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    "# PGE383 - Advanced Geomechanics - Midterm Exam\n",
    "\n",
    "**Instructions:** This exam will be assigned at 11:50 AM Friday, 10/2 via [Canvas](https://utexas.instructure.com/courses/1150226) and is due at 11:00 AM Monday, 10/5.\n",
    "\n",
    "You are allowed any resource (e.g. books, notes, web) to complete the exam except consultation with another student or professor. You are reminded of the [UT Student Honor Code](http://www.utexas.edu/about-ut/mission-core-purpose-honor-code), violations of academic integrity with be [reported](http://deanofstudents.utexas.edu/sjs/acadint_reportingprocedures.php)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 1\n",
    "\n",
    "Valid constitutive models must obey the *Principle of Material Reference Frame Indifference*, i.e. they should be invarient to rigid rotations.  Mathematically, for a Cauchy stress $\\boldsymbold{\\sigma}$ and a orthoganal time-dependent rotation tensor $\\mathbf{R}$ we might say that the stress is invariant to a rotation if\n",
    "\n",
    "\\begin{equation}\n",
    "\\boldsymbol{\\sigma} = \\mathbf{R} \\boldsymbol{\\sigma} \\mathbf{R}^{\\intercal}.\n",
    "\\label{eqn:1}\n",
    "\\end{equation}\n",
    "\n",
    "However, often in plasticity modeling, we use rate-forms of constitutive models.  Is the rate-of-Cauchy stress material reference frame indifferent?  Explain."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Solution**\n",
    "\n",
    "Take the time derivative of both sides of the equation, i.e.\n",
    "\n",
    "\\begin{align*}\n",
    "    \\dot{\\boldsymbol{\\sigma}} = \\dot{\\mathbf{R}} \\boldsymbol{\\sigma} \\mathbf{R}^\\intercal + \\mathbf{R} \\dot{\\boldsymbol{\\sigma}} \\mathbf{R}^\\intercal + \\mathbf{R} \\boldsymbol{\\sigma} \\dot{\\mathbf{R}}^\\intercal\n",
    "\\end{align*}\n",
    "\n",
    "This clearly is not equivalent to $\\ref{eqn:1}$, therefore the rate-of-Cauchy stress **is not** material reference frame indifferent."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 2\n",
    "\n",
    "If the Lagrangian strain components at a point are as shown\n",
    "\n",
    "$$\n",
    "\\mathbf{E} = \n",
    "\\begin{bmatrix}\n",
    "-1 & 0 & 0 \\\\\n",
    " 0 & 1.84 & -0.28 \\\\\n",
    " 0 & -0.28 & 1.84\n",
    "\\end{bmatrix}\n",
    "$$\n",
    "\n",
    "Compute:\n",
    "\n",
    "1. the principle strains\n",
    "\n",
    "2. the maximum shear strain component"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Solution**\n",
    "\n",
    "1. We need to simply solve an eigenvalue problem."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([ 2.12,  1.56, -1.  ])"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import numpy as np\n",
    "import numpy.linalg\n",
    "\n",
    "\n",
    "w, v = np.linalg.eig(np.array([[-1., 0., 0.], [0., 1.84, -0.28], [0., -0.28, 1.84]]));\n",
    "w"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The maximum shear strain is one-half the difference between the maximum and minumum principle strains (from Mohr's circles)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1.5600000000000001"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Emax = 0.5 * (w[0] - w[2]); Emax"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 3\n",
    "\n",
    "For the stress tensor\n",
    "\n",
    "$$\n",
    "\\boldsymbol{\\sigma} =\n",
    "\\begin{bmatrix}\n",
    "2 & 0 & 4 \\\\\n",
    "0 & 3 & 6 \\\\\n",
    "4 & 6 & 0\n",
    "\\end{bmatrix}\n",
    "$$\n",
    "\n",
    "find a direction $\\mathbf{\\hat{n}}$, such that the traction vector on a plane normal to $\\mathbf{\\hat{n}}$ has components $t_1 = t_2 = 0$, and determine $t_3$ on that plane."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Solution**\n",
    "\n",
    "First we will solve for the unknown normal vector using the Cauchy stress equation, i.e. $\\vec{t} = \\sigma^T \\hat{n}$.  Where $\\vec{t}^T$ = [0 0 1], we can use 1 here in place of $t_3$ because any arbrary $t_3$ that appears in the solution, will cancel when dividing the solution by its norm to create the normal vector."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([ 0.66666667,  0.66666667, -0.33333333])"
      ]
     },
     "execution_count": 21,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sigma = np.array([[2., 0, 4],[0, 3, 6], [4, 6, 0]])\n",
    "#First compute the solution\n",
    "n = np.linalg.solve(sigma.T, np.array([0,0,1]))\n",
    "#Now divide by the norm to get a normal vector\n",
    "n_hat = n / np.linalg.norm(n); n_hat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now can use this normal vector to determine $t_3$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([ -4.44089210e-16,   4.44089210e-16,   6.66666667e+00])"
      ]
     },
     "execution_count": 22,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "np.dot(sigma.T, n_hat)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "So the final answers in rational form are\n",
    "\n",
    "$$\\hat{n}^T = \\left[ \\frac{2}{3}, \\frac{2}{3}, -\\frac{1}{3} \\right]$$\n",
    "\n",
    "and\n",
    "\n",
    "$$t_3 = \\frac{20}{3}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 4\n",
    "\n",
    "Evaluate the second invariant of the *deviatoric* stress for the stress tensor\n",
    "\n",
    "$$\n",
    "\\boldsymbol{\\sigma} =\n",
    "\\begin{bmatrix}\n",
    "  2 & -1 & 1 \\\\\n",
    "  -1 & 0 & 1 \\\\\n",
    "  1 & 1 & 2\n",
    "\\end{bmatrix}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Solution**\n",
    "\n",
    "First we compute the deviatoric stress"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 0.66666667, -1.        ,  1.        ],\n",
       "       [-1.        , -1.33333333,  1.        ],\n",
       "       [ 1.        ,  1.        ,  0.66666667]])"
      ]
     },
     "execution_count": 25,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sigma = np.array([[2., -1., 1.], [-1., 0., 1], [1., 1., 2.]])\n",
    "\n",
    "S = sigma - 1./3. * np.trace(sigma) * np.eye(3); S"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now we perform the summation\n",
    "\n",
    "$$\n",
    "J_2 = \\frac{1}{2} S_{ij} S_{ij}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4.333333333333333"
      ]
     },
     "execution_count": 26,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "J_2 = 0.0\n",
    "for i in range(3):\n",
    "    for j in range(3):\n",
    "        J_2 += 0.5 * S[i,j] * S[i, j]\n",
    "        \n",
    "J_2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Or in rational form we have\n",
    "\n",
    "$$J_2 = \\frac{13}{3}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 5\n",
    "\n",
    "Under what condition are these two statements equavalent\n",
    "\n",
    "$$ \\det({\\mathbf{F}}) = 1 $$ \n",
    "\n",
    "$$ \\frac{\\partial v_i}{\\partial x_i} = 0 $$\n",
    "\n",
    "Explain your answer."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Solution**\n",
    "\n",
    "The correct answer to both conditions would be the *incompressibility* condition, from conservation of mass\n",
    "\n",
    "$$\n",
    "\\det(\\mathbf{F}) = 1 = \\frac{\\rho_o}{\\rho}\n",
    "$$\n",
    "\n",
    "which implies that density doesn't change or the motion is *volume preserving* when viewed from the reference configuration.  Likewise, in Eulerian form, conservation of mass is\n",
    "\n",
    "$$\n",
    "\\frac{D \\rho}{Dt} + \\rho \\frac{\\partial v_i}{\\partial x_i} = 0\n",
    "$$\n",
    "\n",
    "When viewed from the position of a moving particle, incompressibility implies $\\frac{D \\rho}{Dt} = 0$ therefore\n",
    "\n",
    "$$\n",
    "\\frac{\\partial v_i}{\\partial x_i} = 0\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Problem 6\n",
    "\n",
    "Show that the *plane stress* equilibrium equations, in the absence of body forces, implies\n",
    "\n",
    "$$\n",
    "2 \\frac{\\partial\\sigma_{12}}{\\partial x_1 \\partial x_2} = -\\frac{\\partial^2 \\sigma_{11}}{\\partial x_1^2} -\\frac{\\partial^2 \\sigma_{22}}{\\partial x_2^2}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Solution**\n",
    "\n",
    "Starting with the momentum equations\n",
    "\n",
    "\\begin{align}\n",
    "    \\rho \\frac{\\partial^2 u_1}{\\partial t^2} &= \\frac{\\partial \\sigma_{11}}{\\partial x_1} + \\frac{\\partial \\sigma_{12}}{\\partial x_2} + \\frac{\\partial \\sigma_{13}}{\\partial x_3} + b_1 \\\\\n",
    "    \\rho \\frac{\\partial^2 u_2}{\\partial t^2} &= \\frac{\\partial \\sigma_{12}}{\\partial x_1} + \\frac{\\partial \\sigma_{22}}{\\partial x_2} + \\frac{\\partial \\sigma_{23}}{\\partial x_3} + b_2 \\\\\n",
    "    \\rho \\frac{\\partial^2 u_3}{\\partial t^2} &= \\frac{\\partial \\sigma_{31}}{\\partial x_1} + \\frac{\\partial \\sigma_{23}}{\\partial x_2} + \\frac{\\partial \\sigma_{33}}{\\partial x_3} + b_3\n",
    " \\end{align}\n",
    " \n",
    "Now we apply the equalibrium condition, i.e. $ \\rho \\frac{\\partial^2 \\vec{u}}{\\partial t^2} = 0$, the plane strain assumptions, i.e. $\\sigma_{33} = \\sigma_{13} = \\sigma_{23} = 0$ and exclude the body forces, we are left with.\n",
    "\n",
    "\\begin{align}\n",
    "    0 &= \\frac{\\partial \\sigma_{11}}{\\partial x_1} + \\frac{\\partial \\sigma_{12}}{\\partial x_2}  \\\\\n",
    "    0 &= \\frac{\\partial \\sigma_{12}}{\\partial x_1} + \\frac{\\partial \\sigma_{22}}{\\partial x_2}\n",
    " \\end{align}\n",
    " \n",
    "Now we differentiate the first equation w.r.t. $x_1$ and the second w.r.t $x_2$ we have \n",
    "\n",
    "\n",
    "\\begin{align}\n",
    "    0 &= \\frac{\\partial^2 \\sigma_{11}}{\\partial x_1^2} + \\frac{\\partial^2 \\sigma_{12}}{\\partial x_1 \\partial x_2}  \\\\\n",
    "    0 &= \\frac{\\partial^2 \\sigma_{12}}{\\partial x_1 \\partial x_2} + \\frac{\\partial^2 \\sigma_{22}}{\\partial x_2^2}\n",
    " \\end{align}\n",
    " \n",
    " Adding the two equations together and rearranging, we have the result\n",
    " \n",
    "$$\n",
    "2 \\frac{\\partial\\sigma_{12}}{\\partial x_1 \\partial x_2} = -\\frac{\\partial^2 \\sigma_{11}}{\\partial x_1^2} -\\frac{\\partial^2 \\sigma_{22}}{\\partial x_2^2}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 7\n",
    "\n",
    "If a ductile material undergoing uniaxial stress, exhibits linear isotropic hardening, with hardening modulus $H$ and yield stress $Y$, write a von Mises yield function for this material as a function of these material constants and equivalent plastic strain $\\varepsilon^p$.\n",
    "\n",
    "**Solution**\n",
    "\n",
    "$$ f = \\sqrt{3 J_2} - Y - H \\varepsilon^p$$"
   ]
  }
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