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{align}

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{align}$$

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{align}

\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{align}$$ or,

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

A k \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) = 2 x_1 \Rightarrow \sum_{n=0}^{\infty} B_n \sin(k_n x_1) = 2 x_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 \ne 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[2 x_1\right] \sin(k_m x_1) dx_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}^{\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$$.