Part 1

Write down expressions for ${\displaystyle N_{1}}$, ${\displaystyle N_{2}}$, and ${\displaystyle N_{3}}$ in terms of the initial configuration (${\displaystyle X,Y}$) ?

In the initial configuration ${\displaystyle t=0}$. Therefore,

${\displaystyle X_{1}=x_{1}(0)=0~;~~Y_{1}=y_{1}(0)=0~;\qquad X_{2}=x_{2}(0)=2~;~~Y_{2}=y_{2}(0)=0~;\qquad X_{3}=x_{3}(0)=0~;~~Y_{3}=y_{3}(0)=1~.}$

Therefore, the initial configuration is given by

${\displaystyle X({\boldsymbol {\xi }})=\sum _{i=1}^{3}X_{i}~N_{i}({\boldsymbol {\xi }})~;\qquad Y({\boldsymbol {\xi }})=\sum _{i=1}^{3}Y_{i}~N_{i}({\boldsymbol {\xi }})~.}$

Substituting the values of ${\displaystyle X_{i}}$ and ${\displaystyle Y_{i}}$, we get

${\displaystyle X({\boldsymbol {\xi }})=2~N_{2}({\boldsymbol {\xi }})~;\qquad Y({\boldsymbol {\xi }})=N_{3}({\boldsymbol {\xi }})~.}$

Hence

${\displaystyle N_{2}({\boldsymbol {\xi }})={\cfrac {X({\boldsymbol {\xi }})}{2}}~;\qquad N_{3}({\boldsymbol {\xi }})=Y({\boldsymbol {\xi }})~.}$

Also, the shape functions must satisfy the partition of unity condition

${\displaystyle \sum _{i=1}^{3}N_{i}({\boldsymbol {\xi }})=1~.}$

Therefore,

${\displaystyle N_{1}({\boldsymbol {\xi }})=1-N_{2}({\boldsymbol {\xi }})-N_{3}({\boldsymbol {\xi }})=1-{\cfrac {X({\boldsymbol {\xi }})}{2}}-Y({\boldsymbol {\xi }})~.}$

The required expressions are

${\displaystyle {N_{1}({\boldsymbol {\xi }})=1-{\cfrac {X({\boldsymbol {\xi }})}{2}}-Y({\boldsymbol {\xi }})~;\qquad N_{2}({\boldsymbol {\xi }})={\cfrac {X({\boldsymbol {\xi }})}{2}}~;\qquad N_{3}({\boldsymbol {\xi }})=Y({\boldsymbol {\xi }})~.}}$

Part 2

Derive expressions for the deformation gradient and the Jacobian determinant for the element as functions of time.

The deformation gradient is given by

${\displaystyle \mathbf {F} =\left[{\begin{aligned}{\frac {\partial x}{\partial X}}&&{\frac {\partial x}{\partial Y}}&&{\frac {\partial x}{\partial Z}}\\{\frac {\partial y}{\partial X}}&&{\frac {\partial y}{\partial Y}}&&{\frac {\partial y}{\partial Z}}\\{\frac {\partial z}{\partial X}}&&{\frac {\partial z}{\partial Y}}&&{\frac {\partial z}{\partial Z}}\end{aligned}}\right]~.}$

Before computing the derivatives, let us express ${\displaystyle x,y,z}$ in terms of ${\displaystyle X,Y,Z}$. Recall

${\displaystyle x({\boldsymbol {\xi }},t)=\sum _{i=1}^{3}x_{i}(t)~N_{i}({\boldsymbol {\xi }})~;\qquad y({\boldsymbol {\xi }},t)=\sum _{i=1}^{3}y_{i}(t)~N_{i}({\boldsymbol {\xi }})}$

Therefore,

${\displaystyle {\begin{aligned}x({\boldsymbol {\xi }},t)&=x_{1}(t)~\left[1-{\cfrac {X({\boldsymbol {\xi }})}{2}}-Y({\boldsymbol {\xi }})\right]+x_{2}(t)~{\cfrac {X({\boldsymbol {\xi }})}{2}}+x_{3}(t)~Y({\boldsymbol {\xi }})\\y({\boldsymbol {\xi }},t)&=y_{1}(t)~\left[1-{\cfrac {X({\boldsymbol {\xi }})}{2}}-Y({\boldsymbol {\xi }})\right]+y_{2}(t)~{\cfrac {X({\boldsymbol {\xi }})}{2}}+y_{3}(t)~Y({\boldsymbol {\xi }})\\z({\boldsymbol {\xi }},t)&=Z({\boldsymbol {\xi }})\end{aligned}}}$

Substituting in the expressions for ${\displaystyle x_{i}}$ and ${\displaystyle y_{i}}$, we get

${\displaystyle {\begin{aligned}x({\boldsymbol {\xi }},t)&=\left[2(1+at)\cos {\cfrac {\pi t}{2}}\right]~{\cfrac {X({\boldsymbol {\xi }})}{2}}-\left[(1+bt)\sin {\cfrac {\pi t}{2}}\right]~Y({\boldsymbol {\xi }})\\y({\boldsymbol {\xi }},t)&=\left[2(1+at)\sin {\cfrac {\pi t}{2}}\right]~{\cfrac {X({\boldsymbol {\xi }})}{2}}+\left[(1+bt)\cos {\cfrac {\pi t}{2}}\right]~Y({\boldsymbol {\xi }})\\z({\boldsymbol {\xi }},t)&=Z({\boldsymbol {\xi }})~.\end{aligned}}}$

In the above expression, the parent coordinates ${\displaystyle {\boldsymbol {\xi }}}$ are no longer useful. Therefore, we write

${\displaystyle {\begin{aligned}x(\mathbf {X} ,t)&=X~(1+at)~\cos {\cfrac {\pi t}{2}}-Y~(1+bt)~\sin {\cfrac {\pi t}{2}}\\y(\mathbf {X} ,t)&=X~(1+at)~\sin {\cfrac {\pi t}{2}}+Y~(1+bt)~\cos {\cfrac {\pi t}{2}}\\z(\mathbf {X} ,t)&=Z\end{aligned}}}$

where ${\displaystyle \mathbf {X} =(X,Y,Z)}$. Taking derivatives, we get

${\displaystyle {\begin{aligned}{\frac {\partial x}{\partial X}}&=(1+at)~\cos {\cfrac {\pi t}{2}}\qquad &{\frac {\partial x}{\partial Y}}&=-(1+bt)~\sin {\cfrac {\pi t}{2}}\qquad &{\frac {\partial x}{\partial Z}}&=0\\{\frac {\partial y}{\partial X}}&=(1+at)~\sin {\cfrac {\pi t}{2}}\qquad &{\frac {\partial y}{\partial Y}}&=(1+bt)~\cos {\cfrac {\pi t}{2}}\qquad &{\frac {\partial y}{\partial Z}}&=0\\{\frac {\partial z}{\partial X}}&=0\qquad &{\frac {\partial z}{\partial Y}}&=0\qquad &{\frac {\partial z}{\partial Z}}&=1\end{aligned}}}$

Therefore,

${\displaystyle {\mathbf {F} (t)=\left[{\begin{aligned}(1+at)~\cos {\cfrac {\pi t}{2}}&&-(1+bt)~\sin {\cfrac {\pi t}{2}}&&0\\(1+at)~\sin {\cfrac {\pi t}{2}}&&(1+bt)~\cos {\cfrac {\pi t}{2}}&&0\\0&&0&&1\end{aligned}}\right]~.}}$

The Jacobian determinant is

${\displaystyle J(t)=\det \mathbf {F} (t)=\left[(1+at)~\cos {\cfrac {\pi t}{2}}\right]\left[(1+bt)~\cos {\cfrac {\pi t}{2}}\right]+\left[(1+bt)~\sin {\cfrac {\pi t}{2}}\right]\left[(1+at)~\sin {\cfrac {\pi t}{2}}\right]~.}$

or

${\displaystyle {J(t)=(1+at)(1+bt)~.}}$

Part 3

What are the values of ${\displaystyle a}$ and ${\displaystyle b}$ for which the motion is isochoric?

For isochoric motion, ${\displaystyle J=1}$. Therefore,

${\displaystyle (1+at)(1+bt)=1~.}$

One possibility is

${\displaystyle {a=b=0}}$

This is a pure rotation.

Another possibility is that

${\displaystyle {b=-{\cfrac {a}{1+at}}}}$

This is a combination of shear and rotation where the volume remains constant.

Part 4

For which values of ${\displaystyle a}$ and ${\displaystyle b}$ do we get invalid motions?

We get invalid motions when ${\displaystyle J\leq 0}$. Let us consider the case where ${\displaystyle J=0}$. Then

${\displaystyle (1+at)(1+bt)=0}$

This is possible when

${\displaystyle at=-1~{\text{or}}~bt=-1~.}$

If ${\displaystyle J<0}$ then

${\displaystyle 1+at<0~{\text{and}}~1+bt>0\qquad {\text{or}}\qquad 1+at>0~{\text{and}}~1+bt<0}$

That is,

${\displaystyle at<-1~{\text{and}}~bt>-1\qquad {\text{or}}\qquad at>-1~{\text{and}}~bt<-1}$

Therefore the values at which we get invalid motions are

${\displaystyle {a\leq -{\cfrac {1}{t}}~~{\text{or}}~~b\leq -{\cfrac {1}{t}}~.}}$

Part 5

Derive the expression for the Green (Lagrangian) strain tensor

for the element as a function of time.

The Green strain tensor is given by

${\displaystyle \mathbf {E} ={\frac {1}{2}}(\mathbf {F} ^{T}\mathbf {F} -\mathbf {I} )}$

Recall

${\displaystyle \mathbf {F} (t)=\left[{\begin{aligned}(1+at)~\cos {\cfrac {\pi t}{2}}&&-(1+bt)~\sin {\cfrac {\pi t}{2}}&&0\\(1+at)~\sin {\cfrac {\pi t}{2}}&&(1+bt)~\cos {\cfrac {\pi t}{2}}&&0\\0&&0&&1\end{aligned}}\right]~.}$

Let us make the following substitutions

${\displaystyle A:=(1+at)~;~~B:=(1+bt)~;~~c:=\cos {\cfrac {\pi t}{2}}~;~~s:=\sin {\cfrac {\pi t}{2}}~.}$

Then

${\displaystyle \mathbf {F} ={\begin{bmatrix}A~c&-B~s&0\\A~s&B~c&0\\0&0&1\end{bmatrix}}~~{\text{and}}~~\mathbf {F} ^{T}={\begin{bmatrix}A~c&A~s&0\\-B~s&B~c&0\\0&0&1\end{bmatrix}}~.}$

Therefore,

${\displaystyle \mathbf {F} ^{T}~\mathbf {F} ={\begin{bmatrix}A^{2}&0&0\\0&B^{2}&0\\0&0&1\end{bmatrix}}={\begin{bmatrix}(1+at)^{2}&0&0\\0&(1+bt)^{2}&0\\0&0&1\end{bmatrix}}~.}$

Hence,

${\displaystyle \mathbf {E} ={\frac {1}{2}}\left({\begin{bmatrix}(1+at)^{2}&0&0\\0&(1+bt)^{2}&0\\0&0&1\end{bmatrix}}-{\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}\right)}$

The Green strain is

${\displaystyle {\mathbf {E} (t)={\frac {1}{2}}{\begin{bmatrix}2at+a^{2}t^{2}&0&0\\0&2bt+b^{2}t^{2}&0\\0&0&0\end{bmatrix}}}}$

Part 6

Derive an expression for the velocity gradient as a function of

time.

The velocity is the material time derivative of the motion.

Recall that the motion is given by

${\displaystyle {\begin{aligned}x(\mathbf {X} ,t)&=X~(1+at)~\cos {\cfrac {\pi t}{2}}-Y~(1+bt)~\sin {\cfrac {\pi t}{2}}\\y(\mathbf {X} ,t)&=X~(1+at)~\sin {\cfrac {\pi t}{2}}+Y~(1+bt)~\cos {\cfrac {\pi t}{2}}\\z(\mathbf {X} ,t)&=Z\end{aligned}}}$

Therefore,

${\displaystyle {\begin{aligned}v_{x}(\mathbf {X} ,t)={\frac {\partial x}{\partial t}}&=X~\left[a~\cos {\cfrac {\pi t}{2}}-(1+at)~{\cfrac {\pi }{2}}~\sin {\cfrac {\pi t}{2}}\right]-Y~\left[b~\sin {\cfrac {\pi t}{2}}+(1+bt)~{\cfrac {\pi }{2}}~\cos {\cfrac {\pi t}{2}}\right]\\v_{y}(\mathbf {X} ,t)={\frac {\partial y}{\partial t}}&=X~\left[a~\sin {\cfrac {\pi t}{2}}+(1+at)~{\cfrac {\pi }{2}}~\cos {\cfrac {\pi t}{2}}\right]+Y~\left[b~\cos {\cfrac {\pi t}{2}}-(1+bt)~{\cfrac {\pi }{2}}~\sin {\cfrac {\pi t}{2}}\right]\\v_{z}\mathbf {X} ,t)={\frac {\partial z}{\partial t}}&=0\end{aligned}}}$

We could compute the velocity gradient using

${\displaystyle \mathbf {l} ={\boldsymbol {\nabla }}\mathbf {v} =\left[{\begin{aligned}{\frac {\partial v_{x}}{\partial x}}&&{\frac {\partial v_{x}}{\partial y}}&&{\frac {\partial v_{x}}{\partial z}}\\{\frac {\partial v_{y}}{\partial x}}&&{\frac {\partial v_{y}}{\partial y}}&&{\frac {\partial v_{y}}{\partial z}}\\{\frac {\partial v_{z}}{\partial x}}&&{\frac {\partial v_{z}}{\partial y}}&&{\frac {\partial v_{z}}{\partial z}}\end{aligned}}\right]}$

after expressing ${\displaystyle \mathbf {v} }$ in terms of ${\displaystyle \mathbf {x} }$. However, that makes the expression quite complicated. Instead, we will use the relation

${\displaystyle \mathbf {l} ={\dot {\mathbf {F} }}~\mathbf {F} ^{-1}~.}$

The time derivative of the deformation gradient is

${\displaystyle {\dot {\mathbf {F} }}=\left[{\begin{aligned}a~\cos {\cfrac {\pi t}{2}}-(1+at)~{\cfrac {\pi }{2}}~\sin {\cfrac {\pi t}{2}}&&-b~\sin {\cfrac {\pi t}{2}}-(1+bt)~{\cfrac {\pi }{2}}~\cos {\cfrac {\pi t}{2}}&&0\\a~\sin {\cfrac {\pi t}{2}}+(1+at)~{\cfrac {\pi }{2}}~\cos {\cfrac {\pi t}{2}}&&b~\cos {\cfrac {\pi t}{2}}-(1+bt)~{\cfrac {\pi }{2}}~\sin {\cfrac {\pi t}{2}}&&0\\0&&0&&0\end{aligned}}\right]~.}$

The inverse of the deformation gradient is

${\displaystyle \mathbf {F} ^{-1}={\cfrac {1}{(1+at)(1+bt)}}\left[{\begin{aligned}(1+bt)~\cos {\cfrac {\pi t}{2}}&&(1+bt)~\sin {\cfrac {\pi t}{2}}&&0\\-(1+at)~\sin {\cfrac {\pi t}{2}}&&(1+at)~\cos {\cfrac {\pi t}{2}}&&0\\0&&0&&(1+at)(1+bt)\end{aligned}}\right]~.}$

Using the substitutions

${\displaystyle A:=(1+at)~;~~B:=(1+bt)~;~~c:=\cos {\cfrac {\pi t}{2}}~;~~s:=\sin {\cfrac {\pi t}{2}}}$

we get

${\displaystyle {\dot {\mathbf {F} }}=\left[{\begin{aligned}a~c-A~s~{\cfrac {\pi }{2}}&&-b~s-B~c~{\cfrac {\pi }{2}}&&0\\a~s+A~c~{\cfrac {\pi }{2}}&&b~c-B~s~{\cfrac {\pi }{2}}&&0\\0&&0&&0\end{aligned}}\right]=\left[{\begin{aligned}a~c&&-b~s&&0\\a~s&&b~c&&0\\0&&0&&0\end{aligned}}\right]+{\cfrac {\pi }{2}}\left[{\begin{aligned}-A~s&&-B~c&&0\\A~c&&-B~s&&0\\0&&0&&0\end{aligned}}\right]}$

and

${\displaystyle \mathbf {F} ^{-1}={\cfrac {1}{A~B}}\left[{\begin{aligned}B~c&&B~s&&0\\-A~s&&A~c&&0\\0&&0&&A~B\end{aligned}}\right]~.}$

The product is

${\displaystyle \mathbf {l} ={\cfrac {1}{A~B}}\left[{\begin{aligned}B~a~c^{2}+A~b~s^{2}&&B~a~c~s-A~b~c~s&&0\\B~a~c~s-A~b~c~s&&B~a~s^{2}+A~b~c^{2}&&0\\0&&0&&0\end{aligned}}\right]+{\cfrac {\pi }{2}}\left[{\begin{aligned}0&&-1&&0\\1&&0&&0\\0&&0&&0\end{aligned}}\right]}$

Note that the first matrix is symmetric while the second is skew-symmetric.

Therefore, the velocity gradient is

${\displaystyle {\mathbf {l} =\left[{\begin{aligned}{\cfrac {a}{1+at}}~\cos ^{2}{\cfrac {\pi t}{2}}+{\cfrac {b}{1+bt}}~\sin ^{2}{\cfrac {\pi t}{2}}&&\left[{\cfrac {a}{1+at}}-{\cfrac {b}{1+bt}}\right]\cos {\cfrac {\pi t}{2}}~\sin {\cfrac {\pi t}{2}}-{\frac {\pi }{2}}&&0\\\left[{\cfrac {a}{1+at}}-{\cfrac {b}{1+bt}}\right]\cos {\cfrac {\pi t}{2}}~\sin {\cfrac {\pi t}{2}}+{\cfrac {\pi }{2}}&&{\cfrac {a}{1+at}}~\sin ^{2}{\cfrac {\pi t}{2}}+{\cfrac {b}{1+bt}}~\cos ^{2}{\cfrac {\pi t}{2}}&&0\\0&&0&&0\end{aligned}}\right]}}$

Part 7

Compute the rate of deformation tensor and the spin tensor.

The rate of deformation is the symmetric part of the velocity gradient:

${\displaystyle {\mathbf {d} =\left[{\begin{aligned}{\cfrac {a}{1+at}}~\cos ^{2}{\cfrac {\pi t}{2}}+{\cfrac {b}{1+bt}}~\sin ^{2}{\cfrac {\pi t}{2}}&&\left[{\cfrac {a}{1+at}}-{\cfrac {b}{1+bt}}\right]\cos {\cfrac {\pi t}{2}}~\sin {\cfrac {\pi t}{2}}&&0\\\left[{\cfrac {a}{1+at}}-{\cfrac {b}{1+bt}}\right]\cos {\cfrac {\pi t}{2}}~\sin {\cfrac {\pi t}{2}}&&{\cfrac {a}{1+at}}~\sin ^{2}{\cfrac {\pi t}{2}}+{\cfrac {b}{1+bt}}~\cos ^{2}{\cfrac {\pi t}{2}}&&0\\0&&0&&0\end{aligned}}\right]}}$

The rate of deformation is the skew-symmetric part of the velocity gradient:

${\displaystyle {\mathbf {w} ={\cfrac {\pi }{2}}\left[{\begin{aligned}0&&-1&&0\\1&&0&&0\\0&&0&&0\end{aligned}}\right]}}$

Part 8

Assume that ${\displaystyle a=1}$ and ${\displaystyle b=2}$. Sketch the undeformed configuration and the deformed configuration at ${\displaystyle t=1}$ and ${\displaystyle t=2}$.  Draw both the deformed and undeformed configurations on the same plot

and label.

Recall that in the initial configuration

${\displaystyle X_{1}=x_{1}(0)=0~;~~Y_{1}=y_{1}(0)=0~;\qquad X_{2}=x_{2}(0)=2~;~~Y_{2}=y_{2}(0)=0~;\qquad X_{3}=x_{3}(0)=0~;~~Y_{3}=y_{3}(0)=1~.}$

Also, the motion is

${\displaystyle {\begin{aligned}x(\mathbf {X} ,t)&=X~(1+at)~\cos {\cfrac {\pi t}{2}}-Y~(1+bt)~\sin {\cfrac {\pi t}{2}}\\y(\mathbf {X} ,t)&=X~(1+at)~\sin {\cfrac {\pi t}{2}}+Y~(1+bt)~\cos {\cfrac {\pi t}{2}}\\z(\mathbf {X} ,t)&=Z\end{aligned}}}$

Plugging in the values of ${\displaystyle a}$ and ${\displaystyle b}$, we get

${\displaystyle {\begin{aligned}x(\mathbf {X} ,t)&=X~(1+t)~\cos {\cfrac {\pi t}{2}}-Y~(1+2t)~\sin {\cfrac {\pi t}{2}}\\y(\mathbf {X} ,t)&=X~(1+t)~\sin {\cfrac {\pi t}{2}}+Y~(1+2t)~\cos {\cfrac {\pi t}{2}}\\z(\mathbf {X} ,t)&=Z\end{aligned}}}$

At ${\displaystyle t=1}$,

${\displaystyle {\begin{aligned}x(\mathbf {X} ,1)&=2X~\cos {\cfrac {\pi }{2}}-3Y~\sin {\cfrac {\pi }{2}}=-3Y\\y(\mathbf {X} ,1)&=2X~\sin {\cfrac {\pi }{2}}+3Y~\cos {\cfrac {\pi }{2}}=2X\\z(\mathbf {X} ,1)&=Z\end{aligned}}}$

At ${\displaystyle t=2}$,

${\displaystyle {\begin{aligned}x(\mathbf {X} ,2)&=3X~\cos \pi -5Y~\sin \pi =-3X\\y(\mathbf {X} ,2)&=3X~\sin \pi +5Y~\cos \pi =-5Y\\z(\mathbf {X} ,2)&=Z\end{aligned}}}$

The deformed and undeformed configurations are shown below.

Deformed and undeformed configurations.

Part 9

Compute the polar decomposition of the deformation gradient with the above values of ${\displaystyle a}$ and ${\displaystyle b}$,

The deformation gradient is

${\displaystyle \mathbf {F} (t)=\left[{\begin{aligned}(1+t)~\cos {\cfrac {\pi t}{2}}&&-(1+2t)~\sin {\cfrac {\pi t}{2}}&&0\\(1+t)~\sin {\cfrac {\pi t}{2}}&&(1+2t)~\cos {\cfrac {\pi t}{2}}&&0\\0&&0&&1\end{aligned}}\right]~.}$

The right Cauchy-Green deformation tensor is

${\displaystyle \mathbf {C} =\mathbf {F} ^{T}~\mathbf {F} ={\begin{bmatrix}(1+t)^{2}&0&0\\0&(1+2t)^{2}&0\\0&0&1\end{bmatrix}}}$

The eigenvalue problem is

${\displaystyle (\mathbf {C} -\lambda ^{2}\mathbf {I} )\mathbf {N} =0~.}$

This problem has a solution if

${\displaystyle \det(\mathbf {C} -\lambda ^{2}\mathbf {I} )=0}$

i.e.,

${\displaystyle \left[\lambda ^{4}-(1+t)^{2}\lambda ^{2}-\lambda ^{2}(1+2t)^{2}+(1+t)^{2}~(1+2t)^{2}\right](1-\lambda ^{2})=0~.}$

The eigenvalues are (as expected)

${\displaystyle \lambda _{1}^{2}=(1+t)^{2}~;~~\lambda _{2}^{2}=(1+2t)^{2}~;~~\lambda _{3}^{2}=1~.}$

The principal stretches are

${\displaystyle \lambda _{1}=(1+t)~;~~\lambda _{2}=(1+2t)~;~~\lambda _{3}=1~.}$

The principal directions are (by inspection)

${\displaystyle \mathbf {N} _{1}=[1~~0~~0]^{T}~;~~\mathbf {N} _{2}=[0~~1~~0]^{T}~;~~\mathbf {N} _{3}=[0~~0~~1]^{T}~.}$

Now, the right stretch tensor is given by

${\displaystyle {\boldsymbol {U}}=\sum _{i=1}^{3}\lambda _{i}{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}}$

Therefore,

${\displaystyle \mathbf {U} =\lambda _{1}~\mathbf {N} _{1}~\mathbf {N} _{1}^{T}+\lambda _{2}~\mathbf {N} _{2}~\mathbf {N} _{2}^{T}+\lambda _{3}~\mathbf {N} _{3}~\mathbf {N} _{3}^{T}~.}$

Hence the right stretch is

${\displaystyle {\mathbf {U} ={\begin{bmatrix}1+t&0&0\\0&1+2t&0\\0&0&1\end{bmatrix}}}}$

At ${\displaystyle t=0,1,2}$, we have

${\displaystyle \mathbf {U} (0)={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}~;~~\mathbf {U} (1)={\begin{bmatrix}2&0&0\\0&3&0\\0&0&1\end{bmatrix}}~;~~\mathbf {U} (2)={\begin{bmatrix}3&0&0\\0&5&0\\0&0&1\end{bmatrix}}~.}$

Now

${\displaystyle \mathbf {R} =\mathbf {F} ~\mathbf {U} ^{-1}}$

and

${\displaystyle \mathbf {U} ^{-1}={\begin{bmatrix}{\cfrac {1}{1+t}}&0&0\\0&{\cfrac {1}{1+2t}}&0\\0&0&1\end{bmatrix}}}$

Therefore, the rotation is

${\displaystyle {\mathbf {R} =\left[{\begin{aligned}\cos {\cfrac {\pi t}{2}}&&-\sin {\cfrac {\pi t}{2}}&&0\\\sin {\cfrac {\pi t}{2}}&&\cos {\cfrac {\pi t}{2}}&&0\\0&&0&&1\end{aligned}}\right]~.}}$

At ${\displaystyle t=0,1,2}$, we have

${\displaystyle \mathbf {R} (0)={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}~;~~\mathbf {R} (1)={\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}}~;~~\mathbf {R} (2)={\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}}~.}$

Part 10

Assume an isotropic, hypoelastic constitutive equation for the material of the element. Compute the material time derivative of the Cauchy stress at ${\displaystyle t=1}$ using (a) the Jaumann rate and (b) the Truesdell rate.

A hypoelastic material behaves according to the relation

${\displaystyle {\overset {\triangle }{\boldsymbol {\sigma }}}={\boldsymbol {\mathsf {C}}}:\mathbf {d} ~.}$

For an isotropic material

${\displaystyle {\boldsymbol {\mathsf {C}}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {I}}}$

Therefore,

${\displaystyle {\overset {\triangle }{\boldsymbol {\sigma }}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}:\mathbf {d} +2\mu ~{\boldsymbol {I}}:\mathbf {d} =\lambda ~{\text{tr}}(\mathbf {d} )~{\boldsymbol {\mathit {1}}}+2\mu ~\mathbf {d} }$

Recall that

${\displaystyle \mathbf {d} =\left[{\begin{aligned}{\cfrac {a}{1+at}}~\cos ^{2}{\cfrac {\pi t}{2}}+{\cfrac {b}{1+bt}}~\sin ^{2}{\cfrac {\pi t}{2}}&&\left[{\cfrac {a}{1+at}}-{\cfrac {b}{1+bt}}\right]\cos {\cfrac {\pi t}{2}}~\sin {\cfrac {\pi t}{2}}&&0\\\left[{\cfrac {a}{1+at}}-{\cfrac {b}{1+bt}}\right]\cos {\cfrac {\pi t}{2}}~\sin {\cfrac {\pi t}{2}}&&{\cfrac {a}{1+at}}~a~\sin ^{2}{\cfrac {\pi t}{2}}+{\cfrac {b}{1+bt}}~\cos ^{2}{\cfrac {\pi t}{2}}&&0\\0&&0&&0\end{aligned}}\right]}$