Membrane analogy

It has been stated earlier that the mathematical solution for the torsion of certain solid and thin-walled sections is complex and beyond the scope of this text. In such cases it is extremely fortunate that an analogy exists known as the membrane analogy, which provides a very convenient mental picture of the way in which stresses build up in such components and allows experimental determination of their values.

It can be shown that the mathematical solution for elastic torsion problems involving partial differential equations is identical in form to that for a thin membrane lightly stretched over a hole. The membrane normally used for visualisation is a soap film. Provided that the hole used is the same shape as the cross-section of the shaft in question and that air pressure is maintained on one side of the membrane, the following relationships exist:

(a) the torque carried by the section is equal to twice the volume enclosed by the membrane;

(b) the shear stress at any point in the section is proportional to the slope of the membrane at that point (Fig. 5.9);

(c) the direction of the shear stress at any point in the section is always at right angles to the slope of the membrane at the same point.

Soap Film Analogy Torsion

Maximum slope

Fig. 5.9. Membrane analog.

Slope a t I Pressurised soap / /film membrane


Maximum slope

Fig. 5.9. Membrane analog.

Application of the above rules to the open sections of Fig. 5.2 shows that each section will carry approximately the same torque at the same maximum shear stress since the volumes enclosed by the membranes and the maximum slopes of the membranes are approximately equal in each case.

The membrane analogy is particularly powerful in the study of the comparative torsional properties of different sections without the need for detailed calculations. For example, it should be evident from the volume relationship (a) above that if two cross-sections have the same area, that which is nearer to circular will be the stronger in torsion since it will produce the greatest enclosed volume.

The analogy also helps to support the theory used for thin-walled open sections in §5.3 when thin rectangular sections are taken to have the same torsional stiffness be they left as a single rectangle or bent into open tubes, angle sections, channel sections, etc.

From the slope relationship (b) the greatest shear stresses usually occur at the boundary of the thickest parts of the section. They are usually high at positions where the boundary is sharply concave but low at the ends of outstanding flanges.

In the preceding paragraphs it has been assumed that the torque is applied at the ends of the member and that all sections are free to warp. In practice, however, there are often cases where one or more sections of a member are constrained in some way so that cross-sections remain plane, i.e. warping is prevented. Whilst this has little effect on the angle of twist of certain solid cross-sections, e.g. rectangular or elliptical sections where the length is significantly greater than the section dimensions, it may have a considerable effect on the twist of open sections. In the latter case the constraint of warping is often accompanied by considerable bending of the flanges. Detailed treatment of warping is beyond the scope of this text^ and it is sufficient to note here that when warping is restrained, angles of twist are generally reduced and hence torsional stiffnesses increased.

5.11. Effect of warping of open sections


Example 5.1

A rectangular steel bar 25 mm wide and 38 mm deep is subjected to a torque of 450 Nm. Estimate the maximum shear stress set up in the material of the bar and the angle of twist, using the experimentally derived formulate stated in §5.1. What percentage error would be involved in each case if the approximate equations are used?

For steel, take G = 80 GN/m2. Solution

The maximum shear stress is given by eqn. (5.1):

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