Question

For a bronze alloy, the stress at which plastic deformation begins is 284 MPa and the modulus of elasticity is 106 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 310 mm2 without plastic deformation? (b) If the original specimen length is 120 mm, what is the maximum length to which it may be stretched without causing plastic deformation?

Answer 1

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