Elasticity/Warping of elliptical cylinder

Example 2 : Elliptical cylinder

Choose warping function

$\psi (x_1,x_2)=k{x}_1{x}_2$

where $k$ is a constant.

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

The traction free BC is

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

Integrating,

$x_1^2+\frac{1-k}{1+k}{x}_2^2=a^2\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

where $a$ is a constant.

This is the equation for an ellipse with major and minor axes $a$ and $b$, where

$b^2=\left(\frac{1+k}{1-k}\right)a^2$

The warping function is

$\psi =-\left(\frac{a^2-b^2}{a^2+b^2}\right)x_1{x}_2$

The torsion constant is

$\tilde{J}=\frac{2b^2}{a^2+b^2}{I}_2+\frac{2a^2}{a^2+b^2}{I}_1=\frac{\pi a^3{b}^3}{a^2+b^2}$

where

$I_1=\underset{S}{\int }{x}_1^{2}dA=\frac{\pi a{b}^3}{4};I_2=\underset{S}{\int }{x}_2^{2}dA=\frac{\pi a^{3}b}{4}$

If you compare $\tilde{J}$ and $J$ for the ellipse, you will find that $\tilde{J}<J$. This implies that the torsional rigidity is less than that predicted with the assumption that plane sections remain plane.

The twist per unit length is

$\alpha =\frac{(a^2+b^2)T}{\mu \pi a^3{b}^3}$

The non-zero stresses are

${\sigma }_{13}=-\frac{2\mu \alpha a^2{x}_2}{a^2+b^2};{\sigma }_{23}=-\frac{2\mu \alpha b^2{x}_1}{a^2+b^2}$

The projected shear traction is

$\tau =\frac{2\mu \alpha }{a^2+b^2}\sqrt{b^4{x}_1^2+a^4{x}_2^2}\Rightarrow {\tau }_{\text{max}}=\frac{2\mu \alpha a^{2}b}{a^2+b^2}(b<a)$

File:Torsion elliptic cs stress.png Shear stresses in the cross section of an elliptical cylinder under torsion

For any torsion problem where $\partial S$ is convex, the maximum projected shear traction occurs at the point on $\partial S$ that is nearest the centroid of $S$.

The displacement $u_3$ is

$u_3=-\frac{(a^2-b^2)T{x}_1{x}_2}{\mu \pi a^3{b}^3}$

File:Torsion elliptic cs disp.png Displacements ($u_3$) in the cross section of an elliptical cylinder under torsion