Design of Shallow Foundation (Isolated Footings)

Objective of challenge:

a) To know the general design procedure of spread footing with hand calculation.

b) To practice how the inputs can be collected and arrived as data for calculation.

c) To understand the concept of bearing capacity of soil, net bearing pressure and safe bearing capacity.

Given parameters:

Column Size-(400mmx400mm)

Axial Load=2000KN

Load Combination= 1.5DL+1.5LL

Safe bearing capacity=150KN/msq

Depth= 2mtr

Grade of concrete=M25

Steps to be carried out:

a) Calculation of gross soil pressure for footing area.

b) Calculation of net soil pressure for flexure & Shear design.

c) Initial depth of footing.

d) Flexural Reinforcement.

e) Check for one way shear and two way shear.

Calculation: Gross Soil Pressure

Unfactored Load,soil surcharge + self wt. of footing/allowable soil pressure

=(2000/1.5x1.1)/150=9.78 sq.mt. ( Considering 10% soil surcharge)

=√9.78=3.12 mtr.

Net Soil Pressure:

Factored Load/ Actual area

=(2000/3.2x3.2)=195.3KN/sq.mtr

Pn=195.3kn/sq.mtr

Assuming depth of footing as 500mm and checking the bearing stress as per IS 456-2000 clause 34.4

√(footing base area/loaded area)≤2

√(2.4x2.4)/(0.4x0.4)=2.4

So we will not use it since it is greater than 2.

Permissible bearing stress= 0.45 fck=0.45x25=11.25N/sqmm

Permissible bearing stress with factored area=11.25x2=22.5N/sqmm

Actual Bearing stress at column base=(2000x1000)/(400x400)=12.5N/sqmm

Actual stress is less than permissible stress. Hence Ok.

Considering 50mm cover Effective depth will be 450mm.

Flexural Reinforcement:

UDL= Net pressurexWidth of footing=195.3x3.2=625KN-mtr.

BM= WLXL/2=625x1.6x1.6/2=800KN-mtr

Reinforcement Calculation as per IS 456-2000

Mu=0.87fyAstd(1-Astfy/bfck) (Fy=500N/sqmm)

Calculating We get Ast=4350 sqmm.

Thus Placing 16MM bar at spacing of 125c/c. Area of rebar =3200/125x201=5145.6 sqmm. Hence ok.

One Way Shear Check:

Shear Force in one way shear area=0.95x3.2x195.3=593.7KN

Design Shear stress at one way cut face

τv=593.7x1000/(3200x450)=0.412N/sqmm<0.425Nsqmm (As per IS 456-2000 Tab -19)

Design Shear stress is less than shear capacity. Hence ok.

Check for two way shear:

Shear force in two way area=((3.2x3.2)-(0.85x0.85))x195.3=1859KN

Design Shear stress at two way face

τv=1859x1000/(3400x450)=1.29 N/sqmm

As per Is Code 456-2000 Shear stress capacity=0.25√fck=1.25N/sqmm

Hence design shear stress is less than shear capacity. Hence ok.