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)](https://img.qammunity.org/2020/formulas/physics/college/twzxsmgjrbgwwicvf0faj4e3l39gc73tgz.png)
![{\Delta L}=(\sigma)/(E){L_o}](https://img.qammunity.org/2020/formulas/physics/college/p4nwjsc1srfgbfhydq2t5j8jwmup3bdere.png)
By putting the values
![{\Delta L}=(\sigma)/(E){L_o}](https://img.qammunity.org/2020/formulas/physics/college/p4nwjsc1srfgbfhydq2t5j8jwmup3bdere.png)
![{\Delta L}=(259)/(116* 1000)* {128}](https://img.qammunity.org/2020/formulas/physics/college/uddtea8cb9le8lfsgem6mzwyr7cntherzx.png)
ΔL=0.285 mm
The final length L
L = ΔL + Lo
L = 0.285 + 128
L=128.285 mm