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.

File:Elliptical cross section.png 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 $\varphi$ 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

$\varphi =C[\frac{x^2}{a^2}+\frac{y^2}{b^2}-1]$

• (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.

$\text{If}f(x)\text{is an even function of}x\text{, then}\underset{-a}{\overset{a}{\int }}f(x)dx=2\underset{0}{\overset{a}{\int }}f(x)dx$

$\frac{1}{2}\underset{-a}{\overset{a}{\int }}(a^2-x^2{)}^{n/2}dx=\frac{1.3.5\dots n}{2.4.6\dots (n+1)}.\frac{\pi }{2}.a^{(n+1)}\text{if}n\text{is odd}$

• (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

${\mathrm{\nabla }}^2\varphi =-2\mu \alpha \Rightarrow {\varphi }_{,11}+{\varphi }_{,22}=-2\mu \alpha$

Plugging in the stress function, we have,

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

or,

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

The torque for a simply connected section is given by

$T=2\underset{𝒮}{\int }\varphi dA$

For the elliptical cross-section, we have

$\begin{array}{cc}T& =2\underset{-a}{\overset{a}{\int }}\left[\underset{-b\sqrt{(1-x^2/a^2)}}{\overset{b\sqrt{(1-x^2/a^2)}}{\int }}C(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1)dy\right]dx\\ & =2C\underset{-a}{\overset{a}{\int }}\left[{|\frac{x^{2}y}{a^2}+\frac{y^3}{3b^2}-y|}_{-b\sqrt{(1-x^2/a^2)}}^{b\sqrt{(1-x^2/a^2)}}\right]dx\\ & =2C\underset{-a}{\overset{a}{\int }}\left[{\left|y[\frac{y^2}{3b^2}-(1-\frac{x^2}{a^2})]\right|}_{-b\sqrt{(1-x^2/a^2)}}^{b\sqrt{(1-x^2/a^2)}}\right]dx\\ & =2C\underset{-a}{\overset{a}{\int }}2b\left(\sqrt{1-\frac{x^2}{a^2}}\right)[\frac{1}{3}(1-\frac{x^2}{a^2})-(1-\frac{x^2}{a^2})]dx\\ & =-\frac{8bC}{3}\underset{-a}{\overset{a}{\int }}\left[{(1-\frac{x^2}{a^2})}^{(3/2)}\right]dx=-\frac{8bC}{3a^3}\underset{-a}{\overset{a}{\int }}\left[{(a^2-x^2)}^{(3/2)}\right]dx\\ & =-\frac{8bC}{3a^3}[(2)\frac{(1)(3)}{(2)(4)}\frac{\pi }{2}{a}^4]=-\pi abC\end{array}$

Therefore,

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

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

$\begin{array}{c}{\sigma }_{13}=\frac{\partial \varphi }{\partial y}=\frac{2Cy}{b^2}\\ {\sigma }_{23}=-\frac{\partial \varphi }{\partial x}=-\frac{2Cx}{a^2}\end{array}$

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

$T=-\pi abC\Rightarrow C=-\frac{T}{\pi ab}$

Therefore,

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