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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
[tex]\sigma = Load, P/ Area, A[/tex]
[tex]\sigma = 4P/\pi d^2[/tex]
substitute the value
[tex]\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi[/tex]