# 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
    ${\begin{aligned}T&=2\int _{-a}^{a}\left[\int _{-b{\sqrt {(1-x^{2}/a^{2})}}}^{b{\sqrt {(1-x^{2}/a^{2})}}}C\left({\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}-1\right)~dy\right]~dx\\&=2C\int _{-a}^{a}\left[\left|{\frac {x^{2}~y}{a^{2}}}+{\frac {y^{3}}{3b^{2}}}-y\right|_{-b{\sqrt {(1-x^{2}/a^{2})}}}^{b{\sqrt {(1-x^{2}/a^{2})}}}\right]~dx\\&=2C\int _{-a}^{a}\left[\left|y\left[{\frac {y^{2}}{3b^{2}}}-\left(1-{\frac {x^{2}}{a^{2}}}\right)\right]\right|_{-b{\sqrt {(1-x^{2}/a^{2})}}}^{b{\sqrt {(1-x^{2}/a^{2})}}}\right]~dx\\&=2C\int _{-a}^{a}2b\left({\sqrt {1-{\frac {x^{2}}{a^{2}}}}}\right)\left[{\frac {1}{3}}\left(1-{\frac {x^{2}}{a^{2}}}\right)-\left(1-{\frac {x^{2}}{a^{2}}}\right)\right]~dx\\&=-{\frac {8bC}{3}}\int _{-a}^{a}\left[\left(1-{\frac {x^{2}}{a^{2}}}\right)^{(3/2)}\right]~dx=-{\frac {8bC}{3a^{3}}}\int _{-a}^{a}\left[\left(a^{2}-x^{2}\right)^{(3/2)}\right]~dx\\&=-{\frac {8bC}{3a^{3}}}\left[(2){\frac {(1)(3)}{(2)(4)}}{\frac {\pi }{2}}a^{4}\right]=-\pi ~a~b~C\end{aligned}}$
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
    ${\begin{aligned}\sigma _{13}={\frac {\partial \phi }{\partial y}}={\frac {2Cy}{b^{2}}}\\\sigma _{23}=-{\frac {\partial \phi }{\partial x}}=-{\frac {2Cx}{a^{2}}}\end{aligned}}$
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,
    ${\tau _{\text{max}}={\frac {2|C|}{b}}={\frac {2T}{\pi ab^{2}}}}$