## 66 Torsion of shafts

(a) Isotropic solids

A torque, T, applied to the ends of an isotropic bar of uniform section, and acting in the plane normal to the axis of the bar, produces an angle of twist B. The twist is related to the torque by the first equation below, in which G is the shear modulus. For round bars and tubes of circular section, the factor K is equal to J, the polar moment of inertia of the section, defined in Section 6.2. For any other section shape K is less than J. Values of K are given in Section 6.2.

If the bar ceases to deform elastically, it is said to have failed. This will happen if the maximum surface stress exceeds either the yield strength, oy, of the material or the stress at which it fractures. For circular sections, the shear stress at any point a distance r from the axis of rotation is

The maximum shear stress, rmax, and the maximum tensile stress, amax, are at the surface and have the values

Td0 GBd0

2K 21

If xmax exceeds ay/2 (using a Tresca yield criterion), or if amax exceeds the MOR, the bar fails, as shown in Figure 6.5. The maximum surface stress for the solid ellipsoidal, square, rectangular and triangular sections is at the

 h-'-h
0 0