Example 9.6 provides the torsion solution of a closed thin-walled section shown in Fig. 9.16.
Investigate the solution of the identical section for the case where a small cut has been introduced as shown in the following figure. This cut creates an open tube and produces significant changes to the stress function (use membrane analogy), stress field, and loadcarrying capacity. The open tube solution can be approximately determined using results (9.5.15) from the thin rectangular solution. For the open tube, develop an equivalent relation as given by (9.6.8) for the closed tube. For identical applied torque, compare the stresses for each case, and justify that the closed tube has much lower stress and is thus much stronger.

Data from example 9.6

Fig 9.16

Equation 9.5.15

Equation 9.6.8

Transcribed Image Text:

The torsion of hollow thin-walled cylinders can be effectively handled using an approximate solution based on the membrane analogy. Consider the general thin-walled tube shown in Fig. 9.16. We assume that thickness t is small, although not necessarily constant. A general section aa is taken through the tube wall at AB, and the expected membrane shape is shown. From our previous theory, the membrane (stress function) will be zero at the outer boundary (point B) and equal to a nonzero constant, say do, on the inner boundary (point A). Because the thickness is small there will be little variation in the membrane slope, and thus shape BC can be approximated by a straight line. Because the membrane slope equals the resultant shear stress, we can write The load-carrying relation (9.3.31) gives t= T=2 t = 2 ] [ pdxdy. pdxdy + 20 Ai (9.6.5) (9.6.6)