RECOMMENDED RESIDENTIAL CONSTRUCTION FOR COASTAL AREAS
Building on Strong and Safe Foundations
FEMA P-550, Second Edition / December 2009
Appendix D. Foundation Analysis and Design Examples
Chapter 3 described the types of loads considered in this manual. This appendix
demonstrates how these loads are calculated using a sample building and
foundation. The reactions from the loads imposed on the example building are
calculated, the loads on the foundation elements are determined, and the total
loads are summed and applied to the foundation.
There is a noteworthy difference between the approach taken for designing the
foundations included in this manual and the analyses that a designer may
undertake for an individual building. The analyses used for the designs in this
manual were based on the “worst case” loading scenario for a “range of building
sizes and weights.” This approach was used to provide designers and contractors
with some flexibility in selecting the home footprint and characteristics for
which these pre-engineered foundations would apply. This also simplifies
application of the pre-engineered foundations, reducing the number of pre-
engineered foundations, and results in conservative designs.
For example, the designs presented were developed to resist uplift and
overturning for a relatively light structure with a flat roof (worst case uplift
and overturning) while gravity loads were based on a relatively heavy structure
to simulate worst case gravity loads. Sliding forces were determined for a
relatively deep home with a steep roof to simulate the largest lateral loads.
The range of building weights and dimensions applicable to the designs are
listed in Appendix C.
Since the designs are inherently conservative for some building dimensions and
weights, a local design professional may be consulted to determine if
reanalyzing to achieve a more efficient design is warranted. If a reanalysis is
determined to be cost-effective, the sample calculations will aid the designer
in completing that analysis.
D.1 Sample Calculations
The sample calculations have been included to show one method of determining and
calculating individual loads, and calculating load combinations.
Type of Building
The sample calculation is based on a two-story home with a 28-foot by 42-foot
footprint and a mean roof height of 26 feet above grade. The home is located on
Little Bay in Harrison County, Mississippi, approximately 1.5 miles southwest of
DeLisle (see Figure D-1). The site is located on the Harrison County Flood
Recovery Map in an area with an Advisory Base Flood Elevation (ABFE) of 18 feet,
located between the 1.5-foot wave contour and the 3-foot wave contour (Coastal A
zone). The local grade elevation is 15 feet North American Vertical Datum
(NAVD). The calculations assume short- and long-term erosion will occur and the
ground elevation will drop 1 foot during the life of the structure. This places
the home in a Coastal A zone with the flood elevation approximately 4 feet above
the eroded exterior grade. Based on ASCE 7-05, the 3-second gust design wind
speed at the site is 128 miles per hour (mph) Exposure Category C. To reduce
possible damage and for greater resistance to high winds, the home is being
designed for a 140-mph design wind speed.
[Begin figure]
[End figure]
The home has a gabled roof with a 3:12 roof pitch. The home is wood-framed,
contains no brick or stone veneer, and has an asphalt shingled roof. It has a
center wood beam supporting the first floor and a center load bearing wall
supporting the second floor. Clear span trusses frame the roof and are designed
to provide a 2-foot overhang. No vertical load path continuity is assumed to
exist in the center supports, but vertical and lateral load path continuity is
assumed to exist elsewhere in the structure.
The proposed foundation for the home is a system of steel pipe piles, a
reinforced concrete grade beam, and concrete columns extending from the grade
beam to the elevated structure.
Methodology
1. Determine the loads based on the building’s parameters (Section D.1.1)
2. Calculate wind and flood loads using ASCE 7-05 (Section D.1.2)
3. Consider the structure as a rigid body, and use force and moment
equilibrium equations to determine reactions at the perimeter foundation
elements (Section D.2)
D.1.1 Determining Individual Loads on a Structure
Building Dimensions and Weights (pulled from the text of the example problem)
B = 42 Building width (ft)
L = 28 Building depth (ft)
F sub 1 = 10 First floor height (ft)
F sub 2 = 10 Second floor height (ft)
R = 3 3:12 roof pitch
W sub ovhg = 2 Width of roof overhang (ft)
Dead Loads
W sub rfDL = 12 Roof dead load including upper level ceiling finish (in pounds
per square foot [psf])
W sub 1stDL = 8 First floor dead load (psf)
W sub 2ndDL = 10 Second floor dead load, including first floor ceiling finish
(psf)
W sub wlDL = 9 Exterior wall weight (psf of wall area)
Live Loads
W sub 1stLL = 40 First floor live load (psf)
W sub 2ndLL = 30 Second floor live load (psf)
W sub rfLL = 20 Roof live load (psf)
Wind Loads
Building Geometry
h/L = 1
L/B = 0.7
Site Parameters (ASCE 7-05, Chapter 6)
K sub zt = 1 Topographic factor (no isolated hills, ridges, or escarpments)
K sub d = 0.85 Directionality factor (for use with ASCE 7-05, Chapter 2, Load
Combinations)
K sub h = 0.94 For simplicity, Velocity Pressure Coefficient used at the 26 foot
mean roof height was applied at all building surfaces. See ASCE 7-05, Table 6-3,
Cases 1 and 2.
I = 1 Importance factor (residential occupancy)
V = 140 3-second gust design wind speed (mph)
G = 0.85 Gust effect factor (rigid structure, ASCE 7-05, Section 6.5.8.1)
External Pressure Coefficients (Cp) (ASCE 7-05, Figure 6-6)
C sub pwwrf = -1.0 Windward roof (side facing the wind)
C sub plwrf = -0.6 Leeward roof
C sub pwwl = 0.8 Windward wall
C sub plwwl = -0.5 Leeward wall (side away, or sheltered, from the wind)
C sub peave = -0.8 Windward eave
Positive coefficients indicate pressures acting on the surface. Negative
coefficients indicate pressures acting away from the surface.
Velocity Pressure (q sub h) (ASCE 7-05, Section 6.5.10)
Velocity pressure at mean roof height:
q sub h = 0.00256 KhKzt Kd V squared I
= 0.00256 (0.94)(1)(0.85)(140) squared (1)
q sub h = 40 psf
Wind Pressures (P)
Determine external pressure coefficients for the various building surfaces.
Internal pressures, which act on all internal surfaces, do not contribute to the
foundation reactions. For sign convention, positive pressures act inward on a
building surface and negative pressures act outward.
P sub wwrfV = q sub h GC sub pwwrf Windward roof
= (40)(0.85)(-1)
P sub wwrfV = -34 psf
Likewise
P sub lwrfV = q sub h GC sub plwrf Leeward roof
P sub lwrfV = -20 psf
P sub wwwl = q sub h GC sub pwwwl Windward wall
P sub wwwl = 27 psf
P sub lwwl = q sub h GC sub plwwl Leeward wall
P sub lwwl = -17 psf
P sub wweave = q sub h GC sub peave Windward roof overhang
P sub wweave = -27 psf - eave
= -34 psf - upper surface
= -61 psf - total
Wind Forces (F) (on a 1-foot wide section of the home)
F sub wwrfV = P sub wwrfV L/2 Windward roof vertical force
= (-34 psf)(14 sf/lf)
= -476 lb/lf
F sub lwrfV = PlwrfV L/2 Leeward roof vertical force
= (-20 psf)(14 sf/lf)
= -280 lb/lf
F sub wwrfH = P sub wwrfH L/2 (r/12) Windward roof horizontal force
= (-34 psf)(14 sf/lf)(3/12)
= -119 lb/lf
F sub lwrfH = P sub lwrfH L/2 (r/12) Leeward roof horizontal force
= (-20 psf)(14 sf/lf)(3/12)
= -70 lb/lf
F sub wwwl1st = P sub wwwl (F1) Windward wall on first floor
= (27 psf)(10 sf/lf)
= 270 lb/lf
F sub wwwl2nd = P sub wwwl (F2) Windward wall on second floor
= (27 psf)(10 sf/lf)
= 270 lb/lf
F sub lwwl1st = P sub lwwl (F1) Leeward wall on first floor
= (-17 psf)(10 sf/lf)
= -170 lb/lf
F sub lwwl2nd = P sub lwwl (F2) Leeward wall on second floor
= (-17 psf)(10 sf/lf)
= -170 lb/lf
F sub wweave = P sub wweave w sub owhg Eave vertical force (lb)
= -(61 psf)(2 sf/lf) horizontal projected areas is negligible so horizontal
force is neglected
= -122 lb/lf
D.1.2 Calculating Reactions from Wind, and Live and Dead Loads
Sum overturning moments (M sub wind) about the leeward corner of the home. For
sign convention, consider overturning moments as negative. Since vertical load
path continuity is assumed not to be present above the center support, the
center support provides no resistance to overturning (see Figure D-2).
Mwind = (-476 lb/lf)(21 ft) + (-280 lb/lf)(7 ft) + (119 lb/lf)(21.75 ft) + (-70
lb/lf)(21.75 ft) + (-270 lb/lf)(5 ft) + (-270 lb/lf)(15 ft) + (-170 lb/lf)(5 ft)
+ (-170 lb/lf)(15 ft) + (-122 lb/lf)(29 ft)
= -23,288 ft-lb/lf
Solving for the windward reaction, therefore:
W sub wind = M sub wind divided by L
W sub wind = -23,228 ft (lb/lf divided 28 ft)
W sub wind = -830 lb/lf
The leeward reaction is calculated by either summing vertical loads or by
summing moments about the windward foundation wall. Leeward reaction equals -48
lb/lf.
Lateral Wind Loads
Sum horizontal loads (Flat) on the elevated structure. (Forces to the left are
positive. See Figure D-2.)
F sub lat = (-119 lb/lf) + (70 lb/lf) + (270 lb/lf) + (270 lb/lf) + (170 lb/lf)
+ (170 lb/lf)
= 831 lb/lf
[Begin figure]
Figure D-2 illustrates paths for wind, live, and dead loads.
[End figure]
Dead Loads
In this example, dead load reactions (W sub dead) are determined by summing
loads over the tributary areas. Since the roof is framed with clear-span trusses
and there is a center support in the home, each exterior foundation wall
supports ½ of the roof load, all of the exterior wall load, and ¼ of the first
and second floor loads. This approach to analysis is somewhat conservative since
it does not consider the entire dead load of the structure to resist
overturning. Standard engineering practice often considers the entire weight of
the structure (i.e., not just the portion supported by the perimeter foundation
walls) available to resist overturning. The closed foundations in this guidance
were developed considering only the tributary dead load to resist uplift. The
open foundations were developed considering all dead loads to resist uplift.
W sub dead = L/2 (w sub rfDL)+ L/4 (w sub 1stDL + w sub 2ndDL) + (F sub 1 + F
sub 2) w sub wlDL
= [14 sf/lf (12 psf)] + [7 sf/lf (8 psf + 10 psf)] + [(10 sf/lf + 10 sf/lf) 9
psf]
= 474 lb/lf
Live Loads
Floor
Live loads (W sub live) are calculated in a similar fashion
W sub live = L/4(W sub 1stLL + W sub 2ndLL)
= (7 sq ft/lf)(40 + 30) psf
= 490 lb/lf
Roof
W sub liveroof = L/2(W sub rfLL)
= (14 sq ft/lf)(20 psf)
= 280 lb/lf
D.2 Determining Load Combinations
Combine loads as specified in Chapter 2 of ASCE 7-05. For this example, an
allowable stress design approach was used. A strength-based design is equally
valid.
Other loads (such as snow, ice, and seismic) are listed in the ASCE 7-05 Load
Combinations, but were considered to be too rare in the Gulf Coast of the United
States to be considered in the design. ASCE 7-05 also lists rain loads that are
appropriate for the Gulf Coast region. Since a minimum roof slope ratio of 3:12
was assumed for the homes, rain loading was not considered. Table D-1 lists
foundation wall reactions for each load case. Critical reactions that contain
the foundation design are highlighted.
[Begin table]
Table D-1. Design Reactions on Base of Elevated Home
ASCE 7-05 Load Combination: #1 D
Vertical (lb/lf): 474
ASCE 7-05 Load Combination: #2 D + L
Vertical (lb/lf): 964
ASCE 7-05 Load Combination: #3 D + L sub r
Vertical (lb/lf): 754
ASCE 7-05 Load Combination: #4 D + 0.75(L) + 0.75(L sub r)
Vertical (lb/lf): {1,052 critical load}
ASCE 7-05 Load Combination: #5 D + W
Vertical (lb/lf): -356
Horizontal (lb/lf): 831
ASCE 7-05 Load Combination: #6 D + 0.75(W) + 0.75(L) + 0.75(L sub r)
Vertical (lb/lf): 1,016
Horizontal (lb/lf): 623
ASCE 7-05 Load Combination: #7 0.6D + W
Vertical (lb/lf): {-546 critical load}
Horizontal (lb/lf):{831 critical load}
ASCE 7-05 Load Combination: #8 0.6D
Vertical (lb/lf): 284
[End table]
Where
D = dead load
L = live load
L sub r = roof live load
W = wind load
Flood Load Effects on a Foundation
In this example, since the foundation selected is a system of steel pipe piles,
the equations used to calculate flood loads are based on open foundations. Some
of the equations used to calculate flood loads will be different if the building
has a closed foundation system.
Many flood calculations depend on the stillwater flooding depth (d sub s). While
not listed on FIRMs, d sub s can be calculated from the BFE by knowing that the
breaking wave height (H sub b) equals 78 percent of the stillwater depth and
that 70 percent of the breaking wave exists above the stillwater depth (see
Figure 11-3 of FEMA 55). Stated algebraically:
BFE = GS + d sub s + 0.70 H sub b
= GS + d sub s + 0.70(0.78 d sub s)
= GS + 1.55 d sub s
GS = 15 ft NAVD (initial elevation) – 1 ft (short- and long- term erosion)
= 14 ft NAVD
d sub s = (BFE – GS) ÷ 1.55
= (18 feet NAVD – 14 feet NAVD) ÷ 1.55
= 4 ft ÷1.55
= 2.6 ft
Hydrostatic Loads
Hydrostatic loads act laterally and vertically on all submerged foundation
elements. On open foundations, lateral hydrostatic loads cancel and do not need
to be considered but vertical hydrodynamic forces (buoyancy) remain. The
buoyancy forces reduce the effective weight of the foundation by the weight of
the displaced water and must be considered in uplift calculations. For example,
normal weight concrete which typically weighs 150 lb/ft3 only weighs 86 lb/ft
cubed when submerged in saltwater (slightly more in freshwater).
In this example, calculations are based on an 18-inch square normal weight
concrete column that extends 4 feet above eroded ground elevation. The column
weighs 1,350 pounds dry ((1.5 ft)(1.5 ft)(4 ft)(150 lb/ft cubed). Under flood
conditions, the column displaces 9 ft3 of saltwater that, at 64 lb/ft cubed3,
weighs 576 pounds so the column weighs only 774 pounds when submerged.
Hydrodynamic Loads
Flood Velocity
Since a Coastal A zone is close to the flood source, flood velocity is
calculated using the ASCE 7-05 Equation C5-2:
V = (g d÷s)superscript 1/2 Upper bound flood velocity
Where
G = Gravitational acceleration (32.2 ft/sec squared)
d sub s = Design stillwater depth (ft)
Hence
V = [(32.2)(2.6)]superscript 1/2
= 9.15 feet per second (fps)
Hydrodynamic Forces
A modified version of ASCE 7-05 Equation C5-4 can be used to calculate the
hydrodynamic force on a foundation element as
F sub dyn = ½ C sub d r V squared A
Where
F sub dyn = Hydrodynamic force (lb) acting on the submerged element
C sub d = 2.0 Drag coefficient (equals 2.0 for a square or rectangular column)
r = 2 Mass density of salt water (slugs/cubic foot)
A = 1.5 d sub s Surface area of obstruction normal to flow (ft squared)
For open foundation, “A” is the area of pier or column perpendicular to flood
direction (calculated for an 18-inch square column).
Hence
F sub dyn = (½) (2)(2)(9.15)2(1.5)(2.6)
= 653 lb/column
The force is assumed to act at a point ds/2 above the eroded ground surface.
The formula can also be used for loads on foundation walls. The drag
coefficient, however, is different. For foundation walls, C sub d is a function
of the ratio between foundation width and foundation height or the ratio between
foundation width and stillwater depth. For a building with dimensions equal to
those used in this example, C sub d for a closed foundation would equal 1.25 for
full submersion (42 feet by 4 feet) or 1.3 if submersed only up to the 2.6-foot
stillwater depth.
Dynamic loads for submersion to the stillwater depth for a closed foundation are
as follows:
F sub dyn = ½ C sub d r V squared A
F sub dyn = (½) (1.3)(2)(9.15) squared (1)(2.6)
= 283 lb/lf of wall
Floodborne Debris Impacts
In this example, the loads imposed by floodborne debris were approximated using
formula 11-9 contained in FEMA 55.
The Commentary of ASCE 7-05 contains a more sophisticated approach for
determining debris impact loading, which takes into account the natural period
of the impacted structure, local debris sources, upstream obstructions that can
reduce the velocity of the floodborne debris, etc.
It is suggested that designers of coastal foundations review the later standards
to determine if they are more appropriate to use in their particular design.
The FEMA 55 Formula 11.9 estimates debris impact loads as follows:
F sub I = wV divided by (g Delta t)
Where
F sub i = impact force (lb)
w = weight of the floodborne debris (lb)
V = velocity of floodborne debris (ft/sec)
g = gravitational constant = 32.2 ft/sec squared
Delta t = impact duration (sec)
Floodborne debris velocity is assumed to equal the velocity of the moving
floodwaters and acting at the stillwater level. For debris weight, FEMA 55
recommends using 1,000 pounds when no other data are available. The impact
duration depends on the relative stiffness of the foundation and FEMA 55
contains suggested impact durations for wood foundations, steel foundations, and
reinforced concrete foundations. For this example, the suggested impact duration
of 0.1 second was used for the reinforced concrete column foundation.
F sub i = wV divided g Delta t
F sub i = [1,000 lb (9.15 ft/sec)] divided by [(32.2 ft/sec squared)(0.1
sec)]
F sub i = 2,842 lb
Breaking Wave Loads
When water is exposed to even moderate winds, waves can build quickly. When
adequate wind speed and upstream fetch exist, floodwaters can sustain wave
heights equal to 78 percent of their stillwater depths. Depending on wind
speeds, maximum wave height for the stillwater depth at the site can be reached
with as little as 1 to 2 miles of upwind fetch.
Breaking wave forces were calculated in this example using ASCE 7-05 formulae
for wave forces on continuous foundation walls (ASCE 7-05 Equations 5-5 and 5-6)
and on vertical piles and columns (ASCE 7-05 Equation 5-4).
The equation for vertical piles and columns from ASCE 7-05 is
F sub brkp = ½ C sub db gama DH sub b squared
Where
F sub brkp = Breaking wave force (acting at the stillwater level) (lb)
C sub db = Drag coefficient (equals 2.25 for square or rectangular
piles/columns)
Gamma = Specific weight of water (64 lb/ft cubed for saltwater)
D = Pile or column diameter in ft for circular sections, or for a square pile or
column, 1.4 times the width of the pile or column (ft). For this example, since
the column is 18 inches square, D = (1.4)(1.5 ft) = 2.1 ft.
H sub b = Breaking wave height (0.78 d sub s)(ft) = (0.78)(2.6) = 2.03 ft
Note: The critical angle of a breaking wave occurs when the wave travels in a
direction perpendicular to the surface of the column. Waves traveling at an
oblique angle (a) to the surface of the waves are attenuated by the factor sin
squared alpha.
F sub brkp = ½ C sub db g DH sub b squared
= ½ (2.25)(64 lb/ft cubed)(2.1 ft)(2.03 ft) squared
= 623 lb
For closed foundations, use equations in Section 5.4.4.2 of ASCE 7-05 to
calculate F sub brkp. FEMA 55 contains the following two equations for
calculating loads on closed foundations:
F sub brkw = 1.1 C sub p rho d sub s squared + 2.4 gamma d sub s squared Case 1
and
F sub brkw = 1.1 C sub p rho d sub s squared 2 + 1.9 gamma d sub s squared Case
2
Where gamma and d sub s are the specific weights of water and design stillwater
depths as before. C sub p is the dynamic pressure coefficient that depends on
the type of structure. C sub p equals 2.8 for residential structures, 3.2 for
critical and essential facilities, and 1.6 for accessory structures where there
is a low probability of injury to human life. The term F sub brkw is a
distributed line load and equals the breaking wave load per foot of wall length
where F sub brkw is assumed to act at the stillwater elevation.
Case 1 of Formula 11.6 represents a condition where floodwaters are not present
on the interior of the wall being designed or analyzed. Case 1 is appropriate
for foundation walls that lack flood vents (see Figure D-3). The less stringent
Case 2 is appropriate for walls where NFIP required flood vents are in place to
equalize hydrostatic loads and reduce forces (see Figure D-4).
[Begin figures]
Figure D-3. Case 1. This is an illustration that shows normally incident
breaking wave pressures against a vertical wall (space behind vertical wall is
dry). Source: ASCE 7-05
Figure D-4. Case 2. This is an illustration that shows normally incident
breaking wave pressures against a vertical wall (stillwater level equal on both
sides of wall). Source: ASCE 7-05
[End figures]
In non-Coastal A Zones, the maximum wave height is 1.5 feet. This corresponds to
a stillwater depth (d sub s) of approximately 2 feet (i.e., 1.5 foot/0.78 for a
depth limited wave). For closed foundations in coastal areas with flood vents, a
1.5-foot breaking wave creates 1,280 lbs per linear foot of wall and 1,400 lbs
per linear foot of wall on foundations that lack flood vents.
Wind Load on Columns
Wind loads have been calculated per ASCE 7-05, Section 6.5.13 (Wind Loads on
Open Buildings and with Monoslope, Pitched, or Troughed Roofs). The velocity
pressure (q sub h) calculated previously was used, although this is a
conservative figure based on the 26-foot mean roof height. The force coefficient
(C sub f) was determined from ASCE 7-05, Figure 6-21 (chimneys, tanks, rooftop
equipment, and similar structures); ASCE 7-05, Figure 6-20 (walls and solid
signs) could have been used as well.
From ASCE 7-05 Equation 6-28:
F sub wind = q sub z G C sub f A sub f
= 40 psf (0.85)(1.33 (interpolated C sub f))(1.5 ft)(4 ft)
= 270 lb
Wind loads on the foundation elements are not considered in combination with
flood loads since the elements are submerged during those events.
Flood Load Combinations
Section 11.6.12 of FEMA 55 provides guidance on combining flood loads. In
Coastal A zones, FEMA 55 suggests two scenarios for combining flood loads. Case
1 involves analyzing breaking wave loads on all vertical supports and impact
loading on one corner or critical support and Case 2 involves analyzing breaking
wave loads applied to the front row of supports (row closest to the flood
source), and hydrodynamic loads applied to all other supports and impact loads
on one corner or critical support.
Depending on the relative values for dynamic and breaking waves, Case 1 often
controls for designing individual piers or columns within a home. Case 2
typically controls for the design of the assemblage of piers or columns working
together to support a home. Because of the magnitude of the load, it is not
always practical to design for impact loads. As an alternative, structural
redundancy can be provided in the elevated home to allow one pier or column to
be damaged by floodborne debris impact without causing collapse or excessive
deflection.
For the sample calculations, Case 1 was used (see Figure D-5) with a breaking
wave load of 623 pounds applied to a non-critical column. The loads were then
determined and summarized. Since the calculations must combine distributed
loads on the elevated structure and discrete loads on the columns themselves, a
column spacing of 7 feet is assumed in the calculations. For lateral loads on
the structure, calculations are based on three rows of columns sharing lateral
loads (Table D-2).
[Begin figure]
Figure D-5 is an illustration showing flood loads.
[End figure]
[Begin table]Table D-2. Loads on Columns Spaced 7 Feet On Center (for three rows
of columns)
ASCE 7-05 Load Combination: #1 D + F
Vertical (lb): 1,350 + [7(474)] = 4,668
ASCE 7-05 Load Combination: #2 D + F + L
Vertical (lb): 1,350 + [7(964)] = 8,098
ASCE 7-05 Load Combination: #3 D + F + L sub r
Vertical (lb): 1,350 + [7(754)] = 6,628
ASCE 7-05 Load Combination: #4 D + F + 0.75(L) + 0.75(L sub r)
Vertical (lb): 774 + 7(1,052) = {8,138}
ASCE 7-05 Load Combination: #5 D + F + W + 1.5(F sub a)
Vertical (lb): 774 + [7(474 - 48)] = 3,756
Horizontal (lb): [(1/3)7(831)] + [1.5(623)] = {2,874}
ASCE 7-05 Load Combination: #6 D + F + 0.75(W) + 0.75(L) + 0.75(L sub r) +
1.5(F sub a)
Vertical (lb): 774 + 7[474 + 0.75(-48+490+280)]= 7,883
Horizontal (lb): [(1/3)7(0.75)(831)] + [1.5(623)]= 2,388
ASCE 7-05 Load Combination: #7 0.6D + W + 1.5(F sub a)
Vertical (lb): 0.6[774 + 7(474)] + [7(-830)]= {-3,555}
Horizontal (lb): [(1/3)7(831)] + [1.5(623)] = {2,874}
ASCE 7-05 Load Combination: #8 0.6D
Vertical (lb): [0.6(1,350)] + [7(306)] = 3,000
[End table]
Where
D = dead load
F = load due to fluids with well-defined pressures and maximum heights (see
Section D.2 for additional information)
F sub a = flood load
L = live load
L sub r = roof live load
W = wind load
Note: Critical loads are in bold with brackets ({ }).
Results
Each perimeter column needs to support the following loads:
Vertical Load = 8,138 lb
Uplift = 3,555 lb
Lateral Load = 2,874 lb
With the critical loads determined, the foundation elements and their
connections to the home can be designed.
The following two examples are to demonstrate designs using information provided
in this manual. The first example is based on a closed foundation; the second
example is based on an open foundation.
D.3 Closed Foundation Example
A structure to be supported by the closed foundation is identical to the
structure analyzed in the example from Section D.1. The site, however, is
different. For the closed foundation design, the structure is to be placed in a
non-Coastal A zone where breaking waves are limited to 1.5 feet. The design
stillwater depth is 2 feet, and the BFE is 3 feet above exterior grade. While
the structure could be placed on a 3-foot foundation, the property owner
requested additional protection from flooding and a 4-foot tall foundation is to
be built. Since the home elevation is identical to that in the example, the
loads and load combinations listed in Table D-1 are identical. However, since
the foundation is closed, flood forces must first be analyzed.
Like the previous analysis example, flood forces consist of hydrodynamic loads,
debris loads, and breaking wave loads. Since the home is located in a non-
Coastal A zone, it is appropriate to use lower bound flood velocities. This will
significantly reduce hydrodynamic and debris loads. From FEMA 55, the following
equation is used:
V sub lower = d sub s
= 2 ft/sec
Hydrodynamic Loads
F sub dyn = ½ C sub d rho V squared A
= ½ C sub d rho Vsquared (1) d sub s
F sub dyn = (½) (1.4)(2)(2) squared(1)(2)
= 11 lb/lf of wall
Where Cd of 1.4 is for a (width of wall/ds) ratio of 21 (42 ft/2 ft)
(From FEMA 55, Table 11.2)
The hydrodynamic load can be considered to act at the mid-depth of the
stillwater elevation. The hydrodynamic load is less than the 27 psf wind load on
the windward wall.
Debris Loads
F sub I = wV divided by gt
F sub I = [1,000 lb (2 ft/sec)] divided [(32.2 ft/sec squared)(0.1 sec)]
F sub I = 620 lb
Due to load distribution, the impact load will be resisted by a section of the
wall. Horizontal shear reinforcement will increase the width of the section of
wall available to resist impact. For this example, a 3-foot section of wall is
considered to be available to resist impact. The debris impact load becomes
F sub iwall = (1/3) 620 lb
F sub iwall = 210 lb/ft
Breaking Wave Loads
The home is to be constructed in an SFHA; hence, the NFIP required flood vents
will be installed. The breaking wave load can be calculated using formulae for
equalized flood depths (Case 2).
F sub brkw = 1.1 C sub p gamma d sub s squared + 1.91 gamma d sub s squared
F sub brkw = gamma d sub s squared (1.1 C sub p + 1.91)
F sub brkw = (64)(2 squared){(1.1)(2.8) + 1.91}
F sub brkw = 1,280 lb/lf
The breaking wave load can be considered to act at the 2-foot stillwater depth
(d sub s) above the base of the foundation wall.
The foundation must resist the loads applied to the elevated structure plus
those on the foundation itself. Chapter 2 of ASCE 7-05 directs designers to
include 75 percent of the flood load in load combinations 5, 6, and 7 for non-
Coastal A zones. Table D-1 lists the factored loads on the elevated structure.
Critical loads from Table D-1 include 546 lb/lf uplift, 1,052 lb/lf gravity, and
831 lb/lf lateral from wind loading. The uplift load needs to be considered when
designing foundation walls to resist wind and flood loads and when sizing
footings to resist uplift; the gravity load must be considered when sizing
footings and the lateral wind and flood loads must be considered in designing
shear walls.
Extending reinforcing steel from the footings to the walls allows the designer
to consider the wall as a propped cantilever fixed at its base and pinned at the
top where it connects to the wood framed floor framing system. The foundation
wall can also be considered simply supported (pinned at top and bottom). The
analysis is somewhat simpler and provides conservative results.
The 1,280 lb/lf breaking wave load is the controlling flood load on the
foundation. The probability that floodborne debris will impact the foundation
simultaneously with a design breaking wave is low so concurrent wave and impact
loading is not considered. Likewise, the dynamic load does not need to be
considered concurrently with the breaking wave load and the 27 lb/sf wind load
can not occur concurrently on a wall submerged by floodwaters.
The breaking wave load is analyzed as a point load applied at the stillwater
level. When subjected to a point load (P), a propped cantilevered beam of length
(L) will produce a maximum moment of 0.197 (say 0.2) PL. The maximum moments
occur when “P” is applied at a distance 0.43L from the base. For the 4-foot tall
wall, maximum moment results when the load is applied near the stillwater level
(ds). In this example, the ASCE 7-05 required flood load of 75 percent of the
breaking wave load will create a bending moment of:
M = (0.2) f sub brkw (L)
= (0.2)(0.75)(1,280 lb/ft) (4 ft)
= 768 ft-lb/lf or
= 9,200 in-lb/lf
The reinforced masonry wall is analyzed as a tension-compression couple with
moment arm “jd,” where “d” is the distance from the extreme compression fiber to
the centroid of the reinforcing steel, and “j” is a factor that depends on the
reinforcement ratio of the masonry wall. While placing reinforcing steel off
center in the wall can increase the distance (d) (and reduce the amount of steel
required), the complexity of off-center placement and the inspections required
to verify proper placement make it disadvantageous to do so. For this design
example, steel is considered to be placed in the center of the wall and “d” is
taken as one half of the wall thickness. For initial approximation, “j” is taken
as 0.85 and a nominal 8-inch wall with a thickness of 7-5/8 inches is assumed.
Solving the moment equation is as follows:
M = T (jd)
T = M/(jd) = Tension force
= M/(j)(t/2) (t = thickness of wall)
T = (9,200 in-lb/lf) ÷ {(0.85)(7.63 in)(0.5)}
= 2,837 lb/lf
For each linear foot of wall, steel must be provided to resist 2,837 pounds of
bending stress and 546 pounds of uplift.
F sub steel = 2,837 lb/lf (bending) + 546 lb/lf (uplift)
= 3,383 lb/lf
The American Concrete Institute (ACI) 530 allows 60 kips per square inch (ksi)
steel to be stressed to 24 ksi so the reinforcement needed to resist breaking
wave loads and uplift is as follows:
A sub steel = 3,383 lb/lf divided 24,000 lb/in squared
= 0.14 in squared /lf
Placing #5 bars (at 0.31 in squared/bar) at 24 inches on centers will provide
the required reinforcement. To complete the analysis, the reinforcement ratio
must be calculated to determine the actual “j” factor and the stresses in the
reinforcing steel need to be checked to ensure the limits dictated in ACI 530
are not exceeded. The wall design also needs to be checked for its ability to
resist the lateral forces from flood and wind.
Footing Sizing
The foundation walls and footings must be sized to prevent overturning and
resist the 546 lb/lf uplift. ASCE 7-05 load combination 6 allows 60 percent of
the dead load to be considered in resisting uplift. Medium weight 8-inch masonry
cores grouted at 24 inches on center weigh 50 lb/sf or, for a 4-foot tall wall,
200 lb/lf. Sixty percent of the wall weight (120 lb/lf) reduces the amount of
uplift the footing must resist to 426 lb/lf. At 90 lb/ft cubed (60 percent of
150 lb/ft cubed for normal weight concrete), the footing would need to have a
cross-sectional area of 4.7 square feet. Grouting all cores increases the dead
load to 68 lb/sf and reduces the required footing area to 4.25 square feet. The
bearing capacity of the soils will control footing dimensions. Stronger soils
can allow narrower footing dimensions to be constructed; weaker soils will
require wider footing dimensions.
The design also needs to be checked to confirm that the footings are adequate to
prevent sliding under the simultaneous action of wind and flood forces. If
marginal friction resistance exists, footings can be placed deeper to benefit
from passive soil pressures.
D.4 Open Foundation Example
For this example, the calculations are based on a two-story home raised 8 feet
above grade with an integral slab-grade beam, mat-type foundation and a 28-foot
by 42-foot footprint. The home is sited approximately 800 feet from the shore in
a Coastal A zone. Subtracting the elevation of the site (determined from a
topographic map or preferably from a survey) from the ABFE and adding estimated
erosion (in feet) determines that the floodwaters during a design event
(including wave effects and runup) will extend 6 feet above the eroded exterior
grade. It is important to note, however, that submittal of an elevation
certificate and construction plans to local building code and floodplain
officials in many jurisdictions will require that the elevation be confirmed by
a licensed surveyor referencing an established benchmark elevation.
The wood framed home has a 3:12 roof pitch with a mean roof height of 30 feet, a
center wood beam supporting the first floor, and a center load bearing wall
supporting the second floor. Clear span trusses frame the asphalt-shingled roof
and are designed to provide a 2-foot overhang. This home is a relatively light
structure that contains no brick or stone veneers.
The surrounding site is flat, gently sloping approximately 1 foot in 150 feet.
The site and surrounding property have substantial vegetation, hardwood trees,
concrete sidewalks, and streets. A four-lane highway and a massive concrete
seawall run parallel to the beach and the established residential area where the
site is located. The beach has been replenished several times in the last 50
years. Areas to the west of the site that have not been replenished have
experienced beach erosion to the face of the seawall. The ASCE 7-05, 3-second
gust design wind speed is 140 mph and the site is in an Exposure Category C.
The proposed foundation for the home incorporates a monolithic carport slab
placed integrally with a system of grade beams along all column lines (see
Figure D-6). The dimensions of the grade beam were selected to provide adequate
bearing support for gravity loads, resistance to overturning and sliding, and
mitigate the potential of undermining of the grade beams and slab due to scour
action. The home is supported by concrete columns, extending from the top of the
slab to the lowest member of the elevated structure, spaced at 14 feet on center
(see Figure D-7).
[Begin figures]
Figure D-6 illustrates a layout of Open Foundation Example.
Figure D-7 is a loading Diagram for Open Foundation Example.
[End figures]
Lateral Wind Loads
Sum horizontal loads (Flat) on the elevated structure (forces to the left are
positive)
F sub lat = (-126 lb/lf) + (74 lb/lf) + (280 lb/lf) + (280 lb/lf) + (180 lb/lf)
+ (180 lb/lf)
= 868 lb/lf
Dead Loads
Dead load reactions (W sub dead) are determined by summing loads over the
tributary areas. For the anterior columns:
W sub dead = L/2 (w sub rfDL) + L/4 (w sub 1stDL + w sub 2ndDL) + (F sub 1 + F
sub 2) w sub wlDL
= [14 sf/lf (12 psf)] + [7 sf/lf (8 psf + 10 psf)] + [(10 sf/lf + 10 sf/lf) 9
psf]
= 474 lb/lf
Live Loads
Floor
Live loads (W sub live) are calculated in a similar fashion
W sub live = L/4(W sub 1stLL + W sub 2ndLL)
= (7 sq ft/lf)(40 + 30) psf
= 490 lb/lf
Roof
W sub liveroof = L/2(W sub rfLL)
= (14 sq ft/lf)(20 psf)
= 280 lb/lf
A minimum roof slope of 3:12 was assumed for the homes; rain loading was not
considered.
Flood Effects
Since the foundation selected is a system of concrete columns, the equations
used to calculate flood loads are based on open foundation. The stillwater
flooding depth (d sub s) is as follows:
d sub s = DFE divided by 1.55
= 6 ft divided by 1.55
= 3.9 ft
Hydrostatic Loads
Calculations are based on a 16-inch square normal weight concrete column that
extends 8 feet above the concrete slab.
The column weighs 2,123 pounds dry ((1.33 ft)(1.33 ft)(8 ft)(150 lb/ft cubed).
Under flood conditions, the column displaces 10.6 ft cubed of saltwater which,
at 64 lb/ft cubed, weighs 679 pounds so the column weighs 1,444 pounds when
submerged.
Hydrodynamic Loads
Flood Velocity
Since a Coastal A zone is close to the flood source, flood velocity is
calculated using the ASCE 7-05 Equation C5-2 as follows:
V = [(32.2 ft/sec squared)(3.9 ft)] superscriopt 1/2
= 11.21 feet per second (fps)
Flood Force
ASCE 7-05 Equation C5-4 is as follows:
F sub dyn = ½ C sub d rho V squared A
= (½) (2)(2)(11.21 fps) squared (1.33 ft)(3.9 ft)
= 1,303 lb/column
Floodborne Debris Impact
The flood debris impact can be estimated, per FEMA 55 Formula 11.9, as follows:
F sub i = wV divided by¸ gt
= [1,000 lb (11.21 ft/sec)] ¸ [(32.2 ft/sec squared)(0.1 sec)]
= 3,478 lb
Breaking Wave Load
The equation for vertical piles and columns from ASCE 7-05 is as follows:
F sub brkp = ½ Cdb gamma DH sub b squared
= ½ (2.25)(64 lb/ft cubed)(1.82 ft)(3.04 ft 9*) squared
= 1,211 lb
* A wave height of 3.04 ft suggests a V zone but, in this example, the depth of
water is increased by erosion, which is not considered in mapping A zones. The
deeper water supports a bigger wave, which in this case exceeds the V-zone wave
height minimum.
Wind Load on Columns
For a load case combining both wind and flood forces, the column would be almost
completely submerged; therefore, the wind load on the column shall not be
included.
Calculating Reactions from Wind, Live, and Dead Loads
Sum overturning moments (M sub wind) and (M sub flood) about the leeward corner
of the mat foundation. For sign convention, consider overturning moments as
negative. Note in this example the home is slightly higher above grade and hence
the wind loads are slightly higher.
M sub wind = (-504 lb/lf)(21 ft) + (-294 lb/lf)(7 ft) + (126 lb/lf)(21.75 ft) +
(-74 lb/lf)(21.75 ft) + (-280 lb/lf)(13 ft) + (-280 lb/lf)(23 ft) + (-180
lb/lf)(13 ft) + (-180 lb/lf)(23 ft) + (-130 lb/lf)(29 ft)
= -31,841 ft-lb/lf
The vertical components of the reaction caused by the wind overturning moment
is:
R sub x = 31,841 lb ÷ 28 ft = +/- 1,137 lb/ft
M sub flood = 1.5[((-1,211 lb)(3.9 ft)) + (2(-1,303 lb)(3.9 ft/2)) + ((-3,478
lb)(3.9 ft))] = 35,053 ft-lb/ft
The vertical component of the reaction caused by the flood overturning moment
is:
R sub x = 35,053 lb ÷ 28 ft = +/- 1,252 lb outboard columns
Load Combinations
Table D-3 summarizes loads on the open foundation example. Loads are listed for
the eight load combinations and critical loads are highlighted.
Table D-3. Loads at Base of Columns Spaced 14 Feet On Center (for three rows of
columns per bay)
ASCE 7-05 Load Combination: #1 D + F
Vertical (lb): 1,444 + 14(474) = 8,080
ASCE 7-05 Load Combination: #2 D + F + L
Vertical (lb): 1,444 + 14(964) = 14,940
ASCE 7-05 Load Combination: #3 D + F + L sub r
Vertical (lb): 1,444 + 14(754) = 12,000
ASCE 7-05 Load Combination: #4 D + F + 0.75(L) + 0.75(sub Lr)
Vertical (lb): 8,080 + (0.75)[(14)(490+280)] = 16,165
ASCE 7-05 Load Combination: #5 D + F + W + 1.5(F sub a)
Vertical (lb): 8,080 +/- 14(1,137) +/- 1,252 = 25,876; -9,716 windward; leeward
Horizontal (lb): wind + (1.5)[F sub dyn + F sub i][(14(868)(1/3)] + (1.5)
[(1,303+3,478)] = 11,222
ASCE 7-05 Load Combination: #6 D + F + 0.75(W) + 0.75(L) + 0.75(L sub r) +
1.5(F sub a)
Vertical (lb): 8,080 +/- (0.75)(14)(1,137) +(0.75)(14)([(490+280)] +/- 1,252 =
2,348; 29,982 windward; leeward
Horizontal (lb): (0.75) wind + (1.5)[F sub dyn + F sub i][(0.75)(14)(868)(1/3)]
+ (1.5)[(1,303+3,478)] = 10,210
ASCE 7-05 Load Combination: #7 0.6D + W + 1.5(F sub a)
Vertical (lb): 0.6 [2,123+14(474)] +/- 14(1,137) +/- (1.5)1,252 = -12,541;
23,051 windward; leeward
Horizontal (lb): wind + (1.5)[F sub dyn + F sub i][(14(868)(1/3)] +
(1.5)[(1,303+3,478)] = 11,222
ASCE 7-05 Load Combination: #8 0.6D
Vertical (lb): [0.6((2,123) + 14(474))] = 5,255
[End table]
Critical loads are in bold.
Where
D = dead load
F = fluid (buoyancy) load
L = live load
L sub r = roof live load
W = wind load
Ww = windward
Lw = leeward
Results
Each perimeter column needs to support the following loads:
Vertical Load = 29,982 lb
Uplift = 12,541 lb
Lateral Load = 11,222 lb
Moment wind + f sub dyn = [(1/3)(14)(1,314)(8) + (1,303)(3.9/2)] ÷ 1,000 lb/kip
= 51.6 ft-kip
Moment wind + f sub brkp = [(1/3)(14)(1,314)(8) + (1,113)(3.9)] ÷ 1,000
lb/kip
= 96.9 ft-kip
Moment debris = (3,478)(3.9) ÷ 1,000 lb/kip
= 13.6 ft-kip
Moment wind + f sub dyn+ debris = 51.6 ft-kip + 13.6 ft-kip
= 65.1 ft-kip
The force is assumed to act at a point ds/2 above the eroded ground surface. For
concrete design, we use load factors per ASCE 7-05.
Ultimate Moment wind + f sub dyn = (48.6)(1.2)+(2.5)(2.0)
= 63.4 ft-kip
Ultimate Moment wind + f sub brkp = (48.6)(1.2)+(4.3)(2.0)
= 66.9 ft-kip
Ultimate Moment wind + f sub dyn + debris = (63.4)+13.6(2)
= 90.6 ft-kip
Foundation Design
Overturning
The overturning moment due to wind with a typical bay of 14 feet wide is as
follows:
M sub wind = (-31,841 ft-lb/lf)(14 ft)
= -445,774 ft-lb
M sub Fa (1.5) = (1.5)[(1,211 lb)(3.9 ft) + (2)(1,303 lb)(3.9 ft/2)]
= -14,707 ft-lb
M sub o = -445,774 ft-lb - 14,707 ft-lb
= -460,481 ft-lb
In this example, it is assumed that the home and the foundation slab are
reasonably symmetrical and uniform; therefore, it is assumed the center of
gravity for the dead loads is at the center of the bay.
Dead load at perimeter columns
D sub ext = (474 lb/ft)(14 ft)(2 columns)
= 13,272 lb
Dead load at an interior column:
D sub int = (14 ft)(14 ft)(8 psf)
= 1,568 lb
Dead load of 3 columns: (3)(8 ft)(1.33 ft x 1.33 ft)(150)lb cubic ft
= 6,368 lb
Assume that only the grade beams are sufficiently reinforced to resist
overturning (neglect weight of slab)
Dead load of the grade beams (area)(depth) (density of concrete)[(28 ft x 3 ft)
+ (3)((11 ft)(3 ft)](4 ft)(150 lb cubic ft) = 109,800 lb
Summing the dead loads = 13,272 + 1,568 + 6,368 + 109,800 = 131,008 lb
Allowable dead load moment of 60 percent
M sub d = (0.6)( 131,008 lb)(14 ft)
= 1,100,476 ft-lb
Since M sub ot = 460,481 ft-lb is very much less than 0.6 M sub d = 1,100,476
ft-lb, the foundation can be assumed to resist overturning.
Sliding
The maximum total lateral load of wind and flood acting on the entire typical
bay is as follows:
L sub wfa = W + 1.5 F sub a
= [(14 ft (868 lb/ ft) + 1.5(7,203 lb)]
= 23,375 lb
Sliding Resistance = (tan phi)N + Passive Resistance at Vertical Foundation
Surfaces
Phi = internal angle of soil friction, assume phi = 25 degrees
N = net normal force (building weight - uplift forces)
N = (131,008) + (14 ft)[((16 ft)(-21 psf )+(14 ft)(-36 psf)+(2 ft)(-65 psf)]
= 117,428 lb
Ignore passive soil pressure
Dead Load Sliding Resistance = (tan 25)(117,428 lb)
= 54,758 lb
Since L sub wfa = 23,371 lb is less than 60 percent of Dead Load Sliding
Resistance = 54,758, the foundation can be assumed to resist sliding.
Soil Bearing Pressure
The simplified approach for this mat foundation assumes that only the grade
beams carry loads to the soil; the slab between grade beams is not considered to
contribute support. It is further assumed that the bearing pressure is uniform
in the absence of wind and flood loading. The areas of the grade beams along the
outboard column lines, in the direction of the flow of wind and flood, are
considered the “critical areas” of the grade beam. The load combination table
below indicates the bearing pressures for the ASCE 7- 05 load combinations for
the critical grade beam area. These load combinations are calculated to ensure
that downward forces of the wind and flood moment couple do not overstress the
soil. The factored dead load moment that resists overturning is of a magnitude
such that there is no net uplift along critical grade beams. Table D-4 presents
foundation bearing pressures for typical bays.
Self Weight of 1 square of foot Grade Beam = (4 ft)(150 lb/cubic ft) = 650 psf
Area of Critical Grade Beam = [(3ft)(14 ft)] + [(3 ft)(5.5 ft)]
= 58.5 ft squared
Weight of Critical Grade Beam = (58.5 ft squared)(4 ft)(150 lb/cubic ft)
= 35,100 lb
Critical Column Uplift = 12,541 lb (load combination 7)
Verification of Uplift Resistance = [35,100 lb(0.6)] – 12,541 lb
= 8,519 lb (positive load, no uplift)
Presumptive Allowable Bearing Pressure = 1,500 psf
[Begin table]
Table D-4. Foundation Bearing Pressures for Typical Bays (for three rows of
columns per bay)
ASCE 7-05 Load Combination: #1 D + F
Combined Loads (lb): 8,080
Bearing Pressures (psf): 8,090 ÷ 58.5 + 650 = 788
ASCE 7-05 Load Combination: #2 D + F + L
Combined Loads (lb): 14,940
Bearing Pressures (psf): 14,950 ÷ 58.5 + 650 = 905
ASCE 7-05 Load Combination: #3 D + F + L sub r
Combined Loads (lb): 12,000
Bearing Pressures (psf): 12,000 ÷ 58.5 + 650 = 855
ASCE 7-05 Load Combination: #4 D + F + 0.75(L) + 0.75(L sub r)
Combined Loads (lb): 16,165
Bearing Pressures (psf): 16,165 ÷ 58.5 + 650 = 926
ASCE 7-05 Load Combination: #5 D + F + W + 1.5(F sub a)
Combined Loads (lb): 28,104
Bearing Pressures (psf): 28,104 ÷ 58.5 + 650 = 1,130
ASCE 7-05 Load Combination: #6 D + F + 0.75(W) + 0.75(L) + 0.75(L sub r) +
1.5(F sub a)
Combined Loads (lb): 29,982
Bearing Pressures (psf): 29,982 ÷ 58.5 + 650 = 1,163
ASCE 7-05 Load Combination: #7 0.6D + W + 1.5(F sub a)
Combined Loads (lb): 23,051
Bearing Pressures (psf): 23,051 ÷ 58.5 + 650 = 1,044
ASCE 7-05 Load Combination: #8 0.6D
Combined Loads (lb): 5,255
Bearing Pressures (psf): 5,255 ÷ 58.5 + 650 = 740
[End table]
Where
D = dead load
F = load due to fluids with well-defined pressures and maximum heights (See
Section D.2 for additional information)
F sub a = flood load
L = live load
L sub r = roof live load
W = wind load
Note: The maximum bearing pressure is in bold (1,163 psf) and is less than the
assumed 1,500 psf bearing pressure.
Design of Concrete members per ACI-318-02 Code and ASCE 7-05; Sections 2.3.2 and
2.3.3-1
Column Design
Verify that 16-inch x 16-inch column design is adequate.
Concrete strength = 4,000 psi
Reinforced with (4) #8 bars, grade 60 reinforcing, with 2½-inch clear cover
Note: 1,000 lb = 1.0 kip
Assume that the total wind load distributed through the floor uniformly to 3
columns.
Check combined axial and bending strength:
Ultimate Moment wind + f sub dyn = (1.6)[(8 ft)((14 ft)(868 lb/ft)/(3))] +
(2.0)[(3.9 ft/2)(1,303) lb)]
= 51,849 ft-lb + 5,082 ft-lb
= 56,931 ft-lb ÷ 1,000 lb/kip = 56.9 ft-kip
Ultimate Moment wind + f sub brkp = (51,849 ft-lb) + (2.0)[(1,211 lb)(3.9 ft)]
= 61,295 ft-lb ÷ 1,000 lb/kip = 61.3 ft-kip
Ultimate Moment wind + f sub brkp + debris = 61.3 ft-kip + (2.0)(3.5 kip)(3.9
ft)
= 88,6 ft-kip
Maximum Factored Moment = 88.6 ft-kip = 1,063 in kip
Refer to Table D-3, conservatively assume flood load factor of 2.0 for all axial
loads
Maximum factored Axial Compression = (2.0)(30.0 kip) = 60.0 kip
Maximum factored Axial Tension = (2.0)(12.5 kip) = 25.0 kip
Based on a chart published by the Concrete Reinforced Steel Institute (CRSI),
the maximum allowable moment for the column = 1,092 in kip for 0 axial load and
1,407 in kip for 102 kip axial load; therefore, the column is adequate.
Check shear strength:
Critical Shear = wind + F sub dyn + F sub i
Ultimate Shear = Vu = [(14 ft)(0.868 kip)(1/3)](1.6) + [(1.3 kip)+(3.5 kip](2.0)
= Vu = 16.0 kip
As the maximum unit tension stress is only 25.0 kip/16 in x 16 in = 0.098 kip/in
squared and the maximum axial compression stress is only 60.0 kip/16 in x 16 in
= 0.234 kip/in squared, we can conservatively treat the column as a flexural
member or beam. The allowable shear of the concrete section then, per ACI-318-02
11.3.1.1, 11.3.1.3, and 11.5.5.1 with minimum shear reinforcing (tie/stirrup),
would be as follows.
Allowable Shear = V sub c = (0.75)(16 in-2.5 in)(16 in )(2)(4,000
psi)superscript 1/2 (1/1,000) = 20.5 kip
The shear strength of the column is adequate with minimum shear reinforcement.
The minimum area of shear, Av, per ACI 318-02, 11.5.5.3 would be:
A sub v = (50)(width of member)(spacing of reinforcing)/yield strength of
reinforcing
= (50)(16 in)(16 in)/(60,000 psi) = 0.21 in squared
2 pieces of #4 bar A sub s = (2)(0.2) = 0.40 in squared
Use of #4 bar for column ties (shear reinforcement) is adequate.
Check spacing per ACI -318-02, 7.10.4
16 diameter of vertical reinforcing bar = (16)(1 in) = 16 in
48 diameter of column tie bar = (48)(1/2 in) = 24 in
Least horizontal dimension = 16 in
Therefore, #4 ties at 16 inches on center are adequate.
The column design is adequate.
Grade Beam Design
The size of the grade beam was configured to provide adequate bearing area,
resistance to uplift, a reasonable measure of protection from damaging scour,
and to provide a factor of redundancy and reserve strength should the foundation
be undermined. A grade beam 36 inches wide and 48 inches deep was selected.
Maximum Bearing Pressure = 1,163 psf = 1.2 ksf (kip/square foot)
Assume a combined load factor of 2.0 (for flood)
Check Shear strength:
Maximum Factored Uniform Bearing Pressure = w sub u = (2.0) (3.0 ft)(1.2 ksf) =
7.2 kip/ft
Maximum Factored Shear = V sub u = (7.2 kip/ft)(14 ft/2) = 50.0 kip
Allowable Shear without minimum shear reinforcing (stirrups) = V sub c/2
V sub c/2= (0.75)(36 in)(48 - 3.5 in)(63 psi)(1/1,000) = 75.7 kip
Use nominal #4 two leg stirrups at 24 inches on center
Check flexural strength: Assume simple span condition
Maximum Factored Moment = M sub u = (7.2 kip/ft)(14 ft) squared (1/8) = 176 ft-
kip
Concrete strength = 4,000 psi
Reinforcement grade = 60,000 psi
Try (4) #6 reinforcing bar continuous top and bottom
A sub s = (4)(0.44 in squared) = 1.76 in squared top or bottom
total A sub s = (2)(1.76) = 3.52 in squared
Reinforcement ratio (r)
Rho = A sub s ÷ [(section width)(section depth- clear cover- ½ bar diameter)]
Rho = A sub s ÷ [(b sub w)(d)]
Rho = (1.76 in squared) ÷ [(36)(48 - 3 - 0.375)]
Rho = 0.01096
One method of calculating the moment strength of a rectangular beam, for a given
section and reinforcement, is illustrated in the 2002 edition of the CRSI Design
Handbook. Referencing page 5-3 of the handbook, the formula for calculating the
moment strength can be written as follows:
Phi M sub n = (phi)[((A sub s)(f sub y)(d)) – (((A sub s)(f sub y)) ÷ ((0.85)(f
'c)(width of member)(2)))]
Phi M sub n = (0.9)[((1.76)(60,000)(44.63)) – (((1.76)(60,000)) ÷
((0.85)(4,000)(36)(2)))]
= 4,241,634 in lb ÷ 12,000 = 353 ft-kip
phi = 0.9
Area of reinforcing steel (A sub s) minimum flexural
= (0.0033)(24)(44.63)
= 3.6 in squared
or
(1.33) (As required by analysis) = F sub Mn is much greater than M sub u
Area of steel for shrinkage and temperature required = (0.0018)(48)(36) = 3.1 in
squared
Total area of steel provided = (8) #6 = (8)(0.44) = 3.52 in squared adequate
Therefore, the grade beam design is adequate, use (4) #6 reinforcing bar
continuous on the top and bottom with #4 stirrups at 24-inch spacing.