Design of Shallow Foundation (Isolated Footings)
AIM: Design a square footing for a column size of 400x400.
INTRODUCTION: The axial compression 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 2 meters below E.G.L.
Given:
Factored axial load: 2000KN
SBC: 150 KN/m2
fck: 25N/mm2
Fy: 500 N/mm2
PROCEDURE:
- First, calculate the area of the footing:
- check permissible bearing stress:
- now calculate flexural reinforcement :
factored net soil pressure:
qn = factored load/area
= 2000/3.2*3.2
= 195.3 KN/m2
= 0.195 N/mm2
moment due to net soil pressure:
mu= force*lever arm
= 0.195*3200*1400*700
=611.52 KN-m
= 620 KN-m(approx)
therefore, as per IS CODE 456-2000, annex-G
after simplifying and putting the value we get a quadratic equation:
(Ast)^2 - 72000 (Ast)+2.3*10^6 =0
therefore Ast= 3320.43 mm2
so 16mm bar, 3320.43/201=16.519
No. od bar = 17
spacing = 3200/17
= 188.23
take spacing = 150mm
no of 16mm bar = 3200/150= 22
area provided= 22*201=4422 mm^2 -----------------------------------------------------------(Trial1)
so, In both directions reinforcement bar is 16 mm @150mm c/c spacing provided.
- One-way shear check:
force on the shaded portion= factored soil pressure*area of shaded
= 0.195*950*3200
= 5902800 N
Shear stress developed due to this force= 592800/3200*450
Vt = 0.412 N/mm2
check design shear stress in concrete from IS code 456:2000 table-19
% reinforcement=100*Ast/(b*d)
=100*4422/(3200*450)
=0.307%
for this % of Ast, for M25
Tc=0.381 N/mm2
so Tc<Tv
therefore, we need to redesign it.
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so I will increase the reinforcement area value.
Take spacing = 125mm
no of 16mm bar = 3200/125= 26
area provided= 26*201=5226 mm^2 -----------------------------------------------------------(Trial 2)
so, In both directions reinforcement bar is 16 mm @125mm c/c spacing provided.
check design shear stress in concrete from IS code 456:2000 table-19
% reinforcement=100*5226/(b*d)
=100*5226/(3200*450)
=0.363%
for this % of Ast, for M25
Tc =0.418 N/mm2
so Tv<Tc
Hence it is safe in one shear.
- Check for two-way shear
force on the shaded portion= factored soil pressure*area of shaded
= 0.195*(3200^2-850^2)
= 1855912 N
Shear stress developed due to this force= 1855912/3400*450
Vt = 1.213 N/mm2
according to IS code 456-2000
Tc=0.25*(25)^1/2
=1.25 N/mm2
Since Tc> Tv
therefore, It is safe in two-way shear.
Result:
