# Introduction to Elasticity/Warping of rectangular cylinder

## Example 3: Rectangular Cylinder

In this case, the form of ${\displaystyle \psi \,}$ is not obvious and has to be derived from the traction-free BCs

${\displaystyle (\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}$

Suppose that ${\displaystyle 2a\,}$ and ${\displaystyle 2b\,}$ are the two sides of the rectangle, and ${\displaystyle a>b\,}$. Also ${\displaystyle a\,}$ is the side parallel to ${\displaystyle x_{1}\,}$ and ${\displaystyle b\,}$ is the side parallel to ${\displaystyle x_{2}\,}$. Then, the traction-free BCs are

${\displaystyle \psi _{,1}=x_{2}~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~\psi _{,2}=-x_{1}~~{\text{on}}~~x_{2}=\pm b\,}$

A suitable ${\displaystyle \psi \,}$ must satisfy these BCs and ${\displaystyle \nabla ^{2}{\psi }=0\,}$.

We can simplify the problem by a change of variable

${\displaystyle {\bar {\psi }}=x_{1}~x_{2}-\psi }$

Then the equilibrium condition becomes

${\displaystyle \nabla ^{2}{\bar {\psi }}=0}$

The traction-free BCs become

${\displaystyle {\bar {\psi }}_{,1}=0~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~{\bar {\psi }}_{,2}=2x_{1}~~{\text{on}}~~x_{2}=\pm b}$

Let us assume that

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=f(x_{1})g(x_{2})}$

Then,

${\displaystyle \nabla ^{2}{\bar {\psi }}={\bar {\psi }}_{,11}+{\bar {\psi }}_{,22}=f^{''}(x_{1})g(x_{2})+g^{''}(x_{2})f(x_{1})=0}$

or,

${\displaystyle {\frac {f^{''}(x_{1})}{f(x_{1})}}=-{\frac {g^{''}(x_{2})}{g(x_{2})}}=\eta }$

### Case 1: ${\displaystyle \eta >0}$ or ${\displaystyle \eta =0}$

In both these cases, we get trivial values of ${\displaystyle C_{1}=C_{2}=0\,}$.

### Case 2: ${\displaystyle \eta <0}$

Let

${\displaystyle \eta =-k^{2}~~~;~~k>0}$

Then,

{\displaystyle {\begin{aligned}f^{''}(x_{1})+k^{2}f(x_{1})=0~~\Rightarrow &~~f(x_{1})=C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\\g^{''}(x_{2})-k^{2}g(x_{2})=0~~\Rightarrow &~~g(x_{2})=C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\end{aligned}}}

Therefore,

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\right]}$

Apply the BCs at ${\displaystyle x_{2}=\pm b\,}$ ~~ (${\displaystyle {\bar {\psi }}_{,2}=2x_{1}}$), to get

{\displaystyle {\begin{aligned}\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\\\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[-C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\end{aligned}}}

or,

${\displaystyle F(x_{1})G^{'}(b)=2x_{1}~~;~~~F(x_{1})G^{'}(-b)=2x_{1}\,}$

The RHS of both equations are odd. Therefore, ${\displaystyle F(x_{1})}$ is odd. Since, ${\displaystyle \cos(kx_{1})\,}$ is an even function, we must have ${\displaystyle C_{1}=0\,}$.

Also,

${\displaystyle F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0}$

Hence, ${\displaystyle G'(b)\,}$ is even. Since ${\displaystyle \sinh(kb)\,}$ is an odd function, we must have ${\displaystyle C_{3}=0\,}$.

Therefore,

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=C_{2}C_{4}\sin(kx_{1})\sinh(kx_{2})=A\sin(kx_{1})\sinh(kx_{2})}$

Apply BCs at ${\displaystyle x_{1}=\pm a\,}$ (${\displaystyle {\bar {\psi }}_{,1}=0}$), to get

${\displaystyle Ak\cos(ka)\sinh(kx_{2})=0\,}$

The only nontrivial solution is obtained when ${\displaystyle \cos(ka)=0}$, which means that

${\displaystyle k_{n}={\frac {(2n+1)\pi }{2a}}~~,~~~n=0,1,2,...}$

The BCs at ${\displaystyle x_{1}=\pm a\,}$ are satisfied by every terms of the series

${\displaystyle {\bar {\psi }}(x_{1},x_{2})=\sum _{n=0}^{\infty }A_{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}$

Applying the BCs at ${\displaystyle x_{1}=\pm b\,}$ again, we get

${\displaystyle \sum _{n=0}^{\infty }A_{n}k_{n}\sin(k_{n}x_{1})\cosh(k_{n}b)=2x_{1}~~\Rightarrow ~~~\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})=2x_{1}}$

Using the orthogonality of terms of the sine series,

${\displaystyle \int _{-a}^{a}\sin(k_{n}x_{1})\sin(k_{m}x_{1})dx_{1}={\begin{cases}0&{\rm {if}}~m\neq n\\a&{\rm {if}}~m=n\end{cases}}}$

we have

${\displaystyle \int _{-a}^{a}\left[\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})\right]\sin(k_{m}x_{1})dx_{1}=\int _{-a}^{a}\left[2x_{1}\right]\sin(k_{m}x_{1})dx_{1}}$

or,

${\displaystyle B_{m}a={\frac {4}{ak_{m}^{2}}}\sin(k_{m}a)}$

Now,

${\displaystyle \sin(k_{m}a)=\sin \left({\frac {(2m+1)\pi }{2}}\right)=(-1)^{m}}$

Therefore,

${\displaystyle A_{m}={\frac {B_{m}}{k_{m}\cosh(k_{m}b)}}={\frac {(-1)^{m}32a^{2}}{(2m+1)^{3}\pi ^{3}\cosh(k_{m}b)}}}$

The warping function is

${\displaystyle \psi =x_{1}x_{2}-{\frac {32a^{2}}{\pi ^{3}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}{(2n+1)^{3}\cosh(k_{n}b)}}}$

The torsion constant and the stresses can be calculated from ${\displaystyle \psi }$.