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fenix001 [56]
3 years ago
10

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 el

asticity of the steel is 29 *106 psi. Determine
Engineering
1 answer:
svetoff [14.1K]3 years ago
3 0

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

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