Design of Shallow Foundation (Isolated Footings)
Answer of the Question
Question:
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.
Answer:
AIM: In This Question We have to 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.
INTRODUCTION:
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
The Following Data Given us in this Question:-
Column size = 400x400mm
SBC = 150 kN/mm^2
Compression axial load = 2000 kN
fck = M25
Steps to design a square footing:
- Size of Footing
- Net upward Pressure
- Bending moment
- Depth of Footing
- Reinforcement
- One way and Two-way shear
PROCEDURE: Compete of the Question We have to follow the following Steps:
STEP-1
Size of Footing:
Load on column = 2000 kN
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 = 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
STEP-4
Depth of footing:
Mu = 0.138 fck.bd^2
= 0.138 x 25 x 4000 x d^2
1336.5 x 106106 = 0.138 x 25 x 4000 x d^2
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-5
Reinforcemet:
Bending moment = 1336.5 x 106106
Percentage of reinforcement, Pt= (50fck/fy)(1-√4.6 x 13336.5x 106106/25x4000x77827782)
Therefore, Pt= 0.1570 %
Area of steel = (Pt/100) x 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 fe 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 spacing = (4000-50-50-8-8)/(No.of bars-1)
= 168.89
∴∴Clear spacing = 168.86-8-8
= 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 x 4 x 1.022
Vu= 843.15 kN
Nominal Stress:
τv= Vu/bd
= (843.15 x103103)/(4000 x 778)
τv= 0.27 N/m
Now compare τv and τc
For the τc value given by IS 456, Table 19
Ptprovided = (100 Ast/bd)
= 100 x Ast/bd
Ast= 24 nos of 16mm dia
= 4823 mm22
∴∴ Pt= (100x 4823)/(4000x778)
= 0.1549
From IS 456, table 19
Therefore, τv< τc
= 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 =42-1.178242-1.1782
= 14.6123mm^2
Vu= Upward pressure x Area of highlighted section
= 206.25 x 14.6123
Vu= 3013.79 kN
Tv= Vu/bd
Peripheral length = 4 x 1178
b= 4712 mm
Tv= (3013.79 x103103)/(4712 x 778)
Tv= 0.8221 N/mm^2
Compare Tv and Tc
For two way shear Tv = Ksx Tc
For Tc value taken from IS 456-2000 page number 58 and 58
Ks= 1 (for square column)
Tc=0.25√fck0.25)
= 1.25 N/mm^2
Therefore,Tc= Ksx Tc
= 1 x 1.25
Tc= 1.25 N/mm^2
∴∴ Tc < Tv
0.8221 < 1.25
Hence Ok
Reinforcement detail drawing:
RESULT:
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