Physics Friday 144

Consider a spherical vessel of (inner) radius r with thin walls of thickness Δr, thin enough so that stress does not vary significantly across that thickness. If we want the tension stress on the material of the walls to not exceed a value of σmax in any direction, then what is the maximum pressure we can have inside the sphere (relative to the pressure outside)?

First, the spherical symmetry tells us that the normal stresses on a small element of the wall must be equal, and there is no shear stress. Next, let us consider an imaginary plane through the center of the sphere, splitting the vessel into two hemispheres. If the (gauge) pressure inside the sphere is P, then the force that pressure exerts on the sphere is πr2P, as shown here. This force on the hemisphere must be countered by the material forces, and so we see that
\sigma{A}=\pi{r^2}P,
where σ is the stress and A the cross-sectional area. Since we have a thin-walled sphere, A=2πrΔr, and so
2\pi{r}\Delta{r}\sigma=\pi{r^2}P,
so the stress is
\sigma=\frac{rP}{2\Delta{r}};
or solving for pressure,
P=\frac{2\Delta{r}}{r}\sigma,
and so our desired pressure range is
P\le{P}_{max}=\frac{2\Delta{r}}{r}\sigma_{max}

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One Response to “Physics Friday 144”

  1. Physics Friday 145 « Twisted One 151's Weblog Says:

    […] extending upon the previous physics post, we now consider a thin-walled cylindrical pressure vessel of (inner) radius r with thin walls of […]

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