User:Eml5526.s11.team7.flater/EGM6611.HW4

=Homework 4=

Description
A material is assumed to have the following constitutive equations:


 * Stress $$\displaystyle \Rightarrow \quad \mathbf{T} = \mathbf{T} (\mathbf{E},\mathbf{\dot{E}},\theta, \nabla \theta)$$
 * Heat flux $$\displaystyle \Rightarrow \quad \mathbf{q} = \mathbf{q} (\mathbf{E},\mathbf{\dot{E}},\theta, \nabla \theta)$$
 * Helmholtz free energy $$\displaystyle \Rightarrow \quad \psi = \psi (\mathbf{E},\mathbf{\dot{E}},\theta, \nabla \theta)$$
 * Entropy $$\displaystyle \Rightarrow \quad s = s (\mathbf{E},\mathbf{\dot{E}},\theta, \nabla \theta)$$


 * where $$\mathbf{E}$$ is Lagrangian finite strain tensor, $$\mathbf{\dot{E}}$$ is time derivative of the Lagrangian finite strain tensor, $$\displaystyle \theta$$ is temperature, and $$\displaystyle \nabla \theta$$ is the gradient of the temperature.

Derive the constitutive equations based on the Clausius-Duhem inequality.

Solution
Clausius-Duhem inequality is defined as


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$$ \frac{ds}{dt} - \frac{r}{\theta} + \frac{1}{\rho \theta} \nabla \cdot \mathbf{q} - \frac{\mathbf{q}}{\rho \theta^2} \cdot \nabla \theta \ge 0 $$     (1.1)
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 * or in terms of the Helmholtz free energy


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$$ \frac{\rho}{\theta}(\dot{\psi} +s \dot{\theta}) + \frac{1}{\theta} \mathbf{T}: \dot{\mathbf{E}} - \frac{1}{\theta^2}Q \cdot \rho \theta \ge 0 $$.     (1.2)
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Solve for $$\displaystyle \dot{\psi}$$ and substitute.


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$$ \frac{\rho}{\theta} \left[ \frac{\partial \psi}{\partial \mathbf{E}} \mathbf{\dot{E}} + \frac{\partial \psi}{\partial \mathbf{\dot{E}}}:\mathbf{\ddot{E}} + \frac{\partial \psi}{\partial \theta} \dot{\theta} + \frac{\partial \psi}{\partial \nabla \theta} \nabla \dot{\theta} +  s \dot{\theta} \right] + \frac{1}{\theta} \mathbf{T}: \dot{\mathbf{E}} - \frac{1}{\theta^2}Q \cdot \rho \theta \ge 0 $$.     (1.3)
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Collect similar terms, and recognize that the above must hold for all values of $$\displaystyle \mathbf{\dot{E}}$$ and $$\displaystyle \mathbf{\ddot{E}}$$. In other words, our constitutive relations are determined.


 * $$\displaystyle \frac{\partial \psi}{\partial \mathbf{\dot{E}}} = 0 \qquad, \qquad \frac{\partial \psi}{\partial \nabla \theta} = 0 \qquad \Rightarrow \qquad \psi = \psi (\mathbf{E}, \theta)$$
 * $$\displaystyle -\frac{\partial \psi}{\partial \mathbf{E}} + \frac{1}{\rho} \mathbf{T} = 0 \qquad \Rightarrow \qquad \mathbf{T}=\rho \frac{\partial \psi}{\partial \mathbf{E}}$$
 * $$\displaystyle \frac{\partial \psi}{\partial \theta} + s = 0 \qquad \Rightarrow \qquad s=- \frac{\partial \psi}{\partial \theta}$$
 * $$\displaystyle -Q \cdot \nabla \theta \ge 0$$

Description
It is often said that for an isotropic material the principle axes of stress coincide with the principle axes of strain. Show that this is true for an elastic solid governed by $$\displaystyle T_{ij}' = 2G \epsilon_{ij}'$$, but not necessarily in a viscoelastic material governed by $$\displaystyle T_{ij}' = 2G \epsilon_{ij}' + 2 \eta \frac{d \epsilon_{ij}'}{dt}$$.

Solution
Principle stresses are defined as


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$$\displaystyle (T_{ij} - T_{(k)} \delta_{ij})n_j^{(k)} = 0 \qquad \text{for} \quad i,j,k = 1,2,3 $$     (2.1)
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 * where $$\displaystyle T_{(k)}$$ is the principle stress and $$\displaystyle n_j^{(k)}$$ is the principle direction. In matrix form,


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$$\displaystyle \begin{bmatrix} T_{11}-T_{(k)} & T_{12} & T_{13}\\ T_{21} & T_{22}-T_{(k)} & T_{23}\\ T_{31} & T_{32} & T_{33}-T_{(k)} \end{bmatrix} \cdot \begin{bmatrix} n_1^{(k)}\\ n_2^{(k)}\\ n_3^{(k)} \end{bmatrix} = 0 $$     (2.2)
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The determinant in the characteristic equation now takes the form


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$$\displaystyle T_{(k)}^3 - \mathrm{I}_T T_{(k)}^2 + \mathrm{II}_T T_{(k)} - \mathrm{III}_T =0 $$     (2.3)
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The stress can be decomposed to


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$$\displaystyle T_{ij}= \delta_{ij} \frac{T_{kk}}{3} + T_{ij}' $$     (2.4)
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 * where $$\displaystyle T_{ij}'$$ is the deviatoric stress and $$\displaystyle T_{kk}$$ is the mean stress.

Therefore, $$\displaystyle T_{ij}$$ and $$\displaystyle T_{ij}'$$ are related. Since the relationship between deviatoric stress and deviatoric strain is linear (see governing equation for an isotropic material above), the axes for principle strain coincide with principle stress. For this case of a rate-dependent viscoelastic material, the relationship between stress and strain is no longer linear. So the axes of principle stress and principle strain do not necessarily coincide.

Description
If the reference axes are principal axes of stress in an isotropic elastic solid governed by $$\displaystyle \sigma_{ij}' = 2G \epsilon_{ij}'$$ and $$\displaystyle p = -K e$$, combine these equations to obtain three equations relating $$\displaystyle \sigma_1, \ \sigma_2, \ \sigma_3, \ \epsilon_1, \ \epsilon_2, \ \epsilon_3$$.

Solution
From Problem 2 it was shown that the principle stress axes coincide with the principle strain axes for an elastic material that follows Hooke's law. So when $$\displaystyle \sigma_{ij}' = 2G \epsilon_{ij}'$$, we recall equation 2.4 such that


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$$\displaystyle \sigma_k - p = 2G \left( \epsilon_k - \frac{1}{3}e \right ) $$     (3.1)
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Substituting $$\displaystyle p = -K e$$ into this equation yields


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$$\displaystyle \sigma_k = 2G \left( \epsilon_k - \frac{1}{3}e \right ) - Ke \qquad \Rightarrow \qquad \sigma_k = 2G \epsilon_k - e \left( \frac{2G}{3} + K\right ) $$     (3.2)
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 * or expanded as


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$$\displaystyle \sigma_k = 2G \epsilon_k - \left( \frac{2G}{3} + K\right ) ( \epsilon_1 +\epsilon_2 + \epsilon_3) $$     (3.3)
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Therefore the three equations relating $$\displaystyle \sigma_1, \ \sigma_2, \ \sigma_3, \ \epsilon_1, \ \epsilon_2, \ \epsilon_3$$ are


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$$\displaystyle \begin{matrix} \sigma_1 = \left( \frac{4G}{3} - K\right ) \epsilon_1 - \left( \frac{2G}{3} + K\right ) ( \epsilon_2 + \epsilon_3) \\ \\ \sigma_2 = \left( \frac{4G}{3} - K\right ) \epsilon_2 - \left( \frac{2G}{3} + K\right ) ( \epsilon_1 + \epsilon_3) \\ \\ \sigma_3 = \left( \frac{4G}{3} - K\right ) \epsilon_3 - \left( \frac{2G}{3} + K\right ) ( \epsilon_1 + \epsilon_2) \end{matrix} $$     (3.4)
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Description
Show that the isotropic Hooke's law may be written as


 * $$\displaystyle T_{ij}=2G \left( E_{ij}+\frac{\nu}{1-2\nu} e \delta_{ij}\right )$$

And hence derive the following form for the equilibrium equations of an elastic body in terms of the displacements, neglecting the distinction between spatial and material coordinates.


 * $$\displaystyle G \left(\nabla^2 u_i + \frac{1}{1-2\nu} \frac{\partial^2 u_j}{\partial x_j \partial x_i} \right ) + \rho b_i = 0$$

Solution
Isotropic Hooke's law is defined as


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$$\displaystyle T_{ij} = \lambda E_{kk} \delta_{ij} + 2 \mu E_{ij} $$.     (4.1)
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 * where Ekk is the volume strain (e).

The two Lamé constants $$ \lambda$$ and $$ \mu$$ are related to the shear and Young's modulii by


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$$\displaystyle \mu = G = \frac{E}{2(1+\nu)} \qquad \text{and} \qquad \lambda = \frac{\nu E}{(1+\nu)(1-2 \nu)} $$.     (4.2)
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By combining equation 4.1 with the equations of 4.2,


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$$\displaystyle T_{ij} = \frac{\nu E}{(1+\nu)(1-2 \nu)} e \delta_{ij} + 2G E_{ij} $$     (4.3)
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 * or


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$$\displaystyle T_{ij} = \quad 2G \left(E_{ij} + \frac{\nu}{1-2\nu} e \delta_{ij} \right) $$     (4.4)
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Now recall the conservation of linear momentum.


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$$\displaystyle \frac{\partial T_{ji}}{\partial x_j} + \rho b_i = \rho \frac{dv_i}{dt} $$     (4.5)
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Combine equations 4.4 and 4.5.


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$$\displaystyle \frac{\partial}{\partial x_j} \left[ 2G \left(E_{ij} + \frac{\nu}{1-2\nu} e \delta_{ij} \right) \right] + \rho b_i = \rho \frac{dv_i}{dt} $$     (4.6)
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Substitute the gradient expression.


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$$\displaystyle G \nabla \left( 2E_{ij} + \frac{2 \nu}{1-2 \nu} e \delta_{ij} \right) + \rho b_i = \rho \frac{dv_i}{dt} $$     (4.7)
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Now move the right-hand side of the equation to the left and simplify.


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$$\displaystyle G \left(\nabla^2 u_i + \frac{1}{1-2\nu} \frac{\partial^2 u_j}{\partial x_j \partial x_i} \right ) + \rho b_i = 0 $$     (4.8)
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Problem 5: Objectivity
Determine the objectivity of the following quantities. Refer to the restrictions for vectors and tensors, respectively:


 * $$\displaystyle a_k' = Q_{kl} a_l$$
 * $$\displaystyle S_{kl}' = Q_{km}S_{mn}Q_{nl}$$

Description
Show that for an orthotropic elastic continuum (three orthogonal planes of elastic symmetry) the elastic coefficient matrix is given as


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$$ [C_{KM}]=\begin{bmatrix} C_{11} & C_{12} & C_{13} & 0 & 0 & 0\\ C_{21} & C_{22} & C_{23} & 0 & 0 & 0\\ C_{31} & C_{32} & C_{33} & 0 & 0 & 0\\ 0 & 0 & 0 & C_{44} & 0 & 0\\ 0 & 0 & 0 & 0 & C_{55} & 0\\ 0 & 0 & 0 & 0 & 0 & C_{66} \end{bmatrix} $$.     (6.1)
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Solution
The generalized Hooke's law includes nine equations expressing the stress components as linear homoegeneous functions of the nine strain components,
 * $$\displaystyle T_{IJ} = C_{IJKM} E_{KM}$$.

The matrix form is shown below. Recall that the stress tensor and strain tensor are symmetric.
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\begin{bmatrix} T_{11}\\ T_{22}\\ T_{33}\\ T_{23}\\ T_{31}\\ T_{12} \end{bmatrix} = \begin{bmatrix} C_{11} & C_{12} & C_{13} & C_{14} & C_{15} & C_{16}\\ C_{21} & C_{22} & C_{23} & C_{24} & C_{25} & C_{26}\\ C_{31} & C_{32} & C_{33} & C_{34} & C_{35} & C_{36}\\ C_{41} & C_{42} & C_{43} & C_{44} & C_{45} & C_{46}\\ C_{51} & C_{52} & C_{53} & C_{54} & C_{55} & C_{56}\\ C_{61} & C_{62} & C_{63} & C_{64} & C_{65} & C_{66}\\ \end{bmatrix} \begin{bmatrix} E_{11}\\ E_{22}\\ E_{33}\\ 2E_{23}\\ 2E_{31}\\ 2E_{12} \end{bmatrix} $$.