Prandtl stress function for circular bar in torsion

[davidwinth]

TL;DR Summary

When applying two different (but equivalent) stress functions for a circular bar, two different results appear.

For a Prandtl stress function to be valid, it must be zero on the boundary. For a circular bar, both of these work:

$$\phi_1 = C\left(\frac{x^2}{r^2}+ \frac{y^2}{r^2} - 1\right)$$

$$\phi_2 = C \left(x^2+ y^2- r^2\right)$$

But performing the integration for the internal torque M gives divergent solutions. Since both functions are legitimate, which one is the "correct" one and why doesn't the other one work?

$$M = 2 \int \int_A \phi_1dxdy = -C \pi r^2$$

$$M = 2 \int \int_A \phi_2 dxdy = -C \pi r^4$$

 

[FEAnalyst]

Have you checked the derivations in Timoshenko's "Theory of Elasticity" ? There you may find the answer to your question.

 

[jack action]

Obviously, the units of $C$ in both $\phi_1$ and $\phi_2$ are different, thus they probably have different values as well.

Since $r$ is a constant, my guess is that $C_{\phi_2} = \frac{C_{\phi_1}}{r^2}$.

$$\phi_1 = C_{\phi_1}\left(\frac{x^2}{r^2}+ \frac{y^2}{r^2} - 1\right) = \frac{C_{\phi_1}}{r^2}\left(x^2+y^2 - r^2\right) = C_{\phi_2}\left(x^2+y^2 - r^2\right) = \phi_2$$

Therefore we can verify this equality:
$$M_{\phi_1} = M_{\phi_2}$$
$$-C_{\phi_1} \pi r^2 = -C_{\phi_2} \pi r^4$$
$$-C_{\phi_1} \pi r^2 = -\frac{C_{\phi_1}}{r^2}\pi r^4$$
$$-C_{\phi_1} \pi r^2 = -C_{\phi_1}\pi r^2$$