Design of Shallow Foundation (Isolated Footings)
AIM : Design a square footing for a column size of 400x400. The compression axial load for the load combination of (1.5 DL + 1.5 LL) is 2000 KN. The safe soil bearing capacity is 150 KN/m2 at a depth of 2 metres below E.G.L. Participants are free to go for either a tapered or stepped footing. Besides the total axial load, also account for the self-weight of the footing and soil above it. Assume M25 concrete grade. Do the following checks:
Introduction :
- footing are the base element of the building so that they can transfer the load from the column to the hard soil strata
- They are common type of footing are strip footing, stepped footing, spread footing , Taped footing , raft footing and combined footing.
Procedure :
Given Data
- Compressive axial load = 2000 KN converting into factoed load = 2000/1.5 = 1333.333
- safe bearing capacity = 150 KN/^m2
- Grade of concrete = M25
- Assuming the thickness of footing is 0.8
- Depth below the 2 m surface ground level
- soil depth is= 2 - 0.8 (depth of footing assumed)
- = 1.2 m depth of soil
- Assumig the clay soil density = 20
- Rcc density = 25 KN/cu.m
Size of footing =
10 % of self weight of footing
load on column = 2000X10 /100
= 200 KN
Axial load on footing = 2000 + 200
= 2200 KN
Area of footing = load on footing / Safe bearing capacity
= 2200/ 150
= 14.66 m^2
B = L = square roof of 14.66 = 3.82
= 3.9 m
Net upword pressure :
factored net soil pressure = load on footing / area of footing
= 2200 / 3.9 x 3.9
= 14.64 converting into factored load
= 14.64 x 1.5
= 216.96 KN/m^2
Depth of footing =
Mu = 0.133fckb^2
d = under root Mu / 0.133fck b
= 650 mm let say
cover = 50 mm
over all depth = Depth + cover + bar dia / 2
= 650 + 50 + 16 /2
= 710 mm
Reinforcement details = Mu / bd^2
= 1328.88X10^6 / 4000 x 700 ^2 = 0.67 say 0.7
Ast = 0.617 X 400 x 700 / 100
= 4676 mm^2
= 4676 /201 mm^2
= 24 numbers
Check for one way shear
shear face = upword soil pressure X (B-b) /2 -d
= 216.96 x (4 -0.4 ) /2- 0.7
= 238. 67 KN
100 Ast / bd = 100 X 4824 X 4000 X 700 = 0.172
Tv = V /bd
= 238.66/ 4000 * 700
= 0.085 N/mm2
Tc = 0.30 N/mm^2 a per table 19 of Is 456
Tv < TC
Check for Two way shear
For two way shear critical section should be at a distance of d/2
Shear face on critical section
V = 238.66 (4 X4 + 1.1 x 1.1)
Pheriphery of section = 2 ( 1.1 +1.1)
= 4.4 m
Tv = Vu / bd
= 3155.81 / 4400 * 700
= 1754.44 KN
The permisible shear stress = Ks Tc
Ks = (0.5 + bc ) > 1
Bc = b/b = 400 / 400
= 1.0
ks = 0.5 + 1
= 1
Permisible shear stress (Tck ) ( N/mm^2 ) = 1.50
0.25 X Fck =
= 1.25
Tv < Tc
1.25 < 1.5
Hence shear is with in the safe limit