## Chapter 1

## Q. 1.27

## Step-by-Step

## Verified Solution

Using Eq. 1.28,

\sigma _{\text{max}}^2 -2\frac{W}{A}\sigma _{\text{max}}-\frac{2WhE}{AL}=0 (1.28)

W\left(h+\frac{\sigma_{\max } L}{E}\right)=\frac{\sigma_{\max }^2}{2 E} \times A L

where, \sigma_{\max }=150 MPa , \quad W=25 N

L = 1000 mm , E = 200000 MPa

A=\frac{\pi}{4} \times 10^2=78.54 mm ^2

or, h=\frac{150^2 \times 78.54 \times 1000}{2 \times 200000 \times 25}-\frac{150 \times 1000}{200000} = 175.96 mm