## R

Therefore maximum elastic torque

XyJ Xy JlR4

If the torque is now increased further it is assumed that, instead of the stress in the outer fibre increasing beyond rv, more and more of the material will yield and take up the stress xy, giving the stress distribution shown in Fig. 3.18(b). Consider the case where the material has yielded to a radius Ri, then: Partially plastic torque

TPP = torque owing to elastic core + torque owing to plastic portion

The first part is obtained directly from eqn. (3.11) with R\ replacing R, nR]

For the second part consider an element of radius r and thickness dr, carrying a stress xy, (see Fig. 3.18(b)), force on element = rv x 2ixr dr contribution to torque = force x radius = (xv x 2itrdr)r = 2nr2 drXy

r total contribution = / Xy2nr2 dr Jr.

2nXy

Therefore, partially plastic torque tpp = -~-rv + — rv[Ä3 - r]}

ttR, 2n

In Fig. 3.18(c) the torque has now been increased until the whole cross-section has yielded, i.e. become plastic. The torque required to reach this situation is then easily determined from eqn. (3.12) since R\ — 0.

0 0