Rectangular Tapered Footing

1. DESIGN OF FOOTING MARKED :- F 1 cl (m) = 0.350
cb (m) = 0.400
l (m) = 0.350
b (m) = 0.400
l' (m) = 0.550
b' (m) = 0.600
L (m) = 2.850
B (m) = 2.850
D1 (m) = 0.750
D2 (m) = 0.200
Dp (m) = 0.000
Depth of foundation 'DF' below
G.L (m) = 1.500
clear cover cx to R/F for forces
about X-axis (m) = 0.050
clear cover cy to R/F for forces
about Y-axis (m) = 0.050
effective depth 'd1' for forces
about X-axis (m) = 0.700
effective depth 'd2' for forces
about Y-axis (m) = 0.700
Unit wt. of Conc.'γc' (t/m3
) = 2.500
Unit wt. of Soil. 'γs' (t/m
3
) = 1.800
Self wt. of footing and weight of soil on footing : fy (N/mm
2
) = 550
Wt. of footing (t) = 8.686 fck (N/mm2
) = 20
Wt. of soil over footing (t) = 10.776
Net Bearing Capacity 'qnet' (t/m
2
)
= 7.9
Footing Wt.+Soil Wt.= (P1 )(t) = 19.463
Gross Bearing Capacity 'qgross'
=(qnet+DF x γs )(t/m
2
) = 10.64
Static case :
P (t) = 59.510 Node 18
Mx (t-m) = -0.590 Load Case 201
My (t-m) = -0.270
Total Load(PT = P + P1 ) (t) = 78.973
partial safety factor 'fs' = 1.5
percent increase in qnet /qgross = 0
Pressure under footing (t/m
2
) = P T /(L x B) + M x x 6 /(L x B
2
) + M y x 6 /(B x L
2
)
9.50 9.81 9.64 9.95
Pressure due to Wt.of footing + Wt.of soil = P 1 /(L x B) (t/m2
) = 2.40
Taking average of pressure and calculating moment at the face of pedestal :
About 1-1 : About 2-2 :
M (t-m) = 15.43 16.20
Mu = (fs x M)(t-m) = 23.15 24.30
ku (N/mm
2
) = Mu /(l' x d1
2
) = 0.859 Mu /(b' x d2
2
) = 0.827
pt = 0.189 0.182
A st (mm
2
) = pt x l' x d1 /100 = 729 pt x b' x d2/100 = 764
Taking average of pressure and calculating one-way shear force at "d" from the face of pedestal :
At 'd1' from 1-1: At 'd2' from 2-2:
V (t) = 10.78 11.40
(α+β)/2
(χ+δ)/2
DF
D1
D2
α
β
χ
δ
cl
cb
X
X
Y
Y
Mx
Mx
My
My
(β+δ)/2
(α+χ)/2
2
2
1 1
P
α β χ δ
L
l'
b
B
l
b'
Dp
(α+β)/2
(χ+δ)/2
DF
D1
D2
α
β
χ
δ
cl
cb
X
X
Y
Y
Mx
Mx
My
MZ
(β+δ)/2
(α+χ)/2
2
2
1 1
P
α β χ δ
L
l'
b
B
l
b'
Dp

2. Vu =(fs x V)(t) = 16.16 17.09
τv(N/mm2
) =Vu/(Lx(D2-cx)+(L+b1')/2x{d1'-(D2-cx)} τv(N/mm2
) =Vu/(Bx(D2-cy)+(BL+b2')/2x{d2'-(D2-cy)}
where b1'=[l'+2x{d1-(b'-b)/2}x(L-l')/(B-b')](m)= 1.777 where b2'=[b'+2x{d2-(l'-l)/2}x(B-b')/(L-l')](m) 1.774
d1'=[(D2-cx)+{d1 -(D2-cx)}/{(B-b')/2}x{(B-b)/2-d1}]= 0.407 d2'=[(D2-cy)+{d2 -(D2-cy)}/{(L-l')/2}x{(L-l)/2-d 0.413
τv(N/mm
2
) = 0.158 0.165
pt = 0.041 0.045
A st (mm
2
) = pt x Agross1 = 1226 pt x Agross2 = 1243
Check for two-way Shear at "d/2" from the face of pedestal :
V(t) = {LxB-(l+d)x(b+d)}xqnet = 55.32
τv = Vu /(po x d) (N/mm
2
) where po = ( l+b+2d )
τv(N/mm2
) = 0.276 τc =(ksx0.25x(fck)1/2
) (N/mm2
) = 1.118
where ks=(1+E/F)but>1.0
Hence OK

3. Seismic/Wind case in X-direction : Node 18
L/C 210
P (t) = 62.720
Mx (t-m) = 0.710
My (t-m) = 4.000
Total Load(PT = P + P1 ) (t) = 82.183
partial safety factor 'fs' = 1.5
percent increase in qnet /qgross = 25
Pressure under footing (t/m
2
) = P T /(L x B) + M x x 6 /(L x B
2
) + M y x 6 /(B x L
2
)
11.34 10.97 9.27 8.90
Pressure due to Wt.of footing + Wt.of soil = P 1 /(L x B) (t/m
2
) = 2.40
Taking average of pressure and calculating moment at the face of pedestal :
About 1-1 : About 2-2 :
M (t-m) = 16.79 18.83
Mu = (fs x M)(t-m) = 25.19 28.24
ku (N/mm
2
) = Mu /(l' x d1
2
) = 0.935 Mu /(b' x d2
2
) = 0.961
pt = 0.207 0.213
A st (mm2
) = pt x l' x d1 /100 = 798 pt x b' x d2/100 = 896
Taking average of pressure and calculating one-way shear force at "d" from the face of pedestal :
At 'd1' from 1-1: At 'd2' from 2-2:
V (t) = 11.78 13.42
Vu =(fs x V)(t) = 17.67 20.12
τv(N/mm
2
) =Vu/(Lx(D2-cx)+(L+b1')/2x{d1'-(D2-cx)} τv(N/mm
2
) =Vu/(Bx(D2-cy)+(BL+b2')/2x{d2'-(D2-cy)}
where b1'=[l'+2x{d1-(b'-b)/2}x(L-l')/(B-b')](m)= 1.777 where b2'=[b'+2x{d2-(l'-l)/2}x(B-b')/(L-l')](m) 1.774
d1'=[(D2-cx)+{d1 -(D2-cx)}/{(B-b')/2}x{(B-b)/2-d1}]= 0.407 d2'=[(D2-cy)+{d2 -(D2-cy)}/{(L-l')/2}x{(L-l)/2-d 0.413
τv(N/mm2
) = 0.173 0.194
pt = 0.049 0.063
A st (mm
2
) = pt x Agross1 = 1226 pt x Agross2 = 1243
Check for two-way Shear at "d/2" from the face of pedestal :
V(t) = {LxB-(l+d)x(b+d)}xqnet = 55.32
τv = Vu /(po x d) (N/mm
2
) where po = ( l+b+2d )
τv(N/mm
2
) = 0.276 τc =(ksx0.25x(fck)
1/2
) (N/mm
2
) = 1.118
where ks=(1+E/F)but>1.0
Hence OK
α β χ δ
α β χ δ

4. Seismic/Wind case in Y-direction : Node 18
L/C 213
P (t) = 63.050
Mx (t-m) = 4.420
My (t-m) = 0.190
Total Load(PT = P + P1 ) (t) = 82.513
partial safety factor 'fs' = 1.5
percent increase in qnet /qgross = 25
Pressure under footing (t/m
2
) = P T /(L x B) + M x x 6 /(L x B
2
) + M y x 6 /(B x L
2
)
11.35 9.06 11.25 8.96
Pressure due to Wt.of footing + Wt.of soil = P 1 /(L x B) (t/m
2
) = 2.40
Taking average of pressure and calculating moment at the face of pedestal :
About 1-1 : About 2-2 :
M (t-m) = 18.35 17.36
Mu = (fs x M)(t-m) = 27.52 26.04
ku (N/mm
2
) = Mu /(l' x d1
2
) = 1.021 Mu /(b' x d2
2
) = 0.886
pt = 0.228 0.196
A st (mm2
) = pt x l' x d1 /100 = 877 pt x b' x d2/100 = 822
Taking average of pressure and calculating one-way shear force at "d" from the face of pedestal :
At 'd1' from 1-1: At 'd2' from 2-2:
V (t) = 13.01 12.23
Vu =(fs x V)(t) = 19.52 18.34
τv(N/mm2
) =Vu/(Lx(D2-cx)+(L+b1')/2x{d1'-(D2-cx)} τv(N/mm2
) =Vu/(Bx(D2-cy)+(BL+b2')/2x{d2'-(D2-cy)}
where b1'=[l'+2x{d1-(b'-b)/2}x(L-l')/(B-b')](m)= 1.777 where b2'=[b'+2x{d2-(l'-l)/2}x(B-b')/(L-l')](m) 1.774
d1'=[(D2-cx)+{d1 -(D2-cx)}/{(B-b')/2}x{(B-b)/2-d1}]= 0.407 d2'=[(D2-cy)+{d2 -(D2-cy)}/{(L-l')/2}x{(L-l)/2-d 0.413
τv(N/mm2
) = 0.191 0.177
pt = 0.061 0.052
A st (mm
2
) = pt x Agross1 = 1226 pt x Agross2 = 1243
Check for two-way Shear at "d/2" from the face of pedestal :
V(t) = {LxB-(l+d)x(b+d)}xqnet = 55.32
τv = Vu /(po x d) (N/mm
2
) where po = ( l+b+2d )
τv(N/mm
2
) = 0.276 τc =(ksx0.25x(fck)
1/2
) (N/mm
2
) = 1.118
where ks=(1+E/F)but>1.0
Hence OK
R/F REQD 1225.51 1242.77
R/F PROVIDED 2147.76 2147.76
Provided 12 dia @ 150 Provided 12 dia @ 150
Ast = 754 mm2
/m Ast = 754 mm2
/m
α β χ δ
α β χ δ