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

Answer:

35.92 kpsi

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

$\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi$