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,

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

Elasticity