Design of Shallow Foundation (Isolated Footings)
AIM:- To design a square footing for a column size of 400*400 using manually. To do the following checks for gross bearing pressure, provide sufficient depth and longitudinal steel, anchorage, and the bearing pressure at the column-footing joint as per IS codes. Also to provide as a simple sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions, and anchorage of steel.
INTRODUCTION:- All the steps are going to be mentioned along with step by step procedure.
PROCEDURE:-
a) Given data:- Axial load, P=2000 KN
Safe bearing capacity of soil = 150 KN/m^2
Size of column = 400 mm * 400 mm
M-25 grade of concrete and assume Fe 500 HYSD bars
b) Size of footing:- Load on column = 2000 KN
Weight of footing and backfill at 10% = (10/100)*2000 = 200 KN
Total load = 2000+200 =2200 KN
Area of footing = 2200/150 = 14.66 mm^2
Size of footing = L =B = √(14.66) = 3.82 m
Adopt 4 m * 4 m size by square footing
Net soil pressure, qu = 2000/(4*4) = 125 KN/m^2 = 0.125 N/mm^2
c) Critical section is at a distance d' from the column face
Factored shear force, Vu1 = 0.125*2400*(4000/2-400/2-d) = 300*(1800-d)
Assume percentage of steel, pt = 0.25% for M25 grade of concrete (Table 19 IS 456-2000) we get τc = 0.36 N/mm^2
One way shear resistance Vc1 = 0.36*4000*d = 1440d
Vu1 less than or equal to Vc1
300 * (1800-d) less than or equal to 1440d
After simplification, we get d greater than or equal to 310 mm
Thickness of footing = 310+75+(12/2) = 391 mm
Provide D = 400 mm and d = 400 - (75-(12/2)) = 331 mm
d) Two way shear:- Assuming the effective depth of slab = 331 mm
Vu2 = 0.125*(4000^2- (400+d)^2) = 0.125*(4000^2-(400+331)^2) = 1933204.875 N
Two way shear resistance Vc2 is computed as Vc2 = Ks*τc * (4*(400+d)*d) where Ks = 1 , τc = 0.25 * √(fck) = 0.25*√(25) = 1.25 N/mm^2
Vc2 = 1*1.25*(4*(400+d)*d) = 1.25d^2+2000*d
Vu2 less than or equal to Vc2
or 1933204.875 less than or equal to 1.25d^2+2000*d
After simplification, we get d = 678.7 mm
Adopt effective depth, d = 331 mm and overall depth, D = 400 mm
e) Design of reinforcement:- The critical section for a moment is halfway between the center line and the edge of the masonry wall ( Clause 34.2.3.2 (b) of IS 456-2000)
d = 4000/2 - 300/4 = 1925 mm
Consider 1 m strip of footing
Mu = 125*1*1.925/2 = 120.31 KN-m
Now Mu/b*d^2 = (120.31*10^6/1000*331^2) = 1.098
Hence percentage of steel, pt = 0.267 (SP-16 Table 3)
Required Ast = pt*b*d/100 = 0.267*1000*331/100 = 883.7 mm^2 per metre length.
Spacing of 12 mm dia bars = 1000*113/883.7 = 127.87 < 300 mm
Provide 12 dia bars at 120 mm c/c
f) Check for development length:- For M25 concrete and Fe 500 HYSD bars, the required development length for 12 mm dia bars from SP-16 Table 65
Ld = 484 mm
Length available = 1600 -75 = 1525 mm > 484 mm which is safe
g) Determine the distribution bars:- Minimum reiforcement along th elength of the footing, As = 0.12% of BD = (0.12/100 )* 400 * 400 = 1920 mm^2
Spacing of 10 mm bars = 4000*78.5/1920 = 163.54
Provide 10 mm dia bars at 150 mm c/c
h) Check for transfer of force at the base of the wall:-
1) At column face:- A1=A2 = 400*400, fck = 25 N/mm^2
√(A1/A2)= √(400/400) = 1, fbr = 0.45 * 25 * 1 = 11.25 N/mm^2
2) At footing face, A1 = 4000 mm^2, A2 = 400 mm^2
√(A1/A2) = √(4000/400) = 3.16>2
Hence adopt a value 2
fbr = 0.45*25*2 = 22.5 N/mm^2
Critical face is the column face
Therefore limiting bearing resistance = 11.25*400*400*10^-3 = 1800 KN < 2000 KN. Hence safe
i) Reinforcement Details:-
RESULTS:- All the results have been illustrated properly which is given below:-
a) Design a square footing for a column size 400 mm * 400 mm manually
b) Checks for bearing pressure
c) Depth and longitudinal steel has been provided
d) Providing bearing and anchorage at the column-footing joint.
e) Reinforcement details