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

$({\psi }_{,1}-x_2){\hat{n}}_1+({\psi }_{,2}+x_1){\hat{n}}_2=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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 ${\mathrm{\nabla }}^2\psi =0$.

We can simplify the problem by a change of variable

$\overline{\psi }=x_1{x}_2-\psi$

Then the equilibrium condition becomes

${\mathrm{\nabla }}^2\overline{\psi }=0$

The traction-free BCs become

${\overline{\psi }}_{,1}=0\text{on}{x}_1=\pm a,\text{and}{\overline{\psi }}_{,2}=2x_1\text{on}{x}_2=\pm b$

Let us assume that

$\overline{\psi }(x_1,x_2)=f(x_1)g(x_2)$

Then,

${\mathrm{\nabla }}^2\overline{\psi }={\overline{\psi }}_{,11}+{\overline{\psi }}_{,22}=f^{{\text{'}}^{\prime }}(x_1)g(x_2)+g^{{\text{'}}^{\prime }}(x_2)f(x_1)=0$

or,

$\frac{f^{{\text{'}}^{\prime }}(x_1)}{f(x_1)}=-\frac{g^{{\text{'}}^{\prime }}(x_2)}{g(x_2)}=\eta$

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,

$\begin{array}{cc}{f}^{{\text{'}}^{\prime }}(x_1)+k^{2}f(x_1)=0\Rightarrow & f(x_1)=C_1\cos (k{x}_1)+C_2\sin (k{x}_1)\\ g^{{\text{'}}^{\prime }}(x_2)-k^{2}g(x_2)=0\Rightarrow & g(x_2)=C_3\cosh (k{x}_2)+C_4\sinh (k{x}_2)\end{array}$

Therefore,

$\overline{\psi }(x_1,x_2)=[C_1\cos (k{x}_1)+C_2\sin (k{x}_1)][C_3\cosh (k{x}_2)+C_4\sinh (k{x}_2)]$

Apply the BCs at $x_2=\pm b$ ~~ (${\overline{\psi }}_{,2}=2x_1$), to get

$\begin{array}{cc}[C_1\cos (k{x}_1)+C_2\sin (k{x}_1)][C_3\sinh (kb)+C_4\cosh (kb)]& =2x_1\\ [C_1\cos (k{x}_1)+C_2\sin (k{x}_1)][-C_3\sinh (kb)+C_4\cosh (kb)]& =2x_1\end{array}$

or,

$F(x_1)G^{\text{'}}(b)=2x_1;F(x_1)G^{\text{'}}(-b)=2x_1$

The RHS of both equations are odd. Therefore, $F(x_1)$ is odd. Since, $\cos (k{x}_1)$ is an even function, we must have $C_1=0$.

Also,

$F(x_1)[G^{\text{'}}(b)-G^{\text{'}}(-b)]=0$

Hence, $G^{\prime }(b)$ is even. Since $\sinh (kb)$ is an odd function, we must have $C_3=0$.

Therefore,

$\overline{\psi }(x_1,x_2)=C_2{C}_4\sin (k{x}_1)\sinh (k{x}_2)=A\sin (k{x}_1)\sinh (k{x}_2)$

Apply BCs at $x_1=\pm a$ (${\overline{\psi }}_{,1}=0$), to get

$Ak\cos (ka)\sinh (k{x}_2)=0$

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

$\overline{\psi }(x_1,x_2)=\sum _{n=0}^{\mathrm{\infty }}{A}_n\sin (k_n{x}_1)\sinh (k_n{x}_2)$

Applying the BCs at $x_1=\pm b$ again, we get

$\sum _{n=0}^{\mathrm{\infty }}{A}_n{k}_n\sin (k_n{x}_1)\cosh (k_{n}b)=2x_1\Rightarrow \sum _{n=0}^{\mathrm{\infty }}{B}_n\sin (k_n{x}_1)=2x_1$

Using the orthogonality of terms of the sine series,

$\underset{-a}{\overset{a}{\int }}\sin (k_n{x}_1)\sin (k_m{x}_1)d{x}_1=\{\begin{array}{cc}0& \mathrm{i}\mathrm{f}m\ne n\\ a& \mathrm{i}\mathrm{f}m=n\end{array}$

we have

$\underset{-a}{\overset{a}{\int }}[\sum _{n=0}^{\mathrm{\infty }}{B}_n\sin (k_n{x}_1)]\sin (k_m{x}_1)d{x}_1=\underset{-a}{\overset{a}{\int }}\left[2x_1\right]\sin (k_m{x}_1)d{x}_1$

or,

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

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

$\psi =x_1{x}_2-\frac{32a^2}{{\pi }^3}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n\sin (k_n{x}_1)\sinh (k_n{x}_2)}{(2n+1{)}^3\cosh (k_{n}b)}$

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