User:EML4500.f08.JAMAMA/Homework4

Methods to Assemble the Global Stiffness Matrix $$K$$


Connecting array "conn," consider the 2-bar truss system (p. 5-6 $$ \rightarrow $$ shown above)



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

Location matrix master array: "lmm" (the same 2-bar truss system (p. 5-6) that is shown above)



lmm(i,j) = equation number (global dof number) for the element stiffness coefficients corresponding to the j -th local dof number.

Alternative Method for Transforming the Local Degrees of Freedom
Method 2 to derive $$ \mathbf{d}_{4 \times 4}^{(e)}: $$

Transform a system with $$ \mathbf{4} $$ dofs into a system with also $$ \mathbf{4} $$ dofs (instead of only $$ \mathbf{2} $$ dofs), so that the transformatin matix is now $$ \mathbf{4 \times 4} $$ and hopefully invertible (local-to-local transformation).

The objective is to transform the set of local degrees of freedom $$\tilde{d}^{(e)}$$ to a different set degrees of freedom $$\tilde{d}^{(e)}_{4X1}$$ using an invertible matrix $$\tilde{T}^{(e)}{_{4X4}}$$.



$$\tilde{d}^{(e)}_{4X1} = \tilde{T}^{(e)}_{4X4}d^{(e)}_{4X1}$$

Lets define equation (1) bellow:

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

$$\tilde{d}_{2}^{(e)} = \overrightarrow{d}_{1}^{(e)}\cdot \overrightarrow{j}$$ (comp. of $$ \overrightarrow{d}_{1}^{(e)}$$ along $$\vec\tilde{j}$$ i.e. y-axis)
 * $$= sin(\theta^{(e)}) + cos(\theta^{(e)}) d_{2}^{(e)}$$

And lets define equation (2) bellow:

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

By putting (1) and (2) in matrix form:

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

$$\underbrace{\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \end{Bmatrix}}_{\tilde{d}_{4X1}^{(e)}} = \underbrace{\begin{bmatrix} R_{2X2}^{(e)}&0_{2X2}\\ 0_{2X2}&R_{2X2}^{(e)}\\ \end{bmatrix}}_{\tilde{T}_{4X4}^{(e)}}\underbrace{\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ \end{Bmatrix}}_{d_{4X1}^{(e)}}$$



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

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

Note: Consider the case: $$\tilde{d}{_{4}}^{(e)}\neq 0$$

$$\tilde{d}{_{1}}^{(e)}=\tilde{d}{_{2}}^{(e)}=\tilde{d}{_{3}}^{(e)}=0$$

$$\tilde{f}^{(e)}=\tilde{k}^{(e)}\tilde{d}^{(e)}=0$$ 4X1  4X4  4X1   4X1

Interp of transverse DOF's.

Recall from Meeting 19: $$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$

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

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

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

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

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

$$\tilde{T}^{(e)}$$=block diagonal matrix (P.19.3)

Consider a general block-diagram matrix:

$$A=\begin{bmatrix} D1 & &  &  &0 \\ & ..&  &  & \\ &  &  ..&  & \\ &  &  &  ..& \\ 0&  &  &  & D5 \end{bmatrix}$$

Question: What is $$A^{-1}$$

Answer using a simpler example: Using a diagonal matrix:

$$B=\begin{bmatrix} d_{11} & &  &  &0 \\ & d_{22}&  &  & \\ & &  d_{33}&  & \\ & &  &  d_{44}& \\ 0& &  &  &d_{mn} \end{bmatrix}= diag\left[d_{11},d_{22},d_{33}....d_{mn} \right]$$

assuming that $$d_{ij}\neq 0 $$ for i=1,....n then $$B^{-1}=diag\left[\frac{1}{d_{11}},\frac{1}{d_{22}},\frac{1}{d_{33}}....\frac{1}{d_{mn}} \right]$$

For a block diagonal matrix A:

$$A=diag\left[D_{1},....D_{5} \right]$$

$$A^{-1}=diag\left[D_{1}^{-1},....D_{5}^{-1} \right]$$

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

P.19-2 $$R^{(e)T}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix}$$

$$R^{(e)T}R^{(e)}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}&l^{(e)} \end{bmatrix}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=I $$ (2x2 identity matrix)

Proof of the Identity Matrix Above

$$R^{(e)T}R^{(e)}=\left[ \begin{matrix} l^{2}+m^{2} & 0 \\ 0 & l^{2}+m^{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\   0 & 1  \\ \end{matrix} \right]=I_{_{2x2}}$$ NOTE: $$\left. \begin{align} & l^{(e)}=\cos \theta \\ & m^{(e)}=\sin \theta \\ \end{align} \right\}\therefore l^{2}+m^{2}=\cos ^{2}\theta +\sin ^{2}\theta =1$$

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

$$\Rightarrow \tilde{T}^{(e)-1}=diag\left[R^{(e)T},R^{(e)T} \right] =\left( diag\left[R^{(e)},R^{(e)} \right]\right)^{T}$$ ...  $$\tilde{T}^{(e)-1}=\tilde{T}^{(e)T}$$

so from earlier it can be shown that: $$\left[ \tilde{T}^{(e)T}\hat{k}^{(e)}\tilde{T}^{(e)}\right]d^{(e)}=f^{(e)}$$

The Meaning of Eigenvectors and Eigenvalues
An eigenvector of an n x n matrix A is a nonzero vector v such that K v =λ v for some scalar λ.

Eigenvalue problem: K v =λ v ( M is the identity matrix)

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

$$K_{_{6x6}}u_{i_{6x1}}=\underbrace{0\cdot u_{i}}_{0_{6x1}},\text{ }i=1,2,3,4$$

The linear combination of {ui, i=1,2,3,4}

$$\sum\limits_{i=1}^{4}{\alpha _{i_{1x1}}u_{i_{6x1}}=:W_{_{6x1}}}=0_{_{1x1}}\cdot W_{_{6x1}}$$

Where $$=:$$ means "equal by definition"

$$\alpha _{i}\text{ is a real number (scalar)}$$

A nonzero solution for d e is possible only if K cannot be inverted. (Because the $$\det K=0$$, the inverse will include a scalar divided by 0, which is an undefined value, therefore making the inverse undefined.) The eigenvalues represent the buckling loads. Substituting each eigenvalue into the global equations, you can compute the nodal displacements. These nodal displacements are the eigenvectors, which are the buckling modes, or mode shapes.

 These mode shapes (corresponding to the zero eigenvalues) may come out as a linear combination of the pure mode shapes. (3 pure rigid body motions and 1 pure mechanism)

For (a), there will be only one zero eigenvalue because the set up is unstable, for instance, you would simply have to push on one of the top nodes to force it out of equilibrium, however, (b) is more stable, and will have more than just one zero eigenvalue.

Justification of the Assembly of the Element Stiffness Matrix $$k^{(e)}$$ into the Global Matrix $$K$$
Consider the 2 bar truss matrix again.

The FD relation for each element is

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

Then using the Euler cut method 2 and look at node 2 with elements 1 and 2.



The sum of the forces at node 2 is zero so

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

$$\sum{F_{y}}=0=-f_{4}^{1}-f_{2}^{2}=0$$

The next step is to utilize the knowledge that the sum of the forces at the global node 2 are equal to zero and substituting each element force with the appropriate force displacement relation: $$k_{4x4}^{(e)}d_{4x1}^{(e)}=f_{4x1}^{(e)}$$


 * For the equations $$f_{3}^{(1)}+f_{1}^{(2)}=0$$ and $$f_{4}^{(1)}+f_{2}^{(2)}=P$$, the substitutions are as follows:

1) Replace $$f_{3}^{(1)}$$, $$f_{1}^{(2)}$$, $$f_{4}^{(1)}$$, and $$f_{2}^{(2)}$$ with the following:

 $$\begin{align} & \text{Row 3: }\underbrace{\left[ k_{31}^{(1)}d_{1}^{(1)}+k_{32}^{(1)}d_{2}^{(1)}+k_{33}^{(1)}d_{3}^{(1)}+k_{34}^{(1)}d_{4}^{(1)} \right]}_{f_{3}^{(1)}}+\underbrace{\left[ k_{11}^{(2)}d_{1}^{(2)}+k_{12}^{(2)}d_{2}^{(2)}+k_{13}^{(2)}d_{3}^{(2)}+k_{14}^{(2)}d_{4}^{(2)} \right]}_{f_{1}^{(2)}}=0 \\ & \\ &  \\ \end{align}$$

$$\text{Row 4: }\underbrace{\left[ k_{41}^{(1)}d_{1}^{(1)}+k_{42}^{(1)}d_{2}^{(1)}+k_{43}^{(1)}d_{3}^{(1)}+k_{44}^{(1)}d_{4}^{(1)} \right]}_{f_{4}^{(1)}}+\underbrace{\left[ k_{21}^{(2)}d_{1}^{(2)}+k_{22}^{(2)}d_{2}^{(2)}+k_{23}^{(2)}d_{3}^{(2)}+k_{24}^{(2)}d_{4}^{(2)} \right]}_{f_{2}^{(2)}}=P$$

2) Now replace all of the local degrees of freedom with their global equivalents.

The equivalent local and global degrees of freedom:



$$\begin{align} & d_{1}=d_{1}^{(1)} \\ & d_{2}=d_{2}^{(1)} \\ & d_{3}=d_{3}^{(1)}=d_{1}^{(2)} \\ & d_{4}=d_{4}^{(1)}=d_{2}^{(2)} \\ & d_{5}=d_{3}^{(2)} \\ & d_{6}=d_{4}^{(2)} \\ \end{align}$$



$$\begin{align} & \text{Row 3: }\underbrace{\left[ k_{31}^{(1)}d_{1}+k_{32}^{(1)}d_{2}+\underbrace{k_{33}^{(1)}d_{3}+k_{34}^{(1)}d_{4}}_{\text{common displacements}} \right]}_{f_{3}^{(1)}}+\underbrace{\left[ \underbrace{k_{11}^{(2)}d_{3}+k_{12}^{(2)}d_{4}}_{\text{common displacements}}+k_{13}^{(2)}d_{5}+k_{14}^{(2)}d_{6} \right]}_{f_{1}^{(2)}}=0 \\ & \\ \end{align}$$

$$\text{Row 4: }\underbrace{\left[ k_{41}^{(1)}d_{1}+k_{42}^{(1)}d_{2}+\underbrace{k_{43}^{(1)}d_{3}+k_{44}^{(1)}d_{4}}_{\text{common displacements}} \right]}_{f_{4}^{(1)}}+\underbrace{\left[ \underbrace{k_{21}^{(2)}d_{3}+k_{22}^{(2)}d_{4}}_{\text{common displacements}}+k_{23}^{(2)}d_{5}+k_{24}^{(2)}d_{6} \right]}_{f_{2}^{(2)}}=P$$

3) Now factor out the common global displacements (noted above):



$$\begin{align} & \text{Row 3: }k_{31}^{(1)}d_{1}+k_{32}^{(1)}d_{2}+\underbrace{\left( k_{33}^{(1)}+k_{11}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 3 } \\ {} & \text{ of the } \\ {} & \text{Global Stiffness Matrix} \\ {} & K \\ \end{matrix}}\left( d_{3} \right)+\underbrace{\left( k_{34}^{(1)}+k_{12}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 4 } \\ {} & \text{of the } \\ {} & \text{Global Stiffness Matrix} \\ {} & K \\ \end{matrix}}\left( d_{4} \right)+k_{13}^{(2)}d_{5}+k_{14}^{(2)}d_{6}=0 \\ & \\ &  \\ \end{align}$$

$$\text{Row 4: }k_{41}^{(1)}d_{1}+k_{42}^{(1)}d_{2}+\underbrace{\left( k_{43}^{(1)}+k_{21}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 3 } \\ {} & \text{          of the }  \\ {} & \text{Global Stiffness Matrix} \\ {} & \text{             }K  \\ \end{matrix}}\left( d_{3} \right)+\underbrace{\left( k_{44}^{(1)}+k_{22}^{(2)} \right)}_{\begin{matrix} {} & \text{Addends for Row 4 } \\ {} & \text{          of the }  \\ {} & \text{Global Stiffness Matrix} \\ {} & \text{             }K  \\ \end{matrix}}\left( d_{4} \right)+k_{23}^{(2)}d_{5}+k_{24}^{(2)}d_{6}=P$$

By factoring these out, the necessity of adding the $$3^{rd}\text{ and }4^{th}$$ rows of the global stiffness matrix $$K$$ has been proven for both equilibrium equations derived above. For a visual of both rows, the global stiffness matrix is reproduced below:

 $$K=\begin{matrix} {} \\   {}  \\   Row\text{ }3  \\ Row\text{ }4 \\ {} \\   {}  \\ \end{matrix}\left[ \begin{matrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 \\ k_{31}^{(1)} & k_{32}^{(1)} & \left( k_{33}^{(1)}+k_{11}^{(2)} \right) & \left( k_{24}^{(1)}+k_{12}^{(2)} \right) & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)} & \left( k_{43}^{(1)}+k_{21}^{(2)} \right) & \left( k_{44}^{(1)}+k_{22}^{(2)} \right) & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \\ \end{matrix} \right]$$

Assembly of the Element Stiffness Matrix
$$k^{(\text{e})}:\text{ e}=1,...,nel$$, into global stiffness matrix $$K$$.

$$K_{_{n\text{ x }n}}=\underset{\text{e}=1}{\overset{nel}{\mathop{A}}}\,k_{_{ned\text{ x }ned}}^{(\text{e})}$$

Where A is the assembly operator. Note: $$\sum – $$ can be used because it implies that $$k^{(\text{e})}$$ has the same dimensions and $$K$$. For example, if each element in a truss has 4 degrees of freedom, and there are 1 million global degrees of freedom, $$k^{(\text{e})}$$ is still a $$4x4$$ matrix.

It is now important to introduce:  $$\begin{array}{*{35}l} {} & n:\text{ the total number of global degrees of freedom before eliminating the boundary conditions}  \\ {} & {} \\   {} & \begin{align} & ned\text{: the number of element degrees of freedom} \\ & \\ & nel:\text{ the number of elements} \\ \end{align} \\ \end{array}$$

In general,

$$ned\ll \text{ }n$$, where $$\ll $$ means "very small compared to".

Principle of Virtual Work (PVW)

The PVW justifies how to eliminate the rows corresponding to the boundary conditions to obtain the global matrix $$K$$. (The columns are eliminated by the boundary conditions alone.)

Recall,

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

and

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


 * We now begin: Deriving the Finite Element Method for Partial Differential Equations

The force displacement relationship for a bar or spring: $$kd=F$$

$$kd=F\underbrace{\Rightarrow }_{implies}kd-F=0$$

However,

$$kd=F\underbrace{\Leftrightarrow }_{equivalent}W(kd-F)=0,\text{ }\underbrace{\text{for all }W}_{\begin{smallmatrix} **\text{Extremely } \\ \text{    Important**} \end{smallmatrix}}$$

Where $$W$$ is the weighting coefficient, and this is also known as the "weak form" $$\equiv $$ PVW

We begin the proof of why $$kd=F\Rightarrow kd-F=0$$ is trivial, while $$kd=F\Leftrightarrow W(kd-F)=0$$ is not.

Because $$W(kd-F)=0,\text{ for all }W$$, you can select any $$W$$:

For example, if $$W=1$$, then $$W(kd-F)=0$$ becomes $$\left[ (1)(kd-F)=0 \right]\Rightarrow \left[ kd-F=0 \right]$$, which is not trivial, because it is specific for $$W=1$$.

Contributing Team Members
Megan Alvarez --EML4500.f08.jamama.megan 23:19, 23 October 2008 (UTC)

Arunas Janulevicius --Eml4500.f08.jamama.jan 09:24, 24 October 2008 (UTC)

Chris Alford --Eml4500.f08.jamama.chris 16:50, 24 October 2008 (UTC)

Cedric Adam --Eml4500.f08.jamama.adam 18:20, 24 October 2008 (UTC)

William Mueller Eml4500.f08.jamama.mueller 19:02, 24 October 2008 (UTC)

Justin McIntire --Eml4500.f08.jamama.justin 19:29, 24 October 2008 (UTC)