Introduction to 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)

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

Displacements (

u_3 \,

) in the cross section of an elliptical cylinder under torsion

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