Elasticity/Rotating rectangular beam

Example : Rotating Rectangular Beam
The body force potential is given by
 * $$\text{(52)}\qquad

V = -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right) $$ Hence,
 * $$ \text{(53)}\qquad

\nabla^2{V} = V_{,11} + V_{,22} = -\cfrac{\rho\dot{\theta}^2}{2} \left(2 + 2\right) $$ or,
 * $$ \text{(54)}\qquad

\nabla^2{V} = -2\rho\dot{\theta}^2 $$ The compatibility condition (in terms of stress) is
 * $$ \text{(55)} \qquad

\nabla^4{\varphi} + \left(2 - \cfrac{1}{\alpha}\right) \nabla^2{V} = 0 $$ Plug $$V$$ in to get
 * $$ \text{(56)} \qquad

\nabla^4{\varphi} - \left(2-\cfrac{1}{\alpha}\right) 2\rho\dot{\theta}^2 = 0 $$ Since $$V$$ is even in $$x_1$$ and $$x_2$$ and BCs are homogeneous, assume
 * $$ \text{(57)} \qquad

\varphi = A x_1^4 + B x_1^2 x_2^2 + C x_2^4 + D x_1^2 + E x_2^2 $$ Hence,
 * $$\begin{align}

\text{(58)}\qquad \sigma_{11} & = \varphi_{,22} + V = 2 B x_1^2 + 12 C x_2^2 + 2 E     -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right)\\ \text{(59)}\qquad\sigma_{22} & = \varphi_{,11} + V = 12 A x_1^2 + 2 B x_2^2 + 2 D      -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right)\\ \text{(60)}\qquad\sigma_{12} & = -\varphi_{,12} = - 4 B x_1 x_2 \end{align}$$ The traction BCs are
 * $$\begin{align}

\text{(61)} \qquad \text{at}~ x_1 = \pm a & &  t_1 = t_2 = 0 \Rightarrow \sigma_{11} = \sigma_{12} = 0 \\ \text{(62)} \qquad \text{at}~ x_2 = \pm b & & t_1 = t_2 = 0 \Rightarrow \sigma_{12} = \sigma_{22} = 0 \end{align}$$ Apply BCs at $$x_2 = \pm b$$.
 * $$\begin{align}

\text{(63}) \qquad \sigma_{22} = 0 & = 12 A x_1^2 + 2 B b^2 + 2 D      -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + b^2\right)\\ \text{(64}) \qquad \sigma_{12} = 0 & = - 4 B b x_1 \end{align}$$ Therefore,
 * $$\begin{align}

\text{(65)} \qquad B & = 0 \\ \text{(66)} \qquad A & = \cfrac{\rho\dot{\theta}^2}{24} \\ \text{(67)} \qquad D & = \cfrac{\rho\dot{\theta}^2b^2}{4} \end{align}$$ We then have,
 * $$ \text{(68)} \qquad

\varphi = \cfrac{\rho\dot{\theta}^2}{24} x_1^4 + C x_2^4 + \cfrac{\rho\dot{\theta}^2b^2}{4} x_1^2 + E x_2^2 $$ Plug into compatibility equation
 * $$ \text{(69)} \qquad

\varphi_{,1111} + 2\varphi_{,1122} + \varphi_{,2222} - \left(2-\cfrac{1}{\alpha}\right) 2\rho\dot{\theta}^2 = 0 $$ to get
 * $$ \text{(70)} \qquad

24 \left(\cfrac{\rho\dot{\theta}^2}{24} + C\right) - \left(2-\cfrac{1}{\alpha}\right) 2\rho\dot{\theta}^2 = 0 $$ or,
 * $$\begin{align}

C & = \left(2-\cfrac{1}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{12} - \cfrac{\rho\dot{\theta}^2}{24} \\ & = \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{24} \qquad \text{(71)} \end{align}$$ Apply BCs at $$x_1 = \pm a$$.
 * $$\begin{align}

\text{(72)} \qquad \sigma_{11} = 0 & = 2 B a^2 + 12 C x_2^2 + 2 E     -\cfrac{\rho\dot{\theta}^2}{2} \left(a^2 + x_2^2\right)\\ & = \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2} x_2^2 + 2 E -\cfrac{\rho\dot{\theta}^2}{2} \left(a^2 + x_2^2\right) \qquad \text{(73)} \end{align}$$ Strong BCs imply that
 * $$ \text{(74)} \qquad

\left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2} -\cfrac{\rho\dot{\theta}^2}{2} = 0 $$ which cannot be true. So weak BCs on $$\sigma_{11}$$ need to be applied at $$x_1 = \pm a$$.
 * $$\begin{align}

\text{(75)} \qquad \text{at}~ x_1 = \pm a & & \int_{-b}^{b} \sigma_{11} dx_2 = 0 \end{align}$$ Hence,
 * $$\text{(76)} \qquad

\left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2}\cfrac{2b^3}{3} + 4 Eb -\cfrac{\rho\dot{\theta}^2}{2} \left(2a^2b + \cfrac{2b^3}{3}\right)=0 $$ or,
 * $$\text{(77)} \qquad

\left(2-\cfrac{2}{\alpha}\right)\cfrac{\rho\dot{\theta}^2b^2}{3} + 4 E -\rho\dot{\theta}^2a^2 = 0 $$ Hence,
 * $$\text{(78)} \qquad

E = \cfrac{\rho\dot{\theta}^2}{4}\left[a^2- \left(2-\cfrac{2}{\alpha}\right)\cfrac{b^2}{3} \right] $$ The stress field is, therefore,
 * $$\begin{align}

\text{(79}) \qquad \sigma_{11} & = \left(3-\cfrac{2}{\alpha}\right) \cfrac{\rho\dot{\theta}^2}{2} x_2^2 + \cfrac{\rho\dot{\theta}^2}{2}\left[a^2- \left(2-\cfrac{2}{\alpha}\right)\cfrac{b^2}{3} \right] -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right)\\ \text{(80}) \qquad \sigma_{22} & = \cfrac{\rho\dot{\theta}^2}{2} x_1^2 + \cfrac{\rho\dot{\theta}^2b^2}{2} -\cfrac{\rho\dot{\theta}^2}{2} \left(x_1^2 + x_2^2\right) \\ \qquad \sigma_{12} & = 0 \end{align}$$ or,
 * $$\begin{align}

\text{(81)} \qquad \sigma_{11} & = \cfrac{\rho\dot{\theta}^2}{2}\left[\left(a^2 - x_1^2\right)+ 2\left(1-\cfrac{1}{\alpha}\right)\left(x_2^2-\cfrac{b^2}{3}\right)\right] \\   \text{(82)} \qquad \sigma_{22} & = \cfrac{\rho\dot{\theta}^2}{2}\left(b^2 - x_2^2\right) \\   \sigma_{12} & = 0 \end{align}$$ The displacements can be found in the standard manner.