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.
