The columns of matrix \(\left [ \Phi \right ] \) are orthogonal w.r.t to the mass matrix. Hence the following two relations will be assumed as given in the derivation that follows\begin {align} \left [ \Phi \right ] ^{T}\left [ M\right ] \left [ \Phi \right ] & =\begin {bmatrix} 1 & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 1 \end {bmatrix} \label {eq:1}\\ \left [ \Phi \right ] ^{T}\left [ K\right ] \left [ \Phi \right ] & =\begin {bmatrix} \omega _{1}^{2} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & \omega _{N}^{2}\end {bmatrix} \nonumber \end {align}
Starting with the coupled EOM given, which is\[ \left [ M\right ] \left \{ q^{\prime \prime }\right \} +\left [ C\right ] \left \{ q^{\prime }\right \} +\left [ K\right ] \left \{ q\right \} =\left \{ Q\right \} \]
Since \(\left \{ q\right \} =\left [ \Phi \right ] \left \{ \eta \right \} ,\) then \(\left \{ q^{\prime \prime }\right \} =\left [ \Phi \right ] \left \{ \eta ^{\prime \prime }\right \} \) and \(\left \{ q^{\prime }\right \} =\left [ \Phi \right ] \left \{ \eta ^{\prime }\right \} \). Substituting these in the above EOM gives\[ \left [ M\right ] \left [ \Phi \right ] \left \{ \eta ^{\prime \prime }\right \} +\left [ C\right ] \left [ \Phi \right ] \left \{ \eta ^{\prime }\right \} +\left [ K\right ] \left [ \Phi \right ] \left \{ \eta \right \} =\left \{ Q\right \} \]
premultiplying by \(\left [ \Phi \right ] ^{T}\) both the LHS and RHS results in\[ \left [ \Phi \right ] ^{T}\left [ M\right ] \left [ \Phi \right ] \left \{ \eta ^{\prime \prime }\right \} +\left [ \Phi \right ] ^{T}\left [ C\right ] \left [ \Phi \right ] \left \{ \eta ^{\prime }\right \} +\left [ \Phi \right ] ^{T}\left [ K\right ] \left [ \Phi \right ] \left \{ \eta \right \} =\left [ \Phi \right ] ^{T}\left \{ Q\right \} \]
Using Eq ?? the above simplifies to \[\begin {bmatrix} 1 & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 1 \end {bmatrix} \left \{ \eta ^{\prime \prime }\right \} +\left [ \Phi \right ] ^{T}\left [ C\right ] \left [ \Phi \right ] \left \{ \eta ^{\prime }\right \} +\begin {bmatrix} \omega _{1}^{2} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & \omega _{N}^{2}\end {bmatrix} \left \{ \eta \right \} =\left [ \Phi \right ] ^{T}\left \{ Q\right \} \]
Replacing \(\left [ C\right ] \) by the expression given in the problem description, the above becomes\begin {align*} \begin {bmatrix} 1 & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 1 \end {bmatrix} \left \{ \eta ^{\prime \prime }\right \} +\left [ \Phi \right ] ^{T}\left ( \left [ M\right ] \left [ \Phi \right ] \begin {bmatrix} \omega _{1}^{2} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & \omega _{N}^{2}\end {bmatrix} \left [ \Phi \right ] ^{T}\left [ M\right ] \right ) \left [ \Phi \right ] \left \{ \eta ^{\prime }\right \} +\begin {bmatrix} \omega _{1}^{2} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & \omega _{N}^{2}\end {bmatrix} \left \{ \eta \right \} & =\left [ \Phi \right ] ^{T}\left \{ Q\right \} \\\begin {bmatrix} 1 & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 1 \end {bmatrix} \left \{ \eta ^{\prime \prime }\right \} +\overset {I}{\overbrace {\left [ \Phi \right ] ^{T}\left [ M\right ] \left [ \Phi \right ] }}\begin {bmatrix} 2\zeta _{1}\omega _{1} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 2\zeta _{N}\omega _{N}\end {bmatrix} \overset {I}{\overbrace {\left [ \Phi \right ] ^{T}\left [ M\right ] \left [ \Phi \right ] }}\left \{ \eta ^{\prime }\right \} +\begin {bmatrix} \omega _{1}^{2} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & \omega _{N}^{2}\end {bmatrix} \left \{ \eta \right \} & =\left [ \Phi \right ] ^{T}\left \{ Q\right \} \end {align*}
Since \(\left [ \Phi \right ] ^{T}\left [ M\right ] \left [ \Phi \right ] \) is the identity matrix, then the above reduces to\[\begin {bmatrix} 1 & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 1 \end {bmatrix} \left \{ \eta ^{\prime \prime }\right \} +\begin {bmatrix} 2\zeta _{1}\omega _{1} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 2\zeta _{N}\omega _{N}\end {bmatrix} \left \{ \eta ^{\prime }\right \} +\begin {bmatrix} \omega _{1}^{2} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & \omega _{N}^{2}\end {bmatrix} \left \{ \eta \right \} =\left [ \Phi \right ] ^{T}\left \{ Q\right \} \]
This is decoupled OEM since there is no coupling in the mass matrix, and no coupling in the damping matrix and no coupling in the stiffness matrix.
QED
EOM is\begin {multline*} \begin {bmatrix} m_{1} & 0 & 0 & 0 & 0\\ 0 & m_{2} & 0 & 0 & 0\\ 0 & 0 & m_{3} & 0 & 0\\ 0 & 0 & 0 & m_{4} & 0\\ 0 & 0 & 0 & 0 & m_{5}\end {bmatrix}\begin {Bmatrix} x_{1}^{\prime \prime }\\ x_{2}^{\prime \prime }\\ x_{3}^{\prime \prime }\\ x_{4}^{\prime \prime }\\ x_{5}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} c_{1}+c_{2} & -c_{2} & 0 & 0 & 0\\ -c_{2} & c_{2}+c_{3} & -c_{3} & 0 & 0\\ 0 & -c_{3} & c_{3}+c_{4} & -c_{4} & 0\\ 0 & 0 & -c_{4} & c_{4}+c_{5} & -c_{5}\\ 0 & 0 & 0 & -c_{5} & c_{5}\end {bmatrix}\begin {Bmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\\ x_{3}^{\prime }\\ x_{4}^{\prime }\\ x_{5}^{\prime }\end {Bmatrix} +\\\begin {bmatrix} k_{1}+k_{2} & -k_{2} & 0 & 0 & 0\\ -k_{2} & k_{2}+k_{3} & -k_{3} & 0 & 0\\ 0 & -k_{3} & k_{3}+k_{4} & -k_{4} & 0\\ 0 & 0 & -k_{4} & k_{4}+k_{5} & -k_{5}\\ 0 & 0 & 0 & -k_{5} & k_{5}\end {bmatrix}\begin {Bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end {Bmatrix} =\begin {Bmatrix} 0\\ 0\\ 0\\ 0\\ F\relax (t) \end {Bmatrix} \end {multline*}
substituting the numerical values gives \(c_{1}=0.2,c_{i}=0.1,i=2,5\), hence EOM becomes\[\begin {bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end {bmatrix}\begin {Bmatrix} x_{1}^{\prime \prime }\\ x_{2}^{\prime \prime }\\ x_{3}^{\prime \prime }\\ x_{4}^{\prime \prime }\\ x_{5}^{\prime \prime }\end {Bmatrix} +\frac {1}{10}\begin {bmatrix} 3 & -1 & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0\\ 0 & -1 & 2 & -1 & 0\\ 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & -1 & 1 \end {bmatrix}\begin {Bmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\\ x_{3}^{\prime }\\ x_{4}^{\prime }\\ x_{5}^{\prime }\end {Bmatrix} +\begin {bmatrix} 3 & -1 & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0\\ 0 & -1 & 2 & -1 & 0\\ 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & -1 & 1 \end {bmatrix}\begin {Bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\end {Bmatrix} =\begin {Bmatrix} 0\\ 0\\ 0\\ 0\\ F\relax (t) \end {Bmatrix} \]
Natural frequency and mass normalized modes are found by solving the eigenvalue problem to find the natural frequencies and the mass normalized modes.
K=[3 -1 0 0 0;-1 2 -1 0 0;0 -1 2 -1 0;0 0 -1 2 -1;0 0 0 -1 1] M=diag(ones(5,1)); [phi,omega]=eig(K,M); omega = sqrt(diag(omega));
\begin {align*} \left [ \Phi \right ] & =\begin {bmatrix} -0.0989 & 0.2871 & -0.4472 & -0.5635 & -0.6247\\ -0.2871 & 0.6247 & -0.4472 & 0.0989 & 0.5635\\ -0.4472 & 0.4472 & 0.4472 & 0.4472 & -0.4472\\ -0.5635 & -0.0989 & 0.4472 & -0.6247 & 0.2871\\ -0.6247 & -0.5635 & -0.4472 & 0.2871 & -0.0989 \end {bmatrix} \\ \omega & =\left \{ 0.3129,0.9080,1.4142,1.7820,1.9754\right \} \text { rad/sec}\\ & =\left \{ 0.0498,0.1445,0.225,0.284,0.314\right \} \text { hz} \end {align*}
in modal coordinates, EOM is decoupled to become\[ I\left \{ \eta ^{\prime \prime }\right \} +\left [ \Phi \right ] ^{T}\left [ C\right ] \left [ \Phi \right ] \left \{ \eta ^{\prime }\right \} +\left [ \Phi \right ] ^{T}\left [ k\right ] \left [ \Phi \right ] \left \{ \eta \right \} =\left [ \Phi \right ] ^{T}\begin {Bmatrix} 0\\ 0\\ 0\\ 0\\ F\relax (t) \end {Bmatrix} \]
EDU>> C = 0.1*K; C = phi.'*C*phi K = phi.'*K*phi syms f(t); F = zeros(5,1); F(5)=1; F = phi.'*F C = 0.0098 0.0000 -0.0000 0.0000 0.0000 0.0000 0.0824 -0.0000 0.0000 -0.0000 -0.0000 -0.0000 0.2000 -0.0000 0.0000 0.0000 0.0000 -0.0000 0.3176 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.3902 K = 0.0979 0.0000 -0.0000 0.0000 0.0000 0.0000 0.8244 -0.0000 0.0000 -0.0000 -0.0000 -0.0000 2.0000 -0.0000 0.0000 0.0000 0.0000 -0.0000 3.1756 -0.0000 0.0000 0 0.0000 -0.0000 3.9021 F = -0.6247 -0.5635 -0.4472 0.2871 -0.0989
Hence EOM in modal coordinates is\begin {multline*} \left [ I\right ] \begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\\ \eta _{3}^{\prime \prime }\\ \eta _{4}^{\prime \prime }\\ \eta _{5}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} 0.0098 & 0 & 0 & 0 & 0\\ 0 & 0.0824 & 0 & 0 & 0\\ 0 & 0 & 0.2 & 0 & 0\\ 0 & 0 & 0 & 0.3176 & 0\\ 0 & 0 & 0 & 0 & 0.3902 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime }\\ \eta _{2}^{\prime }\\ \eta _{3}^{\prime }\\ \eta _{4}^{\prime }\\ \eta _{5}^{\prime }\end {Bmatrix} +\\\begin {bmatrix} 0.0979 & 0 & 0 & 0 & 0\\ 0 & 0.8244 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 3.1756 & 0\\ 0 & 0 & 0 & 0 & 3.9021 \end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\\ \eta _{4}\\ \eta _{5}\end {Bmatrix} =\begin {Bmatrix} -0.6247F\relax (t) \\ -0.5635F\relax (t) \\ -0.4472F\relax (t) \\ 0.2871F\relax (t) \\ -0.0989F\relax (t) \end {Bmatrix} \end {multline*}
Where in the above \(F\relax (t) =\cos \left (\varpi t\right ) \) with \(\varpi \) being the forcing frequency in the range \(0\) to \(1.2\omega _{5}\) where \(\omega _{5}=1.9754\) rad/sec.
Since the equations are now decoupled, the 5\(^{th}\) equation can solved on its own\[ \eta _{5}^{\prime \prime }+0.3902\eta _{5}^{\prime }+3.9021\eta _{5}=\operatorname {Re}\left \{ -0.0989e^{i\varpi t}\right \} \]
Assuming \(\eta _{5}\relax (t) =\operatorname {Re}\left \{ Xe^{i\varpi t}\right \} \) and substituting in the above and simplifying gives\begin {align*} \left (-\varpi ^{2}+i\varpi 0.3902+3.9021\right ) X & =-0.0989\\ X & =\frac {-0.0989}{-\varpi ^{2}+i\varpi 0.3902+3.9021} \end {align*}
Hence \[ \eta _{5}\relax (t) =\operatorname {Re}\left \{ \frac {-0.0989}{-\varpi ^{2}+i\varpi 0.3902+3.9021}e^{i\varpi t}\right \} \]
Similarly, all other \(\eta _{i},i=1,5\) are found. Hence\[\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\\ \eta _{4}\\ \eta _{5}\end {Bmatrix} =\operatorname {Re}\left ( \begin {Bmatrix} \frac {-0.6247}{-\varpi ^{2}+i\varpi 0.0098+0.0979}\\ \frac {-0.5635}{-\varpi ^{2}+i\varpi 0.0824+0.8244}\\ \frac {-0.4472}{-\varpi ^{2}+i\varpi 0.2+2}\\ \frac {0.2871}{-\varpi ^{2}+i\varpi 0.3176+3.176}\\ \frac {-0.0989}{-\varpi ^{2}+i\varpi 0.3902+3.9021}\end {Bmatrix} e^{i\varpi t}\right ) \]
and the solution in physical coordinates is now found from \(\left \{ x\right \} =\left [ \Phi \right ] \left \{ \eta \right \} \). Hence\begin {align*} x_{5} & ={\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) \eta \relax (j) \\ & ={\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) \operatorname {Re}\left \{ X\relax (j) e^{i\varpi t}\right \} \\ & =\operatorname {Re}\left ( {\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) X\relax (j) e^{i\varpi t}\right ) \\ & =\operatorname {Re}\left [ \left (-0.6247X_{1}\relax (t) -0.5635X_{2}\relax (t) -0.4472X_{3}\relax (t) +0.2871X_{4}\relax (t) -0.0989X_{5}\relax (t) \right ) e^{i\varpi t}\right ] \\ & =\operatorname {Re}\left [ \left ( \begin {array} [c]{c}\frac {\left (-0.6247\right ) \left (-0.6247\right ) }{-\varpi ^{2}+i\varpi 0.0098+0.0979}+\frac {\left (-0.5635\right ) \left (-0.5635\right ) }{-\varpi ^{2}+i\varpi 0.0824+0.8244}+\frac {\left (-0.4472\right ) \left ( -0.4472\right ) }{-\varpi ^{2}+i\varpi 0.2+2}+\\ \frac {\left (0.2871\right ) 0.2871}{-\varpi ^{2}+i\varpi 0.3176+3.176}+\frac {\left (-0.0989\right ) \left (-0.0989\right ) }{-\varpi ^{2}+i\varpi 0.3902+3.9021}\end {array} \right ) e^{i\varpi t}\right ] \\ & =\operatorname {Re}\left [ \left ( \begin {array} [c]{c}\frac {0.390\,25}{-\varpi ^{2}+0.0098i\varpi +0.098}+\frac {0.08243}{-\varpi ^{2}+0.3178i\varpi +3.176}+\frac {0.31753}{-\varpi ^{2}+0.0824i\varpi +0.8244}+\\ \frac {0.00978}{-\varpi ^{2}+0.3902i\varpi +3.9021}+\frac {0.19999}{-\varpi ^{2}+0.2i\varpi +2}\end {array} \right ) e^{i\varpi t}\right ] \end {align*}
Therefore \[ x_{5}=\operatorname {Re}\left (Y_{5}e^{i\varpi t}\right ) \] where \begin {multline*} Y_{5}=\frac {0.390\,25}{-\varpi ^{2}+0.0098i\varpi +0.098}+\frac {0.08243}{-\varpi ^{2}+0.3178i\varpi +3.176}+\allowbreak \frac {0.31753}{-\varpi ^{2}+0.0824i\varpi +0.8244}+\\ \frac {0.00978}{-\varpi ^{2}+0.3902i\varpi +3.9021}+\frac {0.19999}{-\varpi ^{2}+0.2i\varpi +2} \end {multline*}
Here is a plot of the magnitude spectrum of \(Y_{5}\) and the phase spectrum for the range of \(\varpi \) of \(0\) to \(1.2\omega _{5}\). This shows that \(x_{5}\left ( t\right ) \) response will have the largest magnitude when the forcing frequency coincides with the first natural frequency (the fundamental frequency). In otherwords when \(\varpi =\omega _{1}\).
The amplitude of \(x_{5}\relax (t) \) at resonance is smaller for the remaining \(4\) natural frequencies. For higher order natural frequencies, resonances at those frequencies produces lower amplitudes than lower order natural frequencies.
Using \(\zeta _{i}=\zeta =0.02\) for \(i=1,5\) the EOM is\[ \left [ I\right ] \begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\\ \eta _{3}^{\prime \prime }\\ \eta _{4}^{\prime \prime }\\ \eta _{5}^{\prime \prime }\end {Bmatrix} +2\zeta \begin {bmatrix} \omega _{1} & 0 & 0 & 0 & 0\\ 0 & \omega _{2} & 0 & 0 & 0\\ 0 & 0 & \omega _{3} & 0 & 0\\ 0 & 0 & 0 & \omega _{4} & 0\\ 0 & 0 & 0 & 0 & \omega _{5}\end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime }\\ \eta _{2}^{\prime }\\ \eta _{3}^{\prime }\\ \eta _{4}^{\prime }\\ \eta _{5}^{\prime }\end {Bmatrix} +\begin {bmatrix} \omega _{1}^{2} & 0 & 0 & 0 & 0\\ 0 & \omega _{2}^{2} & 0 & 0 & 0\\ 0 & 0 & \omega _{3}^{2} & 0 & 0\\ 0 & 0 & 0 & \omega _{4}^{2} & 0\\ 0 & 0 & 0 & 0 & \omega _{5}^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\\ \eta _{4}\\ \eta _{5}\end {Bmatrix} =\begin {Bmatrix} -0.6247F\relax (t) \\ -0.5635F\relax (t) \\ -0.4472F\relax (t) \\ 0.2871F\relax (t) \\ -0.0989F\relax (t) \end {Bmatrix} \]
Hence the solution \[ \eta _{j}=\operatorname {Re}\left \{ X_{j}e^{i\varpi t}\right \} \]
where now \[ X_{j}=\frac {F_{j}}{-\varpi ^{2}+2i\zeta _{j}\varpi \omega _{j}+\omega _{j}^{2}}\]
Hence, since \(\omega =\left \{ 0.3129,0.9080,1.4142,1.7820,1.9754\right \} \) the solutions in modal coordinates is\[\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\\ \eta _{4}\\ \eta _{5}\end {Bmatrix} =\operatorname {Re}\left ( \begin {Bmatrix} \frac {-0.6247}{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left (0.3129\right ) +0.0979}\\ \frac {-0.5635}{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left (0.9080\right ) +0.8244}\\ \frac {-0.4472}{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left (1.4142\right ) +2}\\ \frac {0.2871}{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left (1.7820\right ) +3.176}\\ \frac {-0.0989}{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left (1.9754\right ) +3.9021}\end {Bmatrix} e^{i\varpi t}\right ) \]
and the solution in physical coordinates is now found from \(\left \{ x\right \} =\left [ \Phi \right ] \left \{ \eta \right \} \). Hence\begin {align*} x_{5} & ={\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) \eta \relax (j) \\ & ={\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) \operatorname {Re}\left \{ X\relax (j) e^{i\varpi t}\right \} \\ & =\operatorname {Re}\left ( {\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) X\relax (j) e^{i\varpi t}\right ) \\ & =\operatorname {Re}\left [ \left (-0.6247X_{1}\relax (t) -0.5635X_{2}\relax (t) -0.4472X_{3}\relax (t) +0.2871X_{4}\relax (t) -0.0989X_{5}\relax (t) \right ) e^{i\varpi t}\right ] \\ & =\operatorname {Re}\left [ \left ( \begin {array} [c]{c}\frac {\left (-0.6247\right ) \left (-0.6247\right ) }{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left (0.3129\right ) +0.0979}+\frac {\left ( -0.5635\right ) \left (-0.5635\right ) }{-\varpi ^{2}+2i\varpi \left ( 0.02\right ) \left (0.9080\right ) +0.8244}+\frac {\left (-0.4472\right ) \left (-0.4472\right ) }{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left ( 1.4142\right ) +2}+\\ \frac {\left (0.2871\right ) 0.2871}{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left (1.7820\right ) +3.176}+\frac {\left (-0.0989\right ) \left ( -0.0989\right ) }{-\varpi ^{2}+2i\varpi \left (0.02\right ) \left ( 1.9754\right ) +3.9021}\end {array} \right ) e^{i\varpi t}\right ] \\ & =\operatorname {Re}\left [ \left ( \begin {array} [c]{c}\frac {0.390\,25}{-\varpi ^{2}+1.251\,6\times 10^{-2}i\varpi +0.097\,9}+\frac {0.317\,53}{-\varpi ^{2}+0.036\,32i\varpi +0.824\,4}+\frac {0.19999}{-\varpi ^{2}+5.656\,8\times 10^{-2}i\varpi +2.0}+\\ \frac {8.242\,6\times 10^{-2}}{-\varpi ^{2}+0.071\,28i\varpi +3.176}+\frac {9.781\,2\times 10^{-3}}{-\varpi ^{2}+7.\,\allowbreak 901\,6\times 10^{-2}i\varpi +3.902\,1}\end {array} \right ) e^{i\varpi t}\right ] \end {align*}
Therefore \[ x_{5}=\operatorname {Re}\left (Y_{5}e^{i\varpi t}\right ) \] where \begin {multline*} Y_{5}=\frac {0.390\,25}{-\varpi ^{2}+1.2516\times 10^{-2}i\varpi +0.0979}+\frac {0.317\,53}{-\varpi ^{2}+0.036\,32i\varpi +0.8244}+\frac {0.19999}{-\varpi ^{2}+5.656\,8\times 10^{-2}i\varpi +2.0}+\\ \frac {8.2426\times 10^{-2}}{-\varpi ^{2}+0.07128i\varpi +3.176}+\frac {9.7812\times 10^{-3}}{-\varpi ^{2}+7.9016\times 10^{-2}i\varpi +3.902\,1} \end {multline*}
Here is a plot of the magnitude spectrum of \(Y_{5}\) and the phase spectrum for the range of \(\varpi \) of \(0\) to \(1.2\omega _{5}\) for both part(b) and (c) on the same plot
When \(\zeta =2\%\) was used, the resonance is seen to be higher (part c) compared to part (b). Here is a full range plot of the above.
Comparing the phase between part(b) and (c) gives
Which shows the effect on the phase spectrum.
In structural damping, the damping force is proportional to the elastic force. For example given an EOM \(my^{\prime \prime }+cy^{\prime }+ky=f\), and converting to frequency domain to obtain transfer function \[ Y=\frac {F}{-\varpi ^{2}m+ic\varpi +k}\]
Then structural damping implies replacing \(c\varpi \) with \(\gamma k\) in the above, giving\[ Y=\frac {F}{-\varpi ^{2}m+i\gamma k+k}=\frac {F}{-\varpi ^{2}m+\left ( 1+i\gamma \right ) k}\]
The above method is now applied to the EOM given, and the resulting transfer function for \(x_{5}\) is compared to the last results in order to see the effect of using structural damping on the response. The eigenvalue problem was solved in part (a) where the result was\begin {align*} \left [ \Phi \right ] & =\begin {bmatrix} -0.0989 & 0.2871 & -0.4472 & -0.5635 & -0.6247\\ -0.2871 & 0.6247 & -0.4472 & 0.0989 & 0.5635\\ -0.4472 & 0.4472 & 0.4472 & 0.4472 & -0.4472\\ -0.5635 & -0.0989 & 0.4472 & -0.6247 & 0.2871\\ -0.6247 & -0.5635 & -0.4472 & 0.2871 & -0.0989 \end {bmatrix} \\ \omega & =\left \{ 0.3129,0.9080,1.4142,1.7820,1.9754\right \} \end {align*}
Hence the modal EOM is now\[ \left [ I\right ] \begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\\ \eta _{3}^{\prime \prime }\\ \eta _{4}^{\prime \prime }\\ \eta _{5}^{\prime \prime }\end {Bmatrix} +\left (1+i\gamma \right ) \begin {bmatrix} 0.0979 & 0 & 0 & 0 & 0\\ 0 & 0.8244 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 3.1756 & 0\\ 0 & 0 & 0 & 0 & 3.9021 \end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\\ \eta _{4}\\ \eta _{5}\end {Bmatrix} =\begin {Bmatrix} -0.6247F\relax (t) \\ -0.5635F\relax (t) \\ -0.4472F\relax (t) \\ 0.2871F\relax (t) \\ -0.0989F\relax (t) \end {Bmatrix} \]
Hence the steady state solution now in modal coordinates is Hence the solution \[ \eta _{j}=\operatorname {Re}\left \{ X_{j}e^{i\varpi t}\right \} \]
where now \[ X_{j}=\frac {F_{j}}{-\varpi ^{2}+\left (1+i\gamma \right ) \omega _{j}^{2}}\]
The solutions in modal coordinates are (where \(\gamma =0.04)\)\[\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\\ \eta _{4}\\ \eta _{5}\end {Bmatrix} =\operatorname {Re}\left ( \begin {Bmatrix} \frac {-0.6247}{-\varpi ^{2}+\left (1+i\gamma \right ) 0.0979}\\ \frac {-0.5635}{-\varpi ^{2}+\left (1+i\gamma \right ) 0.8244}\\ \frac {-0.4472}{-\varpi ^{2}+\left (1+i\gamma \right ) 2}\\ \frac {0.2871}{-\varpi ^{2}+\left (1+i\gamma \right ) 3.176}\\ \frac {-0.0989}{-\varpi ^{2}+\left (1+i\gamma \right ) 3.9021}\end {Bmatrix} e^{i\varpi t}\right ) \]
and the solution in physical coordinates is now found from \(\left \{ x\right \} =\left [ \Phi \right ] \left \{ \eta \right \} \). Hence\begin {align*} x_{5} & ={\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) \eta \relax (j) \\ & ={\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) \operatorname {Re}\left \{ X\relax (j) e^{i\varpi t}\right \} \\ & =\operatorname {Re}\left ( {\displaystyle \sum \limits _{j=1}^{5}} \Phi \left (5,j\right ) X\relax (j) e^{i\varpi t}\right ) \\ & =\operatorname {Re}\left [ \left (-0.6247X_{1}\relax (t) -0.5635X_{2}\relax (t) -0.4472X_{3}\relax (t) +0.2871X_{4}\relax (t) -0.0989X_{5}\relax (t) \right ) e^{i\varpi t}\right ] \\ & =\operatorname {Re}\left [ \left ( \begin {array} [c]{c}\frac {0.390\,25}{-\varpi ^{2}+\left (1+i\gamma \right ) 0.0979}+\frac {0.317\,53}{-\varpi ^{2}+\left (1+i\gamma \right ) 0.8244}+\frac {0.199\,99}{-\varpi ^{2}+\left (1+i\gamma \right ) 2}+\\ \frac {8.242\,6\times 10^{-2}}{-\varpi ^{2}+\left (1+i\gamma \right ) 3.176}+\frac {9.781\,2\times 10^{-3}}{-\varpi ^{2}+\left (1+i\gamma \right ) 3.9021}\end {array} \right ) e^{i\varpi t}\right ] \end {align*}
Therefore \[ x_{5}=\operatorname {Re}\left (Y_{5}e^{i\varpi t}\right ) \] where \begin {multline*} Y_{5}=\frac {0.39025}{-\varpi ^{2}+\left (1+i\gamma \right ) 0.0979}+\frac {0.31753}{-\varpi ^{2}+\left (1+i\gamma \right ) 0.8244}+\frac {0.19999}{-\varpi ^{2}+\left (1+i\gamma \right ) 2}+\\ \frac {8.2426\times 10^{-2}}{-\varpi ^{2}+\left (1+i\gamma \right ) 3.176}+\frac {9.7812\times 10^{-3}}{-\varpi ^{2}+\left (1+i\gamma \right ) 3.9021} \end {multline*}
Here is a plot of the magnitude spectrum of \(Y_{5}\) and the phase spectrum for the range of \(\varpi \) of \(0\) to \(1.2\omega _{5}\) using the above transfer function, and superimposed on top of part (c). The magnitude spectrum is identical and no difference can be seen. Looking the phase spectrum there is very small change. Here are the plots. In the following plot, part(d) and (c) can not be distinguished. (the x-axis is drawn using dashed as well, not to be confused with the actual response curve).
To better see the difference, the plot was reproduced by taking the difference of the absolute values from part(d) and part(c) and plotting the \(\log \) to base 20 of this difference. Now the difference can be better seen as very small.
The following the phase difference between case d and c.
The above plots show that using structural damping instead of using the same value of \(\zeta \) for each EOM made very little difference in the result.
Given \(u\left (x,t\right ) \) equations 6.1.1 and 6.1.2 in the text are\begin {align} T_{bar} & =\frac {1}{2}\int _{0}^{L}\dot {u}^{2}\rho Adx\label {eq:6.1.1}\\ V_{bar} & =\frac {1}{2}\int _{0}^{L}EA\left (\frac {\partial u}{\partial x}\right ) ^{2}dx \label {eq:6.1.2} \end {align}
To obtain the mass matrix components \(T_{bar}\) is evaluated and each set of quadratic term are used to generate \(M_{jn}\) as follows. Using Ritz method, Let \(u\left (x,t\right ) ={\displaystyle \sum \limits _{j=1}^{N}} \Psi _{j}\relax (x) q_{j}\relax (t) \). Substituting this in Eq ?? gives\begin {align*} T_{bar} & =\frac {1}{2}\int _{0}^{L}\left (\frac {\partial }{\partial t}{\displaystyle \sum \limits _{j=1}^{N}} \Psi _{j}\relax (x) q_{j}\relax (t) \right ) ^{2}\rho Adx=\frac {1}{2}\int _{0}^{L}\left ( {\displaystyle \sum \limits _{j=1}^{N}} \Psi _{j}\relax (x) q_{j}^{\prime }\relax (t) \right ) ^{2}\rho Adx\\ & =\frac {1}{2}\int _{0}^{L}\left ( {\displaystyle \sum \limits _{j=1}^{N}} \Psi _{j}\relax (x) q_{j}^{\prime }\relax (t) \right ) \left ( {\displaystyle \sum \limits _{n=1}^{N}} \Psi _{n}\relax (x) q_{n}^{\prime }\relax (t) \right ) \rho Adx\\ & =\frac {1}{2}\int _{0}^{L}\left ( {\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} \Psi _{j}\relax (x) \Psi _{n}\relax (x) q_{j}^{\prime }\left ( t\right ) q_{n}^{\prime }\relax (t) \right ) \rho Adx \end {align*}
Replacing order or integration with summation (since both are linear operations) and moving \(q_{j}^{\prime }\relax (t) q_{n}^{\prime }\left ( t\right ) \) outside the integration since it does not depend on \(x\) results in\begin {equation} T_{bar}=\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} \left (\int _{0}^{L}\Psi _{j}\relax (x) \Psi _{n}\relax (x) \rho Adx\right ) q_{j}^{\prime }\relax (t) q_{n}^{\prime }\relax (t) \label {eq:3.1} \end {equation}
Let \[ M_{jn}=\int _{0}^{L}\Psi _{j}\relax (x) \Psi _{n}\relax (x) \rho Adx \]
Then eq ?? becomes Eq 6.1.11 in the textbook\begin {equation} T_{bar}=\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} M_{jn}q_{j}^{\prime }\relax (t) q_{n}^{\prime }\relax (t) \label {eq:6.1.11} \end {equation}
Now, obtain the components of the stiffness matrix. Starting with eq ?? and replacing \(u\left (x,t\right ) \) in this equation gives\begin {align*} V_{bar} & =\frac {1}{2}\int _{0}^{L}EA\left (\frac {\partial }{\partial x}{\displaystyle \sum \limits _{j=1}^{N}} \Psi _{j}\relax (x) q_{j}\relax (t) \right ) ^{2}dx=\frac {1}{2}\int _{0}^{L}EA\left ( {\displaystyle \sum \limits _{j=1}^{N}} \frac {d\Psi _{j}\relax (x) }{dx}q_{j}\relax (t) \right ) ^{2}dx\\ & =\frac {1}{2}\int _{0}^{L}EA\left ( {\displaystyle \sum \limits _{j=1}^{N}} \frac {d\Psi _{j}\relax (x) }{dx}q_{j}\relax (t) \right ) \left ( {\displaystyle \sum \limits _{n=1}^{N}} \frac {d\Psi _{n}\relax (x) }{dx}q_{n}\relax (t) \right ) dx\\ & =\frac {1}{2}\int _{0}^{L}EA\left ( {\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} \frac {d\Psi _{j}\relax (x) }{dx}\frac {d\Psi _{n}\relax (x) }{dx}q_{j}\relax (t) q_{n}\relax (t) \right ) dx \end {align*}
Replacing order of integration with summation and moving \(q_{j}\left ( t\right ) q_{n}\relax (t) \) outside the integration since it does not depend on \(x\) gives\[ V_{bar}=\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} \left (\int _{0}^{L}EA\frac {d\Psi _{j}\relax (x) }{dx}\frac {d\Psi _{n}\relax (x) }{dx}dx\right ) q_{j}\relax (t) q_{n}\left ( t\right ) \]
Let \(K_{jn}=\) \(\int _{0}^{L}EA\frac {d\Psi _{j}\relax (x) }{dx}\frac {d\Psi _{n}\relax (x) }{dx}dx\) then the above becomes\[ V_{bar}=\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} K_{jn}q_{j}q_{n}\]
Which is eq 6.1.13 in the book. QED.
The basic function to use are \(\Psi _{1}=\frac {x}{L},\Psi _{2}=\left (\frac {x}{L}\right ) ^{2},\Psi _{3}=\left (\frac {x}{L}\right ) ^{3}\). Let \(u\left ( x,t\right ) ={\displaystyle \sum \limits _{j=1}^{3}} \Psi _{j}\relax (x) q_{j}\relax (t) \). Now eq 6.1.11 and eq. 6.1.13 are used to obtain the mass matrix and the stiffness matrix components based on the power balance method. \(T_{bar}=\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} M_{jn}q_{j}^{\prime }\relax (t) q_{n}^{\prime }\relax (t) \) where \(M_{jn}=\int _{0}^{L}\Psi _{j}\relax (x) \Psi _{n}\relax (x) \rho Adx\) hence\begin {align*} M_{jn} & =\int _{0}^{L}\Psi _{j}\relax (x) \Psi _{n}\relax (x) \rho Adx\\ & =\int _{0}^{L}\left (\frac {x}{L}\right ) ^{j}\left (\frac {x}{L}\right ) ^{n}\rho Adx\\ & =\int _{0}^{L}\left (\frac {x}{L}\right ) ^{j+n}\rho Adx\\ & =\frac {\rho A}{L^{j+n}}\int _{0}^{L}x^{j+n}dx\\ & =\frac {\rho A}{L^{j+n}}\left [ \frac {x^{j+n+1}}{j+n+1}\right ] _{0}^{L}=\frac {\rho A}{\left (j+n+1\right ) L^{j+n}}L^{j+n+1}\\ & =\frac {\rho AL}{j+n+1} \end {align*}
Therefore, the mass matrix is\[ M=\begin {bmatrix} M_{11} & M_{12} & M_{13}\\ M_{21} & M_{22} & M_{23}\\ M_{31} & M_{32} & M_{33}\end {bmatrix} =\rho AL\begin {bmatrix} \frac {1}{1+1+1} & \frac {1}{1+2+1} & \frac {1}{1+3+1}\\ \frac {1}{2+1+1} & \frac {1}{2+2+1} & \frac {1}{2+3+1}\\ \frac {1}{3+1+1} & \frac {1}{3+2+1} & \frac {1}{3+3+1}\end {bmatrix} =\rho AL\begin {bmatrix} \frac {1}{3} & \frac {1}{4} & \frac {1}{5}\\ \frac {1}{4} & \frac {1}{5} & \frac {1}{6}\\ \frac {1}{5} & \frac {1}{6} & \frac {1}{7}\end {bmatrix} \]
and \(\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} K_{jn}q_{j}q_{n}\) where \(K_{jn}=\) \(\int _{0}^{L}EA\frac {d\Psi _{j}\left ( x\right ) }{dx}\frac {d\Psi _{n}\relax (x) }{dx}dx\), hence\begin {align*} K_{jn} & =\int _{0}^{L}EA\frac {d\left (\frac {x}{L}\right ) ^{j}}{dx}\frac {d\left (\frac {x}{L}\right ) ^{n}}{dx}dx=\int _{0}^{L}EAj\left ( \frac {x^{j-1}}{L^{j}}\right ) n\left (\frac {x^{n-1}}{L^{n}}\right ) dx\\ & =\frac {EAjn}{L^{j}L^{n}}\int _{0}^{L}x^{j-1}x^{n-1}dx\\ & =\frac {EAjn}{L^{j+n}}\int _{0}^{L}x^{j+n-2}dx\\ & =\frac {EAjn}{L^{j+n}}\left [ \frac {x^{j+n-1}}{j+n-1}\right ] _{0}^{L}\\ & =\frac {EAjn}{\left (j+n-1\right ) L^{j+n}}\left [ x^{j+n-1}\right ] _{0}^{L}\\ & =\frac {EAjn}{\left (j+n-1\right ) L^{j+n}}L^{j+n-1}\\ & =\frac {EA}{L}\frac {jn}{j+n-1} \end {align*}
Hence the stiffness matrix is\begin {align*} K & =\begin {bmatrix} K_{11} & K_{12} & K_{13}\\ K_{21} & K_{22} & K_{23}\\ K_{31} & K_{32} & K_{33}\end {bmatrix} =\frac {EA}{L}\begin {bmatrix} \frac {1\relax (1) }{1+1-1} & \frac {1\relax (2) }{1+2-1} & \frac {1\relax (3) }{1+3-1}\\ \frac {2\relax (1) }{2+1-1} & \frac {2\relax (2) }{2+2-1} & \frac {2\relax (3) }{2+3-1}\\ \frac {3\relax (1) }{3+1-1} & \frac {3\relax (2) }{3+2-1} & \frac {3\relax (3) }{3+3-1}\end {bmatrix} \\ & =\frac {EA}{L}\begin {bmatrix} 1 & 1 & 1\\ 1 & \frac {4}{3} & \frac {3}{2}\\ 1 & \frac {3}{2} & \frac {9}{5}\end {bmatrix} \end {align*}
The basic function to use are \(\Psi _{r}=\sin \left (\alpha _{r}\frac {x}{L}\right ) \) where \(\alpha _{r}=\left (\frac {2r-1}{2}\right ) \pi \) for \(r=1,2,3.\)
Let \(u\left (x,t\right ) ={\displaystyle \sum \limits _{j=1}^{3}} \Psi _{j}\relax (x) q_{j}\relax (t) \). Now eq 6.1.11 and eq. 6.1.13 are used to obtain the mass matrix and the stiffness matrix components based on the power balance method. \(T_{bar}=\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} M_{jn}q_{j}^{\prime }\relax (t) q_{n}^{\prime }\relax (t) \) where \(M_{jn}=\int _{0}^{L}\Psi _{j}\relax (x) \Psi _{n}\relax (x) \rho Adx\) hence\begin {align*} M_{jn} & =\int _{0}^{L}\Psi _{j}\relax (x) \Psi _{n}\relax (x) \rho Adx\\ & =\int _{0}^{L}\sin \left (\alpha _{j}\frac {x}{L}\right ) \sin \left ( \alpha _{n}\frac {x}{L}\right ) \rho Adx\\ & =\int _{0}^{L}\sin \left (\left (\frac {2j-1}{2}\right ) \pi \frac {x}{L}\right ) \sin \left (\left (\frac {2n-1}{2}\right ) \pi \frac {x}{L}\right ) \rho Adx \end {align*}
Using \(\sin A\sin B=\frac {1}{2}\left (\cos \left (A-B\right ) -\cos \left ( A+B\right ) \right ) \) the above can be solved.\begin {align*} M_{jn} & =\int _{0}^{L}\frac {1}{2}\left (\cos \left (\left (\frac {2j-1}{2}\right ) \pi \frac {x}{L}-\left (\frac {2n-1}{2}\right ) \pi \frac {x}{L}\right ) -\cos \left (\left (\frac {2j-1}{2}\right ) \pi \frac {x}{L}+\left ( \frac {2n-1}{2}\right ) \pi \frac {x}{L}\right ) \right ) \rho Adx\\ & =\int _{0}^{L}\frac {1}{2}\left [ \cos \left (\frac {\left (2j-1\right ) -\left (2n-1\right ) }{2}\right ) \pi \frac {x}{L}-\cos \left (\frac {\left ( 2j-1\right ) +\left (2n-1\right ) }{2}\right ) \pi \frac {x}{L}\right ] \rho Adx\\ & =\int _{0}^{L}\frac {1}{2}\left [ \cos \left (\frac {2\left (j-n\right ) }{2}\right ) \pi \frac {x}{L}-\cos \left (\frac {2\left (j+n\right ) -2}{2}\right ) \pi \frac {x}{L}\right ] \rho Adx\\ & =\int _{0}^{L}\frac {1}{2}\left [ \cos \left (j-n\right ) \pi \frac {x}{L}-\cos \left (\left (j+n\right ) -1\right ) \pi \frac {x}{L}\right ] \rho Adx \end {align*}
For \(j=1,n=1\) the above gives\[ M_{jn}=\int _{0}^{L}\frac {1}{2}\left [ 1-\cos \pi \frac {x}{L}\right ] \rho Adx=\frac {\rho A}{2}\int _{0}^{L}1-\cos \pi \frac {x}{L}dx=\frac {\rho A}{2}\left ( L-\left (\frac {\sin \pi \frac {x}{L}}{\frac {\pi }{L}}\right ) _{0}^{L}\right ) =\frac {\rho A}{2}L \]
For \(j=1,n=2\)\begin {align*} M_{jn} & =\int _{0}^{L}\frac {1}{2}\left [ \cos \left (-\pi \frac {x}{L}\right ) -\cos 2\pi \frac {x}{L}\right ] \rho Adx=\frac {\rho A}{2}\int _{0}^{L}\cos \left ( \pi \frac {x}{L}\right ) -\cos \left (2\pi \frac {x}{L}\right ) dx\\ & =\frac {\rho A}{2}\left [ \left (\frac {\sin \left (\pi \frac {x}{L}\right ) }{\frac {\pi }{L}}\right ) _{0}^{L}-\left (\frac {\sin \left (2\pi \frac {x}{L}\right ) }{\frac {2\pi }{L}}\right ) _{0}^{L}\right ] \\ & =\frac {\rho A}{2}\left [ 0-0\right ] =0 \end {align*}
The rest of the computation is now done using a small code below to generate the final mass and stiffness matrix
Therefore, the mass matrix is\[ M=\frac {AL\rho }{2}\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} \]
and \(\frac {1}{2}{\displaystyle \sum \limits _{j=1}^{N}} {\displaystyle \sum \limits _{n=1}^{N}} K_{jn}q_{j}q_{n}\) where \(K_{jn}=\) \(\int _{0}^{L}EA\frac {d\Psi _{j}\left ( x\right ) }{dx}\frac {d\Psi _{n}\relax (x) }{dx}dx\), hence\[ K_{jn}=\int _{0}^{L}EA\frac {d\left (\Psi _{j}\right ) }{dx}\frac {d\left ( \Psi _{n}\right ) }{dx}dx \]
From the above code, the result is \[ K=\frac {AE\pi ^{2}}{8L}\begin {bmatrix} 1 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 & 25 \end {bmatrix} \]
Using this set of basis functions produces mass and stiffness matrices that are already decoupled. This is good.
The natural frequencies obtained in problem 2 were\[ problem2\Rightarrow M=\begin {bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end {bmatrix} ,K=\begin {bmatrix} 3 & -1 & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0\\ 0 & -1 & 2 & -1 & 0\\ 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & -1 & 1 \end {bmatrix} \]
\begin {align*} \omega & =\left (0.3129,0.9080,1.4142,1.7820,1.9754\right ) \text { rad/sec}\\ & =\left (0.0498,0.1445,0.225,0.284,0.314\right ) \text { hz} \end {align*}
Now the eigenvalue problem \(\det \left (\left [ k\right ] -\omega ^{2}\left [ M\right ] \right ) \) is solved again using the mass and stiffness matrices in parts b,c above and the natural frequencies are compared with the above result from problem 2. Recall, the \(M\) and \(K\) from part b were\begin {align*} part(b) & \Rightarrow M=\rho AL\begin {bmatrix} \frac {1}{3} & \frac {1}{4} & \frac {1}{5}\\ \frac {1}{4} & \frac {1}{5} & \frac {1}{6}\\ \frac {1}{5} & \frac {1}{6} & \frac {1}{7}\end {bmatrix} ,K=\frac {EA}{L}\begin {bmatrix} 1 & 1 & 1\\ 1 & \frac {4}{3} & \frac {3}{2}\\ 1 & \frac {3}{2} & \frac {9}{5}\end {bmatrix} \\ part(c) & \Rightarrow M=\frac {AL\rho }{2}\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} ,K=\frac {AE\pi ^{2}}{8L}\begin {bmatrix} 1 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 & 25 \end {bmatrix} \end {align*}
First, a numerical values given at end of problem 2 are used, therefore \(\rho AL=m=1\) and \(\frac {EA}{L}=\frac {1}{2}\), hence the \(K\) and \(M\) for part(b) and c become\begin {align*} part(b) & \Rightarrow M=\begin {bmatrix} \frac {1}{3} & \frac {1}{4} & \frac {1}{5}\\ \frac {1}{4} & \frac {1}{5} & \frac {1}{6}\\ \frac {1}{5} & \frac {1}{6} & \frac {1}{7}\end {bmatrix} ,K=\frac {1}{2}\begin {bmatrix} 1 & 1 & 1\\ 1 & \frac {4}{3} & \frac {3}{2}\\ 1 & \frac {3}{2} & \frac {9}{5}\end {bmatrix} \\ part(c) & \Rightarrow M=\frac {1}{2}\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} ,K=\frac {\pi ^{2}}{16}\begin {bmatrix} 1 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 & 25 \end {bmatrix} \end {align*}
The natural frequencies are found. Here is a summary table\[\begin {tabular} [c]{|l|l|l|}\hline & $\omega \ $(rad/sec) & $f$ hz\\\hline problem 2 & $0.3129,0.9080,1.4142,1.7820,1.9754$ & $0.0498,0.1445,0.225,0.284,0.314$\\\hline part(b) & $1.1108,3.4199,7.3872$ & $0.1768,0.5443,1.1757$\\\hline part(c) & $1.1107,3.3322,5.5536$ & $0.1768,0.5303,0.8839$\\\hline \end {tabular} \]
It can be seen that the first three natural frequencies using Ritz basic functions as given for both part b and c are higher than the natural frequencies generated by part b.
The stiffness matrix \(K\) for both parts b and c contains much smaller numerical values than the one used in problem 2. Since \(\omega ^{2}=\frac {k}{m}\) then one expects this result.
The first 3 mode shapes from problem 2 were\[ \left [ \Phi \right ] =\begin {bmatrix} -0.0989 & 0.2871 & -0.4472\\ -0.2871 & 0.6247 & -0.4472\\ -0.4472 & 0.4472 & 0.4472\\ -0.5635 & -0.0989 & 0.4472\\ -0.6247 & -0.5635 & -0.4472 \end {bmatrix} \]
The mode shapes from part(b)\[ \left [ \Phi \right ] =\begin {bmatrix} 2.2642 & -11.2099 & 13.0082\\ -0.2314 & 25.3536 & -47.2984\\ -0.6181 & -12.7003 & 37.5941 \end {bmatrix} \]
The mode shapes from part(c)\[ \left [ \Phi \right ] =\begin {bmatrix} 1.4142 & 0 & 0\\ 0 & 1.4142 & 0\\ 0 & 0 & 1.4142 \end {bmatrix} \]
Here is a plot of the above mode shapes
Here is a plot of the mode shapes overlay.
