DESIGN

OF
FOOTING
 Square Footing # 1:
Column 1: 250mm x 250mm
PDL = 33.57 kN f’c = 21 MPa
PLL = 10.97 kN
PE = 359 kN fy = 415 MPa
q a = 180 kPa

A. Proportion footing dimensions

a. Unfactored Column load
P = DL + LL + E = 33.57 + 10.97 + 359 = 403.54 kN
b. Assume weight of footing to be 20% of column load
W = 0.20(403.54) = 80.708 kN
c. Total load supported by soil
R = 403.54 + 80.708 = 484.25 kN
d. Required footing area
R 484.25 2
A= = =2.69 m
qa 180
e. Footing dimension
B=√ A=√2.45=1.64 m
Use 2m x 2m footing
f. Required footing area
2
A=(2 m)( 2m)=4 m
g. Factored column load
Pu=1.2 ( DL )+1.0E+ f 1 L=1.2 ( 33.57 )+ 1.0 ( 359 )+ 1.0 (10.97 )=410.25 kN
h. Net ultimate upward pressure
P u 410.25
q u= = =¿102.56 kPa
A 4

B. Compute "d" for beam shear

a. Factored shear
V u=(net ultimate upward pressure)(tributary area)
V u=(102.56)(2)(0.85−d)
V u=(205.12)(0.85−d) kN
b. Shear capacity at critical section

∅ V u=∅ ( 16 ) √ f c bd=0.85( 16 ) √ 21(1000)d

'

∅ V u=¿ 649.20d kN

c. Equate factored shear to shear capacity and compute "d"

Vu≤ ∅V u
(205.12)(0.85−d)=¿649.20d kN
d = 0.204
C. Compute "d" for punching shear
a. Factored shear
V u=(net ultimate upward pressure)(tributary area)
2 2
V u=(102.56)[2 −( 0.3+ d ) ]
V u= (102.56 ) [3.91−0.6 d−d 2 ] kN
b. Shear capacity at critical section

∅ V u=∅( 31 ) √ f c b d'
0

∅ V =0.85 ( ) √ 21(1000)(0.3+ d)(d )

1
u
3
∅ V u=¿ 1298.40(0.3d+d2) kN

c. Equate factored shear to shear capacity and compute "d"

Vu≤ ∅V u
( 102.56 ) [ 3.91−0.6 d−d 2 ]=¿ 1298.40(0.3d+d2)
d = 0.38 (controls)
Therefore, use d = 400 mm

D. Compute flexural steel

a. Factored moment
M u=( 102.56 )( 0.85 )( 2 ) ( 0.435 ) =75.84 kN . m
b. Moment resistance coefficient
Mu 75.84 x 10
6
R u= 2
= 2
=0.26
φb d 0.9(2000)( 400)
c. Strength ratio
fy 415
m= '
= =23.25
0.85 f c
0.85(21)
d. Required steel ratio

e.
ρ=
1
m [ √
1− 1−
2 Rum
fy
Minimum steel ratio
=
1
23.25
1− 1− ]
2(0.26)(23.25)
415 [ √ ] = 0.00558

0.25 √ f 'c 0.25 √21

ρmin =
fy
=
415
=0.00276 ≤
1.4
fy
=0.00337
( )
ρmin =0.00337 < ρ=0.0058

Therefore, use ρ ¿ 0.00337

f. Steel area
2
A s=ρbd=0.00 337 ( 2000 ) ( 400 )=42696 mm
 π
Ab = ¿mm2
using 10 mm  bars:
4
A s 2696
n= = =34.33,
A b 78.54
use 34-10 mm  bars (2 layers – 17-10 mm  bars)
g. Check spacing of bars
b−n φ b−2 pc 2000−17 (10)−2(75)
s= = =105 mm
n−1 17−1
desirable spacing, 75 < s < 200mm, OK!!
h. Check development length
α =1.0 β=1.2 γ =1.0 λ=1.0
Ld 3 fyαβγ λ 3(415)(1)(1.2)(1)(1)
= = =65.20
db 5 √f ' c 5 √ 21
Ld =65.20 ( 10 )=652 mm, min Ld =300 mm
Ld =850−75=775 mm>¿ 652 mm, OK!!

E. Check weight of footing

a. thickness of footing
t = 400 + 100 = 500 mm
b. volume of footing
3
V f =2 ( 2 ) ( 0.5 )=2 m
c. weight of footing
W f =24 kN /m3 ( 2 m3 ) =¿48 kN < 80.708 kN, OK!!

F. Details
Square Footing # 2:
Column 2: 275mm x 275mm
PDL = 111.91 kN f’c = 21 MPa
PLL = 36.51 kN
PE = 718 kN fy = 415 MPa
q a = 180 kPa

A. Proportion footing dimensions

a. Unfactored Column load
P = DL + LL + E = 111.91 + 36.51 + 718 = 866.42 kN
b. Assume weight of footing to be 20% of column load
W = 0.20(866.42) = 173.28 kN
c. Total load supported by soil
R = 866.42 + 866.42 = 1039.7 kN
d. Required footing area
R 1039.7 2
A= = =5.78 m
qa 180
e. Footing dimension
B=√ A=√ 5.78=2.4 m
Use 2.6m x 2.6m footing
f. Required footing area
2
A=(2.6 m)(2.6 m)=6.76 m
g. Factored column load
Pu=1.2 ( DL )+1.0E+ f 1 L=1.2 ( 111.91 ) +1.0 ( 718 ) +1.0 ( 36.51 )=888.8 kN
h. Net ultimate upward pressure
P u 888.8
q u= = =¿131.48 kPa
A 6.76

B. Compute "d" for beam shear

a. Factored shear
V u=(net ultimate upward pressure)(tributary area)
V u=(131.48)(2.6)(1.05−d)
V u=(341.85)(1.05−d) kN
b. Shear capacity at critical section

∅ V u=∅ ( 16 ) √ f c bd=0.85( 16 ) √ 21(1000)d

'

∅ V u=¿ 649.20d kN

c. Equate factored shear to shear capacity and compute "d"

Vu≤ ∅V u
(341.85)(1.05−d)=¿649.20d kN
d = 0.36
C. Compute "d" for punching shear
a. Factored shear
V u=(net ultimateupward pressure)(tributary area)
2 2
V u=(131.48)[2.6 −( 0.5+ d ) ]
V u= (131.48 ) [ 6.51−d−d2 ] kN
d. Shear capacity at critical section

∅ V u=∅( 31 ) √ f c b d'
0

∅ V =0.85 ( ) √ 21(1000)(0.5+ d)(d )

1
u
3
∅ V u=¿ 1298.40(0.5d+d2) kN

e. Equate factored shear to shear capacity and compute "d"

Vu≤ ∅V u
( 131.48 ) [ 6.51−d−d 2 ]=¿ 1298.40(0.5d+d2)
d = 0.55 (controls)
Therefore, use d = 550 mm

D. Compute flexural steel

a. Factored moment
M u=( 131.48 )( 1.05 ) ( 2.6 ) ( 0.525 ) =188.44 kN . m
b. Moment resistance coefficient
Mu 188.44 x 10
6
R u= 2
= 2
=0.27
φb d 0.9(2600)(550)
c. Strength ratio
fy 415
m= '
= =23.25
0.85 f c
0.85(21)
d. Required steel ratio

e.
ρ=
1
m [ √
1− 1−
2 Rum
fy
Minimum steel ratio
=
1
23.25
1− 1− ]
2(0.27)(23.25)
415 [ √ ] = 0.00656

0.25 √ f 'c 0.25 √21

ρmin =
fy
=
415
=0.00276 ≤
1.4
fy (
=0.00337
)
ρmin =0.00337 < ρ=0.00656

Therefore, use ρ ¿ 0.00337

f. Steel area
2
A s=ρbd=0.00 337 ( 2600 ) ( 550 )=4819.1 mm
π
using 10 mm  bars: Ab = ¿mm2
4
 A s 4819.1
n= = =61.36,
A b 78.54
use 60-10 mm  bars (3 layers – 20-10 mm  bars)
g. Check spacing of bars
b−n φ b−2 pc 2600−20 (10)−2(75)
s= = =118.42 mm
n−1 20−1
desirable spacing, 75 < s < 200mm, OK!!
h. Check development length
α =1.0 β=1.2 γ =1.0 λ=1.0
Ld 3 fyαβγ λ 3(415)(1)(1.2)(1)(1)
= = =65.20
db 5 √f ' c 5 √ 21
Ld =65.20 ( 10 )=652 mm, min Ld =300 mm
Ld =1050−75=975 mm> ¿ 652 mm, OK!!

E. Check weight of footing

a. thickness of footing
t = 550 + 100 = 650 mm
b. volume of footing
3
V f =2.6 ( 2.6 )( 00.65 )=4.39 m
c. weight of footing
W f =24 kN /m3 ( 4.39 m3 )=¿105.46 kN < 173.28 kN, OK!!

F. Details
Square Footing # 3:
Column 3: 375mm x 375mm
PDL = 223.82 kN f’c = 21 MPa
PLL = 73.02 kN
PE = 1436 kN fy = 415 MPa
q a = 180 kPa

A. Proportion footing dimensions

a. Unfactored Column load
P = DL + LL + E = 223.82 + 73.02 + 1436 = 1732.84 kN
b. Assume weight of footing to be 20% of column load
W = 0.20(1732.84) = 346.57 kN
c. Total load supported by soil
R = 1732.84 + 346.57 = 2079.41 kN
d. Required footing area
R 2079.41 2
A= = =11.55 m
qa 180
e. Footing dimension
B=√ A=√ 11.55=3.4 m
Use 3.4m x 3.4m footing
f. Required footing area
2
A=(3.4 m)(3.4 m)=11.56 m
g. Factored column load
Pu=1.2 ( DL )+1.0E+ f 1 L=1.2 ( 223.82 ) +1.0 ( 1436 )+1.0 ( 73.02 )=1777.6 kN
h. Net ultimate upward pressure
P u 1777.6
q u= = =¿153.77 kPa
A 11.56

B. Compute "d" for beam shear

a. Factored shear
V u=(net ultimate upward pressure)(tributary area)
V u=(153.77)(3.4)(1.4−d)
V u=(522.82)(1.4−d) kN
b. Shear capacity at critical section

∅ V u=∅ ( 16 ) √ f c bd=0.85( 16 ) √ 21(1000)d

'

∅ V u=¿ 649.20d kN

c. Equate factored shear to shear capacity and compute "d"

Vu≤ ∅V u
(522.82)(1.4−d )=¿649.20d kN
d = 0.62
C. Compute "d" for punching shear
a. Factored shear
V u=(net ultimateupward pressure)(tributary area)
2 2
V u=(153.77)[3.4 − ( 0.6+d ) ]
V u= (153.77 ) [11.2−1.2 d−d2 ] kN
b. Shear capacity at critical section

∅ V u=∅( 31 ) √ f c b d'
0

∅ V =0.85 ( ) √ 21(1000)(0.6+ d)(d)

1
u
3
∅ V u=¿ 1298.40(0.6d+d2) kN

c. Equate factored shear to shear capacity and compute "d"

Vu≤ ∅V u
( 153.77 ) [ 11.2−1.2 d−d 2 ] =¿1298.40(0.6d+d2)
d = 0.81 (controls)
Therefore, use d = 800 mm

D. Compute flexural steel

a. Factored moment
M u=( 153.77 )( 1.4 )( 3.4 ) ( 0.7 )=512.36 kN . m
b. Moment resistance coefficient
Mu 512.36 x 10
6
R u= 2
= 2
=0.26
φb d 0.9(3400)(800)
c. Strength ratio
fy 415
m= '
= =23.25
0.85 f c
0.85(21)
d. Required steel ratio

e.
ρ=
1
m [ √
1− 1−
2 Rum
fy
Minimum steel ratio
=
1
23.25
1− 1− ]
2(0.26)(23.25)
415 [ √ ] = 0.0065

0.25 √ f 'c 0.25 √21

ρmin =
fy
=
415
=0.00276 ≤
1.4
fy
=0.00337
( )
ρmin =0.00337 < ρ=0.00656

Therefore, use ρ ¿ 0.00337

f. Steel area
2
A s=ρbd=0.00 337 ( 3400 ) ( 800 ) =9166.4 mm
π
using 20 mm  bars: Ab = ¿mm2
4
 A s 9166.4
n= = =29.18,
A b 314.16
use 30-10 mm  bars both ways
g. Check spacing of bars
b−n φ b−2 pc 3400−30 (20)−2(75)
s= = =91.38 mm
n−1 30−1
desirable spacing, 75 < s < 200mm, OK!!
h. Check development length
α =1.0 β=1.2 γ =1.0 λ=1.0
Ld 3 fyαβγ λ 3(415)(1)(1.2)(1)(1)
= = =65.20
db 5 √f ' c 5 √ 21
Ld =65.20 ( 20 )=1304 mm, min Ld =300 mm
Ld =1400−75=1325 mm>¿ 1304 mm , OK!!

E. Check weight of footing

a. thickness of footing
t = 800 + 100 = 900 mm
b. volume of footing
3
V f =3.4 ( 3.4 ) ( 0.9 )=10.40 m
c. weight of footing
W f =24 kN /m3 ( 10.4 m3 )=¿ 249.70 kN < 346.57 kN, OK!!

F. Details