Use Maple to reduce your manual labor.
The problem becomes easier to solve if we consider numerical values of the parameters. Let the local nodes numbers of element
be
for node
, and
for node
.
Let us assume that the beam is divided into six equal sectors. Then,
![{\displaystyle \theta ={\cfrac {\pi }{4}}~;~~\theta _{1}={\cfrac {\pi }{3}}~;~~\theta _{2}={\cfrac {\pi }{6}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c723d8617a5dec01b4f92a25fbeba0ed46a407d)
Let
and
. Since the blue point is midway between the two,
.
Also, let the components of the stiffness matrix of the composite be
![{\displaystyle C_{11}=10;~~C_{33}=100;~~C_{12}=6;~~C_{13}=60;~~C_{44}=30~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e83dcb08bf6cd8c94380a1aaefce7ab6a577839c)
Let the velocities for nodes
and
of the element be
![{\displaystyle \mathbf {v} _{1}={\begin{bmatrix}v_{1}^{x}\\v_{1}^{y}\\\omega _{1}\end{bmatrix}}={\begin{bmatrix}1\\2\\1\end{bmatrix}}~;~~\mathbf {v} _{2}={\begin{bmatrix}v_{2}^{x}\\v_{2}^{y}\\\omega _{2}\end{bmatrix}}={\begin{bmatrix}2\\1\\1\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8615fd0dc2368a551ddba512b2681b94c2561315)
The
and
coordinates of the master and slave nodes are
![{\displaystyle {\begin{bmatrix}x_{1}\\y_{1}\\x_{2}\\y_{2}\end{bmatrix}}={\begin{bmatrix}r\cos \theta _{1}\\r\sin \theta _{1}\\r\cos \theta _{2}\\r\sin \theta _{2}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d50a992f35a06017c10acca6c8a5a3e3be89ab4)
![{\displaystyle {\begin{bmatrix}x_{1-}\\y_{1-}\\x_{2-}\\y_{2-}\end{bmatrix}}={\begin{bmatrix}r_{1}\cos \theta _{1}\\r_{1}\sin \theta _{1}\\r_{1}\cos \theta _{2}\\r_{1}\sin \theta _{2}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e736366af8f58093f5ea9ff41596a4185fdfb8aa)
![{\displaystyle {\begin{bmatrix}x_{1+}\\y_{1+}\\x_{2+}\\y_{2+}\end{bmatrix}}={\begin{bmatrix}r_{2}\cos \theta _{1}\\r_{2}\sin \theta _{1}\\r_{2}\cos \theta _{2}\\r_{2}\sin \theta _{2}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/beabac4726b48af92f8bc3765eca4b1f05159f4f)
Since there are two master nodes in an element, the parent element is a four-noded quad with shape functions
![{\displaystyle {\begin{aligned}N_{1-}(\xi ,\eta )&={\cfrac {1}{4}}(1-\xi )(1-\eta )&N_{2-}(\xi ,\eta )&={\cfrac {1}{4}}(1+\xi )(1-\eta )\\N_{1+}(\xi ,\eta )&={\cfrac {1}{4}}(1-\xi )(1+\eta )&N_{2+}(\xi ,\eta )&={\cfrac {1}{4}}(1+\xi )(1+\eta )~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63bcb6ac25ebb00dce1ebf966c47aa6296850675)
The isoparametric map is
![{\displaystyle {\begin{aligned}x(\xi ,\eta )&=x_{1-}~N_{1^{-}}(\xi ,\eta )+x_{2-}~N_{2^{-}}(\xi ,\eta )+x_{1+}~N_{1^{+}}(\xi ,\eta )+x_{2+}~N_{2^{+}}(\xi ,\eta )\\y(\xi ,\eta )&=y_{1-}~N_{1^{-}}(\xi ,\eta )+y_{2-}~N_{2^{-}}(\xi ,\eta )+y_{1+}~N_{1^{+}}(\xi ,\eta )+y_{2+}~N_{2^{+}}(\xi ,\eta )~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bdbae2de76496701a623fac7611dc85219d339b)
Therefore, the derivatives with respect to
are
![{\displaystyle {\begin{aligned}x_{,\xi }={\frac {\partial x}{\partial \xi }}\\y_{,\xi }={\frac {\partial y}{\partial \xi }}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51b05dafb60b7c99a2caeb5b1ba4859c0823f972)
Since the blue point is at the center of the element, the values of
and
at that point are zero.
Therefore,
![{\displaystyle \mathbf {x} _{,\xi }=x_{,\xi }\mathbf {e} _{x}-y_{,\xi }\mathbf {e} _{y},~~\lVert \mathbf {x} _{,\xi }\rVert _{}={\sqrt {x_{,\xi }^{2}+y_{,\xi }^{2}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/420f64626043ee83b3f40d198db8b5df5020525b)
The local laminar basis vector
is given by
![{\displaystyle {\widehat {\mathbf {e} }}_{x}={\cfrac {\mathbf {x} _{,\xi }}{\lVert \mathbf {x} _{,\xi }\rVert _{}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45ffe6d2753b5f8720b1b856d4faa66749332523)
The laminar basis vector
is given by
![{\displaystyle {\widehat {\mathbf {e} }}_{y}=\mathbf {e} _{z}\times {\widehat {\mathbf {e} _{x}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/513777252ac737e8fa533784ac3199421ec45334)
To compute the velocity gradient, we have to find the velocities at the slave nodes using the relation
![{\displaystyle {\begin{bmatrix}v_{i-}^{x}\\v_{i-}^{y}\\v_{i+}^{x}\\v_{i+}^{y}\end{bmatrix}}={\begin{bmatrix}1&0&-(y_{i-}-y_{i})\\0&1&(x_{i-}-x_{i})\\1&0&-(y_{i+}-y_{i})\\0&1&(x_{i+}-x_{i})\end{bmatrix}}{\begin{bmatrix}v_{i}^{x}\\v_{i}^{y}\\\omega _{i}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48fb05ee11a7eb65e55673964f210c223beaaf65)
For master node 1 of the element (global node 5), the velocities of the slave nodes are
![{\displaystyle {\begin{bmatrix}v_{1-}^{x}\\v_{1-}^{y}\\v_{1+}^{x}\\v_{1+}^{y}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08c25c6cfaa336e2ef70104eddaa7e8a75e91663)
For master node 2 of the element (global node 6), the velocities of the slave nodes are
![{\displaystyle {\begin{bmatrix}v_{2-}^{x}\\v_{2-}^{y}\\v_{2+}^{x}\\v_{2+}^{y}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/482dba20916aa44e928ff8b1ad405465698ec366)
The interpolated velocity within the element is given by
![{\displaystyle {\begin{aligned}\mathbf {v} ({\boldsymbol {\xi }},t)&={\cfrac {D}{Dt}}[\mathbf {x} ({\boldsymbol {\xi }},t)]\\&=\sum _{i-=1}^{2}{\frac {\partial }{\partial t}}[\mathbf {x} _{i-}(t)]~N_{i^{-}}(\xi ,\eta )+\sum _{i+=1}^{2}{\frac {\partial }{\partial t}}[\mathbf {x} _{i+}(t)]~N_{i^{+}}(\xi ,\eta )\\&=\sum _{i-=1}^{2}\mathbf {v} _{i-}(t)~N_{i-}(\xi ,\eta )+\sum _{i+=1}^{2}\mathbf {v} _{i+}(t)~N_{i+}(\xi ,\eta )~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f78e49ade4daabcf9b49ab545d7d9060936bf55e)
The velocity gradient is given by
![{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} =v_{i,j}={\frac {\partial v_{i}}{\partial x_{j}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/290b69e9ae7a4285b65946d09110ae7c7ca1a482)
The velocities are given in terms of the parent element coordinates (
). We have to convert them to the (
) system in order to compute the velocity gradient. To do that we recall that
![{\displaystyle {\begin{aligned}{\frac {\partial v_{x}}{\partial \xi }}&={\frac {\partial v_{x}}{\partial x}}{\frac {\partial x}{\partial \xi }}+{\frac {\partial v_{x}}{\partial y}}{\frac {\partial y}{\partial \xi }}\\{\frac {\partial v_{x}}{\partial \eta }}&={\frac {\partial v_{x}}{\partial x}}{\frac {\partial x}{\partial \eta }}+{\frac {\partial v_{x}}{\partial y}}{\frac {\partial y}{\partial \eta }}\\{\frac {\partial v_{y}}{\partial \xi }}&={\frac {\partial v_{y}}{\partial x}}{\frac {\partial x}{\partial \xi }}+{\frac {\partial v_{y}}{\partial y}}{\frac {\partial y}{\partial \xi }}\\{\frac {\partial v_{y}}{\partial \eta }}&={\frac {\partial v_{y}}{\partial x}}{\frac {\partial x}{\partial \eta }}+{\frac {\partial v_{y}}{\partial y}}{\frac {\partial y}{\partial \eta }}~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04c273314d3807f76f610f72336f8d0ff67afea9)
In matrix form
![{\displaystyle {\begin{bmatrix}{\frac {\partial v_{x}}{\partial \xi }}\\{\frac {\partial v_{x}}{\partial \eta }}\end{bmatrix}}={\begin{bmatrix}{\frac {\partial x}{\partial \xi }}&{\frac {\partial y}{\partial \xi }}\\{\frac {\partial x}{\partial \eta }}&{\frac {\partial y}{\partial \eta }}\\\end{bmatrix}}{\begin{bmatrix}{\frac {\partial v_{x}}{\partial x}}\\{\frac {\partial v_{x}}{\partial y}}\end{bmatrix}}=\mathbf {J} {\begin{bmatrix}{\frac {\partial v_{x}}{\partial x}}\\{\frac {\partial v_{x}}{\partial y}}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d16df520ecb1cec60f59f16f8f9ccaeede31294b)
and
![{\displaystyle {\begin{bmatrix}{\frac {\partial v_{y}}{\partial \xi }}\\{\frac {\partial v_{y}}{\partial \eta }}\end{bmatrix}}={\begin{bmatrix}{\frac {\partial x}{\partial \xi }}&{\frac {\partial y}{\partial \xi }}\\{\frac {\partial x}{\partial \eta }}&{\frac {\partial y}{\partial \eta }}\\\end{bmatrix}}{\begin{bmatrix}{\frac {\partial v_{y}}{\partial x}}\\{\frac {\partial v_{y}}{\partial y}}\end{bmatrix}}=\mathbf {J} {\begin{bmatrix}{\frac {\partial v_{y}}{\partial x}}\\{\frac {\partial v_{y}}{\partial y}}\end{bmatrix}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96e101e2b08761108ef7663e424d502294bb17ec)
Therefore,
![{\displaystyle {\begin{bmatrix}{\frac {\partial v_{x}}{\partial x}}\\{\frac {\partial v_{x}}{\partial y}}\end{bmatrix}}=\mathbf {J} ^{-1}{\begin{bmatrix}{\frac {\partial v_{x}}{\partial \xi }}\\{\frac {\partial v_{x}}{\partial \eta }}\end{bmatrix}}\qquad {\text{and}}\qquad {\begin{bmatrix}{\frac {\partial v_{y}}{\partial x}}\\{\frac {\partial v_{y}}{\partial y}}\end{bmatrix}}=\mathbf {J} ^{-1}{\begin{bmatrix}{\frac {\partial v_{y}}{\partial \xi }}\\{\frac {\partial v_{y}}{\partial \eta }}\end{bmatrix}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e20672c9f68e97b02302ba45cb319429f20d75fd)
The rate of deformation is given by
![{\displaystyle {\boldsymbol {D}}={\frac {1}{2}}[{\boldsymbol {\nabla }}\mathbf {v} +({\boldsymbol {\nabla }}\mathbf {v} )^{T}]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7497036ca35bfe57c79750ca67a78956012c5965)
The global base vectors are
![{\displaystyle \mathbf {e} _{x}={\begin{bmatrix}1\\0\end{bmatrix}};~\qquad \mathbf {e} _{y}={\begin{bmatrix}0\\1\end{bmatrix}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9b45e3b588b6865a489ce289e85d989d695462f)
Therefore, the rotation matrix is
![{\displaystyle \mathbf {R} _{\text{lam}}={\begin{bmatrix}\mathbf {e} _{x}\bullet {\widehat {\mathbf {e} _{x}}}&\mathbf {e} _{x}\bullet {\widehat {\mathbf {e} _{y}}}\\\mathbf {e} _{y}\bullet {\widehat {\mathbf {e} _{x}}}&\mathbf {e} _{y}\bullet {\widehat {\mathbf {e} _{y}}}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/acea41edd26c5534739ff4b1fbe320214cb38558)
Therefore, the components of the rate of deformation tensor with respect to the laminar coordinate system are
![{\displaystyle \mathbf {D} _{\text{lam}}=\mathbf {R} _{\text{lam}}^{T}\mathbf {D} \mathbf {R} _{\text{lam}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e197e8b855e8a0da9ec573fac17fbe9a83d06b3d)
The rate constitutive relation of the material is given by
![{\displaystyle {\cfrac {D}{Dt}}{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{31}\\\sigma _{12}\end{bmatrix}}={\begin{bmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\C_{12}&C_{11}&C_{13}&0&0&0\\C_{13}&C_{13}&C_{33}&0&0&0\\0&0&0&C_{44}&0&0\\0&0&0&0&C_{44}&0\\0&0&0&0&0&(C_{11}-C_{12})/2\end{bmatrix}}{\begin{bmatrix}D_{11}\\D_{22}\\D_{33}\\D_{23}\\D_{31}\\D_{12}\end{bmatrix}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f4036326466fd4c196e96418b9c875fba43e6e9)
Since the problem is a 2-D one, the reduced constitutive equation is
![{\displaystyle {\cfrac {D}{Dt}}{\begin{bmatrix}\sigma _{11}\\\sigma _{33}\\\sigma _{31}\end{bmatrix}}={\begin{bmatrix}C_{11}&C_{13}&0\\C_{13}&C_{33}&0\\0&0&C_{44}\end{bmatrix}}{\begin{bmatrix}D_{11}\\D_{33}\\D_{31}\end{bmatrix}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d95a932bfe77760aa48f8590b4d6933c23c3e4d)
The laminar
-direction maps to the composite
-direction and the laminar
-directions maps to the composite
-direction. Hence the constitutive equation can be written as
![{\displaystyle {\cfrac {D}{Dt}}{\begin{bmatrix}\sigma _{yy}\\\sigma _{xx}\\\sigma _{xy}\end{bmatrix}}={\begin{bmatrix}C_{11}&C_{13}&0\\C_{13}&C_{33}&0\\0&0&C_{44}\end{bmatrix}}{\begin{bmatrix}D_{yy}\\D_{xx}\\D_{xy}\end{bmatrix}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6be8f6c1b9b70527dcef2f04ffff7e06032290ba)
Rearranging,
![{\displaystyle {\cfrac {D}{Dt}}{\begin{bmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{xy}\end{bmatrix}}={\begin{bmatrix}C_{33}&C_{13}&0\\C_{13}&C_{11}&0\\0&0&C_{44}\end{bmatrix}}{\begin{bmatrix}D_{xx}\\D_{yy}\\D_{xy}\end{bmatrix}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0b932c52c7032a1d38a2a47b371544e1199edc8)
The plane stress condition requires that
in the laminar coordinate system. We assume that the rate of
is also zero. In that case, we get
![{\displaystyle 0={\cfrac {D}{Dt}}(\sigma _{yy})=C_{13}~D_{xx}+C_{11}~D{yy}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e09d128983b6e00c4eec23b2740750865274a18)
or,
![{\displaystyle D_{yy}=-{\cfrac {C_{13}}{C_{11}}}~D_{xx}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b72aaf11beca0a6972fe229d80601063d136a5c)
To get the global stress rate and rate of deformation, we have to rotate the components to the global basis using
![{\displaystyle {\cfrac {D}{Dt}}({\boldsymbol {\sigma }})=\mathbf {R} _{\text{lam}}{\cfrac {D}{Dt}}({\boldsymbol {\sigma }}_{\text{lam}})\mathbf {R} _{\text{lam}}^{T}~;\qquad \mathbf {D} =\mathbf {R} _{\text{lam}}\mathbf {D} _{\text{lam}}\mathbf {R} _{\text{lam}}^{T}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93f4376286236c8711377c76d33b45a181cdb5fb)