Answer:
(a) The stress on the steel wire is 19,000 Psi
(b) The strain on the steel wire is 0.00063
(c) The modulus of elasticity of the steel is 30,000,000 Psi
Explanation:
Given;
length of steel wire, L = 100 ft
cross-sectional area, A = 0.0144 in²
applied force, F = 270 lb
extension of the wire, e = 0.75 in
<u>Part (A)</u> The stress on the steel wire;
δ = F/A
= 270 / 0.0144
δ = 18750 lb/in² = 19,000 Psi
<u>Part (B)</u> The strain on the steel wire;
σ = e/ L
L = 100 ft = 1200 in
σ = 0.75 / 1200
σ = 0.00063
<u>Part (C)</u> The modulus of elasticity of the steel
E = δ/σ
= 19,000 / 0.00063
E = 30,000,000 Psi
