University of Florida/Eml4507/s13 Team 3 Report 4

Problem 4.1

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

Given Mode Shape for system. Mode slope increases and crosses into the positive region

Spring-damper-body arrangement as shown. Two separate forces applied to masses.

$M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}$ $d={\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$ $C={\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}$ $K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}$ $k_{1}=1,k_{2}=3,k_{3}=3$ $K={\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}$ $[K-$ γ$I]x=$ ${\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}$ $-$ γ ${\begin{bmatrix}1&0\\0&1\end{bmatrix}}$ $)$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ $det{\begin{Bmatrix}3-\gamma &-2\\-2&5-\gamma \end{Bmatrix}}=\gamma ^{2}-8\gamma +11=0$ Find

Find the eigenvector $x_{2}$ corresponding to the eigenvalue $\gamma _{2}$ for the spring-mass-damper system on p.53-113. Plot and comment on this mode shape. Verify that the eigenvectors are orthogonal to each other

Solution

Eigenvalues are found
$\gamma _{1}=4+{\sqrt {5}}>0$ $\gamma _{2}=4-{\sqrt {5}}>0$ We find the eigenvectors from $\gamma _{2}$ $\gamma _{2}=4+{\sqrt {5}}$ $[K-\gamma _{2}I]x=$ ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ Set $x_{2}=1$ $(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$ $x_{1}={\frac {2}{-1-{\sqrt {5}}}}$ Eigenvectors are orthogonal to each other:

EDU>> x= [-.8507;-.5257];
EDU>> y= [-.5257;.8507];
EDU>> transpose(y)*x
ans = 0

Problem 4.2

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

Given

Use same given values as in problem 4.1

Find Mode Shape for system. Notice plot is the same even with different initial conditions

Find the eigenvectors for $\gamma _{1}$ and $\gamma _{2}$ when setting $x_{1}=1$ Solution

We find the eigenvectors from $\gamma _{1}$ $\gamma _{1}=4-{\sqrt {5}}$ $[K-\gamma _{1}I]x=$ ${\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ Set $x_{1}=1$ $(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$ $x_{2}={\frac {-1+{\sqrt {5}}}{2}}$ We find the eigenvectors from $\gamma _{2}$ $\gamma _{2}=4+{\sqrt {5}}$ $[K-\gamma _{2}I]x=$ ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$ Set $x_{1}=1$ $(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$ $x_{2}={\frac {-1-{\sqrt {5}}}{2}}$ R4.3 (fead.f08 p.11-3 (Method 1: Square Root Sum of Squares), p.14-3 (Method 2: Transformation Matrix) )

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

Given two methods for finding axial member forces

Method 1:

$P_{1}^{(e)}={\sqrt {(f_{1}^{(e)})^{2}+(f_{2}^{(e)})^{2}}}$ $P_{2}^{(e)}={\sqrt {(f_{3}^{(e)})^{2}+(f_{4}^{(e)})^{2}}}$ Method 2:

$\mathbf {P} ^{(e)}=\mathbf {T} ^{(e)}\mathbf {f} ^{(e)}$ ${\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}}={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}{\begin{Bmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\end{Bmatrix}}$ Find

• Discuss computational efficiency of each method.
• Reconcile analytically using both algebra and geometry

Method 1: Square Root Sum of Squares

Given

$P_{1}^{(e)}={\sqrt {(f_{1}^{(e)})^{2}+(f_{2}^{(e)})^{2}}}$ $P_{2}^{(e)}={\sqrt {(f_{3}^{(e)})^{2}+(f_{4}^{(e)})^{2}}}$ Discussion

The first method uses the Pythagorean Theorem, which is also a distance formula, to find the axial member forces from the nodal forces. This method only requires the two nodal forces on a node to find an axial member force. To use this method, the nodal forces are defined, and then put into the distance formula. Additional axial forces are found by defining additional nodal forces. The formula must be repeated each time to give each axial member force.

Method 2: Transformation Matrix

Given

$\mathbf {P} ^{(e)}=\mathbf {T} ^{(e)}\mathbf {f} ^{(e)}$ ${\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}}={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}{\begin{Bmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\end{Bmatrix}}$ Discussion

The second method uses the transformation matrix to add the projections of the nodal forces along the element. This method requires the two nodal forces and an angle. However, the angle can be applied to both ends of the node. To use this method, the transformation matrix, $\mathbf {T} ^{(e)}$ is created, with $l^{(e)}=\cos {\theta }$ and $m^{(e)}=\sin {\theta }$ , as well as the nodal force matrix $\mathbf {f} ^{(e)}$ .

The matrix multiplication performs the following operations:

$P_{1}=l^{(e)}f_{1}^{(e)}+m^{(e)}f_{2}^{(e)}+0+0$ $P_{1}=f_{1}^{(e)}\cos \theta +f_{2}^{(e)}\sin \theta$ $P_{2}=0+0+l^{(e)}f_{3}^{(e)}+m^{(e)}f_{4}^{(e)}$ $P_{2}=f_{3}^{(e)}\cos \theta +f_{4}^{(e)}\sin \theta$ To find additional axial member forces, new nodal forces and angles can be defined, and the transformation can be expanded. The matrix multiplication only has to be performed once for each element. The multiplication will give a matrix with all the axial forces.

Reconciliation using algebra and geometry

The Pythagorean Theorem is used for the first method. To reconcile this method with the second method, geometry is used to define the nodal forces in terms of $P_{1}$ .

$f_{1}^{(e)}=P_{1}^{(e)}\cos \theta$ $f_{2}^{(e)}=P_{1}^{(e)}\sin \theta$ These nodal force definitions are substituted into the Pythagorean Theorem, and the equation is simplified.

$P_{1}^{(e)}={\sqrt {(f_{1}^{(e)})^{2}+(f_{2}^{(e)})^{2}}}$ $P_{1}^{(e)}={\sqrt {(P_{1}^{(e)}\cos \theta )^{2}+(P_{1}^{(e)}\sin \theta )^{2}}}$ $P_{1}^{(e)}={\sqrt {(P_{1}^{(e)})^{2}(\cos ^{2}\theta +\sin ^{2}\theta )}}$ $P_{1}^{(e)}={\sqrt {(P_{1}^{(e)})^{2}(1)}}$ $P_{1}^{(e)}={\sqrt {(P_{1}^{(e)})^{2}}}$ $P_{1}^{(e)}=P_{1}^{(e)}$ 