User talk:Eml4507.s13.team3.patrick

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=Problem 5.3= On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Given
Letting the coeffients of the eigenvectors look like the following matrices. $$x_1= \begin{Bmatrix} x_{11}\\ x_{12} \end{Bmatrix}$$ $$x_2= \begin{Bmatrix} x_{21}\\ x_{22} \end{Bmatrix}$$ $$ \begin{bmatrix} x_{1} & x_{2} \end{bmatrix} = \begin{bmatrix} x_{ij} \end{bmatrix} = \begin{bmatrix} x_{11}& x_{12}\\ x_{21}& x_{22} \end{bmatrix}$$ Now assume that $$x_{11}=x_{12}=1$$

Find
Find the eigenvectors for $$\gamma_1$$ and $$\gamma_2$$ when setting $$x_{11}=x_{12}=1$$ Plot the modes and create an animation of the moment of the modes

Solution
We find the eigenvectors from $$\gamma_1$$:

$$\gamma_1=4-\sqrt{5}$$ $$[K-\gamma_1I]x=$$ $$ \begin{bmatrix} -1+\sqrt{5} & -2\\ -2 & 1+\sqrt{5} \end{bmatrix}$$ $$ \begin{Bmatrix} x_1\\ x_2 \end{Bmatrix}$$ $$=$$ $$ \begin{Bmatrix} 0\\ 0 \end{Bmatrix}$$ Set $$x_{11}=x_{12}=1$$ $$(-1-\sqrt{5})x_1-(2)x_2=0$$ $$x_2=\frac{-1+\sqrt{5}}{2}$$ $$x_1= \begin{Bmatrix} 1\\ \frac{-1+\sqrt{5}}{2} \end{Bmatrix}$$

We find the eigenvectors from $$\gamma_2$$:

$$\gamma_2=4+\sqrt{5}$$ $$[K-\gamma_2I]x=$$ $$ \begin{bmatrix} -1-\sqrt{5} & -2\\ -2 & 1-\sqrt{5} \end{bmatrix}$$ $$ \begin{Bmatrix} x_1\\ x_2 \end{Bmatrix}$$ $$=$$ $$ \begin{Bmatrix} 0\\ 0 \end{Bmatrix}$$ Set $$x_{11}=x_{12}=1$$ $$(-1-\sqrt{5})x_1-(2)x_2=0$$ $$x_2=\frac{-1-\sqrt{5}}{2}$$ $$x_2= \begin{Bmatrix} 1\\ \frac{-1-\sqrt{5}}{2} \end{Bmatrix}$$ $$ \begin{bmatrix} x_1 & x_2 \end{bmatrix}= \begin{bmatrix} 1 & 1\\ \frac{-1+\sqrt{5}}{2} & \frac{-1-\sqrt{5}}{2} \end{bmatrix}$$



Animation of Mode 1

http://makeagif.com/i/Vi1kwS Animation of Mode 2 http://makeagif.com/i/ULkQrV

= Problem 1: Pb. 3.1, p. 3-8 Section 3 Notes using Matlab and CALFEM = On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Given
Plane truss as shown. A force, F, of 20000 N is applied as shown. The length of each element is 1 meter. Element 2 is vertical and the elements are spaced evenly around node 1. Each element has the same physical properties: Young's modulus of 206 GPa and cross sectional area of .0001 m. Additionally, node 2 experiences an initial displacement of 2 cm in the x direction and -1 cm in the y direction; node 3 experiences an initial displacement of -3 cm in the x direction and 5 cm in the y direction.

Find
Find the unknown displacements and forces at each global node as well as the member forces of each element using Matlab and verify using CALFEM. Plot the deformed and undeformed shape on the same plot.Perform the same tasks once again with zero prescribed displacements, and come compare the outcomes.

Using Matlab
Enter the number of global nodes. >> G=4;

Enter the x, y, and z coordinates (this program was designed solve 1D, 2D, and 3D problems, so the z coordinates must also be entered. As long as all nodes have the same z coordinate, the solution will match the 2D problem). >> x=[0 0 sqrt(3)/2 -sqrt(3)/2]; >> y=[0 1 -.5 -.5]; >> z=[0 0 0 0];

Enter the input force matrix. The first element of each row corresponds to the global degree of freedom; the second element corresponds to the force associated with that degree of freedom. Only the known forces are entered into this matrix. >> F_in=[1 20000/sqrt(2); 2 20000/sqrt(2)];

Enter the input displacement matrix. The first element of each row corresponds to the global degree of freedom; the second element corresponds to the displacement associated with that degree of freedom. Only the known displacements are entered into this matrix. If the force for a particular degree of freedom is unknown, then its displacement is known. >> Q_in=[3 0; 4 .02; 5 -.01; 6 0; 7 -.03; 8 .05; 9 0; 10 0; 11 0; 12 0];

Enter the number of elements. >> E=3;

Enter the global node corresponding to the first local node of each element. The column number corresponds to the element number of the given node. >> l1=[1 1 1];

Enter the global node corresponding to the second local node of each element. >> l2=[3 2 4];

Enter the Young's Modulus associated with each element. The column number corresponds to the element number of the given node. >> Modulus=10^9*[206 206 206];

Enter the Young's Modulus associated with each element. The column number corresponds to the element number of the given node. >> Area=10^-4*[1 1 1];

Run the displacement program. >> [k Q F Q_bar F_bar]=displacement(G, x, y, z, F_in, Q_in, E, l1, l2, Modulus, Area);

The stiffness matrix k: k = 1.0e+07 * 3.0900        0         0         0   -1.5450    0.8920   -1.5450   -0.8920            0    3.0900         0   -2.0600    0.8920   -0.5150   -0.8920   -0.5150            0         0         0         0         0         0         0         0            0   -2.0600         0    2.0600         0         0         0         0      -1.5450    0.8920         0         0    1.5450   -0.8920         0         0       0.8920   -0.5150         0         0   -0.8920    0.5150         0         0      -1.5450   -0.8920         0         0         0         0    1.5450    0.8920      -0.8920   -0.5150         0         0         0         0    0.8920    0.5150

The image to the right shows a plot of the deformed and undeformed truss. The displacement matrix Q: Q = 1.0000  -0.0290       2.0000    0.0108       3.0000         0       4.0000    0.0200       5.0000   -0.0100       6.0000         0       7.0000   -0.0300       8.0000    0.0500       9.0000         0      10.0000         0      11.0000         0      12.0000         0

This shows that the first and second degrees of freedom (x and y directions) for global node 1 experience a displacement of -2.9 cm and 1.08 cm respectively. The remaining displacements were given in the problem.

The Force matrix, F F = 1.0e+05 * 0.1414      0.1414            0            0      -4.2816            0      -3.6562       2.1109            0       3.5148       2.0293            0

This shows the forces at each node. Every 3 rows correspond to 1 node. Global node 1 (rows 1-3) shows the force that was given in the problem.

The internal spring force matrix F_bar F_bar = 1.0e+05 * 4.2219   4.2816    4.0586      -4.2219   -4.2816   -4.0586

Row 1 gives the member forces with respect to the Global nodes and row 2 gives the member forces with respect to the members.

Matlab Code
This code is equivalent to that used in R2.3 except that it uses the given spring constants rather than than the modulus and area to determine the stiffness matrix.

CALFEM Verification
First begin by constructing the Edof  matrix. Column 1 corresponds to the element number; columns 2 and 3 correspond to the global node that corresponds to the first and second local nodes respectively. >> Edof=[1 1 2 5 6 2 1 2 3 4           3 1 2 7 8];

Construct an empty stiffness matrix K that has side lengths equal to the number of degrees of freedom times the number of global nodes. For this problem, there are 4 global nodes and motion is constrained to 2 degrees of freedom. >> K=zeros(8,8);

Construct the force matrix F. The forces corresponding to global node 1 are known; the forces corresponding to global nodes 2, 3, and 4 are unknown. >> F=zeros(8,1); >> F(1)=20000/sqrt(2); >> F(2)=20000/sqrt(2);

Construct the displacement matrix Q. Global nodes 2, 3, and 4 and known displacements as shown. >> Q=[3 .02 4 -.01        5 -.03         6 .05         7 0         8 0];

The function bar2e generates the element stiffness matrices for a bar with 2 degrees of freedom. Its parameters are the x and y coordinates of the element and a vector containing the Young's modulus and cross sectional area. This information is given in the problem statement.

The Vector containing the Young's modulus and cross sectional area. >> E=206e9; >> A=1e-4; >> ep=[E A];

The x and y coordinates for each element. >> ex1=[0 sqrt(3)/2]; >> ey1=[0 -.5]; >> ex2=[0 0]; >> ey2=[0 1]; >> ex3=[0 -sqrt(3)/2]; >> ey3=[0 -.5];

The commands: >> Ke1=bar2e(ex1, ey1, ep); >> Ke2=bar2e(ex2, ey2, ep); >> Ke3=bar2e(ex3, ey3, ep);

Yield the following stiffness matrices: >> Ke1 = 1.0e+07 * 1.5450  -0.8920   -1.5450    0.8920          -0.8920    0.5150    0.8920   -0.5150          -1.5450    0.8920    1.5450   -0.8920           0.8920   -0.5150   -0.8920    0.5150      Ke2 = 0          0           0           0                0    20600000           0   -20600000                0           0           0           0                0   -20600000           0    20600000   >> Ke3 = 1.0e+07 * 1.5450   0.8920   -1.5450   -0.8920          0.8920    0.5150   -0.8920   -0.5150         -1.5450   -0.8920    1.5450    0.8920         -0.8920   -0.5150    0.8920    0.5150

The global stiffness matrix can be assembled with the element stiffness matrices using the following command. >> K=assem(Edof(1,:),K,Ke1); >> K=assem(Edof(2,:),K,Ke2); >> K=assem(Edof(3,:),K,Ke3);

This results in the following stiffness matrix. >> K = 1.0e+07 * 3.0900        0         0         0   -1.5450    0.8920   -1.5450   -0.8920               0    3.0900         0   -2.0600    0.8920   -0.5150   -0.8920   -0.5150               0         0         0         0         0         0         0         0               0   -2.0600         0    2.0600         0         0         0         0         -1.5450    0.8920         0         0    1.5450   -0.8920         0         0          0.8920   -0.5150         0         0   -0.8920    0.5150         0         0         -1.5450   -0.8920         0         0         0         0    1.5450    0.8920         -0.8920   -0.5150         0         0         0         0    0.8920    0.5150

This stiffness matrix is in accord with that computed in Matlab in the previous section.

The function solveq takes the stiffness matrix, known force matrix, and displacement matrix and returns the complete force and displacement matrices. >> [q r]=solveq(K,F,Q) q = -0.0290      0.0108       0.0200      -0.0100      -0.0300       0.0500            0            0   r = 1.0e+05 * 0.0000      0.0000            0      -4.2816      -3.6562       2.1109       3.5148       2.0293

The displacement and force matrices are in accord with those computed in Matlab in the previous section.

The function extract is used to evaluate the displacements of the local nodes for each beam, which is needed to determine the force in each beam. >> ed1=extract(Edof(1,:),q); >> ed2=extract(Edof(2,:),q); >> ed3=extract(Edof(3,:),q);

This results in the following displacements. The number following 'ed' corresponds to the element number. Each column corresponds to the respective degree of freedom for the beam. ed1 = -0.0290   0.0108   -0.0300    0.0500 ed2 = -0.0290   0.0108    0.0200   -0.0100 ed3 = -0.0290   0.0108         0         0

The function bar2s can be used to determine the force in each beam. Its input parameters are the x and y coordinates for the beam (ex1 and ey1), the physical properties of the beam (Young's modulus and cross sectional area - ep) and the displacement (ed). >> es1=bar2s(ex1, ey1,ep,ed1) >> es1=bar2s(ex2, ey2,ep,ed2) >> es1=bar2s(ex3, ey3,ep,ed3)

This results in the following spring forces. es1 = -4.2219e+05 es2 = -4.2816e+05 es3 = -4.0586e+05

This is in accord with the forces computed in Matlab in the previous section.

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 * File:Truss undeformed plot.PNG
 * File:Zero disp plot.PNG

MaintenanceBot (discuss • contribs) 03:45, 11 November 2013 (UTC)