Page 284 - Introduction to Continuum Mechanics

268 Torsion of a Noncircular Cylinder

Because A does turn out to be a constant, we have satisfied both Eq. (5.14.3) and (5.14.4).
Substituting the value of <p into Eq. (5.14.2), we obtain the associated stresses

This distribution of stress gives a surface traction on the end face, xi = I

and the following resultant force system

Denoting Mj = M t and recalling that for an ellipse 733 = n a b/4 and Iii — n b a/4, we
obtain

Similarly the resultant on the other end face x\ — 0 will give rise to a counterbalancing
couple.
In terms of the twisting moment, the stress tensor becomes