5.2 Quiz 2

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5.2.1 Problem description

Consider this 3 DOF system

Suppose a harmonic force \(f(t)=A\cos (\varpi t)\) is applied to the mass in the center.  Use modal analysis to do the following:

1. Find the uncoupled modal equations of motion. Consider the steady state solution for each of these equations. Sketch the modal amplitude (\(X_{j}\) in the book on page 275) for each mode versus frequency.  A hand sketch is sufficient.
2. Use that result to sketch the frequency response of each of the  masses, in other words the complex amplitude \(Y_{n}\) versus \(\omega \)

5.2.2 Answer part (1)

A summary of the steps needed for full modal analysis is first given. In these steps, a column vector is shown as bold letter \(\mathbf {Y}\) and a matrix is shown as \(\left [ M\right ] \). In this summary, the system is assumed to have \(n\) degree of freedom.

The steps are

1. Determine the system of equations of motion and set up \(\left [ M\right ] \mathbf {Y}^{\prime \prime }+\left [ C\right ] \mathbf {Y}^{\prime }+\left [ K\right ] \mathbf {Y}=\mathbf {F}\) in matrix form.
2. Solve the eigenvalue problem \(\det \left (\left [ K\right ] -\omega ^{2}\left [ M\right ] \right ) =0\) in order to determine the \(n\) natural frequencies.
3. For each natural frequency \(\omega _{j}\) determine the corresponding \(j^{th}\) eigenvector \(\mathbf {\varphi }_{j}\) by solving \(\left (\left [ K\right ] -\omega _{j}^{2}\left [ M\right ] \right ) \mathbf {\varphi }_{j}=\mathbf {0}\). In this step, the first component of \(\mathbf {\varphi }_{j}\) is set to \(1\) and the other components are solved relative to it.
4. Obtain the normalized eigenvectors \(\Phi _{j}\) for each \(\mathbf {\varphi }_{j}\) using \(\Phi _{j}=\frac {\mathbf {\varphi }_{j}}{\sqrt {u_{j}}}\) where \(u_{j}=\mathbf {\varphi }_{j}^{T}\left [ M\right ] \mathbf {\varphi }_{j}\). Each \(u_{j}\) will be a scalar.
5. Set up the modal transformation matrix \(\left [ \Phi \right ] =\left [ \Phi _{1}\Phi _{2}\cdots \Phi _{n}\right ] \). This will be an \(n\times n\) matrix.
6. The transformation from normal solution \(y\relax (t) \) to modal \(\eta \relax (t) \) will be \(\mathbf {Y}=\left [ \Phi \right ] \mathbf {\eta }\) and \(\mathbf {\eta }=\left [ \Phi \right ] ^{-1}\mathbf {Y}=\left [ \Phi \right ] ^{T}\left [ M\right ] \mathbf {Y}\)
7. Apply the above transformation on the original equations of motions in matrix form to obtain the equations of motion in modal coordinates \(\left [ \Phi \right ] ^{T}\left [ M\right ] \) \(\left [ \Phi \right ] \mathbf {Y}^{\prime \prime }+\left [ \Phi \right ] ^{T}\left [ C\right ] \left [ \Phi \right ] \mathbf {Y}^{\prime }+\) \(\left [ \Phi \right ] ^{T}\left [ C\right ] \left [ \Phi \right ] \mathbf {Y}=\left [ \Phi \right ] ^{T}\mathbf {F}\). This becomes \(\mathbf {I\eta }^{\prime \prime }\relax (t) +\left [ \tilde {C}\right ] \mathbf {\eta }^{\prime }\relax (t) +\left [ \tilde {K}\right ] \mathbf {\eta }\relax (t) =\left [ \Phi \right ] ^{T}\mathbf {F}\) where \(\mathbf {I}\) is the identity matrix, \(\left [ \tilde {C}\right ] \) is a diagonal damping matrix obtained using a method such as weak damping approximation and \(\left [ \tilde {K}\right ] \) is diagonal matrix with diagonal that contains the natural frequencies squared \(\omega _{j}^{2}\) in each of entries.
8. For steady state solution in modal coordinates, the loading vector \(\left [ \Phi \right ] ^{T}\mathbf {F}\) is assumed to be \(\mathbf {Q}=\left [ \Phi \right ] ^{T}\mathbf {F=}\operatorname {Re}\left (\mathbf {\hat {Q}}e^{i\varpi t}\right ) \) where \(\mathbf {\hat {Q}}\) is the complex amplitude of the loading vector in modal coordinates. Therefore, the steady state solution is \(\mathbf {\eta }_{ss}\relax (t) =\operatorname {Re}\left ( \mathbf {\hat {X}}e^{i\varpi t}\right ) \) where \(\mathbf {\hat {X}}\) is the complex amplitude of each modal response is \(\hat {X}_{j}\mathbf {=}\frac {\mathbf {\Phi }_{j}^{T}\mathbf {F}}{-\varpi ^{2}+i2\zeta _{j}\omega _{j}\varpi +\omega _{j}^{2}}\). For a system with no damping this simplifies to \(\hat {X}_{j}\mathbf {=}\frac {\mathbf {\Phi }_{j}^{T}\mathbf {F}}{-\varpi ^{2}+\omega _{j}^{2}}\). In here, \(\mathbf {\Phi }_{j}^{T}\) represents the transpose of the \(j^{th}\) column of the modal transformation matrix \(\left [ \Phi \right ] \), or the transpose of the \(j^{th}\) mass normalized eigenvector, and \(\omega _{j}\) is the \(j^{th}\) natural frequency.
9. Now the steady state solution in modal coordinate is used to obtain the solution in normal coordinates since \(\mathbf {Y}=\left [ \Phi \right ] \mathbf {\eta }\). Therefore \(\mathbf {Y}_{ss}=\operatorname {Re}\left ( \mathbf {\hat {X}}e^{i\varpi t}\right ) =\operatorname {Re}\left (\left [ \Phi \right ] \mathbf {\hat {X}}e^{i\varpi t}\right ) =\operatorname {Re}\left ( \mathbf {\hat {Y}}e^{i\varpi t}\right ) \). In component form \(\mathbf {Y}_{ss}=\operatorname {Re}\left (\left ( {\displaystyle \sum \limits _{j=1}^{n}} \mathbf {\Phi }_{j}\hat {X}_{j}\right ) e^{i\varpi t}\right ) \)

The EOM are derived in the hand out given. The force \(f\relax (t) \) acting on the second mass is now added, resulting in the following equations of motion for the system\[\begin {bmatrix} m & 0 & 0\\ 0 & m & 0\\ 0 & 0 & m \end {bmatrix}\begin {Bmatrix} q_{1}^{\prime \prime }\\ q_{2}^{\prime \prime }\\ q_{3}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} k_{1}+k_{2} & -k_{2} & 0\\ -k_{2} & k_{1}+2k_{2} & -k_{2}\\ 0 & -k_{2} & k_{1}+k_{2}\end {bmatrix}\begin {Bmatrix} q_{1}\\ q_{2}\\ q_{3}\end {Bmatrix} =\begin {Bmatrix} 0\\ A\cos (\varpi t)\\ 0 \end {Bmatrix} \]

The first step is to obtain the natural frequencies of the system. This is done by solving the eigenvalue problem \(\det \left (\left [ K\right ] -\omega ^{2}\left [ M\right ] \right ) =0\). The solutions are also given in handout. They are \(\omega _{1}^{2}=\frac {k_{1}}{m},\omega _{2}^{2}=\frac {k_{1}+k_{2}}{m},\omega _{3}^{2}=\frac {k_{1}+3k_{2}}{m}\). The non mass normalized eigenvectors associated with these eigenvalues are found as \[ \mathbf {\varphi }_{1}=\begin {Bmatrix} 1\\ 1\\ 1 \end {Bmatrix} ,\mathbf {\varphi }_{2}=\begin {Bmatrix} 1\\ 0\\ -1 \end {Bmatrix} ,\mathbf {\varphi }_{3}=\begin {Bmatrix} 1\\ -2\\ 1 \end {Bmatrix} \]

The next step is to mass normalize the eigenvectors as follows\begin {align*} \mu _{1} & =\mathbf {\varphi }_{1}^{T}\left [ M\right ] \mathbf {\varphi }_{1}=\begin {Bmatrix} 1\\ 1\\ 1 \end {Bmatrix} ^{T}\begin {bmatrix} m & 0 & 0\\ 0 & m & 0\\ 0 & 0 & m \end {bmatrix}\begin {Bmatrix} 1\\ 1\\ 1 \end {Bmatrix} =3m\\ \mu _{2} & =\mathbf {\varphi }_{2}^{T}\left [ M\right ] \mathbf {\varphi }_{2}=\begin {Bmatrix} 1\\ 0\\ -1 \end {Bmatrix} ^{T}\begin {bmatrix} m & 0 & 0\\ 0 & m & 0\\ 0 & 0 & m \end {bmatrix}\begin {Bmatrix} 1\\ 0\\ -1 \end {Bmatrix} =2m\\ \mu _{3} & =\mathbf {\varphi }_{3}^{T}\left [ M\right ] \mathbf {\varphi }_{3}=\begin {Bmatrix} 1\\ -2\\ 1 \end {Bmatrix} ^{T}\begin {bmatrix} m & 0 & 0\\ 0 & m & 0\\ 0 & 0 & m \end {bmatrix}\begin {Bmatrix} 1\\ -2\\ 1 \end {Bmatrix} =6m \end {align*}

Hence the mass normalized eigenvectors are\begin {align*} \mathbf {\Phi }_{1} & =\frac {\mathbf {\varphi }_{1}}{\sqrt {\mu _{1}}}=\frac {1}{\sqrt {3m}}\begin {Bmatrix} 1\\ 1\\ 1 \end {Bmatrix} \\ \mathbf {\Phi }_{2} & =\frac {\mathbf {\varphi }_{2}}{\sqrt {\mu _{2}}}=\frac {1}{\sqrt {2m}}\begin {Bmatrix} 1\\ 0\\ -1 \end {Bmatrix} \\ \mathbf {\Phi }_{3} & =\frac {\mathbf {\varphi }_{3}}{\sqrt {\mu _{3}}}=\frac {1}{\sqrt {6m}}\begin {Bmatrix} 1\\ -2\\ 1 \end {Bmatrix} \end {align*}

Hence the modal transformation matrix \(\left [ \Phi \right ] \) is\[ \left [ \Phi \right ] =\left [ \mathbf {\Phi }_{1}\mathbf {\Phi }_{2}\mathbf {\Phi }_{3}\right ] =\frac {1}{\sqrt {m}}\begin {bmatrix} \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {2}} & \frac {1}{\sqrt {6}}\\ \frac {1}{\sqrt {3}} & 0 & \frac {-2}{\sqrt {6}}\\ \frac {1}{\sqrt {3}} & \frac {-1}{\sqrt {2}} & \frac {1}{\sqrt {6}}\end {bmatrix} =\frac {1}{\sqrt {m}}\begin {bmatrix} 0.577\, & 0.707\, & 0.408\,\\ 0.577\, & 0 & -0.816\,\\ 0.577\, & -0.707\, & 0.408\, \end {bmatrix} \allowbreak \allowbreak \]

The modal EOM’s are now found using the modal transformation matrix \(\left [ \Phi \right ] \) \begin {align*} \left [ \Phi \right ] ^{T}\left [ M\right ] \left [ \Phi \right ] \left \{ \eta ^{\prime \prime }\right \} +\left [ \Phi \right ] ^{T}\left [ K\right ] \left [ \Phi \right ] \left \{ \eta \right \} & =\left [ \Phi \right ] ^{T}\mathbf {Q}\\\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\\ \eta _{3}^{\prime \prime }\end {Bmatrix} +\begin {bmatrix} \omega _{1}^{2} & 0 & 0\\ 0 & \omega _{2}^{2} & 0\\ 0 & 0 & \omega _{3}^{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\end {Bmatrix} & =\left [ \Phi \right ] ^{T}\begin {Bmatrix} 0\\ A\cos (\varpi t)\\ 0 \end {Bmatrix} \\\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\\ \eta _{3}^{\prime \prime }\end {Bmatrix} +\frac {1}{m}\begin {bmatrix} k_{1} & 0 & 0\\ 0 & k_{1}+k_{2} & 0\\ 0 & 0 & k_{1}+3k_{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\end {Bmatrix} & =\frac {1}{\sqrt {m}}\begin {bmatrix} 0.577\, & 0.707\, & 0.408\,\\ 0.577\, & 0 & -0.816\,\\ 0.577\, & -0.707\, & 0.408\, \end {bmatrix} ^{T}\allowbreak \begin {Bmatrix} 0\\ A\cos (\varpi t)\\ 0 \end {Bmatrix} \\\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix}\begin {Bmatrix} \eta _{1}^{\prime \prime }\\ \eta _{2}^{\prime \prime }\\ \eta _{3}^{\prime \prime }\end {Bmatrix} +\frac {1}{m}\begin {bmatrix} k_{1} & 0 & 0\\ 0 & k_{1}+k_{2} & 0\\ 0 & 0 & k_{1}+3k_{2}\end {bmatrix}\begin {Bmatrix} \eta _{1}\\ \eta _{2}\\ \eta _{3}\end {Bmatrix} & =\frac {1}{\sqrt {m}}\begin {Bmatrix} 0.577\,A\cos \left (\varpi t\right ) \\ 0\\ -0.816\,A\cos \left (\varpi t\right ) \end {Bmatrix} \end {align*}

Therefore, the \(3\) uncoupled modal EOM’s are\begin {align*} \eta _{1}^{\prime \prime }\relax (t) +\frac {k_{1}}{m}\eta _{1}\left ( t\right ) & =\frac {0.577\,A}{\sqrt {m}}\cos \left (\varpi t\right ) \\ \eta _{2}^{\prime \prime }\relax (t) +\frac {k_{1}+k_{2}}{m}\eta _{2}\left ( t\right ) & =0\\ \eta _{3}^{\prime \prime }\relax (t) +\frac {k_{1}+3k_{2}}{m}\eta _{3}\relax (t) & =-\frac {0.816\,A}{\sqrt {m}}\cos \left (\varpi t\right ) \end {align*}

To complete the solution, the above EOM are written as follows by using complex form for the loading vector\begin {align*} \eta _{1}^{\prime \prime }\relax (t) +\frac {k_{1}}{m}\eta _{1}\left ( t\right ) & =\operatorname {Re}\left (\frac {0.577\,A}{\sqrt {m}}e^{i\varpi t}\right ) \\ \eta _{2}^{\prime \prime }\relax (t) +\frac {k_{1}+k_{2}}{m}\eta _{2}\left ( t\right ) & =0\\ \eta _{3}^{\prime \prime }\relax (t) +\frac {k_{1}+3k_{2}}{m}\eta _{3}\relax (t) & =\operatorname {Re}\left (\frac {-0.816\,A}{\sqrt {m}}e^{i\varpi t}\right ) \end {align*}

Assuming the steady state solution is\[ \mathbf {\eta }=\operatorname {Re}\left (\mathbf {\hat {X}}e^{i\varpi t}\right ) \]

or in expanded form\begin {align*} \eta _{1}\relax (t) & =\operatorname {Re}\left (\hat {X}_{1}e^{i\varpi t}\right ) \\ \eta _{2}\relax (t) & =\operatorname {Re}\left (\hat {X}_{2}e^{i\varpi t}\right ) \\ \eta _{3}\relax (t) & =\operatorname {Re}\left (\hat {X}_{3}e^{i\varpi t}\right ) \end {align*}

Where\begin {align*} \hat {X}_{1} & =\frac {\frac {0.577\,A}{\sqrt {m}}}{\omega _{1}^{2}+2i\zeta _{1}\omega _{1}\varpi -\varpi ^{2}}\\ \hat {X}_{2} & =0\\ \hat {X}_{3} & =\frac {\frac {-0.816\,A}{\sqrt {m}}}{\omega _{3}^{2}+2i\zeta _{3}\omega _{3}\varpi -\varpi ^{2}} \end {align*}

Dividing the numerator and the denominator by \(\omega _{i}^{2}\) where \(i=1,2,3\) and using \(r_{i}=\frac {\varpi }{\omega _{i}}\) and letting \(\zeta =0\) since no damping exists, results in \begin {align*} \hat {X}_{1} & =\frac {A\sqrt {m}}{k_{1}}\left (\frac {0.577}{1-m\frac {\varpi ^{2}}{k_{1}}}\right ) \\ \hat {X}_{2} & =0\\ \hat {X}_{3} & =\frac {A\sqrt {m}}{k_{1}+3k_{2}}\left (\frac {-0.816\,}{1-m\frac {\varpi ^{2}}{k_{1}+3k_{2}}}\right ) \end {align*}

To sketch these amplitudes, the equations are normalized. This is in effect the same as setting \(m=1,k_{1}=k_{2}=1,A=1\) resulting in

\[ \mathbf {\hat {X}=}\begin {Bmatrix} \hat {X}_{1}\\ \hat {X}_{2}\\ \hat {X}_{3}\end {Bmatrix} =\begin {Bmatrix} \frac {0.577}{1-\varpi ^{2}}\\ 0\\ \frac {1}{4}\left (\frac {-0.816\,}{1-\frac {\varpi ^{2}}{4}}\right ) \end {Bmatrix} \]

Here is a plot of each \(X_{i}\) vs \(\varpi \). The x-axis is the nondimensional forcing frequency \(\Omega \)

Since there is no damping, resonance will occur at \(\Omega =1\) in first mode and at \(\Omega =2\) for mode \(3\).

5.2.3 Answer part (2)

The transformation from modal coordinates to normal coordinates is\[ \mathbf {q}=\left [ \Phi \right ] \mathbf {\eta }\]

In expanded form\[\begin {Bmatrix} q_{1}\\ q_{2}\\ q_{3}\end {Bmatrix} =\begin {Bmatrix} \Phi _{1}^{T}\left \{ \eta \right \} \\ \Phi _{2}^{T}\left \{ \eta \right \} \\ \Phi _{3}^{T}\left \{ \eta \right \} \end {Bmatrix} \]

But \(\left [ \Phi \right ] =\frac {1}{\sqrt {m}}\begin {bmatrix} 0.577\, & 0.707\, & 0.408\,\\ 0.577\, & 0 & -0.816\,\\ 0.577\, & -0.707\, & 0.408\, \end {bmatrix} \) and \(\mathbf {\eta }=\operatorname {Re}\left (\mathbf {\hat {X}}e^{i\varpi t}\right ) \) hence the above becomes\begin {align*} \begin {Bmatrix} q_{1}\\ q_{2}\\ q_{3}\end {Bmatrix} & =\begin {Bmatrix}\begin {Bmatrix} 0.577\,\\ 0.577\,\\ 0.577\, \end {Bmatrix} ^{T}\begin {Bmatrix} \operatorname {Re}\left (\hat {X}_{1}e^{i\varpi t}\right ) \\ \operatorname {Re}\left (\hat {X}_{2}e^{i\varpi t}\right ) \\ \operatorname {Re}\left (\hat {X}_{3}e^{i\varpi t}\right ) \end {Bmatrix} \\\begin {Bmatrix} 0.707\,\,\\ 0\,\\ -0.707\, \end {Bmatrix} ^{T}\begin {Bmatrix} \operatorname {Re}\left (\hat {X}_{1}e^{i\varpi t}\right ) \\ \operatorname {Re}\left (\hat {X}_{2}e^{i\varpi t}\right ) \\ \operatorname {Re}\left (\hat {X}_{3}e^{i\varpi t}\right ) \end {Bmatrix} \\\begin {Bmatrix} 0.408\,\\ -0.816\,\,\\ 0.408\,\, \end {Bmatrix} ^{T}\begin {Bmatrix} \operatorname {Re}\left (\hat {X}_{1}e^{i\varpi t}\right ) \\ \operatorname {Re}\left (\hat {X}_{2}e^{i\varpi t}\right ) \\ \operatorname {Re}\left (\hat {X}_{3}e^{i\varpi t}\right ) \end {Bmatrix} \end {Bmatrix} \\ & =\begin {Bmatrix} 0.577\,\operatorname {Re}\left (X_{1}e^{it\varpi }\right ) +0.577\,\operatorname {Re}\left (X_{2}e^{it\varpi }\right ) +0.577\,\operatorname {Re}\left (X_{3}e^{it\varpi }\right ) \allowbreak \\ 0.707\,\operatorname {Re}\left (X_{1}e^{it\varpi }\right ) -0.707\,\operatorname {Re}\left (X_{3}e^{it\varpi }\right ) \allowbreak \\ 0.408\,\operatorname {Re}\left (X_{1}e^{it\varpi }\right ) -0.816\,\operatorname {Re}\left (X_{2}e^{it\varpi }\right ) +0.408\,\operatorname {Re}\left (X_{3}e^{it\varpi }\right ) \end {Bmatrix} \\ & =\operatorname {Re}\left ( \begin {Bmatrix} 0.577\,X_{1}+0.577\,X_{2}+0.577\,X_{3}\allowbreak \\ 0.707\,X_{1}-0.707\,X_{3}\allowbreak \\ 0.408\,X_{1}-0.816\,X_{2}+0.408\,X_{3}\end {Bmatrix} e^{i\varpi t}\right ) \end {align*}

Comparing the above to \(\mathbf {q}_{ss}=\operatorname {Re}\left ( \mathbf {Y}e^{i\varpi t}\right ) \) shows that

\[ \mathbf {Y}=\begin {Bmatrix} 0.577\,X_{1}+0.577\,X_{2}+0.577\,X_{3}\allowbreak \\ 0.707\,X_{1}-0.707\,X_{3}\allowbreak \\ 0.408\,X_{1}-0.816\,X_{2}+0.408\,X_{3}\end {Bmatrix} \]

To plot each \(Y_{i}\), let \(m=1,k_{1}=1,k_{2}=1,A=1\), and letting \(X_{2}=0\) as found earlier, results in

\[ \mathbf {Y}=\begin {Bmatrix} 0.577\,\frac {0.577}{1-\varpi ^{2}}+\frac {0.577}{4}\,\left (\frac {-0.816\,}{1-\frac {\varpi ^{2}}{4}}\right ) \allowbreak \\ 0.707\frac {0.577}{1-\varpi ^{2}}-\frac {0.707}{4}\,\left (\frac {-0.816\,}{1-\frac {\varpi ^{2}}{4}}\right ) \\ 0.408\,\frac {0.577}{1-\varpi ^{2}}+\frac {0.408}{4}\,\left (\frac {-0.816\,}{1-\frac {\varpi ^{2}}{4}}\right ) \end {Bmatrix} \]

The above shows that when the nondimensional frequency \(\Omega \) is not close to a one of the nondimensional natural frequencies, then the \(Y\) values have comparable magnitudes. For nondimensional frequency \(\Omega \) larger than \(3\) all amplitude are zero, which means the whole system does not oscillate any more in steady state.

5.2.4 Key solution

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