Design of Shallow Foundation (Isolated Footings)
Ans. Step involved in Footing design:
1. Calculation of gross soil pressure of footing area.
2. Calculation of net soil pressure for flexure and shear design .
3. Initial depth of footing calculation.
4. Flextural reinforcement calculation .
5. one way shear check
6. Two way shear check.
Gross soil pressure & footing area:
. Unifactored load with surcharge + self weight of footing / allowable soil pressure
2000 factored load Given.
So i have divide by the 1.5 for the unfactored load.
I dont have any soil surcharge
Assumining 10% acting load will be due surcharge
So,
`=(2000/1.5*1.1)/150=9.78m^2`
Assumining we have squre footing
`sqrt9.78=3.12m` side footing
consider, 3.2 m sided isolated footing
. Net soil pressure & initial footing depth:
. Factored load/actual area of footing
2000/(3.2x3.2)
`P_n=195.3(KN)/m^2`
.Assuming depth of footing 500mm and checking it for bearing stress as per IS 456-2000 clause 34.4
`sqrt`(footing base bearing area/loaded area)`le`2
Net soil pressure & initial footing depth
. Permossible bearing stress `0.45*F_(ck)=11.25N/(mm)^2`
Permissible bearing multiply with area factor =2
So,
11.25x2=22.5 N/mm^2
actual bearing stress at column base
(2000x1000)/400x400=12.5 N/mm^2
12.5 N/mm^2 actual stress is less then permissible 22.5 N/mm^2
Hence okey
we can priceed with 500 mm depth
footing for further design checked
If we assume 50mm cover for footing the effective depth (d) shall be taken as 450mm.
Flexuralo reinforcement
. Conversion of net pressure into uniformly distributed load
Pnx3.2
W= 195.3x 3.2=625 KN/m
. Bending moment due to UDL over a cantilever section
`=(WL^2)/2` L=1.6m
=(625x1.6x1.6)/2
M=800KNm
Reinforcement calculation as per IS 456-2000 Annex-G
`M_u=0.87f_yA_(st)d(1-(A_(st)f_y)/(bdf_(ck)))`
Assume `f_y`=500 N/mm^2 in result `A_(st)`required=4350m^2
if we provide 16mm dia rebar at 125mm spacing the actual rebar provided
=3200/125 X 201
`A_(st)=5146mm^2`
one way shear stress check:
. shear force in the one way shear area =0.95x3.2x195.3=593.7KN
. Design shear syress at one way cut face `tau_v`=593x1000/(3200x450)=0.412N/mm^2
.shear stress capacity of the section as per IS 456-2000 table -19 (include rebar)
=0.425 N/mm^2
design shear stress is less then shear capacity
Hence okey
Two way shear stress check:
Shear force in two way saded area = (3.2x3.2)-(0.85x0.85)x195.3=1859KN
shear5 stress capacity of section as per IS 456-2000 clouse 31.6.3.1
=0.25 `sqrtf_(ck)=1.25N/(mm)^2`
.design shear stress at two way `tau_v`=1859x1000/(3400x3400)=1.215N/mm^2
.design shear stress is less then shear capacity
Hence okey
