Elasticity/Warping of circular cylinder

Example 1: Circular Cylinder
Choose warping function

\psi(x_1,x_2) = 0 \, $$

Equilibrium ($$\nabla^2{\psi} = 0$$) is trivially satisfied.

The traction free BC is

(0 - x_2) \frac{dx_2}{ds} - (0 + x_1) \frac{dx_1}{ds} = 0 ~ \forall (x_1, x_2) \in \partial\text{S} $$ Integrating,

x_2^2 + x_1^2 = c^2 ~ \forall (x_1, x_2) \in \partial\text{S} $$ where $$c$$ is a constant.

Hence, a circle satisfies traction-free BCs.


 * There is no warping of cross sections for circular cylinders

The torsion constant is

\tilde{J} = \int_S (x_1^2 + x_2^2) dA = \int_S r^2 dA = J $$

The twist per unit length is

\alpha = \frac{T}{\mu J} $$

The non-zero stresses are

\sigma_{13} = -\mu\alpha x_2 ~; \sigma_{23} = \mu\alpha x_1 $$

The projected shear traction is

\tau = \mu\alpha\sqrt{(x_1^2 + x_2^2)} = \mu\alpha r $$

Compare results from Mechanics of Materials solution

\phi = \frac{TL}{GJ} \Rightarrow \alpha = \frac{T}{GJ} $$ and

\tau = \frac{Tr}{J} \Rightarrow \tau = G\alpha r $$