1.DATA:

Pu = 1500 KN

column size

b = 300 mm

D = 500 mm

p = 185 KN/m2

fck = 20 KN/mm2

fy = 415 KN/mm2

2.SIZE OF FOOTING:

Load of column= 1500 KN

self weight of footing (10%)= 150 KN

total factored load=Wu = 1650 KN

footing area = Wu /(1.5*p) = 5.95 m2

proportion the footing area in the same proportion as the sides of the column.

hence (b/100)x * (D/100)x = footing area

x = 0.63

short side = 1.89

long side = 3.15

adopt size of footing

x= 2 m

y= 3 m

factored soil pressure at base is computed as :

pu = Pu/(x*y) = 250 <(1.5*p)= 277.5

hence the footing area is adequate since the soil pressure developed at

the base is less than the factored bearing capacity of the soil

3.FACTORED BENDING MOMENT:

cantilver projection from the short side face of the column a =

cantilver projection from the long side face of the column =

bending moment at short side face of column is (0.5*pu*L2)=

bending moment at long side face of column is (0.5*pu*L2)=

4.DEPTH OF FOOTING:

DESIGN OF FOOTING(ISOLATED MAT)

From moment consideration we have :

Mu=0.138fckbd2

d= 266.02 mm

from shear strength considerations we have critical section for one way

shear is located at a distance d from the face of column

shear force per metre width (longer direction)is :

τc = 0.36

Vul = pu(a-d)N pt = 0.25

τc = (Vul/b*d )

by solving d= 512.3 mm

hence adopt d = 550 mm

overall depth D = 600 mm

5.REINFORCEMENT IN FOOTING:

Longer direction:

M u = 0.87. fy. Ast d

Ast= 1023.6 mm2

dia of bar= 16

no of bar= 5.093437 6

Ast=

adopt 16mm dia bar at 16m dia bar at 160mm centres

shorter direction:

by solving Ast = 463.1

ratio of long to short side=β= 1.5

reinforcement in central band width of short span

(y/(β+1)2Ast= 740.99

minimum reinforcement=0.0012*b*D= 1440 > 740.99

dia of bar= 12

no f bar= 7 =7nos

hence provide 12mm diameter bars at 150mm centres

6.CHECK FOR SHEAR STRESS:

The sritical section for one way shear is located at a distance d from the

face of the column .

Ultimate shear force per metre width in the longer direction is:

bdf

Af

ck

stxy

.

.1

Vu=Pu*0.7= 175

100Ast/bd= 0.22

refer table 19 of IS 456-2000

ks= 1

τc= 0.33

ks*τc = 0.33 N/mm2

nominal shear stress = τv= Vu/bd= 0.32 N/mm2

since τv<ks*τc shear stresses are within the safe permissible limits.

KN/m2

1.25 m

0.85 m

195.3 KN.m

90.3 KN.m

DESIGN OF FOOTING(ISOLATED MAT)

calculation

0.148502

0.851498

0.922766

1205.76 mm20.077234

1023.577

calculation

0.068667

0.931333

0.965056

0.034944

463.1163