Introduction to Elasticity/Warping of elliptical cylinder

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[edit] 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

$\tau = \frac{2\mu\alpha}{a^2+b^2}\sqrt{b^4 x_1^2 + a^4 x_2^2} ~~ \Rightarrow ~~ \tau_{\text{max}} = \frac{2\mu\alpha a^2 b}{a^2+b^2} ~~ (b < a)$

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 $S\,$ .