Elasticity/Warping of circular cylinder

Example 1: Circular Cylinder

Choose warping function

$\psi (x_1,x_2)=0$

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

The traction free BC is

$(0-x_2)\frac{d{x}_2}{ds}-(0+x_1)\frac{d{x}_1}{ds}=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

Integrating,

$x_2^2+x_1^2=c^2\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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

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

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

$\varphi =\frac{TL}{GJ}\Rightarrow \alpha =\frac{T}{GJ}$

and

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