Instructions: This exam will be assigned at 11:50 AM Friday, 10/2 via Canvas and is due at 11:00 AM Monday, 10/5.
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, violations of academic integrity with be reported.
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
\begin{equation} \boldsymbol{\sigma} = \mathbf{R} \boldsymbol{\sigma} \mathbf{R}^{\intercal}. \label{eqn:1} \end{equation}However, often in plasticity modeling, we use rate-forms of constitutive models. Is the rate-of-Cauchy stress material reference frame indifferent? Explain.
Solution
Take the time derivative of both sides of the equation, i.e.
\begin{align*} \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 \end{align*}This clearly is not equivalent to $\ref{eqn:1}$, therefore the rate-of-Cauchy stress is not material reference frame indifferent.
If the Lagrangian strain components at a point are as shown
$$ \mathbf{E} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1.84 & -0.28 \\ 0 & -0.28 & 1.84 \end{bmatrix} $$Compute:
the principle strains
the maximum shear strain component
Solution
import numpy as np
import numpy.linalg
w, v = np.linalg.eig(np.array([[-1., 0., 0.], [0., 1.84, -0.28], [0., -0.28, 1.84]]));
w
array([ 2.12, 1.56, -1. ])
The maximum shear strain is one-half the difference between the maximum and minumum principle strains (from Mohr's circles)
Emax = 0.5 * (w[0] - w[2]); Emax
1.5600000000000001
For the stress tensor
$$ \boldsymbol{\sigma} = \begin{bmatrix} 2 & 0 & 4 \\ 0 & 3 & 6 \\ 4 & 6 & 0 \end{bmatrix} $$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.
Solution
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.
sigma = np.array([[2., 0, 4],[0, 3, 6], [4, 6, 0]])
#First compute the solution
n = np.linalg.solve(sigma.T, np.array([0,0,1]))
#Now divide by the norm to get a normal vector
n_hat = n / np.linalg.norm(n); n_hat
array([ 0.66666667, 0.66666667, -0.33333333])
Now can use this normal vector to determine $t_3$.
np.dot(sigma.T, n_hat)
array([ -4.44089210e-16, 4.44089210e-16, 6.66666667e+00])
So the final answers in rational form are
$$\hat{n}^T = \left[ \frac{2}{3}, \frac{2}{3}, -\frac{1}{3} \right]$$and
$$t_3 = \frac{20}{3}$$Evaluate the second invariant of the deviatoric stress for the stress tensor
$$ \boldsymbol{\sigma} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 2 \end{bmatrix} $$Solution
First we compute the deviatoric stress
sigma = np.array([[2., -1., 1.], [-1., 0., 1], [1., 1., 2.]])
S = sigma - 1./3. * np.trace(sigma) * np.eye(3); S
array([[ 0.66666667, -1. , 1. ], [-1. , -1.33333333, 1. ], [ 1. , 1. , 0.66666667]])
Now we perform the summation
$$ J_2 = \frac{1}{2} S_{ij} S_{ij} $$J_2 = 0.0
for i in range(3):
for j in range(3):
J_2 += 0.5 * S[i,j] * S[i, j]
J_2
4.333333333333333
Or in rational form we have
$$J_2 = \frac{13}{3}$$Under what condition are these two statements equavalent
$$ \det({\mathbf{F}}) = 1 $$$$ \frac{\partial v_i}{\partial x_i} = 0 $$Explain your answer.
Solution
The correct answer to both conditions would be the incompressibility condition, from conservation of mass
$$ \det(\mathbf{F}) = 1 = \frac{\rho_o}{\rho} $$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
$$ \frac{D \rho}{Dt} + \rho \frac{\partial v_i}{\partial x_i} = 0 $$When viewed from the position of a moving particle, incompressibility implies $\frac{D \rho}{Dt} = 0$ therefore
$$ \frac{\partial v_i}{\partial x_i} = 0 $$Show that the plane stress equilibrium equations, in the absence of body forces, implies
$$ 2 \frac{\partial\sigma_{12}}{\partial x_1 \partial x_2} = -\frac{\partial^2 \sigma_{11}}{\partial x_1^2} -\frac{\partial^2 \sigma_{22}}{\partial x_2^2} $$Solution
Starting with the momentum equations
\begin{align} \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 \\ \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 \\ \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 \end{align}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.
\begin{align} 0 &= \frac{\partial \sigma_{11}}{\partial x_1} + \frac{\partial \sigma_{12}}{\partial x_2} \\ 0 &= \frac{\partial \sigma_{12}}{\partial x_1} + \frac{\partial \sigma_{22}}{\partial x_2} \end{align}Now we differentiate the first equation w.r.t. $x_1$ and the second w.r.t $x_2$ we have
\begin{align} 0 &= \frac{\partial^2 \sigma_{11}}{\partial x_1^2} + \frac{\partial^2 \sigma_{12}}{\partial x_1 \partial x_2} \\ 0 &= \frac{\partial^2 \sigma_{12}}{\partial x_1 \partial x_2} + \frac{\partial^2 \sigma_{22}}{\partial x_2^2} \end{align}Adding the two equations together and rearranging, we have the result
$$ 2 \frac{\partial\sigma_{12}}{\partial x_1 \partial x_2} = -\frac{\partial^2 \sigma_{11}}{\partial x_1^2} -\frac{\partial^2 \sigma_{22}}{\partial x_2^2} $$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 $\varepsilon^p$.
Solution
$$ f = \sqrt{3 J_2} - Y - H \varepsilon^p$$