# Nonlinear finite elements/Homework 8
## Problem 1: Plate Bending

Consider the flat plate shown in Figure 1.
* Figure 1. Flat plate.
Transverse normals are defined as the straight lines perpendicular to the midsurface. If we assume that the plate is thin, the Kirchhoff-Love hypothesis holds, i.e.,
* Transverse normals remain straight.
* Transverse normals remain normal to the midsurface.
* Transverse normals are inextensible.
Answer the following questions.

### Part 1

The displacement field that satisfies the Kirchhoff-Love hypothesis is given by
    ${\begin{aligned}u_{1}(x_{1},x_{2},x_{3})&=u_{1}^{0}(x_{1},x_{2})-x_{3}~u_{3,1}^{0}\\u_{2}(x_{1},x_{2},x_{3})&=u_{2}^{0}(x_{1},x_{2})-x_{3}~u_{3,2}^{0}\\u_{3}(x_{1},x_{2},x_{3})&=u_{3}^{0}(x_{1},x_{2})\end{aligned}}$
where $u_{i}^{0}$ are the displacements of the midsurface.
Write the above equations (using the standard partial derivative notation with $\partial$~s) after making the following substitutions.
    ${\text{(2)}}\qquad {\begin{aligned}&x:=x_{1}~;~~y:=x_{2}~;~~z:=x_{3}\\&u:=u_{1}~;~~v:=u_{2}~;~~w:=u_{3}\\&u_{0}:=u_{1}^{0}~;~~v_{0}:=u_{2}^{0}~;~~w_{0}:=u_{3}^{0}~.\end{aligned}}$

### Part 2

Assume that the plate undergoes infinitesimal strains and rotations. Then the strain-displacement relations are
    ${\boldsymbol {\varepsilon }}={\frac {1}{2}}\left[{\boldsymbol {\nabla }}\mathbf {u} +({\boldsymbol {\nabla }}\mathbf {u} )^{T}\right]~.$
Show that, in terms of the midsurface displacements, the strain-displacement relations can be written in index notation as
    ${\begin{aligned}\varepsilon _{\alpha \beta }&={\frac {1}{2}}\left(u_{\alpha ,\beta }^{0}+u_{\beta ,\alpha }^{0}\right)-x_{3}~u_{3,\alpha \beta }^{0}\\\varepsilon _{\alpha 3}&=0\\\varepsilon _{33}&=0\end{aligned}}$
where $\alpha =1,2$ and $\beta =1,2$.

### Part 3

The internal virtual work in the plate is given by (using index notation)
    $\delta W_{\text{int}}=\int _{\Omega _{0}}\left(\int _{-h}^{h}\sigma _{\alpha \beta }~\delta \varepsilon _{\alpha \beta }~dx_{3}\right)d\Omega _{0}~.$
Note that if a component of the strain tensor is zero, then the variation of that strain component is also zero.
Show that the internal virtual work can be written as
    ${\text{(3)}}\qquad \delta W_{\text{int}}=\int _{\Omega _{0}}\left(N_{\alpha \beta }\delta \varepsilon _{\alpha \beta }^{0}+M_{\alpha \beta }\delta \varepsilon _{\alpha \beta }^{1}\right)~d\Omega _{0}$
where
    ${\begin{aligned}N_{\alpha \beta }&=\int _{-h}^{h}\sigma _{\alpha \beta }~dx_{3}\\M_{\alpha \beta }&=\int _{-h}^{h}x_{3}~\sigma _{\alpha \beta }~dx_{3}\\\varepsilon _{\alpha \beta }^{0}&={\frac {1}{2}}\left(u_{\alpha ,\beta }^{0}+u_{\beta ,\alpha }^{0}\right)\\\varepsilon _{\alpha \beta }^{1}&=-u_{3,\alpha \beta }^{0}~.\end{aligned}}$
Write down the expanded form of equation (3).

### Part 4

Reddy's book uses the notation of equation (2) to derive the following equilibrium equations for the plate.
    ${\begin{aligned}{\frac {\partial N_{xx}}{\partial x}}+{\frac {\partial N_{xy}}{\partial y}}&=0\\{\frac {\partial N_{xy}}{\partial x}}+{\frac {\partial N_{yy}}{\partial y}}&=0\\{\frac {\partial ^{2}M_{xx}}{\partial x^{2}}}+2{\frac {\partial ^{2}M_{xy}}{\partial x\partial y}}+{\frac {\partial ^{2}M_{yy}}{\partial y^{2}}}+{\mathcal {N}}(u_{0},v_{0},w_{0})+q&=0\end{aligned}}$
where
    ${\mathcal {N}}(u_{0},v_{0},w_{0})={\frac {\partial }{\partial x}}\left(N_{xx}{\frac {\partial w_{0}}{\partial x}}+N_{xy}{\frac {\partial w_{0}}{\partial y}}\right)+{\frac {\partial }{\partial y}}\left(N_{xy}{\frac {\partial w_{0}}{\partial x}}+N_{yy}{\frac {\partial w_{0}}{\partial y}}\right)~.$
Write the above equations in index notation. Then, show that in index-free Gibbs notation, the equilibrium equations for the plate can be written as
    ${\begin{aligned}{\boldsymbol {\nabla }}\bullet {\boldsymbol {N}}&=\mathbf {0} {\text{(4)}}\qquad \\{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\nabla }}\bullet {\boldsymbol {M}})+{\boldsymbol {\nabla }}\bullet ({\boldsymbol {N}}\bullet {\boldsymbol {\nabla u_{3}^{0})}}+q&=0{\text{(5)}}\qquad \end{aligned}}$
where
    ${\begin{aligned}{\boldsymbol {N}}&=N_{\alpha \beta }~\mathbf {e} _{\alpha }\otimes \mathbf {e} _{\beta }~;&{\boldsymbol {M}}&=M_{\alpha \beta }~\mathbf {e} _{\alpha }\otimes \mathbf {e} _{\beta }\\{\boldsymbol {\nabla }}\varphi &=\varphi _{,\alpha }~\mathbf {e} _{\alpha }~;&{\boldsymbol {\nabla }}\mathbf {v} &=v_{\alpha ,\beta }~\mathbf {e} _{\alpha }\otimes \mathbf {e} _{\beta }~;&{\boldsymbol {\nabla }}{\boldsymbol {S}}&=S_{\alpha \beta ,\gamma }~\mathbf {e} _{\alpha }\otimes \mathbf {e} _{\beta }\otimes \mathbf {e} _{\gamma }\\{\boldsymbol {\nabla }}\bullet \mathbf {v} &=v_{\alpha ,\alpha }~;&{\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}&=S_{\alpha \beta ,\beta }~\mathbf {e} _{\alpha }~;&{\boldsymbol {\nabla }}\bullet {\boldsymbol {T}}&=T_{\alpha \beta \gamma ,\gamma }~\mathbf {e} _{\alpha }\otimes \mathbf {e} _{\beta }\end{aligned}}$
and $\alpha =1,2$, $\beta =1,2$, and $\gamma =1,2$.$\varphi$ is a scalar, $\mathbf {v}$ is a vector, ${\boldsymbol {S}}$ is a second-order tensor, and ${\boldsymbol {T}}$ is a third-order tensor.

### Part 5

Derive the symmetric weak forms of the equilibrium equations starting with equations (4) and (5). Use the coordinate-free Gibbs notation in your derivation.
You will find the following identities useful in your derivation
    ${\begin{aligned}{\boldsymbol {\nabla }}\bullet (\varphi \mathbf {v} )&=\varphi ({\boldsymbol {\nabla }}\bullet \mathbf {v} )+\mathbf {v} \bullet ({\boldsymbol {\nabla }}\varphi )\\{\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}^{T}\bullet \mathbf {v} )&=\mathbf {v} \bullet ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}})+{\boldsymbol {S}}:({\boldsymbol {\nabla }}\mathbf {v} )\end{aligned}}$

### Part 6

* Write down the natural boundary conditions in Gibbs notation.
* What are the essential boundary conditions that can be applied to the plate?

## Problem 2: Transient Heat Conduction

Consider the model problem shown in Figure 2.
* Figure 2. Container with burning high energy material.
A 4340 steel casing is used to contain a solid high energy (HE) material. The HE material reacts on the inner (star-shaped) surface and generates gases.Since the gases are confined, the temperature of the gases increases according to the ideal gas law ($PV=nRT$).
We will try to simulate the transient heat transfer and the associated thermal stresses in the HE material and the steel casing.
Since the actual problem is quite complex and involves chemical reactions, equations of state, and fluid-structure interaction, we will simplify the problem.
The assumptions we will make are:
* The burn-surface does not change position.
* The temperature is the same at all points on the burn surface.
* The temperature at the burn surface increases exponentially with time according to the relation
    $T=T_{0}+A~\exp {(B~t)}$
    where $T_{0}$ is the initial temperature (300 K), $T$ is the temperature at time $t$, $A=2$ K, and $B=3$ K/s.
The inner radius of the steel casing is 5 cm. The outer radius of the steel casing is 6 cm.
The inner and outer radii of the star shaped region are 4 cm and 2 cm, respectively. You can make your own assumptions about the shape of the curves in the star shaped region.
Use standard values for the density, Young's modulus, Poisson's ratio, coefficient of thermal expansion, and thermal conductivity for the steel. You can use Matweb to find the properties you need.
Assume that the high energy material has the following properties:
* Density = 1200 kg/m$^{3}$~.
* Young's modulus = 1 GPa~.
* Poisson's ratio = 0.49~.
* Coefficient of thermal expansion = 1.2$\times$10$^{-4}$ /K~.
* Thermal Conductivity = 5 W/m-K (this is 10 times higher than theactual value).
* Specific Heat Capacity = 1000 J/kg-K.
* Assume that the initial temperature of the steel and the HE material is 300 K.Also assume that there is no heat flux on the external surface of the steel casing. Plot the temperature contours and stresses in the two materials at $t=1$ s and $t=2$ s.
* Assume that the initial temperature of the steel and the HE material is 300 K.Also assume that the external surface of the steel casing is maintained at 300 K. Plot the temperature contours and stresses in the two materials at $t=1$ s and $t=2$ s.
You should look at examples given in VM 33 and VM 174 in the ANSYS Verification manual for ideas on how to solve this problem.

## Problem 3: Explicit Finite Elements

Figure 3(a) shows a beam that is built-in at one and loaded at the other. Figure 3(b) shows a plate that is built-in at one end and loaded at the other. The load that is applied is transient and has the form shown in Figure 3(c).
| Figure 3. Built-in cantilever beam and plate. | Figure 3(c). Load curve. |
Assume that the beam and the plate are made of 4340 steel. The beam has dimensions 12 in. $\times$ 0.25 in. $\times$ 0.25 in. The plate has dimensions 12 in. $\times$ 0.25 in. $\times$ 6 in.
Simulate the problem for 10 secs. using LS-DYNA or some other explicit FE code. Use beam elements for the beam and shell elements for the plate.Use an element size of 2 in. for the beam and 2 in. $\times$ 2 in. for the shell.
* Plot the displacement of the end of the beam as a function of time.
* Plot the axial stress at two points on the top of the beam as a function of time.
* Plot the displacement of the end of the plate as a function of time.
* Plot the axial stress at two points on the top of the plate as a function of time.
After you run the LS-DYNA, the program will create a binary file 'd3plot' and ASCII file 'nodout'. To view the result:
* Run LS-Prepost and open the binary file 'd3plot'.
* To view the deflection at free end: click ASCII button.
* In the window below it, select 'nodout', then click LOAD.
* Available nodes will be shown in the next window below.
* Select the node. For shell element, those are 2 nodes at the free end.
* Select 'y-displacement' for beam, and 'z-displacement' for shell, then click PLOT.
You should also try the animation tool to get a feel for how the beam and shell deform over time.