Question

A 20-foot-long W10 x 60 is suspended and hanging from one end. If the modulus of elasticity is 29,000 ksi, determine the following.A. What is the maximum tensile stress?

B. What is the maximum normal strain?

Answer 1

(a) the maximum tensile stress is 68.2 psi

(b) the maximum normal strain is 2.35 x 10⁻⁶

Step-by-step explanation:

Given;

modulus of elasticity, E = 29,000 ksi = 29 x 10⁶ psi

(a) the maximum tensile stress

`\tau = (f)/(A)`

f is the maximum force suspended = 20 x 60 = 1200 lb

A is the area of W10 x 60 = 17.6 in²

`\tau = (1200)/(17.6) \\\\\tau = 68.2 \ psi`

(b) the maximum normal strain.

According to Hook's law stress is directional to strain

τ = Eε

`\epsilon = (\tau)/(E)\\\\\epsilon = (68.2)/(29*10^(6))\\\\\epsilon = 2.35*10^(-6)`