Question

A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in

Answer 1

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, $\sigma=\frac {F}{A}$

where F is applied force and A is original area

Also, engineering strain, $\epsilon=\frac {\triangle l}{l}$ where l is original area and $\triangle l$ is elongation

We also know that Hooke's law states that $E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}$

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

$\triangle l= 0.1 mm= 0.1\times 10^{-3} m$

By substitution we obtain

$E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa$