Answer:
a) the maximum load is 88,040 N
b)
the maximum length to which the specimen may be stretched is 0.12032148 mm
Explanation:
Given the data in the question;
the stress at which plastic deformation begins σ = 284 MPa = 2.84 × 10⁸ Pa
modulus of elasticity E = 106 GPa = 1.06 × 10¹¹ Pa
a)
Area A = 310 mm² = 310 × 10⁻⁶ m ( without plastic deformation )
now, lets consider the equation relating to stress and cross sectional area.
σ = F / A₀
hence, maximum load F = σA₀
so we substitute
F = (2.84 × 10⁸) × (310 × 10⁻⁶)
F = 88,040 N
Therefore, the maximum load is 88,040 N
b)
Initial length specimen l₀ = 120 mm = 120 × 10⁻³ m
using engineering strain, ε = (l₁ - l₀)/l₀
Also from Hooke's law, σ = Eε
so from the equation above;
l₁ = l₀( ε + 1 )
l₁ = l₀( σ/E + 1 )
so we substitute
l₁ = (120 × 10⁻³)( (2.84 × 10⁸)/(1.06 × 10¹¹)) + 1 )
l₁ = (120 × 10⁻³) ( 1.002679 )
l₁ = 0.12032148 mm
Therefore, the maximum length to which the specimen may be stretched is 0.12032148 mm
