A 100 ft long steel wire has a cross-sectional area of 0.0144 in. 2. When a force of 270 lb is applied to the wire, its length increases by 0.75 in. Determine: a. The stress (precision of 1000) b. The strain (precision of 0.00000) c. The modulus of elasticity of the steel (precision of 1,000,000)

(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

Part (A) The stress on the steel wire;

δ = F/A

= 270 / 0.0144

δ = 18750 lb/in² = 19,000 Psi

Part (B) The strain on the steel wire;

σ = e / L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

Part (C) The modulus of elasticity of the steel

E = δ/σ

= 19,000 / 0.00063

E = 30,000,000 Psi