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Design of Footing
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DESIGN OF FOOTING
Footing design (A4) :-
Axial load =427.98 kn
S.B.C =200 kn/m2
Self weight of footing =10%
=1.1 x p =470
Size of footing =1.1 x p/200 =2.35
Area =√2.35 = 1.53
Provide footing of size = 1.70 x 1.50m2
Net load pressure = load/area = 470/1.7 x 1.5 = 184.3 kn/m2
Consider 1m width of footing
Max B.M =184.3 x 0.6252/2
=35.99 kn-m
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)
Mu lim = 0.36 x (xumaxd
) x (1 – 0.42 x (xumaxd
) )x b x d2 x fck
Where (xumaxd
) = 0.48, Fe 415, and M20
= 0.36 x 0.48 x (1 – (0.42 x 0.48)) x 1000 x d2 x 20
35.99 x 106 =2.07 x b x d2
d = 131.85mm
D = 131.85 + 50 + 5 =186.8mm
Say D = 400mm
d = 400 – 50 – 5 = 345mm
Calculation of reinforcement :- (from IS code 456-2000, pg-96)
(a) Reinforcement along short mid span
Ast required :- (from IS code 456-2000, pg-96)
Mu = 0.87 x fy x Ast x d x (1−Ast∗f yb∗d∗f ck
)
35.99 x 106 = 0.87 x 415 x Ast x 345 x (1 – (Ast x 415)/(1000 x 345 x 20) )
35.99 x 106 = 124562 Ast – 0.217 Ast2
Ast = 289mm2
Assume 10mm bar
Spacing:-
A∗∅ 2∗bAst
= π4∗102∗1000
289 = 271.76mm
Say 275mm
Ast = π4∗102∗1000
275 = 285.5mm2
Hence provide 10mm ɸ at 275mm c/c
Shear on one way action :-
Critical section taken at the distance of 345mm away from column
625-345 = 280
Shear force at critical section
Considered 1m width
V = 280/1000 x 1 x 184.3 = 51.6
Vu = 1.5 x 51.6 = 77.4 kn
Nominal shear stress
Jv = vu
b∗deff = 77.4∗103
1000∗345 = 0.22 n/mm2
Percentage of steel = Astb∗d
∗100
= 285.5
1000∗345∗100 =0.08%
Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73
J c = 0.28 n/mm2
Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73
For M20 concrete Jc max = 2.8 n/mm2
Hence Jv < J c Hence it is safe .
Shear two way action :-
Critical action at the distance of 0.5d
= 0.5 x 345 = 172.5mm
Shear force :
624-172.5=452.5
452.5/1000 x 1 x 184.3 = 83.39
Nominal shear stress
Jv = vu
b∗deff = 83.39∗103∗1.5
1000∗345 = 0.36 n/mm2
From shear strength of concrete M20
J c = Ks x J c
Ks =0.5 + length of short column side/long column side
= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J c = 0.25 x √ fck
J c = 1.2 n/mm2
Hence Jv < J c hence ok safe
Load transfer column to footing :-
From code IS 456-2000
Allowable bearing stress = p x 1.5/230 x 450
= 427.98 x 103 x 1.5/230 x 450 = 6.2n/mm2
Allowable bearing pressure = k x 0.45 x fck
K = √ A1A2
= A1 =0.23 x 0.45
A2 = 1.7 x 1.5
K = 0.20 <2 adopt k = 2
2 x 0.45 x 20 = 18 n/mm2
Hence safe
Footing design (A2) :-
Axial load =265.55 kn
S.B.C =200 kn/m2
Self weight of footing =10%
=1.1 x p =292
Size of footing =1.1 x p/200 =1.46
Area =√1.46 = 1.2
Provide footing of size = 1.4 x 1.2m2
Net load pressure = load/area = 292/1.4 x 1.2 = 173.8 kn/m2
Consider 1m width of footing
Max B.M =173.8 x 0.4752/2
=19.6 kn-m
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)
Mu lim = 0.36 x (xumaxd
) x (1 – 0.42 x (xumaxd
) )x b x d2 x fck
Where (xumaxd
) = 0.48, Fe 415, and M20
= 0.36 x 0.48 x (1 – (0.42 x 0.48)) x 1000 x d2 x 20
19.6 x 106 =2.07 x b x d2
d = 97.3mm
D = 97.3 + 50 + 5 =152mm
Say D = 300mm
d = 300 – 50 – 5 = 245mm
Calculation of reinforcement :- (from IS code 456-2000, pg-96)
(b) Reinforcement along short mid span
Ast required :- (from IS code 456-2000, pg-96)
Mu = 0.87 x fy x Ast x d x (1−Ast∗f yb∗d∗f ck
)
19.6 x 106 = 0.87 x 415 x Ast x 245 x (1 – (Ast x 415)/(1000 x 245 x 20) )
19.6 x 106 = 88457 Ast – 0.03 Ast2
Ast = 221.6mm2
Assume 10mm bar
Spacing:-
A∗∅ 2∗bAst
= π4∗102∗1000
221.6 = 345.4mm
Say 300mm
Ast = π4∗102∗1000
300 = 262mm2
Hence provide 10mm ɸ at 300mm c/c
Shear on one way action :-
Critical section taken at the distance of 245mm away from column
475-245 =230
Shear force at critical section
Considered 1m width
V = 230/1000 x 1 x 173.8 = 39.97
Vu = 1.5 x 39.97 = 59.96 kn
Nominal shear stress
Jv = vu
b∗deff = 59.96∗103
1000∗245 = 0.24 n/mm2
Percentage of steel = Astb∗d
∗100
= 262
1000∗245∗100 =0.1%
Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73
J c = 0.28 n/mm2
Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73
For M20 concrete Jc max = 2.8 n/mm2
Hence Jv < J c Hence it is safe .
Shear two way action :-
Critical action at the distance of 0.5d
= 0.5 x 245 = 122.5mm say 125mm
Shear force :
475-125=350
350/1000 x 1 x 173.8 = 60.83
Nominal shear stress
Jv = vu
b∗deff = 60.83∗103∗1.5
1000∗245 = 0.37 n/mm2
From shear strength of concrete M20
J c = Ks x J c
Ks =0.5 + length of short column side/long column side
= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J c = 0.25 x √ fck
J c = 1.2 n/mm2
Hence Jv < J c hence ok safe
Load transfer column to footing :-
From code IS 456-2000
Allowable bearing stress = p x 1.5/230 x 450
= 265.65 x 103 x 1.5/230 x 450 = 3.85n/mm2
Allowable bearing pressure = k x 0.45 x fck
K = √ A1A2
= A1 =0.23 x 0.45
A2 = 1.4 x 1.2
K = 0.248<2 adopt k = 2
2 x 0.45 x 20 = 18 n/mm2
Hence safe
Footing design (A6) :-
Axial load =337 kn
S.B.C =200 kn/m2
Self weight of footing =10%
=1.1 x p =370.7
Size of footing =1.1 x p/200 =1.8
Area =√1.8 = 1.3
Provide footing of size = 1.5 x 1.3m2
Net load pressure = load/area = 370.7/1.5 x 1.3 = 190 kn/m2
Consider 1m width of footing
Max B.M =190 x 0.5252/2
=26.5 kn-m
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)
Mu lim = 0.36 x (xumaxd
) x (1 – 0.42 x (xumaxd
) )x b x d2 x fck
Where (xumaxd
) = 0.48, Fe 415, and M20
= 0.36 x 0.48 x (1 – (0.42 x 0.48)) x 1000 x d2 x 20
26.5 x 106 =2.07 x b x d2
d = 113mm
D = 113 + 50 + 5 =168mm
Say D = 400mm
d = 400 – 50 – 5 = 345mm
Calculation of reinforcement :- (from IS code 456-2000, pg-96)
(c) Reinforcement along short mid span
Ast required :- (from IS code 456-2000, pg-96)
Mu = 0.87 x fy x Ast x d x (1−Ast∗f yb∗d∗f ck
)
26.5 x 106 = 0.87 x 415 x Ast x 345 x (1 – (Ast x 415)/(1000 x 345 x 20) )
26.5 x 106 = 124562 Ast – 0.217 Ast2
Ast = 212mm2
Assume 10mm bar
Spacing:-
A∗∅ 2∗bAst
= π4∗102∗1000
212 = 370.47mm
Say 300mm
Ast = π4∗102∗1000
300 = 262mm2
Hence provide 10mm ɸ at 300mm c/c
Shear on one way action :-
Critical section taken at the distance of 345mm away from column
525-345 =180
Shear force at critical section
Considered 1m width
V = 180/1000 x 1 x 190 = 34.2
Vu = 1.5 x 34.2 = 51.3 kn
Nominal shear stress
Jv = vu
b∗deff = 51.3∗103
1000∗345 = 0.14 n/mm2
Percentage of steel = Astb∗d
∗100
= 262
1000∗3 45∗100 =0.075%
Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73
J c = 0.28 n/mm2
Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73
For M20 concrete Jc max = 2.8 n/mm2
Hence Jv < J c Hence it is safe .
Shear two way action :-
Critical action at the distance of 0.5d
= 0.5 x 345 = 172.5mm
Shear force :
525-172.5=352.5
352.5/1000 x 1 x 190 = 66.97
Nominal shear stress
Jv = vu
b∗deff = 66.97∗103∗1.5
1000∗3 45 = 0.29 n/mm2
From shear strength of concrete M20
J c = Ks x J c
Ks =0.5 + length of short column side/long column side
= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J c = 0.25 x √ fck
J c = 1.2 n/mm2
Hence Jv < J c hence ok safe
Load transfer column to footing :-
From code IS 456-2000
Allowable bearing stress = p x 1.5/230 x 450
= 337 x 103 x 1.5/230 x 450 = 4.88n/mm2
Allowable bearing pressure = k x 0.45 x fck
K = √ A1A2
= A1 =0.23 x 0.45
A2 = 1.5 x 1.3
K = 0.23<2 adopt k = 2
2 x 0.45 x 20 = 18 n/mm2
Hence safe
Footing design (B4) :-
Axial load =1416 kn
S.B.C =200 kn/m2
Self weight of footing =10%
=1.1 x p =1557.6
Size of footing =1.1 x p/200 =7.78
Area =√7.78 = 2.79
Provide footing of size = 3.0 x 2.8m2
Net load pressure = load/area = 1557.6/3.0 x 2.8 = 185.4 kn/m2
Consider 1m width of footing
DEPTH OF FOOTING ;-
VU = QU x B x [B−b
2 - d]
QU = 1.5 x p / area = 1.5 x 1416 /3 x 2.8 = 252 kn/m2
Vu = 0.252 x 2800 x [1250 -d]
Vu = 882000 – 705.6 d------------------------say1
Assume %of steel = 0.2 % & M20
Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73
J c = 0.32 n/mm2
Vu =Jc x B x d
Vu = 0.32 x 2800 x d-----------------------say2
Equating 1&2
d = 251.59
so provide d = 251 + 50 + 5 =306
provide D = 600mm
d = 600 – 50 - 5 = 545mm
chek B.M in long span :-
Mu = QU x B x l2 /2
Mu = 0.252 x 2800 x 12502 / 2
Mu = 55 knm
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)
Mu lim = 0.36 x (xumaxd
) x (1 – 0.42 x (xumaxd
) )x b x d2 x fck
Where (xumaxd
) = 0.48, Fe 415, and M20
Mu lim = 0.36 x 0.48 x (1 – (0.42 x 0.48)) x 1000 x d2 x 20
Mu lim = 203 knm
Mu < Mu lim hence safe
Check for 2 way shear :
Critical action at the distance of 0.5d
= 0.5 x 545 = 272.5mm
Shear force :
1250-272.5=977.5
Vu = 977.5/2800 x 1 x 185.5 = 64.7
Nominal shear stress
Jv = vu
b∗deff = 64.7∗103∗1.5
2800∗545 = 0.063 n/mm2
From shear strength of concrete M20
J c = Ks x J c
Ks =0.5 + length of short column side/long column side
= 0.5 +2.8/3.0
Ks = 0.5+0.5 =1
Permissible stress
J c = 0.25 x √ fck
J c = 1.2 n/mm2
Hence Jv < J c hence ok safe
Calculation of reinforcement :- (from IS code 456-2000, pg-96)
(d) Reinforcement along short mid span
Ast required :- (from IS code 456-2000, pg-96)
Mu = 0.87 x fy x Ast x d x (1−Ast∗f yb∗d∗f ck
)
55 x 106 = 0.87 x 415 x Ast x 545 x (1 – (Ast x 415)/(1000 x 545 x 20) )
55 x 106 = 196772 Ast – 0.00458 Ast2
Ast = 279mm2
Assume 10mm bar
Spacing:-
A∗∅ 2∗bAst
= π4∗102∗1000
279.5 = 786.8mm
Say 300mm
Adopt 10mm bar at 300mm c/c.
Load transfer column to footing :-
From code IS 456-2000
Allowable bearing stress = p x 1.5/500 x 300
= 1416 x 103 x 1.5/500 x 300 = 0.014n/mm2
Allowable bearing pressure = k x 0.45 x fck
K = √ A1A2
= A1 =3.0 x 2.8
A2 = 0.5 x 0.30
K = 7.48<2 adopt k = 2
2 x 0.45 x 20 = 18 n/mm2
Hence safe
Footing design (D3) :-
Axial load =341.7 kn
S.B.C =200 kn/m2
Self weight of footing =10%
=1.1 x p =375.87
Size of footing =1.1 x p/200 =1.8
Area =√1.8 = 1.3
Provide footing of size = 1.5 x 1.3m2
Net load pressure = load/area = 375.87/1.5 x 1.3 = 192.75 kn/m2
Consider 1m width of footing
Max B.M =192.75 x 0.6352/2
=38.86 kn-m
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)
Mu lim = 0.36 x (xumaxd
) x (1 – 0.42 x (xumaxd
) )x b x d2 x fck
Where (xumaxd
) = 0.48, Fe 415, and M20
= 0.36 x 0.48 x (1 – (0.42 x 0.48)) x 1000 x d2 x 20
38.86 x 106 =2.07 x b x d2
d = 137mm
D = 137 + 50 + 5 =192mm
Say D = 400mm
d = 400 – 50 – 5 = 345mm
Calculation of reinforcement :- (from IS code 456-2000, pg-96)
Ast required :- (from IS code 456-2000, pg-96)
Mu = 0.87 x fy x Ast x d x (1−Ast∗f yb∗d∗f ck
)
38.86 x 106 = 0.87 x 415 x Ast x 345 x (1 – (Ast x 415)/(1000 x 345 x 20) )
38.86 x 106 = 124562 Ast – 0.217 Ast2
Ast = 289mm2
Assume 10mm bar
Spacing:-
A∗∅ 2∗bAst
= π4∗102∗1000
289 = 271.76mm
Say 275mm
Ast = π4∗102∗1000
275 = 285.5mm2
Hence provide 10mm ɸ at 275mm c/c
Shear on one way action :-
Critical section taken at the distance of 345mm away from column
635-345 =290
Shear force at critical section
Considered 1m width
V = 290/1000 x 1 x 192.75 = 55.89 kn
Nominal shear stress
Jv = vu
b∗deff = 55.89∗103
1000∗345 = 0.162 n/mm2
Percentage of steel = Astb∗d
∗100
= 285.5
1000∗345∗100 =0.08%
Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73
J c = 0.28 n/mm2
Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73
For M20 concrete Jc max = 2.8 n/mm2
Hence Jv < J c Hence it is safe .
Shear two way action :-
Critical action at the distance of 0.5d
= 0.5 x 345 = 172.5mm
Shear force :
635 -172.5=462.5
462.5/1000 x 1 x 192.75 = 89.14
Nominal shear stress
Jv = vu
b∗deff = 89.14∗103∗1.5
1000∗345 = 0.38 n/mm2
From shear strength of concrete M20
J c = Ks x J c
Ks =0.5 + length of short column side/long column side
= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J c = 0.25 x √ fck
J c = 1.2 n/mm2
Hence Jv < J c hence ok safe
Load transfer column to footing :-
From code IS 456-2000
Allowable bearing stress = p x 1.5/230 x 450
= 341.7 x 103 x 1.5/230 x 450 = 4.9n/mm2
Allowable bearing pressure = k x 0.45 x fck
K = √ A1A2
= A1 =0.23 x 0.45
A2 = 1.5 x 1.3
K = 0.23<2 adopt k = 2
2 x 0.45 x 20 = 18 n/mm2
Hence safe
Footing design (C3) :-
Axial load =1082 kn
S.B.C =200 kn/m2
Self weight of footing =10%
=1.1 x p =1190.75
Size of footing =1.1 x p/200 =5.95
Area =√5.95 = 2.44
Provide footing of size = 2.8 x 2.5m2
Net load pressure = load/area = 1190.75/2.8 x 2.5 = 170 kn/m2
Consider 1m width of footing
DEPTH OF FOOTING ;-
VU = QU x B x [B−b
2 - d]
QU = 1.5 x p / area = 1.5 x 1082.5 /2.8 x 2.5 = 232 kn/m2
Vu = 0.232 x 2500 x [1175 -d]
Vu = 681500 – 580 d------------------------say1
Assume %of steel = 0.2 % & M20
Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73
J c = 0.32 n/mm2
Vu =Jc x B x d
Vu = 0.32 x 2500 x d-----------------------say2
Equating 1&2
d = 493.8
so provide d = 493.8 + 50 + 5 =548
provide D = 550mm
d = 550 – 50 - 5 = 495mm
chek B.M in long span :-
Mu = QU x B x l2 /2
Mu = 0.232 x 2500 x 11752 / 2
Mu = 40 knm
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)
Mu lim = 0.36 x (xumaxd
) x (1 – 0.42 x (xumaxd
) )x b x d2 x fck
Where (xumaxd
) = 0.48, Fe 415, and M20
Mu lim = 0.36 x 0.48 x (1 – (0.42 x 0.48)) x 1000 x d2 x 20
Mu lim = 169 knm
Mu < Mu lim hence safe
Check for 2 way shear :
Critical action at the distance of 0.5d
= 0.5 x 495 = 247.5mm
Shear force :
1175-247.5=927.5
Vu = 927.5/2500 x 1 x 170 = 63
Nominal shear stress
Jv = vu
b∗deff = 94.6∗103∗1.5
2500∗495 = 0.076 n/mm2
From shear strength of concrete M20
J c = Ks x J c
Ks =0.5 + length of short column side/long column side
= 0.5 +2.8/2.5
Ks = 0.5+0.5 =1
Permissible stress
J c = 0.25 x √ fck
J c = 1.2 n/mm2
Hence Jv < J c hence ok safe
Calculation of reinforcement :- (from IS code 456-2000, pg-96)
Ast required :- (from IS code 456-2000, pg-96)
Mu = 0.87 x fy x Ast x d x (1−Ast∗f yb∗d∗f ck
)
40 x 106 = 0.87 x 415 x Ast x 495 x (1 – (Ast x 415)/(1000 x 495 x 20) )
40 x 106 = 178719.7 Ast – 2.99 Ast2
Ast = 224.65mm2
Assume 10mm bar
Spacing:-
A∗∅ 2∗bAst
= π4∗102∗1000
224 .65 = 874mm
Say 300mm
Adopt 10mm bar at 300mm c/c.
Load transfer column to footing :-
From code IS 456-2000
Allowable bearing stress = p x 1.5/450 x 230
= 1082.5 x 103 x 1.5/450 x 230 = 15.68n/mm2
Allowable bearing pressure = k x 0.45 x fck
K = √ A1A2
= A1 =2.8 x 2.5
A2 = 0.45 x 0.23
K = 8.2<2 adopt k = 2
2 x 0.45 x 20 = 18 n/mm2
Hence safe
Footing design (C1) :-
Axial load =354.8 kn
S.B.C =200 kn/m2
Self weight of footing =10%
=1.1 x p =390.28
Size of footing =1.1 x p/200 =1.95
Area =√1.95 = 1.39
Provide footing of size = 1.4 x 1.6m2
Net load pressure = load/area = 390.28/1.4 x 1.6 = 174.23 kn/m2
Consider 1m width of footing
Max B.M =174.23 x 0.5752/2
=28.8 kn-m
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)
Mu lim = 0.36 x (xumaxd
) x (1 – 0.42 x (xumaxd
) )x b x d2 x fck
Where (xumaxd
) = 0.48, Fe 415, and M20
= 0.36 x 0.48 x (1 – (0.42 x 0.48)) x 1000 x d2 x 20
28.8 x 106 =2.07 x b x d2
d = 117.95mm
D = 117.95 + 50 + 5 =172.9mm
Say D = 400mm
d = 400 – 50 – 5 = 345mm
Calculation of reinforcement :- (from IS code 456-2000, pg-96)
Ast required :- (from IS code 456-2000, pg-96)
Mu = 0.87 x fy x Ast x d x (1−Ast∗f yb∗d∗f ck
)
28.8x 106 = 0.87 x 415 x Ast x 345 x (1 – (Ast x 415)/(1000 x 345 x 20) )
28.8 x 106 = 124562 Ast – 0.217 Ast2
Ast = 231.3mm2
Assume 10mm bar
Spacing:-
A∗∅ 2∗bAst
= π4∗102∗1000
231.3 = 339.5mm
Say 300mm
Ast = π4∗102∗1000
300 = 261.7mm2
Hence provide 10mm ɸ at 275mm c/c
Shear on one way action :-
Critical section taken at the distance of 345mm away from column
575 -345 =230
Shear force at critical section
Considered 1m width
V = 230/1000 x 1 x 174.23 = 40 kn
Vu = 40 x 13.5 = 60 kn
Nominal shear stress
Jv = vu
b∗deff = 60∗103
1000∗345 = 0.17 n/mm2
Percentage of steel = Astb∗d
∗100
= 261.79
1000∗345∗100 =0.07%
Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73
J c = 0.28 n/mm2
Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73
For M20 concrete Jc max = 2.8 n/mm2
Hence Jv < J c Hence it is safe .
Shear two way action :-
Critical action at the distance of 0.5d
= 0.5 x 345 = 172.5mm
Shear force :
575 -172.5=402.5
402.5/1000 x 1 x 175.23 = 70
Nominal shear stress
Jv = vu
b∗deff = 70∗103∗1.5
1000∗345 = 0.30 n/mm2
From shear strength of concrete M20
J c = Ks x J c
Ks =0.5 + length of short column side/long column side
= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J c = 0.25 x √ fck
J c = 1.2 n/mm2
Hence Jv < J c hence ok safe
Load transfer column to footing :-
From code IS 456-2000
Allowable bearing stress = p x 1.5/230 x 450
= 354.8 x 103 x 1.5/230 x 450 = 5.14n/mm2
Allowable bearing pressure = k x 0.45 x fck
K = √ A1A2
= A1 =0.23 x 0.45
A2 = 1.6 x 1.4
K = 0.2<2 adopt k = 2
2 x 0.45 x 20 = 18 n/mm2
Hence safe
