Answer:
(a) the maximum tensile stress is 68.2 psi
(b) the maximum normal strain is 2.35 x 10⁻⁶
Explanation:
Given;
modulus of elasticity, E = 29,000 ksi = 29 x 10⁶ psi
(a) the maximum tensile stress
[tex]\tau = \frac{f}{A}[/tex]
f is the maximum force suspended = 20 x 60 = 1200 lb
A is the area of W10 x 60 = 17.6 in²
[tex]\tau = \frac{1200}{17.6} \\\\\tau = 68.2 \ psi[/tex]
(b) the maximum normal strain.
According to Hook's law stress is directional to strain
τ = Eε
[tex]\epsilon = \frac{\tau}{E}\\\\\epsilon = \frac{68.2}{29*10^{6}}\\\\\epsilon = 2.35*10^{-6}[/tex]
