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:-
Aim:- To design the foundation followed by given details by manual method.
Procedure:-
1) Gross soil pressure & footing area :-
Unconfined load with shrinkage + self wt. of footing /allowable soil pressure
=( (2000/1.5)*1.1)/150 = 9.78sqm
0.1 is the 10% of surcharge load assumed.
sqrt9.78 = 3.12m side
so we can consider 3.2m sided isolated footing.
2) Net soil pressure & initial footing depth :-
factored load / actual area of footing
2000/(3.2*3.2)
Pn= 195.3 kn/sqm
Assume depth of footing is 500mm & checking it for bering stress as per IS:456-2000 clause 34.4
Sqrt (footing base beaaring area (loaded area) leaa than or equal to 2
permissible bearing stress 0.45fck = 11.25 N/sqmm
permissible bearing stress multiplied with area
factor 11.25*2 = 22.5N/sqmm
12.5 N/sqmm actual stress is less than permissile 22.5 N/sqmm hence it is okay.
We can proceed with 500mm depth footing for further design checks.
if we assume 50mm cover for footing , the effective depth (d) shall be taken as 450mm
3&4 Flexural reinforcement:-
Conversion of net pressure into UDL
=195.3*3.2=625kn/m
bending moment due to UDL over a cantilever section
M=Wl^2/2 = 625*1.6*1.6/2
M=800KNM.
reiforcement calculation as per IS:456-2000Annex-G
Mu=0.87fyAstd(1-Astfy/bdfck)
Assume fy=500n/mm^2,In result Ast Required = 4350mm^2
If we provide 16m dia rebar at 125mm spacing,the actual rebar provided
3200/125*201
Ast provided = 5146mm^2
5) One way shear check:-
Shear force in part area
0.95*3.2*195.3 = 593.7KN
Design shear stressa at one way cut face
Tv = 593.7*1000/(3200*450)
=0.412N/mm^2
Shear stress capacity of section as per IS:456-2000 table-19 (include rebar)
=0.45N/mm^2
Design shear stress is less than shear capacity hence ok
Note:- if the stress not meet its requirement then increase the depth of the footing and do the design steps from first.
6) Two way shear chek:-
Shear force in two way shear from d
(3.2*3.2)-(0.85*0.85)*195.3 = 1859KN
Design shear stress at two way out face
Tv =1859*1000/(3400*450)=1.215N/mm^2
Shear stress capacity of section as per IS:456-2000 clause 31.6.3.1
0.25 sqrtfck = 1.25 N/mm^2
So the shear stress is less than shear capacity hence t is okay.
Diagram:-
Result:- Finally i got to know how to design a shallow foundation in a manual way with the use of IS:456-2000 codal provision.