Answer:
222.5 Gpa
Explanation:
From definition of engineering stress, [tex]\sigma=\frac {F}{A}[/tex]
where F is applied force and A is original area
Also, engineering strain, [tex]\epsilon=\frac {\triangle l}{l}[/tex] where l is original area and [tex]\triangle l[/tex] is elongation
We also know that Hooke's law states that [tex]E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}[/tex]
Since A=20 mm* 20 mm= 0.02 m*0.02 m
F= 89000 N
l= 100 mm= 0.1 m
[tex]\triangle l= 0.1 mm= 0.1\times 10^{-3} m[/tex]
By substitution we obtain
[tex]E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa[/tex]