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

$\sigma_x = \nu \sigma_y$

Where

σ_{x} = tangential stress

σ_{y} = longitudinal stress

σ_{y} = P_{y} / A = 3140 / (π × 2 × 0.05)

σ_{y} = 31,400/π psi

Thus,

$\sigma_x = 0.30 (31400 / \pi )$

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

$\sigma_x = 2298.5 \, \text{ psi}$ *answer*

Answer seems to be wrong. I think there is some typo here.

The answer should 2998.5 psi.

Since $$\frac{3430}{\pi}=2998.5$$

However we can get more accurate answer by considering the area of cross-section as $$A= \frac{\pi}{4}(d_2^2-d_1^2)$$

Where $$d_2$$is outside diameter and $$d_1$$ is inner diameter.

The formula for area used here is

$$A = \pi d t$$

Where

dis supposed to be mean diameter andtis the thickness. We did not check the extent of the error. Sources of errors are (1) the formula itself is for very thin tube, (2) we use the inner diameter ford, instead of mean diameter.Thank you for pointing it out.