# 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 ${\displaystyle T}$ at the two ends which cause a twist per unit length of ${\displaystyle \alpha }$ in the bar.

The equation of the boundary of the cross section is

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

Since the Prandtl stress function ${\displaystyle \phi }$ 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

${\displaystyle \phi =C\left[{\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}-1\right]}$
• (a) Determine the value of the constant ${\displaystyle C}$.
• (b) Express the twist per unit length (${\displaystyle \alpha }$) in terms of the applied torque (${\displaystyle T}$).
You will find the following results useful in evaluating the integral.
${\displaystyle {\text{If}}~f(x)~{\text{is an even function of}}~x~{\text{, then}}~\int _{-a}^{a}f(x)dx=2\int _{0}^{a}f(x)dx}$
${\displaystyle {\frac {1}{2}}\int _{-a}^{a}(a^{2}-x^{2})^{n/2}dx={\frac {1~.~3~.~5~\dots ~n}{2~.~4~.~6~\dots ~(n+1)}}~.~{\frac {\pi }{2}}~.~a^{(n+1)}~~~{\text{if}}~n~{\text{is odd}}}$
• (c) What is the torsion constant of the section?
• (d) Express the maximum shear stress in the bar in terms of ${\displaystyle T}$, ${\displaystyle a}$ and ${\displaystyle b}$.

## Solution

The Prandtl stress function must satisfy the compatibility condition

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

Plugging in the stress function, we have,

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

or,

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

The torque for a simply connected section is given by

${\displaystyle T=2\int _{\mathcal {S}}\phi ~dA}$

For the elliptical cross-section, we have

{\displaystyle {\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,

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

or,

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

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

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

The stresses in the section are given by

{\displaystyle {\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

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

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

${\displaystyle \tau _{\text{max}}={\frac {2|C|}{b}}}$

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

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

Therefore,

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