User:EML4507.s13.team2.havlock/Report2

=Report 2=

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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$$  \displaystyle results = \left[ \begin{array}{ccccc} 1.7081 & 5.1244 & 8.5406 & 5.1244 & 17.081 \\ 0.6276 & 1.8828 & 3.138 & 1.8828 & 6.276 \\ \end{array} \right] $$ 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.
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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
Problem Statement.... Given:
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$$  \displaystyle math $$     (1.0)
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Solution
test page
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$$  \displaystyle mathstuff $$     (1.1) next is a boxed solution
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$$  \displaystyle mathanswer $$     (1.0)
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Problem Statement
Problem Statement.... Given:
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$$  \displaystyle math $$     (1.0)
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Solution
test page
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$$  \displaystyle mathstuff $$     (1.1) next is a boxed solution
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$$  \displaystyle mathanswer $$     (1.0)
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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
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$$  \displaystyle \omega = 2.0615, y_p(t) = \frac{f_0/m}{\sqrt{(\omega^2 - \bar\omega^2)^2 + (2 \zeta \omega \bar\omega)^2}} e^{i(\bar\omega t - \arctan \frac{2\zeta\omega\bar\omega}{\omega^2 - \bar\omega^2})}, y_p(t)=0.1946e^{i(3.825t - 0.8395)}$$ $$ \mathbb A= 3.512, y(t)=e^{-0.5t}\cos2t+(1/4)e^{-0.5t}\sin2t+0.1946e^{i(3.825t - 0.8395)} $$     (1.29)
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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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