## Example 1: Circular Cylinder
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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

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