Solution to Problem 226 Biaxial Deformation

$\sigma_x = \nu \sigma_y$
 

Where
σ_x = tangential stress
σ_y = longitudinal stress
σ_y = P_y / A = 3140 / (π × 2.05 × 0.05)
σ_y = 9751.15 psi
 

Thus,
$\sigma_x = 0.30 (0.15 )$

$\sigma_x = 9430 / \pi \, \text{ psi}$

$\sigma_x = 2925.34 \, \text{ psi}$           answer