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
=184.3 x 0.6252/2

Max B.M

=35.99 kn-m
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)

Mu lim = 0.36 x

Where

x u max
)
d

x u max
)
d

x (1 0.42 x

x u max
)
d

)x b x d2 x fck

= 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

Astf y
)
bdf 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:2

A b
Ast

1021000
4
289

= 271.76mm

Say 275mm

Ast =

2
10 1000
4
275

= 285.5mm2

Hence provide 10mm 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
bd eff

77.410
1000345

Percentage of steel =

Ast
100
bd

285.5
100
1000345

= 0.22 n/mm2

=0.08%

Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73

J

= 0.28 n/mm2

Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73

J

For M20 concrete

Hence

Jv

<

c max

= 2.8 n/mm2

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
bd eff

83.3910 1.5
1000345

= 0.36 n/mm2

From shear strength of concrete M20

J

= Ks x

Ks =0.5 + length of short column side/long column side

= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J

fck

= 0.25 x

= 1.2 n/mm2

Hence

Jv

<

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=

A1
A2

= 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

Where

x u max
)
d

x u max
)
d

x (1 0.42 x

x u max
)
d

)x b x d2 x fck

= 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

Astf y
)
bdf 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:A2b
Ast

2
10 1000
4
221.6

= 345.4mm

Say 300mm

Ast =

1021000
4
300

= 262mm2

Hence provide 10mm 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
bd eff

59.96103
1000245

Percentage of steel =

= 0.24 n/mm2

Ast
100
bd

262
100
1000245

=0.1%

Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73

J

= 0.28 n/mm2

Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73

For M20 concrete

Hence

Jv

<

c max

= 2.8 n/mm2

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
bd eff

60.8310 1.5
1000245

= 0.37 n/mm2

From shear strength of concrete M20

J

= Ks x

Ks =0.5 + length of short column side/long column side

= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J
J

fck

= 0.25 x

= 1.2 n/mm2

Jv

Hence

<

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=

A1
A2

= 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
=190 x 0.5252/2

Max B.M

=26.5

kn-m

MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)

(

Mu lim = 0.36 x

Where

x u max
)
d

x u max
)
d

x (1 0.42 x

x u max
)
d

)x b x d2 x fck

= 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

Astf y
)
bdf 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:A2b
Ast

1021000
4
212

= 370.47mm

Say 300mm

Ast =

1021000
4
300

= 262mm2

Hence provide 10mm 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
bd eff

51.3103
1000345

= 0.14 n/mm2

Percentage of steel =

Ast
100
bd

262
100
10003 45

=0.075%

Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73

J

= 0.28 n/mm2

Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73

J

For M20 concrete

Hence

Jv

<

c max

= 2.8 n/mm2

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
bd eff

66.971031.5
10003 45

From shear strength of concrete M20

J

= Ks x

= 0.29 n/mm2

Ks =0.5 + length of short column side/long column side

= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J

fck

= 0.25 x

= 1.2 n/mm2

Hence

Jv

<

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=

A1
A2

= 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 [

Bb
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

= 0.32 n/mm2

Vu = J

xBxd

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

Where

x u max
)
d

x u max
)
d

x (1 0.42 x

x u max
)
d

)x b x d2 x fck

= 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
bd eff

64.71031.5
2800545

= 0.063 n/mm2

From shear strength of concrete M20

J

= Ks x

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
J

fck

= 0.25 x

= 1.2 n/mm2

Hence

Jv

<

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

Astf y
)
bdf 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:A2b
Ast

1021000
4
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=

A1
A2

= 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
=192.75 x 0.6352/2

Max B.M

=38.86

kn-m

MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)

Mu lim = 0.36 x

x u max
)
d

x (1 0.42 x

x u max
)
d

)x b x d2 x fck

Where

x u max
)
d

= 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

Astf y
)
bdf 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:A2b
Ast

Say 275mm

2
10 1000
4
289

= 271.76mm

Ast =

1021000
4
275

= 285.5mm2

Hence provide 10mm 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
bd eff

55.8910
1000345

Percentage of steel =

= 0.162 n/mm2

Ast
100
bd

285.5
100
1000345

=0.08%

Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73

J

= 0.28 n/mm2

Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73

For M20 concrete

c max

= 2.8 n/mm2

Hence

Jv

<

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
bd eff

89.1410 31.5
1000345

= 0.38 n/mm2

From shear strength of concrete M20

J

= Ks x

Ks =0.5 + length of short column side/long column side

= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J

fck

= 0.25 x

= 1.2 n/mm2

Hence

Jv

<

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=

A1
A2

= 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 [

Bb
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

= 0.32 n/mm2

Vu = J

xBxd

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

Where

x u max
)
d

x u max
)
d

x (1 0.42 x

x u max
)
d

)x b x d2 x fck

= 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
bd eff

94.610 31.5
2500495

= 0.076 n/mm2

From shear strength of concrete M20

J

= Ks x

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

fck

= 0.25 x

= 1.2 n/mm2

Hence

Jv

<

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

Astf y
)
bdf 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:2

A b
Ast

2
10 1000
4
224 . 6 5

= 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=

A1
A2

= 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
=174.23 x 0.5752/2

Max B.M

=28.8 kn-m
MOMENT OF RESISTENCE :- (from IS code 456-2000, pg-96)

Mu lim = 0.36 x

Where

x u max
)
d

x u max
)
d

x (1 0.42 x

x u max
)
d

)x b x d2 x fck

= 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

Astf y
)
bdf 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:A2b
Ast

2
10 1000
4
231.3

= 339.5mm

Say 300mm

Ast =

1021000
4
300

= 261.7mm2

Hence provide 10mm 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
bd eff

60103
1000345

Percentage of steel =

Ast
100
bd

261.79
100
1000345

= 0.17 n/mm2

=0.07%

Shear strength of concrete:- (from IS code 456-2000) Table-19, pg-73

J

= 0.28 n/mm2

Max Shear strength of concrete:- (from IS code 456-2000) Table-20, pg-73

J

For M20 concrete

Hence

Jv

<

c max

= 2.8 n/mm2

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
bd eff

7010 1.5
1000345

= 0.30 n/mm2

From shear strength of concrete M20

J

= Ks x

Ks =0.5 + length of short column side/long column side

= 0.5 +230/450
Ks = 0.5+0.5 =1
Permissible stress
J

fck

= 0.25 x

= 1.2 n/mm2

Hence

Jv

<

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=

A1
A2

= 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