Challenge

Design of Shallow Foundation (Isolated Footings)

Aim - From the given data calculate the soil pressure for flexure and shear design,depth of footing, reinforcement details and chesk one way amd two way shear

Procedure-

Gross soil pressure and footing area-

Unfactored load with surcharge + self weight / Allowable soil pressure

2000/1.5(factored load)x1.1/150               (1.1=surcharge + self weight)

9.78

Let us assume to take square footing

`sqrt9.78`= 3.12m on either side of footing

Consider it as 3.2m side isolated footing

Net soil pressure and initial footing depth -

Factored load/actual area of footing

2000/3.2x3.2

Soil pressure Pn = 195 Kn/M

Let us assume depth as 500mm and check bearing stress

`sqrtbase shear/loaded area`< 2 

But we got 5.9

Permissible bearing stress= 0.45xFck = 11.25 N/mm2             (Fck= 25)

Actual bearing stress at coloumn base

(2000x1000)/(400x400) = 12.5N/mm2

Here actual is less than the permissible so OK

It is satisfied so we can go to further design checks for 500mm depth

Effective depth = 450mm        (overall depth - cover)

Flexural reinforcement-

Convert the net pressure to UDL

195.3x3.2 = 625Kn/m

BM due to UDL Wl2/2

625x1.6x1.6/2                                          (1.6= span length 

M= 800KnM

Reinforcement area

Mu = 0.87Fy  Ast  d(1-AstFy/bdFck)

We have moment , steel grade as 500 , d= 500 , b =3.2 , and concrete grade as 25

From this we get Ast as 4350mm2

If we use 16mm bars of dia of spacing 125

Its area is 5146 excees for what we got but OK

One way shear check

Cracks develops form one side of the footing because of axial

Shear force in one way

Area x net soil pressure

0.95x3.2x195.3

= 593.7Kn

Design shear

Shear force/face area

593.7x1000/3200x450

=0.412 N/mm2

Shear stress capacity of section from table 19 in IS456

=0.425

Design stress is less than capacity then OK

Two way shear

It is cracks occure around the coloumn

Shear force around 

(3.2X3.2)-(0.85X0.85)X195.3

(Overall area- central area x pressure)

= 1859Kn

Design shear

1859x1000/(3400x450)

= 1.215N/mm2

Shear capacity of section

0.25 `sqrtFck` 

= 0.25 `sqrt25`  = 1.25N/mm2

Design stress is less than capacity Hence OK

We are not doing stepped footing because we have considered d as 500 for step footing it will increase

 

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