Page 281 - Introduction to Continuum Mechanics

Therefore, & (KI , t) must satisfy the equation

where c-p = v/i/p 0.
(b) At the fixed end jti = 0, there is no displacement, therefore,

(c) At the traction-free end x\ = 0, t = -Tej = 0. Thus, ^i \ x =o = 0, Ty^ \ x =g — 0, there-
fore,

Example 5.13.4

A cylindrical bar of square cross-section (see Fig. 5.12) is twisted by end moments. Show
that the displacement field of the torsion of the circular bar does not give a correct solution to
this problem.
Solution. The displacement field for the torsion of circular cylinders has already been shown
to generate an equilibrium stress field. We therefore check if the surface traction of the lateral
surface vanishes. The unit vector on the plane *3 = a is 63, so that the surface traction for the
stress tensor of Eq. (5.13.1) is given by

Similarly, there will be surface tractions in the ej direction on the remainder of the lateral
surface. Thus, the previously assumed displacement field must be altered. To obtain the actual
solution for twisting by end moments only, we must somehow remove these axial surface
tractions. As will be seen in the next section, this will cause the cross-sectional planes to warp.