Design of Shallow Foundation (Isolated Footings)

• Providing isolated footing
• Square footing
• column size=400x400
• Axial load=2000kN
• Soil Bearing capacity= 150KN/m2
• Grade of concrete(fck)=M25

Size of footing:

Assuming self weight of footing equals to 10% of superimposed load

=10% x 2000

=.1x2000

=200KN

Total load=200+2000

=2200KN

Area of footing= W/SBC

=2200/150

=14.66m2

Side of Square= under root area= 3.82 =4m

Size of footing =4mx4m

Area of footing =16m2

Upward pressure:

pressure=total load x factor of safety / size of footing

P= 2200x1.5/16

P=206.25 KN/m2

Depth of footing:

Maximum bending moment at the face of column =pressurex side/8(S-s)2

= 206.25 x4/8 (4-.4)2

1336.5 KN/m

ultimate moment =factor of safety x calculated design moment

Mu= 1.5x 1336.5

Mu=2004.75KN/m

Mu= .138fck bd2

2004.75KN/m=.138x 25x 4000 x d2

d2= 2004.75/.138x25x4000

d2= 145271.73

d= 381.14

d=2.5 x 381.14

d= 952.85mm

D= 952.85 + 50 (clear cover)

D= 1003 mm

Area of Reinforcement :

Ast= 0.5fck/fy [1-(root 1- 4.6 Mu/fck Bd2)]Bd

Ast= 0.5 x 25/ 415 [1- root 1- 4.6x 2004.75x (10)6/25 x 4000 x (953)2] 4000 x 953

Ast= 5985.3 mm2

percentage reinforcement= Ast x 100/ Bd

= 5985.3 x 100 / 4000 x 953

= 0.157%

No. of bars = Total area - c/c / spacing +1 for main bars

Assumin center to center spacing as 200 mm

No. of bars = 5985.3 -50/ 200 + 1

= 29.52

Alternative,

Assuming dia of bar as 20mm

pia/4 x (20)2 = 314.5

no. of bars= 5985.3 /314.5 = 20 bars

spacing = L/ no. of bars

S = 4000/ 20

S= 200mm

20-20mm dia @ 200mm c/c spacing ( Main bars or longitudinal bars)

Check for one way shear:

V= pressure x side {1/2 (S-s) - d}

V= 699.6 KN

Vu = 1.5 x V

Vu= 1049.4 KN

Nominal shear= Vu/Bd

Tu= 1049.4x 10)3/ 4000 x953

Tu= 0.27 N/mm2

So, Tc'= Tc x k (Where k=1)

Tc'= .28

Tu>Tc

safe ok

d= Vu/bxkxTc

d= 937

d(provided)= 953

d (required)= 937

Hence ok

Check for Two way shear:

For two way shear, the section lies at a distance d/2 from the column face

b.= b+d

b.= 400+953

b.= 1353mm

net shear force V=449.1 KN

Vu= 1.5 x V

Vu= 673.65 KN

Tu= Fu/ 4x b.d

Tu= .23 N/mm2

Tc = .25 root fck

Tc= 1.25

ks x Tc= 1.25 (where ks=1)

Tu> Tc

Safe in Two way shear

Hence ok