Question

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

tension with a load of 89,000 N. Under this load it experiences an elongation of 0.10 mm.
Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

Answer 1

222.5 Gpa

Step-by-step 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* 10^(-3) m`

By substitution we obtain

`E=\frac {89000* 0.1}{0.02^(2)* 0.1* 10^(-3)}=2.225* 10^(11)= 225.5 Gpa`