User:Eml4507.s13.team2/Report2

Problem Statement
Given the 2-D truss system with two inclined elements, we want to verify the equilibrium at the second node, or point where the two truss elements meet.

Solution
In order to verify equilibrium at node 2 we must prove the following equations:
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$$  \displaystyle \Sigma Fx={P_{1}}^{(1)}cos\Theta ^{(1)}-{P_{2}}^{(1)}cos\Theta ^{(2)}=0 $$     (1.0)
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$$  \displaystyle \Sigma Fy=P-{P_{1}}^{(1)}sin\Theta ^{(1)}-{P_{2}}^{(1)}sin\Theta ^{(2)}=0 $$     (1.1) Where
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$$  \displaystyle \Theta ^{(1)}=30^{\circ}  and    \Theta ^{(2)}=45^{\circ} $$     (1.2) In problem R1.3 we were able to create a stiffness matrix of the form
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$$  \displaystyle \left[ \begin{array}{cccccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 \\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \\ \end{array} \right] $$ $$\left[ \begin{array}{l} d_1 \\ d_2 \\ d_3  \\ d_4 \\ d_5 \\ d_6 \end{array} \right] = \left[ \begin{array}{l} F_1 \\ F_2  \\ F_3  \\ F_4 \\ F_5 \\ F_6 \end{array} \right] $$     (1.3) plugging in the values that were solved for we get the matrix
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$$  \displaystyle \left[ \begin{array}{cccccc} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\-0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\-0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\\ \end{array} \right] $$     (1.4) Ignoring the columns with values of zeroes, the system can be reduced to:
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$$  \displaystyle \left[ \begin{array}{cccccc} -0.5625 & -0.3248 \\ -0.3248 & -0.1875 \\ 3.0625 & -2.1752 \\ -2.1752 & 2.6875 \\ -2.5 & 2.5 \\2.5 & -2.5 & \\ \end{array} \right]$$ $$\left[ \begin{array}{l}  d_3  \\ d_4 \end{array} \right] = \left[ \begin{array}{l} F_3  \\ F_4  \end{array} \right] $$     (1.5) Given that
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$$  \displaystyle F_{3}=0 and F_{4}=7 $$     (1.7) Given that
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$$  \displaystyle \left[ \begin{array}{l} d_3  \\ d_4 \end{array} \right] = \left[ \begin{array}{cc}3.0625 & -2.1752 \\ -2.1752 & 2.6875\end{array} \right]^{-1} \left[ \begin{array}{l} 0  \\ 7  \end{array} \right]=\left[ \begin{array}{l} 4.325  \\ 6.127  \end{array} \right] $$     (1.8) Now we can solve for the force components from both element 1 and element 2.
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$$  \displaystyle \left[ \begin{array}{cccc} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{array} \right] $$$$\left[ \begin{array}{l} 0 \\ 0 \\ 4.352  \\ 6.127 \end{array} \right] = \left[ \begin{array}{l} -4.438 \\ -2.562  \\ 4.438  \\ 2.562 \end{array} \right] $$     (1.9)
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$$  \displaystyle \left[ \begin{array}{cccc} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{array} \right] $$$$\left[ \begin{array}{l} 4.352 \\ 6.127 \\ 0  \\ 0 \end{array} \right] = \left[ \begin{array}{l} -4.438 \\ 4.438  \\ 4.438  \\ -4.438 \end{array} \right] $$     (1.10) Finally we are ready to solve for our P values in equation 1.1
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$$  \displaystyle P^{(1)}_{1}=\sqrt{(f_{1}^{(1)})^{2}+(f_{2}^{(1)})^{2}}=\sqrt{(-4.438)^{2}+(-2.562)^{2}}=5.124 $$     (1.11)
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$$  \displaystyle P^{(2)}_{2}=\sqrt{(f_{3}^{(2)})^{2}+(f_{4}^{(2)})^{2}}=\sqrt{(4.438)^{2}+(-2.438)^{2}}=6.276 $$     (1.12) Thus,
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$$  \displaystyle \Sigma Fx=5.124cos(30)-6.276cos(45)=0 $$ $$   \displaystyle \Sigma Fy=7-5.124sin(30)-6.276sin(45)=0 $$     (1.13)
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MATLAB with CALFEM
The following MATLAB code was provided by professor Vu-Quoc to a class in a previous semester: http://en.wikiversity.org/wiki/User:Eml4500.f08.delta_6.guzman/MATLAB_Two_Bar_Truss_Example_Modified originally from: http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

In order for the code to run properly, as the past semester report states, there where three other functions needed in order for the code to properly work. The source of those functions is located at http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Once run with the previously mentioned additional functions, the outputs all agree with the work done above. We also get another matrix as follows:
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$$  \displaystyle results = \left[ \begin{array}{ccccc} 1.7081 & 5.1244 & 8.5406 & 5.1244 & 17.081 \\ 0.6276 & 1.8828 & 3.138 & 1.8828 & 6.276 \\ \end{array} \right] $$
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where elements (1,4) and (2,5) are the P values that we previously used when summing the forces in the X and Y direction, confirming that they do in fact equate to 0 and that node 2 is at equilibrium.

Solved By:--Eml4507.s13.team2.rosenberg (talk) 01:43, 6 February 2013 (UTC) This problem was solved using Dr. Vu-Quoc's lecture notes http://upload.wikimedia.org/wikiversity/en/0/04/Eml4500.f08.1.djvu and http://upload.wikimedia.org/wikiversity/en/1/13/Eml4500.f08.2.djvu and was a continuation of our R1.3

The MATLAB code came from the source stated above.

Problem Statement
Consider a linear 2nd-order ODE (L2-ODE-CC) with constant coefficients with the following complex conjugate roots:
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$$  \displaystyle \lambda_{1,2}=-0.5 \pm i2 $$ (2.0)
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Find the homogeneous L2-ODE-CC in standard form in terms of $$ \displaystyle y_h (t) $$ with the above roots and find the damping ratio $$ \displaystyle \zeta $$.

Consider the case with no excitation, and find and plot the solution satisfying these initial conditions:
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$$  \displaystyle y(0)=1, \ y'(0)=0 $$     (2.1)
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Using the plot of the solution, find the approximate amplitudes of oscillation in a cycle. Determine the log decrements for $$ \displaystyle n=1,2,3 $$.

Then, find the average log decrement and verify its value.

Finally, find the quality factor Q and the loss factor $$ \displaystyle \eta $$ for the system described by the above roots.

Solution
The form of a homogeneous L2-ODE-CC in standard form in terms of $$ \displaystyle y_h(t) $$ is:
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$$  \displaystyle y_h'' + ay_h' + by_h = 0 $$     (2.2)
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To find the values $$ \displaystyle a $$ and $$ \displaystyle b $$ and derive the L2-ODE-CC, first find the characteristic equation of the form:
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$     (2.3)
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Start by working backward using the quadratic formula and the given roots.
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$$  \displaystyle \lambda_{1,2}=\frac{-a\pm\sqrt{a^2-4b}}{2}=-0.5 \pm i2 $$ (2.4)
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As a result,
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$$  \displaystyle a=1, \ b=4.25 $$     (2.5)
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Then, the characteristic equation is:
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$$  \displaystyle \lambda^2+\lambda+4.25 = 0 $$     (2.6)
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Thus, the homogeneous L2-ODE-CC is:
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$$  \displaystyle y_h'' + y_h' + 4.25y_h = 0 $$     (2.7)
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To find the damping ratio $$ \displaystyle \zeta $$, first find the undamped natural circular frequency $$ \displaystyle \omega $$ of the system.
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$$  \displaystyle \omega^2:=b=4.25 $$     (2.8)
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$$  \displaystyle \omega = \sqrt{4.25}\approx 2.0616 $$     (2.9)
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The damping ratio $$ \displaystyle \zeta $$ can be found with the equation:
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$$  \displaystyle 2\omega \zeta := a = 1 $$     (2.10)
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Solving for $$ \displaystyle \zeta $$ gives:
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$$  \displaystyle \zeta = \frac{a}{2\omega}=\frac{1}{2*2.0616} $$     (2.11)
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$$  \displaystyle \zeta = 0.2425 $$     (2.12)
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The homogeneous solution for the case of two complex conjugate roots is:
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$$  \displaystyle y_h(t)=c_1e^{mt}\cos{nt}+c_2e^{mt}\sin{nt} $$     (2.13)
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The variables $$ \displaystyle m $$ and $$ \displaystyle n $$ are the components of the complex conjugate roots $$ \displaystyle \lambda_{1,2}=m \pm in $$.

Plugging in the given roots results in:
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$$  \displaystyle y_h(t)=c_1e^{-0.5t}\cos{2t}+c_2e^{-0.5t}\sin{2t} $$     (2.14)
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Differentiating gives:
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$$  \displaystyle y'_h(t)=e^{-0.5t}[(-2c_1-0.5c_2) \sin{2t} + (2c_2-0.5c_1) \cos{2t}] $$     (2.15)
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Applying the initial condition $$ \displaystyle y(0)=1 $$ to Eq. 2.14 results in:
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$$  \displaystyle 1=c_1e^{-0.5(0)} \cos(2*0)+c_2e^{-0.5(0)}\sin(2*0) $$     (2.16)
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$$  \displaystyle 1=c_1\cos(0) + c_2 \sin(0) $$     (2.17)
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$$  \displaystyle 1=c_1 $$     (2.18)
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Applying the initial condition $$ \displaystyle y'(0)=0 $$ to Eq. 2.15 results in:
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$$  \displaystyle 0=e^{-0.5t}[(-2c_1-0.5c_2) \sin{(2*0)} + (2c_2-0.5c_1) \cos{(2*0)}] $$     (2.19)
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$$  \displaystyle 0=(2c_2-0.5c_1) $$     (2.20)
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$$  \displaystyle 0.25=c_2 $$     (2.21)
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Then, the overall homogeneous solution is:
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$$  \displaystyle y_h(t)=e^{-0.5t} \cos(2t) + 0.25e^{-0.5t} \sin(2t) $$     (2.22)
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The solution is plotted below.



Using the MATLAB function "ginput" with the above plot, approximate amplitudes of oscillation were found and used to calculate the log decrements for $$ \displaystyle n=1,2,3 $$ from the following:
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$$  \displaystyle \delta^{(n)} = \log \frac{y_h^{(n)}}{y_h^{(n+1)}} $$     (2.23)
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$$  \displaystyle \delta^{1} = \log \frac{0.2080}{0.0433} \approx 1.5694 $$     (2.24)
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$$  \displaystyle \delta^{2} = \log \frac{0.0433}{0.0089} \approx 1.5821 $$     (2.25)
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$$  \displaystyle \delta^{3} = \log \frac{0.0089}{0.0020} \approx 1.4929 $$     (2.26)
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The average approximated log decrement is:
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$$  \displaystyle \delta \approx \frac13 [\delta^{(1)}+\delta^{(2)} + \delta^{(3)}] $$     (2.27)
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$$  \displaystyle \delta \approx \frac13 [(0.6816)+(0.6871 + (0.6484)] $$     (2.28)
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$$  \displaystyle \delta \approx 1.5481 $$     (2.29)
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To verify the value of $$ \displaystyle \delta $$, the following equation is used:
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$$  \displaystyle \delta = \frac{2 \pi \zeta}{\sqrt{1- \zeta^2}} $$     (2.30)
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$$  \displaystyle \delta = \frac{2 \pi 0.2425}{\sqrt{1- (0.2425)^2}} $$     (2.31)
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$$  \displaystyle \delta \approx 1.5708 $$     (2.32)
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This value is close to the approximated average log decrement above.

The quality factor Q is found by:
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$$  \displaystyle Q:= \frac {\pi}{\delta} $$     (2.33)
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$$  \displaystyle Q:= \frac {\pi}{1.5708}=2 $$     (2.34)
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The loss factor $$ \displaystyle \eta $$ is found by:
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$$  \displaystyle \eta:=\frac{1}{Q} $$     (2.35)
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$$  \displaystyle \eta:=\frac{1}{2}=0.5 $$     (2.36)
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Problem Statement
Problem Statement....

Given:

Solution
Answer:

Problem Statement
(a) Consider the L2-ODE-CC system described by the roots in (1) p.53-5c. Find the system natural circular frequency.

(b) Let the system be excited by a periodic force of the form shown in (1) p.53-6, find the expression for the particular solution.

(c) Find the particular solution for this case with the given information and the amplification factor.

(d) Find the complete solution satisfying the initial conditions in (2) p.53-5c.

(e) Plot the homogeneous, particular and complete equations separate figures.

(f) Plot the homogeneous, particular and complete equations in the same figures.

Given:
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$$  \displaystyle \lambda_{1,2} = -0.5 \pm i2 $$ (1.0)
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Solution
(a) Starting with the characteristic equation and the given information, we can solve for "a" and "b".
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$$  \displaystyle \lambda^2 + a\lambda + b = 0
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$$     (1.1)
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$$  \displaystyle \lambda_{1,2} = \frac{-a \pm {\sqrt{a^2 - 4b}}}{2}
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$$     (1.2)
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$$  \displaystyle -0.5 = -a/2
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$$     (1.3)
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$$  \displaystyle a = 1
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$$     (1.4)
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$$  \displaystyle -2 = \frac{\sqrt{a^2 - 4b}}{2}
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$$     (1.5)
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$$  \displaystyle b = 17/4
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$$     (1.6) After solving for "b" the natural circular frequency can be found
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$$  \displaystyle \omega^2 = b
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$$     (1.7)
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$$  \displaystyle \omega = 2.0615
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$$     (1.8) (b) Using the trial particular solution equation and a few definitions we can solve for the general particular solution
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$$  \displaystyle f(t) = f_0 \cos\bar\omega t
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$$     (1.9)
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$$  \displaystyle y_p(t)=de^{i(\bar\omega t - \phi)}
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$$     (1.10)
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$$  \displaystyle \zeta=\frac{a}{2\omega}
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$$     (1.11)
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$$  \displaystyle d=\frac{f_0/m}{\sqrt{(\omega^2 - \bar\omega^2)^2 + (2 \zeta \omega \bar\omega)^2}}
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$$     (1.12)
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$$  \displaystyle \tan\phi=\frac{2\zeta\omega\bar\omega}{\omega^2 - \bar\omega^2}
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$$     (1.13)
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$$  \displaystyle y_p(t) = \frac{f_0/m}{\sqrt{(\omega^2 - \bar\omega^2)^2 + (2 \zeta \omega \bar\omega)^2}} e^{i(\bar\omega t - \arctan \frac{2\zeta\omega\bar\omega}{\omega^2 - \bar\omega^2})}
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$$     (1.14) Given the following information
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$$  \displaystyle \bar\omega=0.9\omega
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$$     (1.15)
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$$  \displaystyle f_0/m = r_0 = 1
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$$     (1.16) we can solve for the particular solution we are looking to solve
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$$  \displaystyle y_p(t)=0.1946e^{i(3.825t - 0.8395)}
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$$     (1.17) Also, we can solve for the amplification factor
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$$  \displaystyle \rho=\frac{\bar\omega}{\omega}
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$$     (1.18)
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$$  \displaystyle \mathbb A=\frac{1}{\left[(1 - \rho^2)^2 + (2\zeta\rho)^2\right]^{1/2}}
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$$     (1.19)
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$$  \displaystyle \mathbb A= 3.512
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$$     (1.20) (d) Using the homogeneous equation and the initial conditions that are given, we can solve for the homogeneous solution
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$$  \displaystyle y_h(t)=e^{-at/2}[A\cos\tilde\omega t + B\sin\tilde\omega t]
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$$     (1.21)
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$$  \displaystyle y(0)=1
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$$     (1.22)
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$$  \displaystyle y'(0)=0
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$$     (1.23)
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$$  \displaystyle y'_h(t)=A(-0.5e^{-0.5t}\cos2t - 2e^{-0.5 t}\sin2t) + B(-0.5e^{-0.5t}\sin2t + 2e^{-0.5t}\cos2t)
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$$     (1.24)
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$$  \displaystyle A = 1
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$$     (1.25)
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$$  \displaystyle B = 1/4
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$$     (1.26)
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$$  \displaystyle y_h(t)=e^{-0.5t}\cos2t+(1/4)e^{-0.5t}\sin2t
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$$     (1.27) Now we can add the particular solution and the homogeneous solution to finally get the complete solution for the problem
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$$  \displaystyle y(t)=e^{-0.5t}\cos2t+(1/4)e^{-0.5t}\sin2t+0.1946e^{i(3.825t - 0.8395)}
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$$     (1.28)
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(d,e) Here are all the solutions to y(t) on different graphs and one graph altogether

Problem Statement


Given:
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$$ F_{2}=100 \;\; K=1500\;\; u_{1}=u_{3}=0 $$
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Solution



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$$ F_{1}=-40 \;\;\;F_{3}=-60\;\; \; u_{2}=.0133\;
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$$     (1.0)
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Problem Statement
Find all forces on springs, the displacements, and the reaction forces.



Given:

$$ F_3=1000N, k_1=500\frac{N}{mm}, k_2=400\frac{N}{mm}, k_3=600\frac{N}{mm}, k_4=200\frac{N}{mm}, k_5=400\frac{N}{mm}, k_6=300\frac{N}{mm} $$
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$$  \displaystyle
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$$     (1.0)
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Solution
$$ f_1^{(1)}+f_1^{(4)}=F_1=R_1 $$

$$ f_2^{(1)}+f_2^{(2)}+f_2^{(3)}=F_2=0 $$

$$ f_3^{(3)}+f_3^{(4)}+f_3^{(5)}=F_3=1000 $$

$$ f_4^{(2)}+f_4^{(5)}+f_4^{(6)}=F_4=0 $$

$$ f_5^{(6)}=F_5=R_5 $$

$$ 100\begin{bmatrix} k^1+k^4&-k^1&-k^4&0&0 \\ -k^1&k^1+k^2+k^3&-k^3&-k^2&0 \\ -k^4&-k^3&k^3+k^4+k^5&-k^5&0 \\ 0&-k^2&-k^5&k^2+k^5+k^6&-k^6 \\ 0& 0 & 0 & -k^6 & k^6 \end{bmatrix}\begin{bmatrix} u_1\\ u_2\\ u_3\\ u_4\\ u_5 \end{bmatrix}=\begin{bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5 \end{bmatrix} $$

$$ 100\begin{bmatrix} 7&-5&-2&0&0 \\ -5&15&-6&-4&0 \\ -2&-6&12&-4&0 \\ 0&-4&-4&11&-3 \\ 0& 0 & 0 & -3 & 3 \end{bmatrix}\begin{bmatrix} u_1\\ u_2\\ u_3\\ u_4\\ u_5 \end{bmatrix}=\begin{bmatrix} R_1\\ 0\\ 1000\\ 0\\ R_5 \end{bmatrix} $$

Striking the rows and columns corresponding to zero displacement

$$ k^{(e)}=100\begin{bmatrix} 15& -6 & -4\\ -6&12 &-4 \\ -4&-4  &11 \end{bmatrix} $$

$$ f^{(e)}=100\begin{bmatrix} 0\\ 1000\\ 0 \end{bmatrix} $$

$$ q^{(e)}=100\begin{bmatrix} u_2\\ u_3\\ u_4 \end{bmatrix} $$

$$ [k^{(e)}]\left \{ q^{(e)} \right \}=\left \{ f^{(e)} \right \} $$

$$ \left \{ q^{(e)} \right \}=[k^{(e)} ]^{-1}\left \{ f^{(e)} \right \} $$

$$ P^{(e)}=k^{(e)}\Delta ^{(e)} $$

$$ P^{(1)}=500\frac{N}{mm} (0mm-0.8542mm)=-427.1N $$

$$ P^{(2)}=400\frac{N}{mm} (0.8542mm-0.8750mm) =-8.32N $$

$$ P^{(3)}=600\frac{N}{mm}(0.8542mm-1.5521mm)=-418.7N $$

$$ P^{(4)}=200\frac{N}{mm}(0mm-1.5521mm)=-310.4N $$

$$ P^{(5)}=400\frac{N}{mm}(1.5521mm-0.8750mm)=270.8N $$

$$ P^{(6)}=300\frac{N}{mm}(0.8750mm-0mm)=262.5N $$

Confirming with CALFEM


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$$  \displaystyle
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$$     (1.1) next is a boxed solution
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Spring Forces:
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$$  \displaystyle P^{(1)}=-427.1N $$

$$ P^{(2)}=-8.32N $$

$$ P^{(3)}=-418.7N $$

$$ P^{(4)}=-310.4N $$

$$ P^{(5)}=270.8N $$

$$ P^{(6)}=262.5N $$

Reaction Forces: $$ R_1=-737.5N and R_2=-262.5N $$

Displacement: $$ u_2=0.8542mm, u_3=1.5521mm, and u_4=0.8750mm $$

(1.0)
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