User:Eml4500.f08.FEABBQ/HW4

Notes for Oct. 6, 2008
Connectivity Array "conn"

Consider the two bar truss system:

In this diagram the relationship between the local and global nodes can be seen. These relations are stored in the "connectivity array" in our Matlab Code:

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

$$conn(e,j)=$$ global node number of local node $$j$$ at element $$e$$

Location Matrix Master Array "lmm"

The Location Matrix Master Array stores the equation number for the element stiffness coefficient at each local degree of freedom

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

$$lmm(i,j)=$$ equation number (global degree of freedom #) for element stiffness coefficient to the $$j$$th local degree of freedom number $$i$$

Notes for Oct. 8, 2008
GOAL: To find $$\underline{\tilde{T}}^{(e)}$$ that transforms the set of local DOFs, $$\underline{d}^{(e)}$$, to another set of local DOFs, $$\underline{\tilde{d}}^{(e)}$$ , such that $$\underline{\tilde{T}}^{(e)}$$ is invertible.

Figure 2 below shows an element with both the original and transformed local DOFs.



Let:

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

Since,

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

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

we see that,

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

Defining the 2x2 matrix above as $$\underline{R}^{(e)}$$, equation (1) becomes:

$$\left[ \begin{matrix}\tilde{d_1}^{(e)}\\\tilde{d_2}^{(e)}\\\tilde{d_3}^{(e)}\\\tilde{d_4}^{(e)}\end{matrix} \right]=\left[ \begin{matrix} \underline{R}^{(e)} & \underline{0}\\\underline{0} & \underline{R}^{(e)}\end{matrix} \right]\left[ \begin{matrix}{d_1}^{(e)}\\{d_2}^{(e)}\\{d_3}^{(e)}\\{d_4}^{(e)}\end{matrix}\right]$$

Figure 3 below shows an element with its corresponding transformed forces and DOFs.



Here the forces are defined by:

$$\underline{\tilde{f}}^{(e)}=k^{(e)}\left[\begin{matrix}1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{matrix}\right]\underline{\tilde{d}}^{(e)}$$

or

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

Notes for October 10
considering the case where $$ \mathbf{\tilde{d}_{4}^{(e)}} \neq 0     $$,     and       $$  \mathbf{\tilde{d}_{1}^{(e)}}=\mathbf{\tilde{d}_{2}^{(e)}}=\mathbf{\tilde{d}_{3}^{(e)}} = 0   $$

$$ \mathbf{\tilde{f}_{4x1}^{(e)}}= \mathbf{\tilde{k}_{4x4}^{(e)}}\mathbf{\tilde{d}_{4x1}^{(e)}}= \mathbf{0_{4x1}}  $$

The zero matrix $$\mathbf{0_{4x1}} $$ is the fourth col. of matrix$$\mathbf{\tilde{k}_{4x4}^{(e)}}$$

Interpretation of transverse dof's:

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

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

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

$$ \begin{Bmatrix} \tilde{f}_1^{(e)} \\ \tilde{f}_2^{(e)}\\ \tilde{f}_3^{(e)} \\\tilde{f}_4^{(e)} \end{Bmatrix} = \left[ \begin{matrix} R_{2x2}^{(e)} & 0_{2x2} \\  0_{2x2}& R_{2x2}^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$

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

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

if $$\mathbf{\tilde{T}^{(e)}}$$ is invertible then:

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

Where $$\mathbf{\tilde{T}^{(e)}}$$ is block diagonal matrix, consider a general block diagonal matrix A

$$\mathbf{A} = \left[ \begin{matrix} D_{1} &...& 0  \\  ...& D_{2}&...\\ 0&...&D_{n} \end{matrix} \right]$$

$$\mathbf{D}_{1}, \mathbf{D}_{2}, .... , \mathbf{D}_{s}   $$     are matrices

$$\mathbf{A}^{-1}= ?$$

$$\mathbf{B}^{-1}=\left[ \begin{matrix} d_{11} &...& 0 \\  ...& d_22&...\\ 0&...&d_(nn) \end{matrix} \right] $$

$$\mathbf{B}=Diag\left[ \mathbf{d}_{11},\mathbf{d}_{22}, ...... , \mathbf{d}_{nn}  \right] $$

$$\mathbf{B}^{-1}=Diag\left[ \dfrac{1}{d}_{11},\dfrac{1}{d}_{22}, ...... , \dfrac{1}{d}_{nn}  \right] $$

Assuming $$\mathbf{d}_{ii}\neq 0 $$ for i = 1,2, ....,n for a block diagonal matrix A

$$\mathbf{A}=Diag\left[ \mathbf{D}_{1},\mathbf{D}_{2}, ...... , \mathbf{D}_{s}  \right] $$

$$\mathbf{A}^{-1}=Diag\left[ \mathbf{D}_{1}^{-1},\mathbf{D}_{2}^{-1}, ...... , \mathbf{D}_{s}^{-1}  \right] $$

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

$$\mathbf{R}^{(e)^{T}}=\left[ \begin{matrix} l^{(e)} & -m^{(e)} \\  m^{(e)} & l^{(e)} \end{matrix} \right]$$

$$\mathbf{R}_{2x2}^{(e)^{T}}\mathbf{R}_{2x2}^{(e)}=\left[ \begin{matrix} l^{(2)}+m^{(2)} & 0 \\  0 & l^{(2)}+m^{(2)} \end{matrix} \right] = \left[ \begin{matrix} l^{(2)}+m^{(2)} =1 & 0  \\  0 & l^{(2)}+m^{(2) =1} \end{matrix} \right] =\left[ \begin{matrix} 1 & 0  \\  0 & 1 \end{matrix} \right] = \mathbf{I}_{2x2}$$

From this we can conclude that $$\mathbf{R}_{2x2}^{(e)^{(-1)}}=\mathbf{R}_{2x2}^{(e)^{T}}$$

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

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

The above relation agrees with the fundamental relation $$\mathbf{k^{(e)}}\mathbf{d^{(e)}}=\mathbf{f^{(e)}}$$

Where : $$ \mathbf{k^{(e)}}=\mathbf{\tilde{T}^{(e)^{T}}}\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}$$

Notes for October 13, 2008
The four zero-eigenvalues of the $$ \underline{K}$$ matrix correspond to
 * 3 Rigid Body Modes
 * 1 Mechanism

HW 1
Plot the eigenvectors corresponding to the zero eigenvalues of the 2-bar truss system; interpret the results.

Before we do this assignment, it is important to understand the nature of these mode shapes. These mode shapes (that correspond to the zero eigenvalues) may come out as a linear combination of the pure mode shapes (pure rigid body motions and pure mechanism(s)). For example, take a body that has only two rigid body motions: horizontal and directly vertical. The linear combination of these two pure modes would be a diagonal motion. ''In Structural Mechanics, the eigenvalues represent the frequencies while the eigenvectors represent the mode shapes during free vibration. The lower frequencies (eigenvalues) tend to have more prominent vibration.''

Eigenvalue Problem: $$ \underline{K}\underline{v}=\lambda\underline{v} $$

Let {$$ \underline{u}_{1}, \underline{u}_{2}, \underline{u}_{3}, \underline{u}_{4} $$} be the pure eigenvectors corresponding to the four zero eigenvalues.

$$ \underline{K}_{_{6x6}}\underline{u}_{i_{6x1}}=0\cdot\underline{u}_{i_{6x1}}=\underline{0}_{_{6x1}} $$, where i=1,...,4.

A linear combination of {$$ \underline{u}_{i}, i=1,...,4$$}

$$ \sum_{i=1}^4 \alpha_{i_{1x1}} \underline{u}_{i_{6x1}} =: \underline{W}_{_{6x1}} $$, where $$ \alpha_{i} $$ are real numbers.

Note: the $$=:$$ means "equality definition."

$$ \underline{W} $$ is also an eigenvector corresponding to a zero eigenvalue: $$ \underline{K}\underline{W}=\underline{K}(\sum_{i=1}^4\alpha_{i} \underline{u}_{i})=\sum_{i=1}^4 \alpha_{i}(\underline{K}\underline{u}_{i})=\underline{0}=0\cdot\underline{W} $$

From the two-bar truss, we have the following output of $$ \underline{K} $$

And the following is the are the eigenvectors and eigenvalues, using the notation [V,D], where V are the eigenvectors and D are the eigenvalues.

The first four column vectors of V, which correspond to the four zero eigenvalues, are the eigenvectors for those zero eigenvalues. The matrix V is called the modal matrix.

The following are the four column vectors (eigenvectors) plotted as mode shapes. Note: they are shown in the order from the first column to the fourth column of the modal matrix.









HW2
Solve for $$ \underline{\overline{K}}\underline{v}=\lambda\underline{v} $$, the stiffness matrix for the constrained system below. Plot the eigenvectors corresponding to the zero eigenvalues for case (a).



The following is the Matlab code which solves for the global stiffness matrix, $$ \underline{K} $$, for the three bar truss in case (a) above.

The following is the output for the stiffness matrix of the three bar truss in case (a)

Using the method to solve for the eigenvectors of case(a), we get the following. Note: the first five column vectors correspond to zero eigenvalues.

The first five eigenvectors above correspond to zero eigenvalues, therefore their mode shapes are plotted below in order.











The Matlab code which solves for case (a) can be modified to solve for the stiffness matrix, $$ \underline{K} $$ for the four bar truss, case (b)

The following is the output for the stiffness matrix, $$ \underline{K} $$, for the four bar truss system.

Notes for Oct. 15, 2008
Justification of assembly of element stiffness matrices, $$ \mathbf{k^{(e)}} $$ into global stiffness matrices, $$ \mathbf{k} $$.

Consider the example of a 2-bar truss. Recall the element freebody diagram relation of,

$$ \mathbf{k_{4x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}} $$



The equilibrium of global node 2 will be determined using:

1. The Euler cut principle



2. The Free Body Diagrams of element 1 and element 2

3. And for node 2, the identity of global dof's to element dof's for both elements 1 & 2

From these the equilibrium of global node 2 is:



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

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

Notes for Oct. 17, 2008
Rearranging equations 1 and 2 give us the following two conditions:


 * $$ \mbox{Eq (1)}\quad \Rightarrow \quad f_3^{(10} + f_1^{(2)} = 0 \qquad \qquad (1')$$


 * $$ \mbox{Eq (2)}\quad \Rightarrow \quad f_4^{(1)} + f_2^{(2)} = P \qquad \qquad (2')$$

Further expanding each of these equations will prove the correct assembly of the third and fourth rows of the global stiffness matrix. The above equations in terms of local degrees of freedom are


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


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

Now replacing the local displacement degrees of freedom with their global equivalent:


 * $$ (1'):\qquad [k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2 + k_{33}^{(1)}d_3 + k_{34}^{(1)}d_4]\; +\; [k_{11}^{(2)}d_3 + k_{12}^{(2)}d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6] = 0$$


 * $$ (2'):\qquad [k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + k_{43}^{(1)}d_3 + k_{44}^{(1)}d_4]\; +\; [k_{21}^{(2)}d_3 + k_{22}^{(2)}d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6] = P$$

Once again, further expanding and collecting likes terms for each:


 * $$ (1'):\qquad k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2 + (k_{33}^{(1)}+k_{11}^{(2)})d_3 + (k_{34}^{(1)}+k_{12}^{(2)})d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6 = 0$$


 * $$ (2'):\qquad k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + (k_{43}^{(1)}+k_{21}^{(2)})d_3 + (k_{44}^{(1)}+k_{22}^{(2)})d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6 = 0$$

As one can see these two expressions are exactly what would be found in the matrix multiplication of the 3rd and 4th rows of the global force-displacement relation. Therefore, this sufficiently proves the accuracy of the previous assembly of the global stiffness matrix.

In general, the assembly of element stiffness matrices for any number of elements into the global stiffness matrix can be expressed as follows:


 * $$\underline{K}_{\ n\, x\, n} = A_{e=1}^{nel}\ \ \underline{k}_{\ ned\ x\ ned}^{(e)}$$

where
 * $$\quad \underline{K}$$ - Global stiffness matrix
 * n - total number of global DoFs before eliminating b.c.'s
 * $$\underline{k}^{(e)}$$ - element stiffness matrix of element e
 * ned - number of element DoFs    *note*- ned is much smaller than n
 * $$A_{e=1}^{nel}$$ - assembly operator for e = 1,2,...,nel
 * nel - number of elements

One should note that the assembly operator is necessary rather than a simple summation because each element stiffness matrix is smaller in dimension and only a sub-matrix of the final global stiffness matrix. Therefore the assembly operator denotes implementing element stiffness matrices into the proper corresponding locations within the global stiffness matrix.

Principal of Virtual Work (PVW)
PVW is a concept that was previously used to reduce the global stiffness matrix, $$\underline{K}_{\, 6\, x\, 6}$$, to the modified stiffness matrix, $$\underline{\overline{K}}_{\, 2\, x\, 2}$$.

This concept is key in the derivation of finite element methods for solutions to partial differential equations. For an example, consider the force-displacement relation for a simple bar or spring:


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

Implementing the PVW, equation 3 above is equivalent to:


 * $$ W \cdot (Kd - F) = 0 \qquad \mbox{for all W} \qquad \qquad (4)$$

It should be noted that


 * $$ (3) \Rightarrow (4) \qquad \mbox{trivial}$$


 * $$ (4) \Rightarrow (3) \qquad \mbox{NOT trivial}$$

Since equation 4 is assumed to be true for all W, then careful selection of values of W can infer things about the relation that is equation 3. For example, in the case that W = 1, equation 4 becomes


 * $$ 1 \cdot (Kd - F) = 0 \qquad \Rightarrow \qquad Kd - F = 0$$

thus, proving equation 3.

Matlab Homework
HW: Modify the Matlab code for the two bar truss to solve the 3 bar truss system:

--

Back to pg 12: Method 2 to derive k(e)

-transform a system with 4 degrees of freedom into a system also with 4 degrees of freedom (as opposed to 2) so that the transform matrix is now 4x4 and hopefully invertible.

People who contributed to HW4
Andy Koby 14:44, 17 October 2008 (UTC)

Nik Vitt 02:29, 20 October 2008 (UTC)

Esmail Hadman 11:51, 20 October 2008 (UTC)

William Kurth 12:37, 21 October 2008 (UTC)

Stark 19:09, 24 October 2008 (UTC)