Introduction to Elasticity/Warping of rectangular cylinder

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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: $\eta > 0$ or $\eta = 0$

In both these cases, we get trivial values of $C_1 = C_2 = 0\,$ .

Case 2: $\eta < 0$

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 $\psi$ .

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