Section I.3 Elastic Membrane Analogy

The elastic membrane analogy allows the solution of a torsion problem to be determined in a simpler way than that found by the theory of elasticity which requires the availability of the warping function.

Consider a tube which has the same cross-sectional boundary as the bar. For example, if the bar has a solid square cross section of side dimension b, then the tube will have a hollow square cross section of side dimension b as well. Next we stretch an elastic membrane over the tube's cross section and apply internal pressure. The deflected shape of the membrane helps us visualize the stress pattern in the bar under torsion.

The analogy can be viewed as follows:

1. Lines of equal deflection on the membrane (contour lines) correspond to shearing stress lines of the twisted bar.
2. The direction of a particular shear stress resultant at a point is at right angle to the maximum slope of the membrane at the same point.
3. The slope of the deflected membrane at any point, with respect to the edge support plane is proportional in magnitude to the shear stress at the corresponding point on the bar's cross section.
4. The applied torsion on the twisted bar is proportional to twice the volume included between the deflected membrane and plane through the supporting edges.

Keep in mind that the slope at a point on the deflected membrane, and not the displacement from the base, is the parameter that is related to the shear stress in the bar. In all of the following examples, the slope is zero at the very top of the membrane, therefore the stress is zero, not the maximum, at the same location on bar's cross section.

• Figure 1
• Figure 2
• Figure 3 Notice that the stress is maximum at the inside corner and zero at the outside corners.

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