# Introduction to Elasticity/Sample final 4

## Sample Final Exam Problem 4

Consider the torsion of a prismatic bar having an elliptical cross-section as shown in the figure below.

 Torsion of bar with elliptical c.s.

The bar is subjected to equal and opposite torques $T$ at the two ends which cause a twist per unit length of $\alpha$ in the bar.

The equation of the boundary of the cross section is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 = 0$

Since the Prandtl stress function $\phi$ is zero on the boundary of the cross-section of a simply connected prismatic bar, we can choose the Prandtl stress function for the bar with an elliptic cross-section to be

$\phi = C\left[\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1\right]$
• (a) Determine the value of the constant $C$.
• (b) Express the twist per unit length ($\alpha$) in terms of the applied torque ($T$).
You will find the following results useful in evaluating the integral.
$\text{If}~f(x)~\text{is an even function of}~x~\text{, then}~ \int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$
$\frac{1}{2}\int_{-a}^a (a^2 - x^2)^{n/2} dx = \frac{1~.~3~.~5~\dots~n}{2~.~4~.~6~\dots~(n+1)}~.~\frac{\pi}{2}~.~a^{(n+1)} ~~~\text{if}~ n ~\text{is odd}$
• (c) What is the torsion constant of the section?
• (d) Express the maximum shear stress in the bar in terms of $T$, $a$ and $b$.

## Solution

The Prandtl stress function must satisfy the compatibility condition

$\nabla^2{\phi} = -2\mu\alpha ~~\Rightarrow~~ \phi_{,11}+\phi_{,22} = -2\mu\alpha$

Plugging in the stress function, we have,

$C\left[\frac{2}{a^2} + \frac{2}{b^2}\right] = -2\mu\alpha$

or,

${ C = -\frac{\mu\alpha~a^2~b^2}{a^2+b^2} }$

The torque for a simply connected section is given by

$T = 2\int_{\mathcal S} \phi~dA$

For the elliptical cross-section, we have

\begin{align} T & = 2\int_{-a}^a \left[\int_{-b\sqrt{(1-x^2/a^2)}}^{b\sqrt{(1-x^2/a^2)}} C\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1\right)~dy\right]~dx \\ & = 2C\int_{-a}^a \left[ \left|\frac{x^2~y}{a^2} + \frac{y^3}{3b^2} - y \right|_{-b\sqrt{(1-x^2/a^2)}}^{b\sqrt{(1-x^2/a^2)}} \right]~dx \\ & = 2C\int_{-a}^a \left[ \left|y\left[\frac{y^2}{3b^2} - \left(1-\frac{x^2}{a^2}\right)\right] \right|_{-b\sqrt{(1-x^2/a^2)}}^{b\sqrt{(1-x^2/a^2)}} \right]~dx \\ & = 2C\int_{-a}^a 2b\left(\sqrt{1-\frac{x^2}{a^2}}\right) \left[\frac{1}{3}\left(1-\frac{x^2}{a^2}\right) - \left(1-\frac{x^2}{a^2}\right) \right]~dx \\ & = -\frac{8bC}{3}\int_{-a}^a \left[ \left(1-\frac{x^2}{a^2}\right)^{(3/2)} \right]~dx = -\frac{8bC}{3a^3}\int_{-a}^a \left[ \left(a^2-x^2\right)^{(3/2)} \right]~dx\\ & = -\frac{8bC}{3a^3}\left[(2)\frac{(1)(3)}{(2)(4)}\frac{\pi}{2}a^4\right] = -\pi~a~b~C \end{align}

Therefore,

$T = \pi a b \frac{\mu\alpha~a^2~b^2}{a^2+b^2}$

or,

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

The torsion constant $\tilde{J}$ is given by

${ \tilde{J} = \frac{T}{\mu\alpha} = \frac{\pi~a^3~b^3}{a^2+b^2} }$

The stresses in the section are given by

\begin{align} \sigma_{13} = \frac{\partial \phi}{\partial y} = \frac{2Cy}{b^2} \\ \sigma_{23} = -\frac{\partial \phi}{\partial x} = -\frac{2Cx}{a^2} \end{align}

The projected shear traction is

$\tau = \sqrt{\sigma_{13}^2 + \sigma_{23}^2} = 2|C|\sqrt{\frac{y^2}{b^4} + \frac{x^2}{a^4}}$

The maximum projected shear traction is at $(x,y) = (0,\pm b)$. Hence,

$\tau_{\text{max}} = \frac{2|C|}{b}$

To express the magnitude of the maximum shear stress in terms of $T$, $a$, and $b$, we use

$T = -\pi a b C ~~\Rightarrow~~ C = -\frac{T}{\pi a b}$

Therefore,

${ \tau_{\text{max}} = \frac{2|C|}{b} = \frac{2T}{\pi a b^2} }$