Prandtl membrane analogy and narrow sections

Membrane analogy

Prandtl observed that the equation for the torsion stress function, $\nabla^2\phi = -2G\theta$, has the same form as the equation for the small deflection $z$ of a uniformly stretched membrane under pressure $p$ and tension $S$:

$$\nabla^2 z = -\frac{p}{S}$$

So if a thin membrane (soap film) is stretched over a hole of the same shape as the cross-section and inflated slightly, and $p/S$ is made equal to $2G\theta$, the deflection $z$ is equal to $\phi$.

   hole shaped like the section
   +--------------------+
   |   .--------.       |   membrane bulges by z
   |  /  contours \     |   (contour lines = lines of
   |  \          /      |    shear stress direction)
   |   '--------'       |
   +--------------------+

Analogy rules

The analogy gives the stress picture without solving the equation, and explains why the stress is zero at the sharp convex corner of a square and highest at the mid-point of the sides.

Narrow rectangle ($b \gg t$)

The membrane is nearly cylindrical, with a parabolic profile across the thickness $t$. Then

$$\tau_{max} = \frac{3T}{b\,t^2},\qquad \theta = \frac{3T}{G\,b\,t^3},\qquad J = \frac{b\,t^3}{3}$$

The shear stress varies linearly across the thickness, from zero at the mid-plane to $\tau_{max}$ at the long edges.

Open and closed sections

An open thin-walled section (angle, channel, I) behaves as a set of narrow rectangles: $J = \tfrac13\sum b_it_i^3$. The torsional stiffness depends on $t^3$, and the shear flow circulates only within each thin strip.

A closed tube carries a shear flow around the whole wall, enclosing area $A$, with $J = 4A^2/\oint ds/t$, which depends on $t$ only to the first power and on the enclosed area squared. A closed tube is therefore typically tens to hundreds of times stiffer and gives much lower stress than an open section of the same material, which is why box or tubular sections are used where torsion is important.