Page 274 - Introduction to Continuum Mechanics

258 Torsion of a Circular Cylinder

Statics alone does not determine the distribution of the load (a statically indeterminate
problem), so we must consider the deformation induced by the load P. In this problem, there
is no net elongation of the composite bar, therefore

Combining Eqs. (i) and (ii), we obtain

If in particular, Young's moduli are Ey' = 207 GPa (steel) and Ef> = 69 GPa.(aiuminum),
then

5.13 Torsion of a Circular Cylinder
Let us consider the elastic deformation of a cylindrical bar of circular cross-section (of
radius a and length /), twisted by equal and opposite end moments M t (see Fig. 5.8). We choose
the KI axis to coincide with axis of the cylinder and the left and right faces to correspond to the
plane xi = 0 andjcj = / respectively

By the symmetry of the problem, it is reasonable to assume that the motion of each
cross-sectional plane induced by the end moments is a rigid body rotation about the jq axis.
This motion is similar to that of a stack of coins in which each coin is rotated by a slightly
different angle than the previous coin. It is the purpose of this section to demonstrate that for
a circular cross-section, this assumption of the deformation leads to an exact solution within
the linear theory of elasticity.
Denoting the small rotation angle by 0, we evaluate the associated displacement field as

or,

where 6> = 0(jc ] )