Example 1: Circular Cylinder

Choose warping function

${\displaystyle \psi (x_{1},x_{2})=0\,}$

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

The traction free BC is

${\displaystyle (0-x_{2}){\frac {dx_{2}}{ds}}-(0+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

Integrating,

${\displaystyle x_{2}^{2}+x_{1}^{2}=c^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

where ${\displaystyle c}$ is a constant.

Hence, a circle satisfies traction-free BCs.

• There is no warping of cross sections for circular cylinders

The torsion constant is

${\displaystyle {\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2})dA=\int _{S}r^{2}dA=J}$

The twist per unit length is

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

The non-zero stresses are

${\displaystyle \sigma _{13}=-\mu \alpha x_{2}~;~~\sigma _{23}=\mu \alpha x_{1}}$

The projected shear traction is

${\displaystyle \tau =\mu \alpha {\sqrt {(x_{1}^{2}+x_{2}^{2})}}=\mu \alpha r}$

Compare results from Mechanics of Materials solution

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

and

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