61
6
REINFORCED CONCRETE
Finley A. Charney, Ph.D., P.E.
In this chapter, a 12story reinforced concrete office building with some retail shops on the first floor is
designed for both high and moderate seismic loadings. For the more extreme loading, it is assumed that
the structure will be located in Berkeley, California, and for the moderate loading, in Honolulu, Hawaii.
Figure 61 shows the basic structural configuration for each location in plan view and Figure 62, in
section. The building, to be constructed primarily from sandlightweight (LW) aggregate concrete, has
12 stories above grade and one basement level. The typical bays are 30 ft long in the northsouth (NS)
direction and either 40 ft or 20 ft long in the eastwest (EW) direction. The main gravity framing system
consists of seven continuous 30ft spans of pan joists. These joists are spaced 36 in. on center and have
an average web thickness of 6 in. and a depth below slab of 16 in. Due to fire code requirements, a 4in.
thick floor slab is used, giving the joists a total depth of 20 in.
The joists along Gridlines 2 through 7 are supported by variable depth "haunched" girders spanning 40 ft
in the exterior bays and 20 ft in the interior bays. The girders are haunched to accommodate
mechanicalelectrical systems. The girders are not haunched on exterior Gridlines 1 and 8, and the 40ft
spans have been divided into two equal parts forming a total of five spans of 20 ft. The girders along all
spans of Gridlines A and D are of constant depth, but along Gridlines B and C, the depth of the end bay
girders has been reduced to allow for the passage of mechanical systems.
Normal weight (NW) concrete walls are located around the entire perimeter of the basement level. NW
concrete also is used for the first (ground) floor framing and, as described later, for the lower levels of the
structural walls in the Berkeley building.
For both locations, the seismicforceresisting system in the NS direction consists of four 7bay momentresisting
frames. The interior frames differ from the exterior frames only in the end bays where the
girders are of reduced depth. At the Berkeley location, these frames are detailed as special momentresisting
frames. Due to the lower seismicity and lower demand for system ductility, the frames of the
Honolulu building are detailed as intermediate momentresisting frames.
In the EW direction, the seismicforceresisting system for the Berkeley building is a dual system
composed of a combination of frames and framewalls (walls integrated into a momentresisting frame).
Along Gridlines 1 and 8, the frames have five 20ft bays with constant depth girders. Along Gridlines 2
and 7, the frames consist of two exterior 40ft bays and one 20ft interior bay. The girders in each span
are of variable depth as described earlier. At Gridlines 3, 4, 5 and 6, the interior bay has been filled with
a shear panel and the exterior bays consist of 40ftlong haunched girders. For the Honolulu building, the
structural walls are not necessary so EW seismic resistance is supplied by the moment frames along
Gridlines 1 through 8. The frames on Gridlines 1 and 8 are fivebay frames and those on Gridlines 2
through 7 are threebay frames with the exterior bays having a 40ft span and the interior bay having a
20ft span. Hereafter, frames are referred to by their gridline designation (e.g., Frame 1 is located on
FEMA 451, NEHRP Recommended Provisions: Design Examples
62
Figure 62A
102'6"
5 at 20'0"
216'0"
7 at 30'0"
Figure 62B
N
' '
Figure 61 Typical floor plan of the Berkeley building. The Honolulu building is
similar but without structural walls (1.0 ft = 0.3048 m).
Gridline 1). It is assumed that the structure for both the Berkeley and Honolulu locations is founded on
very dense soil (shear wave velocity of approximately 2000 ft/sec).
Chapter 6, Reinforced Concrete
63
R
12
11
10
9
8
7
6
5
4
3
2
G
Story Level
B
1
2
3
4
5
6
7
8
9
10
11
12
11 at 12'6" 18'0" 15'0"
40'0" 20'0" 40'0"
' '
R
12
11
10
9
8
7
6
5
4
3
2
G
Story Level
B
1
2
3
4
5
6
7
8
9
10
11
12
11 at 12'6" 18'0" 15'0"
40'0" 20'0" 40'0"
' '
A. Section at Wall B. Section at Frame
Figure 62 Typical elevations of the Berkeley building; the Honolulu building is
similar but without structural walls (1.0 ft = 0.3048 m).
FEMA 451, NEHRP Recommended Provisions: Design Examples
64
The calculations herein are intended to provide a reference for the direct application of the design
requirements presented in the 2000 NEHRP Recommended Provisions (hereafter, the Provisions) and to
assist the reader in developing a better understanding of the principles behind the Provisions.
Because a single building configuration is designed for both high and moderate levels of seismicity, two
different sets of calculations are required. Instead of providing one full set of calculations for the
Berkeley building and then another for the Honolulu building, portions of the calculations are presented
in parallel. For example, the development of seismic forces for the Berkeley and Honolulu buildings are
presented before structural design is considered for either building. The full design then is given for the
Berkeley building followed by the design of the Honolulu building. Each major section (development of
forces, structural design, etc.) is followed by discussion. In this context, the following portions of the
design process are presented in varying amounts of detail for each structure:
1. Development and computation of seismic forces;
2. Structural analysis and interpretation of structural behavior;
3. Design of structural members including typical girder in Frame 1, typical interior column in Frame 1,
typical beamcolumn joint in Frame 1, typical girder in Frame 3, typical exterior column in Frame 3,
typical beamcolumn joint in Frame 3, boundary elements of structural wall (Berkeley building only)
and panel of structural wall (Berkeley building only).
The design presented represents the first cycle of an iterative design process based on the equivalent
lateral force (ELF) procedure according to Provisions Chapter 5. For final design, the Provisions may
require that a modal response spectrum analysis or time history analysis be used. The decision to use
more advanced analysis can not be made a priori because several calculations are required that cannot be
completed without a preliminary design. Hence, the preliminary design based on an ELF analysis is a
natural place to start. The ELF analysis is useful even if the final design is based on a more sophisticated
analysis (e.g., forces from an ELF analysis are used to apply accidental torsion and to scale the results
from the more advanced analysis and are useful as a check on a modal response spectrum or timehistory
analysis).
In addition to the Provisions, ACI 318 is the other main reference in this example. Except for very minor
exceptions, the seismicforceresisting system design requirements of ACI 318 have been adopted in their
entirety by the Provisions. Cases where requirements of the Provisions and ACI 318 differ are pointed
out as they occur. ASCE 7 is cited when discussions involve live load reduction, wind load, and load
combinations.
Other recent works related to earthquake resistant design of reinforced concrete buildings include:
ACI 318 American Concrete Institute. 1999 [2002]. Building Code Requirements and
Commentary for Structural Concrete.
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for
Buildings and Other Structures.
Fanella Fanella, D.A., and M. Munshi. 1997. Design of LowRise Concrete Buildings for
Earthquake Forces, 2nd Edition. Portland Cement Association, Skokie, Illinois.
ACI 318 Notes Fanella, D.A., J. A. Munshi, and B. G. Rabbat, Editors. 1999. Notes on ACI 31899
Building Code Requirements for Structural Concrete with Design Applications. Portland
Cement Association, Skokie, Illinois.
Chapter 6, Reinforced Concrete
65
ACI SP127 Ghosh, S. K., Editor. 1991. EarthquakeResistant Concrete Structures Inelastic
Response and Design, ACI SP127. American Concrete Institute, Detroit, Michigan.
Ghosh Ghosh, S. K., A. W. Domel, and D. A. Fanella. 1995. Design of Concrete Buildings for
Earthquake and Wind Forces, 2nd Edition. Portland Cement Association, Skokie, Illinois.
Paulay Paulay, T., and M. J. N. Priestley. 1992. Seismic Design of Reinforced Concrete and
Masonry Buildings. John Wiley & Sons, New York.
The Portland Cement Association’s notes on ACI 318 contain an excellent discussion of the principles
behind the ACI 318 design requirements and an example of the design and detailing of a framewall
structure. The notes are based on the requirements of the 1997 Uniform Building Code (International
Conference of Building Officials) instead of the Provisions. The other publications cited above provide
additional background for the design of earthquakeresistant reinforced concrete structures.
Most of the largescale structural analysis for this chapter was carried out using the ETABS Building
Analysis Program developed by Computers and Structures, Inc., Berkeley, California. Smaller portions
of the structure were modeled using the SAP2000 Finite Element Analysis Program, also developed by
Computers and Structures. Column capacity and design curves were computed using Microsoft Excel,
with some verification using the PCACOL program created and developed by the Portland Cement
Association.
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the
changes are described, the design examples and calculations have not been revised to reflect the changes
to the 2003 Provisions.
The changes related to reinforced concrete in Chapter 9 of the 2003 Provisions are generally intended to
maintaining compatibility between the Provisions and the ACI 31802. Portions of the 2000 Provisions
have been removed because they were incorporated into ACI 31802. Other chances to Chapter 9 are
related to precast concrete (as discussed in Chapter 7 of this volume of design examples).
Some general technical changes in the 2003 Provisions that relate to the calculations and/or design in this
chapter include updated seismic hazard maps, revisions to the redundancy requirements, revisions to the
minimum base shear equation, and revisions several of the system factors (R, O0, Cd) for dual systems.
Where they affect the design examples in this chapter, other significant changes to the 2003 Provisions
and primary reference documents are noted. However, some minor changes to the 2003 Provisions and
the reference documents may not be noted.
Note that these examples illustrate comparisons between seismic and wind loading for illustrative
purposes. Wind load calculations are based on ASCE 798 as referenced in the 2000 Provisions, and
there have not been any comparisons or annotations related to ASCE 702.
FEMA 451, NEHRP Recommended Provisions: Design Examples
66
6.1 DEVELOPMENT OF SEISMIC LOADS AND DESIGN REQUIREMENTS
6.1.1 Seismicity
Using Provisions Maps 7 and 8 [Figures 3.33 and 3.34] for Berkeley, California, the short period and
onesecond period spectral response acceleration parameters SS and S1 are 1.65 and 0.68, respectively.
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package.] For the very dense soil conditions, Site Class C is appropriate as described in
Provisions Sec. 4.1.2.1 [3.5.1]. Using SS = 1.65 and Site Class C, Provisions Table 4.1.2.4a [3.31] lists a
short period site coefficient Fa of 1.0. For S1 > 0.5 and Site Class C, Provisions Table 4.1.2.4b [3.32]
gives a velocity based site coefficient Fv of 1.3. Using Provisions Eq. 4.1.2.41 and 4.1.2.42 [3.31 and
3.32], the maximum considered spectral response acceleration parameters for the Berkeley building are:
SMS = FaSS = 1.0 x 1.65 = 1.65
SM1 = FvS1 = 1.3 x 0.68 = 0.884
The design spectral response acceleration parameters are given by Provisions Eq. 4.1.2.51 and 4.1.2.52
[3.33 and 3.34]:
SDS = (2/3) SMS = (2/3) 1.65 = 1.10
SD1 = (2/3) SM1 = (2/3) 0.884 = 0.589
The transition period (Ts) for the Berkeley response spectrum is:
0.589 0.535 sec
1.10
D1
s
DS
T S
S
= = =
Ts is the period where the horizontal (constant acceleration) portion of the design response spectrum
intersects the descending (constant velocity or acceleration inversely proportional to T) portion of the
spectrum. It is used later in this example as a parameter in determining the type of analysis that is
required for final design.
For Honolulu, Provisions Maps 19 and 20 [Figure 3.310] give the shortperiod and 1sec period spectral
response acceleration parameters of 0.61 and 0.178, respectively. For the very dense soil/firm rock site
condition, the site is classified as Site Class C. Interpolating from Provisions Table 4.1.4.2a [3.31], the
shortperiod site coefficient (Fa) is 1.16 and, from Provisions Table 4.1.2.4b [3.32], the interpolated
longperiod site coefficient (Fv) is 1.62. The maximum considered spectral response acceleration
parameters for the Honolulu building are:
SMS = FaSS = 1.16 x 0.61 = 0.708
SM1 = FvS1 = 1.62 x 0.178 = 0.288
and the design spectral response acceleration parameters are:
SDS = (2/3) SMS = (2/3) 0.708 = 0.472
SD1 = (2/3) SM1 = (2/3) 0.288 = 0.192
The transition period (Ts) for the Honolulu response spectrum is:
Chapter 6, Reinforced Concrete
67
0.192 0.407 sec
0.472
D1
s
DS
T S
S
= = =
6.1.2 Structural Design Requirements
According to Provisions Sec. 1.3 [1.2], both the Berkeley and the Honolulu buildings are classified as
Seismic Use Group I. Provisions Table 1.4 [1.3] assigns an occupancy importance factor (I) of 1.0 to all
Seismic Use Group I buildings.
According to Provisions Tables 4.2.1a and 4.2.1b [Tables 1.41 and 1.42], the Berkeley building is
classified as Seismic Design Category D. The Honolulu building is classified as Seismic Design
Category C because of the lower intensity ground motion.
The seismicforceresisting systems for both the Berkeley and the Honolulu buildings consist of momentresisting
frames in the NS direction. EW loading is resisted by a dual framewall system in the
Berkeley building and by a set of momentresisting frames in the Honolulu building. For the Berkeley
building, assigned to Seismic Design Category D, Provisions Sec. 9.1.1.3 [9.2.2.1.3] (which modifies
language in the ACI 318 to conform to the Provisions) requires that all momentresisting frames be
designed and detailed as special moment frames. Similarly, Provisions Sec. 9.1.1.3 [9.2.2.1.3] requires
the structural walls to be detailed as special reinforced concrete shear walls. For the Honolulu building
assigned to Seismic Design Category C, Provisions Sec. 9.1.1.3 [9.2.2.1.3] allows the use of intermediate
moment frames. According to Provisions Table 5.2.2 [4.31], neither of these structures violate height
restrictions.
Provisions Table 5.2.2 [4.31] provides values for the response modification coefficient (R), the system
over strength factor (O0), and the deflection amplification factor (Cd) for each structural system type. The
values determined for the Berkeley and Honolulu buildings are summarized in Table 61.
Table 61 Response Modification, Overstrength, and Deflection Amplification Coefficients
for Structural Systems Used
Location
Response
Direction Building Frame Type R O0 Cd
Berkeley NS Special moment frame 8 3 5.5
EW Dual system incorporating special moment
frame and structural wall
8 2.5 6.5
Honolulu NS Intermediate moment frame 5 3 4.5
EW Intermediate moment frame 5 3 4.5
[For a dual system consisting of a special moment frame and special reinforced concrete shear walls, R =
7, O0 = 2.5, and Cd = 5.5 in 2003 Provisions Table 4.31.]
For the Berkeley building dual system, the Provisions requires that the frame portion of the system be
able to carry 25 percent of the total seismic force. As discussed below, this requires that a separate
analysis of a frameonly system be carried out for loading in the EW direction.
With regard to the response modification coefficients for the special and intermediate moment frames, it
is important to note that R = 5.0 for the intermediate frame is 0.625 times the value for the special frame.
This indicates that intermediate frames can be expected to deliver lower ductility than that supplied by the
more stringently detailed special moment frames.
FEMA 451, NEHRP Recommended Provisions: Design Examples
68
For the Berkeley system, the response modification coefficients are the same (R = 8) for the frame and
framewall systems but are higher than the coefficient applicable to a special reinforced concrete
structural wall system (R = 6). This provides an incentive for the engineer to opt for a framewall system
under conditions where a frame acting alone may be too flexible or a wall acting alone cannot be
proportioned due to excessively high overturning moments.
6.1.3 Structural Configuration
Based on the plan view of the building shown in Figure 61, the only possibility of a plan irregularity is a
torsional irregularity (Provisions Table 5.2.3.2 [4.32]) of Type 1a or 1b. While the actual presence of
such an irregularity cannot be determined without analysis, it appears unlikely for both the Berkeley and
the Honolulu buildings because the lateralforceresisting elements of both buildings are distributed
evenly over the floor. For the purpose of this example, it is assumed (but verified later) that torsional
irregularities do not exist.
As for the vertical irregularities listed in Provisions Table 5.2.3.3 [4.33], the presence of a soft or weak
story cannot be determined without calculations based on an existing design. In this case, however, the
first story is suspect, because its height of 18 ft is well in excess of the 12.5ft height of the story above.
As with the torsional irregularity, it is assumed (but verified later) that a vertical irregularity does not
exist.
6.2 DETERMINATION OF SEISMIC FORCES
The determination of seismic forces requires knowledge of the magnitude and distribution of structural
mass, the short period and long period response accelerations, the dynamic properties of the system, and
the system response modification factor (R). Using Provisions Eq. 5.4.1 [5.21], the design base shear for
the structure is:
V = CSW
where W is the total (seismic) weight of the building and CS is the seismic response coefficient. The upper
limit on CS is given by Provisions Eq. 5.4.1.11 [5.22]:
/DS
S
C S
R I
=
For intermediate response periods, Eq. 5.4.1.12 [5.23] controls:
( / )
D1
S
C S
T R I
=
However, the response coefficient must not be less than that given by Eq. 5.4.1.13 [changed in the 2003
Provisions]:
CS = 0.044SDSI
Note that the above limit will apply when the structural period is greater than SD1/0.044RSDS. This limit is
(0.589)/(0.044 x 8 x 1.1) = 1.52 sec for the Berkeley building and (0.192)/(0.044 x 5 x 0.472) = 1.85 sec
for the Honolulu building. [The minimum Cs value is simply 0.01in the 2003 Provisions, which would
not be applicable to this example as discussed below.]
Chapter 6, Reinforced Concrete
69
In each of the above equations, the importance factor (I) is taken as 1.0. With the exception of the period
of vibration (T), all of the other terms in previous equations have been defined and/or computed earlier in
this chapter.
6.2.1 Approximate Period of Vibration
Requirements for the computation of building period are given in Provisions Sec. 5.4.2 [5.2.2]. For the
preliminary design using the ELF procedure, the approximate period (Ta) computed in accordance with
Provisions Eq. 5.4.2.11 [5.26] could be used:
x
Ta=Crhn
Because this formula is based on lower bound regression analysis of measured building response in
California, it will generally result in periods that are lower (hence, more conservative for use in predicting
base shear) than those computed from a more rigorous mathematical model. This is particularly true for
buildings located in regions of lower seismicity. If a more rigorous analysis is carried out (using a
computer), the resulting period may be too high due to a variety of possible modeling errors.
Consequently, the Provisions places an upper limit on the period that can be used for design. The upper
limit is T = CuTa where Cu is provided in Provisions Table 5.4.2 [5.21].
For the NS direction of the Berkeley building, the structure is a reinforced concrete momentresisting
frame and the approximate period is calculated according to Provisions Eq. 5.4.2.11 [5.26]. Using
Provisions Table 5.4.2.1 [5.22], Cr = 0.016 and x = 0.9. With hn = 155.5 ft, Ta = 1.50 sec. With SD1 >
0.40 for the Berkeley building, Cu = 1.4 and the upper limit on the analytical period is T = 1.4(1.5) = 2.1
sec.
For EW seismic activity in Berkeley, the structure is a framewall system with Cr = 0.020 and x =0.75.
Substituting the appropriate values in Provisions Eq. 5.4.2.11 [5.26], the EW period Ta = 0.88 sec. The
upper limit on the analytical period is (1.4)0.88 = 1.23 sec.
For the Honolulu building, the Ta = 1.5 sec period computed above for concrete moment frames is
applicable in both the NS and EW direction. For Honolulu, SD1 is 0.192g and, from Provisions Table
5.4.2 [5.21], Cu can be taken as 1.52. The upper limit on the analytical period is T = 1.52(1.5) = 2.28 sec.
The period to be used in the ELF analysis will be in the range of Ta to CuTa. If an accurate analysis
provides periods greater than CuTa, CuTa should be used. If the accurate analysis produces periods less
than CuTa but greater than Ta, the period from the analysis should be used. Finally, if the accurate analysis
produces periods less than Ta, Ta may be used.
Later in this chapter, the more accurate periods will be computed using a finite element analysis program.
Before this can be done, however, the building mass must be determined.
6.2.2 Building Mass
Before the building mass can be determined, the approximate size of the different members of the
seismicforceresisting system must be established. For special moment frames, limitations on
beamcolumn joint shear and reinforcement development length usually control. This is particularly true
when lightweight (LW) concrete is used. An additional consideration is the amount of vertical
reinforcement in the columns. ACI 318 Sec. 21.4.3.1 limits the vertical steel reinforcing ratio to 6 percent
for special moment frame columns; however, 4 percent vertical steel is a more practical limit.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1ACI 318 Sec. 21.6.4 [21.7.4] gives equations for the shear strength of the panels of structural walls. In the equations, the term
c appears, but there is no explicit requirement to reduce the shear strength of the concrete when LW aggregate is used. f '
However, ACI 318 Sec. 11.2 states that wherever the term c appears in association with shear strength, it should be f '
multiplied by 0.75 when allLW concrete is used and by 0.85 when sandLW concrete is used. In this example, which utilizes
sandLW concrete, the shear strength of the concrete will be multiplied by 0.85 as specified in ACI 318 Chapter 11.
610
Based on a series of preliminary calculations (not shown here), it is assumed that all columns and
structural wall boundary elements are 30 in. by 30 in., girders are 22.5 in. wide by 32 in. deep, and the
panel of the structural wall is 16 in. thick. It has already been established that pan joists are spaced 36 in.
o.c., have an average web thickness of 6 in., and, including a 4in.thick slab, are 20 in. deep. For the
Berkeley building, these member sizes probably are close to the final sizes. For the Honolulu building
(which has no structural wall and ultimately ends up with slightly smaller elements), the masses computed
from the above member sizes are on the conservative (heavy) side.
In addition to the building structural weight, the following superimposed dead loads (DL) were assumed:
Partition DL (and roofing) = 10 psf
Ceiling and mechanical DL = 15 psf
Curtain wall cladding DL = 10 psf
Based on the member sizes given above and on the other dead load, the individual story weights, masses,
and mass moments of inertia are listed in Table 62. These masses were used for both the Berkeley and
the Honolulu buildings.
As discussed below, the mass and mass moments of inertia are required for the determination of modal
properties using the ETABS program. Note from Table 62 that the roof and lowest floor have masses
slightly different from the typical floors. It is also interesting to note that the average density of this
building is 11.2 pcf. A normal weight (NW) concrete building of the same configuration would have a
density of approximately 14.0 pcf.
The use of LW instead of NW concrete reduces the total building mass by more than 20 percent and
certainly satisfies the minimize mass rule of earthquakeresistant design. However, there are some
disadvantages to the use of LW concrete. In general, LW aggregate reinforced concrete has a lower
toughness or ductility than NW reinforced concrete and the higher the strength, the larger the reduction in
available ductility. For this reason and also the absence of pertinent test results, ACI 318 Sec. 21.2.4.2
allows a maximum compressive strength of 4,000 psi for LW concrete in areas of high seismicity. [Note
that in ACI 31802 Sec. 21.2.4.2, the maximum compressive strength for LW concrete has been increased
to 5,000 psi.] A further penalty placed on LW concrete is the reduction of shear strength. This primarily
affects the sizing of beamcolumn joints (ACI 318 Sec. 21.5.3.2) but also has an effect on the amount of
shear reinforcement required in the panels of structural walls.1 For girders, the reduction in shear strength
of LW aggregate concrete usually is of no concern because ACI 318 disallows the use of the concrete in
determining the shear resistance of members with significant earthquake shear (ACI 318 Sec. 21.4.5.2).
Finally, the required tension development lengths for bars embedded in LW concrete are significantly
greater than those required for NW concrete.
Table 62 Story Weights, Masses, and Moments of Inertia
Story Level Weight (kips)
Mass
(kipssec2/in.)
Mass Moment of Inertia
(in.kipsec2/rad)
Chapter 6, Reinforced Concrete
611
Roof
12
11
10
98765432
Total
2,783
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,169
36,462
7.202
7.896
7.896
7.896
7.896
7.876
7.896
7.896
7.896
7.896
7.896
8.201
4,675,000
5,126,000
5,126,000
5,126,000
5,126,000
5,126,000
5,126,000
5,126,000
5,126,000
5,126,000
5,126,000
5,324,000
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm.
6.2.3 Structural Analysis
Structural analysis is used primarily to determine the forces in the elements for design purposes, compute
story drift, and assess the significance of Pdelta effects. The structural analysis also provides other
useful information (e.g., accurate periods of vibration and computational checks on plan and vertical
irregularities). The computed periods of vibration are addressed in this section and the other results are
presented and discussed later.
The ETABS program was used for the analysis of both the Berkeley and Honolulu buildings. Those
aspects of the model that should be noted are:
1. The structure was modeled with 12 levels above grade and one level below grade. The perimeter
basement walls were modeled as shear panels as were the main structural walls. It was assumed that
the walls were "fixed" at their base.
2. As automatically provided by the ETABS program, all floor diaphragms were assumed to be
infinitely rigid in plane and infinitely flexible outofplane.
3. Beams, columns, and structural wall boundary members were represented by twodimensional frame
elements. Each member was assumed to be uncracked, and properties were based on gross area for
the columns and boundary elements and on effective Tbeam shapes for the girders. (The effect of
cracking is considered in a simplified manner.) The width of the flanges for the Tbeams is based on
the definition of Tbeams in ACI 318 Sec. 8.10. Except for the slab portion of the joists which
contributed to Tbeam stiffness of the girders, the flexural stiffness of the joists was ignored. For the
haunched girders, an equivalent depth of stem was used. The equivalent depth was computed to
provide a prismatic member with a stiffness under equal end rotation identical to that of the
nonprismatic haunched member. Axial, flexural, and shear deformations were included for all
members.
4. The structural walls of the Berkeley building are modeled as a combination of boundary elements and
shear panels.
5. Beamcolumn joints are modeled as 50 percent rigid. This provides effective stiffness for
beamcolumn joints halfway between a model with fully rigid joints (clear span analysis) and fully
flexible joints (centerline analysis).
FEMA 451, NEHRP Recommended Provisions: Design Examples
612
6. Pdelta effects are ignored. An evaluation of the accuracy of this assumption is provided later in this
example.
6.2.4 Accurate Periods from Finite Element Analysis
The computed periods of vibration and a description of the associated modes of vibration are given for the
first 11 modes of the Berkeley building in Table 63. With 11 modes, the accumulated modal mass in
each direction is more than 90 percent of the total mass. Provisions Sec. 5.5.2 [5.3.2] requires that a
dynamic analysis must include at least 90 percent of the actual mass in each of the two orthogonal
directions. Table 64 provides the computed modal properties for the Honolulu building. In this case, 90
percent of the total mass was developed in just eight modes.
For the Berkeley building, the computed NS period of vibration is 1.77 sec. This is between the
approximate period, Ta = 1.5 sec, and CuTa = 2.1 sec. In the EW direction, the computed period is 1.40
sec, which is greater than both Ta = 0.88 sec and CuTa = 1.23 sec.
If cracked section properties were used, the computed period values for the Berkeley building would be
somewhat greater. For preliminary design, it is reasonable to assume that each member has a cracked
moment of inertia equal to onehalf of the gross uncracked moment of inertia. Based on this assumption,
and the assumption that flexural behavior dominates, the cracked periods would be approximately 1.414
(the square root of 2.0) times the uncracked periods. Hence, for Berkeley, the cracked NS and EW
periods are 1.414(1.77) = 2.50 sec, and 1.414(1.4) = 1.98 sec, respectively. Both of these cracked periods
are greater than CuTa, so CuTa can be used in the ELF analysis.
For the Honolulu building, the uncracked periods in the NS and EW directions are 1.78 and 1.87 sec,
respectively. The NS period is virtually the same as for the Berkeley building because there are no walls
in the NS direction of either building. In the EW direction, the increase in period from 1.4 sec to 1.87
sec indicates a significant reduction in stiffness due to the loss of the walls in the Honolulu building. For
both the EW and the NS directions, the approximate period (Ta) for the Honolulu building is 1.5 sec,
and CuTa is 2.28 sec. Both of the computed periods fall within these bounds. However, if cracked section
properties were used, the computed periods would be 2.52 sec in the NS direction and 2.64 sec in the
EW direction. For the purpose of computing ELF forces, therefore, a period of 2.28 sec can be used for
both the NS and EW directions in Honolulu.
A summary of the approximate and computed periods is given in Table 65.
Chapter 6, Reinforced Concrete
613
Table 63 Periods and Modal Response Characteristics for the Berkeley Building
Mode
Period*
(sec)
% of Effective Mass Represented by Mode**
NS EW Description
1
2
3
4
5
6
7
8
9
10
11
1.77
1.40
1.27
0.581
0.394
0.365
0.336
0.230
0.210
0.171
0.135
80.23 (80.2)
0.0 (80.2)
0.0 (80.2)
8.04 (88.3)
0.00 (88.3)
0.00 (88.3)
2.24 (90.5)
0.88 (91.4)
0.00 (91.4)
0.40 (91.8)
0.00 (91.8)
00.00 (0.00)
71.48 (71.5)
0.00 (71.5)
0.00 (71.5)
0.00 (71.5)
14.17 (85.6)
0.00 (85.6)
0.00 (85.6)
0.00 (85.6)
0.00 (85.6)
4.95 (90.6)
First Mode NS
First Mode EW
First Mode Torsion
Second Mode NS
Second Mode Torsion
Second Mode EW
Third Mode NS
Fourth Mode NS
Third Mode Torsion
Fifth Mode NS
Third Mode EW
* Based on gross section properties.
** Accumulated mass in parentheses.
Table 64 Periods and Modal Response Characteristics for the Honolulu Building
Mode
Period*
(sec)
% of Effective Mass Represented by Mode**
NS EW Description
1
2
3
4
5
6
7
8
9
10
11
1.87
1.78
1.38
0.610
0.584
0.452
0.345
0.337
0.260
0.235
0.231
79.7 (79.7)
0.00 (79.7)
0.00 (79.7)
8.79 (88.5)
0.00 (88.5)
0.00 (88.5)
2.27 (90.7)
0.00 (90.7)
0.00 (90.7)
0.89 (91.6)
0.00 (91.6)
0.00 (0.00)
80.25 (80.2)
0.00 (80.2)
0.00 (80.2)
8.04 (88.3)
0.00 (88.3)
0.00 (88.3)
2.23 (90.5)
0.00 (90.5)
0.00 (90.5)
0.87 (91.4)
First Mode EW
First Mode NS
First Mode Torsion
Second Mode EW
Second Mode NS
Second Mode Torsion
Third Mode EW
Third Mode NS
Third Mode Torsion
Fourth Mode EW
Fourth Mode NS
* Based on gross section properties.
** Accumulated mass in parentheses.
Table 65 Comparison of Approximate and "Exact" Periods (in seconds)
Method of Period
Computation
Berkeley Honolulu
NS EW NS EW
Approximate Ta 1.50 0.88 1.50 1.50
Approximate × Cu 2.10* 1.23 2.28 2.28
ETABS (gross) 1.77 1.40 1.78 1.87
ETABS (cracked) 2.50 1.98 2.52 2.64
* Values in italics should be used in the ELF analysis.
6.2.5 Seismic Design Base Shear
FEMA 451, NEHRP Recommended Provisions: Design Examples
614
The seismic design base shear for the Berkeley is computed below.
In the NS direction with W = 36,462 kips (see Table 62), SDS = 1.10, SD1 = 0.589, R = 8, I = 1, and T =
2.10 sec:
,
1.10 0.1375
/ 8/1
DS
S max
C S
R I
= = =
0.589 0.0351
( / ) 2.10(8/1)
D1
S
C S
T R I
= = =
CS,min=0.044SDSI=0.044(1.1)(1)=0.0484
[As noted previously in Sec. 6.2, the minimum Cs value is 0.01 in the 2003 Provisions.]
CS,min = 0.0484 controls, and the design base shear in the NS direction is V = 0.0484 (36,462) = 1,765
kips.
In the stiffer EW direction, CS,max and CS,min are as before, T = 1.23 sec, and
0.589 0.0598
( / ) 1.23(8/1)
D1
S
C S
T R I
= = =
In this case, CS = 0.0598 controls and V = 0.0598 (36,462) = 2,180 kips
For the Honolulu building, base shears are computed in a similar manner and are the same for the NS and
the EW directions. With W = 36,462 kips, SDS = 0.474, SD1 = 0.192, R = 5, I = 1, and T = 2.28 sec:
,
0.472 0.0944
/ 5/1
DS
S max
C S
R I
= = =
0.192 0.0168
( / ) 2.28(5/1)
D1
S
C S
T R I
= = =
CS,min=0.044SDSI=0.044(0.472)(1.0)=0.0207
CS = 0.0207 controls and V = 0.0207 (36,462) = 755 kips
A summary of the Berkeley and Honolulu seismic design parameters are provided in Table 66.
Note that Provisions Sec. 5.4.6 [5.2.6.1] states that for the purpose of computing drift, a base shear
computed according to Provisions Eq. 5.4.1.12 [5.23] (used to compute CS above) may be used in lieu
of the shear computed using Provisions Eq. 5.4.1.13 [5.24] (used to compute CS,min above).
Table 66 Comparison of Periods, Seismic Shears Coefficients, and Base Shears
for the Berkeley and Honolulu Buildings
Location
Response
Direction Building Frame Type
T
(sec) Cs
V
(kips)
Chapter 6, Reinforced Concrete
615
Berkeley NS Special moment frame 2.10 0.0485 1,765
EW Dual system incorporating special moment
frame and structural wall
1.23 0.0598 2,180
Honolulu NS Intermediate moment frame 2.28 0.0207 755
EW Intermediate moment frame 2.28 0.0207 755
1.0 kip = 4.45 kN.
6.2.6 Development of Equivalent Lateral Forces
The vertical distribution of lateral forces is computed from Provisions Eq. 5.4.31 and 5.4.32 [5.210 and
5.211]:
Fx = CvxV
k
x x
vx n
k
i i
i = 1
C = w h
S w h
where
k = 1.0 for T < 0.5 sec
k = 2.0 for T > 2.5 sec
k = 0.75 + 0.5T for 1.0 < T < 2.5 sec
Based on the equations above, the seismic story forces, shears, and overturning moments are easily
computed using an Excel spreadsheet. The results of these computations are shown in Tables 67a and
67b for the Berkeley buildings and in Table 68 for the Honolulu building. A note at the bottom of each
table gives the calculated vertical force distribution factor (k). The tables are presented with as many
significant digits to the left of the decimal as the spreadsheet generates but that should not be interpreted
as real accuracy; it is just the simplest approach. Also, some of the sums are not exact due to truncation
error.
FEMA 451, NEHRP Recommended Provisions: Design Examples
616
Table 67a Vertical Distribution of NS Seismic Forces for the Berkeley Building*
Level
Height h
(ft)
Weight W
(kips) Whk Whk/S
Force Fx
(kips)
Story
Shear Vx
(kips)
Overturning
Moment
Mx (ftk)
R
12
11
10
9
8
7
6
5
4
3
2
Total
155.5
143.0
130.5
118.0
105.5
93.0
80.5
68.0
55.5
43.0
30.5
18.0
2,783
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,169
36,462
24,526,067
23,123,154
19,612,869
16,361,753
13,375,088
10,658,879
8,220,056
6,066,780
4,208,909
2,658,799
1,432,788
575,987
130,821,129
0.187
0.177
0.150
0.125
0.102
0.081
0.063
0.046
0.032
0.020
0.011
0.004
0.998
330.9
311.9
264.6
220.7
180.4
143.8
110.9
81.8
56.8
35.9
19.3
7.8
1764.8
330.9
642.8
907.4
1,128.1
1,308.5
1,452.3
1,563.2
1,645.0
1,701.8
1,737.7
1,757.0
1,764.8
4,136
12,170
23,512
37,613
53,970
72,123
91,663
112,226
133,498
155,219
177,181
208,947
* Table based on T = 2.1 sec and k = 1.8.
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
Table 67b Vertical Distribution of EW Seismic Forces for the Berkeley Building*
Level
Height h
(ft)
Weight W
(kips) Whk Whk/S
Force Fx
(kips)
Story
Shear Vx
(kips)
Overturning
Moment
Mx (ftk)
R
12
11
10
9
8
7
6
5
4
3
2
Total
155.5
143.0
130.5
118.0
105.5
93.0
80.5
68.0
55.5
43.0
30.5
18.0
2,783
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,169
36,462
2,730,393
2,669,783
2,356,408
2,053,814
1,762,714
1,483,957
1,218,579
967,870
733,503
517,758
323,975
163,821
16,982,575
0.161
0.157
0.139
0.121
0.104
0.087
0.072
0.057
0.043
0.030
0.019
0.010
1.000
350.6
342.8
302.5
263.7
226.3
190.5
156.5
124.3
94.2
66.5
41.6
21.0
2180.5
351
693
996
1,260
1,486
1,676
1,833
1,957
2,051
2,118
2,159
2,180
4,382
13,049
25,497
41,242
59,816
80,771
103,682
128,146
153,788
180,260
207,253
246,500
* Table based on T = 1.23 sec and k = 1.365.
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
Chapter 6, Reinforced Concrete
617
Table 68 Vertical Distribution of NS and EW Seismic Forces for the Honolulu Building*
Level
Height h
(ft)
Weight W
(kips) Whk Whk/S
Force Fx
(kips)
Story
Shear Vx
(kips)
Overturning
Moment
Mx (ftk)
R
12
11
10
9
8
7
6
5
4
3
2
Total
155.5
143.0
130.5
118.0
105.5
93.0
80.5
68.0
55.5
43.0
30.5
18.0
2,783
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,051
3,169
36,462
38,626,348
36,143,260
30,405,075
25,136,176
20,341,799
16,027,839
12,210,028
8,869,192
6,041,655
3,729,903
1,948,807
747,115
200,218,197
0.193
0.181
0.152
0.126
0.102
0.080
0.061
0.044
0.030
0.019
0.010
0.004
1.002
145.6
136.2
114.6
94.8
76.7
60.4
46.0
33.4
22.8
14.1
7.3
2.8
754.7
145.6
281.9
396.5
491.2
567.9
628.3
674.3
707.8
730.5
744.6
751.9
754.8
1,820
5,343
10,299
16,440
23,539
31,393
39,822
48,669
57,801
67,108
76,508
90,093
* Table based on T = 2.28 sec and k = 1.89.
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm
The computed seismic story shears for the Berkeley and Honolulu buildings are shown graphically in
Figures 63 and 64, respectively. Also shown in the figures are the story shears produced by ASCE 7
wind loads. For Berkeley, a 3sec gust of 85 mph was used and, for Honolulu, a 3sec gust of 105 mph.
In each case, an Exposure B classification was assumed. The wind shears have been factored by a value
of 1.36 (load factor of 1.6 times directionality factor 0.85) to bring them up to the ultimate seismic
loading limit state represented by the Provisions.
As can be seen from the figures, the seismic shears for the Berkeley building are well in excess of the
wind shears and will easily control the design of the members of the frames and walls. For the Honolulu
building, the NS seismic shears are significantly greater than the corresponding wind shears, but the EW
seismic and wind shears are closer. In the lower stories of the building, wind controls the strength
demands and, in the upper levels, seismic forces control the strength demands. (A somewhat more
detailed comparison is given later when the Honolulu building is designed.) With regards to detailing the
Honolulu building, all of the elements must be detailed for inelastic deformation capacity as required by
ACI 318 rules for intermediate moment frames.
FEMA 451, NEHRP Recommended Provisions: Design Examples
618
0
20
40
60
80
100
120
140
160
0 500 1,000 1,500 2,000 2,500
Shear, kips
Height, ft
EW seismic
NS seismic
EW wind
NS wind
Figure 63 Comparison of wind and seismic story shears for the Berkeley building (1.0
ft = 0.3048 m, 1.0 kip = 4.45 kN).
Chapter 6, Reinforced Concrete
619
0
20
40
60
80
100
120
140
160
0 200 400 600 800 1,000 1,200
Shear, kips
Height, ft
Seismic
EW wind
NS wind
Figure 64 Comparison of wind and seismic story shears for the Honolulu building (1.0 ft
= 0.3048 m, 1.0 kip = 4.45 kN).
6.3 DRIFT AND PDELTA EFFECTS
6.3.1 Direct Drift and PDelta Check for the Berkeley Building
Drift and Pdelta effects are checked according to Provisions Sec. 5.2.8 [5.2.6.1] and 5.4.6 [5.2.6.2],
respectively. According to Provisions Table 5.2.8 [4.51], the story drift limit for this Seismic Use Group
I building is 0.020hsx where hsx is the height of story x. This limit may be thought of as 2 percent of the
story height. Quantitative results of the drift analysis for the NS and EW directions are shown in Tables
69a and 69b, respectively.
With regards to the values shown in Table 69a , it must be noted that cracked section properties were
used in the structural analysis and that 0.0351/0.0484=0.725 times the story forces shown in Table 67a
were applied. This adjusts for the use of Provisions Eq. 5.4.1.13 [not applicable in the 2003 Provisions],
which governed for base shear, was not used in computing drift. In Table 69b, cracked section
FEMA 451, NEHRP Recommended Provisions: Design Examples
620
0
20
40
60
80
100
120
140
160
0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0
Total drift, in.
Height, ft
EW*
6.5 x EW
NS*
5.5 x NS
2% limit
* Elasticlly computed under codeprescribed seismic forces
Figure 65 Drift profile for Berkeley building (1.0 ft =
0.3048 m, 1.0 in. = 25.4 mm).
properties were also used, but the modifying factor does not apply because Provisions Eq. 5.4.1.12 [5.2
3] controlled in this direction.
In neither case does the computed drift ratio (magnified story drift/hsx) exceed 2 percent of the story
height. Therefore, the story drift requirement is satisfied. A plot of the total drift resulting from both the
NS and EW equivalent lateral seismic forces is shown in Figure 65.
An example calculation for drift in Story 5 loaded in the EW direction is given below. Note that the
relevant row is highlighted in Table 69b.
Deflection at top of story = d5e = 1.812 in.
Deflection at bottom of story = d4e = 1.410 in.
Story drift = .5e = d5e  d4e = 1.812  1.410 = 0.402 in.
Deflection amplification factor, Cd = 6.5
Importance factor, I = 1.0
Magnified story drift = .5 = Cd .5e/I = 6.5(0.402)/1.0 = 2.613 in.
Magnified drift ratio = .5/h5 = (2.613/150) = 0.01742 = 1.742% < 2.0% OK
Chapter 6, Reinforced Concrete
621
Table 69a Drift Computations for the Berkeley Building Loaded in the NS Direction
Story
Total Deflection
(in.)
Story Drift
(in.)
Story Drift × Cd *
(in.)
Drift Ratio
(%)
12
11
10
987654321
3.640
3.533
3.408
3.205
2.973
2.697
2.393
2.059
1.711
1.363
0.999
0.618
0.087
0.145
0.203
0.232
0.276
0.305
0.334
0.348
0.348
0.364
0.381
0.618
0.478
0.798
1.117
1.276
1.515
1.675
1.834
1.914
1.914
2.002
2.097
3.397
0.319
0.532
0.744
0.851
1.010
1.117
1.223
1.276
1.276
1.334
1.398
1.573
* Cd = 5.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3
through roof and 216 in. for Level 2.
1.0 in. = 25.4 mm.
Table 69b Drift Computations for the Berkeley Building Loaded in the EW Direction
Story
Total Drift
(in.)
Story Drift
(in.)
Story Drift × Cd
*
(in.)
Drift Ratio
(%)
12
11
10
98765
4321
4.360
4.060
3.720
3.380
3.020
2.620
2.220
1.812
1.410
1.024
0.670
0.362
0.300
0.340
0.340
0.360
0.400
0.400
0.408
0.402
0.386
0.354
0.308
0.362
1.950
2.210
2.210
2.340
2.600
2.600
2.652
2.613
2.509
2.301
2.002
2.353
1.300
1.473
1.473
1.560
1.733
1.733
1.768
1.742
1.673
1.534
1.335
1.089
* Cd = 6.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3
through roof and 216 in. for Level 2.
1.0 in. = 25.4 mm.
When a soft story exists in a Seismic Design Category D building, Provisions Table 5.2.5.1 [4.41]
requires that a modal analysis be used. However, Provisions Sec. 5.2.3.3 [4.3.2.3] lists an exception:
Structural irregularities of Types 1a, 1b, or 2 in Table 5.2.3.3 [4.32] do not apply where no story drift ratio
under design lateral load is less than or equal to 130 percent of the story drift ratio of the next story above. .
. . The story drift ratios of the top two stories of the structure are not required to be evaluated.
FEMA 451, NEHRP Recommended Provisions: Design Examples
622
For the building responding in the NS direction, the ratio of first story to second story drift ratios is
1.573/1.398 = 1.13, which is less than 1.3. For EW response, the ratio is 1.089/1.335 = 0.82, which also
is less than 1.3. Therefore, a modal analysis is not required and the equivalent static forces from Tables
67a and 67b may be used for design.
The Pdelta analysis for each direction of loading is shown in Tables 610a and 610b. The upper limit on
the allowable story stability ratio is given by Provisions Eq. 5.4.6.22 [changed in the 2003 Provisions]
as:
max 0.5 0.50
Cd
.
ß
= =
Taking ß as 1.0 (see Provisions Sec. 5.4.6.2 [not applicable in the 2003 Provisions]), the stability ratio
limit for the NS direction is 0.5/(1.0)5.5 = 0.091, and for the EW direction the limit is 0.5/(1.0)6.5 =
0.077.
[In the 2003 Provisions, the maximum limit on the stability coefficient has been replaced by a
requirement that the stability coefficient is permitted to exceed 0.10 if and only “if the resistance to lateral
forces is determined to increase in a monotonic nonlinear static (pushover) analysis to the target
displacement as determined in Sec. A5.2.3. Pdelta effects shall be included in the analysis.” Therefore,
in this example, the stability coefficient should be evaluated directly using 2003 Provisions Eq. 5.2.16.]
For this Pdelta analysis a (reduced) story live load of 20 psf was included in the total story weight
calculations. Deflections are based on cracked sections, and story shears are adjusted as necessary for use
of Provisions Eq. 5.4.1.13 [5.23]. As can be seen in the last column of each table, the stability ratio (.)
does not exceed the maximum allowable value computed above. Moreover, since the values are less than
0.10 at all levels, Pdelta effects can be neglected for both drift and strength computed limits according to
Provisions Sec. 5.4.6.2 [5.2.6.2].
An example Pdelta calculation for the Level 5 under EW loading is shown below. Note that the relevant
row is highlighted in Table 610b.
Magnified story drift = .5 = 2.613 in.
Story shear = V5 = 1,957 kips
Accumulated story weight P5 = 27,500 kips
Story height = hs5= 150 in.
Cd = 6.5
. = (P5 (.5/Cd)) /(V5hs5) = 27,500(2.613/6.5)/(1957.1)(150) = 0.0377 < 0.077 OK
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions.
The importance factor, I, has been added to 2003 Provisions Eq. 5.216. However, this does not affect
this example because I = 1.0.]
Chapter 6, Reinforced Concrete
623
Table 610a PDelta Computations for the Berkeley Building Loaded in the NS Direction
Level
Story Drift
(in.)
Story Shear *
(kips)
Story Dead
Load
(kips)
Story Live
Load
(kips)
Total Story
Load
(kips)
Accum. Story
Load
(kips)
Stability
Ratio
.
12
11
10
987654321
0.478
0.798
1.117
1.276
1.515
1.675
1.834
1.914
1.914
2.002
2.097
3.397
239.9
466.0
657.8
817.9
948.7
1052.9
1133.3
1192.6
1233.8
1259.8
1273.8
1279.5
2783
3051
3051
3051
3051
3051
3051
3051
3051
3051
3051
3169
420
420
420
420
420
420
420
420
420
420
420
420
3203
3471
3471
3471
3471
3471
3471
3471
3471
3471
3471
3589
3203
6674
10145
13616
17087
20558
24029
27500
30971
34442
37913
41502
0.0077
0.0138
0.0209
0.0257
0.0331
0.0396
0.0471
0.0535
0.0582
0.0663
0.0757
0.0928
* Story shears in Table 67a factored by 0.725. See Sec. 6.3.1.
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
Table 610b PDelta Computations for the Berkeley Building Loaded in the EW Direction
Level
Story Drift
(in.)
Story Shear
(kips)
Story Dead
Load
(kips)
Story Live
Load
(kips)
Total Story
Load
(kips)
Accum. Story
Load
(kips)
Stability
Ratio
.
12
11
10
9876
5
4321
1.950
2.210
2.210
2.340
2.600
2.600
2.652
2.613
2.509
2.301
2.002
2.353
350.6
693.3
995.9
1259.6
1485.9
1676.4
1832.9
1957.1
2051.3
2117.8
2159.4
2180.4
2783
3051
3051
3051
3051
3051
3051
3051
3051
3051
3051
3169
420
420
420
420
420
420
420
420
420
420
420
420
3203
3471
3471
3471
3471
3471
3471
3471
3471
3471
3471
3589
3203
6674
10145
13616
17087
20558
24029
27500
30971
34442
37913
41502
0.0183
0.0218
0.0231
0.0259
0.0307
0.0327
0.0357
0.0377
0.0389
0.0384
0.0361
0.0319
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
6.3.2 Test for Torsional Irregularity for Berkeley Building
In Sec. 6.1.3 it was mentioned that torsional irregularities are unlikely for the Berkeley building because
the elements of the seismicforceresisting system were well distributed over the floor area. This will now
be verified by applying the story forces of Table 63a at an eccentricity equal to 5 percent of the building
dimension perpendicular to the direction of force (accidental torsion requirement of Provisions Sec.
5.4.4.2 [5.2.4.2]). This test is required per Provisions Sec. 5.2.3.2 [4.3.2.2]. Analysis was performed
using the ETABS program.
FEMA 451, NEHRP Recommended Provisions: Design Examples
624
The eccentricity is 0.05(102.5) = 5.125 ft for forces in the NS direction and 0.05(216) = 10.8 ft in the EW
direction.
For forces acting in the NS direction:
Total displacement at center of mass = davg = 3.640 in. (see Table 69a)
Rotation at center of mass = 0.000189 radians
Maximum displacement at corner of floor plate = dmax = 3.640 + 0.000189(102.5)(12)/2 = 3.756 in.
Ratio dmax/davg = 3.756/3.640 = 1.03 < 1.20, so no torsional irregularity exists.
For forces acting in the EW direction:
Total displacement at center of mass = davg = 4.360 in. (see Table 69b)
Rotation at center of mass = 0.000648 radians
Maximum displacement at corner of floor plate = dmax = 4.360 + 0.000648(216)(12)/2 = 5.200 in.
Ratio dmax/davg = 5.200/4.360 = 1.19 < 1.20, so no torsional irregularity exists.
It is interesting that this building, when loaded in the EW direction, is very close to being torsionally
irregular (irregularity Type 1a of Provisions Table 5.2.3.2 [4.32]), even though the building is extremely
regular in plan. The torsional flexibility of the building arises from the fact that the walls exist only on
interior Gridlines 3, 4, 5, and 6.
6.3.3 Direct Drift and PDelta Check for the Honolulu Building
The interstory drift computations for the Honolulu building deforming under the NS and EW equivalent
static forces are shown in Tables 611a and 611b. As with the Berkeley building, the analysis used
cracked section properties. The applied seismic forces, shown previously in Table 63b were multiplied
by the ratio 0.0168/0.0207 = 0.808 to adjust for the use of Provisions Eq. 5.4.1.13. [As noted previously
in Sec. 6.2, the minimum Cs value has been removed in the 2003 Provisions.]
These tables, as well as Figure 66, show that the story drift at each level is less than the allowable
interstory drift of 0.020hsx (Provisions Table 5.2.8 [4.51]). Even though it is not pertinent for Seismic
Design Category C buildings, a soft first story does not exist for the Honolulu building because the ratio
of first story to second story drift does not exceed 1.3.
Chapter 6, Reinforced Concrete
625
EW*
6.5 x EW
NS*
5.5 x NS
2% limit
0
20
40
60
80
100
120
140
160
0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0
Total drift, in.
Height, ft
* Elasticlly computed under codeprescribed seismic forces
Figure 66 Drift profile for the Honolulu building (1.0 ft = 0.3048 m,
1.0 in. = 25.4 mm).
FEMA 451, NEHRP Recommended Provisions: Design Examples
626
Table 611a Drift Computations for the Honolulu Building Loaded in the NS Direction
Story
Total Drift
(in.)
Story Drift
(in.)
Story Drift × Cd
*
(in.)
Drift Ratio
(%)
12
11
10
9
8
7
6
5
4
3
2
1
1.766
1.726
1.656
1.559
1.441
1.306
1.157
0.997
0.829
0.658
0.482
0.297
0.040
0.069
0.097
0.118
0.136
0.149
0.160
0.168
0.171
0.176
0.184
0.297
0.182
0.313
0.436
0.531
0.611
0.669
0.720
0.756
0.771
0.793
0.829
1.338
0.121
0.208
0.291
0.354
0.407
0.446
0.480
0.504
0.514
0.528
0.553
0.619
* Cd = 4.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3
through roof and 216 in. for Level 2.
1.0 in. = 25.4 mm.
Table 611b Drift Computations for the Honolulu Building Loaded in the EW Direction
Story
Total Drift
(in.)
Story Drift
(in.)
Story Drift × Cd *
(in.)
Drift Ratio
(%)
12
11
10
9
8
7
6
5
4
3
2
1
2.002
1.941
1.850
1.734
1.597
1.440
1.269
1.089
0.903
0.713
0.522
0.325
0.061
0.090
0.116
0.137
0.157
0.171
0.179
0.186
0.191
0.191
0.197
0.325
0.276
0.407
0.524
0.618
0.705
0.772
0.807
0.836
0.858
0.858
0.887
1.462
0.184
0.271
0.349
0.412
0.470
0.514
0.538
0.558
0.572
0.572
0.591
0.677
* Cd = 4.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3
through roof and 216 in. for Level 2.
1.0 in. = 25.4 mm.
A sample calculation for Level 5 of Table 611b (highlighted in the table) is as follows:
Deflection at top of story = d5e =1.089 in.
Deflection at bottom of story = d4e = 0.903 in.
Story drift = .5e = d5e  d4e = 1.0890.0903 = 0.186 in.
Deflectiom amplification factor, Cd = 4.5
Importance factor, I = 1.0
Magnified story drift = .5 = Cd .5e/I = 4.5(0.186)/1.0 = 0.836 in.
Magnified drift ratio = .5 / h5 = (0.836/150) = 0.00558 = 0.558% < 2.0% OK
Chapter 6, Reinforced Concrete
627
Therefore, story drift satisfies the drift requirements.
Calculations for Pdelta effects are shown in Tables 612a and 612b for NS and EW loading,
respectively. The stability ratio at the 5th story from Table 612b is computed:
Magnified story drift = .5 = 0.836 in.
Story shear = V5 = 571.9 = kips
Accumulated story weight P5 = 27500 kips
Story height = hs5 = 150 in.
Cd = 4.5
. = [P5 (.5/Cd)]/(V5hs5) = 27500(0.836/4.5)/(571.9)(150) = 0.0596
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions.
The importance factor, I, has been added to 2003 Provisions Eq. 5.216. However, this does not affect
this example because I = 1.0.]
The requirements for maximum stability ratio (0.5/Cd = 0.5/4.5 = 0.111) are satisfied. Because the
stability ratio is less than 0.10 at all floors, Pdelta effects need not be considered (Provisions Sec. 5.4.6.2
[5.2.6.2]). (The value of 0.1023 in the first story for the EW direction is considered by the author to be
close enough to the criterion.)
Table 612a PDelta Computations for the Honolulu Building Loaded in the NS Direction
Level
Story Drift
(in.)
Story Shear *
(kips)
Story Dead
Load
(kips)
Story Live
Load
(kips)
Total Story
Load
(kips)
Accum. Story
Load
(kips)
Stability
Ratio
.
12
11
10
9
8
7
6
5
4
3
2
1
0.182
0.313
0.436
0.531
0.611
0.669
0.720
0.756
0.771
0.793
0.829
1.338
117.7
227.7
320.4
396.9
458.9
507.7
544.9
571.9
590.3
601.6
607.6
609.8
2783
3051
3051
3051
3051
3051
3051
3051
3051
3051
3051
3169
420
420
420
420
420
420
420
420
420
420
420
420
3203
3471
3471
3471
3471
3471
3471
3471
3471
3471
3471
3589
3203
6674
10145
13616
17087
20558
24029
27500
30971
34442
37913
41502
0.0073
0.0136
0.0205
0.0270
0.0337
0.0401
0.0470
0.0539
0.0599
0.0672
0.0766
0.0937
* Story shears in Table 68 factored by 0.808. See Sec. 6.3.3.
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
FEMA 451, NEHRP Recommended Provisions: Design Examples
628
Table 612b PDelta Computations for the Honolulu Building Loaded in the EW Direction
Level
Story Drift
(in.)
Story Shear *
(kips)
Story Dead
Load
(kips)
Story Live
Load
(kips)
Total Story
Load
(kips)
Accum. Story
Load
(kips)
Stability
Ratio
.
12
11
10
9
8
7
6
5
4
3
2
1
0.276
0.407
0.524
0.618
0.705
0.772
0.807
0.836
0.858
0.858
0.887
1.462
117.7
227.7
320.4
396.9
458.9
507.7
544.9
571.9
590.3
601.6
607.6
609.8
2783
3051
3051
3051
3051
3051
3051
3051
3051
3051
3051
3169
420
420
420
420
420
420
420
420
420
420
420
420
3203
3471
3471
3471
3471
3471
3471
3471
3471
3471
3471
3589
3203
6674
10145
13616
17087
20558
24029
27500
30971
34442
37913
41502
0.0111
0.0177
0.0246
0.0314
0.0389
0.0463
0.0527
0.0596
0.0667
0.0728
0.0820
0.1023
* Story shears in Table 68 factored by 0.808. See Sec. 6.3.3.
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
6.3.4 Test for Torsional Irregularity for the Honolulu Building
A test for torsional irregularity for the Honolulu building can be performed in a manner similar to that for
the Berkeley building. However, it is clear that a torsional irregularity will not occur for the Honolulu
building if the Berkeley building is not irregular. This will be the case because the walls, which draw the
torsional resistance towards the center of the Berkeley building, do not exist in the Honolulu building.
6.4 STRUCTURAL DESIGN OF THE BERKELEY BUILDING
6.4.1 Material Properties
For the Berkeley building, sandLW aggregate concrete of 4,000 psi strength is used everywhere except
for the lower two stories of the structural walls where 6,000 psi NW concrete is used. All reinforcement
has a specified yield strength of 60 ksi, except for the panel of the structural walls which contains 40 ksi
reinforcement. This reinforcement must conform to ASTM A706. According to ACI 318 Sec. 21.2.5,
however, ASTM A615 reinforcement may be used if the actual yield strength of the steel does not exceed
the specified strength by more than 18 ksi and the ratio of actual ultimate tensile stress to actual tensile
yield stress is greater than 1.25.
6.4.2 Combination of Load Effects
Using the ETABS program, the structure was analyzed for the equivalent lateral loads shown in Tables
67a and 67b. For strength analysis, the loads were applied at a 5 percent eccentricity as required for
accidental torsion by Provisions Sec. 5.4.4.2 [5.2.4.2]. Where applicable, orthogonal loading effects were
included per Provisions Sec. 5.2.5.2.3 [4.4.2.3]. The torsional magnification factor (Ax) given by
Provisions Eq. 5.4.4.31 [5.213] was not used because the building has no significant plan irregularities.
Provisions Sec. 5.2.7 [4.2.2.1] and Eq. 5.2.71 and 5.2.72 [4.21 and 4.22] require combination of load
effects be developed on the basis of ASCE 7, except that the earthquake load effect, E, be defined as:
Chapter 6, Reinforced Concrete
629
E=.QE+0.2SDSD
when gravity and seismic load effects are additive and
E=.QE0.2SDSD
when the effects of seismic load counteract gravity.
The special load combinations given by Provisions Eq. 5.2.71 and 5.2.72 [4.23 and 4.24] do not apply
to the Berkeley building because there are no discontinuous elements supporting stiffer elements above
them. (See Provisions Sec. 9.6.2 [9.4.1].)
The reliability factor (.) in Eq. 5.2.71 and 5.2.72 [not applicable in the 2003 Provisions] should be
taken as the maximum value of .x defined by Provisions Eq. 5.2.4.2:
2 20
x
x
rmax Ax
. = 
where Ax is the area of the floor or roof diaphragm above the story under consideration and is the rmaxx
largest ratio of the design story shear resisted by a single element divided by the total story shear for a
given loading. The computed value for . must be greater than or equal to 1.0, but need not exceed 1.5.
Special moment frames in Seismic Design Category D are an exception and must be proportioned such
that . is not greater than 1.25.
For the structure loaded in the NS direction, the structural system consists of special moment frames, and
rix is taken as the maximum of the shears in any two adjacent columns in the plane of a moment frame
divided by the story shear. For interior columns that have girders framing into both sides, only 70 percent
of the individual column shear need be included in this sum. In the NS direction, there are four identical
frames. Each of these frames has eight columns. Using the portal frame idealization, the shear in an
interior column will be Vinterior = 0.25 (2/14) V = 0.0357V.
Similarly, the shear in an exterior column will be Vexterior= 0.25 (1/14) V = 0.0179V.
For two adjacent interior columns:
0.7( int int ) 0.7(0.0375 0.0375 ) 0.0525
ix
r V V V V
V V
+ +
= = =
For one interior and one exterior column:
(0.7 int ext ) 0.7(0.0375 ) 0.0179 ) 0.0441
ix
r V V V V
V V
+ +
= = =
The larger of these values will produce the largest value of .x. Hence, for a floor diaphragm area Ax equal
to 102.5 × 216 = 22,140 square ft:
2 20 0.56
0.0525 22,140 . x =  = 
FEMA 451, NEHRP Recommended Provisions: Design Examples
630
As this value is less than 1.0, . will be taken as 1.0 in the NS direction.
For seismic forces acting in the EW direction, the walls carry significant shear, and for the purposes of
computing ., it will be assumed that they take all the shear. According to the Provisions, rix for walls is
taken as the shear in the wall multiplied by 10/lw and divided by the story shear. The term lw represents
the plan length of the wall in feet. Thus, for one wall:
maxx ix 0.25 (10 / 20) 0.125
r r V
V
= = =
Only 80 percent of the . value based on the above computations need be used because the walls are part
of a dual system. Hence, in the EW direction
0.8 2 20 0.740
0.125 22,140 . x
. .
= .. ..=
. .
and as with the NS direction, . may be taken as 1.0. Note that . need not be computed for the columns
of the frames in the dual system, as this will clearly not control.
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a building
assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure beamtocolumn
connections at both ends of a single beam (moment frame system) or failure of a single shear wall with
aspect ratio greater than 1.0 (shear wall system) would not result in more than a 33 percent reduction in
story strength or create an extreme torsional irregularity. Alternatively, if the structure is regular in plan
and there are at least 2 bays of perimeter framing on each side of the structure in each orthogonal
direction, it is permitted to use, . = 1.0. Per 2003 Provisions Sec. 4.3.1.4.3 special moment frames in
Seismic Design Category D must be configured such that the structure satisfies the criteria for . = 1.0.
There are no reductions in the redundancy factor for dual systems. Based on the preliminary design, . =
1.0 for because the structure has a perimeter moment frame and is regular.]
For the Berkeley structure, the basic ASCE 7 load combinations that must be considered are:
1.2D + 1.6L
1.2D + 0.5L ± 1.0E
0.9D ± 1.0E
The ASCE 7 load combination including only 1.4 times dead load will not control for any condition in
this building.
Substituting E from the Provisions, with . taken as 1.0, the following load combinations must be used for
earthquake:
(1.2 + 0.2SDS)D + 0.5L + E
(1.2 + 0.2SDS)D + 0.5L  E
(0.9  0.2 SDS)D + E
(0.9  0.2SDS)D  E
Finally, substituting 1.10 for SDS, the following load combinations must be used for earthquake:
1.42D + 0.5L + E
1.42D + 0.5L  E
Chapter 6, Reinforced Concrete
2The analysis used to create Figures 67 and 68 did not include the 5 percent torsional eccentricity or the 30 percent orthogonal
loading rules specified by the Provisions. The eccentricity and orthogonal load were included in the analysis carried out for
member design.
631
0.68D + E
0.68D  E
It is very important to note that use of the ASCE 7 load combinations in lieu of the combinations given in
ACI Chapter 9 requires use of the alternate strength reduction factors given in ACI 318 Appendix C:
Flexure without axial load f = 0.80
Axial compression, using tied columns f = 0.65 (transitions to 0.8 at low axial loads)
Shear if shear strength is based on nominal axialflexural capacity f = 0.75
Shear if shear strength is not based on nominal axialflexural capacity f = 0.55
Shear in beamcolumn joints f = 0.80
[The strength reduction factors in ACI 31802 have been revised to be consistent with the ASCE 7 load
combinations. Thus, the factors that were in Appendix C of ACI 31899 are now in Chapter 9 of ACI
31802, with some modification. The strength reduction factors relevant to this example as contained in
ACI 31802 Sec. 9.3 are:
Flexure without axial load f = 0.9 (tensioncontrolled sections)
Axial compression, using tied columns f = 0.65 (transitions to 0.9 at low axial loads)
Shear if shear strength is based o nominal axialflexural capacity f = 0.75
Shear if shear strength is not based o nominal axialflexural capacity f = 0.60
Shear in beamcolumn joints f = 0.85]
6.4.3 Comments on the Structure’s Behavior Under EW Loading
Framewall interaction plays an important role in the behavior of the structure loaded in the EW
direction. This behavior is beneficial to the design of the structure because:
1. For frames without walls (Frames 1, 2, 7, and 8), the shears developed in the girders (except for the
first story) do not differ greatly from story to story. This allows for a uniformity in the design of the
girders.
2. For frames containing structural walls (Frames 3 through 6), the overturning moments in the
structural walls are reduced significantly as a result of interaction with the remaining frames (Frames
1, 2, 7, and 8).
3. For the frames containing structural walls, the 40ftlong girders act as outriggers further reducing the
overturning moment resisted by the structural walls.
The actual distribution of story forces developed in the different frames of the structure is shown in
Figure 67.2 This figure shows the response of Frames 1, 2, and 3 only. By symmetry, Frame 8 is similar
to Frame 1, Frame 7 is similar to Frame 2, and Frame 6 is similar to Frame 3. Frames 4 and 5 have a
response that is virtually identical to that of Frames 3 and 6.
As may be observed from Figure 67, a large reverse force acts at the top of Frame 3 which contains a
structural wall. This happens because the structural wall pulls back on (supports) the top of Frame 1. The
deflected shape of the structure loaded in the EW direction (see Figure 65) also shows the effect of
framewall interaction because the shape is neither a cantilever mode (wall alone) nor a shear mode
FEMA 451, NEHRP Recommended Provisions: Design Examples
632
58.4
8.58
8.68
2.88
0.96
4.26
6.76
8.62
8.72
13.28
18.96
107.8
1
2
3
4
5
6
7
8
9
10
1
12
Story force, kips
114.9
30.1
20.18
8.18
1.64
9.56
15.74
19.98
21.48
26.56
29.12
255.7
1
2
3
4
5
6
7
8
9
10
11
12
91.96
31.79
34.77
34.37
36.26
39.88
45.14
52.14
61.29
66.71
120.88
77.67
1
2
3
4
5
6
7
8
9
10
1
12
Story force, kips Story force, kips
Frame 1 Frame 2 Frame 3 (includes wall)
Figure 67 Story forces in the EW direction (1.0 kip = 4.45 kN).
(frame alone). It is the “straightening out” of the deflected shape of the structure that causes the story
shears in the frames without walls to be relatively equal.
A plot of the story shears in Frames 1, 2, and 3 is shown in Figure 68. The distribution of overturning
moments is shown in Figure 69 and indicates that the relatively stiff Frames 1 and 3 resist the largest
portion of the total overturning moment. The reversal of moment at the top of Frame 3 is a typical
response characteristic of framewall interaction.
6.4.4 Analysis of FrameOnly Structure for 25 Percent of Lateral Load
When designing a dual system, Provisions Sec. 5.2.2.1 [4.3.1.1] requires the frames (without walls) to
resist at least 25 percent of the total base shear. This provision ensures that the dual system has sufficient
redundancy to justify the increase from R = 6 for a special reinforced concrete structural wall to R = 8 for
a dual system (see Provisions Table 5.2.2 [4.31]). [Note that R = 7 per 2003 Provisions Table 4.31.]
The 25 percent analysis was carried out using the ETABS program with the mathematical model of the
building being identical to the previous version except that the panels of the structural wall were removed.
The boundary elements of the walls were retained in the model so that behavior of the interior frames
(Frames 3, 4, 5, and 6) would be analyzed in a rational way.
The results of the analysis are shown in Figures 610, 611, and 612. In these figures, the original
analysis (structural wall included) is shown by a solid line and the 25 percent (backup frame) analysis
(structural wall removed) is shown by a dashed line. As can be seen, the 25 percent rule controls only at
the lower level of the building.
Chapter 6, Reinforced Concrete
633
0
20
40
60
80
100
120
140
160
200 100 0 100 200 300 400 500 600
Shear, kips
Height, ft
Frame 1
Frame 2
Frame 3
Figure 68 Story shears in the EW direction (1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN).
0
20
40
60
80
100
120
140
160
10,000 0 10,000 20,000 30,000 40,000 50,000
Bending moment, ftkips
Height, ft
Frame 1
Frame 2
Frame 3
Figure 69 Story overturning moments in the EW direction (1.0 ft = 0.3048 m, 1.0 ftkip = 1.36 kNm).
FEMA 451, NEHRP Recommended Provisions: Design Examples
634
0
20
40
60
80
100
120
140
160
0 50 100 150 200 250 300 350
Shear, kips
Height, ft
25% V analysis for Frame 1
Frame 1
Figure 610 25 percent story shears, Frame 1 EW direction (1.0 ft = 0.3048
m, 1.0 kip = 4.45 kN).
Chapter 6, Reinforced Concrete
635
0
20
40
60
80
100
120
140
160
0 50 100 150 200 250 300 350
Shear, kips
Height, ft
25% V analysis for Frame 2
Frame 2
Figure 611 25 percent story shears, Frame 2 EW direction (1.0 ft = 0.3048
m, 1.0 kip = 4.45 kN).
FEMA 451, NEHRP Recommended Provisions: Design Examples
636
0
20
40
60
80
100
120
140
160
0 50 100 150 200 250 300 350
Shear, kips
Height, ft
25% V analysis for Frame 3
Frame 3 (without panels)
Figure 612 25 percent story shear, Frame 3 EW direction (1.0 ft = 0.3048 m,
1.0 kip = 4.45 kN)..
6.4.5 Design of Frame Members for the Berkeley Building
A sign convention for bending moments is required in flexural design. In this example, when the steel at
the top of a beam section is in tension, the moment is designated as a negative moment. When the steel at
the bottom is in tension, the moment is designated as a positive moment. All moment diagrams are drawn
using the reinforced concrete or tensionside convention. For beams, this means negative moments are
plotted on the top and positive moments are plotted on the bottom. For columns, moments are drawn on
the tension side of the member.
6.4.5.1 Initial Calculations
Before the quantity and placement of reinforcement is determined, it is useful to establish, in an overall
sense, how the reinforcement will be distributed. The preliminary design established that beams would
have a maximum depth of 32 in. and columns would be 30 in. by 30 in. In order to consider the
beamcolumn joints “confined” per ACI 318 Sec. 21.5, it was necessary to set the beam width to 22.5 in.,
which is 75 percent of the column width.
In order to determine the effective depth used for the design of the beams, it is necessary to estimate the
size and placement of the reinforcement that will be used. In establishing this depth, it is assumed that #8
bars will be used for longitudinal reinforcement and that hoops and stirrups will be constructed from #3
deformed bars. In all cases, clear cover of 1.5 in. is assumed. Since this structure has beams spanning in
Chapter 6, Reinforced Concrete
637
2'6"
1.5" cover
#8 bar
#3 hoop
Eastwest
spanning beam
2'8"
2'51
2"
2'41
2"
Northsouth
spanning beam
Figure 613 Layout for beam reinforcement (1.0 ft = 0.3048 m, 1.0 in = 25.4
mm).
two orthogonal directions, it is necessary to layer the flexural reinforcement as shown in Figure 613. The
reinforcement for the EW spanning beams was placed in the upper and lower layers because the strength
demand for these members is somewhat greater than that for the NS beams.
Given Figure 613, compute the effective depth for both positive and negative moment as:
Beams spanning in the EW direction, d = 32  1.5  0.375  1.00/2 = 29.6 in.
Beams spanning in the NS direction, d = 32  1.5  0.375  1.0  1.00/2 = 28.6 in.
For negative moment bending, the effective width is 22.5 in. for all beams. For positive moment, the slab
is in compression and the effective Tbeam width varies according to ACI 318 Sec. 8.10. The effective
widths for positive moment are as follows (with the parameter controlling effective width shown in
parentheses):
20ft beams in Frames 1 and 8 b = 22.5 + 20(12)/12 = 42.5 in. (span length)
Haunched beams b = 22.5 + 2[8(4)] = 86.5 in. (slab thickness)
30ft beams in Frames A, B, C, and D b = 22.5 + [6(4)] = 46.5 in. (slab thickness)
ACI 318 Sec. 21.3.2 controls the longitudinal reinforcement requirements for beams. The minimum
reinforcement to be provided at the top and bottom of any section is:
2
,
200 200(22.5)29.6 2.22 in.
60,000
w
s min
y
A b d
f
= = =
FEMA 451, NEHRP Recommended Provisions: Design Examples
638
This amount of reinforcement can be supplied by three #8 bars with As = 2.37 in.2 Since the three #8 bars
will be provided continuously top and bottom, reinforcement required for strength will include these #8
bars.
Before getting too far into member design, it is useful to check the required tension development length
for hooked bars since the required length may control the dimensions of the columns and the boundary
elements of the structural walls.
From Eq. 216 of ACI 318 Sec. 21.5.4.1, the required development length is:
65
y b
dh
c
f d
l
f
=
'
For NW concrete, the computed length should not be less than 6 in. or 8db. For LW concrete, the
minimum length is the larger of 1.25 times that given by ACI 318 Eq. 216, 7.5 in., or 10db. For fc' =
4,000 psi LW concrete, ACI 318 Eq. 216 controls for #3 through #11 bars.
For straight “top” bars, ld = 3.5ldh and for straight bottom bars, ld = 2.5ldh. These values are applicable
only when the bars are anchored in well confined concrete (e.g., column cores and plastic hinge regions
with confining reinforcement). The development length for the portion of the bar extending into
unconfined concrete must be increased by a factor of 1.6. Development length requirements for hooked
and straight bars are summarized in Table 613.
Where hooked bars are used, the hook must be 90 degrees and be located within the confined core of the
column or boundary element. For bars hooked into 30in.square columns with 1.5 in. of cover and #4
ties, the available development length is 30  1.50  0.5 = 28.0 in. With this amount of available length,
there will be no problem developing hooked bars in the columns. As required by ACI 318 Sec. 12.5,
hooked bars have a 12db extension beyond the bend. ACI 318 Sec. 7.2 requires that #3 through #8 bars
have a 6db bend diameter and #9 through #11 bars have a 8db diameter.
Table 613 is applicable to bars anchored in joint regions only. For development of bars outside of joint
regions, ACI 318 Chapter 12 should be used.
Table 613 Tension Development Length Requirements for Hooked Bars
and Straight Bars in 4,000 psi LW Concrete
Bar Size db (in.) ldh hook (in.) ld top (in.) ld bottom (in.)
#4
#5
#6
#7
#8
#9
#10
#11
0.500
0.625
0.750
0.875
1.000
1.128
1.270
1.410
9.1
11.4
13.7
16.0
18.2
20.6
23.2
25.7
31.9
39.9
48.0
56.0
63.7
72.1
81.2
90.0
22.8
28.5
34.3
40.0
45.5
51.5
58.0
64.2
1.0 in. = 25.4 mm.
Chapter 6, Reinforced Concrete
3See Chapter 1 of the 2nd Edition of the Handbook of Concrete Engineering edited by Mark Fintel (New York: Van Nostrand
Reinhold Company, 1984).
639
6.4.5.2 Design of Members of Frame 1 for EW Loading
For the design of the members of Frame 1, the equivalent lateral forces of Table 67b were applied at an
eccentricity of 10.5 ft together with 30 percent of the forces of Table 67a applied at an eccentricity of 5.0
ft. The eccentricities were applied in such a manner as to maximize torsional response and produce the
largest shears in Frame 1.
For this part of the example, the design and detailing of all five beams and one interior column of Level 5
are presented in varying amounts of detail. The beams are designed first because the flexural capacity of
the asdesigned beams is a factor in the design and detailing of the column and the beamcolumn joint.
The design of a corner column will be presented later.
Before continuing with the example, it should be mentioned that the design of ductile reinforced concrete
moment frame members is dominated by the flexural reinforcement in the beams. The percentage and
placement of beam flexural reinforcement governs the flexural rebar cutoff locations, the size and spacing
of beam shear reinforcement, the crosssectional characteristics of the column, the column flexural
reinforcement, and the column shear reinforcement. The beam reinforcement is critical because the basic
concept of ductile frame design is to force most of the energyabsorbing deformation to occur through
inelastic rotation in plastic hinges at the ends of the beams.
In carrying out the design calculations, three different flexural strengths were used for the beams. These
capacities were based on:
Design strength f = 0.8, tensile stress in reinforcement at 1.00 fy
Nominal strength f = 1.0, tensile stress in reinforcement at 1.00 fy
Probable strength f = 1.0, tensile stress in reinforcement at 1.25 fy
Various aspects of the design of the beams and other members depend on the above capacities as follows:
Beam rebar cutoffs Design strength
Beam shear reinforcement Probable strength of beam
Beamcolumn joint strength Probable strength of beam
Column flexural strength 6/5 × nominal strength of beam
Column shear strength Probable strength of column
In addition, beams in ductile frames will always have top and bottom longitudinal reinforcement
throughout their length. In computing flexural capacities, only the tension steel will be considered. This
is a valid design assumption because reinforcement ratios are quite low, yielding a depth to the neutral
axis similar to the depth of the compression reinforcement (d'/d is about 0.08, while the neutral axis depth
at ultimate ranges from 0.07 to 0.15 times the depth) .3
The preliminary design of the girders of Frame 1 was based on members with a depth of 32 in. and a
width of 22.5 in. The effective depth for positive and negative bending is 29.6 in. and the effective
widths for positive and negative bending are 42.5 and 22.5 in., respectively. This assumes the stress
block in compression is less than the 4.0inch flange thickness.
The layout of the geometry and gravity loading on the three easternmost spans of Level 5 of Frame 1 as
well as the unfactored gravity and seismic moments are illustrated in Figure 614. The seismic moments
are taken directly from the ETABS program output and the gravity moments were computed by hand
FEMA 451, NEHRP Recommended Provisions: Design Examples
640
4,515
4,515
4,708
4,635
4,457
3,988
492
786
562 492
715 715 715 715
492
152
173
242 221 221
152 152
221 221
5,232 5,225
951
4,122 3,453
834
5,834 5,761
4,222 4,149
834
5,641
4,028
5,641
4,028
1.42D + 0.5L + E
0.68D  E
1.2D+1.6L  midspan
(a)
Span layout
and loading
(b)
Earthquake moment
(in.kips)
(c)
Unfactored DL moment
(in.kips)
(d)
Unfactored LL moment
(in.kips)
(e)
Required strength
envelopes (in.kips)
17'6"
20'0" 20'0"
20'0"
'
Figure 614 Bending moments for Frame 1 (1.0 ft = 0.3048 m, 1.0 in.kip = 0.113 kNm).
using the coefficient method of ACI 318 Chapter 8. Note that all moments (except for midspan positive
moment) are given at the face of the column and that seismic moments are considerably greater than those
due to gravity.
Factored bending moment envelopes for all five spans are shown in Figure 614. Negative moment at the
supports is controlled by the 1.42D + 0.5L + 1.0E load combination, and positive moment at the support
is controlled by 0.68D  1.0E. Midspan positive moments are based on the load combination 1.2D + 1.6L.
The design process is illustrated below starting with Span BC.
6.4.5.2.1 Span BC
1. Design for Negative Moment at the Face of the Support
Mu = 1.42(715) + 0.5(221) + 1.0(4515) = 5,641 in.kips
Try two #9 bars in addition to the three #8 bars required for minimum steel:
As = 2(1.0) + 3(0.79) = 4.37 in.2
fc' = 4,000 psi
fy = 60 ksi
Chapter 6, Reinforced Concrete
641
Width b for negative moment = 22.5 in.
d = 29.6 in.
Depth of compression block, a = Asfy/.85fc'b
a = 4.37 (60)/[0.85 (4) 22.5] = 3.43 in.
Design strength, fMn = fAsfy(d  a/2)
fMn = 0.8(4.37)60(29.6  3.43/2) = 5,849 in.kips > 5,641 in.kips OK
2. Design for Positive Moment at Face of Support
Mu = [0.68(715)] + [1.0(4,515)] = 4,028 in.kips
Try two #7 bars in addition to the three #8 bars already provided as minimum steel:
As = [2(0.60)] + [3(0.79)] = 3.57 in. 2
Width b for positive moment = 42.5 in.
d = 29.6 in.
a = [3.57(60)]/[0.85(4)42.5] = 1.48 in.
fMn = 0.8(3.57) 60(29.6  1.48/2) = 4,945 in.kips > 4,028 in.kips OK
3. Positive Moment at Midspan
Mu = [1.2(492)] + [1.6(152)] = 833.6 in.kips
Minimum reinforcement (three #8 bars) controls by inspection. This positive moment reinforcement will
also work for Spans A'B and AA'.
6.4.5.2.2 Span A'B
1. Design for Negative Moment at the Face of Support A'
Mu = [1.42(715)] + [0.5(221)] + [1.0(4,708)] = 5,834 in.kips
Three #8 bars plus two #9 bars (capacity = 5,849 in.kips) will work as shown for Span BC.
2. Design for Negative Moment at the Face of Support B
Mu = [1.42(715)] + [0.5(221)] + [1.0(4,635)] = 5,761 in.kips
As before, use three #8 bars plus two #9 bars.
3. Design for Positive Moment at Face of Support A'
Mu = [0.68(715)] + [1.0(4708)] = 4,222 in.kips
Three #8 bars plus two #7 bars (capacity = 4,945 in.kips) works as shown for Span BC.
4. Design for Positive Moment at Face of Support B'
Mu = [0.68(715)] + [1.0(4,635)] = 4,149 in.kips
As before, use three #8 bars plus two #7 bars.
FEMA 451, NEHRP Recommended Provisions: Design Examples
642
6.4.5.2.3 Span AA'
1. Design for Negative Moment at the Face of Support A
Mu = [1.42(492)] + [0.5(152)] + [1.0(4,457)] = 5,232 in.kips
Try three #8 bars plus two #8 bars:
As = 5 × 0.79 = 3.95 in.2
Width b for negative moment = 22.5 in.
d = 29.6 in.
a = [3.95(60)/[0.85(4)22.5] = 3.10 in.
fMn =[0.8(3.95)60] (29.6  3.10/2) = 5,318 in.kips > 5,232 in.kips OK
2. Design for Negative Moment at the Face of Support A'
Mu = [1.42(786)] + [0.5(242)] + [1.0(3,988)] = 5,225 in.kips
Use three #8 bars plus two #9 bars as required for Support B of Span A'B.
3. Design for Positive Moment at Face of Support A
Mu = [0.68(492)] + [1.0(4,457)] = 4,122 in.kips
Three #8 bars plus two #7 bars will be sufficient.
4. Design for Positive Moment at Face of Support A'
Mu = [0.68(786)] + [1.0(3,988)] = 3,453 in.kips
As before, use three #8 bars plus two #7 bars.
6.4.5.2.4 Spans CC' and C'D
Reinforcement requirements for Spans CC' and C'D are mirror images of those computed for Spans
A'B and AA', respectively.
In addition to the computed strength requirements and minimum reinforcement ratios cited above, the
final layout of reinforcing steel also must satisfy the following from ACI 318 Sec. 21.3.2:
Minimum of two bars continuous top and bottom OK (three #8 bars continuous top and bottom)
Positive moment strength greater than OK (at all joints)
50 percent negative moment strength at a joint
Minimum strength along member greater OK (As provided = three #8 bars is more than
than 0.25 maximum strength 25 percent of reinforcement provided at joints)
The preliminary layout of reinforcement is shown in Figure 615. The arrangement of bars actually
provided is based on the above computations with the exception of Span BC where a total of six #8 top
bars were used instead of the three #8 bars plus two #9 bars combination. Similarly, six #8 bars are used
at the bottom of Span BC. The use of six #8 bars is somewhat awkward for placing steel, but it allows
Chapter 6, Reinforced Concrete
643
'
Note:
Drawing not to scale
'
(2) #8
(3) #8
(2) #9
(2) #7
(3) #8
(2) #7
(3) #8 (3) #8
(2) #8
(2) #9
(2) #7
20'0"
(2) #7
(3) #8
2'8"
2'6"
(typical)
Figure 615 Preliminary rebar layout for Frame 1 (1.0 ft = 03.048 m).
for the use of three #8 continuous top and bottom at all spans. An alternate choice would have been to
use two #9 continuous across the top of Span BC instead of the three of the #8 bars. However, the use of
two #9 bars (. = 0.00303) does not meet the minimum reinforcement requirement .min = 0.0033.
As mentioned above, later phases of the frame design will require computation of the design strength and
the maximum probable strength at each support. The results of these calculations are shown in Table
614.
Table 614 Design and Maximum Probable Flexural Strength For Beams in Frame 1
Item
Location
A A' B C C' D
Negative
Moment
Reinforcement five #8 three #8 +
two #9
six #8 six #8 three #8 +
two #9
five #8
Design Strength
(in.kips) 5,318 5,849 6,311 7,100 5,849 5,318
Probable Strength
(in.kips) 8,195 8,999 9,697 9,697 8,999 8,195
Positive
Moment
Reinforcement three #8 +
two #7
three #8 +
two #7
six #8 six #8 three #8 +
two #7
three #8 +
two #7
Design Strength
(in.kips) 4,945 4,945 6,510 6,510 4,945 4,945
Probable Strength
(in.kips) 7,677 7,677 10,085 10,085 7,655 7,677
1.0 in.kip = 0.113 kNm.
As an example of computation of probable strength, consider the case of six #8 top bars:
As = 6(0.79) = 4.74 in.2
Width b for negative moment = 22.5 in.
d = 29.6 in.
Depth of compression block, a = As(1.25fy)/0.85fc'b
a = 4.74(1.25)60/[0.85(4)22.5] = 4.65 in.
Mpr = 1.0As(1.25fy)(d  a/2)
FEMA 451, NEHRP Recommended Provisions: Design Examples
644
Mpr = 1.0(4.74)1.25(60)(29.6  4.65/2) = 9,697 in.kips
For the case of six #8 bottom bars:
As = 6(0.79) = 4.74 in.2
Width b for positive moment = 42.5 in.
d = 29.6 in.
a = 4.74(1.25)60/(0.85 × 4 × 42.5 ) = 2.46 in.
Mpr = 1.0(4.74)1.25(60)(29.6  2.46/2) = 10,085 in.kips
6.4.5.2.5 Adequacy of Flexural Reinforcement in Relation to the Design of the BeamColumn Joint
Prior to this point in the design process, the layout of reinforcement has been considered preliminary
because the quantity of reinforcement placed in the girders has a direct bearing on the magnitude of the
stresses developed in the beamcolumn joint. If the computed joint stresses are too high, the only
remedies are increasing the concrete strength, increasing the column area, changing the reinforcement
layout, or increasing the beam depth. The option of increasing concrete strength is not viable for this
example because it is already at the maximum (4,000 psi) allowed for LW concrete. If absolutely
necessary, however, NW concrete with a strength greater than 4,000 psi may be used for the columns and
beamcolumn joint region while the LW concrete is used for the joists and beams.
The design of the beamcolumn joint is based on the requirements of ACI 318 Sec. 21.5.3. The
determination of the forces in the joint of the column on Gridline C of Frame 1 is based on Figure 616a,
which shows how plastic moments are developed in the various spans for equivalent lateral forces acting
to the east. An isolated subassemblage from the frame is shown in Figure 616b. The beam shears shown
in Figure 616c are based on the probable moment strengths shown in Table 614.
For forces acting from west to east, compute the earthquake shear in Span BC:
VE = (Mpr
 + Mpr
+ )/lclear = (9,697 + 10,085)/(240  30) = 94.2 kips
For Span CC':
VE = (10,085 + 8,999)/(240  30) = 90.9 kips
With the earthquake shear of 94.2 and 90.9 kips being developed in the beams, the largest shear that
theoretically can be developed in the column above Level 5 is 150.5 kips. This is computed from
equilibrium as shown at the bottom of Figure 616:
94.2(9.83) + 90.9(10.50) = 2Vc(12.5/2)
Vc = 150.4 kips
With equal spans, gravity loads do not produce significant column shears, except at the end column,
where the seismic shear is much less. Therefore, gravity loads are not included in this computation.
The forces in the beam reinforcement for negative moment are based on six #8 bars at 1.25 fy:
T = C = 1.25(60)[(6(0.79)] = 355.5 kips
Chapter 6, Reinforced Concrete
645
'

+

+ +

+
 
9,697 8,999
10,085 10,085
150.4
150.4
90.9
94.2
(a)
Plastic
mechanism
(b)
Plastic
moments (in.
kips) in spans
BC and CC'
(c)
Girder and
column shears
(kips)
'
'
20'0" 20'0"
9'10" 10'6"
Figure 616 Diagram for computing column shears (1.0 ft =
0.3048 m, 1.0 kip = 4.45kN, 1.0 in.kip = 0.113 kNm).
For positive moment, six #8 bars also are used, assuming C = T, C = 355.5 kips.
As illustrated in Figure 617, the joint shear force Vj is computed as:
Vj = T + C  VE
= 355.5 + 355.5  150.4
= 560.6 kips
The joint shear stress is:
2 2
560.5 623 psi
30
j
j
c
V
v
d
= = =
FEMA 451, NEHRP Recommended Provisions: Design Examples
646
C = 355.5 kips
C = 355.5 kips T = 355.5 kips
T = 355.5 kips
Vc = 150.5 kips
V = 2(355.5)150.5
= 560.5 kips
g
Figure 617 Computing joint shear stress (1.0 kip = 4.45kN).
For joints confined on three faces or on two opposite faces, the allowable shear stress for LW concrete is
based on ACI 318 Sec. 21.5.3. Using f = 0.80 for joints (from ACI Appendix C) and a factor of 0.75 as a
modifier for LW concrete:
v j,allowable=0.80(0.75)(15 4,000)=569 psi
[Note that for joints, f = 0.85 per ACI 31802 Sec 9.3 as referenced by the 2003 Provisions. See Sec
6.4.2 for discussion.]
Since the actual joint stress (623 psi) exceeds the allowable stress (569 psi), the joint is overstressed. One
remedy to the situation would be to reduce the quantity of positive moment reinforcement. The six #8
bottom bars at Columns B and C could be reduced to three #8 bars plus two #7 bars. This would require a
somewhat different arrangement of bars than shown in Figure 615. It is left to the reader to verify that
the joint shear stress would be acceptable under these circumstances. Another remedy would be to
increase the size of the column. If the column is increased in size to 32 in. by 32 in., the new joint shear
stress is:
2 2
560.5 547psi < 569 psi
32
j
j
c
V
v
d
= = =
which is also acceptable. For now we will proceed with the larger column, but as discussed later, the final
solution will be to rearrange the bars to three #8 plus two #7.
Joint stresses would be checked for the other columns in a similar manner. Because the combined area of
top and bottom reinforcement used at Columns A, A', C', and D is less than that for Columns B and C,
these joints will not be overstressed.
Given that the joint stress is acceptable, ACI 318 Sec. 21.5.2.3 controls the amount of reinforcement
required in the joint. Since the joint is not confined on all four sides by a beam, the total amount of
transverse reinforcement required by ACI 318 Sec. 21.4.4 will be placed within the depth of the joint. As
shown later, this reinforcement consists of fourleg #4 hoops at 4 in. on center.
Chapter 6, Reinforced Concrete
647
Because the arrangement of steel is acceptable from a joint strength perspective, the cutoff locations of
the various bars may be determined (see Figure 615 for a schematic of the arrangement of
reinforcement). The three #8 bars (top and bottom) required for minimum reinforcement are supplied in
one length that runs continuously across the two end spans and are cut off in the center span. An
additional three #8 bars are placed top and bottom in the center span; these bars are cut off in Spans A'B
and CC'. At Supports A, A', C' and D, shorter bars are used to make up the additional reinforcement
required for strength.
To determine where bars should be cut off in each span, it is assumed that theoretical cutoff locations
correspond to the point where the continuous top and bottom bars develop their design flexural strength.
Cutoff locations are based on the members developing their design flexural capacities (fy = 60 ksi and f =
0.8). Using calculations similar to those above, it has been determined that the design flexural strength
supplied by a section with only three #8 bars is 3,311 in.kips for positive moment and 3,261 in.kips for
negative moment.
Sample cutoff calculations are given first for Span BC. To determine the cutoff location for negative
moment, it is assumed that the member is subjected to earthquake plus 0.68 times the dead load forces.
For positive moment cutoffs, the loading is taken as earthquake plus 1.42 times dead load plus 0.5 times
live load. Loading diagrams for determining cut off locations are shown in Figure 618.
For negative moment cutoff locations, refer to Figure 619a, which is a free body diagram of the west end
of the member. Since the goal is to develop a negative moment capacity of 3,261 in.kips in the
continuous #8 bars summing moments about Point A in Figure 619a:
6,311 0.121 2 73.7 3,261
2
+ x  x=
In the above equation, 6,311 (in.kips) is the negative moment capacity for the section with six #8 bars,
0.121 (kips/in.) is 0.68 times the uniform dead load, 73.3 kips is the end shear, and 3,261 in.kips is the
design strength of the section with three #8 bars. Solving the quadratic equation results in x = 42.9 in.
ACI 318 Sec. 12.10.3 requires an additional length equal to the effective depth of the member or 12 bar
diameters (whichever is larger). Consequently, the total length of the bar beyond the face of the support
is 42.9 + 29.6 = 72.5 in. and a 6 ft1 in. extension beyond the face of the column could be used.
For positive moment cutoff, see Figure 614 and Figure 619b. The free body diagram produces an
equilibrium equation as:
6,510 0.281 2 31.6 3,311
2
 x  x=
where the distance x is computed to be 75.7 in. Adding the 29.6 in. effective depth, the required
extension beyond the face of the support is 76.0 + 29.6 = 105.3 in, or 8 ft9 in. Note that this is exactly at
the midspan of the member.
FEMA 451, NEHRP Recommended Provisions: Design Examples
648
E
E +0.68D
E +1.42D+0.68L
X X+
WL = 0.66 klf
WD = 2.14 klf
Face of
support
17'6"
6,311
3,261
3,311
6,510
Bending moment
(in.kips)
Figure 618 Loading for determination of rebar cutoffs
(1.0 ft = 0.3048 m, 1.0 klf = 14.6 kN/m, 1.0 in.kip =
0.113 kNm).
6,311 in.kips
XA
0.68 WD = 0.121 kip/in.
73.7 kips
X+
A
31.6 kips
1.42D+0.5L = 0.281 kip/in. 6,510 in.kips
(b)
(a)
Figure 619 Free body diagrams (1.0 kip =
4.45kN, 1.0 klf = 14.6 kN/m, 1.0 in.kip =
0.113 kNm).
Clearly, the short bottom bars shown in Figure 615 are impractical. Instead, the bottom steel will be
rearranged to consist of three #8 plus two #7 bars continuous. Recall that this arrangement of
reinforcement will satisfy joint shear requirements, and the columns may remain at 30 in. by 30 in.
As shown in Figure 620, another requirement in setting cutoff length is that the bar being cut off must
have sufficient length to develop the strength required in the adjacent span. From Table 613, the
required development length of the #9 top bars in tension is 72.1 in. if the bar is anchored in a confined
joint region. The confined length in which the bar is developed is shown in Figure 620 and consists of
Chapter 6, Reinforced Concrete
649
dh
F
c
d = 2'8"
d = 2'6"
7'10"
6'1"
Confined
region
Must also check
for force F. Required
length = 3.5 l = 72.1"
#9 bar
Cut off length based
on moments in span
A'B
b '
Figure 620 Development length for top bars (1.0 ft = 0.3048 m, 1.0 in = 25.4 mm).
the column depth plus twice the depth of the girder. This length is 30 + 32 + 32 = 94 in., which is greater
than the 72.1 in. required. The column and girder are considered confined because of the presence of
closed hoop reinforcement as required by ACI 318 Sec. 21.3.3 and 21.4.4.
The bottom bars are spliced at the center of Spans A'B and CC' as shown in Figure 621. The splice
length is taken as the bottom bar Class B splice length for #8 bars. According to ACI 318 Sec. 12.15, the
splice length is 1.3 times the development length. From ACI 318 Sec. 12.2.2, the development length (ld)
is computed from:
'
3
40
d y
b c tr
b
l f
d f c K
d
aß..
=
. + .
. .
. .
using a = 1.0 (bottom bar), ß =1.0 (uncoated), . = 1.0 (#9 bar), . = 1.3 (LW concrete), taking c as the
cover (1.5 in.) plus the tie dimension (0.5 in.) plus 1/2 bar diameter (0.50 in.) = 2.50 in., and using
Ktr = 0, the development length for one #9 bar is:
3 60,000 1 1 1.0 1.3 (1.0) 37.0 in.
40 4,000 2.5 0
1.0
ld
. . × × ×
=... ... .. + .. =
. .
The splice length = 1.3 × 37.0 = 48.1 in. Therefore, use a 48in. contact splice. According to ACI 318
Sec. 21.3.2.3, the entire region of the splice must be confined by closed hoops spaced no closer than d/4
or 4 in.
The final bar placement and cutoff locations for all five spans are shown in Figure 621. Due to the
different arrangement of bottom steel, the strength at the supports must be recomputed. The results are
shown in Table 615.
FEMA 451, NEHRP Recommended Provisions: Design Examples
650
(3) #8
' '
(2) #9 (2) #8
(3) #8
5'0"
4'0"
(3) #8 +
(2) #7
Hoop spacing (from each end):
Typical spans AA', BC, C'D
(4) #3 leg 1 at 2", 19 at 5.5"
Typical spans A'B,CC',
(4) #3 leg 1 at 2",
15 at 5.5", 6 at 4"
Figure 621 Final bar arrangement (1.0 ft = 0.3048 m, 1.0 in = 25.4 mm).
Table 615 Design and Maximum Probable Flexural Strength For Beams in Frame 1 (Revised)
Item
Location
A A' B C C' D
Negative
Moment
Reinforcement five #8 three #8 +
two #9
six #8 six #8 three #8 +
two #9
five #8
Design Moment
(in.kips) 5,318 5,849 6,311 6,311 5,849 5,318
Probable moment
(in.kips) 8,195 8,999 9,696 9,696 8,999 8,195
Positive
Moment
Reinforcement three #8 +
two #7
three #8 +
two #7
three #8 +
two #7
three #8
+ two #7
three #8 +
two #7
three #8 +
two #7
Design Moment
(in.kips) 4,944 4,944 4,944 4,944 4,944 4,944
Probable moment
(in.kips) 7,677 7,677 7,677 7,677 7,677 7,677
1.0 in.kip = 0.113 kNm.
6.4.5.2.6 Transverse Reinforcement
Transverse reinforcement requirements are covered in ACI 318 Sec. 21.3.3 (minimum reinforcement) and
21.3.4 (shear strength).
To avoid nonductile shear failures, the shear strength demand is computed as the sum of the factored
gravity shear plus the maximum probable earthquake shear. The maximum probable earthquake shear is
based on the assumption that f = 1.0 and the flexural reinforcement reaches a tensile stress of 1.25fy. The
probable moment strength at each support is shown in Table 615.
Chapter 6, Reinforced Concrete
651
Figure 622 illustrates the development of the design shear strength envelopes for Spans AA', A'B, and
BC. In Figure 622a, the maximum probable earthquake moments are shown for seismic forces acting to
the east (solid lines) and to the west (dashed lines). The moments shown occur at the face of the supports.
The earthquake shears produced by the maximum probable moments are shown in Figure 622b. For
Span AB, the values shown in the figure are:
pr pr
E
clear
M M
V
l
 + +
=
where lclear = 17 ft6 in. = 210 in.
Note that the earthquake shears act in different directions depending on the direction of load.
For forces acting to the east, VE = (9696 + 7677) / 210 = 82.7 kips.
For forces acting to the west, VE = (8999 + 7677) / 210 = 79.4 kips.
FEMA 451, NEHRP Recommended Provisions: Design Examples
652
'
9,696 9,696
7,677 7,677 7,677 7,677
8,195 8,999
79.4 82.7 82.7
75.6 79.4 82.7
29.5
29.5 29.5
29.5
29.5
29.5
112.2
53.2
112.2
53.2
53.2
112.2
49.9
108.9
49.9
108.9
46.1
105.1
Loading
(a)
Seismic moment
(tension side)
in.kips
kips
positive
kips
positive
kips
positive
(d)
Design shear
seismic + gravity
240"
210"
15" 15"
(c)
Gravity shear
(1.42D + 0.5L)
(b)
Seismic shear
Figure 622 Shear forces for transverse reinforcement (1.0 in = 25.4 mm, 1.0 kip
= 4.45kN, 1.0 in.kip = 0.113 kNm).
Chapter 6, Reinforced Concrete
653
9.3" 54.4" 41.3"
f V s = 132.6 kips s = 5"
s = 6"
s = 7"
f V s = 110.5 kips
f V s = 94.7 kips
112.2 kips
53.2 kips
112.2 kips
53.2 kips
Figure 623 Detailed shear force envelope in Span BC (1.0 in =
25.4 mm, 1.0 kip = 4.45kN).
The gravity shears shown in Figure 622c are:
Factored gravity shear = VG = 1.42Vdead + 0.5Vlive
Vdead = 2.14 × 17.5/2 = 18.7 kips
Vlive = 0.66 × 17.5/2 = 5.8 kips
VG = 1.42(18.7) + 0.5(5.8) = 29.5 kips
Total design shears for each span are shown in Figure 622d. The strength envelope for Span BC is
shown in detail in Figure 623, which indicates that the maximum design shears is 82.7 + 29.5 = 112.2
kips. While this shear acts at one end, a shear of 82.7  29.5 = 53.2 kips acts at the opposite end of the
member.
In designing shear reinforcement, the shear strength can consist of contributions from concrete and from
steel hoops or stirrups. However, according to ACI 318 Sec. 21.3.4.2, the design shear strength of the
concrete must be taken as zero when the axial force is small (Pu/Agf !c
< 0.05) and the ratio VE/Vu is greater
than 0.5. From Figure 622, this ratio is VE/Vu = 82.7/112.2 = 0.73, so concrete shear strength must be
taken as zero. Using the ASCE 7 compatible f for shear = 0.75, the spacing of reinforcement required is
computed as described below. [Note that this is the basic strength reduction factor for shear per ACI 318
02 Sec 9.3. See Sec 6.4.2 for discussion.]
Compute the shear at d = 29.6 in. from the face of the support:
Vu = fVs = 112.2  (29.6/210)(112.2  53.2) = 103.9 kips
Vs = Avfyd/s
Assuming four #3 vertical legs (Av = 0.44 in.2), fv = 60 ksi and d = 29.6 in., compute the required spacing:
s = fAvfyd/Vu = 0.75[4(0.11)](60)(29.6/103.9) = 5.65 in., say 5.5 in.
At midspan, the design shear Vu = (112.2 + 53.2)/2 = 82.7 kips. Compute the required spacing:
s = 0.75[4(0.11)](60)(29.6/82.7) = 7.08 in., say 7.0 in.
Check maximum spacing per ACI 318 Sec. 21.3.3.2:
FEMA 451, NEHRP Recommended Provisions: Design Examples
654
d/4 = 29.6/4 = 7.4 in.
8db = 8(1.0) = 8.0 in.
24dh = 24(3/8) = 9.0 in.
The spacing must vary between 5.5 in. at the support and 7.0 in. at midspan. Due to the relatively flat
shear force gradient, a spacing of 5.5 in. will be used for the full length of the beam. The first hoop must
be placed 2 in. from the face of the support. This arrangement of hoops will be used for Spans AA', BC,
and C'D. In Spans A'B and CC', the bottom flexural reinforcement is spliced and hoops must be placed
over the splice region at d/4 or a maximum of 4 in. on center.
ACI 318 Sec. 21.3.3.1 states that closed hoops are required over a distance of twice the member depth
from the face of the support. From that point on, stirrups may be used. For the girders of Frame 1,
however, stirrups will not be used, and the hoops will be used along the entire member length. This is
being done because the earthquake shear is a large portion of the total shear, the girder is relatively short,
and the economic premium is negligible.
Where hoops are required (first 64 in. from face of support), longitudinal reinforcing bars should be
supported as specified in ACI 318 Sec. 7.10.5.3. Hoops should be arranged such that every corner and
alternate longitudinal bar is supported by a corner of the hoop assembly and no bar should be more than 6
in. clear from such a supported bar. Details of the transverse reinforcement layout for all spans of Level 5
of Frame 1 are shown in Figure 621.
6.4.5.3 Design of a Typical Interior Column of Frame 1
This section illustrates the design of a typical interior column on Gridline A'. The column, which
supports Level 5 of Frame 1, is 30 in. square and is constructed from 4,000 psi LW concrete, 60 ksi
longitudinal reinforcement, and 60 ksi transverse reinforcement. An isolated view of the column is
shown in Figure 624. The flexural reinforcement in the beams framing into the column is shown in
Figure 621. Using simple tributary area calculations (not shown), the column supports an unfactored
axial dead load of 528 kips and an unfactored axial live load of 54 kips. The ETABS analysis indicates
that the maximum axial earthquake force is 84 kips, tension or compression. The load combination used
to compute this force consists of full earthquake force in the EW direction, 30 percent of the NS force,
and accidental torsion. Because no beams frame into this column along Gridline A', the column bending
moment for NS forces can be neglected. Hence, the column is designed for axial force plus uniaxial
bending.
Chapter 6, Reinforced Concrete
655
See Figure 621
for girder
reinforcement
L
30"
'
Level 5
Level 4
20'0" 20'0"
32" 32"
P = 54 kips Includes
PD = 528 kips level 5
12'6"
Figure 624 Layout and loads on column of Frame A (1.0 ft = 0.3048 m,
1.0 in = 25.4 mm, 1.0 kip = 4.45kN).
6.4.5.3.1 Longitudinal Reinforcement
To determine the axial design loads, use the basic load combinations:
1.42D + 0.5L + 1.0E
0.68D  1.0E.
The combination that results in maximum compression is:
Pu = 1.42(528) + 0.5(54) + 1.0(84) = 861 kips (compression)
The combination for minimum compression (or tension) is:
Pu = 0.68(528)  1.0(84) = 275 kips (compression)
The maximum axial compression force of 861 kips is greater than 0.1fc'Ag = 0.1(4)(302) = 360 kips. Thus,
according to ACI 318 Sec. 21.4.2, the nominal column flexural strength must be at least 6/5 of the
nominal flexural strength of the beams framing into the column. Beam moments at the face of the support
are used for this computation. These capacities are provided in Table 615.
Nominal (negative) moment strength at end A' of Span AA' = 5,849/0.8 = 7,311 in.kips
Nominal (positive) moment strength at end A' of Span A' B = 4,945/0.8 = 6,181 in.kips
Average nominal moment framing into joint = 6,746 in.kips
Nominal column design moment = 6/5 × 6746 = 8,095 in.kips.
Knowing the factored axial load and the required design flexural strength, a column with adequate
capacity must be selected. Figure 625 gives design curves for 30 in. by 30 in. columns of 4,000 psi
concrete and reinforcement consisting of 12 #8, #9, or #10 bars. These curves, computed with a
FEMA 451, NEHRP Recommended Provisions: Design Examples
656
(12) #10
(12) #9
(12) #8
0
0
500 1,000 1,500 2,000
1,000
2,000
3,000
4,000
5,000
1,000
2,000
M u (ftkips)
Pu (kips)
Figure 625 Design interaction diagram for column on
Gridline A' (1.0 kip = 4.45kN, 1.0 ftkip = 1.36 kNm).
Microsoft Excel spreadsheet, are based on a f factor of 1.0 as required for nominal strength. At axial
forces of 275 kips and 861 kips, solid horizontal lines are drawn. The dots on the lines represent the
required nominal flexural strength (8095 in.kips) at each axial load level. These dots must lie to the left
of the curve representing the design columns. For both the minimum and maximum axial forces, a
column with 12 #8 bars (with As = 9.48 in.2 and 1.05 percent of steel) is clearly adequate.
6.4.5.3.2 Transverse Reinforcement
ACI 318 Sec. 21.4.4 gives the requirements for minimum transverse reinforcement. For rectangular
sections with hoops, ACI 318 Eq. 213 and 214 are applicable:
0.3 c c g 1
sh
yh ch
A sh f A
f A
. '.. .
= ... ..... ..
0.09 c
sh c
yh
f A shf
'
=
The first of these equations controls when Ag/Ach > 1.3. For the 30in.by30in. columns:
Ach = (30  1.5  1.5)2 = 729 in.2
Ag = 30 (30) = 900 in.2
Ag/Ach = 900/729 = 1.24
ACI 318 Eq. 214 therefore controls.
Chapter 6, Reinforced Concrete
657
For LW concrete, try hoops with four #4 legs and fc' = 4,000 psi:
hc = 30  1.5  1.5  0.25  0.25 = 26.5 in.
s = [4 (0.2) 60,000]/[0.09 (26.5) 4000] = 5.03 in.
However, the maximum spacing of transverse reinforcement is the lesser of onefourth the maximum
column dimension (30/4=7.5 in.), six bar diameters (6 × 1.0 = 6.0 in.), or the dimension sx where:
4 14
3
x
x
s h

= +
and where hx is the maximum horizontal spacing of hoops or cross ties. For the column with twelve #8
bars and #4 hoops and cross ties, hx = 8.833 in. and sx = 5.72 in. The 5.03in. spacing required by ACI Eq.
214 controls, so a spacing of 5 in. will be used. This transverse reinforcement must be spaced over a
distance lo = 30 in. at each end of the member and, according to ACI 318 Sec. 21.5.2, must extend
through the joint at (at most) the same spacing.
ACI 318 Sec. 21.4.4.6 requires a maximum spacing of transverse reinforcement in the region of the
column not covered by Sec. 21.4.4.4. The maximum spacing is the smaller of 6.0 in. or 6db, which for #8
bars is also 6 in. ACI 318 requires transverse steel at this spacing, but it does not specify what the details
of reinforcement should be. In this example, hoops and crossties with the same details as those placed in
the critical regions of the column are used.
6.4.5.3.3 Transverse Reinforcement Required for Shear
The amount of transverse reinforcement computed above is the minimum required. The column also
must be checked for shear with the column shears being based on the maximum probable moments in the
beams that frame into the column. The average probable moment is roughly 1.25/0.8 (f = 0.8) times the
average design moment = (1.25/0.8)(5397) = 8,433 in.kips. With a clear height of 118 in., the column
shear can be estimated at 8433/(0.5x118) = 143 kips. This shear will be compared to the capacity
provided by the 4leg #4 hoops spaced at 6 in. on center. If this capacity is well in excess of the demand,
the columns will be acceptable for shear.
For the design of column shear capacity, the concrete contribution to shear strength may be considered
because Pu > Agf !c
/20. Using a shear strength reduction factor of 0.85 for sandLW concrete (ACI 318
Sec. 11.2.1.2) in addition to the capacity reduction factor for shear, the design shear strength contributed
by concrete is:
fVc=f0.75fc'bcdc=0.75(0.85)( 4,000(30)(27.5)= 33.2 kips
fVs=fAvfyd/s=0.75(4)(0.2)(60)(27.5) / 6=165 kips
fVn=fVc+fVs=33.2+165=198.2 kips > 143 kips OK
The column with the minimum transverse steel is therefore adequate for shear. The final column detail
with both longitudinal and transverse reinforcement is given in Figure 626. The spacing of
reinforcement through the joint has been reduced to 4 in. on center. This is done for practical reasons
only. Column bar splices, where required, should be located in the center half of the of the column and
must be proportioned as (Class B) tension splices.
FEMA 451, NEHRP Recommended Provisions: Design Examples
658
'
Level 7
Level 6
30"
32" 32"
30"
30"
(12) #8 bars
#4 hoops
+ +
2" 7 at 4" 2" 6 at 5" 9 at 6" 6 at 5" 2"
Figure 626 Details of reinforcement for column (1.0 in = 25.4 mm).
6.4.5.4 Design of Haunched Girder
The design of a typical haunched girder of Level 5 of Frame 3 is now illustrated. This girder is of
variable depth with a maximum depth of 32 in. at the support and a minimum depth of 20 in. for the
middle half of the span. The length of the haunch at each end (as measured from the face of the support)
is 8 ft9 in. The width of the web of girder is 22.5 in. throughout.
Based on a tributary gravity load analysis, this girder supports an average of 3.375 kips/ft of dead load
and 0.90 kips/ft of reduced live load. For the purpose of estimating gravity moments, a separate analysis
of the girder was carried out using the SAP2000 program. End A of the girder was supported with
halfheight columns pinned at midstory and End B, which is supported by a shear wall, was modeled as
fixed. Each haunch was divided into four segments with nonprismatic section properties used for each
segment. The loading and geometry of the girder is shown in Figure 627a.
For determining earthquake forces, the entire structure was analyzed using the ETABS program. This
analysis included 100 percent of the earthquake forces in the EW direction and 30 percent of the
Chapter 6, Reinforced Concrete
659
Negative moment hinge
2,000
0
4,000
6,000
8,000
10,000
12,000
14,000
8,000
6,000
4,000
2,000
Range of possible positive
moment hinges
Level 5
4'0" 6'0" 6'0" 4'0"
(d) Potential plastic
hinge locations
Level 5
(c) Flexural
reinforcement
details
(7) #11
(5) #9
6,982 = fMn 4,102 = fMn
13,167 = fMn 6,824 = fMn
1.42D + 0.5L + 0.5E
1.2D + 1.6L
1.42D + 0.5L  0.5E
0.68D + 0.5E
0.68D  E
Strength envelope
Level 5
(a) Span geometry
and loading
(b) Moment envelope
(in.kips)
1'3" 8'9" 10'0" 10'0" 8'9" 1'3"
WL = 0.90 kips/ft
WD = 3.38 kips/ft
Figure 627 Design forces and detailing of haunched girder (1.0 ft = 0.3048 m, 1.0
k/ft = 14.6 kN/m, 1.0 in.kip = 0.113 kNm).
earthquake force in the NS direction, and accidental torsion. Each of these systems of lateral forces was
placed at a 5 percent eccentricity with the direction of the eccentricity set to produce the maximum
seismic shear in the member.
FEMA 451, NEHRP Recommended Provisions: Design Examples
660
6.4.5.4.1 Design and Detailing of Longitudinal Reinforcement
The results of the analysis for five different load combinations are shown in Figure 627b. Envelopes of
maximum positive and negative moment are shown on the figure indicate that 1.42D + 0.5L ± E controls
negative moment at the support, 0.68D ± E controls positive moment at the support, and 1.2D + 1.6L
controls positive moment at midspan. The maximum positive moment at the support is less than 50
percent of the maximum negative moment and the positive and negative moment at midspan is less than
25 percent of the maximum negative moment; therefore, the design for negative moment controls the
amount of reinforcement required at all sections per ACI 318 Sec. 21.3.2.2.
For a factored negative moment of 12,600 in.kips at Support B, try seven #11 bars, and assuming #3
hoops:
As = 7 × 1.54 = 10.92 in.2
d = 32 1.5  3/8  1.41/2 = 29.4 in.
. = 10.92/(29.4 × 22.5) = 0.0165 < 0.025, O.K.
b = 22.5 in.
Depth of compression block, a = [10.92 (60)]/[0.85 (4) 22.5] = 8.56 in.
Design strength, fMn = [0.8 (10.92) 60](29.4  8.56/2) = 13,167 in.kips > 12,600 in.kips OK
For positive moment at the support, try five #9 bars, which supplies about half the negative moment
reinforcement:
As = 5 (1.0) = 5.00 in.2
d = 32  1.5  3/8  1.128/2 = 29.6 in.
. = 5.00/(29.6 × 22.5) = 0.0075 > 0.033, O.K.
b = 86.5 in. (assuming stress block in flange)
a = [5.00 (60)]/(0.85 (4) 86.5] = 1.02 in.
fMn = [0.8 (5.00) 60] (29.6  1.02/2) = 6,982 in.kips.
This moment is larger than the design moment and, as required by ACI 318 Sec. 21.3.2.2, is greater than
50 percent of the negative moment capacity at the face of the support.
For positive moment at midspan the same five #9 bars used for positive moment at the support will be
tried:
As = 5 (1.0) = 5.00 in.2
d = 20  1.5  3/8  1.128/2 = 17.6 in.
. =5.00/(17.6 × 22.5) = 0.0126
b = 86.5 in.
a = [5.00 (60)]/[0.85 (4) 86.5] = 1.02 in.
fMn = [0.8 (5.00) 60] (17.6  1.02/2) = 4,102 in.kips > 3,282 in.kips. OK
The five #9 bottom bars are adequate for strength and satisfy ACI 318 Sec. 21.3.2.2, which requires that
the positive moment capacity be not less than 25 percent of the negative moment capacity at the face of
the support.
For negative moment in the 20ft span between the haunches, four #11 bars (. = 0.016) could be used at
the top. These bars provide a strength greater than 25 percent of the negative moment capacity at the
support. Using four bars across the top also eliminates the possibility that a negative moment hinge will
form at the end of the haunch (8 ft9 in. from the face of the support) when the 0.68D  E load
combination is applied. These four top bars are part of the negative moment reinforcement already sized
Chapter 6, Reinforced Concrete
661
for negative moment at the support. The other three bars extending from the support are not needed for
negative moment in the constant depth region and would be cut off approximately 6 ft beyond the
haunch; however, this detail results in a possible bar cutoff in a plastic hinge region (see below) that is not
desirable. Another alternative would be to extend all seven #11 bars across the top and thereby avoid the
bar cutoff in a possible plastic hinge region; however, seven #11 bars in 20in. deep portion of the girder
provide . = 0.028, which is a violation of ACI 21.3.2.1 (.max = 0.025). The violation is minor and will be
accepted in lieu of cutting off the bars in a potential plastic hinge region. Note that these bars provide a
negative design moment capacity of 6,824 in.kips in the constant depth region of the girder.
The layout of longitudinal reinforcement used for the haunched girder is shown in Figure 627c, and the
flexural strength envelope provided by the reinforcement is shown in Figure 627b. As noted in Table
613, the hooked #11 bars can be developed in the confined core of the columns. Finally, where seven
#11 top bars are used, the spacing between bars is approximately 1.4 in., which is greater than the
diameter of a #11 bar and is therefore acceptable. This spacing should accommodate the vertical column
reinforcement.
Under combined gravity and earthquake load, a negative moment plastic hinge will form at the support
and, based on the moment envelopes from the loading (Figure 627b), the corresponding positive moment
hinge will form in the constant depth portion of the girder. As discussed in the following sections, the
exact location of plastic hinges must be determined in order to design the transverse reinforcement.
6.4.5.4.2 Design and Detailing of Transverse Reinforcement
The design for shear of the haunched girder is complicated by its variable depth; therefore, a tabular
approach is taken for the calculations. Before the table may be set up, however, the maximum probable
strength must be determined for negative moment at the support and for positive moment in the constant
depth region,
For negative moment at the face of the support and using seven #11 bars:
As = 7 (1.56) = 10.92 in.2
d = 32  1.5  3/8  1.41/2 = 29.4 in.
b = 22.5 in.
a = [10.92 (1.25) 60]/[0.85 (4) 22.5] = 10.71 in.
Mpr = 1.0(10.92)(1.25)(60)(29.4  10.71/2) = 19,693 in.kips.
For positive moment in the constant depth region and using five #9 bars:
As = 5 (1.0) = 5.00 in.2
d = 20  1.5  3/8  1.128/2 = 17.6 in.
b = 86.5 in.
a = [5.00 (1.25) 60]/[0.85 (4) 86.5] = 1.28 in.
Mpr = [1.0 (5.00) 1.25 (60)] (17.6  1.28/2) = 6,360 in.kips
Before the earthquake shear may be determined, the location of the positive moment hinge that will form
in the constant depth portion of the girder must be identified. To do so, consider the freebody diagram of
Figure 628a. Summing moments (clockwise positive) about point B gives:
2
0
pr pr 2
M+ +M +Rxwx =
At the positive moment hinge the shear must be zero, thus R – wx = 0
FEMA 451, NEHRP Recommended Provisions: Design Examples
4The equation for the location of the plastic hinge is only applicable if the hinge forms in the constant depth region of the girder.
If the computed distance x is greater than 28 ft  9 in. (345 in.), the result is erroneous and a trial and error approach is required to
find the actual hinge location.
662
By combining the above equations:
2(Mpr Mpr )
x
w
+ + 
=
Using the above equation with Mpr as computed and w = 1.42(3.38) + 0.5(0.90) = 5.25 k/ft = 0.437 k/in.,
x = 345 in., which is located exactly at the point where the right haunch begins.4
The reaction is computed as R = 345 (0.437) = 150.8 kips.
The earthquake shear is computed as VE = R = wL/2 = 150.8(0.437)(450)/2 = 52.5 kips
This earthquake shear is smaller than would have been determined if the positive moment hinge had
formed at the face of support.
The earthquake shear is constant along the span but changes sign with the direction of the earthquake. In
Figure 628a, this shear is shown for the equivalent lateral seismic forces acting to the west. The factored
gravity load shear (1.42VD + 0.5VL) varies along the length of the span as shown in Figure 628b. At
Support A, the earthquake shear and factored gravity shear are additive, producing a design ultimate shear
of 150.8 kips. At midspan, the shear is equal to the earthquake shear acting alone and, at Support C, the
ultimate design shear is 45.8 kips. Earthquake, gravity, and combined shears are shown in Figures 628a
through 628c and are tabulated for the first half of the span in Table 616. For earthquake forces acting
to the east, the design shears are of the opposite sign of those shown in Figure 628.
According to ACI 318 Sec. 21.3.4.2, the contribution of concrete to member shear strength must be taken
as zero when VE/VU is greater than 0.5 and Pu/Agf !c
is less than 0.05. As shown in Table 616, the VE/VU
ratio is less than 0.5 within the first threefourths of the haunch length but is greater than 0.50 beyond this
point. In this example, it is assumed that if VE/VU is less than 0.5 at the support, the concrete strength can
be used along the entire length of the member.
The concrete contribution to the design shear strength is computed as:
fVc=f(0.85)2fc'bwd
where the ASCE 7 compatible f = 0.75 for shear, and the 0.85 term is the shear strength reduction factor
for sandLW concrete. [Note that this is the basic strength reduction factor for shear per ACI 31802 Sec
9.3. See Sec 6.4.2 for discussion.] The remaining shear, fVs = Vu  fVc, must be resisted by closed hoops
within a distance 2d from the face of the support and by stirrups with the larger of 6dh or 3.0 in. hook
extensions elsewhere. The 6dh or 3.0 in. “seismic hook” extension is required by ACI 318 Sec. 21.3.3.3.
Chapter 6, Reinforced Concrete
663
45.8 kips
(a) Location of
plastic hinge
+
PR
'
(b) Earthquake
shear (kips)
(c) Factored gravity
shear (kips)
pos
pos
(d) Earthquake +
factored gravity
shear (kips)
(e)
150.8 kips 98.3 kips
98.3 kips
52.5 kips
face of support
M
"B"
Provide two additional
hoops (detail B) at kink
A B
A B
6 at 6" 13 at 5" 30 at 4"
#3 hoops
6db
6db
R
W
x
Figure 628 Computing shear in haunched girder (1.0 in = 25.4 mm, 1.0 kip = 4.45kN).
FEMA 451, NEHRP Recommended Provisions: Design Examples
664
Table 616 Design of Shear Reinforcement for Haunched Girder
Item
Distance from Center of Support (in.)
Units
15 42.25 67.5 93.75 120 180 240
Ve 52.5 52.5 52.5 52.5 52.5 52.5 52.5
kips
1.42VD + 0.5VL 98.3 86.4 75.4 63.9 52.4 26.2 0.0
Vu 150.8 139.2 127.9 116.6 104.9 78.7 52.5
VE/VU 0.35 0.38 0.41 0.45 0.50 0.67 1.00
d 29.4 26.5 23.5 20.5 17.6 17.6 17.6 in.
fVC 53.3 48.1 42.6 37.2 0.0 0.0 0.0
kips
fVS 97.5 91.2 85.3 79.4 104.9 78.7 52.5
s 5.97 5.78 5.46 5.12 3.32 4.43 6.64
d/4 7.35 6.63 5.88 5.13 4.40 4.40 4.40 in.
Spacing #3 at 6 #3 at 5 #3 at 5 #3 at 5 #3 at 4 #3 at 4 #3 at 4
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
In Table 616, spacings are computed for four #3 vertical leg hoops or stirrups. As an example, consider
four #3 vertical legs at the section at the face of the support:
fVc = f(0.85)2 fc' db = 0.75(0.85)2(4000)0.529.4(22.5) = 53,300 lb = 53.3 kips
fVs = Vu  fVc = 150.8  53.3 = 97.5 kips
fVs = fAvfyd/s = 97.5 kips
s = [0.75(4)0.11(60)29.4]/97.5 = 5.97 in.
The maximum spacing allowed by ACI 318 is shown in Table 616. These spacings govern only in the
center portion of the beam. In the last line of the table, the hoop and stirrup spacing as actually used is
shown. This spacing, together with hoop and stirrup details, is illustrated in Figure 628d. The double
Ushaped stirrups (and cap ties) in the central portion of the beam work well with the #11 top bars and
with the #9 bottom bars.
6.4.5.4.3 Design of BeamColumn Joint
The design of the beamcolumn joint at Support A of the haunched girder is controlled by seismic forces
acting to the west, which produces negative moment at Support A. ACI 318 Sec. 21.5 provides
requirements for the proportioning and detailing of the joint.
A plastic mechanism of the beam is shown in Figure 629a. Plastic hinges have formed at the support
and at the location of the far haunch transition. With a total shear at the face of the support of 150.8 kips,
the moment at the centerline of the column may be estimated as
MCL = Mpr + 15(150.6) = 19,693 + 15(150.6) = 21,955 in.kips.
The total shear in the columns above and below the joint is estimated as 21,955/(150) = 146.3 kips.
Chapter 6, Reinforced Concrete
665
(a)
Plastic mechanism
 +
(b)
Plastic moment
(in.kips)
19,693
10,800
6,360
(c)
Column shears
(kips)
61.1 kips
146.3 kips
146.3 kips
288"
75" 75"
15" 450" 15"
Figure 629 Computation of column shears for use in joint
design (1.0 in = 25.4 mm, 1.0 kip = 4.45 kN).
The stresses in the joint are computed from equilibrium considering the reinforcement in the girder to be
stressed at 1.25fy. A detail of the joint is shown in Figure 630. Compute the joint shear Vj:
Force in the top reinforcement = 1.25Asfy = 1.25(7)1.56(60) = 819 kips
Joint shear = Vj = 819.0  146.3 = 672.7 kips
The joint shear stress vj = Vj/dc
2 = 672.7/[30 (30)] = 0.819 ksi
In the case being considered, all girders framing into the joint have a width equal to 0.75 times the
column dimension so confinement is provided on three faces of the joint. According to ACI 318 Sec.
21.5.3, the allowable joint shear stress = 0.75f(15)2pfc'. The 0.75 term is the strength reduction factor for
LW concrete. Compute the allowable joint shear stress:
vj,allowable = 0.75(0.80)15(4,000)0.5
= 569 psi = 0.569 ksi
FEMA 451, NEHRP Recommended Provisions: Design Examples
666
T = 819 kips
146 kips
V = 819146 = 673 kips
30"
J
C = 819 kips
Figure 630 Computing joint shear force (1.0 kip =
4.45kN).
This allowable stress is significantly less than the applied joint shear stress. There are several ways to
remedy the situation:
1. Increase the column size to approximately 35 × 35 (not recommended)
2. Increase the depth of the haunch so that the area of reinforcement is reduced to seven #10 bars. This
will reduce the joint shear stress to a value very close to the allowable stress.
2. Use 5000 psi NW concrete for the column. This eliminates the 0.75 reduction factor on allowable
joint stress, and raises the allowable stress to 848 psi.
For the remainder of this example, it is assumed that the lower story columns will be constructed from
5000 psi NW concrete.
Because this joint is confined on three faces, the reinforcement within the joint must consist of the same
amount and spacing of transverse reinforcement in the critical region of the column below the joint. This
reinforcement is detailed in the following section.
6.4.5.5 Design and Detailing of Typical Interior Column of Frame 3
The column supporting the west end of the haunched girder between Gridlines A and B is shown in
Figure 631. This column supports a total unfactored dead load of 804 kips and a total unfactored live
load of 78 kips. From the ETABS analysis, the axial force on the column from seismic forces is ±129
kips. The design axial force and bending moment in the column are based on one or more of the load
combinations presented below.
Earthquake forces acting to the west are:
Pu = 1.42(804) + 0.5(78) + 1.0(129)
= 1310 kips (compression)
Chapter 6, Reinforced Concrete
667
Level 5
Level 4
32" 32"
12'6"
20"
P L = 78.4 kips Includes
PD = 803.6 kips level 5
Figure 631 Column loading (1.0 ft = 0.3048 m,
1.0 in = 25.4 mm, 1.0 kip = 4.45kN).
This axial force is greater than 0.1fc'Ag = 360 kips; therefore, according to ACI 318 Sec. 21.4.2.1, the
column flexural strength must be at least 6/5 of the nominal strength (using f = 1.0 and 1.0 fy) of the beam
framing into the column. The nominal beam moment capacity at the face of the column is 16,458
in.kips. The column must be designed for sixfifths of this moment, or 19,750 inkips. Assuming a
midheight inflection point for the column above and below the beam, the column moment at the
centerline of the beam is 19,750/2 = 9,875 in.kips, and the column moment corrected to the face of the
beam is 7,768 in.kips.
Earthquake forces acting to the east are:
Pu = 0.68(804)  1.0(129) = 424 kips (compression)
This axial force is greater than 0.1fc'Ag = 360 kips. For this loading, the end of the beam supported by the
column is under positive moment, with the nominal beam moment at the face of the column being 8,715
in.kips. Because Pu > 0.1fc'Ag, the column must be designed for 6/5 of this moment, or 10,458 in.kips.
Assuming midheight inflection points in the column, the column moment at the centerline and the face of
the beam is 5,229 and 4,113 in.kips, respectively.
Axial force for gravity alone is:
Pu = 1.6(804) + 1.2(78) = 1,380 kips (compression)
FEMA 451, NEHRP Recommended Provisions: Design Examples
668
(12) #10
(12) #9
(12) #8
M u (ftkips)
0
0
500 1,000 1,500 2,000
1,000
2,000
3,000
4,000
5,000
1,000
2,000
Pu (kips)
Figure 632 Interaction diagram and column design forces
(1.0 kip = 4.45kN, 1.0 ftkip = 1.36 kNm).
This is approximately the same axial force as designed for earthquake forces to the west, but as can be
observed from Figure 625, the design moment is significantly less. Hence, this loading will not control.
6.4.5.5.1 Design of Longitudinal Reinforcement
Figure 632 shows an axial forcebending moment interaction diagram for a 30 in. by 30 in. column with
12 bars ranging in size from #8 to #10. A horizontal line is drawn at each of the axial load levels
computed above, and the required flexural capacity is shown by a solid dot on the appropriate line. The
column with twelve #8 bars provides more than enough strength for all loading combinations.
6.4.5.5.2 Design of Transverse Reinforcement
In Sec. 6.4.5.3, an interior column supporting Level 5 of Frame 1 was designed. This column has a shear
strength of 198.2 kips, which is significantly greater than the imposed seismic plus gravity shear of 146.3
kips. For details on the computation of the required transverse reinforcement for this column, see the
“Transverse Reinforcement” and “Transverse Reinforcement Required for Shear” subsections in Sec.
6.4.5.3. A detail of the reinforcement of the column supporting Level 5 of Frame 3 is shown in Figure 6
33. The section of the column through the beams shows that the reinforcement in the beamcolumn joint
region is relatively uncongested.
Chapter 6, Reinforced Concrete
669
Level 6
Level 7
30"
30"
30"
(12) #8 bars
#4 hoops
+ +
6 at 4" 4" 2" 7 at 4" 5" 8 at 6" 5" 7 at 4" 2"
Figure 633 Column detail (1.0 in = 25.4 mm).
6.4.5.6 Design of Structural Wall of Frame 3
The factored forces acting on the structural wall of Frame 3 are summarized in Table 617. The axial
compressive forces are based on a tributary area of 1,800 square ft for the entire wall, an unfactored dead
load of 160 psf, and an unfactored (reduced) live load of 20 psf. For the purposes of this example it is
assumed that these loads act at each level, including the roof. The total axial force for a typical floor is:
Pu = 1.42D + 0.5L = 1,800((1.42×0.16) + 0.50x0.02)) = 427 kips for maximum compression
Pu = 0.68D = 1,800(0.68×0.16) = 196 kips for minimum compression
The bending moments come from the ETABS analysis. Note the reversal in the moment sign due to the
effects of framewall interaction. Each moment contains two parts: the moment in the shear panel and the
couple resulting from axial forces in the boundary elements. For example, at the base of Level 2:
FEMA 451, NEHRP Recommended Provisions: Design Examples
670
ETABS panel moment =162,283 in.kips
ETABS column force = 461.5 kips
Total moment, Mu = 162,283 + 240(461.5) = 273,043 in.kips
The shears in Table 617 also consist of two parts, the shear in the panel and the shear in the column.
Using Level 2 as an example:
ETABS panel shear = 527 kips
ETABS column shear = 5.90 kips
Total shear, Vu = 527 + 2(5.90) = 539 kips
As with the moment, note the reversal in wall shear, not only at the top of the wall but also at Level 1
where the first floor slab acts as a support. If there is some inplane flexibility in the first floor slab, or if
some crushing were to occur adjacent to the wall, the shear reversal would be less significant, or might
even disappear. For this reason, the shear force of 539 kips at Level 2 will be used for the design of
Level 1 as well.
Recall from Sec. 6.2.2 that the structural wall boundary elements are 30 in. by 30 in. in size. The basic
philosophy of this design will be to use these elements as “special” boundary elements where a close
spacing of transverse reinforcement is used to provide extra confinement. This avoids the need for
confining reinforcement in the wall panel. Note, however, that there is no code restriction on extending
the special boundary elements into the panel of the wall.
It should also be noted that preliminary calculations (not shown) indicate that a 12in. thickness of the
wall panel is adequate for this structure. This is in lieu of the 18in. thickness assumed when computing
structural mass.
Table 617 Design Forces for Structural Wall
Supporting
Level
Axial Compressive Force Pu (kips) Moment Mu
(in.kips)
Shear Vu
1.42D + 0.5L 0.68D (kips)
R
12
11
10
987654321
427
854
1,281
1,708
2,135
2,562
2,989
3,416
3,843
4,270
4,697
5,124
5,550
196
392
588
783
979
1,175
1,371
1,567
1,763
1,958
2,154
2,350
2,546
30,054
39,725
49,954
51,838
45,929
33,817
17,847
45,444
78,419
117,975
165,073
273,043
268,187
145
4
62
118
163
203
240
274
308
348
390
539
376 (use 539)
1.0 kip = 4.45 kN, 1.0 in.kip = 0.113 kNm.
Chapter 6, Reinforced Concrete
671
6.4.5.6.1 Design of Panel Shear Reinforcement
First determine the required shear reinforcement in the panel and then design the wall for combined
bending and axial force. The nominal shear strength of the wall is given by ACI 318 Eq. 217:
Vn=Acv(ac fc'+.nfy)
where ac = 2.0 because hw/lw = 155.5/22.5 = 6.91 > 2.0. Note that the length of the wall was taken as the
length between boundary element centerlines (20 ft) plus onehalf the boundary element length (2.5 ft) at
each end of the wall.
Using fc' = 4000 psi, fy = 40 ksi, Acv = (270)(12) = 3240 in.2, and taking f for shear = 0.55, the ratio of
horizontal reinforcement is computed:
Vu = fVn
539.000 (0.85 2 4,000)3,240
0.55 0.0049
. n 3,240(40,000)
.. .. ×
=. . =
Note that the factor of 0.85 on concrete strength accounts for the use of LW concrete. Reinforcement
ratios for the other stories are given in Table 618. This table gives requirements using fc' = 4,000 psi, as
well as 6,000 psi NW concrete. As shown later, the higher strength NW concrete is required to manage
the size of the boundary elements of the wall. Also shown in the table is the required spacing of
horizontal reinforcement assuming that two curtains of #4 bars will be used. If the required steel ratio is
less than 0.0025, a ratio of 0.0025 is used to determine bar spacing.
Table 618 Design of Structural Wall for Shear
Level fc' = 4,000 psi (lightweight) fc' = 6,000 psi (normal weight)
Reinforcement
ratio
Spacing1
(in.)
Reinforcement
ratio
Spacing *
(in.)
R
12
11
10
9
8
7
6
5
4
3
2
1
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00278
0.00487
0.00487
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
12.00 (6.0)
6.84 (6.0)
6.84 (6.0)
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00250
0.00369
0.00369
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (12.0)
13.33 (9.0)
9.03 (9.0)
9.03 (9.0)
* Values in parentheses are actual spacing used.
1.0 in. = 25.4 mm.
For LW concrete, the required spacing is 6.84 in. at Levels 1 and 2. Minimum reinforcement
requirements control all other levels. For the final design, it is recommended to use a 6in. spacing at
FEMA 451, NEHRP Recommended Provisions: Design Examples
672
Levels 1, 2, and 3 and a 12in. spacing at all levels above. The 6in. spacing is extended one level higher
that required because it is anticipated that an axialflexural plastic hinge could propagate this far.
For the NW concrete, the required spacing is 9.03 in. at Levels 1 and 2 and minimum reinforcement
requirements control elsewhere. For the final design, a 9in. spacing would be used at Levels 1, 2, and 3
with a 12in. spacing at the remaining levels.
ACI 318 Sec. 21.6.4.3 [21.7.4.3] requires the vertical steel ratio to be greater than or equal to the
horizontal steel ratio if hwl/lw is less than 2.0. As this is not the case for this wall, the minimum vertical
reinforcement ratio of 0.0025 is appropriate. Vertical steel consisting of two curtains of #4 bars at 12 in.
on center provides a reinforcement ratio of 0.0028, which ill be used at all levels.
6.4.5.6.2 Design for Flexure and Axial Force
The primary consideration in the axialflexural design of the wall is determining whether or not special
boundary elements are required. ACI 318 provides two methods for this. The first approach, specified in
ACI 318 Sec. 21.6.6.2 [21.7.6.2], uses a displacement based procedure. The second approach, described
in ACI 318 Sec. 21.6.6.3 [21.7.6.3], is somewhat easier to implement but, due to its empirical nature, is
generally more conservative. In the following presentation, only the displacement based method will be
used for the design of the wall.
Using the displacement based approach, boundary elements are required if the length of the compression
block, c, satisfies ACI 318 Eq. 218:
600( )
w
u w
c l
d h
=
where du is the total elastic plus inelastic deflection at the top of the wall. From Table 69b, the total
elastic roof displacement is 4.36 in., and the inelastic drift is Cd times the elastic drift, or 6.5(4.36) = 28.4
in. or 2.37 feet. Recall that this drift is based on cracked section properties assuming Icracked = 0.5 Igross and
assuming that flexure dominates. Using this value together with lw = 22.5 ft, and hw = 155.5 ft:
22.5 2.46 ft = 29.52 in.
600( ) 600(2.37 155.5)
w
u w
l
d h
= =
To determine if c is greater than this value, a strain compatibility analysis must be performed for the wall.
In this analysis, it is assumed that the concrete reaches a maximum compressive strain of 0.003 and the
wall reinforcement is elasticperfectly plastic and yields at the nominal value. A rectangular stress block
was used for concrete in compression, and concrete in tension was neglected. A straight line strain
distribution was assumed (as allowed by ACI 318 Sec. 21.6.5.1 [21.7.5.1]). Using this straight line
distribution, the extreme fiber compressive strain was held constant at 0.003, and the distance c was
varied from 100,000 in. (pure compression) to 1 in. (virtually pure tension). For each value of c, a total
cross sectional nominal axial force (Pn) and nominal bending moment (Mn) were computed. Using these
values, a plot of the axial force (Pn) versus neutral axis location (c) was produced. A design value axial
forcebending moment interaction diagram was also produced.
The analysis was performed using an Excel spreadsheet. The concrete was divided into 270 layers, each
with a thickness of 1 in. The exact location of the reinforcement was used. When the reinforcement was
in compression, an adjustment was made to account for reinforcement and concrete sharing the same
physical volume.
Chapter 6, Reinforced Concrete
673
4,000
2,000
0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
16,000
18,000
0 100,000 200,000 300,000 400,000 500,000 600,000 700,000 800,000
Bending moment, in.kips
Axial force, kips
1.42D + 0.5L
0.68 D
6 ksi NW
4 ksi LW
Figure 634 Interaction diagram for structural wall (1.0 kip = 4.45kN, 1.0 in.kip = 0.113 kNm).
Two different sections were analyzed: one with fc' = 4,000 psi (LW concrete) and the other with fc' =
6,000 psi (NW concrete). In each case, the boundary elements were assumed to be 30 in. by 30 in. and
the panel was assumed to be 12 in. thick. Each analysis also assumed that the reinforcement in the
boundary element consisted of twelve #9 bars, producing a reinforcement ratio in the boundary element
of 1.33 percent. Panel reinforcement consisted of two curtains of #4 bars spaced at approximately 12 in.
on center. For this wall the main boundary reinforcement has a yield strength of 60 ksi, and the vertical
panel steel yields at 40 ksi.
The results of the analysis are shown in Figures 634 and 635. The first of these figures is the nominal
interaction diagram multiplied by f = 0.65 for tied sections. Also plotted in the figure are the factored
PM combinations from Table 617. The section is clearly adequate for both 4,000 psi and 6,000 psi
concrete because the interaction curve fully envelopes the design values.
FEMA 451, NEHRP Recommended Provisions: Design Examples
674
3,000
1,000
1,000
3,000
5,000
7,000
9,000
11,000
13,000
15,000
0 50 100 150 200 250
Neutral axis location, in.
Factored axial force, kips
6 ksi NW
4 ksi LW
0
Figure 635 Variation of neutral axis depth with compressive force (1.0 in = 25.4 mm, 1.0 kip = 4.45kN).
Figure 635 shows the variation in neutral axis depth with axial force. For a factored axial force of 5,550
kips, the distance c is approximately 58 in. for the 6,000 psi NW concrete and c is in excess of 110 in. for
the 4,000 psi LW concrete. As both are greater than 29.52 in., special boundary elements are clearly
required for the wall.
According to ACI 318 Sec. 21.6.6.4 [21.7.6.4], the special boundary elements must have a plan length of
c  0.1lw, or 0.5c, whichever is greater. For the 4,000 psi concrete, the first of these values is 110 
0.1(270) = 83 in., and the second is 0.5(110) = 55 in. Both of these are significantly greater than the 30
in. assumed in the analysis. Hence, the 30in. boundary element is not adequate for the lower levels of
the wall if fc' = 4,000 psi. For the 6,000 psi concrete, the required length of the boundary element is
580.1(270) = 31 in., or 0.5(58) = 29 in. The required value of 31 in. is only marginally greater than the
30 in. provided and will be deemed acceptable for the purpose of this example.
The vertical extent of the special boundary elements must not be less than the larger of lw or Mu/4Vu. The
wall length lw = 22.5 ft and, of the wall at Level 1, Mu/4Vu = 273,043/4(539) =126.6 in., or 10.6 ft. 22.5 ft
controls and will be taken as the required length of the boundary element above the first floor. The
special boundary elements will begin at the basement level, and continue up for the portion of the wall
supporting Levels 2 and 3. Above that level, boundary elements will still be present, but they will not be
reinforced as special boundary elements.
Another consideration for the boundary elements is at what elevation the concrete may change from 6,000
psi NW to 4,000 LW concrete. Using the requirement that boundary elements have a maximum plan
dimension of 30 in., the neutral axis depth (c) must not exceed approximately 57 in. As may be seen from
Figure 635, this will occur when the factored axial force in the wall falls below 3,000 kips. From Table
617, this will occur between Levels 6 and 7. Hence, 6,000 psi concrete will be continued up through
Level 7. Above Level 7, 4,000 psi LW concrete may be used.
Chapter 6, Reinforced Concrete
675
(12) #9
#5
at 5" o.c. Alternate
location of 90° bend
4"
4"
#5 at 5" o.c.
4"
#5 x
developed in wall
Figure 636 Details of structural wall boundary element (1.0 in = 25.4 mm).
Where special boundary elements are required, transverse reinforcement must conform to ACI 318 Sec.
21.6.6.4(c) [21.7.6.4(c)], which refers to Sec. 21.4.4.1 through 21.4.4.3. If rectangular hoops are used,
the transverse reinforcement must satisfy ACI 318 Eq. 214:
0.09 c
sh c
yh
f A shf
'
=
If #5 hoops are used in association with two crossties in each direction, Ash = 4(0.31) = 1.24 in.2, and hc =
30  2(1.5)  0.525 = 26.37 in. With fc' = 6 ksi and fyh = 60 ksi:
0.09(12.62.437) 6 5.22
60
s= =
If 4,000 psi concrete is used, the required spacing increases to 7.83 in.
Maximum spacing is the lesser of h/4, 6db, or sx where sx = 4 + (14hx)/3. With hx = 8.83 in., the third of
these spacings controls at 5.72 in. The 5.22in. spacing required by ACI 318 Eq. 214 is less than this, so
a spacing of 5 in. on center will be used wherever the special boundary elements are required.
Details of the panel and boundary element reinforcement are shown in Figures 636 and 637,
respectively. The vertical reinforcement in the boundary elements will be spliced as required using Type
2 mechanical splices at all locations. According to Table 613 (prepared for 4,000 psi LW concrete),
there should be no difficulty in developing the horizontal panel steel into the 30in.by30in. boundary
elements.
FEMA 451, NEHRP Recommended Provisions: Design Examples
676
#5 at 4"
(24) #11
#4 at 12" EF
#4 at 6" EF
Class
B
Class
B
#5 at 4"
#4 at 12" EF
#4 at 6" EF
#5 at 4"
(24) #11
#4 at 12"
See figure EWEF
626
#4 at 12"
EWEF
#4 at 4"
(24) #9
#4 at 12"
EWEF
#4 at 4"
#4 at 12"
EWEF
#4 at 4"
(24) #10
#4 at 12"
EWEF
#4 at 4"
#4 at 12"
EWEF
Class
B
Class
B
See figure
626
(12) #9
#4 at 12" EF
#4 at 12" EF
See figure
626
(12) #9
#4 at 12" EF
#4 at 12" EF
See figure
626
#4 at 12" EF
#4 at 12" EF
See figure
626
#4 at 12" EF
#4 at 12" EF
See figure
626
#4 at 12" EF
#4 at 12" EF
f'c = 4.0ksi
(LW)
f'c =
6.0ksi
(NW)
f'c = 4.0ksi
(LW)
Figure 637 Overall details of structural wall (1.0 in = 25.4 mm).
ACI 318 Sec. 21.6.6.4(d) [21.7.6.4(d)] also requires that the boundary element transverse reinforcement
be extended into the foundation tie beam a distance equal to the tension development length of the #9 bars
used as longitudinal reinforcement in the boundary elements. Assuming the tie beam consists of 6,000 psi
NW concrete, the development length for the #9 bar is 2.5 times the value given by ACI 318 Eq. 216:
2.5 2.5 60,000(1.128) 33.6 in.
65 65 6,000
y b
d
c
f d
l
f
. .
= . .= =
.. ' ..
Hence, the transverse boundary element reinforcement consisting of #5 hoops with two crossties in each
direction, spaced at 5 in. on center, will extend approximately 3 ft into the foundation tie beam.
6.5 STRUCTURAL DESIGN OF THE HONOLULU BUILDING
The structure illustrated in Figure 61 and 62 is now designed and detailed for the Honolulu building.
Because of the relatively moderate level of seismicity, the lateral load resisting system will consist of a
series of intermediate momentresisting frames in both the EW and NS directions. This is permitted for
Seismic Design Category C buildings under Provisions Sec. 9.6 [9.4]. Design guidelines for the
reinforced concrete framing members are provided in ACI 318 Sec. 21.10 [21.12].
Chapter 6, Reinforced Concrete
677
Preliminary design for the Honolulu building indicated that the size of the perimeter frame girders could
be reduced to 30 in. deep by 20 in. wide (the Berkeley building has girders that are 32 in. deep by 22.5 in.
wide) and that the columns could be decreased to 28 in. square (the Berkeley building uses 30in.by30
in. columns). The haunched girders along Frames 2 through 7 have a maximum depth of 30 in. and a
width of 20 in. in the Honolulu building (the Berkeley building had haunches with a maximum depth of
32 in. and a width of 22.5 in.). The Frame 2 through Frame 7 girders in Bays BC have a constant depth
of 30 in. Using these reduced properties, the computed drifts will be increased over those shown in
Figure 66, but will clearly not exceed the drift limits.
6.5.1 Material Properties
ACI 318 has no specific limitations for materials used in structures designed for moderate seismic risk.
For the Honolulu building, 4,000 psi sandLW concrete is used with ASTM A615 Grade 60 rebar for
longitudinal reinforcement and Grade 60 or Grade 40 rebar for transverse reinforcement.
6.5.2 Combination of Load Effects
For the design of the Honolulu building, all masses and superimposed gravity loads generated for the
Berkeley building are used. This is conservative because the members for the Honolulu building are
slightly smaller than the corresponding members for the Berkeley building. Also, the Honolulu building
does not have reinforced concrete walls on Gridlines 3, 4, 5, and 6 (these walls are replaced by infilled,
nonstructural masonry designed with gaps to accommodate frame drifts in the Honolulu building).
Provisions Sec. 5.2.7 [4.2.2] and Eq. 5.2.71 and 5.2.72 [4.21 and 4.22] require a combination of load
effects to be developed on the basis of ASCE 7, except that the earthquake load (E) is defined as:
E=.QE+0.2SDSD
when gravity and seismic load effects are additive and as:
E=.QE0.2SDSD
when the effects of seismic load counteract gravity.
For Seismic Design Category C buildings, Provisions Sec. 5.2.4.1 [4.3.3.1] permits the reliability factor
(.) to be taken as 1.0. The special load combinations of Provisions Eq. 5.2.71 and 5.2.72 [4.23 and
4.24] do not apply to the Honolulu building because there are no discontinuous elements supporting
stiffer elements above them. (See Provisions Sec. 9.6.2 [9.4.1].)
For the Honolulu structure, the basic ASCE 7 load combinations that must be considered are:
1.2D + 1.6L
1.2D + 0.5L ± 1.0E
0.9D ± 1.0E
The ASCE 7 load combination including only 1.4 times dead load will not control for any condition in
this building.
FEMA 451, NEHRP Recommended Provisions: Design Examples
678
Substituting E from the Provisions and with . taken as 1.0, the following load combinations must be used
for earthquake:
(1.2 + 0.2SDS)D + 0.5L + E
(1.2 + 0.2SDS)D + 0.5L  E
(0.9  0.2SDS)D + E
(0.9  0.2SDS)D E
Finally, substituting 0.472 for SDS (see Sec. 6.1.1), the following load combinations must be used for
earthquake:
1.30D + 0.5L + E
1.30D + 0.5L  E
0.80D + E
0.80D  E
Note that the coefficients on dead load have been slightly rounded to simplify subsequent calculations.
As EW wind loads apparently govern the design at the lower levels of the building (see Sec. 6.2.6 and
Figure 64), the following load combinations should also be considered:
1.2D + 0.5L + 1.6W
1.2D + 0.5L  1.6W
0.9D  1.6W
The wind load (W) from ASCE 7 includes a directionality factor of 0.85.
It is very important to note that use of the ASCE 7 load combinations in lieu of the combinations given in
ACI 318 Chapter 9 requires use of the alternate strength reduction factors given in ACI 318 Appendix C:
Flexure without axial load f = 0.80
Axial compression, using tied columns f = 0.65 (transitions to 0.8 at low axial loads)
Shear if shear strength is based on nominal axialflexural capacity f = 0.75
Shear if shear strength is not based on nominal axialflexural capacity f = 0.55
Shear in beamcolumn joints f = 0.80
[The strength reduction factors in ACI 31802 have been revised to be consistent with the ASCE 7 load
combinations. Thus, the factors that were in Appendix C of ACI 31899 are now in Chapter 9 of ACI
31802, with some modification. The strength reduction factors relevant to this example as contained in
ACI 31802 Sec. 9.3 are:
Flexure without axial load f = 0.9 (tensioncontrolled sections)
Axial compression, using tied columns f = 0.65 (transitions to 0.9 at low axial loads)
Shear if shear strength is based o nominal axialflexural capacity f = 0.75
Shear if shear strength is not based o nominal axialflexural capacity f = 0.60
Shear in beamcolumn joints f = 0.85]
6.5.3 Accidental Torsion and Orthogonal Loading (Seismic Versus Wind)
As has been discussed and as illustrated in Figure 64, wind forces appear to govern the strength
requirements of the structure at the lower floors, and seismic forces control at the upper floors. The
seismic and wind shears, however, are so close at the midlevels of the structure that a careful evaluation
Chapter 6, Reinforced Concrete
679
PW
L
PL
PW
Case 1
0.75 PL
Case 3
0.75 PL
0.75 PW
0.75 PW
Case 2 Case 4
0.75 PW
0.56 PW
0.75 PL
0.56 PW 0.56 PL
0.75 PW
L
0.56 PL
PW
0.75 PW 0.75 PL
0.75 PL
0.75 PW
0.75 P
PL
P
Figure 638 Wind loading requirements from ASCE 7.
must be made to determine which load governs for strength. This determination is complicated by the
differing (wind versus seismic) rules for applying accidental torsion and for considering orthogonal
loading effects.
Because the Honolulu building is in Seismic Design Category C and has no plan irregularities of Type 5
in Provisions Table 5.2.3.2 [4.32], orthogonal loading effects need not be considered per Provisions Sec.
5.2.5.2.2 [4.4.2.2]. However, as required by Provisions Sec. 5.4.4.2 [5.2.4.2], seismic story forces must
be applied at a 5 percent accidental eccentricity. Torsional amplification is not required per Provisions
Sec. 5.4.4.3 [5.2.4.3] because the building does not have a Type 1a or 1b torsional irregularity. (See Sec.
6.3.2 and 6.3.4 for supporting calculations and discussion.)
For wind, ASCE 7 requires that buildings over 60 ft in height be checked for four loading cases. The
required loads are shown in Figure 638, which is reproduced directly from Figure 69 of ASCE 7. In
Cases 1 and 2, load is applied separately in the two orthogonal directions. Case 2 may be seen to produce
torsional effects because 7/8 of the total force is applied at an eccentricity of 3.57% the building width.
This is relatively less severe than required for seismic effects, where 100 percent of the story force is
applied at a 5 percent eccentricity.
For wind, Load Cases 3 and 4 require that 75 percent of the wind pressures from the two orthogonal
directions be applied simultaneously. Case 4 is similar to Case 2 because of the torsion inducing pressure
unbalance. As mentioned earlier, the Honolulu building has no orthogonal seismic loading requirements.
FEMA 451, NEHRP Recommended Provisions: Design Examples
680
In this example, only loading in the EW direction is considered. Hence, the following lateral load
conditions were applied to the ETABS model:
100% EW Seismic applied at 5% eccentricity
ASCE 7 Wind Case 1 applied in EW direction only
ASCE 7 Wind Case 2 applied in EW direction only
ASCE 7 Wind Case 3
ASCE 7 Wind Case 4
All cases with torsion are applied in such a manner as to maximize the shears in the elements of Frame 1.
6.5.4 Design and Detailing of Members of Frame 1
In this section, the girders and a typical interior column of Level 5 of Frame 1 are designed and detailed.
For the five load cases indicated above, the girder shears produced from seismic effects control at the fifth
level, with the next largest forces coming from direct EW wind without torsion. This is shown
graphically in Figure 639, where the shears in the exterior bay of Frame 1 are plotted vs. story height.
Wind controls at the lower three stories and seismic controls for all other stories. This is somewhat
different from that shown in Figure 64, wherein the total story shears are plotted and where wind
controlled for the lower five stories. The basic difference between Figures 64 and 639 is that Figure 6
39 includes accidental torsion and, hence, Frame 1 sees a relatively larger seismic shear.
Chapter 6, Reinforced Concrete
681
0
20
40
60
80
100
120
140
160
0 5 10 15 20 25 30 35 40
Girder shear, kips
Height, ft
Seismic (with torsion)
Wind (without torsion)
Figure 639 Wind vs. seismic shears in exterior bay of Frame 1 (1.0 ft = 0.3048 m, 1.0 kip = 4.45kN).
6.5.4.1 Initial Calculations
The girders of Frame 1 are 30 in. deep and 20 in. wide. For positive moment bending, the effective width
of the compression flange is taken as 20 + 20(12)/12 = 40.0 in. Assuming 1.5 in. cover, #3 stirrups and
#8 longitudinal reinforcement, the effective depth for computing flexural and shear strength is 27.6 in.
6.5.4.2 Design of Flexural Members
ACI 318 Sec. 21.10.4 [21.12.4] gives the minimum requirements for longitudinal and transverse
reinforcement in the beams of intermediate moment frames. The requirements for longitudinal steel are
as follows:
1. The positive moment strength at the face of a joint shall be at least onethird of the negative moment
strength at the same joint.
2. Neither the positive nor the negative moment strength at any section along the length of the member
shall be less than onefifth of the maximum moment strength supplied at the face of either joint.
FEMA 451, NEHRP Recommended Provisions: Design Examples
682
2,835
2,835
2,886
2,852
2,796
2,492
502
802
573 502
729 729 729 729
502
155
176
247 225 225
155 155
225 225
3,526 3,658
969
2,394 1,850
850
3,946 3,927
2,302 2,269
850
3,910
2,252
3,910
2,252 1.3D+0.5L+E
0.8D  E
1.2D+1.6L
(a)
Span layout
and loading
(b)
Earthquake moment
(in.kips)
(c)
Unfactored DL moment
(in.kips)
(d)
Unfactored LL moment
(in.kips)
(e)
Required strength
envelopes (in.kips)
17.67'
20.0' 20.0' 20.0'
'
W = 0.66 kips/ft
W = 2.14 kips/ft
L
D
Figure 640 Bending moment envelopes at Level 5 of Frame 1 (1.0 ft = 0.3048 m, 1.0 kip/ft
= 14.6 kN/m, 1.0 in.kip = 0.113 kNm).
The second requirement has the effect of requiring top and bottom reinforcement along the full length of
the member. The minimum reinforcement ratio at any section is taken from ACI 318 Sec. 10.5.1 as 200/fy
or 0.0033 for fy = 60 ksi. However, according to ACI 318 Sec. 10.5.3, the minimum reinforcement
provided need not exceed 1.3 times the amount of reinforcement required for strength.
The gravity loads and design moments for the first three spans of Frame 1 are shown in Figure 640. The
seismic moments are taken directly from the ETABS analysis, and the gravity moments were computed
by hand using the ACI coefficients. All moments are given at the face of the support. The gravity
moments shown in Figures 640c and 640d are slightly larger than those shown for the Berkeley building
(Figure 614) because the clear span for the Honolulu building increases due to the reduction in column
size from 30 in. to 28 in.
Based on preliminary calculations, the reinforcement layout of Figure 641 will be checked. Note that the
steel clearly satisfies the detailing requirements of ACI 318 Sec. 21.10.4 [21.12.4].
Chapter 6, Reinforced Concrete
683
' '
(1) #7
(2) #8
20'0"
30"
28" 48"
#3x stirrups spaced from
each support: 1 at 2", 10 at 6",
5 at 8" (typical each span).
(3) #8
(2) #8
(3) #8
(2) #8
(3) #8
(2) #8
(1) #7 (1) #7
(2) #8
(1) #7
(3) #7 (3) #7
5'0"
(typical)
Figure 641 Preliminary reinforcement layout for Level 5 of Frame 1 (1.0 in = 25.4
mm, 1.0 ft = 0.3048 m).
6.5.4.2.1 Design for Negative Moment at Face of Support A
Mu = 1.3 (502)  0.5 (155)  1.0 (2,796) = 3,526 in.kips
Try three #7 short bars and two #8 long bars.
As = 3 (0.60) + 2 (0.79) = 3.38 in.2
. = 0.0061
Depth of compression block, a = [3.38 (60)]/[0.85 (4) 20] = 2.98 in.
Nominal moment capacity, Mn = Asfy(d  a/2) = [3.38 (60.0)] [27.6  2.98/2] = 5,295 in.kips
Design capacity, fMn = 0.8(5,295) = 4,236 in.kips > 3,526 in.kips OK
6.5.4.2.2 Design for Positive Moment at Face of Support A
Mu = 0.8 (502) + 1.0 (2,796) = 2,394 in.kips
Try three #8 long bars.
Asfy = 3 (0.79) = 2.37 in.2
. = 0.0043
a = 2.37 (60)/[0.85 (4) 40] = 1.05 in.
Mn = Asfy(d  a/2) = [2.37 (60.0)][27.6  1.05/2] = 3,850 in.kips
fMn = 0.8(3850) = 3,080 in.kips > 2,394 in.kips OK
This reinforcement also will work for positive moment at all other supports.
6.5.4.2.3 Design for Negative Moment at Face of Support A'
Mu = 1.3 (729)  0.5 (225)  1.0 (2,886) = 3,946 in.kips
Try four #8 long bars and one #7 short bar:
As = 4 (0.79) + 1 (0.6) = 3.76 in.2
. = 0.0068
FEMA 451, NEHRP Recommended Provisions: Design Examples
684
a = [3.76 (60)]/[0.85 (4) 20] = 3.32 in.
Mn = Asfy(d  a/2) = [3.76 (60.0)][27.6  3.32/2] = 5,852 in.kips
fMn = 0.8(5,852) = 4,681 in.kips > 3,946 in.kips OK
This reinforcement will also work for negative moment at Supports B and C. Therefore, the flexural
reinforcement layout shown in Figure 641 is adequate. The top short bars are cut off 5 ft0 in. from the
face of the support. The bottom bars are spliced in Spans A'B and CC' with a Class B lap length of 48
in. Unlike special moment frames, there are no requirements that the spliced region of the bars in
intermediate moment frames be confined by hoops over the length of the splice.
6.5.4.2.4 Design for Shear Force in Span A'B:
ACI 318 Sec. 21.10.3 [21.12.3] provides two choices for computing the shear strength demand in a
member of an intermediate moment frame:
1. The first option requires that the design shear force for earthquake be based on the nominal moment
strength at the ends of the members. Nominal moment strengths are computed with a flexural
reinforcement tensile strength of 1.0fy and a flexural f factor of 1.0. The earthquake shears computed
from the nominal flexural strength are added to the factored gravity shears to determine the total
design shear.
2. The second option requires that the design earthquake shear force be 2.0 times the factored
earthquake shear taken from the structural analysis. This shear is used in combination with the
factored gravity shears.
For this example, the first option is used. The nominal strengths at the ends of the beam were computed
earlier as 3850 in.kips for positive moment at Support A' and 5,852 in.kips for negative moment at
Support B. Compute the design earthquake shear VE:
5,852 3,850 45.8 kips
VE 212
+
= =
where 212 in. is the clear span of the member. For earthquake forces acting in the other direction, the
earthquake shear is 43.1 kips.
The gravity load shears at the face of the supports are:
2.14(20 2.33) 18.9 kips
VD 2

= =
0.66(20 2.33) 5.83 kips
VL 2

= =
The factored design shear Vu = 1.3(18.9) + 0.5(5.8) + 1.0(45.8) = 73.3 kips. This shear force applies for
earthquake forces coming from either direction as shown in the shear strength design envelope in Figure
642.
The design shear force is resisted by a concrete component (Vc) and a steel component (Vs). Note that the
concrete component may be used regardless of the ratio of earthquake shear to total shear. The required
design strength is:
Chapter 6, Reinforced Concrete
685
Vu # fVc + fVs
where f = 0.75 for shear.
(0.85) (2 4,000)20(27.6) 59.3 kips
Vc= 1,000 =
The factor of 0.85 above reflects the reduced shear capacity of sandLW concrete.
The shear to be resisted by steel, assuming stirrups consist of two #3 legs (Av = 0.22) and fy = 40 ksi is:
73.3 0.75(59.3) 38.4 kips
0.75
u c
s
V V V
f
f
 
= = =
Using VS = Av fyd/s:
(0.22)(40)(27.6) 6.32 in.
38.4
s= =
Minimum transverse steel requirements are given in ACI 318 Sec. 21.10.4.2 [21.12.4.2]. The first stirrup
should be placed 2 in. from the face of the support, and within a distance 2h from the face of the support,
the spacing should be not greater than d/4, eight times the smallest longitudinal bar diameter, 24 times the
stirrup diameter, or 12 in. For the beam under consideration d/4 controls minimum transverse steel, with
the maximum spacing being 27.6/4 = 6.9 in. This is slightly greater, however, than the 6.32 in. required
for strength. In the remainder of the span, stirrups should be placed at a maximum of d/2 (ACI 318 Sec.
21.10.4.3 [21.12.4.3]).
Because the earthquake shear (at midspan) is greater than 50 percent of the shear strength provided by
concrete alone, the minimum requirements of ACI 318 Sec. 11.5.5.3 must be checked:
0.2(40,000) 8.0 in.
smax= 50(20) =
This spacing controls over the d/2 requirement. The final spacing used for the beam is shown in Figure 6
41. This spacing is used for all other spans as well. The stirrups may be detailed according to ACI 318
Sec. 7.1.3, which requires a 90degree hook with a 6db extension. This is in contrast to the details of the
Berkeley building where full hoops with 135degree hooks are required in the critical region (within 2d
from the face of the support) and stirrups with 135degree hooks are required elsewhere.
FEMA 451, NEHRP Recommended Provisions: Design Examples
686
'
5,852 5,852
3,850 3,850 3,850 3,850
5,295 5,852
45.8 45.8 45.8
43.1 45.8 45.8
27.5
27.5 27.5
27.5
27.5
27.5
73.3
18.3
73.3
18.3
Loading
18.3
73.3
18.3
73.3 73.3
18.3
70.6
15.6
(a)
Seismic moment
(tension side)
in.kips
kips
positive
kips
positive
kips
positive
(b)
Seismic shear
(c)
Gravity shear
(1.175D + 1.0L)
(d)
Design shear
seismic + gravity
14" 212" 14"
240"
Figure 642 Shear strength envelopes for Span A'B of Frame 1 (1.0 in =
25.4 mm, 1.0 kip = 4.45kN, 1.0 in.kip = 0.113 kNm).
Chapter 6, Reinforced Concrete
687
'
20'0" 20'0"
12'6"
Level 4
Level 5
30" 30"
28"
See Figure 641
for girder
reinforcement
P = 54 kips Includes
P = 528 kips level 5
Figure 643 Isolated view of column A' (1.0 ft = 0.3048 m, 1.0 kip =
4.45kN).
6.5.4.3 Design of Typical Interior Column of Frame 1
This section illustrates the design of a typical interior column on Gridline A'. The column, which
supports Level 5 of Frame 1, is 28 in. square and is constructed from 4,000 psi LW concrete, 60 ksi
longitudinal reinforcement, and 40 ksi transverse reinforcement. An isolated view of the column is
shown in Figure 643.
The column supports an unfactored axial dead load of 528 kips and an unfactored axial live load of 54
kips. The ETABS analysis indicates that the axial earthquake force is ±33.2 kips, the earthquake shear
force is ±41.9 kips, and the earthquake moments at the top and the bottom of the column are ±2,137 and
±2,708 in.kips, respectively. Moments and shears due to gravity loads are assumed to be negligible.
6.5.4.3.1 Design of Longitudinal Reinforcement
The factored gravity force for maximum compression (without earthquake) is:
Pu = 1.2(528) + 1.6(54) = 720 kips
This force acts with no significant gravity moment.
The factored gravity force for maximum compression (including earthquake) is:
Pu = 1.3(528) + 0.5(54) + 33.2 = 746.6 kips
The factored gravity force for minimum compression (including earthquake) is:
Pu = 0.8(528)  33.2 = 389.2 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
688
Since the frame being designed is unbraced in both the NS and EW directions, slenderness effects
should be checked. For a 28in.by28in. column with a clear unbraced length. lu = 120 in., r = 0.3(28)
= 8.4 in. (ACI 318 Sec. 10.11.3) and lu/r = 120/8.4 = 14.3.
ACI 318 Sec. 10.11.4.2 states that the frame may be considered braced against sidesway if the story
stability factor is less than 0.05. This factor is given as:
u 0
u c
Q P
V l
= S d
which is basically the same as Provisions Eq. 5.4.6.21 [5.216] except that in the ACI equation, the
gravity forces are factored. [Note also that the equation to determine the stability coefficient has been
changed in the 2003 Provisions. The importance factor, I, has been added to 2003 Provisions Eq. 5.216.
However, this does not affect this example because I = 1.0.] ACI is silent on whether or not d0 should
include Cd. In this example, d0 does not include Cd, and is therefore consistent with the Provisions. As
can be seen from earlier calculations shown in Table 612b, the ACI story stability factor will be in excess
of 0.05 for Level 5 of the building responding in the EW direction. Hence, the structure must be
considered unbraced.
Even though the frame is defined as unbraced, ACI 318 Sec. 10.13.2 allows slenderness effects to be
neglected when klu/r < 22. This requires that the effective length factor k for this column be less than
1.54. For use with the nomograph for unbraced columns (ACI 318 Figure R10.12.1b):
(45,000) 187.5
Girder 240
EI E E
L
.. .. = =
. .
According to ACI 318 Sec. 10.12.3:
0.4
(1 )
150
Column
d
Column
EI
EI
L
ß
. .
.. .. =.. + ..
. .
Using the 1.2 and 1.6 load factors on gravity load:
1.2(528) 0.88
ß d= 720 =
3
28 (28) 51,221 in.4
IColumn= 12 =
0.4(51, 221 )
1 0.88 72.7
Column 150
E
EI E
L
.. .. = + =
. .
Because there is a column above and below as well as a beam on either side:
72.7 0.39
.Top =. Bottom=187.5 =
Chapter 6, Reinforced Concrete
5For loading in the NS direction, the column under consideration has no beam framing into it in the direction of loading. If the
stiffness contributed by the joists and the spandrel beam acting in torsion is ignored, the effective length factor for the column in
the NS direction is effectively infinity. However, this column is only one of four in a story containing a total of 36 columns.
Since each of the other 32 columns has a lateral stiffness well in excess of that required for story stability in the NS direction,
the columns on Lines A' and C' can be considered to be laterally supported by the other 32 columns and therefore can be
designed using an effective length factor of 1.0. A Pdelta analysis carried out per the ACI Commentary would be required to
substantiate this.
689
500
1,500
2,500
500
200 400 600 800 1,000
P (kips)
1,000
2,000
3,000
0
M x (ftkips)
0
Figure 644 Interaction diagram for column (1.0 kip =
4.45kN, 1.0 ftkip = 1.36 kNm).
and the effective length factor k = 1.15 (ACI 318 Figure R10.12.1b). As the computed effective length
factor is less than 1.54, slenderness effects need not be checked for this column.5
Continuing with the design, an axialflexural interaction diagram for a 28in.by28in. column with 12
#8 bars (. = 0.0121) is shown in Figure 644. The column clearly has the strength to support the applied
loads (represented as solid dots in the figure).
6.5.4.3.2 Design and Detailing of Transverse Reinforcement
ACI 318 Sec. 21.10.3 [21.12.3] allows the column to be checked for 2.0 times the factored shear force as
derived from the structural analysis. The ETABS analysis indicates that the shear force is 41.9 kips and
the design shear is 2.0(41.9) = 83.8 kips.
The concrete supplies a capacity of:
Vc=0.85(2)fc'bwd=0.85(2) 4,000(28)(25.6)= 77.1 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
690
The requirement for steel reinforcement is:
83.8 0.75(77.1) 34.6 kips
0.75
u c
s
V V V
f
f
 
= = =
Using ties with four #3 legs, s = [4(0.11)] [40.0 (25.6/34.6)] = 13.02 in.
ACI 318 Sec. 21.10.5 [21.12.5] specifies the minimum reinforcement required. Within a region lo from
the face of the support, the tie spacing should not exceed:
8.0db = 8.0 (1.008) = 8.00 in. (using #8 longitudinal bars)
24dtie = 24 (3/8) = 9.0 in. (using #3 ties)
1/2 the smallest dimension of the frame member = 28/2 = 14 in.
12 in.
The 8.0 in. maximum spacing controls. Ties at this spacing are required over a length lo of:
1/6 clearspan of column = 120/6 = 20 in.
maximum cross section dimension = 28 in.
18.0 in.
Given the above, a fourlegged #3 tie spaced at 8 in. over a depth of 28 in. will be used. One tie will be
provided at 4 in. below the beam soffit, the next tie is placed 4 in. above the floor slab, and the remaining
ties are spaced at 8 in. on center. The final spacing is as shown in Figure 645. Note that the tie spacing
is not varied beyond lo.
Chapter 6, Reinforced Concrete
691
'
Level 7
Level 6
28"
30" 30"
28"
28"
(12) #8 bars
4" 14 spaces at 8" o.c. 4"
3" 3 at 8" 3"
Figure 645 Column reinforcement (1.0 in = 25.4 mm).
6.5.4.4 Design of BeamColumn Joint
Joint reinforcement for intermediate moment frames is addressed in ACI 318 Sec. 21.10.5.3 [21.12.5.5],
which refers to Sec. 11.11.2. ACI 318 Sec. 11.11.2 requires that all beamcolumn connections have a
minimum amount of transverse reinforcement through the beamcolumn joints. The only exception is in
nonseismic frames where the column is confined on all four sides by beams framing into the column. The
amount of reinforcement required is given by ACI 318 Eq. 1113:
50 w
v
y
A b s
f
. .
= .. ..
. .
This is the same equation used to proportion minimum transverse reinforcement in beams. Assuming Av
is supplied by four #3 ties and fy = 40 ksi:
4(0.11)(40,000) 12.6 in.
50(28)
s= =
FEMA 451, NEHRP Recommended Provisions: Design Examples
692
1.3D + 0.5L + E
1.2D + 1.6L
1.3D + 0.5L  E
0.8D + E
0.8D  E
Strength envelope
(b) Moment envelope
(in.kips)
L
D
2,000
0
4,000
6,000
8,000
10,000
12,000
8,000
6,000
4,000
2,000
Level 5
Level 7
(a) Span geometry
and loading
48" 48"
1'2"
8'10" 10'0"
1'2"
10'0" 8'10"
W = 0.90 kips/ft
W = 0.90 kips/ft
30"
20"
48"
(3) #10 (3) #10 (2) #10 (2) #10
(4) #9
(4) #9
Figure 646 Loads, moments, and reinforcement for haunched girder (1.0 in = 25.4 mm, 1.0 ft = 0.3048
m, 1.0 kip/ft = 14.6 kN/m, 1.0 in.kip = 0.113 kNm).
This effectively requires only two ties within the joint. However, the first tie will be placed 3 in. below
the top of the beam and then three additional ties will be placed below this hoop at a spacing of 8 in. The
final arrangement of ties within the beamcolumn joint is shown in Figure 645.
6.5.5 Design of Members of Frame 3
6.5.5.1 Design of Haunched Girder
A typical haunched girder supporting Level 5 of Frame 3 is now illustrated. This girder, located between
Gridlines A and B, has a variable depth with a maximum depth of 30 in. at the support and a minimum
depth of 20 in. for the middle half of the span. The length of the haunch at each end (as measured from
the face of the support) is 106 in. The width of the girder is 20 in. throughout. The girder frames into 28
in.by28in. columns on Gridlines A and B. As illustrated in Figure 646c, the reinforcement at Gridline
B is extended into the adjacent span (Span BC) instead of being hooked into the column.
Chapter 6, Reinforced Concrete
693
Based on a tributary gravity load analysis, this girder supports an average of 3.38 kips/ft of dead load and
0.90 kips/ft of reduced live load. A gravity load analysis of the girder was carried out in a similar manner
similar to that described above for the Berkeley building.
For determining earthquake forces, the entire structure was analyzed using the ETABS program. This
analysis included 100 percent of the earthquake forces in the EW direction placed at a 5 percent
eccentricity with the direction of the eccentricity set to produce the maximum seismic shear in the
member.
6.5.5.2 Design of Longitudinal Reinforcement
The results of the analysis are shown in Figure 646b for five different load combinations. The envelopes
of maximum positive and negative moment indicate that 1.2D + 1.6L and 1.3D + 0.5L ± E produce
approximately equal negative end moments. Positive moment at the support is nearly zero under 0.8D 
E, and gravity controls midspan positive moment. Since positive moment at the support is negligible, a
positive moment capacity of at least onethird of the negative moment capacity will be supplied per ACI
318 Sec. 21.10.4.1 [21.12.4.1]. The minimum positive or negative moment strength at any section of the
span will not be less than onefifth of the maximum negative moment strength.
For a factored negative moment of 8,106 in.kips on Gridline A, try six #10 bars. Three of the bars are
short, extending just past the end of the haunch. The other three bars are long and extend into Span BC.
As = 6 (1.27) = 7.62 in.2
d = 30  1.5  0.375  1.27/2 = 27.49 in.
. = 7.62/[20 (27.49)] = 0.0139
Depth of compression block, a = [7.62 (60)]/[0.85 (4) 20.0] = 6.72 in.
Nominal capacity, Mn = [7.62 (60)](27.49  6.72/2) = 11,031 inkips
Design capacity, fMn = 0.8(11,031) = 8,824 in.kips > 8,106 in.kips OK
The three #10 bars that extend across the top of the span easily supply a minimum of onefifth of the
negative moment strength at the face of the support.
For a factored negative moment of 10,641 in.kips on Gridline B, try eight #10 bars. Three of the bars
extend from Span AB, three extend from Span BC, and the remaining two are short bars centered over
Support B.
As = 8 (1.27) = 10.16 in.2
d = 30  1.5  0.375  1.27/2 = 27.49 in.
. = 10.16/[20 (27.49)] = 0.0185
a = [10.16 (60)]/[0.85 (4) 20.0] = 8.96 in.
Mn = [10.16 (60)](27.49  8.96/2) = 13,996 in.kips
fMn = 0.8(13,996) = 11,221 in.kips > 10,641 in.kips OK
For the maximum factored positive moment at midspan of 2,964 inkips., try four #9 bars:
As = 4 (1.0) = 4.00 in.2
d = 20  1.5  0.375  1.128/2 = 17.56 in.
. = 4.0/[20 (17.56)] = 0.0114
a = [4.00 (60)]/[0.85 (4) 84] = 0.84 in. (effective flange width = 84 in.)
Mn = [4.00 (60)](17.56  0.84/2) = 4,113 in.kips
fMn = 0.8(4,113) = 3,290 in.kips > 2,964 OK
FEMA 451, NEHRP Recommended Provisions: Design Examples
694
Even though they provide more than onethird of the negative moment strength at the support, the four #9
bars will be extended into the supports as shown in Figure 646. The design positive moment strength for
the 30in.deep section with four #9 bars is computed as follows:
As = 4 (1.00) = 1.00 in.2
d = 30  1.5  0.375  1.128/2 = 27.56 in.
. = 4.00/[20 (27.56)] = 0.0073
a = [4.0 (60)]/[0.85 (4) 20.0] = 0.84 in.
Mn = [4.00 (60)] (27.56  0.84/2) = 6,514 in.kips
fMn = 0.8(6,514) = 5,211 in.kips
The final layout of longitudinal reinforcement used is shown in Figure 646. Note that the supplied
design strengths at each location exceed the factored moment demands. The hooked #10 bars can easily
be developed in the confined core of the columns. Splices shown are Class B and do not need to be
confined within hoops.
6.5.5.3 Design of Transverse Reinforcement
For the design for shear, ACI 318 Sec. 21.10.3 [21.12.3] gives the two options discussed above. For the
haunched girder, the approach based on the nominal flexural capacity (f = 1.0) of the girder will be used
as follows:
For negative moment and six #10 bars, the nominal moment strength = 11,031 in.kips
For negative moment and eight #10 bars, the nominal strength =13,996 in.kips
For positive moment and four #9 bars, the nominal moment strength = 6,514 in.kips
Earthquake shear when Support A is under positive seismic moment is:
VE = (13,996 + 6,514)/(480  28) = 45.4 kips
Earthquake shear when Support B is under positive seismic moment is:
VE = (11,031 + 6,514)/(480  28) = 38.8 kips
VG = 1.3VD + 0.5VL = 1.3 (63.6) + 0.5(16.9) = 91.1 kips
Maximum total shear occurs at Support B:
Vu = 45.4+91.1 = 136.5 kips
The shear at Support A is 38.8 + 91.9 = 130.1 kips. The complete design shear (demand) strength
envelope is shown in Figure 647a. Due to the small difference in end shears, use the larger shear for
designing transverse reinforcement at each end.
Stirrup spacing required for strength is based on two #4 legs with fy = 60 ksi.
(0.85)(2) 4,000)(20)(27.6) 59.3 kips
Vc= 1,000 =
136.5 0.75(59.3) 122.7 kips
0.75
u c
s
V V V
f
f
 
= = =
Chapter 6, Reinforced Concrete
695
8'10" 20'0" 8'10"
5" o.c.
8" o.c.
6" o.c.
130.1 kips
136.4 kips
42.0 kips
#4 stirrups
(a)
Required shear
strength envelope
(b)
Spacing of
transverse
reinforcement
2"
46.0 kips 52.5 kips
2"
5" o.c.
6" o.c.
Figure 647 Shear force envelope for haunched girder (1.0 ft =
0.3048m, 1.0 in = 25.4 mm, 1.0 kip = 4.45kN).
Using Vs = Av fyd/s:
(0.4)(60)(27.6) 5.39 in.
122.7
s= =
Following the same procedure as shown above, the spacing required for other stations is:
At support, h = 30 in., VU = 136.4 kips s = 5.39 in.
Middle of haunch, h = 25 in., VU = 114.9 kips s = 6.67 in.
End of haunch, h =20 in., VU =93.4 kips s = 7.61 in.
Quarter point of region of 20in. depth, VU = 69.2 kips s = 12.1 in.
Midspan, Vu = 45.1 kips s = 29.7 in.
Within a region 2h from the face of the support, the allowable maximum spacing is d/4 = 6.87 in. at the
support and approximately 5.60 in. at midhaunch. Outside this region, the maximum spacing is d/2 =
11.2 in. at midhaunch and 8.75 in. at the end of the haunch and in the 20in. depth region. At the
haunched segments at either end of the beam, the first stirrup is placed 2 in. from the face of the support
followed by four stirrups at a spacing of 5 in, and then 13 stirrups at 6 in. through the remainder of the
haunch. For the constant 20in.deep segment of the beam, a constant spacing of 8 in. is used. The final
spacing of stirrups used is shown in Figure 647b. Three additional stirrups should be placed at each
bend or “kink” in the bottom bars. One should be located at the kink and the others approximately 2 in.
on either side of the kink.
FEMA 451, NEHRP Recommended Provisions: Design Examples
696
6.5.5.4 Design of Supporting Column
The column on Gridline A which supports Level 5 of the haunched girder is 28 in. by 28 in. and supports
a total unfactored dead load of 803.6 kips and an unfactored reduced live load of 78.4 kips. The layout of
the column is shown in Figure 648. Under gravity load alone, the unfactored dead load moment is 2,603
in.kips and the corresponding live load moment is 693.0 in.kips. The corresponding shears are 43.4 and
11.5 kips, respectively. The factored gravity load combinations for designing the column are as follows:
Bending moment, M = 1.2(2,603) + 1.6(693)
= 4,232 in.kips
This moment causes tension on the outside face of the top of the column and tension on the inside face of
the bottom of the column.
Shear, V = 1.2(43.4) + 0.5(11.5) = 57.8 kips
Axial compression, P = 1.2(803.6) + 1.6(78.4)
= 1,090 kips
For equivalent static earthquake forces acting from west to east, the forces in the column are obtained
from the ETABS analysis as follows:
Moment at top of column = 690 in.kips (tension on inside face subtracts from gravity)
Moment at bottom of column = 874 in.kips (tension on outside face subtracts from gravity)
Shear in column = 13.3 kips (opposite sign of gravity shear)
Axial force = 63.1 kips tension
The factored forces involving earthquake from west to east are:
Moment at top 0.80(2603)  690 = 1,392 in.kips
Moment at bottom = 0.80(2603)  874 = 1,208 in.kips
Shear = 0.80(43.4)  2(13.3) = 8.1 kips (using the second option for computing EQ shear)
Axial force = 0.80(803.6)  63.1 = 580 kips
For earthquake forces acting from east to west, the forces in the column are obtained from the ETABS
analysis as follows:
Moment at top of column = 690 in.kips (tension on outside face adds to gravity)
Moment at bottom of column = 874 in.kips (tension on inside face adds to gravity)
Shear in column = 13.3 kips (same sign of gravity shear)
Axial force = 63.1 kips compression
Chapter 6, Reinforced Concrete
697
P = 78.4 kips Includes
P = 803.6 kips level 5
Level 4
Level 5
L
D
20"
30" 30"
12'6"
28"
Figure 648 Loading for Column A, Frame 3 (1.0 ft =
0.3048 m, 1.0 in = 25.4 mm, 1.0 kip = 4.45kN).
The factored forces involving earthquake from east to west are:
Moment at top 1.3(2,603) + 0.5(693) + 690 = 4,420 in.kips
Moment at bottom = 1.3(2,603) + 0.5(693) + 874 = 4,604 in.kips
Shear = 1.3(43.4) + 0.5(11.5) + 2(13.3) = 94.6 kips (using second option for computing EQ shear)
Axial force = 1.3(803.6) + 0.5(78.4) + 63.1 = 1,147 kips
As may be observed from Figure 649, the column with 12 #8 bars is adequate for all loading
combinations. Since the maximum design shear is less than that for the column previously designed for
Frame 1 and since minimum transverse reinforcement controlled that column, the details for the column
currently under consideration are similar to those shown in Figure 645. The actual details for the column
supporting the haunched girder of Frame 3 are shown in Figure 650.
FEMA 451, NEHRP Recommended Provisions: Design Examples
698
500
1,500
2,500
500
200 400 600 800 1,000
P (kips)
2,000
1,000
0
3,000
M x (ftkips)
0
Figure 649 Interaction diagram for Column A, Frame 3
(1.0 kip = 4.45kN, 1.0 ftkip = 1.36 kNm).
6.5.5.5 Design of BeamColumn Joint
The detailing of the joint of the column supporting Level 5 of the haunched girder is the same as that for
the column interior column of Frame A. The joint details are shown in Figure 650.
Chapter 6, Reinforced Concrete
699
Level 5
Level 4
28"
28"
28"
(12) #8 bars
4" 14 spaces at 8" o.c. 4"
Figure 650 Details for Column A, Frame 3 (1.0
in = 25.4 mm).
71
7
PRECAST CONCRETE DESIGN
Gene R. Stevens, P.E. and James Robert Harris, P.E., Ph.D.
This chapter illustrates the seismic design of precast concrete members using the NEHRP Recommended
Provisions (referred to herein as the Provisions) for buildings in several different seismic design
categories. Very briefly, for precast concrete structural systems, the Provisions:
1. Requires the system (even if the precast carries only gravity loads) to satisfy one of the following two
sets of provisions:
a. Resist amplified chord forces in diaphragms and, if momentresisting frames are used as the
vertical system, provide a minimum degree of redundancy measured as a fraction of available
bays, or
b. Provide a momentresisting connection at all beamtocolumn joints with positive lateral support
for columns and with special considerations for bearing lengths.
(In the authors’ opinion this does not apply to buildings in Seismic Design Category A.)
2. Requires assurance of ductility at connections that resist overturning for ordinary precast concrete
shear walls. (Because ordinary shear walls are used in lower Seismic Design Categories, this
requirement applies in Seismic Design Categories B and C.)
3. Allows special moment frames and special shear walls of precast concrete to either emulate the
behavior of monolithic concrete or behave as jointed precast systems. Some detail is given for special
moment frame designs that emulate monolithic concrete. To validate designs that do not emulate
monolithic concrete, reference is made to a new ACI testing standard (ACI T1.101).
4. Defines that monolithic emulation may be achieved through the use of either:
a. Ductile connections, in which the nonlinear response occurs at a connection between a precast
unit and another structural element, precast or not, or
b. Strong connections, in which the nonlinear response occurs in reinforced concrete sections
(generally precast) away from connections that are strong enough to avoid yield even as the
forces at the nonlinear response location increase with strain hardening.
5. Defines both ductile and strong connections can be either:
a. Wet connections where reinforcement is spliced with mechanical couplers, welds, or lap splices
(observing the restrictions regarding the location of splices given for monolithic concrete) and the
connection is completed with grout, or
FEMA 451, NEHRP Recommended Provisions: Design Examples
72
b. Dry connections, which are defined as any connection that is not a wet connection.
6. Requires that ductile connections be either:
a. Type Y, with a minimum ductility ratio of 4 and specific anchorage requirements, or
b. Type Z, with a minimum ductility ratio of 8 and stronger anchorage requirements.
Many of these requirements have been adopted into the 2002 edition of ACI 318, but some differences
remain. Where those differences are pertinent to the examples illustrated here, they are explained.
The examples in Sec. 7.1 illustrate the design of untopped and topped precast concrete floor and roof
diaphragms of the fivestory masonry buildings described in Sec. 9.2 of this volume of design examples.
The two untopped precast concrete diaphragms of Sec. 7.1.1 show the requirements for Seismic Design
Categories B and C using 8in.thick hollow core precast, prestressed concrete planks. Sec. 7.1.2 shows
the same precast plank with a 2 ½ in.thick composite lightweight concrete topping for the fivestory
masonry building in Seismic Design Category D described in Sec. 9.2. Although untopped diaphragms
are commonly used in regions of low seismic hazard, the only place they are addressed in the Provisions
is the Appendix to Chapter 9. The reader should bear in mind that the appendices of the Provisions are
prepared for trial use and comment, and future changes should be expected.
The example in Sec. 7.2 illustrates the design of an ordinary precast concrete shear wall building in a
region of low or moderate seismicity, which is where most precast concrete seismicforceresisting
systems are constructed. The precast concrete walls in this example resist the seismic forces for a threestory
office building, located in southern New England (Seismic Design Category B). There are very few
seismic requirements for such walls in the Provisions. One such requirement qualifies is that overturning
connections qualify as the newly defined Type Y or Z. ACI 31802 identifies this system as an
“intermediate precast concrete shear wall” and does not specifically define the Type Y or Z connections.
Given the brief nature of the requirements in both the Provisions and ACI 318, the authors offer some
interpretation. This example identifies points of yielding for the system and connection features that are
required to maintain stable cyclic behavior for yielding.
The example in Sec. 7.3 illustrates the design of a special precast concrete shear wall for a singlestory
industrial warehouse building in the Los Angeles. For buildings in Seismic Design Category D,
Provisions Sec. 9.1.1.12 [9.2.2.4] requires that the precast seismicforceresisting system emulate the
behavior of monolithic reinforced concrete construction or that the system’s cyclic capacity be
demonstrated by testing. The Provisions describes methods specifically intended to emulate the behavior
of monolithic construction, and dry connections are permitted. Sec. 7.3 presents an interpretation of
monolithic emulation of precast shear wall panels with ductile, dry connections. Whether this connection
would qualify under ACI 31802 is a matter of interpretation. The design is computed using the
Provisions rules for monolithic emulation; however, the system probably would behave more like a
jointed precast system. Additional clarity in the definition and application of design provisions of such
precast systems is needed.
Tiltup concrete wall buildings in all seismic zones have long been designed using the precast wall panels
as shear walls in the seismicforceresisting system. Such designs have usually been performed using
design force coefficients and strength limits as if the precast walls emulated the performance of castinplace
reinforced concrete shear walls, which they usually do not. In tiltup buildings subject to strong
ground shaking, the inplane performance of the precast panels has rarely been a problem, primarily
because there has been little demand for postelastic performance in that direction. Conventional tiltup
buildings may deserve a unique treatment for seismicresistant design, and they are not the subject of any
of the examples in this chapter, although tiltup panels with large heighttowidth ratios could behave in
the fashion described in design example 7.3.
Chapter 7, Precast Concrete Design
73
In addition to the Provisions, the following documents are either referred to directly or are useful design
aids for precast concrete construction:
ACI 31899 American Concrete Institute. 1999. Building Code Requirements and
Commentary for Structural Concrete.
ACI 31802 American Concrete Institute. 2002. Building Code Requirements and
Commentary for Structural Concrete.
AISC LRFD American Institute of Steel Construction. 2002. Manual of Steel Construction,
Load & Resistance Factor Design, Third Edition.
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for
Buildings and Other Structures.
Hawkins Hawkins, Neil M., and S. K. Ghosh. 2000. “Proposed Revisions to 1997
NEHRP Recommended Provisions for Seismic Regulations for Precast Concrete
Structures, Parts 1, 2, and 3.” PCI Journal, Vol. 45, No. 3 (MayJune), No. 5
(Sept.Oct.), and No. 6 (Nov.Dec.).
Moustafa Moustafa, Saad E. 1981 and 1982. “Effectiveness of ShearFriction
Reinforcement in Shear Diaphragm Capacity of HollowCore Slabs.” PCI
Journal, Vol. 26, No. 1 (Jan.Feb. 1981) and the discussion contained in PCI
Journal, Vol. 27, No. 3 (MayJune 1982).
PCI Handbook Precast/Prestressed Concrete Institute. 1999. PCI Design Handbook, Fifth
Edition.
PCI Details Precast/Prestressed Concrete Institute. 1988. Design and Typical Details of
Connections for Precast and Prestressed Concrete, Second Edition.
SEAA Hollow Core Structural Engineers Association of Arizona, Central Chapter. Design and
Detailing of Untopped HollowCore Slab Systems for Diaphragm Shear.
The following style is used when referring to a section of ACI 318 for which a change or insertion is
proposed by the Provisions: Provisions Sec. xxx (ACI Sec. yyy) where “xxx” is the section in the
Provisions and “yyy” is the section proposed for insertion into ACI 31899.
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect
changes made for the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational
changes (as a result of a reformatting of all chapters for the 2003 Provisions) and substantive technical
changes to the Provisions and its primary reference documents. Although the general conepts of the
changes are described, the design examples and calculations have not been revised to reflect the changes
made for the 2003 Provisions.
The most significant change related to precast concrete in the 2003 Provisions is that precast shear wall
systems are now recognized separately from castinplace systems. The 2003 Provisions recognizes
ordinary and intermediate precast concrete shear walls. The design of ordinary precast shear walls is
based on ACI 31802 excluding Chapter 21 and the design of intermediate shear walls is based on ACI
31802 Sec. 21.13 (with limited modifications in Chapter 9 of the 2003 Provisions). The 2003 Provisions
does not distinguish between precast and castinplace concrete for special shear walls. Special precast
shear walls either need to satisfy the design requirements for special castinplace concrete shear walls
FEMA 451, NEHRP Recommended Provisions: Design Examples
74
(ACI 31802 Sec. 21.7) or most be substantiated using experimental evidence and analysis (2003
Provisions Sec. 9.2.2.4 and 9.6). Many of the design provisions for precast shear walls in the 2000
Provisions have been removed, and the requirements in ACI 31802 are in some ways less specific.
Where this occurs, the 2000 Provisions references in this chapter are simply annotated as “[not applicable
in the 2003 Provisions].” Commentary on how the specific design provision was incorporated into ACI
31802 is included where appropriate.
Some general technical changes for the 2003 Provisions that relate to the calculations and/or designs in
this chapter include updated seismic hazard maps, revisions to the redundancy requirements, and
revisions to the minimum base shear equation. Where they affect the design examples in the chapter,
other significant changes for the 2003 Provisions and primary reference documents are noted. However,
some minor changes may not be noted.
Chapter 7, Precast Concrete Design
1Note that this equation is incorrectly numbered as 5.2.5.4 in the first printing of the 2000 Provisions.
75
7.1 HORIZONTAL DIAPHRAGMS
Structural diaphragms are horizontal or nearly horizontal elements, such as floors and roofs, that transfer
seismic inertial forces to the vertical seismicforceresisting members. Precast concrete diaphragms may
be constructed using topped or untopped precast elements depending on the Seismic Design Category of
the building. Reinforced concrete diaphragms constructed using untopped precast concrete elements are
addressed in the Appendix to Chapter 9 of the Provisions. Topped precast concrete elements, which act
compositely or noncompositely for gravity loads, are designed using the requirements of ACI 31899 Sec.
21.7 [ACI 31802 Sec. 21.9].
7.1.1 Untopped Precast Concrete Units for FiveStory Masonry Buildings Located in
Birmingham, Alabama, and New York, New York
This example illustrates floor and roof diaphragm design for the fivestory masonry buildings located in
Birmingham, Alabama, on soft rock (Seismic Design Category B) and in New York, New York (Seismic
Design Category C). The example in Sec. 9.2 provides design parameters used in this example. The
floors and roofs of these buildings are to be untopped 8in.thick hollow core precast, prestressed concrete
plank. Figure 9.21 shows the typical floor plan of the diaphragms.
7.1.1.1 General Design Requirements
In accordance with the Provisions and ACI 318, untopped precast diaphragms are permitted only in
Seismic Design Categories A through C. The Appendix to Chapter 9 provides design provisions for
untopped precast concrete diaphragms without limits as to the Seismic Design Category. Diaphragms
with untopped precast elements are designed to remain elastic, and connections are designed for limited
ductility. No outofplane offsets in vertical seismicforceresisting members (Type 4 plan irregularities)
are permitted with untopped diaphragms. Static rational models are used to determine shears and
moments on joints as well as shear and tension/compression forces on connections. Dynamic modeling of
seismic response is not required.
The design method used here is that proposed by Moustafa. This method makes use of the shear friction
provisions of ACI 318 with the friction coefficient, µ, being equal to 1.0. To use µ = 1.0, ACI 318
requires grout or concrete placed against hardened concrete to have clean, laitance free, and intentionally
roughened surfaces with a total amplitude of about 1/4 in. (peak to valley). Roughness for formed edges
is provided either by sawtooth keys along the length of the plank or by hand roughening with chipping
hammers. Details from the SEAA Hollow Core reference are used to develop the connection details.
The terminology used is defined in ACI 318 Chapter 21 and Provisions Chapter 9. These two sources
occasionally conflict (such as the symbol µ used above), but the source is clear from the context of the
discussion. Other definitions (e.g., chord elements) are provided as needed for clarity in this example.
7.1.1.2 General InPlane Seismic Design Forces for Untopped Diaphragms
The inplane diaphragm seismic design force (F!px) for untopped precast concrete in Provisions Sec.
9A.3.3 [A9.2.2] “shall not be less than the forcee calculated from either of the following two criteria:”
1. .O0Fpx but not less than .O0Cswpx where
F px is calculated from Provisions Eq. 5.2.6.4.41 [4.63], which also bounds Fpx to be not less than
0.2SDSIwpx and not more than 0.4SDSIwpx. This equation normally is specified for Seismic Design
FEMA 451, NEHRP Recommended Provisions: Design Examples
76
Categories D and higher; it is intended in the Provisions Appendix to Chapter 9 that the same
equation be used for untopped diaphragms in Seismic Design Categories B and C.
. is the reliability factor, which is 1.0 for Seismic Design Categories A through C per Provisions Sec.
5.2.4.1 [4.3.3.1].
O0 is the overstrength factor (Provisions Table 5.2.2 [4.31])
Cs is the seismic response coefficient (Provisions Sec. 5.4.1.1 [5.2.1.1])
wpx is the weight tributary to the diaphragm at Level x
SDS is the spectral response acceleration parameter at short periods (Provisions Sec. 4.1.2 [3.3.3])
I is the occupancy importance factor (Provisions Sec. 1.4 [1.3])
2. 1.25 times the shear force to cause yielding of the vertical seismicforceresisting system.
For the fivestory masonry buildings of this example, the shear force to cause yielding is first
estimated to be that force associated with the development of the nominal bending strength of the
shear walls at their base. This approach to yielding uses the first mode force distribution along the
height of the building and basic pushover analysis concepts, which can be approximated as:
F!px = 1.25KFpx
* where
K is the ratio of the yield strength in bending to the demand, My/Mu. (Note that f = 1.0)
Fpx
* is the seismic force at each level for the diaphragm as defined above by Provisions Eq. 5.2.6.4.4
[4.62] and not limited by the minima and maxima for that equation.
This requirement is different from similar requirements elsewhere in the Provisions. For components
thought likely to behave in a brittle fashion, the designer is required to apply the overstrength factor and
then given an option to check the maximum force that can be delivered by the remainder of the structural
system to the element in question. The maximum force would normally be computed from a plastic
mechanism analysis. If the option is exercised, the designer can then use the smaller of the two forces.
Here the Provisions requires the designer to compute both an overstrength level force and a yield level
force and then use the larger. This appears to conflict with the Commentary.
For Seismic Design Categories B and C, Provisions Sec. 5.2.6.2.6 [4.6.1.9] defines a minimum diaphragm
seismic design force that will always be less than the forces computed above.
For Seismic Design Category C, Provisions Sec. 5.2.6.3.1 [4.6.2.2] requires that collector elements,
collector splices, and collector connections to the vertical seismicforceresisting members be designed in
accordance with Provisions Sec. 5.2.7.1 [4.2.2.2], which places the overstrength factor on horizontal
seismic forces and combines the horizontal and vertical seismic forces with the effects of gravity forces.
Because vertical forces do not normally affect diaphragm collector elements, splices, and connections, the
authors believe that Provisions Sec. 5.2.7.1 [4.2.2.2] is satisfied by the requirements of Provisions Sec.
9A.3.3 [A9.2.2], which requires use of the overstrength factor.
Parameters from the example in Sec. 9.2 used to calculate inplane seismic design forces for the
diaphragms are provided in Table 7.11.
Chapter 7, Precast Concrete Design
77
Table 7.11 Design Parameters from Example 9.2
Design Parameter Birmingham 1 New York City
. 1.0 1.0
Oo 2.5 2.5
Cs 0.12 0.156
wi (roof) 861 kips 869 kips
wi (floor) 963 kips 978 kips
SDS 0.24 0.39
I 1.0 1.0
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
The Provisions Appendix to Chapter 9 does not give the option of using the overstrength factor O0 to
estimate the yield of the vertical system, so Mn for the wall is computed from the axial load moment
interaction diagram data developed in Sec. 9.2. The shape of the interaction diagram between the
balanced point and pure bending is far enough from a straight line (see Figure 9.26) in the region of
interest that simply interpolating between the points for pure bending and balanced conditions is
unacceptably unconservative for this particular check. An intermediate point on the interaction diagram
was computed for each wall in Sec. 9.2, and that point is utilized here. Yielding begins before the
nominal bending capacity is reached, particularly when the reinforcement is distributed uniformly along
the wall rather than being concentrated at the ends of the wall. For lightly reinforced walls with
distributed reinforcement and with axial loads about onethird of the balanced load, such as these, the
yield moment is on the order of 90 to 95 percent of the nominal capacity. It is feasible to compute the
moment at which the extreme bar yields, but that does not appear necessary for design. A simple factor
of 0.95 was applied to the nominal capacity here. Thus, Table 7.12 shows the load information from
Sec. 9.2 (the final numbers in this section may have changed, because this example was completed first).
The factor K is large primarily due to consideration of axial load. The strength for design is controlled by
minimum axial load, whereas K is maximum for the maximum axial load, which includes some live load
and a vertical acceleration on dead load.
FEMA 451, NEHRP Recommended Provisions: Design Examples
78
Table 7.12 Shear Wall Overstrength
Birmingham 1 New York City
Pure Bending, Mn0 963 ftkips 1,723 ftkips
Intermediate Load, MnB 5,355 ftkips 6,229 ftkips
Intermediate Load, PnB 335 kips 363 kips
Maximum Design Load, Pu 315 kips 327 kips
Interpolated Mn 5,092 ftkips 5,782 ftkips
Approximate My 4,837 ftkips 5,493 ftkips
Design Mu 2,640 ftkips 3,483 ftkips
Factor K = My/Mu 1.83 1.58
7.1.1.3 Diaphragm Forces for Birmingham Building 1
The weight tributary to the roof and floor diaphragms (wpx) is the total story weight (wi) at Level i minus
the weight of the walls parallel to the direction of loading.
Compute diaphragm weight (wpx) for the roof and floor as follows:
Roof
Total weight = 861 kips
Walls parallel to force = (45 psf)(277 ft)(8.67 ft/2) = 54 kips
wpx = 807 kips
Floors
Total weight = 963 kips
Walls parallel to force = (45 psf)(277 ft)(8.67 ft) = 108 kips
wpx = 855 kips
Compute diaphragm demands in accordance with Provisions Eq. 5.2.6.4.4 [4.6.3.4]:
n
i
i x
px n px
i
i x
F
F w
w
=
=
S
=
S
Calculations for Fpx are provided in Table 7.13.
Chapter 7, Precast Concrete Design
79
Table 7.13 Birmingham 1 Fpx Calculations
Level
wi
(kips)
n
i
i x
w
= S
(kips)
Fi
(kips)
n
i i
i x
F V
=
S =
(kips)
wpx
(kips)
Fpx
(kips)
Roof
4321
861
963
963
963
963
861
1,820
2,790
3,750
4,710
175
156
117
78
39
175
331
448
527
566
807
855
855
855
855
164
155
137
120
103
1.0 kip = 4.45 kN.
The values for Fi and Vi used in Table 7.13 are listed in Table 9.22.
The minimum value of Fpx = 0.2SDSIwpx = 0.2(0.24)1.0(807 kips) = 38.7 kips (at the roof)
= 0.2(0.24)1.0(855 kips) = 41.0 kips (at floors)
The maximum value of Fpx = 0.4SDSIwpx = 2(38.7 kips) = 77.5 kips (at the roof)
= 2(41.0 kips) = 82.1 kips (at floors)
Note that Fpx by Table 7.13 is substantially larger than the maximum Fpx. This is generally true at upper
levels if the R factor is less than 5. The value of Fpx used for the roof diaphragm is 82.1 kips. Compare
this value to Cswpx to determine the minimum diaphragm force for untopped diaphragms as indicated
previously.
Cswpx = 0.12(807 kips) = 96.8 kips (at the roof)
Cswpx = 0.12(855 kips) = 103 kips (at the floors)
Since Cswpx is larger than Fpx, the controlling force is Cswpx. Note that this will always be true when I =
1.0 and R is less than or equal to 2.5. Therefore, the diaphragm seismic design forces are as follows:
F!px = .O0Cswpx = 1.0(2.5)(96.8 kips) = 242 kips (at the roof)
F!px = .O0Cswpx = 1.0(2.5)(103 kips) = 256 kips (at the floors)
The second check on design force is based on yielding of the shear walls:
F!px = 1.25KFpx* = 1.25(1.85)164 kips = 379 kips (at the roof)
F!px = 1.25KFpx* = 1.25(1.85)155 kips = 358 kips (at the floors)
For this example, the force to yield the walls clearly controls the design. To simplify the design, the
diaphragm design force used for all levels will be the maximum force at any level, 379 kips.
7.1.1.4 Diaphragm Forces for New York Building
The weight tributary to the roof and floor diaphragms (wpx) is the total story weight (wi) at Level i minus
the weight of the walls parallel to the force.
FEMA 451, NEHRP Recommended Provisions: Design Examples
710
Compute diaphragm weight (wpx) for the roof and floor as follows:
Roof
Total weight = 870 kips
Walls parallel to force = (48 psf)(277 ft)(8.67 ft/2) = 58 kips
wpx = 812 kips
Floors
Total weight = 978 kips
Walls parallel to force = (48 psf)(277 ft)(8.67 ft) = 115 kips
wpx = 863 kips
Calculations for Fpx using Provisions Eq. 5.2.6.4.4 [4.6.3.4] are not required for the first set of forces
because Cswpx will be greater than or equal to the maximum value of Fpx = 0.4SDSIwpx when I = 1.0 and R
is less than or equal to 2.5. Compute Cswpx as:
Cswpx = 0.156(812 kips) = 127 kips (at the roof)
Cswpx = 0.156(863 kips) = 135 kips (at the floors)
The diaphragm seismic design forces are:
F!px = .O0Cswpx = 1.0(2.5)(127 kips) = 318 kips (at the roof)
F!px = .O0Cswpx = 1.0(2.5)(135 kips) = 337 kips (at the floors)
Calculations for Fpx using Provisions Eq. 5.2.6.4.4 [4.6.3.4] are required for the second check F!px =
1.25KFpx. Following the same procedure as illustrated in the previous section, the maximum Fpx is 214
kips at the roof. Thus,
1.25KFpx* = 1.25(1.58)214 kips = 423 kips (at the roof)
To simplify the design, the diaphragm design force used for all levels will be the maximum force at any
level. The diaphragm seismic design force (423 kips) is controlled by yielding at the base of the walls,
just as with the Birmingham 1 building.
7.1.1.5 Static Analysis of Diaphragms
The balance of this example will use the controlling diaphragm seismic design force of 423 kips for the
New York building. In the transverse direction, the loads will be distributed as shown in Figure 7.11.
Chapter 7, Precast Concrete Design
711
W2
W1 W1
F F F F
40'0" 3 at 24'0" = 72'0" 40'0"
152'0"
Figure 7.11 Diaphragm force distribution and analytical model (1.0 ft =
0.3048 m).
Assuming the four shear walls have the same stiffness and ignoring torsion, the diaphragm reactions at
the transverse shear walls (F as shown in Figure 7.11) are computed as follows:
F = 423 kips/4 = 105.8 kips
The uniform diaphragm demands are proportional to the distributed weights of the diaphragm in different
areas (see Figure 7.11).
W1 = [67 psf(72 ft) + 48 psf(8.67 ft)4](423 kips / 863 kips) = 3,180 lb/ft
W2 = [67 psf(72 ft)](423 kips / 863 kips) = 2,364 lb/ft
Figure 7.12 identifies critical regions of the diaphragm to be considered in this design. These regions
are:
Joint 1 – maximum transverse shear parallel to the panels at paneltopanel joints
Joint 2 – maximum transverse shear parallel to the panels at the paneltowall joint
Joint 3 – maximum transverse moment and chord force
Joint 4 – maximum longitudinal shear perpendicular to the panels at the paneltowall connection
(exterior longitudinal walls) and anchorage of exterior masonry wall to the diaphragm for outofplane
forces
Joint 5 – collector element and shear for the interior longitudinal walls
FEMA 451, NEHRP Recommended Provisions: Design Examples
712
72'0"
4 1 2 3
5
36'0"
4'0"
24'0"
Figure 7.12 Diaphragm plan and critical design regions (1.0 ft = 0.3048 m).
Provisions Sec. 9.1.1.4 [not applicable in 2003 Provisions] defines a chord amplification factor for
diaphragms in structures having precast gravityload systems. [The chord amplification factor has been
dropped in the 2003 Provisions and does not occur in ASC 31802. See the initial section of this chapter
for additional discussion on changes for the 2003 Provisions.] This amplification factor appears to apply
to buildings with vertical seismicforceresisting members constructed of precast or monolithic concrete.
Because these masonry wall buildings are similar to buildings with concrete walls, this amplification
factor has been included in calculating the chord forces. The amplification factor is:
2
1 0.4
1.0
12
eff
d
d
s
L
b
b
h
. . ..
.+ . ..
.. . ...=
where
Leff = length of the diaphragm between inflection points. Since the diaphragms have no infection
points, twice the length of the 40ftlong cantilevers is used for Leff = 80 ft
hs = story height = 8.67 ft
bd = diaphragm width = 72 ft
The amplification factor = ( ) ( ) = 1.03
2 1 0.4 80
72
72
12 8.67
..+ .. ....
.. . ...
Chapter 7, Precast Concrete Design
713
Joint forces are:
Joint 1 – Transverse forces
Shear, Vu1 = 3.18 kips/ft (36 ft) = 114.5 kips
Moment, Mu1 = 114.5 kips (36 ft/2) = 2,061 ftkips
Chord tension force, Tu1 = M/d = 1.03(2,061 ftkips/71 ft) = 29.9 kips
Joint 2 – Transverse forces
Shear, Vu2 = 3.18 kips/ft (40 ft) = 127 kips
Moment, Mu2 = 127 kips (40 ft/2) = 2,540 ftkips
Chord tension force, Tu2 = M/d = 1.03(2,540 ftkips/71 ft) = 36.9 kips
Joint 3 – Transverse forces
Shear, Vu3 = 127 kips + 2.36 kips/ft (24 ft)  105.8 kips = 78.1 kips
Moment, Mu3 = 127 kips (44 ft) + 56.7 kips (12 ft)  105.8 kips (24 ft) = 3,738 ftkips
Chord tension force, Tu3 = M/d = 1.03(3,738 ftkips/71 ft) = 54.2 kips
Joint 4 – Longitudinal forces
Wall Force, F = 423 kips/8 = 52.9 kips
Wall shear along wall length, Vu4 = 52.9 kips (36 ft)/(152 ft /2) = 25.0 kips
Collector force at wall end, Tu4 = Cu4 = 52.9 kips  25.0 kips = 27.9 kips
Joint 4 – Outofplane forces
The Provisions have several requirements for outofplane forces. None are unique to precast
diaphragms and all are less than the requirements in ACI 318 for precast construction regardless
of seismic considerations. Assuming the planks are similar to beams and comply with the
minimum requirements of Provisions Sec. 5.2.6.1.1 [4.6.1.1] (Seismic Design Category A and
greater) [In the 2003 Provisions, all requirements for Seismic Design Category A are in Sec. 1.5
but they generally are the same as those in the 2000 Provisions. The design and detailing
requirements in 2003 Provisions Sec. 4.6 apply to Seismic Design Category B and greater], the
required outofplane horizontal force is:
0.05(D + L)plank = 0.05(67 psf + 40 psf)(24 ft/2) = 64.2 plf
According to Provisions Sec. 5.2.6.1.2 [4.6.1.2] (Seismic Design Category A and greater), the
minimum anchorage for masonry walls is:
Fp = 400(SDS)I = 400(0.39)1.0 = 156 plf
According to Provisions Sec. 5.2.6.2.7 [4.6.1.3] (Seismic Design Category B and greater),
bearing wall anchorage shall be designed for a force computed as:
0.4(SDS)Wwall = 0.4(0.39)(48 psf)(8.67 ft) = 64.9 plf
Provisions Sec. 5.2.6.3.2 [4.6.2.1] (Seismic Design Category C and greater) requires masonry
wall anchorage to flexible diaphragms to be designed for a larger force. This diaphragm is
FEMA 451, NEHRP Recommended Provisions: Design Examples
714
considered rigid with respect to the walls, and considering that it is designed to avoid yield under
the loads that will yield the walls, this is a reasonable assumption.
Fp = 1.2(SDS)Iwp = 1.2(0.39)1.0[(48 psf)(8.67 ft)] = 195 plf
[In the 2003 Provisions, Eq. 4.61 in Sec. 4.6.2.1 has been changed to 0.85SDSIWp.]
The force requirements in ACI 318 Sec. 16.5 will be described later.
Joint 5 – Longitudinal forces
Wall force, F = 423 kips/8 = 52.9 kips
Wall shear along each side of wall, Vu4 = 52.9 kips [2(36 ft)/152 ft]/2 = 12.5 kips
Collector force at wall end, Tu5 = Cu5 = 52.9 kips  25.0 kips = 27.9 kips
ACI 318 Sec. 16.5 also has minimum connection force requirements for structural integrity of precast
concrete bearing wall building construction. For buildings over two stories there are force requirements
for horizontal and vertical members. This building has no vertical precast members. However, ACI 318
Sec. 16.5.1 specifies that the strengths “. . . for structural integrity shall apply to all precast concrete
structures.” This is interpreted to apply to the precast elements of this masonry bearing wall structure.
The horizontal tie force requirements are:
1. 1,500 lb/ft parallel and perpendicular to the span of the floor members. The maximum spacing of ties
parallel to the span is 10 ft. The maximum spacing of ties perpendicular to the span is the distance
between supporting walls or beams.
2. 16,000 lb parallel to the perimeter of a floor or roof located within 4 ft of the edge at all edges.
ACI’s tie forces are far greater than the minimum tie forces given in the Provisions for beam supports and
anchorage for of masonry walls. They do control some of the reinforcement provided, but most of the
reinforcement is controlled by the computed connections for diaphragm action.
7.1.1.6 Diaphragm Design and Details
Before beginning the proportioning of reinforcement, a note about ACI’s f factors is necessary. The
Provisions cites ASCE 7 for combination of seismic load effects with the effects of other loads. Both
ASCE 7 and the Provisions make it clear that the appropriate f factors within ACI 318 are those
contained within Appendix C of ACI 31899. These factors are about 10% less than the comparable
factors within the main body of the standard. The 2002 edition of ACI 318 has placed the ASCE 7 load
combinations within the main body of the standard and revised the f factors accordingly. This example
uses the f factors given in the 2002 edition of ACI 318, which are the same as those given in Appendix C
of the 1999 edition with one exception. Thus, the f factors used here are:
Tension control (bending and ties) f = 0.90
Shear f = 0.75
Compression control in tied members f = 0.65.
The minimum tie force requirements given in ACI 318 Sec. 16.5 are specified as nominal values, meaning
that f = 1.00 for those forces.
Chapter 7, Precast Concrete Design
715
Splice bars
(2) #7 bars
(chord bars)
3"
2"± 3"± 3"
3"
Grouted
chord / collector
element along exterior
edge of precast plank
Contact
lap splice
Prestressed
hollow core
plank
Artificially roughened
surfaces of void as
required
4"Ø spiral of 1
4" wire
with 2" pitch over each
lap splice may be required
depending on geometry
of specific voids in plank.
Figure 7.13 Joint 3 chord reinforcement at the exterior edge (1.0 in. = 25.4 mm).
7.1.1.6.1 Design and Detailing at Joint 3
Joint 3 is designed first to check the requirements of Provisions Sec. 9A.3.9 [A9.2.4], which references
ACI 318 Sec. 21.7.8.3 [21.9.8.3], which then refers to ACI 318 Sec. 21.7.5.3 [21.9.5.3]. This section
provides requirements for transverse reinforcement in the chords of the diaphragm. The compressive
stress in the chord is computed using the ultimate moment based on a linear elastic model and gross
section properties. To determine the inplane section modulus (S) of the diaphragm, an equivalent
thickness (t) based on the cross sectional area is used for the hollow core precast units as follows.
t = area/width = 215/48 = 4.5 in.
S = td2/6
Chord compressive stress is computed as:
Mu/S = 6Mu3/td2 = 6(3,738 × 12)/(4.5)(72 × 12)2 = 80.1 psi
The design 28day compressive strength of the grout is 4,000 psi. Since the chord compressive stress is
less than 0.2 fc' = 0.2(4,000) = 800 psi, the transverse reinforcement indicated in ACI 318 Sec. 21.4.4.1
through 21.4.4.3 is not required.
Compute the required amount of chord reinforcement as:
Chord reinforcement, As3 = Tu3/ffy = (54.2 kips)/[0.9(60 ksi)] = 1.00 in.2
Use two #7 bars, As = 2(0.60) = 1.20 in.2 along the exterior edges (top and bottom of the plan in Figure
7.1 2). Require cover for chord bars and spacing between bars at splices and anchorage zones by ACI
318 Sec. 21.7.8.3 [21.9.8.3].
Minimum cover = 2.5(7/8) = 2.19 in., but not less than 2.0 in.
Minimum spacing = 3(7/8) = 2.63 in., but not less than 11/2 in.
Figure 7.13 shows the chord element at the exterior edges of the diaphragm. The chord bars extend
along the length of the exterior longitudinal walls and act as collectors for these walls in the longitudinal
direction (see Joint 4 collector reinforcement and Figure 7.17).
FEMA 451, NEHRP Recommended Provisions: Design Examples
716
(2) #6
(collector bars)
33
4"
21 2"
31 2"
2"
#3 x 4'0" (behind)
at each joint
between planks
Figure 7.14 Interior joint reinforcement at the ends of plank
and the collector reinforcement at the end of the interior
longitudinal walls  Joints 1 and 5 (1.0 in. = 25.4 mm).
Joint 3 must also be checked for the minimum ACI tie forces. The chord reinforcement obviously
exceeds the 16 kip perimeter force requirement. The 1.5 kips per foot requirement requires a 6 kip tie at
each joint between the planks, which is satisfied with a #3 bar in each joint (0.11 in.2 at 60 ksi = 6.6 kips).
This bar is required at all bearing walls and is shown in subsequent details.
7.1.1.6.2 Joint 1 Design and Detailing
The design must provide sufficient reinforcement for chord forces as well as shear friction connection
forces as follows:
Chord reinforcement, As1 = Tu1/ffy = (29.9 kips)/[0.9(60 ksi)] = 0.55 in.2 (collector force from Joint 4
calculations at 27.9 kips is not directly additive).
Shear friction reinforcement, Avf1 = Vu1/fµfy = (114.5 kips)/[(0.75)(1.0)(60 ksi)] = 2.54 in.2
Total reinforcement required = 2(0.55 in.2) + 2.54 in.2 = 3.65 in.2
ACI tie force = (3 kips/ft)(72 ft) = 216 kips; reinforcement = (216 kips)/(60 ksi) = 3.60 in.2
Provide four #7 bars (two at each of the outside edges) plus four #6 bars (two each at the interior joint at
the ends of the plank) for a total area of reinforcement of 4(0.60 in2) + 4(0.44 in.2) = 4.16 in.2
Because the interior joint reinforcement acts as the collector reinforcement in the longitudinal direction
for the interior longitudinal walls, the cover and spacing of the two # 6 bars in the interior joints will be
provided to meet the requirements of ACI 318 Sec. 21.7.8.3 [21.9.8.3]:
Minimum cover = 2.5(6/8) = 1.88 in., but not less than 2.0 in.
Minimum spacing = 3(6/8) = 2.25 in., but not less than 11/2 in.
Figure 7.14 shows the reinforcement in the interior joints at the ends of the plank, which is also the
collector reinforcement for the interior longitudinal walls (Joint 5). The two #6 bars extend along the
length of the interior longitudinal walls as shown in Figure 7.18.
Chapter 7, Precast Concrete Design
717
112" 212" 2"
2" (2) #6 anchored 4'0"
into plank at ends.
Figure 7.15 Anchorage region of shear reinforcement for Joint 1 and collector
reinforcement for Joint 5 (1.0 in. = 25.4 mm).
Figure 7.15 shows the extension of the two #6 bars of Figure 7.14 into the region where the plank is
parallel to the bars. The bars will need to be extended the full length of the diaphragm unless
supplemental plank reinforcement is provided. This detail makes use of this supplement plank
reinforcement (two #6 bars or an equal area of strand per ACI 31899 Sec. 21.7.5.2 [21.9.5.2]) and shows
the bars anchored at each end of the plank. The anchorage length of the #6 bars is calculated using ACI
31899 Sec. 21.7.5.4 [21.9.5.4] which references ACI 318 Sec. 21.5.4:
60,000( )
1.6(2.5) 1.6(2.5) 58.2
65 65 4,000
y b b
d b
c
f d d
l d
f
. . . .
= ... '...= .. ..=
The 2.5 factor is for the difference between straight and hooked bars, and the 1.6 factor applies when the
development length is not within a confined core. Using #6 bars, the required ld = 58.2(0.75 in.) = 43.7
in. Therefore, use ld = 4 ft, which is the width of the plank.
7.1.1.6.3 Joint 2 Design and Detailing
The chord design is similar to the previous calculations:
Chord reinforcement, As2 = Tu2/ffy = (36.9 kips)/[0.9(60 ksi)] = 0.68 in.2
The shear force may be reduced along Joint 2 by the shear friction resistance provided by the
supplemental chord reinforcement (2Achord  As2) and by the four #6 bars projecting from the interior
longitudinal walls across this joint. The supplemental chord bars, which are located at the end of the
walls, are conservatively excluded here. The shear force along the outer joint of the wall where the plank
is parallel to the wall is modified as:
( ) ( )( )( ) 2 2 4#6 Mod 127 0.75 60 1.0 4 0.44 47.8 kips
Vu =Vu..ffyµA ..= .. × ..=
This force must be transferred from the planks to the wall. Using the arrangement shown in Figure 7.16,
the required shear friction reinforcement (Avf2) is computed as:
= = 0.79 in.2 ( )
2
2 sin cos
Mod
u
vf
y f f
A V
f f µ a a
=
+ ( )( )
47.8
0.75 60 1.0 sin 26.6° + cos 26.6°
FEMA 451, NEHRP Recommended Provisions: Design Examples
718
Use two #3 bars placed at 26.6 degrees (2to1 slope) across the joint at 4 ft from the ends of the plank
and at 8 ft on center (three sets per plank). The angle (af) used above provides development of the #3 bars
while limiting the grouting to the outside core of the plank. The total shear reinforcement provided is
9(0.11 in.2) = 0.99 in.2
The shear force between the other face of this wall and the diaphragm is:
Vu2  F = 127  106 = 21 kips
The shear friction resistance provided by #3 bars in the grout key between each plank (provided for the
1.5 klf requirement of the ACI) is computed as:
fAvffyµ = (0.75)(10 bars)(0.11 in.2)(60 ksi)(1.0) = 49.5 kips
The development length of the #3 and #4 bars will now be checked. For the 180 degree standard hook
use ACI 318 Sec. 12.5, ldh = lhb times the factors of ACI 318 Sec. 12.5.3, but not less than 8db or 6 in.
Side cover exceeds 21/2 in. and cover on the bar extension beyond the hook is provided by the grout and
the planks, which is close enough to 2 in. to apply the 0.7 factor of ACI 318 Sec. 12.5.3.2. The
continuous #5 provides transverse reinforcement, but it is not arranged to take advantage of ACI 318’s
0.8 factor. For the #3 hook:
0.7(1,200) 0.7(1,200)0.375 = 4.95 in. (6" minimum)
4,000
b
dh
c
d
l
f
= =
'
The available distance for the perpendicular hook is about 51/2 in. The bar will not be fully developed at
the end of the plank because of the 6 in. minimum requirement. The full strength is not required for shear
transfer. By inspection, the diagonal #3 hook will be developed in the wall as required for the computed
diaphragmtoshearwall transfer. The straight end of the #3 will now be checked. The standard
development length of ACI 318 Sec. 12.2 is used for ld.
= 14.2 in. 60,000(0.375)
25 25 4,000
y b
d
c
f d
l
f
= =
'
Figure 7.16 shows the reinforcement along each side of the wall on Joint 2.
Chapter 7, Precast Concrete Design
719
#3x 2'6"
standard hook
grouted into
each key joint
(1) #5
continuous
in joint to
anchor hooks
(2) #5 in
masonry
bond beam
#3 x standard hooks
embedded in grouted
edge cell of plank. Provide
3 sets for each plank.
2
1
71
2"
2'2"
2'2"
2" cover
Vertical
reinforcement
in wall
Figure 7.16 Joint 2 transverse wall joint reinforcement (1.0 in. = 25.4 mm,
1.0 ft = 0.3048 m).
7.1.1.6.4 Joint 4 Design and Detailing
The required shear friction reinforcement along the wall length is computed as:
Avf4 = Vu4/fµfy = (25.0 kips)/[(0.75)(1.0)(60 ksi)] = 0.56 in.2
Based upon the ACI tie requirement, provide #3 bars at each planktoplank joint. For eight bars total, the
area of reinforcement is 8(0.11) = 0.88 in.2, which is more than sufficient even considering the marginal
development length, which is less favorable at Joint 2. The bars are extended 2 ft into the grout key,
which is more than the development length and equal to half the width of the plank.
The required collector reinforcement is computed as:
As4 = Tu4/ffy = (27.9 kips)/[0.9(60 ksi)] = 0.52 in.2
The two #7 bars, which are an extension of the transverse chord reinforcement, provide an area of
reinforcement of 1.20 in.2
The reinforcement required by the Provisions for outofplane force is (195 plf) is far less than the ACI
318 requirement.
Figure 7.17 shows this joint along the wall.
FEMA 451, NEHRP Recommended Provisions: Design Examples
720
#3x 2'6"
standard hook
grouted into
each key joint
(2) #5 in
bond beam
(2) #7 bars
in joint
(chord bars)
Vertical wall
reinforcement
beyond
2"
cover
Figure 7.17 Joint 4 exterior longitudinal walls to diaphragm
reinforcement and outofplane anchorage (1.0 in. = 25.4 mm,
1.0 ft = 0.3048 m).
7.1.1.6.5 Joint 5 Design and Detailing
The required shear friction reinforcement along the wall length is computed as:
Avf5 = Vu5/fµfy = (12.5 kips)/[(0.75)(1.0)(60 ksi)] = 0.28 in.2
Provide #3 bars at each planktoplank joint for a total of 8 bars.
The required collector reinforcement is computed as:
As5 = Tu5/ffy = (27.9 kips)/[0.9(60 ksi)] = 0.52 in.2
Two #6 bars specified for the design of Joint 1 above provide an area of reinforcement of 0.88 in.2 Figure
7.18 shows this joint along the wall.
Chapter 7, Precast Concrete Design
721
4"
#3 x 4'8"
grouted into
each key joint
(2) #5 in
bond beam
(2) #6 bars
in joint
(collector bars)
Vertical wall
reinforcement
beyond
Figure 7.18 Walltodiaphragm reinforcement along interior
longitudinal walls  Joint 5 (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m).
7.1.2 Topped Precast Concrete Units for FiveStory Masonry Building, Los Angeles,
California (see Sec. 9.2)
This design shows the floor and roof diaphragms using topped precast units in the fivestory masonry
building in Los Angeles, California. The topping thickness exceeds the minimum thickness of 2 in. as
required for composite topping slabs by ACI 318 Sec. 21.7.4 [21.9.4]. The topping shall be lightweight
concrete (weight = 115 pcf) with a 28day compressive strength (fc' ) of 4,000 psi and is to act
compositely with the 8in.thick hollowcore precast, prestressed concrete plank. Design parameters are
provided in Sec. 9.2. Figure 9.21 shows the typical floor and roof plan.
7.1.2.1 General Design Requirements
Topped diaphragms may be used in any Seismic Design Category. ACI 318 Sec. 21.7 [21.9]provides
design provisions for topped precast concrete diaphragms. Provisions Sec. 5.2.6 [4.6] specifies the forces
to be used in designing the diaphragms. The amplification factor of Provisions Sec. 9.1.1.4 [not
applicable in the 2003 Provisions] is 1.03, the same as previously computed for the untopped diaphragm.
FEMA 451, NEHRP Recommended Provisions: Design Examples
722
[As noted above, the chord amplification factor has been dropped for the 2003 Provisions and does not
occur in ASC 31802.]
7.1.2.2 General InPlane Seismic Design Forces for Topped Diaphragms
The inplane diaphragm seismic design force (Fpx) is calculated using Provisions Eq. 5.2.6.4.4 [4.62] but
must not be less than 0.2SDSIwpx and need not be more than 0.4SDSIwpx. Vx must be added to Fpx calculated
using Eq. 5.2.6.4.4 [4.62] where:
wpx = the weight tributary to the diaphragm at Level x
SDS = the spectral response acceleration parameter at short periods (Provisions Sec. 4.1.2 [3.3.5])
I = occupancy importance factor (Provisions Sec. 1.4 [1.3])
Vx = the portion of the seismic shear force required to be transferred to the components of the
vertical seismicforceresisting system due to offsets or changes in stiffness of the vertical
resisting member at the diaphragm being designed.
For Seismic Design Category C and higher, Provisions Sec. 5.2.6.3.1 [4.6.2.2] requires that collector
elements, collector splices, and collector connections to the vertical seismicforceresisting members be
designed in accordance with Provisions Sec. 5.2.7.1 [4.2.2.2], which combines the diaphragm forces
times the overstrength factor ( O0
) and the effects of gravity forces. The parameters from example in Sec.
9.2 used to calculate inplane seismic design forces for the diaphragms are provided in Table 7.14.
Table 7.14 Design Parameters from Sec. 9.2
Design Parameter Value
Oo
2.5
wi (roof) 1,166 kips
wi (floor) 1,302 kips
SDS 1.0
I 1.0
Seismic Design Category D
1.0 kip = 4.45 kN.
7.1.2.3 Diaphragm Forces
As indicated previously, the weight tributary to the roof and floor diaphragms (wpx) is the total story
weight (wi) at Level i minus the weight of the walls parallel to the force.
Compute diaphragm weight (wpx) for the roof and floor as:
Roof
Total weight = 1,166 kips
Walls parallel to force = (60 psf)(277 ft)(8.67 ft/2) = 72 kips
wpx = 1,094 kips
Chapter 7, Precast Concrete Design
723
Floors
Total weight = 1,302 kips
Walls parallel to force = (60 psf)(277 ft)(8.67 ft) = 144 kips
wpx = 1,158 kips
Compute diaphragm demands in accordance with Provisions Eq. 5.2.6.4.4 [4.62]:
n
i
i x
px n px
i
i x
F
F w
w
=
=
S
=
S
Calculations for Fpx are provided in Table 7.15. The values for Fi and Vi are listed in Table 9.217.
Table 7.15 Fpx Calculations from Sec. 9.2
Level
wi
(kips)
n
i
i x
w
= S
(kips)
Fi
(kips)
n
i i
i x
F V
=
S =
(kips)
wpx
(kips)
Fpx
(kips)
Roof
4321
1,166
1,302
1,302
1,302
1,302
1,166
2,468
3,770
5,072
6,384
564
504
378
252
126
564
1,068
1,446
1,698
1,824
1,094
1,158
1,158
1,158
1,158
529
501
444
387
331
1.0 kip = 4.45 kN.
The minimum value of Fpx = 0.2SDSIwpx = 0.2(1.0)1.0(1,094 kips) = 219 kips (at the roof)
= 0.2(1.0)1.0(1,158 kips) = 232 kips (at floors)
The maximum value of Fpx = 0.4SDSIwpx = 2(219 kips) = 438 kips (at the roof)
= 2(232 kips) = 463 kips (at floors)
The value of Fpx used for design of the diaphragms is 463 kips, except for collector elements where forces
will be computed below.
7.1.2.4 Static Analysis of Diaphragms
The seismic design force of 463 kips is distributed as in Sec. 7.1.1.6 (Figure 7.11 shows the distribution).
The force is only 9.5 percent higher than that used to design the untopped diaphragm for the New York
design due to the intent to prevent yielding in the untopped diaphragm. Figure 7.12 shows critical
regions of the diaphragm to be considered in this design. Collector elements will be designed for 2.5
times the diaphragm force based on the overstrength factor (O0).
Joint forces taken from Sec. 7.1.1.5 times 1.095 are as:
FEMA 451, NEHRP Recommended Provisions: Design Examples
724
Joint 1 – Transverse forces
Shear, Vu1 = 114.5 kips × 1.095 = 125 kips
Moment, Mu1 = 2,061 ftkips × 1.095 = 2,250 ftkips
Chord tension force, Tu1 = M/d = 1.03 × 2,250 ftkips / 71 ft = 32.6 kips
Joint 2 – Transverse forces
Shear, Vu2 = 127 kips × 1.095 = 139 kips
Moment, Mu2 = 2,540 ftkips × 1.095 = 2,780 ftkips
Chord tension force, Tu2 = M/d = 1.03 × 2,780 ftkips / 71 ft = 39.3 kips
Joint 3 – Transverse forces
Shear, Vu3 = 78.1 kips × 1.095 = 85.5 kips
Moment, Mu2 = 3,738 ftkips × 1.095 = 4,090 ftkips
Chord tension force, Tu3 = M/d = 1.03 × 4,090 ftkips/71 ft = 59.3 kips
Joint 4 – Longitudinal forces
Wall Force, F = 52.9 kips × 1.095 = 57.9 kips
Wall shear along wall length, Vu4 =25 kips × 1.095 = 27.4 kips
Collector force at wall end, O0Tu4 = 2.5(27.9 kips)(1.095) = 76.4 kips
OutofPlane forces
Just as with the untopped diaphragm, the outofplane forces are controlled by ACI 318 Sec. 16.5,
which requires horizontal ties of 1.5 kips per foot from floor to walls.
Joint 5 – Longitudinal forces
Wall Force, F = 463 kips / 8 walls = 57.9 kips
Wall shear along each side of wall, Vu4 = 12.5 kips × 1.095 = 13.7 kips
Collector force at wall end, O0Tu4 = 2.5(27.9 kips)(1.095) = 76.4 kips
7.1.2.5 Diaphragm Design and Details
7.1.2.5.1 Minimum Reinforcement for 2.5 in. Topping
ACI 318 Sec. 21.7.5.1 [21.9.5.1] references ACI 318 Sec. 7.12, which requires a minimum As = 0.0018bd
for welded wire fabric. For a 2.5 in. topping, the required As = 0.054 in.2/ft. WWF 10×10  W4.5×W4.5
provides 0.054 in.2/ft. The minimum spacing of wires is 10 in. and the maximum spacing is 18 in. Note
that the ACI 318 Sec. 7.12 limit on spacing of five times thickness is interpreted such that the topping
thickness is not the pertinent thickness.
7.1.2.5.2 Boundary Members
Joint 3 has the maximum bending moment and is used to determine the boundary member reinforcement
of the chord along the exterior edge. The need for transverse boundary member reinforcement is
reviewed using ACI 318 Sec. 21.7.5.3 [21.9.5.3]. Calculate the compressive stress in the chord with the
ultimate moment using a linear elastic model and gross section properties of the topping. It is
Chapter 7, Precast Concrete Design
725
Figure 7.19 Diaphragm plan and section cuts.
conservative to ignore the precast units, but not necessary. As developed previously, the chord
compressive stress is:
6Mu3/td2 = 6(4,090 × 12)/(2.5)(72 × 12)2 = 158 psi
The chord compressive stress is less than 0.2fc' = 0.2(4,000) = 800 psi. Transverse reinforcement in the
boundary member is not required.
The required chord reinforcement is:
As3 = Tu3/ffy = (59.3 kips)/[0.9(60 ksi)] = 1.10 in.2
7.1.2.5.3 Collectors
The design for Joint 4 collector reinforcement at the end of the exterior longitudinal walls and for Joint 5
at the interior longitudinal walls is the same.
As4 = As5 = O0Tu4/ffy = (76.4 kips)/[0.9(60 ksi)] = 1.41 in.2
Use two #8 bars (As = 2 × 0.79 = 1.58 in.2) along the exterior edges, along the length of the exterior
longitudinal walls, and along the length of the interior longitudinal walls. Provide cover for chord and
collector bars and spacing between bars per ACI 318 Sec. 21.7.8.3 [21.9.8.3].
Minimum cover = 2.5(8/8) = 2.5 in., but not less than 2.0 in.
Minimum spacing = 3(8/8) = 3.0 in., but not less than 11/2 in.
Figure 7.19 shows the diaphragm plan and section cuts of the details and Figure 7.110, the boundary
member and chord/collector reinforcement along the edge. Given the close margin on cover, the
transverse reinforcement at lap splices also is shown.
FEMA 451, NEHRP Recommended Provisions: Design Examples
726
Splice bars
(2) #8 bars
(chord bars)
31
2"
3" 3"
21
2"
Grouted
chord / collector
element along exterior
edge of precast plank
Contact
lap splice
Prestressed
hollow core
plank with
roughened
top surface
Artificially
roughened
edge
WWF bend
down into
chord
21
2" min
(concrete
topping)
41
2"Ø spiral of 1
4" wire
with 2" pitch over each
lap splice.
Figure 7.110 Boundary member, and chord and collector reinforcement (1.0 in. = 25.4
mm).
(2) #8
(collector bars)
3"
21
21 3" 2"
2"
21
2" min
topping
WWF
Figure 7.111 Collector reinforcement at the end
of the interior longitudinal walls  Joint 5 (1.0 in.
= 25.4 mm, 1.0 ft = 0.3048 m).
Figure 7.111 shows the collector reinforcement for the interior longitudinal walls. The side cover of
21/2 in. is provided by casting the topping into the cores and by the stems of the plank. A minimum
space of 1 in. is provided between the plank stems and the sides of the bars.
7.1.2.5.4 Shear Resistance
Chapter 7, Precast Concrete Design
727
Cut out alternate face shells
(16" o.c. each side) and place
topping completely through
wall and between planks
(2) #5 in
masonry
bond beam
#3x4'0" at 16" to
lap with WWF
(2) #8
collector bars
1" clear
Vertical
reinforcement
WWF 10 x 10
W4.5 x W4.5
Figure 7.112 Walltodiaphragm reinforcement along interior
longitudinal walls  Joint 5 (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m).
Thin composite and noncomposite topping slabs on precast floor and roof members may not have reliable
shear strength provided by the concrete. In accordance with ACI 318 Sec. 21.7.7.2 [21.9.7.2], all of the
shear resistance must be provided by the reinforcement (that is, Vc = 0).
fVn = fAcv.nfy = 0.75(0.054 in.2/ft)60 ksi = 2.43 kips/ft
The shear resistance in the transverse direction is:
2.43 kips/ft (72 ft) = 175 kips
which is greater than the Joint 2 shear (maximum transverse shear) of 139 kips. No. 3 dowels are used to
make the welded wire fabric continuous across the masonry walls. The topping is to be cast into the
masonry walls as shown in Figure 7.112, and the spacing of the No. 3 bars is set to be modular with the
CMU.
The required shear reinforcement along the exterior longitudinal wall (Joint 4) is:
Avf4 = Vu4/fµfy = (27.4 kips)/[(0.75)(1.0)(60 ksi)] = 0.61 in.2
7.1.2.5.5 Check OutofPlane Forces
At Joint 4 with bars at 2 ft on center, Fp = 624 plf = 2 ft(624 plf) = 1.25 kips. The required reinforcement,
As = 1.25/(0.9)(60ksi) = 0.023 in.2. Provide #3 bars at 2 ft on center, which provides a nominal strength
of 0.11 x 60 / 2 = 3.3 klf. The detail provides more than required by ACI 318 Sec. 16.5 for the 1.5 klf tie
force. The development length was checked in the prior example. Using #3 bars at 2 ft on center will be
adequate, and the detail is shown in Figure 7.113. The detail at joint 2 is similar.
FEMA 451, NEHRP Recommended Provisions: Design Examples
728
(2) #5 in
masonry
bond beam
(2) #8
(collector bars)
WWF 10 x10
W4.5 x W4.5
2"
Vertical wall
reinforcement
beyond
#3x STD HK
2'6"
at 2'0" o.c.
Cut out face shells
@ 2'0" and place
topping into wall
Figure 7.113 Exterior longitudinal walltodiaphragm
reinforcement and outofplane anchorage  Joint 4 (1.0 in. =
25.4 mm, 1.0 ft = 0.3048 m).
Chapter 7, Precast Concrete Design
729
7.2 THREESTORY OFFICE BUILDING WITH PRECAST CONCRETE SHEAR
WALLS
This example illustrates the seismic design of ordinary precast concrete shear walls that may be used in
regions of low to moderate seismicity. The Provisions has one requirement for detailing such walls:
connections that resist overturning shall be Type Y or Z. ACI 31802 has incorporated a less specific
requirement, renamed the system as intermediate precast structural walls, and removed some of the detail.
This example shows an interpretation of the intent of the Provisions for precast shear wall systems in
regions of moderate and low seismicity, which should also meet the cited ACI 31802 requirements.
[As indicated at the beginning of this chapter, the requirements for precast shear wall systems in the 2003
Provisions have been revised – primarily to point to ACI 31802 by reference. See also Sec. 7.2.2.1 for
more discussion of system requirements.]
7.2.1 Building Description
This precast concrete building is a threestory office building (Seismic Use Group I) in southern New
England on Site Class D soils. The structure utilizes 10ftwide by 18in.deep prestressed double tees
(DTs) spanning 40 ft to prestressed inverted tee beams for the floors and the roof. The DTs are to be
constructed using lightweight concrete. Each of the abovegrade floors and the roof are covered with a 2
in.thick (minimum), normal weight castinplace concrete topping. The vertical seismicforceresisting
system is to be constructed entirely of precast concrete walls located around the stairs and
elevator/mechanical shafts. The only features illustrated in this example are the rational selection of the
seismic design parameters and the design of the reinforcement and connections of the precast concrete
shear walls. The diaphragm design is not illustrated.
As shown in Figure 7.21, the building has a regular plan. The precast shear walls are continuous from
the ground level to 12 ft above the roof. Walls of the elevator/mechanical pits are castinplace below
grade. The building has no vertical irregularities. The storytostory height is 12 ft.
FEMA 451, NEHRP Recommended Provisions: Design Examples
730
25'0" 25'0" 25'0" 25'0" 25'0" 25'0"
150'0"
40'0" 40'0" 40'0"
120'0"
15'0"
8'0"
26 IT 28
precast
beams
18" DT roof and floor
slabs (10 DT 18)
8" precast
shear walls
8'0"
Figure 7.21 Threestory building plan (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m).
The precast walls are estimated to be 8 in. thick for building mass calculations. These walls are normal
weight concrete with a 28day compressive strength, fc' = 5,000 psi. Reinforcing bars used at the ends of
the walls and in welded connectors are ASTM A706 (60 ksi yield strength). The concrete for the
foundations and belowgrade walls has a 28day compressive strength, fc' = 4,000 psi.
7.2.2 Design Requirements
7.2.2.1 Seismic Parameters of the Provisions
The basic parameters affecting the design and detailing of the building are shown in Table 7.21.
Chapter 7, Precast Concrete Design
731
Table 7.21 Design Parameters
Design Parameter Value
Seismic Use Group I I = 1.0
SS (Map 1 [Figure 3.31]) 0.266
S1 (Map 2 [Figure 3.32]) 0.08
Site Class D
Fa 1.59
Fv 2.4
SMS = FaSS 0.425
SM1 = FvS1 0.192
SDS = 2/3 SMS 0.283
SD1 = 2/3 SM1 0.128
Seismic Design Category B
Basic SeismicForceResisting System Bearing Wall System
Wall Type * Ordinary Reinforced Concrete Shear Walls
R 4
O0 2.5
Cd 4
* Provisions Sec. 9.1.1.3 [9.2.2.1.3] provides for the use of ordinary reinforced concrete shear
walls in Seismic Design Category B, which does not require adherence to the special seismic
design provisions of ACI 318 Chapter 21.
[The 2003 Provisions have adopted the 2002 U.S. Geological Survey probabilistic seismic hazard maps
and the maps have been added to the body of the 2003 Provisions as figures in Chapter 3. These figures
replace the previously used separate map package.]
[Ordinary precast concrete shear walls is recognized as a system in Table 4.31 of the 2003 Provisions.
Consistent with the philosophy that precast systems are not expected to perform as well as castinplace
systems, the design factors for the ordinary precast concrete shear walls per 2003 Provisions Table 4.31
are: R = 3, O0 = 2.5, and Cd = 3. Note that while this system is permitted in Seismic Design Category B,
unline ordinary reinforced concrete shear walls, it is not permitted in Seismic Design Category C.
Alternatively, as this example indicates conceptually, this building could be designed incorporating
intermediate precast concrete shear walls with the following design values per 2003 Provisions Table 4.3
1: R = 4, O0 = 2.5, and Cd = 4.]
7.2.2.2 Structural Design Considerations
7.2.2.2.1 Precast Shear Wall System
This system is designed to yield in bending at the base of the precast shear walls without shear slippage at
any of the joints. Although not a stated design requirement of the Provisions or ACI 31802 for this
Seismic Design Category, shear slip could kink the vertical rebar at the connection and sabotage the
intended performance, which counts on an R factor of 4. The flexural connections at the ends of the
FEMA 451, NEHRP Recommended Provisions: Design Examples
732
walls, which are highly stressed by seismic forces, are designed to be the Type Y connection specified in
the Provisions. See Provisions Sec. 9.1.1.2 [9.2.2.1.1] (ACI Sec. 21.1 [21.1]) for the definitions of
ordinary precast concrete structural walls and Provisions Sec. 9.1.1.12 [not applicable for the 2003
Provisions] (ACI Sec. 21.11.6) for the connections. The remainder of the connections (shear connectors)
are then made strong enough to ensure that the inelastic straining is forced to the intended location.
[Per 2003 Provisions Sec. 9.2.2.1.1 (ACI 31802 Sec. 21.1), ordinary precast concrete shear walls need
only satisfy the requirements of ACI 31802 Chapters 118 (with Chapter 16 superceding Chapter 14).
Therefore, the connections are to be designed in accordance with ACI 31802 Sec. 16.6.]
Although it would be desirable to force yielding to occur in a significant portion of the connections, it
frequently is not possible to do so with common configurations of precast elements and connections. The
connections are often unavoidable weak links. Careful attention to detail is required to assure adequate
ductility in the location of first yield and that no other connections yield prematurely. For this particular
example, the vertical bars at the ends of the shear walls act as flexural reinforcement for the walls and are
selected as the location of first yield. The yielding will not propagate far into the wall vertically due to
the unavoidable increase in flexural strength provided by unspliced reinforcement within the panel. The
issue of most significant concern is the performance of the shear connections at the same joint. The
connections are designed to provide the necessary shear resistance and avoid slip without unwittingly
increasing the flexural capacity of the connection because such an increase would also increase the
maximum shear force on the joint. At the base of the panel, welded steel angles are designed to be
flexible for uplift but stiff for inplane shear.
7.2.2.2.2 Building System
No height limitations are imposed (Provisions Table 5.2.2 [4.31]).
For structural design, the floors are assumed to act as rigid horizontal diaphragms to distribute seismic
inertial forces to the walls parallel to the motion. The building is regular both in plan and elevation, for
which, according to Provisions Table 5.2.5.1 [4.44], use of the ELF procedure (Provisions Sec. 5.4 [5.2])
is permitted.
Orthogonal load combinations are not required for this building (Provisions Sec. 5.2.5.2.1 [4.4.2.1]).
Ties, continuity, and anchorage (Provisions Sec. 5.2.6.1 and 5.2.6.2 [4.6.1.1 and 4.6.1.2]) must be
explicitly considered when detailing connections between the floors and roof, and the walls and columns.
This example does not include consideration of nonstructural elements.
Collector elements are required due to the short length of shear walls as compared to the diaphragm
dimensions, but are not designed in this example.
Diaphragms need to be designed for the required forces (Provisions Sec. 5.2.6.2.6 [4.6.1.9]), but that
design is not illustrated here.
The bearing walls must be designed for a force perpendicular to their plane (Provisions Sec. 5.2.6.2.7
[4.6.1.3]), but this requirement is of no real consequence for this building.
The drift limit is 0.025hsx (Provisions Table 5.2.8 [4.51]), but drift is not computed here.
Chapter 7, Precast Concrete Design
733
ACI 318 Sec. 16.5 requires minimum strengths for connections between elements of precast building
structures. The horizontal forces were described in Sec. 7.1; the vertical forces will be described in this
example.
7.2.3 Load Combinations
The basic load combinations (Provisions Sec. 5.2.7 [4.2.2]) require that seismic forces and gravity loads
be combined in accordance with the factored load combinations presented in ASCE 7 except that the
factors for seismic loads (E) are defined by Provisions Eq. 5.2.71 and 5.2.72 [4.21 and 4.22]:
E = .QE ± 0.2SDSD = (1.0)QE ± (0.2)(0.283)D = QE ± 0.0567D
According to Provisions Sec. 5.2.4.1 [4.3.3.1], . = 1.0 for structures in Seismic Design Categories A, B,
and C, even though this seismic resisting system is not particularly redundant.
The relevant load combinations from ASCE 7 are:
1.2D ± 1.0E + 0.5L
0.9D ± 1.0E
Into each of these load combinations, substitute E as determined above:
1.26D + QE + 0.5L
1.14D  QE + 0.5L (will not control)
0.96D + QE (will not control)
0.843D  QE
These load combinations are for loading in the plane of the shear walls.
7.2.4 Seismic Force Analysis
7.2.4.1 Weight Calculations
For the roof and two floors
18 in. double tees (32 psf) + 2 in. topping (24 psf) = 56.0 psf
Precast beams at 40 ft = 12.5 psf
16 in. square columns = 4.5 psf
Ceiling, mechanical, miscellaneous = 4.0 psf
Exterior cladding (per floor area) = 5.0 psf
Partitions = 10.0 psf
Total = 92.0 psf
The weight of each floor including the precast shear walls is:
(120 ft)(150 ft)(92 psf/1,000) + [15 ft(4) + 25 ft(2)](12 ft)(0.10 ksf) = 1,790 kips
Considering the roof to be the same weight as a floor, the total building weight is W = 3(1,790 kips) =
5,360 kips.
7.2.4.2 Base Shear
FEMA 451, NEHRP Recommended Provisions: Design Examples
734
The seismic response coefficient (Cs) is computed using Provisions Eq. 5.4.1.11 [5.22]:
0.283 0.0708
/ 41
DS
S
C S
R I
= = =
except that it need not exceed the value from Provisions Eq. 5.4.1.12 [5.23] computed as:
C
S
S T R I
= D1 = = 0128
029 4 1
0110
( / )
.
. ( / )
.
where T is the fundamental period of the building computed using the approximate method of Provisions
Eq. 5.4.2.11 [5.26]:
T Ch a r n x
= =(0.02)(36)0.75=0.29sec
Therefore, use Cs = 0.0708, which is larger than the minimum specified in Provisions Eq. 5.4.1.13 [not
applicable in the 2003 Provisions]:
Cs = 0.044ISDS = (.044)(1.0)(0.283) = 0 .0125
[The minimum Cs has been changed to 0.01 in the 2003 Provisions.]
The total seismic base shear is then calculated using Provisions Eq. 5.41 [5.21] as:
V = CsW = (0.0708)(5,370) = 380 kips
Note that this force is substantially larger than a design wind would be. If a nominal 20 psf were applied
to the long face and then amplified by a load factor of 1.6, the result would be less than half this seismic
force already reduced by an R factor of 4.
7.2.4.3 Vertical Distribution of Seismic Forces
The seismic lateral force (Fx) at any level is determined in accordance with Provisions Sec. 5.4.3 [5.2.3]:
Fx =CvxV
where
1
k
x x
vx n k
i i
i
C w h
w h
=
=
S
Since the period, T < 0.5 sec, k = l in both building directions. With equal weights at each floor level, the
resulting values of Cvx and Fx are as follows:
Roof Cvr = 0.50 Fr = 190 kips
Third Floor Cv3 = 0.33 F3 = 127 kips
Second Floor Cv2 = 0.17 F2 = 63.0 kips
Chapter 7, Precast Concrete Design
735
12'0" 12'0" 12'0"
95 kips
63.5 kips
31.5 kips
Grade
25'0"
V = SF = 190 kips
Figure 7.22 Forces on the longitudinal walls
(1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m).
7.2.4.4 Horizontal Shear Distribution and Torsion
7.2.4.4.1 Longitudinal Direction
Design each of the 25ftlong walls at the elevator/mechanical shafts for half the total shear. Since the
longitudinal walls are very close to the center of rigidity, assume that torsion will be resisted by the 15ftlong
stairwell walls in the transverse direction. The forces for each of the longitudinal walls are shown in
Figure 7.22.
7.2.4.4.2 Transverse Direction
Design the four 15ftlong stairwell walls for the total shear including 5 percent accidental torsion
(Provisions Sec. 5.4.4.2 [5.2.4.2]). A rough approximation is used in place of a more rigorous analysis
considering all of the walls. The maximum force on the walls is computed as:
V = 380/4 + 380(0.05)(150)/[(100 ft moment arm) × (2 walls in each set)] = 109 kips
Thus
Fr = 109(0.50) = 54.5 kips
F3 = 109(0.33) = 36.3 kips
F2 = 109(0.167) = 18.2 kips
Seismic forces on the transverse walls of the stairwells are shown in Figure 7.23.
FEMA 451, NEHRP Recommended Provisions: Design Examples
736
12'0" 12'0" 12'0"
54.5 kips
36.3 kips
18.2 kips
Grade
15'0"
V = SF = 109 kips
Figure 7.23 Forces on the transverse walls
(1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m).
7.2.5 PROPORTIONING AND DETAILING
The strength of members and components is determined using the strengths permitted and required in
ACI 318 excluding Chapter 21 (see Provisions Sec. 9.1.1.3 [9.2.2.1.3]).
7.2.5.1 Overturning Moment and End Reinforcement
Design shear panels to resist overturning by means of reinforcing bars at each end with a direct tension
coupler at the joints. A commonly used alternative is a threaded posttensioning bar inserted through the
stack of panels, but the behavior is different, and the application of the rules for a Type Y connection to
such a design is not clear.
7.2.5.1.1 Longitudinal Direction
The freebody diagram for the longitudinal walls is shown in Figure 7.24. The tension connection at the
base of the precast panel to the below grade wall is governed by the seismic overturning moment and the
dead loads of the panel and supported floors and roof. In this example, the weights for an elevator
penthouse, with a floor and equipment at 180 psf between the shafts and a roof at 20 psf, are included.
The weight for the floors includes double tees, ceiling and partition (total of 70 psf), but not beams and
columns. Floor live load is 50 psf, except 100 psf is used in the elevator lobby. Roof snow load is 30 psf.
(The elevator penthouse is so small that it was ignored in computing the gross seismic forces on the
building, but it is not ignored in the following calculations.)
Chapter 7, Precast Concrete Design
737
12'0" 12'0" 12'0"
95 kips
63.5 kips
31.5 kips
23'6"
V
12'0"
9" 9"
T
C
12'0"
D
D
D
D
Figure 7.24 Freebody diagram for
longitudinal walls (1.0 kip = 4.45 kN,
1.0 ft = 0.3048 m).
At the base
ME = (95 kips)(36 ft) + (63.5 kips)(24 ft) + (31.5 kips)(12 ft) = 5,520 ftkips
3D = wall + exterior floors (& roof) + lobby floors + penthouse floor + penthouse roof
= (25 ft)(48 ft)(0.1 ksf) + (25 ft)(48 ft/2)(0.070 ksf)(3) + (25 ft)(8 ft/2)(0.070 ksf)(2) +
(25 ft)(8 ft/2)(0.18 ksf) + (25ft )(24 ft/2)(0.02 ksf)
= 120 + 126 + 14 + 18 + 6 = 284 kips
3L = (25 ft)(48 ft/2)(0.05 ksf)(2) + (25 ft)(8 ft/2)(0.1 ksf) = 60 + 10 = 70 kips
3S = (25ft)(48 ft + 24 ft)(0.03 ksf)/2 = 27 kips
Using the load combinations described above, the vertical loads for combining with the overturning
moment are computed as:
Pmax = 1.26 D + 0.5 L + 0.2 S = 397 kips
Pmin = 0.843 D = 239 kips
The axial load is quite small for the wall panel. The average compression Pmax / Ag = 0.165 ksi (3.3
percent of f'c). Therefore, the tension reinforcement can easily be found from the simple couple shown on
Figure 7.24.
The effective moment arm is:
jd = 25  1.5 = 23.5 ft
FEMA 451, NEHRP Recommended Provisions: Design Examples
738
and the net tension on the uplift side is:
min 5320 239 107 kips
u 2 23.5 2
T M P
jd
=  =  =
The required reinforcement is:
As = Tu/ffy = (107 kips)/[0.9(60 ksi)] = 1.98 in.2
Use two #9 bars (As = 2.0 in.2 ) at each end with direct tension couplers for each bar at each panel joint.
Since the flexural reinforcement must extend a minimum distance d (the flexural depth)beyond where it is
no longer required, use both #9 bars at each end of the panel at all three levels for simplicity.
At this point a check of ACI 318 Sec. 16.5 will be made. Bearing walls must have vertical ties with a
nominal strength exceeding 3 kips/ft, and there must be at least two ties per panel. With one tie at each
end of a 25 ft panel, the demand on the tie is:
Tn = (3 kip/ft)(25 ft)/2 = 37.5 kip
The two #9 bars are more than adequate for the ACI requirement.
Although no check for confinement of the compression boundary is required for ordinary precast concrete
shear walls, it is shown here for interest. Using the check from ACI 31899 Sec. 21.6.6.2 [21.7.6.2], the
depth to the neutral axis is:
Total compression force = As fy + Pmax = (2.0)(60) + 397 = 517 kips
Compression block a = (517 kips)/[(0.85)(5 ksi)(8 in. width)] = 15.2 in.
Neutral axis depth c = a/(0.80) = 19.0 in.
The maximum depth (c) with no boundary member per ACI 31899 Eq. 218 [218] is:
c ( )
l
h u w
=
600 d /
where the term (du/hw) shall not be taken less than 0.007. Once the base joint yields, it is unlikely that
there will be any flexural cracking in the wall more than a few feet above the base. An analysis of the
wall for the design lateral forces using 50% of the gross moment of inertia, ignoring the effect of axial
loads, and applying the Cd factor of 4 to the results gives a ratio (du/hw) far less than 0.007. Therefore,
applying the 0.007 in the equation results in a distance c of 71 in., far in excess of the 19 in. required.
Thus, ACI 31899 would not require transverse reinforcement of the boundary even if this wall were
designed as a special reinforced concrete shear wall. For those used to checking the compression stress as
an index:
= 742 psi ( )
( )
( )2 ( )
389 6 5,520
8 25 12 8 25 12
P M
A S
s = + = +
The limiting stress is 0.2fc' , which is 1000 psi, so no transverse reinforcement is required at the ends of
the longitudinal walls.
7.2.5.1.2 Transverse Direction
Chapter 7, Precast Concrete Design
739
12'0" 12'0" 12'0"
18.2 kips
36.3 kips
54.5 kips
12'0"
9" 13'6" 9"
7'0"
V C
T
D
D
D
D
Figure 7.25 Freebody diagram of the transverse walls
(1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m).
The freebody diagram of the transverse walls is shown in Figure 7.25. The weight of the precast
concrete stairs is 100 psf and the roof over the stairs is 70 psf.
The transverse wall is similar to the longitudinal wall.
At the base
ME = (54.5 kips)(36 ft) + (36.3 kips)(24 ft) + (18.2 kips)(12 ft) = 3,052 ftkips
3D = (15 ft)(48 ft)(0.1 ksf) + 2(12.5 ft/2)(10 ft/2)(0.07 ksf)(3) + (15 ft)(8 ft/2)[(0.1 ksf)(3) +
(0.07 ksf)] = 72 + 13 + 18 + 4 = 107 kips
3L = 2(12.5 ft/2)(10 ft/2)(0.05 ksf)(2) + (15 ft)(8 ft/2)(0.1 ksf)(3) = 6 + 18 = 24 kips
3S = [2(12.5 ft/2)(10 ft/2) + (15 ft)(8 ft/2)](0.03 ksf) = 3.7 kips
Pmax = 1.26(107) + 0.5(24) + 0.2(4) = 148 kips
Pmin = 0.843(107) = 90.5 kips
jd = 15  1.5 = 13.5 ft
Tu = (Mnet/jd)  Pmin/2 = (3,052/13.5)  90.5/2 = 181 kips
As = Tu/ffy = (181 kips)/[0.9(60 ksi)] = 3.35 in.2
FEMA 451, NEHRP Recommended Provisions: Design Examples
740
Use two #10 and one #9 bars (As = 3.54 in.2 ) at each end of each wall with a direct tension coupler at each
bar for each panel joint. All three bars at each end of the panel will also extend up through all three levels
for simplicity. Following the same method for boundary member check as on the longitudinal walls:
Total compression force = As fy + Pmax = (3.54)(60) + 148 = 360 kips
Compression block a = (360 kips)/[(0.85)(5 ksi)(8 in. width)] = 10.6 in.
Neutral axis depth c = a/(0.80) = 13.3 in.
Even though this wall is more flexible and the lateral loads will induce more flexural cracking, the
computed deflections are still small and the minimum value of 0.007 is used for the ratio (du/hw). This
yields a maximum value of c = 42.9 in., thus confinement of the boundary would not be required. The
check of compression stress as an index gives:
= 951 psi ( )
( )
( )2 ( )
140 6 2,930
8 15 12 8 15 12
P M
A S
s = + = +
Since s < 1,000 psi, no transverse reinforcement is required at the ends of the transverse walls. Note how
much closer to the criterion this transverse wall is by the compression stress check.
The overturning reinforcement and connection are shown in Figures 7.26. Provisions Sec. 9.1.1.12 [not
applicable in the 2003 Provisions] (ACI 21.11.6.4) requires that this Type Y connection develop a
probable strength of 125% of the nominal strength and that the anchorage on either side of the connection
develop 130% of the defined probable strength. [As already noted, the connection requirements for
ordianry precast concrete shear walls have been removed in the 2003 Provisions and the ACI 31802
requirements are less specific.] The 125% requirement applies to the grouted mechanical splice, and the
requirement that a mechanical coupler develop 125% of specified yield strength of the bar is identical to
the Type 1 coupler defined by ACI 318 Sec. 21.2.6.1. Some of the grouted splices on the market can
qualify as the Type 2 coupler defined by ACI, which must develop the specified tensile strength of the
bar. The development length, ld, for the spliced bars is multiplied by both the 1.25 and the 1.3 factors to
satisfy the Provisions requirement. The bar in the panel is made continuous to the roof, therefore no
calculation of development length is necessary in the panel. The dowel from the foundation will be
hooked, otherwise the depth of the foundation would be more than required for structural reasons. The
size of the foundation will provide adequate cover to allow the 0.7 factor on ACI’s standard development
length for hooked bars. For the # 9 bar:
'
1.3(1.25) (1.6.25)0.7(1200) 1365(1.128) 24.3 in.
4000
b
dh
c
l d
f
= = =
Similarly, for the #10 bar, the length is 27.4 in.
Like many shear wall designs, this design does concentrate a demand for overturning resistance on the
foundation. In this instance the resistance may be provided by a large footing (on the order of 20 ft by 28
ft by 3 ft thick) under the entire stairwell, or by deep piers or piles with an appropriate cap for load
transfer. Refer to Chapter 4 for examples of design of each type of foundation, although not for this
particular example.
Chapter 7, Precast Concrete Design
741
(2) #10 & (1) #9 ea. end,
full ht. of 15' transverse
wall panel
(2) #9 ea. end, full height
of 25' longitudinal
wall panel
8" precast wall
Direct tension
coupler(typical)
8" 8"
Longitudinal Wall Transverse Wall
1" shim and drypack
(typical)
Reinforced foundation
not designed in the
example
3" min 25" min for #9
Development at Foundation
Standard hook to develop
overturning reinforcement
28" min for #10
Figure 7.26 Overturning connection detail at the base of the
walls (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
7.2.5.2 Shear Connections and Reinforcement
Panel joints often are designed to resist the shear force by means of shear friction but that technique is not
used for this example because the joint at the foundation will open due to flexural yielding. This opening
would concentrate the shear stress on the small area of the drypacked joint that remains in compression.
This distribution can be affected by the shims used in construction. Tests have shown that this often leads
to slip of the joint, which could lead to a kink in the principal tension reinforcement at or near its splice
and destroy the integrity of the system. Therefore, the joint will be designed with direct shear connectors
that will prevent slip along the joint. This is the authors’ interpretation of the Provisions text indicating
that “Type Y connections shall develop under flexural, shear, and axial load actions, as required, a
probable strength. . . .” based upon 125 percent of the specified yield in the connection. It would not be
required by the ACI 31802 rules for intermediate precast walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples
742
Welded wire
fabric Plate 3
8x4x1'0"
L4x3x5
16x0'8"
LLH
Plate 1
2x12x1'0"
1
4 8
Drypack
(a) Section through connection
(b) Side elevation
(c) Section through embedded assembly
#5, see (c)
3
4"Ø H.A.S.
1
4
4
C8x18.75
1
4
Figure 7.27 Shear connection at base (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
7.2.5.2.1 Longitudinal Direction
The shear amplification factor is determined as:
2
(1.25) max / 2 (2.0 in. )(1.25)(60 ksi)(23.5 ft) (397 kip)(23.5 ft / 2)
5320 ftkip
capacity s y
demand u
M A fjd P jd
M M
+ +
= =
=1.54
Therefore, the design shear (Vu) at the base is 1.54(190 kips) = 292 kips
The base shear connection is shown in Figure 7.27 and is to be flexible vertically but stiff horizontally in
the plane of the panel. The vertical flexibility is intended to minimize the contribution of these
connections to overturning resistance, which would simply increase the shear demand.
In the panel, provide an assembly with two face plates 3/8 in. × 4 in. × 12 in. connected by a C8x18.75
and with diagonal #5 bars as shown in the figure. In the foundation provide an embedded plate 1/2 × 12 ×
1'6" with six 3/4 in. diameter headed anchor studs. In the field, weld an L 4 × 3 × 5/16 × 0'8", long leg
horizontal, on each face. The shear capacity of this connection is checked:
Chapter 7, Precast Concrete Design
743
Shear in the two loose angles
fVn = f(0.6Fu)tl(2) = (0.75)(0.6)(58 ksi)(0.3125 in.)(8 in.)(2) = 130.5 kip
Weld at toe of loose angles
fVn = f(0.6Fu)tel(2) = (0.75)(0.6)(70 ksi)(0.25 in. / o 2)(8 in.)(2) = 89.1 kip
Weld at face plates, using Table 89 in AISC Manual (3rd edition; same table is 842 in 2nd edition)
fVn = CC1Dl(2 sides)
C1 = 1.0 for E70 electrodes
l = 8 in.
D = 4 (sixteenths of an inch)
k = 2 in. / 8 in. = 0.25
` a = eccentricity, summed vectorially: horizontal component is 4 in.; vertical component is 2.67
in.; thus, al = 4.80 in. and a = 4.8 in./8 in. = 0.6 from the table. By interpolation, C = 1.29
fVn = (1.29)(1.0)(4)(8)(2) = 82.6 kip
Weld from channel to plate has at least as much capacity, but less demand.
Bearing of concrete at steel channel
fc = f(0.85f'c) = 0.65(0.85)(5 ksi) = 2.76 ksi
The C8 has the following properties:
tw = 0.487 in.
bf = 2.53 in.
tf = 0.39 in. (average)
The bearing will be controlled by bending in the web (because of the tapered flange, the critical
flange thickness is greater than the web thickness). Conservatively ignoring the concrete’s
resistance to vertical deformation of the flange, compute the width (b) of flange loaded at 2.76 ksi
that develops the plastic moment in the web:
Mp = fFytw
2/4 = (0.9)(50 ksi)(0.4872 in.2)/4 = 2.67 in.kip/in.
Mu = fc[(btw)2/2  (tw/2)2/2] = 2.76[(b  0.243 in.)2  (0.243 in.)2]/2
setting the two equal results in b = 1.65 in.
Therefore bearing on the channel is
fVc = fc(2  tw)(l) = (2.76 ksi)[(2(1.65)  0.487 in.](6 in.) = 46.6 kip
To the bearing capacity on the channel is added the 4  #5 diagonal bars, which are effective in
tension and compression; f = 0.75 for shear is used here:
fVs = ffyAscosa = (0.75)(60 ksi)(4)(0.31 in.2)(cos 45E) = 39.5 kip
Thus, the total capacity for transfer to concrete is:
fVn = fVc + fVs = 46.6 + 39.6 = 86.1 kip
FEMA 451, NEHRP Recommended Provisions: Design Examples
744
The capacity of the plate in the foundation is governed by the headed anchor studs. The Provisions
contain the new anchorage to concrete provisions that are in ACI 31802 Appendix D. [In the 2003
Provisions, the anchorage to concrete provisions have been removed and replaced by the reference to
ACI 31802.] Capacity in shear for anchors located far from an edge of concrete, such as these, and
with sufficient embedment to avoid the pryout failure mode is governed by the capacity of the steel:
fVs = f n Ase fut = (0.65)(6 studs)(0.44 in.2 per stud)(60 ksi) = 103 kip
Provisions Sec 9.2.3.3.2 (ACI 31802 Sec. D.3.3.3) specifies an additional factor of 0.75 to derate
anchors in structures assigned to Seismic Design Categories C and higher.
In summary the various shear capacities of the connection are:
Shear in the two loose angles: 130.5 kip
Weld at toe of loose angles: 89.1 kip
Weld at face plates: 82.6 kip
Transfer to concrete: 86.1 kip
Headed anchor studs at foundation: 103 kip
The number of embedded plates (n) required for a panel is:
n = 292/82.6 = 3.5
Use four connection assemblies, equally spaced along each side (5'0" on center works well to avoid the
end reinforcement). The plates are recessed to position the #5 bars within the thickness of the panel and
within the reinforcement of the panel.
It is instructive to consider how much moment capacity is added by the resistance of these connections to
vertical lift at the joint. The vertical force at the tip of the angle that will create the plastic moment in the
leg of the angle is:
T = Mp / x = Fylt2/4 / (lk) = (36 ksi)(8 in)(0.31252 in.2)/4]/(4 in.  0.69 in.) = 2.12 kips
There are four assemblies with two loose angles each, giving a total vertical force of 17 kips. The
moment resistance is this force times half the length of the panel, which yields 212 ftkips. The total
demand moment, for which the entire system is proportioned, is 5320 ft  kips. Thus, these connections
will add about 4% to the resistance and ignoring this contribution is reasonable. If a straight plate 1/4 in.
x 8 in., which would be sufficient, were used and if the welds and foundation embedment did not fail first,
the tensile capacity would be 72 kips each, a factor of 42 increase over the angles, and the shear
connections would have the unintended effect of more than doubling the flexural resistance, which could
easily cause failures in the system.
Using ACI 318 Sec. 11.10, check the shear strength of the precast panel at the first floor:
fVc=f2Acv fc'hd=0.85(2) 5,000(8)(23.5)(12) = 271 kips
Because fVc $Vu = 190 kips, the wall is adequate for shear without even considering the reinforcement.
Note that the shear strength of wall itself is not governed by the overstrength required for the connection.
However, since Vu $ fVc/2 = 136 kips, ACI Sec. 11.10.8 requires minimum wall reinforcement in
accordance with ACI 318 Sec. 11.10.9.4 rather than Chapter 14 or 16. For the minimum required .h =
0.0025, the required reinforcement is:
Chapter 7, Precast Concrete Design
745
Plate 5
16x5"x0'8"
1
4
See Figure 7.27
for embedded
plates
Shim and
drypack
Horizontal and
vertical edges
Figure 7.28 Shear connections on each side of the wall at the
second and third floors (1.0 in = 25.4 mm).
Av = 0.0025(8)(12) = 0.24 in.2/ft
As before, use two layers of welded wire fabric, WWF 4×4  W4.0×W4.0, one on each face. Shear
reinforcement provided, Av = 0.12(2) = 0.24 in.2/ft
Next, compute the shear strength at Level 2. Since the end reinforcement at the base extends to the top of
the shear wall, bending is not a concern. Yield of the vertical bars will not occur, the second floor joint
will not open (unlike at the base) and, therefore, shear friction could rationally be used to design the
connections at this level and above. Shear keys in the surface of both panels would be advisable. Also,
because of the lack of flexural yield at the joint, it is not necessary to make the shear connection be
flexible with respect to vertical movement. To be consistent with the seismic force increase from yielding
at the base, the shear at this level will be increased using the same amplification factor as calculated for
the first story.
The design shear, Vu2 = 1.54(95 + 63.5) = 244 kips.
Using the same recessed embedded plate assemblies in the panel as at the base, but welded with a straight
plate, the number of plates, n = 244/82.6 = 2.96. Use three plates, equally spaced along each side.
Figure 7.28 shows the shear connection at the second and third floors of the longitudinal precast concrete
shear wall panels.
FEMA 451, NEHRP Recommended Provisions: Design Examples
746
7.2.5.2.2 Transverse Direction
Use the same procedure as for the longitudinal walls:
2
(1.25) max / 2 (3.54 in. )(1.25)(60 ksi)(13.5 ft) (148 kip)(23.5 ft / 2)
3052 ftkip
capacity s y
demand u
M A fjdP jd
M M
+ +
= =
= 1.50
Design shear, Vu at base is 1.50(105 kips) = 157.5 kips.
Use the same shear connections as at the base of the longitudinal walls (Figure 7.27). The connection
capacity is 82.6 kips and the number of connections required is n = 157.5/82.6 = 1.9. Provide two
connections on each panel.
Check the shear strength of the first floor panel as described previously:
fVc=f2fc'hd=0.85(2) 5,000(8)(13.5)(12) = 156 kips
Similar to the longitudinal direction, fVc $Vu = 142 kips, but Vu $ fVc/2 so provide two layers of welded
wire fabric, WWF 4×4  W4.0×W4.0, one on each face as in the longitudinal walls.
Compute the shear demand at the second floor level joint as indicated below.
The design shear, Vu = 1.50(52.3+ 34.9) = 130.8 kips.
Use the same plates as in the longitudinal walls. The number of plates, n = 130.8/82.6 = 1.6. Use two
plates, equally spaced. Use the same shear connections for the transverse walls as for the longitudinal
walls as shown in Figures 7.27 and 7.28.
Chapter 7, Precast Concrete Design
747
7.3 ONESTORY PRECAST SHEAR WALL BUILDING
This example illustrates precast shear wall seismic design using monolithic emulation as defined in the
Provisions Sec. 9.1.1.12 [not applicable in the 2003 Provisions] (ACI Sec. 21.11.3) for a singlestory
building in a region of high seismicity. For buildings in Seismic Design Category D, Provisions Sec.
9.1.1.12 [not applicable in the 2003 Provisions] (ACI Sec. 21.11.2.1) requires that the precast seismicforce
resisting system emulate the behavior of monolithic reinforced concrete construction or that the
system’s cyclic capacity be demonstrated by testing. This example presents an interpretation of
monolithic emulation design with ductile connections. Here the connections in tension at the base of the
wall panels yield by bending steel angles outofplane. The same connections at the bottom of the panel
are detailed and designed to be very strong in shear and to resist the nominal shear strength of the
concrete panel.
[Many of the provisions for precast concrete shear walls in areas of high seismicity have been moved out
of the 2003 Provisions and into ACI 31802. For structures assigned to Seismic Design Category D,
2003 Provisions Sec. 9.2.2.1.3 (ACI 31802 Sec. 21.21.1.4) permits special precast concrete shear walls
(ACI 31802 Sec. 21.8) or intermediate precast concrete shear walls (ACI 31802 Sec. 21.13). The 2003
Provisions does not differentiate between precast or castinplace concrete for special shear walls. This is
because ACI 31802 Sec. 21.8 essentially requires special precast concrete shear walls to satisfy the same
design requirements as special reinforced concrete shear walls (ACI 31802 Sec. 21.7). Alternatively,
special precast concrete shear walls are permitted if they satisfy experimental and analytical requirements
contained in 2003 Provisions Sec. 9.2.2.4 and 9.6.]
7.3.1 Building Description
The precast concrete building is a singlestory industrial warehouse building (Seismic Use Group I)
located in the Los Angeles area on Site Class C soils. The structure has 8ftwide by 121/2in.deep
prestressed double tee (DT) wall panels. The roof is light gage metal decking spanning to bar joists that
are spaced at 4 ft on center to match the location of the DT legs. The center supports for the joists are
joist girders spanning 40 ft to steel tube columns. The vertical seismicforceresisting system is the
precast/prestressed DT wall panels located around the perimeter of the building. The average roof height
is 20 ft, and there is a 3 ft parapet. The building is located in the Los Angeles area on Site Class C soils.
Figure 7.31 shows the plan of the building, which is regular.
FEMA 451, NEHRP Recommended Provisions: Design Examples
748
15 DT at 8'0" = 120'0"
48'0" 48'0"
12 DT at 8'0" = 96'0"
Steel tube
columns
Joist girder
(typical)
24LH03
at 4'0" o.c.
24LH03
at 4'0" o.c.
3 DT at 8'0" =
24'0"
16'0"
O.H.
door
5 DT at 8'0" = 40'0" 3 DT at 8'0" =
24'0"
16'0"
O.H.
door
Figure 7.31 Singlestory industrial warehouse building plan (1.0 ft = 0.3048 m).
The precast wall panels used in this building are typical DT wall panels commonly found in many
locations but not normally used in Southern California. For these wall panels, an extra 1/2 in. has been
added to the thickness of the deck (flange). This extra thickness is intended to reduce cracking of the
flanges and provide cover for the bars used in the deck at the base. The use of thicker flanges is
addressed later.
Provisions Sec. 9.1.1.5 [9.2.2.1.5.4] (ACI Sec. 21.2.5.1 [21.2.5.1]) limits the grade and type of
reinforcement in boundary elements of shear walls and excludes the use of bonded prestressing tendons
(strand) due to seismic loads. ACI 31899 Sec. 21.7.5.2 [21.9.5.2] permits the use of strand in boundary
elements of diaphragms provided the stress is limited to 60,000 psi. This design example uses the strand
as the reinforcement based on that analogy. The rationale for this is that the primary reinforcement of the
DT, the strand, is not working as the ductile element of the wall panel and is not expected to yield in an
earthquake.
The wall panels are normalweight concrete with a 28day compressive strength, fc ' = 5,000 psi.
Reinforcing bars used in the welded connections of the panels and footings are ASTM A706 (60 ksi).
The concrete for the foundations has a 28day compressive strength, fc ' = 4,000 psi.
Chapter 7, Precast Concrete Design
749
7.3.2 Design Requirements
7.3.2.1 Seismic Parameters of the Provisions
The basic parameters affecting the design and detailing of the building are shown in Table 7.31.
Table 7.31 Design Parameters
Design Parameter Value
Seismic Use Group I I = 1.0
SS (Map 1 [Figure 3.31]) 1.5
S1 (Map 2 [Figure 3.32]) 0.60
Site Class C
Fa 1.0
Fv 1.3
SMS = FaSS 1.5
SM1 = FvS1 0.78
SDS = 2/3 SMS 1.0
SD1 = 2/3 SM1 0.52
Seismic Design Category D
Basic SeismicForceResisting System Bearing Walls System
Wall Type * Special Reinforced Concrete Shear Wall
R 5
O0 2.5
Cd 5
* Provisions Sec. 9.7.1.2 [9.2.2.1.3] requires special reinforced concrete shear walls in Seismic
Design Category D and requires adherence to the special seismic design provisions of ACI 318
Chapter 21.
[The 2003 Provisions have adopted the 2002 U.S. Geological Survey seismic hazard maps and the maps
have been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously
used separate map package).]
7.3.2.2 Structural Design Considerations
7.3.2.2.1 Precast Shear Wall System
The criteria for the design is to provide yielding in a dry connection for bending at the base of each
precast shear wall panel while maintaining significant shear resistance in the connection. The flexural
connection for a wall panel at the base is located in one DT leg while the connection at the other leg is
used for compression. Per Provisions Sec. 9.1.1.12 (ACI Sec. 21.11.3.1) [not applicable in the 2003
Provisions], these connections resist the shear force equal to the nominal shear strength of the panel and
have a nominal strength equal to twice the shear that exists when the actual moment is equal to Mpr
FEMA 451, NEHRP Recommended Provisions: Design Examples
750
(which ACI defines as f = 1.0 and a steel stress equal to 125% of specified yield). Yielding will develop
in the dry connection at the base by bending the horizontal leg of the steel angle welded between the
embedded plates of the DT and footing. The horizontal leg of this angle is designed in a manner to resist
the seismic tension of the shear wall due to overturning and then yield and deform inelastically. The
connections on the two legs of the DT are each designed to resist 50 percent of the shear. The anchorage
of the connection into the concrete is designed to satisfy the Type Z requirements in Provisions Sec.
9.1.1.12 (ACI Sec. 21.11.6.5) [not applicable in the 2003 Provisions.]. Careful attention to structural
details of these connections is required to ensure tension ductility and resistance to large shear forces that
are applied to the embedded plates in the DT and footing.
[Based on the 2003 Provisions, unless the design of special precast shear walls is substantiated by
experimental evidence and analysis per 2003 Provisions Sec. 9.2.2.4 (ACI 31802 Sec. 21.8.2), the design
must satisfy ACI 31802 Sec. 21.7 requirements for special structural walls as referenced by ACI 31802
Sec. 21.8.1. The connection requirements are not as clearly defined as in the 2000 Provisions.]
7.3.2.2.2 Building System
Height limit is 160 ft (Provisions Table 5.2.2 [4.31]).
The metal deck roof acts as a flexible horizontal diaphragm to distribute seismic inertia forces to the walls
parallel to the earthquake motion (Provisions Sec. 5.2.3.1 [4.3.2.1]).
The building is regular both in plan and elevation.
The reliability factor, . is computed in accordance with Provisions Sec. 5.2.4.2 [4.3.3]. The maximum .x
value is given when maxx is the largest value. is the ratio of design story shear resisted by the r maxx r
single element carrying the most shear force to the total story shear. All shear wall elements (8ftwide
panels) have the same stiffness. Therefore, the shear in each element is the total shear along a side
divided by the number of elements (wall panels). The largest maxx value is along the side with the least r
number of panels. Along the side with 11 panels, maxx is computed as: r
maxx = = 0.0455 r
1 2
11
1.0
Ax = 96 ft × 120 ft = 11,520 ft2
2 20 2 20 = 2.10
0.0455 11,520 x
x
rmax Ax
. =  = 
Therefore, use . = 1.0.
[The redundancy requirements have been substantially changed for the 2003 Provisions. For a shear wall
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a single
shear wall with an aspect ratio greater than 1.0 would not result in more than a 33 percent reduction in
story strength or create an extreme torsional irregularity. Based on the design procedures for the walls,
each individual panel should be considered a separate wall with an aspect ratio greater than 1.0.
Alternatively, if the structure is regular in plan and there are at least two bays of perimeter framing on
each side of the structure in each orthogonal direction, the exception in 2003 Provisions Sec. 4.3.3.2
Chapter 7, Precast Concrete Design
751
permits the use of D, . = 1.0. This exception could be interpreted as applying to this example, which is
regular and has more than two wall panels (bays) in both directions.]
The structural analysis to be used is the ELF procedure (Provisions Sec. 5.4 [5.2]) as permitted by
Provisions Table 5.2.5 [4.41].
Orthogonal load combinations are not required for flexible diaphragms in Seismic Design Category D
(Provisions Sec. 5.2.5.2.3 [4.4.2.3]).
This example does not include design of the foundation system, the metal deck diaphragm, or the
nonstructural elements.
Ties, continuity, and anchorage (Provisions 5.2.6.1 through 5.2.6.4 [4.6]) must be explicitly considered
when detailing connections between the roof and the wall panels. This example does not include the
design of these connections, but sketches of details are provided to guide the design engineer.
There are no drift limitations for singlestory buildings as long as they are designed to accommodate
predicted lateral displacements (Provisions Table 5.2.8, footnote b [4.51, footnote c]).
7.3.3 Load Combinations
The basic load combinations (Provisions Sec. 5.2.7) require that seismic forces and gravity loads be
combined in accordance with the factored load combinations as presented in ASCE 7, except that the load
factor for earthquake effects (E) is defined by Provisions Eq. 5.2.71 and 5.2.72 [4.21 and 4.22]:
E = .QE ± 0.2SDSD = (1.0)QE ± (0.2)(1.0)D = QE ± 0.2D
The relevant load combinations from ASCE 7 are:
1.2D ± 1.0E + 0.5L
0.9D ± 1.0E
Note that roof live load need not be combined with seismic loads, so the live load term, L, can be omitted
from the equation.
Into each of these load combinations, substitute E as determined above:
1.4D + QE
1.0D  QE (will not control)
1.1D + QE (will not control)
0.7D  QE
These load combinations are for the inplane direction of the shear walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples
752
7.3.4 Seismic Force Analysis
7.3.4.1 Weight Calculations
Compute the weight tributary to the roof diaphragm
Roofing = 2.0 psf
Metal decking = 1.8 psf
Insulation = 1.5 psf
Lights, mechanical, sprinkler system etc. = 3.2 psf
Bar joists = 2.7 psf
Joist girder and columns = 0.8 psf
Total = 12.0 psf
The total weight of the roof is computed as:
(120 ft × 96 ft)(12 psf/1,000) = 138 kips
The exterior double tee wall weight tributary to the roof is:
(20 ft/2 + 3 ft)[42 psf/1,000](120 ft + 96 ft)2 = 236 kips
Total building weight for seismic lateral load, W = 138 + 236 = 374 kips
7.3.4.2 Base Shear
The seismic response coefficient (Cs) is computed using Provisions Eq. 5.4.1.11 [5.22] as:
1.0 0.20
/ 51
DS
s
C S
R I
= = =
except that it need not exceed the value from Provisions Eq. 5.4.1.12 [5.23] as follows:
( ) ( )
1 0.52 0.55
0.189 5 1
D
s
C S
T R I
= = =
where T is the fundamental period of the building computed using the approximate method of Provisions
Eq. 5.4.2.11 [5.26]:
( )0.75 x (0.02) 20.0 0.189 sec
Ta=Crhn= =
Therefore, use Cs = 0.20, which is larger than the minimum specified in Provisions Eq. 5.4.1.13 [not
applicable in the 2003 Provisions]:
Cs = 0.044ISDS = (0.044)(1.0)(1.0) = 0 .044
[The minimum Cs value has been changed to 0.01 in. the 2003 Provisions.
The total seismic base shear is then calculated using Provisions Eq. 5.41 [5.21] as:
Chapter 7, Precast Concrete Design
753
8'0"
20'0" 3'0"
D1 D1
Vlu
D2
2'0" 2'0" 2'0" 2'0"
DT leg
Foundation
T C
Vlu
Figure 7.32 Freebody diagram of a panel in the
longitudinal direction (1.0 ft = 0.3048 m).
V = CsW = (0.20)(374) = 74.8 kips
7.3.4.3 Horizontal Shear Distribution and Torsion
Torsion is not considered in the shear distribution in buildings with flexible diaphragms. The shear along
each side of the building will be equal, based on a tributary area force distribution.
7.3.4.3.1 Longitudinal Direction
The total shear along each side of the building is V/2 = 37.4 kips. The maximum shear on longitudinal
panels (at the side with the openings) is:
Vlu = 37.4/11 = 3.4 kips
On each side, each longitudinal wall panel resists the same shear force as shown in the freebody diagram
of Figure 7.32, where D1 represents roof joist reactions and D2 is the panel weight.
FEMA 451, NEHRP Recommended Provisions: Design Examples
754
8'0"
20'0" 3'0"
Vtu
D
2'0" 2'0"
DT leg
Foundation
T C
Vtu
Figure 7.33 Freebody diagram of a panel in the
transverse direction (1.0 ft = 0.3048 m).
7.3.4.3.2 Transverse Direction
Seismic forces on the transverse wall panels are all equal and are:
Vtu = 37.4/12 = 3.12 kips
Figure 7.33 shows the transverse wall panel freebody diagram.
Note the assumption of uniform distribution to the wall panels in a line requires that the roof diaphragm
be provided with a collector element along its edge. The chord designed for diaphragm action in the
perpendicular direction will normally be capable of fulfilling this function, but an explicit check should
be made in the design.
7.3.5 Proportioning and Detailing
The strength of members and components is determined using the strengths permitted and required in
ACI 318 including Chapter 21.
Chapter 7, Precast Concrete Design
755
7.3.5.1 Tension and Shear Forces at the Panel Base
Design each precast shear panel to resist the seismic overturning moment by means of a ductile tension
connector at the base of the panel. A steel angle connector will be provided at the connection of each leg
of the DT panel to the concrete footing. The horizontal leg of the angle is designed to yield in bending as
needed in an earthquake. Provisions Sec. 9.1.1.12 [not applicable in the 2003 Provisions] requires that
dry connections at locations of nonlinear action comply with applicable requirements of monolithic
concrete construction and satisfy the following:
1. Where the moment action on the connection is assumed equal to Mpr, the coexisting shear on the
connection shall be no greater than 0.5SnConnection and
2. The nominal shear strength for the connection shall not be less than the shear strengths of the
members immediately adjacent to that connection.
Precisely how ductile dry connections emulate monolithic construction is not clearly explained. The dry
connections used here do meet the definition of a yielding steel element at a connection contained in ACI
31802. For the purposes of this example, these two additional requirements are interpreted as:
1. When tension from the seismic overturning moment causes 1.25 times the yield moment in the angle,
the horizontal shear on this connection shall not exceed onehalf the nominal shear strength of the
connection. For this design, onehalf the total shear will be resisted by the angle at the DT leg in
tension and the remainder by the angle at the DT leg in compression.
2. The nominal shear strength of the connections at the legs need to be designed to exceed the inplane
shear strength of the DT.
Determine the forces for design of the DT connection at the base.
7.3.5.1.1 Longitudinal Direction
Use the freebody diagram shown in Figure 7.32. The maximum tension for the connection at the base
of the precast panel to the concrete footing is governed by the seismic overturning moment and the dead
loads of the panel and the roof. The weight for the roof is 11.2 psf, which excludes the joist girders and
columns.
At the base
ME = (3.4 kips)(20 ft) = 68.0 ftkips
Dead loads
( ) = 1.08 kips 1
11.2 1,000 48 4
2
D = .. ..
. .
D2 = 0.042(23)(8) = 7.73 kips
SD = 2(1.08) + 7.73 = 9.89 kips
1.4D = 13.8 kips
0.7 D = 6.92 kips
Compute the tension force due to net overturning based on an effective moment arm, d = 4.0 ft (distance
between the DT legs). The maximum is found when combined with 0.7D:
FEMA 451, NEHRP Recommended Provisions: Design Examples
756
21 10"
2"
23
8"
4'0"
43
4"
average
M
23
8"
Figure 7.34 Cross section of the DT
drypacked at the footing (1.0 in = 25.4
mm, 1.0 ft = 0.3048 m).
Tu = ME/d  0.7D/2 = 68.0/4  6.92/2 = 13.5 kips
7.3.5.1.2 Transverse Direction
For the transverse direction, use the freebody diagram of Figure 7.33. The maximum tension for
connection at the base of the precast panel to the concrete footing is governed by the seismic overturning
moment and the dead loads of just the panel. No load from the roof is included, since it is negligible.
At the base
ME = (3.12 kips)(20 ft) = 62.4 ftkips
The dead load of the panel (as computed above) is D2 = 7.73 kips, and 0.7D = 5.41.
The tension force is computed as above for d = 4.0 ft (distance between the DT legs):
Tu = 62.4/4  5.41/2 = 12.9 kips
This tension force is less than that at the longitudinal wall panels. Use the tension force of the
longitudinal wall panels for the design of the angle connections.
7.3.5.2 Panel Reinforcement
Check the maximum compressive stress in the DT leg for the
requirement of transverse boundary element reinforcement per ACI
318 Sec. 21.6.6.3 [21.7.6.3]. Figure 7.34 shows the cross section
used. The section is limited by the area of drypack under the DT at
the footing.
The reason to limit the area of drypack at the footing is to locate the
boundary elements in the legs of the DT, at least at the bottom of the
panel. The flange between the legs of the DT is not as susceptible to
cracking during transportation as are the corners of DT flanges
outside the confines of the legs. The compressive stress due to the
overturning moment at the top of the footing and dead load is:
A = 227 in.2
S = 3240 in.3
13,800 12(68,000) 313psi
227 3,240
P ME
A S
s = + = + =
Roof live loads need not be included as a factored axial load in the compressive stress check, but the force
from the prestress steel will be added to the compression stress above because the prestress force will be
effective a few feet above the base and will add compression to the DT leg. Each leg of the DT will be
reinforced with one 1/2in. diameter and one 3/8in. diameter strand. Figure 7.35 shows the location of
these prestressed strands.
Chapter 7, Precast Concrete Design
757
Deck mesh
(1) 1
2" dia. strand
(1) 3
8" dia. strand
Leg mesh
43
4"
average
21
6" 4" 2"
Figure 7.35 Cross section of one DT leg showing the location of the
bonded prestressing tendons or strand (1.0 in = 25.4 mm).
Next, compute the compressive stress resulting from these strands. Note the moment at the height of
strand development above the footing, about 26 in. for the effective stress (fse), is less than at the top of
footing. This reduces the compressive stress by:
(3.4)(26) x 1000 27 psi
3, 240
=
In each leg, use
P = 0.58fpu Aps = 0.58(270 ksi)[0.153 + 0.085] = 37.3 kips
A = 168 in.2
e = yb  CGStrand = 9.48  8.57 = 0.91 in.
Sb = 189 in.3
37,300 0.91(37,300) 402psi
168 189
P Pe
A S
s = + = + =
Therefore, the total compressive stress is approximately 313 + 402  27 = 688 psi.
The limiting stress is 0.2 fc', which is 1000 psi, so no special boundary elements are required in the
longitudinal wall panels.
Reinforcement in the DT for tension is checked at 26 in. above the footing. The strand reinforcement of
the DT leg resisting tension is limited to 60,000 psi. The rationale for using this stress is discussed at the
beginning of this example.
D2 = (0.042)(20.83)(8) = 7.0 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
758
Pmin = 0.7(7.0 + 2(1.08)) = 6.41 kips
ME = (3.4)(17.83) = 60.6 ftkips
Tu = Mnet/d  Pmin/2= 12.0 kips
The area of tension reinforcement required is:
As = Tu/ffy = (12.0 kips)/[0.9(60 ksi)] = 0.22 in.2
The area of one ½ in. diameter and one 3/8 in. diameter strand is 0.153 in.2 + 0.085 in.2 = 0.236 in.2 The
mesh in the legs is available for tension resistance, but not required in this check.
To determine the nominal shear strength of the concrete for the connection design, complete the shear
calculation for the panel in accordance with ACI Sec. 21.6 [21.7]. The demand on each panel is:
Vu = Vlu = 3.4 kips
Only the deck between the DT legs is used to resist the inplane shear (the legs act like flanges, meaning
that the area effective for shear is the deck between the legs). First, determine the minimum required
shear reinforcement based on ACI Sec. 21.6.2.1 [21.7.2]. Since
Acv fc'=2.5(48) 5,000=8.49 kips
exceeds Vu = 3.4 kips, the reinforcement of the deck is per ACI 318 Sec. 16.4.2. Using welded wire
fabric, the required areas of reinforcement are:
Ash = Asv = (0.001)(2.5)(12) = 0.03 in.2/ft
Provide 6 × 6  W2.5 × W2.0 welded wire fabric.
Ash = 0.05 in.2/ft
Asv = 0.04 in.2/ft
The nominal shear strength of the wall panel by ACI 318 Sec. 21.6.4.1 is:
( ) (2.5)(48)2 5,000 0.05(4)(60) 29.0kips
Vn=Acvac fc'+.nfy = 1,000 + =
where ac is 2.0 for hw/lw = 23/4 = 5.75, which is greater than 2.0. Given that the connections will be
designed for a shear of 29 kips, it is obvious that half the nominal shear strength will exceed the seismic
shear demand, which is 3.4 kips.
The prestress force and the area of the DT legs are excluded from the calculation of the nominal shear
strength of the DT wall panel. The prestress force is not effective at the base, where the connection is,
and the legs are like the flanges of a channel, which are not effective in shear.
7.3.5.3 Size the Yielding Angle
The angle, which is the ductile element of the connection, is welded between the plates embedded in the
DT leg and the footing. This angle is a L5 × 31/2 × 3/4 × 0 ft5 in. with the long leg vertical. The steel
for the angle and embedded plates will be ASTM A572, Grade 50. The horizontal leg of the angle needs
to be long enough to provide significant displacement at the roof, although this is not stated as a
Chapter 7, Precast Concrete Design
759
y
x
1" y
4"
5"
Mx
CG
B
21
4"
Tu
L5x31
2x3
4x5
(LLV)
Fillet
weld "t"
Mx
Fillet
weld
k = 11
4"
Location
of plastic
hinge
t
t
Mz
My
y
z
My
Mz
Vu
'
'
Vu'
Tu '
Vu'
Tu '
Vu'
Tu '
Figure 7.36 Freebody of the angle and the fillet weld connecting the embedded plates in
the DT and the footing (elevation and section) (1.0 in = 25.4 mm).
requirement in either the Provisions or ACI 318. This will be examined briefly here. The angle and its
welds are shown in Figure 7.36.
The bending moment at a distance k from the heel of the angle (location of the plastic hinge in the angle)
is:
Mu = Tu(3.5  k) = 13.5(3.5  1.25) = 30.4 in.kips
( ) ( )2 5 0.75
0.9 0.9 50 31.6 in.kips
fbMn FyZ 4
. .
= = . .=
.. ..
Providing a stronger angle (e.g., a shorter horizontal leg) will simply increase the demands on the
remainder of the assembly. Using Provisions Sec. 9.1.1.12 (ACI Sec. 21.11.6.5) [not applicable in the
2003 Provisions], the tension force for the remainder of this connection other than the angle is based upon
a probable strength equal to 140% of the nominal strength. Thus
(1.4) (50)(5)(0.75)2 / 4 1.4 21.9 kips
3.5 3.5 1.25
n
u
T M
k
'= = × =
 
Check the welds for the tension force of 21.9 kips and a shear force (Vu')of 29.0/2 = 14.5 kips, or the
shear associated with Tu', whichever is greater. The bearing panel, with its larger vertical load, will give a
larger shear.
FEMA 451, NEHRP Recommended Provisions: Design Examples
760
1.4D = 13.8 kips, and V = [Tu'(4) + 1.4D(2)]/20 = [21.9/4 + 13.8(2)]/20 = 5.76 kips. Vn for the panel
obviously controls.
But before checking the welds, consider the deformability of the system as controlled by the yielding
angle. Ignore all sources of deformation except the angle. (This is not a bad assumption regarding the
double tee itself, but other aspects of the connections, particularly the plate and reinforcement embedded
in the DT, will contribute to the overall deformation. Also, the diaphragm deformation will overwhelm
all other aspects of deformation, but this is not the place to address flexible diaphragm issues.) The angle
deformation will be idealized as a cantilever with a length from the tip to the center of the corner, then
upward to the level of the bottom of the DT, which amounts to:
L = 3.5 in.  t/2 + 1 in.  t/2 = 3.75 in.
Using an elasticplastic idealization, the vertical deformation at the design moment in the leg is
dv = TL3/3EI = (13.5 kips)(3.75 in.)3/[3(29000 ksi)(5 in.)(0.75 in.)3/12] = 0.047 in.
This translates into a horizontal motion at the roof of 0.24 in. (20 ft to the roof, divided by the 4 ft from
leg to leg at the base of the DT.) With Cd of 4, the predicted total displacement is 0.96 in. These
displacements are not very large, but now compare with the expectations of the Provisions. The
approximate period predicted for a 20fttall shear wall building is 0.19 sec. Given a weight of 374 kips,
as computed previously, this would imply a stiffness from the fundamental equation of dynamics:
T2 W/gK42W/(gT) 42374 /(386 0.19) 201 kip/in.
K
=p . =p =p × =
Now, given the design seismic base shear of 74.8 kips, this would imply an elastic displacement of
dh = 74.8 kip / (201 kip/in.) = 0.37 in.
This is about 50% larger than the simplistic calculation considering only the angle. The bending of angle
legs about their weak axis has a long history of providing ductility and, thus, it appears that this dry
connection will provide enough deformability to be in the range of expectation of the Provisions.
7.3.5.4 Welds to Connection Angle
Welds will be fillet welds using E70 electrodes.
For the base metal, fRn = f(Fy)ABM.
For which the limiting stress is fFy = 0.9(50) = 45.0 ksi.
For the weld metal, fRn = f(Fy)Aw = 0.75(0.6)70(0.707)Aw.
For which limiting stress is 22.3 ksi.
Size a fillet weld, 5 in. long at the angle to embedded plate in the footing:
Using an elastic approach
Resultant force = V2+T2= 14.52+ 21.92= 26.3kips
Chapter 7, Precast Concrete Design
761
Z
X
V
My
V
Mz
Figure 7.37 Freebody of angle with welds, top view,
showing only shear forces and resisting moments.
Aw = 26.3/22.3 = 1.18 in.2
t = Aw/l =1.18 in.2/5 in. = 0.24 in.
For a 3/4 in. angle leg, use a 5/16 in. fillet weld. Given the importance of this weld, increasing the size to
3/8 in. would be a reasonable step. With ordinary quality control to avoid flaws, increasing the strength
of this weld by such an amount should not have a detrimental effect elsewhere in the connection.
Now size the weld to the plate in the DT. Continue to use the conservative elastic method to calculate
weld stresses. Try a fillet weld 5 in. long across the top and 4 in. long on each vertical leg of the angle.
Using the freebody diagram of Figure 7.36 for tension and Figure 7.37 for shear, the weld moments
and stresses are:
Mx = Tu'(3.5) = 21.9(3.5) = 76.7 in.kips
My = Vu'(3.5) = (14.5)(3.5) = 50.8 in.kips
Mz = Vu'(yb + 1.0)
= 14.5(2.77 + 1.0) = 54.7 in.kips
For the weld between the angle and the embedded plate in the DT as shown in Figure 7.37 the section
properties for a weld leg (t) are:
A = 13t in.2
Ix = 23.0t in.4
Iy = 60.4t in.4
FEMA 451, NEHRP Recommended Provisions: Design Examples
762
Ip = Ix + Iy = 83.4t in.4
yb = 2.77 in.
xL = 2.5 in.
To check the weld, stresses are computed at all four ends (and corners). The maximum stress is at the
lower right end of the inverted U shown in Figure 7.36.
14.5 (54)(2.77) 2.93 ksi
13 83.4
21.9 (54.7)(2.5) 0.045 ksi
13 83.4
=  (50.8)(2.5)  (76.7)(2.77) = 11.3 ksi
60.4 23.0t
u zb
x
p
u z L
y
p
y L x b
z
y x
V M y
A I t t t
T M x
A I t t t
M x M y
I I t t
s
s
s
= ' + = + =.. ..
. .
=  ' + =  + =.. ..
. .
=   .. ..
. .
R x2 y2 z21t(2.93)2 (0.045)2 ( 11.3)211.t67ksi
s = s +s +s = + +  =.. ..
. .
Thus, t = 11.67/22.3 = 0.52 in., say 9/16 in. Field welds are conservatively sized with the elastic method
for simplicity and to minimize construction issues.
7.3.5.5 Tension and Shear at the Footing Embedment
Reinforcement to anchor the embedded plates is sized for the same tension and shear, and the
development lengths are lengthened by an additional 30%, per Provisions Sec. 9.1.1.12 (ACI Sec.
21.11.6.5) [not applicable in the 2003 Provisions]. Reinforcement in the DT leg and in the footing will be
welded to embedded plates as shown in Figure 7.38.
The welded reinforcement is sloped to provide concrete cover and to embed the bars in the central region
of the DT leg and footing. The tension reinforcement area required in the footing is:
( )( )
2
,
21.9 0.45 in.
cos 0.9 60 cos26.5
u
s Sloped
y
A T
ff .
'
= = =
o
Use two #5 bars (As = 0.62 in.2 ) at each embedded plate in the footing.
The shear bars in the footing will be two #4 placed on an angle of two (plus)toone. The resultant shear
resistance is:
fVn = 0.75(0.2)(2)(60)(cos 26.5E) = 16.1 kips
Chapter 7, Precast Concrete Design
763
DT
9
1
1
Interior slab 2
L6x4x1
2x10"
(2) #5
5
16 5
L5x31
2x3
4x5
(LLV)
9
16 4 9
16 5
Plate 1
2 x 6 x 0'10"
(2) #4x24"
(see Fig 7.39)
Plate 6x41
2x1
2
C.I.P.
concrete
footing
(2) #4x 1
2
2'6"
2'6"
(2) #4 with
standard hooks
(2) #3 with
standard hooks
(2) #4x48"
(See Fig 7.39)
weld
on #4
Figure 7.38 Section at the connection of the precast/prestressed shear wall panel and
the footing (1.0 in = 25.4 mm).
7.3.5.6 Tension and Shear at the DT Embedment
The area of reinforcement for the welded bars of the embedded plate in the DT, which develop tension as
the angle bends through cycles is:
21.9 0.408 in.2
cos 0.9(60)cos6.3
u
s
y
A T
ff .
'
= = = o
Two #4 bars are adequate. Note that the bars in the DT leg are required to extend upward 1.3 times the
development length, which would be 22 in. In this case they will be extended 22 in. past the point of
development of the effective stress in the strand, which totals about 48 in.
The same embedded plate used for tension will also be used to resist onehalf the nominal shear. This
shear force is 14.5 kips. The transfer of direct shear to the concrete is easily accomplished with bearing
on the sides of the reinforcing bars welded to the plate. Two #5 and two #4 bars (explained later) are
welded to the plate. The available bearing area is approximately Abr = 4(0.5 in.)(5 in.(available)) = 10 in.2
and the bearing capacity of the concrete is fVn = (0.65)(0.85)(5 ksi)(10 in.2) = 27.6 kips > 14.5 kip
demand.
FEMA 451, NEHRP Recommended Provisions: Design Examples
764
The weld of these bars to the plate must develop both the tensile demand and this shear force. The weld
is a flare bevel weld, with an effective throat of 0.2 times the bar diameter along each side of the bar.
(Refer to the PCI Handbook.) For the #4 bar, the weld capacity is
fVn = (0.75)(0.6)(70 ksi)(0.2)(0.5 in.)(2) = 6.3 kips/in.
The shear demand is prorated among the four bars as (14.5 kip)/4 = 3.5 kip. The tension demand is the
larger of 1.25 fy on the bar (15 kip) or Tu/2 (11.0 kip). The vectorial sum of shear and tension demand is
15.4 kip. Thus, the minimum length of weld is 15.4 / 6.3 = 2.4 in.
7.3.5.7 Resolution of Eccentricities at the DT Embedment
Check the twisting of the embedded plate in the DT for Mz.
Use Mz = 54.7 in.kips.
( ) ( )( )
54.7 0.11 in.2
0.9 60 9.0
z
s
y
A M
ff jd
= = =
Use one #4 bar on each side of the vertical embedded plate in the DT as shown in Figure 7.39. This is
the same bar used to transfer direct shear in bearing.
Check the DT embedded plate for My (50.8 in.kips) and Mx (76.7 in.kips) using the two #4 bars welded
to the back side of the plate near the corners of the weld on the loose angle and the two #3 bars welded to
the back side of the plate near the bottom of the DT leg (as shown in Figure 7.39). It is relatively
straightforward to compute the resultant moment magnitude and direction, assume a triangular shaped
compression block in the concrete, and then compute the resisting moment. It is quicker to make a
reasonable assumption as to the bars that are effective and then compute resisting moments about the X
and Y axes. This approximate method is demonstrated here. The #4 bars are effective in resisting Mx,
and one each of the #3 and #4 bars are effective in resisting My. For My assume that the effective depth
extends 1 in. beyond the edge of the angle (equal to twice the thickness of the plate). Begin by assigning
onehalf of the “corner” #4 to each component.
With Asx = 0.20 + 0.20/2 = 0.30 in.2 ,
fMnx = fAs fy jd = (0.9)(0.3 in.2)(60 ksi)(0.95)(5 in.) = 77 in.kips (>76.7).
With Asy = 0.11 + 0.20/2 = 0.21 in.2,
fMny = fAs fy jd = (0.9)(0.21 in.2)(60 ksi)(0.95)(5 in.) = 54 in.kips (>50.8).
Each component is strong enough, so the proposed bars are satisfactory.
Chapter 7, Precast Concrete Design
765
10"
3"
For 1.25 Fy
Plate 41
2"x6"x1
2"
with 5
8" slot
at center
Plate 10"x6"x1
2"
1"
2" 3"
2" 3"
(2) #3 with
standard hook
#4
(2) #4 with
standard hook
Figure 7.39 Details of the embedded plate in the DT at the base (1.0 in = 25.4 mm).
Metal deck
L4x3x1
4x
continuous
DT
Plate at each DT leg
Bar joists
DT corbel at
each leg
Figure 7.311 Sketch of connection of loadbearing DT wall panel at the roof (1.0 in =
25.4 mm).
FEMA 451, NEHRP Recommended Provisions: Design Examples
766
L4x3x1 Bar joist
4
continuous
2'0"
Metal deck
Deck straps
as needed
Plate at each
DT leg
#4 continuous
weld to plates
Figure 7.310 Sketch of connection of nonloadbearing DT wall panel at the
roof (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
7.3.5.8 Other Connections
This design assumes that there is no inplane shear transmitted from panel to panel. Therefore, if
connections are installed along the vertical joints between DT panels to control the outofplane
alignment, they should not constrain relative movement inplane. In a practical sense, this means the
chord for the roof diaphragm should not be a part of the panels. Figures 7.310 and 7.311 show the
connections at the roof and DT wall panels. These connections are not designed here. Note that the
continuous steel angle would be expected to undergo vertical deformations as the panels deform laterally.
Because the diaphragm supports concrete walls out of their plane, Provisions Sec. 5.2.6.3.2 [4.6.2.1]
requires specific force minimums for the connection and requires continuous ties across the diaphragm.
Also, it specifically prohibits use of the metal deck as the ties in the direction perpendicular to the deck
span. In that direction, the designer may wish to use the top chord of the bar joists, with an appropriate
connection at the joist girder, as the continuous cross ties. In the direction parallel to the deck span, the
deck may be used but the laps should be detailed appropriately.
In precast double tee shear wall panels with flanges thicker than 21/2 in., consideration may be given to
using vertical connections between the wall panels to transfer vertical forces resulting from overturning
moments and thereby reduce the overturning moment demand. These types of connections are not
considered here, since the uplift force is small relative to the shear force and cyclic loading of bars in thin
concrete flanges is not always reliable in earthquakes.
81
8
COMPOSITE STEEL AND CONCRETE
James Robert Harris, P.E., Ph.D. and
Frederick R. Rutz, P.E., Ph.D.
This chapter illustrates application of the 2000 NEHRP Recommended Provisions to the design of
composite steel and concrete framed buildings using partially restrained composite connections. This
system is referred to as a “Composite Partially Restrained Moment Frame (CPRMF)” in the Provisions.
An example of a multistory medical office building in Denver, Colorado, is presented. The Provisions set
forth a wealth of opportunities for designing composite steel and concrete systems, but this is the only one
illustrated in this set of design examples.
The design of partially restrained composite (PRC) connections and their effect on the analysis of frame
stiffness are the aspects that differ most significantly from a noncomposite design. Some types of PRC
connections have been studied in laboratory tests and a design method has been developed for one in
particular, which is illustrated in this example. In addition, a method is presented by which a designer
using readily available frame analysis programs can account for the effect of the connection stiffness on
the overall frame.
The example covers only design for seismic forces in combination with gravity, although a check on drift
from wind load is included.
The structure is analyzed using threedimensional static methods. The RISA 3D analysis program, v.4.5
(Risa Technologies, Foothill Ranch, California) is used in the example.
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the
changes are described, the design examples and calculations have not been revised to reflect the changes
to the 2003 Provisions.
Chapter 10 in the 2003 Provisions has been expanded to include modifications to the basic reference
document, AISC Seismic, Part II. These modifications are generally related to maintaining compatibility
between the Provisions and the most recent editions of the ACI and AISC reference documents and to
incorporate additional updated requirements. Updates to the reference documents, in particular AISC
Seismic, have some affect on the calculations illustrated herein.
There are not any general technical changes to other chapters of the 2003 Provisions that have a
significant effect on the calculations and/or design example in this chapter of the Guide with the possible
exception of the updated seismic hazard maps.
FEMA 451, NEHRP Recommended Provisions: Design Examples
82
Where they affect the design examples in this chapter, significant changes to the 2003 Provisions and
primary reference documents are noted. However, some minor changes to the 2003 Provisions and the
reference documents may not be noted.
In addition to the 2000 NEHRP Recommended Provisions (referred to herein as the Provisions), the
following documents are referenced:
ACI 318 American Concrete Institute. 1999. Building Code Requirements for Structural
Concrete, Standard ACI 31899. Detroit: ACI.
AISC LRFD American Institute of Steel Construction. 1999. Load and Resistance Factor Design
Specification for Structural Steel Buildings. Chicago: AISC.
AISC Manual American Institute of Steel Construction. 1998. Manual of Steel Construction, Load
and Resistance Factor Design, Volumes 1 and 2, 2nd Edition. Chicago: AISC.
AISC Seismic American Institute of Steel Construction. 1997. Seismic Provisions for Structural
Steel Buildings, including Supplement No. 2 (2000). Chicago:
AISC SDGS8 American Institute of Steel Construction. 1996. Partially Restrained Composite
Connections, Steel Design Guide Series 8. Chicago: AISC.
ASCE TC American Society of Civil Engineers Task Committee on Design Criteria for
Composite Structures in Steel and Concrete. October 1998. “Design Guide for
Partially Restrained Composite Connections,” Journal of Structural Engineering
124(10)..
ASCE 7 American Society of Civil Engineers. 1998. Minimum Design Loads for Buildings
and Other Structures, ASCE 798. Reston: ASCE.
The shortform designations presented above for each citation are used throughout.
The symbols used in this chapter are from Chapter 2 of the Provisions, the above referenced documents,
or are as defined in the text. Customary U.S. units are used.
Chapter 8, Composite Steel and Concrete
83
W18x35
(typical for
EW beams)
25'0" 25'0" 25'0" 25'0"
12'6" 12'6" 25'0" 25'0" 25'0" 12'6" 12'6"
20K5 at
3'11
2" o.c.
(typical)
25'0"
W10
(typical)
W21x44
(typical for
NS beams)
W E
S
N
Figure 81 Typical floor plan (1.0 ft = 0.3048 m).
8.1 BUILDING DESCRIPTION
This fourstory medical office building has a structural steel framework (see Figures 81 through 83).
The floors and roof are supported by open web steel joists. The floor slab is composite with the floor
girders and the spandrel beams and the composite action at the columns is used to create moment resisting
connections. Figure 84 shows the typical connection. This connection has been studied in several
research projects over the past 15 years and is the key to the building’s performance under lateral loads.
The structure is free of irregularities both in plan and elevation. This is considered a Composite Partially
Restrained Moment Frame (CPRMF) per Provisions Table 5.2.2 and in AISC Seismic, and it is an
appropriate choice for buildings with lowtomoderate seismic demands, which depend on the building as
well as the ground shaking hazard.
FEMA 451, NEHRP Recommended Provisions: Design Examples
84
North and South End Elevation
25'0" 25'0" 25'0"
4 at 13'0" = 52'0"
25'0" 25'0"
2
3
4
Roof
W18x35
(typical)
Figure 82 Building end elevation (1.0 ft = 0.3048 m).
East and West Side Elevation
2
3
4
Roof
12'6" 12'6" 25'0" 25'0" 25'0" 12'6" 12'6"
4 at 13'0" = 52'0"
W21x44
(typical)
Figure 83 Building side elevation (1.0 ft = 0.3048 m).
The building is located in a relatively low hazard region (Denver, Colorado), but some internal storage
loading and Site Class E are used in this example to provide somewhat higher seismic design forces for
purposes of illustration, and to push the example into Seismic Design Category C.
Chapter 8, Composite Steel and Concrete
85
Double angle
web connection
Column
Seat angle
Girder
Rebar
Headed stud
Concrete
Figure 84 Typical composite connection.
There are no foundations designed in this example. For this location and system, the typical foundation
would be a drilled pier and voided grade beam system, which would provide flexural restraint for the
strong axis of the columns at their base (very similar to the foundation for a conventional steel moment
frame). The main purpose here is to illustrate the procedures for the partially restrained composite
connections. The floor slabs serve as horizontal diaphragms distributing the seismic forces, and by
inspection they are stiff enough to be considered as rigid.
The typical bay spacing is 25 feet. Architectural considerations allowed an extra column at the end bay of
each side in the northsouth direction, which is useful in what is the naturally weaker direction. The
exterior frames in the northsouth direction have momentresisting connections at all columns. The
frames in each bay in the eastwest direction have momentresisting connections at all except the end
columns. Composite connections to the weak axis of the column are feasible, but they are not required
for this design. This arrangement is illustrated in the figures.
Material properties in this example are as follows:
1. Structural steel beams and columns (ASTM A992): Fy = 50 ksi
2. Structural steel connection angles and plates (ASTM A36): Fy = 36 ksi
3. Concrete slab (4.5 inches thick on form deck, normal weight): fc' = 3000 psi
4. Steel reinforcing bars (ASTM A615): Fy = 60 ksi
The floor live load is 50 psf, except in 3 internal bays on each floor where medical records storage
imposes 200 psf, and the roof snow load is taken as 30 psf. Wind loads per ASCE 7 are also checked, and
the stiffness for serviceability in wind is a factor in the design. Dead loads are relatively high for a steel
building due to the 4.5" normal weight concrete slab used to control footfall vibration response of the
open web joist system and the precast concrete panels on the exterior walls.
This example covers the following aspects of seismic design that are influenced by partially restrained
composite frame systems:
1. Load combinations for composite design
2. Assessing the flexibility of the connections
3. Incorporating the connection flexibility into the analytical model of the building
FEMA 451, NEHRP Recommended Provisions: Design Examples
86
4. Design of the connections
8.2 SUMMARY OF DESIGN PROCEDURE FOR COMPOSITE PARTIALLY
RESTRAINED MOMENT FRAME SYSTEM
For buildings with low to moderate seismic demands, the partially restrained composite frame system
affords an opportunity to create a seismicforceresisting system in which many of the members are the
same size as would already be provided for gravity loads. A reasonable preliminary design procedure to
develop member sizes for a first analysis is as follows:
1. Proportion composite beams with heavy noncomposite loads based upon the demand for the unshored
construction load condition. For this example, this resulted in W18x35 beams to support the open
web steel joists.
2. Proportion other composite beams, such as the spandrel beams in this example, based upon judgment.
For this example, the first trial was made using the same W18x35 beam.
3. Select a connection such that the negative moment strength is about 75 percent of the plastic moment
capacity of the bare steel beam.
4. Proportion columns based upon a simple portal analogy for either stiffness or strength. If stiffness is
selected, keep the column’s contribution to story drift to no more than onethird of the target. If
strength is selected, an approximate effective column length factor of K = 1.5 is suggested for
preliminary design. Also check that the moment capacity of the column (after adjusting for axial
loads) is at least as large as that for the beam.
Those final design checks that are peculiar to the system are explained in detail as the example is
described. The key difference is that the flexibility of the connection must be taken into account in the
analysis. There are multiple ways to accomplish this. Some analytical software allows the explicit
inclusion of linear, or even nonlinear, springs at each end of the beams. Even for software that does not, a
dummy member can be inserted at each end of each beam that mimics the connection behavior. For this
example another method is illustrated, which is consistent with the overall requirements of the Provisions
for linear analysis. The member properties of the composite beam are altered to become an equivalent
prismatic beam that gives approximately the same flexural stiffness in the sway mode to the entire frame
as the actual composite beams combined with the actual connections. Prudence in the use of this
simplification does suggest checking the behavior of the connections under gravity loads to assure that
significant yielding is confined to the seismic event.
Once an analytic model is constructed, the member and connection properties are adjusted to satisfy the
overall drift limits and the individual strength limits. This is much like seismic design for any other frame
system. Column stability does need to account for the flexibility of the connection, but the AISC LRFD
and the Provisions approaches considering second order moments from the translation of gravity loads are
essentially the same. The further checks on details, such as the strong column rule, are also generally
familiar. Given the nature of the connection, it is also a good idea to examine behavior at service loads,
but there are not truly standard criteria for this.
8.3 DESIGN REQUIREMENTS
8.3.1 Provisions Parameters
The basic parameters affecting the design and detailing of the buildings are shown in Table 8.1 below.
Chapter 8, Composite Steel and Concrete
87
Table 81 Design Parameters
Parameter Value
Ss (Map 1) 0.20
S1 (Map 2) 0.06
Site Class E
Fa 2.5
Fv 3.5
SMS = FaSs 0.50
SM1 = FvS1 0.21
SDS = 2/3SMS 0.33
SD1 = 2/3SM1 0.14
Seismic Design Category C
Frame Type per
Provisions Table 5.2.2
Composite Partially Restrained
Moment Frame
R 6
O0
3
Cd 5.5
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package).]
The frames are designed in accordance with AISC Seismic, Part II, Sec. 8 (Provisions Table 5.2.2). AISC
SDGS8 and ASCE TC describe this particular system in detail. Given the need to determine the
flexibility of the connections, it would be difficult to design such structures without reference to at least
one of these two documents.
8.3.2 Structural Design Considerations Per the Provisions
The building is regular both in plan and elevation. Provisions Table 5.2.5.1 indicates that use of the
Equivalent Lateral Force procedure in accordance with Provisions Sec. 5.4 is permitted.
Nonstructural elements (Provisions Chapter 14) are not considered in this example.
Diaphragms must be designed for the required forces (Provisions Sec. 5.2.6.2.6), however this is not
unique to this system and therefore is not explained in this example.
The story drift limit (Provisions Table 5.2.8) is 0.025 times the story height. Although the Cd factor is
large, 5.5, the seismic forces are low enough that conventional stiffness rules for wind design actually
control the stiffness.
Orthogonal effects need not be considered for Seismic Design Category C, provided the structure does
not have a plan structural irregularity (Provisions Sec. 5.2.5.2.2).
8.3.3 Building Weight and Base Shear Summary
The unit weights are as follows:
FEMA 451, NEHRP Recommended Provisions: Design Examples
88
Noncomposite dead load:
4.5 in. slab on 0.6 in. form deck, plus sag 58 psf
Joist and beam framing 6 psf
Columns 2 psf
66 psf
Composite dead load:
Fire insulation 4 psf
Mechanical and electrical 6 psf
Ceiling 2 psf
Partitions 20 psf
32 psf
Exterior wall:
Precast concrete panels: 0.80 klf
Records storage on 3 bays per floor 120 psf
(50 percent is used for seismic weight; minimum per the Provisions is 25 percent)
The building weight, W, is found to be 8,080 kips. The treatment of the dead loads for analysis is
described in more detail subsequently.
The Seismic Response Coefficient, Cs, is equal to 0.021:
0.14 0.021
1.12 6
1
D1
s
C TS R
I
= = =
. .
. .
. .
The methods used to determine W and Cs are similar to those used elsewhere in this volume of design
examples. The building is somewhat heavy and flexible. The computed periods of vibration in the first
modes are 2.12 and 2.01 seconds in the northsouth and eastwest directions, respectively. These are
much higher than the customary 0.1 second per story rule of thumb, but lowrise frames with small
seismic force demands typically do have periods substantially in excess of the rule of thumb. The
approximate period per the Provisions is 0.66 seconds, and the upper bound for this level of ground
motion is 1.12 seconds.
The total seismic force or base shear is then calculated as follows:
V = CsW = (0.021)(8,080) = 170 kips (Provisions Eq. 5.3.2)
The distribution of the base shear to each floor (again, by methods similar to those used elsewhere in this
volume of design examples) is found to be:
Roof (Level 4): 70 kips
Story 4 (Level 3): 57 kips
Story 3 (Level 2): 34 kips
Story 2 (Level 1): 8 kips
Story 1 (Level 0): 0 kips
S: 169 kips (difference is rounding; total is 170)
Without illustrating the techniques, the gross service level wind force following ASCE 7 is 123 kips.
When including the directionality effect and the strength load factor, the design wind force is somewhat
less than the design seismic base shear. The wind force is not distributed in the same fashion as the
Chapter 8, Composite Steel and Concrete
89
seismic force, thus the story shears and the overturning moments for wind are considerably less than for
seismic.
8.4 DETAILS OF THE PRC CONNECTION AND SYSTEM
8.4.1 Connection M. Relationships
The composite connections must resist both a negative moment and a positive moment. The negative
moment connection has the slab rebar in tension and the leg of the seat angle in compression. The
positive moment connection has the slab concrete in compression (at least the “a” dimension down from
the top of the slab) and the seat angle in tension (which results in flexing of the seat angle vertical leg).
At larger rotations the web angles contribute a tension force that increases the resistance for both negative
and positive bending.
Each of these conditions has a momentrotation relationship available in AISC SDGS8 and ASCE TC.
(Unfortunately there are typographical errors in ASCE TC: A “+” should be replaced by “=” and the
symbol for the area of the seat angle is used where the symbol should be that for the area of the web
angle.) An M. curve can be developed from these equations:
Negative moment connection:
2 (AISC SDGS8, Eq. 1)
1(1 C ) 3
Mn=C e . +C.
where:
C1 = 0.18(4 × AsFyrb + 0.857ALFy)(d + Y3)
C2 = 0.775
C3 = 0.007(AL + AwL)Fy (d + Y3)
. = girder end rotation, milliradians (radians/1000)
d = girder depth, in.
Y3 = distance from top flange of the girder to the centroid of the reinforcement, in.
As = steel reinforcing area, in.2
AL = area of seat angle leg, in.2
AwL = gross area of double web angles for shear calculations, in.2 (For use in these equations AwL is
limited to 150 percent of AL).
Fyrb = yield stress of reinforcing, ksi
Fy = yield stress of seat and web angles, ksi
Positive moment connection:
2 (AISC SDGS8, Eq. 2)
1(1 C ) (3 4)
Mn+=C e . + C +C .
where:
C1 = 0.2400[(0.48AwL ) + AL](d + Y3)Fy
C2 = 0.0210(d + Y3/2)
C3 = 0.0100(AwL + AL )(d + Y3)Fy
C4 = 0.0065 AwL (d + Y3)Fy
From these equations, curves for M. can be developed for a particular connection. Figures 85 and 86
are M. curves for the connections associated with the W18x35 girder and the W21x44 spandrel beam
FEMA 451, NEHRP Recommended Provisions: Design Examples
810
300
200
100
0
100
200
300
25 20 15 10 5 0 5 10 15 20 25
Rotation, milliradians
Moment, ftkip
Positive M
Pos Bilinear
Neg Bilinear
Negative M
Figure 85 M. Curve for W18x35 connection with 6#5 (1.0 ftkip = 1.36 kNm)
respectively, which are used in this example. The selection of the reinforcing steel, connection angles,
and bolts are described in the subsequent section, as are the bilinear approximations shown in the figures.
Among the important features of the connections demonstrated by these curves are:
1. The substantial ductility in both negative and positive bending,
2. The differing stiffnesses for negative and positive bending, and
3. The substantial postyield stiffness for both negative and positive bending.
It should be recognized that these curves, and the equations from which they were plotted, do not
reproduce the line from a single test. They are averages fit to real test data by numerical methods. They
smear out the slip of bolts into bearing. (There are several articles in the AISC Engineering Journal that
describe actual test results. They are in Vol. 24, No.2; Vol. 24, No.4; Vol. 27, No.1; Vol. 27, No. 2; and
Vol 31, No. 2. The typical tests clearly demonstrate the ability of the connection to meet the rotation
capabilities of AISC Seismic, Section 8.4  inelastic rotation of 0.015 radians and total rotation capacity
of 0.030 radians.)
[Based on the modifications to AISC Seismic, Part II, Sec. 8.4 in 2003 Provisions Sec. 10.5.16, the
required rotation capabilities are inelastic rotation of 0.025 radians and total rotation of 0.040 radians.]
Chapter 8, Composite Steel and Concrete
811
500
400
300
200
100
0
100
200
300
400
500
25 20 15 10 5 0 5 10 15 20 25
Rotation, milliradians
Moment, ftkip
Positive M
Pos Bilinear
Neg Bilinear
Negative M
Figure 86 M. Curve for W21x44 connection with 8#5 (1.0 ftkip = 1.36 kNm).
8.4.2 Connection Design and Connection Stiffness Analysis
Table 82 is taken from a spreadsheet used to compute various elements of the connections for this design
example. It shows the typical W18x35 girder and the W21x44 spandrel beam with the connections used
in the final analysis, as well as a W18x35 spandrel beam for the short exterior spans, where a W21x44
was used in the end. Each major step in the table is described in a linebyline description following the
table. [Based on the modifications to AISC Seismic, Part II, Sec. in 2003 Provisions Sec. 10.5.16, the
nominal strength of the connection must be exceed RyMp for the bare steel beam, where Ry is the ratio of
expected yield strength to nominal yield strength per AISC Seismic, Part I, Table I61.]
FEMA 451, NEHRP Recommended Provisions: Design Examples
812
Table 82 Partially Restrained Composite Connection Design
Line Girder Spandrels
Basic Data
2 Beam size W18x35 W21x44 W18x35
3 Span, ft 25 25 12.5
4 Area of beam, in.2 10.3 13 10.3
5 I, of beam alone, in.4 510 843 510
6 Z, plastic modulus of beam, in.3 66.5 95.4 66.5
7 Beam depth, in. 17.7 20.7 17.7
8 Slab thickness, in. 7.0 7.0 7.0
9 Y3 to rebar, in. 5.5 5.5 5.5
10 Column W10x77 W10x88 W10x77
11 Flange width, in. 10.2 10.3 10.2
12 Flange thickness, in. 0.87 0.99 0.87
13 Flange fillet, k1, in. 0.88 0.94 0.88
Basic Negative Moment Capacity
15 Reinforcing bars 6#5 8#5 6#5
16 As, rebar area, in.2 1.86 2.48 1.86
17 Tr, rebar tension, kips 111.6 148.8 111.6
18 Mn

, nominal negative moment, ftkips 215.8 324.9 215.8
19 % Mp (Mn
/beam Mp) 78% 82% 78%
20 Check: > 50%? (75% per ASCE TC) OK OK OK
Seat Demands for Negative Moment
22 Seat angle L7x4x1/2x8 L7x4x5/8x8.5 L7x4x1/2x8
23 Seat Fy, ksi 36 36 36
24 Seat thickness, in. 0.5 0.625 0.5
25 Seat length, in. 8.0 8.5 8.0
26 Leg area, in.2 4.0 5.3125 4.0
27 Minimum area = 1.25 Tr /Fy, in.2 3.875 5.167 3.875
28 Check OK OK OK
29 Leg yield force, kips 144 191.25 144
30 Bolts to beam (4) 1"325X (4) 11/8"490X (4) 1"325X
31 Diameter, in. 1.0 0.875 1.0
32 Bolt design shear capacity, kips (f = 0.75) 141.2 223.6 141.2
33 Check Close enough OK Close enough
Nominal Positive Moment Capacity
35 Seat k, fillet length, in. 1.000 1.125 1.000
36 Mp, vertical leg, in.kips 18.0 29.9 18.0
37 b' (see Figure 87), in. 1.00 0.81 1.00
38 Seat tension from bending, kips 31.5 63.8 31.5
39 Seat tension from shear, kips 86.4 114.75 86.4
40 Tension to bottom flange, kips 31.5 63.8 31.5
41 Nominal Positive Moment, Mn
+, ftkips 67.4 149.9 67.4
42 Percent of Beam Mp 24% 38% 24%
Demand on Tension Bolts at Nominal Capacity
44 a' (see Figure 87), in. 2.0 2.1 2.0
45 Q (prying), kips 6.8 10.7 6.8
46 Bolt tension, kips 38.3 74.5 38.3
47 Bolts to column (2) 1"325X (2) 11/8"490X (2) 1"325X
48 Bolt design tension, kips (f = 0.75) 106 168.4 106
49 Check OK OK OK
Chapter 8, Composite Steel and Concrete
Line Girder Spandrels
813
Compute Total Joint Moment to Column based on Nominal Capacities
51 Connection nominal Mn
 + Mn
+, ftkips 283 475 283
52 Minimum column Mp (125% of sum) 177 297 177
53 Average as percentage of beam 51% 60% 51%
54 Check OK OK OK
Concrete Compression Transfer to Column
56 Rebar Ty + bottom seat Ty, kips 143.10 212.62 143.10
57 0.85 f'c on two flanges, kips 364.14 367.71 364.14
58 Projection for flange Mp, in. 2.72 3.10 2.72
59 Force from flange Mp, kips 225.92 254.88 225.92
60 Ratio, demand / minimum capacity 0.63 0.83 0.63
Web Shear Connection (needed for effective stiffness)
62 Seismic shear demand, kips 11.5 19.9 23.1
63 Web angles L4x4x1/4x8.5 L4x4x1/4x11.5 L4x4x1/4x8.5
64 Aw, area of two legs, in.2 4.25 5.75 4.25
65 Aw, limit based on area of rebar, in.2 2.79 3.72 2.79
66 Aw, used in M. calculation, in.2 2.79 3.72 2.79
Moment Rotation Values for Analysis of Effective Stiffness
68 Mneg at service level (0.0025 rad), ftkip 178.0 267.8 178.0
69 Mneg at maximum capacity(0.020 rad), ftkip 264.5 397.7 264.5
70 Secant stiffness for Mneg at 0.0025 radian 71.2 107.1 71.2
71 Mpos at service level (0.0025 rad), ftkip 73.7 117.3 73.7
72 Mpos at maximum capacity(0.020 rad), ftkip 208.9 313.9 208.9
73 Secant stiffness for Mpos at 0.0025 radian 29.5 46.9 29.5
74 Rotation at nominal Mneg 3.03 3.03 3.03
75 Rotation at nominal Mpos 2.29 3.70 2.29
Beam Moments of Inertia
77 Full composite action force, beam AFy, kips 515.0 650.0 515.0
78 Y2, to plastic centroid in concrete, in. 5.65 5.30 4.31
79 Composite beam inertia for pos. bending, in.4 1,593 2,435 1,402
80 Centroid of all steel for negative bending, in. 6.66 7.81 6.66
81 Composite beam inertia for neg. bending, in.4 834 1366 834
82 Equivalent beam for positive and negative, in.4 1,290 2,008 1,175
83 Weighted connection stiffness, ftkips/radian 61,263 88,105 61,263
84 Eff. prismatic inertia, beam and PRCC, in.4 639 955 412
85 Ratio of eff. prismatic I / I of beam alone 1.25 1.13 0.81
Check Bottom Bolt Tension at Maximum Deformation
87 Rotation at f × (rotation at nominal M pos) × Cd 10.7 14.9 10.7
88 Moment at f × (rot. at nom. M pos) × Cd, ftkips 152.3 268.2 152.3
89 Tension demand, kips 80.5 125.1 80.5
90 Nominal capacity of bolts, kips 141.3 224.5 141.3
Check Positive Moment Capacity as a Percentage of Beam Mp (50% criterion)
92 M pos (at 0.020 radians) / Mp beam 75% 79% 75%
Detailed explanation of the computations in Table 82:
Step 1: Establish nominal negative moment capacity: (This is a step created in this design example; is
not actually an explicit step in the procedures recommended in the references. It appears to be necessary
to satisfy the basic Provisions strength requirement. See Provisions Sec. 5.2.1, Sec. 5.2.7, and ASCE 7
Sec. 2.3.
FEMA 451, NEHRP Recommended Provisions: Design Examples
814
Lines 1518: Mn is taken as a simple couple of rebar in slab and force at connection of bottom flange of
beam; the true maximum moment is larger due to strain hardening in rebar and the bottom connection and
due to tension force in the web connection, so long as the bottom connection can handle the additional
demand. The nominal capacity is plotted in Figures 85 and 86 as the break of the bilinear relation. The
design capacity, using a resistance factor of 0.85, has two requirements:
1. f Mn exceeds demand from seismic load combination: basic Provisions requirement
2. f Mn exceeds demand from total service gravity loads  simply a good idea to maintain reasonable
initial stiffness for lateral loads; by “codes” the factored gravity demand can be checked using plastic
analysis
Lines 1920: Mn exceeds 50 percent (by AISC Seismic, Part II, 8.4) of Mp of the bare steel beam. In this
example, the more stringent recommendation of 75 percent contained within the ASCE TC is followed.
Note that this check is on nominal strength, not design strength. A larger Mn gives a larger stiffness, thus
some drift problems can be addressed by increasing connection capacity.
Step 2: Design bottom seat angle connection for negative moment:
Lines 2228: Provide nominal yield of angle leg at least 125 percent of nominal yield of reinforcing steel.
This allows for increased force due to web shear connection. Strain hardening in the rebar is a factor, but
strain hardening the angle would probably be as large. AISC SDGS8 recommends 120 percent. ASCE
TC recommends 133 percent, but then uses 125 percent to check the bolts. This is a check in
compression, and the authors elected to use 125 percent.
Lines 2933: Provide high strength bolts in normal (not oversized) holes to transfer force between beam
flange and angle by shear; conventional rules regarding threads in the shear plane apply. The references
do apply a resistance factor to the bolts, which may be an inconsistent design methodology. A check
based on overstrength might be more consistent. The capacity at bolt slip could be compared against
service loads, which would be a good idea for designs subject to strong wind forces.
Step 3: Establish nominal positive moment capacity: This connection is less stiff and less linear for
positive moment than for negative moment, and generally weaker. There is not a simple, clear
mechanism for a nominal positive moment. The authors of this example suggest the following procedure
which follows the normal methods of structural engineering and yields a point relevant to the results of
connection tests, in so far as construction of a bilinear approximation is concerned. It significantly
underestimates the ultimate capacity.
Lines 3538: Compute the shear in the vertical leg associated with bending. Figure 87 shows the
mechanics, which is based on methods in the AISC Manual, for computing prying in hangertype
connections. Compute the nominal plastic moment of the angle leg bending out of plane (line 36) and
assume that the location of the maximum moments are at the end of the fillet on the vertical leg (line 35)
and at the edge of the bolt shaft (line 37). The moment near the bolt is reduced for the material lost at the
bolt hole.
Lines 3940: Check the shear capacity, compare with the shear governed by moment, and use the smaller.
Shear will control if the angle is thick.
Line 41: Compute the nominal positive moment as a couple with the force and the distance from the
bottom of the beam to the center of the compression area of the slab on the column. The concrete
compression area uses the idealized Whitney stress block (ACI 318). Note that the capacity to transfer
Chapter 8, Composite Steel and Concrete
815
a' b' k
a b
T
Q
T + Q
Mp
M p
Figure 87 Analysis of seat angle for tension.
concrete compression force to the steel column flange is checked later. The nominal positive moment is
also shown on Figures 85 and 86 at the break point in the bilinear relation.
Step 4: Design the bolts to transfer positive moment tension to the column:
Lines 4445: Compute the prying force following AISC’s recommended method. The moment in the
vertical leg is computed as described above, and the moment arm extends from the edge of the bolt shaft
(closest to the beam) to the bottom edge of the angle. Refer to Figure 87.
Lines 4648: Add the basic tension to the prying force and compare to the factored design capacity of the
bolts. Note that the resistance factor is used here to be consistent with step 2. It is common to use the
same size and grade of bolt as used for the connection to the beam flange, which generally means that
these bolts have excess capacity. Also, for seismic design, another check at maximum positive moment is
recommended (see step 9).
Step 5: Compute the flexural demand on the column: AISC Seismic, Part II, 7 and 8, require that the
flexural resistance of the column be greater than the demand from the connections, but it does not give
any particular margin. ASCE TC recommends a ratio of 1.25.
Lines 5152: The minimum nominal flexural strength of the column, summed above and below as well as
adjusted for the presence of axial load, is set to be 125 percent of the demand from the sum of the nominal
strengths of the connections.
Lines 5354: AISC Seismic, Part II, 8.4 requires that the connection capacity exceed 50 percent of the
plastic moment capacity of the beam. In this example, the negative moment connections are designed for
75 percent of the beam plastic moment, and this check shows that the average of negative and positive
nominal moment capacities for the connection exceeds 50 percent of the plastic moment for the beam. A
later check (step 10) will compare the maximum positive moment resistance to the 50 percent rule.
Step 6: Check the transfer of force from concrete slab to steel column: The tension in the reinforcing
steel and the compression couple from positive bending must both transfer. Both flanges provide
FEMA 451, NEHRP Recommended Provisions: Design Examples
816
resistance if concrete fills the space between the flanges, but full capacity of the second flange has
probably not been exercised in tests.
Line 56: Add the yield force of the reinforcement and the tension yield force of the seat angle, both
previously computed.
Line 57: Compute an upper bound concrete compression capacity as 0.85f'c times the area of concrete
bearing on both flanges.
Lines 5859: Compute the force that would yield the steel column flanges over the thickness of the slab
by computing the projection beyond the web fillet that would yield at a load of 0.85f’c. This ignores the
capacity of the flange beyond the slab thickness and is obviously conservative.
Line 60: Compare the demand with the smaller of the two capacities just computed.
Step 7: Select the web connection:
Line 62: The seismic shear is computed by assuming beam end moments equal to the nominal capacity of
the connections, one in negative moment and one in positive.
Line 63: The gravity demand must be added, and straight gravity demand must also be checked before
selecting the actual connection.
Lines 6466: The web connection influences the overall stiffness and strength of the connection,
especially at large rotation angles. The momentrotation expression include the area of steel in the web
angles, but also places a limitation based upon 150 percent of the area of the leg of the seat angle for use
in the computation.
Step 8: Determine the effective stiffness of the beam and connection system: Determining the equivalent
stiffness for a prismatic beam involves several considerations. Figure 88 shows how the moment along
the beam varies for gravity and lateral loads as well as composite and noncomposite conditions. The
moment of inertia for the composite beam varies with the sense of the bending moment. The end
connections can be modeled as regions with their own moments of inertia, as illustrated in the figure.
Figure 89 shows the effective cross section for each of the four stiffnesses: positive and negative bending
of the composite beam and positive and negative bending of the composite connection. Given a linear
approximation of each connection stiffness expressed as moment per radian, flexural mechanics leads to a
simple expression for a moment of inertia of an equivalent prismatic beam.
Lines 6873: Compute the negative and positive moments at a rotation of 2.5 milliradians, which is the
rotation angle that defines the effective stiffness for lateral analysis (per both AISC SDGS8 and ASCE
TC).
Lines 7475: Using those moments, compute the rotations corresponding to the nominal strength, positive
and negative. (This is useful when idealizing the behavior as bilinear, which is plotted in Figures 85 and
86.)
Lines 7779: Compute the moment of inertia of the composite beam in positive bending. Note that the
system is designed for full composite action, per the recommendations in AISC SDGS8 and ASCE TC,
using the criteria in the AISC manual. The positive bending moment of inertia here is computed using
AISC’s lower bound method, which uses an area of steel in the flange adequate to replace the Whitney
stress block in the concrete flange. This moment of inertia is less than if one used the full concrete area in
Figure 89.
Chapter 8, Composite Steel and Concrete
817
Lines 8081: Compute the moment of inertia of the composite beam in negative bending.
Line 82: Compute an equivalent moment of inertia for the beam recognizing that a portion of the span is
in positive bending and the remainder is in negative bending. Following the recommendations in AISC
SDGS8 and ASCE TC, this is computed as 60 percent of I pos and 40 percent of I neg.
Lines 8384: Compute the moment of inertia of a prismatic beam that will give the same total end
rotation in a sway condition as the actual system. Gravity loads place both connections in negative
moment, so one will be subject to increasing negative moment while the other will be subject to
decreasing negative moment. Thus, initially, the negative moment stiffness is the appropriate stiffness,
which is what is recommended in the AISC SDGS8 and ASCE TC. For this example the positive and
negative stiffnesses are combined, weighted by the nominal strengths in positive and negative bending, to
yield a connection stiffness that is appropriate for analysis up to the nominal strengths defined earlier.
Defining this weighted stiffness as Kconn and the equivalent composite beam moment of inertia as Icomp, the
effective moment of inertia is found by:
I
I
EI
LK
effective
comp
comp
conn
=
1+
6
FEMA 451, NEHRP Recommended Provisions: Design Examples
818
(a)
Typical beam
(b)
Moment
diagram
(c)
Combined
composite
moment
(d)
Variable
moment
of inertia
M
M + M
Seismic
LL
MCombined
MDNC
I2 I1
~ ~
~ ~
~
~
+

I2 I1
Klat 4 Klat 3
DC
Figure 88 Moment diagram for typical beam.
Line 85: compute the ratio of the moment of inertia of the effective prismatic beam to that for the bare
steel beam. When using standard computer programs for analysis that have a library of properties of steel
cross sections, this ratio is a convenient way to adjust the modulus of elasticity and thus easily compute
the lateral drift of a frame. This adjustment could invalidate routines in programs that automatically
check various design criteria that depend on the modulus of elasticity.
Step 9: Check the tension bolts at maximum rotation
Line 87: Compute the rotation at total drift as Cd times the drift at the design positive moment.
Line 88: Compute the positive moment corresponding to that drift.
Line 89: Compute the tension force at the bottom seat angle, ignoring any contribution of the web angles,
from the moment and a moment arm between the center of the slab thickness and the inflection point in
the vertical leg of the seat angle, then add the prying force already calculated for a maximum demand on
the tension bolts.
Line 90: Compare with the nominal capacity of the bolts (set f = 1.0)
Step 10: Check the maximum positive moment capacity:
Chapter 8, Composite Steel and Concrete
819
7"
3
4"
clr cover
6"
1" clear
(6) #5 top bars
in slab
W18
Headed studs
on beam
(2) rows 3
4" Ø H.A.S. spaced at 11" o.c.
4x4x1
4" w/
(3) 3
4" A325 bolts
Bottom flange angle
with (4) 1"Ø A325X bolts
to beam flange and
(2) 1"Ø A325 bolts to
column flange
#4x6'0" @ 10" o.c. transverse.
Alternate above and below #5's.
Figure 810 Elevation of typical connection (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm).
Line 92: The positive moment at 20 milliradians, already calculated, is compared to the plastic moment
capacity of the steel beam. This is the point at which the 50 percent requirement of AISC Seismic, Part
II, 8.4 is checked.
Figure 810 shows many of the details of the connection for the W18x35. The headed studs shown
develop full composite action of the beam between the end and midspan. They do not develop full
composite action between the column and the inflection point, but it may be easily demonstrated that they
are more than capable of developing the full force in the reinforcing steel within that distance. The
transverse reinforcement is an important element of the design, which will be discussed subsequently.
Alternating the position above and below is simply a preference of the authors.
8.5 ANALYSIS
8.5.1 Load Combinations
A 3D model using Risa 3D was developed. Noncomposite dead loads (steel beams, bar joists, form
deck, and concrete) were input as concentrated loads at the columns on each level rather than uniformly
distributed to the beams. This was because we want the model for the seismic load combinations to
address the moments in the PRC connections. The loads subject to composite action are the composite
dead loads, live loads, and seismic loads, not the noncomposite dead loads. But the noncomposite dead
loads still contribute to mass, are subject to ground acceleration, and as such contribute to seismic loads.
This gets confusing; so a detailed look at the load combinations is appropriate.
Let us consider four load cases (illustrated in Figures 811 and 812):
1. Dc  Composite dead load, which is uniformly distributed and applied to beams (based on 32 psf)
2. Dnc  Noncomposite dead load, which is applied to the columns (based on 66 psf)
3. L  Composite live load, which is uniformly distributed to beams, using live load reductions
FEMA 451, NEHRP Recommended Provisions: Design Examples
820
1.2 D nc 1.2 Dc 0.5 L
1.0 QE 0.067 D nc 0.067 D c
Figure 811 Illustration of input for load combination for 1.2D + 0.5L + 1.0QE + 0.2SDSD.
4. E  Earthquake load, which is applied laterally to each level of the building and has a vertical
component applied as a uniformly distributed load to the composite beams
We will investigate two load combinations. Recall that composite loads are applied to beams, while noncomposite
loads are applied to columns. But there is an exception: the 0.2SDSD component, which
represents vertical acceleration from the earthquake is applied to all the dead load on the beams whether it
is composite or noncomposite. This is because even noncomposite dead load contributes to mass, and is
subject to the ground acceleration. Because the noncomposite dead load is not distributed on the beams
in the computer model, an adjustment to the load factor is necessary. The assignment of loads gets a
little complicated, so pay careful attention:
Combination 1 = 1.2D + 0.5L + 1.0E
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.2SDSD
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067 (Dnc + Dc)
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067Dnc(Dc/Dc) + 0.067Dc
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067Dc(Dnc/Dc) + 0.067Dc
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067Dc(66 psf/32 psf) + 0.067Dc
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.138Dc + 0.067Dc
= 1.2Dnc + 1.405Dc + 0.5L + QE
QE will be applied in both the northsouth and the eastwest directions, so this really represents two load
combinations.
Dnc = noncomposite dead load.
Dc = composite dead load
L = live load
QE = horizontal seismic load
Now consider at the second load combination:
Combination 2 = 0.9D + 1.0E
= 0.9Dnc + 0.9Dc + QE  0.2SDSD
= 0.9Dnc + 0.9Dc + QE  0.067 (Dnc + Dc)
= 0.9Dnc + 0.9Dc + QE  0.067 Dnc(Dc/Dc)  0.067Dc
= 0.9Dnc + 0.9Dc + QE  0.067 Dc(Dnc/Dc)  0.067Dc
= 0.9Dnc + 0.9Dc + QE  0.067 Dc(66 psf/32 psf)  0.067Dc
Chapter 8, Composite Steel and Concrete
821
0.9 D nc 0.9 Dc 1.0 Q
0.067 D nc 0.067 D c
E
Figure 812 Illustration of input for load combination for 0.9D + 1.0QE  0.2SDSD.
= 0.9Dnc + 0.9Dc + QE  0.138Dc  0.067Dc
= 0.9Dnc + 0.9Dc + QE  0.205Dc
= 0.9Dnc + 0.695Dc + QE
Again, QE will be applied in both the northsouth and the eastwest directions, so this represents another
two load combinations.
Dnc = noncomposite dead load.
Dc = composite dead load
L = live load
QE = horizontal seismic load
8.5.2 Drift and Pdelta
As defined by the Provisions, torsional irregularity is considered to exist when the maximum
displacement computed including accidental torsion at one end of the structure transverse to an axis is
more than 1.2 times the average of the displacements at the two ends of the structure (Provisions Sec.
5.4.4.3). For this building the maximum displacement at the roof including accidental torsion, is 1.65 in.
The displacement at the other end of the building in this direction is 1.43 in. The average is 1.54 in.
Because 1.65 in. < 1.85 in. = (1.2)(1.54 in.), the structure is not torsionally irregular. Consequently, it is
not necessary to amplify the accidental torsion nor to check the story drift at the corners. A simple check
at the center of the building suffices. [In the 2003 Provisions, the maximum limit on the stability
coefficient has been replaced by a requirement that the stability coefficient is permitted to exceed 0.10 if
and only “if the resistance to lateral forces is determined to increase in a monotonic nonlinear static
(pushover) analysis to the target displacement as determined in Sec. A5.2.3. Pdelta effects shall be
included in the analysis.” Therefore, in this example, the stability coefficient should be evaluated directly
using 2003 Provisions Eq. 5.2.16.]
The elastic story drifts were computed by the RISA 3D analysis for the required load combinations. Like
most modern computer programs for structural analysis, a Pdelta amplification can be automatically
computed, but to illustrate the effect of Pdelta in this structure and to check the limit on the stability
index, two computer runs have been performed, one without the Pdelta amplifier and one with it. The
allowable story drift is taken from Provisions Table 5.2.8. The allowable story drift is 0.025 hsx =
(0.025)(13 ft × 12 in./ft) = 3.9 in. With a Cd of 5.5, this corresponds to a drift 0.71 in. under the
equivalent elastic forces. At this point design for wind does influence the structure. A drift limit of h/400
(= 0.39 in.) was imposed, by office practice, to the service level wind load. In order to achieve the
desired stiffness, the seismic story drift at elastic forces is determined thus:
FEMA 451, NEHRP Recommended Provisions: Design Examples
822
Elastic story drift limit = (wind drift limit)(total seismic force)/(service level wind force)
Elastic story drift limit = (0.39 in.)(170 kip)/123 kip = 0.54 in.
The structure complies with the story drift requirements, but it was necessary to increase the size of the
spandrel beams from the preliminary W18x35 to W21x44 to meet the desired wind stiffness. This is
summarized in Table 83. The structure also complies with the maximum limit on the stability index
(Provisions 5.4.6.22):
.
max ß
. .
. .
= = . .
*
= =
05 05
05 55
018 025
Cd
ß is the ratio of demand to capacity for the story shear, and has not yet been computed. Maximum
demand and design capacity are tabulated in the following section; the average is about twothirds. The
preceding data show that the maximum resistance is higher, especially for positive moment, than the
value suggested here for design capacity. The average ratio of demand to maximum capacity with a
resistance factor is well below 0.5, so that value is arbitrarily used to show that the actual stability index is
within the limits of the Provisions.
Table 83 Story Drift (in.) and Pdelta Analysis
Story
Northsouth (X direction) Eastwest (Z direction)
without
Pdelta
with
Pdelta
Pdelta
amplifier
Stability
index
without
Pdelta
with
Pdelta
Pdelta
amplifier
Stability
index
1 0.358 0.422 1.179 0.152 0.312 0.360 1.154 0.133
2 0.443 0.517 1.167 0.143 0.410 0.471 1.149 0.130
3 0.449 0.513 1.143 0.125 0.402 0.453 1.127 0.113
4 0.278 0.304 1.094 0.086 0.239 0.259 1.084 0.077
8.5.3 Required and Provided Strengths
The maximum beam end moments from the frame analysis for the seismic load combinations are as
follows:
Table 84 Maximum Connection Moments and Capacities (ftkips)
Quantity
W18 Girders W21 Spandrels
Negative Positive Negative Positive
Demand (level 2), Mu
Nominal, Mn
Design capacity, fMn
143
216
184
36.6
67.2
57.1
118
325
276
103
149
127
The capacities, using a resistance factor of 0.85, are well in excess of the demands. The girder member
sizes are controlled by gravity load in the construction condition. All other member and connection
capacities are controlled by the design for drift. The negative moment demands are somewhat larger than
would result from a more careful analysis, because the use of a prismatic member overestimates the end
moments due to distributed load (composite gravity load) along the member. The higher stiffness of the
portion of the beam in positive bending with respect to the connections would result in higher positive
moments at midspan and lower negative moments at the supports. This conservatism has no real effect on
this design example. (The above demands and capacities do not include the girders supporting the storage
Chapter 8, Composite Steel and Concrete
823
bays, which are required to be W18x40 for the gravity load condition. The overall analysis does not take
that larger member into account.)
Snow load is not included in the seismic load combinations. (According to the Provisions, snow load
equal to or less than 30 psf does not have to be included in the mass.) Further, as a designer’s judgment
call, it was considered that the moments from 0.2S (= 6 psf) were so small, considering that the roof was
designed with the same connections as the floors, that it would make no significant difference in the
design analysis.
The maximum column forces are shown in Table 85; the particular column does support the storage load.
The effective length of the columns about their weak axis will be taken as 1.0, because they are braced by
perpendicular frames acting on the strong axes of the columns, and the Pdelta analysis captures the
secondary moments due to the “leaning” column effect. The effective length about their strong axis will
exceed 1.0. The ratio of column stiffness to beam stiffness will use the same effective beam stiffness
computed for the drift analysis, thus for the W10x77 framed into the W18x35 beams:
Icol / Lcol = 455/(13 × 12) = 2.92
Ibeam / Lbeam = 1.25 × 510 / (25 × 12) = 2.12
and the ratio of stiffnesses, G = 2.92 / 2.12 = 1.37
Although the column in the lowest story has greater restraint at the foundation, and thus a lower K factor,
it is illustrative to determine K for a column with the same restraint at the top and bottom. From the
nomographs in the AISC Manual or from equivalent equations, K = 1.45. It turns out that the effective
slenderness about the strong axis is less than that for the weak axis, so the K factor does not really control
this design.
Table 85 Column Strength Check, for W10x77
Seismic Load Combination Gravity Load Combination
Axial force, Pu
Moment, Mu
Interaction equation
391 kip
76.3 ftkip
0.72
557 kip
52.5 ftkip
0.89
8.6 DETAILS OF THE DESIGN
8.6.1 Overview
The requirements in AISC Seismic for CPRMF systems are brief. Some of the requirements are
references to Part I of AISC Seismic for the purely steel components of the system. A few of those detail
checks are illustrated here. For this example, more attention is paid to the details of the joint.
8.6.2 WidthThickness Ratios
The widththickness ratio of the beam flanges, bf /2tf is compared to .p
given in AISC Seismic, Part I,
Table I91. Both beam sizes, W18x35 and W21x44 are found to be acceptable. The W21x44 is
illustrated below:
FEMA 451, NEHRP Recommended Provisions: Design Examples
824
(AISC Seismic, Table I91)
52 52 7.35
50 p
Fy
. = = =
7.22 (AISC Manual)
2
f
f
b
t
=
7.22 < 7.35 OK
The limiting h/t ratios for columns is also given in AISC Seismic, Part I, Table I91. A W10x77 column
from the lower level of an interior bay with storage load is illustrated (the axial load from the seismic load
combination is used):
2 (AISC Seismic, Table I91)
391 kips = 0.385 > 0.125
(0.9)(22.6 in. x 50 ksi)
u
b y
P
f P
=
191 2.33 191[2.33 0.385] 52.5 (AISC Seismic, Table I91)
50
u
p
y b y
P
F P
.
f
. .
= .  .=  =
.. ..
Check:
OK
p 52.5 35.7 253
Fy
. = > =
13.0 (AISC Manual)
w
h
t
=
13.0 < 52.5 OK
8.6.3 Column Axial Strength
AISC Seismic, Part I, 8.2 requires that when Pu/fPn > 0.4 (in a seismic load combination), additional
requirements be met. Selecting the same column as above for our illustration:
2 (AISC Seismic, Part I, 8.2)
391 kips 0.53 > 0.4
(0.85)(22.6 in. )(38.4 ksi)
u
n
P
f P
= =
Therefore the requirements of AISC Seismic, Part I, 8.2a, 8.2b, and 8.2c apply. These necessitate the
calculation of axial loads using the System Overstrength Factor, O0
= 3. Analysis needs to be run for two
additional load combinations:
1.2D + 0.5L + 0.2S + O0QE (AISC Seismic, Part I, Eq. 4.1)
and
0.9D  O0
QE (AISC Seismic, Part I, Eq. 4.2)
Chapter 8, Composite Steel and Concrete
825
Slab
edge
Floor joist
Seat angle
12"
Bottom flange
angle
Figure 813 Detail at column.
The axial seismic force in this column is only 7.5 kips, therefore Pu becomes 397 kips, obviously much
less than fPn. The low seismic axial load is common for a momentresisting frame system. Given that
this requirement is a check ignoring bending moment, it does not control the design.
[The special load combinations have been removed from the 2002 edition of AISC Seismic to eliminate
inconsistencies with other building codes and standards. Therefore, 2003 Provisions Eq. 4.23 and 4.24
should be used in conjunction with the load combinations in ASCE 7.]
8.6.4 Details of the Joint
Figure 813 shows a plan view at an edge column, concentrating on the arrangement of the steel elements.
Figure 814 shows a section at the same location, showing the arrangement of the reinforcing steel. It is
not required that the reinforcing bars be equally distributed on the two sides of the column, but it is
necessary to place at least some of the bars on each side. This means that some overhang of the slab
beyond the column flange is required. This example shows two of the six bars on the outside face.
Figure 815 shows a plan view at a corner column. U shaped bent bars are used to implement the
negative moment connection at such a location. Threaded bars directly attached to the column flange are
also illustrated. Note the close spacing of the headed anchor studs for composite action. The reason for
the close spacing at this location is that the beam span is half the normal span, yet full composite action is
still provided.
FEMA 451, NEHRP Recommended Provisions: Design Examples
826
41
21 2"
2"
12"
Floor joist
#4x4'0" at 10", lap with
1
2"Ø deformed anchor
stud on edge plate
(6) #5 x 15'0"
at each column
L7x4x1
2"x0'8" LLH
(2) 1" dia. A325 bolt
into column
Figure 814 Detail at spandrel.
W18 spandrel girder
(2) #5 x 8'0" with double
nuts at column flange placed
2" below finished floor elevation
Edge of
concrete
slab
(2) rows 3
4" dia. H.A.S. at
41
2" o.c. (12'6" spans only)
W10 column
(3) #5 in slab
W21 spandrel beam
lap with straight bars
Figure 815 Detail at building corner.
The compressive force in the deck is transferred to both flanges of the column. This is shown in Figure 8
16. Note that both flanges can accept compressive forces from the concrete. Also note that the transverse
reinforcement will carry tension as force is transferred from the principal tension reinforcement through
the concrete to bearing on the column flange. Strut and tie models can be used to compute the appropriate
tension.
Chapter 8, Composite Steel and Concrete
827
Column
Longitudinal
Reinforcment
Transverse
Reinforcement
~ ~ ~
~ ~ ~
~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~
Figure 816 Force transfer from deck to column.
AISC SDGS8 and ASCE TC include the following recommendations regarding the reinforcing steel:
1. Place the principal tension reinforcement within a strip of width equal to 7 times the width of the
column flange (or less)
2. Use at least 6 bars for the principal reinforcement, extend it one quarter of the span from the column,
but at least 24 bar diameters beyond the inflection point, and extend at least two of the bars over the
full span
3. Do not use bars larger than number 6 (0.75 in. diameter)
4. Provide transverse reinforcement consistent with a strut and tie model to enable the transfer of forces
(in the authors’ observation such reinforcement is also necessary to preserve the capacity of the
headed studs for shear transfer)
91
9
MASONRY
James Robert Harris, P.E., Ph.D., Frederick R. Rutz, P.E.,
Ph.D. and Teymour Manzouri, P.E., Ph.D.
This chapter illustrates application of the 2000 NEHRP Recommended Provisions (herein after the
Provisions), to the design of a variety of reinforced masonry structures in regions with different levels of
seismicity. Example 9.1 features a singlestory masonry warehouse building with tall, slender walls;
Example 9.2 presents a fivestory masonry hotel building with a bearing wall system designed in areas
with different seismicities; and Example 9.3 covers a twelvestory masonry building having the same plan
as the hotel but located in a region of high seismicity. Selected portions of each building are designed to
demonstrate specific aspects of the design provisions.
Masonry is a discontinuous and heterogeneous material. The design philosophy of reinforced grouted
masonry approaches that of reinforced concrete; however, there are significant differences between
masonry and concrete in terms of restrictions on the placement of reinforcement and the effects of the
joints. These physical differences create significant differences in the design criteria.
All structures were analyzed using twodimensional (2D) static methods. Examples 9.2 and 9.3 use
dynamic analyses to determine the structural periods. Example 9.2 employs the SAP 2000 program,
V6.11 (Computers and Structures, Berkeley, California); Example 9.3 employs the RISA 2D program,
V.5.5 (Risa Technologies, Foothill Ranch, California).
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the
changes are described, the design examples and calculations have not been revised to reflect the changes
to the 2003 Provisions.
The most significant change to the masonry chapter in the 2003 Provisions is the incorporation by
reference of ACI 53002 for strength design in masonry. A significant portion of 2003 Provisions
Chapter 11 has been replaced by a reference to this standard as well as a limited number of modifications
to the standard, similar to other materials chapters. This updated chapter, however, does not result in
significant technical changes as ACI 53002 is in substantial agreement with the strength design
methodology contained in the 2000 Provisions.
Another change to the provisions for masonry structures is the addition of a new lateral system,
prestressed masonry shear walls. This system is not covered in this volume of design examples.
FEMA 451, NEHRP Recommended Provisions: Design Examples
92
Some general technical changes in the 2003 Provisions that relate to the calculations and/or design in this
chapter include updated seismic hazard maps, changes to Seismic Design Category classification for short
period structures, revisions to the redundancy requirements, revisions to the wall anchorage design
requirement for flexible diaphragms, and a new “Simplified Design Procedure” that could be applicable
to some of the examples in this chapter.
Where they affect the design examples in this chapter, other significant changes to the 2003 Provisions
and primary reference documents are noted. However, some minor changes to the 2003 Provisions and
the reference documents may not be noted.
In addition to the Provisions, the following documents are referenced in this chapter:
ACI 318 American Concrete Institute. 1999 [2002]. Building Code Requirements for Concrete
Structures.
ACI 530 American Concrete Institute. 1999 [2002]. Building Code Requirements for Masonry
Structures, ACI 530/ASCE 5/TMS 402.
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for
Buildings and Other Structures.
Amrhein Amrhein, J, and D. Lee. 1994. Tall Slender Walls, 2nd Ed. Masonry Institute of
America.
Drysdale Drysdale R., A. Hamid, and L. Baker. 1999. Masonry Structures, Behavior and Design.
Boulder Colorado: The Boulder Masonry Society.
IBC International Code Council. 2000. International Building Code.
UBC International Conference of Building Officials. 1997. Uniform Building Code.
NCMA National Concrete Masonry Association. A Manual of Facts on Concrete Masonry,
NCMATEK is an information series from the National Concrete Masonry Association,
various dates.
SEAOC Seismology Committee, Structural Engineers Association of California. 1999.
Recommended Lateral Force Requirements and Commentary, 7th Ed.
The short form designations for each citation are used throughout. The citation to the IBC exists for two
reasons. One of the designs employees a tall, slender wall that is partially governed by wind loads and
the IBC provisions are used for that design. Also, the R factors for masonry walls are significantly
different in the IBC than in the Provisions; this is not true for other structural systems.
Chapter 9, Masonry
93
5 bays at 20'0" = 100'0"
5 bays at 40'0" = 200'0"
8" concrete
masonry wall
Typical gluelam
roof beam
Plywood roof
sheathing
Open
12" concrete
masory wall
Figure 9.11 Roof plan (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
9.1 WAREHOUSE WITH MASONRY WALLS AND WOOD ROOF, LOS ANGELES,
CALIFORNIA
This example features a onestory building with reinforced masonry bearing walls and shear walls.
9.1.1 Building Description
This simple rectangular warehouse is 100 ft by 200 ft in plan (Figure 9.11). The masonry walls are 30 ft
high on all sides, with the upper 2 ft being a parapet. The wood roof structure slopes slightly higher
towards the center of the building for drainage. The walls are 8 in. thick on the long side of the building,
for which the slender wall design method is adopted, and 12 in. thick on both ends. The masonry is
grouted in the cells containing reinforcement, but it is not grouted solid. The assumed strength of
masonry is 2,000 psi. Normal weight concrete masonry units (CMU) with type S mortar are assumed.
The long side walls are solid (no openings). The end walls are penetrated by several large doors, which
results in more highly stressed piers between the doors (Figure 9.12); thus, the greater thickness for the
end walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples
94
12'0" 16'0" 2'0"
30'0"
12'0" 8'0"
100'0"
Roof line
Continuous
stem wall
and footing
Door
opening
(typical)
12" masonry
4'0"
Figure 9.12 End wall elevation (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
The floor is concrete slabongrade construction. Conventional spread footings are used to support the
interior steel columns. The soil at the site is a dense, gravelly sand.
The roof structure is wood and acts as a diaphragm to carry lateral loads in its plane from and to the
exterior walls. The roofing is ballasted, yielding a total roof dead load of 20 psf. There are no interior
walls for seismic resistance. This design results in a highly stressed diaphragm with large calculated
deflections. The design of the wood roof diaphragm and the masonry walltodiaphragm connections is
illustrated in Sec. 10.2.
In this example, the following aspects of the structural design are considered:
1. Design of reinforced masonry walls for seismic loads and
2. Computation of Pdelta effects.
9.1.2 Design Requirements
[Note that the new “Simplified Design Procedure” contained in the 2003 Provisions Simplified Alternate
Chapter 4 as referenced by the 2003 Provisions Sec. 4.1.1 is likely to be applicable to this example,
subject to the limitations specified in 2003 Provisions Sec. Alt. 4.1.1.]
9.1.2.1 Provisions Parameters
Site Class (Provisions Sec. 4.1.2.1 [Sec. 3.5]) = C
SS (Provisions Map 5 [Figure 3.33] ) = 1.50
S1 (Provisions Map 6 [Figure 3.34] ) = 0.60
Seismic Use Group (Provisions Sec. 1.3[Sec. 1.2]) = I
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps , and the maps
have been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously
used separate map package).]
Chapter 9, Masonry
95
The remaining basic parameters depend on the ground motion adjusted for site conditions.
9.1.2.2 Response Parameter Determination
The mapped spectral response factors must be adjusted for site class in accordance with Provisions Sec.
4.1.2.4 [3.3.2]. The adjusted spectral response acceleration parameters are computed according to
Provisions Eq. 4.1.2.41 [3.31] and 4.1.2.42 [3.32] for the short period and onesecond period,
respectively, as follows:
SMS = FaSS = 1.0(1.50) = 1.50
SM1 = FvS1 = 1.3(0.60) = 0.78
Where Fa and Fv are site coefficients defined in Provisions Tables 4.1.2.4a [3.31] and 4.1.2.4b [3.32],
respectively. The design spectral response acceleration parameters (Provisions Sec. 4.1.2.5 [Sec. 3.3.3])
are determined in accordance with Provisions Eq. 4.1.2.51 [Eq. 3.33] and 4.1.2.52 [3.34] for the shortperiod
and onesecond period, respectively:
2 2(1.50) 1.00
3 3
SDS= SMS= =
2 2(0.78) 0.52
3 3
SD1= SM1= =
The Seismic Design Category may be determined by the design spectral acceleration parameters
combined with the Seismic Use Group. For buildings assigned to Seismic Design Category D, masonry
shear walls must satisfy the requirements for special reinforced masonry shear walls in accordance with
Provisions Sec. 11.3.8.2 [ACI 530 Sec. 1.13.6.4]. A summary of the seismic design parameters follows:
Seismic Design Category (Provisions Sec. 4.2.1 [1.4]) = D
Seismic Force Resisting System (Provisions Table 5.2.2
[4.31]) = Special Reinforced
Masonry Shear Wall
Response Modification Factor, R (Provisions Table 5.2.2
[4.31]) = 3.5
Deflection Amplification Factor, Cd (Provisions Table 5.2.2
[4.31]) = 3.5
System Overstrength Factor, O0 (Provisions Table 5.2.2
[4.31]) = 2.5
Reliability Factor, . (Provisions Sec. 5.2.4.2 [Sec. 4.3.3]) = 1.0
(Determination of . is discussed in Sec. 9.1.3 below [see Sec. 9.1.3.1 for changes to the reliability factor
in the 2003 Provisions].)
Note that the R factor for this system in the IBC and in ASCE 7 is 4.5. [5.0 in the 2003 IBC and ASCE 7
02] This difference would have a substantial effect on the seismic design; however, the vertical
reinforcement in the tall 8in. walls is controlled by wind loads so it would not change.
9.1.2.3 Structural Design Considerations
With respect to the load path, the roof diaphragm supports the upper 16 ft of the masonry walls (half the
clear span plus the parapet) in the outofplane direction, transferring the lateral force to inplane masonry
shear walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples
96
Soil structure interaction is not considered.
The building is of bearing wall construction.
Other than the opening in the roof, the building is symmetric about both principal axes, and the vertical
elements of the seismic resisting system are arrayed entirely at the perimeter. The opening is not large
enough to be considered an irregularity (per Provisions Table 5.2.3.2[Table 4.32]); thus, the building is
regular, both horizontally and vertically. Provisions Table 5.2.5.1[Table 4.41], permits several analytical
procedures to be used; the equivalent lateral force (ELF) procedure (Provisions Sec. 5.4) is selected for
used in this example. The orthogonality requirements of Provisions Sec. 5.2.5.2 Sec. 4.4.2 are potentially
significant for the piers between the door openings at the end walls. Thus, those walls will be designed
for 100 percent of the forces in one direction plus 30 percent of the forces in the perpendicular direction.
There will be no inherent torsion because the building is symmetric. The effects of accidental torsion,
and its potential amplification, need not be included because the roof diaphragm is flexible. This is the
authors’ interpretation of what amounts to a conflict between Provisions Table 5.2.3.2[Table 4.32], Item
1, and Provisions Sec. 5.4.4.2[Sec. 5.24.2] and Sec. 5.4.4.3[Sec. 5.2.4.3].
The masonry bearing walls also must be designed for forces perpendicular to their plane (Provisions Sec.
5.2.6.2.7)[Sec. 4.6.1.3].
For inplane loading, the walls will be treated as cantilevered shear walls. For outofplane loading, the
walls will be treated as pinned at the bottom and simply supported at the top. The assumption of a pinned
connection at the base is deemed appropriate because the foundation is shallow and narrow which
permits rotation near the base of the wall.
9.1.3 Load Combinations
The basic load combinations (Provisions Sec. 5.2.7 [Sec. 4.2.2]) are the same as specified in ASCE 7 (and
similar to the IBC). The seismic load effect, E, is defined by Provisions Eq. 5.2.71 [4.21] and Eq. 5.2.7
2 [4.22] as:
E = .QE ± 0.2SDSD = (1.0)QE ± 0.2(1.00)D = QE ± 0.2D
This assumes . = 1.0 as will be confirmed in the following section.
9.1.3.1 Reliability Factor
In accordance with Provisions Sec. 5.2.4.2[4.3.3], the reliability factor, ., applies to the inplane load
direction.
For the long direction of building:
10
x
wall
max
story w
r V
V l
. .. .
=... ..... ..
(0.5) 10 0.025
rmaxx= ..200..=
. .
Chapter 9, Masonry
97
2 20 2 20 3.66
20,000 0.025 20,000
. =  =  =
max x r
. = 3.66 < 1.0 = .min, so use . = 1.0.
For the short direction of the building:
10 ( )(0.23) 10 (0.5)(0.23)(1) 0.115
x 8
wall wall
max
story w story
r V V
V l V
=.... ....... ...=.... ....... ...= =
Although the calculation is not shown here, note that a single 8ftlong pier carries approximately 23
percent (determined by considering the relative rigidities of the piers) of the inplane load for each end
wall.
Also, 1.0 was used for the 10/lw term even though 10/8 ft > 1.0. According to Provisions 5.2.4.2, the
10/lw term need not exceed 1.0 only for walls of light frame construction. This example was created based
on a draft version of the 2000 Provisions, which limited the value of the 10/lw term to 1.0 for all shear
walls, a requirement that was later changed for the published edition. Thus, this calculation is not strictly
correct. Using the correct value of would result in . = 1.02 rather than the 0.77 computed below. maxx r
This would result in a slight change in the factor on QE, 1.02 vs. 1.00, which has not been carried through
the remainder of this example.
(When the redundancy factor was developed by the Structural Engineers Association of California in the
wake of the 1994 Northridge earthquake, the upper bound of 1.0 for 10/lw was simply not mentioned. The
1997 Provisions, the UBC, and the IBC were published with no upper bound on 10/lw. However, the
original authors of the concept published their intent with the SEAOC document in 1999 with the upper
bound of 1.0 on 10/lw for all types of shear walls. The same change was adopted within BSSC for the
2000 Provisions. A subsequent change to the 2000 Provisions limited the upper bound of 1.0 to apply
only to light frame walls.)
Therefore,
0.12 rmaxx =
2 20 0.77
0.115 20,000
. =  =
. = 0.77 < 1.0 = .min, so use . = 1.0.
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a shear wall
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear
wall with heighttolengthratio greater than 1.0 would not result in more than a 33 percent reduction in
story strength or create an extreme torsional irregularity. Therefore, the redundancy factor would have to
be investigated only in the transverse direction where the aspect ratios of the piers between door openings
are greater than 1.0. In the longitudinal direct, where the aspect ratio is (significantly) less than 1.0, . =
1.0 by default.]
9.1.3.2 Combination of Load Effects
FEMA 451, NEHRP Recommended Provisions: Design Examples
98
Load combinations for the inplane loading direction from ASCE 7 are:
1.2D + 1.0E + 0.5L + 0.2S
and
0.9D + 1.0E + 1.6H.
L, S, H do not apply for this example so the load combinations become:
1.2D + 1.0E
and
0.9D + 1.0E.
When the effect of the earthquake determined above, 1.2D + 1.0(QE ± 0.2D), is inserted in each of the
load combinations:
1.4D + 1.0 QE
1.0D  1.0 QE
and
0.9D + 1.0(QE ± 0.2D)
which results in:
1.1D + 1.0 QE
and
0.7D  1.0 QE
Thus, the controlling cases from all of the above are:
1.4D + 1.0 QE
when gravity and seismic are additive and
0.7D  1.0 QE
when gravity and seismic counteract.
These load combinations are for the inplane direction of loading. Load combinations for the outofplane
direction of loading are similar except that the reliability coefficient (.) is not applicable. Thus, for
this example (where . = 1.0), the load combinations for both the inplane and the outofplane directions
are:
1.4D + 1.0 QE
and
Chapter 9, Masonry
99
0.7D  1.0 QE.
The combination of earthquake motion (and corresponding loading) in two orthogonal directions must be
considered (Provisions Sec. 5.2.5.2.3) [Sec. 4.4.2.3].
9.1.4 Seismic Forces
9.1.4.1 Base Shear
Base shear is computed using the parameters determined previously. The Provisions does not recognize
the effect of long, flexible diaphragms on the fundamental period of vibration. The approximate period
equations, which limit the computed period, are based only on the height. Since the structure is relatively
short and stiff, shortperiod response will govern the design equations. According to Provisions Sec.
5.4.1 [Sec. 5.2.1.1] and Eq. 5.4.1.11 [Eq. 5.23] (for shortperiod structures):
1.0 0.286
/ 3.5/1
V CsW SRDSI W W W
. . . .
= =.. .. =.. .. =
The seismic weight for forces in the long direction is:
Roof = 20 psf (100)200 = 400 kips
End walls = 103 psf (2 walls)[(30 ft)(100 ft)  5(12 ft)(12 ft)](17.8 ft/28 ft) = 299 kips
Side walls = 65 psf (30ft)(200ft)(2 walls) = 780 kips
Total = 1,479 kips
Note that the centroid of the end walls is determined to be 17.8 ft above the base, so the portion of the
weight distributed to the roof is approximately the total weight multiplied by 17.8 ft/28 ft (weights and
section properties of the walls are described subsequently).
Therefore, the base shear to each of the long walls is:
V = (0.286)(1,479 kips)/2 = 211 kips.
The seismic weight for forces in the short direction is:
Roof = 20 psf (100)200 = 400 kips
Side walls = 65 psf (2 walls)(30ft)(200ft)(15ft/28ft) = 418 kips
End walls = 103 psf (2 walls)[(30ft)(100ft)5(12ft)(12ft)] = 470 kips
Total = 1,288 kips
The base shear to each of the short walls is:
V = (0.286)(1,288 kips)/2 = 184 kips.
9.1.4.2 Diaphragm Force
See Sec. 10.2 for diaphragm forces and design.
9.1.4.3 Wall Forces
because the diaphragm is flexible with respect to the walls, shear is distributed to the walls on the basis of
beam theory ignoring walls perpendicular to the motion (this is the "tributary" basis).
FEMA 451, NEHRP Recommended Provisions: Design Examples
910
The building is symmetric. Given the previously explained assumption that accidental torsion need not
be applied, the force to each wall becomes half the force on the diaphragm.
All exterior walls are bearing walls and, according to Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3], must be
designed for a normal (outofplane) force of 0.4SDSWc. The outofplane design is shown in Sec. 9.1.5.3
below.
9.1.5 Longitudinal Walls
The total base shear is the design force. Provisions Sec. 11.7 [Sec. 11.2] is the reference for design
strengths. The compressive strength of the masonry (fm') is 2,000 psi. Provisions Sec. 11.3.10.2 gives Em
= 750 fm' = (750)(2 ksi) = 1,500 ksi.
[2003 Provisions Sec. 11.2 adopts ACI 530 as a design basis for strength design masonry and provides
some modifications to ACI 530. In general, the adoption of ACI 530 as a reference does not have a
significant effect on this design example. Note that by adopting ACI 530 in the 2003 Provisions, Em =
900f’m per ACI 530 Sec. 1.8.2.2.1, eliminating the conflict discussed below.]
Be careful to use values consistent with the Provisions. Different standards call for different values. To
illustrate this point, the values of Em from different standards are shown in Table 9.11.
Chapter 9, Masonry
911
Table 9.11 Comparison of Em
Standard Em Em for this example
Provisions
IBC
ACI 530
750 fm'
900 fm'
900 fm'
1,500 ksi
1,800 ksi
1,800 ksi
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm.
For 8inch thick CMU with vertical cells grouted at 24 in. o.c. and horizontal bond beams at 48 inch o.c.,
the weight is conservatively taken as 65 psf (recall the CMU are normal weight) and the net bedded area
is 51.3 in.2/ft based on tabulations in NCMATEK 141.
9.1.5.1 Horizontal Reinforcement
As determined in Sec. 9.1.4.1, the design base shear tributary to each longitudinal wall is 211 kips. Based
on Provisions Sec. 11.7.2.2 [ACI 530, Sec. 3.1.3], the design shear strength must exceed either the shear
corresponding to the development of 1.25 times the nominal flexure strength of the wall, which is very
unlikely in this example due to the length of wall, or 2.5 times Vu = 2.5(211) = 528 kips.
From Provisions Eq. 11.7.3.2[ACI 530, Eq. 321] , the masonry component of the shear strength capacity
for reinforced masonry is:
m4.0 1.75 nm0.25 .
M
V Af P
Vd
=  ' + . . ..
.. .. ....
Conservatively treating M/Vd as equal to 1.0 for the long walls and conservatively treating P as the
weight of the wall only without considering the roof weight contribution:
Vm=[4.01.75(1.0)](51.3)(200) 2000+0.25(390)=1130 kips
and
fVm = 0.8(1,130) = 904 kips > 528 kips OK
where f = 0.8 is the resistance factor for shear from Provisions Table11.5.3[ACI 530, sec. 3.1.4] .
Horizontal reinforcement therefore is not required for shear but is required if the wall is to qualify as a
“Special Reinforced Masonry Wall.”
According to Provisions Sec. 11.3.8.3[ACI 530, Sec. 1.13.6.3] , minimum reinforcement is
(0.0007)(7.625 in.)(8 in.) = 0.043 in.2 per course, but it may be wise to use more horizontal reinforcement
for shrinkage in this very long wall and then use minimum reinforcement in the vertical direction (this
concept applies even though this wall requires far more than the minimum reinforcement in the vertical
direction due to its large heighttothickness ratio). Two #5 bars at 48 in. on center provides 0.103 in.2
per course. This amounts to 0.4 percent of the area of masonry plus the grout in the bond beams. The
actual shrinkage properties of the masonry and the grout and local experience should be considered in
deciding how much reinforcement to provide. For long walls that have no control joints, as in this
example, providing more than minimum horizontal reinforcement is appropriate.
FEMA 451, NEHRP Recommended Provisions: Design Examples
912
#7
24" o.c.
8" concrete
masonry unit
Figure 9.13 Trial design for 8in.thick CMU
wall (1.0 in = 25.4 mm).
9.1.5.2 Vertical Reinforcement
Steps for verifying a trial design are noted in the sections that follow.
9.1.5.3 Outof Plane Flexure
As indicated previously, the design demand for seismic outofplane flexure is 0.4SDSWc. For a wall
weight of 65 psf for the 8in.thick CMU side walls, this demand is 0.4(1.00)(65 psf) = 26 psf.
Calculations for outofplane flexure become somewhat involved and include the following:
1. Select a trial design.
2. Investigate to ensure ductility.
3. Make sure the trial design is suitable for wind (or other nonseismic) lateral loadings using the IBC.
Note that many section properties determined in accordance with the IBC are different from those
indicated in the Provisions so section properties will have to be determined multiple times. The
IBC portion of the calculation is not included in this example.
[2003 Provisions and the 2003 IBC both adopt ACI 53002 by reference, so the section properties should
be the same for both documents.]
4. Calculate midheight deflection due to wind by the IBC. (While the Provisions have story drift
requirements, they do not impose a midheight deflection limit for walls).
5. Calculate seismic demand.
6. Determine seismic resistance and compare to demand determined in Step 5.
Proceed with these steps as follows:
9.1.5.3.1 Trial design
A trial design of #7 bars at 24 in. on center is selected. See Figure 9.13.
Chapter 9, Masonry
913
9.1.5.3.2 Investigate to ensure ductility
The critical strain condition corresponds to a strain in the extreme tension reinforcement (which is a
single #7 centered in the wall in this example) equal to 1.3 times the strain at yield stress.
Based on Provisions Sec. 11.6.2.2[ACI 530, Sec. 3.2.3.5.1] for this case:
t = 7.63 in.
d = t/2 = 3.81 in.
em = 0.0025
es = 1.3ey = 1.3(fy/Es) = 1.3(60 ksi /29,000 ksi) = 0.0027
( m ) 1.83 in.
m s
c d
e
e e
. .
=. . =
.. + ..
a = 0.8c = 1.46 in.
The Whitney compression stress block, a = 1.46 in. for this strain distribution, is greater than the 1.25 in.
face shell width. Thus, the compression stress block is broken into two components: one for full
compression against solid masonry (the face shell) and another for compression against the webs and
grouted cells, but accounting for the open cells. These are shown as C1 and C2 in Figure 9.14:
C1 = 0.80fm' (1.25 in.)b = (0.80)(2 ksi)(1.25)(24) = 48 kips (for a 24in. length)
C2 = 0.80 fm' (a1.25 in.)(8 in.) = (0.80)(2 ksi)(1.461.25)(8) = 2.69 kips (for a 24in. length)
The 8in. dimension in the C2 calculation is for combined width of grouted cell and adjacent mortared
webs over a 24in. length of wall. The actual width of one cell plus the two adjacent webs will vary with
various block manufacturers, and may be larger or smaller than 8 in. The 8in. value has the benefit of
simplicity and is correct when considering solidly grouted walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples
914
1.25"
3.81"
1.25"
t = 7.63"
P = P + P
d = 3.81"
1.83" 1.98"
= 0.0025
N.A.
1.3 = 0.0027
1.25"
1.46"
N.A.
T
C
C
0.48"
1.21"
0.8f '
0.21"
#7 at 24" o.c.
1
m
2
y
m
f w
a
c
Figure 9.14 Investigation of outofplane ductility for the 8in.thick CMU side walls
(1.0 in = 25.4 mm).
T is based on 1.25 Fy (Provisions Sec. 11.6.2.2)[ACI 530, Sec. 3.2.3.5.1]:
T = 1.25FyAs = (1.25)(60 ksi)(0.60 in.2) = 45 kips (for a 24in. length)
Use unfactored P (Provisions Sec. 11.6.2.2)[ACI 530, Sec. 3.2.3.5.1]:
P = (Pf + Pw) = (20 psf (10 ft.) + 65 psf (16 ft.)) = 1.24 klf = 2.48 kips (for a 24in. length)
Check C1 + C2 > T + P:
Chapter 9, Masonry
915
T + P = 47.5 kips
C1 + C2 = 50.7 kips > 47.5 kips. OK
The compression capacity is greater than the tension capacity; therefore, the ductile failure mode criterion
is satisfied.
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including
critical strain in extreme tensile reinforcement (1.5 times) and axial force to consider when performing the
ductility check (factored loads).]
9.1.5.3.3 Check for wind load using the IBC
Load factors and section properties are not the same in the IBC and the Provisions (The wind design
check is beyond the scope of this seismic example so it is not presented here.) Both strength and
deflection need to be ascertained in accordance with IBC.
Note that, for comparison, selected properties for the Provisions (and IBC) ductility check, IBC wind
strength check, and Provisions seismic strength check are tabulated below. Keeping track of which
version of a given parameter is used for each of the calculations can get confusing; be careful to apply the
correct property for each analysis.
Table 9.12 Comparison of Variables (explanations in the following text)
Parameter Provisions Ductility
Calculation
Provisions Strength
Calculation
IBC Wind
Calculation
P 1.24 klf 0.87 & 1.74 klf 1.12 klf
Em NA 1,500,000 psi 1,800,000 psi
fr NA 80 psi 112 psi
w NA 26 psf 19 psf (service)
es 0.0027 NA NA
d 3.82 in. 3.82 in. 3.82 in.
c 1.83 in. 1.25 in. 1.25 in.
a 1.46 in. 1.00 in. 1.00 in.
Cres = C1+C2 50.1 kips 52.1 kips 56.4 kips
n = Es/Em NA 19.33 16.11
Ig NA 355 in.4 355 in.4
Sg NA 93.2 in.3 93.2 in.3
Ase NA 0.32 in.2/ft 0.32 in.2/ft
Icr NA 48.4 in.4/ft 48.4 in.4/ft
Mcr = frS NA 7.46 in.kips 10.44 in.kips
dallow NA NA 2.32 in.
NA = not applicable, 1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 in = 25.4 mm, 1.0 ksi = 6.98 MPa,
1.0 in.kip = 0.113 kNm.
FEMA 451, NEHRP Recommended Provisions: Design Examples
916
9.1.5.3.4 Calculate midheight deflection due to wind by the IBC
The actual calculation is not presented here. For this example the midheight deflection was calculated
using IBC Eq. 2141[ACI 530, Eq. 331] with Icr = 47.3 in.4 per ft. Using IBC Eq. 21.41[ACI 530, Eq. 3
31], the calculated deflection is 2.32 in., which is less than 2.35 in. = 0.007h (IBC Eq. 2139[ACI 530,
Eq. 329]).
9.1.5.3.5 Calculate seismic demand
For this case, the two load factors for dead load apply: 0.7D and 1.4D. Conventional wisdom holds that
the lower dead load will result in lower momentresisting capacity of the wall so the 0.7D load factor
would be expected to govern. However, the lower dead load also results in lower Pdelta so both cases
should be checked. (As it turns out, the higher factor of 1.4D governs).
Check moment capacity for 0.7D:
Pu = 0.7(Pf + Pw).
For this example, the iterative procedure for addressing Pdelta from Amrhein will be used, not
Provisions Eq. 11.5.4.3[ACI 530, Commentary Sec. 3.1.5.3] which is intended for inplane deflections:
Roof load, Pf = 0.7(0.2 klf) = 0.14 klf
Eccentricity, e = 7.32 in. (distance from wall centerline to roof reaction centerline)
Modulus of elasticity (Provisions Eq. 11.3.10.2 [ACI 530, 1.8.2.2 ]), Em = 750 fm' = 1,500,000 psi
[Note that by adopting ACI 530 in the 2003 Provisions, Em = 900f’m per ACI 530 Sec. 1.8.2.2.1.]
Modular ratio, s 19.3
m
n E
E
= =
The modulus of rupture (fr) is found in Provisions Table 11.3.10.5.1[ACI 530, Sec. 3.1.7.2.1]. The values
given in the table are for either hollow CMU or fully grouted CMU. Values for partially grouted CMU
are not given; Footnote a indicates that interpolation between these values must be performed. As
illustrated in Figure 9.16, the interpolated value for this example is 80 psi:
(fr  50 psi)/(103 in.2  60 in.2) = (136 psi  50 psi)/(183 in.2  60 in.2)
fr = 80 psi
Ig = 355 in.4/ft
Sg = 93.2 in.3/ft
Mcr = frSg = 7460 inlb/ft
Chapter 9, Masonry
917
16" 8"
b = 24"
1.25"
N.A. N.A.
1.25"
d = 3.81"
nAs
b
c
b = 8.32" inferred from NCMA tabulations
b = 8" used for convenience
Figure 9.17 Cracked moment of inertia (Icr) for 8in.thick CMU side walls (1.0 in = 25.4 mm).
CASE 1
All cells open
A = 60 in.
f = 50 psi
CASE 2
(1) Cell grouted
A = 103 in.
f = 80 psi *
CASE 3
Fully grouted
A = 183 in.
f = 136 psi
* By interpolation
2
r r r
2 2
24" 24" 24"
Figure 9.16 Basis for interpolation of modulus of rupture, fr (1.0 in = 25.4 mm, 1.0 psi = 6.89 kPa).
Refer to Figure 9.17 for determining Icr. The neutral axis shown on the figure is not the conventional
neutral axis by linear analysis; instead it is the plastic centroid, which is simpler to locate, especially when
the neutral axis position results in a T beam crosssection. (For this wall, the neutral axis does not
produce a T section, but for the other wall in this building, a T section does result.) Cracked moments of
inertia computed by this procedure are less than those computed by linear analysis but generally not so
much less that the difference is significant. (This is the method used for computing the cracked section
moment of inertia for slender walls in the standard for concrete structures, ACI 318.) Axial load does
enter the computation of the plastic neutral axis and the effective area of reinforcement. Thus:
P = 1.24 klf
T = ((0.60 in.2)/(2 ft.))(60 ksi) = 18.0 klf
C = T + P = 19.24 klf
a = C/(0.8 f'mb) = (19.24 klf)/(0.8(2.0 ksi)(12 in./ft.)) = 1.002 in.
c = a/0.8 = 1.253 in.
Icr = nAse(dc)2 + bc3/3 = 19.33(0.30 in.2 + (1.24 klf)/60 ksi)(3.81 in.  1.25 in.)2 +
(12 in./ft)(1.25 in.)3/3 = 4.84 in.4/ft
FEMA 451, NEHRP Recommended Provisions: Design Examples
918
Note that Icr could be recomputed for P = 0.7D and P = 1.4D but that refinement is not pursued in this
example.
The standard technique is to compute the secondary moment in an iterative fashion as shown below:
Axial load
Pu = 0.7(Pf + Pw) = 0.7(0.2 klf + 1.04 klf) = 0.868 klf
First iteration
0.868 (0.60)(60) 0.614 in.2/ 2 ft. = 0.312 in.2/ ft
60
u s y
se
y
P Af
A
f
+ +
= = =
2
1 / 8 ( ) u u 0 0 w M =wh +Pe+P+P .
2
1
(26 psf/12)(336 in.) 7.32 in.
(140 plf) (140plf + 728 plf)(0)
8 2 u M= + .. ..+
. .
Mu1 = 31,088 in.lb/lf > Mcr = 7460
( )
2 2
1
5(7460)(336) 5(31,088 7460)(336) 0.165 3.827 3.99 in.
s 48(1,500,000) 355 48(1,500,000)(48.4)

. = + = + =
Second iteration
2 30,576 512 (140 728)(3.99) 34,551 in.lb u M = + + + =
2
2
0.165 5(34,551 7460)(336) 0.165 4.388 4.55 in.
s 48(1,500,000)(48.4)

. = + = + =
Third iteration
Mu3 = 30,576 + 512 + (140 + 728)(4.55) = 35,040 in.lb/lf
( ) 2
3
5 35,040 7460 (336)
0.165 0.165 4.467 4.63 in.
s 48(1,500,000)(48.4)

. = + = + =
Convergence check
4.63 4.55 1.8% 5%
4.55

= <
Mu = 35,040 in.lb (for the 0.7D load case)
Using the same procedure, find Mu for the 1.4D load case. The results are summarized below:
First iteration
P = 7360 plf
Mu1 = 31,601 in.lb/ft
Chapter 9, Masonry
919
.u
1 = 4.08 in.
Second iteration
Mu2 = 38,684 in.lb/ft
.u
2 = 5.22 in.
Third iteration
Mu3 = 40,667 in.lb/ft
.3
= 5.54 in.
Fourth iteration
Mu4 = 41,225 in.lb/ft
.u
4 = 5.63 in.
Check convergence
5.63 5.54 1.7% 5%
5.54

= <
Mu = 41,225 in.lb (for the 1.4D load case)
9.1.5.3.6 Determine flexural strength of wall
Refer to Figure 9.18. As in the case for the ductility check, a strain diagram is drawn. Unlike the
ductility check, the strain in the steel is not predetermined. Instead, as in conventional strength design of
reinforced concrete, a rectangular stress block is computed first and then the flexural capacity is checked.
T = Asfy = (0.30 in.2/ft.)60 ksi = 18.0 klf
The results for the two axial load cases are tabulated below.
Load Case 0.7D + E 1.4D + E
Factored P, klf 0.87 1.74
T + P = C, klf 18.87 19.74
a = C / (0.8f'mb), in. 0.981 1.028
MN = C (d  a/2), in.kip/ft. 62.6 65.1
fMN = 0.85MN, in.kip/ft. 53.2 55.3
MU, in.kip/ft. 35.0 41.2
Acceptance OK OK
FEMA 451, NEHRP Recommended Provisions: Design Examples
920
1.25"
3.81"
1.25"
t = 7.63"
P
d = 3.81"
c
= 0.0025
N.A.
> = 0.00207
a
1.25"
N.A. T = A F + P
C
0.8f '
#7 at 24" o.c.
da/2
u
s
m
m
t
a/2
y
s y
= a/0.80
s
t s t s
Figure 9.18 Outofplane strength for 8in.thick CMU walls (1.0 in = 25.4 mm).
Note that wind actually controls the stiffness and strength outofplane and that this is only a “tentative”
acceptance for seismic. The Provisions requires a check of the combined orthogonal loads in accordance
Chapter 9, Masonry
921
with Provisions Sec. 5.2.5.2, Item a [Sec. 4.4.2.3]. However, as discussed below, a combined orthogonal
load check was deemed unnecessary for this example.
9.1.5.4 InPlane Flexure
Inplane calculations for flexure in masonry walls include two items per the Provisions:
1. Ductility check and
2. Strength check.
It is recognized that this wall is very strong and stiff in the inplane direction. In fact, most engineers
would not even consider these checks necessary in ordinary design. The ductility check is illustrated here
for two reasons: to show a method of implementing the requirement and to point out an unexpected
result. (In the authors’ opinion, the Provisions should reconsider the application of the ductility check
where the M/Vdv ratio is substantially less than 1.0.)
9.1.5.4.1 Ductility check
Provisions Sec. 11.6.2.2 [ACI 530, 3.2.3.5.1] requires that the critical strain condition correspond to a
strain in the extreme tension reinforcement equal to 5 times the strain associated with Fy. This calculation
uses unfactored gravity loads. (See Figure 9.19.)
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing
the ductility check (factored loads).]
P = Pw + Pf = (0.065 ksf (30 ft.) + 0.02 ksf (10 ft.))(200 ft.) = 430 kips
P is at the base of the wall rather than at the midheight.
0.0025 200 ft 38.94 ft
0.0025 0.0103
m
m s
c d
e
e e
=... + ... =... + ... =
a = 0.8c = 31.15 ft = 373.8 in.
Cm = 0.8fm'abavg = 2,560 kips
Where bavg is taken from the average area used earlier, 51.3 in.2/ft.; see Figure 9.19 for locations of
tension steel and compression steel (the rebar in the compression zone will act as compression steel).
From this it can be seen that:
( )( )
(1.25 ) 40.27 (0.60) 453 kips
Ts1 fy 2 2 ft o.c.
. .
= .. .. =
. .
(1.25 ) 120.79 (0.60) 2,718 kips
Ts2= fy .. 2 .. =
. .
6.73 (0.60) 121 kips
Cs1=fy..2 ft. o.c... =
. .
FEMA 451, NEHRP Recommended Provisions: Design Examples
922
= 5E
= 0.0103
c = 38.94' 161.06'
N.A.
61.06' P = P + P
31.15' 7.79'
0.8f '
32.21'
N.A.
21.5'
35.58'
C
C
72.53 ksi
f = 60 ksi
26.84'
100.66'
T
T
1.25 F
= 75 ksi
5 F =
300 ksi
23.37'
C
y
y
s2
s1
40.27' 120.79' y
s1
s2
m
f w
m
s y
m =
0.0025
6.73'
a
Figure 9.19 Inplane ductility check for side walls (1.0 in = 25.4 mm, 1.0 ksi = 6.89 MPa).
( ) 32.21 (0.60) 290 kips
Cs2 fy (2)(2)
. .
= . . =
. .
Note that some authorities would not consider the compression resistance of reinforcing steel that is not
enclosed within ties. The Provisions clearly allows inclusion of compression in the reinforcement.
SC > ST
Chapter 9, Masonry
923
#7 at 24" o.c.
10'4"
All cells grouted
Figure 9.110 Grout cells solid within 10 ft of each end of side walls (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
Cm + Cs1 + Cs2> P + Ts1 + Ts2
2,560 + 121 + 290 = 2,971 < 3,601 = 430 + 453 + 2,718
Therefore, there is not enough compression capacity to ensure ductile failure.
In order to ensure ductile failure with #7 bars at 24 in. on center, one of the following revisions must be
made: either add (3,601 kips  2,971 kips) = 630 kips to Cm or reduce T by reducing As. Since this
amount of reinforcement is needed for outofplane flexure, As cannot be reduced.
Try filling all cells for 10 ft  0 in. from each end of the wall. As shown in Figure 9.110, this results in
10 additional grouted cells.
Area of one grouted cell: (8 in.)(5.13 in.) = 41 in.2
Volume of grout for one cell: (6 in.)(5.13 in.)(30 ft.)/(144 in.2/ft.2) = 6.41 ft.3
Weight of grout for one cell: (0.140 kcf)(6.41) = 0.90 kips/cell
Additional P: (10 additional cells)(0.9) = 9.0 kips
Additional Cm: 0.8 fm' (41 in.2)(10 cells) = 656 kips
Additional Cm  additional P: 656 kips  9 kips = 647 kips
Net additional Cm: 647 kips > 630 kips OK
or, as expressed in terms of the above equation:
SC > ST
2,971 kips + 656 kips = 3,627 kips > 3,610 kips = 3,601 kips + 9 kips OK
Since C > T, the ductile criterion is satisfied.
This particular check is somewhat controversial. In the opinion of the authors, flexural yield is feasible
for walls with M/Vd in excess of 1.0; this criterion limits the compressive strain in the masonry, which
leads to good performance in strong ground shaking. For walls with M/Vd substantially less than 1.0, the
wall will fail in shear before a flexural yield is possible. Therefore, the criterion does not affect
performance. Well distributed and well developed reinforcement to control the shear cracks is the most
important ductility attribute for such walls.
9.1.5.4.2 Strength check
The wall is so long with respect to its height that inplane strength for flexure is acceptable by inspection.
FEMA 451, NEHRP Recommended Provisions: Design Examples
924
9.1.5.5 Combined Loads
Combined loads are not calculated here because the inplane strength is obviously very high. Outofplane
resistance governs the flexural design.
9.1.5.6 Shear in Longitudinal Walls (Side Walls)
Compute outofplane shear at base of wall in accordance with Provisions Sec. 5.2.6.2.7[Sec. 4.6.1.3]:
Fp = 0.4SDSWc = (0.4)(1.00)(65 psf)(28 ft/2) = 364 plf.
Information from the flexural design from Sec. 9.1.5.3 is needed to determine the required shear strength
based upon development of the flexural capacity. The ratio of fMN to MU is the largest for the load case
0.7D + E. The load that would develop the flexural capacity is approximated by ratio (a second Pdelta
analysis does not seem justified for this check):
w w
M
M
N
U
'
/ ./.
.
= × = × = .
f f
26
532 085
350
psf 465 psf
1.25 times this results in a load for shear design of 58 psf. Thus VU = (58 psf)(28 ft. /2) = 818 plf. The
capacity of computed per Provisions Eq.11.7.3.2[ACI 530, Eq. 3.2.1] :
m4.0 1.75 nm0.25
V MAf P
Vd
=...  ... ...... ' +
M/Vd need not be taken larger than 1.0. An is taken as bwd = 8.32(3.81) = 31.7 in.2 per cell from Figure
9.17. Because this shear exists at both the bottom and the top of the wall, conservatively neglect the
effect of P:
[4.0 1.75(1.0)](51.3in.2 / 2 ft.) 2,000 0 1.595 klf m V =  + =
fVm = (0.8)(1.595) = 1.28 klf > 0.81 klf
As indicated in Sec. 9.1.4.1 and Sec. 9.1.5.1, the inplane demand at the base of the wall, Vu = 2.5(211
kips) = 528 kips, and the shear capacity, fVm is larger than 904 kips.
For the purpose of understanding likely behavior of the building somewhat better, Vn is estimated more
accurately for these long walls:
M/Vd = h/l = 28/200 = 0.14
P = 0.7D = 0.7(430) = 301 kip
Vm = [4.0  1.75(0.14)][200(51.3) + 2(10)91.551.3)](0.045) + 0.25(301) = 1870 + 75 = 1945 kip
Vs = 0.5(Av/s)fyd = 0.5(0.62/4.0)(60)(200) = 930 kip
Vn = 1945 + 930 = 2875 kip
Maximum Vn = 6%f'mA = 6(0.045 ksi)(9234 in.2) = 2493 < 2875 kip
fVn = 0.8(2493) = 1994 kip
VE = 211kip
Vn/VE = 11.8 >> R used in design
In other words, it is unlikely that the long masonry walls will yield in either inplane shear or flexure at
the design seismic ground motion. The walls will likely yield in outofplane response and the roof
diaphragm may also yield. The roof diaphragm for this building is illustrated in Sec. 10.2.
Chapter 9, Masonry
925
The combined loads for shear (orthogonal loading, per Provisions Sec. 5.2.5.2.2, Item a)[Sec. 4.4.2.3] are
shown in Table 9.13.
Table 9.13 Combined Loads for Shear in Side Wall
OutofPlane InPlane Total
Case 1
Case 2
1.00(810/1,280)+
0.30(810/1,280)+
0.30(528/1994)=
1.00(528/1994)=
0.71 < 1.00 OK
0.45 < 1.00 OK
Values are in kips; 1.0 kip = 4.45 kN.
9.1.6 Transverse Walls
The transverse walls will be designed in a manner similar to the longitudinal walls. Complicating the
design of the transverse walls are the door openings, which leave a series of masonry piers between the
doors.
9.1.6.1 Horizontal Reinforcement
The minimum reinforcement, per Provisions Sec. 11.3.8.3[ACI 530, Sec.1.13.6.3] , is (0.0007)(11.625
in.)(8 in.) = 0.065 in.2 per course. The maximum spacing of horizontal reinforcement is 48 in., for which
the minimum reinforcement is 0.39 in.2. Two #4 in bond beams at 48 in. on center would satisfy the
requirement. The large amount of vertical reinforcement would combine to satisfy the minimum total
reinforcement requirement. However, given the 100ft length of the wall, a larger amount is desired for
control of restrained shrinkage as discussed in Sec. 9.1.5.1. Two #5 at 48 in. on center will be used.
9.1.6.2 Vertical Reinforcement
The area for each bay subject to outofplane wind is 20 ft wide by 30 ft high because wind load applied
to the doors is transferred to the masonry piers. However, the area per bay subject to both inplane and
outofplane seismic is reduced by the area of the doors. This is because the doors are relatively light
compared to the masonry. See Figures 9.112 and 9.113.
9.1.6.3 OutofPlane Flexure
Outofplane flexure will be considered in a manner similar to that illustrated in Sec. 9.1.5.3 . The design
of this wall must account for the effect of door openings between a row of piers. The steps are the same
as identified previously and are summarized here for convenience:
1. Select a trial design,
2. Investigate to ensure ductility,
3. Make sure the trial design is suitable for wind (or other nonseismic) lateral loadings using IBC,
4. Calculate midheight deflection due to wind by IBC,
5. Calculate the seismic demand, and
6. Determine the seismic resistance and compare to the demand determined in Step 5.
9.1.6.3.1 Trial design
A trial design of 12in.thick CMU reinforced with two #6 bars at 24 in. on center is selected. The selfweight
of the wall, accounting for horizontal bond beams at 4ft on center, is conservatively taken as 103
psf. Adjacent to each door jamb, the vertical reinforcement will be placed into two cells. See Figure 9.1
11.
FEMA 451, NEHRP Recommended Provisions: Design Examples
926
2'0" 2'0" 2'0"
8'0"
(2) #6
6.88"
11.63"
Figure 9.111 Trial design for piers on end walls (1.0 in = 25.4
mm, 1.0 ft = 0.3048 m).
28'0"
20'0" 8'0"
17.8' 20'0"
Inplane
loads
P
V (short wall)
V + V
(long wall)
Area/ bay subject
to seismic (because
masonry walls are
much heavier
than doors)
Area/ bay subject
to wind (because
doors transfer wind
loads to masonry)
Pf = 8 kips f Pf
w
f w
2'0'
Pf
Outofplane loads
applied to bay
Figure 9.112 Inplane loads on end walls (1.0 ft = 0.3048 m).
6'
17.8'
21'
12' 16' 2'
R = 6.82 kips /bay outofplane
4.88 kips /bay inplane
w = (0.4) (103) (8') = 330 plf/bay outofplane
(0.286) (103) (8') = 236 plf/bay inplane
R = 11.97 kips/bay outofplane
8.56 kips/bay inplane
H
H
H
w = (0.4) (103 psf) (20') = 824 plf/bay outofplane
(0.286) (103 psf) (20') = 589 plf/bay inplane
bot
bot
top
2
1
r
top
Figure 9.113 Outofplane load diagram and resultant of lateral loads (1.0 ft = 0.3048 m,
1.0 lb = 4.45 N, 1.0 kip = 4.45 kN).
Next, determine the design loads. The centroid for seismic loads, accounting for the door openings, is
determined to be 17.8 ft above the base. See Figures 9.112 and 9.113.
Chapter 9, Masonry
927
9.1.6.3.2 Investigate to ensure ductility
The critical strain condition is corresponds to a strain in the extreme tension reinforcement (which is a
pair of #6 bars in the end cell in this example) equal to 1.3 times the strain at yield stress. See Figures
9.111 and 9.114.
For this case:
t = 11.63 in.
d = 11.63  2.38 = 9.25 in.
em = 0.0025 (Provisions Sec. 11.6.2.1.b)[ACI 530, Sec. 3.2.2]
es = 1.3 ey = 1.3 (fy/Es) = 1.3 (60 ksi /29,000 ksi) = 0.0027 (Provisions Sec. 11.6.2.2)[ACI 530, Sec.
3.2.3.5.1]
4.45 in.
( )
m
m s
c d
e
e e
. .
=.. + .. =
a = 0.8c = 3.56 in. (Provisions Sec. 11.6.2.2)
FEMA 451, NEHRP Recommended Provisions: Design Examples
928
d = 9.25"
c
4.45"
1.50" 2.06" 0.89" 4.80"
T
0.8f '
1.6 ksi
1.92"
2.98"
3.70"
a = 0.8c = 3.56"
1.3 = 0.0027
C
C
= 0.0025
m
P = (Pf + Pw )
m
2
1
y
11.63"
1.50" 1.50"
(6) #6 (6) #6
N.A.
Cres
Figure 9.114 Investigation of outofplane ductility for end wall (1.0 in = 25.4
mm, 1.0 ksi = 6.89 MPa).
Note that the Whitney compression stress block, a = 3.56 in. deep, is greater than the 1.50in. face shell
thickness. Thus, the compression stress block is broken into two components: one for full compression
against solid masonry (the face shell) and another for compression against the webs and grouted cells but
accounting for the open cells. These are shown as C1 and C2 on Figure 9.115. The values are computed
using Provisions Sec. 11.6.2.1e:[ACI 530, 3.2.2.e];
C1 = 0.80fm' (1.50 in.)b = (0.80)(2 ksi)(1.50)(96) = 230 kips (for full length of pier)
C2 = 0.80fm' (a  1.50 in.)(6(8 in.)) = (0.80)(2 ksi)(3.56  1.50)(48) = 158 kips
The 48 in. dimension in the C2 calculation is the combined width of grouted cell and adjacent mortared
webs over the 96in. length of the pier.
T is based on 1.25Fy (Provisions Sec. 11.6.2.2)[ACI 530, Sec. 3.2.3.5.1]:
Chapter 9, Masonry
929
2'0"
2"
b = 8.31"
d = 9.25"
c = 1.66"
N.A. N.A.
1.25" 6.06" 1" 6.06" 1.25"
(2) #6 at 24"
8.63"
1.5" 1.5"
11.63"
w
Figure 9.115 Cracked moment of inertia (Icr) for end walls (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
Dimension “c” depends on calculations shown for Figure 9.116.
T = 1.25FyAs = (1.25)(60 ksi)(6 × 0.44 in.2) = 198 kips/pier
P = (Pf + Pw) = 8.0 k + (0.103 ksf)(18 ft.)(20 ft.) = 45.1 kips/pier
P is computed at the head of the doors:
C1 + C2 > P + T
388 kip > 243 kips
Since the compression capacity is greater than the tension capacity, the ductility criterion is satisfied.
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including
critical strain in extreme tensile reinforcement (1.5 times) and axial force to consider when performing the
ductility check (factored loads).]
9.1.6.3.3 Check for wind loading using IBC
Note that load factors and section properties are different in the IBC and the Provisions. Note also that
wind per bay is over the full 20 ft wide by 30 ft high bay as discussed above. (The calculations are not
presented here.)
9.1.6.3.4 Calculate midheight deflection due to wind by IBC
Although the calculations are not presented here, note that in Figure 9.115 the neutral axis position and
partial grouting results in a T beam cross section for the cracked moment of inertia. Use of the plastic
neutral axis is a simplification for computation of the cracked moment of inertia. For this example,
midheight outofplane deflection is 1.27 in. < 2.35 in. = 0.007h, which is acceptable.
9.1.6.3.5 Calculate Seismic Demand
For this example, the load combination with 0.7D has been used and, for this calculation, forces and
moments over a single pier (width = 96 in.) are used. This does not violate the “b > 6t” rule (ACI 530
Sec. 7.3.3)[ACI 530, Sec. 3.2.4.3.3] because the pier is reinforced at 24 in. o.c. The use of the full width
of the pier instead of a 24 in. width is simply for calculation convenience.
FEMA 451, NEHRP Recommended Provisions: Design Examples
930
For this example, a Pdelta analysis using RISA2D was run. This resulted in:
Maximum moment, Mu = 66.22 ftkips/bay = 66.22/20 ft = 3.31 klf (does not govern)
Moment at top of pier, Mu = 62.12 ftkips/pier = 62.12 / 8 ft = 7.77 klf (governs)
Shear at bottom of pier, Vu = 6.72 kips/pier
Reaction at roof, Vu = 12.07 kips/bay
Axial force at base, Ru = 54.97 kips/pier
The shears do not agree with the reactions shown in Figure 9.113; because the results in Figure 9.113 do
not include the Pdelta consideration.
9.1.6.3.6 Determine moment resistance at the top of the pier
See Figure 9.116.
As = 6#6 = 2.64 in.2
d = 9.25 in.
T = 2.64(60) = 158.4 kip
C = T + P = 203.5 kip
a = C / (0.8f'mb) = 203.5 / [0.8(2)96] = 1.32 in.
Because a is less than the face shell thickness (1.50 in.), compute as for a rectangular beam. Moments are
computed about the centerline of the wall.
MN = C (t/2  a/2) + P (0) + T (d  t/2)
= 203.5(5.81  1.32/2) + 158.4(9.251.32/2) = 1593 in.kip = 132.7 ft.kip
fMN = 0.85(132.7) = 112.8 ft.kip
Because moment capacity at the top of the pier, fMn = 112.8 ftkips, exceeds the maximum moment
demand at top of pier, Mu = 62.1 ftkips, the condition is acceptable but note that this is only tentative
acceptance.
The Provisions requires a check of the combined loads in accordance with Provisions 5.2.5.2, Item a
[Sec. 4.4.2.3]. See Sec. 9.1.6.5 for the combined loads check.
Chapter 9, Masonry
931
d = 9.25"
a = 4.49" 3.44"
T
0.8f '
= 1.6 ksi
5.15"
m
Pu = (Pf + Pw ) D
11.63"
f
1.50" 1.50"
t tf
1.32"
C
Figure 9.116 Outofplane seismic strength of pier on end wall (1.0 in =
25.4 mm, 1.0 ksi = 6.89 MPa).
9.1.6.4 InPlane Flexure
There are several possible methods to compute the shears and moments in the individual piers of the end
wall. For this example, the end wall was modeled using RISA2D. The horizontal beam was modeled at
the top of the opening, rather than at its midheight. The inplane lateral loads (from Figure 9.112) were
applied at the 12ft elevation and combined with joint moments representing transfer of the horizontal
forces from their point of action down to the 12ft elevation. Vertical load due to roof beams and the selfweight
of the end wall were included. The input loads are shown on Figure 9.117. For this example:
w = (18 ft.)(103 psf) + (20 ft.)(20 psf) = 2.254 klf
H = (184 kip)/5 = 36.8 kip
M = 0.286((400 + 418)(28  12) + 470(17.8  12)) = 452 ftkip
FEMA 451, NEHRP Recommended Provisions: Design Examples
932
H
M
M/2
H/2
w
H/2 H
M
H
M
H
M
M/2
96'0"
12'0"
Figure 9.117 Input loads for inplane end wall analysis (1.0 ft = 0.3048 m).
El. 112'0" = T.O. Pier
8'
12'
V = 43.6 kip
R = 55.0 kip
bot
bot
M b o t = 0
V t o p = 43.6 kip
M t o p = 523 ftkip
P t o p = 45.1 kip
Figure 9.118 Inplane design condition for 8ftwide pier
(1.0 ft = 0.3048 m).
The input forces at the end wall are distributed over all the piers to simulate actual conditions. The RISA
2D frame analysis accounts for the relative stiffnesses of the 4ftand 8ftwide piers. The final
distribution of forces, shears, and moments for an interior pier is shown on Figure 9.118.
As a trial design for inplane pier design, use two #6 bars at 24 in. on center supplemented by adding two
#6 bars in the cells adjacent to the door jambs (see Figure 9.119).
Chapter 9, Masonry
933
8'0' = 96"
11.63"
92"
m = 0.0025
14" 6" 18" 42" 66" 74"
N.A.
s = 5 y
= 0.0103
0.0092
0.0058
0.0025
0.0008
0.0019
c = 18"
P = (Pf + P )
a = 0.8c
14.4"
10.8"
Cm
0.8f 'm = 1.6 ksi 0.8f 'm = 1.6 ksi
Cs1
Cs2
6"
14" 18"
Ts4
42"
Ts3 Ts2 Ts1
66"
74"
1.25Fy
= 75 ksi
5 y
w
4"
Figure 9.119 Inplane ductility check for 8ftwide pier (1.0 in = 25.4 mm, 1.0 ksi = 6.89 MPa).
FEMA 451, NEHRP Recommended Provisions: Design Examples
934
The design values for inplane design at the top of the pier are:
Unfactored 0.7D + 1.0E 1.4D + 1.0E
Axia P = 45.1 kips Pu = 31.6 kips Pu = 63.2 kips
Shea V = 43.6 kips Vu = 43.6 kips Vu = 43.6 kips
Mom M = 523 ftkips Mu = 523 ftkips Mu = 523 ftkips
The ductility check is illustrated in Figure 9.119:
em = 0.0025
es = 5ey = (5)(60/29,000) = 0.0103
d = 92 in.
From the strain diagram, the strains at the rebar locations are:
e66 = 0.0092
e42 = 0.0058
e18 = 0.0025
e6 = 0.0008
e14 = 0.0019
To check ductility, use unfactored loads:
P = Pf + Pw = (0.020 ksf)(20 ft)(20 ft) + (0.103 ksf)(18 ft)(20 ft)
P = 8 kips + 37.1 kips = 45.1 kips
a = 0.8c = 14.4 in.
Cm = (0.8fm' )ab = 1.6 ksi)(14.4 in.)(11.63 in.) = 268.0 kips
Ts1 = Ts2 = Ts3 = Ts4 = (1.25Fy)(As) = (1.25)(60 ksi)(2 × 0.44 in.2) = 66 kips
Cs1 = FyAs(e14/ey) = (60 ksi)(2 × 0.44 in.2)(0.0019/0.00207) = 48.5 kips
Cs2 = FyAs(e6/ey) = (60 ksi)(2 × 0.44 in.2)(0.0008/0.00207) = 20.4 kips
SC > S T + P
Cm + Cs1 + Cs2 > Ts1 + Ts2 + Ts3 + Ts4 + P
268 + 48.5 + 20.4 > 66 + +66 + 66 + 66 + 45.1
336.9 kips > 309.1 kips
Since compression capacity exceeds tension capacity, ductile failure is ensured. Note that 1.25Fy is used
for tension calculations per Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.51] .
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing
the ductility check (factored loads).]
For the strength check, see Figure 9.120.
Chapter 9, Masonry
935
11.63"
m = 0.0025
N.A.
= 0.0017
Cm1 Cm2 Cm3
Ts1 * Ts2
m = 0.0025
Ts4 Ts3 Ts2 Ts1
y = F
E
= 0.00207
y
y = 0.00207
Balanced
Case
P = 0 Case
N.A.
0.8f'm
0.8f'm
y = 0.00207
Cs1
Cs2 = 0.0019
*
96"
48" 44" 4"
a = 11.3"
c = 14.2"
16" 16" 8.3"
40.3"
50.3"
7.7" 2.3"
48" 44"
Center
Line
M
P
Figure 9.120 Inplane seismic strength of pier (1.0 in = 25.4 mm). Strain diagram superimposed on
strength diagram for both cases. Note that low force in reinforcement is neglected in calculations.
To ascertain the strength of the pier, a fPn  fMn curve will be developed. Only the portion below the
“balance point” will be examined as that portion is sufficient for the purposes of this example. Ductile
failures occur only at points on the curve that are below the balance point so this is consistent with the
FEMA 451, NEHRP Recommended Provisions: Design Examples
936
overall approach).
For the P = 0 case, assume all bars in tension reach their yield stress and neglect compression steel (a
conservative assumption):
Ts1 = Ts2 = Ts3 = Ts4 = (2)(0.44 in.2)(60 ksi) = 52.8 kips
Cm = S Ts = (4)(52.8) = 211.2 kips
Cm = 0.8f’mab = (0.8)(2 ksi)a(11.63 in.) = 18.6a
Thus, a = 11.3 in. and c = a/f = 11.3 / 0.8 = 14.2 in.
SMcl = 0:
Mn = 42.35 Cm + 44Ts1 + 36Ts2 + 12Ts3  12Ts4 = 13,168 in.kips
fMn = (0.85)(13,168) = 11,193 in.kips = 933 ftkips
For the balanced case:
d = 92 in.
e = 0.0025
ey = 60/29,000 = 0.00207
m 50.3 in.
m y
c d
e
e e
. .
=... + ... =
a = 0.8c = 40.3 in.
Compression values are determined from the Whitney compression block adjusted for fully grouted cells
or nongrouted cells:
Cm1 = (1.6 ksi)(16 in.)(11.63 in.) = 297.8 kips
Cm2 = (1.6 ksi)(16 in.)(2 × 1.50 in.) = 76.8 kips
Cm3 = (1.6 ksi)(8.3 in.)(11.63 in.) = 154.4 kips
Cs1 = (0.88 in.2)(60 ksi) = 52.8 kips
Cs2 = (0.88 in.2)(60 ksi)(0.0019 / 0.00207) = 48.5 kips
Ts2 = (0.88 in.2)(60 ksi) = 52.8 kips
Ts2 = (0.88 in.2)(60 ksi)(0.0017 / 0/00207) = 43.4 kips
S Fy = 0:
Pn = SC  ST = 297.8 + 76.8 + 154.4 + 52.8 + 48.5 52.8  43.4 = 534 kips
fPn = (0.85)(534) = 454 kips
S Mcl = 0:
Mn = 40Cm1 + 24Cm2 + 11.85Cm3 + 44Cs1 + 36Cs2 + 44Ts1 + 36Ts2 = 23,540 in.kips
fMn = (0.85)(23,540) = 20,009 in.  kips = 1,667 ftkips
The two cases are plotted in Figure 9.121 to develop the fPn  fMn curve on the pier. The demand (Pu ,
Mu) also is plotted. As can be seen, the pier design is acceptable because the demand is within the fPn 
fMn curve. (See the Birmingham 1 example in Sec. 9.2 for additional discussion of fPn  fMn curves.)
By linear interpolation, fMn at the minimum axial load is 968 kip.
Chapter 9, Masonry
937
9.1.6.5 Combined Loads
Combined loads for inplane and outofplane moments in piers at end walls, per Provisions Sec.
5.2.5.2.2, Item a, are shown in Table 9.14.
Table 9.14 Combined Loads for Flexure in End Pier
0.7D
OutofPlane InPlane Total
Case 1
Case 2
1.0(62.12/112.8) +
0.3(62.12/112.8) +
0.3(523/986) =
1.0(523/986) =
0.71 < 1.00 OK
0.70 < 1.00 OK
Values are in kips; 1.0 kip = 4.45 kN.
500
ftkips
200 kips
(P max = 66 kips)
(P min = 33 kips)
fP
fM
M = 986 ftkips
P = 0 Case
(933 ftkips, 0 kips)
Simplified fP  fM curve
100 kips
300 kips
400 kips
500 kips
600 kips
Balance
(1667 ftkips, 454 kips)
1,000
ftkips
1,500
ftkips
2,000
ftkips
M = 523 ftkips
n
n
fPn
fMn
u
u
u
u
u u
Figure 9.121 Inplane fP11  fM11 diagram for pier (1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm).
FEMA 451, NEHRP Recommended Provisions: Design Examples
938
30'
12'
12'
Pf
Vroof + long wall
(upper portion)
Pw
Vshort wall
Vbase
P base
Vbase
Pbase
P
V
M
P pier
Mbase
Figure 9.122 Inplane shear on end wall and pier (1.0 ft =
0.3048 m).
9.1.6.6 Shear at Transverse Walls (End Walls)
The shear at the base of the pier is 43.6 kips/bay. At the head of the opening where the moment demand
is highest, the inplane shear is slightly less (based on the weight of the pier). There, V = 43.6 kips 
0.286(8 ft)(12 ft)(0.103 ksf) = 40.8 kips. (This refinement in shear is not shown in Figure 9.118
although the difference in axial load at the two locations is shown.) The capacity for shear must exceed
2.5 times the demand or the shear associated with 125 percent of the flexural capacity. Using the results
in Table 9.14, the 125 percent implies a factor on shear by analysis of:
125
1
125
1
07
1
085
. . 210
. .
.
1
Demand to capacity ratio
.
. .
.
. .
.
. .
.
. .
=
.. .
.. .
.. .
.. .
=
f
Therefore, the required shear capacities at the head and base of the pier are 91.6 kips and 85.7 kips,
respectively.
The inplane shear capacity is computed as follows where the net area, An, of the pier is the area of face
shells plus the area of grouted cells and adjacent webs:
Chapter 9, Masonry
939
Vm4.0 1.75 VMdAnfm 0.25P
=..  .. .... ' +
. . ..
An = (96 in. × 1.50 in.× 2) + (6 cells × 8 in. × 8.63 in.) = 702 in.2
V
A
s
f d S
V
y V =
.. .
.. .
=
.
. .
.
. .
05
05
062
48
60
2
.
.
.
(
in.
in.
ksi)(96 in.)
= 37.2 kips / bay
At the head of the opening:
Vm = [4.0  1.75(1.0)](702 in.2)(0.0447 ksi) + (0.25)(0.7)(45.1 kips) = 78.5 kips/bay
fVN = (0.8)(78.5 + 37.2) = 92.6 kips/bay
At the base:
Vm = [4.0  1.75(0)](702 in.2)(0.0447 ksi) + (0.25)(0.7)(55.0 kips) = 135.2 kips/bay
fVN = (0.8)(135.2 + 37.2) = 137.9 kips/bay
As discussed previously, M/Vd need not exceed 1.0 in the above equation.
For outofplane shear, see Figure 9.113. Shear at the top of wall is 12.07 kips/bay and shear at the base
of the pier is 6.72 kips/bay. From the values in the figure, the shear at the head of the opening is
computed as 6.72 kips  (12 ft)(0.33 kip/ft) = 2.76 kips. The same multiplier of 2.10 for development of
125 percent of flexural capacity will be applied to outofplane shear resulting in 25.3 kips at the top of
the wall, 5.80 kips at the head of the opening, and 14.11 kips at the base.
Outofplane shear capacity is computed using the same equation. Sbwd is taken as the net area An. Note
that M/Vd is zero at the support because the moment is assumed to be zero; however, a few inches into the
span, M/Vd will exceed 1.0 so the limiting value of 1.0 is used here. This is typically the case when
considering outofplane loads on a wall.
For computing shear capacity at the top of the wall:
An = bwd =((8 in./2 ft.) × 20 ft)(9.25 in.) = 740 in.2
Vm = [4.0  1.75(1)](740 in.2)(0.0447 ksi) + (0.25)(8.0) = 76.9 kips/bay
fVm = (0.8)(76.9) = 61.5 kips/bay
For computing shear capacity in the pier:
An =(8 in./cell)(6 cells)(9.25 in.) = 444 in.2
Vm = [4.0  1.75(1)](444 in.2)(0.0447 ksi) + (0.25)(41.67) = 55.4 kips/bay
fVm = (0.8)(55.4) = 44.3 kips/bay
The combined loads for shear at the end pier (per Provisions 5.2.5.2.2, Item a [Sec. 4.423]) are shown in
FEMA 451, NEHRP Recommended Provisions: Design Examples
940
Table 9.15.
Table 9.15 Combined Loads for Shear in End Wall
InPlane OutofPlane Total
Case 1
Pier base 1.0(91.6/137.9) + 0.3(14.11/44.3) = 0.76 < 1.00 OK
Case 2
Pier base 0.3(91.6/137.9) + 1.0(14.11/44.3) = 0.52< 1.00 OK
Case 1
Pier head 1.0(85.7/92.6) + 0.3(5.80/44.3) = 0.96 > 1.00 OK
Case 2
Pier head 0.3(85.7/92.6) + 1.0(5.80/44.3) = 0.41 < 1.00 OK
Values are in kips; 1.0 kip = 4.45 kN.
9.1.7 Bond Beam
Reinforcement for the bond beam located at the elevation of the roof diaphragm can be used for the
diaphragm chord. The uniform lateral load for the design of the chord is the lateral load from the long
wall plus the lateral load from the roof and is equal to 0.87 klf. The maximum tension in rebar is equal
the maximum moment divided by the diaphragm depth:
M/d = 4,350 ftkips/100 ft = 43.5 kips
The seismic load factor is 1.0. The required reinforcement is:
Areqd = T/fFy = 43.5/(0.85)(60) = 0.85 in.2
This will be satisfied by two #6 bars, As = (2 × 0.44 in.2) = 0.88 in.2
In Sec. 10.2, the diaphragm chord is designed as a wood member utilizing the wood ledger member.
Using either the wood ledger or the bond beam is considered acceptable.
9.1.8 InPlane Deflection
Deflection of the end wall (short wall) has two components as illustrated in Figure 9.123.
Chapter 9, Masonry
941
17.8'
V (short wall)
V + V (long wall)
s
f
12'
4' 12' 8'
18'
16'
28'
5.8'
. 1
. 2
. tot
Figure 9.123 Inplane deflection of end wall (1.0 ft = 0.3048 m).
As obtained from the RISA 2D analysis of the piers, .1 = 0.047 in.:
2
VL
AG
a
. = S
where a is the form factor equal to 6/5 and
G = Em/2(1 + µ) = 1500 ksi / 2(1 + 0.15) = 652 ksi
A = An = Area of face shells + area of grouted cells
= (100 ft × 12 in./ft × 2 × 1.50 in.2) +(50)(8 in.)(8.63 in.) = 7,050 in.2
Therefore:
= 0.0013 + 0.0059 = 0.007 in.
6(67.15)(5.8 12)6(116.9)(16 12)
25(7,050)(652)5(7,050)(652)
. . × . . ×
. = . . + . .
. . . .
and,
.total = Cd(0.047 + 0.007) = 3.5(0.054 in.) = 0.19 in. < 3.36 in.
(3.36 = 0.01hn = 0.01hsx) (Provisions Sec. 11.5.4)
Note that the drift limits for masonry structures are smaller than for other types of structure. It is possible
to interpret Provisions Table 5.2.8 [Table 4.51] to give a limit of 0.007hn for this structure but that limit
also is easily satisfied. The real displacement in this structure is in the roof diaphragm; see Sec. 10.2.
FEMA 451, NEHRP Recommended Provisions: Design Examples
942
6'0"
6'8"
Prestressed
hollow core
slabs
40'0" 24'0" 24'0" 24'0" 40'0"
4'0"
14'0"
24'0" 24'0" 24'0"
152'0"
72'0"
8" concrete
masonry wall
Figure 9.21 Typical floor plan (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
9.2 FIVESTORY MASONRY RESIDENTIAL BUILDINGS IN BIRMINGHAM,
ALABAMA; NEW YORK, NEW YORK; AND LOS ANGELES, CALIFORNIA
9.2.1 Building Description
In plan, this fivestory residential building has bearing walls at 24 ft on center (see Figures 9.21 and 9.2
2). All structural walls are of 8in.thick concrete masonry units (CMU). The floor is of 8in.thick
hollow core precast, prestressed concrete planks. To demonstrate the incremental seismic requirements
for masonry structures, the building is partially designed for four locations: two sites in Birmingham,
Alabama; a site in New York, New York; and a site in Los Angeles, California. The two sites in
Birmingham have been selected to illustrate the influence of different soil profiles at the same location.
The building is designed for Site Classes C and E in Birmingham. The building falls in Seismic Design
Categories B and D in these locations, respectively. For Site Class D soils, the building falls in Seismic
Design Categories C and D for New York and Los Angeles, respectively.
[Note that the method for assigning seismic design category for short period buildings has been revised in
the 2003 Provisions. If the fundamental period, Ta, is less than 0.8Ts, the period used to determine drift is
less than Ts, and the base shear is computed using 2003 Provisions Eq 5.22, then seismic design category
is assigned using just 2003 Provisions Table 1.41 (rather than the greater of 2003 Provisions Tables 1.4
1 and 1.42). This change results in the Birmingham Site Class E building being assigned to Seismic
Design Category C instead of D. The changes to this example based on the revised seismic design
category are not noted in the remainder of the example. The New York building provides an example of
what the Seismic Design Category C requirements would be for the Birmingham Site Class E building.]
Chapter 9, Masonry
943
152'0"
36'0" 80'0" 36'0"
5 at 8'8" = 43'4" 2'0"
45'4"
Figure 9.22 Building elevation (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m).
For the New York and both Birmingham sites, it is assumed that shear friction reinforcement in the joints
of the diaphragm planks is sufficient to resist seismic forces, so no topping is used. For the Los Angeles
site, a castinplace 2 ½in.thick reinforced lightweight concrete topping is applied to all floors. The
structure is free of irregularities both in plan and elevation. The Provisions, by reference to ACI 318,
requires reinforced castinplace toppings as diaphragms in Seismic Design Category D and higher. Thus,
the Birmingham example in Site Class E would require a topping, although that is not included in this
example.
Provisions Chapter 9 has an appendix (intended for trial use and feedback) for the design of untopped
precast units as diaphragms. The design of an untopped diaphragm for Seismic Design Categories A, B,
and C is not explicitly addressed in ACI 318. The designs of both untopped and topped diaphragms for
these buildings are described in Chapter 7 of this volume using ACI 318 for the topped diaphragm in the
Los Angeles building and using the appendix to Provisions Chapter 9 for untopped diaphragms in the
New York building. It is assumed here that the diaphragm for the Birmingham 2 example would be
similar to the New York example, and the extra weight of the Birmingham 2 topping is not included in the
illustration here.
No foundations are designed in this example. However, for the purpose of determining the site class
coefficient (Provisions Sec. 4.1.2.1 [Sec. 3.5]), a stiff soil profile with standard penetration test results of
15 < N < 50 is assumed for Los Angeles and New York sites resulting in a Site Class D for these two
locations. For Birmingham, however, one site has soft rock with N > 50 and the other has soft clay with
N < 15, which results in Site Classes C and E, respectively. The foundation systems are assumed to be
able to carry the superstructure loads including the overturning moments.
The masonry walls in two perpendicular directions act as bearing and shear walls with different levels of
axial loads. The geometry of the building in plan and elevation results in nearly equal lateral resistance in
both directions. The walls are constructed of CMU and are typically minimally reinforced in all
locations. The walls are assumed to act as columns in their planes. Figure 9.23 illustrates the wall
layout.
FEMA 451, NEHRP Recommended Provisions: Design Examples
944
A
B
B
A
A
B
B
A
D
C C D
D
D C C
E
F
G
G
F
E
Wall length
A 36'0"
B 34'0'
C 32'8"
D 32'8"
E 8'0"
F 8'0"
G 8'0"
Figure 9.23 Plan of walls (1.0 ft = 0.3048 m).
The floors serve as horizontal diaphragms distributing the seismic forces to the walls and are assumed to
be stiff enough to be considered rigid. There is little information about the stiffness of untopped precast
diaphragms. The design procedure in the appendix to Provisions Chapter 9 results in a diaphragm
intended to remain below the elastic limit until the walls reach an upper bound estimate of strength,
therefore it appears that the assumption is reasonable.
Material properties are as follows:
The compressive strength of masonry, f!m, is taken as 2,000 psi and the steel reinforcement has a yield
limit of 60 ksi.
The design snow load (on an exposed flat roof) is taken as 20 psf for New York; design for snow does not
control the roof design in the other locations.
This example covers the following aspects of a seismic design:
1. Determining the equivalent lateral forces,
2. Design of selected masonry shear walls for their inplane loads, and
3. Computation of drifts.
See Chapter 7 of this volume for the design and detailing of untopped and topped precast diaphragms.
9.2.2 Design Requirements
9.2.2.1 Provisions Parameters
The basic parameters affecting the design and detailing of the buildings are shown in Table 9.21.
Chapter 9, Masonry
945
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package).]
9.2.2.2 Structural Design Considerations
The floors act as horizontal diaphragms and the walls parallel to the motion act as shear walls for all four
buildings
The system is categorized as a bearing wall system (Provisions Sec. 5.2.2[Sec. 4.3]). For Seismic Design
Category D, the bearing wall system has a height limit of 160 ft and must comply with the requirements
for special reinforced masonry shear walls (Provisions Sec. 11.11.5[Sec. 11.2.1.5]). Note that the
structural system is one of uncoupled shear walls. Crossing beams over the interior doorways (their
design is not included in this example) will need to continue to support the gravity loads from the deck
slabs above during the earthquake, but are not designed to provide coupling between the shear walls.
The building is symmetric and appears to be regular both in plan and elevation. It will be shown,
however, that the building is actually torsionally irregular. Provisions Table 5.2.5 [Table 4.41] permits
use of the equivalent lateral force (ELF) procedure in accordance with Provisions Sec. 5.4 [Sec. 5.2] for
Birmingham 1 and New York City (Seismic Design Categories B and C). By the same table, the
Category D buildings must use a dynamic analysis for design. For this particular building arrangement,
the modal response spectrum analysis does not identify any particular effect of the torsional irregularity,
as will be illustrated.
Table 9.21 Design Parameters
Design Parameter Value for
Birmingham 1
Value for
Birmingham 2
Value for
New York
Value for
Los Angeles
Ss (Map 1) [Figure
3.31]
0.3 0.3 0.4 1.5
S1 (Map 2) [Figure
3.32]
0.12 0.12 0.09 0.6
Site Class C E D D
Fa 1.2 2.34 1.48 1
Fv 1.68 3.44 2.4 1.5
SMS = FaSs 0.36 0.7 0.59 1.5
SM1 = FvS1 0.2 0.41 0.22 0.9
SDS = 2/3 SMS 0.24 0.47 0.39 1
SD1 = 2/3 SM1 0.13 0.28 0.14 0.6
Seismic Design
Category
B D C D
Masonry Wall Type Ordinary
Reinforced
Special
Reinforced
Intermediate
Reinforced
Special Reinforced
Provisions Design Coefficients (Table 5.2.2 [4.31])
R 2.0 3.5 2.5 3.5
FEMA 451, NEHRP Recommended Provisions: Design Examples
Design Parameter Value for
Birmingham 1
Value for
Birmingham 2
Value for
New York
Value for
Los Angeles
946
O0 2.5 2.5 2.5 2.5
Cd 1.75 3.5 2.25 3.5
IBC Design Coefficients (presented for comparison with Provisions coefficients)
R 2.5 5.0 3.5 5.0
O0 2.5 2.5 2.5 2.5
Cd 1.75 3.5 2.25 3.5
The orthogonal effect (Provisions Sec. 5.2.5.2, Item a [Sec. 4.4.2]) applies to structures assigned to
Seismic Design Categories C and D (all of the example buildings except for Birmingham 1). However,
the arrangement of this building is not particularly susceptible to orthogonal effects. This is because the
stresses developed under outofplane loading for shortheight walls (story clear height is 8 ft) are low
and, their contribution to orthogonal effects is minimal.
The walls are all solid and there are no significant discontinuities, as defined by Provisions Sec. 5.2.6.2.3
[Sec. 4.3.2.3], in the vertical elements of the seismicforceresisting system.
Ignoring the short walls at stairs and elevators, there are eight shear walls in each direction, therefore, the
system appears to have adequate redundancy (Provisions Sec. 5.2.6.2.4 [Sec. 4.3.3]). The reliability
factor, however, will be computed. [See Sec. 9.2.3.1 for changes to the reliability factor.]
Tie and continuity requirements (Provisions Sec. 5.2.6.1.2 [Sec. 4.6]) must be addressed when detailing
connections between floors and walls (see Chapter 7 of this volume).
Nonstructural elements (Provisions Chapter 14 [Chapter 6]) are not considered in this example.
Collector elements are required in the diaphragm for longitudinal response (Provisions Sec. 5.2.6.2.5
[Sec. 4.6]). Rebar in the longitudinal direction, spliced into bond beams, will be used for this purpose
(see Chapter 7 of this volume).
Diaphragms must be designed for the required forces (Provisions Sec. 5.2.6.2.6 [Sec. 4.6]).
The bearing walls must be designed for the required force perpendicular to their plane (Provisions Sec.
5.2.6.2.7 [Sec. 4.6.1.3]).
Each wall is a vertical cantilever; there are no coupling beams. The walls are classified as masonry
cantilever shear wall structures in Provisions Table 5.2.8 [Table 4.51], which limits interstory drift to
0.01 times the story height. Provisions Sec.11.5.4.1.1 also limits drift to 0.01 times the wall height for
such a structure.
[The deflection limits have been removed from Chapter 11 of the 2003 Provisions because they were
redundant with the general deflection limits. Based on ACI 530 Sec. 1.13.3.2, the maximum drift for all
masonry structures is 0.007 times the story height. Thus, there appears to be a conflict between ACI 530
and 2003 Provisions Table 4.51.]
Chapter 9, Masonry
947
Vertical accelerations must be considered for the prestressed slabs in Seismic Design Category D
(Provisions Sec. 5.2.6.4.3 [Sec. 4.6.3.1]); refer to Chapter 7 of this volume. The evaluation of such
components involves the earthquake effect determined using Provisions Eq. 5.2.71 [4.21] and 5.2.72
[4.21]. The important load is the vertical effect (0.2SDSD), which reduces the effect of dead loads.
Because the system is prestressed, application of this load might lead to tension where there would
otherwise be no reinforcement. The reinforcement within the topping will control this effect. Refer to
Sec. 7.1 of this volume for the design of precast, prestressed slabs and topping.
Design, detailing, and structural component effects are presented in the chapters of the Provisions that are
relevant to the materials used.
9.2.3 Load Combinations
The basic load combinations (Provisions Sec. 5.2.7 [Sec. 4.2.2]) are the same as those in ASCE 7 (and are
similar to those in the IBC). The seismic load effect, E, is defined by Provisions Eq. 5.2.71 [4.21] and
5.2.72 [4.22] as:
E = .QE ± 0.2SDSD
9.2.3.1 Reliability Factor
Note that . is a multiplier on design force effects and applies only to the inplane direction of the shear
walls. For structures in Seismic Design Categories A, B and C, . = 1.0 (Provisions Sec. 5.2.4.1 [Sec.
4.3.3.1]). For structures in Seismic Design Category D, . is determined per Provisions Sec. 5.2.4.2 [Sec.
4.3.3.2].
For the transverse direction, ignoring accidental torsion:
10 1 10 0.038
x 8 33
wall
max
story w
r V
V l
=.... ....... ....... ...... ...=
and,
2 20 2 20 3.03
0.038 10,944 maxx x r A
. =  =  =
Since the computed . < 1.0 use . = 1.0 for the transverse direction. Accidental torsion does not change
enough to change this conclusion.
maxx r
Based on similar calculations for the longitudinal direction, . is determined to be 1.0.
[The redundancy requirements have been substantially changed in the 2003 Provisions. For structures
assigned to Seismic Design Categories B and C, . = 1.0 in all cases. For a shear wall building assigned to
Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear wall with heighttolength
ratio greater than 1.0 would not result in more than a 33 percent reduction in story strength or
create an extreme torsional irregularity. The intent is that the aspect ratio is based on story height, not
total height. Therefore, the redundancy factor would not have to be investigated (. = 1.0) for the
structure(s) assigned to Seismic Design Category D.]
FEMA 451, NEHRP Recommended Provisions: Design Examples
948
9.2.3.2 Combination of Load Effects
The seismic load effect, E, determined for each of the buildings is:
Birmingham 1 E = (1.0)QE ± (0.2)(0.24)D = QE ± 0.05D
Birmingham 2 E = (1.0)QE ± (0.2)(0.47)D = QE ± 0.09D
New York E = (1.0)QE ± (0.2)(0.39)D = QE ± 0.08D
Los Angeles E = (1.0)QE ± (0.2)(1.00)D = QE ± 0.20D
The applicable load combinations from ASCE 7 are:
1.2D + 1.0E + 0.5L + 0.2S
when the effects of gravity and seismic loads are additive and
0.9D + 1.0E + 1.6H
when the effects of gravity and seismic loads are counteractive. (H is the effect of lateral pressures of soil
and water in soil.)
Load effect H does not apply for this design, and the snow load effect, S, exceeds the minimum roof live
load only at the building in New York. However, even for New York, the snow load effect is only used
for combinations of gravity loading. Consideration of snow loads is not required in the effective seismic
weight, W, of the structure when the design snow load does not exceed 30 psf (Provisions Sec. 5.3 [Sec.
5.2.1]).
The basic load combinations are combined with E as determined above, and the load combinations
representing the extreme cases are:
Birmingham 1 1.25D + QE +0.5L
0.85D  QE
Birmingham 2 1.29D + QE +0.5L
0.81D  QE
New York 1.28D + QE +0.5L +0.2S
0.82D  QE
Los Angeles 1.40D + QE +0.5L
0.70D  QE
These combinations are for the inplane direction. Load combinations for the outofplane direction are
similar except that the reliability coefficient (1.0 in all cases for inplane loading) is not applicable.
It is worth noting that there is an inconsistency in the treatment of snow loads combined with seismic
loads. IBC Sec. 1605.3 clearly deletes the snow term from the ASD combinations where the design snow
load does not exceed 30 psf. There is no similar provision for the strength load combinations in the IBC
for reference standard, ASCE 7.
[The strength design load combinations in the 2003 IBC do have a similar exemption for snow loads, but
ASCE 702 load combinations do not.]
9.2.4 Seismic Design for Birmingham 1
Chapter 9, Masonry
949
9.2.4.1 Birmingham 1 Weights
Use 67 psf for 8in.thick, normal weight hollow core plank plus the nonmasonry partitions. This site is
assigned to Seismic Design Category B, and the walls will be designed as ordinary reinforced masonry
shear walls (Provisions Sec. 11.11.3 [Sec. 4.2.1.3]), which do not require prescriptive seismic
reinforcement. However, both ACI 530 and IBC 2106.1.1.2 stipulate that ordinary reinforced masonry
shear walls have a minimum of vertical #4 bars at 120 in. on center. [By reference to ACI 530, the 2003
Provisions (and 2003 IBC) do have prescriptive seismic reinforcement requirements for ordinary
reinforced masonry shear walls. Refer to ACI 530 Sec. 1.13.2.2.3.] Given the length of the walls,
vertical reinforcement of #4 bars at 8 ft on center works well for detailing reasons and will be used here.
For this example, 45 psf will be assumed for the 8in.thick lightweight CMU walls. The 45 psf value
includes grouted cells and bond beams in the course just below the floor planks.
Story weight, wi, is computed as follows:
For the roof:
Roof slab (plus roofing) = (67 psf) (152 ft)(72 ft) = 733 kips
Walls = (45 psf)(589 ft)(8.67 ft/2) + (45 psf)(4)(36 ft)(2 ft) = 128 kips
Total = 861 kips
Note that there is a 2fthigh masonry parapet on four walls and the total length of masonry wall,
including the short walls, is 589 ft.
For a typical floor:
Slab (plus partitions) = 733 kips
Walls = (45 psf)(589 ft)(8.67 ft) = 230 kips
Total = 963 kips
Total effective seismic weight, W = 861 + (4)(963) = 4,713 kips
This total excludes the lower half of the first story walls, which do not contribute to seismic loads that are
imposed on CMU shear walls.
9.2.4.2 Birmingham 1 Base Shear Calculation
The seismic response coefficient, Cs, is computed using Provisions Sec. 5.4.1.1 [Sec. 5.2.1.1].
Per Provisions Eq. 5.4.1.11 [Eq. 5.22]:
0.24 0.12
/ 21
DS
s
C S
R I = = =
The value of Cs need not be greater than Provisions Eq. 5.4.1.12 [Eq. 5.23]:
( )
0.13 0.192
( / ) 0.338 21
D1
s
C S
T R I
= = =
T is the fundamental period of the building approximated per Provisions Eq. 5.4.2.11[Eq. 5.26] as:
= x=(0.02)(43.330.75 )=0.338 sec
a r n T Ch
FEMA 451, NEHRP Recommended Provisions: Design Examples
950
where Cr = 0.02 and x = 0.75 are from Provisions Table 5.4.2.1 [Table 5.22].
The value for Cs is taken as 0.12 (the lesser of the two computed values). This value is still larger than
the minimum specified in Provisions Eq. 5.4.1.13:
Cs = 0.044ISDS = (0.044)(1.0)(0.24) = 0 0.0106
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value
for longperiod structures, which is not applicable to this example.]
The total seismic base shear is then calculated using Provisions Eq. 5.4.1 [Eq. 5.21]as:
V = CsW = (0.12)(4,713) = 566 kips
9.2.4.3 Birmingham 1 Vertical Distribution of Seismic Forces
Provisions Sec. 5.4.4 [Sec. 5.2.3] stipulates the procedure for determining the portion of the total seismic
load assigned to each floor level. The story force, Fx, is calculated using Provisions Eq. 5.4.31 [Eq. 5.2
10] and 5.4.3.2 [Eq. 5.211], respectively, as:
Fx = CvxV
and
1
k
x x
vx n
k
i i
i
C w h
wh
=
=
S
For T = 0.338 sec < 0.5 sec, k = 1.0.
The seismic design shear in any story is determined from Provisions Eq. 5.4.4[Eq. 5.212]:
Vx= Fi
i=x
nS
The story overturning moment is computed from Provisions Eq. 5.4.5[Eq. 5.214]:
( )
n
x i i x
i x
M F h h
=
=S 
The application of these equations for this building is shown in Table 9.22.
Chapter 9, Masonry
951
Contribution to weight
concentrated at all stories.
Dynamic response to
round motion results
in lateral load at all stories.
Moments are
from S Vh
Weight of entire building
above ground floor
helps to resist moments.
Contribution to weight
concentrated at roof.
Only upper half of walls
out of plane contribute,
but upper half of all walls
used for convenience.
Dynamic response to
ground motion results
in lateral load applied
at roof.
Moment at fifth floor
M = V h
P of roof slab plus
entire height of wall
helps to resist M.
M5 P5
Proof
h
Vroof
5 roof
Figure 9.24 Location of moments due to story shears.
Table 9.22 Birmingham 1 Seismic Forces and Moments by Level
Level
(x)
wx
(kips)
hx
(ft)
wxhx
k
(ftkips)
Cvx Fx
(kips)
Vx
(kips)
Mx
(ftkips)
543213
861
963
963
963
963
4,715
43.34
34.67
26.00
17.33
8.67
37,310
33,384
25,038
16,692
8,346
120,770
0.3089
0.2764
0.2073
0.1382
0.0691
1.0000
175
156
117
78
39
566
1.8e+14 1,515
4,385
8,272
12,836
17,739
1.0 kips = 4.45 kN, 1.0 ft = 0.3048 m.
A note regarding locations of V and M: the vertical weight at the roof (5th level), which includes the
upper half of the wall above the 5th floor (4th level), produces the shear V applied at the 5th level. That
shear in turn produces the moment applied at the top of the 4th level. Resisting this moment is the rebar in
the wall combined with the wall weight above the 4th level. Note that the story overturning moment is
applied to the level below the level thatreceives the story shear. This is illustrated in Figure 9.24.
FEMA 451, NEHRP Recommended Provisions: Design Examples
952
9.2.4.4 Birmingham 1 Horizontal Distribution of Forces
The wall lengths are shown in Figure 9.23. The initial grouting pattern is basically the same for walls A,
B, and C. Because of a low relative stiffness, the effects Walls D, E, and F are ignored in this analysis.
Walls A, B, and C are so nearly the same length that their stiffnesses will be assumed to be the same for
this example.
Torsion is considered according to Provisions Sec. 5.4.4[Sec. 5.2.4]. For a symmetric plan, as in this
example, the only torsion to be considered is the accidental torsion, Mta, caused by an assumed
eccentricity of the mass each way from its actual location by a distance equal to 5 percent of the
dimension of the structure perpendicular to the direction of the applied loads.
Dynamic amplification of the torsion need not be considered for Seismic Design Category B per
Provisions Sec. 5.4.4.3 [Sec. 5.2.4.3].
For this example, the building will be analyzed in the transverse direction only. The evaluation of Wall D
is selected for this example. The rigid diaphragm distributes the lateral forces into walls in both
directions. Two components of force must be considered: direct shear and shear induced by torsion.
The direct shear force carried by Wall D is oneeighth of the total story shear (eight equal walls). The
torsional moment per Provisions Sec. 5.4.4.2 [Sec. 5.2.4.2] is:
Mta = 0.05bVx =(0.05)(152 ft)Vx = 7.6Vx
The torsional force per wall, Vt, is:
2
t
t
V M Kd
Kd
= S
where K is the stiffness (rigidity) of each wall.
Because all the walls in this example are assumed to be equally stiff:
t t 2
V M d
d
. .
= . .
..S ..
where d is the distance from each wall to the center of twisting.
3d2 = 4(36)2 + 4(12)2 + 4(36)2 + 4(12)2 = 11,520
The maximum torsional shear force in Wall D, therefore is:
Vt = 7.6 V(36/11,520) = 0.0238V
Total shear in Wall D is:
=0.125 +0.0238 =0.149 totV V V V
The total story shear and overturning moment may now be distributed to Wall D and the wall proportions
checked. The wall capacity will be checked before considering deflections.
Chapter 9, Masonry
953
9.2.4.5 Birmingham 1 Transverse Wall (Wall D)
The strength or limit state design concept is used in the Provisions. This method was introduced in the
2002 edition of ACI 530, the basic reference standard for masonry design. Because strength design was
not in prior editions of ACI 530, strength design of masonry as defined in the Provisions is illustrated
here.
[The 2003 Provisions adopts by reference the ACI 53002 provisions for strength design in masonry, and
the previous strength design section has been removed. This adoption does not result in significant
technical changes, and the references to the corresponding sections in ACI 530 are noted in the following
sections.]
9.2.4.5.1 Birmingham 1 Shear Strength
Provisions Sec. 11.7.2 [ACI 530, Sec. 3.1.3] states that the ultimate shear loads must be compared to the
design shear strength per Provisions Eq. 11.7.2.1:
Vu # fVn
The strength reduction factor, f, is 0.8 (Provisions Table 11.5.3, ACI 530 [See 3.1.4.3]). The design
shear strength, fVn, must exceed the shear corresponding to the development of 1.25 times the nominal
flexural strength of the member but need not exceed 2.5 times Vu (Provisions Sec. 11.7.2.2 [ACI 530, Sec.
3.1.3]). The nominal shear strength, Vn, is (Provisions Eq. 11.7.3.11 [ACI 530, Eq. 318]):
Vn = Vm + Vs
The shear strength provided by masonry is (Provisions Eq. 11.7.3.2 [ACI 530, Eq. 321]):
m4.0 1.75 nm0.25
V MAf P
Vd
=...  ... ...... ' +
For grouted cells at 8 ft on center:
An = (2 × 1.25 in. × 32.67 ft x 12 in.) + (8 × 5.13 in.2 × 5 cells) = 1,185 in.2
The shear strength provided by reinforcement is given by Provisions Eq. 11.7.3.3 [ACI 530, Sec.
3.2.4.1.2.2] as:
= 0.5.. ..
. .
v
s y v
V AFd
s
The wall will have a bond beam with two #4 bars at each story to bear the precast floor planks and wire
joint reinforcement at alternating courses. Common joint reinforcement with 9 gauge wires at each face
shell will be used; each wire has a crosssectional area of 0.017 in.2 With six courses of joint
reinforcement and two #4 bars, the total area per story is 0.60 in.2 or 0.07 in.2/ft.
Vs = 0.5(0.07 in.2/ft.)(60 ksi)(32.67 ft.) = 68.3 kips
The maximum nominal shear strength of the member (Wall D in this case) for M/Vdv > 1.00 (the
Provisions has a typographical error for the inequality sign) is given by Provisions Eq. 11.7.3.13 [ACI
530, Eq. 322]:
FEMA 451, NEHRP Recommended Provisions: Design Examples
954
(max) = 4 ' n mn V fA
The coefficient 4 becomes 6 for M/Vdv < 0.25. Interpolation between yields the following:
V
M
Vd
f A N
V
m n (max)= 
.
. .
.
. .
(6.67 2.67 ) '
The shear strength of Wall D, based on the equations listed above, is summarized in Table 9.23. Note
that Vx and Mx in this table are values from Table 9.22 multiplied by 0.149 (which represents the portion
of direct and torsional shear assigned to Wall D). P is the dead load of the roof or floor times the
tributary area for Wall D. (Note that there is a small load from the floor plank parallel to the wall.)
Table 9.23 Shear Strength Calculations for Birmingham 1 Wall D
Story Vx
(kips)
Mx
(ftkips)
Mx/Vxd 2.5 Vx
(kips)
P
(kips)
fVm
(kips)
fVs
(kips)
fVn
(kips)
fVn max
(kips)
5 26 225 0.265 65.0 41 158.1 54.6 212.7 252.7
4 49.3 652 0.405 123.3 89 157.3 54.6 211.9 236.9
3 66.7 1230 0.564 166.8 137 155.1 54.6 209.7 218.8
2 78.4 1910 0.746 196.0 184 151.0 54.6 205.6 198.3
1 84.2 2640 0.960 210.5 232 144.8 54.6 199.4 174.1
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
VU exceeds both fVn and fVn max at the first story. It would be feasible to add grouted cells in the first
story to remedy the deficiency. However, it will be shown following the flexural design that the shear to
develop 1.25 times the flexural capacity is 1.94(84.2 kips) = 163 kips, which is OK.
9.2.4.5.2 Birmingham 1 Axial and Flexural Strength
All the walls in this example are bearing shear walls since they support vertical loads as well as lateral
forces. Inplane calculations include:
1. Strength check and
2. Ductility check
9.2.4.5.2.1 Strength check
The wall demands, using the load combinations determined previously, are presented in Table 9.24 for
Wall D. In the table, Load Combination 1 is 1.25D + QE + 0.5L and Load Combination 2 is 0.85D + QE.
Table 9.24 Demands for Birmingham 1 Wall D
Load Combination 1 Load Combination 2
Level PD
(kips)
PL
(kips)
Pu
(kips)
Mu
(ftkips)
Pu
(kips)
Mu
(ftkips)
54321 4.2e+12 8172534 5.111518e+13 2.2565e+17 3.576116e+12 2.256521e+17
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
Strength at the bottom story (where P, V, and M are the greatest) will be examined. (For a real design, all
levels should be examined). The strength design will consider Load Combination 2 from Table 9.24 to
be the governing case because it has the same lateral load as Load Combination 1 but with lower values
of axial force.
Chapter 9, Masonry
955
For the base of the shear walls:
Pumin = 197 kips plus factored weight of lower ½ of 1st story wall = 197 + (0.85)(6.4) = 202 kips
= 307 + (1.25)(6.4) = 315 kips Pumax
Mu = 2,640 ftkips
Try one #4 bars in each end cell and a #4 bar at 8 ft on center for the interior cells. A fPn  fMn curve,
representing the wall strength envelope, will be developed and used to evaluate Pu and Mu determined
above. Three cases will be analyzed and their results will be used in plotting the fPn  fMn curve.
In accordance with Provisions Sec. 11.6.2.1 [ACI 530, Sec. 3.2.2], the strength of the section is reached
as the compressive strains in masonry reach their maximum usable value of 0.0025 for CMU. The force
equilibrium in the section is attained by assuming an equivalent rectangular stress block of 0.8f!m over an
effective depth of 0.8c, where c is the distance of the neutral axis from the fibers of maximum
compressive strain. Stress in all steel bars is taken into account. The strains in the bars are proportional
to their distance from the neutral axis. For strains above yield, the stress is independent of strain and is
taken as equal to the specified yield strength Fy. See to Figure 9.25 for strains and stresses for all three
cases selected.
Case 1 (P = 0)
Assume all tension bars yield (which can be verified later):
Ts1 = (0.20 in.2)(60 ksi) = 12.0 kips
Ts2 = (0.20 in.2)(60 ksi) = 12.0 kips each
Because the neutral axis is close to the compression end of the wall, compression steel, Cs1, will be
neglected (it would make little difference anyway) for Case 1:
SFy = 0:
Cm = ST
Cm =(4)(12.0) = 48.0 kips
The compression block will be entirely within the first grouted cell:
Cm = 0.8 f’mab
48.0 = (0.8)(2.0 ksi)a(7.625 in)
a = 3.9 in. = 0.33 ft
c = a/0.8 = 0.33/0.8 = 0.41 ft
Thus, the neutral axis is determined to be 0.41 ft from the compression end on the wall, which is within
the first grouted cell:
SMcl = 0: (The math will be a little easier if moments are taken about the wall centerline.)
Mn = (16.330.33/2 ft)Cm + (16.00 ft) Ts1 + (0.00 ft)STs2 + (0.00 ft)Pn
Mn = (16.17)(48.0) + (16.00)(12) + 0 + 0 = 968 ftkips
fMn = (0.85)(968) = 823 ftkips
FEMA 451, NEHRP Recommended Provisions: Design Examples
956
y = 0.00207
Ts1
N.A.
= 0.00207
y
Balanced
Case
P = 0 Case
a = 14.15'
c = 17.69'
Center
Line
Ts1
STs2
Cm shell
Cm cells
a = 0.33'
c = 0.41'
m= 0.0025
0.8f'm
Cm tot
32.66'
16.33' 16.00' 0.33'
#4 at 8'0" o.c.
#4 at
end
m= 0.0025
0.8f'm
Cs1
Ts2 Ts2 Ts2
*Ts2
N.A.
*C s2
N.A.
y = 0.00207
Ts1
ST
m= 0.0025
0.8f'm
Cm cell
Ts2 Ts2
Intermediate
Case
Cm shell
y = 0.00207
Cs1
Ts 2 *
a = 6.40'
c = 8.00'
8.00'
M
P
(typical)
fE
y =
16.33' 16.00'
Figure 9.25 Strength of Birmingham 1 Wall D (1.0 ft = 0.3048 m). Strain diagram superimposed on strength
diagram for the three cases. The low force in the reinforcement is neglected in the calculations.
Chapter 9, Masonry
957
To summarize, Case 1:
fPn = 0 kips
fMn = 823 ftkips
Case 2 (Intermediate case between P = 0 and Pbal)
Let c = 8.00 ft.(this is an arbitrary selection). Thus, the neutral axis is defined at 8 ft from the
compression end of the wall:
a = 0.8c = (0.8)(8.00) = 6.40 ft
Cm shells = 0.8f’m(2 shells)(1.25 in. / shell)(6.40 ft. (12 in./ft) = 307.2 kips
Cm cells = 0.8 f’m(41 in.2) = 65.6 kips
Cm tot = Cm shells + Cm cells = 307.2 + 65.6 = 373 kips
Cs1 = (0.20 in.2)(60 ksi) = 12 kips (Compression steel is included in this case)
Ts1 = (0.20 in.2)(60 ksi) = 12 kips
Ts2 = (0.20 in.2)(60 ksi) = 12 kips each
Some authorities would not consider the compression resistance of reinforcing steel that is not enclosed
within ties. The Provisions clearly allows inclusion of compression in the reinforcement.
SFy = 0:
Cm tot + Cs1 = Pn + Ts1 + STs2
373 + 12 = Pn + (3)(12.0)
Pn = 349 kips
fPn = (0.85)(349) = 297 kips
SMcl = 0:
Mn = (13.13 ft)Cm shell + (16.00 ft)(Cm cell + Cs1) + (16.00 ft)Ts1 + (8.00 ft)Ts2
Mn = (13.13)(307.2) + (16.00)(65.6 + 12) + (16.00)(12.0) + (8.00 ft)(12.0) = 5,563 ftkips
fMn = (0.85)(5,563) = 4,729 ftkips
To summarize Case 2:
fPn = 297 kips
fMn = 4,729 ftkips
Case 3 (Balanced case)
In this case, Ts1 just reaches its yield stress:
0.0025 (32.33 ft) = 17.69 ft
(0.0025 0.00207)
c
. .
=.. + ..
a = 0.8c = (0.8)(17.69) = 14.15 ft
Cm shells = 0.8f’m(2 shells)(1.25 in. / shell)(14.15 ft.) (12 in./ft) = 679.2 kips
Cm cells =0.8f’m(2 cells)(41 in.2/cell) = 131.2 kips
Cm tot = Cm shells + Cm cells = 810.4 kips
Cs1 = (0.20 in.2)(60 ksi) = 12.0 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
958
Ts1 = (0.20 in.2)(60 ksi) = 12.0 kips
Cs2 and Ts2 are neglected because they are small, constituting less than 2 percent of the total Pn.
SFy = 0:
Pn = SC  ST
Pn = Cm tot + Cs1  Ts1 = 810.4 + 12.0  12.0 = 810.4 kips
fPn = (0.85)(810.4) = 689 kips
SMcl = 0:
Mn = 9.26 Cm shells + ((16 + 8)/2) Cm cells + 16 Cs1 + 8 Ts2 + 16 Ts1
Mn = (9.26)(679.2) + (12.0)(131.2) + (16.00)(12.0) + (ignore small Ts2) + (16.0)(12.0) = 8,248 kips
fMn = (0.85)(8,248) = 7,011 ftkips
To summarize Case 3:
fPn = 689 kips
fMn = 7,011 ftkips
Using the results from the three cases above, the fPn  fMn curve shown in Figure 9.26 is plotted.
Although the portion of the fPn  fMn curve above the balanced failure point could be determined, it is
not necessary here. Thus, only the portion of the curve below the balance point will be examined. This is
the region of high moment capacity.
Similar to reinforced concrete beamcolumns, inplane compression failure of the cantilevered shear wall
will occur if Pu > Pbal, and tension failure will occur if Pu < Pbal. A ductile failure mode is essential to the
design, so the portion of the curve above the “balance point” is not useable.
As can be seen, the points for Pu min , Mu and Pu max , are within the fPn  fMn envelope; thus, the strength
design is acceptable with the minimum reinforcement. Figure 9.26 shows two schemes for determining
the design flexural resistance for a given axial load. One interpolates along the straight line between pure
bending and the balanced load. The second makes use of intermediate points for interpolation. This
particular example illustrates that there can be a significant difference in the interpolated moment capacity
between the two schemes for axial loads midway between the balanced load and pure bending.
For the purpose of shear design, the value of fMN at the design axial load is necessary. Interpolating
between the intermediate point and the P = 0 point for P = 202 kips yields fMN = 3,480 ftkip. Thus, the
factor on shear to represent development of 125 percent of flexural capacity is:
1.25 / 1.25 3480/ 0.85 1.94
2640
N
U
M
M
f f
= =
Chapter 9, Masonry
959
9.2.4.5.2.2 Ductility check
Provisions Sec.11.6.2.2 [ACI 530, Sec. 3.2.3.5] requires that the critical strain condition correspond to a
strain in the extreme tension reinforcement equal to 5 times the strain associated with Fy. Note that this
calculation uses unfactored gravity axial loads (Provisions Sec.11.6.2.2 [ACI 530, Sec. 3.2.3.5]). Refer to
Figure 9.25 and the following calculations which illustrate this using loads at the bottom story (highest
axial loads). Calculations for other stories are not presented in this example.
5,000 ftkips 10,000 ftkips
500 kips
1,000 kips
P max =
315 kips
P min =
202 kips
fP
fM
M u = 2640 ftkips
Intermediate
(4729 ftkips, 297 kips)
P = 0
(823 ftkips, 0 kips)
Actual fP  fM curve
Balance
(7011 ftkips, 689 kips)
Simplified fP  fM curve
n
n
fPn
fMn
u
u
n
n
n
n
Figure 9.26 fP11  fM11 diagram for Birmingham 1 Wall D (1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm).
FEMA 451, NEHRP Recommended Provisions: Design Examples
960
m = 0.0025
N.A.
s = 5 y
= 0.0103
6.32'
P
Cm
0.8f 'm = 1.6 ksi
Cs1
5.99' 6.51'
Ts3 Ts2 Ts1
Ts4
(5) reinforced cells at 8'0" o.c.
1.25f y
= 75 ksi
5f y = 300 ksi
26.02'
32'8"
3.79'
0.33'
5.06' 1.26'
2.02'
10.02'
18.02'
26.02'
0.33'
f y = 60 ksi
72.86 ksi
23.29 ksi
= 5 ( 60
29,000 )
c
a
Figure 9.27 Ductility check for Birmingham 1 Wall D (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa).
For Level 1 (bottom story), the unfactored axial loads are:
P = 232 kips + weight of half of first story wall = 232 + 6.4 = 238.4 kips
Refer to Figure 9.27:
Cm = 0.8 f’m(ab + Acell) = (1.6 ksi)[(5.06 ft. x 12 in./ft.)(2.5 in.) + 41 in.2] = 308.5 kips (same as above)
Cs1 = FyAs = (60 ksi)(0.20 in.2) = 12.0 kips
Ts1 = Ts2 = Ts3 = (1.25 × 60 ksi)(0.20 in.2) = 15 kips
Chapter 9, Masonry
961
Ts4 = (23.29 ksi)(0.20 in.2) = 4.6 kips
3 C > 3P + T
Cm + Cs1 > P + Ts1 + Ts2 + Ts3 + Ts4
308.5 + 12.0 > 238.4 + 15 + 15 + 15 + 4.6
320.5 kips > 288 kips OK
There is more compression capacity than required so ductile failure condition governs.
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing
the ductility check (factored loads).]
9.2.4.6 Birmingham 1 Deflections
The calculations for deflection involve many variables and assumptions, and it must be recognized that
any calculation of deflection is approximate at best.
Deflections are to be calculated and compared with the prescribed limits set forth by Provisions Table
5.2.8. Deformation requirements for masonry structures are given in Provisions Sec. 11.5.4 [Table 4.51].
The following procedure will be used for calculating deflections:
1. For each story, compare Mx (from Table 9.23) to Mcr = S(fr + Pu min / A) to determine if wall will
crack.
2. If Mcr < Mx, then use cracked moment of inertia and Provisions Eq. 11.5.4.3.
3. If Mcr > Mx, then use In = Ig for moment of inertia of wall.
4. Compute deflection for each level.
Other approximations can be used such as the cubic interpolation formula given in Provisions 11.5.4.3,
but that equation was derived for reinforced concrete members acting as single span beams, not
cantilevers. In the authors’ opinion, all these approximations pale in comparison to the approximation of
nonlinear deformation using Cd.
For the Birmingham 1 building:
be = effective masonry wall width
be = [(2 × 1.25 in.)(32.67 ft × 12) + (5 cells)(41 in.2/cell)]/(32.67 ft × 12) = 3.02 in.
S = be l2/6 = (3.02)(32.67 × 12)2/6 = 77,434 in.3
fr = 0.250 ksi
A = be l = (3.02 in.)(32.67 ft × 12) = 1,185 in.2
Pu is calculated using 1.00D (see Table 9.24). 1.00D is considered to be a reasonable value for axial load
for this admittedly approximate analysis. If greater conservatism is desired, Pu could be calculated using
0.85D.
FEMA 451, NEHRP Recommended Provisions: Design Examples
962
V5
V4
V3
V2
V1
d 5
d 4
d 3
d 2
d1
l
be
Figure 9.28 Shear wall deflections.
The results are shown in Table 9.25.
Table 9.25 Birmingham 1 Cracked Wall Determination
Level Pumin
(kips)
Mcr
(ftkips)
Mu
(ftkips)
Status
54321
41
89
137
185
232
1836
2098
2359
2621
2877
225
652
1230
1910
2640
uncracked
uncracked
uncracked
uncracked
uncracked
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
For uncracked walls:
In = Ig = bl3/12 = (3.02 in.)(32.67 × 12)3 /12 = 1.52 x 107 in.4
The calculation of d will consider flexural and shear deflections. For the final determination of
deflection, a RISA2D analysis was made. The result is summarized Table 9.26 below. Figure 9.28
illustrates the deflected shape of the wall.
Chapter 9, Masonry
963
Table 9.26 Deflections, Birmingham 1
Level F
(kips)
Ieff
(in.4)
dflexural
(in.)
dshear
(in.)
dtotal
(in.)
Cd dtotal
(in.)
.
(in.)
54321 26.0
23.2
17.4
11.7
5.8
1.52 × 107
1.52 × 107
1.52 × 107
1.52 × 107
1.52 × 107
0.108
0.078
0.049
0.024
0.007
0.054
0.049
0.041
0.028
0.015
0.162
0.128
0.090
0.052
0.021
0.284
0.223
0.157
0.091
0.037
0.061
0.066
0.066
0.054
0.037
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm.
The maximum story drift occurs at Level 4 (Provisions Table 5.2.8 [Table 4.51]):
[The specific procedures for computing deflection of shear walls have been removed from the 2003
Provisions. ACI 530 does not contain the corresponding provisions in the text, however, the commentary
contains a discussion and equations that are similar to the procedures in the 2000 Provisions. However,
as indicated previously, there is a potential conflict between the drift limits in 2003 Provisions Table 4.5
1 and ACI 530 Sec. 1.13.3.2.]
. = 0.066 in. < 1.04 in. = 0.01hn OK
9.2.4.7 Birmingham 1 OutofPlane Forces
Provisions Sec 5.2.6.2.7 [Sec. 46.1.3] requires that the bearing walls be designed for outofplane loads
determined as follows:
w = 0.40SDSWc $ 0.1Wc
w = (0.40)(0.24)(45 psf) = 4.3 psf < 4.5 psf = 0.1Wc
The calculated seismic load, w = 4.5 psf, is much less than wind pressure for exterior walls and is also
less than the 5 psf required by IBC Sec. 1607.13 for interior walls. Thus, seismic loads do not govern the
design of any of the walls for loading in the outofplane direction.
9.2.4.8 Birmingham 1 Orthogonal Effects
Orthogonal effects do not have to be considered for Seismic Design Category B (Provisions Sec. 5.2.5.2.1
[Sec. 4.4.2.1]).
This completes the design of Transverse Wall D.
9.2.4.9 Summary of Design for Birmingham 1 Wall D
8 in. CMU
f!m = 2,000 psi
Reinforcement:
One vertical #4 bar at wall end cells
Vertical #4 bars at 8 ft on center at intermediate cells throughout
Bond beam with two  #4 bars at each story just below the floor and roof slabs
FEMA 451, NEHRP Recommended Provisions: Design Examples
964
Horizontal joint reinforcement at 16 inches
Grout at cells with reinforcement and at bond beams.
9.2.5 Seismic Design for New York City
This example focuses on differences from the design for the Birmingham 1 site.
9.2.5.1 New York City Weights
As before, use 67 psf for 8in.thick normal weight hollow core plank plus the nonmasonry partitions.
This site is assigned to Seismic Design Category C, and the walls will be designed as intermediate
reinforced masonry shear walls (Provisions Sec. 11.11.4 [Sec. 11.2.1.4] and Sec. 11.3.7 [Sec. 11.2.1.4]),
which requires prescriptive seismic reinforcement (Provisions Sec. 11.3.7.3 [ACI 530, Sec. 1.13.2.2.4]).
Intermediate reinforced masonry shear walls have a minimum of #4 bars at 4 ft on center. For this
example, 48 psf will be assumed for the 8in. CMU walls. The 48 psf value includes grouted cells and
bond beams in the course just below the floor planks. In Seismic Design Category C, more of the
regularity requirement must be checked. It will be shown that this symmetric building with a seemingly
well distributed lateral force system is torsionally irregular by the Provisions.
Story weight, wi:
Roof
Roof slab (plus roofing) = (67 psf) (152 ft)(72 ft) = 733 kips
Walls = (48 psf)(589 ft)(8.67 ft/2) + (48 psf)(4)(36 ft)(2 ft) = 136 kips
Total = 869 kips
There is a 2ft high masonry parapet on four walls and the total length of masonry wall is 589 ft.
Typical floor
Slab (plus partitions) = 733 kips
Walls = (48 psf)(589 ft)(8.67 ft) = 245 kips
Total = 978 kips
Total effective seismic weight, W = 869 + (4)(978) = 4781 kips
This total excludes the lower half of the first story walls, which do not contribute to seismic loads that are
imposed on CMU shear walls.
9.2.5.2 New York City Base Shear Calculation
The seismic response coefficient, Cs, is computed from Provisions Sec. 5.4.1.1 [Sec. 5.21.1]:
0.39 0.156
/ 2.51
DS
s
C S
R I = = =
The value of Cs need not be greater than:
Chapter 9, Masonry
965
( )
0.14 0.166
( / ) 0.338 2.5 1
D1
s
C S
T R I = = =
where T is the same as found in Sec. 9.2.4.2.
The value for Cs is taken as 0.156 (the lesser of the two computed values). This value is still larger than
the minimum specified in Provisions Eq. 5.3.2.13. Using Provisions Eq. 5.4.1.13:
Cs = 0.044SD1I = (0.044)(0.14)(1) = 0.00616
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value
for longperiod structures, which is not applicable to this example.]
The total seismic base shear is then calculated using Provisions Eq. 5.4.1 [Eq. 5.21]:
V = CsW = (0.156)(4,781) = 746 kips
9.2.5.3 New York City Vertical Distribution of Seismic Forces
The vertical distribution of seismic forces is determined in accordance with Provisions Sec. 5.4.4 [Sec.
5.2.3], which was described in Sec. 9.2.4.3. Note that for Provisions Eq. 5.4.32 [Eq. 5.211], k = 1.0
since T = 0.338 sec (similar to the Birmingham 1 building).
The application of the Provisions equations for this building is shown in Table 9.27:
Table 9.27 New York City Seismic Forces and Moments by Level
Level
(x)
wx
(kips)
hx
(ft)
wxhx
k
(ftkips)
Cvx Fx
(kips)
VX
(kips)
Mx
(ftkips)
543213
869
978
978
978
978
4,781
43.34
34.67
26.00
17.33
8.67
37,657
33,904
25,428
16,949
8,476
122,414
0.3076
0.2770
0.2077
0.1385
0.0692
1.000
229
207
155
103
52
746
2.3e+14 1,985
5,765
10,889
16,907
23,370
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ftkip = 1.36 kNm.
9.2.5.4 New York City Horizontal Distribution of Forces
The initial distribution is the same as Birmingham 1. See Sec. 9.2.4.4 and Figure 9.23 for wall
designations.
Total shear in Wall Type D:
0.125 0.0238 0.149 totV= V+ V= V
Provisions Sec.5.4.4.3 [Sec. 4.3.2.2] requires a check of torsional irregularity using the ratio of maximum
displacement at the end of the structure, including accidental torsion, to the average displacement of the
two ends of the building. For this simple and symmetric structure, the actual displacements do not have
to be computed to find the ratio. Relying on symmetry and the assumption of rigid diaphragm behavior
FEMA 451, NEHRP Recommended Provisions: Design Examples
966
used to distribute the forces, the ratio of the maximum displacement of Wall D to the average
displacement of the floor will be the same as the ratio of the wall shears with and without accidental
torsion:
0.149 1.190
0.125
max
ave
F V
F V
= =
This can be extrapolated to the end of the rigid diaphragm therefore:
1 0.190 152 / 2 1.402
36
max
ave
d
d
= + .. ..=
. .
Provisions Table 5.2.3.2 [Table 4.32] defines a building as having a “Torsional Irregularity” if this ratio
exceeds 1.2 and as having an “Extreme Torsional Irregularity” if this ratio exceeds 1.4. Thus, an
important result of the Seismic Design Category C classification is that the total torsion must be amplified
by the factor:
2 2 1.402 1.365
1.2 1.2
max
x
ave
A
d
d
=.. .. =.. .. =
. . . .
Therefore, the portion of the base shear for design of Wall D is now:
0.125 1.365(0.0238 ) 0.158 DV= V+ V= V
which is a 5.8 percent increase from the fraction before considering torsional irregularity.
The total story shear and overturning moment may now be distributed to Wall D and the wall proportions
checked. The wall capacity will be checked before considering deflections.
9.2.5.5 New York City Transverse Wall D
The strength or limit state design concept is used in the Provisions.
9.2.5.5.1 New York City Shear Strength
Similar to the design for Birmingham 1, the shear wall design is governed by:
Vu=fVn
Vn = Vm + Vs
V fAdepending on M/Vd n mn max =4 to 6 '
m= 41.75( )nm0.25
M
V Af P
Vd
' + . .
.. ..
0.5 v
s y v
V Afd
s
= .. ..
. .
where
Chapter 9, Masonry
967
An = (2 × 1.25 in. × 32.67 ft × 12 in.) + (41 in.2 × 9 cells) = 1,349 in.2
The shear strength of each Wall D, based on the aforementioned formulas and the strength reduction
factor of f = 0.8 for shear from Provisions Table 11.5.3 [ACI 530, Sec. 3.1.4.3], is summarized in Table
9.28. Note that Vx and Mx in this table are values from Table 9.27 multiplied by 0.158 (representing the
portion of direct and indirect shear assigned to Wall D), and P is the dead load of the roof or floor times
the tributary area for Wall D.
FEMA 451, NEHRP Recommended Provisions: Design Examples
968
Table 9.28 New York City Shear Strength Calculation for Wall D
Story Vx
(kips)
Mx
(ftkips)
Mx/Vxd 2.5 Vx
(kips)
P
(kips)
fVm
(kips)
fVs
(kips)
fVn
(kips)
fVn max
(kips)
5 36.1 313 0.265 90.3 42 179.0 54.6 233.6 287.6
4 68.7 908 0.405 171.8 90 176.9 54.6 231.5 269.7
3 93.1 1715 0.564 232.8 139 173.2 54.6 227.8 249.2
2 109.3 2663 0.746 273.3 188 167.7 54.6 222.3 225.8
1 117.5 3680 0.959 293.8 236 159.3 54.6 213.9 198.4
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
Vu exceeds fVn at the lower three stories. As will be shown at the conclusion of the design for flexure, the
factor to achieve 125 percent of the nominal flexural capacity is 1.58. This results in Vu being less than
fVn at all stories. If that were not the case, it would be necessary to grout more cells to increase An or to
increase f’m.
9.2.5.5.2 New York City Axial and Flexural Strength
The walls in this example are all loadbearing shear walls because they support vertical loads as well as
lateral forces. Inplane calculations include:
1. Strength check and
2. Ductility check.
9.2.5.5.2.1 Strength check
Wall demands, using load combinations determined previously, are presented in Table 9.29 for Wall D.
In the table, Load Combination 1 is 1.28D + QE + 0.5L and Load Combination 2 is 0.82D + QE.
Table 9.29 Demands for New York City Wall D
Load Combination 1 Load Combination 2
Level PD
(kips)
PL
(kips)
Pu
(kips)
Mu
(ftkips)
Pu
(kips)
Mu
(ftkips)
54321
42
90
139
188
236
08
17
25
34
54
119
186
253
319
313
908
1715
2663
3680
34
74
114
154
194
313
908
1715
2663
3680
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
As in Sec. 9.2.4.5.2, strength at the bottom story (where P, V, and M are the greatest) will be examined.
The strength design will consider Load Combination 2 from Table 9.29 to be the governing case because
it has the same lateral load as Load Combination 1 but with lower values of axial force. Refer to Fig. 9.2
9 for notation and dimensions.
Chapter 9, Masonry
969
y = 0.00207
Ts1
N.A.
y = 0.00207
Balanced
Case
P = 0 Case
a = 14.15'
c = 17.69'
Center
Line
Ts1
S Ts2
Cm shell
Cm cells
a = 0.66'
c = 0.82'
Cs1
y = 0.00207
m= 0.0025
0.8f'm
Cm tot
32.67'
16.33' 16.00' 0.33'
#4 at 4.00'
on center
(1) #4
y = 0.00207
m= 0.0025
0.8f'm
Cs1
M
P
N.A.
Figure 9.29 Strength of New York City and Birmingham 2 Wall D. Strength diagrams are superimposed over the
strain diagrams for the two cases (intermediate case is not shown) (1.0 ft = 0.3048 m).
Examine the strength of Wall D at Level 1:
= 0.82 D = 0.82 (236 + factored weight of lower half of first story wall) Pumin
= 0.82(236 + 6.4) = 199 kips
= 1.28 D + 0.5 L 319 = 1.28(236 + 6.4) + 0.5(34) = 327 kips Pumax
Mu = 3,680 ftkips
Because intermediate reinforced masonry shear walls are used (Seismic Design Category C), vertical
reinforcement at is required at 4 ft on center in accordance with Provisions Sec. 11.3.7.3 [ACI 530, Sec.
1.13.2.2.4]. Therefore, try one #4 bar in each end cell and #4 bars at 4 ft on center at all intermediate
cells.
FEMA 451, NEHRP Recommended Provisions: Design Examples
970
The calculation procedure is similar to that for the Birmingham 1 building presented in Sec. 9.2.4.5.2.
The results of the calculations (not shown) for the New York building are summarized below.
P = 0 case
fPn = 0
fMn = 1,475 ftkips
Intermediate case
c = 8.0 ft
fPn = 330 kips
fMn = 5,600 ftkips
Balanced case
fPn = 807
fMn = 8,214 ftkips
With the intermediate case, it is simple to use the three points to make two straight lines on the interaction
diagram. Use the simplified fPn  fMn curve shown in Figure 9.210. The straight line from pure
bending to the balanced point is conservative and can easily be used where the design is not as close to
the criterion. It is the nature of lightly reinforced and lightly loaded masonry walls that the intermediate
point is frequently useful.
Use one #4 bar in each end cell and one #4 bar at 4 ft on center throughout the remainder of the wall.
As shown in the design for Birmingham 1,for the purpose of shear design, the value of fMN at the design
axial load is necessary. Interpolating between the intermediate point and the P = 0 point for P = 199 kips
yields fMN = 3,960 ftkip. Thus, the factor on shear to represent development of 125 percent of flexural
capacity is:
125 125
3960 085
3680
. 158
/
.
/ .
.
fM f
M
N
U
= =
Chapter 9, Masonry
971
Figure 9.210 fP11  fM11 Diagram for New York City and Birmingham 2 Wall D (1.0 kip = 4.45 kN, 1.0 ftkip =
1.36 kNm).
9.2.5.5.2.2 Ductility check
Refer to Sec. 9.2.4.5.2, Item 2, for explanation [see Sec. 9.2.4.5.2 for discussion of revisions to the
ductility requirements in the 2003 Provisions.]. For Level 1 (bottom story), the unfactored loads are:
P = 236 + weight of lower ½ of first story wall = 236 + 6.4 = 242.4 kips
M = 3,483 ftkips
Cm = 0.8 fm' [(a)(b) + Acells]
where b = face shells = (2 × 1.25 in.) and Acell = 41 in.2
5,000 ftkips 10,000 ftkips
500 kips
1,000 kips
P max = 342 kips
P min = 206 kips
fP
fM
Birm #2, M = 3283 ftkips
P = 0
(1475 ftkips, 0 kips)
Balance
(8214 ftkips, 807 kips)
P max =
327 kips
P min =
199 kips
NYC, M = 3680 ftkips
3 point fP  fM curve
Intermediate
(5600 ftkips, 330 kips)
Simplified fP  fM curve
n
n
fMn
fPn
u
u
u
u
u
u
n
n
n
n
FEMA 451, NEHRP Recommended Provisions: Design Examples
972
Cm = (1.6 ksi)[(5.03 ft × 12)(2.5 in.) + (2)(41)] = 372.6 kips
Cs1 = FyAs = (60 ksi)(0.20 in.2) = 12 kips
Cs2 = (22.6 ksi)(0.20 in.2) = 4.5 kips
Ts1 = Ts2 = Ts3 = Ts4 = Ts5 = (75 ksi)(0.20 in.2 ) = 15 kips
Ts6 = (69.6 ksi)(0.20 in.2) = 13.9 kips
Ts7= (23.5 ksi)(0.20 sq. in.) = 4.7 kips
3 C > 3 P + T
Cm + Cs1 + Cs2 > P + Ts1 + Ts2 + Ts3 + Ts4 + Ts5 + Ts6 + Ts7
372.6 + 12.0 + 4.5 > 242.5 + 5(15) + 13.9 + 4.7
389 kips > 336 kips OK
Chapter 9, Masonry
973
m = 0.0025
N.A.
s = 5 Ey
= 0.0103
6.29'
P
Cm
0.8f m' = 1.6 ksi
Cs1
5.96' 6.51'
Ts5 Ts3 Ts1
Ts7
(9) #4 at 4'0" o.c.
1.25f y
= 75 ksi
5f y = 300 ksi
26.05'
32'8"
3.78'
0.33'
5.03' 1.26'
26.05'
0.33'
f y = 60 ksi
72.5 ksi
23.5 ksi
= 5 ( 60
29,000 )
c
Ts6 Ts4 Ts2
Cs2
22.6 ksi
69.6 ksi
Figure 9.211 Ductility check for New York City and Birmingham 2 Wall D (1.0 ft = 0.3048 m, 1.0 ksi
= 6.89 MPa).
9.2.5.6 New York Deflections
Refer to 9.2.4.6 for more explanation [see Sec. 9.2.4.6 for discussion of revisions to the deflection
computations and requirements in the 2003 Provisions, as well as the potentially conflicting drift limits].
For the New York City building, the determination of whether the walls will be cracked is:
be = effective masonry wall width
be = [(2 × 1.25 in.)(32.67 ft × 12) + (9 cells)(41 in.2/cell)]/(32.67 ft × 12) = 3.44 in.
A = be l = (3.44 in.)(32.67 × 12) = 1,349 in.2
FEMA 451, NEHRP Recommended Provisions: Design Examples
974
S = be l2/6 = (3.44)(32.67 × 12)2 /6 = 88,100 in.3
fr = 0.250 ksi
Pu is calculated using 1.00D (see Table 9.28 for values and refer to Sec. 9.2.4.6 for discussion). Table
9.210 a summarizes of these calculations.
Table 9.210 New York City Cracked Wall Determination
Level Pu
(kips)
Mcr
(ftkips)
Mx
(ftkips)
Status
54321
42
90
139
188
236
2064
2325
2592
2860
3120
313
908
1715
2663
3680
uncracked
uncracked
uncracked
uncracked
cracked
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
For the uncracked walls:
In = Ig = bl3/12 = (3.44 in.)(32.67 × 12)3/12 = 1.73 × 107 in.4
For the cracked wall, observe that the intermediate point on the interaction diagram is relatively close to
the design point. Therefore, as a different type of approximation, compute a cracked moment of inertia
using the depth to the neutral axis of 8.0 ft:
Icr = bec3/3 + 3nAsd2
Icr = (3.44 in.)(8.0 ft × 12)3/3 + 19.3(0.2)(4.332 + 8.332 + 12.332 + 16.332 + 20.332 + 24.332)144 =
= 1.01 x 106 + 0.84 x 106 = 1.85 x 106 in.4
Per Provisions Eq. 11.5.4.3:
3 3
cr 1 cr
eff n cr n
a a
I IM I M I
M M
. . . . ..
= . .+ .. ..=
. . .. . ...
Ieff = 1.13 x 107 in.4
Provisions 11.5.4.3 would imply that Ieff would be used for the full height. Another reasonable option is
to use Icr at the first story and Ig above that. The calculation of d should consider shear deflections in
addition to the flexural deflections. For this example Ieff will be used over the full height for the final
determination of deflection (a RISA 2D analysis was made). The result is summarized in Table 9.211.
Chapter 9, Masonry
975
Table 9.211 New York City Deflections
Level F
(kips)
Ieff
(in.4)
dflexural
(in.)
dshear
(in.)
dtotal
(in.)
Cd dtotal
(in.)
.
(in.)
54321 34.1
30.9
23.1
15.3
7.7
1.13 × 107
1.13 × 107
1.13 × 107
1.13 × 107
1.13 × 106
0.256
0.189
0.124
0.065
0.020
0.080
0.075
0.064
0.050
0.033
0.336
0.264
0.188
0.115
0.053
0.757
0.593
0.422
0.259
0.118
0.163
0.171
0.163
0.141
0.118
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm
The maximum story drift occurs at Level 4:
.4 = 0.171 in. < 1.04 in. = 0.01 h4 (Provisions Table 5.2.8 [Table 4.51]) OK
The total displacement at the top of the wall is
. = 0.757 in. < 5.2 in. = 0.01 hn (Provisions 11.5.4.1.1) OK
9.2.5.7 New York City OutofPlane Forces
Provisions Sec 5.2.6.2.7 [Sec. 4.4.2.2] requires that the bearing walls be designed for outofplane loads
determined as
w = 0.40 SDS Wc $ 0.1Wc
With SDS = 0.39 , w = 0.156Wc > 0.1Wc , so w = (0.156)(48 psf) = 7.5 psf, which is much less than wind
pressure for exterior walls. Even though Wall D is not an exterior wall, the lateral pressure is sufficiently
low that it is considered acceptable by inspection, without further calculation. Seismic loads do not
govern the design of Wall D for loading in the outofplane direction.
9.2.5.8 New York City Orthogonal Effects
According to Provisions Sec. 5.2.5.2.2, orthogonal interaction effects have to be considered for Seismic
Design Category C when the equivalent lateral force (ELF) procedure is used (as it is here). However, the
outofplane component of only 30 percent of 7.5 psf on the wall will not produce a significant effect
when combined with the inplane direction of loads, so no further calculation will be made.
This completes the design of the transverse Wall D for the New York building.
9.2.5.9 Summary of New York City Wall D Design
8 in. CMU
f!m = 2,000 psi
FEMA 451, NEHRP Recommended Provisions: Design Examples
976
Reinforcement:
Vertical #4 bars at 4 ft on center throughout the wall
Bond beam with two #4 at each story just below the floor or roof slabs
Horizontal joint reinforcement at alternate courses
9.2.6 Birmingham 2 Seismic Design
The emphasis here is on differences from the previous two locations for the same building. Per
Provisions Table 5.2.5.1 [Table 4.41], the torsional irregularity requires that the design of a Seismic
Design Category D building be based on a dynamic analysis. Although not explicitly stated, the
implication is that the analytical model should be threedimensional in order to capture the torsional
response. This example will compare both the equivalent lateral force procedure and the modal response
spectrum analysis procedure and will demonstrate that, as long as the torsional effects are accounted for,
the static analysis could be considered adequate for design.
9.2.6.1 Birmingham 2 Weights
The floor weight for this examples will use the same 67 psf for 8in.thick, normal weight hollow core
plank plus roofing and the nonmasonry partitions as used in the prior examples (see Sec. 9.2.1). This
site is assigned to Seismic Design Category D, and the walls will be designed as special reinforced
masonry shear walls (Provisions Sec. 11.11.5 and Sec. 11.3.8[ACI 530, Sec. 1.13.2.2.5), which requires
prescriptive seismic reinforcement (Provisions Sec. 11.3.7.3). Special reinforced masonry shear walls
have a maximum spacing of rebar at 4 ft on center both horizontally and vertically. Also, the total area of
horizontal and vertical reinforcement must exceed 0.0020 times the gross area of the wall, and neither
direction may have a ratio of less than 0.0007. The vertical #4 bars at 48 in. used for the New York City
design yields a ratio of 0.00055, so it must be increased. Two viable options are #5 bars at 48 in.
(yielding 0.00085) and #4 bars at alternating spaces of 32 in. and 40 in. (12 bars in the wall), which yields
0.0080. The latter is chosen in order to avoid unnecessarily increasing the shear demand. Therefore, the
horizontal reinforcement must be (0.0020  0.0008)(7.625 in.)(12 in./ft.) = 0.11 in.2/ft. or 0.95 in.2 per
story. Two #5 bars in bond beams at 48 in. on center will be adequate. For this example, 56 psf weight
for the 8in.thick CMU walls will be assumed. The 56 psf value includes grouted cells and bond beams.
Story weight, wi:
Roof:
Roof slab (plus roofing) = (67 psf) (152 ft)(72 ft) = 733 kips
Walls = (56 psf)(589 ft)(8.67 ft/2) + (56 psf)(4)(36 ft)(2 ft) = 159 kips
Total = 892 kips
There is a 2fthigh masonry parapet on four walls and the total length of masonry wall is 589 ft.
Typical floor:
Slab (plus partitions) = 733 kips
Walls = (56 psf)(589 ft)(8.67 ft) = 286 kips
Total = 1,019 kips
Total effective seismic weight, W = 892 + (4)(1,019) = 4,968 kips
Chapter 9, Masonry
977
This total excludes the lower half of the first story walls which do not contribute to seismic loads that are
imposed on CMU shear walls.
9.2.6.2 Birmingham 2 Base Shear Calculation
The ELF analysis proceeds as described for the other locations. The seismic response coefficient, Cs, is
computed using Provisions Eq. 5.4.1.11 [Eq. 5.22] and 5.4.1.12 [Eq.5.23]:
(Controls)
0.47 0.134
/ 3.51
DS
s
C S
R I = = =
( )
0.28 0.237
( / ) 0.338 3.51
D1
s
C S
T R I = = =
This is somewhat less than the 746 kips computed for the New York City design due to the larger R
factor.
The fundamental period of the building, based on Provisions Eq. 5.4.2.11 [Eq.5.26], is 0.338 sec as
computed previously (the approximate period, based on building system and building height, will be the
same for all locations). The value for Cs is taken as 0.134 (the lesser of the two values). This value is still
larger than the minimum specified in Provisions Eq. 5.3.2.13 which is:
Cs = 0.044SD1I = (0.044)(0.28)(1) = 0.012
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value for
longperiod structures, which is not applicable to this example.]
The total seismic base shear is then calculated using Provisions Eq. 5.4.1 [Eq.5.21] as:
V = CsW = (0.134)(4,968) = 666 kips
A threedimensional (3D) model was created in SAP 2000 for the modal response spectrum analysis. The
masonry walls were modeled as shell bending elements and the floors were modeled as an assembly of
beams and shell membrane elements. The beams have very little mass and a large flexural moment of
inertia to avoid consideration of models of vertical vibration of the floors. The flexural stiffness of the
beams was released at the bearing walls in order to avoid a wall slab frame that would inadvertently
increase the torsional resistance. The mass of the floors was captured by the shell membrane elements.
Table 9.212 shows data on the modes of vibration used in the analysis.
Provisions Sec. 4.1.2.6 [Sec. 3.3.4] was used to create the response spectrum for the modal analysis. The
key points that define the spectrum are:
TS = SD1/SDS = 0.28/0.47 = 0.60 sec
T0 = 0.2 TS = 0.12 sec
at T = 0, Sa = 0.4 SDS/R = 0.0537 g
from T = T0 to TS, Sa = SDS/R = 0.1343 g
for T > TS, Sa = SD1/(RT) = 0.080/T
The computed fundamental period is less than the approximate period. The transverse direction base
shear from the SRSS combination of the modes is 457.6 kips, which is considerably less than that
obtained using the ELF method.
FEMA 451, NEHRP Recommended Provisions: Design Examples
978
Provisions Sec. 5.5.7 [Sec. 5.3.7] requires that the modal base shear be compared with the ELF base shear
computed using a period somewhat larger than the approximate fundamental period (CuTa). Per Sec.
9.2.4.2, Ta = 0.338 sec. and per Provisions Table 5.4.2 [Table 5.21] Cu = 1.4. Thus, CuTa = 0.48 sec.,
which is less that SD1/SDS. Therefore, the ELF base shear for comparison is 666 kips as just computed.
Because 85 percent of 666 kips = 566 kips, Provisions Sec. 5.5.7 [Sec. 5.3.7] dictates that all the results
of the modal analysis be factored by:
0.85 566 1.24
458
ELF
Modal
V
V
= =
Both analyses will be carried forward as discussed in the subsequent sections.
Table 9.212 Birmington 2 Periods, Mass Participation Factors, and Modal Base Shears in the
Transverse Direction for Modes Used in Analysis
Mode Period, Individual mode (percent) Cumulative sum (percent) Trans.
number (seconds) Long. Trans. Vert. Long. Trans. Vert. base shear
1 0.2467 0.00 0.00 0.00 0.00 0.00 0.00 0.0
2 0.1919 0.00 70.18 0.00 0.00 70.18 0.00 451.1
3 0.1915 70.55 0.00 0.00 70.55 70.18 0.00 0.0
4 0.0579 0.00 18.20 0.00 70.55 88.39 0.00 73.9
5 0.0574 17.86 0.00 0.00 88.41 88.39 0.00 0.0
6 0.0535 0.00 4.09 0.00 88.41 92.48 0.00 16.1
7 0.0532 4.17 0.00 0.00 92.58 92.48 0.00 0.0
8 0.0413 0.00 0.01 0.00 92.58 92.48 0.00 0.0
9 0.0332 1.50 0.24 0.00 94.08 92.72 0.00 0.8
10 0.0329 0.30 2.07 0.00 94.38 94.79 0.00 7.1
11 0.0310 1.28 0.22 0.00 95.66 95.01 0.00 0.8
12 0.0295 0.22 1.13 0.00 95.89 96.14 0.00 3.8
13 0.0253 1.97 0.53 0.00 97.86 96.67 0.00 1.7
14 0.0244 0.53 1.85 0.00 98.39 98.52 0.00 5.9
15 0.0190 1.05 0.36 0.00 99.44 98.89 0.00 1.1
16 0.0179 0.33 0.94 0.00 99.77 99.82 0.00 2.8
17 0.0128 0.19 0.07 0.00 99.95 99.90 0.00 0.2
18 0.0105 0.03 0.10 0.00 99.99 99.99 0.00 0.3
1 kip = 4.45 kN.
9.2.6.3 Birmingham 2 Vertical Distribution of Seismic Forces
The dynamic analysis will be revisited for the horizontal distribution of forces in the next section but as
demonstrated there, the ELF procedure is considered adequate to account for the torsional behavior in this
example. The dynamic analysis can certainly be used to deduce the vertical distribution of forces. This
analysis was constructed to study amplification of accidental torsion. It would be necessary to integrate
the shell forces to find specific story forces, and it is not necessary to complete the design. Therefore, the
vertical distribution of seismic forces for the ELF analysis is determined in accordance with Provisions
Sec. 5.4.4 [Sec. 5.2.3], which was described in Sec. 9.2.4.3. For Provisions Eq. 5.4.32 [Sec. 5.211], k =
1.0 since T = 0.338 sec (similar to the Birmingham 1 and New York City buildings). It should be noted
that the response spectrum analysis may result in moments that are less than those calculated using the
ELF method; however, because of its relative simplicity, the ELF is used in this example.
Application of the Provisions equations for this building is shown in Table 9.213:
Chapter 9, Masonry
979
Table 9.213 Birmingham 2 Seismic Forces and Moments by Level
Level
(x)
wx
(kips)
hx
(ft)
wxhx
(ftkips)
Cvx Fx
(kips)
Vx
(kips)
Mx
(ftkips)
543213
892
1,019
1,019
1,019
1,019
4,968
43.34
34.67
26.00
17.33
8.67
38,659
35,329
26,494
17,659
8,835
126,976
0.3045
0.2782
0.2086
0.1391
0.0695
1.000
203
185
139
93
46
666
203
388
527
620
666
1,760
5,124
9,693
15,068
20,843
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ftkip = 1.36 kNm.
9.2.6.4 Birmingham 2 Horizontal Distribution of Forces
For the ELF analysis, this is the same as that for New York City location; see Sec. 9.2.5.4.
Total shear in wall type D:
0.125 Vtot= V+1.365(0.0238)V=0.158V=104.9kips
The dynamic analysis shows that the fundamental mode is a pure torsional mode. The fact that the
fundamental mode is torsional does confirm, to an extent, that the structure is torsionally sensitive. This
modal analysis does not show any significant effect of the torsion, however. The pure symmetry of this
structure is somewhat idealistic. Real structures usually have some real eccentricity between mass and
stiffness, and dynamic analysis then yields coupled modes, which contribute to computed forces.
The Provisions does not require that the accidental eccentricity be analyzed dynamically. For illustration,
however, this was done by adjusting the mass of the floor elements to generate an eccentricity of 5
percent of the 152ft length of the building. Table 9.214 shows the results of such an analysis.
(Accidental torsion could also be considered using a linear combination of the dynamic results and a
statically applied moment equal to the accidental torsional moment.)
The transverse direction base shear from the SRSS combination of the modes is 403.8 kips, significantly
less than the 457.6 kips for the symmetric model. The amplification factor for this base shear is 566/404
= 1.4. This smaller base shear from modal analysis of a model with an artificially introduced eccentricity
is normal. For two primary reasons. First, the mass participates in more modes. The participation in the
largest mode is generally less, and the combined result is dominated by the largest single mode. Second,
the period for the fundamental mode generally increases, which will reduce the spectral response except
for structures with short periods (such as this one).
The base shear in Wall D was computed by adding the inplane reactions. For the symmetric model the
result was 57 kips, which is 12.5 percent of the total of 458 kips, as would be expected. Amplifying this
by the 1.24 factor yields 71 kips The application of a static horizontal torsion equal to the 5 percent
eccentricity times a base shear of 566 kips (the “floor”) adds 13 kips, for a total of 84 kips. If the static
horizontal torsion is amplified by 1.365, as found in the analysis for the New York location, the total
becomes 89 kips, which is less than the 99 kips and 105 kips computed in the ELF analysis without and
with, respectively, the amplification of accidental torsion. The Wall D base shear from the eccentric
model was 66 kips; with the amplification of base shear = 1.4, this becomes 92 kips. Note that this value
is less than the direct shear from the symmetric model plus the amplified static torsion. The obvious
conclusion is that more careful consideration of torsional instability than actually required by the
FEMA 451, NEHRP Recommended Provisions: Design Examples
980
Provisions does not indicate any more penalty than already given by the procedures for the ELF in the
Provisions. Therefore the remainder of the example designs for this building are completed using the
ELF.
Table 9.214 Birmingham periods, Mass Participation Factors, and Modal Base
Shears in the Transverse Direction for Modes Used in Analysis
Mode Period Individual mode (percent) Cumulative sum (percent) Trans.
Number (sec) Long. Trans. Vert. Long. Trans. Vert. Base Shear
1 0.2507 0.0 8.8 0.0 0.0 8.8 0.0 56.3
2 0.1915 70.5 0.0 0.1 70.5 8.8 0.1 0.0
3 0.1867 0.0 61.4 0.0 70.5 70.2 0.1 394.9
4 0.0698 0.0 2.9 0.0 70.5 73.1 0.1 12.7
5 0.0613 1.1 0.0 23.0 71.6 73.1 23.1 0.0
6 0.0575 19.2 0.0 0.0 90.9 73.1 23.2 0.0
7 0.0570 0.0 13.7 0.0 90.9 86.8 23.2 55.5
8 0.0533 0.0 5.6 0.0 90.9 92.4 23.2 22.0
9 0.0480 1.2 0.0 12.8 92.0 92.4 35.9 0.0
10 0.0380 1.4 0.0 0.0 93.5 92.4 35.9 0.0
11 0.0374 0.0 0.4 0.0 93.5 92.8 35.9 1.3
12 0.0327 1.7 0.0 0.2 95.2 92.8 36.1 0.0
13 0.0322 0.0 3.1 0.0 95.2 95.9 36.1 10.4
14 0.0263 2.8 0.0 0.1 98.0 95.9 36.2 0.0
15 0.0243 0.0 3.0 0.0 98.0 98.8 36.2 9.5
16 0.0201 1.6 0.0 0.1 99.6 98.8 36.3 0.0
17 0.0164 0.0 1.1 0.0 99.6 100.0 36.3 3.4
18 0.0141 0.4 0.0 0.1 100.0 100.0 36.3 0
The total story shear and overturning moment (from the ELF analysis) may now be distributed to Wall D
and the wall proportions checked. The wall capacity will be checked before considering deflections.
The “extreme torsional irregularity” has an additional consequence for Seismic Design Category D:
Provisions 5.6.2.4.2 [Sec. 4.6.3.2] requires that the design forces for connections between diaphragms,
collectors, and vertical elements (walls) be increased by 25 percent above the diaphragm forces given in
Provisions 5.4.1 [Sec. 4.6.3.4]. For this example, the diaphragm of precast elements is designed using the
different requirements of the appendix to Provisions Chapter 9 (see Chapter 7 of this volume).
9.2.6.5 Birmingham 2 Transverse Wall (Wall D)
The design demands are slightly smaller than for the New York City design, yet there is more
reinforcement, both vertical and horizontal in the walls. This illustration will focus on those items where
the additional reinforcement has special significance.
9.2.6.5.1 Birmingham 2 Shear Strength
Refer to Sec. 9.2.5.5.1 for most quantities. The additional horizontal reinforcement raises Vs and the
additional grouted cells raises An and, therefore both Vm and Vn max.
Av/s = (4)(0.31 in.2)/(8.67 ft.) = 0.1431 in.2/ft
Vs = 0.5(0.1431)(60 ksi)(32.67 ft) = 140.2 kips
An = (2 × 1.25 in. × 32.67 ft x 12 in.) + (41 in.2 × 12 cells) = 1,472 in.2
Chapter 9, Masonry
981
The shear strength of Wall D is summarized in Table 9.215 below. (Note that Vx and Mx in this table are
values from Table 9.213 multiplied by 0.158, the portion of direct and torsional shear assigned to the
wall). Clearly, the dynamic analysis would make it possible to design this wall for smaller forces, but the
minimum configuration suffices. The 1.96 multiplier on Vx to determine Vu is explained in the subsequent
section on flexural design.
Table 9.215 Shear Strength Calculations for Wall D, Birmingham 2
Level
(x)
Vx
(kips)
Mx
(ftkips)
Mx/Vxd 1.98Vx
(kips)
P
(kips)
fVm
(kips)
fVs
(kips)
fVn
(kips)
fVn max
(kips)
5 32.0 277 0.265 63.4 42 194.6 112.2 306.8 313.9
4 61.1 907 0.454 121 90 186.8 112.2 299 287.3
3 83.0 1527 0.563 164.3 139 186.6 112.2 298.8 272.0
2 97.7 2373 0.743 193.4 188 179.7 112.2 291.9 246.7
1 104.9 3283 0.958 207.7 236 169.6 112.2 281.8 216.6
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
Note that Vn max is less than Vn at all levels except the top story. The capacity is greater than the demand
at all stories, therefore, the design is satisfactory for shear.
9.2.6.5.2 Birmingham 2 Axial and Flexural Strength
Once again, the similarities to the design for the New York City location will be exploited. Normally, the
inplane calculations include:
1. Strength check
2. Ductility check
9.2.6.5.2.1 Strength check
The wall demands, using the load combinations determined previously, are presented in Table 9.216 for
Wall D. In the table, Load Combination 1 is 1.29D + QE + 0.5L and Load Combination 2 is 0.81D + QE.
Table 9.216 Birmingham 2 Demands for Wall D
Load Combination 1 Load Combination 2
Level PD
(kips)
PL
(kips)
Pu
(kips)
Mu
(ftkips)
Pu
(kips)
Mu
(ftkips)
54321
43
94
145
196
247
0
8
17
25
34
55
125
196
265
336
277
807
1527
2373
3283
36
76
117
159
200
277
807
1527
2373
3283
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm
Strength at the bottom story (where P, V, and M are the greatest) are less than required for the New York
City design. The demands are plotted on Figure 9.210, showing that the design for New York City has
sufficient axial and flexural capacity for this Birmingham 2 location. For this design, the interaction
capacity line will be shifted to the right, due to the presence of additional reinforcing bars. The only
calculation here will be an estimate of the factor to develop the flexural capacity at the design axial load.
FEMA 451, NEHRP Recommended Provisions: Design Examples
982
The flexural capacity for lightly load walls is approximately proportional to the sum of axial load plus the
yield of the reinforcing steel:
Birmingham #2 capacity
NewYorkCapacity
kips 12 0.20 in. ksi
kips 9 0.20 in. ksi
2
2 =
+ × ×
+ × ×
= =
200 60
199 60
344
307
1.12
Therefore the factor by which the walls shears must be multiplied to represent 125 percent of flexural
capacity, given that the factor was 1.58 for the New York design is:
158 112 177
746
666
.×. =.× =1.98
New York base shear
Birmingham #2 base shear
9.2.6.5.2.2 Ductility check
The Provisions requirements for ductility are described in Sec. 9.2.4.5.2 and 9.2.5.5.2. Since the wall
reinforcement and loads are so similar to those for the New York City building, the computations are not
repeated here.
[Refer to Sec. 9.2.4.5.2 for discussion of revisions to the ductility requirements in the 2003 Provisions.]
9.2.6.6 Birmingham 2 Deflections
The calculations for deflection would be very similar to that for the New York City location. Ironically,
that procedure will indicate that the wall is not cracked at the design load. The Cd factor is larger, 3.5 vs.
2.25. However, the calculation is not repeated here; refer to Sec. 9.2.4.6 and Sec. 9.2.5.6.
[Refer to Sec. 9.2.4.6 for discussion of revisions to the deflection computations and requirements in the
2003 Provisions, as well as the potentially conflicting drift limits.]
9.2.6.7 Birmingham 2 OutofPlane Forces
Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3] requires that the bearing walls be designed for outofplane loads
determined:
w = 0.40 SDS Wc $ 0.1Wc
w = (0.40)(0.47)(56 psf) = 10.5 psf $ 0.1Wc
The calculated seismic load, w = 10.5 psf, is less than wind pressure for exterior walls. Even though Wall
D is not an exterior wall, the lateral pressure is sufficiently low that it is considered acceptable by
inspection without further calculation. Seismic loads do not govern the design of Wall D for loading in
the outofplane direction.
9.2.6.8 Birmingham 2 Orthogonal Effects
According to Provisions Sec. 5.2.5.2.2 [Sec. 4.4.2.3], orthogonal interaction effects have to be considered
for Seismic Design Category D when the ELF procedure is used (as it is here). However, the outofplane
component of only 30 percent of 10.5 psf on the wall will not produce a significant effect when combined
with the inplane direction of loads so no further calculation will be made.
This completes the design of the Transverse Wall D.
Chapter 9, Masonry
983
9.2.6.9 Birmingham 2 Summary of Wall Design for Wall D
8in. CMU
f'm = 2,000 psi
Reinforcement:
12 vertical #4 bars per wall (spaces alternate at 32 and 40 in. on center)
Two bond beams with 2  #5 at each story, at bearing for the planks, and at 4 ft above each floor.
Horizontal joint reinforcement at alternate courses is recommended, but not required.
9.2.7 Seismic Design for Los Angeles
Once again, the differences from the designs for the other locations will be emphasized. As explained for
the Birmingham 2 building, the Provisions would require a dynamic analysis for design of this building.
For the reasons explained in Sec. 9.2.6.4, this design is illustrated using the ELF procedure.
9.2.7.1 Los Angeles Weights
Use 91 psf for 8in.thick, normal weight hollow core plank, 2.5 in. lightweight concrete topping (115
pcf), plus the nonmasonry partitions. This building is Seismic Design Category D, and the walls will be
designed as special reinforced masonry shear walls (Provisions Sec. 11.11.5 and Sec. 11.3.8
[Sec.11.2.1.5]), which requires prescriptive seismic reinforcement (Provisions Sec. 11.3.8.3 [ACI 530,
Sec. 1.13.2.2.5]). Special reinforced masonry shear walls have a minimum spacing of vertical
reinforcement of 4 ft on center. For this example, 60 psf weight for the 8in. CMU walls will be assumed.
The 60 psf value includes grouted cells and bond beams in the course just below the floor planks and in
the course 4 ft above the floors. A typical wall section is shown in Figure 9.212.
FEMA 451, NEHRP Recommended Provisions: Design Examples
984
8 " p recast
concrete p lank
8 " con crete
m asonry w all
See F igure 7 .1 8
fo r g ro u ted
sp a c e d e ta il.
V ertic a l b ar
at 4 '0" o .c.
B o n d b eam
w / (2) # 5 (ty p )
9 g a . horizo n tal
jo in t re in fo rc em en t
at 1 '4" o .c.
8"
4'0"
7'91 8"
2"
4'0"
21
2" 21
2"
L ig h tw eig h t
concrete
to p p ing
Figure 9.212 Typical wall section for the Los Angeles location (1.0 in. = 25.4
mm, 1.0 ft = 0.3048 m)
Story weight ,wi:
Roof weight:
Roof slab (plus roofing) = (91 psf) (152 ft)(72 ft) = 996 kips
Walls = (60 psf)(589 ft)(8.67 ft/2) + (60 psf)(4)(36 ft)(2 ft) = 170 kips
Total = 1,166 kips
There is a 2fthigh masonry parapet on four walls and the total length of masonry wall is 589 ft.
Typical floor:
Slab (plus partitions) = 996 kips
Walls = (60 psf)(589 ft)(8.67 ft) = 306 kips
Total = 1,302 kips
Total effective seismic weight, W = 1,166 + (4)(1,302) = 6,374 kips
This total excludes the lower half of the first story walls, which do not contribute to seismic loads that are
not imposed on the CMU shear walls.
Chapter 9, Masonry
985
9.2.7.2 Los Angeles Base Shear Calculation
The seismic response coefficient, Cs, is computed using Provisions Eq. 5.4.1.11 [Eq. 5.22] and 5.4.1.12
[Eq. 5.23]:
Controls
1.00 0.286
/ 3.51
DS
s
C S
R I = = =
( )
0.60 0.507
( / ) 0.338 3.5 1
D1
s
C S
T R I = = =
where T is the fundamental period of the building, which is 0.338 sec as computed previously (the
approximate period, based on building system and building height, will be the same for all locations).
The value for Cs is taken as 0.286 (the lesser of these two). This value is still larger than the minimum
specified in Provisions Eq. 5.3.2.13 which is:
Cs = 0.044SD1I = (0.044)(0.60)(1) = 0.026
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value
for longperiod structures, which is not applicable to this example.]
The total seismic base shear is then calculated Provisions Eq. 5.4.1 [Eq.5.21]:
V = CsW = (0.286)(6,374) = 1,823 kips
9.2.7.3 Los Angeles Vertical Distribution of Seismic Forces
The vertical distribution of seismic forces is determined in accordance with Provisions Sec. 5.4.4 [Sec.
5.2.3], which as described in Sec. 9.2.4.3. Note that for Provisions Eq. 5.4.32 [Eq. 5.211], k = 1.0 since
T = 0.338 sec (similar to the previous example buildings).
The application of the Provisions equations for this building is shown in Table 9.217:
Table 9.217 Los Angeles Seismic Forces and Moments by Level
Level
(x)
wx
(kips)
hx
(ft)
wxhx
k
(ftkips)
Cvx Fx
(kips)
Vx
(kips)
Mx
(ftkips)
543213
1,166
1,302
1,302
1,302
1,302
6,374
43.34
34.67
26.00
17.33
8.67
50,534
45,140
33,852
22,564
11,288
163,378
0.309
0.276
0.207
0.138
0.069
1.000
564
504
378
252
126
1,824
564
1,608
1,446
1,698
1,824
4,890
14,150
26,686
41,409
57,222
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ftkip = 1.36 kNm
9.2.7.4 Los Angeles Horizontal Distribution of Forces
This is the same as for the Birmingham 2 design; see Sec. 9.2.6.4.
Total shear in Wall Type D:
FEMA 451, NEHRP Recommended Provisions: Design Examples
986
Vtot = 0.125V + 1.365(0.0238)V = 0.158V
The total story shear and overturning moment may now be distributed to each wall and the wall
proportions checked. The wall capacity will be checked before considering deflections.
9.2.7.5 Los Angeles Transverse Wall D
The strength or limit state design concept is used in the Provisions.
9.2.7.5.1 Los Angeles Shear Strength
The equations are the same as for the prior locations for this example building. Looking forward to the
design for flexural and axial load, the amplification factor on the shear is computed as:
125 125
9156 085
9012
. . 149
/ .
.
M
M
n
u
= = (which is less than the 2.5 upper bound)
Therefore, the demand shear is 1.49 times the value from analysis. (This design continues to illustrate the
ELF analysis and; as explained for the Birmingham 2 design, smaller demands could be derived from the
dynamic analysis.) All other parameters are similar to those for Birmingham 2 except that:
An = (2 × 1.25 in. × 32.67 ft x 12 in.) + (41 in.2 × 15 cells) = 1,595 in.2
The shear strength of each Wall D, based on the aforementioned formulas and data, are summarized in
Table 9.218.
Table 9.218 Los Angeles Shear Strength Calculations for Wall D
Story
Vx
(kips)
Mx
(ftkips)
Mx/Vxd 1.49Vx
(kips)
P
(kips)
fVm
(kips)
fVs
(kips)
fVn
(kips)
fVn max
(kips)
5 88.8 770 0.265 132.3 42 210.1 112.2 322.3 340
4 168.2 2229 0.406 250.6 90 205.7 112.2 317.9 318.7
3 227.7 4203 0.565 339.3 139 199.6 112.2 311.8 294.5
2 267.4 6522 0.747 398.4 188 191.3 112.2 303.8 266.8
1 287.2 9012 0.960 427.9 236 179.5 112.2 291.7 234.3
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm
Just as for the Birmingham 2 design, the maximum on Vn controls over the sum of Vm and Vs at all stories
except the top. Unlike the prior design, the shear capacity is inadequate in the lower three stories. The
solution is to add grout. At the first story, solid grouting is necessary:
An = (7.625 in.)(32.67 ft.)(12 in./ft.) = 2989 in.2
fVn max = 0.8(4.11)(0.0447 ksi)(2989 in.2) = 439 kips > 428 kips OK
At the third story, six additional cells are necessary, and at the second story, approximately two out of
three cells must be grouted. The additional weight adds somewhat to the demand but only about 2
percent. If the entire building were grouted solid (which would be common practice in the hypothetical
location), the weight would increase enough that the shear strength criterion might be violated.
Chapter 9, Masonry
987
9.2.7.5.2 Los Angeles Axial and Flexural Strength
The basics of the flexural design have been demonstrated for the previous locations. The demand is much
higher at this location, however, which introduces issues about the amount and distribution of
reinforcement in excess of the minimum requirements. Therefore, the strength and ductility checks will
both be examined.
9.2.7.5.2.1 Strength check
Load combinations, using factored loads, are presented in Table 9.219 for Wall D. In the table, Load
Combination 1 is 1.4D + QE + 0.5L, and Load Combination 2 is 0.7D + QE.
Table 9.219 Los Angeles Load Combinations for Wall D
Load Combination 1 Load Combination 2
Level
(x)
PD
(kips)
PL
(kips)
Pu
(kips)
Mu
(ftkips)
Pu
(kips)
Mu
(ftkips)
54321
63
126
189
251
314
0
8
17
25
34
88
180
273
364
456
770
2229
4203
6522
9012
44
88
132
176
220
770
2229
4203
6522
9012
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm
Strength at the bottom story (where P, V, and M are the greatest) is examined. This example considers
Load Combination 2 from Table 9.2.19 to be the governing case, because it has the same lateral load as
Load Combination 1 but lower values of axial force.
Refer to Figure 9.213 for notation and dimensions.
FEMA 451, NEHRP Recommended Provisions: Design Examples
988
Figure 9.213 Los Angeles: Strength of wall D (1.0 ft = 0.3048 m).
Strength diagrams superimposed on strain diagrams for the two cases.
Examine the strength of Wall D at Level 1:
Pumin = 220 kips
= 456 kips max u P
Mu = 9,012 ftkips
Because special reinforced masonry shear walls are used (Seismic Design Category D), vertical
reinforcement at 4 ft. on center and horizontal bond beams at 4 ft on center are prescribed (Provisions
Sec. 11.3.7.3 [ACI 530, Sec. 1.13.2.2.5]). (Note that the wall is 43.33 ft high, not 8 ft high, for purposes
of determining the maximum spacing of vertical and horizontal reinforcement.)
y = 0.00207
Ts1
N.A.
y = 0.00207
P = 0 Case
a = 14.15'
c = 17.69'
Center
Line
S Ts2
Cm cell 1 Cm shell
a = 1.86'
c = 2.33'
y = 0.00207
m= 0.0025
0.8f'
C m + C
32.67'
16.33' 16.00' 0.33'
y = 0.00207
m= 0.0025
0.8f'
Cs1
M
P
(7) #5 at 4.00' on center
(4) #8 at 0.67' on center
Cs2 Cs2 Cs2
Ts1
Ts2 Ts2 Ts2
= 0.00179
Cm cell 2
Ts1
N.A.
Balanced
Case
s
m
m
Chapter 9, Masonry
989
For this bending moment, the minimum vertical reinforcement will not suffice. For reinforcement
uniformly distributed, a first approximation could be taken from a simple model using an effective
internal moment arm of 80 percent of the overall length of the wall:
2 2
ksi
0.8 / 2 9012 0.8 220 32.67 / 2 7.8 in 0.24 in /ft.
60( )(0.8 / 2) 60 0.8 32.67 / 2 s
A M Pl
l
  × ×
= = = =
× ×
The minimum vertical steel is 0.0007 times the gross area, which is 0.064 in.2/ft. At the maximum
spacing of 4 ft, a #5 bar is slightly above the minimum. Experimental evidence indicates that uniformly
distributed reinforcement will deliver good performance. This could be accomplished with a #9 bar at 48
in. or a #8 bar at 40 in. This design will work well in a wall that is solidly grouted; however, for walls
that are grouted only at cells containing reinforcement, it will be found that this wall fails the ductility
check (which can be remedied by placing several extra grouted cells near each end of the wall as was
shown in Sec. 9.1.5.4). The flexural design was completed before the shear design (described in the
previous section) discovered the need for solid grout in the first story. The remainder of this flexural
design check is carried out without consideration of the added grout. (It is unlikely that the interaction
line will be affected near the design points, but the balanced point will definitely change.)
It has long been common engineering practice to concentrate flexural reinforcement near the ends of the
wall. (This a normal result of walls that intersect to form flanges with reinforcement in both web and
flange.) For this design, if one uses the minimum #5 bar at 48 inches, then the extra steel at the ends of
the walls is approximately:
(7.8 7 0.31) / 2 2.8 in2 s end A =  × =
Try #8 bars in each of the first four end cells and #5 bars at 4 ft on center at all intermediate cells.
The calculation procedure is similar to that presented in Sec. 9.2.4.5.2. The strain and stress diagrams are
shown in Figure 9.213 for the Birmingham 1 building and the results are as follows:
P = 0 case
fPn = 0
fMn = 6,636 ftkips
Intermediate case, setting c = 4.0 ft
fPn = 223 kips
fMn = 9,190 ftkips
Balanced case
fPn = 1049 kips
fMn = 14,436 ftkips
The simplified fPn  fMn curve is shown in Figure 9.214 and indicates the design with #8 bars in the first
four end cells and #5 bars at 4 ft on center throughout the remainder of the wall is satisfactory.
FEMA 451, NEHRP Recommended Provisions: Design Examples
990
9.2.7.5.2.2 Ductility check
Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.5] has been illustrated in the prior designs. Recall that this
calculation uses unfactored gravity axial loads (Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.5]). Refer
to Figure 9.215 and the following calculations which illustrate this using loads at the bottom story
(highest axial loads). The extra grout required for shear is also ignored here. More grout gives higher
compression capacity, which is conservative.
5,000 ftkips 10,000 ftkips
500 kips
1,000 kips
P max = 456 kips
P min = 220 kips
fP
fM
P = 0
(6636 ftkips, 0 kips)
Balance
(15,320 ftkips, 989 kips)
M = 9012 ftkips
15,000 ftkips
Simplified fP  fM curve
u
3 point fP  fM curve
Intermediate
(9190 ftkips, 223 kips)
n
n
fPn
fMn
u
u
n
n
n
n
Figure 9.214 fP11  fM11 diagram for Los Angeles Wall D (1.0 kip = 4.45 kN, 1.0 kipft = 1.36 kNm).
Chapter 9, Masonry
991
m = 0.0025
N.A.
s = 5 y
= 0.0103
6.29'
P
Cm
0.8f 'm = 1.6 ksi
Cs1
5.96' 6.51'
Ts4 Ts1
(7) #5 at 4'0" o.c.
1.25f y
= 75 ksi
5f y = 300 ksi
26.05'
32'8"
3.78'
0.33'
5.03' 1.26'
26.05'
0.33'
f y = 60 ksi
72.86 ksi
23.5 ksi
= 5 ( 60
29,000 )
c
S T s2
Cs2
22.6 ksi
(4) #8 at
0.67' o.c.
1.08' 5.21'
(4) #8 at
0.67' o.c.
53.2 ksi
Ts3
69.6 ksi
45.6 ksi
Figure 9.215 Ductility check for Los Angeles Wall D (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa)
For Level 1 (bottom story), the unfactored loads are:
P = 314 kips
Cm = 0.8f’m[(a)(b) + Acells]
where b = flange width = (2 × 1.25 = 2.5 in.) and Acells = 41 in.2
Cm = (1.6 ksi)[(5.03 ft × 12)(2.5 in.) + (5 cells)(41)] = 569.4 kips
Cs1 = 0.79(2 x 60 + 53.2 + 45.6) = 172.9 kips
Cs2 = (22.6 ksi)(0.31 in.2) = 7.0 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
992
3C = 749 kips
3Ts1 = (4 × 0.79 in.2)(75 ksi) = 237 kips
3Ts2 = (4 × 0.31 in.2)(75 ksi ) = 93.0 kips
Ts3 = (0.31 in.2)(69.6 ksi) = 21.6 kips
Ts4 = (0.31 in.2)(23.5 ksi) = 7.3 kips
3T = 359 kips
3 C > 3 P + T
749 kips > 673 kips
If a solution with fully distributed reinforcement were used, the tension from reinforcement would
increase while the compression from grout at the end of the wall, as well as compression of steel at the
compression would also decrease. The criterion would not be satisfied. Adding grout would be required.
[Refer to Sec. 9.2.4.5.2 for discussion of revisions to the ductility requirements in the 2003 Provisions.]
9.2.7.6 Los Angeles Deflections
Recall the assertion that the calculations for deflection involve many variables and assumptions and that
any calculation of deflection is approximate at best. The requirements and procedures for computing
deflection are provided in Sec. 9.2.4.6. [Refer to Sec. 9.2.4.6 for discussion of revisions to the deflection
computations and requirements in the 2003 Provisions, as well as the potentially conflicting drift limits.]
For the Los Angeles building, the determination of whether the walls will be cracked is as follows:
be = effective masonry wall width
be = [(2 × 1.25 in.)(32.67 ft × 12) + (15 cells)(41 in.2/cell)]/32.67 ft × 12) = 4.07 in.
A = be l = (4.07 in.)(32.67 × 12) = 1595 in.2
S = be l2/6 = (7.07)(32.67 × 12)2/6 = 104,207 in.3
fr = 0.250 ksi
Pu is calculated using 1.00D (See Table 9.218 for values, and refer to Sec. 9.2.4.6 for discussion). Table
9.220 provides a summary of these calculations. (The extra grout required for shear strength is also not
considered here; the revision would slightly reduce the computed deflections by raising the cracking
moment.)
Table 9.220 Los Angeles Cracked Wall Determination
Level Pumin
(kips)
Mcr
(ftkips)
Mx
(ftkips)
Status
54321
63
126
189
251
314
2514
2857
3200
3538
3880
770
2229
4203
6522
9012
uncracked
uncracked
cracked
cracked
cracked
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
For the uncracked walls (Levels 4 and 5):
In = Ig = bel3/12 = (4.07 in.)(32.67 × 12)3/12 = 2.04 × 107 in.4
Chapter 9, Masonry
993
For the cracked walls, the transformed cross section will computed by classic methods. Assuming the
neutral axis to be about 10 ft in from the compression face gives five #5 bars in tension. The tension
reinforcement totals:
As = 4(0.79) + 5(0.31) = 3.16 + 1.55 = 4.71 in.2
The axial compression stiffens the wall. The effect is approximated with an equivalent area of tension
reinforcement equal to half the compression. Thus, the total reinforcement becomes:
Ase = 4.71 + 0.5(314)/60 = 4.71 + 2.62 = 7.33 in.2
The centroid of this equivalent reinforcement is 29.5 ft from the compression face. Following the classic
method for transformed cracked cross sections and with n = 19.3:
. = 7.33/(4.04 x 29.5 x 12) = 0.0051
.n = 0.0051(19.3) = 0.099
k = sqrt(.n2 + 2.n)  .n = 0.36
kd = c = 10.5 feet (which is close enough to the assumed 10 feet)
Icr = bc3/3 + 3nAsd2 = 4.04(29.5 x 12)3 + 19.3(7.33)(29.5  10.5)2(144) = 1.01 x 107 in.4
The Provisions encourages the use of the cubic interpolation formula illustrated for the previous
locations. For the values here, this yields Ieff = 1.09 x 107 in.4, which is about half the gross moment of
inertia (which in itself is not a bad approximation for a cracked and well reinforced cross section). For
this example, the deflection computation will instead use the cracked moment of inertia in the lower three
stories and the gross moment of inertia in the upper two stories. The results from a RISA 2D analysis are
shown in Table 9.221, and are about 5 percent higher than use of Ieff over the full height.
Table 9.221 Los Angeles Deflections
Level F
(kips)
Ieff
(in.4)
dflexural
(in.)
dshear
(in.)
dtotal
(in.)
Cd dtotal
(in.)
.
(in.)
54321
84.0
75.1
56.3
37.6
18.8
2.04 × 107
2.04 × 107
1.01 × 107
1.01 × 107
1.01 × 107
0.491
0.370
0.237
0.118
0.033
0.127
0.118
0.096
0.068
0.035
0.618
0.488
0.333
0.186
0.068
2.163
1.708
1.166
0.651
0.238
0.455
0.543
0.515
0.413
0.238
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm.
The maximum drift occurs at Level 4 per Provisions Table 5.2.8 is:
. = 0.543 in. < 1.04 in. = 0.01 hn (Provisions Table 5.2.8 [Table 4.51]) OK
9.2.7.7 Los Angeles OutofPlane Forces
Provisions Sec 5.2.6.2.7 [Sec. 4.6.1.3] requires that the bearing walls be designed for outofplane loads
determined as follows:
w = 0.40 SDS Wc $ 0.1Wc
w = (0.40)(1.00)(60 psf) = 24psf $ 0.1Wc
FEMA 451, NEHRP Recommended Provisions: Design Examples
994
The outofplane bending moment, using the strength design method for masonry, for a pressure, w = 24
psf and considering the Pdelta effect, is computed to be 2,232 in.lb/ft. This compares to a computed
strength of the wall of 14,378 in.lb/ft, considering only the #5 bars at 4 ft on center. Thus, the wall is
loaded to about 16 percent of its capacity in flexure in the outofplane direction. (See Sec. 9.1 for a more
detailed discussion of strength design of masonry walls, including the Pdelta effect.)
9.2.7.8 Los Angeles Orthogonal Effects
According to Provisions Sec. 5.2.5.2.2 [Sec. 4.4.2.3], orthogonal interaction effects have to be considered
for Seismic Design Category D when the ELF procedure is used (as it is here).
The outofplane effect is 16 percent of capacity, as discussed in Sec. 9.2.7.7 above. When considering
the 0.3 combination factor, the outofplane action adds about 5 percent overall to the interaction effect.
For the lowest story of the wall, this could conceivably require a slight increase in capacity for inplane
actions. In the authors’ opinion, this is on the fringe of requiring real consideration (in contrast to the end
walls of Example 9.1).
This completes the design of the transverse Wall D.
9.2.7.9 Los Angeles Summary of Wall Design for Wall D
8in. CMU
f!m = 2,000 psi
Reinforcement:
Four vertical #8 bars, one bar in each cell for the four end cells
Vertical #5 bars at 4 ft on center at intermediate cells
Two bond beams with two #5 bars at each story, at floor bearing and at 4 ft above each floor
Horizontal joint reinforcement at alternate courses recommended, but not required
Grout:
All cells with reinforcement and bond beams, plus solid grout at first story, at two out of three
cells in the second story, and at six extra cells in the third story
Table 9.222 compares the reinforcement and grout for Wall D designed for each of the four locations.
Table 9.222 Variation in Reinforcement and Grout by Location
Birmingham 1 New York City Birmingham 2 Los Angeles
Vertical bars 5  #4 9  #4 12  #4 8  #8 + 7  #5
Horizontal bars 10  #4 + jt. reinf 10  #4 + jt. reinf 20  #5 20  #5
Grout (cu. ft.) 91 122 189 295
1 cu. ft. = 0.0283 m3.
Chapter 9, Masonry
995
40'0" 24'0" 24'0" 24'0" 40'0"
4'0"
72'0"
24'0" 24'0" 24'0'
152'0"
A
B
B
A
Joists
at 2'6"
A
B
B
A
C C
C C
D D
D D
Bulbs at ends
of walls
101
2" grouted
brick walls at
lower 6 stories.
8" brick in
upper 6 stories.
Figure 9.31 Floor plan (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm)
9.3 TWELVESTORY RESIDENTIAL BUILDING IN
LOS ANGELES, CALIFORNIA
9.3.1 Building Description
This 12story residential building has a plan form similar to that of the fivestory masonry building
described in Sec. 9.2. The floor plan and building elevation are illustrated in Figures 9.31 and 9.32,
respectively. The floors are composed of 14in.deep open web steel joists spaced at 30 in. that support a
3in. concrete slab on steel form deck. A firerated ceiling is included at the bottom chord of the joists.
Partitions, including the shaft openings, are gypsum board on metal studs, and the exterior nonstructural
curtain walls are glass and aluminum.
All structural walls are of grouted brick. For purposes of illustration, two styles of wall are included. The
lower six stories have 101/2in.thick walls consisting of two wythes of 4in. (nominal) brick and a 31/4
in. grout space. The upper six stories have 8in. (nominal) brick, hollow unit style, with the vertical
reinforcing in the cells and the horizontal reinforcing in bond beams. (In actual construction, however, a
single style wall might be used throughout: either a twowythe grouted wall or a throughthewall unit of
an appropriate thickness). The walls are subject to high overturning moments and have a reinforced
masonry column at each end. The column concentrates the flexural reinforcement and increases resistance
to overturning. (Similar concentration and strength could be obtained with transverse masonry walls
serving as flanges for the shear walls had the architectural arrangement been conducive to this approach.)
Although there is experimental evidence of improved performance of walls with all vertical reinforcement
uniformly distributed, concentration at the ends is common in engineering practice and the flexural
demands are such for this tall masonry building that the concentration of masonry and reinforcement at
the ends is simply much more economical.
The compressive strength of masonry, f!m, used in this design is 2,500 psi for Levels 1 through 6 and
3,000 psi for Levels 7 through 12.
FEMA 451, NEHRP Recommended Provisions: Design Examples
996
152'0"
36'0" 80'0" 36'0"
12 at 10'0"=120'0" 2'0"
122'0"
Figure 9.32 Elevation (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm)
This example illustrates the following aspects of the seismic design of the structure:
1. Development of equivalent lateral forces
2. Reinforced masonry shear wall design
3. Check for building deflection and story drift
4. Check of diaphragm strength.
9.3.2 Design Requirements
9.3.2.1 Provisions Design Parameters
Table 9.31 shows the design parameters for building design.
Chapter 9, Masonry
997
Table 9.31 Design Parameters
Design Parameter Value
Ss (Map 1 [Figure 3.33]) 1.5
S1 (Map 2 [Figure 3.34]) 0.6
Site Class C
Fa 1
Fv 1.3
SMS = FaSs 1.5
SM1 = FvS1 0.78
SDS = 2/3 SMS 1
SD1 = 2/3 SM1 0.52
Seismic Design Category D
Masonry Wall Type Special Reinforced
R 3.5
O0 2.5
Cd 3.5
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package).]
9.3.2.2 Structural Design Requirements
The load path consists of the floors acting as horizontal diaphragms and the walls parallel to the motion
acting as shear walls.
Soilstructure interaction is not considered.
The building is a bearing wall system (Provisions Table 5.2.2 [4.31]) .
Deformational compatibility must be assured (Provisions Sec. 5.2.2.4.3 [Sec. 4.5.3]). The structural
system is one of noncoupled shear walls. Crossing beams over the halls (their design is not included in
this example) will need to continue to support the gravity loads from the floors and roof during an
earthquake but will not provide coupling between the shear walls.
The building is symmetric in plan but has the same torsional irregularity described in Sec. 9.2.5.4. The
vertical configuration is regular except for the change in wall type between the sixth and seventh stories,
which produces a significant discontinuity in stiffness and strength, both for shear and flexure (Provisions
Sec. 5.2.3.3 [Sec. 4.3.2.3] and 5.2.6.2.3 [Sec. 4.6.1.6]). There is no weak story because the strength does
not increase as one goes upward. The stiffness discontinuity will be shown to qualify as regular.
Provisions Table 5.2.5.1 [Table 4.41] would not permit the use of the ELF procedure of Provisions Sec.
5.4 [Sec. 5.2]; instead a dynamic analysis of some type is required. As will be illustrated, this particular
building does not really benefit from this requirement.
FEMA 451, NEHRP Recommended Provisions: Design Examples
998
The design and detailing must comply with the requirements of Provisions Sec. 5.2.6 [Sec. 4.6].
The walls must resist forces normal to their plane (Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3]). These forces
will be used when addressing the orthogonal effects (Provisions Sec.5.2.5.2.2 [Sec. 4.4.2.3]).
With eight walls in each direction, the system is expected to be redundant.
Tie and continuity requirements for anchorage of masonry walls must be considered when detailing the
connections between floors and walls (Provisions Sec. 5.2.6.1.2 [Sec. 4.6.2.1] and 5.2.6.1.3).
Openings in walls and diaphragms need to be reinforced (Provisions Sec. 5.2.6.2.2 [Sec. 4.6.1.4]).
Diaphragms need to be designed to comply with Provisions Sec. 5.2.6.2.6 [Sec. 4.6.3.4].
The story drift limit is 0.01hsx (Provisions Sec. 5.2.8 [Sec. 4.5.1]) and the overall drift limit is 0.01hsn
(Provisions Sec. 11.5.4.1). For this structure the difference between these two is significant, as will be
shown.
[The deflection limits have been removed from Chapter 11 of the 2003 Provisions because they were
redundant with the general deflection limits. Based on ACI 530 Sec. 1.13.3.2, the maximum drift for all
masonry structures is 0.007 times the story height. Thus, there appears to be a conflict between ACI 530
and 2003 Provisions Table 4.51.]
9.3.2.3 Load Combinations
The basic load combinations (Provisions Sec. 5.2.7 [Sec. 4.2.2]) are the same as those in ASCE 7 except
that the seismic load effect, E, is defined by Provisions Eq. 5.2.71 [Eq. 4.21] and 5.2.72 [Eq. 4.22] as:
E = .QE ± 0.2SDSD
Based on the configuration of the shear walls and the results presented in Sec. 9.2, the reliability factor, .,
is treated as equal to 1.0 for both directions of loading. Refer to Sec. 9.2.3.1 for additional information.
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a shear wall
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear
wall with heighttolengthratio greater than 1.0 would not result in more than a 33 percent reduction in
story strength or create an extreme torsional irregularity. The intent is that the aspect ratio is based on
story height, not total height. Therefore, the redundancy factor would not have to be investigated (. =
1.0) for this building.]
The discussion on load combinations for the Los Angeles site in Sec.9.2 is equally applicable to this
example. Refer to Sec. 9.2.3.2 for determination of load combinations.
The load combinations representing the extreme cases are:
1.4D + QE + 0.5L
0.7D + QE
9.3.3 Seismic Force Analysis
The analysis is performed using the ELF procedure of Provisions Sec. 5.4 [Sec. 5.2] and checked with a
modal response spectrum (MRS) analysis in conformance with Provisions Sec. 5.5 [Sec. 5.3]. This
Chapter 9, Masonry
999
example illustrates an analysis for earthquake motions acting in the transverse direction only. Earthquake
motions in all directions will need to be addressed for an actual project.
9.3.3.1 Building Weights
For the ELF analysis, the masses are considered to be concentrated at each floor level whereas, for the
MRS analysis, it is distributed on both wall and floor elements. Note that the term “level” corresponds to
the slab above each story. Thus Level 1 is the second floor; Level 12 is the roof.
Lower Levels:
Slab, joists, partitions, ceiling, mechanical/electrical (M/E), curtain wall at 53 psf
(0.053 ksf)(152 ft)(72 ft) = 580 kips/story
Walls: 10.5 in. at 114 psf (brick at 73 psf + grout at 41 psf)
(0.114 ksf)(10 ft)[(8)(29 ft) + (4)(30 ft) + (4)(32 ft)] = 547 kips/story
Bulbs at ends of walls: (24 in. × 24 in. bulb)
Brick: 2.01 ft2 /bulb; Grout: 1.99 ft2/bulb
[(2.01 ft2)(0.120 kcf) + (1.99 ft2)(0.150 kcf)] (10 ft)(32 bulbs) = 173 kips/story
Upper Levels:
Slab, joists, partitions, ceiling, M/E, curtain wall at 53 psf
(0.053 ksf)(152 ft)(72 ft) = 580 kips/story
Walls: 8 in. Partially grouted brick at 48 psf
(0.048 ksf)(10 ft)[(8)(29.67 ft) + (4)(30.67 ft) + (4)(32.67 ft)] = 236 kips/story
Bulbs at ends of walls (grouted 20 in. × 20 in. brick bulb)
(0.315 klf/bulb)(10 ft)(32 bulbs) = 101 kips/story
Roof:
Slab, roofing, joists,, ceiling, M&E, curtain wall at 53 psf :
(0.053 ksf)(152 ft)(72 ft) = 580 kips
Walls
(238 kips/story + 101 kips/story)/2 = 170 kips
Parapet
(4 parapets)(2 ft)[(0.048 kips/lf)(33 ft) +(3.15 kips/bulb)(2 bulbs)/(10 ft)] = 18 kips
Preliminary design indicates a 101/2in. wall with bulbs for the six lower stories and an 8in. wall with
bulbs for the six upper stories. Therefore, effective seismic weight, W, is computed as follows:
Levels 15 (5)(580 + 547 + 173) = (5)(1,300 kips/level) = 6,500 kips
Level 6 580 + (547 + 236)/2 + (173 + 101)/2 = 1,109 kips = 1,109 kips
Levels 711 (5)(580 + 236 + 101) = (5)(917 kips/level) = 4,585 kips
Level 12 (roof) (580 + 170 + 18) = 768 kips = 768 kips
Total = 12,962 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
9100
The weight of the lower half of walls for the first story are not included with the walls for Level 1 because
the walls do not contribute to the seismic loads.
9.3.3.2 Base Shear
The seismic coefficient, Cs, for the ELF analysis is computed as:
1.00 0.286
s / 3.5/1.0
C SDS
R I = = =
The value of Cs need not be greater than:
0.52
0.198
( / ) (0.75)(3.5 /1)
= = = D1
s
S
C
T R I
The value of the fundamental period, T, was determined from a dynamic analysis of the building modeled
as a cantilevered shear wall. RISA 2D was used for this analysis, with cracked sections taken into
account. From this analysis, a period of T = 0.75 sec was determined. See Sec. 9.3.4. This value is also
obtained from the 3D dynamic analysis (described subsequently) for the first translational mode in the
transverse direction when using a reduced modulus of elasticity to account for cracking in the masonry
(approximately 60 percent of the nominal value for E). Provisions Sec. 5.4.2 [Sec. 5.22] requires that the
fundamental period, T, established in a properly substantiated analysis be no larger than the approximate
period, Ta , multiplied by Cu, determined from Provisions Table 5.4.2 [Table 5.21]. The approximate
period of the building, Ta, is calculated based as:
= 3/ 4=(0.02)(120)0.75=0.725 sec a r n T Ch
where Cr = 0.02 from Provisions Table 5.4.2.1 [Table 5.22], and hn = 120 ft
TaCu = (0.725)(1.4) = 1.015 sec > 0.75 sec = T
(Note that T = 0.75 sec will be verified later when deflections are examined).
The value for Cs is taken to be 0.198 (the minimum of the two values computed above). This value is
still larger than the minimum specified:
=0.044 =(0.044)(1.0)(0.60)=0.0264 Cs ISD1
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value
for longperiod structures, which is not applicable to this example.]
The total seismic base shear is then calculated by Provisions Eq. 5.4.1 [Eq. 5.21]:
V = CsW = (0.198)(12,962 kips) = 2,568 kips
A 3D model was created in SAP 2000 for the MRS analysis. Just as for the fivestory building described
in Sec. 9.2, the masonry walls were modeled as shell bending elements and the floors were modeled as an
assembly of beams and shell membrane elements. See Sec. 9.2.6.2 for further description. The difference
in f’m between upper and lower stories was not modeled; the value of Em used was 1,100 ksi, which is 59
percent of the value from Provisions Eq. 11.3.10.2 for the lower stories. [Note that by adopting ACI 530
in the 2003 Provisions, Em = 900f’m per ACI 530 Sec. 1.8.2.2.1.] As mentioned, this value was selected as
Chapter 9, Masonry
9101
an approximation of the effects of flexural cracking. Unlike the fivestory building, the difference in
length between the longitudinal and transverse walls was modeled. However, to simplify construction of
the model, wall types A and B are the same length. Because this example illustrates design in the
transverse direction, this liberty has little effect. Table 9.32 shows data on the modes of vibration used in
the analysis.
FEMA 451, NEHRP Recommended Provisions: Design Examples
9102
Table 9.32 Periods, mass participation ratios, and modal base shears in the transverse
direction for modes used in analysis
Mode Period Individual mode (percent) Cumulative sum (percent) Trans.
number (seconds) Long. Trans. Vert. Long. Trans. Vert. base shear
1 0.9471 0.00 0.00 0.00 0.00 0.00 0.00 0.0
2 0.7469 0.00 59.12 0.00 0.00 59.12 0.00 1528.0
3 0.6941 59.16 0.00 0.00 59.16 59.12 0.00 0.0
4 0.2247 0.00 0.00 0.00 59.16 59.12 0.00 0.0
5 0.1763 0.00 24.38 0.00 59.16 83.50 0.00 896.2
6 0.1669 24.57 0.00 0.00 83.73 83.50 0.00 0.0
7 0.1070 0.00 0.01 0.00 83.73 83.51 0.00 0.5
8 0.1059 0.00 0.00 0.28 83.74 83.51 0.28 0.0
9 0.1050 0.00 0.00 29.48 83.74 83.51 29.76 0.0
10 0.0953 0.00 0.00 0.00 83.74 83.51 29.76 0.0
11 0.0900 0.00 0.00 1.51 83.74 83.51 31.27 0.0
12 0.0858 0.00 0.03 0.01 83.74 83.54 31.28 1.1
13 0.0832 0.00 7.25 0.00 83.74 90.79 31.28 234.4
14 0.0795 7.11 0.00 0.00 90.85 90.79 31.28 0.0
15 0.0778 0.04 0.00 0.19 90.88 90.79 31.48 0.0
16 0.0545 0.00 4.47 0.00 90.88 95.26 31.48 117.5
17 0.0526 4.44 0.00 0.00 95.32 95.26 31.48 0.0
18 0.0413 0.01 1.24 0.00 95.33 96.51 31.48 29.1
19 0.0392 1.66 0.05 0.00 96.99 96.55 31.48 1.1
20 0.0358 0.07 0.87 0.00 97.06 97.43 31.48 19.5
21 0.0288 1.59 0.33 0.01 98.66 97.76 31.49 7.0
22 0.0278 0.33 1.40 0.00 98.99 99.16 31.49 28.9
23 0.0191 0.76 0.23 0.01 99.75 99.39 31.50 4.3
24 0.0186 0.23 0.60 0.00 99.98 99.98 31.50 11.1
1 kip = 4.45kN.
The combined modal base shear is 1,791 kips
The fundamental mode captures no translation of mass; it is a pure torsional response. This is a
confirmation of the intent of the torsional irregularity provision. The first translational mode has a period
of 0.75 sec, confirming the earlier statements. Also note that the base shear is only about 70 percent of
the ELF base shear (2,568 kips) even though the fundamental period is the same. The ELF analysis
assumes that all the mass participates in the fundamental mode whereas the dynamic analysis does not.
The absolute sum of modal base shears is higher than the ELF but the statistical sum is not. Provisions
Sec. 5.5.7 requires that the modal base shear be compared with 85 percent of the ELF base shear. The
comparison value is 0.85(2,568 kips), which is 2,183 kips. Because this is greater than the value from the
modal analysis, the modal analysis results would have to be factored upwards by the ratio 2,183/1,791 =
1.22. The period used for this comparison cannot exceed CuTa, which is 1.015 sec as described
previously. Note that the period used is from Mode 2, because Mode 1 is a purely torsional mode. The
1.22 factor is very close to the factor for the five story building computed in Sec. 9.2.6.2; an additional
comparison will follow.
9.3.3.3 Vertical Distribution of Seismic Forces
Carrying forward with the ELF analysis, Provisions Sec. 5.4.3 [Sec. 5.2.3] provides the procedure for
determining the portion of the total seismic loads assigned to each floor level. The story force, Fx, is
calculated as:
Chapter 9, Masonry
9103
Fx=CvxV
and
=1
=
S
k
x x
vx n
k
i i
i
C w h
wh
For T = 0.75 sec, which is between 0.5 sec and 2.5 sec, the value of k is determined to be 1.125 based on
interpolation (Provisions Sec. 5.4.3 [Sec. 5.2.3]).
The seismic design shear in any story shall be determined from:
n
x i
i x
V F
=
= S
The story overturning moment is computed from:
( )
n
x i i x
i x
M F h h
=
=S 
Table 9.33 shows the application of these equations for this building.
Table 9.33 Seismic Forces and Moments by Level
Level
(x)
wx
(kips)
hx
(kips)
1.125
wxhx
(ftkips)
Cvx Fx
(kips)
Vx
(kips)
Mx
(kips)
12
11
10
9
8
7
6
5
4
3
2
1
768
917
917
917
917
917
1,109
1,300
1,300
1,300
1,300
1,300
120
110
100
90
80
70
60
50
40
30
20
10
167,700
181,500
163,100
144,800
126,900
109,200
111,000
106,000
82,500
59,700
37,800
17,300
1,307,400
0.128
0.139
0.125
0.111
0.097
0.084
0.085
0.081
0.063
0.046
0.029
0.013
1.00
329
357
320
284
249
214
218
208
162
117
74
34
2,568
329
686
1,006
1,291
1,540
1,754
1,972
2,181
2,342
2,460
2.534
2,568
3,300
10,200
20,200
33,100
48,500
66,000
85,800
107,600
131,000
155,600
181,000
206,600
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ftkip = 1.36 kNm.
The dynamic modal analysis does give a direct output for the gross overturning moment, of about 66
percent of the moment from the ELF analysis. Because the model is built with shell elements, there is no
direct value for the variation of moment with height.
FEMA 451, NEHRP Recommended Provisions: Design Examples
9104
9.3.3.4 Horizontal Distribution
For the ELF analysis, the approach is essentially the same as used for the fivestory masonry building
described in Sec. 9.2.4.4:
Direct shear: All transverse walls have the same properties, except axial load. Axial load affects cracking
but, each wall considered has the same stiffness. Therefore, each will resist an equivalent amount in
direct shear:
V = V/8 = 0.125Vx
Torsion: The center of mass corresponds with the center of resistance; therefore, the only torsion is due
to the 5 percent accidental eccentricity in accordance with Provisions Sec. 5.4.4.2 [Sec. 5.2.4.2]:
Mta = 0.05 bV = (0.05)(152 ft)V = 7.6V
The longitudinal walls are slightly longer than the transverse walls. Unlike the example in Sec. 9.2, this
difference will be illustrated here in a simple fashion. The diaphragm is assumed to be rigid. When the
walls are not identical, a measure of the actual stiffness is necessary; for masonry walls, this involves both
flexural and shear deformations. The conventional technique is an application of the following equation
for deformation of a simple cantilever wall without bulbs or flanges at the ends:
3 6
3 5 wall
m m
Vh Vh
E I G A
. = +
Considering Gm = 0.4 Em, A = Lt, and I = L3t/12, this can be simplified to :
3
4 3 wall
V h h
Et L L
. = .... .. + .. ....
... . . ...
Rigidity, K, is inversely proportional to deflection. Considering E and t as equal for all walls:
3
1
4 3
K
h h
L L
=
.. .. + .. ..
. . . .
Figure 9.33 identifies the walls.
For a multistory building, the quantity h is not easy to pin down. For this example, the authors suggest
the following approach: use h= 10 ft (one story) to evaluate the shear in the wall at the base and also use h
= 80 ft (two thirds of total height) to evaluate the moments in the walls. Table 9.34 shows some of the
intermediate steps for these two assumptions.
(d, as used here, is the distance of the wall to the centroid of the building, not the length of the
wall, as used elsewhere)
Chapter 9, Masonry
9105
Table 9.34 Relative Rigidities
Wall Length
(ft)
Arm, d
(ft)
for shear, h = 10 ft for moment, h = 80 ft
h/d K Kd2 (ft2) h/d K Kd2 (ft2)
ABCD
36
34
33
33
36
12
12
36
0.278
0.294
0.303
0.303
1.088
1.017
0.980
0.980
1,410
146
141
1,270
2,967
2.22
2.35
2.42
2.42
0.01978
0.01690
0.01556
0.01556
25.63
2.43
2.24
20.17
50.47
1.0 ft = 03.048 m
The total torsional rigidity is four times the amount in Table 9.34, since there are four walls of each type.
When considering torsion, Wall D is the critical member (shortest length, greatest d).
For shear due to accidental torsion:
2
(0.980)(36)
7.6 0.0226 or 7.6
4(2,967)
0.01556(36) 0.0211
4(50.47) t
MKd
V V V V
Kd
= = = V
S
... ... ... ...=
FEMA 451, NEHRP Recommended Provisions: Design Examples
9106
When considering the approximations involved, the remainder of the ELF example will simply use
0.0226V for Vt. Because a 3D analytical model exists, a simplistic load case with a static horizontal
torsion at each level was defined. The couple varied directly with height, so the variation of mass with
height was ignored. Examining the base reactions for Wall D yields a torsional shear equal to 0.0221V
and an overturning moment corresponding to 0.0191V. Therefore, the hand computations illustrated are
somewhat conservative.
Total shear for Wall D is equal to the direct shear plus shear due to accidental torsion, which is computed
as:
0.125V + 0.0226V = 0.148V
2'0" 1'8"
1'8"
2'0"
d
d
Wall at Stories 7 to 12
Wall at Stories 1 to 6
8"
101
2"
Wall length
Wall d
A 36'0"
B 34'0"
C 33'0"
D 33'0"
Figure 9.33 Wall dimensions (1.0 in. = 25.4 mm, 1.0 ft = 0.0348 m).
Chapter 9, Masonry
9107
The resulting shears and overturning moments for Wall D are shown in Table 9.35.
Table 9.35 Shear for Wall D
Level Story Shear
(kips)
Wall Shear
(kips)
Story Moment
(ftkips)
Wall Moment
(ftkips)
12
11
10
9
8
7
6
5
4
3
2
1
329
686
1,006
1,291
1,540
1,754
1,972
2,181
2,342
2,460
2,534
2,568
49
102
149
191
228
260
292
323
347
364
375
380
3,300
10,200
20,200
33,100
48,500
66,000
85,800
107,600
131,000
155,600
181,000
206,600
500
1,500
3,000
4,900
7,200
9,800
12,700
15,900
19,400
23,000
26,800
30,60
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
When considering accidental torsion, a check for torsional irregularity must be made. First consider the
case used for design: a direct shear of 0.125V and a torsional shear of 0.0226V. The ratio of extreme
displacement to average displacement can be found from these values and the dimensions, considering
symmetry:
Average displacement is proportional to 0.125V
Torsional displacement at Wall D is proportional to 0.0226V
Torsional displacement at the corner is proportional to (0.0226V)((152 ft./2)/36 ft. = 0.0447V
Ratio of corner to average displacement = (0.125 + 0.0447)/0.125 = 1.38
If the lower value of torsional shear, 0.0191V, found from the 3D computer analysis for the static torsion
is used, the ratio becomes 1.32. In either case, the result is a torsional irregularity (ratio exceeds 1.2) but
not an extreme torsional irregularity (ratio does not exceed 1.4). The reason for the difference from the
fivestory building, in which the ratio exceeded 1.4, is that the longer walls in the longitudinal directions.
For the ELF analysis, Provisions Sec. 5.4.4.3 [Sec. 5.2.4.3] requires the accidental torsion to be
amplified:
2
3.0
1.2 x
A Max displacement
Ave displacement
. .
=. .=
. .
If one uses the ratio of 1.32 based on the 3D computer analysis, the amplifier is 1.21 and the torsional
shear becomes 1.32(0.0191V) = 0.0231V. This is close enough to the unamplified 0.0226V that the ELF
analysis will simply proceed with a torsional shear of 0.0226V.
As described in Sec. 9.2.6.4 for the fivestory building, the 3D analytical model was altered to offset the
center of mass from the center of rigidity. The modal periods, mass participation ratios, and base shears
are given in Table 9.36. The total base shear is 1620 kips, down from the 1,791 kips found without the
eccentricity. The 1,620 kips still slightly exceeds the minimum for design of 1,613 kips described earlier.
Thus, MRS analysis can be used directly in the load combinations and can be considered to include the
FEMA 451, NEHRP Recommended Provisions: Design Examples
9108
amplified accidental torsion.
Table 9.36 Periods, Mass Participation Ratios, and Modal Base Shears in the Transverse
Direction for Modes Used in Analysis of Building with Deliberate Eccentricity
Mode Period Individual mode (percent) Cumulative sum (percent) Trans.
number (seconds) Long. Trans. Vert. Long. Trans. Vert. base shear
1 0.965 0.0 8.5 0.0 0.0 8.5 0.0 169.4
2 0.723 0.0 50.6 0.0 0.0 59.1 0.0 1352.7
3 0.694 59.2 0.0 0.0 59.2 59.1 0.0 0.0
4 0.229 0.0 3.3 0.0 59.2 62.5 0.0 122.7
5 0.171 0.0 21.0 0.0 59.2 83.5 0.0 772.7
6 0.167 24.6 0.0 0.0 83.7 83.5 0.0 0.0
7 0.120 0.0 0.0 20.3 83.7 83.5 20.3 0.0
8 0.108 0.0 1.0 0.0 83.7 84.4 20.3 35.3
9 0.105 0.0 0.0 0.0 83.7 84.4 20.3 0.1
10 0.097 0.0 0.0 0.0 83.7 84.4 20.3 0.1
11 0.090 0.0 0.0 11.4 83.8 84.4 31.7 0.0
12 0.081 0.0 6.2 0.0 83.8 90.7 31.7 198.6
13 0.079 7.1 0.0 0.1 90.9 90.7 31.8 0.0
14 0.074 0.0 0.0 3.1 90.9 90.7 34.8 0.1
15 0.072 0.0 0.6 0.1 90.9 91.3 34.9 17.5
16 0.061 0.0 0.5 0.1 90.9 91.7 34.9 12.8
17 0.053 4.3 0.0 0.0 95.2 91.7 34.9 0.0
18 0.052 0.0 4.1 0.0 95.2 95.8 34.9 104.8
19 0.043 0.7 0.0 0.0 95.9 95.8 34.9 0.0
20 0.037 1.5 0.0 0.0 97.4 95.8 35.0 0.1
21 0.035 0.0 2.5 0.0 97.4 98.3 35.0 54.7
22 0.027 1.8 0.0 0.0 99.2 98.3 35.0 0.0
23 0.022 0.0 1.7 0.0 99.2 100.0 35.0 32.8
24 0.018 0.8 0.0 0.0 100.0 100.0 35.0 0
1 kip = 4.45 kN
The combined modal base shear is 1620 kips.
Mode 1 now includes a translational component, and the comparison to an ELF base shear would be
performed with its period. For T = 0.965 sec, the ELF base shear becomes 1,996 kips, and the
comparison value is 0.85 (1996) = 1,696 kips. This is 78 percent of the value for the symmetric model
and illustrates one of the problems in handling accidental torsion in a consistent fashion.
Without factoring the modal results up to achieve a base shear of 1,696 kips (a factor of 1.047), the
reactions indicate that the base shear for wall D is 266.5 kips, or 0.1645 times the total base shear. If one
takes the direct shear as oneeighth (0.125V), that leaves 0.0395V for the dynamically amplified shear due
to accidental torsion, which could be interpreted to be an amplification of 1.79 over the shear of 0.0221V
found for the static torsion. Thus, it is clear that the amplification value of 1.21 from the equation given
for the ELF analysis underestimates the dynamic amplification of accidental torsion. The bottom line is
that the Wall D shear of 266.5 kips from the dynamic analysis is significantly less than the shear of 380
kips found in the ELF analysis without amplification of accidental torsion. The example will proceed
based upon the shear of 380 kips.
9.3.3.5 Transverse Wall (Wall D)
Chapter 9, Masonry
9109
The strength or limit state design concept is used in the Provisions.
[The 2003 Provisions adopts by reference the ACI 53002 provisions for strength design in masonry, and
the previous strength design section has been removed. This adoption does not result in significant
technical changes, and the references to the corresponding sections in ACI 530 are noted in the following
sections.]
9.3.3.5.1 Axial and Flexural Strength General
The walls in this example are all bearing shear walls since they support vertical loads as well as lateral
forces.
The demands for the representative design example, Wall D, are presented in this section. The design of
the lower and upper portions of Wall D is presented in the next two sections. For both locations, inplane
calculations include:
1. Strength check and
2. Ductility check.
The axial and flexural demands for Wall D, using the load combinations identified in Sec. 9.3.2.3, are
presented in Table 9.37. In the table, Load Combination 1 represents 1.4D + 1.0 QE + 0.5L, and Load
Combination 2 represents 0.7D + 1.0 QE.
Table 9.37 Load Combinations for Wall D
Load Combination 1 Load Combination 2
Level PD
(kips)
PL
(kips)
Pu
(kips)
Mu
(ftkips)
Pu
(kips)
Mu
(ftkips)
12
11
10
9
8
7
6
5
4
3
2
1
37
80
124
168
212
255
308
370
432
494
556
618
0
8
17
25
34
43
50
59
67
76
84
92
51
117
182
247
313
379
457
548
639
730
821
912
500
1,500
3,000
4,900
7,200
9,800
12,700
15,900
19,400
23,000
26,800
30,600
26
56
87
117
148
179
216
259
303
346
389
433
500
1,500
3,000
4,900
7,200
9,800
12,700
15,900
19,400
23,000
26,800
30,60
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
Strength at the lowest story (where P, V, and M are the greatest) for both the lower wall (Level 1) and the
upper wall (Level 7) constructions will be examined. The design for both locations is based on the values
for Load Combination 2 in Table 9.37.
FEMA 451, NEHRP Recommended Provisions: Design Examples
9110
2'0"
2'0"
1012"
358" 358"
(16) #9
with #3 ties
at 1'4" o.c.
Figure 9.34 Bulb reinforcement at lower levels
(1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m).
9.3.3.5.2 Axial and Flexural Strength Lower Levels
Examine the strength of Wall D at Level 1:
9.3.3.5.2.1 Strength Check (Level 1)
Pumin = 433 kips + factored weight of half of 1st story wall = 433 + (0.7)(21.9) = 448 kips
Mu = 30,600 ftkips
For this Seismic Design Category D building, the special reinforced masonry shear walls must have
vertical and horizontal reinforcement spaced at no more than 4 ft on center. The minimum in either
direction is 0.0007(10.5 in.) = 0.0074 in.2/in. (vertical #5 at 42 in. on center). That will be used as the
vertical reinforcement (although some of the subsequent calculations of flexural resistance are based upon
a spacing of 48 in. on center); the shear strength demands for horizontal reinforcement will be greater and
will satisfy the total amount of 0.0020(10.5 in.) = 0.021 in.2/in.(horizontal #5 at 22 in. on center will
suffice).
Try 16 #9 bars in each bulb. Refer to Figure 9.34 for the placement of the reinforcement in the bulb. In
some of the strength calculations, the #5 bars in the wall will be neglected as a conservative
simplification.
Chapter 9, Masonry
9111
Cm bulb
N.A.
P = 0 Case
Intermediate
Case
Balanced
Case
m = 0.0035
m = 0.0035
m = 0.0035
y = 0.00207
y = 0.00207
Ts
y = 0.00207
y = 0.00207
Ts
Ts
y = 0.00207
Cs
Cm wall
Cs
N.A.
Cm bulb Cm wall
Cm bulb
N.A.
2.00' 2.00'
33.00'
29.00'
M
P
15.50'
a = 0.8c
= 1.37'
c = 1.71'
2.00'
c = 3.00'
a = 0.8c
= 2.40'
a = 0.8c = 16.09'
c = 20.11'
S Ts2
S Ts2
Figure 9.35 Strength of Wall D, Level 1 (1.0 ft = 0.3048 m)
For evaluating the capacity of the wall, a fPn  fMn curve will be developed to represent the wall strength
envelope. The demands (Pu and Mu determined above) will then be compared to this curve. Several cases
will be analyzed and their results used in plotting the fPn  fMn curve. Refer to Figure 9.35 for notation
and dimensions.
FEMA 451, NEHRP Recommended Provisions: Design Examples
9112
Case 1 (P = 0)
The neutral axis will be within the compression bulb, so assume that only the bars closest to the
compression face are effective in compression. (Also recall that the Provisions clearly endorses the use of
compression reinforcement in strength computations.)
Ts1 = (16 bars)(1.00 in.2)(60 ksi) = 960 kips
Ts2 = (7 bars)(0.31 in.2)(60 ksi) = 130 kips
Cs = (5 bars)(1.00 in.2)(60 ksi) = 300 kips
SC = ST + P
Cm + Cs = Ts1 + Ts2 + P
Cm = 960 + 130 +0  300 = 790 kips
Cm = 790 kips = ff’m(24 in.)a = (0.8)(2.5 ksi)(24 in.)a
a = 16.46 in. = 1.37 ft
c = a/0.8 = 1.37/0.8 = 1.71 ft = 20.7 in.
Check strain in compression steel
es = 0.0035(20.7 in.  6 in.)/(20.7 in.) = 0.0025 > yield; assumption OK
SMcl = 0
Mn = (790 kips)(16.5 ft 1.37 ft/2) + (300 + 960 kips)(15.5 ft) + (130 kips)(0 ft.) = 32,170 ftkips
fMn = (0.85)(32,170) = 27,340 ftkips
Case 2 (Intermediate case between P = 0 and balanced case):
Select an intermediate value of c. Let c = 3.0 ft, and determine Pn and Mn for this case.
a = 0.8c = 2.4 ft
Cm bulb = (0.8)(2.5 ksi)(24 in.)2 = 1152 kips
Cm wall = (0.8)(2.5 ksi)(10.5 in.)(0.4 ft. × 12) = 101 kips
Cs = (16 bars)(1.00 in.2)(60 ksi) = 960 kips (approximate; not all bars reach full yield)
SC = (1152 + 101 + 960) = 2213 kips
ST = 960 +130 kips = 1090 kips
SFy = 0
Pn = GC  GT = 2213  1090 = 1123 kips
fPn = (0.85)(1123) = 955 kips
SMcl = 0
Mu = (1152 +960 kips)(15.5 ft) + (101 kips)(14.1 ft) + (960 kips)(15.5 ft) = 49,040 ftkips
fMn = (0.85)(49,040) = 41,680 ftkips
Case 3 (Balanced case):
0.0035 (32.00 ft) 20.11 ft
0.0035 0.00207
=... + ... =
c
a = (0.8)c = 16.09 ft
Ignore the distributed #5 bars for this case.
Chapter 9, Masonry
9113
Cm bulb = 1152 kips
Cm wall = (0.8)(2.5 ksi)(10.5 in.)(14.09 ft. × 12) = 3,550 kips
Cs = 960 kips
SC = (1152 + 3550 + 960) = 5,662 kips
Ts = 960 kips
SFy = 0
Pn = GC  GT = 5662  960 = 4,702 kips
fPn = (0.85)(4,702) = 3,997 kips
SMcl = 0
Mu = (1152 +960 kips)(15.5 ft) + (3550 kips)(7.46 ft) + (960 kips)(15.5 ft) = 74,100 ftkips
fMn = (0.85)(74,100) = 62,980 ftkips
The actual design strength, fMn, at the level of minimum axial load can be found by interpolation to be
33,840 ft.kip.
FEMA 451, NEHRP Recommended Provisions: Design Examples
9114
9.3.3.5.2.2 Ductility check (Level 1)
Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.5] requires that the critical strain condition correspond to a
strain in the extreme tension reinforcement equal to 5 times the strain associated with Fy. Note that this
calculation uses unfactored gravity axial loads (Provisions 11.6.2.2 [ACI 530, Sec. 3.2.3.5]). See Figure
9.37 and the following calculations.
1,000 kips P max = 912 kips
P min = 433 kips
fP
fM
M = 30,600 ftkips
Balance
(62,980 ftkips,
3,997 kips)
Intermediate
(41,680 ft kips, 955 kips)
20,000
ftkips
30,000
ftkips
40,000
ftkips
50,000
ftkips
60,000
ftkips
2,000 kips
3,000 kips
4,000 kips
(27,340 ft kips, 0 kips)
n
n
fPn
fMn
u
u
u
Figure 9.36 fP11  fM11 Diagram for Level 1 (1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm).
Chapter 9, Masonry
9115
2.00'
33.00'
2.00'
24.88'
P
8.12'
s = 5
= 0.0103
1.25 f =
Ts1
4.49'
6.49'
0.8f 'm =
2.00'
Cs1
m = 0.0035
c
2.0 ksi
Cm bulb Cm wall
N.A.
y
102 ksi
300 ksi
5f =
23.88' 1.00'
1.00'
N.A.
75 ksi
a
2.00'
f y = 60 ksi
y
y
1.63'
S Ts2
Figure 9.37 Ductility check for Wall D, Level 1 (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa).
(0.0003.50003.50103) (32.00 ft) = 8.12 ft
. .
. .
.. ..
=
c +
a = 0.8 c = 6.49 ft
For Level 1, the unfactored loads are:
P = 618 kips
M = 30,600 ftkips
=0.8 ' =1,152 kips
Cmbulb fmAbulb
=0.8 ' (10.5 in.)(4.49 ft × 12)=1131 kips
Cmwall fm
FEMA 451, NEHRP Recommended Provisions: Design Examples
9116
The distributed wall rebar that is near the neutral axis is divided between tension and compression, and
therefore it will not have much effect on the result of this check, so it will be neglected.
Cs1 = (60 ksi)(16 × 1.00 in.2) = 960 kips
Ts1 = (16 × 1.00 in.2)(75 ksi) = 1,200 kips
Ts2 = (4 × 0.31 in.2)(75 ksi) = 93 kips
P = 618 kips (unfactored dead load)
3C > 3P + 3T
1152 +1131 + 960 > 618 + 1200 + 93
3,243 kips > 1,818 kips
There is more compression capacity than tension capacity, so a ductile failure condition governs.
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing
the ductility check (factored loads).]
9.3.3.5.3 Axial and Flexural Strength Upper Levels
Examine the strength of Wall D at Level 7.
Pumin = 179 kips + factored weight of ½ of 7th story wall = 179 + (0.7)(11.4) =190 kips
Mu = 9,800 ftkips
This is a point, however, where some of the reinforcement in the lower wall will be terminated. Although
not required by the Provisions, most design standards require the longitudinal reinforcement to be
extended a distance d beyond the point where it could theoretically be terminated. (The ASD chapter of
ACI 530 has such a requirement.) Therefore, the reinforcement at level 6 (7th floor) should be capable of
resisting the moment d below. d is approximately three stories for this wall, therefore, take Mu = 19,400
ftkip (and P = 303 kip) from Level 3.
Try eight #9 in each bulb and vertical #5 bars at 4 ft on center in the wall. Refer to Figure 9.38 for the
placement of the bulb reinforcement.
Chapter 9, Masonry
9117
8"
1'8"
1'8"
Figure 9.38 Bulb reinforcement at upper levels
(1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm).
For evaluating the capacity of the wall, a fPn  fMn curve will be developed to represent the wall strength
envelope for Level 7. The demands (Pu and Mu determined above) will then be compared to this curve.
Several cases will be analyzed and their results used in plotting the fPn  fMn curve. Refer to Figure. 9.3
9 for notation and dimensions.
FEMA 451, NEHRP Recommended Provisions: Design Examples
9118
T
N.A.
C
m = 0.0035
m bulb Cm wall
s
Cs
y = 0.00207
Balance
Case
P = 0 Case
N.A.
Cm bulb
m = 0.0035
Ts
y = 0.00207
y = 0.00207
1.67' 1.67'
33.00'
29.67'
b e = 2.96"
a = 1.16'
c = 1.45'
1.67' 14.50'
16.17'
c = 20.21'
15.67'
M
P
S Ts2
+ Cs
Figure 9.39 Strength of Wall D at Level 7 (1.0 ft = 0.3048 m)
Case 1 (P = 0)
Tension forces:
Ts1 = (8 bars)(1.00 in.2)(60 ksi) = 480 kips
Ts2 = (7 bars)(0.31 in.2)(60 ksi) = 130 kips
Equilibrium:
SC = ST + P
Chapter 9, Masonry
9119
SC = 480 + 130 + 0 = 610 kips
Assume bars closest to compression face yield:
SC = Cs + Cm
Cs = (3 bars)(1.00 in.2)(60 ksi) = 180 kips
Cm = 610  180 = 430 kips
Locate equivalent stress block and neutral axis:
430 = ff’m(20 in.)a = (0.8)(3 ksi)(20 in.)a
a = 8.96 in. = 0.75 ft
c = a/0.8 = 0.75/0.8 = 0.93 ft = 11.2 in.
Verify strain in compression steel:
At the outside layer, e = (0.0035)(7.2 in. / 11.2 in.) = 0.0023 > yield,
At the central layer, e = (0.0035)(1.2 in. / 11.2 in.) = 0.0004 => fs = 11 ksi
Resultant moment:
SMcl = 0:
Mn = (430 kips)(16.5 ft 0.75 ft/2) + (180 kips)(16.5  0.33 ft) + (480 kips)(16.5  0.83 ft)
+ (130 kips)(0 ft.) = 17,370 ftkips
fMn = (0.85)(17,280) = 14,760 ftkips
Case 2 (Intermediate)
Assume the neutral axis at the face of the bulb, c = 1.67 ft
a = 0.8c = 1.33 ft. = 16 in.
Cm = (2.4 ksi)(16 in.)(20 in.) = 768 kip
For the compression steel, it is necessary to compute the strains:
At the outside layer, e = (0.0035)(16 in. / 20 in.) = 0.0028 > yield
At the central layer, e = (0.0035)(10 in. / 20 in.) = 0.00175 => fs = 50 ksi
At the inside layer, e = (0.0035)(4 in. / 20 in.) = 0.0007 => fs = 20 ksi
Cs = (3.0 x 60 ksi + 2.0 x 50 ksi + 3.0 x 20 ksi) = 340 kips
Ts1 = (8 bars)(1.00 in.2)(60 ksi) = 480 kips
Ts2 = (7 bars)(0.31 in.2)(60 ksi) = 130 kips
Pn = 768 + 340  480 130 = 498 kips
fPn = (0.85)(498) = 423 kips
Mn = (768 kips)(16.5 ft 1.33 ft/2) + (340 + 480 kips)(16.5  1.67/2 ft) + (130 kips)(0 ft)
= 25,010 ftkips
fMn = (0.85)(25,010) = 21,260 ftkips
At P = 303 kips, fMn = 19,990 ft.kips by interpolation (exceeds 19,400 ft.kips, OK)
FEMA 451, NEHRP Recommended Provisions: Design Examples
9120
Case 3 (Balanced Case):
0.0035 (32.17 ft) = 20.21 ft
0.0035 0.00207
. .
. .
.. ..
=
+ c
a = (0.8)c = 16.17 ft
be = effective width of wall
be = [(2)(1.3125 in.)(29.67 ft x 12) + (8 cells)(15 in.2/cell)] = 2.96 in./ft
Cm bulb = 960 kips
Cm wall = (0.8)(3 ksi)(2.96 in.)(14.17 ft x 12) = 101 kips
Cs = 480 kips
SC = (960 + 101 + 480) = 1,541 kips
Ts = 480 kips, ignoring the distributed #5 bars
SFy = 0
Pn = GC  GT = 1,541  480 = 1,061 kips
fPn = (0.85)(1,061) = 902 kips
SMcl = 0
Mu = (480 +960 kips)(15.67 ft) + (101 kips)(7.42 ft) + (480 kips)(15.67 ft) = 30,830 ftkips
fMn = (0.85)(30,830) = 26,210 ftkips
Chapter 9, Masonry
9121
The ductility check is performed similar to that for the wall at Level 1. See Figure 9.311 and the
following calculations.
20,000 ftkips 30,000 ftkips
500 kips
1,000 kips
379 kips
179 kips
fP
fM
M = 9,800 ftkips
Balance
(26,210 ftkips, 902 kips)
Simplified fP  fM curve
10,000 ftkips
P = 0 Case
(14,760 ftkips, 0 kips)
Intermediate
(21,260 ftkips, 423 kips)
Extension of demand
a distance "d"
(19,400 ftkips, 303 kips)
n
n
fMn
fPn
u u
u
Figure 9.310 fP11  fM11 Diagram for Level 7 (1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm).
FEMA 451, NEHRP Recommended Provisions: Design Examples
9122
1.67'
33.00'
1.67'
24.63'
P
8.37'
s = 5
= 0.0103
1.67'
25.2 ksi
1.25 fy =
Ts1
5.03'
6.70'
8.37'
0.8f 'm =
1.67'
Cs1
m= 0.0035
c
2.4 ksi
Cm wall
Cm bulb
N.A.
y
102 ksi
300 ksi
5fy =
24.63'
0.84'
N.A.
0.84'
75 ksi
21.9 ksi
fy = 60 ksi
Cs2
Ts2
Ts3
Cs3
a
0.67'
1.67'
c
Figure 9.311 Ductility check for Wall D, Level 7 (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa)
For Level 7, the unfactored loads are:
P = 255 kips
M = 11,600 ftkips
Chapter 9, Masonry
9123
Cmb = 0.8f’mAb = 0.8(3.0 ksi)(400 in.2) = 960 kips
Cmw = 0.8f’mAw = 0.8(3.0 ksi)[2(1.3125 in.)(5.03 ft.)(12 in/ft) + (1 cell)(25.6 in.2/cell)]
= 442 kips
Cs1 = (60 ksi)(8× 1.00 in.2) = 480 kips
Cs2 = (60 ksi)(0.31 in.2) = 19 kips
Ts1 = (8 × 1.00 in.2)(75 ksi) = 600 kips
Ts2 = (4 × 0.31 in.2)(75 ksi ) = 116 kips
Ts3 = (0.31 in.2)(25.2 ksi) = 7 kips
P = 255 kips (unfactored dead load)
GC > P + GT
960 + 442 + 480 + 19 > 255 + 600 + 116 + 7 ?
1901 kips > 978 kips OK
There is more compression capacity than tension capacity, so a ductile failure condition governs.
9.3.3.5.4 Shear Strength
The first step is to determine the net area, An, for Wall D. The definition of An in the Provisions, however,
does not explicitly address bulbs or flanges at the ends of walls. Following an analogy with reinforced
concrete design, the area is taken as the thickness of the web of the wall times the overall length. In
partially grouted walls this is not extended to the point of ignoring the grouted cores because the
implication is that grouted cores are intended to be included. (However, if the spacing of bond beams
greatly exceeds the spacing of grouted cores, even that assumption might be questionable.)
For Levels 1 through 6:
An = (10.5 in.(33 ft × 12))= 4,158 in.2
For Levels 7 through 12 (using 8 x 8 x 12 clay brick units):
An = (2)(1.3125 in.)(12 in.)(33 ft) + (7 + 2 × 3 cells)(25.6 in.2 /cell with adjacent webs)
= 1,372 in.2
Shear strength is determined as described in Sec. 9.2 using Provisions Eq. 11.7.2.1 [ACI 520, Sec. 3.13]
and Provisions Eq. 11.7.3.11 [ACI 530, Eq. 318], respectively:
Vu # fVn
Vn = Vm + Vs
For Levels 1 through 6 using Provisions Eq. 11.7.3.13 where x > 1.0 :
x
M
V d
(max)=4 ' = (4)( 2,500)(4,806) = 961 kips (P3) n mn V fA
FEMA 451, NEHRP Recommended Provisions: Design Examples
9124
For Levels 7 through 12 where varies from 0.30 to 1.14: x
x
M
V d
Vn (max) varies from 5.87 m'n to 4 m'n f A f A
Therefore, Vn (max) varies from 5.87 3,000(1836) = 590 kips to 4 3,000(1836) = 402 kips
depending on the value of . The masonry shear strength is computed as: x
x
M
V d
m= 4  1.75 nm0.25
V MAf P
Vd
... ... ...... ' +
The shear strength of Wall D, based on the aforementioned formulas and the strength reduction factor of
f = 0.8 for shear from Provisions Table 11.5.3 [ACI 530, Sec. 3.1.4.3] , is summarized in Table 9.38. Vx
and Mx in this table are values from Table 9.32 multiplied by 0.148 (the portion of direct and torsional
shear assigned to Wall D). P is the dead load of the roof or floor multiplied by the tributary area for Wall
D, and d is the wall length, not height (d = 32.67 ft for Wall D).
The demand shear, Vu, is found by amplifying the loads to a level that produces a moment of 125 percent
of the nominal flexural strength at the base of the wall. Given the basic flexural demand of 30,600 ftkips,
a design resistance of 33,840 ftkips, f = 0.85, and the 1.25 factor, the overall amplification of design
load is 1.63.
Table 9.38 Shear Strength for Wall D
Level Vx/wall
(kips)
Mx/wall
(ftkips)
Mx/Vxd 1.63Vx fVn max
(kips)
fVm
(kips)
P
(kips)
fVm
(kips)
Req’d fVs
(kips)
12 49 500 0.309 80 351.2 OK 37 215 
11 102 1500 0.446 166 329.3 OK 80 210 
10 149 3000 0.610 243 303.0 OK 124 201 42
9 191 4900 0.777 311 276.2 NG 168 192 119
8 228 7200 0.957 372 247.4 NG 212 182 190
7 260 9800 1.142 424 240.5 NG 255 186 238
6 292 12700 1.318 476 665.3 OK 308 436 40
5 323 15900 1.492 526 665.3 OK 370 448 78
4 347 19400 1.694 566 665.3 OK 432 461 106
3 364 23000 1.915 593 665.3 OK 494 473 120
2 375 26800 2.166 611 665.3 OK 556 485 126
1 380 30600 2.440 619 665.3 OK 618 498 121
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
Note that 1.63 Vx exceeds Vn max at Levels 7, 8, and 9. The next most economical solution appears to be to
add grout to increase An and, therefore, both Vm and Vn max. Check Level 7 using solid grout:
An = (7.5 in.)( 33 ft)(12 in./ft) = 2970 in.2
(max) 4 (4)( 2,500)(2,970) = 650 kips n mn V = f'A=
fVn max = 0.8(650) = 520 kips > 424 kips OK
Chapter 9, Masonry
9125
fVm =(0.80)..(4.01.75(1.0))2,970 3000 /1000+0.25(255)..=344 kips
fVs = Vu  fVm = 424  344 = 80 kips
Check minimum reinforcement for capacity. With vertical #5 at 48 in., a reinforcement ratio of 0.00086
is provided. Thus the horizontal reinforcement must exceed (0.0020  0.00086)(7.5 in.)(12 in.) = 0.1025
in.2/ft. With story heights of 10 ft., bond beams at 40 in. on center are convenient, which would require
0.34 in.2 Therefore, for 2  #4 at 40 in. on center:
fVs = 0.80(0.5)(0.4 / 40) (60 ksi)(33 ft.)(12 in./ft.) = 95 kips > 80 kips OK
The largest demand for fVx in the lower levels is 126 kips at level 2. As explained in the design of the
lower level walls for flexural and axial loads (Sec. 9.3.3.5.1), horizontal #5 at 22 in. are required to satisfy
minimum reinforcement. Given the story height, check for horizontal #5 at 20 in.:
fVs = 0.8(0.5)(0.31 / 20) (60 ksi)(33 ft.)(12 in./ft.) = 147 kips > 126 kips OK
In summary, for shear it is necessary to grout the hollow units at story 7 solid, and to add some grout at
stories 8 and 9. Horizontal reinforcement is 2  #4 in bond beams at 40 in. on center in the upper stories
and one #5 at 20 in. on center in the grouted cavity of the lower stories.
9.3.4 Deflections
The calculations for deflection involve many variables and assumptions, and it must be recognized that
any calculation of deflection is approximate at best.
Deflections are to be calculated and compared with the prescribed limits set forth by Provisions Table
5.2.8 [Table 4.51]. Deformation requirements for masonry structures are discussed in Provisions Sec.
11.5.4.
The following procedure will be used for calculating deflections:
1. Determine if the wall at each story will crack by comparing Mx (see Table 9.36) to Mcr where
= ( + )
cr r umin M S f P A
2. If Mcr < Mx, then use cracked moment of inertia and Provisions Eq. 11.5.4.3.
3. If Mcr > Mx, then use In = Ig for moment of inertia of wall.
4. Compute deflection for each level.
5. dmax = 3 story drift
[The specific procedures for computing deflection of shear walls have been removed from the 2003
Provisions. ACI 530 does not contain the corresponding provisions in the text, however, the commentary
contains a discussion and equations that are similar to the procedures in the 2000 Provisions. Based on
ACI 530 Sec. 1.13.3.2, the maximum drift for all masonry structures is 0.007 times the story height.
Thus, there appears to be a conflict between ACI 530 and 2003 Provisions Table 4.51.]
For the upper levels (the additional grout required for shear strength is not considered here):
be = effective masonry wall width
be = [(2 × 1.3125 in.)(356) + (7 cells)(15 in.2/cell)]/(356) = 2.92 in.
A = Awall + 2Abulb = (2.92 in.)(356 in.) + (2)(400 in.2) = 1,840 in.2
FEMA 451, NEHRP Recommended Provisions: Design Examples
9126
[Note that by adopting ACI 530 in the 2003 Provisions, Em = 900f’m per ACI 530 Sec. 1.8.2.2.1]
Per Provisions Eq. 11.3.10.2 [ACI 530, 1.8.2.2]:
E = 750 f’m = 2,250 ksi (n = 12.89)
Ig = Iwall + I bulb
3 2
6 4 (2.96)(356) 376
(2 bulbs)(20 20) 39.4 10 in.
12 2
= + × . .= ×
. .
. . g I
S = Ig/c = 39.4 × 106 /(198) = 199,000 in.3
fr = 0.250 ksi
= 1.00D (see Table 9.36.) Pumin
For the lower levels:
A = A wall + 2Abulb = (10.5 in.)(348 in.) + (2)(576 in.2) = 4,806 in.2
E = 750 f’‘m = 1,875 ksi (n = 15.47)
I = Iwall + I bulb
3 2
6 4 (10.5)(348) 372
(2 bulbs)(24 24) 76.7 10 in.
12 2
= + × . .= ×
. .
. . g I
S = Ig / c = 76.7 × 106 /(198) = 387,000 in.3
fr = 0.250 ksi
= 1.00D (see Table 9.36.) Pumin
Table 9.39 provides a summary of these calculations.
Table 9.39 Cracked Wall Determination
Level Pumin
(kips)
Mcr
(ftkips)
Mx
(ftkips)
Status
12
11
10
987654321
37
80
124
168
212
255
308
370
432
494
556
618
7,620
8,820
8,950
9,620
10,300
11,000
15,400
16,000
16,700
17,300
18,000
18,600
500
1,500
3,000
4,900
7,200
9,800
12,700
15,900
19,400
23,000
26,800
30,600
uncracked
uncracked
uncracked
uncracked
uncracked
uncracked
uncracked
uncracked
cracked
cracked
cracked
cracked
1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
For the uncracked walls at the upper levels:
Chapter 9, Masonry
9127
In = Ig = 39.4 × 106 in.4
For the uncracked walls at the lower levels:
In = Ig = 76.7 × 106 in.4
For the cracked walls at the lower levels, the determination of Icr will be for the load combination of 1.0D
+ 0.5L. The 0.5L represents an average condition of live load. Making reference to Figure 9.36, it can
be observed that at this level of Pu, the point on the fPn  fMn curve is near the “intermediate point”
previously determined. This is where c = 3.0 ft. (The actual c dimension will be very close to 3.0 ft).
For this case, and referring to Figure 9.35, the cracked moment of inertia is:
I cr = I bulb + I wall + I nAs
= [244 + (24 × 24)(24)2 ] + [10.5 × 123/3 ] + [(15.47 × 16 in.2)(29 ft × 12)2]
= 30.3 × 106 in4.
Note that 98.9 percent of the value comes from one term: the reinforcement in the tension bulb. If the
distributed #5 bars is added to this computation, the value becomes 31.5 × 106 in.4
With the other masonry examples, the interpolation between gross and cracked section properties was
used. The application of that is less clear here, where the properties step at midheight, so two analyses are
performed. First, each story is considered to be cracked or uncracked. Second is the author’s
interpretation of the effective moment of inertia equation as:
For all the cracked walls (Provisions Eq. 11.5.4.3 [ACI 530, Commentary Sec. 3.1.5.3]):
3 3
= + 1 =
. . . . ..
. . . . ..
. . .. . ...
cr cr
eff n cr n
a a
M M
I I I I
M M
( ) ( ) 3 3
618, 600 6 18,600 64
1
30,600 30,600
=76.7 10 +30.6 10  41.0 10 in.
× .. .. × .. .. ....= ×
. . . . ..
eff I
The entire 12story Wall D will be treated as a stepped, vertical masonry cantilever shear wall. For the
lower step (Levels 16), Ieff = 41.0 × 106 in.4 Even though the upper walls are uncracked, Ieff of the upper
step (Levels 712), will be In reduced in the same proportion as the lower levels:
(upper levels)
6
6 6
6
(39.4 10 ) 41.0 10 = 21.1 10
76.7 10
. × .
= × .. × .. × eff I
Both the deflections and the fundamental period can now be found. Two RISA 2D analyses were run,
and the deflections shown in Table 9.310 were obtained. The deflection from the RISA 2D analysis at
each level is multiplied by Cd (= 3.5) to determine the inelastic deflection at each level. From these, the
story drift, ., at each level can be found.
The periods shown in the table validate the period of T = 0.75 sec previously used to determine the base
shear in Sec. 9.3.3.2.
Table 9.310 Deflections for ELF Analysis (inches)
Using gross and cracked properties, Using effective moment of inertia
FEMA 451, NEHRP Recommended Provisions: Design Examples
9128
story by story (T = 0.798 sec) (T = 0.755 sec)
Level Elastic Total Drift Ratio Elastic Total Drift Ratio
12 3.40 11.90 3.14 10.99
11 3.04 10.65 1.25 2.77 9.71 1.28
10 2.68 9.38 1.27 1.01 2.40 8.41 1.30 1.01
9 2.32 8.10 1.27 1.00 2.04 7.13 1.28 0.99
8 1.95 6.84 1.26 0.99 1.68 5.87 1.26 0.98
7 1.60 5.60 1.24 0.98 1.34 4.68 1.19 0.95
6 1.26 4.41 1.19 0.96 1.02 3.58 1.10 0.92
5 0.95 3.34 1.07 0.90 0.76 2.65 0.93 0.85
4 0.66 2.31 1.03 0.96 0.52 1.82 0.83 0.90
3 0.40 1.40 0.91 0.88 0.32 1.11 0.71 0.85
2 0.20 0.68 0.71 0.78 0.16 0.55 0.56 0.79
1 0.06 0.20 0.48 0.67 0.05 0.17 0.38 0.68
0 0 0.00 0.20 0.42 0 0.00 0.17 0.44
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm
The two methods give comparable results. The maximum building deflection is compared to the
maximum deflection permitted by Provisions Sec. 11.5.4.1.1 as follows:
Cddmax = 11.90 in. < 14.4 in. = 0.1hn OK
The maximum story drift occurs at Story 11 and is compared to the maximum story drift permitted by
Provisions Table 5.2.8 [Table 4.51] as follows:
. = 1.30 in. > 1.20 in. = 0.01hsx NG
Although this indicates a failure to satisfy the Provisions, in the author’s opinion the drift is satisfactory
for two reasons. First the MRS analysis shows smaller drifts (Table 9.311) that are within the criteria.
On a more fundamental level, however, the authors believe the basic check for drift of a masonry wall is
performed according to Provisions Sec. 11.5.4.1.1, which applies only to the total displacement at the top
of the wall, and that the story drift for any particular story is more properly related to the values for nonmasonry
buildings. That limit is 0.020 hsx, or 2.4 in. per story. For a building with a torsional
irregularity, Provisions Sec. 5.4.6.1 [Sec. 4.5.1] requires that the story drift be checked at the plan
location with the largest drift, which would be a corner for this building. That limit is satisfied for this
building, both by ELF and MRS analyses.
Chapter 9, Masonry
9129
Table 9.311 Displacements from Modal Analysis, inches
At corner of floor plate with
maximum displacements. Story
drift would be pertinent,
although not at 0.010
At wall with maximum inplane
displacement. Roof limit for
masonry would be pertinent.
Level Elastic Total Approx.
Drift
Elastic Total Approx.
Drift
12 2.48 8.67 1.91 6.70
11 2.20 7.69 0.98 1.69 5.90 0.79
10 1.91 6.70 0.99 1.48 5.17 0.74
9 1.63 5.70 1.00 1.26 4.40 0.77
8 1.35 4.72 0.98 1.04 3.64 0.76
7 1.08 3.77 0.95 0.83 2.91 0.73
6 0.83 2.90 0.88 0.64 2.23 0.68
5 0.62 2.16 0.74 0.48 1.67 0.57
4 0.43 1.50 0.66 0.33 1.16 0.51
3 0.27 0.93 0.57 0.21 0.72 0.44
2 0.14 0.48 0.46 0.11 0.37 0.35
1 0.05 0.16 0.32 0.04 0.12 0.25
0 0.00 0.00 0.16 0.00 0.00 0.12
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm.
The drifts in Table 9.311 are not the true modal drifts. The values are computed from the modal sum
maximum displacements, rather than being a modal sum of drifts in each mode. The values in the table
are less than the true value.
Both tables also confirm that the change in stiffness at midheight does not produce a stiffness irregularity.
Provisions Sec. 5.2.3.3, Exception 1 [Sec. 4.3.2.3, Exception 1], clarifies that if the drift in a story never
exceeds 130 percent of the drift in the story above, then there is no vertical stiffness irregularity. Note
that the inverse does not apply; even though the drift in Story 2 is more than double that in Story 1, it
does not constitute a stiffness irregularity.
9.3.5 OutofPlane Forces
Provisions Sec 5.2.6.2.7 [Sec. 4.6.1.3] states that the bearing walls shall be designed for outofplane
loads equal to:
w = 0.40 SDS Wc $ 0.1Wc
w = (0.40)(1.00)(114 psf) = 45.6 psf $ 0.1Wc
Therefore, w = 45.6 psf. Outofplane bending, using the strength design method for masonry, for a load
of 45.6 psf acting on a 10 ft story height is approximated as 456 ft.lb. per linear ft of wall. This
compares to a computed strength of the wall of 1,600 in.lb per linear foot of wall, considering only the
#5 bars at 4 ft on center. Thus the wall is loaded to 28.5 percent of its capacity in flexure in the outofplane
direction. The upper wall has the same reinforcement, about 42 percent of the load and about 71
percent of the thickness. Therefore, it will be loaded to a smaller fraction of its capacity. (Refer to
Example 9.1 for a more detailed discussion of strength design of masonry walls, including the Pdelta
effect.)
FEMA 451, NEHRP Recommended Provisions: Design Examples
9130
9.3.6 Orthogonal Effects
In accordance with Provisions Sec. 5.2.5.2.2 [Sec. 4.4.2.3], orthogonal interaction effects must be
considered for buildings in Seismic Design Category D when the ELF procedure is used. Any outofplane
effect on the heavily reinforced bulbs is negligible compared to the inplane effect, so orthogonal
effects on the bulbs need not be considered further. Considering only the #5 bars and the load
combination of 100 percent of inplane load plus 30 percent of the outofplane load, yields a result that
0.3(0.285), or 8.6 percent of the capacity of the #5 bars is not available for inplane resistance. Given that
the #5 bars contribute about 12 percent to the tension resistance (130 kips, vs 960 kips for the bulb
reinforcement), the overall effect is a change of about 1 percent in inplane resistance, which is negligible.
This completes the design of the transverse Wall D.
9.3.7 Wall Anchorage
The anchorage for the bearing walls must be designed for the force, Fp, determined in accordance with
Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3] as:
Fp = 0.4SDSWc = (0.4)(1.00)(10 ft)(114 psf) = 456 lb/ft
Minimum force = 0.10Wc = (0.10)(10 ft)(114 psf) = 114 lb/ft
Provisions Sec. 5.2.6.3.2 [Sec. 4.6.2.1] references Provisions Sec. 6.1.3 [Sec. 6.2.2] for anchorage of
walls where diaphragms are not flexible. For the lower wall:
0.4 (1 2 / ) 0.4(1.0)(1.0)(10ft. 114 psf )(1 2(0.5)) 364 lb./ft.
/ 2.5/1.0
p DS p
p
p p
a S W
F zh
R I
×
= + = + =
Therefore, design for 456 lb/ft. For a 2 ft6 in. joist spacing, the anchorage force at each joist is Fp =
1,140 lb.
Refer to Figure 9.312 for the connection detail. A 3/16in. fillet, weld 2 in. long on each side of the joist
seat to its bearing plate will be more than sufficient. Two1/2 in.diameter headed anchor studs on the
bottom of the bearing plate also will be more than sufficient to transfer 4,560 lb into the wall.
9.3.8 Diaphragm Strength
See Example 7.1 for a more detailed discussion on the design of horizontal diaphragms.
To compute the story force associated with the diaphragm on each level, use Provisions Eq. 5.2.6.44 [Eq.
4.62]:
=
=
=
S
S
n
i
i x
px n px
i
i x
F
F w
w
Chapter 9, Masonry
9131
The results are shown in Table 9.312. Note that wi is approximately the same as wpx for this case, the
only difference being the weight of the walls perpendicular to the force direction, so the wi values were
used for both.
Table 9.312 Diaphragm Seismic Forces
Level wi
(kips)
Fi
(kips)
Fpx
(kips)
12
11
10
987654321
768
917
917
917
917
917
1109
1300
1300
1300
1300
1300
329
357
320
284
249
214
218
208
162
117
74
34
329
373
355
336
318
301
338
365
336
309
282
258
1.0 kip = 4.45 kN
The maximum story force is 373 kips. Therefore, use 373 kips/152 ft = 2.45 kips/ft in the transverse
direction. The shear in the diaphragm is shown in Figure 9.313b. The reaction, R, at each wall pair is
373/4 = 93.25 kips. The diaphragm force at each wall pair is 93.25 kips/(2 × 33 ft) = 1.41 kips/ft.
Joist to
plate
Slab reinforcement at
perimeter of building
is continuous through
the wall. Use
#4 x 4'0" at 16" o.c.
dowels elsewhere
(center on wall).
3
16 2"
Plate 3
8"x 6" x 0'10"
at each joist (2'6" o.c.)
with (2) 1
2" dia. x 6" H.A.S.
Figure 9.312 Floor anchorage to wall (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m).
FEMA 451, NEHRP Recommended Provisions: Design Examples
9132
The maximum diaphragm shear stress is v = V/td = 1410 plf/(2.5 in.)(12 in.) = 47 psi. This compares to
an allowable shear of
fvc=(0.85)(2)fc'=(0.85)(2) 3,000=93 psi
for 3,000 psi concrete. Thus, no shear reinforcing is necessary. Provide . = 0.0018 as minimum
reinforcement, so As = 0.054 in.2/ft. Use WWF 6 × 62.9/2.9, which has As = 0.058 in.2/ft.
The moment in the diaphragm is shown in Figure 9.313c. The maximum moment is 2,460 ftkips.
Perimeter reinforcement in the diaphragm is determined from:
T = M/d = (2,460 ftkips)/(72 ft) = 34.1 kips
As = T/fFy = 34.1 kips / (0.85)(60 ksi) = 0.67 in.2
Boundary elements of diaphragms may also serve as collectors. The collector force is not usually the
same as the chord force. Provisions Sec. 5.2.6.4.1 [Sec. 4.6.2.2] requires that collector forces be
amplified by O0. Collector elements are required in this diaphragm for the longitudinal direction. A
similar design problem is illustrated in Chapter 7 of this volume. Where reinforcing steel withing a
topping slab is used for chords or collectors, ACI 318, Sec. 21.9.8 (2002 edition) imposes special spacing
and cover requirements. Given the thin slab in this building, the chord reinforcement will have to be
limited to bars with couplers at the splices or a thickened edge will be required.
Chapter 9, Masonry
9133
Figure 9.313 Shears and moments for diaphragm
(1.0 kip/ft = 14.6 kN/m, 1.0 kip = 4.45 kN, 1.0 kipft = 1.36kNm)
Fp = 2.45 kips/ft
2,460 ftkips
93.25 kips
a) Load
b) Shear
c) Moment
Figure 9.313.
101
10
WOOD DESIGN
Peter W. Somers, P.E. and Michael Valley, P.E.
This chapter examines the design of a variety of wood building elements. Section10.1 features a BSSC
trial design prepared in the early 1980s as a starting point. Section 10.2 completes the roof diaphragm
design for the building featured in Section 9.1. In both cases, only those portions of the designs necessary
to illustrate specific points are included.
Typically, the weak links in wood systems are the connections, but the desired ductility must be
developed by means of these connections. Wood members have some ductility in compression
(particularly perpendicular to grain) but little in tension. Nailed plywood shear panels develop
considerable ductility through yielding of nails and crushing of wood adjacent to nails. Because wood
structures are composed of many elements that must act as a whole, the connections must be considered
carefully to ensure that the load path is complete. “Tying the structure together ,” which is as much an art
as a science, is essential to good earthquakeresistant construction.
Wood elements often are used in lowrise masonry and concrete buildings. The same basic principles
apply to the design of wood elements, but certain aspects of the design (for example, walltodiaphragm
anchorage) are more critical in mixed systems than in allwood construction.
Wood structural panel sheathing is referred to as “plywood” in this chapter. As referenced in the 2000
NEHRP Recommended Provisions, wood structural panel sheathing includes plywood and other products,
such as orientedstrand board (OSB), that conform to the materials standards of Chapter 12. According to
Provisions Chapter 12, panel materials other than wood structural panel sheathing do not have a
recognized capacity for seismicforce resistance in engineered construction.
In addition to the 2000 NEHRP Recommended Provisions and Commentary (hereafter, the Provisions and
Commentary), the documents edited below are either referenced directly, or are useful design aids for
wood construction.
AF&PA Manual American Forest & Paper Association 1996. Manual for Engineered Wood
Construction (LRFD), including supplements, guidelines, and ASCE 1695,
AF&PA.
[AF&PA Wind American Forest and Paper Association. 2001. Special Design Provisions for
& Seismic Wind and Seismic, ADS/LRFD Supplement. AF&PA.]
FEMA 451, NEHRP Recommended Provisions: Design Examples
102
ANSI/AITC A190.1 American National Standards Institute/American Institute of Timber
Construction. 1992. American National Standard for Wood Products:
Structural GluedLaminated Timber, A190.1. AITC.
APA PDS American Plywood Association. 1998. Plywood Design Specification, APA.
APA 138 American Plywood Association. 1998. Plywood Diaphragms, APA Research
Report 138. APA.
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for
Buildings and Other Structures. ASCE.
ASCE 16 American Society of Civil Engineers. 1995. Standard for Load and Resistance
Factor Design (LRFD) for Engineered Wood Construction. ASCE.
UBC Std 232 International Conference of Building Officials. 1997. UBC Standard 232
Construction and Industrial Plywood, Uniform Building Code. ICBO.
Roark Roark, Raymond. 1985. Formulas for Stress and Strain, 4th Ed. McGrawHill.
USGS CDROM United States Geological Survey. 1996. Seismic Design Parameters, CDROM.
USGS.
WWPA Rules Western Wood Products Association. 1991. Western Lumber Grading Rules.
WWPA.
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect
changes made to the 2003 Provisions. Annotations within brackets,[ ], indicated both organizational
changes (as a result of a reformat of all the chapters of the 2003 Provisions) and substantive technical
changes to the 2003 Provisions and it’s primary reference documents. While the general concepts of the
changes are described, the design examples and calculations have not been revised to reflect the changes
to the 2003 Provisions.
The most significant change to the wood chapter in the 2003 Provisions is the incorporation by reference
of the AF&PA, ADS/LRFD Supplement, Special Design Provisions for Wind and Seismic for design of
the engineered wood construction. A significant portion of the 2003 Provisions Chapter 12, including the
diaphragm and shear wall tables, has been replaced by a reference to this document. This updated
chapter, however, does not result in significant technical changes , as the Supplement, (referred to herein
as AF&PA Wind&Seismic) is in substantial agreement with the 2000 Provisions. There are, however,
some changes to the provisions for perforated shear walls, which are covered in Section 10.1
Some general technical changes in the 2003 Provisions that relate to the calculations and/or design in this
chapter include updated seismic hazard maps, changes to the Seismic Design Category classification for
short period structures, revisions to the redundancy requirements, revisions to the wall anchorage design
requirement for flexible diaphragms, and a new “Simplified Design Procedure” that could be applicable
to the examples in this chapter.
Where the affect the design examples in this chapter, other significant changes to the 2003 Provisions and
primary reference documents are noted. However, some minor changes to the 2003 Provisions and the
reference documents may not be noted.
Chapter 10, Wood Design
103
10.1 THREESTORY WOOD APARTMENT BUILDING; SEATTLE, WASHINGTON
This example features a wood frame building with plywood diaphragms and shear walls. It is based on a
BSSC trial design by Bruce C. Olsen, Structural Engineer, Seattle, Washington.
10.1.1 BUILDING DESCRIPTION
This threestory, wood frame apartment building has plywood floor diaphragms and shear walls. The
building has a doubleloaded central corridor. Figure 10.11 shows a typical floor plan, and Figure 10.12
shows a longitudinal section and elevation. The building is located in a neighborhood a few miles north
of downtown Seattle. The site coordinates for determining the seismic design parameters are 47.69° N,
122.32° W.
The shear walls in the longitudinal direction are located on the exterior faces of the building and along the
corridor. (In previous versions of this volume of design examples, the corridor walls were gypsum
wallboard sheathed shear walls; however, gypsum wallboard sheathing, is no longer recognized for
engineered design of shear walls per Provisions Sec. 12.3.5. Therefore, plywood shear walls are provided
at the corridors.) The entire solid (nonglazed) area of exterior walls plywood sheathing, but only a
portion of the corridor walls will require sheathing. For the purposes of this example, assume that each
corridor wall will have a net of 55 ft of plywood (the reason for this is explained later). In the transverse
direction, the end walls and one line of interior shear walls provide lateral resistance. (In previous
versions of this example, only the end walls were shear walls. The interior walls now are required for
control of diaphragm deflections given the increased seismic ground motion design parameters for the
Seattle area.)
The floor and roof systems consist of wood joists supported on bearing walls at the perimeter of the
building, the corridor lines, plus one postandbeam line running through each bank of apartments.
Exterior walls are framed with 2×6 studs for the full height of the building to accommodate insulation.
Interior bearing walls require 2×6 or 3×4 studs on the corridor line up to the second floor and 2×4 studs
above the second floor. Apartment party walls are not load bearing; however, they are double walls and
are constructed of staggered, 2×4 studs at 16 in. on center. Surfaced, dry (seasoned) lumber, is used for
all framing to minimize shrinkage. Floor framing members are assumed to be composed of Douglas
FirLarch material, and wall framing is Hemfir No. 2, as graded by the WWPA. The material and
grading of other framing members associated with the lateral design is as indicated in the example. The
lightweight concrete floor fill is for sound isolation, and is interrupted by the party walls, corridor walls,
and bearing walls.
The building is founded on interior footing pads, continuous strip footings, and grade walls (Figure
10.13). The depth of the footings, and the height of the grade walls, are sufficient to provide crawl space
clearance beneath the first floor.
The building is typical of apartment construction throughout the western United States, and has the
weight necessary to balance potential overturning forces in the transverse direction. If the ground floor
were a slabongrade, however, the resulting shallower grade wall might well require special attention,
due to the possibility of overturning on some of the shear wall units.
FEMA 451, NEHRP Recommended Provisions: Design Examples
104
Typical apartment
partitions
Post &
beam
lines
148'0"
28'0" 8'0"
56'0"
9'0" 9'0"
25'0" 6'0" 25'0"
11
2" Lightweight concrete
over plywood deck on
joists at 16" o.c.
Figure 10.11 Typical floor plan (1.0 ft = 0.3048 m).
148'0"
15'0" 26'0" 30'0" 26'0" 30'0" 13'0"
Sheathed
wall
Glazed
wall
4' 9' 9' 9'
Doors
Corridor
walls Stairs
8'0"
Figure 10.12 Longitudinal section and elevation (1.0 ft = 0.3048 m).
Chapter 10, Wood Design
105
Pad
footings
Continuous footing
and grade wall
Figure 10.13 Foundation plan.
10.1.1.1 Scope
In this example, the structure is designed and detailed for forces acting in the transverse and longitudinal
directions. However, greater attention is paid to the transverse direction, because of diaphragm
flexibility. The example includes the following
1. Development of equivalent static loads, including torsional effects on plywood diaphragms,
2. Design and detailing of transverse plywood walls for shear and overturning moment,
3. Design and detailing of plywood floor and roof diaphragms,
4. Design and detailing of wall and diaphragm chord members,
5. Detailed deflection and Pdelta calculations, and
6. Design and detailing of longitudinal plywood walls using the requirements for perforated shear walls.
[Note that the new “Simplified Design Procedure” contained in 2003 Provisions Simplified Alternate
Chapter 4 as referenced by 2003 Provisions Sec. 4.1.1 is likely to be applicable to this example, subject to
the limitations specified in 2003 Provisions Sec. Alt. 4.1.1.]
10.1.2 BASIC REQUIREMENTS
10.1.2.1 Provisions Parameters
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I
Occupancy Importance Factor, I (Provisions Sec. 1.4 [1.3]) = 1.0
Site Coordinates = 47.69° N, 122.32° W
Short Period Response, SS (USGS CDROM) = 1.34
One Second Period Response, S1 (USGS CDROM) = 0.46
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = D
Seismic Design Category (Provisions Sec. 4.2 [1.4]) = D
SeismicForceResisting System (Provisions Table 5.2.2 [4.31]) = Wood panel shear wall
Response Modification Coefficient, R (Provisions Table 5.2.2 [4.31) = 6.5
System Overstrength Factor, O0 (Provisions Table 5.2.2 [4.31) = 3
Deflection Amplification Factor, Cd (Provisions Table 5.2.2 [4.31) = 4
FEMA 451, NEHRP Recommended Provisions: Design Examples
106
10.1.2.2 Structural Design Criteria
10.1.2.2.1 Ground Motion (Provisions Sec. 4.1.2 [3.3])
Based on the site location, the spectral response acceleration values can be obtained from either the
seismic hazard maps accompanying the Provisions or from the USGS CDROM. For site coordinates
47.69° N, 122.32° W, the USGS CDROM returns short period response, SS = 1.34 and one second period
response, S1 = 0.46.
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package). A CDROM containing the site response parameters based on the 2002 maps is
also available.]
The spectral response factors are then adjusted for the site class (Provisions Sec. 4.1.2.4 [3.5]). For this
example, it is assumed that a site class recommendation was not part of the soils investigation, which
would not be uncommon for this type of construction. When soil properties are not known, Provisions
Sec. 4.1.2.1 [3.5] defaults to Site Class D, provided that soft soils (Site Class E or F) are not expected to
be present at the site (a reasonable assumption for soils sufficient to support a multistory building on
shallow spread footings). The adjusted spectral response acceleration parameters are computed,
according to Provisions Eq. 4.1.2.41 and 4.1.2.42 [3.31 and 3.34], for the short period and one second
period, respectively, as:
SMS = FaSS = 1.0(1.34) = 1.34
SM1 = FvS1 = 1.54(0.46) = 0.71
where Fa and Fv are site coefficients defined in Provisions Tables 4.1.2.4a and 4.1.2.4b [3.31 and 3.32],
respectively. Note that Straight line interpolation was used for Fv.
Finally, the design spectral response acceleration parameters (Provisions Sec. 4.1.2.5 [3.3.3]) are
determined in accordance with Provisions Eq. 4.1.2.51 and 4.1.2.52 [3.33 and 3.34], for the short
period and one second period, respectively, as:
2 2(1.34) 0.89
SDS=3SMS=3 =
2 2(0.71) 0.47
SD1=3SM1=3 =
10.1.2.2.3 Seismic Design Category (Provisions Sec. 4.2 [1.4])
Based on the Seismic Use Group and the design spectral response acceleration parameters, the Seismic
Design Category is assigned to the building based on Provisions Tables 4.2.1a and 4.2.1b [1.41 and 1.4
2]. For this example, the building is assigned Seismic Design Category D.
[Note that the method for assigning seismic design category for short period buildings has been revised in
the 2003 Provisions. If the fundamental period, Ta, is less than 0.8Ts, the period used to determine drift is
less than Ts, and the base shear is computed using 2003 Provisions Eq 5.22, then seismic design category
is assigned using just 2003 Provisions Table 1.41 (rather than the greater of 2003 Provisions Tables 1.4
1 and 1.42). The change does not affect this example.]
Chapter 10, Wood Design
107
148'0"
56'0"
25'0" 25'0"
A
15'0" 30'0" 30'0"
15'0" 30'0" 30'0"
A: Another solution would be to use the full
end for the shear wall, not just 25 feet.
Solid
interior
wall
Solid end
wall
55 feet of net shear wall
length, each side of
corridor, distributed over
length of building.
Perforated
56'0" 84'0" exterior wall
Figure 10.14 Load path and shear walls (1.0 ft = 0.3048 m).
10.1.2.2.4 Load Path (Provisions Sec. 5.2.1 [14.21])
See Figure 10.14. For both directions, the load path for seismic loading consists of plywood floor and
roof diaphragms and plywood shear walls. Because the lightweight concrete floor topping is
discontinuous at each partition and wall, it is not considered to be a structural diaphragm.
10.1.2.2.5 Basic SeismicForceResisting Systems (Provisions Sec. 5.2.2 [4.3])
Building Class (Provisions Table 5.2.2 [4.31]): Bearing Wall System
SeismicForceResisting System (Provisions Table 5.2.2 [4.31]): Light frame walls with shear panels
with R = 6.5, O0 = 3, and Cd = 4 for both directions
10.1.2.2.6 Structure Configuration (Provisions Sec. 5.2.3 [5.3.2])
Diaphragm Flexibility (Provisions Sec. 5.2.3.1 [4.3.2.1]): Rigid (wood panel diaphragm in light frame
structure with structural panels for shear resistance). Provisions Sec. 12.4.1.1 [4.3.2.1]defines a structural
panel diaphragm as flexible, if the maximum diaphragm deformation is more than two times the average
story drift. Due to the central shear wall, this is not expected to be the case in this building.
Plan Irregularity (Provisions Sec. 5.2.3.2 [4.3.2.2]): Since the shear walls are not balanced for loading in
the transverse direction (see Figure 10.14), there will be some torsional response of the system. The
potential for torsional response combined with the rigid diaphragm requires that the building be checked
for a torsional irregularity (Provisions Table 5.2.3.2 [4.32]). This check will be performed following the
initial determination of seismic forces, and the final seismic forces will be modified if required.
Vertical Irregularity (Provisions Sec. 5.2.3.3 [4.3.2.3]): Regular
FEMA 451, NEHRP Recommended Provisions: Design Examples
108
10.12.2.7 Redundancy (Provisions Sec. 5.2.4 [4.3.3])
For Seismic Design Category D, the reliability factor, ., is computed per Provisions Eq. 5.2.4.2 [4.3.3.2].
Since the computation requires more detailed information than is known prior to the design, assume . =
1.0 for the initial analysis and verify later. If, in the engineer's judgement, the initial design appears to
possess relatively few lateral elements, the designer may wish to use an initial . greater than 1.0 (but no
greater than 1.5).
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a shear wall
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear
wall with heighttolengthratio greater than 1.0 would not result in more than a 33 percent reduction in
story strength or create an extreme torsional irregularity. The intent is that the aspect ratio is based on
story height, not total height. Therefore, the redundancy factor would have to be investigated only in the
longitudinal direction where the aspect ratios of the perforated shear walls would be interpreted as having
aspect ratios greater than 1.0 at individual piers. In the longitudinal direct, where the aspect ratio is less
than 1.0, . = 1.0 by default.]
10.1.2.2.8 Analysis Procedure (Provisions Sec. 5.2.5 [4.4.1])
Design in accordance with the equivalent lateral force (ELF) procedure (Provisions Sec. 5.4 [5.2]): No
special requirements. In accordance with Provisions Sec. 5.2.5.2.3 [4.4.2], the structural analysis must
consider the most critical load effect due to application of seismic forces in any direction for structures
assigned to Seismic Design Category D. For the ELF procedure, this requirement is commonly satisfied
by applying 100 percent of the seismic force in one direction, and 30 percent of the seismic force in the
perpendicular direction; as specified in Provisions Sec. 5.2.5.2.2, Item a [4.4.2.2, Item 1]. For a lightframed
shear wall building, with shear walls in two orthogonal directions, the only element affected by
this directional combination would be the design of the shear wall end post and tiedown, located where
the ends of two perpendicular walls intersect. In this example, the requirement only affects the shear wall
intersections at the upper left and lower left corners of Figure 10.14. The directional requirement is
satisfied using a twodimensional analysis in the design of the remainder of the shear wall and diaphragm
elements.
10.1.2.2.9 Design and Detailing Requirements (Provisions Sec. 5.2.6 [4.6])
See Provisions Chapters 7 and 12 for special foundation and wood requirements, respectively. As
discussed in greater detail below, Provisions Sec. 12.2.1, now utilizes Load and Resistance Factor Design
(LRFD) for the design of engineered wood structures. Therefore, the design capacities are consistent with
the strength design demands of Provisions Chapter 5 [4 and 5].
10.1.2.2.10 Combination of Load Effects (Provisions Sec. 5.2.7 [4.2.2])
The basic design load combinations are as stipulated in ASCE 7 and modified by the Provisions
Eq. 5.2.71 and 5.2.72 [4.21 and 4.21]. Seismic load effects according to the Provisions are:
E = .QE + 0.2SDSD
and
E = .QE  0.2SDSD
when seismic and gravity are additive and counteractive, respectively.
Chapter 10, Wood Design
109
For SDS = 0.89 and assuming . = 1.0 (both discussed previously), the design load combinations are as
stipulated in ASCE 7:
1.2D + 1.0E + 0.5L + 0.2S = 1.38D + 1.0QE + 0.5L + 0.2S
and
0.9D  1.0E = 0.72D  1.0QE
10.1.2.2.11 Deflection and Drift Limits (Provisions Sec. 5.2.8 [4.5.1])
Assuming that interior and exterior finishes have not been designed to accommodate story drifts, then the
allowable story drift is (Provisions Table 2.5.8 [4.51]):
.a = 0.020 hsx
where interstory drift is computed from story drift as (Provisions Eq. 5.4.6.1 [5.215]):
( 1)
1
d xe x e
x x x
C
I
d d
d d 

. =  = ..  ..
where Cd is the deflection amplification factor, I is the occupancy importance factor, and dxe is the total
elastic deflection at Level x.
10.1.2.3 Basic Gravity Loads
Roof:
Live/Snow load (in Seattle, snow load governs over roof live = 25 psf
load; in other areas this may not be the case)
Dead load (including roofing, sheathing, joists, = 15 psf
insulation, and gypsum ceiling)
Floor:
Live load = 40 psf
Dead load (1½ in. lightweight concrete, sheathing, joists, and = 20 psf
gypsum ceiling. At first floor, omit ceiling but add insulation.)
Interior partitions, corridor walls (8 ft high at 11 psf) = 7 psf distributed floor
load
Exterior frame walls (wood siding, plywood sheathing, = 15 psf of wall surface
2×6 studs, batt insulation, and 5/8in. gypsum drywall)
Exterior double glazed window wall = 9 psf of wall surface
Party walls (doublestud sound barrier) = 15 psf of wall surface
Stairways = 20 psf of horiz.
projection
FEMA 451, NEHRP Recommended Provisions: Design Examples
1010
Perimeter footing (10 in. by 1 ft4 in.) and grade beam = 562 plf
(10 in. by 3 ft2 in.)
Corridor footing (10 in by 1 ft4 in.) and grade wall = 292 plf
(8 in. by 1 ft3 in.); 18 in. minimum crawl space under first floor
Applicable seismic weights at each level = 182.8 kips
Wroof = area (roof dead load + interior partitions + party walls)
+ end walls + longitudinal walls
W3 = W2 = area (floor dead load + interior partitions + party walls) = 284.2 kips
+ end walls + longitudinal walls
Effective total building weight, W = 751 kips
For modeling the structure, the first floor is assumed to be the seismic base, because the short crawl space
with concrete foundation walls is quite stiff compared to the superstructure.
10.1.3 SEISMIC FORCE ANALYSIS
The analysis is performed manually following a stepbystep procedure for determining the base shear
(Provisions Sec. 5.4.1 [5.2.1]), and the distribution of vertical (Provisions Sec. 5.4.3 [5.2.3]) and
horizontal (Provisions Sec. 5.4.4 [5.2.4]) shear forces. Since there is no basic irregularity in the building
mass, the horizontal distribution of forces to the individual shear walls is easily determined. These forces
need only be increased to account for accidental torsion (see subsequent discussion).
No consideration is given to soilstructure interaction since there is no relevant soil information available;
a common situation for a building of this size and type. The soil investigation ordinarily performed for
this type of structure is important, but is not generally focused on this issue. Indeed, the cost of an
investigation sufficiently detailed to permit soilstructure interaction effects to be considered, would
probably exceed the benefits to be derived.
10.1.3.1 Period Determination and Calculation of Seismic Coefficient (Provisions Sec. 5.4.1 [5.2.1])
Using the values for SD1, SDS, R, and I from Sec. 10.1.2.1, the base shear is computed per Provisions Sec.
5.4.1 [5.2.1]. The building period is based on Provisions Eq. 5.4.2.11 [5.26]:
x 0.237
Ta=Crhn=
where Cr = 0.020, hn = 27 ft, and x = 0.75.
According to Provisions Eq. 5.4.1.11 [5.22]:
0.89 0.137
/ 6.5/1.0
DS
s
C S
R I
= = =
but need not exceed Provisions Eq. 5.4.1.12 [5.23]:
0.47 0.305
( / ) (0.237)(6.5 /1.0)
D1
s
C S
T R I
= = =
Chapter 10, Wood Design
1011
Ground floor
Roof
56'0"
9'0" 9'0" 9'0"
2 floor
3 floor F3rd
Froof
F2nd
nd
rd
Ground floor
Roof
56'0"
9'0" 9'0" 9'0"
2 floor
3 floor F3rd
Froof
F2nd
nd
rd
Figure 10.15 Vertical shear distribution (1.0 ft = 0.3048 m).
Although it would probably never govern for this type of structure, also check minimum value according
to Provisions Eq. 5.4.1.13:
Cs = 0.044ISDS = (0.044)(1.0)(0.89) = 0.039
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value for
longperiod structures, which is not applicable to this example.]
The calculation of actual T as based on the true dynamic characteristics of the structure would not affect
Cs; thus, there is no need to compute the actual period because the Provisions does not allow a calculated
period that exceeds CuTa where Cu = 1.4 (see Provisions Sec. 5.4.2 [5.2.2]). Computing the base shear
coefficient, per Provisions Eq. 5.4.1.11 [5.22], using this maximum period, would give Cs = 0.218,
which still exceeds 0.137. Therefore, short period response governs the seismic design of the structure,
which is common for lowrise buildings.
10.1.3.2 Base Shear Determination
According to Provisions Eq. 5.4.1 [5.21]:
V = CsW = 0.137(751) = 103 kips (both directions)
where effective total weight is W = 751 kips as computed previously.
10.1.3.3 Vertical Distribution of Forces
Forces are distributed as shown in Figure 10.15, where the story forces are calculated according to
Provisions Eq. 5.4.31 and 5.4.32 [5.210 and 5.211]:
1
k
x x
x vx n k
i i
i
F C V w h V
w h
=
. .
. .
= =. .
..S ..
. .
FEMA 451, NEHRP Recommended Provisions: Design Examples
1012
For T < 0.5, k = 1.0 and k 182.2(27) 284.2(18) 284.2(9) 12,610
Swihi = + + =
Froof = [182.8(27)/12,608](103.2) = 40.4 kips
F3rd = [284.2(18)/12,608](103.2) = 41.9 kips
F2nd = [284.2(9)/12,608](103.2) = 20.9 kips
S = 103 kips
It is convenient and common practice, to perform this calculation along with the overturning moment
calculation. Such a tabulation is given in Table 10.11.
Table 10.11 Seismic Coefficients, Forces, and Moments
Level
x
Wx
(kips)
hx
(ft)
wxh k
x
(k = 1)
Cvx Fx
(kips)
Vx
(kips)
Mx
(ftkips)
Roof 182.8 27 4,936 0.391 40.4
40.4 364
3 284.2 18 5,115 0.406 41.9
82.3 1,104
2 284.2 9 2,557 0.203 20.9
103.2 2,033
S 751.2 12,608
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ftkip= 1.36 kNm.
10.1.3.4 Horizontal Distribution of Shear Forces to Walls
Since the diaphragms are defined as “rigid” by the Provisions (Sec 5.2.3.1 [4.3.2.1] and 12.4.1.1), the
horizontal distribution of forces must account for relative rigidity of the shear walls, and horizontal
torsion must be included. As discussed below, for buildings with a Type 1a or 1b torsional irregularity
(per Provisions Sec. 5.2.3.2 [4.3.2.2]) the torsional amplification factor (Provisions Sec. 5.4.4.3 [5.2.4.3])
must be calculated.
It has been common practice for engineers to consider wood diaphragms as flexible, regardless of the
relative stiffness between the walls and the diaphragms. Under the flexible diaphragm assumption, loads
are distributed to shear walls based on tributary area, without taking diaphragm continuity, or relative
wall rigidity into account. Recognizing that diaphragm stiffness should not be ignored (even for wood
structural panel sheathing), the Provisions provides limits on when the flexible diaphragm assumption
may be used and when it may not be used.
The calculation of horizontal force distribution for rigid diaphragms, can be significantly more laborious
than the relatively simple tributary area method. Therefore, this example illustrates some simplifications
that can be used as relatively good approximations (as confirmed by using more detailed calculations).
The design engineer is encouraged to verify all simplifying assumptions that are used to approximate
rigid diaphragm force distribution.
For this example, forces are distributed as described below.
Chapter 10, Wood Design
1013
10.1.3.4.1 Longitudinal Direction
Based on the rigid diaphragm assumption, force is distributed based on the relative rigidity of the
longitudinal walls, and the transverse walls are included for resisting torsion. By inspection, however, the
center of mass coincides with the center of rigidity and , thus, the torsional demand is just the accidental
torsional moment resulting from a 5 percent eccentricity of force from the center of mass (Provisions Sec.
5.4.4.2 [5.2.4.2]).
In this direction, there are four lines of resistance and the total torsional moment is relatively small.
Although the walls are unequal in length, the horizontal distribution of the forces can be simplified by
making two reasonable assumptions. First, for plywood shear walls, it is common to assume that stiffness
is proportional to net inplane length of sheathing (assuming sheathing thickness, nailing, and chord
elements are roughly similar). Second, for this example, assume that all of the torsional moment is
resisted by the end walls in the transverse directions. This is a reasonable assumption because the walls
have a greater net length than the longitudinal exterior shear walls and are located much farther from the
center of rigidity (and thus contribute more significantly to the rotational resistance).
Therefore, direct shear is distributed in proportion to wall length and torsional shear is neglected. Each
exterior wall has 45 ft net length and each corridor wall has 55 ft net length for a total of 200 ft of net
shear wall. The approximate load to each exterior wall is (45/200)Fx = 0.225Fx, and the load to each
corridor wall is (55/200)Fx = 0.275Fx. (The force distribution was also computed using a complete
rigidity model, including accidental torsion, with all transverse and longitudinal wall segments. The
resulting distribution is 0.231Fx to the exterior walls and 0.276 Fx to the corridor walls, for a difference of
2.7 percent and 0.4 percent, respectively).
10.1.3.4.2 Transverse Direction
Again, based on the rigid diaphragm assumption, force is to be distributed based on relative rigidity of the
transverse walls, and the longitudinal walls are included for resisting torsion because the center of mass
does not coincide with the center of rigidity. The torsional demand must be computed. Assuming that all
six transverse wall segments have the same rigidity, the distance from the center of rigidity (CR) to the
center of mass (CM) can be computed as:
148/ 2 4 60 144 4.67ft
3
CM CR
+ +
 =  =
The accidental torsional moment resulting from a 5 percent eccentricity of force from the center of mass
(Provisions Sec. 5.4.4.2 [5.2.4.2]) must also be considered.
As in the longitudinal direction, the force distribution in the transverse direction can be computed with
reasonable accuracy by utilizing a simplified model. This simplification is made possible largely because
the transverse wall segments are all of the same length and, thus, the same rigidity (assuming nailing and
chord members are also the same for all wall segments). First, although the two wall segments at each
line of resistance are offset in plan (Figure 10.14), assume that the wall segments do align, and are
located at the centroid of the two segments. Second, assume that the longitudinal walls do not resist the
torsional moment. Third, since interior transverse wall segments are located close to the center of rigidity,
assume that they do not contribute to the resistance of the torsional moment.
To determine the forces on each wall, split the seismic force into two parts: that due to direct shear and
that due to the torsional moment. Because their rigidities are the same, the direct shear is resisted by all
six wall segments equally and is computed as 0.167Fx. The torsional moment is resisted by the four end
wall segments. Including accidental torsion, the total torsional moment is computed as:
FEMA 451, NEHRP Recommended Provisions: Design Examples
1014
Total Torsional Moment, Mt + Mta = Fx × (4.67 ft + 0.05 × 148 ft) = 12.07Fx
Torsional Shear to End Walls = 12.07Fx/140 ft = 0.086Fx
Therefore, the simplified assumption yields a total design force of 0.167Fx + 0.086Fx/2= 0.210Fx, at each
of the end wall segments on the right side of Figure 10.41, 0.167Fx  0.086Fx/2= 0.124Fx at each of the
end wall segments on the left side of the figure, and 0.167Fx at each interior wall segment.
Next, determine the relationship between the maximum and average story drifts, to determine if a
torsional irregularity exists, as defined in Provisions Table 5.2.3.2 [4.32]. Assuming that all of the shear
walls will have the same plywood and nailing, the deflection of each wall will be proportional to the force
in that wall. Therefore, the maximum drift for any level, will be proportional to 0.210Fx, and the average
drift proportional to (0.210Fx + 0.124Fx)/2 = 0.167Fx. The ratio of maximum to average deflection is
0.210Fx/0.167Fx = 1.26. Since the ratio is greater than 1.2, the structure is considered to have a torsional
irregularity Type 1a, in accordance with Provisions Table 5.2.3.2 [4.32]. Therefore, the horizontal force
distribution must be computed again using the torsional amplification factor, Ax, from Provisions Sec.
5.4.4.3 [5.2.4.3]. (Diaphragm connections and collectors, not considered in this example, must satisfy
Provisions Sec. 5.2.6.4.2. [4.6.3.2])
In accordance with Provisions Eq 5.4.4.31 [5.213]:
( )
2 2
0.210 1.10
1.2 1.2 0.167
max x
x
avg x
A F
F
d
d
. . . .
=. . =. .=
.. .. .. ..
Therefore, recompute the torsional moment and forces as:
Total torsional moment, Ax(Mt + Mta) = 1.10[Fx × (4.67 ft + 0.05 × 148 ft)] = 13.28Fx
Torsional shear to End walls = 13.28Fx/140 ft = 0.095Fx
The revised simplified assumption yields a maximum total design force of 0.167Fx + 0.095Fx/2= 0.214Fx
at the end walls (right side of Figure 10.41), which will be designed in the following sections.
(The force distribution was also computed using a complete rigidity model including all transverse and
longitudinal wall segments. Relative rigidities were based on length of wall and the torsional moment
was resisted based on polar moment of inertia. The resulting design forces to the shear wall segments
were 0.208Fx, 0.130Fx, and 0.168Fx for the right end, left end, and interior segments, respectively. The
structure is still torsionally irregular (dmax / dave = 1.23), but the irregularity is less substantial and the
resulting Ax = 1.05. This more rigorous analysis, has shown that the simplified approach provides
reasonable results in this case. Since the simplified method is more likely to be used in design practice,
the values from that approach will be used for the remainder of this example.)
10.1.3.5 Verification of Redundancy Factor
Once the horizontal force distribution is determined, the assumed redundancy factor must be verified.
Provisions Eq. 5.2.4.2, is used to compute the redundancy factor, ., for each story as:
2 20
x
x
rmax Ax
. = 
Chapter 10, Wood Design
1015
where Ax is the floor area in square feet and rmaxx is the ratio of the design story shear resisted by the
single element carrying the most shear force in the story to the total story shear. As defined for shear
walls in Provisions Sec. 5.2.4.2, shall be taken as the shear in the most heavily loaded wall, rmaxx
multiplied by 10/lw, where lw is the wall length in feet.
The most heavily loaded wall is the end wall, farthest away from the interior shear wall. Since the force
distribution is the same for all three levels, the redundancy factor need only be determined for one level,
in this case, at the first floor. The redundancy factor is computed as:
Floor area = (140)(56) = 7,840 ft2
Max load to wall = 0.214V
Wall length = 25 ft
= 0.214V(10/25)/V = 0.0856 rmaxx
Therefore:
2 20 0.64
0.0856 7,840 . x =  = 
Because the calculated redundancy factor is less than the minimum permitted value of 1.0, the initial
assumption of . = 1.0 is correct, and the design can proceed using the previously computed lateral forces.
[See Sec. 10.1.2.2 for discussion of the changes to the redundancy requirements in the 2003 Provisions.]
10.1.3.6 Diaphragm Design Forces
As specified in Provisions Sec. 5.2.6.4.4 [4.6.3.4], floor and roof diaphragms must be designed to resist a
force, Fpx, in accordance with Provisions Eq. 5.2.6.4.4 [4.62]:
n
i
i x
px n px
i
i x
F
F w
w
=
=
S
=
S
plus any force due to offset walls (not applicable for this example). The diaphragm force as computed
above need not exceed 0.4SDSIwpx, but shall not be less than 0.2SDSIwpx. This latter value often governs at
the lower levels of the building, as it does here. The maximum required diaphragm demand does not
govern in this example.
The weight tributary to the diaphragm, wpx, need not include the weight of walls parallel to the force. For
this example, however, since the shear walls in both directions are relatively light compared to the total
tributary diaphragm weight, the diaphragm force is computed based on the total story weight, for
convenience.
Transverse direction
Roof Level
SFi = 40.4
Swi = 182.8
FP,roof = (40.4/182.8)(182.8) = 40.4 kips (controls for roof)
0.2SDSIwpx = (0.2)(0.89)(1.0)(182.8) = 32.7 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
1016
Third Floor
SFi = 40.4 + 41.9 = 82.3
Swi = 182.8 + 284.2 = 447.0
FP,3rd =(82.3/447.0)(284.2) = 50.1 kips
0.2SDSIwpx = (0.2)(0.89)(1.0)(284.2) = 50.8 kips (controls for 3rd floor)
Second Floor
SFi = 40.4 + 41.9 + 20.9 = 103,231
Swi = 182.8 + 284.2 + 284.2 = 751.2
FP,2nd = (103.2/751.2)(284.2) = 39.1 kips
0.2SDSIwpx = (0.2)(0.89)(1.0)(284.2) = 50.8 kips (controls for 2nd floor)
Diaphragm forces in the longitudinal direction are computed in a similar manner. Since the weight of the
exterior walls is more significant in the longitudinal direction, the designer may wish to subtract this
weight from the story force in order to compute the diaphragm demands.
10.1.4 Basic Proportioning
Designing a plywood diaphragm and plywood shear wall building, principally involves the determination
of sheathing thicknesses and nailing patterns to accommodate the applied loads. In doing so, some
successive iteration of steps may be needed to satisfy the deflection limits.
Nailing patterns in diaphragms and shear walls have been established on the basis of tabulated
requirements included in the Provisions. It is important to consider the framing requirements for a given
nailing pattern and capacity as indicated in the notes following the tables. In addition to strength
requirements, Provisions Sec. 12.4.1.2 [12.4.1.1] places aspect ratio limits on plywood diaphragms
(length/width shall not exceed 4/1 for blocked diaphragms), and Provisions Sec. 12.4.2.3 [12.4.2.3] places
similar limits on shear walls (height/width shall not exceed 2/1 for full design capacities). However, it
should be taken into consideration that compliance with these aspect ratios does not guarantee that the
drift limits will be satisfied.
Therefore, diaphragms and shear walls have been analyzed for deflection as well as for shear capacity. A
procedure for computing diaphragm and shear wall deflections is provided in Commentary Sec. 12.4.1.
This procedure is illustrated below. The Commentary does not indicate how to compute the nail
deformation (nail slip) factor, but there is a procedure contained in the commentary of ASCE16.
In the calculation of diaphragm deflections, the chord slip factor can result in large additions to the total
deflection. This can be overcome by using “neat” holes for bolts, and proper shimming at butt joints.
However, careful attention to detailing and field inspection are essential, to ensure that they are provided.
[AF&PA Wind & Seismic also contains procedures for computing diaphragm and shear wall deflections.
The equations are slightly different from the more commonly used equations that appear in the
Commentary and AF&PA LRFD Manual. In AF&PA Wind & Seismic, the shear and nail slip terms are
combined using an “apparent shear stiffness” parameter. However, the apparent shear stiffness values are
only provided for OSB. Therefore, the deflection equations in the Commentary or AF&PA LRFD
Manual must be used in this example which has plywood diaphragms and shear walls. The apparent
shear stiffness values for plywood will likely be available in future editions of AF&PA Wind & Seismic.]
10.1.4.1 Strength of Members and Connections
Chapter 10, Wood Design
1017
The 2000 Provisions has adopted Load and Resistance Factor Design (LRFD) for engineered wood
structures. The Provisions includes the ASCE 16 standard by reference and uses it as the primary design
procedure for engineered wood construction. Strength design of members and connections is based on
the requirements of ASCE 16. The AF&PA Manual and supplements contain reference resistance values
for use in design. For convenience, the Provisions contain design tables for diaphragms and shear walls
that are identical to those contained in the AF&PA StructuralUse Panels Supplement. However, the
modification of the tabulated design resistance for shear walls and diaphragms with framing other than
Douglas FirLarch or Southern Pine is different between the two documents. This example illustrates the
modification procedure contained in the Provisions tables, which is new to the 2000 edition.
Throughout this example, the resistance of members and connections subjected to seismic forces, acting
alone, or in combination with other prescribed loads, is determined in accordance with ASCE 16 and the
AF&PA Manual; with the exception of shear walls and diaphragms for which design resistance values are
taken from the Provisions. The LRFD standard incorporates the notation D’, T’, Z’, etc. to represent
adjusted resistance values, which are then modified by a time effect factor, ., and a capacity reduction
factor, f, to compute a design resistance, which is defined as “factored resistance” in the Provisions and
ASCE 16. Additional discussion on the use of LFRD is included in Commentary Sec. 12.2 and 12.3, and
in the ASCE 16 commentary. It is important to note that ACSE 16 and the AF&PA Manual use the term,
“resistance,” to refer to the design capacities of members and connections while “strength” refers to
material property values.
It is worth noting that the AF&PA Manual contains a Preengineered Metal Connections Guideline for
converting allowable stress design values for cataloged metal connection hardware (for example, tiedown
anchors) into ultimate capacities for use with strength design. This procedure is utilized in this
example.
[The primary reference for design of wood diaphragms and shear walls in the 2003 Provisions is AF&PA
Wind & Seismic. Much of the remaining text in the 2003 Provisions results from differences between
AF&PA Wind & Seismic and Chapter 12 of the 2000 Provisions as well as areas not addressed by
AF&PA Wind & Seismic. Because the AF&PA Wind & Seismic tabulated design values for diaphragms
and shear walls do not completely replace the tables in the 2000 Provisions, portions of the tables remain
in the 2003 Provisions. Therefore, some diaphragm and shear wall design values are in the 2003
Provisions and some are in AF&PA Wind & Seismic. The design values in the tables are different
between the two documents. The values in the 2003 Provisions represent factored shear resistance
(.fD’), while the values in AF&PA Wind & Seismic represent nominal shear resistance that must then be
multiplied by a resistance factor, f, (0.65) and a time effect factor ,., (1.0 for seismic loads). Therefore,
while the referenced tables may be different, the factored resistance values based on the 2003 Provisions
should be the same as those in examples based on the 2000 Provisions. The calculations that follow are
annotated to indicate from which table the design values are taken.]
10.1.4.2 Transverse Shear Wall Nailing
The design will focus on the more highly loaded end walls; interior walls are assumed to be similar.
10.1.4.2.1 Load to Any One of Four 25ft End Walls
Froof = 0.214(40.4) = 8.65 kips
F3rd = 0.214(41.9) = 8.97 kips
F2nd = 0.214(20.9) = 4.47 kips
S = 22.09 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
1018
9' 9' 9'
25'0"
27'0"
22.09 kips
v = 0.883 kip/ft
v = 0.704 kip/ft
v = 0.346 kip/ft
Fr o o f = 8.65 kips
F3 rd = 8.97 kips
F2 n d = 4.47 kips
9' 9' 9'
25'0"
27'0"
22.09 kips
v = 0.883 kip/ft
v = 0.704 kip/ft
v = 0.346 kip/ft
Fr o o f = 8.65 kips
F3 rd = 8.97 kips
F2 n d = 4.47 kips
Figure 10.16 Transverse section: end wall (1.0 ft = 0.3048 m, 1.0
kip = 4.45 kN, 1.0 kip/ft = 14.6 kN/m).
10.1.4.2.2 Roof to Third Floor Design
V = 8.65 kips
v = 8.65/25 = 0.346 klf
Try a ½in. (15/32) plywood rated sheathing (not Structural I) on blocked 2in. Douglas firLarch
members at 16 in. on center with 10d common nails at sixin. on center at panel edges and 12 in. on center
at intermediate framing members. According to Note b of Provisions Table 12.4.32a [12.43a], the
design shear resistance must be adjusted for Hemfir wall framing. The specific gravity adjustment factor
equals 1(0.5SG) where SG is the specific gravity of the framing lumber. From Table 12A of the
AF&PA Structural Connections Supplement, the SG = 0.43 for Hemfir. Therefore, the adjustment factor
is 1(0.50.43) = 0.93.
From Provisions Table 12.4.32a [AF&PA Wind/Seismic Table 4.3A], .fD’ = 0.40 klf
Adjusted shear resistance = 0.93(0.40) = 0.37 klf > 0.346 klf OK
8d nails could be used at this level but, because 10d nails are required below, 10d nails are used here for
consistency.
Deflection of plywood shear panels depends, in part, on the slip at the nails which, in turn, depends on the
load per nail. See Sec. 10.1.4.3 for a detailed discussion of the appropriate slip for use with the
Provisions and see Table 10.12 for fastener slip equations used here.
Load per nail = 0.346(6/12)(1000) = 173 lb
Nail slip en = 1.2(173/769)3.276 = 0.00904 in.
In the above equation, 1.2 = factor for plywood other than Structural I, and 769 and 3.276 are constants
explained in Sec. 10.1.4.3.
[As indicated previously, AF&PA Wind & Seismic does not provide apparent shear stiffness values for
plywood sheathing. Therefore, the deflection equations in the Commentary or AF&PA LRFD Manual
must be used in this case.]
10.1.4.2.3 Third Floor to Second Floor
V = 8.65 + 8.97 = 17.62 kips
Chapter 10, Wood Design
1019
v = 17.62/25 = 0.704 klf
Try ½in. (15/32) plywood rated sheathing (not Structural I) on blocked 2in. Douglas firLarch members
at 16 in. on center with 10d nails at 3in. on center at panel edges, and at 12 in. on center at intermediate
framing members.
From Provisions Table 12.4.32a [AF&PA Wind&Seismic Table 4.3A], .fD’ = 0.78 klf
Adjusted shear resistance = 0.93(0.78) = 0.73 klf > 0.704 klf OK
Table Note e [AF&PA Wind&Seismic Sec. 4.3.7.1] requires 3in. framing at adjoining panel edges.
Load per nail = 0.704(3/12)(1000) = 176 lb
Nail slip en = 1.2(176/769)3.276 = 0.00960 in.
10.1.4.2 Second Floor to First Floor
V = 17.62 + 4.47 = 22.09 kips
v = 22.09/25 = 0.883 klf
Try a ½in. (15/32) plywood rated sheathing (not Structural I) on blocked 2in. Douglas firLarch
members at 16 in. on center with 10d common nails at 2in. on center at panel edges and 12 in. on center
at intermediate framing members.
From Provisions Table 12.4.32a [AF&PA Wind&Seismic Table 4.3A], .fD’ = 1.00 klf
Adjusted shear resistance = 0.93(1.00) = 0.93 klf > 0.883 klf OK
Table Notes d and e [AF&PA Wind&Seismic Sec. 4.3.7.1]require 3in. framing at adjoining panel edges.
Load per nail = 0.883(2/12)(1000) = 147 lb
Nail slip en = 1.2(147/769)3.276 = 0.00534 in.
10.1.4.3 General Note on the Calculation of Deflections for Plywood Shear Panels
The commonly used expressions for deflection of diaphragms and shear walls, which are contained in
Commentary Sec. 12.4.1, and the ASCE 16 commentary standard (see also UBC Std 232 or APA 138),
include a term for nail slip. The ASCE 16 commentary includes a procedure for estimating nail slip that
is based on LRFD design values.
The values for en used in this example (and in Sec. 10.2) are calculated according to Table 10.12, which
is taken from Table C9.51 of ASCE 16.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1020
Table 10.12 Fastener Slip Equations
Fastener
Minimum
Penetration
(in.)
For Maximum
Loads Up to
(lb.)
Approximate Slip, en
*
Green/Dry Dry/Dry
6d common nail 11/4 180 (Vn/434)2.314 (Vn/456)3.144
8d common nail 17/16 220 (Vn/857)1.869 (Vn/616)3.018
10d common nail 15/8 260 (Vn/977)1.894 (Vn/769)3.276
14ga staple 1 to 2 140 (Vn/902)1.464 (Vn/596)1.999
14ga staple 2 170 (Vn/674)1.873 (Vn/361)2.887
*Fabricated green/tested dry (seasoned); fabricated dry/tested dry. Vn = fastener load in pounds. Values
based on Structural I plywood fastened to Group II lumber. Increase slip by 20 percent when plywood is not
Structural I.
1.0 in. = 25.4 mm, 1.0 lb = 4.45 N.
This example is based on the use of surfaced dry lumber so the “dry/dry” values are used. The
appearance of the equations for both shear wall and diaphragm deflection, implies greater accuracy than is
justified. The reader should keep in mind that the deflections calculated are only rough estimates.
10.1.4.4 Transverse Shear Wall Deflection
From Commentary Sec. 12.4, modified as described below, shear wall deflection is computed as:
8 3 0.75 n a
vh vh he h d
wEA Gt w
d = + + +
The above equation produces displacements in inches and the individual variables must be entered in the
force or length units as described below:
v = V/w where V is the total shear on the wall and v is in units of pounds/foot
8vh3/wEA = bending deflection, as derived from the formula db = Vh3/3EI, where V is the total shear
in pounds, acting on the wall and I = Aw2/2 (in.4)
vh/Gt = shear deflection, as derived from the formula dv = Vh/GA', where A' = wt (in.2)
0.75 hen = nail slip, in inches. Note that with h being given in feet, the coefficient 0.75 carries units
of 1/ft.
(h/w)da = deflection due to anchorage slip, in inches. Note that for use in the deflection equation
contained in the Commentary, the term da represents the vertical deflection due to anchorage details.
In the deflection equation contained in the commentary in the AF&PA Manual, there is no h/w factor,
so it should be assumed that the term da represents the horizontal deflection at the top of the wall due
to anchorage details.
For this example:
E = 1,600,000 psi
G = 75,000 psi
A (area) = 2(2.5)5.5 = 27.5 in.2 for assumed double 3×6 end posts
Chapter 10, Wood Design
1021
Roof
2 floor
3 floor
nd
rd
Roof
2 floor
3 floor
nd
rd
Figure 10.17 Force
distribution for flexural
deflections.
w (shear wall length) = 25 ft
h (story height) = 9 ft
t (effective thickness) = 0.298 in. for ½in. unsanded plywood (not Structural I)
en = nail deformation factor from prior calculations, inches.
This equation is designed for a onestory panel and some modifications are in order for a multistory
panel. The components due to shear distortion and nail slip are easily separable (see Table 10.13).
Table 10.13 Wall Deflection (per story) Due to Shear and Nail Slip
Story v
(plf)
vh/Gt
(in.)
en
(in.)
0.75enh
(in.)
Roof 346 0.139 0.00904 0.061
3 704 0.284 0.00960 0.065
2 883 0.356 0.00534 0.036
1.0 in. = 25.4 mm, 1.0 plf = 14.6 N/m.
Likewise, the component due to anchorage slip is easily separable; it is a rigid body rotation. If a 1/8in.
upward slip is assumed (on the tension side only), the deflection per story is (9/25)(1/8) = 0.045 in. (Table
10.14).
The component due to bending is more difficult to separate. For this example, a grossly simplified,
distributed triangular load on a cantilever beam is used (Figure 10.17).
The total load V is taken as 21.3 kips, the sum of the story forces on the wall. The equation for the deflection,
taken from Roark's Formulas for Stress and Strain, is:
5 4 4 5
2 2
2 (11 15 5 )
5 x
V h hx hx x
EAb h
d=  + 
where x is the distance (ft) from the top of the building to the story in question, h is the total height (ft), b is
the wall width (ft), A is the chord cross sectional area (in.2), E is the modulus of elasticity (psi), and the
resulting displacement, d, is in inches.
This somewhat underestimates the deflections, but it is close enough for design. The results are shown in
Table 10.14.
The total wall deflections, shown in Table 10.15, are combined with the diaphragm deflections (see Sec.
10.1.4.7, 10.1.4.8, and 10.1.4.9). Drift limits are checked after diaphragm deflections are computed.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1022
Table 10.14 Wall Deflection (per story) Due to Bending and Anchorage Slip
Level Effective
8vh3/wEA (in.)
(h/w)da
(in.)
Roof 0.031 0.045
3 0.027 0.045
2 0.012 0.045
1.0 in. = 25.4 mm.
Table 10.15 Total Elastic Deflection and Drift of End Wall
Level Shear
(in.)
Bending
(in.)
Nail Slip
(in.)
Anchor
Slip (in.)
Drift .e
(in.)
Total de
(in.)
Roof 0.139 0.031 0.061 0.045 0.277 1.145
3 0.284 0.027 0.065 0.045 0.420 0.869
2 0.356 0.012 0.036 0.045 0.448 0.448
1.0 in. = 25.4 mm.
10.1.4.5 Transverse Shear Wall Anchorage
Provisions Sec. 12.4.2.4 [12.4.2.4.1] requires tiedown (holddown) anchorage at the ends of shear walls
where net uplift is induced. Net uplift is computed as the combination of the seismic overturning moment
and the dead load counterbalancing moment using the load combination indicated in Provisions Eq.
5.2.72 [4.22].
Traditionally, tiedown devices were designed to resist this net overturning demand. However,
Provisions Sec. 12.4.2.4 [12.4.2.4.1] requires that the nominal strength of tiedown devices exceeds the
expected maximum vertical force that the shear panels can deliver to the tensionside end post.
Specifically, the nominal strength (f = 1.0) of the tiedown device must be equal to, or greater than the
net uplift forces resulting from O0/1.3 times the factored shear resistance of the shear panels, where O0 is
the system overstrength factor. These uplift forces are cumulative over the height of the building. Also,
dead load forces are not used in these calculations to offset the uplift forces for the design of the tiedown
anchorage.
[The requirements for uplift anchorage are slightly different in the 2003 Provisions. The tie down force is
based on the “nominal strength of the shear wall” rather than the O0/1.3 times the factored resistance as
specified in the 2000 Provisions. This change is primarily intended to provide consistent terminology and
should not result in a significant impact on the design of the tie down.]
An additional requirement of Provisions Sec. 12.4.2.4 [12.4.2.4.1], is that end posts must be sized such
that failure across the net section does not control the capacity of the system. That is, the tensile strength
of the net section must be greater than the tiedown strength. Note that the tiedowns designed in the
following section are not located at wall intersections where the directional combination requirements of
Provisions Sec. 5.2.5.2.3 [4.4.2.3] would apply (see also the “Analysis Procedure” in Sec. 10.1.2.2).
10.1.4.5.1 Tiedown Anchors at Third Floor
For the typical 25ft end wall segment, the overturning moment at the third floor is:
M0 = 9(8.65) = 77.8 ftkip = QE
Chapter 10, Wood Design
1023
The counterbalancing moment, 0.72 QD = 0.72D(25 ft/2). The width of the floor contributing to this, is
taken as half the span of the exterior window header equal to 6.5 feet (see Figures 10.11 and 10.12).
For convenience, the same length is used for the longitudinal walls, the weight of which (interior and
exterior glazed wall) is assumed to be 9 psf.
End wall self weight = 9 ft (25ft )15 psf/1000 = 3.4 kips
Tributary floor = 6.5 ft (25 ft) (15) psf/1000 = 2.4 kips
Tributary longitudinal walls = 9 ft (6.5 ft) 9 psf (2)/1000 = 1.1 kips
S = 6.9 kips
0.72QD = 0.72(6.9)12.5 = 62.1 ftkip
M0 (net) = 77.8  62.1 = 15.7 ftkip
Therefore, uplift anchorage is required. Using the procedure described above, nominal tiedown strength
must be equal to or exceed the tension force, T, computed as:
T = 0.37 klf (3/1.3)(8ft) = 6.83 kips
where
0.37 klf = factored shear wall resistance at the third floor
3/1.3 = O0/1.3
8 ft = net third floor wall height (9ft story height minus approximately 1 ft of framing).
Use a double tiedown device to eliminate the eccentricity associated with a single tiedown. Also, use
the same size end post over the full height of the wall to simplify the connections and alignments. As will
be computed below, a 6×6 (Douglas firLarch) post is required at the first floor. At the third floor, try
two sets of double tiedown anchors connected through the floor with a 5/8in. threaded rod to the end
posts with two 5/8in. bolts similar to Figure 10.18.
According to Provisions Sec. 12.4.2.4 [12.4.2.4.1], the nominal tiedown strength is defined as the
“maximum test load the device can resist under cyclic testing without connection failure by either metal
or wood failure.” [Note that this definition for nominal tie down strength has been removed in the 2003
Provisions. This change is primarily intended to provide consistent terminology and should not result in
a significant impact on the design of the tie down.] For a single tiedown device as described above, the
cataloged allowable uplift capacity is 2.76 kips and the cataloged average ultimate load is 12.15 kips.
However, according to the documentation in this particular supplier’s product catalog, the ultimate values
are based on static testing and the type of failure is not indicated. Therefore, for this example, the
cataloged ultimate value is not considered to satisfy the requirements of the Provisions. As an alternate
approach, this example will utilize the methodology contained in the AF&PA Manual for converting
cataloged allowable stress values to strength values.
Based on the procedure in the AF&PA Manual, PreEngineered Metal Connectors Guide, the strength
conversion factor for a connector for which the catalog provides an allowable stress value with a 1.33
factor for seismic is 2.88/1.33.
Since the typical product catalogs provide design capacities only for single tiedowns, the design of
double tiedowns requires two checks. First, consider twice the capacity of one tiedown, and second, the
capacity of the bolts in double shear. f = 1.0 for both these calculations.
For the double tiedown, the nominal strength is computed as:
FEMA 451, NEHRP Recommended Provisions: Design Examples
1024
2.fZ’ = 2(1.0)(1.0)(2.76)(2.88/1.33) = 12.0 kips > 6.83 kips OK
For the two bolts through the end post in double shear, the AF&PA Manual gives:
2.fZ’ = 2(1.0)(1.0)(4.90) = 9.80 kips > 6.83 kips OK
The factored capacity of the tiedowns must also be checked for the design loads, which in this case will
not govern the design.
10.1.4.5.2 Tiedown Anchors at Second Floor
Since tiedowns are required at the third floor, it would be common practice to provide tiedowns at the
second and first floors, whether or not calculations indicate that they are required. Nevertheless, the
overturning calculations are performed for illustrative purposes. The overturning moment at the second
floor is:
M0 = 18(8.65) + 9(8.97) = 236 ftkip.
The counterbalancing moment, 0.72 QD = 0.72(DL)12.5.
End wall self weight = 18 ft (25 ft) 15 psf/1000 = 6.75 kips
Tributary floor = 6.5 ft (25 ft) (15 + 27) psf/1000 = 6.83 kips
Tributary longitudinal walls = 18 ft (6.5 ft) 9 psf (2)/1000 = 2.10 kips
S = 15.7 kips
0.72QD = 0.72(15.7)12.5 = 141 ftkips
M0 (net) = 236141 = 95 ftkips
As expected, uplift anchorage is required. The design uplift force is computed using an adjusted factored
shear resistance of 0.73 klf at the second floor, and a net length of wall height equal to 8 ft. Note that 8 ft.
is appropriate for this calculation given the detailing for this structure. As shown in Figure 10.110, the
plywood sheathing is not detailed as continuous across the floor framing, which results in a net sheathing
height of about 8ft. If the sheathing were detailed across the floor framing, then 9 ft would be the
appropriate wall height for use in computing tiedown demands. Combined with the uplift force at the
third floor, the total design uplift force at the second floor is:
T = 6.83 kips + 0.73 klf (3/1.3)(8ft) = 20.3 kips
Use two sets of double tiedown anchors to connect the 6×6 end posts. Using the same procedure as for
the third floor, tiedowns with a seveneighthsin. threaded rod and three seveneighthsin. bolts are
computed to be adequate. See Figure 10.18.
Chapter 10, Wood Design
1025
Threaded rod connected
to tiedown anchors
with thru bolts 7
diameters from
end of stud
Post and
tie down
Floor framing
Solid blocking
at post
Corner
of wall
Figure 10.18 Shear wall tie down at suspended floor framing.
10.1.4.5.3 Tiedown Anchors at First Floor
The overturning moment at the first floor is:
M0 = 27(8.65) + 18(8.97) + 9(4.47) = 435 ftkip = QE.
The counterbalancing moment, 0.72 QD = 0.72D(25 ft/2).
End wall self weight = 27 ft (25ft )15 psf/1000 = 10.1 kips
Tributary floor = 6.5 ft (25 ft) (15 + 27 + 27) psf/1000 = 11.2 kips
Tributary longitudinal walls = 27 ft (6.5 ft) 9 psf (2)/1000 = 3.2 kips
S = 24.5 kips
0.72QD = 0.72(24.5)12.5 = 221 ftkip
M0 (net) = 435  221 = 214 ftkip
As expected, uplift anchorage is required. The design uplift force is computed using an adjusted factored
shear resistance of 0.93 klf at the first floor, and a net wall height of 8ft. Combined with the uplift force
at the floors above, the total design uplift force at the first floor is:
T = 20.3 kips + 0.93 klf (3/1.3)(8ft.) = 37.5 kips.
Use a double tiedown anchor that extends through the floor with an anchor bolt into the foundation.
Tiedowns with a 7/8in. threaded rod, and four 7/8in. bolts are adequate.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1026
The strength of the end post, based on failure across the net section, must also be checked (Provisions
Sec. 12.4.2.4 [12.4.2.4.1]). For convenience, the same size post has been used over the height of the
building, so the critical section is at the first floor. A reasonable approach to preclude net tension failure
from being a limit state would be to provide an end post, whose factored resistance exceeds the nominal
strength of the tiedown device. (Note that using the factored resistance rather than nominal strength of
the end post provides an added margin of safety that is not explicitly required by the Provisions.) The
nominal strength of the first floor double tiedown is 42.9 kips, as computed using the procedure
described above. Therefore, the tension capacity at the net section must be greater than 42.9 kips.
Try a 6×6 Douglas FirLarch No. 1 end post. In some locations, the shear wall end post also provides
bearing for the window header, so this size is reasonable. Accounting for 1in. bolt holes, the net area of
the post is 24.75 in.2 According to the AF&PA Manual, Structural Lumber Supplement:
.fT’ = (1.0)(0.8)(2.23ksi)(24.75in2) = 44.2 kips > 42.9 kips OK
For the maximum compressive load at the end post, combine maximum gravity load, plus the seismic
overturning load. In the governing condition, the end post supports the header over the glazed portion of
the exterior wall (end wall at right side of Figure 10.11). Assume that the end post at the exterior side of
the wall supports all the gravity load from the header, and resists onehalf of the seismic overturning load.
Compute gravity loads based on a 6.5foot tributary length of the header:
Tributary DL = ((27 ft)(9 psf) + (8 ft)(15 + 27 + 27)psf)(6.5ft)/1000 = 5.17 kips
Tributary LL = 8 ft (7 ft) (40 + 40) psf/1000 = 4.48 kips
Tributary SL = 8 ft (6.5 ft) (25 psf)/1000 = 1.30 kips
The overturning force is based on the seismic demand (not wall capacity as used for tension anchorage)
assuming a moment arm of 23 ft:
Overturning, QE = 435/23 ft = 18.9 kips
Per load combination associated with Provisions Eq. 5.2.7.11 [4.21]:
Maximum compression = 1.38(5.17) + 1.0(18.9/2) + 0.5(4.48) + 0.2(1.30) = 19.1 kips.
Due to the relatively short clear height of the post, the governing condition is bearing perpendicular to the
grain on the bottom plate. Check bearing of the 6×6 end post on a 3×6 Douglas firLarch No. 2 plate, per
the AF&PA Manual, Structural Lumber Supplement:
.fP’= (1.0)(0.8)(1.30ksi)(30.25 in.2) = 31.5 kips > 19.1 kips. OK
The 6×6 end post is slightly larger than the double 3×6 studs assumed above for the shear wall deflection
calculations. Therefore, the computed shear wall deflection is slightly conservative, but the effect is
minimal.
10.1.4.5.4 Check Overturning at the Soil Interface
A summary of the overturning forces is shown in Figure 10.19. To compute the overturning at the soil
interface, the overturning moment must be increased for the 4ft foundation height:
M0 = 435 + 22.09(4.0) = 523 ftkip
Chapter 10, Wood Design
1027
However, it then may be reduced in accordance with Provisions Sec. 5.3.6:
M0 = 0.75(523) = 392 ftkip
To determine the total resistance, combine the weight above with the dead load of the first floor and
foundation.
Load from first floor = 25 ft (6.5 ft) (27  4 + 1)psf /1000 = 3.9 kips
where 4 psf is the weight reduction due to the absence of a ceiling, and 1 psf is the weight of insulation.
The length of the longitudinal foundation wall included, is a conservative approximation of the amount
carried by minimum nominal reinforcement in the foundation.
Foundation weight = (562 plf (13 ft + 25 ft) + 292 plf (13 ft))/1000 = 25.1 kips
First floor = 3.9 kips
Structure above = 24.5 kips
S = 53.5 kips
Therefore, 0.72D  1.0QE = 0.72(53.5)12.5 ft  1.0(392) = 89.5 ftkips, which is greater than zero, so the
wall will not overturn.
10.1.4.5.5 Anchor Bolts for Shear
At the first floor:
v = 0.883 klf.
A common anchorage for nonengineered construction is a ½in. bolt at 4ft0 in. For a ½in. bolt in a
Douglas firLarch 3×6 plate, in single shear, parallel to the grain:
.fZ’/4ft = (1.0)(0.65)(2.38 kips)/4 = 0.39 klf < 0.883 klf NG
Try a larger bolt and tighter spacing. For this example, use a 5/8in. bolt at 32 in. on center:
.fZ’/2.67ft = (1.0)(0.65)(3.72 kips)/2.67 = 0.91 klf > 0.883 klf OK
Provisions Sec. 12.4.2.4 [12.4.2.4.2]requires plate washers at all shear wall anchor bolts. A summary of
the transverse shear wall elements is shown in Figure 10.19.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1028
10.1.4.6 Remarks on Shear Wall Connection Details
In normal platform frame construction, details must be developed that will transfer the lateral loads
through the floor system and, at the same time, accommodate normal material sizes and the crossgrain
shrinkage in the floor system. The connections for wall overturning in Sec. 10.1.4.5 are an example of
one of the necessary force transfers. The transfer of diaphragm shear to supporting shear walls is another
important transfer as is the transfer from a shear wall on one level to the level below.
The floortofloor height is nineft with about oneft occupied by the floor framing. Using standard 8ftlong
plywood sheets for the shear walls, a gap occurs over the depth of the floor framing. It is common to
use the floor framing to transfer the lateral shear force. Figures 10.110 and 10.111 depict this
accomplished by nailing the plywood to the bottom plate of the shear wall, which is nailed through the
floor plywood to the double 2×12 chord in the floor system.
9' 9' 9'
Provide tiedown
anchor at each end
bolted to end post
at each level.
F 3 r d = 8.97 kips
25'0"
F r o o f = 8.65 kips
F 2 n d = 4.47 kips
Provide 5
8" anchor
bolts at 2'8" o.c.
WD
Figure 10.19 Transverse wall: overturning
(1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN).
Chapter 10, Wood Design
1029
Wall and floor nailing
as in bearing wall
detail (Figure 10.110)
Framing clip or
alternate toe nailing
as in bearing wall
detail (Figure 10.110)
8d at 6" o.c.
into blocking
2x12 joists
Sheet metal strap
connecting blocks
across bottom of
joists
2x12 blocking between joists at 4'0" o.c. extending
inward from wall to third joist and toenailed to
each joist with 28d and to wall plate with 216d.
Connect across joist with sheet metal strap and
210d each side. Remainder of blocking at
plywood joists to be 2x3 lumber.
Figure 10.111 Nonbearing wall (1.0 in. = 25.4 mm).
Joists at 16" o.c.
toenailed to wall
plate with 38d
2x12 blocking between
joists toenailed to
wall plate with 216d
and to 22x12 chord
with 216d
Provide sheet metal framing
clips (under blocking) to
connect 22x12
chord to wall top plate
Where shear forces are
small, sheet metal
framing clips may be
replaced by toenailing
the 22x12 chord
to the wall top plate
Floor edge
nailing
Wall edge
nailing
Bottom plate nailing to
develop capacity of shear wall
22x12 chord
for diaphragm
Figure 10.110 Bearing wall (1.0 in = 25.4 mm).
FEMA 451, NEHRP Recommended Provisions: Design Examples
1030
Bottom plate nailing
2x12 joist toenailed
to sill with 316d
2x12 blocking between
joists. Connect to sill
with sheet metal
framing clips.
3x6 sill, dense, treated
Wall edge nailing
Flashing, sheathing,
and finish siding
not shown
5
8" dia. anchor
bolt at 2'8" o.c.
w/ plate washer
Figure 10.112 Foundation wall detail (1.0 in. = 25.4 mm).
The top plate of the lower shear wall also is connected to the double 2×12 by means of sheet metal
framing clips to the double 2×12 to transfer the force back out to the lower plywood. (Where the forces
are small, using toe nails between the double 2×12 and the top plate may be used for this connection.)
This technique leaves the floor framing free for crossgrain shrinkage. Although some designers in the
past may have used a short tier of plywood nailed to the plates of the stud walls to accomplish the
transfer, Provisions Sec. 12.4.2.6 prohibits this type of detailing.
The floor plywood is nailed directly to the framing at the edge of the floor, before the plate for the upper
wall is placed. Also, the floor diaphragm is connected directly to framing that spans over the openings
between shear walls. The axial strength, and the connections of the double 2×12 chords, allow them to
function as collectors to move the force from the full length of the diaphragm to the discrete shear walls.
(According to Provisions Sec. 5.2.6.4.1 [4.6.2.2], the design of collector elements in wood shear wall
buildings in Seismic Design Category D need not consider increased seismic demands due to
overstrength.)
The floor joist is toe nailed to the wall below for forces normal to the wall. Likewise, fulldepth blocking
is provided adjacent to walls that are parallel to the floor joists, as shown in Figure 10.111. (Elsewhere
the blocking for the floor diaphragm only need be small pieces, flat 2×4s, for example.) The connections
at the foundation are similar (see Figure 10.112).
The particular combinations of nails and bent steel framing clips shown in Figures 10.110, 10.111, and
10.112, to accomplish the necessary force transfers, are not the only possible solutions. A great amount
Chapter 10, Wood Design
1031
of leeway exists for individual preference, as long as the load path has no gaps. Common carpentry
practices often will provide most of the necessary transfers but, just as often, a critical few will be missed.
As a result, careful attention to detailing and inspection is an absolute necessity.
10.1.4.7 Roof Diaphragm Design
While it has been common practice to design plywood diaphragms as simply supported beams spanning
between shear walls, the diaphragm design for this example must consider the continuity associated with
rigid diaphragms. The design will be based on the maximum shears and moments that occur over the
entire diaphragm. From Sec. 10.1.3.5, the diaphragm design force at the roof is, FP,roof = 40.4 kips.
As discussed previously, the design force computed in this example includes the internal force due to the
weight of the walls parallel to the motion. Particularly for onestory buildings, it is common practice to
remove that portion of the design force. It is conservative to include it, as is done here.
10.1.4.7.1 Diaphragm Nailing
The maximum diaphragm shear occurs at the end walls. From Sec. 10.1.3.4, each 25ft end wall segment
resists 21 percent of the total story (diaphragm, in this case) force. Distributing the diaphragm force at the
same rate, the diaphragm shear over the entire diaphragm width at the end walls is:
V = (0.214)(2)(40.4) = 17.3 kips
v = 17.3/56 ft. = 0.308 klf
Try ½in. (15/32) plywood rated sheathing (not Structural I) on blocked 2in. Douglas firLarch members
at 16 in. on center, with 8d nails at 6 in. on center at panel edges and 12 in. on center at intermediate
framing members.
From Provisions Table 12.4.31a [AF&PA Wind&Seismic Table 4.2A], .fD’ = 0.35 klf > 0.308 klfOK
The determination of nail slip for diaphragms is included below.
10.1.4.7.2 Chord and Splice Connection
Diaphragm continuity is an important factor in the design of the chords. The design must consider the
tension/compression forces, due to positive moment at the middle of the span, as well as negative moment
at the interior shear wall. It is reasonable (and conservative) to design the chord for the positive moment
assuming a simply supported beam and for the negative moment accounting for continuity. The positive
moment is wl2/8, where w is the unit diaphragm force, and l is the length of the governing diaphragm
span. For a continuous beam of two unequal spans, under a uniform load, the maximum negative moment
is:
3 3
1 2
8(1 2)
M wl wl
l l
 +
=
+
where w is the unit diaphragm force, and l1 and l2 are the lengths of the two diaphragm spans. For w =
40.4 kips /140 ft = 0.289 klf, the maximum positive moment is:
0.289(84)2 / 8 = 255 ftkip
and the maximum negative moment is:
FEMA 451, NEHRP Recommended Provisions: Design Examples
1032
Stagger joints in
opposite faces. Use
16'0" minimum
lengths.
Drive shim in joint
22x12 chord  Douglas
FirLarch, No. 1
7" 7"
4" dia. split ring.
3
4" dia. machine
bolt. 3
4" dia. hole. 1'2"
Figure 10.113 Diaphragm chord splice (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm).
0.289(84)30.289(56)3198 ftkip
8(84 56)
+
=
+
The positive moment controls, and the design chord force is 255/56 = 4.55 kips. Try a double 2×12
Douglas firLarch No. 2 chord. Due to staggered splices, compute the tension capacity based on a single
2×12, with a net area of An = 15.37 in.2 (accounting for 1in. bolt holes). Per the AF&PA Manual,
Structural Lumber Supplement:
.fT’ = (1.0)(0.8)(1.55psi)(15.37in2) = 19.1 kips > 4.55 kips OK
For chord splices, use 4 in. diameter splitring connectors with 3/4in. bolts. For split rings is single
shear, the capacity of one connector is:
.fT’ = (1.0)(0.65)(17.1) = 11.1 kips > 4.55 kips. OK
This type of chord splice connection, shown in Figure 10.113, is generally used only for heavily loaded
chords and is shown here for illustrative purposes. A typical chord splice connection for less heavily
loaded chords can be accomplished more easily by using 16d nails to splice the staggered chord members.
10.1.4.7.3 Diaphragm Deflection
The procedure for computing diaphragm deflections, contained in Commentary Sec. 12.4 (similar to the
commentary of ASCE 16), is intended for the single span, “flexible” diaphragm model that has been used
in common practice. The actual deflections of multiple span “rigid” diaphragms may, in general, be
similar to those of singlespan diaphragms because shear deflection and nail slip (both based on shear
demand) tend to dominate the behavior. As multiple span deflection computations tend to be
cumbersome, it is suggested that the design engineer compute diaphragm deflections based on the single
Chapter 10, Wood Design
1033
span model. This will result with a reasonable (and conservative), estimation of overall displacements. If
these displacements satisfy the drift criteria, then the design is assumed to be adequate; if not, then a more
rigorous computation of displacements could be performed.
It is the authors’ opinion that for diaphragm displacement computations, the diaphragm loading should be
according to Provisions Eq. 5.2.6.4.4 [5.211]; the minimum design demand (0.2SDSIwpx) need not be
considered. Although the Provisions does not provide a specific requirement either way, this
interpretation is consistent with Provisions Sec. 5.4.6.1 [5.2.6.1], which indicates that the minimum base
shear equation (Provisions Eq. 5.4.1.13) need not be used for calculation of story drift.
[See Sec. 10.1.3.1 for a discussion of diaphragm deflection computation using the 2003 Provisions and
AF&PA Wind & Seismic reference.]
From Commentary Sec. 12.4, the diaphragm deflection is computed as:
.
5 3 0.188 ( )
8 4 2
c
n
vL vL Le X
wEA Gt w
d
S .
= + + +
The equation produces the midspan diaphragm displacement in inches, and the individual variables must
be entered in the force or length units as described below. For the single span approximation, consider
the longer, 84 ft long span so that:
(84/140)
2
Fpx
v
w
=
where Fpx is the story diaphragm force at Level x, w is the diaphragm width, and v is in pounds/ft. For
this calculation, the unit shear will be the same at both ends of the diaphragm. Therefore, it
underestimates the actual unit shear at the end wall but overestimates the actual unit shear at the interior
wall. These inaccuracies are assumed to be roughly offsetting. The individual terms of the above
equation represent the following:
5vL3/8wEA = bending deflection, as derived from the formula db = 5vL4/384EI, where v is the
diaphragm unit force in pounds per ft and I = Aw2/2 (in2)
vL/4Gt = shear deflection as derived from the formula dv = vL2/8GA, where v is the diaphragm unit
force in pounds per ft and A = wt (in2)
0.188Len = nail slip deflection in inches
S(.cX)/2w = deflection due to chord slip in inches
For this example:
v = 40.4(84/140)/[(2(56))(1000)] = 216 plf (ignoring torsion)
L = 84 ft and w = 56 ft, diaphragm length and width
A = 2(1.5) 11.25 = 33.75 in.2 (double 2×12)
t = 0.298 in. for ½in. unsanded plywood
E = 1,600,000 psi
G =75,000 psi
Nail slip is computed using the same procedure as for shear walls (Sec. 10.1.4.3), with the load per nail
based on the diaphragm shear from this section. The diaphragm nailing is 8d at 6 in. on center.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1034
Load per nail = 216(6/12) = 108 lb
Nail slip en = 1.2(108/616)3.018 = 0.0063 in.
In the above equation, 1.2 is the factor for plywood other than Structural I and 616 and 3.018 are
coefficients for seasoned lumber.
Although a tight chord splice has been specified, the chord slip is computed for illustrative purposes.
Assume chord slip occurs only at the side tension since the compression splices are tightly shimmed as
shown in Figure 10.113. For this example:
.c = chord splice slip (1/16 in. used for this example)
X = distance from chord splice to nearest support
Assuming splices at 20 ft on center, along the 84ft diaphragm length (ignore diaphragm continuity for
this term), the sum of the chord splice slip is:
S(.cX) = (1/16)(20 + 40 + 24 + 4) = 5.5
Thus:
5(216)(843 ) 216(84) 0.188(84)(0.0063) 5.5
8(1,600,000)(33.75)(56) 4(75,000)(0.298) 2(56)
d= + + +
= 0.027 + 0.203 + 0.100 + 0.055 = 0.385 in.
Wall and floor drifts are added, and checked in Sec. 10.1.4.9.
10.1.4.8 Second and Third Floor Diaphragm Design
The design of the second and third floor diaphragms follows the same procedure as the roof diaphragm.
From Sec. 10.1.3.6, the diaphragm design force for both floors is Fp,3rd = Fp,2nd = 50.8 kips.
10.1.4.8.1 Diaphragm Nailing
The maximum diaphragm shear occurs at the end walls. From Sec. 10.1.3.4, each end wall segment
resists 21 percent of the total story (diaphragm, in this case) force. Distributing the diaphragm design
force at the same rate, the diaphragm shear at the end walls is:
V = (0.214)(2)(50.8) = 21.7 kips
v = 21.7/56 ft = 0.388 klf
Try ½in. (15/32) plywood rated sheathing (not Structural I) on blocked 2in. Douglas firLarch members
at 16 in. on center, with 8d nails at 4 in. on center at boundaries and continuous panel edges, at 6 in. on
center at other panel edges, and 12 in. on center at intermediate framing members.
From Provisions Table 12.4.31a [AF&PA Wind&Seismic Table 4.2A], .fD' = 0.47 klf > 0.388 klfOK
10.1.4.8.2 Chord and Splice Connection
Computed as described above for the roof diaphragm, the maximum positive moment is 320 kips and the
design chord force is 5.71 kips.
Chapter 10, Wood Design
1035
By inspection, a double 2×12 chord spliced with 4in. diameter split ring connectors, as at to the roof
level, is adequate. A typical chord splice connection is shown in Figure 10.113.
10.1.4.8.3 Diaphragm Deflection
Using the same procedure as before:
L = 84 ft and w = 56 ft, diaphragm length and width
A = 2(1.5) 11.25 = 33.75 in.2 (double 2×12)
t = 0.298 in. for ½in. unsanded plywood
E = 1,600,000 psi
G = 75,000 psi
As discussed previously, diaphragm deflection computations need not consider the minimum diaphragm
design forces.
Therefore, at the third floor:
v = 50.1(84/140)/[(2(56))(1000)] = 268 plf
Load per nail = 268(4/12) = 89 lb
Nail slip en = 1.2(89/616)3.018 = 0.0035 in.
5(268)(843 ) 268(84) 0.188(84)(0.0035) 5.5
8(1,600,000)(33.75)(56) 4(75,000)(0.298) 2(56)
d= + + +
= 0.033 + 0.252 + 0.056 + 0.055 = 0.396 in.
At the second floor:
v = 39.1(84/140)/(2(56))(1000) = 209 plf
Load per nail = 209(4/12) = 70 lb
Nail slip en = 1.2(70/616)3.018 = 0.0017 in.
5(209)(843 ) 209(84) 0.188(84)(0.0017) 5.5
8(1,600,000)(33.75)(56) 4(75,000)(0.298) 2(56)
d= + + +
= 0.026 + 0.197 + 0.026 + 0.055 = 0.304 in.
10.1.4.9 Transverse Deflections, Drift, and Pdelta Effects
Transverse deflections for walls and diaphragms were calculated above. The diaphragm deflections for
the second and third floors are based on the seismic force analysis and not the minimum diaphragm
design forces (Provisions Sec. 5.2.6.4.4 [5.2.3]).
To determine the maximum story deflections and drifts (see Section 10.1.2.2), the midspan diaphragm
deflection is combined with the wall deflection. This is summarized in Table 10.16, which shows the
drift below the level considered.
Table 10.16 Total Deflection and Drift
Level Wall (in.) Diaphragm (in.) Total de (in.) d = Cdde/I (in.) .,drift (in.)
FEMA 451, NEHRP Recommended Provisions: Design Examples
1036
Roof 1.14 0.38 1.52 6.08 1.00
3 0.87 0.40 1.27 5.08 2.08
2 0.45 0.30 0.75 3.00 3.00
1.0 in. = 25.4 mm.
The maximum permissible drift is 0.020 hs = 2.16 in. Therefore, the drift limitations are satisfied at the
third floor and roof. The drift in the first story is about 38 percent too large so the design must be revised.
One beneficial aspect of the design that has been ignored in these calculations is the lightweight concrete
floor fill. Although it is discontinuous at the stud walls, it will certainly stiffen the diaphragm. No
studies that would support a quantitative estimate of the effect have been found. Because the shear
deformation of the ½in. plywood diaphragm represents a significant portion of the total drift at the first
and second levels, any significant increase in shear stiffness that might be provided by the concrete would
further reduce the expected drifts. In this case, about 60 percent of the drift is contributed by the shear
walls. The diaphragm stiffness would need to be tripled to meet the drift criteria, so the shear walls will
be revised.
If Structural I plywood is used for the shear walls in this direction, the drift criteria are satisfied at all
levels. In Sec. 10.1.4.4, G becomes 90,000 ksi, and t (effective thickness) becomes 0.535 in. The
deflections due to shear and nail slip (the 1.2 factor no longer applies) are reduced. The resulting total
elastic deflection of the shear walls at Levels 2, 3, and Roof are, 0.242 in., 0.488 in., and 0.669 in.,
respectively. Table 10.16b shows the revised results, which satisfy the drift criteria.
Table 10.16b Total Deflection and Drift (Structural I Plywood Shear Walls)
Level Wall (in.) Diaphragm (in.) Total de (in.) d = Cdde/I (in.) .,drift (in.)
Roof 0.67 0.38 1.05 4.20 0.64
3 0.49 0.40 0.89 3.56 1.40
2 0.24 0.30 0.54 2.16 2.16
1.0 in. = 25.4 mm.
Because the tiedown calculations (Sec. 10.1.4.5) depend on the tabulated capacities of the shear wall
panels and Structural I panels have slightly higher capacities, the connection designs must be verified.
Additional calculations (not shown here) confirm that the hardware that was previously selected still
works. It is also worth noting that although many designers fail to perform deflection calculations for
wood construction, the drift criteria control the selection of sheathing grade in this example.
The Pdelta provision also must be examined. Following Provisions Sec. 5.4.6.2 [5.2.6.2] and assuming
the total mass deflects twothirds of the maximum diaphragm deflection, the Pdelta coefficient is as
shown in Table 10.17.
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions.
The importance factor, I, has been added to 2003 Provisions Eq. 5.216. However, this does not affect
this example because I = 1.0.]
Table 10.17 Pdelta Stability Coefficient
Level PD
(kips)
PL
(kips)
SP
(kips)
.
(in.)
V
(kips) . = P./VhCd
Chapter 10, Wood Design
1037
Roof 183 204 387 0.67 40.4 0.015
3 284 130 801 1.27 82.3 0.029
2 284 130 1,215 1.76 103.2 0.048
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
For example, to compute the effective Pdelta drift at the roof:
droof = Cd[dwall + 2/3(ddiaphragm)] = 4[0.67 + 2/3(0.38)] = 3.693 in.
d3 = 4[0.49 + 2/3(0.40)] = 3.027 in.
d = 3.693  3.027 = 0.67 in.
The story dead load is the same as shown in Table 10.11, and the story live load is based on 25 psf snow
load for the roof and 16 psf reduced live load (0.4×40 psf) acting over the entire area of Levels 2 and 3.
For . < 0.10, no deflection amplification due to Pdelta effects is necessary.
10.1.4.10 Longitudinal Direction
Only one exterior shear wall section will be designed here. The design of the corridor shear walls would
be similar to the transverse walls. (This example has assumed a greater length of corridor wall than
would likely be required to resist the design forces. This increased length is intended to reduce the
demand to the exterior walls, assuming rigid diaphragm distribution, to a level below the maximum
permitted inplane shear for the perforated shear wall design procedure as discussed below.) For loads in
the longitudinal direction, diaphragm stresses and deflections are negligible.
The design of the exterior wall will utilize the guidelines for perforated shear walls (Provisions Sec.
12.4.3 [AF&PA Wind&Seismic Sec. 4.3]), which are new to the 2000 Provisions. [The provisions for
perforated shear walls are contained in AF&PA Wind & Seismic and therefore have been removed from
the 2003 Provisions. The design provisions are spread throughout AF&PA Wind & Seismic Sec. 4.3.3,
but are not substantially different for the provisions contained in the 2000 Provisions except for the
revisions to the tie down requirements as noted below.] The procedure for perforated shear walls applies
to walls with openings that have not been specifically designed and detailed for forces around the
openings. Essentially, a perforated wall is treated in its entirety, rather than as a series of discrete wall
piers. The use of this design procedure is limited by several conditions (Provisions Sec. 12.4.3.2
[AF&PA Wind&Seismic Sec. 4.3.5.2]), the most relevant to this example is that the factored design shear
resistance shall not exceed 0.64 klf. This requirement essentially limits the demand on perforated shear
walls such that the required factored design shear resistance is less than 0.64 klf. If the configuration
required higher design values, then walls must be added in order to reduce the demand.
The main aspects of the perforated shear wall design procedure are as follows. The design shear capacity
of the shear wall is the sum of the capacities of each segment (all segments shall have the same sheathing
and nailing) reduced by an adjustment factor that accounts for the geometry of the openings. Uplift
anchorage (tiedown) is required only at the ends of the wall (not at the ends of all wall segments), but all
wall segments must resist a specified tension force (using anchor bolts at the foundation and with
strapping or other means at upper floors). Requirements for shear anchorage and collectors (drag struts)
across the openings are also specified. It should be taken into account that the design capacity of a
perforated shear wall, is less than a standard segmented wall with all segments restrained against
overturning. However, the procedure is useful in eliminating interior hold downs for specific conditions
and, thus, is illustrated in this example.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1038
The portion of the story force resisted by each exterior wall was computed previously as 0.225Fx. The
exterior shear walls are composed of three separate perforated shear wall segments (two at 30 ft long and
one at 15 ft long, all with the same relative length of full height sheathing), as shown in Figure 10.12.
This section will focus on the design of a 30ft section. Assuming that load is distributed to the wall
sections based on relative length of shear panel, then the total story force to the 30ft section is
(30/75)0.225Fx = 0.090Fx per floor. The load per floor is
Froof = 0.090(40.4) = 3.64 kips
F3rd = 0.090(41.9) = 3.77 kips
F2nd = 0.090(20.9) = 1.88 kips
S = 9.29 kips
10.1.4.10.1 Perforated Shear Wall Resistance
The design shear capacity (Provisions Sec. 12.4.3.3 [AF&PA Wind&Seismic Table 4.3.3.4]) is computed
as the factored shear resistance for the sum of the wall segments, multiplied by an adjustment factor that
accounts for the percentage of full height (solid) sheathing and the ratio of the maximum opening to the
story height. At each level, the design shear capacity, Vwall, is:
Vwall = (vC0)SLi
where
v = unadjusted factored shear resistance (Provisions Table 12.4.32a [Table 12.4.3a or AF&PA
Wind&Seismic Table 4.3A).
C0 = shear capacity adjustment factor (Provisions Table 12.4.31 [AF&PA Wind&Seismic Table
4.3.3.4]) = 0.83
The percent of full height sheathing is (4 + 10 + 4)/30 = 0.60, and the maximum opening height ratio is 4
ft /8 ft = 0.5. Per Provisions Table 12.4.31 [AF&PA Wind&Seismic Table 4.3.3.4], C0 = 0.83.
SLi = sum of widths of perforated shear wall segments = 4 + 10 + 4 = 18 ft
Chapter 10, Wood Design
1039
Froof
4'0" 6'0" 10'0" 6'0" 4'0"
F3rd
F2nd
2'6" 2'6" 4'0"
Floor
framing
Perforated
shear wall
h = 8'0"
Provide
tiedown
for uplift
at ends
Perforated
shear wall
segment
Figure 10.114 Perforated shear wall at exterior (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm)
The wall geometry (and thus the adjustment factor and total length of wall segments) is the same at all
three levels, as shown in Figure 10.114. Perforated shear wall plywood and nailing are determined
below.
Roof to third floor:
V = 3.64 kips
Required v = 3.64/0.83/18 = 0.244 klf
Try ½in. (15/32) Structural I plywood rated sheathing on blocked 2in. Douglas firLarch members at 16
in. on center with 10d common nails at 6 in. on center at panel edges and 12 in. on center at intermediate
framing members. (Structural I plywood would not be required to satisfy the strength requirements.
However, to minimize the possibility for construction errors, the same grade of sheathing is used on walls
in both directions.)
From Provisions Table 12.4.32a [AF&PA Wind&Seismic Table 4.3A], .fD’ = 0.44 klf > 0.244 klf OK
Third floor to second floor:
V = 3.64 + 3.77 = 7.41 kips
Required v = 7.41/0.83/18 = 0.496 klf
FEMA 451, NEHRP Recommended Provisions: Design Examples
1040
Try ½in. (15/32) Structural I plywood rated sheathing on blocked 2 in. Douglas firlarch members at 16
in. on center with 10d common nails at four in. on center at panel edges and 12 in. on center at
intermediate framing members.
From Provisions Table 12.4.32a [AF&PA Wind&Seismic Table 4.3A]:
.fD’ = 0.66 klf (0.64 klf max) > 0.496 klf OK
As discussed above, Provisions Sec. 12.4.3.2, Item b [12.4.3], limits the factored shear resistance in
Provisions Table 12.4.32a [AF&PA Wind&Seismic Table 4.3A] to 0.64 klf, which still exceeds the
demand at this level, so the limitation is satisfied.
Second floor to first floor:
V = 7.41 + 1.88 = 9.29 kips
Required v = 9.29/0.83/18 = 0.622 klf
By inspection, the same plywood and nailing from above will work.
10.1.4.10.2 Perforated Shear Wall Uplift Anchorage
According to Provisions Sec. 12.4.3.4.1 [AF&PA Wind&Seismic Table 4.3A], the requirements for uplift
anchorage must be evaluated at the ends of the wall only. Uplift at each wall segment is treated
separately as described later. Uplift forces, based on the strength of the shear panels (per Provisions Sec.
12.4.2.4 [12.4.2.4.1]), are to be computed as discussed in Sec. 10.1.4.5. For this example, calculations
involving seismic overturning and counterbalancing moments are assumed not to be applicable for
perforated shear walls, as they are not expected to act as rigid bodies in resisting global overturning.
The tiedown design force is determined as O0/1.3 times the factored shear resistance of the shear panels,
as discussed previously. For this example, the tiedown will be designed at the first floor only; the other
floors would be computed similarly, and tiedown devices, as shown in Figure 10.18, would be used.
[The requirements for uplift anchorage are slightly different in the 2003 Provisions. The tie down force is
based on the “nominal strength of the shear wall” rather than the O0/1.3 times the factored resistance as
specified in the 2000 Provisions. This change is primarily intended to provide consistent terminology and
should not result in a significant impact on the design of the tie down.]
The uplift forces are computed as:
Roof: T = 0.44 klf (3.0/1.3) (8 ft) = 8.12 kips
Third floor: T = 0.66 klf (3.0/1.3) (8 ft) = 12.18 kips
Second floor: T = 0.66 klf (3.0/1.3) (8 ft) = 12.18 kips
S = 32.48 kips
Since the chord member supports the window header as well, use a 6×6 Douglas firLarch No. 1 similar
to the transverse walls. Try a double tiedown device with a 7/8in. anchor bolt and three 7/8in. stud
bolts. Using the method described above for computing the strength of a double tiedown, the nominal
design strength is 34.3 kips, which is greater than the demand of 32.48 kips.
The design of the tiedowns at the second and third floors is similar.
10.1.4.10.3 Perforated Shear Wall Compression Chords
Chapter 10, Wood Design
1041
Provisions Sec. 12.4.3.4.4 [AF&PA Wind&Seismic Sec. 4.3.6.1], requires each end of a perforated shear
wall to have a chord member designed for the following compression force from each story:
0S i
C Vh
C L
=
where
V = design shear force in the shear wall (not wall capacity as used for uplift)
h = shear wall height (per floor)
C0 = shear capacity adjustment factor
SLi = sum of widths of perforated shear wall segments
For h = 8 ft, C0 = 0.83 and SLi = 18 ft, the compression force is computed as:
Third floor: C = 3.64(8)/0.83/18 = 1.95 kips
Second floor: C = (3.64 + 3.77)(8)/0.83/18 = 3.96 kips
First floor: C = (3.64 + 3.77 + 1.88)(8)/0.83/18 = 4.98 kips
S = 10.89 kips
Again, just the chord at the first floor will be designed here; the design at the upper floors would be
similar. Although not explicitly required by Provisions Sec. 12.4.3.4.4 [AF&PA Wind&Seismic Sec.
4.3.6.1], it is rational to combine the chord compression with gravity loading (using the load combination
1.38D + 1.0QE + 0.5L + 0.2S in accordance with Provisions Eq. 5.2.7.11 [4.21]), in order to design the
chord member. The end post of the longitudinal shear wall supports the same tributary weight at the end
post of the transverse shear walls. Using the weights computed previously in Sec. 10.1.4.5, the design
compression force is:
1.38(5.17) + 1.0(10.89) + 0.5(4.48) + 0.2(1.30) = 20.5 kips.
The bearing capacity on the bottom plate was computed previously as 31.5 kips, which is greater than
20.5 kips. Where end posts are loaded in both directions, orthogonal effects must be considered in
accordance with Provisions Sec. 5.2.5.2 [4.4.2.3].
10.1.4.10.4 Anchorage at Shear Wall Segments
The anchorage at the base of a shear wall segment (bottom plate to floor framing or foundation wall), is
designed per Provisions Sec. 12.4.3.4.2 [AF&PA Wind&Seismic Sec. 4.3.6.4]. While this anchorage
need only be provided at the full height sheathing, it is usually extended over the entire length of the
perforated shear wall to simplify the detailing and reduce the possibility of construction errors.
0 i
v V
C L
=
S
where
V = design shear force in the shear wall
C0 = shear capacity adjustment factor
SLi = sum of widths of perforated shear wall segments
FEMA 451, NEHRP Recommended Provisions: Design Examples
1042
This equation is the same as was previously used to compute unit shear demand on the wall segments.
Therefore, the inplane anchorage will be designed to meet the following unit, inplane shear forces:
Third floor: v = 0.244 klf
Second floor: v = 0.496 klf
First floor: v = 0.622 klf.
In addition to resisting the inplane shear force, Provisions Sec. 12.4.3.4.3 [AF&PA Wind&Seismic Sec.
4.3.6.4] requires that the shear wall bottom plates be designed to resist a uniform uplift force, t, equal to
the unit inplane shear force. Per Provisions Sec. 12.4.3.4.5 [AF&PA Wind&Seismic Sec. 4.3.6.4], this
uplift force must be provided with a complete load path to the foundation. That is, the uplift force at each
level must be combined with the uplift forces at the levels above (similar to the way overturning moments
are accumulated down the building).
At the foundation level, the unit inplane shear force, v, and the unit uplift force, t, are combined for the
design of the bottom plate anchorage to the foundation wall. The design unit forces are:
Shear: v = 0.622 klf
Tension: t = 0.244 + 0.496 + 0.622 = 1.36 klf
Assuming that stresses on the wood bottom plate govern the design of the anchor bolts, the anchorage is
designed for shear (single shear, woodtoconcrete connection) and tension (plate washer bearing on
bottom plate). The interaction between shear and tension need not be considered in the wood design for
this configuration of loading. As for the transverse shear walls, try a 5/8in. bolt at 32 in. on center with a
3in. square plate washer (Provisions Sec. 12.4.2.4 [12.4.2.4.2] requires 1/4×3×3 in. plate washer for
5/8in. anchor bolts). As computed previously, the shear capacity is 0.91 klf so the bolts are adequate.
For anchor bolts at 32 in. on center, the tension demand per bolt is 1.36 klf (32/12) = 3.63 kips. Bearing
capacity of the plate washer (using a Douglas fir No. 2 bottom plate) is computed per the AF&PA
Manual, Structural Connections Supplement, as:
.fP’ = (1.0)(0.8)(1.30 ksi)(9 in.2) = 9.36 kips > 3.63 kips OK
The anchor bolts themselves must be designed for combined shear, and tension in accordance with
Provisions Sec. 9.2 [11.2].
In addition to designing the anchor bolts for uplift, a positive load path must be provided to transfer the
uplift forces into the bottom plate. One method for providing this load path continuity, is to use metal
straps nailed to the studs and lapped around the bottom plate, as shown in Figure 10.115. Attaching the
studs directly to the foundation wall (using embedded metal straps) for uplift and using the anchor bolts
for shear only is an alternative approach.
Chapter 10, Wood Design
1043
1
4x3x3 steel plate
washer at anchor bolt
Metal strap used
to resist uplift,
lapped under
sill plate.
5
8" dia. anchor
bolt at 2'8" o.c.
(Used to resist
shear and tension)
Nail to
stud
Figure 10.115 Perforated shear wall detail at foundation (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm).
Nail to stud
(typical)
Nails to resist shear
Sheet metal framing
clips (Under blocking)
used to resist shear.
Metal strap used
to resist uplift
Figure 10.116 Perforated shear wall detail at floor framing.
At the upper floors, the load transfer for inplane shear is accomplished by using nailing or framing clips
between the bottom plates, rim joists, and top plates in a manner similar to that for standard shear walls.
The uniform uplift force can be resisted either by using the nails in withdrawal (for small uplift demand)
or by providing vertical metal strapping between studs above and below the level considered. This type
of connection is shown in Figure 10.116. For this type of connection (and the one shown in Figure 10.1
15) to be effective, shrinkage of the floor framing must be minimized using dry or manufactured lumber.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1044
~
~
Typ. articulated
gluelam beams
Concrete
masonry walls
Typ. roof joists
at 24" o.c.
Steel column on
spread footing
Plywood roof
sheathing
Open
5 bays at 40'0" = 200'0"
5 bays at 20'0" = 100'0"
Figure 10.21 Building plan (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm).
For example, consider the second floor. The required uniform uplift force, t = 0.244 + 0.496 = 0.740 klf.
Place straps at every other stud, so the required strap force is 0.740 (32/12) = 1.97 kips. Provide an 18
gauge strap with 1210d nails at each end.
10.2 WAREHOUSE WITH MASONRY WALLS AND WOOD ROOF, LOS ANGELES,
CALIFORNIA
This example features the design of the wood roof diaphragm, and walltodiaphragm anchorage for the
onestory masonry building, described in Sec. 9.1 of this volume of design examples. Refer to that
example for more detailed building information and the design of the masonry walls.
10.2.1 Building Description
This is a very simple rectangular warehouse, 100 ft by 200 ft in plan (Figure 10.21), with a roof height of
28 ft. The wood roof structure slopes slightly, but it is nominally flat. The long walls (side walls) are 8
in. thick and solid, and the shorter end walls are 12 in. thick and penetrated by several large openings.
Based on gravity loading requirements, the roof structure consists of wood joists, supported by 8¾in.
wide, by 24in. deep, gluedlaminated beams, on steel columns. The joists span 20 ft and the beams span
40 ft, as an articulated system. Typical roof framing is assumed to be Douglas firLarch No.1 as graded
by the WWPA. The gluedlaminated beams meet the requirements of combination 24FV4 per
ANSI/AITC A190.1.
The plywood roof deck acts as a diaphragm to carry lateral loads to the exterior walls. There are no
interior walls for seismic resistance. The roof contains a large opening that interrupts the diaphragm
continuity.
The diaphragm contains continuous cross ties in both principal directions. The details of these cross ties
and the masonry walltodiaphragm anchorage are substantially different from those shown in previous
versions of this example. This is primarily due to significant revisions to the Provisions requirements for
anchorage of masonry (and concrete) walls to flexible wood diaphragms.
Chapter 10, Wood Design
1045
The following aspects of the structural design are considered in this example:
1. Development of diaphragm forces based on the equivalent lateral force procedure used for the
masonry wall design ( Sec. 9.1)
2. Design and detailing of a plywood roof diaphragm with a significant opening
3. Computation of drift and Pdelta effects
4. Anchorage of diaphragm and roof joists to masonry walls and
5. Design of cross ties and subdiaphragms
[Note that as noted in Sec. 9.1, the new “Simplified Design Procedure” contained in 2003 Provisions
Simplified Alternate Chapter 4 as referenced by 2003 Provisions Sec. 4.1.1 is likely to be applicable to
this example, subject to the limitations specified in 2003 Provisions Sec. Alt. 4.1.1.]
10.2.2 Basic Requirements
10.2.2.1 Provisions Parameters
SS (Provisions Maps [Figure 3.3.3]) = 1.50
S1 (Provisions Maps [Figure 3.3.4]) = 0.60
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = C
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I
Seismic Design Category (Provisions Sec. 4.2 [1.4]) = D
Seismic Force Resisting System (Provisions Table 5.2.2 [4.31]) = Special reinforced
masonry shear wall
Response Modification Factor, R (Provisions Table 5.2.2 [4.31]) = 3.5
System Overstrength Factor, O0 (Provisions Table 5.2.2 [4.31]) = 2.5
Deflection Amplification Factor, Cd (Provisions Table 5.2.2 4.31]) = 3.5
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package).]
10.2.2.2 Structural Design Criteria
A complete discussion on the criteria for ground motion, seismic design category, load path, structural
configuration, redundancy, analysis procedure, and shear wall design, is included in Sec. 9.1 of this
volume of design examples.
10.2.2.2.1 Design and Detailing Requirements (Provisions Sec. 5.2.6 [4.6])
See Provisions Chapter 12, for wood design requirements. As discussed in greater detail in Sec. 10.1,
Provisions Sec. 12.2.1 utilizes load and resistance factor design (LRFD) for the design of engineered
wood structures. The design capacities are therefore, consistent with the strength design demands of
Provisions Chapter 5.
The large opening in the diaphragm must be fitted with edge reinforcement (Provisions Sec. 5.2.6.2.2
[4.6.1.4]). However, the diaphragm does not require any collector elements that would have to be
designed for the special load combinations (Provisions Sec. 5.2.6.4.1 [4.6.2.2]).
The requirements for anchoring of masonry walls to flexible diaphragms (Provisions Sec. 5.2.6.3.2
[4.6.2.1]) are of great significance in this example.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1046
10.2.2.2.2 Combination of Load Effects (Provisions Sec. 5.2.7 [4.2.2])
The basic design load combinations for the lateral design, as stipulated in ASCE 7, and modified by the
Provisions Eq. 5.2.71 and 5.2.72 [4.21 and 4.22], were computed in Sec. 9.1 of this volume of design
examples as:
1.4D + 1.0QE
and
0.7D  1.0QE.
The roof live load, Lr, is not combined with seismic loads, and the design snow load is zero for this Los
Angeles location.
10.2.2.2.3 Deflection and Drift Limits (Provisions Sec. 5.2.8 [4.5.1])
Inplane deflection and drift limits for the masonry shear walls are considered in Sec. 9.1.
As illustrated below, the diaphragm deflection is much greater than the shear wall deflection. According
to Provisions Sec. 5.2.6.2.6 [4.5.2], inplane diaphragm deflection shall not exceed the permissible
deflection of the attached elements. Because the walls are essentially pinned at the base, and simply
supported at the roof, they are capable of accommodating large deflections at the roof diaphragm.
For illustrative purposes, story drift is determined and compared to the requirements of Provisions Table
5.2.8 [4.51]. However, according to this table, there is essentially no drift limit for a single story
structure as long as the architectural elements can accommodate the drift (assumed to be likely in a
warehouse structure with no interior partitions). As a further check on the deflection, Pdelta effects
(Provisions Sec. 5.4.6.2 [5.2.6.2]) are evaluated.
10.2.3 Seismic Force Analysis
Building weights and base shears are as computed in Sec. 9.1 of this volume of design examples. (The
building weights used in this example are based on a preliminary version of Example 9.1 and, thus, minor
numerical differences may exist between the two examples). Provisions Sec. 5.2.6.4.4 [4.6.3.4]specifies
that floor and roof diaphragms be designed to resist a force, Fpx, in accordance with Provisions Eq.
5.2.6.4.4 [4.62]as follows:
n
i
i x
px n px
i
i x
F
F w
w
=
=
S
=
S
plus any force due to offset walls (not applicable for this example). For onestory buildings, the first term
of this equation will be equal to the seismic response coefficient, Cs, which is 0.286. The effective
diaphragm weight, wpx, is equal to the weight of the roof, plus the tributary weight of the walls
perpendicular to the direction of the motion. The tributary weights are:
Roof = 20(100)(200) = 400 kips
Side walls = 2(65)(28/2+2)(200) = 416 kips
End walls = 2(103)(28/2+2)(100) = 330 kips
Chapter 10, Wood Design
1047
The diaphragm design force is computed as:
Transverse Fp,roof = 0.286(400 + 416) = 233 kips
Longitudinal Fp,roof = 0.286(400 + 330) = 209 kips
These forces exceed the minimum diaphragm design forces given in Provisions Sec. 5.2.6.4.4 [4.6.3.4],
because Cs exceeds the minimum factor of 0.2SDS.
10.2.4 Basic Proportioning of Diaphragm Elements
The design of plywood diaphragms primarily involves the determination of sheathing sizes and nailing
patterns to accommodate the applied loads. Large openings in the diaphragm and wall anchorage
requirements, however, can place special requirements on the diaphragm capacity. Diaphragm deflection
is also a consideration.
Nailing patterns for diaphragms are established on the basis of tabulated requirements included in the
Provisions. It is important to consider the framing requirements for a given nailing pattern and capacity
as indicated in the notes following the tables. In addition to strength requirements, Provisions Sec.
12.4.1.2 places aspect ratio limits on plywood diaphragms (lengthtowidth shall not exceed 4/1 for
blocked diaphragms). However, it should be taken into consideration that compliance with this aspect
ratio does not guarantee that drift limits will be satisfied.
While there is no specific limitation on deflection for this example, the diaphragm has been analyzed for
deflection as well as for shear capacity. A procedure for computing diaphragm deflections is illustrated in
detail, in Sec. 10.1.4.7.
In the calculation of diaphragm deflections, the chord splice slip factor can result in large additions to the
total deflection. This chord splice slip, however, is often negligible where the diaphragm is continuously
anchored to a bond beam in a masonry wall. Therefore, chord splice slip is assumed to be zero in this
example.
10.2.4.1 Strength of Members and Connections
The 2000 Provisions have adopted Load and Resistance Factor Design (LRFD) for engineered wood
structures. The Provisions includes the ASCE 16 standard by reference and uses it as the primary design
procedure for engineered wood construction. Strength design of members and connections is based on
the requirements of ASCE 16. The AF&PA Manual and supplements contain reference resistance values
for use in design. For convenience, the Provisions contains design tables for diaphragms that are
identical to those contained in the AF&PA StructuralUse Panels Supplement. Refer to Sec. 10.1.4.1 for
a more complete discussion of the design criteria.
[The primary reference for design of wood diaphragms in the 2003 Provisions is AF&PA Wind &
Seismic. Much of the remaining text in the 2003 Provisions results from differences between AF&PA
Wind & Seismic and Chapter 12 of the 2000 Provisions as well as areas not addressed by AF&PA Wind
& Seismic. Because the AF&PA Wind & Seismic tabulated design values for diaphragms do not
completely replace the tables in the 2000 Provisions, portions of the tables remain in the 2003 Provisions.
Therefore, some diaphragm design values are in the 2003 Provisions and some are in AF&PA Wind &
Seismic. The design values in the tables are different between the two documents. The values in the
2003 Provisions represent factored shear resistance (.fD’), while the values in AF&PA Wind & Seismic
represent nominal shear resistance that must then be multiplied by a resistance factor, f, (0.65) and a time
effect factor ,., (1.0 for seismic loads). Therefore, while the referenced tables may be different, the
factored resistance values based on the 2003 Provisions should be the same as those in examples based on
FEMA 451, NEHRP Recommended Provisions: Design Examples
1048
the 2000 Provisions. The calculations that follow are annotated to indicate from which table the design
values are taken.]
10.2.4.2 Roof Diaphragm Design for Transverse Direction
10.2.4.2.1 Plywood and Nailing
The diaphragm design force, Fp,roof = 233 kips. Accounting for accidental torsion (Provisions Sec. 5.4.4.2
[5.2.4.2]), the maximum end shear = 0.55Fp,roof = 128 kips. This corresponds to a unit shear force v =
(128/100) = 1.28 klf. Although there is not a specific requirement in the Provisions, it is the authors’
opinion that accidental torsion should be considered, even for “flexible” diaphragms, to account for the
possibility of a nonuniform mass distribution in the building.
Because the diaphragm shear demand is relatively high, the plywood sheathing and roof framing at the
ends of the building, must be increased in size over the standard roof construction for this type of
building. Assuming 3in. nominal framing, try blocked 3/4in. (23/32) Structural I plywood rated
sheathing with two lines of 10d common nails at 21/2 in. on center at diaphragm boundaries, continuous
panel edges, and two lines at 3 in. on center at other panel edges.
From Provisions Table 12.4.31a [12.41a], .fD’ = 1.60 klf > 1.28 klf OK
Because the diaphragm shear decreases towards the midspan of the diaphragm, the diaphragm capacity
may be reduced towards the center of the building. Framing size and plywood thickness are likely to
have a more significant impact on cost than nail spacing, determine a reasonable location to transition to 2
in. nominal roof joists and ½ in. (15/32) plywood. A reasonable configuration for the interior of the
building utilizes ½ in. (15/32) Structural I plywood rated sheathing with a single line of 10d at 2 ½ in. on
center nailing at diaphragm boundaries, continuous panels edges, and 4 in. on center nailing at other panel
edges. Using 2×4 flat blocking at continuous panel edges, the requirements found in Notes f, and g of
Provisions Table 12.4.31a [AF&PA Wind&Seismic Sec. 4.2.7.1] are met. Determine the distance, X,
from the end wall where the transition can be made as:
.fD’ = 0.83 klf (Provisions Table 12.4.31a [AF&PA Wind&Seismic Table 4.2A]),
Shear Capacity = 0.83(100) = 83.0 kips
Uniform Diaphragm Demand = 233/200 = 1.165 klf
X = (128  83)/1.165 = 38.6 ft, say 40 ft from the diaphragm edge
[Here is an example where both the Provisions tables and the AF&PA Wind & Seismic tables are
required to complete the design. The design value for this plywood thickness and nailing pattern is
contained in AF&PA Wind & Seismic, but the design value for the highercapacity diaphragm at the ends
is contained in the Provisions.]
In a building of this size, it may be beneficial to further reduce the diaphragm nailing towards the middle
of the roof. However, due to the requirements for subdiaphragms, (see below) and diaphragm capacity, in
the longitudinal direction and for simplicity of design, no additional nailing pattern is used.
Table 10.21 contains a summary of the diaphragm framing and nailing requirements (All nails are 10d
common). See Figure 10.22 for designation of framing and nailing zones, and Figure 10.23 for typical
plywood layout.
Table 10.21 Roof Diaphragm Framing and Nailing Requirements
Zone* Framing Plywood Nail Spacing (in.) Capacity
(kip/ft)
Chapter 10, Wood Design
1049
40'0" 100'0" 20'0" 40'0"
100'0"
Zone a Zone b Zone a
Extended zone a due to
diaphragm opening.
Figure 10.22 Diaphragm framing and nailing layout (1.0 ft = 0.3048 m).
Boundaries
and Cont.
Panel Edges
Other Panel
Edges
Intermediate
Framing
Members
a 3×12 ¾ in. 2½ (2 lines) 3 (2 lines) 12 (1 line) 1.60
b 2×12 ½ in. 2½ (1 line) 4 (1 line) 12 (1 line) 0.83
1.0 in. = 25.4 mm, 1.0 kip/ft = 14.6 kN/m.
* Refer to Figure 10.22 for zone designation.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1050
Longitudinal
Transverse
2x12 or 3x12 joists at 24" o.c.
2x4 flat blocking at edges
of plywood panels
4'x8' plywood with
grain to joists and
end joints staggered
Figure 10.23 Plywood layout (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm).
10.2.4.2.2 Chord Design
Although the bond beam at the masonry wall could be used as a diaphragm chord, this example illustrates
the design of the wood ledger member as a chord. Chord forces are computed using a simply supported
beam analogy, where the design force is the maximum moment divided by the diaphragm depth.
Diaphragm moment, M = wL2/8 = Fp,roofL/8 = 233(200/8) = 5,825 ftkips
Chord force, T = C = 5,825/(100  16/12) = 59.0 kips
Try a select structural Douglas firlarch 4×12, for the chord. Assuming two 1/16 in. bolt holes (for 1in.
bolts) at splice locations, the net chord area is 31.9 in.2 Tension strength (parallel to wood grain), per the
AF&PA Manual, Structural Lumber Supplement:
.fT’ (1.0)(0.8)(2.70)(31.9) = 68.9 kips > 59.0 kips OK
Design the splice for the maximum chord force of 59.0 kips. Try bolts with steel side plates using 1 in.
A307 bolts, with a 3½ in. length in the main member. The capacity, according to the AF&PA Manual,
Structural Connections Supplement, is:
.fZ’ = (1.0)(0.65)(16.29) = 10.6 kips per bolt.
Number of bolts required = 59.0/10.6 = 5.6
Use two rows of three bolts. The reduction (group action factor) for multiple bolts is negligible. Net area
of the 4×12 chord with two rows of 11/16 in. holes is 31.9 in.2 as assumed above. Therefore, use six 1
in. A307 bolts on each side of the chord splice (Figure 10.24). Although it is shown for illustration, this
type of chord splice may not be the preferred splice against a masonry wall since the bolts, and side plate,
would have to be recessed into the wall.
Chapter 10, Wood Design
1051
7" 4"
61" A307 bolts on
each side of chord.
Splice in tight hole.
4x12 Select
Structural
Doug. fir chord
Drive shim in joint
2 plates
1
4"x7"x2'9"
31
2"
4"
31
41 2"
31 4"
2"
Figure 10.24 Chord splice detail (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm).
10.2.4.2.3 Diaphragm Deflection and Pdelta Check
The procedure for computing diaphragm deflection is described in Sec. 10.1.4.7.
[AF&PA Wind & Seismic also contains procedures for computing diaphragm deflections. The equations
are slightly different from the more commonly used equations that appear in the Commentary and
AF&PA LRFD Manual. In AF&PA Wind & Seismic, the shear and nail slip terms are combined using an
“apparent shear stiffness” parameter. However, the apparent shear stiffness values are only provided for
OSB. Therefore, the deflection equations in the Commentary or AF&PA LRFD Manual must be used in
this example which has plywood diaphragms. The apparent shear stiffness values for plywood will likely
be available in future editions of AF&PA Wind & Seismic.]
As stated in Commentary Sec. 12.4, the diaphragm deflection is computed as:
5 3 0.188 ( )
8 4 2
c
n
vL vL Le X
wEA Gt w
d
S .
= + + +
The equation produces the midspan diaphragm displacement in inches, and the individual variables must
be entered in the force or length units as described below. A small increase in diaphragm deflection due
to the large opening is neglected. An adjustment factor, F, for nonuniform nailing is applied to the third
term of the above equation for this example.
Bending deflection = 5vL3/8wEA = 0.580 in.
Shear deflection = vL/4Gt = 1.210 in.
Effective nails slip deflection = 0.188 LenF = 0.265 in.
Deflection due to chord slip at splices = S(.cX)/2w = 0.000 in.
(The chord slip deflection is assumed to be zero because the chord is connected to the continuous bond
beam at the top of the masonry wall.)
The variables above and associated units used for computations are:
v = (233/2)/100 = 1,165 plf (shear per foot at boundary, ignoring torsion)
FEMA 451, NEHRP Recommended Provisions: Design Examples
1052
40'0" 60'0"
End Center
0.146 kips
0.233 kips
0.088 kips
1
2
Figure 10.25 Adjustment for nonuniform nail spacing (1.0 ft = 0.3048 m, 1.0 kip =
4.45 kN)
L = 200 ft and w = 100 ft (diaphragm length and depth)
A = effective area of 4×12 chord and two#6 rebars assumed to be in the bond beam
= 39.38 in.2 + 2(0.44)(29,000,000/1,900,000) = 52.81 in.2
t = 0.535 in. (effective, for ½ in. Structural I plywood, unsanded; neglect ¾ in. plywood at edge)
E = 1,900,000 psi (for Douglas firlarch select structural chord)
G = 90,000 psi (Structural I plywood)
en = nail slip for 10d nail at end of diaphragm (use one and onehalfin. nail spacing, two lines at 3 in.,
not the 11/4 in. perimeter spacing)
Design nail load = 1,165/(12/1.5) = 146 lb/nail
Using the procedure in Sec. 10.1.4.3 and the value for dry/dry lumber from Table 10.12, en =
(146/769)3.276 = 0.00433 in.
F = adjustment for nonuniform nail spacing
The 0.188 coefficient assumes a uniform nail spacing, which implies an average load per nail of onehalf the
maximum. One common practice for calculating deformation is simply to use the load per nail that would
result from the larger spacing should that spacing be used throughout. However, this gives a large estimate
for deformation due to nail slip. Figure 10.25 shows the graphic basis for computing a more accurate nail
slip term. The basic amount is taken as that for the smaller nail spacing (11/2 in.), which gives 146 lbs per
nail.
Using a larger nail spacing at the interior increases the deformation. The necessary increase may be
determined as the ratio of the areas of the triangles in Figure 10.25, which represents the load per nail
along the length of the diaphragm. The ratio is for Triangle 2 representing zone b compared to Triangle 1
representing zone a, where zones a and b are as shown Figure 10.22.
Ratio = (233  88)60 / 146(100) = 0.63
thus, F = 1 + 0.63 = 1.63.
Total for diaphragm:
d = 0.580 + 1.210 + 0.265 + 0.00 = 2.055 in.
End wall deflection = 0.037 in. (see Sec. 9.1 of this volume of design examples)
Chapter 10, Wood Design
1053
Therefore, the total elastic deflection dxe = 2.055 + 0.037 = 2.092 in.
Total deflection, dx = Cd dxe/I = 3.5(2.092)/1.0 = 7.32 in. = .
Pdelta effects are computed according to Provisions Sec. 5.4.6.2 [5.2.6.2]using the stability coefficient
computed per Provisions Eq. 5.4.6.21 [5.216]:
x
x sx d
P
V h C
.
.
=
Because the midspan diaphragm deflection is substantially greater than the deflection at the top of the
masonry end walls, it would be overly conservative to consider the entire design load at the maximum
deflection. Therefore, the stability coefficient is computed by splitting the Pdelta product into two terms
one for the diaphragm and one for the end walls.
For the diaphragm, consider the weight of the roof and side walls at the maximum displacement. (This
overestimates the Pdelta effect. The computation could consider the average displacement of the total
weight, which would lead to a reduced effective delta. Also, the roof live load need not be included.)
P = 400 + 416 = 816 kips
.= 7.32 in.
V = 233 kips (diaphragm force)
For the end walls, consider the weight of the end walls at the wall displacement:
P = 330 kips
. = (3.5)(0.037) = 0.13 in.
V = 264 kips (additional base shear for wall design)
For story height, h = 28 feet, the stability coefficient is:
816(7.32) 330(0.13) (28)(12)(3.5) 0.022
d 233 264
P P hC
V V
.
=.. .+ ... =.. + .. =
. . . .
For . < 0.10, no deflection amplification due to Pdelta effects is necessary.
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions.
The importance factor, I, has been added to 2003 Provisions Eq. 5.216. However, this does not affect
this example because I = 1.0.]
Drift index =./hsx = 7.32/[28(12)] = 0.022.
This is slightly less than the limiting drift ratio of 0.025 applied for most lowrise buildings in Seismic
Use Group I (Provisions Sec. 5.2.8 [4.5.1] and Table 5.2.8 [4.51]). However, for onestory buildings,
Table 5.2.8, Note b [4.51, Note c], and Provisions Sec. 5.2.6.2.6 [4.5.2], permit unlimited drift provided
that the structural elements and finishes can accommodate the drift. The limit for masonry cantilever
shear wall structures (0.007) should only be applied to the inplane movement of the end walls (0.13/h =
0.0004 << 0.007). The construction of the outofplane walls allows them to accommodate very large
drifts. It is further expected that the building does not contain interior elements that are sensitive to drift.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1054
Open
20'
Gluedlaminated
beams
20' 20' 40' to
diaphragm
boundary
Open
20'
Gluedlaminated
beams
20' 20'
Open
20'
Gluedlaminated
beams
20' 20' 40' to
diaphragm
boundary
Figure 10.26 Diaphragm at roof opening (1.0 ft = 0.3048 m).
Given the above conditions, and the fact that Pdelta effects are not significant for this structure, the
computed diaphragm deflections appear acceptable.
10.2.4.2.4 Detail at Opening
Consider diaphragm strength at the roof opening (Figure 10.26), as required by Provisions Sec. 5.2.6.2.2
[4.6.1.4].
Check diaphragm nailing for
required shear area (shear in diaphragm at edge of opening):
Shear = 128  [40(1.165)] = 81.4 kips
v = 81.4/(100  20) = 1.02 klf
Because the opening is centered in the width of the diaphragm, half the force to the diaphragm must be
distributed on each side of the opening.
Diaphragm capacity in this area = 0.830 klf as computed previously (see Table 10.21 and Figure 10.22).
Because the diaphragm demand at the reduced section exceeds the capacity, the extent of the Zone a
nailing and framing should be increased. For simplicity, extend the Zone a nailing to the interior edge of
the opening (60 ft from the end wall). The diaphragm strength is now adequate for the reduced overall
width at the opening.
10.2.4.2.5 Framing around Opening
The opening is located 40 ft from one end of the building and is centered in the other direction (Figure
10.26). This does not create any panels with very high aspect ratios.
In order to develop the chord forces, continuity will be required across the gluedlaminated beams in one
direction and across the roof joists in the other direction.
10.2.4.2.6 Chord Forces at Opening
To determine the chord forces on the edge joists, model the diaphragm opening as a Vierendeel truss and
assume the inflection points will be at the midpoint of the elements (Figure 10.27). Compute forces at
the opening using a uniformly distributed diaphragm demand of 233/200 = 1.165 klf.
For Element 1 (shown in Figure 10.27):
Chapter 10, Wood Design
1055
3
1 2
Element 1
109 kips
1.165 kip/ft
w1
M1
V1A
M1
V1 B
40'0"
20'0" 40'0"
Figure 10.27 Chord forces and Element 1 freebody diagram (1.0 ft = 0.3048
m, 1.0 kip = 4.45 kN, 1.0 kip/ft = 14.6 kN/m).
w1 = 1.165/2 = 0.582 kips/ft
V1B = 0.5[128(40)(1.165)] = 40.7 kips
V1A = 40.7  20(0.582) = 29.1 kips
M1 due to Vierendeel action = (½)[40.7(10) + 29.1(10)] = 349 ftkips
Chord force due to M1 = 349/40 = 8.72 kips. This is only 41 psi on the gluedlaminated beam on the edge
of the opening. This member is adequate by inspection. On the other side of this diaphragm element, the
chord force is much less than the maximum global chord force (59.0 kips), so the ledger and ledger splice
are adequate.
For Element 3, analyze Element 2 (shown in Figure 10.28):
w2 = 1.165(40/100) = 0.466 kips/ft
V3 = 128(40/100) = 51.2 kips
V1B = 40.7 kips
M1 = 349 ftkips.
T1B is the chord force due to moment on the total diaphragm:
M = 128(40)  1.165(402/2) = 4,188 ftkips
T1B = 4,188/100 = 41.9 kips
SM0: M3 = M1 + 40V3  40T1B  w2402/2 = 348 ftkips
Therefore, the chord force on the roof joist = 348/40 = 8.7 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
1056
M1
M 3 = M
V3
2
O w
2
V1B
T1B
2
Figure 10.28 Freebody
diagram for Element 2.
Alternatively, the chord design should consider the wall anchorage force interrupted by the opening. As
described in Sec. 10.2.4.3, the edge members on each side of the opening are used as continuous crossties,
with maximum crosstie force of 25.0 kips. Therefore, the crosstie will adequately serve as a chord
at the opening.
10.2.4.3 Roof Diaphragm Design for Longitudinal Direction
Force = 209 kips
Maximum end shear = 0.55(209) = 115 kips
Diaphragm unit shear v = 115/200 = 0.58 klf
For this direction, the plywood layout is Case 3 in Provisions Table 12.4.31a [AF&PA Wind&Seismic
Table 4.2A]. Using ½ in. Structural I plywood rated sheathing, blocked, with 10d common nails at 2 ½
in. on center at diaphragm boundaries and continuous panel edges parallel to the load (ignoring the
capacity of the extra nails in the outer zones):
.fD’ = 0.83 plf > 0.58 plf, Provisions Table 12.4.31a [AF&PA Wind&Seismic Table 4.2A] OK
Therefore, use the same nailing designed for the transverse direction. Compared with the transverse
direction, the diaphragm deflection and Pdelta effects will be satisfactory.
10.2.4.4 Masonry Wall Anchorage to Roof Diaphragm
As stipulated in Provisions Sec. 5.2.6.3.2 [4.6.2.1], masonry walls must be anchored to flexible
diaphragms to resist outofplane forces computed per Provisions Eq. 5.2.6.3.2 [4.61]as:
FP = 1.2SDSIWp = 1.2(1.0)(1.0)Wp = 1.2 Wp
Side walls, FP = 1.2(65psf)(2+28/2)/1000 = 1.25 klf
End walls, FP = 1.2(103psf)(2 + 28/2)/1000 = 1.98 klf
[In the 2003 Provisions the anchorage force for masonry walls connected to a flexible diaphragm has
been reduced to 0.8SDSIWp.]
10.2.4.3.1 Anchoring Joists Perpendicular to Walls (Side Walls)
Chapter 10, Wood Design
1057
Because the roof joists are spaced at 2 ft on center, provide a connection at each joist that will develop
2(1.25) = 2.50 kip/joist.
A common connection for this application is metal tension tie or holddown device that is anchored to the
masonry wall with an embedded bolt and is either nailed or bolted to the roof joist. Other types of
anchors include metal straps that are embedded in the wall and nailed to the top of the joist. The ledger is
not used for this force transfer because the eccentricity between the anchor bolt and the plywood creates
tension perpendicular to the grain in the ledger (crossgrain bending), which is prohibited. Also, using
the edge nails to resist tension perpendicular to the edge of the plywood, is not permitted.
Try a tension tie with a 3/4 in. headed anchor bolt, embedded in the bond beam and 18 10d nails into the
side of the joist (Figure 10.210). Modifying the allowable values using the procedure in Sec. 10.1.4.5
results in a design strength of:
.fZ’ = (1.0)(0.65)(4.73) = 3.07 kips per anchor > 2.50 kips OK
The joists anchored to the masonry wall must also be adequately connected to the diaphragm sheathing.
Determine the adequacy of the typical nailing for intermediate framing members. The nail spacing is 12
in. and the joist length is 20 ft, so there are 20 nails per joist. From the AF&PA Manual, Structural
Connections Supplement, the strength of a single 10d common nail is:
.fZ’ = (1.0)(0.65)(0.298) = 0.194 kips per nail
20(0.194) = 3.88 kips > 2.50 kips OK
The embedded bolt also serves as the ledger connection, for both gravity loading, and inplane shear
transfer at the diaphragm. Therefore, the strength of the anchorage to masonry, and the strength of the
bolt in the wood ledger, must be checked.
For the anchorage to masonry, check the combined tension and shear, resulting from the outofplane
seismic loading (2.50 kips/bolt), and the vertical gravity loading. Assuming 20 psf dead load (roof live
load is not included), and a 10foot tributary roof width, the vertical load per bolt = (20 psf)(10 ft)(2
ft)/1000 = 0.40 kip. Using the load combination described previously, the design horizontal tension and
vertical shear on the bolt are:
ba = 1.0QE = 2.50 kips
bv = 1.4D = 1.4(0.40) = 0.56 kip
The anchor bolts in masonry, are designed according to Provisions Sec. 11.3.12. Using 3/4 in. headed
anchor bolts, both axial strength, Ba, and shear strength, Bv, will be governed by masonry breakout. Per
Provisions Eq. 11.3.12.31, and 11.3.12.11, respectively, the design strengths are:
Ba= 4f Ap fm'
Bv=1750f 4fm' Ab
where f = 0.5, f’m = 2,000 psi, Ab = tensile area of bolt = 0.44 in.2, and Ap = projected area on the masonry
surface of a right circular cone = 113 in.2 (assuming 6 in. effective embedment). Therefore,
Ba = 10.1 kips and Bv = 4.8 kips. Shear and tension are combined per Provisions Eq. 11.3.12.4 as:
2.50 0.56 0.36 1.0 OK
10.1 4.8
a v
a v
b b
B B
+ = + = <
FEMA 451, NEHRP Recommended Provisions: Design Examples
1058
Bond beam at top
Vert reinf. to top
Bond beam
with 2#7 cont.
for chord
4x12 ledger
2x12 or 3x12 joist
with joist hanger
Structural sheathing
(See plan for thickness
and nailing)
Tension tie (e.g. Simpson
LTT131) at each joist
3
4" dia. bolt at tension
tie (24" o.c.)
Figure 10.29 Anchorage of masonry wall perpendicular to joists (1.0 in. = 25.4 mm).
[The 2003 Provisions refers to ACI 530, Sec. 3.1.6 for strength design of anchorage to masonry, as
modified by 2003 Provisions Sec. 11.2. In general, the methodology for anchorage design using ACI 530
should be comparable to the 2000 Provisions, though some of the equations and reduction factors may be
different. In addition, the 2003 Provisions require that anchors are either controlled by a ductile failure
mode or are designed for 2.5 times the anchorage force.]
Figure 10.29 summarizes the details of the connection. Inplane seismic shear transfer (combined with
gravity) and orthogonal effects are considered in a subsequent section.
According to Provisions Sec. 5.2.6.3.2 [4.6.2.1, diaphragms must have continuous crossties to distribute
the anchorage forces into the diaphragms. Although the Provisions do not specify a maximum spacing,
20 ft is common practice for this type of construction and Seismic Design Category.
For crossties at 20 ft on center, the wall anchorage force per crosstie is:
(1.25 klf)(20 ft) = 25.0 kips
Try a 3×12 (Douglas firLarch No. 1) as a crosstie. Assuming one row of 15/16 in. bolt holes, the net
area of the section is 25.8 in.2 Tension strength (parallel to wood grain) per the AF&PA Structural
Lumber Supplement, is:
.fT’ = (1.0)(0.8)(1.82)(25.8) = 37.5 kips > 25.0 kips OK
At the splices, try a double tiedown device with four 7/8 in. bolts in double shear through the 3×12
(Figure 10.210). Product catalogs provide design capacities for single tiedowns only; the design of
double hold downs requires two checks. First, consider twice the capacity of one tiedown and, second,
consider the capacity of the bolts in double shear.
For the double tiedown, use the procedure in Sec. 10.1.4.5 to modify the allowable values:
2.fZ’ = 2(1.0)(0.65)(20.66) = 26.9 kips > 25.0 kips OK
Chapter 10, Wood Design
1059
3x12 with
joist hanger
Gluelam beam
3x12 with
joist hanger
7
8" dia. threaded
rod at hold downs
Plywood with boundary
nailing along 3x12
Tiedown on both sides
of 3x12 with 47
8" dia. thru
bolts. (Each side of beam)
Figure 10.210 Chord tie at roof opening (1.0 in. = 25.4 mm).
The reduction (group action) factor for multiple bolts, Cg = 0.97.
For the four bolts, the AF&PA Manual, Structural Connections Supplement, gives:
4.fCgZ’ = 4(1.0)(0.65)(0.97)(10.17) = 25.6 kips > 25.0 kips
OK
In order to transfer the wall anchorage forces into the crossties, the subdiaphragms between these ties
must be checked per Provisions Sec. 5.2.6.3.2 [4.6.2.1]. There are several ways to perform these
subdiaphragm calculations. One method is illustrated in Figure 10.211. The subdiaphragm spans
between crossties and utilizes the gluedlaminated beam and ledger as it’s chords. The 1to1 aspect
ratio meets the requirement of 2½ to 1 for subdiaphragms per Provisions Sec. 5.2.6.3.2 [4.2.6.1].
For the typical subdiaphragm (Figure 10.211):
Fp = 1.25 klf
v = (1.25)(20/2)/20 = 0.625 klf.
The subdiaphragm demand is less than the minimum diaphragm capacity (0.83 klf along the center of the
side walls). In order to develop the subdiaphragm strength, and boundary nailing must be provided along
the crosstie beams.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1060
Fp
V V
Subdiaphragm
F p = 1.25 kips/ft
Subdiaphragm
20'0" 20'0"
20'0"
Glued 
laminated
beam
Spliced
continuous
cross ties
Masonry
wall
Typical joist,
anchored to
masonry wall
~
~
Figure 10.211 Cross tie plan layout and subdiaphragm freebody
diagram for side walls (1.0 ft = 0.3048 m, 1.0 kip/ft = 14.6 kN/m).
10.2.4.3.2 Anchorage at Joists Parallel to Walls (End Walls)
Where the joists are parallel to the walls, tied elements must transfer the forces into the main body of the
diaphragm, which can be accomplished by using either metal strapping and blocking or metal rods and
blocking. This example uses threaded rods that are inserted through the joists and coupled to the anchor
bolt (Figure 10.212). Blocking is added on both sides of the rod to transfer the force into the plywood
sheathing. The tension force in the rod causes a compression force on the blocking through the nut, and
bearing plate at the innermost joist.
Chapter 10, Wood Design
1061
Bond beam
with chord
reinf.
4x12 ledger
3x12 joist
at 24" o.c.
3
4" dia. threaded rod
at 4'0" o.c. extend
through joists
3
4" Structural I
plywood
3
4" dia. bolt
at 2'0" o.c.
Coupler
12'0"
10d nails at 6"
o.c. at blocking
Plate washer
and nut at
last joist
2x12 blocking on
each side of
threaded rod
Bond beam
at top
Vertical
reinforcement
Figure 10.212 Anchorage of masonry wall parallel to joists (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm)
The anchorage force at the end walls is 1.98 klf. Space the connections at 4 ft on center so that the wall
need not be designed for flexure (Provisions Sec. 5.2.6.1.2 [4.6.1.2]). Thus, the anchorage force is 7.92
kips per anchor.
Try a 3/4 in. headed anchor bolt, embedded into the masonry. In this case, gravity loading on the ledger
is negligible and can be ignored, and the anchor can be designed for tension only. (Inplane shear transfer
and orthogonal effects are considered later.)
As computed for 3/4 in. headed anchor bolts (with 6 in. embedment), the design axial strength Ba = 10.1
kips > 7.92 kips. Therefore, the bolt is acceptable.
Using couplers rated for 125 percent of the strength of the rod material, the threaded rods are then
coupled to the anchor bolts and extend six joist spaces (12 ft) into the roof framing. (The 12 ft are
required for the subdiaphragm force transfer as discussed below.)
Nailing the blocking to the plywood sheathing is determined, using the AF&PA Structural Connections
Supplement. As computed previously, the capacity of a single 10d common nail, .fZ’ = 0.194 kips.
Thus, 7.92/0.194 = 41 nails are required. This corresponds to a nail spacing of about 7 in. for two 12ft
rows of blocking. Space nails at 6 in. for convenience.
Use the gluedlaminated beams (at 20 ft on center) to provide continuous crossties, and check the
subdiaphragms between the beams to provide adequate load transfer to the beams per Provisions Sec.
5.2.6.3.2 [4.6.2.1].
Design tension force on beam = (1.98 klf)(20 ft) = 39.6 kips
The stress on the beam is ft = 39.600/[8.75(24)] = 189 psi, which is small. The beam is adequate for
combined moment due to gravity loading and axial tension.
At the beam splices, try onein. bolts with steel side plates. Per the AF&PA Manual, Structural
Connections Supplement:
FEMA 451, NEHRP Recommended Provisions: Design Examples
1062
Fp
V
V
Subdiaphragm
Fp = 1.98 kips/ft
Subdiaphragm
20'0"
Glued  laminated
beam, spliced as
continuous cross tie
Masonry
wall
Typical wall,
anchor at
4'0" o.c.
12'0"
~
~
Roof joist for
subdiaphragm
chord
~
Figure 10.213 Cross tie plan layout and subdiaphragm freebody diagram for end
walls (1.0 ft = 0.3048 m, 1.0 kip/ft = 14.6 kN/m).
lfZ’ = (1.0)(0.65)(18.35) = 11.93 kips per bolt.
The number of bolts required = 39.6/11.93 = 3.3.
Use four bolts in a single row at midheight of the beam, with 1/4in. by 4in. steel side plates. The
reduction (group action factor) for multiple bolts is negligible. Though not included in this example, the
steel side plates should be checked for tension capacity on the gross and net sections. There are
preengineered hinged connectors for gluedlaminated beams that could provide sufficient tension
capacity for the splices.
In order to transfer the wall anchorage forces into the crossties, the subdiaphragms between these ties
must be checked per Provisions Sec. 5.2.6.3.2 [4.6.2.1]. The procedure is similar to that used for the side
walls as described previously. The end wall condition is illustrated in Figure 10.213. The subdiaphragm
spans between beams and utilizes a roof joist as its chord. In order to adequately engage the
subdiaphragm, the wall anchorage ties must extend back to this chord. The maximum aspect ratio for
subdiaphragms is 2½ to 1 per Provisions Sec. 5.2.6.3.2 [4.6.2.1]. Therefore, the minimum depth is 20/2.5
= 8 ft.
For the typical subdiaphragm (Figure 10.213):
Chapter 10, Wood Design
1063
Fp = 1.98 klf
v = (1.98)(20/2)/8 = 2.48 klf
As computed previously (see Table 10.21 and Figure 10.2.2), the diaphragm strength in this area is 1.60
klf < 2.48 klf. Therefore, increase the subdiaphragm depth to 12 ft (six joist spaces):
v = (1.98)(20/2)/12 = 1.65 klf .1.60 klf OK
In order to develop the subdiaphragm strength, boundary nailing must be provided along the crosstie
beams. There are methods of refining this analysis using multiple subdiaphragms so that all of the tension
anchors need not extend 12 ft into the building.
10.2.4.3.3 Transfer of Shear Wall Forces
The inplane diaphragm shear must be transferred to the masonry wall by the ledger, parallel to the wood
grain. The connection must have sufficient capacity for the diaphragm demands as:
Side walls = 0.575 klf
End walls = 1.282 klf
For each case, the capacity of the bolted wood ledger and the capacity of the anchor bolts embedded into
masonry must be checked. Because the wall connections provide a load path for both inplane shear
transfer and outofplane wall forces, the bolts must be checked for orthogonal load effects in accordance
with Provisions Sec. 5.2.5.2.3 [4.4.2.3]. That is, the combined demand must be checked for 100 percent
of the lateral load effect in one direction (e.g., shear) and 30 percent of the lateral load effect in the other
direction (e.g., tension).
At the side walls, the wood ledger with 3/4in. bolts (Figure 10.29) must be designed for gravity loading
(0.56 kip per bolt as computed above) as well as seismic shear transfer. The seismic load per bolt (at 2 ft
on center) is 0.575(2) = 1.15 kips.
Combining gravity shear and seismic shear, produces a resultant force of 1.38 kips at an angle of 26
degrees from the axis of the wood grain. Using the formulas for bolts at an angle to the grain per the
AF&PA Structural Connections Supplement gives
lfZ’ = (1.0)(0.65)(3.47) = 2.26 kips > 1.38. OK
This bolt spacing satisfies the load combination for gravity loading only.
For the check of the embedded anchor bolts, the factored demand on a single bolt is 1.15 kips in
horizontal shear (inplane shear transfer), 2.50 kips in tension (outofplane wall anchorage), 0.56 kip in
vertical shear (gravity). Orthogonal effects are checked, using the following two equations:
(0.3)(2.5) 1.152 0.562 0.34
19.1 4.8
+
+ =
and
(controls) < 1.0 OK
(2.5) [0.3(1.15)]2 0.562 0.38
10.1 4.8
+
+ =
FEMA 451, NEHRP Recommended Provisions: Design Examples
1064
At the end walls, the ledger with 3/4in. bolts (Figure 10.212) need only be checked for inplane seismic
shear because gravity loading is negligible. For bolts spaced at 4 ft on center, the demand per bolt is
1.282(4) = 5.13 kips parallel to the grain of the wood. Per the AF&PA Structural Manual, Connections
Supplement:
Z’ = (1.0)(0.65)(5.37) = 3.49 kips < 5.13 kips NG
Therefore, add 3/4in. headed bolts evenly spaced between the tension ties such that the bolt spacing is 2
ft on center and the demand per bolt is 1.282(2) = 2.56 kips. These added bolts are used for inplane
shear only and do not have coupled tension tie rods.
For the check of the embedded bolts, the factored demand on a single bolt is 2.56 kips in horizontal shear
(inplane shear transfer), 7.92 kips in tension (outofplane wall anchorage), 0 kip in vertical shear
(gravity is negligible). Orthogonal effects are checked using the following two equations:
(0.3)(7.92) 2.56 0.77
10.1 4.8
+ =
and
(controls) < 1.0 OK
7.92 0.3(2.56) 0.94
10.1 4.8
+ =
Therefore, the wall connections satisfy the requirements for combined gravity, and seismic loading,
including orthogonal effects.
111
11
SEISMICALLY ISOLATED STRUCTURES
Charles A. Kircher, P.E., Ph.D.
Chapter 13 of the 2000 NEHRP Recommended Provisions addresses the design of buildings that
incorporate a seismic isolation system. The Provisions provides essentially a stand alone set of design
and analysis criteria for an isolation system. Chapter 13 defines load, design, and testing requirements
specific to the isolation system and interfaces with the appropriate materials chapters for design of the
structure above the isolation system and of the foundation and structural elements below.
A discussion of background, basic concepts, and analysis methods is followed by an example that
illustrates the application of the Provisions to the structural design of a building with an isolation system.
In this example, the building is a threestory emergency operations center (EOC) with a steel
concentrically braced frame above the isolation system. Although the facility is hypothetical, it is of
comparable size and configuration to actual baseisolated EOCs, and is generally representative of
baseisolated buildings.
The EOC is located in San Francisco and has an isolation system that utilizes elastomeric bearings, a type
of bearing commonly used for seismic isolation of buildings. The example comprehensively describes
the EOC’s configuration, defines appropriate criteria and design parameters, and develops a preliminary
design using the equivalent lateral force (ELF) procedure of Chapter 13. It also includes a check of the
preliminary design using dynamic analysis as required by the Provisions and specifies isolation system
design and testing criteria.
Located in a region of very high seismicity, the building is subject to particularly strong ground motions.
Large seismic demands pose a challenge for the design of baseisolated structures in terms of the capacity
of the isolation system and the configuration of the structure above the isolation system. The isolation
system must accommodate large lateral displacements (e.g., in excess of 2 ft). The structure above the
isolation system should be configured to produce the smallest practical overturning loads (and uplift
displacements) on the isolators. The example addresses these issues and illustrates that isolation systems
can be designed to meet the requirements of the Provisions, even in regions of very high seismicity.
Designing an isolated structure in a region of lower seismicity would follow the same approach. The
isolation system displacement, overturning forces, and so forth would all be reduced, and therefore, easier
to accommodate using available isolation system devices.
The isolation system for the building in the example is composed of highdamping rubber (HDR)
elastomeric bearings. HDR bearings are constructed with alternating layers of rubber and steel plates all
sheathed in rubber. The first baseisolated building in the United States employed this type of isolation
system. Other types of isolation systems used to base isolate buildings employed leadcore elastomeric
bearings (LR) and sliding isolators, such as the friction pendulum system (FPS). In regions of very high
seismicity, viscous dampers have been used to supplement isolation system damping (and reduce
displacement demand). Using HDR bearings in this example should not be taken as an endorsement of
this particular type of isolator to the exclusion of others. The concepts of the Provisions apply to all types
FEMA 451, NEHRP Recommended Provisions: Design Examples
112
of isolations systems, and other types of isolators (and possible supplementary dampers) could have been
used equally well in the example.
In addition to the 2000 NEHRP Recommended Provisions and Commentary (hereafter, the Provisions and
Commentary), the following documents are either referenced directly or are useful aids for the analysis
and design of seismically isolated structures.
ATC 1996 Applied Technology Council. 1996. Seismic Evaluation and Retrofit of
Buildings, ATC40.
Constantinou Constantinou, M. C., P. Tsopelas, A. Kasalanati, and E. D. Wolff. 1999.
Property Modification Factors for Seismic Isolation Bearings, Technical Report
MCEER990012. State University of New York.
CSI Computers and Structures, Inc. (CSI). 1999. ETABS Linear and Nonlinear
Static and Dynamic Analysis and Design of Building Systems.
FEMA 273 Federal Emergency Management Agency. 1997. NEHRP Guidelines for the
Seismic Rehabilitation of Buildings, FEMA 273.
FEMA 222A Federal Emergency Management Agency. 1995. NEHRP Recommended
Provisions for Seismic Regulations for New Buildings, FEMA 222A.
91 UBC International Conference of Building Officials. 1991. Uniform Building Code.
94 UBC International Conference of Building Officials. 1994. Uniform Building Code.
Kircher Kircher, C. A., G. C. Hart, and K. M. Romstad. 1989. "Development of Design
Requirements for Seismically Isolated Structures" in Seismic Engineering and
Practice, Proceedings of the ASCE Structures Congress, American Society of
Civil Engineers, May 1989.
SEAOC 1999 Seismology Committee, Structural Engineers Association of California. 1999.
Recommended Lateral Force Requirements and Commentary, 7th Ed.
SEAOC 1990 Seismology Committee, Structural Engineers Association of California. 1990.
Recommended Lateral Force Requirements and Commentary, 5th Ed.
SEAONC Isolation Structural Engineers Association of Northern California. 1986. Tentative
Seismic Isolation Design Requirements.
Although the guide is based on the 2000 Provisions , it has been annotated to reflect changes made to the
2003 Provisions. Annotations within brackets, [ ], indicate both organizational changes (as a result of a
reformat of all of the chapters of the 2003 Provisions) and substantiative technical changes to the 2003
Provisions and its primary reference documents. While the general changes to the document are
described , the deign examples and calculations have not been revised to reflect the changes to the 2003
Provisions.
In the 2003 edition of the Provisions, Chapter 13 has been restructured so that it is better integrated into
the Provisions as a whole and is less of a stand alone set of requirements. Where they affect the design
examples in this chapter, other significant changes to the 2003 Provisions and primary reference
documents may be noted.
Chapter 11, Seismically Isolated Structures
113
FEMA 451, NEHRP Recommended Provisions: Design Examples
114
11.1 BACKGROUND AND BASIC CONCEPTS
Seismic isolation, commonly referred to as base isolation, is a design concept that presumes a structure
can be substantially decoupled from potentially damaging earthquake ground motions. By decoupling the
structure from ground shaking, isolation reduces the level of response in the structure that would
otherwise occur in a conventional, fixedbase building. Conversely, baseisolated buildings may be
designed with a reduced level of earthquake load to produce the same degree of seismic protection. That
decoupling is achieved when the isolation scheme makes the fundamental period of the isolated structure
several times greater than the period of the structure above the isolation system.
The potential advantages of seismic isolation and the advancements in isolation system products led to the
design and construction of a number of isolated buildings and bridges in the early 1980s. This activity, in
turn, identified a need to supplement existing seismic codes with design requirements developed
specifically for such structures. These requirements assure the public that isolated buildings are safe and
provide engineers with a basis for preparing designs and building officials with minimum standards for
regulating construction.
Initial efforts developing design requirements for baseisolated buildings began with ad hoc groups of the
Structural Engineers Association of California (SEAOC), whose Seismology Committee has a long
history of contributing to codes. The northern section of SEAOC was the first to develop guidelines for
the use of elastomeric bearings in hospitals. These guidelines were adopted in the late 1980s by the
California Office of Statewide Health Planning and Development (OSHPD) and were used to regulate the
first baseisolated hospital in California. At about the same time, the northern section of SEAOC
published SEAONC Isolation, first set of general requirements to govern the design of baseisolated
buildings. Most of the basic concepts for the design of seismically isolated structures found in the
Provisions can be traced back to the initial work by the northern section of SEAOC.
By the end of the 1980s, the Seismology Committee of SEAOC recognized the need to have a more
broadly based document and formed a statewide committee to develop design requirements for isolated
structures Kircher. The "isolation" recommendations became an appendix to the 1990 SEAOC Blue
Book. The isolation appendix was adopted with minor changes as a new appendix in the 1991 Uniform
Building Code and has been updated every three years, although it remains largely the same as the
original 91 UBC appendix. (SEAOC 1990 and 1999 are editions of SEAOC’s Recommended Lateral
Force Requirements and Commentary, which is also known as the Blue Book.)
In the mid1990s, the Provisions Update Committee of the Building Seismic Safety Council incorporated
the isolation appendix of the 94 UBC into the 1994 Provisions (FEMA 222A). Differences between the
Uniform Building Code (UBC) and the Provisions were intentionally minimized and subsequent editions
of the UBC and the Provisions are nearly identical. Additional background may be found in the
commentary to the 1999 SEAOC Blue Book.
The Provisions for designing the isolation system of a new building were used as the starting point for the
isolation system requirements of the NEHRP Guidelines for Seismic Rehabilitation of Buildings (FEMA
273). FEMA 273 follows the philosophy that the isolation system for a rehabilitated building should be
comparable to that for a new building (for comparable ground shaking criteria, etc.). The superstructure,
however, could be quite different, and FEMA 273 provides more suitable design requirements for
rehabilitating existing buildings using an isolation system.
11.1.1 Types of Isolation Systems
Chapter 11, Seismically Isolated Structures
115
The Provisions requirements are intentionally broad, accommodating all types of acceptable isolation
systems. To be acceptable, the Provisions requires the isolation system to:
1. Remain stable for maximum earthquake displacements,
2. Provide increasing resistance with increasing displacement,
3. Have limited degradation under repeated cycles of earthquake load, and
4. Have wellestablished and repeatable engineering properties (effective stiffness and damping).
The Provisions recognizes that the engineering properties of an isolation system, such as effective
stiffness and damping, can change during repeated cycles of earthquake response (or otherwise have a
range of values). Such changes or variability of design parameters are acceptable provided that the design
is based on analyses that conservatively bound (limit) the range of possible values of design parameters.
The first seismic isolation systems used in buildings in the United States were composed of elastomeric
bearings that had either a highdamping rubber compound or a lead core to provide damping to isolated
modes of vibration. Other types of isolation systems now include sliding systems, such as the friction
pendulum system (FPS), or some combination of elastomeric and sliding isolators. Some applications at
sites with very strong ground shaking use supplementary fluidviscous dampers in parallel with either
sliding or elastomeric isolators to control displacement. While generally applicable to all types of
systems, certain requirements of the Provisions (in particular, prototype testing criteria) were developed
primarily for isolation systems with elastomeric bearings.
Isolation systems typically provide only horizontal isolation and are rigid or semirigid in the vertical
direction. A rare exception to this rule is the full isolation (horizontal and vertical) of a building in
southern California isolated by large helical coil springs and viscous dampers. While the basic concepts
of the Provisions can be extended to full isolation systems, the requirements are only for horizontal
isolation systems. The design of a full isolation system requires special analyses that explicitly include
vertical ground shaking and the potential for rocking response.
Seismic isolation is commonly referred to as base isolation because the most common location of the
isolation system is at or near the base of the structure. The Provisions does not restrict the plane of
isolation to the base of the structure but does require the foundation and other structural elements below
the isolation system to be designed for unreduced (RI = 1.0) earthquake forces.
11.1.2 Definition of Elements of an Isolated Structure
The design requirements of the Provisions distinguish between structural elements that are either
components of the isolation system or part of the structure below the isolation system (e.g., foundation)
and elements of the structure above the isolation system. The isolation system is defined by the
Provisions as:
The collection of structural elements that includes all individual isolator units, all structural elements that
transfer force between elements of the isolation system, and all connections to other structural elements.
The isolation system also includes the windrestraint system, energydissipation devices, and/or the
displacement restraint system if such systems and devices are used to meet the design requirements of
Chapter 13.
Figure 11.11 illustrates this definition and shows that the isolation system consists not only of the
isolator units but also of the entire collection of structural elements required for the system to function
properly. The isolation system typically includes segments of columns and connecting girders just above
the isolator units because such elements resist moments (due to isolation system displacement) and their
yielding or failure could adversely affect the stability of isolator units.
FEMA 451, NEHRP Recommended Provisions: Design Examples
116
Structure Below The
Isolation System
Structural Elements That Transfer
Force Between Isolator Units
Isolator
Unit
Isolator
Unit
Structure Above The
Isolation System
Isolation
Interface
Figure 11.11 Isolation system terminology.
The isolation interface is an imaginary boundary between the upper portion of the structure, which is
isolated, and the lower portion of the structure, which is assumed to move rigidly with the ground.
Typically, the isolation interface is a horizontal plane, but it may be staggered in elevation in certain
applications. The isolation interface is important for design of nonstructural components, including
components of electrical and mechanical systems that cross the interface and must accommodate large
relative displacements.
The windrestraint system is typically an integral part of isolator units. Elastomeric isolator units are very
stiff at very low strains and usually satisfy drift criteria for wind loads, and the static (breakaway) friction
force of sliding isolator units is usually greater than the wind force.
11.1.3 Design Approach
The design of isolated structures using the Provisions (like the UBC and SEAOC’s Blue Book) has two
objectives: achieving life safety in a major earthquake and limiting damage due to ground shaking. To
meet the first performance objective, the isolation system must be stable and capable of sustaining forces
and displacements associated with the maximum considered earthquake and the structure above the
isolation system must remain essentially elastic when subjected to the design earthquake. Limited
ductility demand is considered necessary for proper functioning of the isolation system. If significant
inelastic response was permitted in the structure above the isolation system, unacceptably large drifts
could result due to the nature of longperiod vibration. Limiting ductility demand on the superstructure
has the additional benefit of meeting the second performance objective of damage control.
The Provisions addresses the performance objectives by requiring:
Chapter 11, Seismically Isolated Structures
117
1. Design of the superstructure for forces associated with the design earthquake, reduced by only a
fraction of the factor permitted for design of conventional, fixedbase buildings (i.e., RI = 3/8 R #2.0).
2. Design of the isolation system and elements of the structure below the isolation system (e.g.,
foundation) for unreduced design earthquake forces.
3. Design and prototype testing of isolator units for forces (including effects of overturning) and
displacements associated with the maximum considered earthquake.
4. Provision of sufficient separation between the isolated structure and surrounding retaining walls and
other fixed obstructions to allow unrestricted movement during the maximum considered earthquake.
11.1.4 Effective Stiffness and Effective Damping
The Provisions utilizes the concepts of effective stiffness and damping to define key parameters of
inherently nonlinear, inelastic isolation systems in terms of amplitudedependent linear properties.
Effective stiffness is the secant stiffness of the isolation system at the amplitude of interest. Effective
damping is the amount of equivalent viscous damping described by the hysteresis loop at the amplitude of
interest. Figure 11.12 shows the application of these concepts to both hysteretic isolator units (e.g.,
friction or yielding devices) and viscous isolator units and shows the Provisions equations used to
determine effective stiffness and damping from tests of prototypes. Ideally, the effective damping of
velocitydependent devices (including viscous isolator units) should be based on the area of hysteresis
loops measured during cyclic testing of the isolation system at fullscale earthquake velocities. Tests of
prototypes are usually performed at lower velocities (due to test facility limitations), resulting in
hysteresis loops with less area, which produce lower (conservative) estimates of effective damping.
11.2 CRITERIA SELECTION
As specified in the Provisions the design of isolated structures must be based on the results of
the equivalent lateral force (ELF) procedure, response spectrum analysis, or (nonlinear) time
history analysis. Because isolation systems are typically nonlinear, linear methods (ELF
procedure and response spectrum analysis) use effective stiffness and damping properties to
model nonlinear isolation system components.
The ELF procedure is intended primarily to prescribe minimum design criteria and may be used
for design of a very limited class of isolated structures (without confirmatory dynamic analyses).
The simple equations of the ELF procedure are useful tools for preliminary design and provide a
means of expeditious review and checking of more complex calculations. The Provisions also
uses these equations to establish lowerbound limits on results of dynamic analysis that may be
used for design. Table 11.21 summarizes site conditions and structure configuration criteria
that influence the selection of an acceptable method of analysis for designing of isolated
structures. Where none of the conditions in Table 11.21 applies, all three methods are
permitted.
FEMA 451, NEHRP Recommended Provisions: Design Examples
118
Force
Hysteretic Isolator Viscous Isolator
Force
Disp. Disp.
Eloop Eloop
eff
F F
k
. .
+ 
+ 
+
=
+
( )2
2 loop
eff
eff
E
k
ß
p .+ .
. .
= .. ..
.. + ..
. + . +
.  . 
F+ F+
F F
Figure 11.12 Effective stiffness and effective damping.
Table 11.21 Acceptable Methods of Analysis*
Site condition or
Structure Configuration Criteria
ELF
Procedure
Response
Spectrum
Analysis
Time
History
Analysis
Site Conditions
Nearsource (S1 > 0.6) NP P P
Soft soil (Site Class E or F) NP NP P
Superstructure Configuration
Flexible or irregular superstructure
(height > 4 stories, height > 65 ft, or
TM > 3.0 sec., or TD #3T)
NP P P
Nonlinear superstructure (requiring explicit
modeling of nonlinear elements;
Provisions Sec. 13.2.5.3.1) [13.4.1.2]
NP NP P
Isolation System Configuration
Highly nonlinear isolation system or
system that otherwise does not meet the
criteria of Provisions Sec. 13.2.5.2, Item 7
[13.2.4.1, Item 7]
NP NP P
* P indicates permitted and NP indicates not permitted by the Provisions.
Chapter 11, Seismically Isolated Structures
119
Seismic criteria are based on the same site and seismic coefficients as conventional, fixedbase structures
(e.g., mapped value of S1 as defined in Provisions Chapter 4 [3]). Additionally, sitespecific design
criteria are required for isolated structures located on soft soil (Site Class E of F) or near an active source
such that S1 is greater than 0.6, or when nonlinear time history analysis is used for design.
11.3 EQUIVALENT LATERAL FORCE PROCEDURE
The equivalent lateral force (ELF) procedure is a displacementbased method that uses simple equations
to determine isolated structure response. The equations are based on ground shaking defined by 1 second
spectral acceleration and the assumption that the shape of the design response spectrum at long periods is
inversely proportional to period as shown in Provisions Figure 4.1.2.6 [3.315]. [In the 2003 edition of
the Provisions , there is also a 1/T2 portion of the spectrum at periods greater than TL. However, in most
parts of the Unites States TL is longer than the period of typical isolated structures.] Although the ELF
procedure is considered a linear method of analysis, the equations incorporate amplitudedependent
values of effective stiffness and damping to implicitly account for the nonlinear properties of the isolation
system. The equations are consistent with the nonlinear static procedure of FEMA 273 assuming the
superstructure is rigid and lateral displacements to occur primarily in the isolation system.
11.3.1 Isolation System Displacement
The isolation system displacement for the design earthquake is determined by using Provisions Eq.
13.3.3.1 [13.31]:
4 2
D1 D
D
D
D g S T
p B
=.. ..
. .
where the damping factor ,BD, is based on effective damping, ßD, using Provisions Table 13.3.3.1 [13.3
1]. This equation describes the peak (spectral) displacement of a singledegreeoffreedom (SDOF)
system with period, TD, and damping, ßD, for the design earthquake spectrum defined by the seismic
coefficient, SD1. SD1 corresponds to 5 percent damped spectral response at a period of 1 second. BD,
converts 5 percent damped response to the level of damping of the isolation system. BD is 1.0 when
effective damping, ßD, is 5 percent of critical. Figure 11.31 illustrates the underlying concepts of
Provisions Eq. 13.3.3.1 [13.31] and the amplitudedependent equations of the Provisions for effective
period, TD, and effective damping, ßD.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1110
Mass .. Design earthquake
spectral acceleration
(W/g .. SD1/T)
Mass .. Spectral
acceleration
Spectral
displacement
Maximum
stiffness
curve
Minimum
stiffness
curve
Response reduction, BD
(effective damping,
ßD > 5% of critical)
kDmax
kDmin
4 2 D1 D
g S T
p
. .
. .
. .
D 2
Dmin
T W
k g
= p
2
1
2
D
D
Dmax D
E
k D
ß
p
. .
= .. ..
. .
S
SED DD
Vb
Figure 11.31 Isolation system capacity and earthquake demand.
The equations for maximum displacement, DM, and design displacement, DD, reflect differences due to the
corresponding levels of ground shaking. The maximum displacement is associated with the maximum
considered earthquake (characterized by SM1) whereas the design displacement corresponds to the design
earthquake (characterized by SD1). In general, the effective period and the damping factor (TM and BM,
respectively) used to calculate the maximum displacement are different from those used to calculate the
design displacement (TD and BD) because the effective period tends to shift and effective damping may
change with the increase in the level of ground shaking.
As shown in Figure 11.31, the calculation of effective period, TD, is based on the minimum effective
stiffness of the isolation system, kDmin, as determined by prototype testing of individual isolator units.
Similarly, the calculation of effective damping is based on the minimum loop area, ED, as determined by
prototype testing. Use of minimum effective stiffness and damping produces larger estimates of effective
period and peak displacement of the isolation system.
The design displacement, DD, and maximum displacement, DM, represent peak earthquake displacements
at the center of mass of the building without the additional displacement, that can occur at other locations
due to actual or accidental mass eccentricity. Equations for determining total displacement, including the
effects of mass eccentricity as an increase in the displacement at the center of mass, are based on the plan
dimensions of the building and the underlying assumption that building mass and isolation stiffness have
a similar distribution in plan. The increase in displacement at corners for 5 percent mass eccentricity is
about 15 percent if the building is square in plan, and as much as 30 percent if the building is long in plan.
Figure 11.32 illustrates design displacement, DD, and maximum displacement, DM, at the center of mass
of the building and total maximum displacement, DTM, at the corners of an isolated building.
Chapter 11, Seismically Isolated Structures
1111
Total Maximum Displacement
(maximum considered earthquake
corner of building)
Plan View of Building
Maximum Displacement
(maximum considered earthquake
center of building)
Design Displacement
(design earthquake center
of building)
DTM
DM
DD
Figure 11.32 Design, maximum, and total maximum displacement.
11.3.2 Design Forces
Forces required by the Provisions for design of isolated structures are different for design of the
superstructure and design of the isolation system and other elements of the structure below the isolation
system (e.g., foundation). In both cases, however, use of the maximum effective stiffness of the isolation
system is required to determine a conservative value of design force.
In order to provide appropriate overstrength, peak design earthquake response (without reduction) is used
directly for design of the isolation system and the structure below. Design for unreduced design
earthquake forces is considered sufficient to avoid inelastic response or failure of connections and other
elements for ground shaking as strong as that associated with the maximum considered earthquake (i.e.,
shaking as much as 1.5 times that of the design earthquake). The design earthquake base shear, Vb, is
given by Provisions Eq. 13.3.4.1 [13.37]:
Vb = kDmaxDD,
where kDmax is the maximum effective stiffness of the isolation system at the design displacement, DD.
Because the design displacement is conservatively based on minimum effective stiffness, Provisions Eq.
13.3.4.1 implicitly induces an additional conservatism of a worst case combination mixing maximum and
minimum effective stiffness in the same equation. Rigorous modeling of the isolation system for dynamic
FEMA 451, NEHRP Recommended Provisions: Design Examples
1112
analyses precludes mixing of maximum and minimum stiffness in the same analysis (although separate
analyses are typically required to determine bounding values of both displacement and force).
Design earthquake response is reduced by a modest factor for design of the superstructure above the
isolation interface, as given by Provisions Eq. 13.3.4.2 [13.38]:
b Dmax D
s
I I
V V k D
R R
= =
The reduction factor, RI, is defined as threeeighths of the R factor for the seismicforceresisting system
of the superstructure, as specified in Provisions Table 5.2.2 [4.31], with an upperbound value of 2.0. A
relatively small RI factor is intended to keep the superstructure essentially elastic for the design
earthquake (i.e., keep earthquake forces at or below the true strength of the seismicforceresisting
system). The Provisions also impose three limits on design forces that require the value of Vs to be at
least as large as each of:
1. The shear force required for design of a conventional, fixedbase structure of period TD.
2. The shear force required for wind design, and/or
3. A factor of 1.5 times the shear force required for activation of the isolation system.
These limits seldom govern design but reflect principles of good design. In particular, the third limit is
included in the Provisions to ensure that isolation system displaces significantly before lateral forces
reach the strength of the seismicforceresisting system.
For designs using the ELF procedure, the lateral forces, Fx, must be distributed to each story over the
height of the structure, assuming an inverted triangular pattern of lateral load (Provisions Eq.13.3.5 [13.3
9]):
1
s x x
x n
i i
i
F V w h
w h
=
=
S
Because the lateral displacement of the isolated structure is dominated by isolation system displacement,
the actual pattern of lateral force in the isolated mode of response is distributed almost uniformly over
height. The Provisions require an inverted triangular pattern of lateral load to capture possible
highermode effects that might be missed by not modeling superstructure flexibility. Rigorous modeling
of superstructure flexibility for dynamic analysis would directly incorporate highermode effects in the
results.
Example plots of the design displacement, DD, total maximum displacement, DTM, and design forces for
the isolation system, Vb, and the superstructure, Vs (RI = 2), are shown in Figure 11.33 as functions of the
effective period of the isolation system. The figure also shows the design base shear required for a
conventional building, V (R/I = 5). The example plots are for a building assigned to Seismic Design
Category D with a onesecond spectral acceleration parameter, SD1, equal to 0.6, representing a stiff soil
site (Site Class D) located in a region of high seismicity but not close to an active fault. In this example,
the isolation system is assumed to have 20 percent effective damping (at all amplitudes of interest) and
building geometry is assumed to require 25 percent additional displacement (at corners/edges) due to the
requisite 5 percent accidental eccentricity.
Chapter 11, Seismically Isolated Structures
1113
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
1 1.5 2 2.5 3 3.5 4
Effective period of isolation system (seconds)
Design base shear (fraction of W)
0
5
10
15
20
25
30
35
40
Isolation system displacement (in.)
SD1=0.6 BD=1.5 DTMDM=1.25
Vb
Vs(RI =2)
V(R I =5)
DTM
DD
Figure 11.33 Isolation system displacement and shear force as function of period (1.0 in. = 25.4 mm).
The plots in Figure 11.33 illustrate the fundamental trade off between displacement and force as a
function of isolation system displacement. As the period is increased, design forces decrease and design
displacements increase linearly. Plots like those shown in Figure 11.33 can be constructed during
conceptual design once site seismicity and soil conditions are known (or are assumed) to investigate trial
values of effective stiffness and damping of the isolation system. In this particular example, an isolation
system with an effective period falling between 2.5 and 3.0 seconds would not require more than 22 in. of
total maximum displacement capacity (assuming TM # 3.0 seconds). Design force on the superstructure
would be less than about eight percent of the building weight (assuming TD $ 2.5 seconds and RI $ 2.0).
11.4 DYNAMIC LATERAL RESPONSE PROCEDURE
While the ELF procedure equations are useful tools for preliminary design of the isolations system, the
Provisions requires a dynamic analysis for most isolated structures. Even where not strictly required by
the Provisions, the use of dynamic analysis (usually time history analysis) to verify the design is
common.
11.4.1 Minimum Design Criteria
The Provisions encourages the use of dynamic analysis but recognize that along with the benefits of more
complex models and analyses also comes an increased chance of design error. To avoid possible under
design, the Provisions establishes lowerbound limits on results of dynamic analysis used for design.
The limits distinguish between response spectrum analysis (a linear, dynamic method) and time history
analysis (a nonlinear, dynamic method). In all cases, the lowerbound limit on dynamic analysis is
established as a percentage of the corresponding design parameter calculated using the ELF procedure
equations. Table 11.41 summarizes the percentages that define lowerbound limits on dynamic analysis.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1114
Table 11.41 Summary of Minimum Design Criteria for Dynamic Analysis
Design Parameter Response
Spectrum Analysis
Time History
Analysis
Total design displacement, DTD 90% DTD 90% DTD
Total maximum displacement, DTM 80% DTM 80% DTM
Design force on isolation system, Vb 90% Vb 90% Vb
Design force on irregular superstructure, Vs 100% Vs 80% Vs
Design force on regular superstructure, Vs 80% Vs 60% Vs
The Provisions permits more liberal drift limits when the design of the superstructure is based on dynamic
analysis. The ELF procedure drift limits of 0.010hsx are increased to 0.015hsx for response spectrum
analysis and to 0.020hsx for time history analysis (where hsx is the story height at level x). Usually a stiff
system (e.g., braced frames) is selected for the superstructure and drift demand is typically less than about
0.005hsx. Provisions Sec. 13.4.7.4 [13.4.4] requires an explicit check of superstructure stability at the
maximum considered earthquake displacement if the design earthquake story drift ratio exceeds 0.010/RI.
11.4.2 Modeling Requirements
As for the ELF procedure, the Provisions requires the isolation system to be modeled for dynamic
analysis using stiffness and damping properties that are based on tests of prototype isolator units.
Additionally, dynamic analysis models are required to account for:
1. Spatial distribution of individual isolator units,
2. Effects of actual and accidental mass eccentricity,
3. Overturning forces and uplift of individual isolator units, and
4. Variability of isolation system properties (due to rate of loading, etc.).
The Provisions requires explicit nonlinear modeling of elements if time history analysis is used to justify
design loads less than those permitted for ELF or response spectrum analysis. This option is seldom
exercised and the superstructure is typically modeled using linear elements and conventional methods.
Special modeling concerns for isolated structures include two important and related issues: the uplift of
isolator units, and the Pdelta effects on the isolated structure. Isolator units tend to have little or no
ability to resist tension forces and can uplift when earthquake overturning (upward) loads exceed factored
gravity (downward) loads. Local uplift of individual elements is permitted (Provisions Sec. 13.6.2.7
[13.2.5.7]) provided the resulting deflections do not cause overstress or instability. To calculate uplift
effects, gap elements may be used in nonlinear models or tension may be released manually in linear
models.
The effects of Pdelta loads on the isolation system and adjacent elements of the structure can be quite
significant. The compression load, P, can be large due to earthquake overturning (and factored gravity
loads) at the same time that large displacements occur in the isolation system. Computer analysis
programs (most of which are based on smalldeflection theory) may not correctly calculate Pdelta
moments at the isolator level in the structure above or in the foundation below. Figure 11.41 illustrates
moments due to Pdelta effects (and horizontal shear loads) for an elastomeric bearing isolation system
and a sliding isolation system. For the elastomeric system, the Pdelta moment is split onehalf up and
onehalf down. For the sliding system, the full Pdelta moment is applied to the foundation below (due to
the orientation of the sliding surface). A reverse (upside down) orientation would apply the full Pdelta
moment on the structure above.
Chapter 11, Seismically Isolated Structures
1115
C
A
B D
.
H2
H1
H4
H3
Elastomeric Isolator Sliding Isolator
V
P
V
P
P
V
V
P
.
Figure 11.41 Moments due to horizontal shear and Pdelta effects.
1
2
3
4
2
2
A
B
C
D
M P VH
M P VH
M VH
M P VH
.
= +
.
= +
=
= . +
For time history analysis, nonlinear forcedeflection characteristics of isolator units are explicitly modeled
(rather than using effective stiffness and damping). Forcedeflection properties of isolator units are
typically approximated by a bilinear, hysteretic curve whose properties can be accommodated by
commercially available nonlinear structural analysis programs. Such bilinear hysteretic curves should
have approximately the same effective stiffness and damping at amplitudes of interest as the true
forcedeflection characteristics of isolator units (as determined by prototype testing).
Figure 11.42 shows a bilinear idealization of the response of a typical nonlinear isolator unit. Figure
11.42 also includes simple equations defining the yield point (Dy, Fy) and end point (D, F) of a bilinear
approximation that has the same effective stiffness and damping as the true curve (at a displacement, D).
FEMA 451, NEHRP Recommended Provisions: Design Examples
1116
Force
Displacement
Eloop
F
F
D
D
4( )
eff
y y loop
F k
D
DF F D E
˜
 ˜
0.32 (using in inches)
0.637
eff
y y
eff
T DW D
F
DF F D
F D
ß
=
.  .
= . .
. .
At the displacement of interest:
Dy
Fy
Figure 11.42 Bilinear idealization of isolator unit behavior.
11.4.3 Response Spectrum Analysis
Response spectrum analysis requires that isolator units be modeled using amplitudedependent values of
effective stiffness and damping that are the same as those for the ELF procedure. The effective damping
of the isolated modes of response is limited to 30 percent of critical. Higher modes of response are
usually assumed to have five percent damping—a value of damping appropriate for the superstructure,
which remains essentially elastic. As previously noted, maximum and minimum values of effective
stiffness are typically used to individually capture maximum displacement of the isolation system and
maximum forces in the superstructure. Horizontal loads are applied in the two orthogonal directions, and
peak response of the isolation system and other structural elements is determined using the 100 percent
plus 30 percent combination method.
11.4.4 Time History Analysis
Time history analysis with explicit modeling of nonlinear isolator units is commonly used for the
evaluation of isolated structures. Where at least seven pairs of time history components are employed, the
values used in design for each response parameter of interest may be the average of the corresponding
analysis maxima. Where fewer pairs are used (with three pairs of time history components being the
minimum number permitted), the maximum value of each parameter of interest must be used for design.
The time history method is not a particularly useful design tool due to the complexity of results, the
number of analyses required (e.g., to account for different locations of eccentric mass), the need to
combine different types of response at each point in time, etc. It should be noted that while Provisions
Chapter 5 does not require consideration of accidental torsion for either the linear or nonlinear response
history procedures, Chapter 13 does require explicit consideration of accidental torsion, regardless of the
analysis method employed. Time history analysis is most useful when used to verify a design by
Chapter 11, Seismically Isolated Structures
1117
Figure 11.51 Threedimensional model of the structural system.
checking a few key design parameters, such as: isolation displacement, overturning loads and uplift, and
story shear force.
11.5 EMERGENCY OPERATIONS CENTER USING ELASTOMERIC BEARINGS,
SAN FRANCISCO, CALIFORNIA
This example features the seismic isolation of a hypothetical emergency operations center (EOC), located
in the center of San Francisco, California, an area of very high seismicity. Using highdamping rubber
bearings, other types of isolators could be designed to have comparable response properties. Isolation is
an appropriate design strategy for EOCs and other buildings where the goal is to limit earthquake damage
and protect facility function. The example illustrates the following design topics:
1. Determination of seismic design parameters,
2. Preliminary design of superstructure and isolation systems (using the ELF procedure),
3. Dynamic analysis of seismically isolated structures, and
4. Specification of isolation system design and testing criteria.
While the example includes development of the entire structural system, the primary focus is on the
design and analysis of the isolation system. Examples in other chapters may be referred to for more
indepth descriptions of the provisions governing detailed design of the superstructure (i.e., the structure
above the isolation system) and the foundation.
11.5.1 System Description
This EOC is a threestory, steelbraced frame structure with a large, centrally located mechanical
penthouse. Story heights of 15 ft at all floors accommodate computer access floors and other
architectural and mechanical systems. The roof and penthouse roof decks are designed for significant live
load to accommodate a helicopterlanding pad and meet other functional requirements of the EOC. Figure
11.51 shows the threedimensional model of the structural system.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1118
30'0" 30'0" 30'0" 30'0"
30'0" 30'0" 30'0" 30'0" 30'0" 30'0"
Figure 11.52 Typical floor framing plan (1.0 ft = 0.3048 m).
30'0" 30'0" 30'0" 30'0"
30'0" 30'0" 30'0" 30'0" 30'0" 30'0"
Figure 11.53 Penthouse roof framing plan (1.0 ft = 0.3048 m).
The structure (which is regular in configuration) has plan dimensions of 120 ft. by 180 ft. at all floors
except for the penthouse, which is approximately 60 ft by 120 ft in plan. Columns are spaced at 30 ft in
both directions. This EOC’s relatively large column spacing (bay size) is used to reduce the number of
isolator units for design economy and to increase gravity loads on isolator units for improved earthquake
performance. Figures 11.52 and 11.53, are framing plans for the typical floor levels (1, 2, 3, and roof)
and the penthouse roof.
Chapter 11, Seismically Isolated Structures
1119
Penthouse Roof
Roof
Third Floor
Second Floor
First Floor
Base
Figure 11.54 Longitudinal bracing elevation (Column Lines B and D).
Penthouse Roof
Roof
Third Floor
Second Floor
First Floor
Base
a) b)
Figure 11.55 Transverse bracing elevations: (a) on Column Lines 2 and 6 and ( b) on Column Line 4.
The vertical load carrying system consists of concrete fill on steel deck floors, roofs supported by steel
beams at 10 ft on center, and steel girders at column lines. Isolator units support the columns below the
first floor. The foundation is a heavy mat (although the spread footings or piles could be used depending
on the soil type, depth to the water table, and other site conditions).
The lateral system consists of a symmetrical pattern of concentrically braced frames. These frames are
located on Column Lines B and D in the longitudinal direction, and on Column Lines 2, 4 and 6 in the
transverse direction. Figures 11.54 and 11.55, respectively, show the longitudinal and transverse
elevations. Braces are specifically configured to reduce the concentration of earthquake overturning, and
uplift loads on isolator units by:
1. Increasing the continuous length of (number of) braced bays at lower stories,
2. Locating braces at interior (rather than perimeter) column lines (more holddown weight), and
3. Avoiding common end columns for transverse and longitudinal bays with braces.
The isolation system is composed of 35 identical elastomeric isolator units, located below columns. The
first floor is just above grade, and the isolator units are approximately 3 ft below grade to provide
FEMA 451, NEHRP Recommended Provisions: Design Examples
1120
clearance below the first floor, for construction and maintenance personnel. A short retaining wall
borders the perimeter of the facility and provides 30 in. of “moat” clearance for lateral displacement of
the isolated structure. Access to the EOC is provided at the entrances by segments of the first floor slab,
which cantilever over the moat.
Girders at the firstfloor column lines are much heavier than the girders at other floor levels, and have
momentresisting connections to columns. These girders stabilize the isolator units by resisting moments
due to vertical (Pdelta effect) and horizontal (shear) loads. Column extensions from the first floor to the
top plates of the isolator units are stiffened in both horizontal directions, to resist these moments and to
serve as stabilizing haunches for the beamcolumn moment connections.
11.5.2 Basic Requirements
11.5.2.1 Specifications
General: 1997 Uniform Building Code (UBC)
Seismic: NEHRP Recommended Provisions
11.5.2.2 Materials
Concrete:
Strength fc ' = 3 ksi
Weight (normal) . =150 pcf
Steel:
Columns Fy = 50 ksi
Primary firstfloor girders (at column lines) Fy = 50 ksi
Other girders and floor beams Fy = 36 ksi
Braces Fy = 46 ksi
Steel deck: 3in.deep, 20gauge deck
Seismic isolator units (highdamping rubber):
Maximum longtermload (1.2D + 1.6L) face pressure, sLT 1,400 psi
Maximum shorttermload (1.5D + 1.0L + QMCE) face pressure, sST 2,800 psi
Minimum bearing diameter (excluding protective cover) 1.25DTM
Minimum rubber shear strain capacity (isolator unit), .max 300 percent
Minimum effective horizontal shear modulus, Gmin 65  110 psi
Third cycle at . = 150 percent (after scragging/recovery)
Maximum effective horizontal shear modulus, Gmax 1.3 × Gmin
First cycle at . = 150 percent (after scragging/recovery)
Minimum effective damping at 150 percent rubber shear strain, ßeff 15 percent
11.5.2.3 Gravity Loads
Dead loads:
Main structural elements (slab, deck, and framing) self weight
Chapter 11, Seismically Isolated Structures
1121
Miscellaneous structural elements (and slab allowance) 10 psf
Architectural facades (all exterior walls) 750 plf
Roof parapets 150 plf
Partitions (all enclosed areas) 20 psf
Suspended MEP/ceiling systems and supported flooring 15 psf
Mechanical equipment (penthouse floor) 50 psf
Roofing 10 psf
Reducible live loads:
Floors (13) 100 psf
Roof decks and penthouse floor 50 psf
Live load reduction:
The 1997 UBC permits areabased live load reduction, of not more than 40 percent for elements with
live loads from a single story (e.g., girders), and not more than 60 percent for elements with live loads
from multiple stories (e.g., axial component of live load on columns at lower levels and isolator
units).
EOC weight (dead load) and live load:
Penthouse roof WPR = 965 kips
Roof (penthouse) WR = 3,500 kips
Third floor W3 = 3,400 kips
Second floor W2 = 3,425 kips
First floor W1 = 3,425 kips
Total EOC weight W = 14,715 kips
(See Chapter 1 for a discussion of live load contributions to the seismic weight.)
Live load (L) above isolation system L = 7,954 kips
Reduced live load (RL) above isolation system RL = 3,977 kips
Table 11.51 Gravity Loads on Isolator Units*
Dead/live loads (kips)
Column
line 1 2 3 4
ABC
182/73
336/166
295/149
349/175
570/329
520/307
303/153
606/346
621/356
345/173
539/309
639/358
1.0 kip = 4.45 kN.
* Loads at Column Lines 5, 6 and 7 (not shown) are the same as those at Column Lines 3, 2, and 1,
respectively; loads at Column Lines D and E (not shown) are the same as those at Column Lines B
and A, respectively.
FEMA 451, NEHRP Recommended Provisions: Design Examples
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11.5.2.4 Provisions Parameters
11.5.2.4.1 Performance Criteria (Provisions Sec. 1.3 [1.2 and 1.3])
Designated Emergency Operation Center Seismic Use Group III
Occupancy Importance Factor I = 1.5 (conventional)
Occupancy Importance Factor (Provisions Chapter 13) I = 1.0 (isolated)
Chapter 13 does not require use of the occupancy importance factor to determine the design loads on
the structural system of an isolated building (that is, I = 1.0), but the component importance factor is
required by Chapter 6, to determine seismic forces on components (Ip = 1.5 for some facilities).
11.5.2.4.2 Ground Motion (Provisions Chapter 4 [3])
Site soil type (assumed) Site Class D
Maximum considered earthquake (MCE) spectral acceleration at SS = 1.50
short periods (Provisions Map 7)
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package.]
Site coefficient (Provisions Table 4.1.2.4b [3.31]) Fa = 1.0
MCE spectral acceleration adjusted for site class (FaSS) SMS = 1.50
Design earthquake (DE) spectral acceleration (2/3SMS) SDS = 1.0
MCE spectral acceleration at a period of 1 second (Provisions Map 8) S1 = 0.9
On Map 8, the contour line of S1 = 0.9 runs approximately through the center of the San Francisco
peninsula, with other San Francisco peninsula contour lines ranging from 0.6 (greatest distance from
San Andreas Fault), to about 1.6 (directly on top of the San Andreas Fault).
Site coefficient (Provisions Table 4.1.2.4b [3.32]) Fv = 1.5
MCE spectral acceleration adjusted for site class (FvS1) SM1 = 1.35
Design earthquake (DE) spectral acceleration (2/3SM1) SD1 = 0.9
Seismic Design Category (Provisions Table 4.2.1b [1.42]) Seismic Design Category F
[In the 2003 edition of the Provisions, the ground motion trigger for Seismic Design Categories E and F
have been changed to S1 $0.60. No change would result for this example.]
Chapter 11, Seismically Isolated Structures
1123
MCE
80%
of MCE
DE
80%
of DE
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Period (seconds)
Spectral Acceleration (g)
Figure 11.56 Example design spectra (5 percent damping).
11.5.2.4.3 Design Spectra (Provisions Sec. 13.4.4.1)
Figure 11.56 plots design earthquake and maximum considered earthquake response spectra as
constructed in accordance with the procedure of Provisions Sec. 13.4.4.1 [3.4] using the spectrum shape
defined by Provisions Figure 4.1.2.6 [3.315]. [In the 2003 edition of Provisions, the shape of the design
spectrum changes at period beyond TL, but no change would result for this example.] Provisions Sec.
13.4.4.1[13.2.3.1] requires sitespecific design spectra to be calculated for sites with S1 greater than 0.6
(e.g., sites near active sources). Sitespecific design spectra may be taken as less than 100 percent but not
less than 80 percent of the default design spectra of Provisions Figure 4.1.2.6 [3.315].
For this example, sitespecific spectra for the design earthquake and the maximum considered earthquake,
were assumed to be 80 percent (at long periods) and 100 percent (at short periods) of the respective
spectra shown in Figure 11.56. The 80 percent factor reflects a reduction in demand that could be
achieved through a detailed geotechnical investigation of site soil conditions. In general, sitespecific
spectra for regions of high seismicity, with well defined fault systems (like those of the San Francisco
Bay Area), would be expected to be similar to the default design spectra of Provisions Figure 4.1.2.6 [3.3
15].
11.5.2.4.4 Design Time Histories (Provisions Sec. 13.4.4.2 [13.2.3.2])
For time history analysis, Provisions Sec. 13.4.4.7 [13.4.2.3] requires no less than three pairs of
horizontal ground motion time history components to be selected from actual earthquake records, and
scaled to match either the design earthquake (DE) or the maximum considered earthquake (MCE)
spectrum. The selection and scaling of appropriate time histories is usually done by an earth scientist, or
a geotechnical engineer experienced in the seismology of the region; and takes the earthquake
FEMA 451, NEHRP Recommended Provisions: Design Examples
1124
magnitudes, fault distances, and source mechanisms that influence hazards at the building site into
consideration.
For this example, records from the El Centro Array Station No. 6, the Newhall Fire Station, and the
Sylmar Hospital are selected from those recommended by ATC40 [ATC, 1996], for analysis of buildings
at stiff soil sites with ground shaking of 0.2g or greater. These records, and the scaling factors required to
match the design spectra are summarized in Table 11.52. These three records represent strong ground
shaking recorded relatively close to fault rupture; and contain longperiod pulses in the direction of
strongest shaking, which can cause large displacement of the isolated structures. MCE scaling factors for
the El Centro No. 6 and Sylmar records are 1.0, indicating that these records are used without
modification for MCE time history analysis of the EOC.
Provisions Sec. 13.4.4.2 [13.2.3.2] provides criteria for scaling earthquake records to match a target
spectrum over the period range of interest, defined as 0.5TD to 1.25TM. In this example, TD and TM are
both assumed to be 2.5 seconds, so the period range of interest is from 1.25 seconds to 3.125 seconds.
For each period in this range, the average of the squarerootofthesumofthesquares (SRSS)
combination of each pair of horizontal components of scaled ground motion, may not fall below the target
spectrum by more 10 percent. The target spectrum is defined as 1.3 times the design spectrum of interest
(either the DE or the MCE spectrum).
Figure 11.57 compares the spectra of scaled time history components with the spectrum for the design
earthquake. Rather than comparing the average of the SRSS spectra with 1.3 times the design spectrum,
as indicated in the Provisions, Figure 11.57 shows the (unmodified) design spectrum and the average of
the SRSS spectra divided by 1.3. The effect is the same, but the method employed eliminates one level of
obscurity and permits direct comparison of the calculated spectra without additional manipulation. A
comparison of maximum, considered earthquake spectra would look similar (with values that were 1.5
times larger). The fit is very good at long periods  the period range of interest for isolated structures. At
2.5 seconds, the spectrum of the scaled time histories has the same value as the target spectrum. Between
1.25 seconds and 3.125 seconds, the spectrum of the scaled time histories is never less than the target
spectrum by more than 10 percent. At short periods, the spectrum of scaled time history motions are
somewhat greater than the target spectrum  a common result of scaling real time histories to match the
longperiod portion of the design spectrum.
Figure 11.57 also includes plots of upperbound (maximum demand) and lowerbound (minimum
demand) envelopes of the spectra of the six individual components of scaled time histories. The
maximum demand envelope illustrates that the strongest direction (component) of shaking of at least one
of the three scaled time histories is typically about twice the sitespecific spectrum in the period range of
the isolated structure  consistent with strong ground shaking recorded near sources in the fault normal
direction.
Table 11.52 Earthquake Time History Records and Scaling Factors
Record
No.
Earthquake Source and Recording Station Scaling factor
Year Earthquake Station (owner) DE MCE
1 1979 Imperial Valley, CA El Centro Array Station 6 (USGS) 0.67 1.0
2 1994 Northridge, CA Newhall Fire Station (CDMG) 1.0 1.5
3 1994 Northridge, CA Sylmar Hospital (CDMG) 0.67 1.0
Chapter 11, Seismically Isolated Structures
1125
0.0
0.5
1.0
1.5
2.0
2.5
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Period (seconds)
Spectral Acceleration (g)
Target spectrum
Average of SRSS spectra divided by 1.3
Maximum and minimum demands
Period range of interest for
scaling time history records
Figure 11.57 Comparison of design earthquake spectra.
11.5.2.5 Structural Design Criteria
11.5.2.5.1 Design Basis (Provisions Sec. 5.2 and 13.2 [4.3 and 13.2])
Seismicforceresisting system. Special steel concentrically
braced frames (height < 100 ft)
Response modification factor, R (Provisions Table 5.2.2 [4.31] 6 (conventional)
Response modification factor for design of the superstructure, RI 2 (isolated)
( Provisions Sec. 13.3.4.2 [13.3.3.2], 3/8R # 2)
Plan irregularity (of superstructure)  (Provisions Table 5.2.3.2) None
Vertical irregularity (of superstructure)  from Table 5.2.3.3 None
Lateral response procedure (Provisions Sec. 13.2.5.2 [13.2.4.1], S1 > 0.6) Dynamic analysis
Redundancy/reliability factor  (Provisions Sec. 5.2.4.2 [4.3.3]) . > 1.0 (conventional)
. = 1.0 (isolated)
Provisions Sec. 5.2.4.2 [4.3.3] requires the use of a calculated . value, which would be greater than
1.0 for a conventional structure with a brace configuration similar to the superstructure of the
baseisolated EOC. However, in the author's opinion, the use of RI equal to 2.0 (rather than R equal
FEMA 451, NEHRP Recommended Provisions: Design Examples
1126
to 6) as required by Provisions Sec. 13.3.4.2 precludes the need to further increase superstructure
design forces for redundancy/reliability. Future editions of the Provisions should address this issue.
[The redundancy factor is changed substantially in the 2003 Provisions. However, the rationale set forth
above by the author still holds; use of . = 1.0 is reasonable since RI = 2.0.]
11.5.2.5.2 Horizontal Earthquake Loads and Effects (Provisions Chapters 5 and 13)
Design earthquake (acting in either the X or Y direction) DE (site specific)
Maximum considered earthquake (acting in either the X or Y direction) MCE (site specific)
Mass eccentricity  actual plus accidental 0.05b = 6 ft (z to X axis)
0.05d = 9 ft (z to Y axis)
The superstructure is symmetric about both principal axes, however, the placement of the braced
frames results in a ratio of maximum corner displacement to average displacement of 1.25, exceeding
the threshold of 1.2 per the definition of the Provisions. If the building were not on isolators, the
accidental torsional moment would need to be increased from 5 percent to 5.4 percent of the building
dimension. The input to the superstructure is controlled by the isolation system, and it is the author’s
opinion that the amplification of accidental torsion is not necessary for such otherwise regular
structures. Future editions of the Provisions should address this issue. Also refer to the discussion of
analytical modeling of accidental eccentricities in Guide Chapter 1.
Superstructure design (reduced DE response) QE = QDE/2 = DE/2.0
Isolation system and foundation design (unreduced DE response) QE = QDE = DE/1.0
Check of isolation system stability (unreduced MCE response) QE = QMCE = MCE/1.0
11.5.2.5.3 Combination of Horizontal Earthquake Load Effects (Provisions Sec. 5.25.2.3 [4.4.2.3] and
13.4.6.3 [13.4.2.2])
Response due earthquake loading in the X and Y directions QE = Max (1.0QEX+0.3QEY,
0.3QEX+1.0QEY)
In general, the horizontal earthquake load effect, QE, on the response parameter of interest is
influenced by only one direction of horizontal earthquake load, and QE = QEX or QE = QEY.
Exceptions include vertical load on isolator units due to earthquake overturning forces.
11.5.2.5.4 Combination of Horizontal and Vertical Earthquake Load Effects (Provisions Sec. 5.2.7
[4.2.2.1])
Design earthquake (.QE ± 0.2SDSD) E = QE ± 0.2D
Maximum considered earthquake (.QE ± 0.2SMSD) E = QE ± 0.3D
11.5.2.5.5 Superstructure Design Load Combinations (UBC Sec. 1612.2, using RI = 2)
Gravity loads (dead load and reduced live load) 1.2D + 1.6L
Gravity and earthquake loads (1.2D + 0.5L + 1.0E) 1.4D + 0.5L + QDE/2
Chapter 11, Seismically Isolated Structures
1127
Gravity and earthquake loads (0.9D  1.0E) 0.7D  QDE/2
11.5.2.5.6 Isolation System and Foundation Design Load Combinations (UBC Sec. 1612.2)
Gravity loads (for example, long term load on isolator units) 1.2D + 1.6L
Gravity and earthquake loads (1.2D + 0.5L + 1.0E) 1.4D + 0.5L + QDE
Gravity and earthquake loads (0.9D  1.0E) 0.7D  QDE
11.5.2.5.7 Isolation System Stability Load Combinations (Provisions Sec. 13.6.2.6 [13.2.5.6])
Maximum short term load on isolator units (1.2D + 1.0L + E) 1.5D + 1.0L + QMCE
Minimum short term load on isolator units (0.8D  E) 0.8D  QMCE
Note that in the above combinations, the vertical earthquake load (0.2SMSD) component of E is
included in the maximum (downward) load combination, but excluded from the minimum (uplift)
load combination. It is the author's opinion that vertical earthquake ground shaking is of a dynamic
nature, changing direction too rapidly to affect appreciably uplift of isolator units and need not be
used with the load combinations of Provisions Sec. 13.6.2.6 [13.2.5.6] for determining minimum
(uplift) vertical loads on isolator units due to the MCE. Future editions of the Provisions should
explicitly address this issue.
11.5.3 Seismic Force Analysis
11.5.3.1 Basic Approach to Modeling
To expedite calculation of loads on isolator units, and other elements of the seismicforceresisting
system, a 3dimensional model of the EOC is developed and analyzed using the ETABS computer
program (CSI, 1999). While there are a number of programs to choose from, ETABS (version 7.0) is
selected for this example since it permits the automated release of tension in isolator units subject to uplift
and has builtin elements for modeling other nonlinear properties of isolator units. Arguably, all of the
analyses performed by the ETABS program could be done by hand, or by spreadsheet calculation (except
for confirmatory time history analyses).
The ETABS model is used to perform the following types of analyses and calculations:
1. Dead load weight and live loads for the buildingcalculate maximum longterm load on isolator units
(Guide Table 11.51)
2. ELF procedure for gravity and reduced design earthquake loadsdesign superstructure (ignoring
uplift of isolator units)
3. Nonlinear static analysis with ELF loads for gravity and unreduced design earthquake loadsdesign
isolation system and foundation (considering uplift of isolator units)
4. Nonlinear static analysis with ELF loads  gravity and unreduced design earthquake loadscalculate
maximum shortterm load (downward force) on isolator units (Guide Table 11.55) and calculate
minimum shortterm load (downward force) of isolator units (Guide Table 11.56)
FEMA 451, NEHRP Recommended Provisions: Design Examples
1128
5. Nonlinear static analysis with ELF loads for gravity and unreduced MCE loadscalculate maximum
shortterm load (downward force) on isolator units (Guide Table 11.57) and calculate minimum
shortterm load (uplift displacement) of isolator units (Guide Table 11.58)
6. Nonlinear time history analysis  gravity and scaled DE or MCE time historiesverify DE
displacement of isolator units and design story shear (Guide Table 11.510), verify MCE
displacement of isolator units (Guide Table 11.511), verify MCE shortterm load (downward force)
on isolator units (Guide Table 11.512), and verify MCE shortterm load (uplift displacement) of
isolator units (Guide Table 11.513).
The Provisions requires a response spectrum analysis for the EOC (see Guide Table 11.21). In general,
the response spectrum method of dynamic analysis is considered sufficient for facilities that are located at
a stiff soil site, which have an isolation system meeting the criteria of Provisions Sec. 13.2.5.2.7, Item 7
[13.2.4.1, Item 7]. However, nonlinear static analysis is used for the design of the EOC, in lieu of
response spectrum analysis, to permit explicit modeling of uplift of isolator units. For similar reasons,
nonlinear time history analysis is used to verify design parameters.
11.5.3.2 Detailed Modeling Considerations
Although a complete description of the ETABS model is not possible, key assumptions and methods used
to model elements of the isolation system and superstructure are described below.
11.5.3.2.1 Mass Eccentricity
Provisions Sec. 13.4.5.2 [13.4.1.1] requires consideration of mass eccentricity. Because the building in
the example is doubly symmetric, there is no actual eccentricity of building mass (but such would be
modeled if the building were not symmetric). Modeling of accidental mass eccentricity would require
several analyses, each with the building mass located at different eccentric locations (for example, four
quadrant locations in plan). This is problematic, particularly for dynamic analysis using multiple time
history inputs. In this example, only a single (actual) location of mass eccentricity is considered, and
calculated demands are increased moderately for the design of the seismicforceresisting system, and
isolation system (for example, peak displacements calculated by dynamic analysis are increased by 10
percent for design of the isolation system).
11.5.3.2.2 Pdelta Effects
Pdelta moments in the foundation and the first floor girders just above isolator units due to the large
lateral displacement of the superstructure are explicitly modeled. The model distributes half of the
Pdelta moment to the structure above, and half of the Pdelta moment to the foundation below the
isolator units. ETABS (version 7.0) permits explicit modeling of the Pdelta moment, but most computer
programs (including older versions of ETABS) do not. Therefore, the designer must separately calculate
these moments and add them to other forces for the design of affected elements. The Pdelta moments are
quite significant, particularly at isolator units that resist large earthquake overturning loads along lines of
lateral bracing.
11.5.3.2.5 Isolator Unit Uplift
Provisions Sec. 13.6.2.7 [13.2.5.7] permits local uplift of isolator units provided the resulting deflections
do not cause overstress of isolator units or other structural elements. Uplift of some isolator units is likely
(for unreduced earthquake loads) due to the high seismic demand associated with the site. Accordingly,
isolator units are modeled with gap elements that permit uplift when the tension load exceeds the tensile
capacity of an isolator unit. Although most elastomeric bearings can resist some tension stress before
Chapter 11, Seismically Isolated Structures
1129
yielding (typically about 150 psi), isolator units are assumed to yield as soon as they are loaded in
tension, producing larger estimates of uplift displacement and overturning loads on isolator units that are
in compression.
The assumption that isolator units have no tension capacity is not conservative, however, for design of the
connections of isolator units to the structure above and the foundation below. The design of anchor bolts
and other connection elements, must include the effects of tension in isolator units (typically based on a
maximum stress of 150 psi).
11.5.3.2.4 Bounding Values of Bilinear Stiffness of Isolator Units
The design of elements of the seismicforceresisting system is usually based on a linear, elastic model of
the superstructure. When such models are used, Provisions Sec. 13.4.5.3.2 [13.4.1.2] requires that the
stiffness properties of nonlinear isolation system components be based on the maximum effective stiffness
of the isolation system (since this assumption produces larger earthquake forces in the superstructure). In
contrast, the Provisions require that calculation of isolation system displacements be based on the
minimum effective stiffness of the isolation system (since this assumption produces larger isolation
system displacements).
The concept of bounding values, as discussed above, applies to all of the available analysis methods. For
the ELF procedure, the Provisions equations are based directly on maximum effective stiffness, kDmax and
kMmax, are used for calculating design forces; and on minimum effective stiffness, kDmin and kDmax, are used
for calculating design displacements. Where (nonlinear) time history analysis is used, isolators are
explicitly modeled as bilinear hysteretic elements with upper or lowerbound stiffness curves.
Upperbound stiffness curves are used to verify the forces used for the design of the superstructure and
lowerbound stiffness curves are used to verify design displacements of the isolation system.
11.5.4 Preliminary Design Based on the ELF Procedure
11.5.4.1 Calculation of Design Values
11.5.4.1.1 Design Displacements
Preliminary design begins with the engineer's selection of the effective period (and damping) of the
isolated structure, and the calculation of the design displacement, DD. In this example, the effective
period of the EOC facility (at the design earthquake displacement) is TD = 2.5 seconds; and the design
displacement is calculated as follows, using Provisions Eq. 13.3.3.1:
2
(9.8) 0.9(2.5) 16.3 in.
4 1.35
D1 D
D
D
D g S T
p B
=.. .. = =
. .
The 1.35 value of the damping coefficient, BD, is given in Provisions Table 13.3.3.1 [13.31] assuming 15
percent effective damping at 16.3 in. of isolation system displacement. Effective periods of 2 to 3
seconds and effective damping values of 10 to 15 percent are typical of highdamping rubber (and other
types of) bearings.
Stability of the isolation system must be checked for the maximum displacement, DM, which is calculated
using Provisions Eq. 13.3.3.3 as follows:
( ) 2
9.8 1.35(2.5) 24.5 in.
4 1.35
M1 M
M
M
Dg S T
p B
=.. .. = =
. .
FEMA 451, NEHRP Recommended Provisions: Design Examples
1130
For preliminary design, the effective period and effective damping at maximum displacement are assumed
to be the same as the values at the design displacement. While both the effective period and damping
values may reduce slightly at larger rubber strains, the ratio of the two parameters tend to be relatively
consistent.
The total displacement of specific isolator units (considering the effects of torsion) is calculated based on
the plan dimensions of the building, the total torsion (due to actual, plus accidental eccentricity), and the
distance from the center of resistance of the building to the isolator unit of interest. Using Provisions Eq.
13.3.3.51 and 13.3.3.52 [13.35 and 13.36], the total design displacement, DTD, and the total maximum
displacement, DTM, of isolator units located on Column Lines 1 and 7 are calculated for the critical
(transverse) direction of earthquake load as follows:
2 2 2 2
1 12 16.3 1 90 12(0.05)(180) 16.3(1.21) 19.7 in.
120 180 TD D
D D y e
b d
= ...+ ...+ ......= ...+ ... + ......= =
2 2 2 2
1 12 24.5 1 90 12(0.05)(180) 24.5(1.21) 29.6 in.
120 180 TM M
D D y e
b d
= ...+ ...+ ......= ...+ ... + ......= =
The equations above assume that mass is distributed in plan in proportion to isolation system stiffness and
shifted by 5percent, providing no special resistance to rotation of the building on the isolation system. In
fact, building mass is considerably greater toward the center of the building, as shown by the schedule of
gravity loads in Table 11.51. The stiffness of the isolation system is uniform in plan (since all isolators
are of the same size) providing significant resistance to dynamic earthquake rotation of the building.
While the Provisions permit a reduction in the total displacements calculated using the ELF procedure
(with proper substantiation of resistance to torsion), in this example the 21 percent increase is considered
to be conservative for use in preliminary design and for establishing lowerbound limits on dynamic
analysis results.
11.5.4.1.2 Minimum and Maximum Effective Stiffness
Provisions Eq. 13.3.3.2 [13.32] expresses the effective period at the design displacement in terms of
building weight (dead load) and the minimum effective stiffness of the isolation system, kDmin.
Rearranging terms and solving for minimum effective stiffness:
2
2 2
4 114,715240 kips/in.
9.8 2.5 Dmin
D
k W
g T
=.. p .. =.. .. =
. . . .
This stiffness is about 6.9 kips/in. for each of 35 identical isolator units. The effective stiffness can vary
substantially from one isolator unit to another and from one cycle of prototype test to another. Typically,
an isolator unit’s effective stiffness is defined by a range of values for judging acceptability of prototype
(and production) bearings. The minimum value of the stiffness range, kDmin, is used to calculate isolation
system design displacements; the maximum value of the stiffness, kDmax, is used to define design forces.
The variation in effective stiffness depends on the specific type of isolator, elastomeric compound,
loading history, etc., but must, in all cases, be broad enough to comply with the Provisions Sec. 13.9.5.1
[13.6.4.1]requirements that define maximum and minimum values of effective stiffness based on testing
of isolator unit prototypes. Over the three required cycles of test at DD, the maximum value of effective
stiffness (for example, at the first cycle) should not be more than about 30 percent greater than the
minimum value of effective stiffness (for example, at the third cycle) to comply with Provisions Sec.
Chapter 11, Seismically Isolated Structures
1131
13.9.4 [13.6.3]. On this basis, the maximum effective stiffness, kDmax, of the isolation system in this
example, is limited to 312 kips/in. (that is, 1.3 × 240 kips/in.).
The range of effective stiffness defined by kDmin and kDmax (as based on cyclic tests of prototype isolator
units) does not necessarily bound all the possible variations in the effective stiffness of elastomeric
bearings. Other possible sources of variation include: stiffness reduction, due to postfabrication
"scragging" of bearings (by the manufacturer), and partial recovery of this stiffness over time.
Temperature and aging effects of the rubber material, and other changes in properties that can also occur
over the design life of isolator units. (Elastomeric bearings are typically “scragged” immediately
following molding and curing to loosen up the rubber molecules by application of vertical load.)
Provisions Sec. 13.6.2.1 [13.2.5.1] requires that such variations in isolator unit properties be considered in
design, but does not provide specific criteria. A report  Property Modification Factors for Seismic
Isolation Bearings (Constantinou, 1999)  provides guidance for establishing a range of effective stiffness
(and effective damping) properties that captures all sources of variation over the design life of the isolator
units. The full range of effective stiffness has a corresponding range of effective periods (with different
levels of spectral demand). The longest effective period (corresponding to the minimum effective
stiffness) of the range would be used to define isolation system design displacements; and the shortest
effective period (corresponding to the maximum effective stiffness) of the range would be used to define
the design forces on the superstructure.
11.5.4.1.3 Lateral Design Forces
The lateral force required for the design of the isolation system, foundation, and other structural elements
below the isolation system, is given by Provisions Eq. 13.3.4.1 [13.37]:
Vb = kDmax DD = 312(16.3) = 5,100 kips
The lateral force required for checking stability and ultimate capacity of elements of the isolation system,
may be calculated as follows:
VMCE = kDmax DM = 312(24.5) = 7,650 kips
The (unreduced) base shear of the design earthquake is about 35 percent of the weight of the EOC, and
the (unreduced) base shear of the MCE is just over 50 percent of the weight. In order to design the
structure above the isolation system, the design earthquake base shear is reduced by the RI factor in
Provisions Eq. 13.3.4.2 [13.38]:
312(16.3) 2,550 kips
2.0
Dmax D
s
1
Vk D
R
= = =
This force is about 17 percent of the dead load weight of the EOC, which is somewhat less than, but
comparable to, the force that would be required for the design of a conventional, fixedbase building of
the same size and height, seismicforceresisting system, and site seismic conditions. Story shear forces
on the superstructure are distributed vertically over the height of the structure in accordance with
Provisions Eq. 13.3.5 [13.39], as shown in Table 11.53.
Table 11.53 Vertical Distribution of Reduced Design Earthquake Forces (DE/2)
FEMA 451, NEHRP Recommended Provisions: Design Examples
1132
Story level, x Weight, wx
(kips)
Height above
isolation
system, hx (ft)
Force,
x x s
x
i i
i
F w h V
w h
=
S
(kips)
Cumulative
force (kips)
Force divided
by weight,
x
x
F
w
Penthouse roof 965 64 370 370 0.38
Roof 3,500 49 1,020 1,390 0.29
Third floor 3,400 34 690 2,080 0.20
Second floor 3,425 19 390 2,470 0.11
First floor 3,425 4 80 2,550 0.023
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN.
Provisions Eq. 13.3.5 distributes lateral seismic design forces (DE/2) over the height of the building in an
inverted, triangular, pattern as indicated by the ratio of Fx/wx, shown in Table 11.53. Because the
superstructure is much stiffer laterally, than the isolation system, it tends to move as a rigid body in the
first mode with a pattern of lateral seismic forces that is more uniformly distributed over the height of the
building. The use of a triangular load pattern for design is intended to account for highermode response
that may be excited due to flexibility of the superstructure. Provisions Eq. 13.3.5 [13.39] is also used to
distribute forces over the height of the building for unreduced DE and MCE forces, as summarized in
Table 11.54.
Table 11.54 Vertical Distribution of Unreduced DE and MCE Forces
Story level, x
Design earthquake (DE) Maximum considered earthquake (MCE)
Force
(kips)
Cumulative
force (kips)
Force
divided by
weight
Force
(kips)
Cumulative
force (kips)
Force
divided by
weight
Penthouse roof 740 740 0.76 1,110 1,110 1.14
Roof 2,040 2,780 0.59 3,060 4,170 0.88
Third floor 1,380 4,160 0.41 2,070 6,240 0.61
Second floor 780 4,940 0.23 1,170 7,410 0.34
First floor 160 5,100 0.047 240 7,650 0.071
1.0 kip = 4.45 kN.
11.5.4.1.4 Design Earthquake Forces for Isolator Units
Tables 11.55 and 11.56 show the maximum and minimum downward forces for design of the isolator units.
These forces are a result from the simultaneous application of unreduced design earthquake story forces,
summarized in Table 11.54 and appropriate gravity loads to the model of the EOC. (See Guide Sec. 11.5.2.5
for the design load combinations.) As described in Guide Sec. 11.5.2.5, loads are applied simultaneously in
two horizontal directions. The tables report the results for both of the orientations: 100 percent in the X
direction, plus 30 percent in the Y direction, and 30 percent in the X direction, plus 100 percent in the Y
direction. Where the analyses indicate that certain isolator units could uplift during peak DE response (as
indicated by zero downward force), the amount of uplift is small and would not appreciably affect the
distribution of earthquake forces in the superstructure.
Chapter 11, Seismically Isolated Structures
1133
Table 11.55 Maximum Downward Force (kips) for Isolator Design (1.4D + 0.5L + QDE)*
Maximum downward force (kips)
100%(X) ± 30%(Y) / 30%(X) ± 100%(Y)
Column
line 1 2 3 4
A 347/348 661/892 522/557 652/856
B 833/634 1,335/1,519 1,418/1,278 984/1,186
C 537/502 939/890 1,053/1,046 1,074/1,070
1.0 kip = 4.45 kN
.*Forces at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively; loads at
column lines D and E (not shown) are the same as those at column lines B and A, respectively.
Table 11.56 Minimum Downward Force (kips) for Isolator Design (0.7D  QDE)*
Maximum downward force (kips)
100%(X) ± 30%(Y) / 30%(X) ± 100%(Y)
Column
line 1 2 3 4
A 87/84 165/0 200/169 163/0
B 0/123 11/0 23/69 280/59
C 169/196 299/243 428/427 447/442
1.0 kip = 4.45 kN
* Forces at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively; loads at
column lines D and E (not shown) are the same as those at column lines B and A, respectively.
11.5.4.1.5 Maximum Considered Earthquake Forces and Displacements for Isolator Units
Simultaneous application of the unreduced MCE story forces, as summarized in Table 11.54 and appropriate
gravity loads to the model of the EOC, result in the maximum downward forces on isolator units shown in
Guide Table 11.57, and the maximum uplift displacements shown in Table 11.58. The load orientations
and MCE load combinations, are described in Guide Sec. 11.5.2.5. The tables report the results for both of
the load orientations: 100 percent in the X direction, plus 30 percent in the Y direction, and 30 percent in the
X direction, plus 100 percent in the Y direction. Since the nonlinear model assumes that the isolators have
no tension capacity, the values given in Guide Table 11.58 are upper bounds on uplift displacements.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1134
Table 11.57 Maximum Downward Force (kips) on Isolator Units (1.5D + 1.0L + QMCE)*
Maximum downward force (kips)
100%(X) ± 30%(Y) / 30%(X) ± 100%(Y)
Column
line 1 2 3 4
A 444/445 829/1,202 645/710 819/1,146
B 1,112/794 1,739/2,006 1,848/1,635 1,219/1,521
C 680/618 1,171/1,091 1,298/1,282 1,316/1,307
1.0 kip = 4.45 kN.
* Forces at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively; loads at
column lines D and E (not shown) are the same as those at column lines B and A, respectively.
Table 11.58 Maximum Uplift Displacement (in.) of Isolator Units (0.8D  QMCE)*
Maximum uplift displacement (in.)
100%(X) ± 30%(Y) / 30%(X) ± 100%(Y)
Column
line 1 2 3 4
A No uplift 0.00/0.94 No uplift 0.00/0.50
B 0.54/0.00 0.19/0.45 0.13/0.00 0.00/0.14
C No uplift No uplift No uplift No uplift
1.0 in. = 25.4 mm.
* Displacements at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively;
displacements at column lines D and E (not shown) are the same as those at column lines B and A, respectively.
11.5.4.1.6 Limits on Dynamic Analysis
The displacements and forces determined by the ELF procedure provide a basis for expeditious assessment
of size and capacity of isolator units and the required strength of the superstructure. The results of the ELF
procedure also establish limits on design parameters when dynamic analysis is used as the basis for design.
Specifically, the total design displacement, DTD, and the total maximum displacement of the isolation system,
DTM, determined by dynamic analysis cannot be less than 90 percent and 80 percent, respectively, of the
corresponding ELF procedure values:
DTD, dynamic $0.9DTD, ELF = 0.9(19.7) = 17.7 in.
DTM, dynamic $0.8DTM, ELF = 0.8(29.6) = 23.7 in.
The superstructure, if regular, can also be designed for less base shear, but not less than 80 percent of the base
shear from the ELF procedure:
Vs, dynamic $0.8Vs, ELF = 0.8(2,550) = 2,040 kips (= 0.14W)
As an exception to the above, design forces less than 80 percent of the ELF results are permitted if justified
by time history analysis (which is seldom, if ever, the case).
Chapter 11, Seismically Isolated Structures
1135
Third Floor
Base
First Floor
Second Floor
Roof
W24x84 W24x84 Penthouse Roof
W24x104 W24x104 W24x104 W24x104
W24x104 W24x104 W24x104 W24x104
W24x104 W24x104 W24x104 W24x104
W24x192 W24x131 W24x131 W24x192
TS12x12x3
4 TS12x12x3
4 TS12x12x3
4
TS12x12x3
4
TS12x12x3
4
W12x120 W12x120
W12x120 W12x120
W12x96 W12x65
W12x65 W12x65 W12x53
W12x65 W12x65
W12x53 W12x53
W12x65 W12x65
W12x65 W12x65 W12x53
TS12x12x3
4
TS12x12x3
4 TS12x12x3
4 TS12x12x3
4
TS12x12x3
4
Figure 11.58 Elevation of framing on Column Line 2 (Column Line 6 is similar).
11.5.4.2 Design of the Superstructure
The lateral forces, developed in the previous section, in combination with gravity loads, provide a basis for
the design of the superstructure, using methods similar to those used for a conventional building. In this
example, selection of member sizes were made based on the results of ETABS model calculations. Detailed
descriptions of the design calculations are omitted, since the focus of this section is on design aspects unique
to isolated structures (i.e., design of the isolation system, which is described in the next section).
Figures 11.58 and 11.59 are elevation views at Column Lines 2 and B, respectively. Figure 11.510 is a
plan view of the building that shows the framing at the first floor level.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1136
W24 x146
W24 x131
W24 x131 W24 x131 W24 x131 W24 x131 W24 x146
W24 x192 W24 x192 W24 x192 W24 x192 W24 x192 W24 x192
W24 x162 W24 x146 W24 x146 W24 x146 W24 x146 W24 x162
W24 x192 W24 x192 W24 x192 W24 x192 W24 x192 W24 x192
W24 x146 W24 x131 W24 x117 W24 x117 W24 x131 W24 x146
W24 x104 W24 x104 W24 x131
W18x40
W24 x192 W24 x131 W24 x131 W24 x192
W24 x146 W24 x117 W24 x117 W24 x146
W24 x176 W24 x131 W24 x131 W24 x176
W24 x146 W24 x117 W24 x117 W24 x146
W24 x192 W24 x131 W24 x131 W24 x192
W24 x131 W24 x104 W24 x104 W24 x131
W18x40 W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
W18x40
Figure 11.510 First floor framing plan.
Penthouse Roof
Roof
Third Floor
Second Floor
First Floor
Base
W24x68 W24x68 W24x68 W24x68
W24x104 W24x131 W24x146 W24x146 W24x131
W24x117 W24x117 W24x131 W24x131 W24x117
W24x117 W24x131 W24x131 W24x131 W24x131 W24x117
W24x192 W24x192 W24x192 W24x192 W24x192 W24x192
W12x65 W12x65 W12x53
W12x120 W12x120 W12x65
W12x120 W12x120 W12x53
W12x120 W12x120 W12x53
W12x120 W12x120 W12x53
W12x120 W12x120 W12x65
W12x65 W12x65 W12x53
W12x65
W12x53
W12x53
W12x53
W12x65
TS12x12x3
4
TS12x12x3
4
TS12x12x3
4
TS12x12x3
4 TS12x12x3
4
TS12x12x3
4
TS12x12x3
4 TS12x12x3
4
TS12x12x3
4
TS12x12x3
4
TS12x12x3
4
TS12x12x3
4 TS12x12x3
4 TS12x12x3
4
Figure 11.59 Elevation of framing on Column Line B (Column Line D is similar).
Chapter 11, Seismically Isolated Structures
1137
As shown in the elevations (Figures 11.58 and 11.59), fairly large (12×12×¾ in.) tubes are consistently used
throughout the structure for diagonal bracing. A quick check of these braces indicates that stresses will be
at, or below yield for design earthquake loads. The six braces at the third floor, on lines 2, 4, and 6 (critical
floor and direction of bracing) carry a reduced design earthquake force of about 400 kips each (= 2,080 kips/6
braces × cos(30°)). The corresponding stress is about 12.5 ksi for reduced design earthquake forces, or about
25 ksi for unreduced design earthquake forces; indicating that the structure is expected to remain elastic
during the design earthquake.
As shown in Figure 11.510, the first floor framing has heavy, W24 girders along lines of bracing (lines B,
D, 2, 4, and 6). These girders resist Pdelta moments, as well as other forces. A quick check of these girders
indicates that only limited yielding is likely, even for the MCE loads (up to about 2 ft of MCE displacement).
Girders on Line 2 that frame into the column at Line B (critical columns and direction of framing), resist a
Pdelta moment due to the MCE of about 1,000 kipft (2,000 kips/2 girders × 2 ft/2). Moment in these girders
due to MCE shear force in isolators is about 450 kipft (8.9 kips/in. × 24 in. × 4 ft/2 girders). Considering
additional moment due to gravity loads, the plastic capacity of the first floor girders (1,680 kipft) would not
be reached until isolation system displacements exceed about 2 ft. Even beyond this displacement, postyield
deformation would be limited (due to the limited extent and duration of MCE displacements beyond 2 ft),
and the first floor girders would remain capable of stabilizing the isolator units.
11.5.4.3 Design of the Isolation System
The displacements and forces calculated in Guide Sec. 11.5.4.1 provide a basis to either:
1. Develop a detailed design of the isolator units or
2. Include appropriate design properties in performancebased specifications.
Developing a detailed design of an elastomeric bearing, requires a familiarity with rubber bearing technology,
that is usually beyond the expertise of most structural designers; and often varies based on the materials used
by different manufacturers. This example, like most recent isolation projects, will define design properties
for isolator units that are appropriate for incorporation into a performance specification (and can be bid by
more than one bearing manufacturer). Even though the specifications will place the responsibility for meeting
performance standards with the supplier, the designer must still be knowledgeable of available products and
potential suppliers, to ensure success of the design.
11.5.4.3.1 Size of Isolator Units
The design properties of the seismic isolator units are established based on the calculations of ELF demand,
recognizing that dynamic analysis is required to verify these properties (and will likely justify slightly more
lenient properties). The key parameters influencing size are:
1. Peak displacement of isolator units,
2. Average longterm (gravity) load on all isolator units,
3. Maximum longterm (gravity) load on individual isolator units, and
4. Maximum shortterm load on individual isolator units (gravity plus MCE loads) including maximum
uplift displacement.
These parameters are summarized in Guide Table 11.59. Loads or displacements on individual isolator units
are taken as the maximum load or displacement on all isolator units (since the design is based on only one
size of isolator unit). Reduced live load is used for determining longterm loads on isolators.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1138
Table 11.59 Summary of Key Design Parameters [isolator unit location]
Key Design Parameter ELF Procedure
or Gravity Analysis
Dynamic Analysis
Limit
DE displacement at corner of building (DTD) 19.7 in. [A1] > 17.7 in. [A1]
MCE displacement at corner of building (DTM) 29.6 in. [A1] > 23.7 in. [A1]
Average longterm load (1.0D + 0.5L) 477 kips [ALL] N/A
Maximum longterm load (1.2D + 1.6L) 1,053 kips [C4] N/A
DE max shortterm load (1.4D + 0.5L + QDE) 1,519 kips [B2] N/A
DE minimum shortterm load (0.7D  QDE) 150 kips uplift [A2] N/A
MCE max shortterm load (1.5D + 1.0L + QMCE) 2,006 kips [B2] N/A
MCE minimum shortterm load (0.8D  QMCE) 0.94 in. uplift [A2] N/A
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
As rule of thumb, elastomeric isolators should have a diameter, excluding the protective layer of cover, of
no less than 1.25 times maximum earthquake displacement demand. In this case, the full displacement
determined by the ELF procedure would require an isolator diameter of:
iISO $ 1.25(29.6) = 37 in. (0.95 m)
If justified by dynamic analysis, then 80 percent of the ELF procedure displacement would require an isolator
diameter of:
iISO $ 1.25(23.7) = 30 in. (0.75 m)
For this example, a single size of isolator unit is selected with a nominal diameter of no less than 35.4 in.
(0.90 m). Although the maximum vertical loads vary enough to suggest smaller diameter of isolator units
at certain locations (such as at building corners), all of the isolator units must be large enough to sustain MCE
displacements; which are largest at building corners, due to torsion. An isolator unit with a diameter of 35.4
in., has a corresponding bearing area of about Ab = 950 square in. The maximum longterm face pressure is
about 1,100 psi (i.e., 1,053 kips/950 in.2), which is less than the limit for most elastomeric bearing
compounds. Average longterm face pressure is about 500 psi (i.e., 477 kips/950 in.2) indicating the
reasonably good distribution of loads among all isolator units.
The minimum effective stiffness is 6.9 kips/in. per isolator unit at the design displacement (i.e., about 16 in.).
The height of the isolator unit is primarily a function of the height of the rubber, hr, required to achieve this
stiffness, given the bearing area, Ab, and the effective stiffness of the rubber compound. The EOC design
accommodates rubber compounds with minimum effective shear modulus (at 150 percent shear strain)
ranging from G150% = 65 psi to 110 psi. Numerous elastomeric bearing manufacturers have rubber compounds
with a shear modulus that falls within this range. Since rubber compounds (and in particular, highdamping
rubber compounds) are nonlinear, the effective stiffness used for design must be associated with a shear strain
that is close to the strain level for the design earthquake (e.g., 150 percent shear strain). For a minimum
effective shear modulus of 65 psi, the total height of the rubber, hr, would be:
Chapter 11, Seismically Isolated Structures
1139
W12x120
W24x192
Figure 11.511 Typical detail of the isolation system at columns (for clarity,
some elements not shown).
2 2
150% 65 lb/in. 950 in. 8.9 in. 9 in.
6,900 lb/in.
b
r
eff
hG A
k
×
= = = .
The overall height of the isolator unit, H, including steel shim and flange plates, would be about 15 in. For
compounds with an effective minimum shear modulus of 110 psi, the rubber height would be proportionally
taller (about 15 in.), and the total height of the isolator unit would be about 24 in.
11.5.4.3.2 Typical Isolation System Detail
For the EOC design, the isolation system has a similar detail at each column, as shown in Figure 11.511.
The column has an extra large base plate that bears directly on the top of the isolator unit. The column base
plate is circular, with a diameter comparable to that of the top plate of the isolator unit. Heavy, first floor
girders frame into, and are moment connected to the columns (moment connections are required at this floor
only). The columns and base plates are strengthened by plates that run in both horizontal directions, from
the bottom flange of the girder to the base. Girders are stiffened above the seat plates, and at temporary
jacking locations. The top plate of the isolator unit is bolted to the column base plate, and the bottom plate
of the isolator unit is bolted to the foundation.
The foundation connection accommodates the removal and replacement of isolator units, as required by
Provisions Sec. 13.6.2.8 [13.2.5.8]. The bottom plate of the isolator unit bears on a steel plate, that has a
shear lug at the center grouted to the reinforced concrete foundation. Anchor bolts pass through holes in this
plate and connect to threaded couplers that are attached to deeply embedded rods.
With the exception of the portion of the column above the first floor slab, each element shown in Figure
11.511 is an integral part of the isolation system (or foundation), and is designed for the gravity and
FEMA 451, NEHRP Recommended Provisions: Design Examples
1140
unreduced design earthquake loads. In particular, the first floor girder, the connection of the girder to the
column, and the connection of the column to the base plate, are designed for gravity loads and forces caused
by horizontal shear and Pdelta effects due to the unreduced design earthquake load (as shown earlier in
Figure 11.41).
11.5.4.3.3 Design of Connections of Isolator Units
Connection of the top plate of the isolator unit, to the column base plate; and the connection of the bottom
plate to the foundation, are designed for load combinations that include maximum downward forces (1.4D
+ 0.5L + QDE), and minimum downward (uplift) forces (0.7D  QDE). The reactions to bolts at the top and
bottom plates of isolator units (ignoring shear friction and shear capacity of the lug at the base) are
approximately equal, and include shear, axial load, and moment, in the most critical direction of response for
individual bolts. Moments include the effects of Pdelta and horizontal shear across the isolator unit, as
described by the equations shown in Figure 11.41:
2 2
M P VH
.
= +
In this case, H1 = H2 = H/2, where H is the height of the isolator unit (assumed to be 24 in., the maximum
permissible height of isolator units). For maximum downward acting loads, maximum tension on any one
of N (12) uniformly spaced bolts located in a circular pattern of diameter, Db, must be equal to 43 in. and is
estimated as follows:
b 4 b14 12( )
b b b b b b
F M PA M PA MP P VHP
S A N A D A N N D N D
. . . . . . . . + .
=..  .. =.. ·  .. = ..  ..= ..  ..
1 2(1,519 kips)(19.7 in.) 175 kips(24 in.) 1,519 kips 1,587 1,519 6 kips
12 43 in. 12
=.. +  ..=  .
. .
In this calculation, the vertical load, P = 1,519 kips, is the maximum force occurring at Column B2, the
deflection is based on DTD = 19.7 in., and the shear force is calculated as V = 8.9 kips/in. × 19.7 in. The
calculation indicates only a modest amount of tension force in anchor bolts, due to the maximum downward
loads on the isolator units. However, the underlying assumption of the plane section’s remaining plane is not
conservative if top and bottom flange plates yield during a large lateral displacement of the isolator units.
Rather than fabricate bearings with overly thick flange plates, manufacturers usually recommend anchor bolts
that can carry substantial amounts of shear and tension, which can arise due to local bending of flanges.
Tension forces in anchor bolts can occur due to local uplift of isolator units. The ETABS model did not
calculate uplift forces, since it was assumed that the isolator units would yield in tension. An estimate of the
uplift load may be based on a maximum tension yield stress of about 150 psi, which produces a tension force,
Ft, of no more than 150 kips for isolator units with a bearing area of 950 in.2 Applying uplift and shear loads
to the isolator unit, the maximum tension force in each bolt is estimated at:
b 1 2( )
t
b b
FM P A P VHP
S A N N D
. . . . + .
=. + . = . + .
. . . .
1 2(150 kips(19.7 in.) +175 kips(24 in.) 150 kips 333 150 40 kips
12 43 in. 12
=.. + ..=+.
. .
The maximum shear load per bolt is:
Chapter 11, Seismically Isolated Structures
1141
175 15 kips
v 12
FV
N
= = .
Twelve 1¼in. diameter anchor bolts of A325 material are used for these connections. Alternatively, eight
1½in. diameter bolts are permitted for isolator units manufactured with eight, rather than twelve, bolt holes.
Bolts are tightened as required for a slipcritical connection (to avoid slip in the unlikely event of uplift).
Design of stiffeners and other components of the isolation system detail are not included in this example, but
would follow conventional procedures for design loads. Design of the isolator unit including top and bottom
flange plates is the responsibility of the manufacturer. Guide Sec. 11.5.6 includes example performance
requirements of specifications that govern isolator design (and testing) by the manufacturer.
11.5.5 Design Verification Using Nonlinear Time History Analysis
The design is verified (and in some cases isolation system design properties are improved) using time history
analysis of models that explicitly incorporate isolation system nonlinearity, including lateral forcedeflection
properties and uplift of isolator units, which are subject to net tension loads. Using three sets of horizontal
earthquake components the EOC is analyzed separately for design earthquake (DE) and MCE ground
shaking,. All analyses are repeated for two models of the EOC, one with upperbound (UB) stiffness
properties and the other with lowerbound (LB) stiffness properties of isolator units.
11.5.5.1 Ground Motion
Each pair of horizontal earthquake components is applied to the model in three different orientations relative
to the principal axes of the building. Time histories are first applied to produce the maximum response along
the X axis of the EOC model. The analyses are then repeated with the time histories, they are rotated 90° to
produce the maximum response along the Y axis of the EOC model. Additionally, the time histories are
rotated 45° to produce the maximum response along an XY line (to check if this orientation would produce
greater response in certain elements than the two principal axis orientations). A total of 18 analyses (2 models
× 3 component orientations × 3 sets of earthquake components) are performed separately for both DE and
the MCE loads. Results are based on the maximum response of the parameter of interest calculated by each
DE or MCE analysis, respectively. Parameters of interest include:
Design earthquake
1. Peak isolation system displacement
2. Maximum story shear forces (envelope over building height)
Maximum considered earthquake
1. Peak isolation system displacement
2. Maximum downward load on any isolator unit (1.5D + 1.0L + QMCE)
3. Maximum uplift displacement of any isolator unit (0.8L  QMCE).
11.5.5.2 Bilinear Stiffness Modeling of Isolator Units
Nonlinear forcedeflection properties of the isolator units are modeled using a bilinear curve; whose hysteretic
behavior is a parallelogram, which is supplemented by a small amount of viscous damping. A bilinear curve
is commonly used to model the nonlinear properties of elastomeric bearings; although other approaches are
sometimes used, including a trilinear curve that captures stiffening effects of some rubber compounds at very
high strains.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1142
0
100
200
300
400
0 5 10 15 20 25 30
Displacement (in.)
Force (kip)
Upperbound stiffness
Lowerbound stiffness
Property Lowerbound Upperbound
Effective s tiffnes s (kip/in.) 7 11
Initial s tiffnes s (kip/in.) 13.0 20.5
Yield force (kip) 52 82
Pos t/preyield ratio 0.408 0.408
Supplementary damping 3% 3%
Figure 11.512 Stiffness and damping properties of EOC isolator units (1.0 in. = 25.4 mm, 1.0 kip =
4.45 kN).
Using engineering judgement, the initial stiffness, the yield force, the ratio of postyield to preyield stiffness
for the bilinear curves, and the amount of supplementary viscous damping, ßv, are selected. The effective
stiffness and effective damping values are used for preliminary design. The lowerbound bilinear stiffness
curve is based on kDmin = 6.9 kips/in. (rounded to 7 kips/in.). The upperbound bilinear stiffness curve is based
on an effective stiffness of kDmax = 8.9 kips/in. times 1.2 (which the result is rounded to 11 kips/in.). The 1.2
factor is taken into account for the effects of aging over the design life of the isolators. Both upperbound
and lowerbound stiffness curves are based on effective damping (combined hysteretic and supplementary
viscous) of ßD – 15 percent. Figure 11.512 illustrates upperbound and lowerbound bilinear stiffness curves,
and summarizes the values of the parameters that define these curves.
For isolators with known properties, parameters defining bilinear stiffness may be based on test data provided
by the manufacturer. For example, the bilinear properties shown in Figure 11.512, are compared with
effective stiffness and effective damping test data that are representative of a HK090H6 highdamping rubber
bearing manufactured by Bridgestone Engineered Products Company, Inc. Bridgestone is one of several
manufacturers of highdamping rubber bearings and provides catalog data on design properties of standard
isolators. The HK090H6 bearing has a rubber height of about 10 in., a diameter of 0.9 m, without the cover,
and a rubber compound with a relatively low shear modulus (less than 100 psi at high strains).
Plots of the effective stiffness and damping of the HK090H6 bearing, are shown in Figure 11.513 (solid
symbols), and are compared with plots of the calculated effective stiffness and damping (using the equations
from Figure 11.42 and the bilinear stiffness curves shown in Figure 11.512). The effective stiffness curve
of the HK090H6 bearing, is based on data from the third cycle of testing; and therefore, best represents the
minimum effective (or lowerbound) stiffness. The plots indicate the common trend in effective stiffness of
highdamping elastomeric bearings– significant softening up to 100 percent shear strain, fairly stable stiffness
from 100 percent to 300 percent shear strain, and significant stiffening beyond 300 percent shear strain. The
trend in effective damping is a steady decrease in damping with amplitude beyond about 150 percent shear
strain.
Chapter 11, Seismically Isolated Structures
1143
0
5
10
15
20
25
30
0 5 10 15 20 25 30
Displacement (in.)
Stiffness (kip/in.) or
Damping (% of critical)
Effective damping  Test Data
Effective damping
Upperbound effective stiffness
Effective stiffness  Test Data
Lowerbound effective stiffness
Figure 11.513 Comparison of modeled isolator properties to test data (1.0 in. = 25.4 mm, 1.0
kip/in. = 0.175 kN/mm).
The effective stiffness and damping of the bilinear stiffness curves capture the trends of the HK090H6 bearing
reasonably well, as shown in Figure 11.513. Lowerbound effective stiffness fits the HK090H6 data well
for response amplitudes of interest (i.e., the displacements in the range of about 15 in. through 25 in.).
Likewise, the effective damping of the bilinear stiffness curves, is conservatively less than the effective
damping based on test data for the same displacement range. These comparisons confirm that the bilinear
stiffness properties of isolator units used for nonlinear analysis, are valid characterizations of the
forcedeflection properties of the Bridgestone HK090H6 bearing, also presumably, the bearings of other
manufacturers that have comparable values of effective stiffness and damping.
11.5.5.3 Summary of Results for Time History Analyses
Tables 11.510 and 11.511, compare the results of the design earthquake and maximum considered
earthquake time history analyses, respectively, with the corresponding values calculated using the ELF
procedure. The peak response values noted for the time history analyses are, the maxima of Xaxis and
Yaxis directions of response. Because torsional effects were neglected in the time history analyses for this
example, the displacement at the corner of the EOC was assumed to be 1.1 times the center displacement; this
assumption may not be conservative. The reported shears occur below the indicated level.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1144
Table 11.510 Design Earthquake Response Parameters  Results of ELF Procedure and Nonlinear
Time History Analysis
Response parameter ELF
procedure
Time history analysis
Peak response Model Record
Isolation system displacement (in.)
Displacement at center of EOC, DD 16.3 in. 14.3 in. LB
stiffness Sylmar
Displacement at corner of EOC, DTD 19.7 in. 15.7 in.*
Superstructure forces – story shear (kips)
Penthouse roof 740 kips 400 kips
UB stiffness Newhall
Roof (penthouse) 2,780 kips 1,815 kips
Third floor 4,160 kips 3,023 kips
Second floor 4,940 kips 4,180 kips
First floor 5,100 kips 5,438 kips
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
* Displacement includes an arbitrary 10 percent increase for possible torsional response.
Table 11.511 Maximum Considered Earthquake Response Parameters – Results of ELF Procedure
and Nonlinear Time History Analysis
Response parameter
ELF
procedure
Time history analysis
Peak response Model Record
Displacement at center of EOC, DM 24.5 in. 26.4 in. LB
stiffness
El Centro
Displacement at corner of EOC, DTM 29.7 in. 29.0 in.1 No. 6
1.0 in. = 25.4 mm.
* Displacement includes an arbitrary 10 percent increase for possible torsional response.
The results of the design earthquake time history analyses, verify that the results of the ELF procedure are
generally conservative. A design displacement of 16 in., and total design displacement of 20 in., are
conservative bounds on calculated displacements, even if significant torsion should occur. The story shears
calculated using the ELF procedure, are larger than those from the dynamic analyses at all superstructure
elevations. The higher the elevation, the larger the margin between the story shears, which is an indication
that the inverted triangular pattern (the results from Provisions Eq. 13.3.5 [13.39]), produces conservative
results. At the second story (below the third floor), the critical level for the brace design the story shear from
the dynamic analyses (3,023 kips) is only about threequarters of the story shear from the ELF procedure
(4,160 kips). The results of the MCE time history analyses show that the ELF procedure can underestimate
the maximum displacement. Accordingly, a maximum displacement of 27 in., and total maximum
displacement of 30 in., are used for the design specifications.
Tables 11.512 and 11.513, respectively, report the maximum downward forces on isolator units and the
maximum uplift displacements determined from the maximum considered earthquake time history analyses.
Chapter 11, Seismically Isolated Structures
1145
These tables report two values for each isolator location representing both the Xaxis and Yaxis orientations
of the strongest direction of shaking of the time history record.
Table 11.512 Maximum Downward Force (kips) on Isolator Units (1.5D + 1.0L + QMCE)*
Upperbound stiffness model – Newhall record – Xaxis/Yaxis orientations
Column
line 1 2 3 4
A 417/411 930/1,076 651/683 889/1,027
B 1,048/881 1,612/1,766 1,696/1,588 1,308/1,429
C 656/628 1,145/1,127 1,301/1,300 1,319/1,318
1.0 kip = 4.45 kN.
* Forces on Column Lines 5, 6 and 7 (not shown) are enveloped by those on Column Lines 3, 2, and 1, respectively; forces
on Column Lines D and E (not shown) are enveloped by those on Column Lines B and A, respectively.
Table 11.513 Maximum Uplift Displacement (in.) Of Isolator Units (0.8D  QMCE)1
Upperbound stiffness model – Newhall record – Xaxis/Yaxis orientations
Column
line 1 2 3 4
A No uplift 0.00/0.39 No uplift 0.00/0.17
B 0.26/0.00 0.19/0.12 No uplift No uplift
C No uplift No uplift No uplift No uplift
1.0 in. = 25.4 mm.
* Displacement on Column Lines 5, 6 and 7 (not shown) enveloped by those on Column Lines 3, 2, and 1, respectively;
displacement on Column Lines D and E (not shown) are enveloped by those on Column Lines B and A, respectively.
Table 11.512 indicates a maximum downward force of 1,766 kips (at column B2), which is somewhat less
than the value predicted using the ELF procedure (2,006 kips in Table 11.57). Table 11.513 includes the
results from the controlling analysis, and indicates a maximum uplift displacement of 0.39 in. (at column A2),
which is significantly less than the value predicted using the ELF procedure (0.94 in. as noted in Table
11.58). Based on these results, the design specification (Guide Sec. 11.5.2.10) uses values of 2,000 kips
maximum downward force and ½ in. of maximum uplift displacement.
11.5.6 Design and Testing Criteria for Isolator Units
Detailed design of the isolator units for the EOC facility, is the responsibility of the bearing manufacturer
subject to the design and testing (performance)of the criteria included in the construction documents
(drawings and/or specifications). Performance criteria typically includes a basic description, and
requirements for isolator unit sizes, the design life and durability, the environmental loads and fireresistance
criteria, Quality Assurance and Quality Control (including QC testing of production units), the design forces
and displacements, and prototype testing requirements. This section summarizes key data and performance
criteria for the EOC, including the criteria for prototype testing of isolator units, as required by Provisions
Sec. 13.9.2 [13.6.1].
FEMA 451, NEHRP Recommended Provisions: Design Examples
1146
d b
D b
H
tf tf
D
D + 2t
D
o
f
o o
Figure 11.514 Isolator dimensions.
11.5.6.1 Dimensions and Configuration
Steel shim plate diameter, Do $ 35.4 in.
Rubber cover thickness, to $ 0.5 in.
Rubber layer thickness, tr # 0.375 in.
Gross rubber diameter, Do + 2to # 38 in.
Bolt pattern diameter, Db = 43 in.
Bolt hole diameter, db = 1.5 in.
Steel flange plate diameter, Df = 48 in.
Steel flange plate thickness, tf $ 1.5 in.
Total height, H # 24 in.
11.5.6.2 Prototype Stiffness and Damping Criteria
Minimum effective stiffness (third cycle of test, typical vertical load) kDmin, kMmin $ 7.0 kips/in.
Maximum effective stiffness (first cycle test, typical vertical load) kDmax, kMmax # 9.0 kips/in.
The kDmin, kDmax, kMmin, and kMmax, are properties of the isolation system as a whole (calculated from the
properties of individual isolator units using Provisions Eq. 13.9.5.11 through 13.9.5.14 [13.63 through
13.66]). Individual isolator units may have stiffness properties that fall outside the limits (by, perhaps,
10 percent), provided the average stiffness of all isolator units complies with the limits.
Chapter 11, Seismically Isolated Structures
1147
Effective damping
(minimum of three cycles of test, typical vertical load) ßD $ 15 percent
(minimum of three cycles of test, typical vertical load) ßM $ 12 percent
Vertical Stiffness
(average of three cycles of test, typical vertical load kv $ 5,000 kips/in.
± 50 percent)
11.5.6.3 Design Forces (Vertical Loads)
Maximum longterm load (individual isolator) 1.2D + 1.6L = 1,200 kips
Typical load  cyclic load tests (average of all isolators) 1.0D + 0.5L = 500 kips
Upperbound load  cyclic load tests (all isolators) 1.2D + 0.5L + E = 750 kips
Lowerbound load  cyclic load tests (all isolators) 0.8D  E = 250 kips
Maximum shortterm load (individual isolator) 1.2D + 1.0L + E = 2,000 kips
Minimum shortterm load (individual isolator) 0.8D  E = tension force due
to ½ in. of uplift
11.5.6.4 Design Displacements
Design earthquake displacement DD = 16 in.
Total design earthquake displacement DTD = 20 in.
Maximum considered earthquake displacement DM = 27 in.
Total maximum considered earthquake displacement DTM = 30 in.
11.5.6.5 Prototype Testing Criteria
Table 11.514 summarizes the prototype test criteria found in Provisions Sec. 13.9.2 [13.6.1] as well as the
corresponding loads on isolator units of the EOC.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1148
Table 11.514 Prototype Test Requirements
Provisions criteria EOC criteria
No. of cycles Vertical load Lateral load Vertical load Lateral load
Vertical stiffness test
3 cycles Typical ± 50% None 500 ± 250 kips None
Cyclic load tests to check wind effects (Provisions Sec. 13.9.2.3 [13.6.1.2])
20 cycles Typical Design force 500 kips ± 20 kips
Cyclic load tests to establish effective stiffness and damping (Provisions Sec. 13.9.2.3 [13.6.1.2])
3 cycles Typical 0.25DD 500 kips ± 4 in.
3 cycles Upperbound* 0.25DD 750 kips ± 4 in.
3 cycles Lowerbound* 0.25DD 250 kips ± 4 in.
3 cycles Typical 0.5DD 500 kips ± 8 in.
3 cycles Upperbound* 0.5DD 750 kips ± 8 in.
3 cycles Lowerbound* 0.5DD 250 kips ± 8 in.
3 cycles Typical 1.0DD 500 kips ± 16 in.
3 cycles Upperbound* 1.0DD 750 kips ± 16 in.
3 cycles Lowerbound* 1.0DD 250 kips ± 16 in.
3 cycles Typical 1.0DM 500 kips ± 27 in.
3 cycles Upperbound* 1.0DM 750 kips ± 27 in.
3 cycles Lowerbound* 1.0DM 250 kips ± 27 in.
3 cycles Typical 1.0DTM 500 kips ± 30 in.
Cyclic load tests to check durability (Provisions Sec. 13.9.2.3 [13.6.1.2])
30SD1/SDSBD(= 20)** Typical load 1.0DTD 500 kips ± 20 in.
Static load test of isolator stability (Provisions Sec. 13.9.2.6 [13.6.1.5])
N/A Maximum 1.0DTM 2,000 kips 30 in.
N/A Minimum 1.0DTM ½in. of uplift 30 in.
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
* Tests with upperbound and lowerbound vertical loads are required by Provisions Sec. 13.9.2.3 [13.6.1.2] for isolator units
that are verticalloadcarrying elements.
** The Provisions contains a typographical error where presenting this expression. The errata to the Provisions correct the
error.
121
12
NONBUILDING STRUCTURE DESIGN
Harold O. Sprague Jr., P.E.
Chapter 14 of the 2000 NEHRP Recommended Provisions and Commentary (hereafter, the Provisions and
Commentary) is devoted to nonbuilding structures. Nonbuilding structures comprise a myriad of
structures constructed of all types of materials with markedly different dynamic characteristics and a wide
range of performance requirements.
Nonbuilding structures are a general category of structure distinct from buildings. Key features that
differentiate nonbuilding structures from buildings include human occupancy, function, dynamic
response, and risk to society. Human occupancy, which is incidental to most nonbuilding structures, is
the primary purpose of most buildings. The primary purpose and function of nonbuilding structures can
be incidental to society or the purpose and function can be critical for society.
In the past, many nonbuilding structures were designed for seismic resistance using building code
provisions developed specifically for buildings. These code provisions were not adequate to address the
performance requirements and expectations that are unique to nonbuilding structures. For example
consider secondary containment for a vertical vessel containing hazardous materials. Nonlinear
performance and collapse prevention, which are performance expectations for buildings, are inappropriate
for a secondary containment structure, which must not leak.
Traditionally, the seismic design of nonbuilding structures depended on the various trade organizations
and standards development organizations that were disconnected from the building codes. The Provisions
have always been based upon strength design and multiple maps for seismic ground motion definition,
whereas most of the industry standards were based on allowable stress design and a single zone map. The
advent of the 1997 Provisions exacerbated the problems of the disconnect for nonbuilding structures with
direct use of seismic spectral ordinates, and with the change to a longer recurrence interval for the seismic
ground motion. It became clear that a more coordinated effort was required to develop appropriate
seismic design provisions for nonbuilding structures.
This chapter develops examples specifically to help clarify Chapter 14 of the Provisions. The solutions
developed are not intended to be comprehensive but instead focus on interpretation of Provisions Chapter
14 (Nonbuilding Structure Design Requirements). Complete solutions to the examples cited are beyond
the scope of this chapter.
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the
changes are described, the design examples and calculations have not been revised to reflect the changes
to the 2003 Provisions.
FEMA 451, NEHRP Recommended Provisions: Design Examples
122
Several noteworthy changes were made to the nonbuilding structures requirements of the 2003
Provisions. These include clearer definition of the scopes of Chapters 6 and 14, expanded, direct
definition of structural systems (along with design parameters and detailing requirements) in Chapter 14,
and a few specific changes for particular nonbuilding structural systems.
In addition to changes Provisions Chapter 14, the basic earthquake hazard maps were updated, the
redundancy factor calculation was completely revised, and the minimum base shear equation for areas
without nearsource effects was eliminated.
Where they affect the design examples in this chapter, significant changes to the 2003 Provisions and
primary reference documents are noted. However, some minor changes to the 2003 Provisions and the
reference documents may not be noted.
In addition to the Provisions and Commentary, the following publications are referenced in this chapter:
United States Geological Survey, 1996. Seismic Design Parameters (CDROM) USGS.
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the
maps have been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the
previously used separate map package). The CDROM also has been updated.]
American Water Works Association. 1996. Welded Steel Tanks for Water Storage. AWWA.
American Petroleum Institute (API), Welded steel tanks for oil storage. API 650, 10th Edition,
November 1998.
12.1 NONBUILDING STRUCTURES VERSUS NONSTRUCTURAL COMPONENTS
Many industrial structures are classified as either nonbuilding structures or nonstructural components.
This distinction is necessary to determine how the practicing engineer designs the structure. The intent of
the Provisions is to provide a clear and consistent design methodology for engineers to follow regardless
of whether the structure is a nonbuilding structure or a nonstructural component. Central to the
methodology is how to determine which classification is appropriate.
The design methodology contained in Provisions Chapter 6, Architectural, Mechanical, and Electrical
Components Design Requirements, focuses on nonstructural component design. As such, the
amplification by the supporting structure of the earthquakeinduced accelerations is critical to the design
of the component and its supports and attachments. The design methodology contained in Provisions
Chapter 14 focuses on the direct effects of earthquake ground motion on the nonbuilding structure.
Table 121 Applicability of the Chapters of the Provisions
Supporting
Structure
Supported Item
Nonstructural Component Nonbuilding Structure
Building
Chapter 5 [4 and 5]for supporting
structure
Chapter 6 for supported item
Chapter 5 [4 and 5]for supporting
structure
Chapter 14 for supported item
Nonbuilding Chapter 14 for supporting structure
Chapter 6 for supported item
Chapter 14 for both supporting
structure and supported item
Chapter 12, Nonbuilding Structure Design
123
30' 30' 30' 30'
25'
80'
Inlet filter
Figure 121 Combustion turbine building (1.0 ft = 0.3048 m).
The example shown in Figure 121 is a combustion turbine, electricpowergenerating facility with four
bays. Each bay contains a combustion turbine and supports an inlet filter on the roof. The uniform
seismic dead load of the supporting roof structure is 30 psf. Each filter weighs 34 kips.
The following two examples illustrate the difference between nonbuilding structures that are treated as
nonstructural components, using Provisions Chapter 6, and those which are designed in accordance with
Provisions Chapter 14. There is a subtle difference between the two chapters:
6.1: “. . .if the combined weight of the supported components and nonbuilding structures with flexible dynamic
characteristics exceeds 25 percent of the weight of the structure, the structure shall be designed considering
interaction effects between the structure and the supported items.”
14.4: “If the weight of a nonbuilding structure is 25 percent or more of the combined weight of the nonbuilding
structure and the supporting structure, the design seismic forces of the nonbuilding structure shall be
determined based on the combined nonbuilding structure and supporting structural system.. . . ”
The difference is the plural components and the singular nonbuilding structure, and that difference is
explored in this example.
[The text has been cleaned up considerably in the 2003 edition but some inconsistencies persist. Sec.
14.1.5 indicates the scopes of Chapters 6 and 14. Both chapters consider the weight of an individual
supported component or nonbuilding structure in comparison to the total seismic weight. Where the
weight of such an individual item does not exceed 25 percent of the seismic weight, forces are determined
in accordance with Chapter 6. Where a nonbuilding structure’s weight exceeds 25 percent of the seismic
weight, Sec. 14.1.5 requires a combined system analysis and the rigidity or flexibility of the supported
nonbuilding structure is used in determining the R factor. In contrast, Sec. 6.1.1 requires consideration of
interaction effects only where the weight exceeds 25 percent of the seismic weight and the supported item
has flexible dynamic characteristics.]
12.1.1 Nonbuilding Structure
FEMA 451, NEHRP Recommended Provisions: Design Examples
124
For the purpose of illustration assume that the four filter units are connected in a fashion that couples their
dynamic response. Therefore, the plural components used in Provisions Sec. 6.1 is apparently the most
meaningful provision.
[The text no longer contains a plural, but conceptually the frame could be considered a single item in this
instance (just as the separate items within a single rooftop unit would be lumped together).]
12.1.1.1 Calculation of Seismic Weights
All four inlet filters = WIF = 4(34 kips) = 136 kips
Support structure = WSS = 4 (30 ft)(80 ft)(30 psf) = 288 kips
The combined weight of the nonbuilding structure (inlet filters) and the supporting structural system is
Wcombined = 136 kips + 288 kips = 424 kips
12.1.1.2 Selection of Design Method
The ratio of the supported weight to the total weight is:
136 0.321 25%
424
IF
Combined
W
W
= = >
Because the weight of the inlet filters is 25 percent or more of the combined weight of the nonbuilding
structure and the supporting structure (Provisions Sec. 14.4 [14.1.5]), the inlet filters are classified as
“nonbuilding structures” and the seismic design forces must be determined from analysis of the combined
seismicresistant structural systems. This would require modeling the filters, the structural components of
the filters, and the structural components of the combustion turbine supporting structure to determine
accurately the seismic forces on the structural elements as opposed to modeling the filters as lumped
masses. [See the discussion added to Sec. 12.1.]
12.1.2 NONSTRUCTURAL COMPONENT
For the purpose of illustration assume that the inlet filters are independent structures, although each is
supported on the same basic structure. In this instance, one filter is the nonbuilding structure. The
question is whether it is heavy enough to significantly change the response of the combined system.
12.1.2.1 Calculation of Seismic Weights
One inlet filter = WIF = 34 kips
Support structure = WSS = 4 (30 ft)(80 ft)(30 psf) = 288 kips
The combined weight of the nonbuilding structures (all four inlet filters) and the supporting structural
system is
Wcombined = 4 (34 kips) + 288 kips = 424 kips
12.1.2.2 Selection of Design Method
The ratio of the supported weight to the total weight is:
Chapter 12, Nonbuilding Structure Design
125
W
W
IF
Combined
= 34 = <
424
0.08 25%
Because the weight of an inlet filter is less than 25 percent of the combined weight of the nonbuilding
structures and the supporting structure (Provisions Sec. 14.4 [14.1.5]), the inlet filters are classified as
“nonstructural components” and the seismic design forces must be determined in accordance with
Provisions Chapter 6. In this example, the filters could be modeled as lumped masses. The filters and the
filter supports could then be designed as nonstructural components.
12.2 PIPE RACK, OXFORD, MISSISSIPPI
This example illustrates the calculation of design base shears and maximum inelastic displacements for a
pipe rack using the equivalent lateral force (ELF) procedure.
12.2.1 Description
A twotier, 12bay pipe rack in a petrochemical facility has concentrically braced frames in the
longitudinal direction and ordinary moment frames in the transverse direction. The pipe rack supports
four runs of 12in.diameter pipe carrying naphtha on the top tier and four runs of 8in.diameter pipe
carrying water for fire suppression on the bottom tier. The minimum seismic dead load for piping is 35
psf on each tier to allow for future piping loads. The seismic dead load for the steel support structure is
10 psf on each tier.
Pipe supports connect the pipe to the structural steel frame and are designed to support the gravity load
and resist the seismic and wind forces perpendicular to the pipe. The typical pipe support allows the pipe
to move in the longitudinal direction of the pipe to avoid restraining thermal movement. The pipe support
near the center of the run is designed to resist longitudinal and transverse pipe movement as well as
provide gravity support; such supports are generally referred to as fixed supports.
Pipes themselves must be designed to resist gravity, wind, seismic, and thermally induced forces,
spanning from support to support.
If the pipe run is continuous for hundreds of feet, thermal/seismic loops are provided to avoid a
cumulative thermal growth effect. The longitudinal runs of pipe are broken up into sections by providing
thermal/seismic loops at spaced intervals. In Figure 122, it is assumed thermal/seismic loops are
provided at each end of the pipe run.
FEMA 451, NEHRP Recommended Provisions: Design Examples
126
20'0"
6 bays @ 20'0"
= 120'0"
PLAN
ELEVATION SECTION
10'0" 8'0"
15'0" 3'0"
20'0"
5 bays @ 20'0"
= 100'0"
Expansion loop
breaks the
continuity
Horizontal bracing
at braced bay only
Figure 122 Pipe rack (1.0 ft = 0.3048 m).
12.2.2 Provisions Parameters
12.2.2.1 Ground Motion
The spectral response acceleration coefficients at the site are
SDS = 0.40
SD1 = 0.18.
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package).]
12.2.2.2 Seismic Use Group and Importance Factor
The upper piping carries a hazardous material (naphtha) and the lower piping is required for fire
suppression. The naphtha piping is included in Provisions Sec. 1.3.1, Item 11 [Sec. 1.2.1, Item 11],
therefore, the pipe rack is assigned to Seismic Use Group III.
According to Provisions Sec. 14.5.1.2 [14.2.1], the importance factor, I, is 1.5 based on Seismic Use
Group, Hazard, and Function. If these three measures yield different importance factors, the largest factor
applies.
12.2.2.3 Seismic Design Category
For this structure assigned to Seismic Use Group III with SDS = 0.40 and SD1 = 0.18, the Seismic Design
Category is D according to Provisions Sec. 4.2.1 [1.4].
Chapter 12, Nonbuilding Structure Design
127
12.2.3 Design in the Transverse Direction
[Chapter 14 has been revised so that it no longer refers to Table 4.31. Instead values for design
coefficients and detailing requirements are provided with the chapter.]
12.2.3.1 Design Coefficients
Using Provisions Table 14.5.1.1 [14.42] (which refers to Provisions Table 5.2.2 [4.31]), the parameters
for this ordinary steel moment frame are
R = 4
O0 = 3
Cd = 3½
[In the 2003 Provisions, R factor options are presented that correspond to required levels of detailing. R =
3.5, O = 3; Cd = 3.]
Ordinary steel moment frames are retained for use in nonbuilding structures such as pipe racks because
they allow greater flexibility for accommodating process piping and are easier to design and construct
than special steel moment frames.
12.2.3.2 Seismic Response Coefficient
Using Provisions Eq. 5.4.1.11 [5.22]:
C S
s R I
= DS = 04 =
4 15
. 015
.
.
From analysis, T = 0.42 sec. For nonbuilding structures, the fundamental period is generally
approximated for the first iteration and must be verified with final calculations. For many nonbuilding
structures the maximum period limit contained in the first paragraph of Provisions Sec. 5.4.2 [5.2.2] is not
appropriate. As a result, the examples in this chapter neglect that limit. Future editions of the Provisions
will clarify that this limit does not apply to nonbuilding structures. [In the 2003 Provisions, Sec. 14.2.9
makes clear that the approximate period equations do not apply to nonbuilding structures.]
Using Provisions Eq. 5.4.1.12 [5.23], CS does not need to exceed
( ) ( )
0.18 0.161
0.42 4 1.5
D1
s
C S
T R I
= = =
Using Provisions Eq. 5.4.1.13, Cs shall not be less than
Cs = 0.044ISDS = 0.044(1.5)(0.4) = 0.0264
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value for
longperiod structures, which is not applicable to this example.]
Provisions Eq. 5.4.1.11 [5.22] controls; Cs = 0.15.
FEMA 451, NEHRP Recommended Provisions: Design Examples
128
12.2.3.3 Seismic Weight
W = 2(20 ft)(20 ft)(35 psf + 10 psf) = 36 kips
12.2.3.4 Base Shear (Provisions Sec. 5.3.2 [5.2.1])
V = CsW = 0.15(36 kips) = 5.4 kips
12.2.3.5 Drift
Although not shown here, drift of the pipe rack in the transverse direction was calculated by elastic
analysis using the design forces calculated above. The calculated lateral drift, dxe = 0.328 in.
Using Provisions Eq. 14.3.2.1 [5.215],
3.5(0.328 in.)
0.765 in.
1.5
d xe
x
C
I
d
d = = =
The lateral drift must be checked with regard to acceptable limits. The acceptable limits for nonbuilding
structures are not found in codes. Rather, the limits are what is acceptable for the performance of the
piping. In general, piping can safely accommodate the amount of lateral drift calculated in this example.
Pdelta effects must also be considered and checked as required in Provisions Sec. 5.4.6.2 [5.2.6.2].
12.2.3.6 Redundancy Factor
Some nonbuilding structures are designed with parameters from Provisions Table 5.2.2 [4.31]; if they are
termed “nonbuilding structures similar to buildings”. For such structures the redundancy factor applies,
if the structure is in Seismic Design Category D, E, or F. Pipe racks, being fairly simple moment frames
or braced frames, are in the category similar to buildings. Because this structure is assigned to Seismic
Design Category D, Provisions Sec. 5.2.4.2 [4.3.3.2] applies. The redundancy factor is calculated as
. = 2  20
rmaxx Ax
where r is the fraction of the seismic force at a given level resisted by one component of the vertical maxx
seismicforceresisting system at that level, and Ax is defined as the area of the diaphragm immediately
above the story in question. Some interpretation is necessary for the pipe rack. Considering the
transverse direction, the seismicforceresisting system is an ordinary moment resisting frame with only
two columns in a single frame. The frames repeat in an identical pattern. The “diaphragm” is the pipes
themselves, which are not rigid enough to make one consider the 240 ft length between expansion joints
as a diaphragm. Therefore, for the computation of . in the transverse direction, each 20by20 ft bay will
be considered independently.
The maximum of the sum of the shears in the two columns equals the story shear, so the ratio rmax is 1.0.
The diaphragm area is simply the bay area:
Ax = 20 ft × 20 ft = 400 ft2,
therefore,
. = 2 =
20
10
10
.
.
400
Chapter 12, Nonbuilding Structure Design
129
[The redundancy requirements have been changed substantially in the 2003 Provisions.]
12.2.3.7 Determining E
E is defined to include the effects of horizontal and vertical ground motions as follows:
E = .QE ± 0.2 SDS D
where QE is the effect of the horizontal earthquake ground motions, which is determined primarily by the
base shear just computed, and D is the effect of dead load. By putting a simple multiplier on the effect of
dead load, the last term is an approximation of the effect of vertical ground motion. For the moment
frame, the joint moment is influenced by both terms. E with the “+” on the second term when combined
with dead and live loads will generally produce the largest negative moment at the joints, while E with the
“”on the second term when combined with the minimum dead load (0.9D) will produce the largest
positive joint moments.
The Provisions also requires the consideration of an overstrength factor, O0, on the effect of horizontal
motions in defining E for components susceptible to brittle failure.
E = . O0 QE ± 0.2 SDS
The pipe rack does not appear to have components that require such consideration.
12.2.4 Design in the Longitudinal Direction
[In the 2003 Provisions, Chapter 14 no longer refers to Table 4.31. Instead, Tables 14.22 and 14.23
have design coefficient values and corresponding detailing requirements for each system.]
12.2.4.1 Design Coefficients
Using Provisions Table 14.5.1.1 [14.22] (which refers to Provisions Table 5.2.2 [4.31]), the parameters
for this ordinary steel concentrically braced frame are:
R = 4
O0 = 2
Cd = 4½
[The 2003 Provisions allow selection of appropriate design coefficients and corresponding detailing for
several systems. In the case of this example, R would equal 5, but the calculations that follow are not
updated.]
Where Provisions Table 5.2.2 [4.31] is used to determine the values for design coefficients, the detailing
reference sections noted in the table also apply. A concentric braced frame has an assigned R of 5, but an
R of 4 is used to comply with Provisions Sec. 5.2.2.2.1 [4.3.1.2.1].
[In the 2003 Provisions, Chapter 14 no longer refers to Table 4.31. Instead, Tables 14.22 and 14.23
have design coefficient values and corresponding detailing requirements for each system. Chapter 14
contains no requirements corresponding to that found in Sec. 4.3.1.2.1 (related to R factors for systems in
orthogonal directions).]
12.2.4.2 Seismic Response Coefficient
Using Provisions Eq. 5.4.1.11 [5.22]:
FEMA 451, NEHRP Recommended Provisions: Design Examples
1210
C S
s R I
= DS = 04 =
4 15
. 015
.
.
From analysis, T = 0.24 seconds. The fundamental period for nonbuilding structures, is generally
approximated for the first iteration and must be verified with final calculations. For many nonbuilding
structures the maximum period limit contained in the first paragraph of Provisions Sec. 5.4.2 [5.2.2] is not
appropriate. As a result, the examples in this chapter neglect that limit. Future editions of the Provisions
are expected to clarify that this limit does not apply to nonbuilding structures. [In the 2003 Provisions,
Sec. 14.2.9 makes clear that the approximate period equations do not apply to nonbuilding structures.]
Using Provisions Eq. 5.4.1.12 [5.23], CS does not need to exceed:
( ) ( ) C S
s T R I
= D1 = 018 =
024 4 15
. 0281
. .
.
Provisions Sec. 14.5.1 [14.2.8] provides equations for minimum values of Cs that replace corresponding
equations in Provisions Sec. 5.4.1.1 [5.2.1.1]. However, according to Item 2 of Sec. 14.5.1 [14.2.8,
replacement of Chapter 5 equations for minima occurs only “for nonbuilding systems that have an R value
provided in Table 14.5.1.1” [14.42]. In the present example the R values are taken from Table 5.2.2 so
the minima defined in Sec. 5.4.1.1 apply. [In the 2003 Provisions this is no longer the case as reference
to Table 4.31 has been eliminated. Since the example structure would satisfy exception 1 of Sec. 14.2.8
and the minimum base shear equation in Chapter 5 was removed, no additional minimum base shear must
be considered.]
Using Provisions Eq. 5.4.1.13, Cs shall not be less than:
Cs = 0.044ISDS = 0.044(1.5)(0.4) = 0.0264
Provisions Eq. 5.4.1.11 [5.22] controls; Cs = 0.12.
12.2.4.3 Seismic Weight
W = 2(240 ft)(20 ft)(35 psf + 10 psf) = 432 kips
12.2.4.4 Base Shear
Using Provisions Eq. 5.3.2 [5.21]:
V = CsW = 0.15(432 kips) = 64.8 kips
12.2.4.5 Redundancy Factor
For the longitudinal direction, the diaphragm is the horizontal bracing in the bay with the braced frames.
However, given the basis for the redundancy factor, it appears that a more appropriate definition of Ax
would be the area contributing to horizontal forces in the diagonal braces. Thus Ax = 20(240) = 4800 ft2.
The ratio rx is 0.25; each of the four braces has the same stiffness, and each is capable of tension and
compression. Therefore:
. = 2  20 = <
025
085 10 10
.
. ., .
4800
use
Chapter 12, Nonbuilding Structure Design
1211
8'0"
3'0" 3'0" 3'0" 3'0"
3'0"
8'0" 8'0" 8'0" 8'0"
N
W E
S
Figure 123 Steel storage rack (1.0 ft = 0.3048 m).
[The redundancy requirements have been changed substantially in the 2003 Provisions.]
12.3 STEEL STORAGE RACK, OXFORD, MISSISSIPPI
This example uses the equivalent lateral force (ELF) procedure to calculate the seismic base shear in the
eastwest direction for a steel storage rack.
12.3.1 Description
A fourtier, fivebay steel storage rack is located in a retail discount warehouse. There are concentrically
braced frames in the northsouth and eastwest directions. The general public has direct access to the
aisles and merchandise is stored on the upper racks. The rack is supported on a slab on grade. The design
operating load for the rack contents is 125 psf on each tier. The weight of the steel support structure is
assumed to be 5 psf on each tier.
12.3.2 Provisions Parameters
12.3.2.1 Ground Motion
FEMA 451, NEHRP Recommended Provisions: Design Examples
1212
The spectral response acceleration coefficients at the site are as follows:
SDS = 0.40
SD1 = 0.18
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package).]
12.3.2.2 Seismic Use Group and Importance Factor
Use Provisions Sec. 1.3 [1.2]. The storage rack is in a retail facility. Therefore the storage rack is
assigned to Seismic Use Group I. According to Provisions Sec. 14.6.3.1 and 6.1.5 [14.3.5.2], I = Ip = 1.5
because the rack is in an area open to the general public.
12.3.2.3 Seismic Design Category
Use Provisions Tables 4.2.1a and 4.2.1b [1.41 and 1.42]. Given Seismic Use Group I, SDS = 0.40, and
SD1 = 0.18, the Seismic Design Category is C.
12.3.2.4 Design Coefficients
According to Provisions Table 14.5.1.1 [14.23], the design coefficients for this steel storage rack are
R = 4
O0 = 2
Cd = 3½
12.3.3 Design of the System
12.3.3.1 Seismic Response Coefficient
Provisions Sec. 14.6.3 [14.3.5]allows designers some latitude in selecting the seismic design
methodology. Designers may use the Rack Manufacturer’s Institute specification if they modify the
equations to incorporate the seismic spectral ordinates contained in the Provisions; or they may use an R
of 4 and use Provisions Chapter 5 according to the exception in Provisions Sec. 14.6.3.1. The exception
is used in this example. [In the 2003 Provisions these requirements have been restructured so that the
primary method is use of Chapter 5 with the design coefficients of Chapter 14; racks designed using the
RMI method of Sec. 14.3.5.6 are deemed to comply.]
Using Provisions Eq. 5.4.1.11 [5.23]:
0.4 0.15
4 1.5
DS
s
C S
R I
= = =
From analysis, T = 0.24 seconds. For this particular example the short period spectral value controls the
design. The period, for taller racks, however, may be significant and will be a function of the operating
weight. Using Provisions Eq. 5.4.1.12 [5.23], CS does not need to exceed
( ) ( )
0.18 0.281
0.24 4 1.5
D1
s
C S
T R I
= = =
Chapter 12, Nonbuilding Structure Design
1213
Provisions Sec. 14.5.1 [14.2.8]provides equations for minimum values of Cs that replace corresponding
equations in Sec. 5.4.1.1 [5.2.1.1]. The equations in Sec. 14.5.1 [14.2.8] are more conservative than those
in Sec. 5.4.1.1 [5.2.1.1] because nonbuilding structures generally lack redundancy and are not as highly
damped as building structures. These equations generally govern the design of systems with long periods.
According to Item 2 of Sec. 14.5.1 [14.2.8], replacement of the Chapter 5 equations for minima occurs
only “for nonbuilding systems that have an R value provided in Table 14.5.1.1” [14.22]. In the present
example the R value is taken from Table 14.5.1.1 [14.22]and the Seismic Design Category is C so Eq.
14.5.11 [14.22] applies. Using that equation, Cs shall not be less than the following:
Cs = 0.14SDSI = 0.14(0.4)(1.5) = 0.084
Provisions Eq. 5.4.1.11[5.22] controls; Cs = 0.15.
12.3.3.2 Condition “a” (each rack loaded)
12.3.3.2.1 Seismic Weight
In accordance with Provisions Sec. 14.6.3.2 [14.3.5.3], Item a:
Wa = 4(5)(8 ft)(3 ft)[0.67(125 psf)+5 psf] = 42.6 kips
12.3.3.2.2 Design Forces and Moments
Using Provisions Eq. 5.4.1 [5.21], the design base shear for condition “a” is calculated
Va = CsW = 0.15(42.6 kips) = 6.39 kips
In order to calculate the design forces, shears, and overturning moments at each level, seismic forces must
be distributed vertically in accordance with Provisions Sec. 14.6.3.3 [14.3.5.4]. The calculations are
shown in Table 12.31.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1214
Table 12.31 Seismic Forces, Shears, and Overturning Moments
Level
x
Wx
(kips)
hx
(ft)
wxh k
x
(k = 1)
Cvx Fx
(kips)
Vx
(kips)
Mx
(ftkips)
5 10.65 12 127.80 0.40 2.56
2.56 7.68
4 10.65 9 95.85 0.30 1.92
4.48 21.1
3 10.65 6 63.90 0.20 1.28
5.76 38.4
2 10.65 3 31.95 0.10 0.63
6.39 57.6
S 42.6 319.5
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ftkip = 1.36 kNm.
12.3.3.2.3 Resisting Moment at the Base
MOT, resisting = Wa (1.5 ft) = 42.6(1.5 ft) = 63.9 ftkips
12.3.3.3 Condition “b” (only top rack loaded)
12.3.3.3.1 Seismic Weight
In accordance with Provisions Sec. 14.6.3.2 [14.3.5.3], Item b:
Wb = 1(5)(8 ft)(3 ft)(125 psf) + 4(5)(8 ft)(3 ft)(5 psf) = 17.4 kips
12.3.3.3.2 Base Shear
Using Provisions Eq. 5.4.1 [5.21], the design base shear for condition “b” is calculated as follows:
Vb = CsW = 0.15(17.4 kips) = 2.61 kips
12.3.3.3.3 Overturning Moment at the Base
Although the forces could be distributed as shown above for condition “a”, a simpler, conservative
approach for condition “b” is to assume that a seismic force equal to the entire base shear is applied at the
top level. Using that simplifying assumption,
MOT = Vb (12 ft) = 2.61 kip (12 ft) = 31.3 ftkips
12.3.3.3.4 Resisting Moment at the Base
MOT, resisting = Wb (1.5 ft) = 17.4(1.5 ft) = 26.1 ftkips
12.3.3.4 Controlling Conditions
Condition “a” controls shear demands at all but the top level.
Although the overturning moment is larger under condition “a ,” the resisting moment is larger than the
overturning moment. Under condition “b” the resistance to overturning is less than the applied
Chapter 12, Nonbuilding Structure Design
1215
overturning moment. Therefore, the rack anchors must be designed to resist the uplift induced by the
base shear for condition “b”.
12.3.3.5 Torsion
It should be noted that the distribution of eastwest seismic shear will induce torsion in the rack system
because the eastwest brace is only on the back of the storage rack. The torsion should be resisted by the
northsouth braces at each end of the bay where the eastwest braces are placed. If the torsion were to be
distributed to each end of the storage rack, the engineer would be required to calculate the transfer of
torsional forces in diaphragm action in the shelving, which may be impractical.
12.4 ELECTRIC GENERATING POWER PLANT, MERNA, WYOMING
This example highlights some of the differences between the design of nonbuilding structures and the
design of building structures. The boiler building in this example illustrates a solution using the
equivalent lateral force (ELF) procedure. Due to mass irregularities, the boiler building would probably
also require a modal analysis. For brevity, the modal analysis is not illustrated.
12.4.1 Description
Large boilers in coalfired electric power plants are generally suspended from support steel near the roof
level. Additional lateral supports (called buck stays) are provided near the bottom of the boiler. The
buck stays resist lateral forces but allow the boiler to move vertically. Lateral seismic forces are resisted
at the roof and at the buck stay level. Close coordination with the boiler manufacturer is required in order
to determine the proper distribution of seismic forces.
In this example, a boiler building for a 950 mW coalfired electric power generating plant is braced
laterally with ordinary concentrically braced frames in both the northsouth and the eastwest directions.
The facility is part of a grid and is not for emergency back up of a Seismic Use Group III facility.
The dead load of the structure, equipment, and piping, WDL, is 16,700 kips.
The weight of the boiler in service, WBoiler, is 31,600 kips.
The natural period of the structure (determined from analysis) is as follows:
NorthSouth, TNS = 1.90 seconds
EastWest, TEW = 2.60 seconds
FEMA 451, NEHRP Recommended Provisions: Design Examples
1216
N S
E
W
Buck stays
(Typical)
BOILER
Plate girder to
support boiler
(Typical)
240'0"
185'0"
230'0"
Section AA
A A
Chapter 12, Nonbuilding Structure Design
1217
12.4.2 Provisions Parameters
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = II
(for continuous operation, but not for emergency
back up of a Seismic Use Group III facility)
Occupancy Importance Factor, I (Provisions Sec. 1.4 [14.2.1]) = 1.25
Site Coordinates = 42.800° N, 110.500° W
Short Period Response, SS (Seismic Design Parameters) = 0.966
One Second Period Response, S1 (Seismic Design Parameters) = 0.278
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = D (default)
Accelerationbased site coefficient, Fa (Provisions Table 4.1.2.4a
[3.31]) = 1.11
Velocitybased site coefficient, Fv (Provisions Table 4.1.2.4b
[3.32]) = 1.84
Design spectral acceleration response parameters
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.11)(0.966) = 0.715
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.84)(0.278) = 0.341
Seismic Design Category (Provisions Sec. 4.2 [1.4]) = D
SeismicForceResisting System (Provisions Table 14.5.1.1
[14.22]) = Steel concentrically braced
frame (Ordinary)
Response Modification Coefficient, R (Provisions Table 5.2.2) = 5
System Overstrength Factor, O0 (Provisions Table 5.2.2) = 2
Deflection Amplification Factor, Cd (Provisions Table 5.2.2) = 4½
Height limit (Provisions Table 14.5.1.1) = None
Note: If the structure were classified as a “building,” its height would be limited to 35 ft for a Seismic
Design Category D ordinary steel concentrically braced frame, according to the Provisions Table 5.2.2.
The structure is, however, defined as a nonbuilding structure according to Provisions Sec. 14.6.3.4.
Provisions Table 14.5.1.1 does not restrict the height of a nonbuilding structure using an ordinary steel
concentrically braced frame.
[Changes in the 2003 Provisions would affect this example significantly. Table 14.22 would be used to
determine design coefficients and corresponding levels of detailing. For structures of this height using an
ordinary concentrically braced frame system, R = 1.5, O0
= 1, and Cd = 1.5. Alternatively, a special
concentrically braced frame system could be employed.]
FEMA 451, NEHRP Recommended Provisions: Design Examples
1218
12.4.3 Design in the NorthSouth Direction
12.4.3.1 Seismic Response Coefficient
Using Provisions Eq. 5.4.1.11[5.22]:
0.715 0.179
5 1.25
DS
s
C S
R I
= = =
From analysis, T = 1.90 seconds. Using Provisions Eq. 5.4.1.12 [5.23], CS does not need to exceed
( ) ( )
0.341 0.045
1.90 5 1.25
D1
s
C S
T R I
= = =
but using Provisions Eq. 5.4.1.13, Cs shall not be less than:
Cs = 0.044ISDS = 0.044(1.25)(0.715) = 0.0393
[Under the 2003 Provisions no additional minimum base shear must be considered since the example
structure would satisfy exception 1 of Sec. 14.2.8 and the minimum base shear equation in Chapter 5 was
removed.]
Provisions Eq. 5.4.1.12 [5.23] controls; Cs = 0.045.
12.4.3.2 Seismic Weight
Calculate the total seismic weight, W, as:
W = WDL + WBoiler = 16,700 kips + 31,600 kips = 48,300 kips
12.4.3.3 Base Shear
Using Provisions Eq. 5.4.1 [5.21]:
V = CsW = 0.045(48,300 kips) = 2170 kips
12.4.3.4 Redundancy Factor
Refer to Sec. 12.2.3.6 for an explanation of the application of this factor to nonbuilding structures similar
to buildings. The seismic force resisting system is an ordinary concentric braced frame with five columns
in a single line of framing. The number of bays of bracing diminishes near the top, and the overall plan
area is large. For the purposes of this example, it will be assumed that the structure lacks redundancy and
. = 1.5.
[The redundancy requirements have been substantially changed in the 2003 Provisions. If it is assumed
that the structure would fail the redundancy criteria, . = 1.3.]
12.4.3.5 Determining E
See Sec. 12.2.3.7.
Chapter 12, Nonbuilding Structure Design
1219
12.4.4 Design in the EastWest Direction
12.4.4.1 Seismic Response Coefficient
Using Provisions Eq. 5.4.1.11 [5.22]:
0.715 0.179
5 1.25
DS
s
C S
R I
= = =
From analysis, T = 2.60 seconds. Using Provisions Eq. 5.4.1.12 [5.23], CS does not need to exceed:
( ) ( )
0.341 0.0328
2.60 5 1.25
D1
s
C S
T R I
= = =
Using Provisions Eq. 5.4.1.13, Cs shall not be less than:
Cs = 0.044ISDS = 0.044(1.25)(0.715) = 0.0393
[Under the 2003 Provisions no additional minimum base shear must be considered since the example
structure would satisfy exception 1 of Sec. 14.2.8 and the minimum base shear equation in Chapter 5 was
removed.]
Provisions Eq. 5.4.1.13 controls; Cs = 0.0393. [Under the 2003 Provisions, Eq. 5.23 would control the
base shear coefficient for this example.]
12.4.4.2 Seismic Weight
Calculate the total seismic weight, W, as
W = WDL + WBoiler = 16,700 kips + 31,600 kips = 48,300 kips
12.4.4.3 Base Shear
Using Provisions Eq. 5.4.1 [5.21]:
V = CsW = 0.0393(48,300 kips) = 1900 kips
12.5 PIER/WHARF DESIGN, LONG BEACH, CALIFORNIA
This example illustrates the calculation of the seismic base shear in the eastwest direction for the pier
using the ELF procedure.
12.5.1 Description
A private shipping company is developing a pier in Long Beach, California, to service container vessels.
In the northsouth direction, the pier is tied directly to an abutment structure supported on grade. In the
eastwest direction, the pier resists seismic forces using moment frames.
The design live load for container storage is 1000 psf.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1220
15'0"
PLAN N
ELEVATION
3'0" 3'0"
S
W
E
30'0" 20'0" 10'0"
Mean Sea Level
Mud
Dense
sand
10'0" 10'0" 10'0" 10'0" 3'0"
Figure 125 Pier plan and elevation (1.0 ft = 0.3048 m).
12.5.2 Provisions Parameters
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I
(The pier serves container vessels that carry no
hazardous materials.)
Importance Factor, I (Provisions Sec. 14.5.1.2 [14.2.1]) = 1.0
Short Period Response, SS = 1.75
One Second Period Response, S1 = 0.60
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = D (dense sand)
Accelerationbased Site Coefficient, Fa (Provisions Table 4.1.2.4a
[3.31]) = 1.0
Chapter 12, Nonbuilding Structure Design
1221
Velocitybased Site Coefficient, Fv (Provisions Table 4.1.2.4b
[3.32]) = 1.5
Design spectral acceleration response parameters
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.0)(1.75) = 1.167
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.5)(0.60) = 0.60
Seismic Design Category (Provisions Sec. 4.2) = D
SeismicForceResisting System (Provisions Table 14.5.1.1
[14.22]) = Intermediate concrete
moment frame
Response Modification Coefficient, R (Provisions Table 5.2.2) = 5
(The International Building Code and the 2002 edition of
ASCE 7 would require an R value of 3.)
System Overstrength Factor, O0 (Provisions Table 5.2.2) = 3
Deflection Amplification Factor, Cd (Provisions Table 5.2.2) = 4½
Height limit (Provisions Table 14.5.1.1) = 50 ft
If the structure was classified as a building, an intermediate reinforced concrete moment frame
would not be permitted in Seismic Design Category D.
[Changes in the 2003 Provisions would affect this example significantly. Table 14.22 would be used to
determine design coefficients and corresponding levels of detailing. For structures of this height using an
intermediate concrete moment frame system, R = 3, O0
= 2, and Cd = 2.5.]
12.5.3 Design of the System
12.5.3.1 Seismic Response Coefficient
Using Provisions Eq. 5.4.1.11 [5.22]:
1.167 0.233
5 1.0
DS
s
C S
R I
= = =
From analysis, T = 0.596 seconds. Using Provisions Eq. 5.4.1.12 [5.23], CS does not need to exceed:
( ) ( )
0.60 0.201
0.596 5 1.0
D1
s
C S
T R I
= = =
Using Provisions Eq. 5.4.1.13 , Cs shall not be less than:
Cs = 0.044ISDS = 0.044(1.0)(1.167) = 0.0513
FEMA 451, NEHRP Recommended Provisions: Design Examples
1222
[Under the 2003 Provisions no additional minimum base shear must be considered since the example
structure would satisfy exception 1 of Sec. 14.2.8 and the minimum base shear equation in Chapter 5 was
removed.]
Provisions Eq. 5.4.1.12 [5.23] controls; Cs = 0.201.
12.5.3.2 Seismic Weight
In accordance with Provisions Sec. 5.3 [5.2.1] and 14.6.6 [14.2.6], calculate the dead load due to the deck,
beams, and support piers, as follows:
WDeck = 1.0 ft(43 ft)(21 ft)(0.150 kip/ft3) = 135.5 kips
WBeam = 4(2 ft)(2 ft)(21 ft)(0.150 kip/ft3) = 50.4 kips
WPier = 8[p(1.25 ft)2][(10 ft  3 ft) + (20 ft)/2](0.150 kip/ft3) = 100.1 kips
WDL = WDeck + WBeams + WPiers = 135.5 + 50.4 + 100.1 = 286.0 kips
Calculate 25 percent of the storage live load
W1/4 LL = 0.25(1000 psf)(43 ft)(21 ft) = 225.8 kips
Calculate the weight of the displaced water (Provisions Sec. 14.6.6 [14.3.3.1])
WDisp. water = 8[p(1.25 ft)2](20 ft)(64 pcf) = 50.27 kips
Therefore, the total seismic weight is
W = WDL + W1/4LL + WDisp. water = 286.0 + 225.8 + 50.27 = 562.1 kips
12.5.3.3 Base Shear
Using Provisions Eq. 5.3.2 [5.21]:
V = CsW = 0.201(562.1 kips) = 113.0 kips
12.5.3.4 Redundancy Factor
This structure is small in area and has a large number of piles. Following the method described in Sec.
12.2.3.6, yields . = 1.0.
12.6 TANKS AND VESSELS, EVERETT, WASHINGTON
The seismic response of tanks and vessels can be significantly different from that of buildings. For a
structure composed of interconnected solid elements, it is not difficult to recognize how ground motions
accelerate the structure and cause inert forces within the structure. Tanks and vessels, when empty,
respond in a similar manner.
When there is liquid in the tank, the response is much more complicated. As earthquake ground motions
accelerate the tank shell, the shell applies lateral forces to the liquid. The liquid, which responds to those
lateral forces. The liquid response may be amplified significantly if the period content of the earthquake
ground motion is similar to the natural sloshing period of the liquid.
Chapter 12, Nonbuilding Structure Design
1223
H = 10'0" 5'0"
H R = 15'0"
D = 20'0"
Freeboard
d s
Figure 126 Storage tank section (1.0 ft = 0.3048 m).
Earthquakeinduced impulsive fluid forces are those calculated assuming that the liquid is a solid mass.
The convective fluid forces are those that result from sloshing in the tank. It is important to account for
the convective forces on columns and appurtenances inside the tank, because they are affected by
sloshing in the same way that waves affect a pier in the ocean.
The freeboard considerations are critical. Often times, the roof acts as a structural diaphragm. If a tank
does not have sufficient freeboard, the sloshing wave can rip the roof from the wall of the tank. This
could result in the failure of the wall and loss of the liquid within.
The nature of seismic design for liquid containing tanks and vessels is complicated. The fluid mass that is
effective for impulsive and convective seismic forces is discussed in the literature referenced in the
NEHRP Provisions and Commentary.
12.6.1 FlatBottom Water Storage Tank
12.6.1.1 Description
This example illustrates the calculation of the design base shear using the equivalent lateral force (ELF)
procedure for a steel water storage tank used to store potable water for a process within a chemical plant
(Figure 126).
The tank is located away from personnel working within the facility.
The weight of the tank shell, roof, and equipment is 15,400 lb.
12.6.1.2 Provisions Parameters
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I
Importance Factor, I (Provisions Sec. 14.5.1.2 [14.2.1]) = 1.0
Site Coordinates = 48.000° N, 122.250° W
Short Period Response, SS = 1.236
FEMA 451, NEHRP Recommended Provisions: Design Examples
1224
One Second Period Response, S1 = 0.406
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = C (per geotech)
Accelerationbased Site Coefficient, Fa (Provisions Table 4.1.2.4a
[3.31]) = 1.0
Velocitybased Site Coefficient, Fv (Provisions Table 4.1.2.4b
[3.32]) = 1.39
Design spectral acceleration response parameters
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.0)(1.236) = 0.824
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.39)(0.406) = 0.376
SeismicForceResisting System (Provisions Table 14.5.1.1
[14.23]) = Flatbottom, groundsupported,
anchored, bolted
steel tank
Response Modification Coefficient, R (Provisions Table 14.5.1.1
[14.23]) = 3
System Overstrength Factor, O0 (Provisions Table 5.2.2
[14.23]) = 2
Deflection Amplification Factor, Cd (Provisions Table 5.2.2
[14.23]) = 2½
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package). The CDROM also has been updated.]
12.6.1.3 Calculations for Impulsive Response
12.6.1.3.1 Natural Period for the First Mode of Vibration
Based on analysis, the period for impulsive response of the tank and its contents is Ti = 0.14 sec.
12.6.1.3.2 Spectral Acceleration
Based on Provisions Figure 14.7.3.61 [14.41]:
0.376 0.456 seconds
0.824
D1
s
DS
T S
S
= = =
Using Provisions Sec. 14.7.3.6.1 [114.4.7.5.1] with Ti < Ts,:
Sai = SDS = 0.824
12.6.1.3.3 Seismic (Impulsive) Weight
Chapter 12, Nonbuilding Structure Design
1225
Wtank = 15.4 kips
W1water = p(10 ft)2(10 ft)(0.0624 kip/ft3) (W1/WT)= 196.0 (0.75) kips = 147 kips
The ratio W1/WT (= 0.75) was determined from AWWA D100 (it depends on the ratio of height
to diameter)
Wi = Wtank + W1water = 15.4 + 147 = 162.4 kips
12.6.1.3.4 Base Shear
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]:
( )
V S W
i R
= ai i = =
0824 162 4
3
44 6
. .
.
kips
kips
12.6.1.4 Calculations for Convective Response Natural Period for the First Mode of Sloshing
12.6.1.4.1 Natural Period for the First Mode of Sloshing
Using Provisions Section 14.7.3.6.1 [14.4.7.5.1]:
( )2
ft
s
2 2 20 ft 2.58 s
3.68 tanh 3.68 3.68 32.174 tanh 3.68(10 ft)
10 ft
c
T D
g H
D
=p =p =
. . . .
. . . .
. . . .
12.6.1.4.2 Spectral Acceleration
Using Provisions Sec. 14.7.3.6.1 [14.4.7.5.1] with Tc < 4 seconds:
1.5 1.5(0.376) 0.219
2.58
D1
ac
c
S S
T
= = =
12.6.1.4.3 Seismic (Convective) Weight
Wc = Wwater (W2/WT) = 196 (0.30) = 58.8 kips
The ratio W2/WT (= 0.30) was determined from AWWA D100.
12.6.1.4.4 Base Shear
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]:
( )
V S W
c R
= ac c = =
0 219 588
3
4 29
. .
.
kips
kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
1226
12.6.1.5 Design Base Shear
Although Item b of Provisions Sec. 14.7.3.2 [14.4.7.1] indicates that impulsive and convective
components may, in general, be combined using the SRSS method, Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]
requires that the direct sum be used for groundsupported storage tanks for liquids. Using Provisions Eq.
14.7.3.6.1 [14.41]:
V = Vi + Vc = 44.6 + 4.29 = 48.9 kips
[In the 2003 Provisions, use of the SRSS method is also permitted for groundsupported storage tanks for
liquids.]
12.6.2 FLATBOTTOM GASOLINE TANK
12.6.2.1 Description
This example illustrates the calculation of the base shear and the required freeboard using the ELF
procedure for a petrochemical storage tank in a refinery tank farm near a populated city neighborhood.
An impoundment dike is not provided to control liquid spills.
The tank is a flatbottom, groundsupported, anchored, bolted steel tank constructed in accordance with
API 650. The weight of the tank shell, roof, and equipment is 15,400 lb.
12.6.2.2 Provisions Parameters
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = III
(The tank is used for storage of hazardous material.)
Importance Factor, I (Provisions Sec. 14.5.1.2 [14.2.1]) = 1.5
Site Coordinates = 48.000° N, 122.250° W
Short Period Response, SS = 1.236
One Second Period Response, S1 = 0.406
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = C (per geotech)
Accelerationbased Site Coefficient, Fa (Provisions Table 4.1.2.4a
[3.31]) = 1.0
Velocitybased Site Coefficient, Fv (Provisions Table 4.1.2.4b
[3.32]) = 1.39
Design spectral acceleration response parameters
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.0)(1.236) = 0.824
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.39)(0.406) = 0.376
Chapter 12, Nonbuilding Structure Design
1227
SeismicForceResisting System (Provisions Table 14.5.1.1
[14.23]) = Flatbottom, groundsupported,
anchored, bolted
steel tank
Response Modification Coefficient, R (Provisions Table 14.5.1.1
[14.23]) = 3
System Overstrength Factor, O0 (Provisions Table 5.2.2
[14.23]) = 2
Deflection Amplification Factor, Cd (Provisions Table 5.2.2
[14.23]) = 2½
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package). The CDROM also has been updated.]
12.6.2.3 Calculations for Impulsive Response
12.6.2.3.1 Natural Period for the First Mode of Vibration
Based on analysis, the period for impulsive response of the tank and its contents is Ti = 0.14 sec.
12.6.2.3.2 Spectral Acceleration
Based on Provisions Figure 14.7.3.61 [14.41]:
0.376 0.456 seconds
0.824
D1
s
DS
T S
S
= = =
Using Provisions Sec. 14.7.3.6.1 [ 14.4.7.5.1] with Ti < Ts,:
Sai = SDS = 0.824
12.6.2.3.3 Seismic (Impulsive) Weight
Wtank = 15.4 kips
WGas = p(10 ft)2(10 ft)(0.046 kip/ft3)(W1/WT) = 144.5 kips (0.75) = 108.4 kips
Note: The ratio W1/WT was determined from AWWA D100, but API 650 should be used.
Wi = Wtank + WGas = 15.4 + 108.4 = 123.8 kips
12.6.2.3.4 Base Shear
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]:
( )( )
V
S IW
i R
= ai i= =
0824 15 1238
3
510
. . .
.
kips
kips
FEMA 451, NEHRP Recommended Provisions: Design Examples
1228
12.6.2.4 Calculations for Convective Response
12.6.2.4.1 Natural Period for the First Mode of Sloshing
The dimensions are the same as those used for the water tank in Sec. 12.6.1; therefore, Tc = 2.58 sec.
12.6.2.4.2 Spectral Acceleration
Likewise, Sac = 0.219.
12.6.2.4.3 Seismic (Convective) Weight
Wc = WLNG (W2/WT) = 144.5 ( 0.30) = 43.4 kips
The ratio W2/WT was determined from AWWA D100.
12.6.2.4.4 Base shear
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]:
( )( )
V
S IW
c R
= ac c= =
0824 15 435
3
17 9
. . .
.
kips
kips
12.6.2.5 Design Base Shear
Using Provisions Eq. 14.7.3.6.1 [14.41]:
V = Vi + Vc = 51.0 + 17.9 = 68.9 kips
12.6.2.6 Minimum Freeboard
Provisions Table 14.7.3.6.1.2 [14.42] indicates that a minimum freeboard equal to ds is required for this
tank. Using Provisions Eq. 14.7.3.6.1.2 [14.49]:
ds = 0.5DISac = 0.5(20 ft)(1.5)(0.219) = 3.29 ft
The 5 ft freeboard provided is adequate.
12.7 EMERGENCY ELECTRIC POWER SUBSTATION STRUCTURE, ASHPORT,
TENNESSEE
The main section addressing electrical transmission, substation, and distribution structures is in the
appendix to Chapter 14 of the Provisions. The information is in an appendix so that designers can take
time to evaluate and comment on the seismic design procedures before they are included in the main text
of the Provisions.
[In the 2003 Provisions Sections A14.2.1 and A14.2.2 were removed because the appropriate industry
standards had been updated to include seismic design criteria and earthquake ground motions consistent
with the Provisions. Therefore, all references to the Provisions in Sec. 12.7 of this chapter are obsolete.]
Chapter 12, Nonbuilding Structure Design
1229
20'0" 15'0"
7'0"
15'0"
15'0"
cg
PLAN
ELEVATION
Figure 127 Platform for elevated transformer (1.0 ft = 0.3048 m).
12.7.1 Description
This example illustrates the calculation of the base shear using the ELF procedure for a braced frame that
supports a large transformer (Figure 127). The substation is intended to provide emergency electric
power to the emergency control center for the fire and police departments of a community. There is only
one center designed for this purpose.
The weight of the transformer equipment is 17,300 lb.
The weight of the support structure is 12,400 lb.
The period of the structure is T = 0.240 sec.
Although the ratio of the supported structure over the total weight is greater than 25 percent, experience
indicates that the transformer will behave as a lumped rigid mass.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1230
12.7.2 Provisions Parameters
12.7.2.1 Ground Motion
The design response spectral accelerations are defined as
SDS = 1.86
SD1 = 0.79
12.7.2.2 Seismic Use Group and Importance Factor
The structure is for emergency electric power for a Seismic Use Group III facility. Therefore, the
platform is assigned to Seismic Use Group III, as required by Provisions Sec. 1.3 [1.2]. Using Provisions
Table 14.5.1.2 [14.21], the Importance Factor, I, is equal to 1.5.
12.7.2.3 Response Modification Coefficient
From Provisions Table 14A.2.1, R is 3.
12.7.3 Design of the System
12.7.3.1 Seismic Response Coefficient
Provisions Sec. 14A.2.2 defines Cs in a manner that is not consistent with the rest of the Provisions. This
inconsistency will be eliminated in future editions of the Provisions. In this example, the equations are
applied in a manner that is consistent with Chapters 5 [4 and 5] and 14 – that is, R is applied in the
calculation of Cs rather than in the calculation of V.
Using Provisions Section 14A.2.2:
1.86 0.93
3 1.5
DS
s
C S
R I
= = =
but Cs need not be larger than:
( ) ( )
0.79 1.646
0.24 3 1.5
D1
s
C S
T R I
= = =
Therefore, Cs = 0.93.
12.7.3.2 Seismic Weight
W = WTransformer + WSupport structure = 17.3 + 12.4 = 29.7 kips
12.7.3.3 Base Shear
Using Provisions Section 14A.2.2: V = CsW = 0.93(29.7 kips) = 27.6 kips
131
13
DESIGN FOR
NONSTRUCTURAL COMPONENTS
Robert Bachman, P.E. and Richard Drake, P.E.
Chapter 6 of the 2000 NEHRP Recommended Provisions and Commentary (hereinafter, the Provisions
and Commentary) addresses architectural, mechanical, and electrical components of buildings. Two
examples are presented here to illustrate many of the requirements and procedures. Design and anchorage
are illustrated for exterior precast concrete cladding, and for a roofmounted HVAC unit. The rooftop
unit is examined in two common installations: directly attached, and isolated with snubbers. This chapter
also contains an explanation of the fundamental aspects of the Provisions, and an explanation of how
piping, designed according to the ASME Power Piping code, is checked for the force and displacement
requirements of the Provisions.
A large variety of materials and industries are involved with nonstructural components is large, and
numerous documents define and describe methods of design, construction, manufacture, installation,
attachment, etc. Some of the documents address seismic issues but many do not. Provisions Sec. 6.1.1
[6.1.2] contains a listing of approved standards for various nonstructural components.
Although the Guide is based on the 2000 Provisions, it has been annotated to reflect changes made to the
2003 Provisions. Annotations within brackets, [ ], indicate both organizational changes (as a result of a
reformat of all of the chapters of the 2003 Provisions) and substantive technical changes to the 2003
Provisions and its primary reference documents. While the general concepts of the changes are
described, the design examples and calculations have not been revised to reflect the changes to the 2003
Provisions.
A few noteworthy changes were made to the nonstructural components requirements of the 2003
Provisions. These include explicit definition of load effects (including vertical seismic forces) within the
chapter and revised classification of nonductile anchors (based on demonstrated ductility or
prequalification rather than embedmentlengthtodiameter ratio).
In addition to changes Provisions Chapter 6, the basic earthquake hazard maps were updated and the
concrete design reference was updated to ACI 31802 (with a significant resulting changes to the
calculations for anchors in concrete).
Where they affect the design examples in this chapter of the Guide, significant changes to the 2003
Provisions and primary reference documents are noted. However, some minor changes to the 2003
Provisions and the reference documents may not be noted.
In addition to the Provisions, the following are referenced in this chapter:
FEMA 451, NEHRP Recommended Provisions: Design Examples
132
ACI 318 American Concrete Institute. 1999 [2002]. Building Code Requirements and
Commentary for Reinforced Concrete.
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for
Buildings and Other Structures.
ASHRAE APP IP American Society of Heating, Refrigeration, and AirConditioning Engineers
(ASHRAE). 1999. Seismic and Wind Restraint Design, Chapter 53.
ASME B31.1 American Society of Mechanical Engineers. Power Piping Code.
IBC International Code Council. 2000. International Building Code.
The symbols used in this chapter are drawn from Chapter 2 of the Provisions or reflect common
engineering usage. The examples are presented in U.S. customary units.
[In the 2003 Provisions, definitions and symbols specific to nonstructural components appear in Sec.
6.1.3 and 6.1.4, respectively.]
13.1 DEVELOPMENT AND BACKGROUND OF THE PROVISIONS FOR
NONSTRUCTURAL COMPONENTS
13.1.1 Approach to Nonstructural Components
The Provisions requires that nonstructural components be checked for two fundamentally different
demands placed upon them by the response of the structure to earthquake ground motion: resistance to
inertial forces and accommodation of imposed displacements. Building codes have long had
requirements for resistance to inertial forces. Most such requirements apply to the component mass an
acceleration that vary with the basic ground motion parameter and a few broad categories of components.
The broad categories are intended distinguish between components whose dynamic response couples with
that of the supporting structure in such a fashion as to cause the component response accelerations to be
amplified above the accelerations of the structure and those components that are rigid enough with respect
to the structure so that the component response is not amplified over the structural response. In recent
years, a coefficient based on the function of the building or of the component have been introduced as
another multiplier for components important to life safety or essential facilities.
The Provisions includes an equation to compute the inertial force that involves two additional concepts:
variation of the acceleration with relative height within the structure, and reduction in design force based
upon available ductility in the component, or its attachment. The Provisions also includes a quantitative
measure for the deformation imposed upon nonstructural components. The inertial force demands tend to
control the seismic design for isolated or heavy components, whereas, the imposed deformations are
important for the seismic design for elements that are continuous through multiple levels of a structure, or
across expansion joints between adjacent structures, such as cladding or piping.
The remaining portions of this section describe the sequence of steps and decisions prescribed by the
Provisions to check these two seismic demands on nonstructural components.
13.1.2 Force Equations
The following seismic force equations are prescribed for nonstructural components:(Provisions Eq.
6.1.31 [6.21], 6.1.32 [6.23], and 6.1.33 [6.24]):
Chapter 13, Nonstructural Components
133
0.4
p DS p 1 2
p
p
p
F aRS W hz
I
= ..+ ..
. .
1.6 Fpmax = SDSIpWp
0.3 Fpmin = SDSIpWp
where:
Fp = horizontal equivalent static seismic design force centered at the component’s center of gravity
and distributed relative to the component’s mass distribution.
ap = component amplification factor (either 1.0 or 2.5) as tabulated in Provisions Table 6.2.2 [6.31]
for architectural components and Provisions Table 6.3.2 [6.41] for mechanical and electrical
components (Alternatively, may be computed by dynamic analysis)
SDS = five percent damped spectral response acceleration parameter at short period as defined in
Provisions Sec. 4.1.2 [3.3.3]
Wp = component operating weight
Rp = component response modification factor (varies from 1.0 to 5.0) as tabulated in Provisions
Table 6.2.2 [6.31]for architectural components and Provisions Table 6.3.2 [6.41] for
mechanical and electrical components
Ip = component importance factor (either 1.0 or 1.5) as indicated in Provisions Sec. 6.1.5 [6.22]
z = elevation in structure of component point of attachment relative to the base
h = roof elevation of the structure or elevation of highest point of the seismicforceresisting system
of the structure relative to the base
The seismic design force, Fp, is to be applied independently in the longitudinal, and transverse directions.
The effects of these loads on the component are combined with the effects of static loads. Provisions Eq.
6.1.32 [6.23 and 6.24] and 6.1.33, provide maximum and minimum limits for the seismic design force.
For each point of attachment, a force, Fp, should be determined based on Provisions Eq. 6.1.31 [6.21].
The minima and maxima determined from Provisions Eq. 6.1.32 and 6.1.33 [6.21] must be considered
in determining each Fp. The weight, Wp, used to determine each Fp should be based on the tributary
weight of the component associated with the point of attachment. For designing the component, the
attachment force, Fp, should be distributed relative to the component’s mass distribution over the area
used to establish the tributary weight. With the exception of the bearing walls, which are covered by
Provisions Sec. 5.2.6.2.7 [4.6.1.3], and outofplane wall anchorage to flexible diaphragms, which is
covered by Provisions Sec. 5.2.6.3.2 [4.6.2.1], each anchorage force should be based on simple statics
determined by using all the distributed loads applied to the complete component. Cantilever parapets that
are part of a continuous element, should be separately checked for parapet forces.
13.1.3 Load Combinations and Acceptance Criteria
FEMA 451, NEHRP Recommended Provisions: Design Examples
134
13.1.3.1 Seismic Load Effects
When the effects of vertical gravity loads and horizontal earthquake loads are additive, Provisions Eq.
5.2.7.11 [4.21] is used:
E = .QE + 0.2SDSD
When the effects of vertical gravity load counteract those of horizontal earthquake loads, Provisions Eq.
5.2.7.12 [4.22] is used:
E = .QE  0.2SDSD
where:
E = effect of horizontal and vertical earthquakeinduced forces
. = redundancy factor (= 1.0 for nonstructural components)
QE = effect of horizontal seismic forces (due to application of Fp for nonstructural components)
D = effect of dead load
0.2SDSD = effect of vertical seismic forces
13.1.3.2 Strength Load Combinations
Provisions Sec. 5.2.7 [4.2.2] requires the use of ASCE 7 factored load combinations. The combinations
from ASCE 7 Sec. 2.3.2 that include earthquake effects are:
U = 1.2D + 1.0E + 0.5L + 0.2S
U = 0.9D + 1.0E + 1.6H
13.1.4 Component Amplification Factor
The component amplification factor, ap, found in Provisions Eq. 6.1.31 [6.21] represents the dynamic
amplification of the component relative to the maximum acceleration of the component support point(s).
Typically, this amplification is a function of the fundamental period of the component, Tp, and the
fundamental period of the support structure, T. It is recognized that at the time the components are
designed or selected, the effective fundamental period of the structure, T, is not always available. It is
also recognized that for a majority of nonstructural components, the component fundamental period, Tp,
can be accurately obtained only by expensive shaketable or pullback tests. As a result, the determination
of a component’s fundamental period by dynamic analysis, considering T/Tp ratios, is not always
practicable. For this reason, acceptable values of ap have been provided in the Provisions tables.
Therefore, component amplification factors from either these tables or a dynamic analysis may be used.
Values for ap are tabulated for each component based on the expectation that the component will behave
in either a rigid or a flexible manner. For simplicity, a step function increase based on input motion
amplifications is provided to help distinguish between rigid and flexible behavior. If the fundamental
period of the component is less than 0.06 seconds, no dynamic amplification is expected and ap may be
taken to equal 1.00. If the fundamental period of the component is greater than 0.06 seconds, dynamic
amplification is expected, and ap is taken to equal 2.50. In addition, a rational analysis determination of
Chapter 13, Nonstructural Components
135
ap is permitted if reasonable values of both T and Tp are available. Acceptable procedures for determining
ap are provided in Commentary Chapter 6.
13.1.5 Seismic Coefficient at Grade
The short period design spectral acceleration, SDS, considers the site seismicity and local soil conditions.
The site seismicity is obtained from the design value maps (or CDROM), and SDS is determined in
accordance with Provisions Sec. 4.1.2.5 [3.3.3]. The coefficient SDS is the used to design the structure.
The Provisions approximates the effective peak ground acceleration as 0.4SDS, which is why 0.4 appears
in Provisions Eq. 6.1.31 [6.21].
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package). The CDROM also has been updated.]
13.1.6 Relative Location Factor
The relative location factor, 1 2z , scales the seismic coefficient at grade, resulting in values linearly
h
.. + ..
. .
varying from 1.0 at grade to 3.0 at roof level. This factor approximates the dynamic amplification of
ground acceleration by the supporting structure.
13.1.7 Component Response Modification Factor
The component response modification factor, Rp, represents the energy absorption capability of the
component’s construction and attachments. In the absence of applicable research, these factors are based
on judgment with respect to the following benchmark values:
1. Rp = 1.0 or 1.5, brittle or buckling failure mode is expected
2. Rp = 2.5, some minimal level of energy dissipation capacity
3. Rp = 3.5 or 5.0, highly ductile materials and detailing
13.1.8 Component Importance Factor
The component importance factor, Ip, represents the greater of the life safety importance and/or the hazard
exposure importance of the component. The factor indirectly accounts for the functionality of the
component or structure by requiring design for a lesser amount of inelastic behavior (or higher force
level). It is assumed that a lesser amount of inelastic behavior will result in a component that will have a
higher likelihood of functioning after a major earthquake.
13.1.9 Accommodation of Seismic Relative Displacements
The Provisions requires that seismic relative displacements, Dp, be determined in accordance with several
equations. For two connection points on Structure A (or on the same structural system), one at Level x
and the other at Level y, Dp is determined from Provisions Eq. 6.1.41 [6.25] as:
Dp =d xAd yA
Because the computed displacements are frequently not available to the designer of nonstructural
components, one may use the maximum permissible structural displacements per Provisions Eq. 6.1.42
[6.26]:
FEMA 451, NEHRP Recommended Provisions: Design Examples
136
( ) max
aA
p
sx
D X Yh
.
= 
For two connection points on Structures A and B (or on two separate structural systems), one at Level x,
and the other at Level y, DP and DPmax are determined from Provisions Eq. 6.1.43 [6.27] and 6.1.44
[6.28], respectively, as:
Dp =d xA +d yB
max
aA aB
p
sx sx
D X Y
h h
. .
= +
where:
Dp = seismic relative displacement that the component must be designed to accommodate
dxA = deflection of building Level x of Structure A, determined by an elastic analysis as defined
in Provisions Sec. 5.4.6.1 or 5.5.5 [5.2.6, 5.3.5, or 5.4.3] and multiplied by the Cd factor
dyA = deflection of building Level y of Structure A, determined in the same fashion as dxA
X = height of upper support attachment at Level x as measured from the base
Y = height of lower support attachment at Level y as measured from the base
.aA = allowable story drift for Structure A as defined in Provisions Table 5.2.8 [4.51]
hsx = story height used in the definition of the allowable drift, .a, in Provisions Table 5.2.8
[4.51]
dyB = deflection of building Level y of Structure B, determined in the same fashion as dxA
.aB = allowable story drift for Structure B as defined in Provisions Table 5.2.8 [4.51]
The effects of seismic relative displacements must be considered in combination with displacements
caused by other loads as appropriate. Specific methods for evaluating seismic relative displacement
effects of components and associated acceptance criteria are not specified in the Provisions. However,
the intention is to satisfy the purpose of the Provisions. Therefore, for nonessential facilities,
nonstructural components can experience serious damage during the design level earthquake provided
they do not constitute a serious life safety hazard. For essential facilities, nonstructural components can
experience some damage or inelastic deformation during the design level earthquake provided they do not
significantly impair the function of the facility.
13.1.10 Component Anchorage Factors and Acceptance Criteria
Design seismic forces in the connected parts, Fp, are prescribed in Provisions Sec. 6.1.3 [6.2.6].
When component anchorage is provided by expansion anchors or shallow anchors, a value of Rp = 1.5 is
used. Shallow anchors are defined as those with embedment lengthtodiameter ratios of less than 8.
Anchors embedded in concrete or masonry are proportioned to carry the least of the following:
1. The design strength of the connected part
Chapter 13, Nonstructural Components
137
2. 1.3 times the prescribed seismic design force or
3. The maximum force that can be transferred to the connected part by the component structural system
Determination of design seismic forces in anchors must consider installation eccentricities, prying effects,
multiple anchor effects, and the stiffness of the connected system.
Use of powderdriven fasteners is not permitted for seismic design tension forces in Seismic Design
Categories D, E, and F unless approved for such loading.
[In the 2003 Provisions reference is made to “poweractuated” fasteners so as to cover a broader range of
fastener types than is implied by “powderdriven.”]
The design strength of anchors in concrete is determined in accordance with, Provisions Chapter 9, which
is basically the same as IBC Sec. 1913, and has been updated somewhat in Appendix D of ACI 3182002.
(These rules for anchors in concrete will probably be deleted from the next edition of the Provisions in
favor of a reference to ACI 318.)
[The 2003 Provisions refer to Appendix D of ACI 31802 rather than providing specific, detailed
requirements.]
13.1.11 Construction Documents
Construction documents must be prepared by a registered design professional and must include sufficient
detail for use by the owner, building officials, contractors, and special inspectors; Provisions Table 6.1.7
[6.21] includes specific requirements.
13.2 ARCHITECTURAL CONCRETE WALL PANEL
13.2.1 Example Description
In this example, the architectural components are a 4.5in.thick precast normal weight concrete spandrel
panel and a column cover supported by the structural steel frame of a fivestory building as shown in
Figures 13.21 and 13.22.
FEMA 451, NEHRP Recommended Provisions: Design Examples
138
5 at 13'6" = 67'6"
24'0"
Structural
steel frame
Typical window
frame system
Spandrel panel
under consideration
Column cover
under consideration
Figure 13.21 Fivestory building elevation showing panel location (1.0 ft = 0.3048 m)
Chapter 13, Nonstructural Components
139
13'6" story height
T.O.S
T.O.S
24'0"
6'6" panel 7'0" column cover
3'0"
Spandrel panel
under consideration
Typical window
frame system
Column cover
under consideration
~
~
Figure 13.22 Detailed building elevation (1.0 ft = 0.3048 m).
The columns, at the third level of the fivestory office building, support the spandrel panel under
consideration. The columns between the third and fourth levels of the building support the column cover
under consideration. The building, located near a significant active fault in Los Angeles, California, is
assigned to Seismic Use Group I. Wind pressures normal to the building are 17 psf determined in
accordance with ASCE 7. The spandrel panel supports glass windows weighing 10 psf.
This example develops prescribed seismic forces for the selected spandrel panel and prescribed seismic
displacements for the selected column cover.
It should be noted that details of precast connections vary according to the preferences and local practices
of the precast panel supplier. In addition, some connections may involve patented designs. As a result,
this example will concentrate on quantifying the prescribed seismic forces and displacements. After the
prescribed seismic forces and displacements are determined, the connections can be detailed and designed
according to the appropriate AISC and ACI codes and PCI (Precast/Prestressed Concrete Institute)
recommendations.
13.2.2 Design Requirements
13.2.2.1 Provisions Parameters and Coefficients
ap = 1.0 for wall panels (Provisions Table 6.2.2 [6.31])
ap = 1.25 for connection fasteners (Provisions Table 6.2.2 [6.31])
FEMA 451, NEHRP Recommended Provisions: Design Examples
1310
SDS = 1.487 (Design Values CDROM for the selected location and site class)
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package). The CDROM also has been updated.]
Seismic Design Category = D (Provisions Table 4.2.1a [1.41] for S1 < 0.75)
[In the footnote to 2003 Provisions Table 1.41, the value of S1 used to trigger assignment to Seismic
Design Category E or F was changed from 0.75 to 0.6.]
Spandrel Panel Wp = (150 lb/ft3)(24 ft)(6.5 ft)(0.375 ft) = 8775 lb
Glass Wp = (10 lb/ft2)(21 ft)(7 ft) = 1470 lb (supported by spandrel panel)
Column Cover Wp = (150 lb/ft3)(3 ft)(7 ft)(0.375 ft) = 1181 lb
Rp = 2.5 for wall panels (Provisions Table 6.3.2 [6.31])
Rp = 1.0 for connection fasteners ((Provisions Sec. 6.1.6.1 [6.2.8.1])
[In the 2003 Provisions component anchorage is designed using Rp = 1.5 unless specific ductility or
prequalification requirements are satisfied.]
Ip = 1.0 (Provisions Sec. 6.1.5 [6.2.2])
40.5 ft 0.6 (at third floor)
67.5 ft
z
h
= =
. = 1.0 (Provisions Sec. 6.1.3)
[The 2003 Provisions indicate that the redundancy factor does not apply to the design of nonstructural
components. Although the effect is similar to stating that . = 1, there is a real difference since load
effects for such components and their supports and attachments are now defined in Chapter 6 rather than
by reference to Chapter 4.]
13.2.2.2 Performance Criteria
Component failure should not cause failure of an essential architectural, mechanical, or electrical
component (Provisions Sec. 6.1 [6.2.3]).
Component seismic attachments must be bolted, welded, or otherwise positively fastened without
considering the frictional resistance produced by the effects of gravity (Provisions Sec. 6.1.2 [6.2.5]).
The effects of seismic relative displacements must be considered in combination with displacements
caused by other loads as appropriate (Provisions Sec. 6.1.4 [6.2.7]).
Exterior nonstructural wall panels that are attached to or enclose the structure shall be designed to resist
the forces in accordance with Provisions Eq. 6.1.31 or 6.1.32 [6.2.6] and must be able to accommodate
movements of the structure resulting from response to the design basis ground motion, Dp, or temperature
changes (Provisions Sec. 6.2.4 [6.3.2]).
Chapter 13, Nonstructural Components
1311
A
A1
A
A1
B
B
3rd Story beam
6'6"
panel
7'0"
column cover /
window frame
24'0"
Panel C.G.
Figure 13.23 Spandrel panel connection layout from interior (1.0 ft = 0.3048 m).
13.2.3 Spandrel Panel
13.2.3.1 Connection Details
Figure 13.23 shows the types and locations of connections that support one spandrel panel.
The connection system must resist the weight of the panel and supported construction including the
eccentricity between that load and the supports as well as inertial forces generated by response to the
seismic motions in all three dimensions. Furthermore, the connection system must not create undue
interaction between the structural frame and the panel, such as restraint of the natural shrinkage of the
panel or the transfer of floor live load from the beam to the panel. The panels are usually very stiff
compared to the frame, and this requires careful release of potential constraints at connections. PCI’s
Architectural Precast Concrete (2nd Ed. 1989), provides an extended discussion of important design
concepts for such panels.
For this example, the basic gravity load, and vertical accelerations are resisted at Points A, which
provides the recommended simple and statically determinant system for the main gravity weight. The
eccentricity of vertical loads is resisted by a force couple at the two pairs of A1 and A connections.
Horizontal loads parallel to the panel are resisted by the A connections. Horizontal loads perpendicular to
the panel are resisted by all six connections. The A connections, therefore, restrain movement in three
dimensions while the A1 and B connections restrain movement in only one dimension, perpendicular to
the panel. Connection components can be designed to resolve some eccentricities by bending of the
element; for example, the eccentricity of the horizontal inplane force with the structural frame can be
resisted by bending the A connection.
The practice of resisting the horizontal inplane force at two points varies with seismic demand and local
industry practice. The option is to resist all of the inplane horizontal force at one connection in order to
avoid restraint of panel shrinkage. The choice made here depends on local experience indicating that
precast panels of this length have been restrained at the two ends without undue shrinkage restraint
problems.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1312
The A and A1 connections are often designed to take the loads directly to the columns, particularly on
steel moment frames where attachments to the flexural hinging regions of beams are difficult to
accomplish. The lower B connection often require an intersecting beam to provide sufficient stiffness and
strength to resist the loads.
The column cover is supported both vertically and horizontally by the column, transfers no loads to the
spandrel panel, and provides no support for the window frame.
The window frame is supported both vertically and horizontally along the length of the spandrel panel
and transfers no loads to the column covers.
13.2.3.2 Prescribed Seismic Forces
Lateral forces on the wall panels and connection fasteners include seismic loads in accordance with the
Provisions and wind loads in accordance with ASCE 7 as indicated in the problem statement. Wind
forces are not illustrated here.
13.2.3.2.1 Panels
D = Wp = 8775 lb + 1470 lb = 10245 lb (vertical gravity effect)
(Provisions Eq. 6.1.31 [6.21])
( )( )( )
0.4 1.0 1(.487 )10245 lb( ( ))
1 2 0.6 5362 lb
2.5
1.0
Fp= + =
1.6(1.487)(1.0)(10245 lb) = 24375 lb (Provisions Eq. 6.1.32 [6.23]) Fpmax =
0.3(1.487)(1.0)(10245 lb) = 4570 lb (Provisions Eq. 6.1.33 [6.24]) Fpmin =
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be
considered in design are indicated directly and the redundancy factor does not apply, so the following five
steps would be cast differently; the result is the same.]
QE (due to application of Fp) = 5362 lb (Provisions Sec. 6.1.3)
.QE = (1.0)(5362 lb) = 5362 lb (horizontal earthquake effect)
0.2SDSD = (0.2)(1.487)(10245 lb) = 3047 lb (vertical earthquake effect)
E = .QE +0.2SDSD (Provisions Eq. 5.2.7.11 [4.21])
E = .QE  0.2SDSD (Provisions Eq. 5.2.7.12 [4.22])
13.2.3.2.2 Connection Fasteners
The Provisions specifies a reduced RP, and an increased aP for “Fasteners,” which is intended to prevent
premature failure in those elements of connections that are inherently brittle, such as embedments that
depend on concrete breakout strength, or are simply too small to adequately dissipate energy inelastically,
such as welds or bolts. The net effect more than triples the design seismic force.
(Provisions Eq. 6.1.31 [6.21])
( )( )( )
( ) 0.4 1.25 1.487 10245 lb ( ( ))
1 2 0.6 16757 lb
1.0
1.0
Fp= + =
Chapter 13, Nonstructural Components
1313
1.6(1.487)(1.0)(10245 lb) = 24375 lb (Provisions Eq. 6.1.32 [6.23]) Fpmax =
0.3(1.487)(1.0)(10245 lb) = 4570 lb (Provisions Eq. 6.1.33 [6.24]) Fpmin =
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be
considered in design are indicated directly and the redundancy factor does not apply, so the following five
steps would be cast differently; the result is the same.]
QE (due to application of Fp) = 16757 lb (Provisions Sec. 6.1.3)
.QE = (1.0)(16757 lb) = 16757 lb (horizontal earthquake effect)
0.2SDSD = (0.2)(1.487)(10245 lb) = 3047 lb (vertical earthquake effect)
E = .QE +0.2SDSD (Provisions Eq. 5.2.7.11 [4.21])
E = .QE  0.2SDSD (Provisions Eq. 5.2.7.12 [4.22])
13.2.3.3 Proportioning and Design
13.2.3.3.1 Panels
The wall panels should be designed for the following loads in accordance with ACI 318. The design of
the reinforced concrete panel is standard and is not illustrated in this example. Spandrel panel moments
are shown in Figure 13.24. Reaction shears (Vu), forces (Hu), and moments (Mu) are calculated for
applicable strength load combinations.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1314
L = Column spacing
V = Dead and/or earthquake load
H = Wind or earthquake load
Strong
axis
M
Weak
axis
M
u
ux
uy
Vu L
8
32
H u L
u
Figure 13.24 Spandrel panel moments.
Chapter 13, Nonstructural Components
1315
U = 1.4D
Vu = 1.4(10245 lb) = 14343 lb (vertical load downward)
(strong axis moment)
(14343 lb)(24 ft)
43029 ftlb
Mux= 8 =
U = 1.2D + 1.0E
1.2(10245 lb) 1.0(3047 lb) 15341 lb (vertical load downward) Vumax= + =
.Hu =1.0(5362 lb)=5362 lb (horizontal load parallel to panel)
.Hu =1.0(5362 lb)=5362 lb (horizontal load perpendicular to panel)
(strong axis moment) (15341 lb)(24 ft)
46023 ftlb
Muxmax= 8 =
(weak axis moment)
(5362 lb)(24 ft)
4022 ftlb
Muy= 32 =
U = 0.9D + 1.0E
0.9(10245 lb) 1.0(3047 lb) 6174 lb (vertical load downward) Vumin=  =
.Hu =1.0(5362 lb)=5362 lb (horizontal load parallel to panel)
.Hu =1.0(5362 lb)=5362 lb (horizontal load perpendicular to panel)
(strong axis moment)
(6174 lb)(24 ft)
18522 ftlb
Muxmin= 8 =
(weak axis moment)
(5362 lb)(24 ft)
4022 ftlb
Muy= 32 =
13.2.3.3.2 Connection Fasteners
The connection fasteners should be designed for the following loads in accordance with ACI 3182002
(Appendix D) and the AISC specification. There are special reduction factors for anchorage in high
seismic demand locations, and the parameters for this project would invoke those reduction factors. The
design of the connection fasteners is not illustrated in this example. Spandrel panel connection forces are
shown in Figure 13.25. Reaction shears (Vu), forces (Hu), and moments (Mu) are calculated for applicable
strength load combinations.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1316
D = Dead
E = Earthquake
W = Wind
R = Reaction
Panel C.G.
RA
RA1
RA
E, W
D
E
1'6"
8" 1'4"
1'4"
Figure 13.25 Spandrel panel connection forces.
U = 1.4D
(vertical load downward at Point A and A1)
1.4(10245 lb)
7172 lb
VuA= 2 =
MuA = (7172 lb)(1.5 ft) = 10758 ftlb (moment resisted by paired Points A and A1)
Horizontal couple from moment at A and A1 = 10758 / 1.33 = 8071 lb
U = 1.2D + 1.0E
1.2(10245 lb) 1.0(3047 lb) 7671 lb (vertical load downward at Point A)
uAmax 2 V
+
= =
1.0(16757 lb) 3 3142 lb (horizontal load perpendicular to panel at Points A and A1)
16 uA .H = =
HAin = (7671 lb)(1.5 ft) / (1.33 ft) + (3142 lb)(2.0 ft) / (1.33 ft) = 13366 lb (inward force at Point A)
HA1out = (7671 lb)(1.5 ft) / (1.33 ft) + (3142 lb)(0.67) / (1.33 ft) = 10222 lb(outward force at Point A1)
1.0(16757 lb) 8378 lb (horizontal load parallel to panel at Point A)
2 uA .H = =
Mu2A = (8378 lb)(1.5 ft) = 12568 ftlb (flexural moment at Point A)
1.0(16757 lb) 5 10473 lb (horizontal load perpendicular to panel at Points B and B1)
8 uB .H = =
Chapter 13, Nonstructural Components
1317
1'3" 6'0"
1'6"
24'0"
Cover C.G.
C E
D E
column cover /
window frame
6'6" 7'0"
panel
Figure 13.26 Column cover connection layout (1.0 ft = 0.3048 m).
HB = (10743 lb)(2.0 ft) / (1.33 ft) = 15714 lb (inward or outward force at B)
HB1 = (10473 lb)(0.67 ft) / (1.33 ft) = 5237 lb (inward or outward force at B1)
U = 0.9D + 1.0E
0.9(10245 lb) 1.0(3047 lb) 3086 lb (vertical load downward at Point A)
uAmin 2 V

= =
Horizontal forces are the same as combination 1.2 D + 1.0 E. No uplift occurs; the net reaction at A is
downward. Maximum forces are controlled by prior combination. It is important to realize that inward
and outward acting horizontal forces generate different demands when the connections are eccentric to the
center of mass as it is in this example. Only the maximum reactions are computed above.
13.2.3.4 Prescribed Seismic Displacements
Prescribed seismic displacements are not applicable to the building panel because all connections are
essentially at the same elevation.
13.2.4 Column Cover
13.2.4.1 Connection Details
Figure 13.26 shows the key to the types of forces resisted at each column cover connection.
Vertical loads, horizontal loads parallel to the panel, and horizontal loads perpendicular to the panel are
resisted at Point C. The eccentricity of vertical loads is resisted by a force couple at Points C and D. The
horizontal load parallel to the panel eccentricity between the panel and the support is resisted in flexure of
the connection. The connection is designed to take the loads directly to the columns.
Horizontal loads parallel to the panel and horizontal loads perpendicular to the panel are resisted at Point
D. The vertical load eccentricity between the panel and the support is resisted by a force couple of Points
FEMA 451, NEHRP Recommended Provisions: Design Examples
1318
C and D. The eccentricity of horizontal loads parallel to the panel is resisted by flexure at the connection.
The connection must not restrict vertical movement of the panel due to thermal effects or seismic input.
The connection is designed to take the loads directly to the columns.
Horizontal loads perpendicular to the panel are resisted at Points E. The connection is designed to take
the loads directly to the columns.
There is no load eccentricity associated with the horizontal loads perpendicular to the panel.
In this example, all connections are made to the sides of the column because there usually is not enough
room between the outside face of the column and the inside face of the cover to allow a feasible
loadcarrying connection.
13.2.4.2 Prescribed Seismic Forces
Calculation of prescribed seismic forces for the column cover are not shown in this example. They
should be determined in the same manner as illustrated for the spandrel panels.
13.2.4.3 Prescribed Seismic Displacements
The results of an elastic analysis of the building structure are not usually available in time for use in the
design of the precast cladding system. As a result, prescribed seismic displacements are usually
calculated based on allowable story drift requirements:
hsx = story height = 13 ft 6 in
X = height of upper support attachment = 47 ft 9 in
Y = height of lower support attachment = 41 ft 9 in
.a = 0.020hsx (Provisions Table 5.2.8 [4.51])
( ) (72 in.) 0.020 1.44 in. (Provisions Eq. 6.1.42 [6.26]) max
a sx
p
sx sx
D X Y h
h h
.
=  = =
The joints at the top and bottom of the column cover must be designed to accommodate an inplane
relative displacement of 1.44 inches. The column cover will rotate somewhat as these displacements
occur, depending on the nature of the connections to the column. If the supports at one level are “fixed”
to the columns while the other level is designed to “float,” then the rotation will be that of the column at
the point of attachment.
13.2.5 Additional Design Considerations
13.2.5.1 Window Frame System
The window frame system is supported by the spandrel panels above and below. Assuming that the
spandrel panels move rigidly inplane with each floor level, the window frame system must accommodate
a prescribed seismic displacement based on the full story height.
(Provisions Eq. 6.1.42 [6.26]) ( ) (162 in.) 0.020 3.24 in. max
a sx
p
sx sx
D X Y h
h h
.
=  = =
Chapter 13, Nonstructural Components
1319
The window frame system must be designed to accommodate an inplane relative displacement of 3.24 in.
between the supports. This is normally accommodated by a clearance between the glass and the frame.
Provisions Sec. 6.2.10.1 [6.3.7], prescribes a method of checking such a clearance. It requires that the
clearance be large enough so that the glass panel will not fall out of the frame unless the relative seismic
displacement at the top and bottom of the panel exceeds 125 percent of the value predicted amplified by
the building importance factor. If hp and bp are the respective height and width of individual panes and if
the horizontal and vertical clearances are designated c1 and c2, respectively, then the following expression
applies:
2
1
1
2 1 p 1.25
clear p
p
h c
D c D
b c
. .
= .. + ..=
. .
For hp = 7 ft, bp = 5 ft, and Dp = 3.24 in., and setting c1 = c2, the required clearance is 0.84 in.
13.2.5.2 Building Corners
Some thought needs to be given to seismic behavior at external building corners. The preferred approach
is to detail the corners with two separate panel pieces, mitered at a 45 degree angle, with high grade
sealant between the sections. An alternative choice of detailing Lshaped corner pieces, would introduce
more seismic mass and load eccentricity into connections on both sides of the corner column.
13.2.5.3 Dimensional Coordination
It is important to coordinate dimensions with the architect and structural engineer. Precast concrete
panels must be located a sufficient distance from the building structural frame to allow room for the
design of efficient load transfer connection pieces. However, distances must not be so large as to
unnecessarily increase the load eccentricities between the panels and the frame.
13.3 HVAC FAN UNIT SUPPORT
13.3.1 Example Description
In this example, the mechanical component is a 4fthigh, 5ftwide, 8ftlong, 3000lb HVAC fan unit
that is supported on the two long sides near each corner (Figure 13.31). The component is located at the
roof level of a fivestory office building, near a significant active fault in Los Angeles, California. The
building is assigned to Seismic Use Group I. Two methods of attaching the component to the 4,000 psi,
normalweight roof slab, are considered as follows:
1. Direct attachment to the structure with 36 ksi, carbon steel, castinplace anchors and
2. Support on vibration isolation springs, that are attached to the slab with 36 ksi carbon steel
postinstalled expansion anchors.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1320
Concrete Concrete
a = 7'0" b = 5'6"
4'0"
2'0"
Centerofmass
Elevation
Plan
8'0"
5'0"
Attachment location (typical).
HVAC fan unit Direct attachment shown
W = 3,000 lbs
Elevation
Figure 13.31 Air handling fan unit (1.0 ft = 0.3048 m, 1.0 lb = 4.45 N).
13.3.2 Design Requirements
13.3.2.1 Provisions Parameters and Coefficients
ap = 1.0 for direct attachment (Provisions Table 6.3.2 [6.41])
ap = 2.5 for vibration isolated (Provisions Table 6.3.2 [6.41])
SDS = 1.487 (Design Values CDROM)
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used
separate map package).]
Seismic Design Category = D (Provisions Table 4.2.1a [14.41])
Wp = 3000 lb (given)
Rp = 2.5 for HVAC system equipment (Provisions Table 6.3.2 [6.41])
Rp = 1.5 for expansion anchors and shallow,
castinplace anchors* ((Provisions Sec. 6.1.6.1 [6.2.8.1])
Shallow anchors are defined by Provisions Sec. 2.1 as anchors having embedmenttodiameter
ratios of less than 8.
Chapter 13, Nonstructural Components
1321
[In the 2003 Provisions component anchorage is treated differently. Rather than making distinctions
based on an anchor being “shallow,” component anchorage is designed using Rp = 1.5 unless specific
ductility or prequalification requirements are satisfied.]
Ip = 1.0 (Provisions Sec. 6.1.5 [6.2.2])
z/h = 1.0 (for roof mounted equipment)
. = 1.0 (Provisions Sec. 6.1.3 [6.2.6])
[The 2003 Provisions indicate that the redundancy factor does not apply to the design of nonstructural
components. Although the effect is similar to stating that . = 1, there is a real difference since load
effects for such components and their supports and attachments are now defined in Chapter 6 rather than
by reference to Chapter 4.]
13.3.2.2 Performance Criteria
Component failure should not cause failure of an essential architectural, mechanical, or electrical
component (Provisions Sec. 6.1 [6.2.3]).
Component seismic attachments must be bolted, welded, or otherwise positively fastened without
consideration of frictional resistance produced by the effects of gravity (Provisions Sec. 6.1.2 [6.2.5]).
Anchors embedded in concrete or masonry must be proportioned to carry the least of: (a) the design
strength of the connected part, (b) 1.3 times the force in the connected part due to the prescribed forces, or
(c) the maximum force that can be transferred to the connected part by the component structural system
(Provisions Sec. 6.1.6.2 [6.2.8.2]).
Attachments and supports transferring seismic loads must be constructed of materials suitable for the
application and must be designed and constructed in accordance with a nationally recognized structural
standard (Provisions Sec. 6.3.13.2.a [6.4.4, item 6]).
Components mounted on vibration isolation systems must have a bumper restraint or snubber in each
horizontal direction. Vertical restraints must be provided where required to resist overturning. Isolator
housings and restraints must also be constructed of ductile materials. A viscoelastic pad, or similar
material of appropriate thickness, must be used between the bumper and equipment item to limit the
impact load (Provisions Sec. 6.3.13.2.e). Such components must also resist an amplified design force.
13.3.3 Direct Attachment to Structure
This section illustrates design for castinplace concrete anchors with embedmentlengthtodiameters
ratios of 8 or greater; thus, the use of Rp = 2.5 is permitted. [In 2003 Provisions Sec. 6.2.8.1, the value of
Rp used in designing component anchorage is no longer based on the embedment depthtodiameter ratio.
Instead Rp = 1.5 unless specific ductility or prequalification requirements are satisfied.]
13.3.3.1 Prescribed Seismic Forces
See Figure 13.32 for freebody diagram for seismic force analysis.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1322
(Provisions Eq. 6.1.31 [6.21]) ( )( )( )
0.4 1.0 (1.487 )3000 lb ( ( ))
1 21 2141 lb
2.5
1.0
Fp= + =
1.6(1.487)(1.0)(3000 lb) 7138 lb (Provisions Eq. 6.1.32 [6.23]) Fpmax= =
0.3(1.487)(1.0)(3000 lb) 1338 lb (Provisions Eq. 6.1.33 [6.24]) Fpmin= =
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be
considered in design are indicated directly and the redundancy factor does not apply, so the following six
steps would be cast differently; the result is the same.]
QE (due to application of Fp) = 2141 lb (Provisions Sec. 6.1.3)
.QE = (1.0)(2141 lb) = 2141 lb (horizontal earthquake effect)
0.2SDSD = (0.2)(1.487)(3000 lb) = 892 lb (vertical earthquake effect)
D = Wp = 3000 lb (vertical gravity effect)
E = .QE + 0.2SDSD (Provisions Eq. 5.2.7.11 [4.21])
E = .QE  0.2SDSD (Provisions Eq. 5.2.7.12 [4.22])
U = 1.2D + 1.0E + 0.5L + 0.2S
1.0(2141 lb)
535 lb/bolt
Vu= 4 bolts =
no tension
( )( ) ( )( ) ( )( )
( )( )
1.2 3000 lb 2.75 ft 1.0 2141 lb 2 ft 1.0 892 lb 2.75 ft
288 lb/bolt
Tu 5.5 ft 2 bolts
 + +
= =
U = 0.9D  (1.3W or 1.0E)
R
V V
T
D
.QE
0.2S D S D
2'0"
5'6"
u
u
u
u
Figure 13.32 Freebody diagram for seismic force analysis
(1.0 ft = 0.348 m).
Chapter 13, Nonstructural Components
1323
Component base
Tu
Vu
Figure 13.33 Anchor for direct
attachment to structure.
1.0(2141 lb)
535 lb/bolt
Vu= 4 bolts =
no tension
( )( ) ( )( ) ( )( )
( )( )
0.9 3000 lb 2.75 ft 1.0 2141 lb 2 ft 1.0 892 lb 2.75 ft
63 lb/bolt
Tu 5.5 ft 2 bolts
 + +
= =
13.3.3.2 Proportioning and Design
See Figure 13.33 for anchor for direct attachment to structure.
Check one ¼in.diameter castinplace anchor embedded 2 in. into the concrete slab with no transverse
reinforcing engaging the anchor and extending through the failure surface. Although there is no required
tension strength on these anchors, design strengths and tension/shear interaction acceptance relationships
are calculated to demonstrate the use of the Provisions equations.
5.3.3.2.1 Design Tension Strength on Isolated Anchor in Slab, Away from Edge, Loaded Concentrically
[The 2003 Provisions refer to Appendix D of ACI 31802 rather than providing specific, detailed
requirements. Note also that some of the resistance factors, f, are different.]
Following Provisions Sec. 9.2, for a headed bolt or a rod with a nut at the bottom:
(not shallow, Rp = 2.5 is permitted)
2 in. 8.0
0.25 in.
L
d
= =
[In 2003 Provisions selection of Rp is no longer based on the L/d ratio; see Sec. 6.2.8.1.]
Tension capacity of steel, f = 0.80:
Ns = AseFy = (0.049 in.2)(36000 psi) = 1764 lb (Provisions Sec. 9.2.5.1.2 [ACI 31802 Eq. D3])
[In Appendix D of ACI 31802 this capacity calculation is based on fut rather than Fy, since “the large
majority of anchor materials do not exhibit a welldefined yield point.”]
Tension capacity of concrete, f = 0.70, with no eccentricity, no pullthrough, and no edge or group effect:
(Provisions Eq. 9.2.5.2.21 [ACI 31802 Eq. D7]) 1.5 24 4000 psi(81.5 ) 4293 lb c cef N =k f'h = =
FEMA 451, NEHRP Recommended Provisions: Design Examples
1324
[In Appendix D of ACI 31802 this item is defined as Nb rather than Nc.]
The steel controls but, with no tension demand, the point is moot.
5.3.3.2.2 Design Shear Strength on Isolated Anchor, Away from Edge
Shear capacity of steel, f = 0.80:
Vs = AseFy = (0.049 in.2)(36000 psi) = 1764 lb (Provisions Eq. 9.2.6.1.21 [ACI 31802 Eq. D17)
[In Appendix D of ACI 31802 this capacity calculation is based on fut rather than Fy, since “the large
majority of anchor materials do not exhibit a welldefined yield point.”]
Shear capacity of concrete, far from edge, limited to pryout, f = 0.70:
Vcp = 2Nc = 2(4293) = 8493 lb (Provisions Eq. 9.2.6.3.1 [ACI 31802 Eq. D28)
The steel controls, with fVN = 0.8(1764) = 1411 lb
Per Provisions Sec. 6.1.6.2 [6.2.8.2], anchors embedded in concrete or masonry are to be proportioned to
carry at least 1.3 times the force in the connected part due to the prescribed forces. Thus, Vu = 1.3(535) =
696 lb and the anchor is clearly adequate.
5.3.3.2.3 Combined Tension and Shear
The Provisions gives a new equation (Eq. 9.2.7.3 [ACI 31802 Eq. D29]) for the interaction of tension
and shear on an anchor or a group of anchors:
u u 1.2 , which applies when either term exceeds 0.2
N N
N V
fN fV
+ =
5.3.3.2.3 Summary
At each corner of the component, provide one ¼in.diameter castinplace anchor embedded 2 in. into
the concrete slab. Transverse reinforcement engaging the anchor and extending through the failure
surface is not necessary.
13.3.4 Support on Vibration Isolation Springs
13.3.4.1 Prescribed Seismic Forces
Design forces for vibration isolation springs are determined by an analysis of earthquake forces applied in
a diagonal horizontal direction as shown in Figure 13.34. Terminology and concept are taken from
ASHRAE APP IP.
Angle of diagonal loading:
tan1 b (ASHRAE APP IP. Eq. 17)
a
. = .. ..
. .
Tension per isolator:
Chapter 13, Nonstructural Components
1325
h
1 a 2 2 b 3
FP sin (.)
WP
FPV FPV
WP
FP cos (.)
1 2
X
Plan view
X
Y Y
F P sin (.)
. F P cos (.)
FP
4 3
F P = Seismic horizontal force
WP = Operating weight of equipment
F P V = Seismic vertical force
a = Distance between vibration isolators along YY
b = Distance between vibration isolators along XX
h = Height of center of gravity
= Vibration isolator location
Figure 13.34 ASHRAE diagonal seismic force analysis for vibration isolation springs.
(ASHRAE APP IP. Eq. 18)
cos sin
4 2
p pv p
u
W F Fh
T
b a
=   .. .+ ...
. .
Compression per isolator:
(ASHRAE APP IP. Eq. 19)
cos sin
4 2
p pv p
u
W F Fh
C
b a
= + + .. .+ ...
. .
Shear per isolator:
(ASHRAE APP IP. Eq. 20)
4
p
u
F
V =
Select worst case assumption: Design for postinstalled expansion anchors, requiring the use of Rp = 1.5.
(Provisions Eq. 6.1.31 [6.21]) ( )( )( )
( ) 0.4 2.5 1.487 3000 lb ( ( ))
1 2 1 8922 lb
1.5
1.0
Fp= + =
1.6(1.487)(1.0)(3000 lb) 7138 lb (Provisions Eq. 6.1.32 [6.23]) Fpmax= =
0.3(1.487)(1.0)(3000 lb) 1338 lb (Provisions Eq. 6.1.33 [6.24]) Fpmin= =
FEMA 451, NEHRP Recommended Provisions: Design Examples
1326
Components mounted on vibration isolation systems shall have a bumper restraint or snubber in each
horizontal direction. Per Provisions Table 6.3.2 [6.41], Footnote B, the design force shall be taken as
2Fp.
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be
considered in design are indicated directly and the redundancy factor does not apply, so the following
steps would be cast differently; the result is the same.]
QE = Fp = 2(7138 lb) = 14276 lb (Provisions Sec. 6.1.3)
.QE = (1.0)(14276 lb) = 14276 lb (horizontal earthquake effect)
Fpv(ASHRAE) = 0.2SDSD = (0.2)(1.487)(3000 lb) = 892 lb (vertical earthquake effect)
D = Wp = 3000 lb (vertical gravity effect)
E = .QE + 0.2SDSD (Eq. 5.2.7.11 [4.21])
E = .QE  0.2SDSD (Eq. 5.2.7.12 [4.22])
U = 1.2D + 1.0E + 0.5L + 0.2S
tan17 ft 51.8o
5.5 ft
.=  .. ..=
. .
1.2(3000 lb) (892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 2624 lb
4 2 7 ft 5.5 ft u T
=   .. °+ °..=
. .
1.2(3000 lb)+(892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 4424 lb
4 2 7 ft 5.5 ft u C
= + .. °+ °..=
. .
14276 lb 3569 lb
4 u V = =
U = 0.9D + 1.0E
tan17 ft 51.8o
5.5 ft
.=  .. ..=
. .
0.9(3000 lb) (892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 2849 lb
4 2 7 ft 5.5 ft u T
=   .. °+ °..=
. .
0.9(3000 lb) (892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 4199 lb
4 2 7 ft 5.5 ft u C = + + .. °+ °..=
. .
14276 lb 3569 lb
4 u V = =
13.3.4.2 Proportioning and Details
Chapter 13, Nonstructural Components
1327
T
W = 4"
V
"O"
T
V
H = 5"
Vibration isolator
w/ seismic housing
Equipment frame
u
u
b
b
Figure 13.35 Anchor and snubber loads for
support on vibration isolation springs (1.0 in. = 25.4
mm).
Anchor and snubber loads for support on vibration isolation springs are shown in Figure 13.35.
Check vibration isolation system within housing anchored with two 1in.diameter postinstalled
expansion anchors embedded 9 inches into the concrete.
The Provisions does not provide a basis for determining the design strength of postinstalled expansion
anchors. Although manufacturers provide ultimate strength shear and tension loads for their products, the
Provisions does not provide resistance or quality values, f, to allow determination of design shear and
tension strengths. The 2002 edition of ACI 318 contains provisions for postinstalled anchors which
depend on anchor testing per the ACI 355.201 testing standard. Prior to the adoption of this method, the
best course available was to use allowable stress loads published in evaluation reports prepared by model
building code agencies. These allowable stress loads were then multiplied by 1.4 to convert to design
strengths. This technique will be illustrated in the following.
[The 2003 Provisions refer to Appendix D of ACI 31802.]
Allowable stress combinations are included in the 2000 IBC. The 1.4 strength conversion factor is
unnecessary when using allowable stress loads published in evaluation reports prepared by model
building code agencies.
13.3.4.2.1 Design Tension Strength on Isolated Anchor in Slab, Away from Edge
Allowable stress tension values are obtained from ICBO Evaluation Services, Inc., ER4627 for Hilti
Kwik Bolt II concrete anchors. Similar certified allowable values are expected with anchors from other
manufacturers.
anchor diameter = 1 in.
anchor depth = 9 in.
fc' = 4000 psi
FEMA 451, NEHRP Recommended Provisions: Design Examples
1328
with special inspection
Tallow = 8800 lb
Ps = fPc = 1.4(8800 lb) = 12320 lb
13.3.4.2.2 Design Shear Strength on Isolated Anchor in Slab, Away from Edge
Allowable stress shear values are obtained from ICBO Evaluation Services, Inc., ER4627 for Hilti Kwik
Bolt II concrete anchors. Similar certified allowable values are expected with anchors from other
manufacturers.
anchor diameter = 1 in.
anchor depth = 9 in.
fc' = 4000 psi
Vallow = 8055 lb
Vs = fVc = 1.4(8055 lb) = 11277 lb
13.3.4.2.3 Combined Tension and Shear
Per Provisions Sec. 6.1.6.2 [6.2.8.2], anchors embedded in concrete or masonry shall be proportioned to
carry at least 1.3 times the force in the connected part due to the prescribed forces.
In the Provisions and in the 2000 IBC, the factor of 2.0 is reduced to 1.3. This will greatly reduce the
prescribed seismic forces.
Interaction relationships for combined shear and tension loads are obtained from ICBO Evaluation
Services, Inc. ER4627 for Hilti Kwik Bolt II concrete anchors. Similar results are expected using other
anchors.
As stated in ICBO ES evaluation report:
5 5
3 3
s s 1
t t
P V
P V
. . . .
. . +. . =
. . . .
Using Provisions terminology:
5 5
3 3
b b 1
c c
T V
fP fV
. . . .
. . +. . =
. . . .
OK
5 5
2 4462 lb32 1785 lb30.68 0.13 0.81 1
11277 lb 12075 lb
..× .. +..× .. = + = <
. . . .
13.3.4.2.4 Summary
Chapter 13, Nonstructural Components
1329
Gap Gap
Equipment
frame
Impact
material
Steel bushing
bolted or welded
to equipment frame
Figure 13.36 Lateral restraint required to resist seismic forces.
At each corner of the HVAC fan unit, provide a vibration isolation system within a housing anchored
with two 1in.diameter postinstalled expansion anchors embedded 9 in. into the concrete slab. Special
inspection is required. A raised concrete pad is probably required to allow proper embedment of the
postinstalled expansion anchors.
Other postinstalled anchors, such as chemical (adhesive) or undercut postinstalled anchors also could be
investigated. These anchors may require more involved installation procedures, but they may allow the
use of Rp = 2.5 if they have an embedment depthtodiameter ratio of at least 8 (i.e., are not shallow
anchors). The higher Rp value will result in much smaller prescribed seismic forces and, therefore, a
much reduced embedment depth.
[Again note that in 2003 Provisions selection of Rp is no longer based on the L/d ratio; see Sec. 6.2.8.1.]
13.3.5 Additional Considerations for Support on Vibration Isolators
Vibration isolation springs are provided for equipment to prevent vibration from being transmitted to the
building structure. However, they provide virtually no resistance to horizontal seismic forces. In such
cases, some type of restraint is required to resist the seismic forces. Figure 13.36 illustrates one concept
where a bolt attached to the equipment base is allowed to slide a controlled distance (gap) in either
direction along its longitudinal axis before it contacts resilient impact material.
Design of restraints for vibrationisolated equipment varies for different applications and for different
manufacturers. In most cases, restraint design incorporates all directional capability with an air gap, a
soft impact material, and a ductile restraint or housing.
Restraints should have alldirectional restraint capability to resist both horizontal and vertical motion.
Vibration isolators have little or no resistance to overturning forces. Therefore, if there is a difference in
height between the equipment center of gravity and the support points of the springs, rocking is inevitable
and vertical restraint is required.
FEMA 451, NEHRP Recommended Provisions: Design Examples
1330
An air gap between the restraint device and the equipment prevents vibration from transmitting to the
structure during normal operation of the equipment. Air gaps are generally no greater than ¼ in.
Dynamic tests indicate a significant increase in acceleration for air gaps larger than ¼ in.
A soft impact material often an elastomer such as bridge bearing neoprene reduces accelerations and
impact loads by preventing steeltosteel contact. The thickness of the elastomer can significantly reduce
accelerations to both the equipment and the restraint device and should be specifically addressed for life
safety applications.
A ductile restraint or housing is critical to prevent catastrophic failure. Unfortunately, housed isolators
made of brittle materials such as cast iron often are assumed to be capable of resisting seismic loads and
continue to be installed in seismic zones.
Overturning calculations for vibration isolated equipment must consider a worst case scenario as
illustrated in Guide Sec. 13.3.4.1. However, important variations in calculation procedures merit further
discussion. For equipment that is usually directly attached to the structure, or mounted on housed
vibration isolators, the weight can be used as a restoring force since the equipment will not transfer a
tension load to the anchors until the entire equipment weight is overcome at any corner. For equipment
installed on any other vibration isolated system (such as the separate spring and snubber arrangement
shown in Figure 13.35), the weight cannot be used as a restoring force in the overturning calculations.
As the foregoing illustrates, design of restraints for resiliently mounted equipment is a specialized topic.
The Provisions sets out only a few of the governing criteria. Some suppliers of vibration isolators in the
highest seismic zones are familiar with the appropriate criteria and procedures. Consultation with these
suppliers may be beneficial.
13.4 ANALYSIS OF PIPING SYSTEMS
13.4.1 ASME Code Allowable Stress Approach
Piping systems are typically designed to satisfy national standards such as ASME B31.1. Piping required
to be designed to other ASME piping codes use similar approaches with similar definition of terms.
13.4.1.1 Earthquake Design Requirements
ASME B31.1 Sec. 101.5.3 requires that the effects of earthquakes, where applicable, be considered in the
design of piping, piping supports, and restraints using data for the site as a guide in assessing the forces
involved. However, earthquakes need not be considered as acting concurrently with wind.
13.4.1.2 Stresses Due to Sustained Loads
The effects of pressure, weight, and other sustained loads must meet the requirements of ASME B31.1
Eq. 11A:
0.75 1.0
4
o A
L h
n
SPD iM S
t Z
= + =
where:
SL = sum of the longitudinal stresses due to pressure, weight, and other sustained loads
P = internal design pressure, psig
Chapter 13, Nonstructural Components
1331
Do = outside diameter of pipe, in.
tn = nominal pipe wall thickness, in.
i = stress intensification factor from ASME Piping Code Appendix D, unitless
= 1.0 for straight pipe
$ 1.0 for fittings and connections
MA = resultant moment loading on cross section due to weight and other sustained loads, in.lb
Z = section modulus, in.3
Sh = basic material allowable stress at maximum (hot) temperature from ASME Piping Code
Appendix A
For example, ASTM A53 seamless pipe and tube, Grade B: Sh = 15.0 ksi for 20 to 650 degrees F.
13.4.1.3 Stresses Due to Occasional Loads
The effects of pressure, weight, and other sustained loads, and occasional loads including earthquake
must meet the requirements of ASME B31.1 Eq. 12A:
0.75 0.75
4
o A B
h
n
PD iM iM kS
t Z Z
+ + =
where:
MB = resultant moment loading on crosssection due to occasional loads, such as from thrust loads,
pressure and flow transients, and earthquake. Use onehalf the earthquake moment range.
Effects of earthquake anchor displacements may be excluded if they are considered in Eq. 13A,
in.lb
k = duration factor, unitless
= 1.15 for occasional loads acting less than 10% of any 24 hour operating period
= 1.20 for occasional loads acting less than 1% of any 24 hour operating period
= 2.00 for rarely occurring earthquake loads resulting from both inertial forces and anchor
movements (per ASME interpretation)
13.4.1.4 Thermal Expansion Stress Range
The effects of thermal expansion must meet the requirements of ASME B31.1 Eq. 13A:
C ( )
E A h L
S iM S f S S
Z
= = + 
where:
SE = sum of the longitudinal stresses due to thermal expansion, ksi
FEMA 451, NEHRP Recommended Provisions: Design Examples
1332
MC = range of resultant moments due to thermal expansion. Also includes the effects of earthquake
anchor displacements if not considered in Eq. 12A, in.lb
SA = allowable stress range, ksi (per ASME B31.1 Eq. 1, (1.25 0.25 ) A c h S =f S + S
f = stress range reduction factor for cyclic conditions from the ASME Piping Code Table 102.3.2.
Sc = basic material allowable stress at minimum (cold) temperature from the ASME Piping Code
Appendix A
13.4.1.5 Summary
In the ASME B31.1 allowable stress approach, the earthquake’s effects only appear in the MB and MC
terms.
Earthquake inertial effects  MB term
Earthquake displacement effects  MC term
13.4.2 Allowable Stress Load Combinations
ASME B31.1 utilizes an allowable stress approach; therefore, allowable stress force levels and allowable
stress load combinations should be used. While the Provisions are based on strength design, the IBC
provides the following two sets of allowable stress loads and load combinations. The IBC load
combinations are appropriate for use for piping systems when considering earthquake effects. When
earthquake effects are not considered, load combinations should be taken from the appropriate piping
system design code.
13.4.2.1 IBC Basic Allowable Stress Load Combinations
No increases in allowable stress are permitted for the following set of load combinations:
D (IBC Eq. 167)
D + L + (Lr or S or R) (IBC Eq. 169)
D + (W or 0.7 E) (IBC Eq. 1610)
0.6D  0.7 E (IBC Eq. 1612)
13.4.2.2 IBC Alternate Basic Allowable Stress Load Combinations
Increases in allowable stress (typically 1/3) are permitted for the following alternate set of load
combinations that include W or E:
D + L + (Lr or S or R) (IBC Eq. 1613)
D + L + S + E/1.4 (IBC Eq. 1617)
0.9D + E/1.4 (IBC Eq. 1618)
13.4.2.3 Modified IBC Allowable Stress Load Combinations
Chapter 13, Nonstructural Components
1333
It is convenient to define separate earthquake load terms to represent the separate inertial and
displacement effects.
EI = Earthquake inertial effects  MB term
E. = Earthquake displacement effects  MC term
It is also convenient to use the IBC Alternate Basic Allowable Stress Load Combinations modified to use
ASME Piping Code terminology, deleting roof load effects (Lr or S or R) and multiplying by 0.75 to
account for the 1.33 allowable stress increase when W or E is included. Only modified IBC Eq. 1617
and 1618 will be considered in the discussion that follows.
0.75[D + L + S + (EI + E.)/1.4] (modified IBC Eq. 1617)
0.75[0.9D + (EI + E.)/1.4] (modified IBC Eq. 1618)
13.4.3 Application of the Provisions
13.4.3.1 Overview
Provisions Sec. 6.3.11 [6.4.2, item 4] requires that, in addition to their attachments and supports, piping
systems assigned an Ip greater than 1.0 must themselves be designed to meet the force and displacement
requirements of Provisions Sec. 6.1.3 and 6.1.4 [6.2.6 and 6.2.7] and the additional requirements of this
section.
13.4.3.1 Forces
Provisions Sec. 6.1.3 [6.2.6] provides specific guidance regarding the equivalent static forces that must be
considered. In computing the earthquake forces for piping systems, the inertial portion of the forces
(noted as EI in this example) are computed using Provisions Eq. 6.1.31, 6.1.32, and 6.1.33 [6.21, 6.2
3, and 6.24] for Fp with ap = 1 and Rp = 3.5. For anchor points with different elevations, the average
value of the Fp may be used with minimum and maximums observed. In addition, when computing the
inertial forces, the vertical seismic effects (± 0.2SDSWp) should be considered.
The term EI can be expressed in terms of the forces defined in the Provisions converted to an allowable
stress basis by the 1.4 factor:
0.2
1.4 1.4
p DSp
I
horizontal vertical
F SW
E
. . . .
=. . ±. .
. . . .
It is convenient to designate the term by the variable ß.
1 0.2
1.4
DS .. ± S ..
. .
The vertical component of EI can now be defined as ßMa and applied to all load combinations that include
EI.
MB can now be defined as the resultant moment induced by the design force Fp/1.4 where Fp is as defined
by Provisions Eq. 6.1.31, 6.1.32, or 6.1.33.
13.4.3.3 Displacements
FEMA 451, NEHRP Recommended Provisions: Design Examples
1334
Provisions Sec. 6.1.4 [6.2.7] provides specific guidance regarding the relative displacements that must be
considered. Typically piping systems are designed considering forces and displacements using elastic
analysis and allowable stresses for code prescribed wind and seismic equivalent static forces in
combination with operational loads.
However, no specific guidance is provided in the Provisions except to say that the relative displacements
should be accommodated. The intent of the word "accommodate" was not to require that a piping system
remain elastic. Indeed, many types of piping systems typically are very ductile and can accommodate
large amounts of inelastic strain while still functioning quite satisfactorily. What was intended was that
the relative displacements between anchor and constraining points that displace significantly relative to
one another be demonstrated to be accommodated by some rational means. This accommodation can be
made by demonstrating that the pipe has enough flexibility and/or inelastic strain capacity to
accommodate the displacement by providing loops in the pipeline to permit the displacement or by adding
flex lines or articulating couplings which provide free movement to accommodate the displacement.
Sufficient flexibility may not exist where branch lines, may be forced to move with a ceiling or other
structural system are connected to main lines. Often this “accommodation” is done by using engineering
judgment, without calculations. However, if relative displacement calculations were required for a piping
system, a flexibility analysis would be required. A flexibility analysis is one in which a pipe is modeled
as a finite element system with commercial pipe stress analysis programs (such as Autopipe or CAESAR
II) and the points of attachment are displaced by the prescribed relative displacements. The allowable
stress for such a condition may be significantly greater than the normal allowable stress for the pipe.
The internal moments resulting from support displacement may be computed by means of elastic analysis
programs using the maximum computed relative displacements as described earlier and then adjusted.
As with elastic inertial forces, the internal moments caused by relative displacement can be divided by Rp
to account for reserve capacity. A reduction factor of 1.4 may be used to convert them for use with
allowable stress equations. Therefore, MC can now be defined as the resultant moment induced by the
design relative seismic displacement Dp/1.4Rp where Dp is defined by Provisions Eq. 6.1.41, 6.1.42,
6.1.43, or 6.1.44 [6.25, 6.26, 6.27, or 6.28].
13.4.3.4 Load Combinations
Combining ASME B31.1 Eq. 12A and 13A with modified IBC Eq. 1605.3.2.5 and 1605.3.2.6 yields the
following:
For modified IBC Eq. 1617
0.75 0.75
4
o A B
h
n
PD iM iM kS
t Z Z
+ß.. ..± =
. .
C ( )
A h L
iM S f S S
Z
= + 
and for modified IBC Eq. 1618
0.9 0.75 0.75
4
o A B
h
n
PD iM iM kS
t Z Z
+ ß.. .. =
. .
C ( )
A h L
iM S f S S
Z
= + 
Chapter 13, Nonstructural Components
1335
where:
1 0.2
1.4
DS ß =.. ± S ..
. .
MA = the resultant moment due to weight
MB = the resultant moment induced by the design force Fp/1.4 where Fp is as defined by Provisions
Eq. 6.1.31, 6.1.32, or 6.1.33 [6.21, 6.23, or 6.24].
MC = the resultant moment induced by the design relative seismic displacement Dp/1.4Rp where Dp is
as defined by Provisions Eq. 6.1.41, 6.1.42, 6.1.43, or 6.1.44 [6.25, 6.26, 6.27, or 6.28].
SDS, Wp, and Rp are as defined in the Provisions.
P, Do, tn, I, Z, k, Sh, SA, f, and SL are as defined in ASME B31.1.
A1
Appendix A
THE BUILDING SEISMIC
SAFETY COUNCIL
The purpose of the Building Seismic Safety Council is to enhance the public's safety by providing a
national forum to foster improved seismic safety provisions for use by the building community. For the
purposes of the Council, the building community is taken to include all those involved in the planning,
design, construction, regulation, and utilization of buildings.
To achieve its purposes, the Council shall conduct activities and provide the leadership needed to:
1. Promote development of seismic safety provisions suitable for use throughout the United States;
2. Recommend, encourage, and promote adoption of appropriate seismic safety provisions in
voluntary standards and model codes;
3. Assess implementation progress by federal, state, and local regulatory and construction agencies;
4. Identify opportunities for the improvement of seismic regulations and practices and encourage
public and private organizations to effect such improvements;
5. Promote the development of training and educational courses and materials for use by design
professionals, builders, building regulatory officials, elected officials, industry representatives,
other members of the building community and the public.
6. Provide advice to governmental bodies on their programs of research, development, and
implementation; and
7. Periodically review and evaluate research findings, practice, and experience and make
recommendations for incorporation into seismic design practices.
The scope of the Council's activities encompasses seismic safety of structures with explicit consideration
and assessment of the social, technical, administrative, political, legal, and economic implications of its
deliberations and recommendations.
Achievement of the Council's purpose is important to all in the public and private sectors. Council
activities will provide an opportunity for participation by those at interest, including local, State, and
Federal Government, voluntary organizations, business, industry, the design professions, the construction
industry, the research community and the public. Regional and local differences in the nature and
magnitude of potentially hazardous earthquake events require a flexible approach adaptable to the relative
risk, resources and capabilities of each community. The Council recognizes that appropriate earthquake
hazard reduction measures and initiatives should be adopted by existing organizations and institutions and
incorporated into their legislation, regulations, practices, rules, codes, relief procedures and loan
A2
requirements, whenever possible, so that these measures and initiatives become part of established
activities rather than being superposed as separate and additional.
The Council is established as a voluntary advisory, facilitative council of the National Institute of
Building Sciences, a nonprofit corporation incorporated in the District of Columbia, under the authority
given the Institute by the Housing and Community Development Act of 1974, (Public Law 93383), Title
VIII, in furtherance of the objectives of the Earthquake Hazards Reduction Act of 1977 (Public Law 95
124) and in support of the President's National Earthquake Hazards Reduction Program, June 22, 1978.
20052006 BSSC BOARD OF DIRECTION
Chair  Jim W. Sealy, FAIA, Architect/Consultant, Dallas, TX
Vice Chair  David Bonneville, Degenkolb Engineers, San Francisco, California
Secretary  Jim Rinner, Project Manager II, Kitchell CEM, Sacramento, California
ExOfficio Member  Charles Thornton, Chairman/Principal, ThorntonTomasetti Group, Inc., New
York, New York
Members
Edwin Dean, Nishkian Dean, Portland, Oregon
Bradford K. Douglas, Director of Engineering, American Forest and Paper Association, Washington,
D.C.
Cynthia J. Duncan, Director of Specifications, American Institute of Steel Construction, Chicago, Illinois
Henry Green, Executive Director, Bureau of Construction Codes and Fire Safety, State of Michigan,
Department of Labor and Economic Growth, Lansing, Michigan (representing the National Institute of
Building Sciences)
Jay W. Larson, American Iron and Steel Institute, Betlehem, Pennsylvania
Joseph Messersmith, Coordinating Manager, Regional Code Services, Portland Cement Association,
Rockville, Virginia (representing the Portland Cement Association)
Ronald E. Piester, Assistant Director for Code Development, New York State, Department of State,
Kinderhook, New York
James Rossberg, Manager, Technical Activities for the Structural Engineering Institute, American Society
of Civil Engineers, Reston Virginia
W. Lee Shoemaker, Director, Engineering and Research, Metal Building Manufacturers Association,
Cleveland, Ohio
Howard Simpson, Simpson Gumpertz and Heger, Arlington, Massachusetts (representing National
Council of Structural Engineers Associations)
Shyam Sunder, Deputy Director, Building Fire Research Laboratory, National Institute of Standards and
Technology, Gaithersburg, Maryland (representing Interagency Committee on Seismic Safety in
Construction)
Charles A. Spitz, Architect/Planner/Code Consultant, Wall New Jersey (representing the American
Institute of Architects)
Robert D. Thomas, Vice President Engineering, National Concrete Masonry Association, Herndon,
Virginia
BSSC Staff
Claret M. Heider, Vice President for BSSC Programs
A3
Bernard F. Murphy, PE, Director, Special Projects
Carita Tanner, Communications Manager
BSSC MEMBER ORGANIZATIONS
Voting Members
AFLCIO Building and Construction Trades Department
American Concrete Institute
American Consulting Engineers Council
American Forest and Paper Association
American Institute of Architects
American Institute of Steel Construction
American Iron and Steel Institute
American Society of Civil Engineers
ASCE, Kansas City Chapter
American Society of Heating, Refrigeration, and AirConditioning Engineers
American Society of Mechanical Engineers
American Welding Society
APA  The Engineered Wood Association
Applied Technology Council
Associated General Contractors of America
Association of Engineering Geologists
Association of Major City Building Officials
Brick Industry Association
California Geotechnical Engineers Association
California Division of the State Architect, Office of Regulation Services
Canadian National Committee on Earthquake Engineering
Concrete Masonry Association of California and Nevada
Concrete Reinforcing Steel Institute
Earthquake Engineering Research Institute
General Services Administration Seismic Program
Hawaii State Earthquake Advisory Board
Institute for Business and Home Safety
Interagency Committee on Seismic Safety in Construction
International Code Council
International Masonry Institute
Masonry Institute of America
Metal Building Manufacturers Association
MidAmerica Earthquake Center
National Association of Home Builders
National Concrete Masonry Association
National Conference of States on Building Codes and Standards
National Council of Structural Engineers Associations
National Elevator Industry, Inc.
National Fire Sprinkler Association
National Institute of Building Sciences
National Ready Mixed Concrete Association
A4
Portland Cement Association
Precast/Prestressed Concrete Institute
Rack Manufacturers Institute
Santa Clara University
Seismic Safety Commission (California)
Steel Deck Institute, Inc.
Structural Engineers Association of California
Structural Engineers Association of Central California
Structural Engineers Association of Colorado
Structural Engineers Association of Illinois
Structural Engineers Association of Kentucky
Structural Engineers Association of Northern California
Structural Engineers Association of Oregon
Structural Engineers Association of San Diego
Structural Engineers Association of Southern California
Structural Engineers Association of Texas
Structural Engineers Association of Utah
Structural Engineers Association of Washington
The Masonry Society
U.S. Army CERL
Western States Clay Products Association
Wire Reinforcement Institute
Affiliate Members
Bay Area Structural, Inc.
Building Technology, Incorporated
City of Hayward, California
Felten Engineering Group, Inc.
H&H Group
HLM Design
LaPay Consulting, Inc.
Niehoff, Dennis, PE
Square D Company
Steel Joist Institute
Vibration Mountings and Controls
York, a Johnson Controls Company
A5
BSSC PUBLICATIONS
Available free from the Federal Emergency Management Agency at 18004802520 ( by FEMA
Publication Number). For detailed information about the BSSC and its projects, contact: BSSC,
1090 Vermont Avenue, N.W., Suite 700, Washington, D.C. 20005 Phone 2022897800; Fax 202
2891092; email ctanner@nibs.org
NEW BUILDINGS PUBLICATIONS
NEHRP (National Earthquake Hazards Reduction Program) Recommended Provisions for Seismic
Regulations for New Buildings, 2003 Edition, 2 volumes and maps, FEMA 450 (issued as a CD with only
limited print copies available)
NEHRP (National Earthquake Hazards Reduction Program) Recommended Provisions for Seismic
Regulations for New Buildings, 2000 Edition, 2 volumes and maps, FEMA 368 and 369
NEHRP Recommended Provisions: Design Examples, 2006, FEMA 451 (issued as a CD)
A Nontechnical Explanation of the NEHRP Recommended Provisions, Revised Edition, 1995, FEMA 99
Homebuilders= Guide to EarthquakeResistant Design and Construction, 2006, FEMA 232
Seismic Considerations for Steel Storage Racks Located in Areas Accessible to the Public. 2005, FEMA
460
Seismic Considerations for Communities at Risk, Revised Edition, 1995, FEMA 83
Seismic Considerations: Apartment Buildings, Revised Edition, 1996, FEMA 152
Seismic Considerations: Elementary and Secondary Schools, Revised Edition, 1990, FEMA 149
Seismic Considerations: Health Care Facilities, Revised Edition, 1990, FEMA 150
Seismic Considerations: Hotels and Motels, Revised Edition, 1990, FEMA 151
Seismic Considerations: Office Buildings, Revised Edition, 1996, FEMA 153
Societal Implications: Selected Readings, 1985, FEMA 84
EXISTING BUILDINGS
NEHRP Guidelines for the Seismic Rehabilitation of Buildings, 1997, FEMA 273
NEHRP Guidelines for the Seismic Rehabilitation of Buildings: Commentary, 1997, FEMA 274
Case Studies: An Assessment of the NEHRP Guidelines for the Seismic Rehabilitation of Buildings, 1999,
FEMA 343
Planning for Seismic Rehabilitation: Societal Issues, 1998, FEMA 275
Example Applications of the NEHRP Guidelines for the Seismic Rehabilitation of Buildings, 1999, FEMA
276
A6
NEHRP Handbook of Techniques for the Seismic Rehabilitation of Existing Buildings, 1992, FEMA 172
NEHRP Handbook for the Seismic Evaluation of Existing Buildings, 1992, FEMA 178
An Action Plan for Reducing Earthquake Hazards of Existing Buildings, 1985, FEMA 90
MULTIHAZARD
An Integrated Approach to Natural Hazard Risk Mitigation, 1995, FEMA 261/295
LIFELINES
Abatement of Seismic Hazards to Lifelines: An Action Plan, 1987, FEMA 142
Abatement of Seismic Hazards to Lifelines: Proceedings of a Workshop on Development of An Action
Plan, 6 volumes:
Papers on Water and Sewer Lifelines, 1987, FEMA 135
Papers on Transportation Lifelines, 1987, FEMA 136
Papers on Communication Lifelines, 1987, FEMA 137
Papers on Power Lifelines, 1987, FEMA 138
Papers on Gas and Liquid Fuel Lifelines, 1987, FEMA 139
Papers on Political, Economic, Social, Legal, and Regulatory Issues and General Workshop
Presentations, 1987, FEMA 143
(August 2006)