Solution to Problem 228 Biaxial Deformation

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

$\sigma_x = \nu \, \sigma_y = \sigma_l$       → longitudinal stress

$\sigma_t = \sigma_y$       → tangential stress
 

$\sigma_t = \dfrac{pD}{2t} = \dfrac{6000(3)}{2(0.10)}$

$\sigma_t = 90\,000 \, \text{ psi}$           answer
 

$\sigma_l = \nu \, \sigma_y = \frac{1}{3} (90\,000)$

$\sigma_l = 30\,000 \, \text{ psi}$           answer