# Elasticity/Warping of rectangular cylinder
## Example 3: Rectangular Cylinder

In this case, the form of $\psi \,$ is not obvious and has to be derived from the traction-free BCs
    $(\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
Suppose that $2a\,$ and $2b\,$ are the two sides of the rectangle, and $a>b\,$. Also $a\,$ is the side parallel to $x_{1}\,$ and $b\,$ is the side parallel to $x_{2}\,$. Then, the traction-free BCs are
    $\psi _{,1}=x_{2}~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~\psi _{,2}=-x_{1}~~{\text{on}}~~x_{2}=\pm b\,$
A suitable $\psi \,$ must satisfy these BCs and $\nabla ^{2}{\psi }=0\,$.
We can simplify the problem by a change of variable
    ${\bar {\psi }}=x_{1}~x_{2}-\psi$
Then the equilibrium condition becomes
    $\nabla ^{2}{\bar {\psi }}=0$
The traction-free BCs become
    ${\bar {\psi }}_{,1}=0~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~{\bar {\psi }}_{,2}=2x_{1}~~{\text{on}}~~x_{2}=\pm b$
Let us assume that
    ${\bar {\psi }}(x_{1},x_{2})=f(x_{1})g(x_{2})$
Then,
    $\nabla ^{2}{\bar {\psi }}={\bar {\psi }}_{,11}+{\bar {\psi }}_{,22}=f^{''}(x_{1})g(x_{2})+g^{''}(x_{2})f(x_{1})=0$
or,
    ${\frac {f^{''}(x_{1})}{f(x_{1})}}=-{\frac {g^{''}(x_{2})}{g(x_{2})}}=\eta$

### Case 1: η > 0 or η = 0

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

### Case 2: η < 0

Let
    $\eta =-k^{2}~~~;~~k>0$
Then,
    ${\begin{aligned}f^{''}(x_{1})+k^{2}f(x_{1})=0~~\Rightarrow &~~f(x_{1})=C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\\g^{''}(x_{2})-k^{2}g(x_{2})=0~~\Rightarrow &~~g(x_{2})=C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\end{aligned}}$
Therefore,
    ${\bar {\psi }}(x_{1},x_{2})=\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\right]$
Apply the BCs at $x_{2}=\pm b\,$ ~~ (${\bar {\psi }}_{,2}=2x_{1}$), to get
    ${\begin{aligned}\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\\\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[-C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\end{aligned}}$
or,
    $F(x_{1})G^{'}(b)=2x_{1}~~;~~~F(x_{1})G^{'}(-b)=2x_{1}\,$
The RHS of both equations are odd. Therefore, $F(x_{1})$ is odd. Since, $\cos(kx_{1})\,$ is an even function, we must have $C_{1}=0\,$.
Also,
    $F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0$
Hence, $G'(b)\,$ is even. Since $\sinh(kb)\,$ is an odd function, we must have $C_{3}=0\,$.
Therefore,
    ${\bar {\psi }}(x_{1},x_{2})=C_{2}C_{4}\sin(kx_{1})\sinh(kx_{2})=A\sin(kx_{1})\sinh(kx_{2})$
Apply BCs at $x_{1}=\pm a\,$ (${\bar {\psi }}_{,1}=0$), to get
    $Ak\cos(ka)\sinh(kx_{2})=0\,$
The only nontrivial solution is obtained when $\cos(ka)=0$, which means that
    $k_{n}={\frac {(2n+1)\pi }{2a}}~~,~~~n=0,1,2,...$
The BCs at $x_{1}=\pm a\,$ are satisfied by every terms of the series
    ${\bar {\psi }}(x_{1},x_{2})=\sum _{n=0}^{\infty }A_{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})$
Applying the BCs at $x_{1}=\pm b\,$ again, we get
    $\sum _{n=0}^{\infty }A_{n}k_{n}\sin(k_{n}x_{1})\cosh(k_{n}b)=2x_{1}~~\Rightarrow ~~~\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})=2x_{1}$
Using the orthogonality of terms of the sine series,
    $\int _{-a}^{a}\sin(k_{n}x_{1})\sin(k_{m}x_{1})dx_{1}={\begin{cases}0&{\rm {if}}~m\neq n\\a&{\rm {if}}~m=n\end{cases}}$
we have
    $\int _{-a}^{a}\left[\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})\right]\sin(k_{m}x_{1})dx_{1}=\int _{-a}^{a}\left[2x_{1}\right]\sin(k_{m}x_{1})dx_{1}$
or,
    $B_{m}a={\frac {4}{ak_{m}^{2}}}\sin(k_{m}a)$
Now,
    $\sin(k_{m}a)=\sin \left({\frac {(2m+1)\pi }{2}}\right)=(-1)^{m}$
Therefore,
    $A_{m}={\frac {B_{m}}{k_{m}\cosh(k_{m}b)}}={\frac {(-1)^{m}32a^{2}}{(2m+1)^{3}\pi ^{3}\cosh(k_{m}b)}}$
The warping function is
    $\psi =x_{1}x_{2}-{\frac {32a^{2}}{\pi ^{3}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}{(2n+1)^{3}\cosh(k_{n}b)}}$
The torsion constant and the stresses can be calculated from $\psi$.