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)