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Template:Engineering/Intro

Engineering is the application of scientific and technical knowledge to solve human problems. Engineers use imagination, judgment and reasoning to apply science, technology, mathematics, and practical experience. The result is the design, production, and operation of useful objects or processes.

A school is a large organizational structure which can contain various departments and divisions. The departments and divisions should be listed in the departments and divisions section. The school should not contain any learning resources. The school can contain projects for developing learning resources.

For specific questions, why not consult the Engineering Help Desk? We also have an engineering specific discussion portal to help get those technical details straight or just enjoy shooting the breeze with other technically interested people.

title: Template:Engineering/Intro


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Template:Engineering/Departments

Divisions and Departments of the School exist on pages in "topic" namespace. Start the name of departments with the "Topic:" prefix; departments reside in the Topic: namespace. Departments and divisions link to learning materials and learning projects. Divisions can link subdivisions or to departments. For more information on schools, divisions and departments look at the Naming Conventions. To add additional Divisions or Departments within the School of Engineering, edit the list of "Topic:" namespaces on the Engineering Template. '''The list of Engineering Departments below has been restructured under popularly established major engineering branches and the new "Tree" categorization system has been implemented. Please edit all Engineering pages to reflect these changes.'''

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Template:Engineering/Departments


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Template:Engineering/News and events

 * November 9, 2007 - Began using the Template:Engineering-list.
 * Aug 4, 2007 - Reformat of main page.
 * Jan 2007 - School founded!

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School:Engineering/Active participants

The histories of Wikiversity pages indicate who the active participants are. If you are an active participant in this school, you can list your name here (this can help small schools grow and the participants communicate better; for large schools it is not needed).

 * 1) Since 9 February 2012 --Marshallsumter (discuss • contribs) 22:25, 7 April 2018 (UTC)
 * 2) Joshua (discuss • contribs) 19:24, 23 August 2018 (UTC)

Inactive participants
See: List of participants
 * 1) Learning !Wasami007 (discuss • contribs) 14:52, 19 November 2014 (UTC)
 * 2) TargetAquired 11:25, 16 March 2012 (UTC)User:TargetAquired
 * 3) abhishek mishra 21:01, 22 January 2007 (UTC)kishu
 * 4) Markemark1005 16:34, 22 February 2007 (UTC)
 * 5) Dionysios (talk), a Participant in the Wikiversity School of Advanced General Studies, Date: 2007-04-25 (April 25, 2007) Time: 1239301 UTC
 * 6) Chocoman 31 August,2007
 * 7) Gustable - 03:47, 17 December 2008 (UTC)
 * 8) Theornamentalist 01:07, 25 January 2009 (UTC)
 * 9) Khoshhal Ahmad Stanikzai
 * 10) Savannah

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Template:Engineering/Research projects

 * Unmanned aerial systems
 * If you read news sites like Physorg.com there are many stories of breakthrough technologies. It seems however, that these breakthrough technologies just don't appear on the market or take several years to do so. Why do these technologies take so long to appear? Why do these technologies sometimes not appear? Do these news sites hype the story to present an unrealistic picture of the technology or is there some obstacle that prevents these technologies from appearing on the market? What could be done to increase the speed at which these technologies enter the market?

title: Template:Engineering/Research projects


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School:Engineering/Wikipedia articles

 * Engineering
 * List of engineering topics
 * List of basic engineering topics
 * Fields of Engineering
 * Engineering vehicle
 * Hydraulic machinery
 * Pneumatics
 * Hydraulic fluid
 * Category:Engineering vehicles

title: School:Engineering/Wikipedia articles


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Template:Engineering/Open source software

 * avoCADo - Collaborative, intuitive CAD tools
 * freeCAD
 * Topic:Engineering/Free or open source design tools


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Template:Engineering/External links

 * NASA Centennial Challenges - Cash for engineering feats.
 * Oilgae - Open source biodiesel from Algae
 * Appleseed Reactor - Open source biofuel reactor
 * Cornell Computational Synthesis Lab
 * "14 great engineering challenges"

title: Template:Engineering/External links


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School:Engineering/Textbooks

 * Fluid Mechanics
 * Soil mechanics
 * Circuit Theory
 * Engineering Tables
 * Communication Systems
 * Nanotechnology
 * Statics
 * Jet Propulsion

 * Rocket Propulsion
 * Automotive Systems
 * Introduction to Materials
 * Structural engineering
 * Semiconductors
 * Electronics
 * Microtechnology
 * Electric Vehicle Conversion
 * Engineering Acoustics

 * Signals and Systems
 * Engineering Mechanics
 * Heat Transfer
 * House Construction
 * Automatic Control Engineering
 * Engineering Analysis
 * Introduction to Materials Science
 * SensorNetworks
 * Structured Analysis of Systems
 * Probability for Engineers

Introduction to Engineering (Christian Liberty Press)

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Talk:Principles of marketing

Practice Opportunity
Participants studying marketing are welcome to practice or develop their skills at Lunar Boom Town. Many or most of the business ideas or plans there will need some market analysis to refine the business models. Mirwin 08:35, 30 September 2007 (UTC)

Product Management Sections
These sections seem a bit out of place. They do not really address the marketeer's recommended responsibilities or participation methods in the product development cycle. Some augmentation from this viewpoint of typical practice in various industries would be nice. Mirwin 03:55, 20 December 2007 (UTC)

NEED HELP!!!!!!
can anyone describe to me how the marketing principles are relevant to businesses???

^^^^ -- I don't think it's reasonable to throw out these kinds of questions to a wiki.

Marketing
What is all about marketing Ianzulu (discuss • contribs) 20:47, 17 December 2017 (UTC)

On my own I think these are activities which take place in-between the customers and the produers Ianzulu (discuss • contribs) 20:50, 17 December 2017 (UTC)

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User:Anderswohin

I have an interest in learning more about Philosophy and Critical Thinking. I look forward to having a discussion with anyone while I work at acquiring a better understanding myself.

Personally, I am a filmmaker and sometimes musician. My work includes the internet show, Anderecast. . I shall be working on a new show called, "Andereplatz" which will be an open reading and discussion forum via radio podcasting.

title: User:Anderswohin


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Nonlinear finite elements/Homework 1/Solutions

Problem 1: Mathematical Model Development
 Given:

 Assumptions:

 * 1) The rod and the mass at the end are rigid.
 * 2) The mass of the rod is negligible compared to the mass of the bob.
 * 3) There is no friction at the pivot.

Part 1
''Find the equation of motion of the bob (angular displacement as a function of time). State all other assumptions.'' The component gravitational force in the direction tangent to the motion is

F_x = - mg\sin\theta ~. $$ The tangential velocity of the bob is

v_x = \cfrac{ds}{dt} = \cfrac{d}{dt}(l\theta) = l\cfrac{d\theta}{dt} $$ where $$s$$ is an arc length along the motion of the bob.

Therefore, the tangential acceleration of the bob is

a_x = \cfrac{d}{dt}\left(l\cfrac{d\theta}{dt}\right) = l \cfrac{d^2\theta}{dt^2} ~. $$ From Newton's second law (balance of momentum) we have

F_x = m a_x \qquad \implies \qquad -mg\sin\theta = ml\cfrac{d^2\theta}{dt^2} ~. $$ Therefore  the equation of motion of the bob is

{       \cfrac{d^2\theta}{dt^2} + \lambda^2\sin\theta = 0, ~\qquad \lambda^2 = \cfrac{g}{l} ~. }     $$ Additional assumptions are lack of viscosity in the surrounding medium, no wind load, and so on.

Part 2
 Why is the equation of motion "nonlinear"?

Let us try the technique we learned in class. Let

A := \cfrac{d^2\theta }{dt^2} ~; \lambda = 1 ~. $$ Then, the equation of motion can be written as

{       A + \sin\theta = 0 ~. }     $$ Without making further assumptions, this equation cannot be expressed as the equation of a line of the form

y = m\theta ~. $$ Hence the equation of motion is nonlinear.

Part 4
''Assume that $$\theta$$ is "small". Derive the equation of motion for this situation.''

From elementary calculus, for small values of $$\theta$$ we have

\sin\theta \approx \theta ~. $$ Therefore, for small values of $$\theta$$, the equation of motion becomes

{       \cfrac{d^2\theta}{dt^2} + \lambda^2\theta = 0~. }     $$

Part 5
Why is the small $$\theta$$ equation of motion "linear"? In this case we can write the differential equation as

{       A + \theta = 0 ~. }     $$ This equation  can be expressed as the equation of a line of the form

y = m\theta ~. $$ Hence the equation of motion is linear.

Part 6
Derive a finite element formulation for the linear equation of motion.

We know that if we can formulate the finite element system of equations for one element, we can assemble the element equations into a global system.

Let us find the finite element system of equations for one element. In this problem, we discretize the time variable into elements. Let the element extend from time $$t_1$$ to time $$t_2$$.

The first step in the process is the derivation of a approximate weak form of the differential equation.

Let $$w(t)$$ be a weighting function. After multiplying the ODE by the weighting function and integrating the product over the element, we get

\int_{t_1}^{t_2} \left(\cfrac{d^2\theta}{dt^2} + \lambda^2\theta\right)~w(t)~dt = 0~. $$ After integrating the first term by parts, we get

\left.\left(\cfrac{d\theta}{dt}~w\right)\right|_{t_1}^{t_2} - \int_{t_1}^{t_2} \cfrac{d\theta}{dt}\cfrac{dw}{dt}~dt + \lambda^2\int_{t_1}^{t_2}\theta~w~dt = 0~. $$ The rearranged weak form is

{       \int_{t_1}^{t_2} \cfrac{d\theta}{dt}\cfrac{dw}{dt}~dt - \lambda^2\int_{t_1}^{t_2}\theta~w~dt = \left.\left(\cfrac{d\theta}{dt}~w\right)\right|_{t_1}^{t_2} ~. }     $$ Let $$\theta_h(t)$$ be an approximate solution. To get the approximate weak form, we plug $$\theta_h$$ into the exact weak form. Notice that the right hand side of the exact weak form contains a weighted boundary lux term. We do not approximate this term because it is presumed to be specified or known exactly. In this problem, this term represents the angular velocity $$\omega = d\theta/dt$$. Then the approximate weak form is

{       \int_{t_1}^{t_2} \cfrac{d\theta_h}{dt}\cfrac{dw}{dt}~dt - \lambda^2\int_{t_1}^{t_2}\theta_h~w~dt = \left.\left(\omega~w\right)\right|_{t_1}^{t_2} ~. }     $$ At this stage we choose the form of the approximate solution. Let us assume that each element has two nodes and the approximate solution within an element has the form

{       \theta_h(t) = \theta_1 N_1(t) + \theta_2 N_2(t) = \sum_{j=1}^2 \theta_j N_j(t) }     $$ where $$\theta_1$$ is the nodal rotation at node $$1$$ of the element and      $$\theta_2$$ is the nodal rotation at node $$2$$ of the element and $$N_1$$ and $$N_2$$ are the corresponding shape functions.

We have seen that instead of thinking of the weighting function as a series of the same form as $$\theta_h(t)$$, we can just consider these functions to have the form $$N_i$$ - where the index $$i$$ stands for a row of the finite element system of equations. Therefore, we write

{       w = N_i(t) ~. }     $$ Plug the approximate solution and the weighting function into the weak form to get

\int_{t_1}^{t_2} \left(\sum_{j=1}^2 \theta_j\cfrac{dN_j}{dt}\right)\cfrac{dN_i}{dt}~dt - \lambda^2 \int_{t_1}^{t_2}\left(\sum_{j=1}^2\theta_j N_j\right)N_i~dt = \left.\left(\omega~N_i\right)\right|_{t_1}^{t_2} ~. $$ Take the summation sign and the constants outside the integral and get

\sum_{j=1}^2 \left(\theta_j          \int_{t_1}^{t_2}\cfrac{dN_j}{dt}\cfrac{dN_i}{dt}~dt\right) - \lambda^2 \sum_{j=1}^2 \left(\theta_j          \int_{t_1}^{t_2} N_j N_i~dt\right) = \left.\left(\omega~N_i\right)\right|_{t_1}^{t_2} ~. $$ Collect terms on the left hand side

\sum_{j=1}^2 \left[\theta_j \int_{t_1}^{t_2}\left(\cfrac{dN_j}{dt}\cfrac{dN_i}{dt} -                         \lambda^2 N_j N_i\right)~dt\right] = \left.\left(\omega~N_i\right)\right|_{t_1}^{t_2} ~. $$ Rearrange and get

\sum_{j=1}^2 \left[ \int_{t_1}^{t_2}\left(\cfrac{dN_i}{dt}\cfrac{dN_j}{dt} -                         \lambda^2 N_i N_j\right)~dt\right]\theta_j = \left.\left(\omega~N_i\right)\right|_{t_1}^{t_2} ~. $$ The above has the form

\sum_{j=1}^2 K_{ij} \theta_j = f_i $$ where

{        K_{ij} =  \int_{t_1}^{t_2}\left(\cfrac{dN_i}{dt}\cfrac{dN_j}{dt} -                          \lambda^2 N_i N_j\right)~dt } \qquad \leftarrow ~\text{Element stiffness matrix.} $$ and

{        f_i = \left.\left(\omega~N_i\right)\right|_{t_1}^{t_2} ~. } \qquad \leftarrow ~\text{Element load vector.} $$ Given the element stiffness matrix and the element load vector, we can assemble a global system of equations and solve it for the unknown values of $$\theta_i$$.

Let us assume that the shape functions are

{       N_1(t) = \cfrac{t_2 - t}{\Delta t} \qquad \text{and} \qquad N_2(t) = \cfrac{t - t_1}{\Delta t}        } $$ where $$\Delta t = t_2 - t_1$$.

Then, the element stiffness matrix terms are
 * $$\begin{align}

K_{11} & = \int_{t_1}^{t_2}\left(\cfrac{dN_1}{dt}\cfrac{dN_1}{dt} -                     \lambda^2 N_1 N_1\right)~dt = \int_{t_1}^{t_2}\left[\cfrac{1}{\Delta t^2} - \lambda^2\left(\cfrac{t_2-t}{\Delta t}\right)^2\right]~dt = \cfrac{1}{3\Delta t} (\lambda^2\Delta t^2 - 3) \\ K_{12} & = \int_{t_1}^{t_2}\left(\cfrac{dN_1}{dt}\cfrac{dN_2}{dt} -                     \lambda^2 N_1 N_2\right)~dt = \int_{t_1}^{t_2}\left[-\cfrac{1}{\Delta t^2} - \lambda^2\cfrac{(t-t_1)(t_2-t)}{\Delta t^2}\right]~dt = \cfrac{1}{6\Delta t} (\lambda^2\Delta t^2 + 6) \\ K_{22} & = \int_{t_1}^{t_2}\left(\cfrac{dN_2}{dt}\cfrac{dN_2}{dt} -                     \lambda^2 N_2 N_2\right)~dt = \int_{t_1}^{t_2}\left[\cfrac{1}{\Delta t^2} - \lambda^2\left(\cfrac{t-t_1}{\Delta t}\right)^2\right]~dt = \cfrac{1}{3\Delta t} (\lambda^2\Delta t^2 - 3) \end{align}$$ The load vector terms are
 * $$\begin{align}

f_1 & = \left.\left(\omega~N_1\right)\right|_{t_1}^{t_2} = \omega(t_2)~N_1(t_2) - \omega(t_1)~N_1(t_1) = -\omega(t_1) =: -\omega_1 \\ f_2 & = \left.\left(\omega~N_2\right)\right|_{t_1}^{t_2} = \omega(t_2)~N_2(t_2) - \omega(t_1)~N_2(t_1) = \omega(t_2) =: \omega_2\\ \end{align}$$ In matrix form:

{       \cfrac{1}{3\Delta t}        \begin{bmatrix} (\lambda\Delta t)^2 - 3 & (\lambda\Delta t)^2/2 + 3 \\ (\lambda\Delta t)^2/2 + 3 & (\lambda\Delta t)^2 - 3 \end{bmatrix} \begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix} =        \begin{bmatrix} -\omega_1 \\ \omega_2 \end{bmatrix} }     $$ The Maple script for getting the components of the stiffness matrix is shown below.

Part 7
''Use ANSYS or some other tool of your choice (including your own code) to solve the linear equation of motion via finite elements. Compare with the exact solution. You can use any reasonable values for time ($$t$$), mass ($$m$$), and length ($$l$$).'' The Matlab script for solving the problem is given below:

Comparisons are shown below between the FEM solution and the exact solution for three sets of initial conditions.

 * When the initial angular velocity is 0 and the initial angle is nonzero, the FEM solution is close to the exact solution.  This is a lucky coincidence.  The total time interval has been taken to be 2.0 secs.  It turns out that at exactly 2 secs, the angular velocity goes to zero and hence the problem appears as if a angular velocity boundary condition has been applied at time $$t = 2$$ secs.  You can see the problem if you increase the time interval to  $$t = 2.2$$ secs (see Figure 3).
 * In problems that involve time, we usually know only the initial state but not the final state. Therefore, integrals over the total time interval cannot be evaluated easily.  That is one of the reasons why we don't use finite elements to discretize time.  Instead, we use techniques that directly discretize the differential operator, i.e., we work from the strong form.
 * You cannot solve a problem without specifying BCs at all points on the boundary. Applying two BCs at a single point does not work unless they are for different variables.

Part 8
What happens when you try to solve the nonlinear equation of motion with finite elements? In this case, the approximate weak form becomes

\int_{t_1}^{t_2} \cfrac{d\theta_h}{dt}\cfrac{dw}{dt}~dt - \lambda^2\int_{t_1}^{t_2}\sin(\theta_h)~w~dt = \left.\left(\omega~w\right)\right|_{t_1}^{t_2} ~. $$ If we plug in the approximate solution, we get

\sum_{j=1}^2 \left(\int_{t_1}^{t_2} \cfrac{dN_i}{dt}\cfrac{dN_j}{dt}~dt\right) \theta_j - \lambda^2\int_{t_1}^{t_2}\sin(\theta_h)~N_i~dt = \left.\left(\omega~N_i\right)\right|_{t_1}^{t_2} ~. $$ To get a linear system of equations, we can take the $$\sin\theta$$ term to the right hand side.

\sum_{j=1}^2 \left(\int_{t_1}^{t_2} \cfrac{dN_i}{dt}\cfrac{dN_j}{dt}~dt\right) \theta_j = \lambda^2\int_{t_1}^{t_2}\sin(\theta_h)~N_i~dt + \left.\left(\omega~N_i\right)\right|_{t_1}^{t_2} ~. $$ At this stage we have to start with an initial guess for $$\theta_1$$ and   $$\theta_2$$, plug them into the $$\sin(\theta)$$ term, integrate       numerically and form a right hand side. We can then solve for the $$\theta$$ values and get a new set. We can then plug these back into the RHS and repeat the process. A straightforward linear solution is no longer possible.

Problem 2: Numerical Exercise
 Given:

Mini-centrifuge problem. Temperature range: -80 C to 100 C., RPM: 25000.

 Assumptions: Assume that the current material is a light metal (any other material of your choice will also do). The only requirement is that it should be cheap and stiff. Ignore bending and the weight of the sample tubes.

 Find:

Find the thinnest beam of the material of your choice that can stand the axial load that is generated due to rotation at the new speed. Use ANSYS or your own code to solve the problem. State all simplifying assumptions.

Solution
Let us consider one of the spokes and assume that it's a one-dimensional bar. The governing equation equation for a 1-D bar is

AE \cfrac{d^2u}{dx^2} + q = 0 ~. $$ where $$q$$ is a body force which has units of force per unit length.

For a rotating bar, the axial acceleration is

a(x) = \omega^2~x $$ where $$\omega$$ is the angular velocity of rotation (assumed constant in this case). The acceleration of the fixed end is assumed to be zero (but may not be zero for the actual centrifuge).

Therefore, the body force per unit length due to the rotation of the bar is

q(x) = \rho~A~a(x) = \rho A\omega^2 x $$ where $$\rho$$ is the density of the material, and $$A$$ is the area of  cross-section. Therefore, the governing differential equation is

AE \cfrac{d^2u}{dx^2} + \rho A \omega^2 x = 0 $$ or,

{   E \cfrac{d^2u}{dx^2} + \rho \omega^2 x = 0 ~. } $$ Clearly, the solution for the displacement does not depend on the area of cross-section of the material. Hence, the strains do not depend on the area of cross-section and therefore, neither do the stresses. However, the stresses do depend on the length.

'''If we assume that the width and the length of the spokes have to be maintained constant in order to hold the tubes, the thickness of the spokes cannot be optimized on the basis of inertial forces. We have to consider bending.'''

If we work out the finite element equations over an element for this problem, we will get
 * $$\begin{align}

K^e_{ij} & = \int_{x_1}^{x_2} \cfrac{dN_j}{dx} E \cfrac{dN_i}{dx}~dx \\ f^e_j & = \rho\omega^2\int_{x_1}^{x_2} x~N_j~dx \end{align}$$ The second equation above shows you how to apply the body forces (assuming ANSYS does not allow that - which is surprising.)

At this stage, an ANSYS solution becomes a formality that may or may not be performed.

For the ANSYS solution, let us assume that Here's a sample ANSYS script for the problem
 * 1) The material is 6061-T6 aluminum alloy.
 * 2) The density is 2790 kg/m$$^3$$.
 * 3) The Young's modulus at 100 C is 68.8 GPa.
 * 4) The Poisson's ratio is 0.35.
 * 5) The initial yield stress at -80 C is 350 MPa.
 * 6) The initial yield stress at 100 C is 250 MPa.
 * 7) The fracture strain at -80 C is 0.05.
 * 8) The fracture strain at 100 C is 0.50.
 * 9) The length is 140 mm.
 * 10) The width is 40 mm.
 * 11) The thickness is 10 mm.

Results from ANSYS
Ideally you should run a convergence study at this stage to see how close your results are to the actual solution and show plots of the results.

Numbers that have five decimal places or are close to machine precision are usually useless when presented in tables.

From the axial bar problem in the learning resources, you know the exact solution for this problem, which is

{   \begin{align} u(x) & = \cfrac{\rho\omega^2}{6E}(3L^2x - x^3) \\ \sigma(x) & = \cfrac{\rho\omega^2}{2}(L^2 - x^2) \end{align} } $$

Here's a comparison between the exact solution and the finite element solution The stresses are lower than the yield stress of Aluminum, but they are close. The strains are also relatively small (approximately 0.002). The structure should hold up under the new conditions.


title: File:PendulumSol1.png


title: File:PendulumSol2.png


title: File:PendulumSol3.png


title: File:PendulumSol4.png


text: File:NFE HW1Prob2Sol.png

title: File:NFE HW1Prob2Sol.png


text:
Nonlinear finite elements/Quiz 1/Solutions

Quiz 1: Given
Heat conduction in an isotropic material with a constant thermal conductivity and no internal heat sources is described by Laplace's equation

\nabla^2 T = 0 ~\qquad~ \text{or,} \qquad \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} = 0 ~. $$

Part 1
Derive a symmetric weak form for the Laplace equation in 1-D (an insulated rod).

The 1-D version of Laplace's equation is

\frac{\partial^2 T}{\partial x^2} = 0 ~. $$ To derive the symmetric weak form we multiply the equation by a weighting function ($$w$$) and integrate by parts. Thus,

\int_{\Omega} w~\frac{\partial^2 T}{\partial x^2}~dx = 0 $$ or

{ \int_{\Omega} \frac{\partial w}{\partial x}~\frac{\partial T}{\partial x}~dx = \left.w~\left(\frac{\partial T}{\partial x}\right)\right|_{\Gamma} ~ \qquad \leftarrow \qquad \text{Symmetric Weak Form} ~. } $$

Part 2
What are the expressions for the components of the finite element stiffness matrix ($K_{ij}$) and the load vector ($f_i$) for this 1-D problem?

The stiffness matrix terms are

{ K^e_{ij} = \int_{\Omega^e} \frac{\partial N^e_i}{\partial x}\frac{\partial N^e_j}{\partial x}~dx } $$ The load vector terms are

{ f^e_{i} = \left.N^e_i\left(\frac{\partial T}{\partial x}\right)\right|_{\Gamma^e} ~ } $$

Part 3
Assume that the one of ends of the rod is maintained at a temperature of $T_1$ (which is nonzero) and the other end has a prescribed heat flux of $Q_2$. If we discretize the rod into two elements, what does the reduced finite element system of equations look like?You do not have to work out the terms of the stiffness matrix - just use generic labels.

The finite element system of equations for a two element mesh (with linear shape functions) is

\begin{bmatrix} K_{11} & K_{12} & 0\\ K_{12} & K_{22} & K_{23} \\ 0 & K_{23} & K_{33} \end{bmatrix} \begin{bmatrix} T_1 \\ T_2 \\ T_3 \end{bmatrix} = \begin{bmatrix} f_1 \\ 0 \\ Q_2 \end{bmatrix} $$ If $$T_1$$ is not zero, the reduced system of equations will be

{ \begin{bmatrix} K_{22} & K_{23} \\ K_{23} & K_{33} \end{bmatrix} \begin{bmatrix} T_2 \\ T_3 \end{bmatrix} = \begin{bmatrix} -K_{12}T_1 \\ Q_2 \end{bmatrix} ~. } $$

Part 4
Now, assume that the thermal conductivity of the material varies with temperature. What form does the governing equation take? (We will call this the modified problem.)

If the thermal conductivity ($$\kappa$$) is a function of temperature, the governing equation takes the form

\frac{\partial }{\partial x}\left(\kappa(T)\frac{\partial T}{\partial x}\right) = 0 ~. $$ Since $$\kappa$$ is a function only of temperature, we can take it outside the derivative to get

\kappa(T)\frac{\partial^2 T}{\partial x^2} = 0 \qquad \implies \qquad { \frac{\partial^2 T}{\partial x^2} = 0 ~.} $$ The equation does not change!

Part 5
List the steps needed to solve the modified problem using finite elements.

The standard steps for a linear ODE are applicable.

 * 1) Derive the symmetric weak form.
 * 2) Substitute the approximate solution into the weak form and find the symmetric element stiffness matrix and element load vector.
 * 3) Assemble global stiffness matrix and load vector.
 * 4) Apply boundary conditions.
 * 5) Solve.


text:
Finite elements

Welcome to this learning project about !

Learning Project Summary

 * Project code:
 * Suggested Prerequisites:
 * Linear algebra
 * Partial differential equations
 * Time investment: 6 months
 * Portal:Engineering and Technology
 * School:Engineering
 * Department:Mechanical Engineering
 * Stream:Applied Mechanics
 * Level: Upper-division undergraduate/First year graduate

Content summary
This course will introduce you to the topic of finite element analysis. The course will cover linear finite elements and the analysis of simple solid mechanics and heat transfer problems.

Goals
This course aims to:
 * Introduce you to the finite element method
 * Show you how finite element formulations are arrived at
 * Show you how some engineering problems are solved numerically

Syllabus and Learning Materials

 * 1) Mathematical Preliminaries
 * 2) Set notation
 * 3) Functions
 * 4) Vectors
 * 5) Matrices
 * 6) Tensors
 * 7) Partial differential equations
 * 8) Variational calculus
 * 9) Tensor exercises and Physical Principles in Differential Form
 * 10) Quick indicial notation review
 * 11) Conservation of mass
 * 12) Conservation of linear momentum
 * 13) Conservation of angular momentum
 * 14) Conservation of energy
 * 15) Finite element basics in one-dimension
 * 16) An example: Axially loaded bar
 * 17) * Strong form: governing differential equation
 * 18) * Weak form: integral equation
 * 19) * Approximate solution: Galerkin method
 * 20) * Approximate solution: Finite element method
 * 21) * Approximate solution: Finite element method as a lower bound solution
 * 22) More examples: Some model problems
 * 23) * Weak form: Weighted residual methods
 * 24) * Choosing a weight function
 * 25) * Bubnov Galerkin methods
 * 26) * Finite element basis functions
 * 27) * Example of a finite element approximation
 * 28) A time-dependent problem: the heat equation
 * 29) * Steady state heat conduction
 * 30) * Weak form of the steady state heat equation
 * 31) * Weak form of the time-dependent heat equation
 * 32) * Finite element approximation of Poisson equation
 * 33) * Finite element approximation of the heat equation
 * 34) * Time integration of the heat equation.

Assignments

 * Homework 1 Problem set
 * Solutions
 * Homework 2 Problem set
 * Solutions
 * Homework 3 Problem set
 * Solutions
 * Homework 4 Problem set
 * Solutions
 * Homework 5 Problem set
 * Solutions
 * Homework 6 Problem set
 * Solutions
 * Homework 7 Problem set
 * Solutions
 * Homework 8 Problem set
 * Solutions
 * Homework 9 Problem set
 * Solutions
 * Homework 10 Problem set
 * Solutions
 * Homework 11 Problem set
 * Solutions

Tests and Quizzes

 * Quiz 1
 * Solutions

Textbooks and References
 Textbooks
 * The Finite Element Method: Linear Static and Dynamic Finite Element Analysis by T. J. R. Hughes, Dover Publications, 2000.
 * K-J. Bathe (1996), Finite Element Procedures, Prentice-Hall. Useful repository of information on nonlinear finite elements.
 * J. N. Reddy (1993), An Introduction to the Finite Element Method, McGraw-Hill. This book is referred to a number of times in one of the texts.
 * O. C. Zienkiewicz and R. L. Taylor (2000), The Finite Element Method: Volume 2 Solid Mechanics, Butterworth-Heinemann. Another excellent repository of information of nonlinear finite elements geared toward the Civil Engineers.

References
 * Mathematics:
 * R. M. Brannon (2004), Elementary Vector and Tensor Analysis for Engineers. This free online book is the best introduction I have seen for vector and tensor analysis for nonlinear mechanics.
 * B. Daya Reddy (1998), Introductory Functional Analysis: With applications to boundary value problems and finite elements., Springer-Verlag.  Excellent book for engineers who want to understand the terminology used in the finite element literature and how error analysis is done.
 * A.P.S. Selvadurai (2000), Partial Differential Equations in Mechanics 1,2. Springer. Excellent introductory text on partial differential equations with engineers in mind.

title: Finite elements


text: Finite elements/Sets

title: Finite elements/Sets


text: Finite elements/Functions

title: Finite elements/Functions


text: Finite elements/Matrices

title: Finite elements/Matrices


text: Finite elements/Partial differential equations


text: Finite elements/Calculus of variations

title: Finite elements/Calculus of variations


text: Finite elements/Axially loaded bar

title: Finite elements/Axially loaded bar


text: Finite elements/Axial bar strong form

title: Finite elements/Axial bar strong form


text: Finite elements/Axial bar weak form

title: Finite elements/Axial bar weak form


text: Finite elements/Axial bar approximate solution


text: Finite elements/Axial bar finite element solution


text: Finite elements/Model problems

title: Finite elements/Model problems


text: Finite elements/Weighted residual methods


text: Finite elements/Choosing weight function


text: Finite elements/Bubnov Galerkin method

title: Finite elements/Bubnov Galerkin method


text: Finite elements/Finite element basis functions


text: Finite elements/Model finite element approximation


text: Finite elements/Linear heat equation

title: Finite elements/Linear heat equation


text: Finite elements/Steady state heat conduction


text: Finite elements/Weak form of Poisson equation


text: Finite elements/Weak form of heat equation


text: Finite elements/Solution of Poisson equation


text: Finite elements/Solution of heat equation


text: Finite elements/Heat equation time integration


text:
Continuum mechanics

Welcome to this learning project about continuum mechanics!

Introduction
Topic of continuum mechanics deals with the basics equations of motion that governs the mechanics of both solid and fluid objects in nature.

Project metadata

 * Suggested Prerequisites:
 * Linear algebra
 * Partial differential equations
 * Time investment: 6 months
 * Portal: Engineering and Technology
 * School: Engineering
 * Department: Mechanical engineering, Civil engineering, Aeronautical engineering, Applied mechanics
 * Level: Senior year undergraduate and graduate students

Content summary
This is an introductory course on the continuum mechanics of materials. Small deformation theory is generalized for finite deformation scenarios and applied for both solid and fluids.

Goals
This learning project aims to.
 * provide the mathematical foundations of continuum mechanics
 * expose students to extension of small deformation theory into the finite deformation regime in solids and fluids

Contents
{|width=100%
 * valign=top width=50%|

Syllabus and Learning Materials

 * valign=top width=50%|

Assignments

 * }

Syllabus and Learning Materials

 * 1) Mathematical Preliminaries
 * 2) Set notation
 * 3) Functions
 * 4) Vectors
 * 5) Matrices
 * 6) Tensors
 * 7) Useful tensor algebra identities
 * 8) Useful relations between tensors and vectors
 * 9) Curl of the gradient of a vector
 * 10) Relations between surface and volume integrals
 * 11) Leibniz formula in one dimension
 * 12) Leibniz formula in three dimensions
 * 13) Partial differential equations
 * 14) Variational calculus
 * 15) Kinematics
 * 16) Motion, displacement, velocity, acceleration
 * 17) Strains and deformations
 * 18) Polar decomposition
 * 19) Spectral decompositions of kinematic quantities
 * 20) Volume change and area change
 * 21) Time derivatives and rate quantities
 * 22) Objectivity of kinematic quantities
 * 23) Stress measures and stress rates
 * 24) Stress measures
 * 25) Deviatoric and volumetric stress
 * 26) Objective stress rates
 * 27) Balance laws
 * 28) Governing equations and thermodynamics
 * 29) Balance of mass
 * 30) Balance of linear momentum
 * 31) Balance of angular momentum
 * 32) Balance of energy
 * 33) Clausius-Duhem inequality
 * 34) Constitutive relations
 * 35) Thermoelasticity
 * 36) Relation between Cauchy stress and Green strain
 * 37) Maxwell relations for thermoelasticity
 * 38) Balance of energy
 * 39) Clausius-Duhem inequality
 * 40) Specific heat relations
 * 41) Nonlinear Elasticity
 * 42) Plasticity
 * 43) Viscoplasticity
 * 44) Viscoelasticity.

Tests and Quizzes

 * Quiz 1
 * Solutions
 * /Kinematics Test/

Textbooks

 * Continuum Mechanics by A.J.M Spencer, Dover Publications, 2004
 * Introduction to the Mechanics of a Continuous Medium by L.E. Malvern, Prentice-Hall, 1969

title: Continuum mechanics


text: Continuum mechanics/Sets

title: Continuum mechanics/Sets


text:
Category:Continuum mechanics

This category contains content related to the mechanics of solid and fluid continua in the condensed state.

title: Category:Continuum mechanics


text: Continuum mechanics/Functions

title: Continuum mechanics/Functions


text: Continuum mechanics/Matrices

title: Continuum mechanics/Matrices


text: Continuum mechanics/Partial differential equations


text: Continuum mechanics/Calculus of variations


text:
User talk:MysticSong

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title: User talk:MysticSong


text:
File:Noblerinthemind SBTDS thumbnail Storyboard.png

Used with the WikiU Film School Lesson on how to do thumbnail storyboarding


text:
File:Noblerinthemind sample thumbnail storyboard.png

Used with the WikiU Film School Lesson on how to do thumbnail storyboarding


text:
Socrates, Plato and Aristotle

The dialogues of Socrates were written by his student Plato. Both are frequently quoted by Plato's student Aristotle, who also refuted and incorporated aspects of the materialist philosophy of Democritus into his hylomorphic theory of ontology. The main concept that Socrates pioneered is also a method named after him - The Socratic Method

Socratic Method
At its base, the Socratic method is deceptively simple: question all weaknesses of an argument until a gap in reasoning is revealed. That gap is an opportunity to learn.

Now in practice, this is a much more difficult concept to implement as it takes exceptional mental awareness and a willingness to face the truth. These alone are what drive the citizens of Athens into accusing Socrates of poisoning the minds of the youth because he was revealing the priesthood and professional sophists as fools who believed in ideas they could not support with reason.

The oracle told Socrates that he was the wisest of all people. In typical Socratic fashion, Socrates decided to test this premise. He went to the artisans in town and started asking them questions about their craft and what they believed in, hoping to find someone without a flaw in their beliefs. Over time, he exposed each of the artisans as people who really didn't know as much as they thought they did, so Socrates proceeded to move on to the merchants in the town. Again, he met the same situation time and again - exposing each in turn. Finally, he turned to those in the government...surely these wise men who ran Athens would have at least one among them with no flaw in his beliefs? Sadly, the same results as the previous two attempts. By this point in time Socrates has turned most of the town against him for exposing visible members of the community as fools. Socrates, realizing that the oracle was right then set upon himself, coming up with the conclusion that he was the wisest man because he knew that he knew nothing whereas others were deluding themselves that they knew something but upon closer inspection did not. The climax of this tale ends with Socrates being brought up on charges and in his defense the court asked him what he thought his punishment should be. Socrates replied that he should be given room and board, at government expense, for the rest of his life (an honor usually given to Olympians/heroes). Incensed at not only being made fools of, but of this ridiculous request, the court made judgement that Socrates either be banished from Athens or made to drink hemlock (poison). Since being banished from Athens was considered a cop-out and would be considered today as being exiled to Siberia or Antarctica, Socrates chose the hemlock. In another of Plato's writings, there is a discussion about Socrates possibly going on a jailbreak; however he refuses, citing personal morals and conviction as cornerstones of humanity.

=Plato=

The Platonic Cave
The Platonic cave is analogy that Plato makes in 'The Republic' to demonstrate that the truth of particular things in the known world lie in their abstract form, for example that the beauty of a painting participates in the objective idea of beauty and not the opinions of a person. The analogy is as follows; imagine a group of prisoners are imprisoned within a cave and cannot even see each other. The only knowledge they gain is from objects moving across a bridge whose shadows are illuminated by a fire. Thus all that the prisoners know are the shadows of real things. Now imagine one of the prisoners, let us call him Socrates, breaks free one day. Socrates goes out into the world outside the cave and sees things as they really are. He sees not a shadow of a tree, or a person but actual trees and people. Socrates rejoices in the fact that he has seen the true forms of real things and after a time studying objects in the world returns to the cave to tell the other prisoners. Upon doing so the prisoners cannot comprehend the true reality of things, they know the shadows on the cave wall and cannot accept the reality of objects. They kill Socrates for trying to convince them otherwise. As an analogy the cave works as a metaphor for the world in which each individual human being lives. Socrates is the only person able to escape and is a metaphor for the philosophers who are able to glimpse the true reality of things. In other texts Plato describes the world of the forms or ideas, that is the world in which things exist as they actually are, as what always is but is never becoming. In other words such a reality is eternal in nature due to its abstract form. In contrast the particular world, the cave, is the world in which everything is always becoming but never is. Meaning that all objects of the world are always trying to become like the forms but due to the fact that they are not objective can never be exactly like the objective forms which they represent. A beautiful painting can never be actually beautiful, it can only ever be a close approximation to beauty, hence why not everyone finds it beautiful. Some debate is held as to whether Plato actually meant that there are two separate worlds, one of forms and one of particular things, or whether it is all part of one world in which the abstract forms exist eternally and particular thing finitely. For example formulas gained from studying geometry are abstract and always true in nature but the actual shapes in nature only participate in a likeness of such a formula.

Forms as Cause
Plato suggests that the cause of all particular things in the perceived world are the ideas or forms. This is due to the need of particular things to participate in something abstract that is in perfect being. The existence of the forms themselves is due to the fact that they are perfect and good in themselves. As the worst thing for a perfect existence is its non-existence it must exist always out of necessity. Hence there is always something rather than nothing and no thing can be caused by nothing what so ever.

Plato's Influences
Plato was influenced by two accounts of nature that are seemingly apposed. The one, Hericlitus, suggested that everything in the world is in a constant state of flux, that is always in motion and ever changing. Thus no one can stand in the same river twice. In contrast Parmenides has an account of nature suggests nothing is in movement, the world we perceive is an illusion of movement. Plato's abstract forms constitute a state where nothing is in movement, whilst at the same time the things participating in them are in flux out of necessity.

=Aristotle= See main course: Aristotle.

=Aristotle's Epistemology= Aristotle was a pupil of Plato's academy. Upon not becoming the head teacher of the school after Plato's death he set up his own school of thought that differed in its way of thinking. Aristotle was an epistemologist believing that the truth of objects in the world was to be found in objects themselves, not in their separate abstract form as Plato had suggested. This claim can be seen as follows; Socrates never thinks without an image present to him. All truth is present within objects in the world. Thus any truth of the objects must be contained in the images Socrates has present to him. The study of metaphysics, the study of being qua being, (Qua meaning the capacity or role of an item) or first philosophy reveils truths to us present in the objects. =Aristotle's Rejection of the Platonic forms- The Third Man Argument= Aristotle formalizes in his 'Metaphysics' a rejection of Plato's theory of the forms known as the Third Man Argument, although it is found in one of Plato's own works 'Parmenides'. It is as follows: if particular things participate in the likeness of forms then they cannot be completely different, Socrates cannot be completely different from the form man else he cannot be participating in a likeness of man. If they are the same then Socrates is equivalent to the form man and nothing is added to the argument by the form. Thus Socrates must be similar to the form man, not equivalent to nor completely different to the form man. But such likeness, Man 3, is a third thing between the form and the particular, again this likeness cannot be equivalent or completely different to the form, due to the above arguments and so a fourth likeness is needed to explain the similarity, Man four. There is an infinite regress in which there is always a need for another likeness. The effectiveness of the third man argument can be disputed. If Socrates participates in the form man his similarity is not one of similarity given to Socrates but by trying to be like it. This is something Aristotle himself posits in his concept of matter and God, when all matter is trying to be like the first cause, God but can never actually be God itself. Never the less for a long time, particularly the medieval period when Aristotle was known as 'The Philosopher', this argument was seen as logical proof for the non-existence of abstract forms as apart from particular things. This is not to say that there are not forms in which all things participate.

title: Socrates, Plato and Aristotle


text:
Nonlinear finite elements/Homework 10/Solutions

Problem 1: Kinematics and Stress Rates
 Given:

Figure 1 shows a linear three-noded triangular element in the reference configuration.

The motion of the nodes is given by:
 * $$\begin{align}

\text{Node}~1:&& x_1(t) & = 0& \qquad y_1(t) & = 0 \\ \text{Node}~2:&& x_2(t) & = 2(1 + at)\cos\cfrac{\pi t}{2} &\qquad y_2(t) & = 2(1 + at)\sin\cfrac{\pi t}{2} \\ \text{Node}~3:&& x_3(t) & = -(1 + bt)\sin\cfrac{\pi t}{2} &\qquad y_3(t) & =(1 + bt)\cos\cfrac{\pi t}{2} \end{align}$$ The configuration ($$x,y$$) of the element at time $$t$$ is given by

x(\boldsymbol{\xi}, t) = \sum_{i=1}^3 x_i(t)~N_i(\boldsymbol{\xi}) ~;\qquad y(\boldsymbol{\xi}, t) = \sum_{i=1}^3 y_i(t)~N_i(\boldsymbol{\xi}) $$

Part 1
Write down expressions for $N_1$, $N_2$, and $N_3$ in terms of the initial configuration ($X,Y$) ?

In the initial configuration $$t = 0$$. Therefore,

X_1 = x_1(0) = 0 ~; Y_1 = y_1(0) = 0 ~;\qquad X_2 = x_2(0) = 2 ~; Y_2 = y_2(0) = 0 ~;\qquad X_3 = x_3(0) = 0 ~; Y_3 = y_3(0) = 1 ~. $$ Therefore, the initial configuration is given by

X(\boldsymbol{\xi}) = \sum_{i=1}^3 X_i~N_i(\boldsymbol{\xi}) ~;\qquad Y(\boldsymbol{\xi}) = \sum_{i=1}^3 Y_i~N_i(\boldsymbol{\xi})~. $$ Substituting the values of $$X_i$$ and $$Y_i$$, we get

X(\boldsymbol{\xi}) = 2~N_2(\boldsymbol{\xi}) ~;\qquad Y(\boldsymbol{\xi}) = N_3(\boldsymbol{\xi})~. $$ Hence

N_2(\boldsymbol{\xi}) = \cfrac{X(\boldsymbol{\xi})}{2} ~;\qquad N_3(\boldsymbol{\xi}) = Y(\boldsymbol{\xi}) ~. $$ Also, the shape functions must satisfy the partition of unity condition

\sum_{i=1}^3 N_i(\boldsymbol{\xi}) = 1 ~. $$ Therefore,

N_1(\boldsymbol{\xi}) = 1 - N_2(\boldsymbol{\xi}) - N_3(\boldsymbol{\xi}) = 1 - \cfrac{X(\boldsymbol{\xi})}{2} - Y(\boldsymbol{\xi}) ~. $$ The required expressions are

{ N_1(\boldsymbol{\xi}) = 1 - \cfrac{X(\boldsymbol{\xi})}{2} - Y(\boldsymbol{\xi})~;\qquad N_2(\boldsymbol{\xi}) = \cfrac{X(\boldsymbol{\xi})}{2} ~;\qquad N_3(\boldsymbol{\xi}) = Y(\boldsymbol{\xi}) ~. } $$

Part 2
Derive expressions for the deformation gradient and the Jacobian determinant for the element as functions of time.

The deformation gradient is given by

\mathbf{F} = \left[ \begin{align} \frac{\partial x}{\partial X} & & \frac{\partial x}{\partial Y} & & \frac{\partial x}{\partial Z} \\ \frac{\partial y}{\partial X} & & \frac{\partial y}{\partial Y} & & \frac{\partial y}{\partial Z} \\ \frac{\partial z}{\partial X} & & \frac{\partial z}{\partial Y} & & \frac{\partial z}{\partial Z} \end{align} \right]~. $$ Before computing the derivatives, let us express $$x,y,z$$ in terms of $$X,Y,Z$$. Recall

x(\boldsymbol{\xi}, t) = \sum_{i=1}^3 x_i(t)~N_i(\boldsymbol{\xi}) ~;\qquad y(\boldsymbol{\xi}, t) = \sum_{i=1}^3 y_i(t)~N_i(\boldsymbol{\xi}) $$ Therefore,

\begin{align} x(\boldsymbol{\xi}, t) & = x_1(t)~\left[1 - \cfrac{X(\boldsymbol{\xi})}{2} - Y(\boldsymbol{\xi})\right] + x_2(t)~\cfrac{X(\boldsymbol{\xi})}{2} + x_3(t)~Y(\boldsymbol{\xi}) \\ y(\boldsymbol{\xi}, t) & = y_1(t)~\left[1 - \cfrac{X(\boldsymbol{\xi})}{2} - Y(\boldsymbol{\xi})\right] + y_2(t)~\cfrac{X(\boldsymbol{\xi})}{2} + y_3(t)~Y(\boldsymbol{\xi}) \\ z(\boldsymbol{\xi}, t) & = Z(\boldsymbol{\xi}) \end{align} $$ Substituting in the expressions for $$x_i$$ and $$y_i$$, we get

\begin{align} x(\boldsymbol{\xi}, t) & = \left[2(1 + at)\cos\cfrac{\pi t}{2}\right]~ \cfrac{X(\boldsymbol{\xi})}{2} - \left[(1 + bt)\sin\cfrac{\pi t}{2}\right]~Y(\boldsymbol{\xi}) \\ y(\boldsymbol{\xi}, t) & = \left[2(1 + at)\sin\cfrac{\pi t}{2}\right]~ \cfrac{X(\boldsymbol{\xi})}{2} + \left[(1 + bt)\cos\cfrac{\pi t}{2}\right]~Y(\boldsymbol{\xi}) \\ z(\boldsymbol{\xi}, t) & = Z(\boldsymbol{\xi}) ~. \end{align} $$ In the above expression, the parent coordinates $$\boldsymbol{\xi}$$ are no longer useful. Therefore, we write

\begin{align} x(\mathbf{X}, t) & = X~(1 + at)~\cos\cfrac{\pi t}{2} - Y~(1 + bt)~\sin\cfrac{\pi t}{2} \\ y(\mathbf{X}, t) & = X~(1 + at)~\sin\cfrac{\pi t}{2} + Y~(1 + bt)~\cos\cfrac{\pi t}{2} \\ z(\mathbf{X}, t) & = Z \end{align} $$ where $$\mathbf{X} = (X, Y, Z)$$. Taking derivatives, we get

\begin{align} \frac{\partial x}{\partial X} & =(1 + at)~\cos\cfrac{\pi t}{2}\qquad & \frac{\partial x}{\partial Y} & = -(1 + bt)~\sin\cfrac{\pi t}{2}\qquad & \frac{\partial x}{\partial Z} & = 0 \\ \frac{\partial y}{\partial X} & = (1 + at)~\sin\cfrac{\pi t}{2}\qquad& \frac{\partial y}{\partial Y} & = (1 + bt)~\cos\cfrac{\pi t}{2}\qquad & \frac{\partial y}{\partial Z} & = 0 \\ \frac{\partial z}{\partial X} & = 0\qquad & \frac{\partial z}{\partial Y} & = 0\qquad & \frac{\partial z}{\partial Z} & = 1 \end{align} $$ Therefore,

{ \mathbf{F}(t) = \left[ \begin{align} (1 + at)~\cos\cfrac{\pi t}{2} & & -(1 + bt)~\sin\cfrac{\pi t}{2} & & 0 \\ (1 + at)~\sin\cfrac{\pi t}{2} & & (1 + bt)~\cos\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 1 \end{align} \right]~. } $$ The Jacobian determinant is

J(t) = \det\mathbf{F}(t) = \left[(1 + at)~\cos\cfrac{\pi t}{2}\right] \left[(1 + bt)~\cos\cfrac{\pi t}{2}\right] + \left[(1 + bt)~\sin\cfrac{\pi t}{2}\right] \left[(1 + at)~\sin\cfrac{\pi t}{2}\right] ~. $$ or

{ J(t) = (1+at)(1+bt) ~. } $$

Part 3
What are the values of $a$ and $b$ for which the motion is isochoric?

For isochoric motion, $$J = 1$$. Therefore,

(1 + at)(1 + bt) = 1 ~. $$ One possibility is

{ a = b = 0 } $$ This is a pure rotation.

Another possibility is that

{ b = -\cfrac{a}{1+at} } $$ This is a combination of shear and rotation where the volume remains constant.

Part 4
For which values of $a$ and $b$ do we get invalid motions?

We get invalid motions when $$J \le 0$$. Let us consider the case where $$J = 0$$. Then

(1 + at) (1 + bt) = 0 $$ This is possible when

at = -1 ~\text{or}~ bt = -1 ~. $$ If $$J < 0$$ then

1 + at < 0 ~\text{and}~ 1 + bt > 0\qquad\text{or}\qquad 1 + at > 0 ~\text{and}~ 1 + bt < 0 $$ That is,

at < -1 ~\text{and}~ bt > -1 \qquad\text{or}\qquad at > -1 ~\text{and}~ bt < -1 $$ Therefore the values at which we get invalid motions are

{ a \le -\cfrac{1}{t} \text{or} b \le -\cfrac{1}{t} ~. } $$

Part 5
Derive the expression for the Green (Lagrangian) strain tensor for the element as a function of time.

The Green strain tensor is given by

\mathbf{E} = \frac{1}{2}(\mathbf{F}^T \mathbf{F} - \mathbf{I}) $$ Recall

\mathbf{F}(t) = \left[ \begin{align} (1 + at)~\cos\cfrac{\pi t}{2} & & -(1 + bt)~\sin\cfrac{\pi t}{2} & & 0 \\ (1 + at)~\sin\cfrac{\pi t}{2} & & (1 + bt)~\cos\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 1 \end{align} \right]~. $$ Let us make the following substitutions

A := (1+at) ~; B := (1+bt) ~; c := \cos\cfrac{\pi t}{2} ~; s := \sin\cfrac{\pi t}{2} ~. $$ Then

\mathbf{F} = \begin{bmatrix} A~c & -B~s & 0 \\ A~s &B~c & 0 \\ 0 &0 & 1 \end{bmatrix} \text{and} \mathbf{F}^T = \begin{bmatrix} A~c & A~s &0 \\ -B~s & B~c &0 \\ 0 & 0 & 1 \end{bmatrix}~. $$ Therefore,

\mathbf{F}^T~\mathbf{F} = \begin{bmatrix} A^2 & 0 & 0 \\ 0 & B^2 & 0 \\ 0 &0& 1 \end{bmatrix} = \begin{bmatrix} (1+at)^2 & 0 & 0 \\ 0 & (1+bt)^2 & 0 \\ 0 &0& 1 \end{bmatrix} ~. $$ Hence,

\mathbf{E} = \frac{1}{2} \left(\begin{bmatrix} (1+at)^2 & 0 & 0 \\ 0 & (1+bt)^2 & 0 \\ 0 &0& 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) $$ The Green strain is

{ \mathbf{E}(t) = \frac{1}{2} \begin{bmatrix} 2at + a^2 t^2 & 0 & 0 \\ 0 & 2bt + b^2 t^2 & 0 \\ 0 &0& 0 \end{bmatrix} } $$

Part 6
Derive an expression for the velocity gradient as a function of time.

The velocity is the material time derivative of the motion.

Recall that the motion is given by

\begin{align} x(\mathbf{X}, t) & = X~(1 + at)~\cos\cfrac{\pi t}{2} - Y~(1 + bt)~\sin\cfrac{\pi t}{2} \\ y(\mathbf{X}, t) & = X~(1 + at)~\sin\cfrac{\pi t}{2} + Y~(1 + bt)~\cos\cfrac{\pi t}{2} \\ z(\mathbf{X}, t) & = Z \end{align} $$ Therefore,

\begin{align} v_x(\mathbf{X},t) = \frac{\partial x}{\partial t} & = X~\left[ a~\cos\cfrac{\pi t}{2} - (1+at)~\cfrac{\pi}{2}~\sin\cfrac{\pi t}{2} \right] - Y~\left[ b~\sin\cfrac{\pi t}{2} + (1 + bt)~\cfrac{\pi}{2}~\cos\cfrac{\pi t}{2} \right] \\ v_y(\mathbf{X},t) = \frac{\partial y}{\partial t} & = X~\left[ a~\sin\cfrac{\pi t}{2} + (1 + at)~\cfrac{\pi}{2}~\cos\cfrac{\pi t}{2} \right]+ Y~\left[ b~\cos\cfrac{\pi t}{2} - (1 + bt)~\cfrac{\pi}{2}~\sin\cfrac{\pi t}{2} \right]\\ v_z\mathbf{X},t) = \frac{\partial z}{\partial t} & = 0 \end{align} $$ We could compute the velocity gradient using

\mathbf{l} = \boldsymbol{\nabla} \mathbf{v} = \left[ \begin{align} \frac{\partial v_x}{\partial x} & &\frac{\partial v_x}{\partial y} & & \frac{\partial v_x}{\partial z} \\ \frac{\partial v_y}{\partial x} & &\frac{\partial v_y}{\partial y} & & \frac{\partial v_y}{\partial z} \\ \frac{\partial v_z}{\partial x} & &\frac{\partial v_z}{\partial y} & & \frac{\partial v_z}{\partial z} \end{align} \right] $$ after expressing $$\mathbf{v}$$ in terms of $$\mathbf{x}$$. However, that makes the expression quite complicated. Instead, we will use the relation

\mathbf{l} = \dot{\mathbf{F}}~\mathbf{F}^{-1} ~. $$ The time derivative of the deformation gradient is

\dot{\mathbf{F}} = \left[ \begin{align} a~\cos\cfrac{\pi t}{2} - (1+at)~\cfrac{\pi}{2}~\sin\cfrac{\pi t}{2} & & -b~\sin\cfrac{\pi t}{2} - (1 + bt)~\cfrac{\pi}{2}~\cos\cfrac{\pi t}{2} & & 0 \\ a~\sin\cfrac{\pi t}{2} + (1 + at)~\cfrac{\pi}{2}~\cos\cfrac{\pi t}{2} & & b~\cos\cfrac{\pi t}{2} - (1 + bt)~\cfrac{\pi}{2}~\sin\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right]~. $$ The inverse of the deformation gradient is

\mathbf{F}^{-1} = \cfrac{1}{(1+at)(1+bt)}\left[ \begin{align} (1 + bt)~\cos\cfrac{\pi t}{2} & & (1 + bt)~\sin\cfrac{\pi t}{2} & & 0 \\ -(1 + at)~\sin\cfrac{\pi t}{2} & & (1 + at)~\cos\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & (1+at)(1+bt) \end{align} \right]~. $$ Using the substitutions

A := (1+at) ~; B := (1+bt) ~; c := \cos\cfrac{\pi t}{2} ~; s := \sin\cfrac{\pi t}{2} $$ we get

\dot{\mathbf{F}} = \left[ \begin{align} a~c - A~s~\cfrac{\pi}{2} & & -b~s - B~c~\cfrac{\pi}{2} & & 0 \\ a~s + A~c~\cfrac{\pi}{2} & &b~c - B~s~\cfrac{\pi}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] = \left[ \begin{align} a~c & & -b~s & & 0 \\ a~s & &b~c & & 0 \\ 0 & & 0 & & 0 \end{align} \right] + \cfrac{\pi}{2} \left[ \begin{align} - A~s & &- B~c & & 0 \\ A~c & &- B~s & & 0 \\ 0 & & 0 & & 0 \end{align} \right] $$ and

\mathbf{F}^{-1} = \cfrac{1}{A~B}\left[ \begin{align} B~c & & B~s & & 0 \\ -A~s & & A~c & & 0 \\ 0 & & 0 & & A~B \end{align} \right]~. $$ The product is

\mathbf{l} = \cfrac{1}{A~B} \left[ \begin{align} B~a~c^2 + A~b~s^2 & & B~a~c~s - A~b~c~s & & 0 \\ B~a~c~s - A~b~c~s & &B~a~s^2 + A~b~c^2 & & 0 \\ 0 & & 0 & & 0 \end{align} \right] + \cfrac{\pi}{2} \left[ \begin{align} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{align} \right] $$ Note that the first matrix is symmetric while the second is skew-symmetric.

Therefore, the velocity gradient is

{ \mathbf{l} = \left[ \begin{align} \cfrac{a}{1+at}~\cos^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\sin^2\cfrac{\pi t}{2} & & \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} - \frac{\pi}{2} & & 0 \\ \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} + \cfrac{\pi}{2} & & \cfrac{a}{1+at}~\sin^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\cos^2\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] } $$

Part 7
Compute the rate of deformation tensor and the spin tensor.

The rate of deformation is the symmetric part of the velocity gradient:

{ \mathbf{d} = \left[ \begin{align} \cfrac{a}{1+at}~\cos^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\sin^2\cfrac{\pi t}{2} & & \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} & & 0 \\ \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} & & \cfrac{a}{1+at}~\sin^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\cos^2\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] } $$ The rate of deformation is the skew-symmetric part of the velocity gradient:

{ \mathbf{w} = \cfrac{\pi}{2} \left[ \begin{align} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{align} \right] } $$

Part 8
Assume that $$a=1$$ and $$b=2$$. Sketch the undeformed configuration and the deformed configuration at $$t=1$$ and $$t=2$$. Draw both the deformed and undeformed configurations on the same plot and label.

Recall that in the initial configuration

X_1 = x_1(0) = 0 ~; Y_1 = y_1(0) = 0 ~;\qquad X_2 = x_2(0) = 2 ~; Y_2 = y_2(0) = 0 ~;\qquad X_3 = x_3(0) = 0 ~; Y_3 = y_3(0) = 1 ~. $$ Also, the motion is

\begin{align} x(\mathbf{X}, t) & = X~(1 + at)~\cos\cfrac{\pi t}{2} - Y~(1 + bt)~\sin\cfrac{\pi t}{2} \\ y(\mathbf{X}, t) & = X~(1 + at)~\sin\cfrac{\pi t}{2} + Y~(1 + bt)~\cos\cfrac{\pi t}{2} \\ z(\mathbf{X}, t) & = Z \end{align} $$ Plugging in the values of $$a$$ and $$b$$, we get

\begin{align} x(\mathbf{X}, t) & = X~(1 + t)~\cos\cfrac{\pi t}{2} - Y~(1 + 2t)~\sin\cfrac{\pi t}{2} \\ y(\mathbf{X}, t) & = X~(1 + t)~\sin\cfrac{\pi t}{2} + Y~(1 + 2t)~\cos\cfrac{\pi t}{2} \\ z(\mathbf{X}, t) & = Z \end{align} $$ At $$t = 1$$,

\begin{align} x(\mathbf{X}, 1) & = 2X~\cos\cfrac{\pi}{2} - 3Y~\sin\cfrac{\pi}{2} = -3Y \\ y(\mathbf{X}, 1) & = 2X~\sin\cfrac{\pi}{2} + 3Y~\cos\cfrac{\pi}{2} = 2X\\ z(\mathbf{X}, 1) & = Z \end{align} $$ At $$t = 2$$,

\begin{align} x(\mathbf{X}, 2) & = 3X~\cos\pi - 5Y~\sin\pi = -3X \\ y(\mathbf{X}, 2) & = 3X~\sin\pi + 5Y~\cos\pi = -5Y \\ z(\mathbf{X}, 2) & = Z \end{align} $$ The deformed and undeformed configurations are shown below.

Part 9
Compute the polar decomposition of the deformation gradient with the above values of $a$ and $b$,

The deformation gradient is

\mathbf{F}(t) = \left[ \begin{align} (1 + t)~\cos\cfrac{\pi t}{2} & & -(1 + 2t)~\sin\cfrac{\pi t}{2} & & 0 \\ (1 + t)~\sin\cfrac{\pi t}{2} & & (1 + 2t)~\cos\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 1 \end{align} \right]~. $$ The right Cauchy-Green deformation tensor is

\mathbf{C} = \mathbf{F}^T~\mathbf{F} = \begin{bmatrix} (1+t)^2 & 0 & 0 \\ 0 & (1+2t)^2 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ The eigenvalue problem is

(\mathbf{C} - \lambda^2\mathbf{I})\mathbf{N} = 0 ~. $$ This problem has a solution if

\det(\mathbf{C} - \lambda^2\mathbf{I}) = 0 $$ i.e.,

\left[\lambda^4-(1+t)^2\lambda^2-\lambda^2 (1+2t)^2+(1+t)^2~(1+2t)^2\right](1-\lambda^2) = 0 ~. $$ The eigenvalues are (as expected)

\lambda_1^2 = (1+t)^2 ~; \lambda_2^2 = (1+2t)^2 ~; \lambda_3^2 = 1~. $$ The principal stretches are

\lambda_1 = (1+t) ~; \lambda_2 = (1+2t) ~; \lambda_3 = 1~. $$ The principal directions are (by inspection)

\mathbf{N}_1 = [1 0 0]^T ~; \mathbf{N}_2 = [0 1 0]^T ~; \mathbf{N}_3 = [0 0 1]^T ~. $$ Now, the right stretch tensor is given by

\boldsymbol{U} = \sum_{i=1}^3 \lambda_i \boldsymbol{N}_i\otimes\boldsymbol{N}_i $$ Therefore,

\mathbf{U} = \lambda_1~\mathbf{N}_1~\mathbf{N}_1^T + \lambda_2~\mathbf{N}_2~\mathbf{N}_2^T + \lambda_3~\mathbf{N}_3~\mathbf{N}_3^T ~. $$ Hence the right stretch is

{ \mathbf{U} = \begin{bmatrix} 1+t & 0 & 0 \\ 0 & 1+2t & 0 \\ 0 & 0 & 1 \end{bmatrix} } $$ At $$t=0,1,2$$, we have

\mathbf{U}(0) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}~; \mathbf{U}(1) = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}~; \mathbf{U}(2) = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix}~. $$ Now

\mathbf{R} = \mathbf{F}~\mathbf{U}^{-1} $$ and

\mathbf{U}^{-1} = \begin{bmatrix} \cfrac{1}{1+t} & 0 & 0 \\ 0 & \cfrac{1}{1+2t} & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Therefore, the rotation is

{ \mathbf{R} = \left[ \begin{align} \cos\cfrac{\pi t}{2} & & -\sin\cfrac{\pi t}{2} & & 0 \\ \sin\cfrac{\pi t}{2} & & \cos\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 1 \end{align} \right]~. } $$ At $$t=0,1,2$$, we have

\mathbf{R}(0) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}~; \mathbf{R}(1) = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}~; \mathbf{R}(2) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}~. $$

Part 10
Assume an isotropic, hypoelastic constitutive equation for the material of the element. Compute the material time derivative of the Cauchy stress at $t = 1$ using (a) the Jaumann rate and (b) the Truesdell rate.

A hypoelastic material behaves according to the relation

\overset{\triangle}{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}}:\mathbf{d} ~. $$ For an isotropic material

\boldsymbol{\mathsf{C}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2\mu~\boldsymbol{I} $$ Therefore,

\overset{\triangle}{\boldsymbol{\sigma}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}}:\mathbf{d} + 2\mu~\boldsymbol{I}:\mathbf{d} = \lambda~\text{tr}(\mathbf{d})~\boldsymbol{\mathit{1}} + 2\mu~\mathbf{d} $$ Recall that

\mathbf{d} = \left[ \begin{align} \cfrac{a}{1+at}~\cos^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\sin^2\cfrac{\pi t}{2} & & \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} & & 0 \\ \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} & & \cfrac{a}{1+at}~a~\sin^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\cos^2\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] $$ Using the values of $$a$$ and $$b$$ from the previous part, at $$t = 1$$,

\mathbf{d} = \left[ \begin{align} \cfrac{1}{2}~\cos^2\cfrac{\pi}{2} + \cfrac{2}{3}~\sin^2\cfrac{\pi}{2} & & \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} & & 0 \\ \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} & & \cfrac{1}{2}~\sin^2\cfrac{\pi}{2} + \cfrac{2}{3}~\cos^2\cfrac{\pi}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] = \begin{bmatrix} \cfrac{2}{3} & 0 & 0 \\ 0 & \cfrac{1}{2} & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ Therefore, the trace of the rate of defromation is

\text{tr}(\mathbf{d}) = \cfrac{7}{6} $$ Therefore,

\overset{\triangle}{\boldsymbol{\sigma}} = \cfrac{7\lambda}{6}~\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + 2\mu~\begin{bmatrix} \cfrac{2}{3} & 0 & 0 \\ 0 & \cfrac{1}{2} & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} $$ For the Jaumann rate

\overset{\triangle}{\boldsymbol{\sigma}} = \dot{\boldsymbol{\sigma}} + \boldsymbol{\sigma}\bullet\mathbf{w} - \mathbf{w}\bullet\boldsymbol{\sigma} $$ where the spin is

\mathbf{w} = \cfrac{\pi}{2} \left[ \begin{align} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{align} \right] $$

Therefore,

\dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} - \cfrac{\pi}{2}\begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} \begin{bmatrix} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{bmatrix} + \cfrac{\pi}{2}\begin{bmatrix} 0 & & - 1 & & 0 \\ 1 & & 0 & & 0 \\ 0 & & 0 & & 0 \end{bmatrix} \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} $$ or,

{ \dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} + \cfrac{\pi}{2} \begin{bmatrix} -\sigma_{xy} & \sigma_{xx}-\sigma_{yy} & -\sigma_{yz} \\ \sigma_{xx}-\sigma_{yy} & 2\sigma_{xy} & \sigma_{xz} \\ -\sigma_{yz} & \sigma_{xz} & 0 \end{bmatrix} } $$ For the Truesdell rate

\overset{\triangle}{\boldsymbol{\sigma}} = \overset{\circ}{\boldsymbol{\sigma}} = \dot{\boldsymbol{\sigma}} - \boldsymbol{\sigma}\bullet\mathbf{l}^T - \mathbf{l}\bullet\boldsymbol{\sigma} + \text{tr}(\mathbf{l})\boldsymbol{\sigma} ~. $$ Therefore,

\dot{\boldsymbol{\sigma}} = \overset{\circ}{\boldsymbol{\sigma}} + \boldsymbol{\sigma}\bullet\mathbf{l}^T + \mathbf{l}\bullet\boldsymbol{\sigma} - \text{tr}(\mathbf{l})\boldsymbol{\sigma} ~. $$ Recall,

\mathbf{l} = \left[ \begin{align} \cfrac{a}{1+at}~\cos^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\sin^2\cfrac{\pi t}{2} & & \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} - \frac{\pi}{2} & & 0 \\ \left[\cfrac{a}{1+at} - \cfrac{b}{1+bt}\right] \cos\cfrac{\pi t}{2}~\sin\cfrac{\pi t}{2} + \cfrac{\pi}{2} & & \cfrac{a}{1+at}~\sin^2\cfrac{\pi t}{2} + \cfrac{b}{1+bt}~\cos^2\cfrac{\pi t}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] $$ For $$a=1$$, $$b=2$$, $$t=1$$, we have

\mathbf{l} = \left[ \begin{align} \cfrac{1}{2}~\cos^2\cfrac{\pi}{2} + \cfrac{2}{3}~\sin^2\cfrac{\pi}{2} & & \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} - \frac{\pi}{2} & & 0 \\ \left[\cfrac{1}{2} - \cfrac{2}{3}\right] \cos\cfrac{\pi}{2}~\sin\cfrac{\pi}{2} + \cfrac{\pi}{2} & & \cfrac{1}{2}~\sin^2\cfrac{\pi}{2} + \cfrac{2}{3}~\cos^2\cfrac{\pi}{2} & & 0 \\ 0 & & 0 & & 0 \end{align} \right] = \begin{bmatrix} \cfrac{2}{3} & -\cfrac{\pi}{2} & 0 \\ \cfrac{\pi}{2} & \cfrac{1}{2} & 0 \\ 0 &0 &0 \end{bmatrix} ~. $$ Therefore,

\text{tr}(\mathbf{l}) = \cfrac{7}{6} = \text{tr}(\mathbf{d}) $$ and

\dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} + \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} \begin{bmatrix} \cfrac{2}{3} & \cfrac{\pi}{2} & 0 \\ -\cfrac{\pi}{2} & \cfrac{1}{2} & 0 \\ 0 &0 &0 \end{bmatrix} + \begin{bmatrix} \cfrac{2}{3} & -\cfrac{\pi}{2} & 0 \\ \cfrac{\pi}{2} & \cfrac{1}{2} & 0 \\ 0 &0 &0 \end{bmatrix} \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} - \cfrac{7}{6} \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix} $$ Hence,

{ \dot{\boldsymbol{\sigma}} = \begin{bmatrix} \cfrac{7\lambda}{6} + \cfrac{4\mu}{3} & 0 & 0 \\ 0 & \cfrac{7\lambda}{6} + \mu & 0 \\ 0 & 0 & \cfrac{7\lambda}{6} \end{bmatrix} + \begin{bmatrix} \cfrac{1}{6}\sigma_{xx} + \pi\sigma_{xy} & \cfrac{\pi}{2}(-\sigma_{xx} + \sigma_{yy}) & \cfrac{1}{2}(-\sigma_{xz} + \pi\sigma_{yz}) \\ \cfrac{\pi}{2}(-\sigma_{xx} + \sigma_{yy}) & -\cfrac{1}{6}\sigma_{yy} - \pi\sigma_{xy} & -\cfrac{2}{3}\sigma_{yz} -\cfrac{\pi}{2}\sigma_{xz} \\ \cfrac{1}{2}(-\sigma_{xz} + \pi\sigma_{yz}) & -\cfrac{2}{3}\sigma_{yz} -\cfrac{\pi}{2}\sigma_{xz} & -\cfrac{7}{6}\sigma_{zz} \end{bmatrix} } $$

Problem 2: Hyperelastic Pinched Cylinder Problem
Read the following paper on shells:

Buchter, N., Ramm, E., and Roehl, D., 1994, "Three-dimensional extension of non-linear shell formulation based on the enhanced assumed strain concept," '' Int. J. Numer. Meth. Engng.'',  37, pp. 2551-2568.

Answer the following questions:

Part 1
What do the authors mean by "enhanced assumed strain"?

See Wikipedia article on [w:Enhanced assumed strain|enhanced assumed strain]].

Part 2
Example 8.2 (and Figures 3 and 4 and Table III) in the paper discusses the simulation of a hyperelastic cylinder. Perform the same simulation using ANSYS for a shell thickness of 0.2 cm. Use shell elements and the Neo-Hookean hyperelastic material model that ANSYS provides.

The following material properties are used:

$$E=16800$$ kN/cm$$^2$$, $$\nu=0.4$$, $$\mu=6000$$ kN/cm$$^2$$, and $$d=2/K=7.14\times 10^{-5}$$ cm$$^2$$/kN. Symmetry is used and only half of the model is meshed. At the support, the model is constraint in all directions. The load of 36 kN is applied on the top of the cylinder (Fig~2. ANSYS input listing is shown Fig~6 of Appendix.

Part 3
Compare the total load needed to achieve an edge displacement of 16 cm with the results given in Table III. Comment on your results.

The plot of force vs. edge displacement (vertical) and the deformed model are shown in Fig 3. From the plot, one sees that a load of 35.1 kN is required to deform the edge by 16 cm. This is less than 1% difference compared to the result given in the paper using 7-parameter shell theory.

Problem 3: Elastic-Plastic Punch Indentation
Read the following paper on elastic-viscoplastic FEA:

Rouainia, M. and Peric, D., 1998, "A computational model for elasto-viscoplastic solids at finite strain with reference to thin shell applications," '' Int. J. Numer. Meth. Engng.'',  42, pp. 289-311.

Answer the following questions:

Part 1
Example 5.4 of the paper shows a simulation of the deformation of a thin sheet by a square punch. Perform the same simulation for 6061-T6 aluminum. Assume linear isotropic hardening and no rate dependence.

The following data are used for the thin plate:$$E=70$$ GPa, $$\nu=0.3$$, $$\sigma_Y=240$$ Mpa, $$\sigma_t=7800$$ Mpa, see Fig 4 for the stress-strain data. Symmetry is used and only half of the model is meshed (Fig 4). At the support, the model is constrained in all directions. A plastic finite strained shell (SHELL43) is chosen for this simulation.

The following material properties are used for the punch and die:$$E=100\times 10^6$$ GPa, $$\nu=0.3$$. The value of Young's modulus is arbitrary chosen so long as it is high enough to remain rigid during the simulation.

The meshed model is illustrated in Fig 4. A load of 10 kN is applied on the top of the punch. Load steps are split into two steps. First load step the punch is moved close to the plate to activate the contact elements (CONTAC49). The second load step, 30 kN is applied on top of the punch. ANSYS input listing is shown Fig 7 of the Appendix.

Part 2
Draw a plot of the punch force vs. punch travel and compare your result with the results shown in Figure 13 of the paper (qualitative comparison only).

The punch force vs. punch travel and the plot of the deformation at the final load step is shown in Fig 5. The curve displays similar pattern as those shown in the paper.


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User:Prometeus~enwikiversity

The debate about if America is an Empire or not is useless, until "empire" be defined in some usefull and meaning way.

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Nonlinear finite elements/Homework 11/Solutions/Problem 3

Problem 3: Taylor impact test
 Given:

 * Cylindrical impact specimen strikes rigid target.
 * Frictionless contact between cylinder and target.
 * $$l_0 $$ = 30 mm; $$d_0$$ = 6 mm.
 * $$v_0 $$ = 188 m/s; $$T_0$$ = 718 K.
 * $$\mu$$ = 46 GPa; $$K$$ = 129 GPa;$$\alpha$$ = 1.76e-5 /K.
 * $$\rho_0$$ = 8930 kg/m$$^3$$;$$\kappa$$ = 386 W/m-K; $$C_v$$ = 414 J/kg-K.
 * Yield stress given by the Johnson-Cook model.
 * $$A$$ = 90 MPa; $$B$$ = 292 MPa; $$C$$ = 0.025; $$n$$ = 0.31;$$m$$ = 1.09.
 * 1) $$\dot{\varepsilon}_{0}^p$$ = 1.0 /s; $$T_0$$ = 294 K; $$T_m$$ = 1356 K.
 * Experimental deformed profile:

Part 1
Plot of final deformed shape of specimen and comparison with experimental data.

This problem is solved using LS-DYNA. Due to symmetry, only a quarter of the cylinder is modelled. The plot of the final deformed shape of the specimen is shown in Fig 6.

Part 2
Comment of differences between simulation and experiment.

The error in the predicted final width of the base of the cylinder is less than 1%

However, the width of the bulge is underestimated by the simulation by almost 10% not predicting enough strain hardening at 718 K or that there is some volumetric locking during the simulation.

The final length of the cylinder is also underestimated by the simulation. This could because the Johnson-Cook model predicts a higher yield stress at 718 K than is observed in experiments. It could also be because our mesh is too coarse.

Overall, agreement with experiment is not very good and could be improved with improvements in the plasticity model and the finite element model.

Part 3
Plot of energy. Is energy conserved?

Though the total energy at the end of the simulation is less than the the energy at the beginning of the simulation, the variation in the energy is small. Hence, we can say that the energy is reasonably well conserved during the simulation.


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Nonlinear finite elements/Homework 11/Solutions/Problem 2

Problem 2: Billet upset forging
 Given:

 * Compression of cylindrical billet with rigid dies.
 * No-slip contact between billet and dies.
 * $$r_0 $$ = 10 mm; $$l_0$$ = 30 mm.
 * $$\mu$$ = 384.6 MPa; $$K$$ = 833.3 MPa.
 * Yield stress given by

\sigma_y = \sigma_0 + H~\varepsilon_{p}~. $$
 * $$\sigma_0$$ = 1 MPa; $$H$$ = 3 MPa.
 * Compressive force of 1 KN applied to die.

Part 1
Plot of final shape of billet.

This problem is solved using ANSYS 8.0. Plane182 (2-D 4-node axisymmetric element) is used to model a quarter of the billet and the die. Contac48 (2-D contact element) is used for the node to surface contact. The final shape of the billet is shown in Fig 4.

Part 2
Compare final shape with that given in Simo and Hughes.

The die moves less in the ANSYS simulation than it does in the plot shown in Simo and Hughes (p. 325, Fig. 9.8 (b)). In Simo and Hughes, the die stroke is around 20.5 mm. In our simulation, the die stroke is around 19 mm.

The top of the billet shows an undulation that indicates that contact is not being done correctly. The billet material appears to be moving through the die during deformation. Simo and Hughes do not show such an effect.

Part 3
Plot die force (N) vs. die stroke (mm).

A plot of the die force vs. the displacement at the center of the billet is shown in Fig 5. The die stroke is (30 - $$u$$) mm where $$u$$ is the displacement shown in the plot.

Part 4
Compare plot of force vs. stroke with Simo and Hughes.

See Fig 9.8 (c) in Simo and Hughes. Our simulation shows a smaller displacement for the same load when compared with Simo and Hughes.

Part 5
Comment on volumetric locking.

The smaller displacement observed in our simulation may be an indication of volumetric locking when fully integrated PLANE182 elements are used. Underintegrated elements or assumed strain elements may alternatively be used to confirm that possibility.


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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 16

Problem 1: Part 16: Newton iterations
Let the nonlinear equations be $$g(\Delta\gamma) = 0$$. Recall that the Newton method requires that we iterate using the formula

\Delta\gamma_{r+1} = \Delta\gamma_r - \cfrac{g(\Delta\gamma_r)}{\cfrac{dg(\Delta\gamma_r)}{d\Delta\gamma}} $$ where $$r$$ is the Newton iteration number. Derive an expression for the derivative of $$g$$ that is required in the above formula.

Let us find the derivatives term by term. For the first term

\cfrac{d}{d\Delta\gamma}\left[9~\mu^2~(\Delta\gamma)^2\right] = 18~\mu^2~\Delta\gamma $$ For the second term

\cfrac{d}{d\Delta\gamma}\left[ 6~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n \right]

= 6~\mu~\sqrt{\cfrac{3}{2}}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n $$ For the fourth term

\cfrac{d}{d\Delta\gamma}\left[ \cfrac{3}{2}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^{\text{trial}} \right] = 0 $$ For the third term

\cfrac{d}{d\Delta\gamma}\left[ \left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}^2 \left\{1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\}^2 \right] = \cfrac{d}{d\Delta\gamma}[P_n~Q_n] = \cfrac{dP_n}{d\Delta\gamma}~Q_n + \cfrac{dQ_n}{d\Delta\gamma}~P_n $$ Now,

\begin{align} \cfrac{dP_n}{d\Delta\gamma} & = \cfrac{d}{d\Delta\gamma}\left[ \left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}^2\right] \\ & = 2~\left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\} \cfrac{d}{d\Delta\gamma}\left[ \sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right] \\ & = 2~\left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}\left[n~B~ \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \right] \cfrac{d}{d\Delta\gamma}\left( \alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right) \\ & = 2~\left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}\left[n~B~ \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \right] \left(\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol{\varepsilon}

^p_n\rVert_{}}\right) \\ & = 2~n~B~ \left(\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol{\varepsilon}

^p_n\rVert_{}}\right) \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \left[\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right] \end{align} $$ Similarly,

\begin{align} \cfrac{dQ_n}{d\Delta\gamma} & = \cfrac{d}{d\Delta\gamma}\left[ \left\{1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\}^2 \right] \\ & = 2~\left\{1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\} \cfrac{d}{d\Delta\gamma}\left[ - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right) \right] \\ & = -2~\sqrt{\cfrac{3}{2}}~\left\{1 - \left( \cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p}~ \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\} ~\left(\cfrac{\chi~\lVert\mathbf{s}_n\rVert_{}}{\rho_n~C_p~(T_m-T_0)}\right)\\ & = -\sqrt{6} ~\left(\cfrac{\chi~\lVert\mathbf{s}_n\rVert_{}}{\rho_n~C_p}\right) ~\left[\cfrac{T_m - T_n - \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p}~\lVert\mathbf{s}_n\rVert_{}} {(T_m -T_0)^2}\right] \end{align} $$ Therefore, the full expression for the derivative is

{ \begin{align} \cfrac{dg(\Delta\gamma_r)}{d\Delta\gamma} & = 18~\mu^2~\Delta\gamma - 6~\mu~\sqrt{\cfrac{3}{2}}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n \\ & \qquad - 2~n~B~ \left(\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol{\varepsilon}

^p_n\rVert_{}}\right) \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \left[\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right] \left[ 1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right]^2 \\ & \qquad + \sqrt{6} ~\left(\cfrac{\chi~\lVert\mathbf{s}_n\rVert_{}}{\rho_n~C_p}\right) ~\left[\cfrac{T_m - T_n - \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p}~\lVert\mathbf{s}_n\rVert_{}} {(T_m -T_0)^2}\right] \left[\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right]^2 \end{align} } $$


text:
Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 15

Problem 1: Part 15: Finding the plastic flow parameter
The discretized form of the Kuhn-Tucker conditions in conjunction with the consistency condition gives us

f(\boldsymbol{\sigma}_{n+1}, \alpha_{n+1}, T_{n+1}) = 0 ~. $$ Use this condition and the relations you have derived in the previous sections to arrive at a nonlinear equation in $$\Delta\gamma$$ that can be solved using Newton iterations.

The yield function is

f = \sqrt{\cfrac{3}{2}}~\sqrt{\mathbf{s}:\mathbf{s}} - \left[\sigma_0 + B \alpha^n\right] \left[1 - \left(\cfrac{T - T_0}{T_m -T_0}\right)\right] $$ Therefore the discretized form of Kuhn-Tucker + consistency is

\sqrt{\cfrac{3}{2}}~\sqrt{\mathbf{s}_{n+1}:\mathbf{s}_{n+1}} - \left[\sigma_0 + B \alpha_{n+1}^n\right] \left[1 - \left(\cfrac{T_{n+1} - T_0}{T_m -T_0}\right)\right] = 0 $$ or,

\mathbf{s}_{n+1}:\mathbf{s}_{n+1} = \cfrac{2}{3}\left[\sigma_0 + B \alpha_{n+1}^n\right]^2 \left[1 - \left(\cfrac{T_{n+1} - T_0}{T_m -T_0}\right)\right]^2 $$ Now,

\mathbf{s}_{n+1} = \mathbf{s}_{n+1}^{\text{trial}} - 2~\mu~\sqrt{\cfrac{3}{2}}

~\Delta\gamma~\mathbf{n}_n $$ Therefore,

\begin{align} \mathbf{s}_{n+1}:\mathbf{s}_{n+1} & = \mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^

{\text{trial}} - 4~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n + 4~\mu^2~\cfrac{3}{2}~(\Delta\gamma)^2~\mathbf{n}_n:\mathbf{n}_n \\ & = \mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^{\text{trial}} - 4~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n + 6~\mu^2~(\Delta\gamma)^2 \end{align} $$ Plugging into the discretized yield condition, we have

\cfrac{2}{3}\left[\sigma_0 + B \alpha_{n+1}^n\right]^2 \left[1 - \left(\cfrac{T_{n+1} - T_0}{T_m -T_0}\right)\right]^2 = \mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^{\text{trial}} - 4~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n + 6~\mu^2~(\Delta\gamma)^2 $$ or,

\left[\sigma_0 + B \alpha_{n+1}^n\right]^2 \left[1 - \left(\cfrac{T_{n+1} - T_0}{T_m -T_0}\right)\right]^2 = \cfrac{3}{2}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^{\text{trial}} - 6~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n + 9~\mu^2~(\Delta\gamma)^2 $$ Also

\begin{align} \alpha_{n+1}& = \alpha_n + \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol {\varepsilon}^p_n\rVert_{}}\\ T_{n+1} & = T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} ~\lVert\mathbf{s}_n\rVert_{} \end{align} $$ Therefore,

\left[\sigma_0 + B \left\{\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right\}^n \right]^2 \left[1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} ~\lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right]^2 = \cfrac{3}{2}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^{\text{trial}} - 6~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n + 9~\mu^2~(\Delta\gamma)^2 $$ The nonlinear equation in $$\Delta\gamma$$ is

{ \begin{align} g(\Delta\gamma) & = 0 \\ & = 9~\mu^2~(\Delta\gamma)^2 - 6~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n \\ & \qquad - \left[\sigma_0 + B \left\{\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right\}^n \right]^2 \left[1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right]^2 \\ &\qquad + \cfrac{3}{2}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^{\text{trial}} \end{align} } $$


text:
Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 14

Problem 1: Part 14: Return mapping
Show that

\mathbf{s}_{n+1} = \mathbf{s}_{n+1}^{\text{trial}} - 2~\mu~\Delta\gamma~\mathbf{n}_n ~. $$

We have

\begin{align} \boldsymbol{\sigma}_{n+1} & = \boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon}^e_{n+1} = \boldsymbol{\mathsf{C}}:(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_

{n+1}) \\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol

{\mathsf{I}}): (\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_{n+1}) \\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol

{\mathsf{I}}): (\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac{3}{2}}

~\Delta\gamma~\mathbf{n}_n) \\ & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} - \text{tr}(\boldsymbol{\varepsilon}^p_n)~\boldsymbol{\mathit{1}} - \sqrt{\cfrac{3}{2}}~\Delta\gamma~\text{tr}(\mathbf{n}_n)~\boldsymbol{\mathit{1}}\right] + 2~\mu~\left[\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac

{3}{2}}~\Delta\gamma~\mathbf{n}_n \right] \\ & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol

{\varepsilon}^p_n) - \sqrt{\cfrac{3}{2}}~\Delta\gamma~\text{tr}(\mathbf{n}_n)\right]\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac

{3}{2}}~\Delta\gamma~\mathbf{n}_n \right] \\ & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol

{\varepsilon}^p_n)\right]\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac

{3}{2}}~\Delta\gamma~\mathbf{n}_n \right] \qquad~(\text{since}~ \text{tr}(\mathbf{n}) = 0) \end{align} $$ Now

\mathbf{s}_{n+1} = \boldsymbol{\sigma}_{n+1} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma}_

{n+1})~\boldsymbol{\mathit{1}} $$ The trace of the stress is given by

\begin{align} \text{tr}(\boldsymbol{\sigma}_{n+1}) & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol

{\varepsilon}^p_n)\right]\text{tr}(\boldsymbol{\mathit{1}}) + 2~\mu~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol{\varepsilon}

^p_n) - \sqrt{\cfrac{3}{2}}~\Delta\gamma~\text{tr}(\mathbf{n}_n) \right] \\ & = 3~\lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol

{\varepsilon}^p_n)\right] + 2~\mu~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol{\varepsilon}

^p_n)\right] \qquad~(\text{since}~ \text{tr}(\mathbf{n}) = 0) \\ & = (3~\lambda + 2~\mu)\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - (3~\lambda + 2~\mu)\text{tr}(\boldsymbol{\varepsilon}^p_n) \end{align} $$ Therefore,

\begin{align} \mathbf{s}_{n+1} & = \boldsymbol{\sigma}_{n+1} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma}_{n+1})~\boldsymbol{\mathit{1}}\\ & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol{\varepsilon}^p_n)\right]\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n \right] \\ & \qquad - (\lambda + \cfrac{2}{3}~\mu)\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} + (\lambda + \cfrac{2}{3}~\mu)\text{tr}(\boldsymbol{\varepsilon}^p_n)~\boldsymbol{\mathit{1}} \\ & = 2~\mu\left[\boldsymbol{\varepsilon}_{n+1} - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}}\right] - 2~\mu\left[\boldsymbol{\varepsilon}^p_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^p_n)~\boldsymbol{\mathit{1}}\right] - 2~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n \\ & = 2~\mu\mathbf{e}_{n+1} - 2~\mu\left[\boldsymbol{\varepsilon}_n - \boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}_n - \boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}}\right] - 2~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n \\ & = 2~\mu\mathbf{e}_{n+1} - 2~\mu\left[\boldsymbol{\varepsilon}_n - \boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}_n)~\boldsymbol{\mathit{1}} + \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}}\right] - 2~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n \\ & = 2~\mu\left[\mathbf{e}_{n+1} - \mathbf{e}_n\right] + 2~\mu\left[\boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}}\right] - 2~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n \\ & = \mathbf{s}_{n+1}^{\text{trial}} - \mathbf{s}_n + 2~\mu\left[\boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}}\right] - 2~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n \end{align} $$ The stress-strain relation is

\boldsymbol{\sigma}_n = \lambda~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol

{\mathit{1}} + 2~\mu~\boldsymbol{\varepsilon}^e_n $$ Hence,

\begin{align} \mathbf{s}_n & = \lambda~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}} +

2~\mu~\boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\lambda~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\text{tr}(\boldsymbol{\mathit

{1}})~\boldsymbol{\mathit{1}} - \cfrac{2}{3}~\mu~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}} \\ & = 2~\mu\left[\boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol

{\varepsilon}^e_n)~\boldsymbol{\mathit{1}}\right] \end{align} $$ Plugging into expression for $$\mathbf{s}_{n+1}$$, we get

\mathbf{s}_{n+1} = \mathbf{s}_{n+1}^{\text{trial}} - \mathbf{s}_n + \mathbf{s}_n - 2~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n $$ Therefore,

{ \mathbf{s}_{n+1} = \mathbf{s}_{n+1}^{\text{trial}} - 2~\mu~\sqrt{\cfrac{3}{2}}

~\Delta\gamma~\mathbf{n}_n } $$

Remark: If we write the yield function as

f = \sqrt{\mathbf{s}:\mathbf{s}} - \sqrt{\cfrac{2}{3}}~\sigma_y ~. $$ then the above equation takes the form

{ \mathbf{s}_{n+1} = \mathbf{s}_{n+1}^{\text{trial}} - 2~\mu~\Delta\gamma~\mathbf{n}_n } $$ These are equivalent.


text:
Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 13

Problem 1: Part 13: Trial elastic stress
Starting from equation (3) show that

\mathbf{s}_{n+1}^{\text{trial}} = \mathbf{s}_n + 2~\mu~(\mathbf{e}_{n+1} - \mathbf{e}_n) $$ where $$\mathbf{s}$$ is the deviatoric part of $$\boldsymbol{\sigma}$$ and $$\mathbf{e}$$ is the deviatoric part of $$\boldsymbol{\varepsilon}$$.

From equation (3) we have

\begin{align} \boldsymbol{\sigma}_{n+1}^{\text{trial}} & = [\lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} +

2~\mu~\boldsymbol{\varepsilon}_{n+1}] - [\lambda~\text{tr}(\boldsymbol{\varepsilon}^p_n)~\boldsymbol{\mathit{1}} +

2~\mu~\boldsymbol{\varepsilon}^p_n] \\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n)\\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n + \boldsymbol{\varepsilon}^e_n)\\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):\boldsymbol{\varepsilon}^e_n + (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) \\ & = \boldsymbol{\sigma}_n + (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) \end{align} $$ The deviatoric parts of the stress and strain are

\mathbf{s}_{n+1}^{\text{trial}} = \boldsymbol{\sigma}_{n+1}^{\text{trial}} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma}_{n+1}^{\text{trial}})~\boldsymbol{\mathit{1}}

~; \mathbf{e}_{n+1} = \boldsymbol{\varepsilon}_{n+1} - \frac{1}{3}~\text{tr}(\boldsymbol

{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} ~; \mathbf{e}_n = \boldsymbol{\varepsilon}_n - \frac{1}{3}~\text{tr}(\mathbf{e}_n)~\boldsymbol

{\mathit{1}} $$ Therefore,

\begin{align} \mathbf{s}_{n+1}^{\text{trial}} & = \boldsymbol{\sigma}_{n+1}^{\text{trial}} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma}_{n+1}^{\text{trial}})~\boldsymbol{\mathit{1}} \\ & = \boldsymbol{\sigma}_n + (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) - \frac{1}{3}~\text{tr}{\boldsymbol{\sigma}_n}~\boldsymbol{\mathit{1}} - \frac{1}{3}~\text

{tr}[ \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n)] ~\boldsymbol{\mathit{1}} \end{align} $$ Now,

(\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) = \lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\varepsilon}_{n+1} - \lambda~\text{tr}(\boldsymbol{\varepsilon}_n)~\boldsymbol{\mathit{1}} - 2~\mu~\boldsymbol{\varepsilon}_n $$ Therefore,

\begin{align} \text{tr}[ (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n)] & = \lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\text{tr}(\boldsymbol{\mathit{1}}) + 2~\mu~\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \lambda~\text{tr}(\boldsymbol{\varepsilon}_n)~\text{tr}(\boldsymbol{\mathit{1}}) - 2~\mu~\text{tr}(\boldsymbol{\varepsilon}_n) \\ & = 3~\lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1}) + 2~\mu~\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - 3\lambda~\text{tr}(\boldsymbol{\varepsilon}_n) - 2~\mu~\text{tr}(\boldsymbol{\varepsilon}_n) \end{align} $$ Hence

\begin{align} \mathbf{s}_{n+1}^{\text{trial}} = &~ \boldsymbol{\sigma}_n - \frac{1}{3}~\text{tr}{\boldsymbol{\sigma}_n} + \lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} +

2~\mu~\boldsymbol{\varepsilon}_{n+1} - \lambda~\text{tr}(\boldsymbol{\varepsilon}_n)~\boldsymbol{\mathit{1}} - 2~\mu~\boldsymbol

{\varepsilon}_n \\ & - \lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} - \cfrac{2}{3}~\mu~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} + \lambda~\text{tr}(\boldsymbol{\varepsilon}_n)~\boldsymbol{\mathit{1}} + \cfrac{2}{3}

~\mu~\text{tr}(\boldsymbol{\varepsilon}_n)~\boldsymbol{\mathit{1}} \\ = & ~ \mathbf{s}_n + 2~\mu~\left(\boldsymbol{\varepsilon}_{n+1} - \frac{1}{3}~\text{tr}

(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}}\right) - 2~\mu~\left(\boldsymbol{\varepsilon}_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}

_n)~\boldsymbol{\mathit{1}}\right) \\ = & ~ \mathbf{s}_n + 2~\mu~\mathbf{e}_{n+1} - 2~\mu~\mathbf{e}_n \end{align} $$ This shows that

{ \mathbf{s}_{n+1}^{\text{trial}} = \mathbf{s}_n + 2~\mu~(\mathbf{e}_{n+1} - \mathbf{e}_n) } $$


text:
Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 12

Problem 1: Part 12: Trial elastic stress
In the radial return algorithm, we define a trial elastic state as
 * $$\text{(2)} \qquad

\boldsymbol{\sigma}_{n+1}^{\text{trial}} = \boldsymbol{\sigma}_n + \boldsymbol{\mathsf{C}}: (\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) $$ where $$\boldsymbol{\sigma}_n, \boldsymbol{\varepsilon}_n$$ are the stress and strain at $$t = t_n$$ and $$\boldsymbol{\sigma}_{n+1}, \boldsymbol {\varepsilon}_{n+1}$$ are the values at $$t = t_{n+1}$$. Show that, if the elastic response of the material is linear, equation (2) can be written as
 * $$\text{(3)} \qquad

\boldsymbol{\sigma}_{n+1}^{\text{trial}} = [\lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\varepsilon}_{n+1}] - [\lambda~\text{tr}(\boldsymbol{\varepsilon}^p_n)~\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\varepsilon}^p_n] ~. $$

From equation (2)

\boldsymbol{\sigma}_{n+1}^{\text{trial}} = \boldsymbol{\sigma}_n + \boldsymbol{\mathsf{C}}:

(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) = \boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon}^e_n + \boldsymbol{\mathsf{C}}:

(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^e_n - \boldsymbol{\varepsilon}

^p_n) = \boldsymbol{\mathsf{C}}:(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n) $$ Therefore,

{ \boldsymbol{\sigma}_{n+1}^{\text{trial}} = = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol

{\mathsf{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n) = \lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1}) + 2~\mu~\boldsymbol{\varepsilon}_{n+1} - \lambda~\text{tr}(\boldsymbol{\varepsilon}^p_n) - 2~\mu~\boldsymbol{\varepsilon}^p_n~. } $$ Hence shown.


text:
Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 11

Problem 1: Part 11: Discrete Kuhn-Tucker
The Kuhn-Tucker loading-unloading conditions are

\dot{\gamma} \ge 0 ~, f(\boldsymbol{\sigma}, \alpha, T) \le 0 ~; \dot{\gamma} f(\boldsymbol{\sigma},\alpha,T) = 0 ~. $$ Write down a discrete form of the Kuhn-Tucker conditions.

The discrete form of the Kuhn-Tucker conditions is

{ \begin{align} \Delta\gamma & \ge 0 \\ f(\boldsymbol{\sigma}_{n+1}, \alpha_{n+1}, T_{n+1}) & \le 0 \\ \Delta\gamma~f(\boldsymbol{\sigma}_{n+1}, \alpha_{n+1}, T_{n+1}) & = 0 \end{align} } $$


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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 10

Problem 1: Part 10: Discrete evolution equations
Discretize the equations for $\dot{\mathbf{e}}^p$ (equation 1), $\dot{\alpha}$ (from part 1), and $\dot{T}$ (from part 2) using Forward Euler. |undefined

The relevant equations are

\begin{align} f_{\boldsymbol{\sigma}} & = \sqrt{\cfrac{3}{2}}~\mathbf{n} \\ \dot{\boldsymbol{\varepsilon}}^p & = \dot{\gamma}~f_{\boldsymbol{\sigma}} = \sqrt{\cfrac{3}{2}}~\dot{\gamma}~\mathbf{n} \\ \dot{\alpha} & = \sqrt{\cfrac{2}{3}}~\dot{\gamma}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol

{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} = \dot{\gamma}~\cfrac{\boldsymbol{\varepsilon}^p:\mathbf{n}}{\lVert\boldsymbol

{\varepsilon}^p\rVert_{}} \\ \dot{T} & = \cfrac{\chi~\dot{\gamma}}{\rho~C_p} ~ \boldsymbol{\sigma}:\frac{\partial f}{\partial \boldsymbol{\sigma}} = \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\dot{\gamma}}{\rho~C_p} ~ \boldsymbol{\sigma}:\mathbf{n} = \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\dot{\gamma}}{\rho~C_p}~\lVert\mathbf{s}\rVert_{} \end{align} $$ A Forward Euler time discretization gives us

\begin{align} \boldsymbol{\varepsilon}^p_{n+1} & = \boldsymbol{\varepsilon}^p_n + \sqrt{\cfrac{3}{2}}

~\Delta t~\dot{\gamma}_n~\mathbf{n}_n \\ \alpha_{n+1}& = \alpha_n + \Delta t~\dot{\gamma}_n~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}

{\lVert\boldsymbol{\varepsilon}^p_n\rVert_{}} \\ T_{n+1} & = T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta t~\dot{\gamma}_n}{\rho_n~C_p} ~\lVert\mathbf{s}_n\rVert_{} \end{align} $$ or

{ \begin{align} \boldsymbol{\varepsilon}^p_{n+1} & = \boldsymbol{\varepsilon}^p_n + \sqrt{\cfrac{3}{2}}

~\Delta\gamma~\mathbf{n}_n \\ \alpha_{n+1}& = \alpha_n + \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}} \\ T_{n+1} & = T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} ~\lVert\mathbf{s}_n\rVert_{} \end{align} } $$