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User:Eml4500.f08.bottle.butler/HW 4

Before we continue with our analysis of the 3-bar truss system we will make a quick comment on modifying a 2-bar truss program in MATLAB to solve a 3-bar truss system.

First, we define the connection array as "conn". Consider the 2-bar truss system:

$$ conn = \begin{bmatrix} 1 & 2\\ 2& 3 \end{bmatrix}$$ where column 1 corresponds to local node $$ 1 \! $$ and column 2 corresponds to local node $$ 2 \! $$. Similarly, row 1 represents element $$ 1 \! $$ and row 2 represents element $$ 2 \! $$. This array can then be represented by conn(e,j) = global node # of local node $$ j \! $$ of element $$ e \! $$

We denote the location master array by "lmm". Therefore

$$ lmm = \begin{bmatrix} 1 & 2 & 3 & 4\\ 3& 4 & 5 & 6 \end{bmatrix} $$

where row 1 represents element $$ 1 \! $$ and row 2 represents element $$ 2 \! $$, each column represents a local degree of freedom number with each number in the matrix representing the global degree of freedom number (or equation number) in $$ K \! $$.

This can be rewritten as limm(i,j)= equation number (global degree of freedom) for the element stiffness matrix coefficient corresponding to the $$jth \! $$ local degree of freedom number

Returning to our analysis of FEA, our goal is to find $$ \tilde{T}^{(e)}_{4X4} $$ that transforms the set of local element dof's $$ d^{(e)}_{4X1} $$ to another set of local element $$ \tilde{d}^{(e)}_{4X1}$$ such that $$ \tilde{T}^{(e)} $$ is invertible.

image

From previous lectures we know that:

$$ \underbrace {\tilde{d}_{1}^{(e)}}_{q_{1}^{(e)}} = \begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

and $$ \tilde{d}_{2}^{(e)}= \vec{d}^{(e)}_{1}\circ \vec{\tilde{j}} $$ $$= -\sin \theta ^{(e)}d_{1}^{(e)}+\cos \theta ^{(e)}d_{2}^{(e)}$$

therefore:

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

If we combine the two we get the following equation in matrix form: $$ \begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

where $$ \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} = R^{(e)} $$

This can be expanded to : $$\underbrace{\begin{Bmatrix} \tilde{d}^{(e)}_{1}\\ \tilde{d}^{(e)}_{2}\\ \tilde{d}^{(e)}_{3}\\ \tilde{d}^{(e)}_{4} \end{Bmatrix}}_{\tilde{d}^{(e)}_{4X1}}= \underbrace{\begin{bmatrix} R^{(e)} & 0\\ 0 & R^{(e)} \end{bmatrix}}_{\tilde{T}^{(e)}_{4X4}}\underbrace{\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}}_{d^{(e)}_{4X1}}$$

Examining the element below

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

where $$ \tilde{f}^{(e)}_{4X1}=\tilde{k}^{(e)}_{4X4}\tilde{d}^{(e)}_{4X1} $$

In previous meetings we found the eigenvalues associated with a 2-bar truss system. Four eigenvalues were found, three of which were due to rigid body motion and the forth resulting from mechanisms. To understand the concept of a mechanism, we imagine that the support on a 2-bar truss system is cut and the structure can now move freely. The two bars can now move in a "scissor" motion about the pin joint. This is possible because there is no stored energy in the pin.

For the eigenvalue problem we have: $$ Kv=\lambda v \! $$, where $$ Kv $$ is the mass matrix and $$ \lambda v $$ is the identity matrix.

Let $$ \begin{Bmatrix} (u_{1})_{6X1} & (u_{2})_{6X1} & (u_{3})_{6X1} & (u_{4})_{6X1} \end{Bmatrix} $$ be pure eigenvalues corresponding to the four eigenvalues.

$$ K_{6X6}(u_{i})_{6X1}= \underbrace {0 u_{i}}_{0_{6X1}}, i = 1, ... ,4 $$

We can write the linear combination of $$ \begin{Bmatrix} u_{i}, i = 1, ... , 4 \end{Bmatrix}$$ as $$ \Sigma (\alpha _{i})_{1X1}(u_{i})_{6X1} = W_{6X1} $$

where $$\alpha _{i} = real numbers $$

Physically this means that the linear combination of movement in the  x \! and $$ y \! $$ directions and rotation.

$$ W \! $$ is also an eigenvalue that corresponds to a zero eigenvalue $$ KW=K(\sum_{i = 1}^{4} \alpha_{i}u_{i} ) $$ $$= \sum_{i = 1}^{4}{\alpha _{i}\underbrace{(Ku_{i})}_{0_{6X1}}} = 0 $$ $$ = 0_{1x1} W_{6X1} \!$$ We will now begin the justification of the assembly of the elemental stiffness matrix $$ k^{(e)} \!$$ into the global stiffness matrix.

Consider the example of the 2-bar truss and recall the following: 1. The elemental FD relationship can be written as $$ k^{(e)}d^{(2)}=f^{(e)} \!$$ 2. We used the Euler cut principle to find the equivalent of global node $$ 2 \! $$ 3. We drew the FBD's of element $$ 1 \! $$ and $$ 2 \! $$ to find the elemental degrees of freedom $$ d^{(e)}_{4X1} $$ 4. We identified the global degrees of freedom to elemental degrees of freedom for node $$ 2 \! $$  for element  $$ 1 \! $$ and $$ 2 \! $$

We can draw the following FBD to represent the equilibrium for node $$ 2 \! $$

We can now see that the addition of factors in the global stiffness matrix is because of the equilibrium position of node $$ 2 \! $$. The equations of equilibrium are $$ \sum{F_{x}}= 0 = -f^{(1)}_{3}-f^{(2)}_{1}$$ $$ \sum{F_{y}}= 0 = P-f^{(1)}_{4}-f^{(2)}_{2} $$

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User:Eml4500.f08.ateam/hw 4

A-Team Teampage

Homework Report #4

This page is intended for Homework Report #4 of the A-Team.

Due date: Friday, 24 October 08, 5 pm EDT = 17:00 EDT (or 21:00 UTC).

Connectivity and LMM Arrays

From the diagram of the two bar truss above, we can see the relationships between the global and local nodes of elements 1 and 2. From these relationships, we can derive the connectivity array. In MATLAB, this is denoted as. In this array, the rows represent the elements (e) and the columns represent the local node numbers (j). The values inside the array are the global node numbers. Using the two bar truss diagram above, the connectivity array becomes:

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

For example, in this case the local node of element 1 corresponds to the global node 1. Therefore, the value of in the first row and column is 1.

Another array that we will need to derive is the location matrix master array or "LMM". This array, will represent the relationships between the local degrees of freedom (DOF's) and the global DOF's numbers. As in the connectivity array, the rows in the LMM represent the elements (e), but the columns represent the local DOF's (j). In MATLAB this command is.

The above diagrams show the elements of the two bar truss and their local DOF's. From these diagrams, we will populate our location matrix master array. The values in the array correspond to the global DOF's. The array then becomes:

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

Transforming Local DOFs to Another Local Set
The goal of this next step is to find $$ \tilde{T}_{4x4}^{(e)}$$ that transforms the set of local, or elemental, degrees of freedom, $${d_{4x1}^{(e)}}$$, to another set of local degrees of freedom, $$ \tilde{d}_{4x1}^{(e)}$$, so that the matrix $$ \tilde{T}_{4x4}^{(e)}$$ is invertible.

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

$$ \tilde{d}_1^{(e)} = [ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$

The right side of the equation has been seen before in previous lectures when trying to find the axial displacement, or the orthogonal projection, of the node. That means we can also write the equation as:

$$ {q}_1^{(e)} = [ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$

Derivations
Derivation of $$ {d_1^{(e)}} $$:

$$\tilde {d}_1^{(e)} = {\vec d_1^{(e)}} \cdot \vec \tilde{i} $$

$$ {d_1^{(e)}} = d_1^{(e)} \vec i + d_2^{(e)} \vec j $$.

$$\tilde {d_1^{(e)}} = (d_1^{(e)} \tilde{i} + d_2^{(e)} \tilde{j}) \cdot \vec \tilde{i} $$

$$\tilde {d_1^{(e)}} = d_1^{(e)} (\vec i \cdot \vec \tilde{i}) + d_2^{(e)} (\vec j \cdot \vec \tilde{i}) $$

Where:

$$ l^{(e)} = {\tilde{i}} \cdot {i} = cos(\theta^{(e)}) $$

$$ m^{(e)} = {\tilde{i}} \cdot {j} = sin(\theta^{(e)}) $$

Therefore:

$$ \tilde{d}_1^{(e)} = [ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$ (1)

This process is similar when finding $$ \tilde{d}_2^{(e)}$$. The only difference is that the components of $$ \overrightarrow {d}_1^{(e)}$$ are found along the $$ \tilde {j}$$ or the $$ \tilde {y} $$ axis.

Derivation of $$ {d_2^{(e)}} $$:

$$ {d_1^{(e)}} = d_1^{(e)} \vec i + d_2^{(e)} \vec j $$.

$$\tilde {d}_2^{(e)} = \vec d_1^{(e)} \cdot \vec \tilde{j} $$

$$\tilde {d}_2^{(e)} = (d_1^{(e)} \tilde{i} + d_2^{(e)} \tilde{j}) \cdot \vec \tilde{j} $$

$$\tilde {d}_2^{(e)} = d_1^{(e)} (\vec i \cdot \vec \tilde{j}) + d_2^{(e)} (\vec j \cdot \vec \tilde{j}) $$

$$(\vec i \cdot \vec \tilde{j}) = -sin(\theta^{(e)})$$ and

$$(\vec j \cdot \vec \tilde{j}) = cos(\theta^{(e)})$$

$$\tilde {d}_2^{(e)} = -m^{(e)} d_1^{(e)} +  l^{(e)} d_2^{(e)} $$

$$\tilde {d}_2^{(e)} = [ \begin{matrix} -m^{(e)} & l^{(e)} \end{matrix} ] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$ (2)

Equations (1) and (2) can now be put into a matrix form

$$ \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} \\  -m^{(e)}& l^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix} $$

where $$ \left[ \begin{matrix} l^{(e)} & m^{(e)} \\  -m^{(e)}& l^{(e)} \end{matrix} \right] $$ can be written as $$ {R_{2x2}^{(e)}} $$

Now the equation $$ \tilde{d}_{4x1}^{(e)} = \tilde{T}_{4x4}^{(e)} {d_{4x1}^{(e)}}$$ can be formed from these equations.

$$ \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)} \\\tilde{d}_4^{(e)} \end{Bmatrix} = \left[ \begin{matrix} R_{2x2}^{(e)} & 0_{2x2} \\  0_{2x2}& R_{2x2}^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$

Transformed Forces and Degrees of Freedom
The element above has been rotated for an easier understanding.

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

If you try to move this node transversely, nothing will happen. That is why columns 2 and 4 are all zeros, because they do not stretch the string.

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

This gives us the final equation of:

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

Changing global degrees of freedom to a local coordinate system aligned along the truss
The tilde vectors represent the axial and normal forces from the axis of the bar. The same equation is used to start the derivation.

$$ \tilde\mathbf {f}^{(e)}= \tilde\mathbf {k}^{(e)} \tilde\mathbf {d}^{(e)} = 0 $$

A transformation vector will be used to get $$ \tilde\mathbf {d} $$ as well as $$ \tilde {f} $$.

$$ \tilde\mathbf {d}^{(e)} = \tilde\mathbf {T}^{(e)} \mathbf{d}^{(e)} $$

$$ \tilde\mathbf {f}^{(e)} = \tilde\mathbf {T}^{(e)} \mathbf{f}^{(e)} $$

By substituting the values for $$ \tilde \mathbf{T} $$ it can be proven.

$$ \tilde \mathbf{T} = \begin{bmatrix} cos (\Theta) & sin (\Theta) & 0 & 0\\ -sin (\Theta) & cos (\Theta) & 0 & 0\\0 & 0 & cos (\Theta) & sin (\Theta)\\0 & 0 & -sin (\Theta) & cos (\Theta)\end{bmatrix}=\begin{bmatrix} \mathbf{R}^{(e)} & \mathbf{0}\\\mathbf{0} & \mathbf{R}^{(e)}\end{bmatrix}$$

$$ \tilde \mathbf{T}^{(1)} \mathbf{f}^{(1)}= \begin{bmatrix} cos (30) & sin (30) & 0 & 0\\ -sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & sin (30)\\0 & 0 & -sin (30) & cos (30)\end{bmatrix}\begin{bmatrix}-4.378\\-2.5622\\4.4378\\2.5622\end{bmatrix} = \tilde\mathbf{f}^{(1)}$$

$$\tilde \mathbf{f}^{(1)} = \begin{bmatrix} -5.1243\\0\\5.1243\\0\end{bmatrix} $$

This confirms the previous statics work for $$ \tilde \mathbf{f}^{(1)} $$.

By substituting the previous relationships into $$ \tilde \mathbf{f}^{(e)}= \tilde \mathbf{k}^{(e)} \tilde \mathbf{d}^{(e)}$$, this equation can be rewritten as follows.

$$ \tilde \mathbf{k}^{(e)} \tilde \mathbf{T}^{(e)}{d}^{(e)}= \tilde \mathbf{T}^{(e)} \tilde \mathbf{f}^{(e)}$$

Verify transformation matrix transpose theorem
If the transformation matrix, $$\tilde \mathbf{T}^{(e)}$$, is invertible then the equation can be changed into the following form.

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

Now since only a square matrix can be inverted, the following process was investigated. Since $$\tilde \mathbf{T}^{(e)}$$ is composed of R matrices which are sin and cos functions, when the transpose of $$\tilde \mathbf{R}^{(e)}$$ is multiplied by $$\tilde \mathbf{R}^{(e)}$$, the result is the identity matrix.

$$ \mathbf {R}^{(e)T} \ \mathbf {R}^{(e)} = \begin{bmatrix} \sin^2 {\Theta} + \cos^2 {\Theta} & 0 \\ 0 & \sin^2 {\Theta} + \cos^2 {\Theta} \end{bmatrix}=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$

Therefore if both sides are divided by $$ \mathbf {R}^{(e)}$$, then the following relationship can be made.

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

This simplifies the math involved significantly and allows for rectangular matrices to be used instead of solely square matrices. Since $$\tilde \mathbf{T}^{(e)}$$ is comprised of a diagonal $$\tilde \mathbf{R}^{(e)}$$ matrix, the previous relationship between the transpose and inverse can be applied to $$\tilde \mathbf{T}^{(e)}$$ as well.

To verify that $$ [\tilde \mathbf{T}^{(e)-1} \ \tilde \mathbf{k}^{(e)} \ \tilde\mathbf {T}^{(e)}] $$, the values from the 2-bar truss will be inserted. The first element's values will be used.

$$ [\tilde \mathbf{T}^{(e)-1} \ \tilde \mathbf{k}^{(e)} \ \tilde\mathbf {T}^{(e)}] = \begin{bmatrix} cos (30) & -sin (30) & 0 & 0\\ sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & -sin (30)\\0 & 0 & sin (30) & cos (30)\end{bmatrix}\begin{bmatrix}0\\0\\4.4352\\6.1271\end{bmatrix} \begin{bmatrix} cos (30) & sin (30) & 0 & 0\\ -sin (30) & cos (30) & 0 & 0\\0 & 0 & cos (30) & sin (30)\\0 & 0 & -sin (30) & cos (30)\end{bmatrix}$$

$$= \begin{bmatrix} 0.5625 & 0.325&-0.5625&-0.325\\0.325&0.1875&-0.325&-0.1875\\-0.5625&-0.325&0.5625&0.325\\-0.325&-0.1875&0.325&0.1875\end{bmatrix}= \mathbf{k}^{(1)}$$

This verifies the derivation made earlier.

Eigenvalues of the Matrix
When the eigenvalues of the 6x6 matrix $$ \underline{K}$$ are calculated, there are 4 zero values. Three of these values correspond to the three rigid body motions and the other corresponds to the mechanism.

Eigenvalue problem: For this problem we will use the following equation: $$ \underline{K}\underline{v}=\lambda\underline{v} $$

For this matrix, we will let {$$ \underline{u}_{1}, \underline{u}_{2}, \underline{u}_{3}, \underline{u}_{4} $$} be the pure eigenvectors corresponding to the four zero eigenvalues. The equation above now becomes:

$$ \underline{K}\underline{u}_{i}=0\cdot\underline{u}_{i}=\underline{0} $$

Here $$ \underline{K}$$ is a 6x6 matrix, $$ \underline{u}$$ is a 6x1 matrix, and i=1,...,4.

We can now compute a linear combination on the values of $$ \underline{u}_{i}, i=1,...,4$$ with a summation:

$$ \sum_{i=1}^4 \alpha_{i} \underline{u}_{i} =: \underline{W} $$.

In this summation, $$ \alpha_{i}$$ is a 1x1 matrix, $$ \underline{u}_{i}$$ is a 6x1 matrix, $$ \underline{W}$$ is a 6x1, and $$ \alpha_{i}$$ are real numbers. The $$ =$$ sign here means "equal by definition" and $$ \underline{W}$$ is a zero eigenvector corresponding to a zero eigenvalue. Our original equation for the eigenvalue problem now becomes:

$$ \underline{K}\underline{W}=\underline{K}(\sum_{i=1}^4\alpha_{i} \underline{u}_{i})=\sum_{i=1}^4 \alpha_{i}(\underline{K}\underline{u}_{i})=\underline{0}=0\cdot\underline{W} $$

Using the equation $$ \underline{K}\underline{v}=\lambda\underline{v} $$ we will now evaluate a three and four bar truss.

Given the values in the diagram, we can compute the stiffness matrix $$ \underline{K}$$.

HW: Plot the eigenvectors corresponding to the zero eigenvalues of the two-bar truss system.

For the two bar truss system, our matrix $$ \underline{K}$$ is defined as:

With this matrix, we can now plot the eigenvalues and eigenvectors using the Matlab code [V,D]=eig(K). This will display the eigenvectors (V) and eigenvalues (D):

HW2: Plot the eigenvectors corresponding to the zero eigenvalues of the three-bar truss system above. The matrix $$ \underline{K}$$ has been determined to be:

Using the same command from earlier ([V,D]=eig(A)) we can determine the eigenvalues and eigenvectors of this matrix:

Justification of Assembly of Elemental Stiffness Matrices into the Global Stiffness Matrix
k(e) = element stiffness matrix k = global stiffness matrix

Consider the example of the 2 bar truss: Recall the element force displacement (FD) relationship:

$$k^{(e)}*d=f^{(e)}$$

Use the Euler cut principle (Method 2) to solve for the forces. The Euler cut principle states that node 2 should be in equilibrium. We take this and solve for the forces in each of the 2 bars.

Free Body Diagrams of elements 1 and 2.

The element degrees of freedom are given in the matrix: d(e). This is a 4x1 matrix.

For node 2, identify the global degrees of freedom to element degrees of freedom for both of the bars.

Now we can use the equilibrium concept from the Euler cut method to solve for the forces.

Sum of forces in x direction: F = 0 = -f(1)3 - f(2)1 Sum of forces in y direction: F = 0 = P -f (1)4 - f(2)2

Next we use the element force displacement relationship to obtain the displacement of node 2:

$$k^{(e)}*d=f^{(e)}$$

The free body diagram at Node 2 yields the following force balance equations.

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

$$f_4^{(1)}+f_2^{(2)}=\mathbf{P}\rightarrow (2)$$

Next, the elemental Force-Displacement relation,

$$k^{(e)}d^{(e)}=f^{(e)}$$

can be used to rewrite Equation (1) as follows.

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

Similarly, Equation (2) can be written as follows.

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

Now, Equations (1) and (2) can be transformed into the global coordinate system as follows.

$$(1)\rightarrow [k_{31}^{(1)}d_1+k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4]+ [k_{11}^{(2)}d_3+k_{12}^{(2)}d_4+k_{13}^{(1)}d_5+k_{14}^{(1)}d_6]$$

$$(2)\rightarrow [k_{41}^{(1)}d_1+k_{42}^{(1)}d_2+k_{43}^{(1)}d_3+k_{44}^{(1)}d_4]+ [k_{21}^{(2)}d_3+k_{22}^{(2)}d_4+k_{23}^{(1)}d_5+k_{24}^{(1)}d_6]$$

Now, the elemental stiffness matrices, k(e), can be assembled into the global stiffness matrix, K, through

$$\mathbf{K}=\mathbf{A}\mathbf{k}$$

where K is an n x n matrix, A is nel x 1 matrix and k is an ned x ned matrix. Here, n is the total number of degrees of freedom, nel is the total number of elements, ned is the total number of elemental degrees of freedom, and A is the assembly operator.

As we continue to justify the assembly of the elemental stiffness matrices into the global stiffness matrix, we must use the Principal of Virtual Work, as described previously, to obtain $$\mathbf{\overline{K}}$$ by eliminating the corresponding boundary conditions.

Once the Principal of Virtual Work is applied, the resulting elemental axial force equation and the elemental elemental stiffness equation can be written as follows.

$$\mathbf{q^{(e)}}=\mathbf{T^{(e)}}\mathbf{d^{(e)}}$$

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

Now, it is important to derive the Finite Element Method for Partial Differential Equations. Consider the Force-Displacement relation for a bar or spring with

$$kd=F$$

This implies that

$$kd-F=0\rightarrow (3)$$

or equivalently

$$W(kd-F)=0\rightarrow (4)$$

for all W.

Upon investigation, $$(3)\Rightarrow(4)$$ is trivial, but $$(4)\Rightarrow(3)$$ is not trivial.

Since (4) is valid for all W, select W=1. Now, (4) becomes $$1(kd-F)=0\Rightarrow(3)$$.

Five-Bar Truss System: MATLAB Dissection
Analysis of the following five-bar truss system MATLAB code is important before attempting to write code for the three-bar truss system. The following code is from the  file found at the website for the class textbook, Fundamental Finite Element Analysis and Applications.

Additional Functions
The following functions are essential for the MATLAB program above to run. They are also found at the course textbook website. These functions include,  , and. The details of these functions are outlined below.

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

MATLAB Results
The following is a printout of the exact results of the MATLAB analysis. It includes the complete Global Stiffness matrix (K), the Applied Force matrix (R), the Displacement matrix (d), the reaction forces, and the resulting stresses and strains.

K = Columns 1 through 6 32600       76067       -32600       -76067            0            0        76067  2.9749e+005       -76067 -1.7749e+005            0    -1.2e+005 -32600      -76067  2.4309e+005  1.1914e+005       -32998        32998 -76067 -1.7749e+005 1.1914e+005  2.4309e+005        32998       -32998 0           0       -32998        32998    1.53e+005       -32998 0   -1.2e+005        32998       -32998       -32998    1.53e+005 0           0 -1.7749e+005       -76067    -1.2e+005            0 0           0       -76067       -32600            0            0

Columns 7 through 8 0           0            0            0 -1.7749e+005       -76067 -76067      -32600    -1.2e+005            0 0           0  2.9749e+005        76067 76067       32600

R = 0          0           0     -150000           0           0           0           0

d = 0           0      0.53895     -0.95306       0.2647      -0.2647            0            0

reactions = 54927 1.5993e+005 -54927     -9926.7

results = -0.0001743     -34.859 -1.3944e+005 -3.15e-005     -6.2999       -25200 -5.2941e-005     -10.588       -31764 -5.2941e-005     -10.588       -31764 0.00032087      22.461        44922

The d matrix represents the displacements at each node.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\ 0 \\ 0.53895 \\ -0.95306 \\ 0.2647 \\ -0.2647 \\ 0 \\ 0 \\ \end{Bmatrix}$$

Reaction forces at Global Node 1 and Global Node 4:

Resulting Stresses and Strains:

Three-Bar Truss System: MATLAB Dissection
The following code solves the three-bar truss system presented in class and shown below. The two-bar truss system code provided by Dr. Vu-Quoc on the EML4500 website was modified to account for the additional element. The original code can be found here. After analysis, the resulting deformation was scaled by a factor of 1/500 for presentation.

MATLAB Results
n_nodes =

4

n_elem =

3

k =

0.9     0.51962         -0.9     -0.51962      0.51962          0.3     -0.51962         -0.3         -0.9     -0.51962          0.9      0.51962     -0.51962         -0.3      0.51962          0.3

k =

0.6     0.34641         -0.6     -0.34641      0.34641          0.2     -0.34641         -0.2         -0.6     -0.34641          0.6      0.34641     -0.34641         -0.2      0.34641          0.2

k =

0.3         0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3          0.3          0.3

K =

Columns 1 through 6

0.9     0.51962         -0.9     -0.51962            0            0      0.51962          0.3     -0.51962         -0.3            0            0         -0.9     -0.51962          1.8        1.166         -0.6     -0.34641     -0.51962         -0.3        1.166          0.8     -0.34641         -0.2            0            0         -0.6     -0.34641          0.6      0.34641            0            0     -0.34641         -0.2      0.34641          0.2            0            0         -0.3         -0.3            0            0            0            0         -0.3         -0.3            0            0

Columns 7 through 8

0           0            0            0         -0.3         -0.3         -0.3         -0.3            0            0            0            0          0.3          0.3          0.3          0.3

R =

0    0     0    30     0     0     0     0

d = 0           0      -435.17       671.77            0            0            0            0

reactions = 42.588      24.588       28.392       16.392      -70.981      -70.981

Unstable Three-Bar Truss: MATLAB Dissection
The following code is the three-bar truss system code modified for the unstable three-bar truss system shown at the left of the figure below. To stabilize this truss, an additional member is added at shown at the right of the figure. A force, P, of magnitude 0.5 is applied as shown to evaluate the truss system stability.

Breakdown of FEM
Running this MATLAB code results in a breakdown of the FEM. The system above is unstable and results in a singular stiffness matrix. FEM cannot be used to solve for the displacements and forces at each node. This breakdown is shown below.

EDU>> three_unstable

n_nodes =

4

n_elem =

3

k =

0    0     0     0     0     6     0    -6     0     0     0     0     0    -6     0     6

k =

6    0    -6     0     0     0     0     0    -6     0     6     0     0     0     0     0

k =

0    0     0     0      0     6     0    -6      0     0     0     0      0    -6     0     6

K =

0    0     0     0     0     0     0     0      0     6     0    -6     0     0     0     0      0     0     6     0    -6     0     0     0      0    -6     0     6     0     0     0     0      0     0    -6     0     6     0     0     0      0     0     0     0     0     6     0    -6      0     0     0     0     0     0     0     0      0     0     0     0     0    -6     0     6

R =

0     0      0      0      1      0      0      0

Warning: Matrix is singular to working precision. > In NodalSoln at 12 In three_unstable at 67

d =

0     0    Inf NaN NaN NaN 0     0

reactions =

NaN NaN NaN NaN

Stabilization Due to Additional Member
To add the cross-bar member to the truss the MATLAB code is modified to the following.

This code produces the following results.

EDU>> three_stable

n_nodes =

4

n_elem =

4

k =

2.2496e-032 3.6739e-016 -2.2496e-032 -3.6739e-016 3.6739e-016           6 -3.6739e-016           -6 -2.2496e-032 -3.6739e-016 2.2496e-032  3.6739e-016 -3.6739e-016          -6  3.6739e-016            6

k =

6           0           -6            0            0            0            0            0           -6            0            6            0            0            0            0            0

k =

0    0     0     0     0     6     0    -6     0     0     0     0     0    -6     0     6

k =

2.1213      2.1213      -2.1213      -2.1213       2.1213       2.1213      -2.1213      -2.1213      -2.1213      -2.1213       2.1213       2.1213      -2.1213      -2.1213       2.1213       2.1213

K =

Columns 1 through 7

2.1213      2.1213 -2.2496e-032 -3.6739e-016      -2.1213      -2.1213            0 2.1213      8.1213 -3.6739e-016           -6      -2.1213      -2.1213            0 -2.2496e-032 -3.6739e-016           6  3.6739e-016           -6            0            0 -3.6739e-016          -6  3.6739e-016            6            0            0            0 -2.1213     -2.1213           -6            0       8.1213       2.1213            0      -2.1213      -2.1213            0            0       2.1213       8.1213            0            0            0            0            0            0            0            0            0            0            0            0            0           -6            0

Column 8

0           0            0            0            0           -6            0            6

R =

0           0          0.5            0            0            0            0            0

d =

0           0      0.40237 -2.4638e-017 0.31904   -0.083333            0            0

reactions =

-0.5        -0.5            0          0.5

Contributing Team Members
The following students contributed to this report:

Mark Barry Eas4200c.f08.blue.a 16:16, 24 October 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 23:02, 19 October 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 03:49, 24 October 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 17:09, 24 October 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 22:47, 23 October 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 21:21, 23 October 2008(UTC)

title: User:Eml4500.f08.ateam/hw 4


text: File:DOG TRAINING TRAIL II.JPG

title: File:DOG TRAINING TRAIL II.JPG


text:
Earth-impact events

What objects exist in space near the Earth? How often do they collide with our planet? What are the risks of such an impact? Is this something that we should be worried about? Have scientists and journalists done a good job of informing the public? Please read the information on this page and visit the links, and then join the discussion.

The science of Near-Earth objects

The Solar System consists of the Sun, the planets and their moons, and many smaller objects. These smaller objects include asteroids, comets and large meteoroids. Most of these objects are distant from the Earth and never pass close to our planet. For example, the vast majority of asteroids orbit within the main asteroid belt. But there are objects that are known to approach the Earth very closely, and many have collided with our planet thoughout its history. These Near-Earth objects include some that are potentially hazardous.

Near-Earth objects
Near-Earth objects (NEO) are asteroids, comets and large meteoroids whose orbits bring them close to Earth's orbit, and which may therefore pose a collision danger.

 * On October 7, 2008 a meteoroid called 2008 TC3 which was 2 - 5 meters (7 - 16 ft) in diameter entered Earth's atmosphere and burned up before it reached the ground. The object had just been discovered about a day before the impact.  It exploded tens of kilometers above the ground with the energy of around 1 - 2 kiloton of TNT, causing a large fireball in the early morning sky over northern Sudan.  An impact like this happens 2 - 4 times per year, but this is the first time that such an object has been observed and tracked before entering the atmosphere.

Potentially Hazardous Asteroids
Potentially Hazardous Asteroids (PHA) are Near-Earth asteroids that are big enough to cause unprecedented regional devastation if they were to hit the Earth.

 * On September 29, 2004 the asteroid 4179 Toutatis made a particularly close approach (within 4 LD, or lunar distances) from Earth. The next close approach is November 9, 2008, but it will be five times more distant. It is about 4.6 km by 2.4 km in size.

 * The asteroid 2007 CN26 passed within 0.6 LD of the Earth on September 2, 1985. The size of this object is in the range of 170 - 380 meters in diameter.  It was not discovered until February 2007, more than 20 years after the close approach to the Earth.

 * There is some concern that the asteroid 99942 Apophis could impact earth on April 13, 2036. It is about 210 - 330 meters in size.  The probability of a collision is currently esitmated to be 1 in 45,000.  These calculations are uncertain due to another close encounter in 2029 which will modify the orbit of the asteroid, making a collision either more or less likely.  New measurements possible in 2011-2013 will likely confirm that the asteroid will miss the earth.

 * The asteroid 2007 VK184 is currently estimated to have a probability of 1 in 3,030 of impacting Earth on June 3, 2048. This 130 meter object has the potential of releasing 150 megatons of energy if it were to collide.

Close-Approaches

1 LD = Lunar Distance = ~384,000 kilometers

See the NASA/JPL Near Earth Object Program for up-to-date information.

In fiction

 * Fictional impact events
 * Fictional asteroid collisions with Earth
 * Off on a comet (Jules Verne novel)
 * Meteor (1979 film)
 * Armageddon (1998 film)
 * Deep Impact (1998 film)

In the news

 * Near-Earth Asteroid 2007 TU24 to Pass Close to Earth on Jan. 29 - Should be Observable with Modest Sized Telescopes - January 22, 2008
 * Asteroid to narrowly miss Earth - January 24, 2008
 * Small Asteroid Predicted to Cause Brilliant Fireball over Northern Sudan - October 6, 2008
 * Boulder-Size Asteroid to Be Fireball Over Earth Monday - October 6, 2008
 * Impact of Asteroid 2008 TC3 Confirmed - October 7, 2008
 * Incoming Asteroid Burned Up in Earth's Atmosphere Right on Time - October 8, 2008
 * Little Asteroid Makes a Big Splash - October 9, 2008
 * For first time, scientists predict arrival of asteroid - October 11, 2008
 * Who's watching our skies? - October 12, 2008
 * Asteroid 2008 TC3 Strikes Earth: Predictions and Observations Agree - November 4, 2008

Commentary

 * Asteroids ahead - don't panic - June 6, 1998
 * Asteroid False Alarm Shows Limits of Alert Systems - March 8, 2004
 * Impact! - March 7, 2007
 * The Sky Is Falling. Really. - March 16, 2007
 * The Sky Is Falling - June 2008
 * What 'Deep Impact' might an asteroid make on Earth, astronauts ask - September 25, 2008
 * Will an asteroid destroy Earth? Time for U.N. to keep tabs, say experts - September 25, 2008
 * Death by meteorite - October 13, 2008

title: Earth-impact events


text:
User:Unirivercity

Unirivercity is the name of a Wikiversity user, contributor, and editor.

Unirivercity is an instructional editor at a university town with two rivers.

title: User:Unirivercity


text:
User:Eml 4500.f08.delta 6.specht/HW4

 Comment: This page is a revision from the HW that was turned in before the deadline. Here is the comparison between these two versions. --Eml 4500.f08.delta 6.specht 17:47, 4 November 2008 (UTC)

Lecture Notes (October 6th - October 17th, 2008)

 * BEGINNING OF OCT 6 LECTURE

Connectivity & Location Matrix Master Arrays
Consider the two-bar truss below:

and its respective bar element:

We can notice that there is a common global node, and its corresponding dofs, when comparing both bar element diagrams. Since the FEA method can be applied to a large system with multiple common nodes an elements, then it is important to organize these element notations so as to visualize which elements share which nodes and bounded by which dofs.

The Connectivity Array, "conn"
This array is set up as follows:

 * $$\begin{matrix}

1 & 2 \end{matrix}$$   <-- Local node numbers

 * conn = $$\left.\ \begin{vmatrix}

1 & 2\\ 2 & 3 \end{vmatrix} \right\}$$ <-- Global node numbers, each row for each element

In short-hand notation, we say that conn = (e,j): global node number of local node j of element e.

The Location Matrix Master Array, "lmm"
This array is set up as follows:

 * $$\begin{matrix}

1 & 2 & 3 & 4 \end{matrix}$$   <-- Local dofs numbers

 * lmm = $$\left.\ \begin{vmatrix}

1 & 2 & 3 & 4\\ 3 & 4 & 5 & 6 \end{vmatrix} \right\}$$ <-- Global dofs numbers, each row for each element

In short-hand notation, we say that lmm = (i,j): equation number (global dof number) for the element stiffness coefficient corresponding to the jth local dof number.

 * END OF OCT 6 LECTURE

Development of Invertible Local DOF Transfer Function
Goal: Find $$ \tilde{T}_{4x4}^{(e)} $$ that transforms the set of local element d.o.f.'s $$ d^{(e)} $$ to another set of local element d.o.f.'s $$ \tilde{d}_{4x1}^{(e)} $$ such that $$ \tilde{T}^{(e)} $$ is invertible.

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

$$ (1) \hat{d}_{1}^{(e)} = \begin{bmatrix} l^{(e)}\ m^{(e)}\

\end{bmatrix}

\begin{Bmatrix}d_1^{(e)} \\ d_{2}^{(e)} \end{Bmatrix} $$

Where,

$$ \hat{d}_{1}^{(e)} = q_1^{(e)} $$

$$\hat{d}_{2}^{(e)} = \vec{d}_1^{(e)} \cdot \tilde{j}$$

Note: $$\hat{d}_{2}^{(e)}$$ is the component of $$\vec{d}_1^{(e)}$$ along $$\tilde{j}$$, i.e., $$\tilde{y}$$ axis.

Thus,

$$\tilde{d}_2^{(e)} = -\sin \theta^{(e)}d_1^{(e)} + \cos \theta^{(e)} d_2^{(e)}$$

Equation (2) is as follows:

$$(2) \hat{d}_{2}^{(e)} = \begin{bmatrix} -m^{(e)}\ l^{(e)}\

\end{bmatrix}

\begin{Bmatrix}d_1^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

Next, put (1) & (2) into matrix form,

$$\begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & -l^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}$$

Where,

$$R^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & -l^{(e)} \end{bmatrix}$$

Thus,

$$\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)}\\ \tilde{d}_4^{(e)} \end{Bmatrix} = \begin{bmatrix} R_{2x2}^{(e)} & 0_{2x2}\\ 0_{2x2} & R_{2x2}^{(e)} \end{bmatrix} \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)} \end{Bmatrix}$$

which represents,

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

Next, rotate the element achieving the following

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

Note: all zero values are zero because transverse force does not stretch the spring.

Next, we can conclude:

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

Note: this is the stopping point of transforming the FD relation back to original (x,y) coordinate system.

Derivation of the Global Force Vector from the Axial Displacements and Forces
$$\tilde{f}^{(e)}_{4x1} = \tilde{k}_{4x4} ^{(e)} * \tilde{d}_{4x1}^{(e)}$$ Note: Consider the case where $$ \tilde{d}_{4} \neq 0 $$ then $$ \tilde{d}_1^{(e)} = \tilde{d}_2^{(e)} = \tilde{d}_3^{(e)} = 0 $$

you will get the following relationship $$ \tilde{f}_{4x1} ^{(e)} = \tilde{k}_{4x4} ^{(e)} * \tilde{d}_{4x1} ^{(e)} = 0_{4x1} $$ where the zero matrix is the 4th column of matrix $$ k^{(e)} $$ (interpretation of the transverse dofs)

Recall from 19-3 that $$ \tilde{d}^{(e)} = \tilde{T}^{(e)} * d^{(e)} $$ similarly there is the same relationship for f where $$\tilde{f}^{(e)} = \tilde{T}^{(e)} * f^{(e)} $$

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

The lesson here is that we should use the transverse dofs instead of just using the inverse dofs therefore finding the relationship for the global matrix $$ k=d*f $$ $$\tilde{k}^{(e)} *\tilde{T}^{(e)} * d^{(e)}= \tilde{T}^{(e)} * f^{(e)}$$

If $$ T^{(e)}$$ is invertible, then the matrix that we are looking for is:

$$f^{(e)} = \begin{bmatrix} \tilde{T}^{(e)^{-1}} * \tilde{k}^{(e)} * \tilde{T}^{(e)}*d^{(e)} \end{bmatrix}$$ where $$T^{(e)}$$ is the block diagram matrix.

Considering a general block-diagram matrix A:

$$ A=

\begin{bmatrix}

D_{1} & 0 & 0 & 0\\

0 & . & 0 & 0\\

0 & 0 & . & 0\\

0 & 0 & 0 & D_{s}

\end{bmatrix} $$

so what will be the inverse of matrix A? This can be answered by observing the simple example of the diagonal matrix B:

$$ B=

\begin{bmatrix}

d_{11} & 0 & 0 & 0\\

0 & d_{22} & 0 & 0\\

0 & 0 & . & 0\\

0 & 0 & 0 & d_{mm}

\end{bmatrix} $$

$$ B= Diag \begin{bmatrix} d_{11}, d_{22}, ... , d_{mm} \end{bmatrix} $$

$$ B^{-1} = Diag\begin{bmatrix} \frac{1}{d_{11}}, \frac{1}{d_{22}}, ... , \frac{1}{d_{mm}} \end{bmatrix} $$

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

For a block diagonal matrix A

$$ A = Diag\begin{bmatrix} D_{11}, D_{22}, ... , D_{s} \end{bmatrix} $$

$$ A^{-1} = Diag\begin{bmatrix} D_{11}^{-1}, D_{22}^{-1}, ... , D_{s}^{-1} \end{bmatrix} $$

$$\tilde{T}^{(e)^{-1}} = Diag \begin{bmatrix} R^{(e)^{-1}}, ... , R^{(e)^{-1}} \end{bmatrix}$$

$$R^{(e)^{T}} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix}$$

Since $$l^{2}*m^{2} = 1 $$ the matrix becomes where $$ I_{2x2} $$ is the identity matrix.

$$R_{2x2}^{(e)^{T}}*R_{2x2}^{(e)} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}_{2x2} = I_{2x2} $$

$$ R^{(e)^{T}}= R^{(e)^{-1}} $$ $$ \tilde{T}^{(e)^{-1}} = Diag\begin{bmatrix} R^{(e)^{T}}, R^{(e)^{T}} \end{bmatrix}$$ where $$ \tilde{T}^{(e)} = \begin{pmatrix} Diag \begin{bmatrix} R^{(e)}, R^{(e)} \end{bmatrix} \end{pmatrix}^{T} $$ $$ \tilde{T}^{(e)^{-1}} = \tilde{T}^{(e)^{T}} $$ $$ \begin{bmatrix} \tilde{T}^{(e)^{T}} * \tilde{k}^{(e)}* \tilde{T}^{(e)} \end{bmatrix} *d^{(e)}= f^{(e)} $$ where $$ k^{e} = \begin{bmatrix} \tilde{T}^{(e)^{T}} * \tilde{k}^{(e)}* \tilde{T}^{(e)} \end{bmatrix} $$

Mechanisms
The dynamics eigenvalue problem is given as: $$Kv=\lambda M v$$ In this equation, the mass matrix M is given as the identity matrix. This problem can then be reduced to: $$Kv=\lambda v$$

The global K matrix for the two bar truss system is:

$$K= \ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix} 0.5625& 0.3248& -0.5625 & -0.3248 & 0 & 0\\  0.3248& 0.1875 & -0.3248 & -0.1875 & 0 & 0\\ -0.5625 & -0.3248 &3.0625  & -2.1752 & -2.5 & 2.5\\  -0.3248& -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\  0& 0 & -2.5 & 2.5 & 2.5 & -2.5\\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5 \end{bmatrix}_{6X6}\,$$

When this matrix is solved for the eigenvalues, the values found are given as: $$\lambda = \begin{bmatrix} 0&0 &0  &0  & 1.4707 & 10.0294 \end{bmatrix}$$

Notice that there are four eigenvalues that are equal to zero. These correspond to the three rigid body displacements and one mechanism of the two bar truss system. Two of the three rigid body motions are the translational displacements in the two dimension cartesian coordinate system. The other rigid body motion is the rotation of the system around global node 2. The mechanism of the system can be described as the motion of each element as they pivot about node 2. The mechanism can also be described as the change in the angle between elements 1 and 2 at node 2, due to the applied force, and the displacement of node 2. If the row vector consisting of the eigenvectors that correspond to the four zero eigenvalues is given as $$\begin{Bmatrix}u_{1} & u_{2} & u_{3} & u_{4}\end{Bmatrix}$$, this equation can be expressed as (for i=1:4):

$$K_{6x6}u_{i,6x1}=0_{6x1}u_{i,6x1}$$ The linear combination of each of the eigenvectors can be equated to: $$ \sum_{i=1}^{4}{\alpha _{i}u_{i,6x1}}=:W_{6x1} $$

The value of $$\alpha_{i}$$ can be any real scalar value. The symbol used in the equation( =:) makes the right hand side equal to the left hand side by definition. The column vector $$W$$ is the eigenvector symbolizing the sum of all four of the eigenvectors corresponding to the eigenvalues ($$u_{i}$$). Thus the dynamics eigenvalue problem can be expressed as the following equation for the eigenvectors corresponding to the zero eigenvalues.

$$ K_{6x6}W_{6x1}=K_{6x6}(\sum_{i=1}^{4}{\alpha _{i}u_{i,6x1}}) $$ $$ K_{6x6}W_{6x1}=\sum_{i=1}^{4}{\alpha _{i}(K_{6x6}u_{i,6x1}}) $$ Because the values in parentheses is proven to be equal to a 6x1 column vector of zeros, the dynamics eigenvalue problem can be written as the dot product of the zero vector and the column vector $$W$$, which is the sum of the eigenvectors corresponding to the zero eigenvalues. $$ K_{6x6}W_{6x1}=0_{6x1}\cdot W_{6x1} $$

Justification of Assembly of Elemental Stiffness Matrix into Global Stiffness Matrix
Consider the example of the 2-bar truss system discussed earlier. If we recall the elemental Force-Discplacement relationship:

$$ k^{(e)}d^{(e)} = f^{(e)} $$

The equilibrium of node 2 can be found using three methods we have discussed earlier this semester. 1. The Euler Cut Principle 2. The Free Body Diagrams of each element 3. The Identity of Global D.O.F.'s to Element D.O.F.'s

At node 2, the elquilibrium can be found by simple summing the forces in each direction.

$$ \sum{F_{x}} = 0 = -f_{3}^{(1)} - f_{1}^{(2)} = 0 $$ $$ \sum{F_{y}} = 0 = P -f_{4}^{(1)} - f_{2}^{(2)} = 0 $$

When equations one and two, from the equilibrium of node two, are rewritten they become: $$f^{(1)}_{3} + f^{(2)}_{1}= 0$$ $$f^{(1)}_{4} + f^{(2)}_{2}= 0$$ The element forces can then be written in terms of the element stiffness and displacements. $$f_{3}^{(1)} = \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}$$ $$f_{1}^{(2)} = \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}$$ The forces written in terms of the elements stiffness and displacements are then substituted into equation one and equation one becomes. $$ \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix} + \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}=0$$ Now recall the relationship between the global degrees of freedom and the element degrees of freedom: $$d_{1}=d_{1}^{(1)},d_{2}=d_{2}^{(1)},d_{3}=d_{3}^{(1)}=d_{1}^{(2)},d_{4}=d_{4}^{(1)}=d_{2}^{(2)},d_{5}=d_{3}^{(2)},d_{6}=d_{4}^{(2)}$$ The global degrees of freedom can now be substituted into equation one for the element degrees of freedom and equation one becomes. $$ \begin{bmatrix} k_{31}^{(1)}d_{1} + k_{32}^{(1)}d_{2} + k_{33}^{(1)}d_{3} + k_{34}^{(1)}d_{4} \end{bmatrix} + \begin{bmatrix} k_{11}^{(2)}d_{3} + k_{12}^{(2)}d_{4} + k_{13}^{(2)}d_{5} + k_{14}^{(2)}d_{6} \end{bmatrix}=0$$ Now the common terms are grouped and factored (i.e. The terms that include $$d_{3}$$ and $$d_{4}$$). The same process is repeated for equation two. Equation one results in the third row of the global stiffness matrix and equation two results in the fourth row of the global stiffness matrix. The work is shown in the homework for this section.

Assembly of the global stiffness matrix
Before starting the assembly of the global stiffness matrix a quick review of the goals and the end that is being worked towards is necessary. The first goal is the use of The Principle of Virtual Work (PVW) which allows the formation of the global stiffness matrixK by eliminating rows that correspond to the boundary conditions. The ultimate goal is to use partial differential equations (PDE's) to derive the Finite Element Method (FEM). Recall now two equations that have already been derived in the process of the FEM. $$q^{(e)}_{2x1}$$ = $$T^{(e)}_{2x4}$$ $$d^{(e)}_{4x1}$$ $$K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

The assembly of the global stiffness matrix (K) is accomplished using the element stiffness matrices ($$k^{(e)}$$) and an assemble operator (A). Where e= 1,....,nel, and nel is the number of elements. The assembly operator is introduced because the sigma operator is not correct. The global stiffness matrix is assembled using a combination of element stiffness matrices. The element stiffness matrices are small when compared to the global stiffness matrix. The global stiffness matrix could be orders of magnitude larger than the element stiffness matrices. The sum of $$k_{nedxned}^{(e)}$$ doesn't equal to a $$K_{nxn}$$ matrix. Therefore the assembly operator (A) is necessary for the notation of the process to be correct. $$K=Ak^{(e)}$$ With this, new notation needs to be introduced. n is equal to the number of global degrees of freedom before eliminating the boundary conditions. ned is the number of element degrees of freedom. Where ned is very small when compared to n. Recall that the force displacement relationship for a bar or spring is kd=F. After some algebra: $$ kd-F=0$$ This will be called equation three and is equivalent to: w(kd-F)=0 This will be called equation four. This holds true for all values of w. Therefore it is trivial to imply that equation three is equal to equation four but is extremely important to understand that equation four is equal to equation three. This is true because equation four is valid for all w. So letting w=1, equation four is converted back into equation three.

Five Bar Truss Structure (MATLAB)
The code that was used to run the 5-Bar Truss System is listed as follows The first section of this code defines the material properties for each element. Since some of the values are used for more than one section, it was only necessary to declare them once. The variable nodes describes the position of each of the global nodes, which are all multiplied by a scalar value (1000) to get the actual position. The two functions that follow it are the Connectivity array (conn) and the Location Master Matrix array (lmm). The connectivity array holds the number of the global node that is being described when the following is called: conn(i,j) = global node number This function will give the global node number for the ith element, and the jth local node of that element. This is necessary to reduce the amount of storage space for the program, and to allow for the use of for loops in the body of the program. The Location Master Matrix array returns the global dof number for the ith element and the jth local dof for the element. This can be called in the same manner that the Connectivity array is called. The Connectivity array and the Location Master Matrix array can both be used to assemble the global stiffness matrix for the structure. The code in this program is divided into two sections in order to construct the global K matrix. This is done in order to use for loops to construct the matrix, since multiple matrices have the same values for the modulus of elasticity (e in the program) and for the area (a). There are three sets of values for the area, which is the maximum, so this is the limiting factor of the program. Therefore construction of this matrix must take place in at least three parts, which is done in the program for each set of corresponding area values and modulus of elasticity. The function PlaneTrussElement is discussed later. After the global stiffness matrices are constructed, the applied load that is imposed on the structure is defined. Once the load is defined, the last several lines in the program call the functions that are discussed below (PlaneTrussElement, PlaneTrussResults, NodalSoln) in order to compute the displacements and reactions at each node. There are several functions which are also called by the program in order to compute the values for the displacements, the reactions, and the results. In order to do this, the functions must be called several times, based on the fact that there are three different values for area, and two different values for the modulus of elasticity. The first function is called PlaneTrussElement, which is used to generate the element stiffness matrices for each element in the structure. In order to use this function in the body of the program values for the modulus of elasticity, area, and the coordinates at the endpoints of the element must all be included in the declaration. This then computes the length of the element, which is in turn followed by the computation of the element values ls and ms. These values are the horizontal and vertical components of the element, respectively.

The second function is the PlaneTrussResults function, and requires that the modulus, area, coordinates of the endpoints, and displacements of the endpoints be declared when the function is called. This function computes the axial displacement for the element and then uses this value in order to compute the axial force on the element. In order to do this the length is computed first, followed by the horizontal and vertical components, similar to the manner described for the first function. The transfer function (T) is then computed, which allows the displacements to be translated from the cartesian coordinate system, to the coordinate system in which the x axis is pointed from local node 1 to node 2 along the length of the element. The strain is then computed, and multiplied by the modulus in order to obtain the stress on the element. By multiplying the stress on the element by the area of the element, the axial force on the element is found, and returned to the body of the program. The last function that is called by the program is the NodalSoln function, which requires that the global stiffness matrix, the applied force vector, the element dofs, and the values for these dofs all be declared when the function is called. The number of global dofs is set to the length of the applied force vector. The value df is then set to the difference between the global dofs and the number of dofs for the element, which is iterated from 1 to the total number of global dofs. The value in the global stiffness matrix at the (df, df) position is then found. A scalar force value is established by the value in the df position of the applied force matrix minus the value of the value in the global stiffness matrix at the (df,debc) position, which is multiplied by the value of the dof before being subtracted. The displacement vector for the element is then established by setting the value in the debcth row of the specified value for the displacement, and the value in dof-debcth row to the value of the scalar force value divided by the value from the global stiffness matrix that was found earlier. The resultant force for each node is then computed by multiplying the debcth row of the global stiffness matrix by the displacement vector that was constructed, and then subtracting the value from the debcth row in the applied force vector. This value is then returned to the main body of the program.

The results from the code that is given above are listed below. First the global stiffness matrix is computed and returned. The second value given in the value of the applied force vector. After that, the displacement vector for the system is given, followed by the reactions at the fixed nodes, and the resultant forces through each element. The code that is listed below was generated in order to plot the undeformed and deformed shapes of the 5-Bar Truss Structure. This first calculates the original position of each node. Then the original position of each node is added to the vector d from the result of the code above, which is given as the displacement of each node. In order to better display the effect of the force on the global node 2, the displacement vector, d, has been multiplied by a weighting factor of 500. If this weighting factor were not present, the shape of the deformed structure would very closely coincide with the shape of the undeformed structure. The weighting factor thus allows the effect of the applied force on the system to be better understood visually, while the true structure would not deform as much as is shown. The last part of the code is necessary to include the labels at the nodes and on the elements of the structure, as well as including the key, which gives the linetype for the undeformed and deformed structures.

Three Bar Truss Example (MATLAB)
It is important to understand how to solve a bar truss system employing a finite element MATLAB code in order to compare the resulting matrices with those obtained from hand calculations.

Take this truss example with the following:

Data

MATLAB Code (Analysis of the system)

MATLAB Functions

In order to run this MATLAB code the following functions are needed:

Note: this MATLAB code was obtained from the MATLAB.zip Chapter 9 folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Note: this MATLAB code was obtained from the MATLAB.zip Common folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Note: this MATLAB code was obtained from the MATLAB.zip Chapter 9 folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Results MATLAB Code (Plot of the system)

Derivation of $$\tilde{d}_{2}^{(e)}$$
The vector $$\tilde{d}_{2}^{(e)}$$ is the translational vector of node 1 for each element, while the vector $$\tilde{d}_{1}^{(e)}$$ is the axial displacement of the node. The value for the first vector can be found by multiplying the displacement of node 1 (given in standard cartesian coordinates) by the unit vector in the $$\tilde{y}$$ direction ($$\vec{\tilde{j}}$$). The general form of this equation is given as: $$ \tilde{d}_{2}^{(e)}=\vec{d}_{Local Node 1}^{(e)}\cdot \vec{\tilde{j}} $$

The equations for the variables on the right hand side of the equation are given as the following two equations. (1)

$$ \vec{d}_{Local Node 1}^{(e)}=d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j} $$

(2)

$$ \vec{\tilde{j}}=-sin\theta ^{(e)}\vec{i}+cos\theta ^{(e)}\vec{j} $$

Plugging the values for equations (1) and (2) into the general equation given above yields the equation for $$\tilde{d}_{2}^{(e)}$$ in terms of $$d_{1}^{(e)}$$ and $$d_{2}^{(e)}$$ for each element.

$$ \tilde{d}_{2}^{(e)}=(-sin\theta ^{(e)}\vec{i}\cdot d_{1}^{(e)}\vec{i})+(cos\theta ^{(e)}\vec{j}\cdot d_{2}^{(e)}\vec{j} ) $$

This can be reduced by noting the following:

$$ \vec{i}\cdot \vec{i}= 1 $$

$$ \vec{j}\cdot \vec{j}= 1 $$

When inputting the values for these dot products the equation reduces to: $$ \tilde{d}_{2}^{(e)}=(-sin\theta ^{(e)}\cdot d_{1}^{(e)})+(cos\theta ^{(e)}\cdot d_{2}^{(e)}) $$

Since the values for $$-sin\theta^{(e)}$$ and $$cos\theta^{(e)}$$ can be given as $$-m^{(e)}$$ and $$l^{(e)}$$ respectively, this can be reduced to the following equation:

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

Verify the element stiffness matrix
Start with equation 1: $$k^{(e)}={\tilde{T}^{(e)T}}{\tilde{k}^{(e)}}{\tilde{T}^{(e)}}$$ The substitute in with: equation 2 $$R^{(e)T}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix}$$ and equation 3 $$R^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix}$$ and equation 4 $${\tilde{T}^{(e)T}}=Diag\begin{bmatrix} R^{(e)T} & R^{(e)T}\end{bmatrix}$$ and equation 5 $${\tilde{T}^{(e)}}=Diag\begin{bmatrix} R^{(e)} & R^{(e)}\end{bmatrix}$$ also $${\tilde{k}^{(e)}}=k^{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$ From previous work $$l^{(1)}=cos(30)=.866025$$ and $$m^{(1)}=sin(30)=.5$$. so $$R^{(e)T}=\begin{bmatrix} .866025 & -.5\\ .5 & .866025 \end{bmatrix}$$ and $$R^{(e)}=\begin{bmatrix} .866025 & .5\\ -.5 & .866025 \end{bmatrix}$$ and $$k^{(e)}=\begin{bmatrix} .5625 & 3248 & -.5625 & -.3248\\ .3248 & .1875 & -.3248 & -.1875\\ -.5625 & -.3248 & .5625 & .3248\\ -.3248 & -.1875 & .3248 &.1875 \end{bmatrix}$$ Now the known matrices are substituted into equations 2, 3, 4 and 5. Then equations 2, 3, 4 and 5 are substituted into equation 1 to verify that: $$k^{(e)}=\begin{bmatrix} .5625 & 3248 & -.5625 & -.3248\\ .3248 & .1875 & -.3248 & -.1875\\ -.5625 & -.3248 & .5625 & .3248\\ -.3248 & -.1875 & .3248 &.1875 \end{bmatrix}$$

Proof of Global Transfer Function Use with Element Force Matrix
Known formulas: equation 2 $${\tilde{f}^{(e)}}={\tilde{k}^{(e)}}{\tilde{d}^{(e)}}$$ equation 3 $${\tilde{d}^{(e)}}={\tilde{T}^{(e)}}d^{(e)}$$ equation 4, where I is the identity matrix $${\tilde{k}^{(e)}}=k^{(e)}I$$ equation 5 $$f^{(e)}=k^{(e)}d^{(e)}$$ Substituting equation 2 into equation 1: $${\tilde{f}^{(e)}}={\tilde{k}^{(e)}}{\tilde{T}^{(e)}}d^{(e)}$$ The substituting equation 3 and equation 4 into equation 1: $${\tilde{f}^{(e)}}=k^{(e)}I{\tilde{T}^{(e)}}d^{(e)}$$ Finally, equation 5 is substituted into equation 1 and the Identity matrix is dropped. This gives the wanted results. $${\tilde{f}^{(e)}}={\tilde{T}^{(e)}}f^{(e)}$$

Plot of the Eigenvectors Corresponding to the Zero Eigenvalues
Using MATLAB, the output of the Global Stiffness Matrix is:

The V and D matrices below correspond to the eigenvectors and eignevalues, respectively.

As can be seen by the D matrix, the first four columns of the V matrix correspond to the eigenvectors of the zero eigenvalues.

Columns 1 & 2

Columns 3 & 4

Completion of $$f_1^{(2)} $$
The force component in the x direction of the local node 1 of element 2, $$f_1^{(2)}$$ can be expressed as:

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

Derivation of Fourth Row of Global Force Vector
$$f_4^{(1)} + f_2^{(2)} = P$$ =F_4

Where,

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

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

Thus,

$$f_{3}^{(1)}+f_{1}^{(2)}=[k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}]+ [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}]$$ $$f_{3}^{(1)}+f_{1}^{(2)}=[k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + k_{43}^{(1)}d_3 + k_{44}^{(1)}d_4]+ [k_{21}^{(2)}d_3 + k_{22}^{(2)}d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6]$$

Observation: Last two items in 1. & first two items in 2. correspond to the 4th row in the global stiffness matrix K6x6.

Hence,

$$[(k_{43}^{(1)} + k_{21}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)})]$$

Three-Bar Truss Eigenvalues
We have two truss systems, one 3-bar and 5-bar. They are illustrated below:

Where $$E^{(1)}=E^{(2)}=E^{(3)}=1$$; $$A^{(1)}=A^{(2)}=A^{(3)}=1$$; $$L^{(1)}=L^{(2)}=L^{(3)}=1$$. It is possible to analyze the eigenvalues and eigen vectors according to $$\mathbf{K}v=\mathbf{\lambda} v$$ for the 3-bar truss case.

We construct the stffness matrix from each individual element stiffness matrix. It was found to be:
 * K=$$\begin{bmatrix}

0&0&0&0&0&0&0&0\\ 0&1&0&-1&0&0&0&0\\ 0&0&1&0&-1&0&0&0\\ 0&-1&0&2&0&0&0&-1\\ 0&0&-1&0&1&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&-1&0&0&0&1 \end{bmatrix}$$ This matrix was then inputed into MATLAB and its eigenvalues and eigenvectors were determined. The MATLAB results are as follow:

V = -0.0000   0.0000    0.0000   -0.4740    0.7103   -0.2504    0.4082    0.0000   -0.7071   -0.4674   -0.1769    0.2838   -0.0000   -0.7071         0   -0.2136   -0.3585   -0.5702   -0.8165   -0.0000    0.0000   -0.4674   -0.1769    0.2838    0.0000    0.7071    0.0000   -0.2136   -0.3585   -0.5702   -0.0000    0.0000   -0.0000   -0.0556    0.1402   -0.1014   -0.0000    0.0000    0.0000   -0.1604    0.3534   -0.1873    0.4082    0.0000    0.7071   -0.4674   -0.1769    0.2838 D = 3.0000        0         0         0         0         0         0    2.0000         0         0         0         0         0         0    1.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0   -0.0000

Contributing Team Members
Iain Specht--Eml 4500.f08.delta 6.specht 23:57, 16 October 2008 (UTC) Micheal Krueger--Eml4500.f08.delta 6.krueger 20:29, 19 October 2008 (UTC) Nicolas Castrillon--Eml4500.f08.delta 6.castrillon 15:56, 21 October 2008 (UTC) Diana Guzman--Eml4500.f08.delta 6.guzman 06:26, 24 October 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 07:51, 24 October 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 21:03, 24 October 2008 (UTC)

title: User:Eml 4500.f08.delta 6.specht/HW4


text:
User:Juandev/Total disappearance

This is a personal lerning project, how to total disappear. I.e. from the sight of your frinds, isntitutions, school, work whatever. You sudenley left, without leaving any traits.

title: User:Juandev/Total disappearance


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text:
User:Eas4200c.f08.ZYX/HW Week 7-8

 Comment: This page is identical to the previous one, except that the MATLAB Basic section has been added. Here is a comparison between the two versions. Eas4200c.f08.gator.reger 03:15, 19 November 2008 (UTC)

title: User:Eas4200c.f08.ZYX/HW Week 7-8


text:
User:Uf.team.wiki/HW4

= Method 2 To Derive $${\underset{4 \times 4}\mathbf k^{(e)}}$$ =

 * Key to Method 2: Transform a system with 4 degrees of freedom into a system with also 4 degrees of freedom (instead of just 2), so that the transform matrix is now 4x4 and (hopefully) invertible
 * Goal: To find $${\underset{4 \times 4}\mathbf \tilde T^{(e)}}$$ that transforms the set of local (element-level) degrees of freedom $${\underset{4 \times 1}\mathbf d^{(e)}}$$ to another set of local (element-level) degrees of freedom $${\underset{4 \times 1}\mathbf \tilde d^{(e)}}$$ such that $${\underset{4 \times 4}\mathbf \tilde T^{(e)}}$$ is invertible

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

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

$$ \tilde d_2^{(e)} = \overrightarrow{d_{[1]}^{(e)}} \cdot \overrightarrow{\tilde j} $$

 * NOTE- This expression translates to: "the component of $$\overrightarrow{d_{[1]}^{(e)}}$$  along the unit vector   $$\overrightarrow{\tilde j} $$ (ie: the y-axis)." Thus,

$$ \Rightarrow \tilde d_2^{(e)} = -sin \theta^{(e)} d^{(e)}_1 + cos \theta^{(e)} d^{(e)}_2$$

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

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

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

Thus,

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

= Example #1 =

Horizontal Element:

$$ \tilde \mathbf f^{(e)} = k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \tilde \mathbf d^{(e)} \Rightarrow {\underset{4 \times 1} \tilde \mathbf f^{(e)}} ={\underset{4 \times 4}\tilde \mathbf k^{(e)}}{\underset{4 \times 1} \tilde \mathbf d^{(e)}}$$

In this case,

 * $$ \tilde \mathbf d^{(e)}_4 \ne 0 $$
 * $$ \tilde \mathbf d^{(e)}_1 = \tilde \mathbf d^{(e)}_2 = \tilde \mathbf d^{(e)}_3 = 0 $$
 * $${\underset{4 \times 1} \tilde \mathbf f^{(e)}} ={\underset{4 \times 4}\tilde \mathbf k^{(e)}}{\underset{4 \times 1} \tilde \mathbf d^{(e)}} = \underbrace{\underset{4 \times 1}0}_{4th column of \tilde \mathbf k^{(e)}}$$

= Interpretation of Transverse/Transform Degrees of Freedom =

Recall:
 * $$\tilde \mathbf d^{(e)}=\mathbf \tilde T^{(e)}\mathbf d^{(e)} $$

Thus,
 * $$\tilde \mathbf f^{(e)}=\mathbf \tilde T^{(e)}\mathbf f^{(e)} $$
 * $$\tilde \mathbf k^{(e)}\mathbf \tilde d^{(e)}=\mathbf f^{(e)} $$

Substitution leads to:

$$\tilde \mathbf k^{(e)}\mathbf \tilde T^{(e)}\mathbf d^{(e)}=\mathbf \tilde T^{(e)}\mathbf f^{(e)} $$

If $$\mathbf \tilde T^{(e)}$$ is invertible $$\Rightarrow[\tilde \mathbf T^{(e)^{-1}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}] \mathbf d^{(e)}=\mathbf f^{(e)}$$

Recall the block diagonal matrix: $$\tilde\mathbf T^{(e)}=\begin{bmatrix} {\underset{2 \times 2}\mathbf R^{(e)}} & {\underset{2 \times 2}\mathbf 0^{(e)}} \\ {\underset{2 \times 2}\mathbf 0^{(e)}} & {\underset{2 \times 2}\mathbf R^{(e)}} \\ \end{bmatrix}$$

Consider a general block diagonal matrix: $$\mathbf A = \begin{bmatrix} \mathbf D_1    &  & \underline 0      \\ & \ddots  \\ \underline 0    &  & \mathbf D_s\end{bmatrix}$$

Question: What is $$\mathbf A^{-1}$$?

Demonstrating with a simpler example (diagonal matrix):

$$\mathbf B = \begin{bmatrix} d_{11} & & &\underline 0   \\ & d_{22} & & &  \\ & &  \ddots  & \\ \underline 0 & & & d_{nn}\end{bmatrix}= Diag[d_{11},d_{22},...d_{nn}]$$

$$\mathbf B^{(-1)}=Diag[\frac{1}{d_{11}},\frac{1}{d_{22}},...\frac{1}{d_{nn}}]$$(Assuming dii ǂ 0 for i=1,2,...n)

For block-diagonal matrix A:

$$ \mathbf A = Diag[\mathbf D_1,...D_s]$$

Thus the inverse of A is written as:

$$ \mathbf A^{-1} = Diag[\mathbf D_1^{-1},...D_s^{-1}] $$

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

$$\mathbf R^{(e)T} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \\ \end{bmatrix}$$

$${\underset{2 \times 2}\mathbf R^{(e)T}}{\underset{2 \times 2}\mathbf R^{(e)}}= \begin{bmatrix} l^{(e)^2}-m^{(e)^2} & 0\\ 0 & l^{(e)^2}-m^{(e)^2} \\ \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \\ \end{bmatrix} = {\underset{2 \times 2}\mathbf I} (identity  matrix)$$

This verifies the relationship $$\mathbf R^{(e)^{-1}}=\mathbf R^{(e)^{T}}$$

$$\Rightarrow \tilde \mathbf T^{(e)-1} = Diag[\mathbf R^{(e)^T},\mathbf R^{(e)^T}] = \underbrace{(Diag[\mathbf R^{(e)},\mathbf R^{(e)}])^T}_{\tilde \mathbf T^{(e)^{-1}}}$$

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

$$\underbrace{[\tilde \mathbf T^{(e)^{-1}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}]}_{\mathbf k^{(e)}}\mathbf d^{(e)}=\mathbf f^{(e)}$$

= Eigenvalue Problem =

$$\mathbf K \mathbf v = \mathbf \lambda \mathbf v$$

Let {u1,u2,u3,u4} be the pure eigenvalues corresponding to the 4 zero eigenvalues:

$${\underset{6 \times 6}\mathbf K} {\underset{6 \times 1}\mathbf {u_i}} = \underbrace{0 \cdot \mathbf {u_i}}_ \quad i=1,2,3,4$$

Linear combination of {ui, i=1,2,3,4}:

$$\sum_{i=1}^4 {\underset{1 \times 1}\mathbf {\alpha_i}}{\underset{6 \times 1}\mathbf {u_i}} =:{\underset{6 \times 1}\mathbf W} $$

Where αi is:
 * A scalar (ie: a 1x1 matrix is a scalar)
 * Real numbers

Note: W is also an eigenvector corresponding to a zero eigenvalue.

$$\mathbf {KW} = \mathbf K(\sum_{i=1}^4 \mathbf \alpha_i \mathbf u_i) = \sum_{i=1}^4 \mathbf \alpha_i(\mathbf K \mathbf u_i) = {\underset{6 \times 1}\mathbf 0^{(e)}} = {\underset{1 \times 1} 0}\cdot{\underset{6 \times 1} \mathbf W}$$

= Three Bar Truss Example Problem =

Given:
 * a = b = 1m
 * E = 2
 * A = 3

Solve: $$\quad \overline \mathbf K \mathbf v = \mathbf \lambda \mathbf v \qquad $$ (where $$ \quad \overline \mathbf K \quad $$ is the stiffness matrix for constrained systems)

= Useful Matlab Arrays=

 The Connectivity Array
 * Matlab variable "conn"

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

 * Decoding the matrix
 * Each column corresponds to a local node number
 * Each row corresponds to an element number
 * Each value inside the matrix corresponds to the global node number

 * When using the connectivity array in Matlab code
 * conn(e,j) = The global number of element e
 * j = The local node of element e

 The Location Matrix Master Array
 * Matlab variable "lmm"

lmm= $$\begin{bmatrix} 1&2&3&4\\ 3&4&5&6 \end{bmatrix}$$

 * Decoding the matrix
 * Each column corresponds to a local DOF number
 * Each row corresponds to an element number
 * Each value inside the matrix corresponds to the global DOF number
 * (or equivalent number in  K )

 * When using the location matrix master array in Matlab code
 * lmm(i,j) = The equation number (global DOF number)
 * the stiffness coefficient corresponds to the $$j^t$$ local DOF number

= Three Bar Truss System = The following code calculates the displacement and reaction forces for the Three Bar Truss System found in HW2. This program also plots the undeformed and deformed system. Note: The first three functions found in the Five Barr Truss System are required to run the following program

The following code are the results of the above program.

The following figure is the plot of the undeformed and deformed state of the Three Bar Truss System. Note: The displacement in the deformed plot is magnified by a factor of 0.2. This is due to the actual large displacement and will make it more visible.

= Five Bar Truss System = Before running the Five Bar Truss System program, the following three Matlab program functions must be in the same directory. They are obtained from the textbook website (See HW1 for textbook information). Matlab Truss Files

The following code calculates the displacement and reaction forces for the Five Bar Truss System found in p25 of the textbook. This program also plots the undeformed and deformed system.

The following code are the results of the above program.

The following figure is the plot of the undeformed and deformed state of the Five Bar Truss System. Note: The displacement in the deformed plot is magnified by a factor of 2000. This is due to the actual small displacement and will make it more visible.

= Justification of Elemental Stiffness Matrix ( k (e)) Into Global Matrix ( K ) =

Recall element FD relation: $$\underset{4 \times 4}{k^{(e)}}\underset{4 \times 1}{d^{(e)}} = \underset{4 \times 1}{f^{(e)}} $$     {e = 1, 2}

For two bar truss system:

$$\sum{F_{x}}=0=-f_{3}^{(1)}-f_{1}^{(2)}=0\;\;(1)$$

$$\sum{F_{y}}=0=P-f_{4}^{(1)}-f_{2}^{(2)}=0\;\;(2)$$

We can re-write equation (1) in the form of $$\mathbf{Kd=F}$$:

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

Hw: now we re-write equation (2) as

$$\left[k_{41}^{(1)}d_1^{(1)} +k_{42}^{(1)}d_2^{(1)}+k_{43}^{(1)}d_3^{(1)}+k_{44}^{(1)}d_4^{(1)}\right]$$ $$+\left[k_{21}^{(2)}d_1^{(2)} +k_{22}^{(2)}d_2^{(2)}+k_{23}^{(2)}d_3^{(2)}+k_{24}^{(2)}d_4^{(2)}\right]=P\; \; \;(2)$$

We can now substitute the global DOFs which are related to the local DOFs

$$d_1^{(1)}=d_1$$ $$d_2^{(1)}=d_2$$ $$d_3^{(1)}=d_2^{(2)}=d_3$$ $$d_4^{(1)}=d_1^{(2)}=d_4$$ $$d_3^{(2)}=d_5$$ $$d_4^{(2)}=d_6$$

The new equations with the global DOFs sub substituted in are

$$\left[k_{31}^{(1)}d_1 +k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4\right]$$ $$+\left[k_{11}^{(2)}d_3 +k_{12}^{(2)}d_4+k_{13}^{(2)}d_5+k_{14}^{(2)}d_6\right]=0\;\;(Eq.1)$$

Hw: now we re-write equation (2) as

$$\left[k_{41}^{(1)}d_1 +k_{42}^{(1)}d_2+k_{43}^{(1)}d_3+k_{44}^{(1)}d_4\right]$$ $$+\left[k_{21}^{(2)}d_3 +k_{22}^{(2)}d_4+k_{23}^{(2)}d_5+k_{24}^{(2)}d_6\right]=P\;\;(Eq.2)$$

= Assembly operator =

We assemble the Global stiffness matrix with the assembly operator $$\mathbf{A}$$. We use this operator to assemble the element stiffness matrices into the global stiffness matrix.

Assembly of $$\mathbf{K^{(e)}}, e=1,...,n_{el}$$ is done by

$$\mathbf{K}_{nxn}=A\mathbf{K}_{n_{el}xn_{el}}^{(e)}$$

Where

 * nel: number of elements:
 * A: assembly operation
 * n: total number of global DOFs before elimination with boundary conditions
 * ned: number of element degrees of freedom
 * (ned << n)

FD relationship for a bar or spring is: kd=F , which implies

$$\Rightarrow kd-F=0 (Eq.3)$$  which is equivalent to $$\Leftrightarrow w(kd-F)=0  (Eq.4)$$ Eq.4 is considered "weak form", also known as PVW (Principle of Virtual Work)

= Deriving Finite Element Method for PDEs =

We use the principles of virtual work to eliminate rows and columns from the global stiffness, DOF, and force matrix.

from $$kd=f$$ we can derive a relationship

$$kd-f=0\;\;(3)$$

$$W(kd-f)=0\;\;(4)$$

where

 * W: weighting coefficient
 * $$(kd-F)$$: known as the weak form

From this we can say that going from

(3) $$\Rightarrow$$  (4) is trivial

(4) $$\Rightarrow$$  (3) is not trivial

= Contributing Members = The following Team Wiki members contributed to this report.

Rodney Dagulo --Eml4500.f08.wiki.dagulo.r 17:22, 23 October 2008 (UTC)

Cortland Glowacki --Eml4500.f08.wiki.cort 14:13, 24 October 2008 (UTC)

Ricardo Lopez --Eml4500.f08.wiki.Lopez 15:13, 24 October 2008 (UTC)

Thomas Kehoe -- EML4500.f08.Wiki.Kehoe 15:18, 24 October 2008 (UTC)

Renee Hood -- EML4500.f08.Wiki.Hood 19:35, 24 October 2008 (UTC)

Gonzalo Barcia -- EML4500.f08.Wiki.Barcia 16:50, 24 October 2008 (UTC)

title: User:Uf.team.wiki/HW4


text:
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)

title: User:Eml4500.f08.qwiki.bishop/HW4


text:
User talk:Annareihl

good to see you here, Anna. Take a look at Introduction_to_Russian when you get a chance!
 * Stevenarntson 02:33, 17 October 2008 (UTC)

Welcome!
 Hello Annareihl, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 09:05, 30 October 2008 (UTC)

You may wish to visit Topic:Russian, Russian language, or Russian to help you in your efforts to learn Russian and about it. Good luck! The Jade Knight (d'viser) 09:05, 30 October 2008 (UTC)

title: User talk:Annareihl


text:
Talk:Fruit Clock

Definitely important to distinguish between ripe/unripe (harvestable is a good idea, too). I also think it will be very important to distinguish between cultivated and uncultivated varieties. The Jade Knight (d'viser) 03:04, 17 October 2008 (UTC)

 * The cultivated vs uncultivated can probably be done on the regional data sections (like on the bloom clock), as will the ripe/unripe/harvestable (many fruits (like summer squashes) are actually harvested well before they are technically ripe). Edibility and toxicity will be another issue. --SB_Johnny talk 22:48, 18 October 2008 (UTC)

title: Talk:Fruit Clock


text:
User talk:Jbrannigan

 Hello Jbrannigan, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 03:27, 17 October 2008 (UTC)

title: User talk:Jbrannigan


text: File:Simplified airfoil.svg

title: File:Simplified airfoil.svg


text: File:Simplified multicellular airfoil.svg


text:
User:Eml4500.f08.group/Homework4

=Homework 4=

 * Lecture Notes (current version)
 * Matlab (current version)
 * Contributing Members (current version)

title: User:Eml4500.f08.group/Homework4


text:
User:Eas4200c.f08.carbon.orear/HW4/Contributors

Jeff Wright Eas4200c.f08.carbon.w 06:14, 17 October 2008 (UTC)

Brian Taylor Eas4200c.f08.carbon.taylor 01:44, 19 October 2008 (UTC)

Jeff O'Rear Eas4200c.f08.carbon.orear 21:03, 23 October 2008 (UTC)

Beau Guidry Eas4200c.f08.carbon.guidry 23:42, 23 October 2008 (UTC)

Abby Booth Eas4200c.f08.carbon.booth 13:56, 24 October 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 14:36, 27 October 2008 (UTC)