Date : 2016-06-23 12:24:21


The simplest torsion problem is that of the twisting of a uniform thin circular tube.



The tube shown in Figure is of thickness f, and the mean radius of the wall is r, L is the length of the tube.

Shearing stresses T are applied around the circumference of the tube at each end, and in opposite directions.

Figure:Torsion of a thin-walled circular tube.

If the stresses T are uniform around the boundary, the total torque Tat each end of the tube is

Thus the shearing stress around the Circumference due to an applied torque Tis

We consider next the strains caused by these shearing stresses. We note firstly that complementary shearing stresses are set up in the wall parallel to the longitudinal axis of the tube.

If δs is a small length of the circumference then an element of the wall ABCD, is in a state of pure shearing stress. If the remote end of the tube is assumed not to twist, then the longitudinal element ABCD is distorted into the parallelogram ABC'D', Figure 16.1, the angle of shearing strain being

if the material is elastic, and has a shearing (or rigidity) modulus G. But if 8 is the angle of twist of the near end of the tube we have

It is sometimes more convenient to defme the twist of the tube as the rate of change of twist per unit length; this is given by (θ/L)

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