User:Eml4500.f08.echo.mott/HW4

= Homework Report 4=
 * This homework report contains lecture notes from 10/6/08 to 10/17/08 and Matlab solutions requested in class.
 * The previous version of the page was submitted into the HW report table.

Lecture Notes
Connectivity Array


 * Consider the 2-bar truss system described earlier.


 * $$\mathbf {conn}=

\begin{Bmatrix} 1 & 2\\ 2 & 3 \end{Bmatrix}$$

Where
 * Columns represent local nodes (ie. column 1 = local node 1)
 * Rows represent elements (ie. row 1 = element 1)
 * Numbers within matrix represent global node numbers

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

Location Matrix Master Array

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

Where
 * Columns represent the local degree of freedom number (ie. 1 = 1)
 * Rows represent the element numbers (ie. 1= 1)
 * Numbers within matrix represent the global degree of freedom number, or the equivalent number, in k

lmm (i,j)= equivalent number (global degree of freedom number) for the element stiffness coefficient corresponding to the $$ j^{th} $$ local degree of freedom number.

Goal: Want to find $$\mathbf{\tilde{T}}_{4x4}^{(e)}$$ that transforms the set of local elements dof's $$\mathbf d_{4x1}^{(e)}$$ to another set of local element dof's $$\mathbf{\tilde{d}}_{4x1}^{(e)}$$ such that $$\mathbf{\tilde{T}}_{ }^{(e)}$$ is invertible. The tilde symbol refers to displacements and forces in local axial directions.



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

$$\tilde{d_1^{(e)}}=q_1^{(e)}= \begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix} \qquad \qquad (1)$$

$$\tilde{d}_{2}^{(e)}=d_1^{(e)}\cdot\vec{\tilde{j}}$$ (component of $$\vec{d}_1^{(e)}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis)

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

For HW, derive the following: $$\tilde{d}_{2}^{(e)}=\begin{vmatrix} -m_{ }^{(e)} & l_{ }^{(e)} \end{vmatrix} \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix} \qquad \qquad (2)$$

Beginning with the transformation equation,

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

$$ \begin{bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{e)} \\ \tilde{d}_3^{(e)} \\ \tilde{d}_4^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{bmatrix} $$

$$ \begin{align} \tilde{d}_2^{(e)} &= \begin{bmatrix} -m^{(e)} & l^{(e)} & 0 & 0 \end{bmatrix} \begin{bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{bmatrix} \\ &= \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{bmatrix} \end{align}                    $$

Put (1) & (2) in matrix form: $$\begin{Bmatrix} \tilde{d_1^{(e)}} \\ \tilde{d_2^{(e)}} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}$$

$$\begin{Bmatrix} \tilde{d_1^{(e)}} \\ \tilde{d_2^{(e)}} \\ \tilde{d_3^{(e)}} \\ \tilde{d_4^{(e)}} \end{Bmatrix}=\begin{bmatrix} \mathbf{R}_{2x2}^{(e)} & \mathbf{0}_{2x2} \\ \mathbf{0}_{2x2} & \mathbf{R}_{2x2}^{(e)} \end{bmatrix}\begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)}\end{Bmatrix}$$

In simpler terms, this can be notated as:

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

For an example, take the following diagram:



This can be described as:

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

This can be simplified and summarized as:

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

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

and

$$\tilde{d}_1^{(e)}=\tilde{d}_2^{(e)}=\tilde{d}_3^{(e)}=0$$

Starting with the element axial F-D relation,

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

The zero vector being the 4th column of $$\tilde{\mathbf{k}}^{(e)}$$

Interpretation of the transverse dof's:

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

We have derived that $$\tilde{\mathbf{d}}^{(e)} = \tilde{\mathbf{T}}^{(e)} {\mathbf{d}}^{(e)}$$.

Similarly we would have the same relation for $$\tilde{\mathbf{f}}$$ and $${\mathbf{f}}$$:

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

We can write the displacement relation differently as

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

By substitution,

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

If $$\tilde{\mathbf{T}}^{(e)}$$ is invertible, then the inverse can be used and the force matrix becomes:

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

$$\tilde{\mathbf{T}}^{(e)}$$ is a block diagonal matrix.

Let's consider a more general block diagonal matrix consisting of s individual matrices :

$$\mathbf {A} = \begin{bmatrix} \mathbf{D}_1 & 0 & 0 \\ 0 & \ \ddots \quad & 0  \\ 0 & 0 & \mathbf{D}_{s} \end{bmatrix}$$

What would be the inverse of this matrix $$\mathbf{A}^{-1}$$?

Let us look at a simpler diagonal matrix:

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

Assuming that the values of $$d_{11}, d_{22} , ... , d_{nn}$$ do not equal zero, the inverse of $$\mathbf{B}$$ can be written as:

$$\mathbf{B}^{-1} = diag [ 1/d_{11}, 1/d_{22} , ... , 1/d_{nn} ] $$

For the general block diagonal matrix $$\mathbf{A}$$:

$$\mathbf{A} = diag [ \mathbf{D}_{11}, \mathbf{D}_{22} , ... , \mathbf{D}_{ss} ] $$

and

$$\mathbf{A}^{-1} = diag [ {\mathbf{D}_{11}}^{-1}, {\mathbf{D}_{22}}^{-1} , ... , {\mathbf{D}_{ss}}^{-1} ] $$

Applying this theory to $$ \tilde{\mathbf{T}}$$,

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

From previous notes it is shown that:

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

The identity matrix is found when R is multiplied by its transpose, concluding that the transpose is equivalent to the inverse.

$${\mathbf R}^{(e)^{T}}{\mathbf R}^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)^{2}} + {m}^{(e){2}}&0 \\0& l^{(e)^{2}} + {m}^{(e)^{2}} \end{bmatrix} = \begin{bmatrix} 1&0 \\0& 1 \end{bmatrix} $$

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

Now $$ \tilde{\mathbf{T}}$$ can be written as:

$$ \begin{align} {\tilde{\mathbf{T}}}^{(e)^{-1}} &= diag [ {\mathbf{R}}^{(e)^{T}} ,{\mathbf{R}}^{(e)^{T}}] \\ &= (diag [ {\mathbf{R}}^{(e)} ,{\mathbf{R}}^{(e)}])^{T}\\ &= {\tilde{\mathbf{T}}}^{(e)^{T}} \\ \end{align} $$

This means that:

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

HW - Verify that $$ {\mathbf{k}}^{(e)} = [{\tilde{\mathbf{T}}}^{(e)^{T}} {\tilde{\mathbf{k}}}^{(e)} {\tilde{\mathbf{T}}}^{(e)}]$$

$$ \begin{align} {\mathbf{k}}^{(e)} &= \begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 \\ m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & -m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \\ &= \begin{bmatrix} l^{(e)} & 0 & -l^{(e)} & 0 \\ m^{(e)} & 0 & -m^{(e)} & 0 \\ -l^{(e)} & 0 & l^{(e)} & 0 \\ -m^{(e)} & 0 & m{(e)} & 0 \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \\ &= \begin{bmatrix} l^{(e)^2} & l^{(e)}m^{(e)} & -l^{(e)^2} & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & m^{(e)^2} & -l^{(e)}m^{(e)} & -m^{(e)^2} \\ -l^{(e)^2} & -l^{(e)}m^{(e)} & l^{(e)^2} & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -m^{(e)^2} & l^{(e)}m^{(e)} & m^{(e)^2}\\ \end{bmatrix} \\ \end{align} $$

Reference [this page]

Eigenvalues problem: $$ \vec{k} \vec{v}=\lambda \vec{v}$$

let $$ \begin{bmatrix} \vec{u}_1 & \vec{u}_2 & \vec{u}_3 & \vec{u}_4 \end{bmatrix}$$ be the pure eigenvalues corresponding to 4 zero eigenvalues

$$ \vec{k}_{6x6} \vec{u}_{6x1}=0 \cdot \vec{u}_i, i=1 , ...4$$

The zero dot product with $$u_i$$ will yield a 6x1 matrix. In this case the zero is a scalar.

linear combination of $$ {\vec{u}_i, i=1, ...4}$$

$$\sum_{i=1}^4 \alpha_i \vec{u}_i=: \vec {W}_{6x1} $$

where $$ \alpha_i $$ represents real numbers.

$$ \vec {W}$$ is also an eigenvector corresponding to a zero eigenvalue:

$$ \vec {k} \vec{W} = \vec {k} (\sum_{i=1}^4 \alpha_i \vec {u}_i)$$


 * $$= \sum_{i=1}^4 \alpha_i ( \vec {k} \vec{u}_i) = \vec{0}_{6x1}$$


 * $$= 0_{1x1} \times \vec{W}_{6x1}$$

HW:



$$\vec{k} \vec{v} = \lambda \vec{v}$$

$$ \vec {k} $$ stiffness matrix for constrained system

Plot eigenvector corresponding to zero eigenvalue for case (a).

Assume $$L_{}^{1} = L_{}^{2} = L_{}^{3} = 1$$, $$E_{}^{1} = E_{}^{2} = E_{}^{3} = 1$$, and $$A_{}^{1} = A_{}^{2} = A_{}^{3} = 1$$

For case A the global Stiffness matrix was found to be the Following:

$$K = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 \end{bmatrix}$$

When run through MATLAB the following results were recieved:

V = 1.0000        0         0         0         0         0         0         0         0   -0.7071         0         0         0         0         0   -0.7071         0         0   -0.7071         0         0   -0.7071         0         0         0   -0.7071         0         0         0         0         0    0.7071         0         0   -0.7071         0         0    0.7071         0         0         0         0         0         0   -0.7071         0   -0.7071         0         0         0         0    1.0000         0         0         0         0         0         0         0         0   -0.7071         0    0.7071         0

D = 0    0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     2     0     0     0     0     0     0     0     0     2     0     0     0     0     0     0     0     0     2

With the diagonal of D being the eigenvectors.

Eigenvector Plot for Case A:

Justification of assembly of element stiffness matrix ($$\mathbf{k}^{(e)}$$) into global stiffness matrix ($$\mathbf{K}$$) where e=1...number of elements

Consider the example of the 2-bar truss from Homework 1. Recall the element force displacement relation: $$\mathbf{k}_{4x4}^{(e)}\mathbf{d}_{4x1}^{(e)}=\mathbf{f}_{4x1}^{(e)}$$

The methods and steps previously examined were:

- The Euler cut principle, method 2 (Equilibrium of global node 2), shown near the end of Homework 2.

- Free body diagrams of element 1 & 2. Element dof's $$\mathbf{d}_{4x1}^{(e)}$$, shown in Homework 1.

- For node 2, identifying the global dof's to element dof's for both element 1 & 2. This is shown in detail in the below illustration. Notice that the forces are equal and opposing.



Using the summation of force and FD relations:

$$\sum F_X=0=-f_3^{(1)}-f_1^{(2)}=0$$

$$\sum F_Y=0=P-f_4^{(1)}-f_2^{(2)}=0$$

Next, use element force displacement relations such that $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

Taking the summation of forces for Node 2 and rearranging:

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

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

Writing the forces in terms of displacements and element stiffness:

$$f_3^{(1)}=\begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}$$

$$f_1^{(2)}=\begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}$$

Substituting the previous two equations into Equation 1:

$$\begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}+\begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}=0$$

Thus we obtain the 3rd row of the Global Stiffness Matrix. Follow the same link for $$\mathbf{k}_{6x6}\mathbf{d}_{6x1}=\mathbf{F}_{6x1}$$

Homework: Derive details of Equation 2 above & the assembly of row 4 in $$\mathbf{K}\mathbf{d}=\mathbf{F}$$

$$f_4^{(1)}=\begin{bmatrix} k_{41}^{(1)}d_{1}^{(1)} + k_{42}^{(1)}d_{2}^{(1)} + k_{43}^{(1)}d_{3}^{(1)} + k_{44}^{(1)}d_{4}^{(1)} \end{bmatrix}$$

$$f_2^{(2)}=\begin{bmatrix} k_{21}^{(2)}d_{1}^{(2)} + k_{22}^{(2)}d_{2}^{(2)} + k_{23}^{(2)}d_{3}^{(2)} + k_{24}^{(2)}d_{4}^{(2)} \end{bmatrix}$$

Substituting $$f_4^{(1)}$$ and $$f_2^{(2)}$$ into Equation 2:

$$\begin{bmatrix} k_{41}^{(1)}d_{1}^{(1)} + k_{42}^{(1)}d_{2}^{(1)} + k_{43}^{(1)}d_{3}^{(1)} + k_{44}^{(1)}d_{4}^{(1)} \end{bmatrix}+\begin{bmatrix} k_{21}^{(2)}d_{1}^{(2)} + k_{22}^{(2)}d_{2}^{(2)} + k_{23}^{(2)}d_{3}^{(2)} + k_{24}^{(2)}d_{4}^{(2)} \end{bmatrix}=P$$

The assembly of $$\mathbf{K}^{(e)} $$ where e goes from 1 to the number of elements.

Global Stiffness Matrix K:

$$\mathbf{K}_{nxn}=A_{e=1}\mathbf{k}^{(e)}$$

n: Total number of global dof's before eliminating boundary conditions

ned: Number of element dof's.

Note: ned<<n

A: Assembly Operator

Principle of Virtual Work (PVW) - here


 * -Elimination of rows corresponding to boundary conditions to obtain $$\mathbf{\bar k}_{2x2}$$


 * -$$\mathbf{q}_{2x1}^{(e)}=\mathbf{T}_{2x4}^{(e)}\mathbf{d}_{4x1}^{(e)}$$

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


 * Deriving FEM for Partial Differential Equations (PDE's)


 * FD relation for a bar or spring $$ Kd=F $$ implies:

$$K_{ }^{ }d-F=0 \qquad (3)$$


 * Equivalently, with a weighting coefficient it can be simplified to:

$$w(K_{ }^{ }d-F)=0 \qquad (4)$$  ←for all W


 * Proof that (3) ↔ (4):

A) Going from equation (3)→(4)

If equation (3) = 0, equation (4) must also equal 0 (trivial)

B) Going from equation (4)→(3)

If equation (4) = 0, equation (4) must also equal 0. (not trivial)

Since equation (4) is valid for all w, we select w=1.

Equation (4) becomes $$ (Kd-F)=0 $$

Contributing Team Members
Chris Mott 17:04, 24 October 2008 (UTC)

Brandon Sell 17:06, 24 October 2008 (UTC)

Charles Marshall 18:49, 24 October 2008 (UTC)

Chris Sell 19:10, 24 October 2008 (UTC)

Jason Bruce 19:10, 24 October 2008 (UTC)

Bradley LaCroix 20:42, 24 October 2008 (UTC)

Past Homeworks
[Homework 3]

[Homework 2]

[Homework 1]