Design of Shallow Foundation (Isolated Footings)

Design of Shallow Foundation (Isolated Footings)

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.

Aim:

To design the square footing for use by the given data and also shown in the simple sketch, the size of the footing.

Steps:

1. Calculation of footing area
2. Calculation of gross soil pressure for footing area and net soil pressure for flexure and shear design
3. Intial dept of footing calcluation
4. One way shear check
5. Two way shear check

Given data:

• Column Size- 400x400 mm
• Soli Bearing Capacity- 150KN/m2
• Comprission axial load- 2000KN
• Grade of concrete- M25

Solution:

Step 1: Size of footing

Assume self wight of footing equal to 10% of imposed load

`underset (w) W= 10%xx2000`

= 0.1x2000

= 200 KN

Total load, W= `underset (L) W+ underset (w) W`

= 2000+200

= 2200 KN

Therefore area of footing= W/SBC

= 2200/150 =14.66 `m^2`

B=`sqrt A`

= `sqrt 14.66`

= 3.82

Hence provide the size of footing is 4mx4m

The area of footing is 16 `m^2`

Step 2: Net upward pressure (`underset (0) rho`)

`underset (0) rho=`wx1.5/size of footing

= 2200x1.5/4x4

`underset (0) rho =`206.25 KN/`M^2`

Step 3: Depth of footing on the basis of BM

The maximum BM at the face of the coulumn and is given by,

`M= underset (0) rho B/8(B-b)^2`

`= 206.25xx4/8(4-0.4)^2`

= 1336.5 KN/M

Therefore ultimate moment (Mu) =1.5XM

= 1.5X1336.5

= 2004.75 KN/M

Depth of footing:

Mu= 0.138 Fck `Bd^2`

2004.75=0.138x25x4000x `d^2`

`d^2`= 2004.75x `10^6`/0.138x25x4000

`d^2`= 145271.73

d= `sqrt145271.73`

d= 381.14m

By use the thumb rule take the 2.5 times higher value of d for check shear

d=381.14x2.5

d=952.85 m

D= d+50

= 952.85+50

=1003 m

Step 4:Area of Reinforcement:

`Ast= 0.5(fck)/(fy) (1-sqrt(1-(4.6Mu)/(fck B d^2) )) Bd`

= `(0.5xx25)/(415) (1-sqrt(1-(4.6xx2004.75xx10^6)/(25xx4000xx953^2) ))xx4000xx953`

Ast= 5985.3 `mm^2`

Therefore % of reinforcement

P= Ast/ Bd x100

= (5985.3/4000x953) x 100

P= 0.157%

No. of reinforcement required for footing,

Assume 20mm dia of bars in Ast= 5985.3 `mm^2`

`pi/4xx20^2=314.5`

No. of bars= 5985.3/314.5 = 20

Required no. of reinforcement is 20.

Spacing between the bars = B/ No. of bars

= 4000/20

= 200 mm @ c/c

Step 5: Check for One way Shear

Shear force (v)= `underset (0) rho B {1/2(B-b)-d}`

= 206.25x4{1/2(4-0.4)-0.952}

= 699.6 KN

Ultimate Shear force (Vu)= 1.5(v)

= 1.5x699.9

= 1049.4 KN

Nominal Shear Stree `(Tv)=(Vu)/(Bd)`

= 1049.4x`10^3`/4000x952

= 0.27N/`mm^2`

Shear Strength of concrete (Tc)'

(Tc)'= Tc k ( As per IS456-2000, Page.No. 72)

K=1, depth>mm

Tc for 0.157% of P is 0.28 N/`mm^2`

Tc=0.28 N/`mm^2`

Therefore, Tv<Tc

Hence Ok.

Now, Tv<Tc'

Vu/Bd=K.Tc

d=Vu/B.K.Tc

d=1049.4x`10^3`/4000x1x0.28

d=937mm

d provided=953mm>937mm, Hence ok.

Step 6: Check for Two way Shear

Width=`underset(0)b=b+d`

=400+953

=1353 mm

Shear force acting on a perimeter,

`F= underset (0) rho[B^2-underset(0)b^2]`

`=underset(0)rho[B^2-(b+b)^2]`

=206.25[4-`(1.35)^2`]

=449.10KN

Therefore, Ultimate Shear force, Fu=1.5(F)

=1.5x449.10

Fu=673.65KN

from Page.No.57, Clause 31.6.2 (IS456-2000)

Tv=Fu/4`underset(0)b`d

=673.65x`10^3`/4x1353x953

=0.23N/`mm^2`

Therefore, Tc=0.25`sqrtfck`

=`0.25 sqrt25`

= 1.25

Ks Tc=1x1.25

=1.25

Tv<Tc

Hence safe in Two way shear

Result:

The desing and size of the footing is designed successfully by using the given details.