Answer:
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
[tex]\dfrac{\Delta L}{L}=\dfrac{\sigma}{E}[/tex]
[tex]{\Delta L}=\dfrac{\sigma}{E}{L_o}[/tex]
By putting the values
[tex]{\Delta L}=\dfrac{\sigma}{E}{L_o}[/tex]
[tex]{\Delta L}=\dfrac{259}{116\times 1000}\times {128}[/tex]
ΔL=0.285 mm
The final length L
L = ΔL + Lo
L = 0.285 + 128
L=128.285 mm