User:Eml4500.f08.wiki1.aguilar/TeamHW4

Continued Assembly Instructions for a Three-Bar Truss
To continue, we're going to need to be familiar with a "connectivity array" and a "location matrix master array."

A Connectivity array, "conn" is shown below. (Consider the two-bar truss system from your class notes, p.5-6)



Conn(e,j) = global node number of local node j of element e

We also want to establish a Location Matrix Master array: "lmm".



lmm (i,j) = equation number (global degree of freedom number) for element stiffness coefficient corresponding to the jth local degree of freedom number.

Going back to p.12-1, Method 2 to derive $$\mathbf{k}^{(e)}_{4x4} $$

Finding a $$\tilde {\mathbf{T}}^{(e)}$$ that is Invertible
Goal: Our goal is to find a $$\mathbf{\tilde{T}}^{(e)}_{4x4}$$ that can transform the set of local element degrees of freedom, $$\mathbf{d}^{(e)}_{4x1} $$ to another set of local element degrees of freedom, $$\mathbf{\tilde{d}}^{(e)}_{4x1} $$, such that $$\mathbf{\tilde{T}}^{(e)}_{4x4}$$ is invertible.



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

We illustrate this below:

$${\tilde{d}^{(e)}}_{1}={\vec{d}^{(e)}}_{[1]}\cdot \vec{\tilde{j}}$$ This is the component of vector $${\vec{d}^{(e)}}_{[1]}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis. $${\tilde{d}^{(e)}}_{1}=-\sin \theta ^{(e)} d^{(e)}_{1}+\cos\theta ^{(e)} d^{(e)}_{2}$$

This is also done for $${\tilde{d}^{(e)}}_{2}$$. $${\tilde{d}^{(e)}}_{2}= \begin{bmatrix} l^{(e)} &m^{(e)} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\end{Bmatrix}(2)$$

$${\tilde{d}^{(e)}}_{2}={\vec{d}^{(e)}}_{[2]}\cdot \vec{\tilde{j}}$$ This is the component of vector $${\vec{d}^{(e)}}_{[2]}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis. $${\tilde{d}^{(e)}}_{2}=-\sin \theta ^{(e)} d^{(e)}_{1}+\cos\theta ^{(e)} d^{(e)}_{2}$$

If we put equations (1) and (2) in matrix form:

 $$ \begin{Bmatrix}{\tilde{d}^{(e)}}_{1} \\ {\tilde{d}^{(e)}}_{2} \\ {\tilde{d}^{(e)}}_{3} \\ {\tilde{d}^{(e)}}_{4}\end{Bmatrix}= \begin{bmatrix} \mathbf{R}^{(e)}_{2x2} & 0 \\ 0 & \mathbf{R}^{(e)}_{2x2} \end{bmatrix} \begin{Bmatrix}d^{(e)}_{1} \\ d^{(e)}_{2} \\ d^{(e)}_{3} \\ d^{(e)}_{4}\end{Bmatrix} $$

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



$$\mathbf{\tilde f}^{(e)}=k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\mathbf{\tilde d}^{(e)}$$ Note: the zeroes arise because there is no stretching in the transverse direction.

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

Note: Consider the case: $${\tilde{d}^{(e)}}_{4}\neq 0$$ $${\tilde{d}^{(e)}}_{1}={\tilde{d}^{(e)}}_{2}={\tilde{d}^{(e)}}_{3}=0$$ $$\mathbf{\tilde f}^{(e)}_{4x1}=\mathbf{\tilde k}^{(e)}_{4x4} \mathbf{\tilde d}^{(e)}_{4x1}=\mathbf{0}$$ Here, $$\mathbf{0}_{4x1}$$ corresponds to the 4th column of $$\mathbf{\tilde k}^{(e)}$$ This is the interpolation of transverse degrees of freedom.

From our notes on p.19-3, we recall that:

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

We use the same argument for:

<p style="text-align:center;">$${\tilde{\mathbf{f}}^{(e)}}={\tilde{\mathbf{T}}^{(e)}} {\mathbf{f}}^{(e)}$$

Also:

<p style="text-align:center;">$${\tilde{\mathbf{k}}^{(e)}} {\tilde{\mathbf{d}}^{(e)}} ={\tilde{\mathbf{f}}^{(e)}}$$

Which leads to:

<p style="text-align:center;">$${\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}} {\mathbf{d}}^{(e)}= {\tilde{\mathbf{T}}^{(e)}} {\mathbf{f}}^{(e)}$$

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

<p style="text-align:center;">$$[{\tilde{\mathbf{T}}^{(e)^{-1}}}{\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}}]{\mathbf{d}}^{(e)}={\mathbf{f}}^{(e)}$$

$${\tilde{\mathbf{T}}^{(e)}}$$ is a block diagonal matrix, as you can see by referring back to your notes on p. 19-3.

Consider a general block diagonal matrix:

<p style="text-align:center;">

<p style="text-align:center;">Q: What is $$\mathbf{A}^{-1}$$?

Simple example:

<p style="text-align:center;">

<p style="text-align:center;">$$\mathbf{B}=Diag[d_{11},d_{22},...,d_{nn}]$$

So, <p style="text-align:center;">$$\mathbf{B}^{-1}=Diag[\frac{1}{d_{11}},\frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$ Assuming $$d_{ii}\neq 0$$ for i=1,...,n.

For a block diagonal matrix A:

<p style="text-align:center;">$$\mathbf{A}=Diag[\mathbf{D}_{1},\mathbf{D}_{2},...,\mathbf{D}_{s}]$$

So, <p style="text-align:center;">$$\mathbf{A}^{-1}=Diag[\mathbf{D}_{1}^{-1},\mathbf{D}_{2}^{-1},...,\mathbf{D}_{s}^{-1}]$$

<p style="text-align:center;">$$\tilde{\mathbf{T}}^{(e)^{-1}} = Diag[\mathbf{R}^{(e)^{-1}} ,\mathbf{R}^{(e)^{-1}}  ]$$ And from p19-2, <p style="text-align:center;">$$\tilde{\mathbf{R}}^{(e)^{T}}=\begin{bmatrix}l^{(e)} &-m^{(e)} \\ m^{(e)} &l^{(e)} \end{bmatrix}$$

And, <p style="text-align:center;">$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}l^{(e)} &-m^{(e)} \\ m^{(e)} &l^{(e)} \end{bmatrix}\begin{bmatrix}l^{(e)} &m^{(e)} \\ -m^{(e)} &l^{(e)} \end{bmatrix}$$ Thus:$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}l^{(e)^{2}} m^{(e)^{2}} &0 \\ 0 &l^{(e)^{2}} m^{(e)^{2}} \end{bmatrix}$$

Which simplifies to:<p style="text-align:center;">$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}1&0 \\ 0 &1\end{bmatrix}$$

Which gives us $$\mathbf{I}_{2x2}$$, an identity matrix.

We can see that:<p style="text-align:center;">$$ \mathbf{R}^{(e)^{-1}} = \tilde{\mathbf{R}}^{(e)^{T}}$$

<p style="text-align:center;">$$\tilde{\mathbf{T}}^{(e)^{-1}} = Diag[\mathbf{R}^{(e)^{T}} ,\mathbf{R}^{(e)^{T}} ]$$

And then, <p style="text-align:center;">$$\tilde{\mathbf{T}}^{(e)^{-1}} = (Diag[\mathbf{R}^{(e)} ,\mathbf{R}^{(e)} ])^{T}$$

And then we get:

<p style="text-align:center;">$$\tilde{\mathbf{T}}^{(e)^{-1}} = \tilde{\mathbf{T}}^{(e)^{T}}$$

Going back to our notes on p.20-1: $${\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}} {\mathbf{d}}^{(e)}= {\tilde{\mathbf{T}}^{(e)}} {\mathbf{f}}^{(e)}$$ which is equal to :$$[{\tilde{\mathbf{T}}^{(e)^{T}}}{\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}}]{\mathbf{d}}^{(e)}={\mathbf{f}}^{(e)}$$

and comparing it with $${\tilde{\mathbf{k}}^{(e)}} {\tilde{\mathbf{d}}^{(e)}} ={\tilde{\mathbf{f}}^{(e)}}$$,

we finally get:

<p style="text-align:center;">$$\mathbf{k}^{e}=\tilde{\mathbf{T}}^{(e)^{T}} \tilde{\mathbf{k}}^{(e)} \tilde{\mathbf{T}}^{(e)}$$

The roles of Eigenvectors and Eigenvalues
Eigenvalue problem: $$\mathbf{kv} = \lambda \mathbf{v}$$

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

<p style="text-align:center;">$$\mathbf{ku_i} = 0\mathbf{u_i} = \mathbf{0}\left(i = 1,2,3,4 \right)$$

<p style="text-align:center;">Linear combination of $$\left\{u_i, i = 1,2,3,4 \right\} $$

<p style="text-align:center;"> = \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W} <p style="text-align:center;">$$ \sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \equiv \mathbf{W} $$ (equal by definition)

<p style="text-align:center;">$$ \alpha_i \Rightarrow $$(real numbers)

W is also an eigenvector corresponding to a zero eigenvalue:

<p style="text-align:center;">$$ \mathbf{kW} = \mathbf{k}\left(\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \right) $$

<p style="text-align:center;">$$= \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}$$

Justification of assembly of element stiffness matrix $$\left(\mathbf{k^{(e)}}, e = 1,..., nel \right)\Rightarrow nel =$$ number of element

into global stiffness matrix $$\left(\mathbf{K}\right)$$

Consider example of 2-bar truss, (Mtg. 9):

Recall elemental FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

Mtg. 11: Euler cut principle, Method 2 (equilibrium of global node 2)

Mtg. 4: Free Body Diagrams of element 1 and element 2 with element degrees of freedom $$\mathbf{d}^{(e)}_{4\times 1}$$

Mtg. 8: For node 2, identify the relationships between the global degrees of freedom and the element degrees of freedom for elements 1 and 2

Equilibrium of node 2:

<p style="text-align:center;"> <p style="text-align:center;"> Free Body Diagram Using Euler Cut Principle

$$(1) \sum{F_x} = 0 = -f^{(1)}_3 - f^{(2)}_1 = 0$$

$$(2) \sum{F_y} = 0 = P - f^{(1)}_4 - f^{(2)}_2 = 0$$

Next we use the FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

<p style="text-align:center;">$$\mathbf{Eq. (1) \ }\Rightarrow f^{(1)}_3 + f^{(2)}_1 = 0  \mathbf{\left( 1\right)}$$

<p style="text-align:center;">$$\mathbf{Eq. (2) \ }\Rightarrow f^{(1)}_4 + f^{(2)}_2 = P \mathbf{\left( 2\right)}$$

<p style="text-align:center;">$$\mathbf{(1):}\left[k^{(1)}_{31}d^{(1)}_1 + k^{(1)}_{32}d^{(1)}_2 + ... + k^{(1)}_{34}d^{(1)}_4\right] \equiv f^{(1)}_3$$ <p style="text-align:center;">$$+ \left[k^{(1)}_{11}d^{(2)}_1 + ... + k^{(2)}_{14}d^{(2)}_4\right] \equiv f^{(2)}_1 = 0$$

Mtg. 8: Rewrite local degrees of freedom to global degrees of freedom:

<p style="text-align:center;">$$\mathbf{\left(1 \right):} \left[k^{(1)}_{31}d_1 + k^{(1)}_{32}d_2 + k^{(1)}_{33}d_3 + k^{(1)}_{34}d_4\right]$$

<p style="text-align:center;">$$+ \left[k^{(2)}_{11}d_3 + k^{(2)}_{12}d_4 + k^{(2)}_{13}d_5 + k^{(2)}_{14}d_6\right] = 0$$

$$\Rightarrow $$ Thus obtain the 3rd row of $$\mathbf{k}$$ (Mtg. 9) (See also Mtg. 8)

Assembly of $$\mathbf{k^{(e)}}, e = 1,...,nel $$, into global stiffness matrix $$\mathbf{K}$$:

<p style="text-align:center;">$$\mathbf{K_{n\times n}= }A^{nel}_{e=1}\mathbf{k^{(e)}_{ned\times ned}}$$

<p style="text-align:center;">A: assembly operator <p style="text-align:center;">n: total number of global dof's before eliminating boundary conditions <p style="text-align:center;">ned: number of element dof's <p style="text-align:center;">(ned<<n)

Principle of Virtual Work (PVW)
Mtg. 10: Elimination of rows corresponding to the boundary conditions to obtain $$\mathbf{\overline{k}_{2\times 2}} $$

Mtg. 12: $$\mathbf{q^{(e)}_{2\times 1}=T^{(e)}_{2\times 4}d^{(e)}_{4\times 1}}$$

Mtg. 14: $$\mathbf{k^{(e)}=T^{(e)^T}\hat{k}^{(e)}T^{(e)}}$$

Deriving FEM for partial differential equations (PDE's):

FD relation for a bar or a spring: $$kd = F$$

<p style="text-align:center;">$$\Rightarrow kd - F = 0\;\;\;\;\; (3)$$

<p style="text-align:center;">$$\Leftrightarrow W(kd - F) = 0\;\;\;\left(for \, all\, W\, \right) \;\;\;\;\;(4)$$

<p style="text-align:center;">W: weighting coefficient <p style="text-align:center;">"weak form" $$\equiv $$ PVW

 Proof :

$$A) \; \left(3 \right)\Rightarrow (4): trivial$$

$$B) \; \left(4 \right)\Rightarrow (3): \mathbf{NOT\; TRIVIAL}$$

Since $$\left(4 \right)$$ is valid for all $$W$$, select $$W=1$$ Then $$\left(4 \right)$$ becomes:

<p style="text-align:center;">$$1\times (kd - F) = 0 \Rightarrow \left(3 \right)$$

Other Resources
ADINA FEA Eigen Frequency Models and Animations Eigenvalues and Eigenvectors on Wikipedia

Contributors
Eml4500.f08.wiki1.schaet 20:09, 24 October 2008 (UTC) Eml4500.f08.wiki1.aguilar 19:03, 23 October 2008 (UTC) Eml4500.f08.wiki1.oatley 20:37, 21 October 2008 (UTC) Eml4500.f08.wiki1.ambrosio 20:40, 23 October 2008 (UTC) Eml4500.f08.wiki1.brannon 01:52, 24 October 2008 (UTC) Eml4500.f08.wiki1.handy 21:27, 23 October 2008 (UTC)