Part (a)
Tangential stress (*longitudinal section*):

$F = 2T$

$pDL = 2(\sigma_t tL)$

$\sigma_t = \dfrac{pD}{2t} = \dfrac{4.5(400)}{2(20)}$

$\sigma_t = 45 \, \text{MPa}$ *answer*

Longitudinal Stress (*transverse section*):

$F = P$

$\frac{1}{4} \pi D^2 p = \sigma_l (\pi Dt)$

$\sigma_l = \dfrac{pD}{4t} = \dfrac{4.5(400)}{4(20)}$

$\sigma_l = 22.5 \, \text{MPa}$ *answer*

Part (b)

From (a), $\sigma_t = \dfrac{pD}{2t}$ and $\sigma_l = \dfrac{pD}{4t}$ thus, $\sigma_t = 2\sigma_l$, this shows that tangential stress is the critical.

$\sigma_t = \dfrac{pD}{2t}$
$120 = \dfrac{p(400)}{2(20)}$

$p = 12 \, \text{MPa}$ *answer*

The bursting force will cause a stress in the longitudinal section that is twice to that of the transverse section. Thus, fracture is expected as shown.