Design of Shallow Foundation (Isolated Footings)

Aim: To design a square footing for a column size of 400x400

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, W_w = 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 = A^1/2 = 14.66^1/2 = 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

• P_o = (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, M_max = P_oB (B-b)²/8 = 206.25 *4*(4-0.4)²/8 = 1336.5 KN/m

• Ultimate Moment(M_u) = 1.5M = 1.5* 1336.5 = 2004.75

• We know, M_u = 0.138f_ck*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 = A_st *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 = A_st /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 = P_OB [(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, V_u = 1.5V =1.5* 699.6 = 1049.4 KN

• As per IS 456: 2000,

• Nominal Shear Stress, T_v= V_u/ Bd

  = 1049.4*10³/ (4000*952)

  = 0.27 N/mm²

• Design Shear strength of Concrete, T_C^’ = T_C*K

  But we take K=1, Thus , T_C^’ = T_C

• T_C for 0.157% of reinforcement is 0.28 N/mm²

• Since T_C > T_v, Hence OK

• From the equation, V_U/Bd =K*T_C

  We get, d= V_u /(B*K*T_C)

  = (1049.4* 10³) / (4000*1*0.28) = 937mm

• Thus, d_provided > d, Hence OK

Step 6: Check for 2 way shear

• Width,b_o = b+d = 400 + 953 = 1353mm

• Net Shear Force acting on the perimeter, F = P_o (B²-b_o²)

  = 206.25 ( 4² -1.353² )

  = 449.10 KN

• Ultimate Shear Force, F_u = 1.5F

  =1.5* 449.10

  =673.65 KN

• From IS 456:2000, we know,

• T_v = F_u / (4b_o*d)

  = (673.65 *10³)/(4*1353*953)

  = 0.23 N/mm²

• T_c = 0.25 f_ck^1/2

  = 0.25*25^1/2

  = 1.25

• T_c >T_v , Hence OK