Page 278 - Introduction to Continuum Mechanics

262 Torsion of a Circular Cylinder

In terms of the twisting couple M t, the stress tensor becomes

In reality, when a bar is twisted the exact distribution of the applied forces is rarely, if ever
known. Invoking St. Venant's principle, we conclude that as long as the resultants of the
applied forces on the two ends of a slender bar are equal and opposite couples of strength
M t, the state of stress inside the bar is given by Eq, (5.13.11).

Example 5.13.1
For a circular bar of radius a in torsion (a) find the magnitude and location of the greatest
normal and shearing stresses throughout the bar; (b) find the principal direction at the position
*2 = 0» *3 = a -
Solution, (a) We first evaluate the principal stresses as a function of position by solving the
characteristic equation

Thus, the principal values at any point are

where r is the distance from the axis of the bar.
In this case, the magnitude of the maximum shearing and normal stress at any point are
equal and are proportional to the distance r. Therefore, the greatest shearing and normal stress
both occur on the boundary, r = a with

(b) For the principal value A = M t a/I p at the boundary points (xi, 0, a) the eigenvector
equation becomes