Nonlinear finite elements/Homework 8/Solutions

Problem 1: Plate Bending
 Given:

Consider the flat plate shown in Figure 1.

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.

Part 1
The displacement field that satisfies the Kirchhoff-Love hypothesis is given by
 * $$\begin{align}

u_1(x_1,x_2,x_3) & = u^0_1(x_1,x_2) - x_3~u^0_{3,1} \\ u_2(x_1,x_2,x_3) & = u^0_2(x_1,x_2) - x_3~u^0_{3,2} \\ u_3(x_1,x_2,x_3) & = u^0_3(x_1,x_2) \end{align}$$ where $$u^0_i$$ are the displacements of the midsurface.

Write the expanded form of the above equations after making the following substitutions.
 * $$\text{(2)} \qquad

\begin{align} & x := x_1 ~; y := x_2 ~; z := x_3 \\ & u := u_1 ~; v := u_2 ~; w := u_3 \\ & u_0 := u^0_1 ~; v_0 := u^0_2 ~; w_0 := u^0_3 ~. \end{align} $$

After making the substitutions we get
 * $$\begin{align}

u(x,y,z) & = u_0(x,y) - z~w_{0,1} \\ v(x,y,z) & = v_0(x,y) - z~w_{0,2} \\ w(x,y,z) & = w_0(x,y) \end{align}$$ The expanded form is

\begin{align} u(x,y,z) & = u_0(x,y) - z~\frac{\partial w_0}{\partial x} \\ v(x,y,z) & = v_0(x,y) - z~\frac{\partial w_0}{\partial y} \\ w(x,y,z) & = w_0(x,y) \end{align} $$

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{align}

\varepsilon_{\alpha\beta} & = \frac{1}{2} \left(u^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right) - x_3~u^0_{3,\alpha\beta} \\ \varepsilon_{\alpha 3} & = 0 \\ \varepsilon_{33} & = 0 \end{align}$$ where $$\alpha = 1,2$$ and $$\beta = 1,2$$.

The strain-displacement relations in index notation are

\varepsilon_{ij} = \frac{1}{2}\left[u_{i,j} + u_{j,i}\right] $$ where $$i,j = 1,2,3$$. If the indices $$\alpha$$ and $$\beta$$ run from 1 to 2, we can write the above strain-displacement relations as
 * $$\begin{align}

\varepsilon_{\alpha\beta} & = \frac{1}{2}\left[u_{\alpha,\beta} + u_{\beta,\alpha}\right] \\ \varepsilon_{\alpha 3}= \varepsilon_{3\alpha} & = \frac{1}{2}\left[u_{\alpha,3} + u_{3,\alpha}\right] \\ \varepsilon_{33}& = \frac{1}{2}\left[u_{3,3} + u_{3,3}\right] = u_{3,3} \end{align}$$ The displacements are given by
 * $$\begin{align}

u_1 & = u^0_1 - x_3~u^0_{3,1} \\ u_2 & = u^0_2 - x_3~u^0_{3,2} \\ u_3 & = u^0_3 \end{align}$$ If the index $$\alpha$$ runs from 1 to 2, the displacements can be written as
 * $$\begin{align}

u_\alpha & = u^0_{\alpha} - x_3~u^0_{3,\alpha} \\ u_3 & = u^0_3 \end{align}$$ The partial derivatives of the displacements are (since $$u^0_1, u^0_2, u^0_3$$ are functions only of $$x_1$$ and $$x_2$$)
 * $$\begin{align}

u_{\alpha,\beta} & = u^0_{\alpha,\beta} - x_3~u^0_{3,\alpha\beta} \\ u_{\alpha, 3} & = u^0_{\alpha, 3} - x_3~u^0_{3,\alpha 3} = 0 \\ u_{3,\alpha} & = u^0_{3,\alpha} = 0 \\ u_{3,3} & = u^0_{3,3} = 0 \end{align}$$ Therefore,

\begin{align} \varepsilon_{\alpha\beta} & = \frac{1}{2}\left[u^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right] - x_3~u^0_{3,\alpha\beta} \\ \varepsilon_{\alpha 3}= \varepsilon_{3\alpha} & = 0 \\ \varepsilon_{33}& = 0 \end{align} $$

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{align}

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^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right) \\ \varepsilon_{\alpha\beta}^1 & = - u^0_{3,\alpha\beta} ~. \end{align}$$ Write down the expanded form of equation (3).

The strains are given by

\varepsilon_{\alpha\beta} = \frac{1}{2}\left[u^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right] - x_3~u^0_{3,\alpha\beta} = \varepsilon_{\alpha\beta}^0 + x_3~\varepsilon_{\alpha\beta}^1 $$ Therefore, the variations of the strains are

\delta \varepsilon_{\alpha\beta} = \delta \varepsilon_{\alpha\beta}^0 + x_3~\delta \varepsilon_{\alpha\beta}^1 $$ The internal virtual work is
 * $$\begin{align}

\delta W_{\text{int}} & = \int_{\Omega_0}\left[\int_{-h}^{h} \sigma_{\alpha\beta}~\left( \delta \varepsilon_{\alpha\beta}^0 + x_3~\delta \varepsilon_{\alpha\beta}^1\right) ~dx_3\right] d\Omega_0\\ & = \int_{\Omega_0}\left[\delta \varepsilon_{\alpha\beta}^0\int_{-h}^{h} \sigma_{\alpha\beta}~dx_3 + \delta \varepsilon_{\alpha\beta}^1\int_{-h}^{h} x_3~\sigma_{\alpha\beta}~dx_3\right]~d\Omega_0 \\ & = \int_{\Omega_0}\left[\delta \varepsilon_{\alpha\beta}^0~N_{\alpha\beta} + \delta \varepsilon_{\alpha\beta}^1~M_{\alpha\beta} \right]~d\Omega_0 \end{align}$$ Therefore,

\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 $$ The expanded form of the above equation is

\begin{align} \delta W_{\text{int}} = \int_{\Omega_0}( & N_{11}\delta \varepsilon_{11}^0 + N_{12}\delta \varepsilon_{12}^0 +  N_{21}\delta \varepsilon_{21}^0 + N_{22}\delta \varepsilon_{22}^0 + \\ & M_{11}\delta \varepsilon_{11}^1 + M_{12}\delta \varepsilon_{12}^1 +  M_{21}\delta \varepsilon_{21}^1 + M_{22}\delta \varepsilon_{22}^1)~d\Omega_0 \end{align} $$ or,

\begin{align} \delta W_{\text{int}} = \int_{\Omega_0}( & N_{11}\delta \varepsilon_{11}^0 + 2 N_{12}\delta \varepsilon_{12}^0 +  N_{22}\delta \varepsilon_{22}^0 + \\ & M_{11}\delta \varepsilon_{11}^1 + 2 M_{12}\delta \varepsilon_{12}^1 +  M_{22}\delta \varepsilon_{22}^1)~d\Omega_0 \end{align} $$

Part 4
The  equilibrium equations for the plate can be written as
 * $$\begin{align}

\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{align}$$ 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.

The equilibrium equations are (written using indices $$1, 2, 3$$):
 * $$\begin{align}

& N_{11,1} + N_{12,2} = 0 \\ & N_{21,1} + N_{22,2} = 0 \\ & M_{11,11} + 2M_{12,12} + M_{22,22} + \left(N_{11}~u^0_{3,1} + N_{12}~u^0_{3,2}\right)_{,1} + \left(N_{21}~u^0_{3,1} + N_{22}~u^0_{3,2}\right)_{,2} + q = 0 \end{align}$$ Using the summation convention, we get
 * $$\begin{align}

& N_{1\beta, \beta} = 0 \\ & N_{2\beta, \beta} = 0 \\ & M_{\alpha\beta,\alpha\beta} + \left(N_{1\beta}~u^0_{3,\beta}\right)_{,1} + \left(N_{2\beta}~u^0_{3,\beta}\right)_{,2} + q = 0 \end{align}$$ The first two equations can then be written as

N_{\alpha\beta, \beta} = 0 $$ and the third equation can be written as

M_{\alpha\beta,\alpha\beta} + \left(N_{\alpha\beta}~u^0_{3,\beta}\right)_{,\alpha} + q = 0 $$ Therefore the equilibrium equations in index notation are

\begin{align} & N_{\alpha\beta, \beta} = 0 \\ & M_{\alpha\beta,\alpha\beta} + \left(N_{\alpha\beta}~u^0_{3,\beta}\right)_{,\alpha} + q = 0 \end{align} $$

Part 5
Show that in index-free  Gibbs notation the equilibrium equations for the plate can be written as
 * $$\begin{align}

\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^0_3)} + q & = 0 \text{(5)} \qquad \end{align}$$ where

\begin{align} \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{align} $$ and $$\alpha=1,2$$, $$\beta=1,2$$, and $$\gamma=1,2$$.

Equation (4) in index notation is

\boldsymbol{\nabla} \bullet \boldsymbol{N} = N_{\alpha\beta,\beta}~\mathbf{e}_{\alpha} = \mathbf{0}\qquad \implies N_{\alpha\beta,\beta} = 0 $$ The above equations have been shown to be equivalent to the first two equilibrium equations in the previous part.

Let us now look at equation (5). Let

\mathbf{p} := \boldsymbol{\nabla} \bullet \boldsymbol{M} \qquad \implies p_{\alpha}\mathbf{e}_{\alpha} = M_{\alpha\beta,\beta} \mathbf{e}_{\alpha} $$ Then

\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M}) = \boldsymbol{\nabla} \bullet \mathbf{p} = p_{\alpha,\alpha} = M_{\alpha\beta,\alpha\beta} $$ For the next term, let

\mathbf{w} := \boldsymbol{\nabla} u^0_3 \qquad \implies w_{\alpha}\mathbf{e}_\alpha = u^0_{3,\alpha}\mathbf{e}_\alpha \implies w_{\alpha} = u^0_{3,\alpha} $$ Then,

\mathbf{q} := \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3 = \boldsymbol{N}\bullet\mathbf{w} \qquad \implies q_{\alpha}~\mathbf{e}_\alpha = N_{\alpha\beta}~w_{\beta}~\mathbf{e}_\alpha = N_{\alpha\beta}~u^0_{3,\beta}~\mathbf{e}_\alpha $$ Hence

\boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)} = \boldsymbol{\nabla} \bullet \mathbf{q} = q_{\alpha,\alpha} = (N_{\alpha\beta}~u^0_{3,\beta})_{,\alpha} $$ Therefore, equation (5) can be written as

M_{\alpha\beta,\alpha\beta}+(N_{\alpha\beta}~u^0_{3,\beta})_{,\alpha} + q = 0 $$ The above equation has been shown to be equivalent to the third equilibrium equations in the previous part.

Part 6
Derive the symmetric weak forms of the equilibrium equations starting with equations (4) and (5). Use the coordinate-free Gibbs notation in your derivation.

For equation 4, we multiply the equation by a vector weighting function $$\mathbf{w}_1$$ and integrate over the midsurface $$\Omega_0$$. Then,
 * $$\begin{align}

\int_{\Omega_0} \mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N})~d\Omega_0 & = 0 \end{align}$$ Using the identity

\boldsymbol{\nabla} \bullet (\boldsymbol{S}^T\bullet\mathbf{v}) = \mathbf{v}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{S}) + \boldsymbol{S}:(\boldsymbol{\nabla} \mathbf{v}) $$ we get

\mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N}) = \boldsymbol{\nabla} \bullet (\boldsymbol{N}^T\bullet\mathbf{w}_1) - \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1) $$ From the symmetry to the stress tensor $$\boldsymbol{N} = \boldsymbol{N}^T$$. Hence,

\mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N}) = \boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\mathbf{w}_1) - \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1) $$ Therefore
 * $$\begin{align}

\int_{\Omega_0} \mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N})~d\Omega_0 & = \int_{\Omega_0} \boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\mathbf{w}_1)~d\Omega_0 - \int_{\Omega_0} \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1)~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet(\boldsymbol{N}\bullet\mathbf{w}_1)~d\Gamma_0 - \int_{\Omega_0} \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1)~d\Omega_0 \end{align}$$ where $$\mathbf{n}$$ is the normal to the boundary $$\Gamma_0$$ of the midsurface.

Hence the symmetric weak form of the first equation is

\int_{\Omega_0} \boldsymbol{N}:\boldsymbol{\nabla} \mathbf{w}_1~d\Omega_0 = \int_{\Gamma_0} \mathbf{n}\bullet(\boldsymbol{N}\bullet\mathbf{w}_1)~d\Gamma_0 $$

For equation 5, we multiply the equation by a scalar weighting function $$w_2$$ and integrate over the midsurface $$\Omega_0$$. Then,

\int_{\Omega_0} w_2\left[\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M}) + \boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)} + q\right] ~d\Omega_0 = 0 $$ Using the identity,

\boldsymbol{\nabla} \bullet (\varphi\mathbf{v}) = \varphi (\boldsymbol{\nabla} \bullet \mathbf{v}) + \mathbf{v}\bullet(\boldsymbol{\nabla} \varphi) $$ we get

w_2[\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M})] = \boldsymbol{\nabla} \bullet [w_2(\boldsymbol{\nabla}\bullet\boldsymbol{M})] - (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2) $$ Therefore,
 * $$\begin{align}

\int_{\Omega_0} w_2[\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M})]~d\Omega_0 & = \int_{\Omega_0} \boldsymbol{\nabla} \bullet [w_2(\boldsymbol{\nabla}\bullet\boldsymbol{M})]~d\Omega_0 - \int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet[w_2(\boldsymbol{\nabla} \bullet \boldsymbol{M})]~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 \\ & = \int_{\Gamma_0} w_2~[\mathbf{n}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{M})]~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 \end{align}$$ In order to have natural BCs directly in terms of moments, we have to perform a similar set of operations the last term above. Thus,
 * $$\begin{align}

\int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 & = \int_{\Omega_0} \boldsymbol{\nabla} \bullet [\boldsymbol{M}\bullet(\boldsymbol{\nabla w_2)]}~d\Omega_0 - \int_{\Omega_0} \boldsymbol{M}:[\boldsymbol{\nabla} (\boldsymbol{\nabla}{w_2)}]~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet[\boldsymbol{M}\bullet{(\boldsymbol{\nabla} w_2)}]~d\Gamma_0 - \int_{\Omega_0} \boldsymbol{M}:[\boldsymbol{\nabla} (\boldsymbol{\nabla} w_2)]~d\Omega_0 \end{align}$$ Hence,

\int_{\Omega_0} w_2~\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M})~d\Omega_0= \int_{\Gamma_0} \left\{w_2~\mathbf{n}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{M})- \mathbf{n}\bullet\boldsymbol{M}\bullet(\boldsymbol{\nabla} w_2)\right\} d\Gamma_0 + \int_{\Omega_0} \boldsymbol{M}:\left(\boldsymbol{\nabla} \boldsymbol{\nabla}{w_2}\right)~d\Omega_0 $$ Similarly,
 * $$\begin{align}

\int_{\Omega_0} w_2\left[\boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)}\right]~d\Omega_0&= \int_{\Omega_0} \boldsymbol{\nabla} \bullet [w_2(\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)]}~d\Omega_0 - \int_{\Omega_0} (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet(\boldsymbol{\nabla} w_2) ~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet[w_2(\mathbf{d}ot{\boldsymbol{N}{\boldsymbol{\nabla} u^0_3})]}~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet(\boldsymbol{\nabla} w_2) ~d\Omega_0 \\ & = \int_{\Gamma_0} w_2~[\mathbf{n}\bullet(\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)]~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet(\boldsymbol{\nabla} w_2) ~d\Omega_0 \end{align}$$ Therefore, the symmetric weak form of the second equation is
 * $$\begin{align}

\int_{\Omega_0} & \left[ (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet\boldsymbol{\nabla} w_2 - \boldsymbol{M}:\left(\boldsymbol{\nabla} \boldsymbol{\nabla} w_2\right) \right]~d\Omega_0 = \\ & \int_{\Omega_0} w_2~q~d\Omega_0 + \int_{\Gamma_0} \left[w_2\left\{ \mathbf{n}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{M}) + \mathbf{n}\bullet\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right\} - \mathbf{n}\bullet(\boldsymbol{M}\bullet\boldsymbol{\nabla} w_2) \right]~d\Gamma_0 \end{align}$$ Using the distributive law for vectors

(\mathbf{a}+\mathbf{b})\bullet\mathbf{c} = \mathbf{a}\bullet\mathbf{c} + \mathbf{b}\bullet\mathbf{c} $$ we can also write the symmetric weak form for the second equation as:

\begin{align} \int_{\Omega_0} & \left[ (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet{\boldsymbol{\nabla} w_2} - \boldsymbol{M}:\left(\boldsymbol{\nabla} \boldsymbol{\nabla} w_2\right) \right]~d\Omega_0 = \\ & \int_{\Omega_0} w_2~q~d\Omega_0 + \int_{\Gamma_0} \mathbf{n}\bullet\left( \boldsymbol{\nabla} \bullet \boldsymbol{M} + \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right)~w_2~d\Gamma_0 - \int_{\Gamma_0} \mathbf{n}\bullet(\boldsymbol{M}\bullet\boldsymbol{\nabla} w_2)~d\Gamma_0 \end{align} $$

Part 7
Write down the natural boundary conditions in Gibbs notation.

From the expressions for the weak form, the weighted natural boundary conditions are

\begin{align} & \int_{\Gamma_0} \mathbf{n}\bullet(\mathbf{d}ot{\boldsymbol{N}{\mathbf{w}_1})}~d\Gamma_0 \\ & \int_{\Gamma_0} \mathbf{n}\bullet\left( \boldsymbol{\nabla} \bullet \boldsymbol{M} + \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right)~w_2~d\Gamma_0 \\ & - \int_{\Gamma_0} \mathbf{n}\bullet(\mathbf{d}ot{\boldsymbol{M}{\boldsymbol{\nabla} w_2})}~d\Gamma_0 \end{align} $$ Therefore, the natural boundary conditions are:

\begin{align} \text{Boundary Membrane Forces}~ & & \mathbf{n}\bullet\boldsymbol{N} \\ \text{Boundary Transverse Forces}~ & & \mathbf{n}\bullet\left(\boldsymbol{\nabla} \bullet \boldsymbol{M} + \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right)\\ \text{Boundary Moments}~ & & \mathbf{n}\bullet\boldsymbol{M} \end{align} $$

Part 8
What are the essential boundary conditions that can be applied to the plate?

The essential boundary conditions that can be applied are

\begin{align} \text{Boundary Membrane Displacements}~ & & u^0_1, u^0_2 \\ \text{Boundary Transverse Displacements}~ & & u^0_3 \\ \text{Boundary Out-of-plane Rotations}~ & & \boldsymbol{\nabla} u^0_3 \end{align} $$

Problem 2: Transient Heat Conduction
 Given:

The model problem shown in Figure 6.

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 ($$PV = nRT$$).

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.

The high energy material has 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, Specific Heat Capacity = 1000 J/kg-K.

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.

Solution
The properties of normalized 4340 steel from Matweb are

Part 1
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.

The temperature contours for the two materials are shown in Figure 7. The low thermal conductivity of the HE material leads to a slow rate of diffusion of heat into the material. Therefore, the steel casing sees no change in temperature. However, the casing does deform due to the deformation of the HE material.

The stress contours for the two materials are shown in Figures 8, 9, 10, and 11. The principal stress $$\sigma_1$$ is tensile in the steel casing and reaches a values of around 200 MPa (close to its yield point). In the HE material, $$\sigma_1$$ is compressive in some regions and tensile in others and reaches values well beyond the ulimate tensile strength of the material inside some elements.

The ANSYS input file used for this problem is given below. The preprocessing stage is shown in the PDF file NFE_HW8Prob2Ans.pdf. The solution stage is shown below:

Part 2
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.

The temperature distribution from the previous problem suggests that the heat flux at the external surface is zero if the outside is maintained at 300 K (because of the low thermal conductivity of the HE material.) We don't expect to see any significant difference in the temperature and stress distributions.

Problem 3: Explicit Finite Elements
 Given:

Figure 13(a) shows a beam that is built-in at one and loaded at the other. Figure 13(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 13(c).

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.

Part 1
Plot the displacement of the end of the beam as a function of time.

Figure 14 shows the displacement of the end of the beam as a function of time. Some numerical damping appears to occur if Hughes-Liu beam elements are used with 2$$\times$$2 quadrature.

Part 2
Plot the axial stress at two points on the top of the beam as a function of time.

Figure 15 shows the axial stress at a Gauss point near the top of the beam for elements 2 and 5.

The LS-Dyna keyword file for the beam is shown below.

Part 3
Plot the displacement of the end of the plate as a function of time.

Figure 16 shows the displacement of the end of the plate as a function of time. Hughes-Liu shell elements with 3 integration points are used.

Part 4
Plot the axial stress at two points on the top of the plate as a function of time.

Figure 17 shows the axial stress at a Gauss point near the top of the plate for elements 8 and 11.

The LS-Dyna keyword file for the beam is shown below.