Longitudinal stress:

$\sigma_y = \dfrac{pD}{4t} = \dfrac{4(80)}{4(3)}$

$\sigma_y = \frac{80}{3} \, \text{MPa}$

The strain in the x-direction is:

$\varepsilon_x = \dfrac{\sigma_x}{E} - \nu \dfrac{\sigma_y}{E} = 0$

$\sigma_x = \nu \, \sigma_y$ → tangential stress

$\sigma_x = \frac{1}{3}(\frac{80}{3})$

$\sigma_x = 8.89 \, \text{ MPa}$ *answer*

## Comments

## In the solution

In the solution

you used the formula for longitudinal stress and found it. Yet you didnt simply used the formula for tangential stress. Why?