Longitudinal stress is half of circumferential stress

A thin cylinder of internal diameter $d$, wall thickness $t$ and internal pressure $p$ is subjected to two stresses: circumferential (hoop) stress $\sigma_c$ acting on longitudinal sections, and longitudinal stress $\sigma_l$ acting on circular cross-sections. We show that $\sigma_l = \tfrac12\sigma_c$.

   Longitudinal section        Transverse section
   (hoop stress)               (longitudinal stress)

     ______________             _________
    |<----- d ---->|           /  p  p   \
  sc|   p  p  p    |sc        | p  p  p   |-> sl
    |______________|           \_________/
       length L                 area = pi*d*t

Circumferential stress (cut along a diameter)

Cut the cylinder by a longitudinal plane through its axis. The pressure on the projected area $d\times L$ is balanced by the tension in the two walls:

$$\text{Bursting force} = p\times d\times L, \qquad \text{Resisting force} = 2\times\sigma_c\times L\times t$$ $$p\,d\,L = 2\sigma_c\,L\,t \Rightarrow \sigma_c = \frac{pd}{2t}$$

Longitudinal stress (cut across the cylinder)

Cut the cylinder by a transverse plane. The pressure acts on the end area $\frac{\pi}{4}d^2$. This force is resisted by the metal ring of area $\pi d t$ (thin wall):

$$p\times\frac{\pi}{4}d^2 = \sigma_l\times\pi d\,t \Rightarrow \sigma_l = \frac{pd}{4t}$$

Relation

$$\frac{\sigma_l}{\sigma_c} = \frac{pd/4t}{pd/2t} = \frac{1}{2} \quad\Rightarrow\quad \sigma_l = \frac{\sigma_c}{2}$$

So the longitudinal stress is half of the circumferential stress. This is why a boiler shell fails along a longitudinal seam first, and why the longitudinal joint needs more strength than the circumferential joint.