User:Eml4500.f08.Ateam.paquette/Homework5

 Lecture notes 24 October 20, 2008 

Justification of eliminating rows 1,2,5,6.
The goal of eliminating rows 1,2,5 and 6 in a stiffness matrix is to find $$\mathbf {K_{2x2}}$$ in the original 2 bar truss shown below:



The Force Displacement relationship can be described as: $$\mathbf {K_{6x6}d_{6x1} = F_{6x1}}$$

This can be rearranged into the equation $$\mathbf {K_{6x6}d_{6x1} - F_{6x1} = 0_{6x1}}$$ (1)

For the Principle of Virtual Work (PVW):

$$\mathbf {W_{6x1}\cdot (K_{6x6}d_{6x1} - F_{6x1}) = 0_{1x1}}$$ (2)

This is for all $$\mathbf {W_{6x1}}$$, also known as the Weighting Matrix.

As seen above, it is trivial when going from equation 1 to equation 2, but it is very important to understand how to get from equation 2 to equation 1. The next part will be verifying the steps in going from equation 2 to equation 1.

Eq. 1 $$\Leftrightarrow $$ Eq. 2
When proving Eq. 2 $$\Leftrightarrow $$ Eq. 1, choices have to be made for what the Weighting Coefficient will equal.

*Choice 1

Let $$\mathbf {W_1 = 1}$$ and $$\mathbf {W_2 = W_3 = W_4 = W_5 = W_6 = 0}$$

This gives the Weighting Matrix $$ W = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$1\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The only equation that is used is the first one since the rest become zero.

$$ 1\cdot\left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}}\right]$$

or

$$ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}}$$

*Choice 2

Next $$\mathbf W_2$$ will be equal to 1 while the rest of the weighting matrix is equal to 0.

$$ W = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

The same process will be taken to find the next part of the equation.

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

Again all of the equations are going to become zero, but this time the second equation will be the only one that matters since does not have a zero coefficient. The final equation for choice 2 is:

$$ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}}$$

*Choice 3

$$\mathbf W_3$$ for this choice will be the factor equal to 1.

$$ W = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 3 becomes:

$$ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}$$

*Choice 4

$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 4 becomes:

$$ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}$$

*Choice 5

$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 5 becomes:

$$ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}$$

*Choice 6

Choice 6 is not any different and it ends the process of trying to prove that going from the equation with a weighting coefficient to the force displacement relationship is not trivial.

$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 6 becomes:

$$ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}$$

Boundary Conditions
For the Principle of Virtual Work, the boundary conditions need to be accounted for. In the 2-bar truss, $$\mathbf {d_1 = d_2 = d_5 = d_6 = 0 }$$. The only two displacements the matter are $$\mathbf d_3$$ and $$\mathbf d_4$$ in the two bar truss. Weighting coefficients need to be kinematically admissible, which means they cannot violate the boundary conditions. This concludes that $$ \mathbf {W_1 = W_2 = W_5 = W_6 = 0 }$$. According to the calculus of variations, the weighting coefficients are equivalent to the virtual displacements.

The original equation $$\mathbf({W)\cdot({K} {d} - {F}) = 0}$$ can now be expressed in matrix form and can be further reduced.

$$ (W)_{6x1}K_{6x6} \begin{bmatrix} d_1=0 \\ d_2=0 \\ d_3=x  \\ d_4=y \\ d_5=0 \\ d_6=0  \end{bmatrix}_{6x1}- F_{6x1} = 0$$

Note that x and y are arbitrary values not equal to zero in this problem. The principle of virtual work allows the equation to be condensed to only what is needed and all of the zeros to be removed. This new equations will be noted as equation 3 and is shown below.

$$(W)_{6x1}\cdot \begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix}_{6x2} \begin{bmatrix} d_3 \\ d_4 \end{bmatrix}_{2x1} - F_{2x1} = 0 $$

Or: $$ \mathbf{W \cdot(Kd - F)}$$

This equation can be even further reduced to

$$\begin{Bmatrix} W_3\\W_4 \end{Bmatrix}\cdot\left(\begin{bmatrix} K_{33} &K_{34} \\ K_{43}& K_{44}\end{bmatrix}_{2x2} \begin{Bmatrix} d_3\\d_4 \end{Bmatrix}_{2x1} - \begin{Bmatrix} F_3\\F_4 \end{Bmatrix}_{2x1}\right) = 0 $$

This equation can be rewritten as

$$ \mathbf {\overline{K} \overline{d} = \overline{F}}$$

Where

$$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

$$\mathbf{\overline{d}}= \begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix} $$

$$\overline{F } = \begin{Bmatrix}F_{3} \\ F_{4} \end{Bmatrix} $$

for any $$\begin{Bmatrix}W_3\\W_4 \end{Bmatrix}$$

 Lecture notes 26 October 27, 2008 

Matlab Homework
In Homework 5, the 5-bar truss code needs to be reworked into a 6-bar truss. Also the 2-bar code needs to be corrected. The 6-bar truss will have the same E and A as the 5-bar truss. Note that the 2-bar truss has different E and A. $${d_{4x1}^{(i)}}$$ are the element degrees of freedom in the global (x,y) coordinate system for element i. This can be related to the axial displacement in the equation $$ {q}_{2x1}^{(i)} = {T}_{2x4} {d_{4x1}^{(i)}}$$. Also for the 2-bar truss, the complete degrees of freedom in global coordinate system $${d_{6x1}}$$ needs to be changed to d(lmm(i.;)).

Back to Principle of Virtual Work
An example everyone can understand for PVW is the equation for our course grade.

Course Grade = $$\alpha_{o}\,$$ x (Homework grade) + $$\sum_{i=1}^3 \alpha_{i}\,$$ x Exami

where $$\alpha_{o}\,$$ and $$\alpha_{i}\,$$ are the weighting coefficients.

For deriving $$ {k}^{(e)} = {T}^{(e)T}\quad\hat{k}^{(e)} {T^{(e)}}$$ we need to recall the force displacement relationship with the axial degrees of freedoms $$ q^{(e)} $$.

$$\quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}}=  {p}_{2x1}^{(e)}$$  can be rearranged to the formula

$$\quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}} - {p}_{2x1}^{(e)} = 0_{2x1} $$ (1) Note that 0 is not a scalar but the 2x1 matrix $$ 0 $$.

With the weighting coefficient added, the PVW equation becomes $$W_{2x1}\cdot \quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}} - {p}_{2x1}^{(e)} = 0 $$.

Note that since $$W$$ is a 2x1 matrix, the 0 is a scalar. This works for all $$W_{2x1}$$ because equation 1 is equal to zero. Now we have again shown that equation 1 is equivalent to equation 2, like from slide 24-1.

Again recall $$ {q}_{2x1}^{(e)} = {T}_{2x4}^{(e)} {d_{4x1}^{(e)}}$$. (3) Real displacement is represented in this equation.

$$\quad\hat{W}_{2x1} = {T_{2x4}^{(e)}} {W}_{4x1}$$ (4)

The equation above shows that $$\quad\hat{W}$$ relates to the axial $$q$$, and it is also the virtual displacement.

 Lecture notes 27 October 29, 2008 

For the next lectures, Chapter 2 sections 1,2,6 and 7 need to be read for a better understanding of what we will be taught.

As a reminder, $$\quad\hat{W}_{2x1}$$ is the virtual axial displacement corresponding to $$ {q}_{2x1}^{(e)} $$. $$ {W}_{4x1} $$ is the virtual displacement in the global coordinate system corresponding to $$ {d}_{4x1}^{(e)} $$.

Our goal is to replace equations 3 and 4 into equation 2. The result is shown below.

$$ (T^{(e)}W)\cdot [\quad\hat{k}^{(e)}(T^{(e)}d^{(e)}) - p^{(e)}] = 0 $$ for all $$W_{4x1}$$. (5)

Transposing the first matrix and multiplying it by the second matrix is equivalent to taking the dot product of two matrices.

Recall that Recall $$\left(AB \right)^{T}=B^{T}A^{T}$$ (6)

Homework: Proof $$\left(AB \right)^{T}=B^{T}A^{T}$$

$${A}_{2x3} = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}$$

$${B}_{3x3} = \begin{bmatrix} 7 & 8 & 9\\1 & 2 & 3 \\4 & 5 & 6 \end{bmatrix}$$

$${AB}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix} $$

$${AB}^T=\begin{bmatrix} 21 & 57 \\ 27 & 72 \\33 & 87 \\ \end{bmatrix} $$

$${B}^T_{3x3}=\begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6\end{bmatrix}$$

$${A}^T_{2x3}=\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \\ \end{bmatrix}$$

$${B}^T{A}^T=\begin{bmatrix} 21 & 57 \\ 27 & 72 \\ 33 & 87 \\ \end{bmatrix} $$

Therefore, $$\mathbf{{B}^T{A}^T = {(AB)}^T}$$

Also recall that $${a_{nx1} \cdot b_{nx1}=a^T_{1xn}b_{nx1}=S_{1x1}}$$ (7) Now equations 7 and 6 need to be applied into equation 5.

$${\left( T^{(e)}w \right)^T} \left[ {\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0_{1x1}$$

$${ w^TT^{(e)T}} \left[ {\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0_{1x1}$$ Next the transposed T matrix is distributed throughout the rest of the equation.

$${ w^T}\cdot \left[ {\left(T^{(e)T}\hat{k}^{(e)}T^{(e)}\right)d^{(e)} - T^{(e)T}p^{(e)}} \right]=0_{1x1}$$

We can set $$\mathbf{(T^{(e)T}\hat{k}^{(e)}T^{(e)}) = k^{(e)}}$$ and $$\mathbf {T^{(e)T}p^{(e)} = f^{(e)}} $$ to shorten the equation to:

$${ w^T}\cdot \left[{{k}^{(e)}d^{(e)} - f^{(e)}} \right]=0_{1x1}$$

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

As shown before, it is important to find the steps going from the equation with the weighting coefficient to the one basic equation.

So far we have only done the discrete case, or the non continuous case, and now we need to learn about the continuous case, or Partial Differential Equations (PDEs). A motivational model problem is an elastic bar with varying A(x) and E(x) subject to varying axial load (distributed) and concentrated loads. The axial and the concentrated loads are both time dependent.