# Introduction to Elasticity/Rotating rectangular beam

## Example : Rotating Rectangular Beam

 A rotating rectangular beam

The body force potential is given by

${\displaystyle {\text{(52)}}\qquad V=-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)}$

Hence,

${\displaystyle {\text{(53)}}\qquad \nabla ^{2}{V}=V_{,11}+V_{,22}=-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(2+2\right)}$

or,

${\displaystyle {\text{(54)}}\qquad \nabla ^{2}{V}=-2\rho {\dot {\theta }}^{2}}$

The compatibility condition (in terms of stress) is

${\displaystyle {\text{(55)}}\qquad \nabla ^{4}{\varphi }+\left(2-{\cfrac {1}{\alpha }}\right)\nabla ^{2}{V}=0}$

Plug ${\displaystyle V}$ in to get

${\displaystyle {\text{(56)}}\qquad \nabla ^{4}{\varphi }-\left(2-{\cfrac {1}{\alpha }}\right)2\rho {\dot {\theta }}^{2}=0}$

Since ${\displaystyle V}$ is even in ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ and BCs are homogeneous, assume

${\displaystyle {\text{(57)}}\qquad \varphi =Ax_{1}^{4}+Bx_{1}^{2}x_{2}^{2}+Cx_{2}^{4}+Dx_{1}^{2}+Ex_{2}^{2}}$

Hence,

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle x_{2}=\pm b}$.

{\displaystyle {\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,

{\displaystyle {\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,

${\displaystyle {\text{(68)}}\qquad \varphi ={\cfrac {\rho {\dot {\theta }}^{2}}{24}}x_{1}^{4}+Cx_{2}^{4}+{\cfrac {\rho {\dot {\theta }}^{2}b^{2}}{4}}x_{1}^{2}+Ex_{2}^{2}}$

Plug into compatibility equation

${\displaystyle {\text{(69)}}\qquad \varphi _{,1111}+2\varphi _{,1122}+\varphi _{,2222}-\left(2-{\cfrac {1}{\alpha }}\right)2\rho {\dot {\theta }}^{2}=0}$

to get

${\displaystyle {\text{(70)}}\qquad 24\left({\cfrac {\rho {\dot {\theta }}^{2}}{24}}+C\right)-\left(2-{\cfrac {1}{\alpha }}\right)2\rho {\dot {\theta }}^{2}=0}$

or,

{\displaystyle {\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 ${\displaystyle x_{1}=\pm a}$.

{\displaystyle {\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

${\displaystyle {\text{(74)}}\qquad \left(3-{\cfrac {2}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}}{2}}-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}=0}$

which cannot be true. So weak BCs on ${\displaystyle \sigma _{11}}$ need to be applied at ${\displaystyle x_{1}=\pm a}$.

{\displaystyle {\begin{aligned}{\text{(75)}}\qquad {\text{at}}~x_{1}=\pm a&&\int _{-b}^{b}\sigma _{11}dx_{2}=0\end{aligned}}}

Hence,

${\displaystyle {\text{(76)}}\qquad \left(3-{\cfrac {2}{\alpha }}\right){\cfrac {\rho {\dot {\theta }}^{2}}{2}}{\cfrac {2b^{3}}{3}}+4Eb-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(2a^{2}b+{\cfrac {2b^{3}}{3}}\right)=0}$

or,

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

Hence,

${\displaystyle {\text{(78)}}\qquad E={\cfrac {\rho {\dot {\theta }}^{2}}{4}}\left[a^{2}-\left(2-{\cfrac {2}{\alpha }}\right){\cfrac {b^{2}}{3}}\right]}$

The stress field is, therefore,

{\displaystyle {\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,

{\displaystyle {\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 (${\displaystyle \sigma _{11}}$) in a rotating rectangular beam
 Stresses (${\displaystyle \sigma _{22}}$) in a rotating rectangular beam