Nonlinear finite elements/Homework 10/Solutions

Problem 1: Kinematics and Stress Rates
 Given:

Figure 1 shows a linear three-noded triangular element in the reference configuration.

The motion of the nodes is given by:
 * $$\begin{align}

\text{Node}~1:&& x_1(t) & = 0& \qquad y_1(t) & = 0 \\ \text{Node}~2:&& x_2(t) & = 2(1 + at)\cos\cfrac{\pi t}{2} &\qquad y_2(t) & = 2(1 + at)\sin\cfrac{\pi t}{2} \\ \text{Node}~3:&& x_3(t) & = -(1 + bt)\sin\cfrac{\pi t}{2} &\qquad y_3(t) & =(1 + bt)\cos\cfrac{\pi t}{2} \end{align}$$ The configuration ($$x,y$$) of the element at time $$t$$ is given by

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}) $$

Part 1
Write down expressions for $N_1$, $N_2$, and $N_3$ in terms of the initial configuration ($X,Y$) ?

In the initial configuration $$t = 0$$. Therefore,

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

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})~. $$ Substituting the values of $$X_i$$ and $$Y_i$$, we get

X(\boldsymbol{\xi}) = 2~N_2(\boldsymbol{\xi}) ~;\qquad Y(\boldsymbol{\xi}) = N_3(\boldsymbol{\xi})~. $$ Hence

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

\sum_{i=1}^3 N_i(\boldsymbol{\xi}) = 1 ~. $$ Therefore,

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

{ 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

\mathbf{F} = \left[ \begin{align} \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{align} \right]~. $$ Before computing the derivatives, let us express $$x,y,z$$ in terms of $$X,Y,Z$$. Recall

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,

\begin{align} 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{align} $$ Substituting in the expressions for $$x_i$$ and $$y_i$$, we get

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

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

\begin{align} \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{align} $$ Therefore,

{ \mathbf{F}(t) = \left[ \begin{align} (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{align} \right]~. } $$ The Jacobian determinant is

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

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

Part 3
What are the values of $a$ and $b$ for which the motion is isochoric?

For isochoric motion, $$J = 1$$. Therefore,

(1 + at)(1 + bt) = 1 ~. $$ One possibility is

{ a = b = 0 } $$ This is a pure rotation.

Another possibility is that

{ b = -\cfrac{a}{1+at} } $$ This is a combination of shear and rotation where the volume remains constant.

Part 4
For which values of $a$ and $b$ do we get invalid motions?

We get invalid motions when $$J \le 0$$. Let us consider the case where $$J = 0$$. Then

(1 + at) (1 + bt) = 0 $$ This is possible when

at = -1 ~\text{or}~ bt = -1 ~. $$ If $$J < 0$$ then

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

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

{ a \le -\cfrac{1}{t} \text{or} b \le -\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

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

\mathbf{F}(t) = \left[ \begin{align} (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{align} \right]~. $$ Let us make the following substitutions

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

\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,

\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,

\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

{ \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

\begin{align} 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{align} $$ Therefore,

\begin{align} 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{align} $$ We could compute the velocity gradient using

\mathbf{l} = \boldsymbol{\nabla} \mathbf{v} = \left[ \begin{align} \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{align} \right] $$ after expressing $$\mathbf{v}$$ in terms of $$\mathbf{x}$$. However, that makes the expression quite complicated. Instead, we will use the relation

\mathbf{l} = \dot{\mathbf{F}}~\mathbf{F}^{-1} ~. $$ The time derivative of the deformation gradient is

\dot{\mathbf{F}} = \left[ \begin{align} 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{align} \right]~. $$ The inverse of the deformation gradient is

\mathbf{F}^{-1} = \cfrac{1}{(1+at)(1+bt)}\left[ \begin{align} (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{align} \right]~. $$ Using the substitutions

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

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

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

\mathbf{l} = \cfrac{1}{A~B} \left[ \begin{align} 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{align} \right] + \cfrac{\pi}{2} \left[ \begin{align} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{align} \right] $$ Note that the first matrix is symmetric while the second is skew-symmetric.

Therefore, the velocity gradient is

{ \mathbf{l} = \left[ \begin{align} \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{align} \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:

{ \mathbf{d} = \left[ \begin{align} \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{align} \right] } $$ The rate of deformation is the skew-symmetric part of the velocity gradient:

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

Part 8
Assume that $$a=1$$ and $$b=2$$. Sketch the undeformed configuration and the deformed configuration at $$t=1$$ and $$t=2$$. Draw both the deformed and undeformed configurations on the same plot and label.

Recall that in the initial configuration

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

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

\begin{align} 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{align} $$ At $$t = 1$$,

\begin{align} 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{align} $$ At $$t = 2$$,

\begin{align} 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{align} $$ The deformed and undeformed configurations are shown below.

Part 9
Compute the polar decomposition of the deformation gradient with the above values of $a$ and $b$,

The deformation gradient is

\mathbf{F}(t) = \left[ \begin{align} (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{align} \right]~. $$ The right Cauchy-Green deformation tensor is

\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

(\mathbf{C} - \lambda^2\mathbf{I})\mathbf{N} = 0 ~. $$ This problem has a solution if

\det(\mathbf{C} - \lambda^2\mathbf{I}) = 0 $$ i.e.,

\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)

\lambda_1^2 = (1+t)^2 ~; \lambda_2^2 = (1+2t)^2 ~; \lambda_3^2 = 1~. $$ The principal stretches are

\lambda_1 = (1+t) ~; \lambda_2 = (1+2t) ~; \lambda_3 = 1~. $$ The principal directions are (by inspection)

\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

\boldsymbol{U} = \sum_{i=1}^3 \lambda_i \boldsymbol{N}_i\otimes\boldsymbol{N}_i $$ Therefore,

\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

{ \mathbf{U} = \begin{bmatrix} 1+t & 0 & 0 \\ 0 & 1+2t & 0 \\ 0 & 0 & 1 \end{bmatrix} } $$ At $$t=0,1,2$$, we have

\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

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

\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

{ \mathbf{R} = \left[ \begin{align} \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{align} \right]~. } $$ At $$t=0,1,2$$, we have

\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 $t = 1$ using (a) the Jaumann rate and (b) the Truesdell rate.

A hypoelastic material behaves according to the relation

\overset{\triangle}{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}}:\mathbf{d} ~. $$ For an isotropic material

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

\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

\mathbf{d} = \left[ \begin{align} \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{align} \right] $$ Using the values of $$a$$ and $$b$$ from the previous part, at $$t = 1$$,

\mathbf{d} = \left[ \begin{align} \cfrac{1}{2}~\cos^2\cfrac{\pi}{2} + \cfrac{2}{3}~\sin^2\cfrac{\pi}{2} & & \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} & & 0 \\ \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} & & \cfrac{1}{2}~\sin^2\cfrac{\pi}{2} + \cfrac{2}{3}~\cos^2\cfrac{\pi}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] = \begin{bmatrix} \cfrac{2}{3} & 0 & 0 \\ 0 & \cfrac{1}{2} & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ Therefore, the trace of the rate of defromation is

\text{tr}(\mathbf{d}) = \cfrac{7}{6} $$ Therefore,

\overset{\triangle}{\boldsymbol{\sigma}} = \cfrac{7\lambda}{6}~\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + 2\mu~\begin{bmatrix} \cfrac{2}{3} & 0 & 0 \\ 0 & \cfrac{1}{2} & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} $$ For the Jaumann rate

\overset{\triangle}{\boldsymbol{\sigma}} = \dot{\boldsymbol{\sigma}} + \boldsymbol{\sigma}\bullet\mathbf{w} - \mathbf{w}\bullet\boldsymbol{\sigma} $$ where the spin is

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

Therefore,

\dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} - \cfrac{\pi}{2}\begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} \begin{bmatrix} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{bmatrix} + \cfrac{\pi}{2}\begin{bmatrix} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{bmatrix} \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} $$ or,

{ \dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} + \cfrac{\pi}{2} \begin{bmatrix} -\sigma_{xy} & \sigma_{xx}-\sigma_{yy} & -\sigma_{yz} \\ \sigma_{xx}-\sigma_{yy} & 2\sigma_{xy} & \sigma_{xz} \\ -\sigma_{yz} & \sigma_{xz} & 0 \end{bmatrix} } $$ For the Truesdell rate

\overset{\triangle}{\boldsymbol{\sigma}} = \overset{\circ}{\boldsymbol{\sigma}} = \dot{\boldsymbol{\sigma}} - \boldsymbol{\sigma}\bullet\mathbf{l}^T - \mathbf{l}\bullet\boldsymbol{\sigma} + \text{tr}(\mathbf{l})\boldsymbol{\sigma} ~. $$ Therefore,

\dot{\boldsymbol{\sigma}} = \overset{\circ}{\boldsymbol{\sigma}} + \boldsymbol{\sigma}\bullet\mathbf{l}^T + \mathbf{l}\bullet\boldsymbol{\sigma} - \text{tr}(\mathbf{l})\boldsymbol{\sigma} ~. $$ Recall,

\mathbf{l} = \left[ \begin{align} \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{align} \right] $$ For $$a=1$$, $$b=2$$, $$t=1$$, we have

\mathbf{l} = \left[ \begin{align} \cfrac{1}{2}~\cos^2\cfrac{\pi}{2} + \cfrac{2}{3}~\sin^2\cfrac{\pi}{2} & & \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} - \frac{\pi}{2} & & 0 \\ \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} + \cfrac{\pi}{2} & & \cfrac{1}{2}~\sin^2\cfrac{\pi}{2} + \cfrac{2}{3}~\cos^2\cfrac{\pi}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] = \begin{bmatrix} \cfrac{2}{3} & -\cfrac{\pi}{2} & 0 \\ \cfrac{\pi}{2} & \cfrac{1}{2} & 0 \\ 0 &0 &0 \end{bmatrix} ~. $$ Therefore,

\text{tr}(\mathbf{l}) = \cfrac{7}{6} = \text{tr}(\mathbf{d}) $$ and

\dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} + \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} \begin{bmatrix} \cfrac{2}{3} & \cfrac{\pi}{2} & 0 \\ -\cfrac{\pi}{2} & \cfrac{1}{2} & 0 \\ 0 &0 &0 \end{bmatrix} + \begin{bmatrix} \cfrac{2}{3} & -\cfrac{\pi}{2} & 0 \\ \cfrac{\pi}{2} & \cfrac{1}{2} & 0 \\ 0 &0 &0 \end{bmatrix} \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} - \cfrac{7}{6} \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} $$ Hence,

{ \dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} + \begin{bmatrix} \cfrac{1}{6}\sigma_{xx} + \pi\sigma_{xy} & \cfrac{\pi}{2}(-\sigma_{xx} + \sigma_{yy}) & \cfrac{1}{2}(-\sigma_{xz} + \pi\sigma_{yz}) \\ \cfrac{\pi}{2}(-\sigma_{xx} + \sigma_{yy}) & -\cfrac{1}{6}\sigma_{yy} - \pi\sigma_{xy} & -\cfrac{2}{3}\sigma_{yz} -\cfrac{\pi}{2}\sigma_{xz} \\ \cfrac{1}{2}(-\sigma_{xz} + \pi\sigma_{yz}) & -\cfrac{2}{3}\sigma_{yz} -\cfrac{\pi}{2}\sigma_{xz} & -\cfrac{7}{6}\sigma_{zz} \end{bmatrix} } $$

Problem 2: Hyperelastic Pinched Cylinder Problem
Read the following paper on shells:

Buchter, N., Ramm, E., and Roehl, D., 1994, "Three-dimensional extension of non-linear shell formulation based on the enhanced assumed strain concept," '' Int. J. Numer. Meth. Engng.'',  37, pp. 2551-2568.

Answer the following questions:

Part 1
What do the authors mean by "enhanced assumed strain"?

See Wikipedia article on [w:Enhanced assumed strain|enhanced assumed strain]].

Part 2
Example 8.2 (and Figures 3 and 4 and Table III) in the paper discusses the simulation of a hyperelastic cylinder. Perform the same simulation using ANSYS for a shell thickness of 0.2 cm. Use shell elements and the Neo-Hookean hyperelastic material model that ANSYS provides.

The following material properties are used:

$$E=16800$$ kN/cm$$^2$$, $$\nu=0.4$$, $$\mu=6000$$ kN/cm$$^2$$, and $$d=2/K=7.14\times 10^{-5}$$ cm$$^2$$/kN. Symmetry is used and only half of the model is meshed. At the support, the model is constraint in all directions. The load of 36 kN is applied on the top of the cylinder (Fig~2. ANSYS input listing is shown Fig~6 of Appendix.

Part 3
Compare the total load needed to achieve an edge displacement of 16 cm with the results given in Table III. Comment on your results.

The plot of force vs. edge displacement (vertical) and the deformed model are shown in Fig 3. From the plot, one sees that a load of 35.1 kN is required to deform the edge by 16 cm. This is less than 1% difference compared to the result given in the paper using 7-parameter shell theory.

Problem 3: Elastic-Plastic Punch Indentation
Read the following paper on elastic-viscoplastic FEA:

Rouainia, M. and Peric, D., 1998, "A computational model for elasto-viscoplastic solids at finite strain with reference to thin shell applications," '' Int. J. Numer. Meth. Engng.'',  42, pp. 289-311.

Answer the following questions:

Part 1
Example 5.4 of the paper shows a simulation of the deformation of a thin sheet by a square punch. Perform the same simulation for 6061-T6 aluminum. Assume linear isotropic hardening and no rate dependence.

The following data are used for the thin plate:$$E=70$$ GPa, $$\nu=0.3$$, $$\sigma_Y=240$$ Mpa, $$\sigma_t=7800$$ Mpa, see Fig 4 for the stress-strain data. Symmetry is used and only half of the model is meshed (Fig 4). At the support, the model is constrained in all directions. A plastic finite strained shell (SHELL43) is chosen for this simulation.

The following material properties are used for the punch and die:$$E=100\times 10^6$$ GPa, $$\nu=0.3$$. The value of Young's modulus is arbitrary chosen so long as it is high enough to remain rigid during the simulation.

The meshed model is illustrated in Fig 4. A load of 10 kN is applied on the top of the punch. Load steps are split into two steps. First load step the punch is moved close to the plate to activate the contact elements (CONTAC49). The second load step, 30 kN is applied on top of the punch. ANSYS input listing is shown Fig 7 of the Appendix.

Part 2
Draw a plot of the punch force vs. punch travel and compare your result with the results shown in Figure 13 of the paper (qualitative comparison only).

The punch force vs. punch travel and the plot of the deformation at the final load step is shown in Fig 5. The curve displays similar pattern as those shown in the paper.