A rotating rectangular beam
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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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01a3c6d22634b54b778c99088821daf111d7ada7)
Hence,
![{\displaystyle {\text{(53)}}\qquad \nabla ^{2}{V}=V_{,11}+V_{,22}=-{\cfrac {\rho {\dot {\theta }}^{2}}{2}}\left(2+2\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3cb5226410e786756491054e1ca5db066296b21)
or,
![{\displaystyle {\text{(54)}}\qquad \nabla ^{2}{V}=-2\rho {\dot {\theta }}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f19fa298a83a00e5e1758554b51d0ae1d05eb2c)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c3e48e08d48804dd75963c51e7763f7692996fb)
Plug
in to get
![{\displaystyle {\text{(56)}}\qquad \nabla ^{4}{\varphi }-\left(2-{\cfrac {1}{\alpha }}\right)2\rho {\dot {\theta }}^{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4d51493d95a5f066b99b94ff5428f41e37fbafe)
Since
is even in
and
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3e7b79e0db74e5431c1c8ae4f45aa8106c70f3d)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3617fa46b0ada8dedeba9ee831a644d1b9d00e6d)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b505bd625668343cdf6f3495ad4fda17ed6706c)
Apply BCs at
.
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb5db6e4e4f6b53271142b71d992fef44eafd049)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f872116f11d343cfdf4c3ae9feca716eba5400ab)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/706ac805621b89c5d77fa5d969da03803db76f59)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3bcb112d2d39353ceb0bc27bc1c01f2af98a8df)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec3f454b26c8b1ea4692d399fed0f9f2ff6625ff)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01595348f4bf93806eb22aff4dc9e031eb8ea8d6)
Apply BCs at
.
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb7f7f7eece78cc2a44a8e4d64251ee063e6abaf)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b8f77b00748403397a50f042bed4fd1ace22e95)
which cannot be true. So weak BCs on
need to be applied at
.
![{\displaystyle {\begin{aligned}{\text{(75)}}\qquad {\text{at}}~x_{1}=\pm a&&\int _{-b}^{b}\sigma _{11}dx_{2}=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/589c19c996970bb9dbb780e1d579f3917ed06daf)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5d67f759d76a1f35aef80aa93fde735be132bf0)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d92f51bbf909d6e3980cf223422af68174d2fd1)
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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97fe7c9ef0e981d3a14a40ba56b0b219be2206a5)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3190a13ae7886a97334d71f05015e4e5f9b4167e)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57eba5cd415efb24a76ee11c9604bde31d3719b4)
The displacements can be found in the standard manner.
Stresses ( ) in a rotating rectangular beam
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Stresses ( ) in a rotating rectangular beam
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