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. 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} } $$