User:Eml4500.f08.qwiki.bishop/HW4

 Comment: The pictures that were previously used in this homework submission have been changed. The previous images ( and ) were not originally authored by Team Qwiki, however I did not intentionally plagiarize Team Echo's work. While working on Homework 2, both Team Qwiki and Team Echo created two images, with the same name: and . Due to the nature of MediaWiki, if two different teams create an image with the same name, only the latest version is displayed when called upon in an article. Therefore, even though Team Qwiki uploaded two of our own images (Element1.jpg and Element2.jpg) during homework 2, Team Echo's images were displayed because their upload was the most recent. In the future, we will make sure to upload files that have a unique name. For Homework 4, I created and  on October 23 because MediaWiki would not display Element1.jpg properly. Since I assumed that Element1.jpg was the work of Team Qwiki in Homework 2, I navigated to our HW2 article, right clicked on Element1.jpg and saved it to my computer. I then uploaded the image as Element1_Qwiki.jpg and Element1Qwiki.jpg, but only used Element1_Qwiki.jpg in Homework 4. The problem has been fixed and Team Qwiki files are now displayed here. In the future I will be sure to: A comparison between the current version of Homework 4 and our prior submission can be found here. Lee Harris Eml4500.f08.qwiki.harris 19:03, 12 November 2008 (UTC) =Lecture 18 - Monday, October 6, 2008=
 * Create unique names when uploading files
 * Check the file history of all images before I right click and save to my computer

Explanation of Matlab Variables
There are two arrays in finite element analysis, the connectivity array and the location matrix master array, which can be represented in Matlab as the variables 'conn' and 'lmm'.

Connectivity Array
Conn is the variable used for the Connectivity array. Each model must have a connectivity array. The array shows how the nodes are connected together. The equation for the connectivity array is: $$N_{element}\times N_{node}$$

Considering the two bar truss system from session 5, the matrix becomes: $$conn=\begin{bmatrix} 1 & 2\\ 2 & 3 \end{bmatrix}$$

In Matlab code: conn(i,j)=global node number of local node j of element i.

Location Matrix Master Array
The variable lmm can be used for the location matrix master array. The equation for the lmm array is: $$N_{element}\times N_{D.O.F.}$$

Considering the two bar truss system again, the matrix becomes: $$lmm=\begin{bmatrix} 1 & 2 & 3 & 4\\ 3 & 4 & 5 & 6 \end{bmatrix}$$

In Matlab code: lmm(i, j) = global degree of freedom number for the element stiffness coefficients corresponding to the local (jth) d.o.f. number.

Five-Bar Truss
From problem 1.4 on page 25 of the text book, the following five-bar truss is examined.

Detailed Explanation
The Matlab code starts by defining K as a 10 x 10 matrix, and R as a 10 x 1 matrix.

Next we have to input element 1's k and r matrix.

The location master matrix (lmm) is defined by the degrees of freedom: 1, 2, and 5. Thus yielding:

Next, the elements 1, 2 and 5 of the R vector must be extracted and added to the r vector.

The matrix k can be assembled into the global matrix K as follows.

Next we have to input element 2's k and r matrix.

The location master matrix (lmm) is defined by the degrees of freedom: 1, 2, and 5. Thus yielding:

Next, the elements 1, 2 and 5 of the R vector must be extracted and added to the r vector.

The matrix k can be assembled into the global matrix K as follows.

Results and Graphs
After running the above program, the results are as follows.

Three-Bar Truss

 * The force, P, has a value of 50


 * The first step in the code is to define each element's length, cross-sectional area, modulus, and k value. Then each of the element stiffness matrices are found (where km1 is the stiffnes matrix for element one, km2 for element 2, and km3 for element 3:


 * Then the global stiffness matrix, K, was found by combining all the elemental stiffness matrices according their association with the lobal nodes:


 * The displacements of the only unconstrained global node were:
 * d3=-11.6737 and d4=44.4051


 * In the plot below the the undeformed truss is denoted by the dashed line and the deformed truss is denoted by the solid line. Also, the original, undeformed global node number 2 is the origin of the plot.

=Lecture 19 - Wednesday, October 8, 2008=

Lecture Goal
The goal of this lecture is to find the matrix $$\tilde{T}_{(4x4)}^{(e)}$$  that can transform the set of local (element) degree of freedom matrix,  $$d_{(4x4)}^{(e)}$$, to another set of local (element) degree of freedom matrix, $$\tilde{d}_{(4x1)}^{(e)}$$, such that  $$\tilde{T}_{(4x4)}^{(e)}$$  is invertible.

Consider the following diagram:



The T matrix for this diagram will be found using the following expression:

$$\mathbf{\tilde{d}_{(4x1)}^{(e)}}=\mathbf{\tilde{T}_{(4x4)}^{(e)}}\mathbf{d_{(4x4)}^{(e)}}$$

First, the element degree of freedom matrix for nodes 1 and 2 must be defined:

$$\tilde{d}_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{pmatrix}$$

The R Matrix
Putting (1) and (2) into matrix form yields:

$$\begin{pmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \end{pmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{pmatrix}$$

where the matrix $$\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)}\\ \end{bmatrix}$$ equals  $$\mathbf{R^{(e)}}$$

Therefore, the following relation is true for element (e):

$$\begin{pmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \end{pmatrix}=\begin{bmatrix} \mathbf{R}^{(e)} & \mathbf{0}\\ \mathbf{0} & \mathbf{R}^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ \end{pmatrix}$$

Following the expression $$\mathbf{\tilde{d}_{(4x1)}^{(e)}}=\mathbf{\tilde{T}_{(4x4)}^{(e)}}\mathbf{d_{(4x4)}^{(e)}}$$, the T matrix is now found.

$$\mathbf{\tilde{T}_{(4x4)}^{(e)}}=\begin{bmatrix} \mathbf{R}^{(e)} & \mathbf{0}\\ \mathbf{0} & \mathbf{R}^{(e)}\\ \end{bmatrix}$$

Force Displacement Relation
The diagram can now be represented entirely in the new tilde coordinate system.



The force displacement relation can now be applied.

$$\mathbf{\tilde{f}_{(4x1)}^{(e)}}=\mathbf{\tilde{k}_{(4x4)}^{(e)}}\mathbf{\tilde{d}_{(4x1)}^{(e)}}$$

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

=Lecture 20 - Friday, October 10, 2008=

Note

 * Consider case: $$\tilde{d}^{(e)}_{4}\neq 0$$
 * $$\tilde{d}^{(e)}_{1}=\tilde{d}^{(e)}_{2}=\tilde{d}^{(3)}_{4}=0$$
 * $$\tilde{f}^{(e)}=\tilde{k}^{(e)}\tilde{d}^{(e)}$$=0
 * (4x1)=(4x4)(4x1)=(4x1)
 * The zero matrix represents the fourth column of the elemental stiffness matrix, k

T matrix application to the elemental Force matrix
\tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m_{(e)}&l^{(e)} \end{bmatrix}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{e}_{2}\\ \end{Bmatrix}=\textbf{R}^{(e)}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{(e)}_{2}\\ \end{Bmatrix}$$ \tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2}\\ \tilde{f}^{(e)}_{3}\\ \tilde{f}^{(e)}_{4} \end{Bmatrix}=\begin{bmatrix} \mathbf{R}^{(e)}_{2x2} & \mathbf{0}_{2x2}\\ \mathbf{0}_{2x2} & \mathbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{(e)}_{2}\\ {f}^{(e)}_{3}\\ {f}^{(e)}_{4} \end{Bmatrix}\Rightarrow \tilde{\mathbf{f}}^{(e)}_{4x1}=\tilde{\mathbf{T}}^{(e)}_{4x4}\mathbf{f}^{(e)}_{4x1}$$
 * p.19-3, $$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$
 * Similarly, $$\tilde{f}^{(e)}=\tilde{T}^{(e)}f^{(e)}$$
 * $$\begin{Bmatrix}
 * $$\begin{Bmatrix}
 * The set of R and zero matrices can be represented by the T matrix.
 * If we look at the force-displacement relationship, we can apply the T matrix conversion between global and axial coordiantes.
 * $$\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{d}}^{(e)}=\tilde{\mathbf{f}}^{(e)}\Rightarrow \tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}\mathbf{d}^{(e)}=\tilde{\mathbf{T}}^{(e)}\mathbf{f}^{(e)}$$
 * But if the T matrix can be inverted, then:
 * $$[\tilde{\mathbf{T}}^{(e) -1}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

General block diagonal matrix
d_{11}&0 &0 \\ 0& d_{22}&0 \\ 0& 0 &d_{33} \end{bmatrix}=diag[d_{11}, d_{22}, d_{33}]$$ l^{(e)} & -m^{(e)} \\ m_{(e)}&l^{(e)} \\ \end{bmatrix}$$ 1 & 0 \\ 0&1 \\ \end{bmatrix}=\mathbf{I}$$
 * $$\tilde{\mathbf{T}}^{(e)}$$ is a block diagonal matrix.
 * To denote a block diagonal matrix in shorthand use the following notation:
 * $$\mathbf{B}=\begin{bmatrix}
 * The inverse of B is:
 * $$\mathbf{B^{(-1)}}=diag[\frac{1}{d_{11}},\frac{1}{d_{22}}, \frac{1}{d_{33}}]=diag[d_{11}^{(-1)}, d_{22}^{(-1)}, d_{33}^{(-1)}]$$
 * Assuming that no value of d is zero
 * In applying this to the T-matrix:
 * $$\tilde{\mathbf{T}}^{(e)-1}=diag[\mathbf{R}^{(e)-1},\mathbf{R}^{(e)-1}]$$
 * In reference to p.19-2:
 * $$\mathbf{R}^{(e)T}=\begin{bmatrix}
 * In order to prove that the transpose of R is equivalent to the inverse of R:
 * $$\mathbf{R}^{(e)T}\mathbf{R}^{(e)}=\begin{bmatrix}
 * So, $$\mathbf{R}^{(e)T}=\mathbf{R}^{(e)-1}$$
 * The inverse of the T-matrix can be written as:
 * $$\tilde{\mathbf{T}}^{(e)-1}=diag[\mathbf{R}^{(e)T},\mathbf{R}^{(e)T}]=\tilde{\mathbf{T}}^{(e)T}$$

Force-Displacement Relationship with T

 * $$[\tilde{\mathbf{T}}^{(e)T}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$
 * To show that $$[\tilde{\mathbf{T}}^{(e)T}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]=\mathbf{k}^{(e)}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l^{(e)} &-m^{(e)} & 0 &0 \\ m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&-m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}\begin{bmatrix} 1 & 0 &-1 &0 \\ 0& 0 & 0 & 0\\ -1&0  &1  &0 \\ 0&  0&  0& 0 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & 0 &0 \\ -m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l &0 &-l  & 0\\ m& 0& -m & 0\\ -l & 0& l & 0\\ -m& 0 & m&0 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & 0 &0 \\ -m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l^2 &lm &-l^2  & -ml\\ ml& m^2& -ml & -m^2\\ -l^2 & -lm& l^2 & lm\\ -ml& -m^2 & ml&m^2 \end{bmatrix}$$
 * So, the relationship between the elemental k matrix in global coordinates and the k matrix in axial coordinates using the T matrix is confirmed.

=Lecture 21 - Monday, October 13, 2008=

Comments on Homework 3: Using infinitesimal displacement to close the loop as previously mentioned. The zero eigenvalues from the eigenvalue problem refer to the rigid body motion and mechanisms of the system. Three zero eigenvalues correspond to the rigid body motion including and a x-axis displacement, a y-axis displacement, and a rotation. The fourth zero eigenvalue corresponds to mechanisms of the body.

Homework 4: Plot the eigenvectors corresponding to the zero eigenvalues of the two-bar-truss system, and interpret the results (mode shapes). These mode shapes, corresponding to zero eigenvalues, may come out as a linear combination of the pure mode shapes.



In the above diagram a = b = 1 m, and for case (a) E = 2 and A = 3.

Case (a) will be solved using the equation: $${\mathbf{\bar{k}}}\mathbf{v}=\lambda \mathbf{v}$$

=Lecture 22 - Wednesday, October 15, 2008=

Formal Justification of Assembly Process of Element Stiffness Matrix into Global Stiffness Matrix
For a formal justification, we will consider the two-bar truss discussed in meeting 9. We must also recall that the Force-Displacement relationship for each element is $$\mathbf{k}^{(e)}_{(4x4)}\mathbf{d}^{(e)}_{(4x1)}=\mathbf{f}^{(e)}_{(4x1)}$$. This relationship allows us to compute the nodal displacements and forces, unlike statics which only considers element displacements and forces. Therefore, if we follow the Finite Element Method we can isolate and analyze each node and element in the system according to the Euler Cut Principle.

In terms of the DOFs and their naming for this justification, we refer to the Free Body Diagrams of Element 1 and 2 from Meeting 4. Notice that global node 2 DOFs correspond to node 2 of element 1 and node 1 of element 2.


 * Global Node 1:
 * $$d_1=d_1^{(1)}$$
 * $$d_2=d_2^{(1)}$$


 * Global Node 2:
 * $$d_3=d_3^{(1)}=d_1^{(2)}$$
 * $$d_4=d_4^{(1)}=d_2^{(2)}$$


 * Global Node 3:
 * $$d_5=d_3^{(2)}$$
 * $$d_6=d_4^{(2)}$$

Since global node 2 is the intersecting node of element 1 and 2, we can examine the equilibrium of global node 2



The next step of the justification process is to invoke the element FD relationship $$\mathbf{k}_{(4x4)}^{(e)}\mathbf{d}_{(4x1)}^{(e)} = \mathbf{f}_{(4x1)}^{(e)}$$ for each element. We will replace each internal element force in equations 1 and 2 shown above with its respective internal displacements, where $$f_i^{(e)} = \sum_{j=1}^{4}{k_{ij}^{(e)}d_{j}^{(e)}}$$
 * $$\sum{F}_x = -f_3^{(1)} - f_1^{(2)} = 0$$ Eqn. 1
 * $$\sum{F}_y = \mathbf{P} - f_1^{(4)} - f_2^{(2)} = 0$$   Eqn. 2

=Lecture 23 - Friday, October 17, 2008=

From Lecture on 10/15/2008

Equation 1

$$f_3^{(1)} + f_1^{(2)} = 0$$

Equation 2

$$f_4^{(1)} + f_2^{(1)} = \underline{P}$$

and

$$f_3^{(1)} = k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)} + k_{33}^{(1)}d_3^{(1)} + k_{34}^{(1)}d_4^{(1)}$$

$$f_1^{(2)} = k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)} + k_{13}^{(2)}d_3^{(2)} + k_{14}^{(2)}d_4^{(2)}$$

so. ..

$$(1): [k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)} + k_{33}^{(1)}d_3^{(1)} + k_{34}^{(1)}] + [k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)} + k_{13}^{(2)}d_3^{(2)} + k_{14}^{(2)}d_4^{(2)}] = 0$$

Thus we obtain the 3rd row of K, the golobal stiffness matrix

HW: Write details of Eq.2 and the assembly of row 4 in $$\underline{K}_{6x6}\underline{d}_{6x1}=\underline{F}_{6x1}$$

$$f_4^{(1)} = k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}$$

$$f_2^{(2)} = k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}$$

so. ..

$$(2): [k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}] + [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}] = 0$$

Assembly of $$K^{(e)}$$ where e goes from 1 to the number of elements. into the global stiffness matrix $$\underline{K}$$

$$K_{nxn} = A_{e=1}k^{(e)}$$

n = the total number of global degree of freedoms before eliminating boundary conditions

n$$_{ed} $$= the number of element degree of freedoms

$$n_{ed} << n $$

A = assembly operator

Principle of Virtual Work (PVW)

Elimination of rows corresponding to boundary conditions to obtain $$\underline{K}$$

$$q_{2x1}^{(e)} = T_{2x4}^{(e)}d_{4x1}^{(e)}$$

$$k^{(e)} = T^{(e)T}\hat{k}^{(e)}T^{(e)}$$

Deriving FEM for Partial Differential Equations (PDE)

FD relation for a bar or spring: Kd=F

implies. ..

$$Kd-F = 0 (3)$$

equivalently. ..

$$w(Kd-F) = 0$$ (for all w) (4)

=Contributing Team Members= The following students contributed to this report
 * Chris Bishop Eml4500.f08.qwiki.bishop 18:51, 23 October 2008 (UTC)
 * Michael Berry Eml4500.f08.qwiki.berry 20:04, 23 October 2008 (UTC)
 * Andrea Booher EML4500.f08.qwiki.booher 00:13, 24 October 2008 (UTC)
 * Lee Harris Eml4500.f08.qwiki.harris 02:03, 24 October 2008 (UTC)
 * Eric Gatch Eml4500.f08.qwiki.gatch 13:51, 24 October 2008 (UTC)
 * David Nobles Eml4500.f08.qwiki.nobles 14:00, 24 October 2008 (UTC)