Design of Shallow Foundation (Isolated Footings)

• AIM: To design a square footing for a clumn size for 400 X 400.

• PROCEDDURE:

1. Design of square footing
2. Net soil Upword Pressure
3. Depth of footing
4. Gross bearing capacity
5. Reinforcement
6. Check For one way shear
7. Check for two way shear

• Design Of square Footing

• Column size : 400 X 400 mm
• Soil bearing capacity : 150 KN/M^2
• Compression axial load:2000KN
• Compressive Strength: M25

• Determine the size of the footing

• Assume self weight of footing equal to 10% of superimposed load

• Wd = 10% x 2000 = 0.1X 2000 KN

= 200KN

• Total Wd = 2200KN
• Area of Footing = W /SBC
• =14.66 MM^2
• Now , take (B) =`root`A
• Now, we provide the size of square footing 4m X 4m

Area of footing = 16m^2

• Net Upword Pressure

Po = 1.5XW /Size of Footing

= 1.5 X 2200 /4 X 4

= 206.25 KN/M^2

• Depth of the Footing on the basis of BM

At critical section of the footing face of the column from the edge of the footing

(4000/2) -(400/2) = 1800 mm

Load on Critical Solution = 206.25 X 4

Load = 825 KN/M

Bending Moment = Wl^2/2

= 825 X 1.8^2/2

= 1336.5 KN-M

Mu =0.138Fckbd^2

1336.5 X 106 =0.138 X 25 X4000X D^2

d = 311.20mm

D = d + Effective cover = Bar diam + Hall of bar dia

= 778 + 50 + 20 +10

= 860 mm

• Area of Reinforcement

Bending Moment = !336.5 X 106

Percentage of reinforcement,Pt= ( 50 Fck/Fy)(1- root 4.6 X 1335.6*106/25 X4000 X7782)

Pt = 0.1570%

Area of steel (As) = (Pt/100)Xb xd

=(0.157/100) X 4000 X778

=4885 mm^2

Take Number of Bars = 20mm

Number of bar = 4885/314.5

=16 nos

Provide 16 no. of 20mm bar

Ast(provided) = 16 X 314.5

=5032 mm^2

Ast> Ast provided

4885 > 5032 Hence OK

For Is 456 -Table 15 for fe415 the clear spacing between the bars should be less than 180mm

C/C spacing = (4000-50-50-8-8)(No of bar-1)

= 168.89

Clear spacing = 168.88 -8-8

= 152.86 mm

So 152.86< 180

Hence OK

• Check For one way shear

One way shear check at the critical section for one way shear at the distance d from face at the column

Area of highjlighted Section = (4000/2)-(400/2)-778

=1.022

Now, Vu = Upword Prssure X Area of highlighted Section

= 206.25 X 4X 1.022

=843.15KN

Nominal Stress

tv = Vu/bd

= 843.15 X10^3/4000 X778

= 0.27N/m

Now compare tv amd tc

Pt = (100Ast/bd)

Ast = 24 No of 16 mm dia =0.1549

From IS 456,table 19

Therefore tv<tc

0.15<0.28

Hence Ok

• Check For two way shear

Area of highlighted section = 400+(778/2)+(778/2)

=1178mm^2

THerefore , area = 42-1.178

=14.6133

Vu = Upword Pressure X Area of highlighted section

Vu = 206.25 X 14.6123

Vu = 3013.79KN

tv =Vu/bd

Peripheral length = 4 X1178

b = 4712 mm

tv = (3013.79X 103)/(4712 X778)

tv = 0.8221 N/mm^2

Compare tv and tc

For two way shear (tc) = Ks X tc

= 1X1.25

=1.25 N/mm^2

Tv<Tc

0.8821 <1.25 Hence Ok

• Reinforcement Drawing

• RESULT : From above details square footig has been designed succesfully