Gr

Therefore partially plastic torque, from eqn. (3.12),

Example 3 J

A 50 mm diameter steel shaft is case-hardened to a depth of 2 mm. Assuming that the inner core remains elastic up to a yield stress in shear of 180 MN/m2 and that the case can also be assumed to remain elastic up to failure at the shear stress of 320 MN/m2, calculate:

(a) the torque required to initiate yielding at the outside surface of the case;

(b) the angle of twist per metre length at this stage.

Take G — 85 GN/m2 for both case and core whilst they remain elastic. Solution

Since the modulus of rigidity G is assumed to be constant throughout the shaft whilst elastic, the angle of twist 6 will be constant.

The stress distribution throughout the shaft cross-section at the instant of yielding of the outside surface of the case is then as shown in Fig. 3.41, and it is evident that whilst the failure stress of 320 MN/m2 has only just been reached at the outside of the case, the yield stress of the core of 180 NM/m2 has been exceeded beyond a radius r producing a fully plastic annulus and an elastic core.

\Yielded portion of core

Stress distribution

\Yielded portion of core

Stress distribution

By proportions, since Gcase

320 ~25 180

0 0