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. 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 :
Footings are structural members used to support columns and walls and transmit their loads to the underlying soils. Reinforced concrete is a material admirably suited for footings and is used as such for both reinforced concrete and structural steel buildings, bridges, towers, and other structures.The permissible pressure on a soil beneath a footing is normally a few tons per square foot. The compressive stresses in the walls and columns of an ordinary structure may run as high as a few hundred tons per square foot. It is therefore necessary to spread these loads over sufficient soil areas to permit the soil to support the loads safely.Not only is it desired to transfer the superstructure loads to the soil beneath in a manner that will prevent excessive or uneven settlements and rotations, but it is also necessary to provide sufficient resistance to sliding and overturning.
Given data:
Size of the Column = 400*400
Compression axial load = 2000 KN
Safe Soil Bearing Capacity (Safe SBC) = 150 KN/m2
Grade of concrete = M25
Procedure:
Step 1: Determination of the size of the footing
Lets assume the Self weight of the designing footing is equal to 10% of the Super imposed Load
Thus, Ww = 10% * 2000 = 200 KN
Therefore, Total load, W = 2000 +200 = 2200 KN
Area of the footing, A = W/ Safe SBC = 2200/ 150 = 14.66 m2
Since a square footing, Breadth of the footing , B = A1/2 = 14.661/2 = 3.82 m
Thus, we provide a square footing of 4m *4m
Thus, Area of the footing = 16 m2
Step 2: Determination of Net upward Pressure
Po = (W*1.5)/ Size of the footing = (2200*1.5)/(4*4) = 206.25 KN/m2
Step 3: Depth of the Footing
Maximum Bending Moment at the face of the column, Mmax = PoB (B-b)2/8 = 206.25 4(4-0.4)2/8 = 1336.5 KN/m
Ultimate Moment(Mu) = 1.5M = 1.5* 1336.5 = 2004.75
We know, Mu = 0.138fck*Bd2
On substituting, 2004.75* 106 = 0.138*25*4000*d2
Thus, d2 = 145271.739
d = 381.14 mm
For shear check, take 2.5 times higher the values of d, thus, d= 381.14 *2.5 = 952.85mm
Assuming a clearence of 50mm
Thus, Depth of the footing, D = d + 50 = 952.85 + 50 = 1002.85mm ~ 1003 mm
Step 4: Determination of Area of Reinforcement
Percentage of reinforcement required = Ast *100 / Bd
On substituting, we get, Percentage of reinforcement required = 0.157%
Now, Assume the bars of 20mm dia, thus, Area of 20mm dia bar = 314.5mm2
Number of bars = Ast /Area of 20mm dia bar = 19.03
Thus we provide 20 number of 20 mm dia bars
Now spacing between the bars = B / number of bars = 4000/20 = 200
Therefor, we provide, 20 number of 20 mm dia bars at 200mm c/c
Step 5: Check for 1 way shear
Shear force, V = POB [(B-b)/2 - d]
On substituting the values, V= 206.25 4[ (4-0.4)/2 - 0.952]
V= 699.6 KN
Ultimate Shear force, Vu = 1.5V =1.5* 699.6 = 1049.4 KN
As per IS 456: 2000,
Nominal Shear Stress, Tv= Vu/ Bd
= 1049.4*103/ (4000*952)
= 0.27 N/mm2
Design Shear strength of Concrete, TC’ = TC*K
But we take K=1, Thus , TC’ = TC
TC for 0.157% of reinforcement is 0.28 N/mm2
Since TC > Tv, Hence OK
From the equation, VU/Bd =K*TC
We get, d= Vu /(B*K*TC)
= (1049.4* 103) / (4000*1*0.28) = 937mm
Thus, dprovided > d, Hence OK
Step 6: Check for 2 way shear
Width,bo = b+d = 400 + 953 = 1353mm
Net Shear Force acting on the perimeter, F = Po (B2-bo2)
= 206.25 ( 42 -1.3532 )
= 449.10 KN
Ultimate Shear Force, Fu = 1.5F
=1.5* 449.10
=673.65 KN
From IS 456:2000, we know,
Tv = Fu / (4bo*d)
= (673.65 *103)/(4*1353*953)
= 0.23 N/mm2
Tc = 0.25 fck1/2
= 0.25*251/2
= 1.25
Tc >Tv , Hence OK