Design of Shallow Foundation (Isolated Footings)
Given:
Column size = 400 * 400mm
SBC = 150kN / m * m ^ 2
Compression axial load = 2000 kN
fck = M25
Steps to design a square footing:
1) Size of Footing
2) Net upward Pressure
3) Bending moment
4) Depth of Footing
5) Reinforcement
6) One way and Two-way shear
Step 1:
Size of Footing:
Load on column = 2000kN
Add 10% of column axial load for footing = 10% (2000)
= 2000-200
= 2200 KN
Area of Footing:
Area of Footing = Load on footing/Safe bearing capacity
= 2200/150
= 14.66 m^2
√14.66=3.82
= 4m x 4m
Therefore, provided area = 16m^2
Step 2:
Net upward pressure:
σ = Force/Area
=1.5 x Total load on footing/Area
=1.5 x 2200/4 x 4
σ = 206.25 kN/m^2
Step 3:
Bending Moment:
At the critical section of the footing face of the column from the edge of footing
(4000/2)-(400/2) = 1800mm
Therefore, load on critical section = σx breadth of footing
= 206.25 x 4
Load = 825 kN/m
So, the bending moment = wl^2/2
= (825 x 1.8^2)/2
= 1336.5 kN-m
Step 4:
Depth of footing:
Mu = 0.138fckbd^2
= 0.138x25x4000xd^2
1336.5*10^6 = 0.138*25*4000*d^2
d = 311.20mm
By the thumb rule, we can take 2 to 2.5 times higher value of d for shear check
d = 778mm
D = d+ effective cover = bar dia + half of bar dia
= 778 + 50+20 +10 (Assume bar dia as 20mm)
D = 860mm
Step 5:
Reinforcemet.
Bending moment = 1336.5*10^6
Percentage of reinforcement, Pt= (50 ƒck/fy)(1-√4.6x13336.5*10^6/25x4000x778^2)
Therefore, Pt = 0.1570 %
Area of steel = (Pt/100)*bd
= (0.157/100) x 4000 x 778
Ast= 4885 mm^2
No of bars:
Take 20mm bar
No of bar = 4885/314.5
= 15.55 = 16 nos
Provide 16 nos of 20mm bar
Ast provide = 16 x 314.5
= 5032 mm^2
Ast > Ast provided
4885 > 5032
Hence ok.
Step 6:
Check for clear spacing between bars:
As per IS 456-table 15, for fc 415, the clear spacing between the bars should be less than 180mm
c/c spacing= (4000-50-50-10-10)/(no of bars-1)
= 258.66 mm
clear spacing = 242.5- half of dia - half of dia bar
=242.5-10-10
clear spacing= 238 mm
As per IS code value, our value is greater
so obtained clear spacing is greater than IS code value.
In order, the minimize the value reduce the bar size.
Take 16mm bar = 4885/200.96
No. of bar = 24.30
= 24
C/C sin g = (4000 - 50 - 50 - 8 - 8) /(No.of bars-1)
=168.89
Clear spacing = 168.86 - 8
= 152.86mm
So, Clear spacing 152.86 < 180
Hence Ok
Step 7:
Check for Oneway Shear:
One way shear check at the critical section for one way shear at the distance, d from face of the column
Vu=Upward pressure x Area of the highlighted section
= 206.25 x 4 x?
Area of highlighted section = (4000/2) - (400/2) - 778
= 1.022
Now, Vu = 206.25 * 4 * 1.022
Vu = 843.15kN
Nominal Stress:
Tv = Vu/bd
= (843.15 x 10³)/(4000 x 778)
Tv = 0.27N/m
Now compare Tv and Tc
For the value given by IS 456, Table 19
Pt provided = (100 Ast/bd)
= 100 x Ast/bd
Ast = 24 nos of 16mm dia
= 4823 mm^2
Pt = (100x 4823)/(4000x778)
= 0.1549
From IS 456, table 19
100Ast/bdvalueofgreaterthan or equal 0.15is0.28N/mm^2
Therefore, Tv<Tc
= 0.15<0.28
Hence Ok
Step 8:
Check for Two way shear:
At peripheral d/2 distance from face of the column
Area of highlighted section = 400+(778/2)+(778/2)
= 1178 mm^2
Therefore, area = 4^2 - 1.178^2
= 14.6123 mm^2
Vu = Upward pressure x Area of highlighted section
= 206.25 x 14.6123
Vu = 3013.79 kN
Tv= Vu/bd
Peripheral length = 4 * 1178
b = 4712mm
tv = (3013.79*10^3) / (4712 * 778)
tv= 0.8221N /mm^2
Compare Tv and Tc.
For two way shear Tc = Ks*Tc
For Tc value taken from IS 456-2000 page number 58 and 58
Ks = 1 (for square column)
Tc =0.25√fck
= 1.25 N/mm²
Therefore, Tc = Ks*Tc
= 1x 1.25
Tc = 1.25N /mm^2
0.8221<1.25
Hence Ok
Reinforcement detail drawing
