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For a bronze alloy, the stress at which plastic deformation begins is 259 MPa and the modulus of elasticity is 116 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 302 mm2 without plastic deformation? (b) If the original specimen length is 128 mm, what is the maximum length to which it may be stretched without causing plastic deformation?

1 Answer

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Answer:

a)P = 78.21 KN

b)L=128.285 mm

Step-by-step 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


(\Delta L)/(L)=(\sigma)/(E)


{\Delta L}=(\sigma)/(E){L_o}

By putting the values


{\Delta L}=(\sigma)/(E){L_o}


{\Delta L}=(259)/(116* 1000)* {128}

ΔL=0.285 mm

The final length L

L = ΔL + Lo

L = 0.285 + 128

L=128.285 mm

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