Design of Shallow Foundation (Isolated Footings)

Given Data :

Column Size: (400x400)mm

Compression axial load for the load combination of (1.5 DL + 1.5 LL) : 2000 KN

Safe soil bearing capacity at a depth of 2 meters below E.G.L is 150 kN/m²

Grade of concrete(M₂₅)= 25 kN/m²

Assuming f_y =415 N/mm²

Assuming the self-weight of the footing and soil above it as a 10% load

Calculations:

1.) Size of Footing:

Un-factored Load = 2000/1.5 = 1333.33kN

Safe Bearing Capacity of Soil (SBC) = 150kN/M²

Area of Footing = (1333.33 x 1.1)/150 = 9.78M²

For Square footing = (9.78)^0.5 = 3.5M

Area of Footing = 12.25M²

2.) Net Soil Pressure:

Net Soil Pressure = (Factored load x 1.5)/Size of Footing\

=(2000)/12.25

=163.26kN/M²

3.) Depth of Footing:

Max Bending is at the face of the column

Length of the critical section = 3.5-0.2= 1.55M

Load on Critical Section = Net Soil pressure X Length

= 163.26 X 3.5 = 571.76kN/M

Moment, M = (wl²)/2

=(572 x 1.55²)/2 = 684kN/M

Ultimate Moment, M_u = 1.5 x 684 = 1026kN/M

Depth of Footing, D = M_u=0.138f_ck xBxd²

= 1026x 10⁶ = 0.138 x 0.25 x 3500 x d²

d =287mm

For Shear Check(taking values by 2.5 times)

d = 2.5 x 287 = 717mm ~ 720 mm

d = (720+ 50) = 770mm ~ 800mm (Providing 50mm As cover)

4.) Gross bearing capacity

Assuming the unit weight of the soil to be 18kN/m³

Weight of the footing, g = 3.5 X 3.5 X 25 X 0.8 = 245kN

Weight of the soil= 3.5 X 3.5 X (2 - 0.8) X 18 = 261.6kN

Load on the footing= 1333.3kN

Total load= 1852kN

Gross bearing capacity= 1852/3.5 X 3.5 =257KN/m2

As 257 kN/m² >150kN/m², hence we need to increase the size of the footing to 3.8m X 3.8m (Section Redesign Required)

Now Total load=1934kN

Gross bearing capacity=1934/3.8x3.8=133.9KN/mm²

133.9 kN/m² <150kN/m², hence safe

Corrected net soil pressure = 2000/3.8X3.8= 138.5kN

5.) Reinforcement

Area of Steel Provided ,A_st = 0.5f_ck/f_y[1-{(1-4.6M_u/F_ckBd2)}^0.5]Bd

= 3640mm²

Percentage of Steel = (Ast/Bd)x 100 = 0.119%

Numbers of Steel Bars Required = 3640/3.145x16x16/4 = 18

Spacing Between Bars = 3800/18 =211 ~210 mm

Lets Provide 18nos of 16mm Bars @ 160mm c/c

6.) Check for One -Way Shear:

Shear Force τ_u, = Net Soil Pressure x B{0.5 x(B-b)-d}

= 138.5 x 3.8 x{0.5 x(3.8-0.4)-0.8}

= 473.67kN

Now from IS-456: 2000 Table 19 We can get value of τ_c = 0.29

Since τ_v < τ_c shear stress is less than permissible stress hence ok.

7.) Check for Two-way Shear ( Punching Shear):

Net Shear force, V_u = Net Soil Pressure x( B² - b₀²)

= 138.5 x { 3.8² - (0.8+0.4)²}

= 1800.5kN

Pheriphery of Section, B₀ = 2(1.2+1.2) = 4.8 m

τ_v =Vu /Bd = (1800.5x 10³)/4800x4800 =0.46N/mm2

Permissible Shear Stress = K_s x τ_c = 1 x 0.25 x(25)^0.5 = 1.25N/mm2 (from IS-456: 2000)

τv<τc Hence ok.

8.) Development Length: ( 0.87 x f_s x Φ)/4 x τ x b x d

= ( 0.87 x 25 x 16)/4 x 1.6 x 1.4 = 677mm

Length Available = 1600-50 = 1550mm

RESULT: Sloped isolated footing for the square column is designed for the given conditions and also checked.

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