For a bronze alloy, the stress at which plastic deformation begins is 259 mpa and

a)P = 78.21 KN

b)L=128.285 mm

Explanation:

Given that

E= 116 GPa

σ = 259 MPa

a)

A= 302 mm²

Lets take load P can applied without plastic deformation

We know that

P = σ A

P = 259 x 302 N

P = 78218 N

P = 78.21 KN

b)

Lo= 128 mm

Lets take final length = L

Change in length ΔL

Strain ε = ΔL/L

ε = σ/E

By putting the values

ΔL=0.285 mm

The final length L

L = ΔL + Lo

L = 0.285 + 128

L=128.285 mm

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