Design of Shallow Foundation (Isolated Footings)

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:

• If the gross bearing pressure under the footing is within safe bearing capacity
• Provide sufficient depth and longitudinal steel for resisting bending and shear
• Sufficient anchorage for column steel into the footing
• If the bearing pressure at column-footing joint is within 0.45 times the maximum concrete compressive strength

Provide a simple sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions and anchorage of column steel.

Please do go through the solved examples in the document provided with the challenge.

ans.

given data.

column size= 400X400

load=2000kN

safe bearing capacity =150KN/m2

depth =2m

grade of concrete grade=m25

Steps involved in footing Design.

1. Calculated of gross soil pressure for footing area
2. calculation of net soil pressure for flexure and shear design
3. initial depth of footing calculation.
4. flexural reinforcement calculation
5. one way shear check
6. two way shear check

Step 01.

gross soil pressure & footing area

• Unfactored load with surcharge + self weight of footing /allowable soil pressure

Note. the load which we have given is factored load which means it with 1.5Dl+1.5LL and accouring to IS Code we need unfactored load for calculating gross soil pressure

so 2000\1.5=1333.33

self weight of footing is consider 10 persent of the total column load

2000X10/100=200kN

area of the footing = unfactored load +self weight of the footing / allowable soil pressure

so 1333.33+200/150 = 10.22

assume we have square footing √10.22 =3.12

consider 3.2 m sided footing

Step.02

Net soil pressure & initial footing depth

Note (for net soil pressure now we use factored load )

factored load /actual area of footing

factored load =2000

actual area of footing =3.2X 3.2=10.24

so 2000/10.24 =195.3KN/m2

assuming depth of the footing is 500mm

Step.03

check for bearing stress as per is -456-2000-34-4

√area on the base of footing /loaded area at column base (which is column size ) is not greater then 2

in over case size of the column is 400

depth we assume is 500mm so accourding to is 456-2000-34.4 base bearing size is 1:2 which means if depth is 500 then base is double the depth 1000mm

so total base area is 400+1000+1000=2400mm

permissible bearing stress accouring to IS- 456-2000

for working stress method 0.25fck

for limit state method 0.45fck

we take 0.45fck so for M25

0.45X25=11.25

now we multiply it with area factor which is also given in code =2

11.25X2=22.5N/mm2

actual bearing stress on column base is

2000X1000/400X400=12.5N/mm2

actual stress is less then permissible stress hence ok

if we assume 50mm cover for footing then effective depth of footing is 450mm

Step.04

flexural Reinforcment

convert net pressure into udl

195.3X 3.2 =625KN/M

bending moment due to udl over a cantiliver is

WL2/2 625X1.6X1.6/2

M=800KN/M

Rainforcment calculation as per IS-456-2000 is

Mu=0.87fy Ast d(1-AstXfy/bdXfck)

assume fy is 500

area of steel required is = 4700mm2

Step 05. one way shear check

shaded area is find out =total lenth /2-half of column size- effective depth

shaded area is =3200/2-200-450

=950

shear force in one way sheaded area =0.95X3.2X195.3=593.7KN

Design shear stress at one way cut face is

shear force /bd

593.7X1000/(3200X450)=0.412N/mm2

shear stress capasity as per IS 456-2000

0.412N/mm2

design shear stress is less then shear capacity hence ok

Step 05.

two way shear check

shear force in the two way shaded area

Shear force in the way shaded area

=3.2X3.2-o.85X0.85X195.3=1859KN

Design Shear Stress at Two Way cut face

1859X1000/(3400X450)=1.25N/mm2

Shear stress Capacity of Section as per Is-456-2000

0.25X√fck=1.25N/mm2

design shear stress is less then shear capacity hence ok.