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SPREAD FOOTING DESIGN
GIVEN: 300 kips
250 kips
115 pcf Cx = 30fy = 60,000 psif'c = 4,000 psi Ly = 10 Cy = 12
Allowable Soil Bearing pressure ( gross ) Qs = 4 ksf
F0OTING DIMENSION Lx = 16
Ly= 10 ft FOOTING PLANLx = 16 ftCy = 12 in PCx = 30 in
h = 4.5 ftM = 6
M = 6 inD = 2 ftT = 2.5 ft D = 2
h = 4.5
T = 2.5
FOOTING ELEVATIONDetermine Base Area Required
Total weight of surcharge, Qsoil = Ws x h
= 0.52 ksf
Net permissible soil bearing pressure, Qs = Qs - Qsoil= 3.48 ksf
Required base area of footing,
= 157.9
157.9 O.K
160
Factored loads and soil reaction
Pu = = 760 kips
Qu = = 4.75 ksf
PD =
PL =
SOIL WEIGHT (WS ) =
Af = ( PD + PL ) / Qs
ft2
Use 10 ft x 16 ft footing ( Af = 160 ft2 ) >
Use Af = ft2
1.2 PD + 1.6 PL
Pu / Af
Check beam shear ( wide beam shear) --- see figure AAssume footing thickness, T = 2.5 ft 30 in
And average effective thickness, d = 2.21 ft 26.5 in
Vn ≥ Vuф --------- ACI - Eq. 11-1where;
Vu = Qu * (tributary area)
Tributary area, = Ly ( Lx/2 - T/2 - d )= 45.42
Vu = Qu x tributary area= 215.73 kips
Vn =ф --------- ACI - Eq. 11-3
= 301.68 kips
therfore, Vn =ф 301.68 > Vu = 215.73 O.Kratio = Vu / Ø Vn
= 0.72 < 1.0 O.K
Check two-way action( punching shear) --- see figure Acheck punching shear along tributary perimeter area
where;Vu = Qu x (tributary area)
Tributary area, = (Lx * Ly) - ((Cx + d) (Cy + d)/144 )= 144.89
Vu = Qu x tributary area= 688.25 kips
Vn ≥ Vuф ( Vn=Vc ) --------- ACI - Eq. 11-3
Vc shall be the smallest of the following:
Vc = --------- ACI - Eq. 11-33
Vc = --------- ACI - Eq. 11-34
Vc = --------- ACI - Eq. 11-35
2 ( T + d ) + 2 ( Cy + d ) = 190 in
Cx / 12 = 2.5 ft
ft2
Ø( 2 (f'c)½ bw d )
ft2
(2 + 4/ßc ) (f'c)1/2 bo d
αs d / bo + 2 (f'c)1/2 bo d
4 (f'c)1/2 bo d
bo =
ßc =
ratio = = 0.14 in
for interior column = 40 in
Vc = = 3.6 kips
Vc = = 7.58 kips
Vc = = 4.0 kips
therfore, Vc = 3.6 kips governs
Vc =ф 0.75 * (2 + 4/ßc ) (f'c)1/2 bo d= 859.79 kips
therfore, Vn =ф 859.79 kips > 688.25 kips O.Kfooting thickness, T = 30 in is adequate
ratio = Vu / Ø Vn= 0.80 < 1.0 O.K
Beam Shear and Punching Shear in Footingd
punching shearbeam shear
Ly = 10Lx = 16
Cx + d = 4.71Cy + d = 3.21 Ly Cy + d
d = 2.21Lx-Cx-2d = 4.54
Cx + d
LxLx - (Cx + d)-2d
Figure A
Design of Footing Reinforcement Cx
W = Qu * Lx= 47.5 ksf
I == 6.75 ft # 7 # 8
Mu =
= 1082.109 k - ft critical sectionMu = Mnф (Lx - Cx) / 2Mn = Mu/ ф Lx
d / bo
αs =
(2 + 4ßc ) (f'c)1/2 bo d
αs d / bo + 2 (f'c)1/2 bo d
4 (f'c)1/2 bo d
Ultimate moment, Mu = WI2 / 2
(Lx-Cx)/2
WI2 / 2
= 1202.344 k - ft
F = b = Ly
= 7.02Kn = Mn / F
= 171.21From Table 6;
0.003 > 0.0018
p x b x d b = Ly
= 9.54
RSB to be provided ( trial );No. of bars, n = 14
# 8, db = 1.0 in
= 11.0 > 9.54 O.K.Required Spacing
bd2 / 12,000
in3
preqd = pmin =
As =
in2
Diameter of bar, db =
As = (πd2/4) x n
in2 in2
SPREAD FOOTING DESIGN
GIVEN: 300 kN
250 kN
16 Cx = 800
fy = 60,000
f'c = 4,000 Ly = 3.1 Cy = 300
Allowable Soil Bearing pressure ( gross )
Qs = 200
F0OTING DIMENSION Lx = 4.8
Ly= 3.1 m FOOTING PLANLx = 4.8 mCy = 300 mm PCx = 800 mm
h = 1.4 mM = 6
M = 6 inD = 0.6 mT = 0.8 m D = 0.6
h = 1.4T = 0.8
FOOTING ELEVATIONDetermine Base Area Required
Total weight of surcharge, Qsoil = Ws x h = 0.02 ksf
Net permissible soil bearing pressure, Qs = Qs - Qsoil= 199.98 ksf
Required base area of footing,
= 2.8
2.8 O.K
14.88
Factored loads and soil reactionPu = = 760 kips
Qu = = 51.08 ksf
PD =
PL =
SOIL WEIGHT (WS ) = kN/m3
kN/m2
kN/m2
kN/m3
Af = ( PD + PL ) / Qs
ft2
Use 10 ft x 16 ft footing ( Af = 160 ft2 ) >
Use Af = ft2
1.2 PD + 1.6 PL
Pu / Af
Check beam shear ( wide beam shear) --- see figure AAssume footing thickness, T = 0.8 ft 9.6 in
And average effective thickness, d = 0.51 ft 6.1 in
Vn ≥ Vuф --------- ACI - Eq. 11-1where;
Vu = Qu * (tributary area)
Tributary area, = Ly ( Lx/2 - T/2 - d )= 4.62
Vu = Qu x tributary area= 236.18 kips
Vn =ф --------- ACI - Eq. 11-3
= 21.53 kips
therfore, Vn =ф 21.53 > Vu = 236.18 O.Kratio = Vu / Ø Vn
= 10.97 < 1.0 O.K
Check two-way action( punching shear) --- see figure Acheck punching shear along tributary perimeter area
where;Vu = Qu x (tributary area)
Tributary area, = (Lx * Ly) - ((Cx + d) (Cy + d)/144 )= -1698.64
Vu = Qu x tributary area= ### kips
Vn ≥ Vuф ( Vn=Vc ) --------- ACI - Eq. 11-3
Vc shall be the smallest of the following:
Vc = --------- ACI - Eq. 11-33
Vc = --------- ACI - Eq. 11-34
Vc = --------- ACI - Eq. 11-35
2 ( T + d ) + 2 ( Cy + d ) = 643.6 in
Cx / 12 = 66.667 ft
ratio = = 0.01 in
ft2
Ø( 2 (f'c)½ bw d )
ft2
(2 + 4/ßc ) (f'c)1/2 bo d
αs d / bo + 2 (f'c)1/2 bo d
4 (f'c)1/2 bo d
bo =
ßc =
d / bo
for interior column = 40 in
Vc = = 2.1 kips
Vc = = 2.38 kips
Vc = = 4.0 kips
therfore, Vc = 2.06 kips governs
Vc =ф 0.75 * (2 + 4/ßc ) (f'c)1/2 bo d= 383.62 kips
therfore, Vn =ф 383.62 kips > -86758.61 kips O.Kfooting thickness, T = 9.6 in is adequate
ratio = Vu / Ø Vn= -226.16 < 1.0 O.K
Beam Shear and Punching Shear in Footingd
punching shearbeam shear
Ly = 3.1Lx = 4.8
Cx + d = 67.18Cy + d = 25.51 Ly Cy + d
d = 0.51Lx-Cx-2d = -31.44
Cx + d
LxLx - (Cx + d)-2d
Figure A
Design of Footing Reinforcement Cx
W = Qu * Lx= 158.3333 ksf
I == -30.9333 ft # 7 # 8
Mu =
= 75752.3 k - ft critical sectionMu = Mnф (Lx - Cx) / 2Mn = Mu/ ф Lx
αs =
(2 + 4ßc ) (f'c)1/2 bo d
αs d / bo + 2 (f'c)1/2 bo d
4 (f'c)1/2 bo d
Ultimate moment, Mu = WI2 / 2
(Lx-Cx)/2
WI2 / 2
= 84169.22 k - ft
