Solution to Problem 225 Biaxial Deformation

σ_y = longitudinal stress
$\sigma_y = \dfrac{pD}{4t} = \dfrac{1.5(1200)}{4(10)}$

$\sigma_y = 45 \, \text{MPa}$
 

σ_x = tangential stress
$\sigma_y = \dfrac{pD}{2t} = \dfrac{1.5(1200)}{2(10)}$

$\sigma_y = 90 \, \text{MPa}$
 

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

$\varepsilon_x = \dfrac{90}{200\,000} - 0.3 \left( \dfrac{45}{200\,000} \right)$

$\varepsilon_x = 3.825 \times 10^{-4}$
 

$\varepsilon_x = \dfrac{\Delta D}{D}$

$\Delta D = \varepsilon_x \, D = (3.825 \times 10^{-4})(1200)$

$\Delta D = 0.459 \, \text{ mm}$           answer