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.
AIM :
To design a square footing for a column size of 400x400 .
PROCEDURE :
Design of square footing :
Given:
* Column size = 400*400 mm
* Soil bearing capacity SBC = 150 KN/mm2
* Compression axial load = 2000KN
* Compressive strength fck = M25
Steps involved in design of square footing :
* Determine the size of footing
* Upward pressure P0
* Depth of footing on the basis of BM
* Area of R/F
* Check for one way shear
* Check for two way shear
* Developmnet length
* R/F detailed
Step 1 : Determine the size of footing
- Assume self weight of footing equal to 10% of superimposed load
Wd = 10% * 2000 = 0.1 * 2000 = 200 KN
Now,
- Total load W = WL + WD = 2000 + 200 = 2200 KN
Then,
- Area of footing (A) = W/SBC = 2200/150 = 14.66 m2
- let us take B = √A = √14.66 = 3.82 m
- Hence, we provide the size of square footing is 4M * 4M
- The area of footing is 16 m2
Step 2 : Net upward pressure
P0 = 1.5 * W / size of footing
= 1.5 * 2200 / 4*4
= 206.25 KN/m2
Step 3: Depth of the footing on the basis of BM
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 = σ x 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
Mu = 0.138 * fck * bd2
= 0.138 x 25 x 4000 x d2
1336.5*106 = 0.138 x 25 x 4000 x d2
d = 311.20 mm
By the thumb rule, we can take 2 to 2.5 times higher value of d for shear check
∴ d= 778 mm
D = d + effective cover = bar dia + half of bar dia
= 778 + 50+ 20 +10 ( Assume bar dia as 20mm)
D = 860 mm
Step 4: Area of R/F
Bending moment = 1336.5 x 106
Percentage of reinforcement, Pt= (50fck/fy)(1-√4.6 x 13336.5*106/25x4000x7782)
Therefore, Pt = 0.1570 %
Area of steel = (Pt/100) x bd
= (0.157/100) x 4000 x 778
Ast= 4885 mm2
No of bars:
Take 20mm bar
No of bar = 4885/314.5
= 15.55 = 16 nos
Provide 16 nos of 20mm bar
Astprovide = 16 x 314.5
= 5032 `mm^2
∴Ast > Astprovided
4885 > 5032
Hence ok.
Check for clear spacing between bars:
As per IS 456 - table 15, for fefe415, 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 = 242.5-10-10= 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 spacing = (4000-50-50-8-8)/(No.of bars-1) = 168.89 mm
Clear spacing = 168.86-8-8 = 152.86mm
So, Clear spacing 152.86 < 180>
Hence Ok
Step 5: Check for one way 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 x 4 x 1.022 = 843.15 kN
Nominal Stress:
τv = Vu/bd
= (843.15 x 103)/(4000 x 778) = 0.27 N/m
Now compare τv and τc
For the τc value given by IS 456, Table 19
Ptprovided = (100 Ast/bd)
Ast = 24 nos of 16mm dia = 4823 mm2
Pt= (100x 4823)/(4000x778) = 0.1549
From IS 456, table 19
Therefore, τv< τc
= 0.15<0>
Hence Ok
Step 6: 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 = 42-1.1782
= 14.6123 mm2
Vu= Upward pressure x Area of highlighted section
= 206.25 x 14.6123
Vu = 3013.79 kN
τv= Vu/bd
Peripheral length = 4 x 1178
b= 4712 mm
τv= (3013.79 x 103)/(4712 x 778)
τv = 0.8221 N/mm2mm2
Compare τv and τc,
For two way shear τc = Ksx τc
For τc value taken from IS 456-2000 page number 58 and 58
Ks= 1 (for square column)
τc= 0.25o√fck0.25fck)
= 1.25 N/mm2
Therefore, τc= Ks x τc = 1 x 1.25= 1.25 N/mm2mm2
τv < τc
0.8221 < 1>
Hence Ok
Reinforcement detail drawing:
RESULT:-
Using the given details the square sloped footing has been designed and check successfully.