User:Eml4500.f08.ateam/hw 5

A-Team Teampage

Homework Report #5

This page is intended for Homework Report #5 of the A-Team.

Due date: Friday, 07 November 08, 5 pm EST = 22:00 UTC.

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



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

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

For the Principle of Virtual Work (PVW):

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

This is for all $$ {W_{6x1}}$$, 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: Weighting Coefficient Choices
When proving Eq. 2 $$\Leftrightarrow $$ Eq. 1, choices have to be made for what the Weighting Coefficient will equal.

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

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

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

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

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

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

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 $$ W_2$$ 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.

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

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

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

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

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

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

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

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

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

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

The final equation for choice 3 becomes:

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

Choice 4
$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

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

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

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

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

The final equation for choice 4 becomes:

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

Choice 5
$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}_{6x1}$$

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

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

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

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

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.

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

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

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

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

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

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

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.

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)

Proof
$$\left(AB \right)^{T}=B^{T}A^{T}$$ will be verified with an example.

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

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

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

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

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

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

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.



A(x), E(x), f(x,t), and p(x) are all given.

Example of a Composite Material
Some composite materials are examples of how the modulus of elasticity can be a function of x. One example is reinforced concrete walls. The modulus of the concrete is significantly different than that of the reinforcing steel. In order to accurately model this material the modulus would change as one moved through each material.



source: Rebar

Another material that has a modulus that changes with x is an aluminum composite panel which consists of two aluminum sheets sandwiched around a non-aluminum material.

source: (Aluminium_composite_panel)

Partial Differential Method


A very small slice of the beam shall be considered of width dx. The force balance in the x direction gives the following result.

$$ \sum F_x=0= -N(x,t) + N(x+dx,t) +f(x,t)dx - m(x)\ddot{u}dx$$

$$=\frac{\partial {N}}{\partial{x}}(x,t)dx + higher\ order\ terms + f(x,t)dx-m(x)\ddot{u}dx\quad (Eqn. 1)$$

These higher order terms come from the Taylor Series expansion, shown in the general form here.

$$f(x+dx)=f(x)+\frac{df(x)}{dx}dx+\frac{1}{2}\frac{d^2f(x)}{dx^2}dx^2...$$

Where all the $$dx^2...$$ and higher terms are neglected due to their small effect on the result. After those simplifications are made, Eqn. 1 can be written in the following form.

$$\frac{\partial{N}}{\partial{x}}+f(x,t)=m(x)\ddot{u}\quad (Eqn. 2)$$

In this form, it is easy to see that this is the equation of motion. In order to get to a more usable form, N(x,t) needs to be rewritten into its constituent components.

$$N(x,t)=A(x)\ \sigma(x,t)$$

$$\sigma(x,t)=E(x)\ \epsilon(x,t)$$

$$\epsilon(x,t)=\frac{\partial{u}}{\partial{x}}(x,t)$$

Using these relationships, N(x,t) can be written as follows.

$$N(x,t)=[A(x)\ E(x)\ \frac{\partial{u}}{\partial{x}}(x,t)]$$

Now Eqn. 1 can be rewritten in the final form.

$$\frac{\partial{ }}{\partial{x}}[A(x)\ E(x)\ \frac{\partial{u}}{\partial{x}}(x,t)]+f(x,t)=m(x)\ddot{u}$$

This is the final form of the equation of motion that will be used to solve finite element problems with the partial derivative method.

Initial and Boundary Conditions
In order to solve these problems, certain aspects of the conditions must be known. For the following case, two boundary conditions are known.

 $$u(0,t)=0\ and\ u(L,t)=0$$

In another example, two initial conditions are known.



$$u(0,t)=0\ and\ N(L,t)=A(L)\sigma(L,t)=F(t)$$

Axial DOF's and Forces/Element DOF's and Forces in Global xyz-Coordinate System for 6x1 Matrices


Using the previous diagram, we can write the following relationships.

$$\underline{k}^{(e)} \underline{q}^{(e)} = \underline{p}^{(e)}$$

$$\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix} \begin{bmatrix}q_{1}^{(e)}\\ q_{2}^{(e)}\end{bmatrix} = \begin{bmatrix}p_{1}^{(e)}\\ p_{2}^{(e)}\end{bmatrix}$$



From the diagram above, we can write the element DOF's and forces for a 3-D system.

$$\underline{f}^{(e)} = \begin{vmatrix}F_1\\F_2 \\F_3 \\ F_4\\ F_5\\ F_6\end{vmatrix}_{6x1}$$ and $$\underline{d}^{(e)} = \begin{vmatrix} d_1\\d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

We can now use the transformation matrix $$\underline{T}$$ to transform the 6 element DOF's in the global xyz-coordinate system to the 2 axial DOF's $$p$$ and $$q$$.

$$\underline{q}^{(e)} = \underline{T} \underline{d}^{(e)}$$

This equation expands to:

$$\begin{vmatrix}q_{1}^{(e)}\\q_{2}^{(e)} \end{vmatrix}_{2x1} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}_{2x6} \begin{vmatrix}d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{vmatrix}_{6x1}$$

Similarly, this can be applied to the equation:

$$\underline{p} = \underline{T} \underline{f}$$

From either of the above equations, to solve for the element forces $$\underline{f}$$ or displacements $$\underline{d}$$ we can use the transpose of the transformation matrix $$\underline{T}^{T}$$.

$$\underline{f} = \underline{T}^{T} \underline{p}$$

$$\underline{d} = \underline{T}^{T} \underline{q}$$

Derivation of FD Relation in Global Coordinates Using the PVW
Derive the transformation matrix based on the lecture for the 2-D truss elements:

$$a_1^{(e)} = d_1^{(e)} \cdot \vec\tilde{i} = ( d_1 \tilde{i} + d_2 \tilde{j} + d_3 \tilde{k} ) \cdot \vec\tilde{i} = d_1 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_2 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_3 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_1^{(e)} = l^{(e)}d_1 + m^{(e)}d_2 + n^{(e)}d_3$$

$$a_1^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_1\\ d_2\\ d_3 \end{vmatrix}$$

$$a_2^{(e)} = d_2^{(e)} \cdot \vec\tilde{i} = ( d_4 \tilde{i} + d_5 \tilde{j} + d_6 \tilde{k} ) \cdot \vec\tilde{i} = d_4 \cdot ( \tilde{i} \cdot \vec\tilde{i} ) + d_5 \cdot ( \tilde{j} \cdot \vec\tilde{i} ) + d_6 \cdot ( \tilde{k} \cdot \vec\tilde{i} )$$

where $$( \tilde{i} \cdot \vec\tilde{i} ) = l^{(e)}, ( \tilde{j} \cdot \vec\tilde{i} ) = m^{(e)}, and ( \tilde{k} \cdot \vec\tilde{i} ) = n^{(e)}$$

$$a_2^{(e)} = l^{(e)}d_4 + m^{(e)}d_5 + n^{(e)}d_6$$

$$a_2^{(e)} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix}\begin{vmatrix} d_4\\ d_5\\ d_6 \end{vmatrix}$$

$$\begin{vmatrix} a_1^{(e)}\\ q_2^{(e)} \end{vmatrix} = \begin{vmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix}$$

Note: The transformation matrix is in terms of the three director cosines.

Recall the relation between axial DOF's and element DOF's in the global coordinate system:

$$\underline{q} = \underline{T} \underline{d}$$

and also the relation between axial forces and element forces in the global coordinate system:

$$\underline{P} = \underline{T} \underline{f}$$

Using the principle of virtual work (PVW) we can derive the force-displacement relationship in the global coordinate system.

$$\underline{k} \underline{d} = \underline{F}$$

therefore,

$$\underline{k} \underline{d} - \underline{F} = 0$$

With the PVW, this becomes:

$$\underline{W} \cdot ( \underline{k} \underline{d} - \underline{F} ) = 0$$

where $$\underline{W}$$ is the weighting matrix. For this system we have 6 choices for $$\underline{W}$$:

Choice 1:

$$ W = \begin{vmatrix}1 & 0 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j - F_1 = 0}$$

$$\sum_{j=1}^{6}{k_{1,j}d_j = F_1}$$

Choice 2:

$$ W = \begin{vmatrix}0 & 1 & 0 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j - F_2 = 0}$$

$$\sum_{j=1}^{6}{k_{2,j}d_j = F_2}$$

Choice 3:

$$ W = \begin{vmatrix}0 & 0 & 1 & 0 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j - F_3 = 0}$$

$$\sum_{j=1}^{6}{k_{3,j}d_j = F_3}$$

Choice 4:

$$ W = \begin{vmatrix}0 & 0 & 0 & 1 & 0 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j - F_4 = 0}$$

$$\sum_{j=1}^{6}{k_{4,j}d_j = F_4}$$

Choice 5:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 1 & 0\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j - F_5 = 0}$$

$$\sum_{j=1}^{6}{k_{5,j}d_j = F_5}$$

Choice 6:

$$ W = \begin{vmatrix}0 & 0 & 0 & 0 & 0 & 1\end{vmatrix}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j - F_6 = 0}$$

$$\sum_{j=1}^{6}{k_{6,j}d_j = F_6}$$

$$\begin{vmatrix} k_{1,1} & k_{1,2} & k_{1,3} & k_{1,4} & k_{1,5} & k_{1,6}\\ k_{2,1} & k_{2,2} & k_{2,3} & k_{2,4} & k_{2,5} & k_{2,6}\\ k_{3,1} & k_{3,2} & k_{3,3} & k_{3,4} & k_{3,5} & k_{3,6}\\ k_{4,1} & k_{4,2} & k_{4,3} & k_{4,4} & k_{4,5} & k_{4,6}\\ k_{5,1} & k_{5,2} & k_{5,3} & k_{5,4} & k_{5,5} & k_{5,6}\\ k_{6,1} & k_{6,2} & k_{6,3} & k_{6,4} & k_{6,5} & k_{6,6}\\ \end{vmatrix} \begin{vmatrix} d_1\\d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6

\end{vmatrix} = \begin{vmatrix} F_1\\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6

\end{vmatrix}$$

Derivation of Elemental Stiffness Matrix in Global Coordinates Using PVW
The following is an expression of the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ in terms of the transformation matrix $$\underline{T}^{(e)}$$ and the element stiffness matrix $$\underline\hat{k}^{(e)}$$ from the axial force displacement relationship. We will show that by using the priciple of virtual work, we can derive the equation for the element stiffness matrix in global coordinates.

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

Recall the force displacement relationship with axial DOF's $$ q^{(e)}$$

$$\underline\hat{k}_{2x2}^{(e)} \underline{q}_{2x1}^{(e)} = \underline{p}_{2x1}^{(e)}$$

Rearranging this equation yields:

$$ \underline\hat{k}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)} = \underline{0}_{2x1}$$    (1)

We can now apply the principle of virtual work (PVW) to get:

$$\underline\hat{W} \cdot ( \underline\hat{k}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)} ) = 0^{'}_{1x1}$$   (2)  for all $$\underline\hat{W}_{2x1}$$

We now can recall the equation:

$$\underline{q}_{2x1}^{(e)} = \underline{T}_{2x6}^{(e)} \underline{d}_{6x1}^{(e)}$$    (3)

Similarly,

$$\hat{W}_{2x1} = \underline{T}^{(e)}_{2x6} \underline{W}_{6x1}$$ (4)

Now, substitute equations (3) & (4) into equation (2) which yields:

$$( \underline{T}^{(e)} \underline{W} ) \cdot [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$  for all $$ \underline{W}_{6x1}$$   (5)

Recall that:

$$ ( \underline{A}_{pxq} \underline{B}_{qxr} )^{T} = \underline{B}^{T} \underline{A}^{T}$$  (6)

And also that:

$$\underline{a}_{nx1} \cdot \underline{b}_{nx1} = \underline{a}^{T}_{1xn} \underline{b}_{nx1}$$   (7)

Now we can apply equations (6) and (7) into (5) to get:

$$ ( \underline{T}^{(e)} \underline{W} )^{T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - ( \underline{p}^{(e)} ] = 0 $$ for all $$\underline{W}_{6x1}$$

In this equation, $$\underline\hat{W}_{2x1}$$ represents the virtual axial displacement and from eq. (4) and the matrix $$\underline{W}_{6x1}$$ represents the virtual displacement in the global coordinate system corresponding to $$\underline{d}^{(e)}$$

The previous equation now becomes:

$$\underline{W}^{T} \underline{T}^{(e)T} [ \underline\hat{k}^{(e)} ( \underline{T}^{(e)} \underline{d}^{(e)} ) - \underline{p}^{(e)} ] = 0$$

$$\underline{W} \cdot [ ( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} ) \underline{d}^{(e)} - ( \underline{T}^{(e)T} \underline{p}^{(e)}) ] = 0$$ for all $$\underline{W}_{6x1}$$

In the equation above, the element stiffness matrix $$\underline{K}_{6x6}^{(e)}$$ is represented by $$( \underline{T}^{(e)T} \underline\hat{k}^{(e)} \underline{T}^{(e)} )$$and the force matrix $$\underline{f}_{6x1}^{(e)}$$ is represented by $$\underline{T}^{(e)T} \underline{p}^{(e)}$$

The equation now becomes:

$$\underline{W} \cdot [ \underline{K}^{(e)} \underline{d}^{(e)} - \underline{f}^{(e)} ] = 0$$ for all $$\underline{W}$$

We have the original expression for the elemental stiffness matrix in terms of the axial force displacement relationship.

$$\underline{K}^{(e)} \underline{d}^{(e)} = \underline{f}^{(e)}$$

Expression for element forces in terms of transformation matrix and element axial forces
From the previous derivations, it was determined that the element axial force was related to the element forces in global coordinates by

$$p^{(e)}_{2x1}=T^{(e)}_{2x6}f^{(e)}_{6x1}$$

Therefore, the expression for the element forces in global coordinates in terms of the transformation matrix and the element axial forces is

$$f^{(e)}_{6x1}=T^{(e)T}_{6x2}p^{(e)}_{2x1}$$

Element Stiffness Matrix in Global Coordinates
The equation for the elemental stiffness matrix in global coordinates is

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

This equation can be combined with the director cosines to yield

$$ \mathbf{k}_{6\times 6}^{(e)} = k^{(e)}\begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} \\ m^{(e)}l^{(e)}& m^{(e)^{2}} & m^{(e)}n^{(e)} & -m^{(e)}l^{(e)} & -m^{(e)^{2}} & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)}& m^{(e)}n^{(e)} & n^{(e)^{2}} & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -n^{(e)^{2}} \\ -l^{(e)^{2}}& -m^{(e)}l^{(e)} & -n^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)} & n^{(e)}l^{(e)} \\ -m^{(e)}l^{(e)}& -m^{(e)^{2}} & -m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & m^{(e)^{2}} & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)}& -m^{(e)}n^{(e)} & -n^{(e)^{2}} & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & n^{(e)^{2}} \end{bmatrix} $$

where $$l^{(e)}=\frac{x^{(e)}_2-x^{(e)}_1}{L}$$, $$m^{(e)}=\frac{y^{(e)}_2-y^{(e)}_1}{L}$$ , $$n^{(e)}=\frac{z^{(e)}_2-z^{(e)}_1}{L}$$.

Essential MATLAB Functions for Finite Element Analysis
Similar to the five-bar truss system, all of the following MATLAB programs require additional functions to run. These functions are found at the course textbook website, Fundamental Finite Element Analysis and Applications. They include,  , and. The details of these functions are outlined below.

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

Two-Bar Truss System: MATLAB Code Debugging
In Homework Report #2, the team analyzed the MATLAB code for a two-bar truss system. This code was provided by Dr. Vu-Quoc and can be found here. After the homework report submission deadline, a bug was found in the provided code. This bug occurs during the building of the  array. Instead of passing the proper Young's modulus,, and cross-sectional area,  , indicies into the   function, the program passes in the entire   and   arrays. This is sufficient for cases in which all truss elements have an identical cross-sectional area and Young's modulus, but fails when these properties vary among elements.



Note: Applied load P = 7

To fix the bug, a simple modification was made to the following program segment:

Where the  array is assembled the variables   and   are replaced by the array indices   and   respectively.

was modified to be

The resulting MATLAB program code is as follows:

Running
Running this code produces the following output: >> twobar_debug k_local = 0.75       -0.75         -0.75         0.75 k = 0.5625     0.32476      -0.5625     -0.32476       0.32476       0.1875     -0.32476      -0.1875        -0.5625     -0.32476       0.5625      0.32476      -0.32476      -0.1875      0.32476       0.1875 k_local = 5   -5     -5     5 k = 2.5        -2.5         -2.5          2.5          -2.5          2.5          2.5         -2.5          -2.5          2.5          2.5         -2.5           2.5         -2.5         -2.5          2.5 K = 0.5625     0.32476      -0.5625     -0.32476            0            0       0.32476       0.1875     -0.32476      -0.1875            0            0       -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5      -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5             0            0         -2.5          2.5          2.5         -2.5             0            0          2.5         -2.5         -2.5          2.5 R = 0     0      0      7      0      0 d = 0            0         4.352        6.1271             0             0 reactions = -4.4378      -2.5622        4.4378       -4.4378 results = 1.7081      5.1244       5.1244        0.6276        3.138        6.276

Compared to the previously obtained  array shown below, this new   array makes much more sense. The previous array contained additional meaningless values along with the desired values.

Previous  array:

results = 1.7081      5.1244       8.5406       5.1244       17.081        0.6276       1.8828        3.138       1.8828        6.276

Verifying Array Accuracy
The new  array values correspond to the table shown below.

The  array for finite element analysis of the truss system is verified using principles of statics by simply summing the forces in the X and Y directions and setting these equations equal to zero. This produces two equations with two unknown values from which the axial forces of each element are derived. From here, using the element properties, the element stresses and strains can be determined. This is shown below.

$$\sum{F_{y}} = 7 - F_1\cdot sin(30^{\circ} ) - f_2\cdot sin(45^{\circ}) = 0$$

$$\sum{F_{x}} = F_1\cdot cos(30^{\circ} ) - F_2\cdot cos(45^{\circ}) = 0$$

Where $$F_1$$ is the axial force of Element 1 and $$F_2$$ is the axial force of Element 2.

Solving these equations simultaneously yields the element axial forces.

$$F_1 = 5.1244$$

$$F_2 = 6.2760$$

Now the axial stress for each element can be computed using the axial force and cross-sectional area for each element.

$$\sigma_1 = \frac{F_1}{A_1} = \frac{5.1244}{1.00} = 5.1244$$

$$\sigma_2 = \frac{F_2}{A_2} = \frac{6.2760}{2.00} = 3.1380$$

The values for axial stress can now be used along with the Young's modulus to determine the axial strain for each element.

$$\epsilon_1 = \frac{\sigma_1}{E_1} = \frac{5.1244}{3.00} = 1.7081$$

$$\epsilon_2 = \frac{\sigma_2}{E_2} = \frac{3.1380}{5.00} = 0.6276$$

These values are almost identical to the values obtained through finite element analysis. The MATLAB results have been verified to be accurate. For this case, principles of statics proved to be a much more efficient method to solve the problem. This is not the case as mesh complexity increases.

Six-Bar Truss System: MATLAB Finite Element Analysis
The problem to be evaluated in this section is found as Example 4.1 on pg.226 of the course textbook. The MATLAB code used to evaluate the problem is from the  file found at the Student Companion Site for the class textbook.

Additionally, a modified version of the problem is analyzed accordingly in this section.

The load $$P = 20 kN$$ acts at an angle $$\alpha = 30^\circ$$ for each version of the problem.

Original Problem: Example 4.1
Analysis of the original six-bar truss system is important before comparing to the modified six-bar truss system. The following code is from the  file found at the website for the class textbook.

Running
Running this code produces the following results.

>> SixBarTrussEx K = Columns 1 through 6 85355       35355       -50000            0            0            0         35355        35355            0            0            0            0        -50000            0        85355       -35355            0            0             0            0       -35355  1.0202e+005            0            0 0           0            0            0        71554       -35777             0            0            0            0       -35777        17889             0            0            0            0            0            0             0            0            0       -66667            0            0        -35355       -35355       -35355        35355       -71554        35777        -35355       -35355        35355       -35355        35777       -17889   Columns 7 through 10 0           0       -35355       -35355             0            0       -35355       -35355             0            0       -35355        35355             0       -66667        35355       -35355             0            0       -71554        35777             0            0        35777       -17889         71554        35777       -71554       -35777         35777        84555       -35777       -17889        -71554       -35777  2.1382e+005            0 -35777      -17889            0  1.0649e+005 R = 0            0         10000         17321             0             0             0             0             0             0 d  = 0            0       0.21311       0.24998             0             0             0             0    -0.0060971      0.012242 reactions = -10873      -217.27        874.27       -437.13       -1.7279        -16666 results = 5.3276e-005      10.655        10655 -4.6334e-006    -0.92669      -926.69 -4.8873e-006    -0.97746      -977.46 -8.3326e-005     -16.665       -16665 1.5363e-006     0.30727       307.27 -9.659e-009  -0.0019318      -1.9318

The  array values correspond to the table shown below.

The original displacements are shown in the following matrix. It is important to note that the nodal positions were magnified by a factor of 1000 in the original  file.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\           0 \\   0.21311 \\   0.24998 \\            0 \\            0 \\            0 \\            0 \\   -0.0060971 \\     0.012242 \\ \end{Bmatrix}$$

Plotting Example 4.1
To plot the undeformed and deformed shapes, the following code was appended to the original  file. For viewing, the nodal magnification factor in the original file was removed and the resulting nodal displacements were multiplied by a factor of 2500.



Modified Problem: Varying Young's Modulus
This modification of Example 4.1 involves varying the Young's modulus, E, of the elements as shown in the table below. This requires re-writing the  code to allow for an array of E values. The subsequent MATLAB file is called.

To allow for the passing of array indices, the MATLAB code was modified as shown. Much like fixing the bug in the two-bar truss system code,  replaces   inside the   loop that builds the stiffness matrices and the results array.

Running
The results obtained by running the code are as follows. Again, it is important to note that the nodal positions are multiplied by a factor of 1000. >> sixbarmod K = Columns 1 through 6 76391       38891       -37500            0            0            0         38891        38891            0            0            0            0        -37500            0        69320       -31820            0            0             0            0       -31820        98486            0            0             0            0            0            0        71554       -35777             0            0            0            0       -35777        17889             0            0            0            0            0            0             0            0            0       -66667            0            0        -38891       -38891       -31820        31820       -71554        35777        -38891       -38891        31820       -31820        35777       -17889   Columns 7 through 10 0           0       -38891       -38891             0            0       -38891       -38891             0            0       -31820        31820             0       -66667        31820       -31820             0            0       -71554        35777             0            0        35777       -17889         89443        44721       -89443       -44721         44721        89027       -44721       -22361        -89443       -44721  2.3171e+005        16015 -44721      -22361        16015  1.1096e+005 R = 0            0         10000         17321             0             0             0             0             0             0 d = 0            0       0.26485       0.26083             0             0             0             0    0.00063864     -0.001246 reactions = -9908.3       23.619       -90.274        45.137       -1.4008        -17389 results = 6.6213e-005      9.9319       9931.9 5.347e-007    0.096245       96.245 5.0465e-007     0.10093       100.93 -8.6943e-005     -17.389       -17389 -1.5183e-007   -0.033402      -33.402 -6.2645e-009  -0.0015661      -1.5661

The  array values correspond to the table shown below.

The original displacements are shown in the following matrix. It is important to note that the nodal positions were magnified by a factor of 1000 in the  file.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\           0 \\      0.26485 \\      0.26083 \\            0 \\            0 \\            0 \\            0 \\   0.00063864 \\    -0.001246 \\ \end{Bmatrix}$$

Plotting Example 4.1 and the Modified Six-Bar Truss System
To plot the undeformed and deformed shapes, the following code was appended to the original  file. The subsequent file was plotted over the existing plot for Example 4.1 to show how varying Young's modulus throughout the truss system can affect deformation. Subsequent deformation clearly differed from that of elements with identical Young's modulus. However, deformation did not vary greatly.

The undeformed truss system is represented by the blue dashed line. The deformed truss system of Example 4.1 is represented by the solid red line. The deformed modified truss system is represented by the green dotted line. For viewing, the nodal magnification factor in the original files was removed and the resulting nodal displacements were multiplied by a factor of 2500.



Analysis of Three-Bar Space Truss System: Example 4.2
This problem is found on page 230 of the course textbook.



The truss system in Example 4.2 is a pin-jointed structure with three elements. Much like problems solved earlier, the element properties vary in this truss system and are displayed below. This system adds another dimension to the FEA and results in a total of 12 degrees of freedom. Nodal displacements for this problem are denoted by,    and   in the  ,    and   directions, respectively.

Before implementing FEA to solve the problem, it is important to properly set up the  and   arrays. $$= \begin{bmatrix} 1&4\\ 2&4\\ 3&4 \end{bmatrix}$$

$$= \begin{bmatrix} 1 & 2 & 3 & 10 & 11 & 12\\ 4 & 5 & 6 & 10 & 11 & 12\\ 7 & 8 & 9 & 10 & 11 & 12 \end{bmatrix} $$

The  file for this problem was found on the Student Companion Site for the course textbook. It is as follows.

Additional MATLAB Functions
These functions were found on the Student Companion Site for the course textbook. They behave similarly to their corresponding functions for plane trusses detailed in the previous section.

Note: The program also uses, detailed in the previous section.

Running
Running this code produces the following results. Note that the nodal positions are multiplied by a factor of 1000 for simplicity. >> ThreeBarSpaceTrussEx K = Columns 1 through 6 1459.7      2919.3      -3040.9            0            0            0        2919.3       5838.6      -6081.9            0            0            0       -3040.9      -6081.9       6335.3            0            0            0             0            0            0       3566.7      -3566.7       4953.8             0            0            0      -3566.7       3566.7      -4953.8             0            0            0       4953.8      -4953.8       6880.3             0            0            0            0            0            0             0            0            0            0            0            0             0            0            0            0            0            0       -1459.7      -2919.3       3040.9      -3566.7       3566.7      -4953.8       -2919.3      -5838.6       6081.9       3566.7      -3566.7       4953.8        3040.9       6081.9      -6335.3      -4953.8       4953.8      -6880.3   Columns 7 through 12 0           0            0      -1459.7      -2919.3       3040.9             0            0            0      -2919.3      -5838.6       6081.9             0            0            0       3040.9       6081.9      -6335.3             0            0            0      -3566.7       3566.7      -4953.8             0            0            0       3566.7      -3566.7       4953.8             0            0            0      -4953.8       4953.8      -6880.3             0            0            0            0            0            0             0            0            0            0            0            0             0            0        60000            0            0       -60000             0            0            0       5026.4      -647.45       1912.9             0            0            0      -647.45       9405.4       -11036             0            0       -60000       1912.9       -11036        73216 d  = 0            0             0             0             0             0             0             0             0      -0.18705        -2.592       -0.3858 reactions = 6666.7        13333        -13889       -6666.7        6666.7       -9259.3             0             0         23148 results = 0.00050936      101.87        20375    0.00033036       66.072        13214    -0.0001929       -38.58       -23148

The  array values correspond to the table shown below.

The displacement matrix is as follows.

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \\ d_{11} \\ d_{12} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\        0 \\         0 \\         0 \\         0 \\         0 \\         0 \\         0 \\         0 \\   -0.1871 \\   -2.5920 \\   -0.3858 \\ \end{Bmatrix}$$

The reactions are as follows.

$$\begin{Bmatrix} R_1 \\ R_2 \\ R_3 \\ R_4 \\ R_5 \\ R_6 \\ R_7 \\ R_8 \\ R_9 \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 6667 \\  13333 \\  -13889 \\   -6667 \\    6667 \\   -9259 \\       0 \\       0 \\   23148 \\ \end{Bmatrix}$$

Static Determinacy
The three bar space truss was found to be statically determinate. There are three unknowns and three equations to use, summing the forces in x, y, and z directions. By using these three equations, the following relationships are made.

$$\sum{F_{x}}= F_{1}l^{(1)} + F_{2}l^{(2)} = 0$$

$$\sum{F_{y}}= F_{1}m^{(1)} + F_{2}m^{(2)} = -P$$

$$\sum{F_{z}}= F_{1}n^{(1)} + F_{2}n^{(2)} + F_{3}n^{(3)} = 0$$

Plugging in P = 20,000 N, and solving the set of linear equations gives the following forces along each element.

$$F_{1} = 20375\ N$$

$$F_{2} = 13124\ N$$

$$F_{3} = -23149\ N$$

In order to solve for the axial stress and strain, the following equations will be used.

$$\sigma_{n} = \frac{F_{n}}{A_{n}}$$

$$\epsilon_{n} = \frac{\sigma_{n}}{E_{n}}$$

Plugging in the given values for the Young's Modulus and area for each element yields these values. (Sigma is expressed in MPa and epsilon is expressed in mm.)

$$\sigma_{1} = \frac{20375 N}{200\ mm^2}= 101.875\qquad \qquad \epsilon_{1} = \frac{\sigma_{1}}{200\ GPa}= 0.5094$$

$$\sigma_{2} = \frac{13214 N}{200\ mm^2}= 66.07\qquad \qquad \ \ \ \epsilon_{2} = \frac{\sigma_{2}}{200\ GPa}= 0.33035$$

$$\sigma_{3} = \frac{-23149 N}{600\ mm^2}= -38.58\qquad \qquad \epsilon_{3} = \frac{\sigma_{3}}{200\ GPa}= -0.1929$$

These values for the displacement of each element matches the results from the MATLAB confirming the method.

MATLAB Plots of Example 4.2
To plot the undeformed and deformed states of the three-bar space truss system the following code was appended to the original  file. Each plotting view was manipulated within the MATLAB plot editor.

Perspective View: (1, 1, 1)
To make the deformation more clear the following perspective view was included.



Contributing Team Members
The following students contributed to this report:

Mark Barry Eml4500.f08.Ateam.barry 03:17, 7 November 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 00:37, 3 November 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 05:22, 7 November 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 00:31, 7 November 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 14:13, 3 November 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 01:11, 7 November 2008 (UTC)