#!/usr/bin/env python
# coding: utf-8

# Homework Assignment 2
# =====================
# 
# 
# ## Problem 1
# 
# 1. Write out the conservation of linear momentum equations without using the
# summation convection, i.e. all components.
# 
# 1. For the plane stress state $\sigma_{11} = 12 A x^2 y, \sigma_{22} = B y^2,
# \sigma_{12} = -C x y$, determine the constants $B$ and $C$ in terms of $A$ for
# equilibrium, i.e. $\rho \frac{D\vec{v}}{Dt} = 0$ with negligible body forces.
# 

# **Solution**
# 
# Writing out the components, using an $x,y,z$ component convention
# 
# \begin{align*}
#   \rho \frac{Dv_x}{Dt} &= \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z} + \rho b_x \\
#   \rho \frac{Dv_y}{Dt} &= \frac{\partial \sigma_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{yz}}{\partial z} + \rho b_y \\
#   \rho \frac{Dv_z}{Dt} &= \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} + \rho b_z
# \end{align*}
# 
# Substituting in the stress components
# 
# \begin{align*}
#   0 &=  \frac{\partial}{\partial x}(12 A x^2 y) - \frac{\partial}{\partial y}(C x y) \\
#   0 &= -\frac{\partial}{\partial x}(C x y)      + \frac{\partial}{\partial y}(B y^2) 
# \end{align*}
# 
# 
# Evaluating the partial derivatives
# 
# \begin{align*}
#   0 &=  24 A x y - C x &\implies C = 24 A y \\
#   0 &= -C y      + 2 B y &\implies B = 12 A y
# \end{align*}
# 

# ## Problem 2
# 
# For most of the class, we've used a solid mechanics setting to motivate the
# physics of interest; however, the principles we've derived are general enough to
# apply to fluids as well.  In an ideal nonviscous fluid there can be no shear
# stress.  Hence, the stress tensor is entire hydrostatic, $\sigma_{ij} = -p
# \delta_{ij}$.  Show that this leads to the following form of the momentum
# equation, known as Euler's equation of motion for a frictionless fluid:
# 
# $$
# -\frac{1}{\rho} \nabla p + \vec{b} = \frac{\partial \vec{v}}{\partial t} +
# \vec{v} \cdot \nabla \vec{v}
# $$
# 

# **Solution**
# 
# Starting with the conservation of momentum equation
# 
# \begin{align*}
# \rho \frac{D \vec{v}}{D t} &= \nabla \cdot \boldsymbol{\sigma} + \rho \vec{b} \\
# \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} &= \frac{1}{\rho} \nabla \cdot \boldsymbol{\sigma} + \vec{b} \\
# \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} &= -\frac{1}{\rho} \nabla \cdot (p \mathbf{I}) + \vec{b} \\
# \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} &= -\frac{1}{\rho} \nabla p + \vec{b}
# \end{align*}

# ## Problem 3
# 
# Using the different forms of conservation of mass, derive an expression for $\frac{d J}{dt}$ in terms of $\vec{v}$, where $J = \det(\mathbf{F})$
# 
# **Solution**
# 
# Using the material form of conservation of mass
# 
# \begin{align*}
#   J &= \frac{\rho_o}{\rho} \\
#   \rho J &= \rho_o \\
#   \frac{d}{dt}(\rho J) &= \frac{d}{dt}(\rho_o) \\
#   J \frac{d \rho}{dt} + \rho \frac{d J}{dt} &= 0 \\
#   J  \nabla \cdot (\rho \vec{v}) &= \rho \frac{d J}{dt} \\
#   \frac{\rho_o}{\rho} \nabla \cdot (\rho \vec{v}) &= \rho \frac{d J}{dt} \\
#   \frac{\rho_o}{\rho^2} \nabla \cdot (\rho \vec{v}) &= \frac{d J}{dt} 
# \end{align*}

# ## Problem 4
# 
# Assume that the internal-energy density can be given as $u = u(\epsilon, T)$,
# that the heat flux is governed by Fourier's law $\vec{q} = −k(T)\nabla T$, and
# that $r = 0$. Defining the specific heat $C = \frac{\partial u}{\partial T}$,
# write the equation resulting from combining these assumptions with the energy-
# balance equation.
# 

# **Solution**
# 
# Starting with the energy equation, for small displacements $D_{ij} \approx \dot{\varepsilon}_{ij}$
# 
# \begin{align*}
# \rho \frac{D u}{Dt} &= \boldsymbol{\sigma} : \boldsymbol{\dot\varepsilon} + \nabla \cdot \vec{q} \\
# \rho \frac{D u}{DT}\frac{DT}{Dt} + \rho \frac{D u}{D\epsilon} \frac{D\epsilon}{Dt} &= \boldsymbol{\sigma} : \boldsymbol{\dot\varepsilon} - \nabla \cdot (k(T) \nabla T) \\
# \rho C \dot{T} + \rho \frac{D u}{D\epsilon} \dot{\epsilon} &= \boldsymbol{\sigma} : \boldsymbol{\dot\varepsilon} - \nabla k(T)\nabla T - k(T)\nabla^2 T \\
# \rho C \dot{T} + \rho \frac{D u}{D\epsilon} \dot{\epsilon} &= \boldsymbol{\sigma} : \boldsymbol{\dot\varepsilon} - k(T)\nabla^2 T \\
# \end{align*}
# 
# where it is understood that $\epsilon$ is a scalar internal state variable.

# ## Problem 5
# 
# Show that, in an isotropic linearly elastic solid, the principal stress and
# principal strain directions coincide.
# 
# **Solution**
# 
# There are several ways to demonstrate this, first, let's assume that there is an orthonormal tensor $\mathbf{V}$ that diagonalizes the *strain tensor* $\boldsymbol{\varepsilon}$, i.e. it consists of eigenvectors of the strain tensor and produces a diagonal tensor, $\boldsymbol{\varepsilon'}$ with the principle strains on the diagonal.
# 
# $$\boldsymbol{\varepsilon'} = \mathbf{V}^{-1} \boldsymbol{\varepsilon} \mathbf{V}$$
# 
# We'll use the eigenvectors from the strain tensor to transform the stress tensor, if the result is diagonal,  we know they are also eigenvectors of the stress tensor, that is the principle directions of both tensors are in the same direction.
# 
# \begin{align*}
# \sigma_{ij}' &= V_{ik}^{-1} \sigma_{kl} V_{lj} \\
#              &= V_{ik}^{-1} \left(2 \mu \varepsilon_{kl} + \lambda \varepsilon_{mm} \delta_{kl} \right) V_{lj} \\
#              &= 2 \mu V_{ik}^{-1} \varepsilon_{kl} V_{lj} + \lambda \varepsilon_{mm} V_{il}^{-1} V_{lj} \\
#              &= 2 \mu V_{ik}^{-1} \varepsilon_{kl} V_{lj} + \lambda \varepsilon_{mm} \delta_{ij} \\
#              &= 2 \mu \varepsilon'_{ij} + \lambda \varepsilon_{mm} \delta_{ij} \\
# \end{align*}
# 
# By inspection we can see that both the first and second terms of the resulting expressions are diagonal by definition, therefore $\boldsymbol{\sigma}'$ is also diagonal and $\mathbf{V}$ also contains the eigenvectors of $\boldsymbol{\sigma}$

# ## Problem 6
# 
# Write the elastic modulus matrix $C_{IJ}$ for an isotropic linearly elastic
# solid in terms of the Young’s modulus $E$ and the Poisson’s ratio $\nu$.
# 
# 
# **Solution**
# 
# 
# $$
# C_{IJ} = \frac{E}{2 (1+\nu)}
# \begin{bmatrix}
# \frac{2(1-\nu)}{1-2\nu} & \frac{2\nu}{1-2\nu}     & \frac{2\nu}{1-2\nu} & 0 & 0 & 0 \\
# \frac{2\nu}{1-2\nu}     & \frac{2(1-\nu)}{1-2\nu} & \frac{2\nu}{1-2\nu} & 0 & 0 & 0 \\
# \frac{2\nu}{1-2\nu}     & \frac{2\nu}{1-2\nu}     & \frac{2(1-\nu)}{1-2\nu} & 0 & 0 & 0 \\
# 0 & 0 & 0 & 1 & 0 & 0 \\
# 0 & 0 & 0 & 0 & 1 & 0 \\
# 0 & 0 & 0 & 0 & 0 & 1
# \end{bmatrix}
# $$

# ## Problem 7
# 
# Combine the generalized Hooke's law for an isotropic linearly elastic solid with
# the equations of motion and the definition of small strain in order to derive
# the equations of motion for such a solid entirely in terms of displacement,
# using
# 
# 1. $\lambda$ and $\mu$
# 
# 2. $G$ and $\nu$
# 
# **Solution**
# 
# 
# \begin{align*}
# \rho \frac{D^2\vec{u}}{Dt^2} &= \nabla \cdot \boldsymbol{\sigma} + \rho \vec{b} \\
# &= \nabla \cdot \left(2 \mu \boldsymbol{\varepsilon} + \lambda \text{tr}(\boldsymbol{\varepsilon}) \mathbf{I} \right)+ \rho \vec{b} \\
# &= \mu \frac{\partial}{\partial x_j} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right) + \lambda \frac{\partial^2 u_k} {\partial x_j \partial x_k} \delta_{ij}\\
# &= \mu \frac{\partial^2 u_i}{\partial x_j^2} + \mu \frac{\partial^2 u_j}{\partial x_i\partial x_j} + \lambda \frac{\partial^2 u_k} {\partial x_i \partial x_k} \\ 
# &= \mu \frac{\partial^2 u_i}{\partial x_j^2} + \mu \frac{\partial^2 u_j}{\partial x_i\partial x_j} + \lambda \frac{\partial^2 u_j} {\partial x_i \partial x_j} \\
# &= \mu \frac{\partial^2 u_i}{\partial x_j^2} + (\mu + \lambda) \frac{\partial^2 u_j} {\partial x_i \partial x_j} \\
# &= \mu \nabla^2 \vec{u} + (\mu + \lambda) \nabla (\nabla \cdot \vec{u}) \\
# &= G \nabla^2 \vec{u} + \frac{G}{1 - 2 \nu}\nabla (\nabla \cdot \vec{u})
# \end{align*}

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