User:Eml4500.f08.bike.sweeting

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Homework 1-Individual Part
EML4500 Summary: Aug. 25, 2008 – Aug. 29, 2008

Finite Element Analysis with Dr. Loc Vu-Quoc

Lectures are held at the University of Florida on Monday, Wednesday, and Friday from 4:05-4:55 PM in Florida Gym 270.

EML4500 teaches the fundamentals of the finite elemental method used in the analysis of truss, beam, and solid structures. The finite element method is taught through matrix algebra with the derivation of the stiffness matrix. Local and global coordinate systems are emphasized, as well as 3-D truss and beam structures and their linear stability. Coding and plotting is another area of interest in this course, as Matlab programming will be used in order to solve many of the problems. Applications of the finite element method are discussed, as well as its uses in mechanical, aerospace, and biomedical engineering fields.

The text book that will be used throughout the semester is: M. Asghar Bhatti, Finite Element Analysis With Mathematica and Matlab Computations and Practical Applications: Fundamental Concepts, John Wiley & Sons, Inc. Copyright 2005

Students will be required to work in groups of 5 to 6 under a team name that is 4 to 6 characters long. Cooperative learning techniques will be used, and group member evaluations will be held at the end of the semester. These evaluations will directly impact the homework grade of that individual, approximately 31% of his/her overall grade. The rest of the student’s individual grade is comprised of 3 exams, each accounting for 23%.

Each group will be required to use Wikiversity in order to submit homework and projects. Each member must create his/her own personal class account where his/her portion of the work will be posted. Each members name and UFID number, as well as his/her Wikiversity class username, is to be submitted to the instructor on hard copy.

Homework and project load will be divided evenly among the team members at their discretion. Once each group member has completed his/her assigned task the assignment will be posted as a whole on the group leader’s user namespace. This should be done close to the homework deadline in order to keep other groups from stealing ideas or plagiarizing. Another safety concern is that users may change information on a students Wikiversity page. In order to prevent this type of fraud from occurring the student must save an archived version of the assignment and email that address, not the address of the updated page, to the TA.

Homework 2-Individual Part(9/17)
Step 4 in the finite element analysis process begins with the reduction of the global force-displacement relations. This reduction is done by first eliminating the known degrees of freedom(DOF’s). $$ d_{1}=d_{2}=d_{5}=d_{6}=0 $$

Deleting the corresponding columns in the global stiffness matrix and the force matrix results in the following relationship.

$$ \left[ \begin{array}{cc} k_{33} & k_{34}\\ k_{43}& k_{44} \end{array} \right]_{2x2} $$ $$\begin{Bmatrix} d_{3}\\d_{4} \end{Bmatrix}_{2x1}=\begin{Bmatrix}F_{3}\\F_{4} \end{Bmatrix}_{2x1}$$

In order to find the inverse of the stiffness matrix one must first calculate the determinant as follows. det $$k = k_{33}k_{44}-k_{34}k_{43} $$

The inverse of the stiffness matrix can then be calculated as shown below.

$$k^{-1} = \frac {1}{det(k)}$$$$ \left[ \begin{array}{cc} k_{44} & -k_{34}\\ -k_{43}& k_{33} \end{array} \right] $$

If done correctly the stiffness matrix multiplied by its inverse will yield the identity matrix, I. $$I= \left[ \begin{array}{cc} 1 & 0\\ 0& 1 \end{array} \right] $$

For the two beam truss system shown below, the displacement matrix is calculated by multiplying the inverse matrix by $$\begin{Bmatrix}{0} \\ {P} \end{Bmatrix}$$, where P=7.



$$\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=k^{-1}\begin{Bmatrix}{0}\\{7}\end{Bmatrix}$$

det $$k=(3.0625)(2.6875)-(-2.175)(-2.175)=3.50$$

Therefore, $$ k^{-1}=\frac {1}{3.50}$$$$ \left[ \begin{array}{cc}2.6875&2.175\\ 2.175&3.0625\end{array} \right]=\left[ \begin{array}{cc}.7678&.6214\\ .6214&.8750\end{array} \right]$$

so $$\begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix}=\left[ \begin{array}{cc}.7678&.6214\\ .6214&.8750\end{array} \right]\begin{Bmatrix}{0}\\{7}\end{Bmatrix}=\begin{Bmatrix}{4.352}\\{6.1271}\end{Bmatrix}$$

Step 5 consists of computing the reactions, which can be done by two different methods. Method 1 is to use the element force-displacement relations and method 2 is the Euler cut principle. The force-displacement relation corresponding to the two beam truss system is shown below.

$$k_{4x4}^{(e)}d_{4x1}^{(e)}=f_{4x1}^{(e)}$$ where $$e=1,2$$

The free body diagram of element 1(e=1) is shown below, along with the corresponding force-displacement relation. $$k^{(1)}d^{(1)}=f^{(1)}$$ where $$k^{(1)}$$ and $$d^{(1)}$$ are known.

$$d^{(1)}=\begin{Bmatrix}{0}\\{0}\\{4.352}\\{6.1271}\end{Bmatrix}$$ and $$f^{(1)}=\begin{Bmatrix}{f_{1}^{(1)}}\\{f_{2}^{(1)}}\\{f_{3}^{(1)}}\\{f_{4}^{(1)}}\end{Bmatrix}$$

The free body diagram of element 2(e=2) is shown below, along with the corresponding force-displacement relation. $$k^{(2)}d^{(2)}=f^{(2)}$$ where $$k^{(2)}$$ and $$d^{(2)}$$ are known.

$$d^{(2)}=\begin{Bmatrix}{4.352}\\{6.1271}\\{0}\\{0}\end{Bmatrix}$$ and $$f^{(2)}=\begin{Bmatrix}{f_{1}^{(2)}}\\{f_{2}^{(2)}}\\{f_{3}^{(2)}}\\{f_{4}^{(2)}}\end{Bmatrix}$$

Homework 4-Individual Part(10/6)
The elements of a two bar truss system are shown below by Figures 1 and 2.

The connectivity array, denoted "conn," is shown below for the two bar truss system.

conn = $$ \left[ \begin{array}{cc} {1} & {2} \\ {2}& {3} \end{array} \right]$$

Each row of the connectivity array represents an element(i.e. row 1 = element 1, row 2 = element 2), and each column represents a local node number(i.e. column 1 = local node 1, column 2 = local node 2). The numbers within the connectivity array correspond to the global node numbers, such that conn(e,j) = global node number of local node j of element e.

The location matrix master array, denoted "lmm," is shown below for the two bar truss system.

lmm = $$ \left[ \begin{array}{cccc} {1} & {2} & {3} & {4} \\ {3} & {4} & {5} & {6} \end{array} \right]$$

Somewhat similar to the connectivity array, the location matrix master array has rows that correspond to the element number and columns that represent the local degree of freedom(dof) number. The numbers within the location matrix master array correspond to the global dof number, or equation number, in the matrix K. Generally, lmm(i,j) = global dof number(or equation number) for the element stiffness coefficient corresponding to the jth local dof number.

Method 2 to derive the stiffness matrix $$K_{4x4}^{e}$$ is to transform a system with 2 dof's into a system with 4 dof's so that the transformed matrix is a 4x4, and hopefully invertible.

Homework 4-Individual Part(10/15)
For the justification of assembly of the element stiffness matrix into the global stiffness matrix, consider the two-bar truss system.

Let the global stiffness matrix be represented by K and the element stiffness matrix be represented by $$k^{(e)}$$, where e = 1,...,n and n is any number of element.

Recall the element force-displacement relationship $$k^{(e)}_{4x4}d^{(e)}_{4x1}=f^{(e)}_{4x1}$$ as well as the second method of the Euler cut principle shown below. This method involves an equilibrium analysis of node 2.

Also recall the free body diagram of the two bar truss system in question. The free body diagrams of elements 1 and 2 are shown below, taking note of the naming process for the element degrees of freedom.

It is possible to relate the global degrees of freedom to element degrees of freedom for both element 1 and element 2. This is done below for node 2.

$$

\textrm{Node\ 2:\ }d_{3}=d_{3}^{(1)}=d_{1}^{(2)}\textrm{,\ }d_{4}=d_{4}^{(1)}=d_{2}^{(2)}

$$

The equilibrium of node 2 can then be analyzed by cutting each element and relating the forces as shown below.

$$\sum F_x = 0 = -f_3^{(1)} - f_1^{(2)} = 0 $$

$$\sum F_y = 0 = P - f_4^{(1)} - f_2^{(2)} = 0 $$

Homework 5-Individual Part(10/31)
The free body diagram of section dx can be seen below in Figure 1.

Summing the forces in the x direction yields

$$\sum F_x = 0 = -N(x,t) + N(x+dx,t) + f(x,t)dx - m(x){\ddot{u}}dx $$

$$ = \frac{\partial N(x,t)}{\partial x}dx + h.o.t + f(x,t)dx - m(x){\ddot{u}}dx $$ ⇨ $$ Equation \left(1\right)$$

The dx terms are canceled out of this equation, as well as the high order terms(h.o.t). Let this equation be call equation (1).

Why are the high order terms neglected?

Recall Taylor series expansion: $$ f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^{2}f(x)}{dx^{2}}dx^{2} + ... $$ where $$\frac{1}{2}\frac{d^{2}f(x)}{dx^{2}}dx^{2}$$ is considered a h.o.t and is therefore neglected.

From equation (1) the equation of motion(EOM) of the system can be derived. Let the EOM be called equation (2).

$$ \frac{\partial N}{\partial x} + f = m{\ddot{u}} $$ ⇨ $$ Equation \left(2\right)$$

Equation (3), the constitutive relation, can then be determined.

$$ N\left(x,t\right) = A(x)\sigma(x,t) $$ ⇨ $$ Equation \left(3\right)$$ where $$ \sigma\left(x,t\right) = E(x)\epsilon(x,t) $$ and $$ \epsilon(x,t) = \frac{\partial u(x,t)}{\partial x} $$

Substituting equation (3) into equation (2) yields the partial differential equation of motion, equation (4) shown below.

$$ \frac{\partial}{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}] + f(x,t) = m(x){\ddot{u}} $$  ⇨   $$ Equation \left(4\right)$$

There are 2 initial conditions(displacement and velocity), and therefore 2 boundary conditions(2nd order derivative with respect to x).

For Figure 2 the boundary conditions are $$ u\left(0,t\right) = 0 = u(L,t) $$

For Figure 3 the boundary condition at 0 is also $$ u\left(0,t\right) = 0 $$, but the boundary condition at L can be derived as follows

$$ N\left(L,t\right) = F(t) $$ where $$ N = A\left(L\right)\sigma(L,t)$$ and $$\sigma\left(L,t\right) = E(L)\epsilon(L,t)$$ and $$ \epsilon(L,t) = \frac{\partial u(L,t)}{\partial x} $$

The boundary condition at L can therefore be simplified to

$$ \frac{\partial u(L,t)}{\partial x} = \frac{F(t)}{A(L)E(L)} $$

Homework 6-Individual Part(11/3)
Figure 1 shows the loaded elastic bar system in question.



The initial conditions for the system in Figure 1 at t = 0 are

$$ u\left(x,t=0\right)=\overline{u}(x) $$ for the known displacement function

and

$$ \frac{\partial u}{\partial t}(x,t=0)={\ddot{u}}(x,t=0)=\overline{V}(x) $$ for the known velocity function.

The partial differential equation for the continuous principle of virtual work(PVW) of dynamics for the elastic bar in Figure 1 can be seen below in Equation 1.

$$ PDE: \frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f=m{\ddot{u}} $$ ⇨ $$ Equation \left(1\right)$$

The discrete equation of motion can then be shown by Equation 2, where $$ \underline{M} $$ is the mass matrix. Equation 2 represents a multiple defree of freedom system(MDOF). Note that a single degree of freedom system(SDOF) has a very similar equation of motion, shown be Equation 3.

MDOF System ⇨ $$ \underline{M}\underline{\ddot{d}}+\underline{k}\underline{d}=\underline{F} $$ ⇨ $$ Equation \left(2\right)$$

SDOF System ⇨ $$ m\ddot{d}+kd=F $$ ⇨ $$ Equation \left(3\right)$$

The integral shown in Equation 4 was used to derive Equation 2 from Equation 1. The term $$ w(x) $$ represents the weighting function.

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$ for all possible $$ w(x) $$ ⇨ $$ Equation \left(4\right)$$

To go from Equation 1 to Equation 4 is a trivial task, but going from Equation 4 to Equation 1 is not trivial. This process is shown below.

For simplicity, let $$ \left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)=g(x) $$

Equation 4 then becomes $$ \int_{x=0}^{x=L} w(x)g(x)dx=0 $$ for all $$ w\left(x\right) $$

Since Equation 4 holds for all $$ w\left(x\right) $$, select $$ w\left(x\right)=g\left(x\right) $$,

then Equation 4 becomes $$ \int_{x=0}^{x=L}g^2dx=0 $$ ⇨ $$ g\left(x\right)=0 $$

Homework 6-Individual Part(11/5)
Recall the steps for and integration by parts:

$$ \left(rs\right)^{'}=r^{'}s+rs^{'} $$

$$ r^{'}=\frac{dr}{dx} $$ and $$ s^{'}=\frac{ds}{dx} $$

$$ \int \left(rs\right)^{'}=\int r^{'}s + \int rs^{'} $$

Which yields a final result of $$ \int r^{'}s=rs-\int rs^{'} $$

Now Recall the Cont. Principle of Virtual Work(PVW) shown below.

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$ for all possible $$ w(x) $$

In the first term, let $$ s\left(x\right)=w(x) $$ and $$ r\left(x\right)=(EA)\frac{\partial u}{\partial x} $$

Solving using integration by parts yields

$$ \int_{x=0}^{x=L} w\left(x\right)\frac{\partial}{\partial x}\left[\left(EA\right)\frac{\partial u}{\partial x}\right]dx = \left[w(EA)\frac{\partial u}{\partial x}\right]_{x=0}^{x=L} - \int_{x=0}^{x=L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

$$ = w\left(L\right)(EA)(L)\frac{\partial u}{\partial x}(L,t)-w(0)(EA)(0)\frac{\partial u}{\partial x}(0,t)-\int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

where $$ \left(EA\right)(L)\frac{\partial u}{\partial x}(L,t) = N(L,t)=F $$

and $$ (EA)(0)\frac{\partial u}{\partial x}(0,t)=N(0,t) $$

Now lets consider the model problem with 2 boundary conditions shown below.

At $$ x=0 $$ select $$ w(x) $$ such that $$ w(0)=0 $$. This means that the system is kinematically admissible.

Motivation: discrete PVW applied to the equation shown below.

$$ \underline{W}_{6x1}\left(\left[ \underline K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1}-\underline{F}_{6x1}\right)=0_{1x1} $$ for all $$ \underline{W} $$

where $$ \underline{F}^{T}=\left[F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}\right] $$

$$ F_{3}, F_{4} $$ are known reactions and $$ F_{1}, F_{2}, F_{5}, F_{6} $$ are unknown reactions.

Since $$ \underline{w} $$ can be selected arbitrarily, select $$ \underline{w} $$ such that $$ w_{1}=w_{2}=w_{5}=w_{6}=0 $$ so to eliminate equations involving unknown reactions. Simply put, rows 1, 2, 5, and 6 are eliminated. The result is shown below in Equation 1.

$$ \underline{W}\left(\underline{K}\underline{d}-\underline{F}\right)=0 $$ for all $$ \underline{W} $$

$$ \underline{K}_{2x2}\underline{d}_{2x1}=\underline{F}_{2x1} $$ ⇨ $$ Equation \left(1\right) $$

Back to the principle of virtual work, the unknown reaction is

$$ N(0,t)=(EA)(0)\frac{\partial u}{\partial x}(0,t) $$

The equation then becomes

$$ W\left(L\right)F(t)-\int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx+\int w(x)\left[f-m\ddot{u}\right]dx=0 $$ for all $$ w\left(x\right) $$ such that $$ w\left(0\right)=0 $$

The final equation can be written as

$$ \int_{0}^{L}w\left(m\ddot{u}\right)dx+\int_{0}^{L} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx=W(L)F(t)+\int_{0}^{L}wfdx $$ for all $$ w\left(x\right) $$ such that $$ w\left(0\right)=0 $$

Homework 7-Individual Part(11/21)
$$ \left[\epsilon\right] = \frac{[du]}{[dx]} = \frac{L}{L} = 1 $$

$$ \left[\sigma\right] = \left[E\right] = \frac{F}{L^2} $$

$$ \left[A\right] = L^2, [I] = L^4 $$

$$ \left[\frac{EA}{L}\right] = [\tilde{k}_{11}] = \frac{\frac{F}{L^2}{L^2}}{L} = \frac{F}{L} $$

$$ \left[\tilde{k}_{11}\tilde{d_1}\right] = [\tilde{k}_{11}\tilde{d_1}] = F $$

$$ \left[\tilde{k}_{23}\tilde{d_3}\right] = [\tilde{k}_{23}][\tilde{d_3}] $$ where $$ \tilde{d_3} = 1 $$

$$ = \frac{\left[6\right][E][I]}{L^2} $$

$$ = \frac{\left(1\right)\left(\frac{F}{L^2}\right)(L^4)}{L^2} = F $$

The element force displacement relation in global coordinates from the element force displacement relation in local coordinates yields:

$$ \underline{k}^{(e)}_{6x6}\underline{d}^{(e)}_{6x1} = \underline{f}^{(e)}_{6x1} $$

with $$ \underline{k}^{(e)}_{6x6} = {\underline{\tilde{T}}^{(e)}_{6x6}}^T\underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{T}}^{(e)}_{6x6} $$

from $$ \underline{\tilde{k}}^{(e)}_{6x6}\underline{\tilde{d}}^{(e)}_{6x1} = \underline{\tilde{f}}^{(e)}_{6x1} $$



The derivation of $$ \tilde{K}^{(e)} $$ from PVW results in the equation shown below, focusing on the bending effects.

$$ \frac{\partial^{2}}{\partial x^{2}}\left((EI)\frac{\partial^{2}v}{\partial x^{2}}\right) - f_{t}(x) = m(x)\ddot{v} $$

Homework 7-Individual Part(12/1)
The interpolations of the applied internal forces $$ N_{5}\left(\tilde{x}\right) $$ and $$ N_{6}\left(\tilde{x}\right) $$ are shown below.

Recall the equation of the displacement matrix in the tilde coordinate system shown below, where $$ \underline{d}_{6x1}^{(e)} $$ is known after solving the finite element system.

$$ \underline{\tilde{d}}_{6x1}^{(e)} = \underline{\tilde{T}}_{6x6}^{(e)}\underline{d}_{6x1}^{(e)}$$

Now we will look at the computation of $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$.

$$ \underline{u}\left(\tilde{x}\right) = u(\tilde{x})\tilde{i} + v(\tilde{x})\tilde{j} = u_{x}(\tilde{x})\tilde{i} + u_{y}(\tilde{x})\tilde{j} $$

Computing $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$ required using the equations shown below.

$$u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4$$

$$v(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$

The computation of $$ u_{x}\left(\tilde{x}\right) $$ and $$ u_{y}\left(\tilde{x}\right) $$ from $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$ is shown below.

$$\begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \underline{R}^{T}\begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix}$$

$$ \begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix} = \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} $$ where $$ \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix} $$ is represented by $$ \underline{N}\left(\tilde{x}\right) $$ and $$\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix}$$ is denoted $$ \underline{\tilde{d}}^{(e)} $$

$$\begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \underline{R}^{T}\underline{N}\left(\tilde{x}\right)\underline{\tilde{T}}^{(e)}\underline{d}^{(e)} $$