User:EGM4313.f14.Team1.Bishop/Report 2

Solution

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$$9y''-30y'+25y=0$$
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$$y(x)=y_1(x)+y_2(x)$$
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$$y_1(x)=C_1 e^{r x}$$
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$$y_2(x)=C_2 e^{r x}$$
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$$9r^2-30r+25=0$$
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$$r=\frac{30 \pm \sqrt{30^2-4(9)(725)}}{2(9)}$$
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$$r=\frac{30 \pm \cancelto{0}\sqrt{900-900}}{18}$$
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$$r=\frac{5}{3}$$
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$$y_1(x)=C_1 e^{\frac{5x}{3}}$$
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$$y_2(x)=u(x)y_1(x)=C_1 u(x) e^{\frac{5x}{3}}$$
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$$y_2'(x)=C_1 u'(x) e^{\frac{5x}{3}} + \frac{5}{3} C_1 u(x) e^{\frac{5x}{3}}$$
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$$y_2(x)=C_1 u(x) e^{\frac{5x}{3}} + \frac{10}{3} C_1 u'(x) e^{\frac{5x}{3}} + \frac{25}{9} C_1 u(x) e^{\frac{5x}{3}}$$
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$$9y_2''(x)-30y_2'(x)+25y_2(x)=0$$
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$$9C_1 u''(x) e^{\frac{5x}{3}} + 30C_1 u'(x) e^{\frac{5x}{3}} + 25C_1 u(x) e^{\frac{5x}{3}} - 30C_1 u'(x) e^{\frac{5x}{3}} - 50C_1 u(x) e^{\frac{5x}{3}} + 25u(x)y_1(x)=C_1 u(x) e^{\frac{5x}{3}}=0$$
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$$9C_1 u''(x) e^{\frac{5x}{3}}=0$$
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$$u''(x)=0$$
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$$\int \! u''(x)\,dx = \int \! 0\,dx$$
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$$u'(x)=C_2$$
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$$\int \! u'(x)\,dx = \int \! C_2\,dx$$
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$$u(x)=C_2 x + C_3$$
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$$y(x)=C_1 e^{\frac{5x}{3}} + C_2 x e^{\frac{5x}{3}} + C_3 e^{\frac{5x}{3}}$$
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$$y(x)=C_4 e^{\frac{5x}{3}} + C_2 x e^{\frac{5x}{3}}, C_4=C_1+C_3$$
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Solution
$$z_1=-11+10i$$

$$z_2=-1+4i$$

$$\overline{z_1}+\overline{z_2}=\overline{z_1+z_2}$$

Solution
Element 1




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$$ f_A^{(1)} = k_1 ( -d_B + d_A ) + c_1 ( -d'_B + d'_A ) = 5 ( -d_B + d_A ) + 2 ( -d'_B + d'_A ) $$
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$$ f_B^{(1)} = k_1 ( d_B - d_A ) + c_1 ( d'_B - d'_A ) = 5 ( d_B - d_A ) + 2 ( d'_B - d'_A ) $$
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Rearrange Terms in Equations to Prepare to Change to Matrix Form


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$$ f_A^{(1)} = k_1 ( d_A - d_B ) + c_1 ( d'_A - d'_B ) = 5 ( d_A - d_B ) + 2 ( d'_A - d'_B ) $$
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$$ f_B^{(1)} = k_1 ( -d_A + d_B ) + c_1 ( -d'_A + d'_B ) = 5 ( -d_A + d_B ) + 2 ( -d'_A + d'_B ) $$
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Equations in Matrix Form


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$$ \begin{Bmatrix} f_A^{(1)} \\ f_B^{(1)} \end{Bmatrix} = \begin{bmatrix} k_1 & - k_1 \\ -k_1 &  k_1 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} c_1 & - c_1 \\ -c_1 & c_1 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} = \begin{bmatrix} 5 & - 5 \\ -5 &  5 \end{bmatrix} \begin{Bmatrix} d_A \\ d_B \end{Bmatrix} + \begin{bmatrix} 2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_A \\ d'_B \end{Bmatrix} $$
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Element 2 Start with FBD:



Force matrix follows same topology as in element 1.


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$$ \begin{Bmatrix} f_B^{(2)} \\ f_C^{(2)} \end{Bmatrix} = \begin{bmatrix}   k_2 & - k_2 \\ -k_2 &   k_2 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  c_2 & - c_2 \\ -c_2 & c_2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} + \begin{bmatrix} \frac{EA}{L} & -\frac{EA}{L} \\ -\frac{EA}{L} & \frac{EA}{L} \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} = \begin{bmatrix}  5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_B \\ d'_C \end{Bmatrix} + \begin{bmatrix} \frac{1}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{1}{5} \end{bmatrix} \begin{Bmatrix} d_B \\ d_C \end{Bmatrix} $$
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Element 3

Start with FBD:



Force matrix follows same topology as in elements 1 and 2.


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$$ \begin{Bmatrix} f_C^{(3)} \\ f_D^{(3)} \end{Bmatrix} = \begin{bmatrix}   k_3 & - k_3 \\ -k_3 &   k_3 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  c_3 & - c_3 \\ -c_3 & c_3 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} = \begin{bmatrix}   5 & - 5 \\ -5 &   5 \end{bmatrix} \begin{Bmatrix} d_C \\ d_D \end{Bmatrix} + \begin{bmatrix}  2 & - 2 \\ -2 & 2 \end{bmatrix} \begin{Bmatrix} d'_C \\ d'_D \end{Bmatrix} $$
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Mass B



$$ \sum_{} f = F_B(t)- m_Bd''-f_B^{(1)}-f_B^{(2)} = 0 $$


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$$ f_B^{(1)} = \begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix} = \begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix} $$
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$$ f_B^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
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Sub in For Forces


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$$ F_B(t) = m_Bd''_B + (\begin{bmatrix} -k_1d_A + k_1d_B \end{bmatrix} + \begin{bmatrix} -c_1d'_A + c_1d'_B \end{bmatrix}) + (\begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_B - c_2d'_C \end{bmatrix}) $$  (1)
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$$ = 1.3d''_B + (\begin{bmatrix} -5d_A + 5d_B \end{bmatrix} + \begin{bmatrix} -2d'_A + 2d'_B \end{bmatrix}) + (\begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_B - 2d'_C \end{bmatrix}) $$  (1)
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Mass C



$$ \sum_{} f = 0 = F_C(t)- m_Cd''_C- f_C^{(2)} - f_C^{(3)} $$


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$$ f_C^{(2)} = \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} = \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} $$
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$$ f_C^{(3)} = \begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix} = \begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix} $$
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Sub in For Forces


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$$ F_C(t) = m_Cd''_C + (\begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix}) + (\begin{bmatrix} k_3d_C - k_3d_D \end{bmatrix} + \begin{bmatrix} c_3d'_C - c_3d'_D \end{bmatrix}) $$ (2)
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$$ = 1.3d''_C + (\begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix}) + (\begin{bmatrix} 5d_C - 5d_D \end{bmatrix} + \begin{bmatrix} 2d'_C - 2d'_D \end{bmatrix}) $$ (2)
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Boundary Conditions for Both Equations of Motion


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$$ d_A(t) = d'_A(t) = 0 $$
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$$ d_D(t) = d'_D(t) = 0 $$
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Equations of Motion After Substituting in Boundary Conditions


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$$ F_B(t) = m_Bd_B + k_1d_B + c_1d'_b + \begin{bmatrix} k_2d_B - k_2d_C \end{bmatrix} + \begin{bmatrix} c_2d'_b - c_2d'_C \end{bmatrix} = 1.3d_B + 5d_B + 2d'_b + \begin{bmatrix} 5d_B - 5d_C \end{bmatrix} + \begin{bmatrix} 2d'_b - 2d'_C \end{bmatrix} $$   (3)
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$$ F_C(t) = m_Cd_C + \begin{bmatrix} -k_2d_B + k_2d_C \end{bmatrix} + \begin{bmatrix} -c_2d'_B + c_2d'_C \end{bmatrix} + k_3d_C + c_3d'_C = 1.3d_C + \begin{bmatrix} -5d_B + 5d_C \end{bmatrix} + \begin{bmatrix} -2d'_B + 2d'_C \end{bmatrix} + 5d_C + 2d'_C $$   (4)
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Into Matrix Form


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$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   m_B & 0 \\ 0 &  m_C \end{bmatrix} \begin{bmatrix} d_B \\ d_C \end{bmatrix} + \begin{bmatrix}  c_1 + c_2 & - c_2 \\ -c_2 & c_2 + c_3 \end{bmatrix} \begin{bmatrix} d'_B \\ d'_C \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} d_b \\ d_c \end{bmatrix} $$
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Sub In Values Below into Equation of Motion
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$$ k_1 = k_2 = k_3 = 5 $$
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$$ m_1 = m_2 = 1.3 $$
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$$ c_1 = c_2 = c_3 = 2 $$
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$$ m_B = m_1 $$
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$$ m_C = m_2 $$
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Final Equation of Motion with All Values Substituted In


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$$ \begin{Bmatrix} F_B(t) \\ F_C(t) \end{Bmatrix} = \begin{bmatrix}   1.3 & 0 \\ 0 &  1.3 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} + \begin{bmatrix}  4 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} d'_1 \\ d'_2 \end{bmatrix} + \begin{bmatrix} 10 & -5 \\ -5 & 10 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$
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