Physics Friday 145

Now, extending upon the previous physics post, we now consider a thin-walled cylindrical pressure vessel of (inner) radius r with thin walls of thickness d. Now, we have two different directions for the stress, which need not be equal: the longitudinal stress (parallel to the cylinder axis), and the hoop stress. With these directions chosen, there is no shear stress; these stresses, σl and σh, are the principal stresses. Making an imaginary cut perpendicular to the cylinder axis, and examining the force balance as we did for the sphere, we see the net force (along the axis) on one piece of the vessel due to pressure must be the pressure times the cross sectional area (due to similar arguments as here and here), or F=P\cdot\pi{r^2}. This must be balanced by the longitudinal stress force, which is the longitudinal stress σl times the cross-section of the thin wall, which gives
F=\sigma_l\cdot2\pi{r}d.
Equating, we find
2\pi{r}d\sigma_l=\pi{r^2}P,
and so
\sigma_l=\frac{rP}{2d};
equivalent to the hoop stress in the sphere of radius r and wall thickness d.

Now, for the hoop stress, let us consider a cylindrical segment of length Δz. If we split it into two semi-circular halves, we see the pressure force on one half is F=P\cdot\Delta{z}\cdot2r, while the stress force is
F=\sigma_h\cdot2\cdot\Delta{z}d.
Equating,
2d\Delta{z}\sigma_h=2r\Delta{z}P,
so
\sigma_h=\frac{rP}{d},
twice the longitudinal stress. Note that both stresses are independent of the vessel length.

The fact that the hoop stress is larger than the longitudinal stress can be most easily seen in cooking hot dogs; when a cooking hot dog breaks open, due to internal steam pressure, it usually does so along a longitudinal split, indicating failure to withstand the hoop stress.

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