User:Eml4500.f08.ateam.nobrega/Homework Report Three

Axial Force-Displacement






The equation for a two-force element,


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

can also be expressed with axial forces and displacements where,



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

This can also be expressed as,



K^{(e)} \cdot \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} $$

where,



K^{(e)} = \frac{EA}{L} $$

The relationship between q(e) and d(e) can be expressed in the form of the equation,



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

where q(e) is a 2x1 matrix, T(e) is a 2x4 matrix, and d(e) is a 4x1 matrix.

Consider the displacement vector of local node one, it is denoted by d[1](e) and can be seen at right where,



d_{[1]}^{(e)} = d_{1}^{(e)}\overrightarrow{i} + d_{2}^{(e)}\overrightarrow{j} $$

relating q to d,



q_{1}^{(e)} = d_{[1]}^{(e)}\overrightarrow{\tilde{i}} $$



q_{1}^{(e)} = (d_{1}^{(e)}\overrightarrow{i} + d_{2}^{(e)}\overrightarrow{j})\cdot \overrightarrow{\tilde{i}} $$



q_{1}^{(e)} = d_{1}^{(e)}(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}}) + d_{2}^{(e)}(\overrightarrow{j}\cdot \overrightarrow{\tilde{i}}) $$

Where,



\overrightarrow{i}\cdot \overrightarrow{\tilde{i}} = \cos\theta = l^{(e)} $$



\overrightarrow{j}\cdot \overrightarrow{\tilde{i}} = \sin\theta = m^{(e)} $$



q_{1}^{(e)} = l^{(e)}d_{1}^{(e)} + m^{(e)}d_{2}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

For node 2, denoted by d[2](e),



d_{[2]}^{(e)} = d_{3}^{(e)}\overrightarrow{i} + d_{4}^{(e)}\overrightarrow{j} $$

relating q to d,



q_{2}^{(e)} = d_{[2]}^{(e)}\overrightarrow{\tilde{i}} $$



q_{2}^{(e)} = (d_{3}^{(e)}\overrightarrow{i} + d_{4}^{(e)}\overrightarrow{j})\cdot \overrightarrow{\tilde{i}} $$



q_{2}^{(e)} = d_{3}^{(e)}(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}}) + d_{4}^{(e)}(\overrightarrow{j}\cdot \overrightarrow{\tilde{i}}) $$



q_{2}^{(e)} = l^{(e)}d_{3}^{(e)} + m^{(e)}d_{4}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{bmatrix} $$

Therefore,



\begin{bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)} \end{bmatrix} $$

Here q(e) is a 2x1 matrix, T(e) is a 2x4 matrix, and d(e) is a 4x1 matrix.

Note: On Wednesday September 24 of 2008, the class took the first exam. Dr. Vu-Quoc posted the solutions to the exam on his course website but the direct link is available below.

Exam 1 Solutions.



\begin{bmatrix} P_1^{(e)}\\ P_2^{(e)}\\ \end{bmatrix} = T^{(e)} \cdot \begin{bmatrix} f_1^{e}\\ f_2^{e}\\ f_3^{e}\\ f_4^{e}\\ \end{bmatrix} $$

Where T(e) is a 2x4 transformation matrix which can be used to transform distance and force from one coordinate system to another.



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



P^{(e)} = T^{(e)} \cdot f^{(e)} $$

Recall



P^{(e)} = q^{(e)} \cdot k^{(e)} $$

Where q and P are axial components with k being a 2x2 overlap stiffness matrix and q being 2x1 distance matrix.

Thus

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



T^{(e)} \cdot f^{(e)} = P^{(e)} $$



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

The goal is to have a FD distance relationship as described by:

k(e) * d(e) = f(e)

Move the transformation matrix from the RHS to the LHS by premultiplying the equation by Y(e)-1. (inverse of the T matrix)

The T matrix is a rectangular 2x4 matrix, thus it cannot be inverted. However, the T matrix can be transposed to achieve similar results:

[T^{(e)T} \cdot k^{(e)} \cdot T^{(e)}] \cdot d^{(e)} = (T^{(e)} \cdot f^{(e)}) $$

Where the transpose of T is a 4x2 matrix and the overall resultant matrix of the transpose of TT * k * T leads to a 4x4 matrix.

The following problem has been worked out to verify the above statement.

$$     \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} \cdot \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} \cdot \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}^T $$

>> k=[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 5/2]

k =

2.5000  -2.5000   -2.5000    2.5000    2.5000    2.5000    2.5000   -2.5000   -2.5000    2.5000    2.5000   -2.5000    2.5000   -2.5000   -2.5000    2.5000

>> T=[sqrt(2)/2 -sqrt(2)/2 0 0; 0 0 sqrt(2)/2 -sqrt(2)/2]

T =

0.7071  -0.7071         0         0         0         0    0.7071   -0.7071

>> TT=T'

TT =

0.7071        0   -0.7071         0         0    0.7071         0   -0.7071

>> K=k*TT

K =

3.5355  -3.5355         0    3.5355   -3.5355    3.5355    3.5355   -3.5355

>> K=K*T

K =

2.5000  -2.5000   -2.5000    2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000    2.5000   -2.5000    2.5000   -2.5000   -2.5000    2.5000

The justification of this concept can be provided using the principal virtual work theorem. This theorem will be introduced later in the semester.


 * See Lecture 10 for the first example of PVW/Reduction of global FD relationship.

REMARK The k equation from the exam could not be solved using: k-1 * F = d because  k  is a singular matrix.

 K  is a singular matrix, which means the determinant of  k  is 0 and thus  k  cannot be inverted.

RECALL: You need to compute the (det)-1 to find the inverse of the  k  matrix.


 * A famous quote narrated by the professor:

" Ask what not your country can do for you               -ask what you can do for your country." JFK For an unconstrained structure system, there are three possible rigid body motions in 2-D (2 truss - 1 rotation).

Find the eigenvalues of the K matrix, and make observations about the number of 0 eigenvalues.

A =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.2500   -0.3248   -0.2500         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.2500    2.8248    1.9375    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

>> eig(A)

ans =

8.4194   2.5879   -0.1729   -0.0219   -0.0000   -0.0000 Observation: Two of the six eigenvalues are zero.

Dynamic Evaluation Problem:
 * k * v = λ * M * V

Where k*v is the stiffness matrix, λ is the eigenvalue related to the vibrational frequency, M is the mass of the system and V is the velocity matrix. Thus, zero eigenvalues lead to zero stored elastic energy meaning the system is rigid.

Infinitesimal Displacement
An alternative way of solving for the reactions is by using the Global Force Displacement relation. Since one knows the values for the displacement column matrix ( d : 6x1) and the global stiffness matrix ( K : 6x6). One can solve for the force column matrix ( F : 6x1).



\begin{bmatrix} &  &   &   &   &  & \\         &   &   &   &   &  & \\         &   &   &   K   &  & \\ &  &   &   &   &  & \\         &   &   &   &   &  & \\         &   &   &   &   &  & \\     \end{bmatrix} \cdot \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} $$

Example Calculation:



\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0.00 & 0.00\\   0.3248 & 0.1875 & -0.3248 & -0.1875 & 0.00 & 0.00\\    -0.5625 & -0.3248 & 3.0625 & -2.1752 & 2.5 & -2.5\\    -0.3248 & -0.1875 & -1.9375 & 2.6875 & -2.5 & 2.5\\    0.00 & 0.00 & 2.5 & -2.5 & 2.5 & -2.5\\    0.00 & 0.00 & -2.5 & 2.5 & -2.5 & 2.5\\ \end{bmatrix} \cdot \begin{bmatrix} 0\\      0\\      4.352\\      6.127\\       0\\       0\\     \end{bmatrix} =      \begin{bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{bmatrix} $$

The form above can be reduced to the form below :



\begin{bmatrix} -0.5625 & -0.3248\\    -0.3248 & -0.1875\\       3.0625& -2.1752\\     -1.9375 &  2.6875\\      -2.5   &   2.5  \\      -2.5   &   2.5  \\ \end{bmatrix} \cdot \begin{bmatrix} 4.352\\     6.127\\     \end{bmatrix} =      \begin{bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{bmatrix} $$

How to solve in MATLAB:

>> K = [-0.5625,-0.32476;-0.32476,-0.1875;3.0625,-2.1752;-2.1752,2.6875;-2.5,2.5;2.5,-2.5]

K =

-0.5625  -0.3248   -0.3248   -0.1875    3.0625   -2.1752   -2.1752    2.6875   -2.5000    2.5000    2.5000   -2.5000

>> d = [4.352;6.1271]

d =

4.3520   6.1271

>> F = K*d

F =

-4.4378  -2.5622    0.0003    7.0001    4.4377   -4.4377

Note: F3 and F4 return the given input values for P. Therefore, to make computations easier there is no      need to compute rows 3, 4 since row 3 and 4 will return the values for F3 and F4.

Results Comparison:

Method 1: Using the element force displacement relationship to calculate the force reactions.
 * F 1 = -4.4378
 * F 2 = -2.5622
 * F 5 = 4.4378
 * F 6 = -4.4378

Method 2: Using the global force displacement relationship to calculate the force reactions.
 * F 1 = -4.4378
 * F 2 = -2.5622
 * F 5 = 4.4377
 * F 6 = -4.4377

Method 3: Using statics to calculate the force reactions.
 * F 1 = -5.124cos(30o)= -4.4375
 * F 2 = -5.124sin(30o)= -2.5622
 * F 5 = 6.276cos(-45o)= 4.4378
 * F 6 = 6.276sin(-45o)= -4.4378



The image to the left is a visual representation of closing the loop between the finite element method and statics, to be further described below.To be able to close the loop we need to know how to go from the computed force reaction to an unknown displacement. By the statics method described above, we know the reactions of each element. Therefore, we also know the axial force being applied to that member. If we know both the axial load P(e) and the element stiffness k(e). Then we can calculate the axial displacement of the element. See calculations below for further explanation.

Recall the Equations for Axial Displacement :

$$ q_{2}^{(1)} = \frac{|P_{1}^{(1)}|}{k^{(1)}} =\frac{5.1243}{3/4} = 6.8324 $$

$$ q_{1}^{(2)} = \frac{|P_{2}^{(2)}|}{k^{(2)}} =\frac{6.276}{5} = 1.2552 $$



The image on the left is a geometric representation of the deformed shape of a two bar truss. Length AB is the geometric representation of the axial displacement of element 2. Length AC is the geometric representation of the axial displacement of element 1. The intersection of the perpendicular projection of line segment AB and the perpendicular projection of line segment AC, would be the new location of node 2 after the initial load is applied. If you connect a line from node 1 and node 3 to the new location of node 2 this would give you what is known as the deformed shape of the truss after being loaded.

The infinitesimal displacements are very closely related to the virtual displacement from the principal of virtual work. It is important to note in the figure at right that the coordinate system (in red) is taken to be at point A which correlates to the Global node 2 of the two bar truss system. The black vector ending at point D is the virtual displacement vector.

The length of the displacements AC and AB of the deformed shape are then solved for using the following equations.



AC = \frac{|P_{1}^{(1)}|}{k^{(1)}} =\frac{5.1243}{3/4} = 6.8324 $$



AB = \frac{|P_{1}^{(2)}|}{k^{(2)}} =\frac{6.276}{5} = 1.2552 $$

The next step is to solve for the (x,y) coordinates at points B and C using the following equations.



X_{B} = AB \cdot \cos{(135)} = 1.2552\cos{(135)} = -.88756 $$



Y_{B} = AB \cdot \sin{(135)} = 1.2552\sin{(135)} = .88756 $$



X_{C} = AC \cdot \cos{(30)} = 6.8324\cos{(30)} = 5.91703 $$



Y_{C} = AC \cdot \sin{(30)} = 6.8324\sin{(30)} = 3.4162 $$

The final two unknowns are XD and YD which can be solved for with the following equations.

To do so an equation is needed for the line AB and the line BC



PQ = (PQ)i^{~} = (X-X_{P})i + (Y-Y_{P})j $$



(PQ)i^{~} = PQ[ \cos{\theta}i + \sin{\theta}j ] $$

Therefore



X - X_{P} = PQ\cdot \cos{\theta} $$



Y - Y_{P} = PQ\cdot \sin{\theta} $$



\frac {Y-Y_{P}}{X-X_{P}} = \tan{\theta} $$



Y-Y{P} = (\tan{\theta})\cdot(X-X_{P}) $$

The equation for a line perpendicular and passing through P is as follows,



Y-Y_{P} = \tan{(\theta + \frac{\pi}{2})}(X-X{P}) $$

By definition and from the Finite Element Method in which we solved for d3 and d4



AD = d_{3}i + d_{4}j $$

Continuing, $$ X_{D} = 4.35 $$ and $$ Y_{D} = 6.125 $$

A Three Bar Truss
At right the global diagram of a three bar truss can seen.

Here,the force acting on the system is given to be,


 * $$ P = 30 $$

The properties of each bar are given below,



\begin{matrix} E^{(1)}=2 & E^{(2)}=4 & E^{(3)}=3\\ A^{(1)}=3 & A^{(2)}=1 & A^{(3)}=2\\ L^{(1)}=5 & L^{(2)}=5 & L^{(3)}=10 \end{matrix}

$$

The corresponding global and element free body diagrams are as follows.





Recall the equations used to solve static problems:



\sum{F_x=0} $$



\sum{F_y=0} $$



\sum{M_z=0} $$

Summing the moments about point A is trivial as all forces pass through the point and have a zero length moment arm. This action will give us no new information about the system. Therefore, there are 2 equations and 4 unknown forces. This problem is statically indeterminate.

Breaking Down the Element Coordinate Systems




The three diagrams above describe the element coordinate systems for the Three Bar Truss System.

The Moment About Point B
If the sum of the moments are computed about point B, will the sum still be zero? Since point B does not exist on any line of action, will this strategy allow us to solve for sum of the moments?




 * $$\overrightarrow{BA'}=\overrightarrow{BA} + \overrightarrow{AA'}$$

3D- Explanation:


 * $$\sum \overrightarrow{M_B}={\overrightarrow{BA}} \times {\overrightarrow{F}} ={\overrightarrow{BA'}} \times {\overrightarrow{F}}$$

For all values of A', on the line of action of the vector F:


 * $$\sum \overrightarrow{M_B}=({\overrightarrow{BA}} + {\overrightarrow{AA'}}) \times {\overrightarrow{F}}$$
 * $$\sum \overrightarrow{M_B}={\overrightarrow{BA}} \times {\overrightarrow{F}} + {\overrightarrow{AA'}} \times {\overrightarrow{F}} $$

Where:


 * $${\overrightarrow{AA'}} \times {\overrightarrow{F}} = \overrightarrow{O} $$

Three Bar Truss (cont)
Point A is in equilibrium: $$ \sum_{i=0}^{3}{\overrightarrow {Fi}}=\overrightarrow O $$

Note that the above equation is the same as the two scalar equations summing forces in the X and Y directions.

If we sum moments about B:


 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA'_i} \times \overrightarrow{F_i} $$


 * $$A'_{i}=$$ any point on line of action of $$\overrightarrow{F_i}$$


 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA} \times \overrightarrow{F_i} = \overrightarrow {BA}\times \sum_{i}\overrightarrow {F_i}$$

Since $$\sum_{i}\overrightarrow {F_i}=0$$, the sum of the moments about point B is zero.


 * $$ \sum_{i}\overrightarrow {M}_{B,i}=\sum_{i}\overrightarrow {BA} \times \overrightarrow{O} $$

Construction of Global Stiffness Matrix for a Three Bar Truss
The Three Bar Truss problem has 8 Degrees of Freedom (Thus an 8x8 matrix):



$$k_3$$ is in total a 4 x 4 matrix.

$$K_{33}=k^{(1)}_{33}+k^{(2)}_{11}+k^{(3)}_{11}$$

$$K_{34}=k^{(1)}_{34}+k^{(2)}_{12}+k^{(3)}_{12}$$

$$K_{43}=k^{(1)}_{43} + k^{(2)}_{21} + k^{(3)}_{21}$$

$$K_{44}=k^{(1)}_{44} + k^{(2)}_{22} + k^{(3)}_{22}$$

$$

K_{global}= \left [\begin{array}{cccccccc} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} & 0 & 0 & 0 & 0 \\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} & 0 & 0 & 0 & 0\\ k^{(1)}_{31} & k^{(1)}_{32} & (k^{(1)}_{33} + k^{(2)}_{11} + k^{(3)}_{11})& (k^{(1)}_{34} + k^{(2)}_{12}+k^{(3)}_{12})& k^{(2)}_{13} & k^{(2)}_{14} & k^{(3)}_{13} & k^{(3)}_{14} \\ k^{(1)}_{41} & k^{(1)}_{42} & (k^{(1)}_{43} + k^{(2)}_{21} + k^{(3)}_{21}) & (k^{(1)}_{44} + k^{(2)}_{22} + k^{(3)}_{22}) & k^{(2)}_{23}) & k^{(2)}_{24} & k^{(3)}_{23} & k^{(3)}_{24} \\ 0 & 0 & k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34} & 0 & 0 \\ 0 & 0 & k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} & 0 & 0 \\ 0 & 0 & k^{(3)}_{31} & k^{(3)}_{32} & 0 & 0 & k^{(3)}_{33} & k^{(3)}_{34} \\ 0 & 0 & k^{(3)}_{41} & k^{(3)}_{42} & 0 & 0 & k^{(3)}_{43} & k^{(3)}_{44}\end{array} \right] $$

MATLAB: Two Bar Truss Deformation
The undeformed truss is represented with the dotted lines. The deformed truss is the solid lines.

% HW 3 - Two Bar Truss System Deformation Plot % Description---This code plots the initial undeformed truss system and the resulting deformed truss % Created by: %          Michael Carr %          Akash Shah %          Taylor Boggs %          Pedro Rivero %          Josh McNally %          Tatiana Nobrega % Due --08OCT08

clear; %clear screen close; %close open windows

dof = 2; %degree of freedoms n_elem = 4; %number of elements (both deformed and original) n_node = 4; %number of nodes total_dof = 2 * n_node; %total dofs

position(:, 1) = [ 0; 0]; position(:, 2) = [ 3.464; 2]; position(:, 3) = [ 4.8783; 0.58579]; position(:, 4) = [ 7.8161; 8.1271];

% set up the nodal coordinate arrays x, y

for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); end

% setting up the element connectivity array node_connect, defined as follows: % node_connect(local node number, element number) = global node number

node_connect(1, 1) = 1; node_connect(2, 1) = 2; node_connect(1, 4) = 4; node_connect(2, 4) = 3; node_connect(1, 2) = 2; node_connect(2, 2) = 3; node_connect(1, 3) = 1; node_connect(2, 3) = 4;

% plot the whole truss system % loop over the elements

for i = 1 : 2 % for element i, do the following: % node_1 = global node number corresponding to the local node 1 node_1 = node_connect(1,i); % node_2 = global node number corresponding to the local node 1 node_2 = node_connect(2,i); % xx : 1x2 array containing x coordinates of node_1 and node_2 xx = [x(node_1),x(node_2)]; % yy : 1x2 array containing y coordinates of node_1 and node_2 yy = [y(node_1),y(node_2)]; % plot the element i in 3-D using command plot3 % use axis command to avoid having the plot window too tight axis([0 10 0 10]) % solid line plot(xx,yy,'--') %z axis removed from Crab.m code % hold the current figure for the next element hold on  end %The following loop connects the remaining elements for i = 3 : 4 node_3 = node_connect(1,i); node_4 = node_connect(2,i); xx = [x(node_3),x(node_4)]; yy = [y(node_3),y(node_4)]; axis([-2 10 -2 10]) plot(xx,yy,'-g') hold on  end

%This lables the nodes and elements

text(x(node_connect(1,1)),y(node_connect(1,1)), 'Global Node 1', 'HorizontalAlignment', 'center') text(x(node_connect(2,1)),y(node_connect(2,1)), 'Global Node 2', 'HorizontalAlignment', 'center') text(x(node_connect(2,3)),y(node_connect(1,4)), 'Global Node 2','HorizontalAlignment', 'center') text(x(node_connect(2,2)),y(node_connect(2,2)), 'Global Node 3', 'HorizontalAlignment', 'left') text(x(node_connect(2,1))/2,y(node_connect(2,1))/2, 'Element 1', 'HorizontalAlignment', 'center') text(x(node_connect(2,3))/2,y(node_connect(2,3))/2, 'Element 1','HorizontalAlignment', 'center') text(x(node_connect(1,2))/2 + x(node_connect(2,2))/2,y(node_connect(1,2))/2+y(node_connect(2,2))/2, 'Element 2', 'HorizontalAlignment', 'center') text(x(node_connect(2,3))/2 + x(node_connect(2,4))/2,y(node_connect(2,3))/2+y(node_connect(2,4))/2, 'Element 2', 'HorizontalAlignment', 'center')

%Labeling axis and titling graph title('Two Bar Truss---Original Configuration & Resulting Deformation') xlabel('x') ylabel('y')

Contributing Mediawiki Writers:

Eml4500.f08.ateam.mcnally 17:33, 7 October 2008 (UTC)

Eml4500.f08.ateam.nobrega 21:28, 7 October 2008 (UTC)

Eml4500.f08.ateam.boggs.t 21:30, 7 October 2008 (UTC)

Eml4500.f08.ateam.shah 21:29, 7 October 2008 (UTC)

Eml4500.f08.ateam.carr 21:47, 7 October 2008 (UTC)

Eml4500.f08.ateam.rivero 22:50, 7 October 2008 (UTC)