# Elasticity/Rotating rectangular beam
## Example : Rotating Rectangular Beam

* A 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
    ${\text{(57)}}\qquad \varphi =Ax_{1}^{4}+Bx_{1}^{2}x_{2}^{2}+Cx_{2}^{4}+Dx_{1}^{2}+Ex_{2}^{2}$
Hence,
    ${\begin{aligned}{\text{(58)}}\qquad \sigma _{11}&=\varphi _{,22}+V=2Bx_{1}^{2}+12Cx_{2}^{2}+2E-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)\\{\text{(59)}}\qquad \sigma _{22}&=\varphi _{,11}+V=12Ax_{1}^{2}+2Bx_{2}^{2}+2D-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)\\{\text{(60)}}\qquad \sigma _{12}&=-\varphi _{,12}=-4Bx_{1}x_{2}\end{aligned}}$
The traction BCs are
    ${\begin{aligned}{\text{(61)}}\qquad {\text{at}}~x_{1}=\pm a&&t_{1}=t_{2}=0\Rightarrow \sigma _{11}=\sigma _{12}=0\\{\text{(62)}}\qquad {\text{at}}~x_{2}=\pm b&&t_{1}=t_{2}=0\Rightarrow \sigma _{12}=\sigma _{22}=0\end{aligned}}$
Apply BCs at $x_{2}=\pm b$.
    ${\begin{aligned}{\text{(63}})\qquad \sigma _{22}=0&=12Ax_{1}^{2}+2Bb^{2}+2D-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+b^{2}\right)\\{\text{(64}})\qquad \sigma _{12}=0&=-4Bbx_{1}\end{aligned}}$
Therefore,
    ${\begin{aligned}{\text{(65)}}\qquad B&=0\\{\text{(66)}}\qquad A&={\cfrac {\rho {\dot {\theta }}^{2}}{24}}\\{\text{(67)}}\qquad D&={\cfrac {\rho {\dot {\theta }}^{2}b^{2}}{4}}\end{aligned}}$
We then have,
    ${\text{(68)}}\qquad \varphi ={\cfrac {\rho {\dot {\theta }}^{2}}{24}}x_{1}^{4}+Cx_{2}^{4}+{\cfrac {\rho {\dot {\theta }}^{2}b^{2}}{4}}x_{1}^{2}+Ex_{2}^{2}$
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,
    ${\begin{aligned}C&=\left(2-{\cfrac {1}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}}{12}}-{\cfrac {\rho {\dot {\theta }}^{2}}{24}}\\&=\left(3-{\cfrac {2}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}}{24}}\qquad {\text{(71)}}\end{aligned}}$
Apply BCs at $x_{1}=\pm a$.
    ${\begin{aligned}{\text{(72)}}\qquad \sigma _{11}=0&=2Ba^{2}+12Cx_{2}^{2}+2E-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(a^{2}+x_{2}^{2}\right)\\&=\left(3-{\cfrac {2}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}}{2}}x_{2}^{2}+2E-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(a^{2}+x_{2}^{2}\right)\qquad {\text{(73)}}\end{aligned}}$
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 $x_{1}=\pm a$.
    ${\begin{aligned}{\text{(75)}}\qquad {\text{at}}~x_{1}=\pm a&&\int _{-b}^{b}\sigma _{11}dx_{2}=0\end{aligned}}$
Hence,
    ${\text{(76)}}\qquad \left(3-{\cfrac {2}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}}{2}}{\cfrac {2b^{3}}{3}}+4Eb-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(2a^{2}b+{\cfrac {2b^{3}}{3}}\right)=0$
or,
    ${\text{(77)}}\qquad \left(2-{\cfrac {2}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}b^{2}}{3}}+4E-\rho {\dot {\theta }}^{2}a^{2}=0$
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,
    ${\begin{aligned}{\text{(79}})\qquad \sigma _{11}&=\left(3-{\cfrac {2}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}}{2}}x_{2}^{2}+{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left[a^{2}-\left(2-{\cfrac {2}{\alpha }}\right){\cfrac {b^{2}}{3}}\right]-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)\\{\text{(80}})\qquad \sigma _{22}&={\cfrac {\rho {\dot {\theta }}^{2}}{2}}x_{1}^{2}+{\cfrac {\rho {\dot {\theta }}^{2}b^{2}}{2}}-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)\\\qquad \sigma _{12}&=0\end{aligned}}$
or,
    ${\begin{aligned}{\text{(81)}}\qquad \sigma _{11}&={\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left[\left(a^{2}-x_{1}^{2}\right)+2\left(1-{\cfrac {1}{\alpha }}\right)\left(x_{2}^{2}-{\cfrac {b^{2}}{3}}\right)\right]\\{\text{(82)}}\qquad \sigma _{22}&={\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(b^{2}-x_{2}^{2}\right)\\\sigma _{12}&=0\end{aligned}}$
The displacements can be found in the standard manner.
* Stresses ($\sigma _{11}$) in a rotating rectangular beam
* Stresses ($\sigma _{22}$) in a rotating rectangular beam