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

The traction free BC is

(kx_2 - x_2) \frac{dx_2}{ds} - (kx_1 + x_1) \frac{dx_1}{ds} = 0 ~ \forall (x_1, x_2) \in \partial\text{S} $$ Integrating,

x_1^2 + \frac{1-k}{1+k}x_2^2 = a^2 ~ \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 = \int_S x_1^2 dA = \frac{\pi a b^3}{4} ; I_2 = \int_S 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) $$

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)Tx_1x_2}{\mu\pi a^3b^3} $$