
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