Question

13. An 8 inch by 8 inch reinforced concrete bar needs to be designed to carry a compressive axial force of 235,000 pounds. Assume the normal stress in concrete to be a uniform maximum of 3ksi and the normal stress in steel bars to be 20ksi, determine the minimum number of 0.25" diameter steel bars needed

Answer 1

Here we have,

Dimensions of bar = 8 in × 8 in

Gross Area of cross section of bar,A = 64 in²

Design compressive axial load,P = 235000 lb

Maximum normal stress in concrete, σ_c = 3 ksi = 3000 psi

Maximum normal stress in steel,σ_{​​​​​​s}​​​​​ = 20 ksi = 20000 psi

Diameter of bar,D =0.25 in

Cross section area of steel bar = (π/4)D² = 0.0490873852 in²

Now we know that Load taken by concrete and steel should be equal to the design load.

Thus, let the area of steel be A_{​​​​​​s} and area of concrete be A_{​​​​​​c}

1.) A_{​​​​​​s} + A_{​​​​​​c} = A

2.) σ_{​​​​​​c}.A_c + σ_{​​​​​​s}​​​​​.A_s = P

>> 3000(A - As ) + 20000A_{​​​​​​s} = 235000

>> 3000 × 64 + 17000.A_s = 235000

>> 17000.A_s = 43000

>> A_{​​​​​​s} = 2.52941176 in²

Now, let the number of 0.25 in bar required be N,thus we have

N.a = A_{​​​​​​s}

N = 2.52941176/0.0490873852 = 51.5287533

Thus, atleast 52 bar will be required.