User:EML4500.f08.A-team.VandenBerg/HW4

 Comment: This is HW4 located in the user namespace of EML4500.f08.A-team.VandenBerg 20:28, 24 October 2008 (UTC)

HW4
18-1

Connecting Array
Considering the two-bar truss system from the the earlier page 5-6 for connecting array "conn".

conn: $$ \begin{bmatrix} 1     & 2 \\  2     & 3 \end{bmatrix}$$

Where:
 * Row one represents element one
 * Row 2 represents element two
 * The numbers within the matrix represent the global node numbers

conn(e,j) = This is the global node number of local node "j" of element "e".

Location Matrix Master Array
The following location matrix master is for "1mm":

1mm = $$\begin{vmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 5 & 6 \end{vmatrix}$$

Where:
 * Row one represents element one
 * Column one represents local degree of freedom number one
 * Column two represents local degree of freedom number two
 * Column three represents local degree of freedom number three
 * Column four represents local degree of freedom number four
 * The numbers within the matrix represent the global degree of freedom number (or eq. number) in K

18-2 1mm(i,j) = This represents the eq. number (global degree of freedom number) to the j th local degree of freedom number for the element stiffness coefficient.

HW

 * The homework for this meeting is to solve the three bar truss system from page 16-4 by modifying the Matlab code for the two bar truss system.
 * Also, make a transformation matrix that is a 4x4 and (hopefully) invertible by transforming a four degree of freedom system to a four degree of freedom system from page 12-1.

Meeting 19: Wednesday, 8 October 2008. EML4500

 * Goal: We want to find the transformation matrix; $${\underset{4 \times 4}\mathbf \tilde T^{(e)}}$$, that transforms the set of local elemental degrees of freedom; $${\underset{4 \times 1}\mathbf d^{(e)}}$$ which is set on the cartesian cordinate system as seen on the figure below, to the another set of local elemental degrees of freedom $${\underset{4 \times 1}\mathbf \tilde d^{(e)}}$$, such that $${\underset{4 \times 4}\mathbf \tilde T^{(e)}}$$ is inversible.

$$\underset{4 \times 1}\tilde \mathbf d^{(e)}=\underset{4 \times 4}\mathbf \tilde T^{(e)}\underset{4 \times 1} \mathbf d^{(e)} $$

$$\underset{4 \times 1}\tilde \mathbf d^{(e)}=\underset{4 \times 4}\mathbf \tilde T^{(e)}\underset{4 \times 1} \mathbf d^{(e)} $$

$$\underbrace_{q_1^{(e)}} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$ {1}

$${\underset{4 \times 1}\mathbf \tilde {d^{(e)}_2}} = \begin{bmatrix} -m^{(e)} & l^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$ {2}

$$\mathbf \tilde d^{(e)}_2 = \begin{bmatrix} -sin \theta^{(e)} & cos \theta^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix} = \begin{bmatrix} -m^{(e)} & l^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$

Put {1} and {2} in matrix form:

$$\begin{pmatrix} \tilde d^{(e)}_1 \\ \tilde d^{(e)}_2 \end{pmatrix} = \underbrace{\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \\ \end{bmatrix}}_{\mathbf R^{(e)}}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$

$$\underset{4 \times 1}\tilde \mathbf d^{(e)}=\underset{4 \times 4}\mathbf \tilde T^{(e)}\underset{4 \times 1} \mathbf d^{(e)} \Rightarrow \underbrace{\begin{pmatrix} \tilde d^{(e)}_1 \\ \tilde d^{(e)}_2 \\ \tilde d^{(e)}_3 \\ \tilde d^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1}\tilde \mathbf d^{(e)}} = \underbrace{\begin{bmatrix} {\underset{2 \times 2}\mathbf R^{(e)}} & {\underset{2 \times 2}\mathbf 0^{(e)}} \\ {\underset{2 \times 2}\mathbf 0^{(e)}} & {\underset{2 \times 2}\mathbf R^{(e)}} \\ \end{bmatrix}}_{\underset{4 \times 4}\mathbf \tilde T^{(e)}}\underbrace{\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1} \mathbf d^{(e)}}$$


 * Shown in the figure below, the element is represented in matrix form below:



$$ \tilde \mathbf f^{(e)} = k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \tilde \mathbf d^{(e)}$$


 * $${\underset{4 \times 1} \tilde \mathbf f^{(e)}} ={\underset{4 \times 4}\tilde \mathbf k^{(e)}}{\underset{4 \times 1} \tilde \mathbf d^{(e)}}$$

Meeting 20: Friday, 10 October 2008. EML4500
Note: consider case: $$\tilde{d}^{(e)}_4 \neq 0$$

$$ \tilde{d}^{(e)}_1 {=} \tilde{d}^{(e)}_2 {=} \tilde{d}^{(e)}_3 {=} 0$$

$$ \underline{\tilde{f}}^{(e)}_{4x1} {=} \underline{\tilde{k}}^{(e)}_{4x4} \cdot \underline{\tilde{d}}^{(e)}_{4x1} {=} \underline{0}_{4x1} \leftarrow 4^{th}$$ column of $$\underline{\tilde{k}}^{(e)}$$

Interpretation of trans. dofs

p. 19-3: $$ \underline{\tilde{d}}^{(e)} {=} \underline{\tilde{T}}^{(e)} \cdot \underline{d}^{(e)}$$

Similarly  $$ \underline{\tilde{f}}^{(e)} {=} \underline{\tilde{T}}^{(e)} \cdot \underline{f}^{(e)}$$

Also:  $$ \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{d}}^{(e)} {=} \underline{\tilde{f}}^{(e)} $$

$$ \rightarrow \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)} \cdot \underline{\tilde{d}}^{(e)} {=} \underline{\tilde{T}}^{(e)} \cdot \underline{\tilde{f}}^{(e)}$$

If $$\underline{\tilde{T}}$$ is invertible, then:

$$ [\underline{\tilde{T}}^{(e)-1} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)}] \cdot \underline{d}^{(e)} {=} \underline{f}^{(e)} $$

$$ \underline{\tilde{T}}^{(e)}$$ block diag. matrix (p. 19-3)

Consider a general block-diag. matrix:



Question: What is A-1

Simpler example: Diag. matrix:



$$ {=} diag[d_{11}, d_{22} , d_{33} , ... , d_{nn}] $$

$$ \underline{B}^{-1} = Diag[\frac{1}{d_{11}}, \frac{1}{d_{22}} , ... , \frac{1}{d_{nn}}]$$

Assuming $$ d_{ii} \neq 0 $$ for $$ i {=} 1, ... , n$$

For a block-diag matrix A:

$$ \underline{A} {=} Diag[\underline{D}_1, ... , \underline{D}_s]$$

$$ \underline{A}^{-1} {=} Diag[\underline{D}_1^{-1}, ... , \underline{D}^{-1}_{s}]$$

$$ \underline{\tilde{T}}^{(e)-1} {=} Diag[\underline{R}^{(e) -1}, ... , \underline{R}^{(e) -1}] $$

p. 19-2: $$ \underline{R}^{(e) T} {=} \begin{vmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{vmatrix} $$

$$ \underline{R}^{(e)T}_{2x2} \cdot \underline{R}^{(e)}_{2x2} {=} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}_{2x2} {=} \underline{I}_{2x2} $$

$$ \rightarrow \underline{R}^{(e)-1} = \underline{R}^{(e)T} $$

$$ \rightarrow \underline{\tilde{T}}^{(e)-1} = Diag[\underline{R}^{(e)T}, \underline{R}^{(e)T}] $$

$$ {=} \underbrace{ (Diag[ \underline{R}^{(e)}, \underline{R}^{(e)}]) }_{\underline{\tilde{T}}^{(e)}} $$

$$ \underline{\tilde{T}}^{(e)-1} {=} \underline{\tilde{T}}^{(e)T} $$

p. 20-1: $$ \underbrace{ [\underline{\tilde{T}}^{(e)T} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)}]}_{\underline{k}^{(e)}} \cdot \underline{d}^{(e)} {=} \underline{f}^{(e)} $$

HW: Verify: $$ \underline{k}^{(e)} {=} \underline{\tilde{T}}^{(e)T} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)} $$

EML4500.f08.A-team.kirley 03:46, 21 October 2008 (UTC) 23-1

Meeting 21: Monday, 13 October 2008. EML4500
Eigenvalue problem: $$\mathbf{kv} = \lambda \mathbf{v}$$

Let $$\left\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}, \mathbf{u_4} \right\}$$ be the pure eigenvectors corresponding to the 4 zero eigenvalues:

$$\mathbf{ku_i} = 0\mathbf{u_i} = \mathbf{0}\left(i = 1,2,3,4 \right)$$

Linear combination of $$\left\{u_i, i = 1,2,3,4 \right\} $$

$$\sum_{i=1}^{4}{\alpha_{i_{1x1}}\mathbf{u}_{i_{6x1}}=:\mathbf{W}_{_{6x1}}}=\mathbf{0}_{_{1x1}}\cdot \mathbf{W}_{_{6x1}}$$

Where $$=:$$ means "equal by definition"

$$\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \equiv \mathbf{W} $$ (equal by definition)

$$ \alpha_i \Rightarrow $$(real numbers)

W is also an eigenvector corresponding to a zero eigenvalue:

$$ \mathbf{kW} = \mathbf{k}\left(\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \right) $$

$$= \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}$$

Eml4500.f08.a-team.robinson 02:58, 24 October 2008 (UTC)

Meeting 22: Wednesday, 15 October 2008. EML4500
Justification of assembly of $$ \underbrace{ element \ stiffness \ matrix }_{\underline{k}^{(e)} } $$ into  $$ \underbrace{ global \ stiffness \ matrix }_{\underline{K}} $$

e = 1,...,n (# of elements)

Consider the example of a two-bar truss.

Recall element FD relation: k(e) d(e) = f(e)

p.11-3: Euler cut principle, method 2 (equilibrium of global node 2)

p.4-2, 4-3: FBDs of element 1 and element 2, element dofs. d(e)

p.8-3: For node 2, identify global dofs to element dofs for both element 1 and 2.



$$ \sum Fx = 0 $$ = -f3(1) - f1(2) = 0

$$ \sum Fy = 0 $$ = P - f4(1) - f2(2) = 0

Next, use element FD relation k(e) d(e) = f(e)

Meeting 23: Friday, 17 October 2008. EML4500
From page 22-2:
 * Equation (1) $$ \rightarrow f_3^{1} + f_1^(2) $$ = 0
 * Equation (2) $$ \rightarrow f_4^{1} + f_2^(2) $$ = P

(1): $$ \underbrace {k_{31}^{(1)} d_1^{(1)} + k_{32}^{(1)} d_2^{(1)} + k_{33}^{(1)} d_3^{(1)} + k_{34}^{(1)} d_4^{(1)}}_{f_3^{(1)}} $$ + $$ \underbrace {k_{11}^{(2)} d_1^{(2)} + k_{12}^{(2)} d_2^{(2)} + k_{13}^{(2)} d_3^{(2)} + k_{14}^{(2)} d_4^{(2)}}_{f_1^{(2)}} $$

For the conversion between global and local nodes please refer to page 8-3 and 8-4: e.g.... $$ k_{31}^{(1)} d_1 $$

From equation (1), using the third and fourth values from the first set of brackets and the first and second values from the second set of brackets you can obtain the third row of K as seen on page 9-1. For applications of K see page 8-4 for the equation:  $$ k_{6x6}d_{6x1} = F_{6x1} $$ 23-2

Assembly of k(e), e = 1,...., ned, into global stiffness matrix K:

$$ \underset {nxn}{\mathbf K} $$ = $$ \overset {ned} {\underset {e=1}{\mathbf A}} $$ $$ {\mathbf {k}}_{nel \ x \ nel}^{(e)} $$

where:
 * "n" is the total number of global degrees of freedom before the boundary conditions are eliminated.
 * "ned" is the number of element degrees of freedom where ned << n (where "<<" means "very small compared to).
 * "A" is the assembly operator

Principle of Virtual Work
The principle of virtual work is also known as $$ \mathbf{P} \mathbf{V} \mathbf{W} $$.

Page 10-1 states that to obtain $$ \overline {\mathbf{K}_{2 \ x \ 2}} $$, the corresponding rows of boundary conditions need to be eliminated. 23-3 Page 12-2: $$ \mathbf{q}_{2x1}^{(e)} $$ = $$ \mathbf{T}_{2x4}^{(e)} \ $$ $$ \mathbf{d}_{4x1}^{(e)} $$


 * $$ \mathbf{k}^{(e)} $$ = $$ \mathbf{T}^{(e)^{T}} \ $$ $$ \hat{\mathbf{k}}^{(e)} \ $$ $$ \mathbf{T}^{(e)} $$: which is from page 14-3

Deriving FEM for PDE's
Where:
 * FEM stands for Finite Element Method
 * PDE stands for Partial Differential Equation
 * FD stands for Force Displacement

The force displacement for a spring or bar is known as kd = F


 * $$ \Rightarrow kd - F = 0 \qquad $$ - (3)


 * $$ \ \Leftrightarrow w(kd - F) = 0 \qquad $$ - for all w  - (4)

Where:
 * "$$ \Rightarrow $$" stands for implies
 * "$$ \Leftrightarrow $$" stands for equivalent

Proof:

$$ \qquad (3) \ \Rightarrow \ (4) \qquad $$ - trivial

$$ \qquad (4) \ \Rightarrow \ (3) \qquad $$ - not trivial

Since (4) is valid for all w, we can select any arbitrary w (i.e. w = 1), and the (4) becomes:


 * $$ \qquad 1.(kd - F) = 0 \ \Rightarrow \ (3) $$

MatLab Problems:

Three Bar Truss System:

The program for the graph is as follows:

Five Bar Truss Evaluation:

Eigen plot of 2-bar truss:

Contributions
--EML4500.f08.A-team.rieth 18:31, 20 October 2008 (UTC)

--EML4500.f08.A-team.kirley 03:47, 21 October 2008 (UTC)

--Eml4500.f08.a-team.robinson 01:03, 24 October 2008 (

--EML4500.f08.A-team.melvin 18:12, 24 October 2008 (UTC)

--EML4500.f08.A-team.VandenBerg 20:28, 24 October 2008 (UTC)

--EML4500.f08.A-team.morford 20:33, 24 October 2008 (UTC)