User:Eml4500.f08.wiki1.aguilar/TeamHW5

 See my comments below. Eml4500.f08 12:42, 10 November 2008 (UTC)

 Comment: This page is identical to the previous one, except that the failure to parse error has been corrected. LaTeX Formula needed to be modified before working within Wikipedia. Here is a comparison between these two versions. Eml4500.f08.wiki1.aguilar 22:13, 7 November 2008 (UTC)

 Comment: The Two Bar Truss Results Code was not copied from Team Delta_6. There are two sites that the MATLAB files can be downloaded from ( and ). Each site has a different code for the Two Bar Truss Results. Our team and Team Delta_6 happened to use the same website, but we modified the final code so we could obtain the correct results. Eml4500.f08.wiki1.ambrosio 22:01, 10 November 2008 (UTC)

=Team Homework 5= This consists of course notes from 10/20 - 1031.

Two Bar Truss
The following code was used to re-run the two bar truss system.

 Note: This team had to modify the original code to print out the correct array "results" because they used the incomplete code from Team Delta_6; see my comments further below and in "The best of HW5". Eml4500.f08 12:48, 10 November 2008 (UTC)

The following codes were used in the two bar truss code.

 Note: This team used the incomplete code PlaneTrussResults from Team Delta_6 (the last line was missing); see my comments above and in "The best of HW5". Eml4500.f08 12:48, 10 November 2008 (UTC)

When the MATLAB code is ran, the following results are obtained.

Row 1 of the "results" represents element 1, while row 2 represents element 2. Column 1 is the strain, column 2 the stress, and column 3 is the axial force for each element.

Six Bar Truss Example
When the MATLAB code is ran, the following results are obtained.

As with the "results" matrix from the two bar truss code, the row indicates the element while column 1 is the strain, cloumn 2 is the stress, and column 3 is the axial force.

The figure below shows the undeformed truss (dashed blue line) and deformed truss (solid red line) with a weighting factor of 10000.



Six Bar Truss With Different Young's Modulus
When the MATLAB code is ran, the following results are obtained.

As with the "results" matrix from the two bar and six bar truss code, the row indicates the element while column 1 is the strain, cloumn 2 is the stress, and column 3 is the axial force.

The figure below shows the undeformed truss (dashed blue line), deformed truss where all elements have the same Young's Modulus (solid red line), and deformed truss where all elements have a different Young's Modulus (solid green line) with a weighting factor of 10000.



Justification for Eliminating Rows 1,2,5, and 6 to Obtain $$\overline{K}$$ in 2-bar Truss
Force-Displacement Relationship: K*d=F K*d-F=0 $$\rightarrow Equation(1)$$ By the Principle of Virtual Work:W•(Kd-F)=0 for all W $$\rightarrow Equation(2)$$ $$Equation(1)\Leftrightarrow Equation(2)$$ Proof: $$(1) \rightarrow (2)$$: Trivial $$(2) \rightarrow (1)$$: Not Trivial We want to show $$(2) \rightarrow (1)$$. Choice 1: Select W such that W1=1, W2=0, W3=0, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{Bmatrix}$$ W•(Kd-F)= 1.) $$[\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot [\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{1j}d_j=F_1$$   $$\rightarrow (1st Equation)$$ Choice 2: Select W such that W1=0, W2=1, W3=0, W4=0, W5=0, W6=0  W=$$ \begin{Bmatrix} 0\\1\\0\\0\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot [\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{2j}d_j=F_2 \rightarrow (2nd Equation)$$ Choice 3: Select W such that W1=0, W2=0, W3=1, W4=0, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\1\\0\\0\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{3j}d_j=F_3 \rightarrow (3rd Equation)$$ Choice 4: Select W such that W1=0, W2=0, W3=0, W4=1, W5=0, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\1\\0\\0\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+[\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{4j}d_j=F_4 \rightarrow (4th Equation)$$ Choice 5: Select W such that W1=0, W2=0, W3=0, W4=0, W5=1, W6=0 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\1\\0\end{Bmatrix}$$ 2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+[\sum_{j=1}^6 K_{5j}d_j-F_5]+0\cdot [\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{5j}d_j=F_5 \rightarrow (5th Equation)$$ Choice 6: Select W such that W1=0, W2=0, W3=0, W4=0, W5=0, W6=1 W=$$ \begin{Bmatrix} 0\\0\\0\\0\\0\\1\end{Bmatrix}$$  2.) $$0\cdot [\sum_{j=1}^6 K_{1j}d_j-F_1]+0\cdot[\sum_{j=1}^6 K_{2j}d_j-F_2]+0\cdot[\sum_{j=1}^6 K_{3j}d_j-F_3]+0\cdot [\sum_{j=1}^6 K_{4j}d_j-F_4]$$$$+0\cdot [\sum_{j=1}^6 K_{5j}d_j-F_5]+[\sum_{j=1}^6 K_{6j}d_j-F_6]$$ $$\sum_{j=1}^6 K_{6j}d_j=F_6 \rightarrow (6th Equation)$$ $$\begin{bmatrix} K_{11}&\cdots&K_{16}\\ \vdots&\ddots&\vdots\\ K_{61}&\cdots&K_{66}\end{bmatrix}* \begin{Bmatrix} d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{Bmatrix}= \begin{Bmatrix} F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{Bmatrix}$$ Hence, Kd=F (or Equation (1) from above)

Principle of Virtual Work, Relating to W
The Principle of Virtual Work accounts for the boundary conditions.

For the two-bar truss system, d1=d2=d5=d6=0 Weighting coefficients must be "kinematically admissible". In other words, the weighting coefficients cannot violate the boundary conditions.

Weighting Coefficients$$\equiv$$Virtual Displacements $$W_1=W_2=W_5=W_6=0$$ W•(Kd-F) $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}\cdot (\bar{\mathbf{K}} \bar{\mathbf{d}} -\bar{\mathbf{F}})=0$$ for all $$\begin{Bmatrix}W_3\\W_4\end{Bmatrix}$$ (3) $$\bar{\mathbf{K}}=\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix}\qquad \bar{\mathbf{d}}=\begin{Bmatrix}d_3\\d_4\end{Bmatrix}\qquad \bar{\mathbf{F}}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$

$$\begin{Bmatrix}W_3=n\\W_4=n\end{Bmatrix}\cdot (\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix})=0$$

Because the $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}$$ term is equal to $$\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$, $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix} -\begin{Bmatrix}F_3\\F_4\end{Bmatrix}=\begin{Bmatrix}0\\0\end{Bmatrix}$$ $$\begin{bmatrix}K_{33}&K_{34}\\K_{43}&K_{44}\end{bmatrix} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}=\begin{Bmatrix}F_3\\F_4\end{Bmatrix}$$(4)

More Derivations
We want to derive $$\mathbf{k}^{(e)}_{\,4x4}= \mathbf{T}^{(e)T}_{4x2} \,\hat{\mathbf{k}}^{(e)}_{2x2}\, \mathbf{T}^{(e)}_{2x4}$$, an equation that can be found on p14.3 and p23.3 of our class notes.

Recall the force-displacement relationship with axial degrees of freedom, $$q^{(e)}$$.

$$\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} = \mathbf{p}^{(e)}$$

Moving some things around we can get

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

Plugging this in for the Principle of Virtual Work:

$$ PVW\Rightarrow \mathbf{\hat{w}}\cdot \underbrace{(\hat{\mathbf{k}}^{(e)}\, \mathbf{q}^{(e)} - \mathbf{p}^{(e)})}_{2x1\mathrm{\;for\;all\;} \mathbf{\hat{w}}_{2x1}} = {0}_{1x1} $$         (2)

We showed that equations (1) and (2) are equivalent on p24.1 of our class notes.

Recall,

Here we can see the similarities between the equation for the real displacement and the virtual displacement.



$${\hat{\mathbf{W}}}_{2x1}$$ is the virtual axial displacement which corresponds to $${\mathbf{q}}^{(e)}_{2x1}$$. On the other hand $${\mathbf{W}}_{4x1}$$ is the virtual displacement in the global coordinate system which corresponds to $${\mathbf{d}}^{(e)}_{4x1}$$. If we combine equations (2), (3) and (4) we get

$$(\mathbf{T^{(e)}}\mathbf{W})[\hat{\mathbf{k^{(e)}}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}})-\mathbf{P^{(e)}}]=0$$ (5)

Which applies for all $${\mathbf{W}}_{4x1}$$. If we recall that

$$\mathbf{(A_{pxq}}\mathbf{B_{qxr})^T}=\mathbf{B^TA^T}$$ (6)

and

$$\mathbf{(a_{nx1}}\mathbf{b_{nx1})}=\mathbf{a_{1xn}^Tb_{nx1}}$$ (7)

we can manipulate (5) using (7). This gives us

<p style="text-align:center;">$$\mathbf{(T^{(e)}W)^T[\hat{k^{(e)}}(T^{(e)}d^{(e)})-P^{(e)}}]=0 $$

Then by substituting in (6) we get

<p style="text-align:center;">$$\mathbf{(T^{(e)T}W^T)[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}]}=0$$

which can be simplified to

<p style="text-align:center;">$$\mathbf{W[k^{(e)}d^{(e)}-f^{(e)}}]=0$$

and finally

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

Up until now we have dealt with the discrete case with matrices. From here forward we will consider the continuous case which involves partial differential equations. The model problem we will use consists of an elastic bar with varying area and modulus. It will be subjected to varying concentrated and axial loads, as well as inertial forces.



Perform literature search for composite materials such that E(x) (varying Young's modulus) (doping, etc.)

→One example includes polyester-50% glass fiber ply. See []



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

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

Neglect higher order terms

Recall Taylor Series Expansion:

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

$$\Rightarrow \frac{\partial N}{\partial x} + f = m\ddot{u}\; \mathbf{(2)}$$ Equation of Motion (EOM)

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

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

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

$$\Rightarrow N(x,t) = A(x)E(x)\frac{\partial u}{\partial x}(x,t)\; \mathbf{(3)}$$ Constitutive Relation

(3) in (2) yields:

$$\frac{\partial }{\partial x}\left[A(x)E(x)\frac{\partial u}{\partial x} \right] + f(x,t) = m(x)\ddot{u}\; \mathbf{(4)}$$

$$\ddot{u} = \frac{\partial^2 u}{\partial t^2}$$ PDE of Motion

Need 2 boundary conditions (2nd order derivative with respect to x)

Need 2 initial conditions (2nd derivative with respect to t) → (initial displacement, initial velocity)



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



1) $$u(0,t) = 0$$

2) $$N(L,t) = F(t)$$

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

$$\sigma(L,t) = E(L)\varepsilon(L,t)$$

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

Boundary Conditions: $$\Rightarrow \frac{\partial u}{\partial x}(L,t) = \frac{F(t)}{A(L)E(L)}$$

Initial Conditions: (at t=0 prescribed)

$$u(x,t=0) = \bar{u}(x)$$ known function (displacement)

$$\frac{\partial u}{\partial t}(x,t=0) = \dot{u}(x,t=0) = \bar{v}(x)$$ known function (velocity)

Problem Statement
3-D Three Bar Truss Problem p230 in Bhatti Consider the three bar truss problem shown below. The cross sectional are of elements 1 and 2 is 200mm2 and that of element 3 is 600mm2. All elements are made of the same material with E=200Gpa. The applied load is P=20kN. We were also asked to determine if this truss system was statically determinate. If it was, we were to solve it using statics and compare the solution to the FEA solution generated with matlab. Below is the picture, similar to the one given in the book. The nodal coordinates are as follows: The solution given in the book is as follows:

Matlab Code
Below is the code for the 3-D Three Bar Truss System: <p style="text-align:center;"> Three Bar Space Truss

Needed m-files for Space Truss Problem
This code uses three m-files. The first mfile, SpaceTrussElement.m, defines the element stiffness matrix. The second m-file, SpaceTrussResults.m, computes the element strain, stress and axial forces. The third m-file, NodalSoln.m, was given in an earlier homework assignment. It computes the nodal solutions. <p style="text-align:center;"> SpaceTrussElement.m <p style="text-align:center;"> SpaceTrussResults.m <p style="text-align:center;"> NodalSoln.m

conn and lmm Arrays of Space Truss Problem
Below are the arrays lmm and conn used to solve the 3-D three-bar truss system: conn= $$ \begin{bmatrix} 1 & 4 \\ 2 & 4 \\ 3 & 4\end{bmatrix} $$ lmm= $$ \begin{bmatrix} 1 & 2 & 3 & 10 & 11 & 12 \\ 4 & 5 & 6 & 10 & 11 & 12 \\ 7 & 8 & 9 & 10 & 11 & 12 \end{bmatrix}$$

Images of Space Truss Problem
Below are the four images of the undeformed shape vs. the deformed shape. The undeformed shape is the blue dotted line and the deformed shape is the red solid line. Their perspective plot is from the point (-2, -2, 3) as suggested. <p style="text-align:center;">From point (-2,-2,3) <p style="text-align:center;">YZ plane <p style="text-align:center;">XZ plane <p style="text-align:center;">XY plane

Matlab Code for the Images of Space Truss Problem
Below is the matlab code used to obtain the previous images: <p style="text-align:center;"> Image Code

Statics Solution of the Space Truss Problem
This problem, unfortunately for us, is statically determinate. In order to solve this problem, we first note that there will be three reaction forces at node 1, node 2, and node 3. Each one of these reaction forces can be divided into the x, y, and z components. Doing so gives us the following equation: $$ R_1 = R_{1x}\hat i + R_{1y}\hat j + R_{1z}\hat k $$ $$ R_2 = R_{2x}\hat i + R_{2y}\hat j + R_{2z}\hat k $$ $$ R_3 = R_{3x}\hat i + R_{3y}\hat j + R_{3z}\hat k $$ In order to solve for these reactions, we need to find what each reaction components are. Noting that R3x and R3y are both zero, we have the following equations to work off of: $$\mathbf{R_{1x}} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1}$$ $$\mathbf{R_{1y}} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1}$$ $$\mathbf{R_{1z}} = \mathbf{R_1}\sin{\phi_1}$$ $$\mathbf{R_{2x}} = \mathbf{R_2}\cos{\phi_2}\cos{\theta_2}$$ $$\mathbf{R_{2y}} = \mathbf{R_2}\cos{\phi_2}\sin{\theta_2}$$ $$\mathbf{R_{2z}} = \mathbf{R_2}\sin{\phi_2}$$ $$\mathbf{R_{3z}} = \mathbf{R_3}$$ Using the pythagorean theorem and trigonometry the following calculations were made in order to find the values for θ and Φ. $$\theta_1 = 63.435^\circ$$ $$\theta_2 = -45^\circ$$ $$\phi_1 = 42.975^\circ$$ $$\phi_2 = 44.482^\circ$$ Using these equations, we set up the equations for the forces in the following steps: $$ \sum F = 0 $$ $$\sum \mathbf{F_x} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1} + \mathbf{R_2}\cos{\phi_2}\cos{\theta_2} = 0$$ $$\sum \mathbf{F_y} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1} + \mathbf{R_2}\cos{\phi_2}\sin{\theta_2} - \mathbf{P} = 0$$ $$\sum \mathbf{F_z} = \mathbf{R_1\sin{\phi_1}} + \mathbf{R_2\sin{\phi_2}} + \mathbf{R_3} = 0$$ Plugging in all of the known values, given and calculated, the following results are obtained: R1 = 20374.6588 R2 = 13214.4037 R3 = -23148.1172 These values are just like the ones received using Matlab and the FEA method.

3-Dimensional Derivations
Some of the equations that we have learned over the course of the semester need to be updated for use with 3-dimensional systems or space trusses.

With the local $$\tilde{x}$$ axis along the axis of the element, the local element equations are still the same as the 2-node axial deformation element:

$$\frac{EA}{L}\begin{pmatrix}\;\; 1&-1\\-1&\;\;1\end{pmatrix}\begin{pmatrix}q_1\\q_2\end{pmatrix}=\begin{pmatrix}P_1\\P_2\end{pmatrix}$$

Which gives us our familiar axial displacement relationship: $$\mathbf {\hat {k}}^{(e)} \mathbf q^{(e)} = \mathbf P^{(e)} $$



In the global coordinates, the nodal degrees of freedom and the nodal applied forces for each element are:

$$\mathbf q = \begin{pmatrix}d_1\\d_2\\d_3\\d_4\\d_5\\d_6\end{pmatrix}$$     and       $$\mathbf p = \begin{pmatrix}f_1\\f_2\\f_3\\f_4\\f_5\\f_6\end{pmatrix}$$

In order to derive the transformation matrix, we must do the following:

Consider the displacement vector of the local node 1 denoted by $$\vec{d}_{1}^{(e)}$$.



$$\vec{d}_{[1]}^{(e)}={d}_{1}^{(e)}\vec{i}+{d}_{1}^{(e)}\vec{j}+{d}_{1}^{(e)}\vec{k}$$

$${q}_{1}^{(e)}$$ = axial displacement of node 1 is the orthagonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of element 'e'.

$${q}_{1}^{(e)}=\vec{d}_{1}^{(e)}\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}=({d}_{1}^{(e)}\vec{i}+{d}_{2}^{(e)}\vec{j}+{d}_{3}^{(e)}\vec{k})\cdot\vec{\tilde{i}}$$

$${q}_{1}^{(e)}={d}_{1}^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+{d}_{2}^{(e)}(\vec{j}\cdot\vec{\tilde{i}})+{d}_{3}^{(e)}(\vec{k}\cdot\vec{\tilde{i}})$$

$$L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\frac{x_2-x_1}{L}=l^{(e)}$$

$$(\vec{j}\cdot\vec{\tilde{i}})=\frac{y_2-y_1}{L}=m^{(e)}$$

$$(\vec{k}\cdot\vec{\tilde{i}})=\frac{z_2-z_1}{L}=m^{(e)}$$

$${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}+n^{(e)}{d}_{3}^{(e)}$$

$${q}_{1}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}&n^{(e)} \end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}  \end{Bmatrix}_{3x1}$$

Here we can see that $${q}_{1}^{(e)}$$ is a 1x1 scalar.

We do this for node 2 as well. $${q}_{2}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)} & n^{(e)}\end{bmatrix}_{1x3}\begin{Bmatrix}{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)}  \end{Bmatrix}_{3x1}$$

which leads us to the transformation between global to local degrees of freedom which are as follows:

$$\begin{Bmatrix} {q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}= \begin{bmatrix} l^{(e)}& m^{(e)}& n^{(e)}&0&0&0\\0&0&0& l^{(e)}& m^{(e)}& n^{(e)} \end{bmatrix} \begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)} \end{Bmatrix}\Rightarrow \textbf{q}_{2x1}^{(e)}=\textbf{T}_{2x6}^{(e)}\textbf{d}_{6x1}^{(e)}$$

$$\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)}\\{d}_{5}^{(e)}\\{d}_{6}^{(e)} \end{Bmatrix}= \begin{bmatrix} l^{(e)}&0\\m^{(e)}&0\\ n^{(e)}&0\\0&l^{(e)}\\0&m^{(e)}\\0& n^{(e)}\end{bmatrix}\begin{Bmatrix}{q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}

\Rightarrow \textbf{d}_{6x1}^{(e)} = {\textbf{T}_{6x2}^{(e)}}^{T} \textbf{q}_{2x1}^{(e)} $$

The conversion from axial forces and element forces in global coordinates is as follows:

$$\mathbf{f}^{(e)}=\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}$$ in global coordinates $$\mathbf{\hat{k}^{(e)}q^{(e)}=p^{(e)}}={EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} q_1^{(e)}\\q_2^{(e)} \end{pmatrix}= \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ in axial coordinates $$\mathbf{q^{(e)}=T^{(e)}d^{(e)}}$$ $$\mathbf{\hat{k}^{(e)}T^{(e)}d^{(e)}=p^{(e)}}$$ $${EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix}\begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$ Multiplying both sides by T(e)T as completed in the 2D case, $$\mathbf{T^{(e)T}\hat{k}^{(e)}T^{(e)}d^{(e)}=T^{(e)T}p^{(e)}}$$ $$\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} {EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix} \begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)}  & m^{(e)} & n^{(e)} \end{pmatrix} \begin{pmatrix} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)}\\ d_5^{(e)}\\ d_6^{(e)} \end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix} \begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix} $$ and with $$\mathbf{k^{(e)}=T^{(e)T}\hat{k}^{(e)}T^{(e)}},\qquad\mathbf{f^{(e)}=T^{(e)T}p^{(e)}}$$ We have $$\mathbf{(T^{(e)T}\hat{k}^{(e)}T^{(e)})d^{(e)}=(T^{(e)T}p^{(e)})}$$ or $$\mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$ $$\begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix}=\begin{pmatrix} l^{(e)} & 0\\m^{(e)} & 0\\n^{(e)}  & 0\\ 0 &l^{(e)}\\0&m^{(e)}\\ 0&n^{(e)} \end{pmatrix}\begin{pmatrix}p_1^{(e)}\\p_2^{(e)}\end{pmatrix}$$

We have already shown that $$\mathbf{k^{(e)}} = \mathbf{T}^{(e)T} \mathbf{{\hat{k}}^{(e)}} \mathbf{{T}^{(e)}} $$ .By using the transformation matrix we get the following.

$$\begin{align} \mathbf{k^{(e)}} & = \mathbf{T}^{(e)T} \mathbf{{\hat{k}}^{(e)}} \mathbf{{T}^{(e)}} \\

\mathbf{{k}^{(e)}} & =

\begin{pmatrix} l^{(e)} & 0 \\ m^{(e)} & 0 \\ n^{(e)} & 0 \\ 0 & l^{(e)} \\ 0 & m^{(e)} \\ 0 & n^{(e)} \end{pmatrix}

{EA \over L} \begin{bmatrix}1&-1\\-1&1\end{bmatrix}

\begin{pmatrix} l^{(e)} & m^{(e)} & n^{(e)}  & 0 & 0 & 0\\ 0 & 0 &0 & l^{(e)} & m^{(e)}  & n^{(e)} \end{pmatrix}

\end{align} $$

$$ \mathbf{k^{(e)}}= \frac{EA}{L} $$ $$ \begin{pmatrix} {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{pmatrix} $$ $$ \begin{pmatrix} d_1^{(e)}\\d_2^{(e)}\\d_3^{(e)}\\d_4^{(e)}\\d_5^{(e)}\\d_6^{(e)}\end{pmatrix}= \begin{pmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\\f_{5}^{(e)}\\f_{6}^{(e)}\end{pmatrix} $$

=Contributing Team Members=

Eml4500.f08.wiki1.oatley 17:05, 7 November 2008 (UTC) Eml4500.f08.wiki1.ambrosio 15:02, 7 November 2008 (UTC) Eml4500.f08.wiki1.aguilar 19:36, 7 November 2008 (UTC) Eml4500.f08.wiki1.brannon 18:40, 7 November 2008 (UTC) Eml4500.f08.wiki1.schaet 16:55, 7 November 2008 (UTC) Eml4500.f08.wiki1.handy 17:44, 7 November 2008 (UTC)