Question

A steel bar with a diameter of .875 inches and a length of 15.0 ft is axially loaded with a force of 21.6 kip. The modulus of elasticity of the steel is 29 *106 psi. Determine

Answer 1

35.92 kpsi

Step-by-step explanation:

Given data:

diameter of the steel bar d = 0.875 in

Area A = πd^2/4 = π(0.875)^2/4

length L = 15.0 ft

Load P = 21.6 kip

Modulus of elesticity E = 29×10^6 Psi

Assume we are asked to determine axial stress in the bar which is given as

`\sigma = Load, P/ Area, A`

`\sigma = 4P/\pi d^2`

substitute the value

`\sigma = (4* 21.6)/(\pi * (0.875)^2) \\=35.92\ kpsi`