Design of Shallow Foundation (Isolated Footings)

Aim of challenge- To know the general design procedure of stepped footing with manual calculation and to practice how the inputs can be collaborated and arrived as data for calculations.

Introduction- The design consideration and design data has been given below,

Column size of square footing= 400×400mm

Factored compressive axial load= 2000 kN

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

Grade of concrete= M25

Grade of steel =Fe500

Calculation of Gross soil pressure for footing area

Assuming we have the square footing

Footing area= (un-factored load with surcharge + self-weight of the footing)/ allowable soil pressure

= ((2000/1.5)×1.1)/150 kN/m² =9.75 m²

As we have assumed square footing, so the side of the square footing= square root (9.78)= 3.12m

Considering 3.2 m of side of the isolated footing.

Calculation of Net soil pressure and initial footing depth

• Net soil pressure= (Factored load/ actual area of footing)

=(2000/(3.2×3.2))= 195.3 kN/m²

P_n= 195.3 kN/m²

Assuming depth of footing 500mm and checking it for bearing stress as per 456-2000 clause 34.4

Square root (footing base bearing area/loaded area) is less than or equal to 2.

• Permissible bearing stress for limit state method as per IS 456 clause 34.4

Permissible bearing stress= 0.45× Fck = 11.25 N/mm²

Permissible bearing stress multiplied with area factor = 11.25 ×2= 22.5 N/mm²

• Actual bearing stress at column base= ((2000×1000)/(400×400))= 12.5 N/mm²

Actual stress of 12.5 N/mm² is less than permissible i.e. 22.5 N/mm². Hence ok. We can proceed with 500mm depth of footing for further checks.

Assuming cover for footing is 50mm that means effective depth (d) shall be taken as 450mm.

Calculation of Flexural reinforcement

• Conversion of net pressure into uniformly distributed load= 195.3 ×3.2= 624.96 kN/m which is nearly equal to 625 kN/m.
• Bending moment due to UDL over a cantilever section = ((wL^2/2)= (625× 1.6²)/2)= 800 kNm
• Reinforcement calculation as per IS 456-2000, Annex G

Where, M_u= 800 kNm

d= 450mm

Fck= 25 n/mm²

b= 3200mm

Assuming Fy= 500 N/mm²

M_u=0.87×500×A_st×450 (1-((A_st×500)/(3200×450×25)))

=>800×10⁶N-mm= 195750×A_st-2.71875×A_st²

=>A_st= 4349.61mm² OR 67650.38904 mm²

=>A_st required = 4350mm²

If we provide 16mm diameter of rebar at 125mm spacing, the actual rebar provided= (3200/125)× ((Pie/4)×15m²)= (3200/125)×201.06= 5145.6mm²

A_st provided= 5146 mm²

• Checking of one way shear stress

Shear force in the one way shaded area= 0.95×3.2×195.3= 593.7 kN

Design shear stress at one way cut face

τ_v= 593.7×1000 /(3200×450)= 0.412 N/mm²

Shear stress capacity of section as per IS 456-2000 table-19 = 0.425 N/mm²

Design shear stress is less than shear capacity, Hence ok.

• Checking of two way shear stress

Shear force in the two way shaded area = (3.2×3.2)-(0.85×0.85) ×195.3= 1895kN

Design shear stress at two way cut face,

τ_v= 1895×1000 /(3200×450)= 1.215 N/mm²

Shear stress capacity of section as per IS 456-2000 clause31.6.3.1,

=0.25×square root (Fck) = 0.25×square root (25) = 1.25 N/mm²

Design shear stress is less than shear capacity, Hence ok.

Result-

The detail design reforcement for footing has been given below,

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