(a) Determine the natural frequencies for the three degree-of-freedom system shown in Figure 8.13.
(b) Calculate and graphically illustrate the normalized mode shape vectors.
(c) Demonstrate mode shape orthogonality

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Accepted Answer

The differential equations governing the system may be formulated using Newton’s law or Lagrange’s equations (the full set or via influence coefficients as)

[latex]\begin{bmatrix} m & 0 & 0 \0 & m & 0 \0 & 0 & 2m \end{bmatrix}\begin{bmatrix} \ddot{x}_{1}\ \ddot{x}_{2}\ \ddot{x}_{3} \end{bmatrix} +\begin{bmatrix} 2k & -2k & 0 \-2k & 3k & -k \ 0 & -k & k \end{bmatrix} \begin{bmatrix}x_{1}\x_{2}\x_{3} \end{bmatrix} =\begin{bmatrix}0 \ 0\0 \end{bmatrix} [/latex] (a)

(a) The natural frequencies are the square roots of the eigenvalues of [latex] \pmb{M^{-1}K}[/latex],

[latex]\det(\pmb{M^{-1}K}-\lambda \pmb{I})=\pmb{0}\Rightarrow \begin{bmatrix}\frac{1}{m} & 0 & 0 \ 0 & \frac{1}{m} & 0 \0 & 0 & \frac{1}{2m} \end{bmatrix}\begin{bmatrix} 2k & -2k & 0 \ -2k & 3k & -k \ 0 & -k & k \end{bmatrix} - \lambda \begin{bmatrix}1 & 0 & 0\0 & 1 & 0\0 & 0 & 1 \end{bmatrix} [/latex]

[latex] =\begin{bmatrix} 4\phi -\lambda & -4\phi & 0 \-4\phi & 6\phi -\lambda & -2\phi \0 & -\phi & \phi -\lambda \end{bmatrix} [/latex] (b)

where [latex]\phi =\frac{k}{2m} [/latex]. Evaluation of the determinant in Equation (b) leads to

[latex]-\lambda ^{3}+11\phi \lambda ^{2}-16\phi ^{2}\lambda =0 [/latex] (c)

The smallest toot of Equation (c) is [latex]\lambda=0[/latex]. The system is unrestrained. The other roots are obtained by solving

[latex]\lambda ^{2}-11\phi \lambda +16\phi ^{2}=0 [/latex] (d)

The solutions of Equation (b) are

[latex]\lambda = 0,1.725 \phi ,9.275\phi [/latex] (e)

[latex]\omega _{1}=0 [/latex] [latex] \omega _{2}=\sqrt{\frac{1.725k}{2m} }=0.928\sqrt{\frac{k}{m} } [/latex] [latex]\omega _{3}=\sqrt{\frac{9.275k}{2m} } =2.15\sqrt{\frac{k}{m} } [/latex] (f)

(b) The mode shape vectors are determined from

[latex]\begin{bmatrix} 4\phi -\lambda & -4\phi & 0 \-4\phi & 6\phi -\lambda & -2\phi \0 & -\phi &\phi -\lambda \end{bmatrix}\begin{bmatrix}X_{1}\X_{2}\X_{3}\end{bmatrix} =\begin{bmatrix} 0\0\0 \end{bmatrix} [/latex] (g)

The first equation gives

[latex]X_{1}=\frac{4\phi }{4\phi -\lambda }X_{2} [/latex] (h)

while the third equation gives

[latex] X_{3}=\frac{\phi }{\phi -\lambda }X_{2} [/latex] (i)

When evaluated for the values of which are the eigenvalues of [latex]\pmb{M^{-1}K}.[/latex] and keeping [latex]X_{2}=C [/latex], arbitrary Equations (h) and (i) yield

[latex]\pmb{X}_{1}=C\begin{bmatrix} 1\1\1 \end{bmatrix} [/latex] [latex]\pmb{X}_{2}=C\begin{bmatrix} 1.758 \ 1 \-1.379 \end{bmatrix} [/latex] [latex]\pmb{X}_{3}=C\begin{bmatrix} -0.758 \ 1 \ -0.121 \end{bmatrix} [/latex] (j)

The mode shapes are normalized by requiring [latex]\pmb{(X_{i,X_{i}})_{M}=X^{T}_{i}MX_{i}} =1[/latex]. For example, the normalization of [latex]\pmb{X_{2}} [/latex] chooses C such that

[latex]\pmb{X^{T}_{2}MX_{2}}=1\Rightarrow 1=C\begin{bmatrix} 1.758 &1 &-1.3679 \end{bmatrix} \begin{bmatrix} m &0&0 \0&m&0\ 0&0&2m\end{bmatrix} C\begin{bmatrix} 1.758 \1\ -1.379 \end{bmatrix} [/latex]

[latex]=C^{2}\begin{bmatrix} 1.758 & 1& -1.3679 \end{bmatrix} \begin{bmatrix} 1.75m\m\-2.758m \end{bmatrix} [/latex]

[latex] =C^{2}\begin{bmatrix} 1.758(1.758m)+1(m)-1.479(-2.758m)=C^{2}(8.939m) \end{bmatrix} [/latex] (k)

or [latex] C=\frac{0.334}{\sqrt{m} } [/latex]. Similar calculations are performed yielding the normalized mode-shape vectors as

[latex]\pmb{X}_{1}=\frac{1}{\sqrt{m} }\begin{bmatrix} 0.5 \ 0.5 \ 0.5 \end{bmatrix} [/latex] [latex] \pmb{X}_{2}=\frac{1}{\sqrt{m} }\begin{bmatrix} 0.587 \ 0.334 \-0.461 \end{bmatrix} [/latex] [latex] \pmb{X}_{3}=\frac{1}{\sqrt{m} }\begin{bmatrix} -.599 \0.790 \-0.096\end{bmatrix} [/latex] (l)

The normalized mode shapes are illustrated in Figure 8.12(b). The first mode is a rigidbody mode corresponding to the natural frequency of zero. There is a node for the second mode in the spring connecting the second and third masses. Two nodes mark the third mode. One is in the spring connecting the first two masses; the second is in the spring connecting the second and third mass, but not in the same location as the node for the second mode.
(a) Mode-shape orthogonality implies [latex]\pmb{(X_{i},X_{j})}_{M}=\pmb{X}^{T}_{j}\pmb{MX}_{i}=0 [/latex] for [latex] i eq j[/latex]. The demonstration of this relation follows

[latex]\pmb{(X_{1},X_{2})}_{M}=\pmb{X}^{T}_{2}\pmb{MX}_{1}=\frac{1}{\sqrt{m} }\begin{bmatrix} 0.587&0.334 &-0.461 \end{bmatrix} \begin{bmatrix} m & 0 & 0 & \0 & m & 0\0 & 0 & 2m\end{bmatrix} \frac{1}{\sqrt{m} } \begin{bmatrix}0.5\0.5\0.5 \end{bmatrix} [/latex]

[latex]= \frac{1}{m }\begin{bmatrix} 0.587 & 0.334 &-0.461 \end{bmatrix} \begin{bmatrix} 0.5\0.5\m \end{bmatrix} [/latex]

[latex] =0.587(0.5)+0.334(0.5)=0.461(1)=0[/latex] (m)

[latex]\pmb{(X_{2},X_{3})}_{M}=\pmb{X}^{T}_{3}\pmb{MX}_{2}=\frac{1}{\sqrt{m} }\begin{bmatrix} 0.599 & 0.790 & -0.096 \end{bmatrix} \begin{bmatrix} m & 0 & 0 & \0 & m & 0\0 & 0 & 2m \end{bmatrix} \frac{1}{\sqrt{m} } \begin{bmatrix} -0.587\0.334\ -0.461\end{bmatrix} [/latex]

[latex] =\frac{1}{m}\begin{bmatrix} -0.599 & 0.790 & 0.096 \end{bmatrix} \begin{bmatrix} 0.587m \ 0.334 m\0.922m \end{bmatrix} [/latex]

[latex]=-0599(0.587)+0.790(0.334)-0.096(-0.922)=0[/latex] (n)

[latex]\pmb{(X_{3},X_{1})}_{M}=\pmb{X}^{T}_{2}\pmb{MX}_{1}=\frac{1}{\sqrt{m} }\begin{bmatrix} 0.5 & 0.5 & -0.5 \end{bmatrix} \begin{bmatrix} m & 0 & 0 & \0 & m & 0\0 & 0 & 2m \end{bmatrix} \frac{1}{\sqrt{m} } \begin{bmatrix} -00.599\00.790\-0.096 \end{bmatrix} [/latex]

[latex]=\frac{1}{m}\begin{bmatrix} 0.5 & 0.5 & -0.5 \end{bmatrix} \begin{bmatrix} -0.599m\0.790m\-0.192m \end{bmatrix} [/latex]

[latex]=0.5(-0.599)+0.5(0.790)+0.5(-0.192)=0 [/latex] (o)

Question 8.17: (a) Determine the natural frequencies for the three degree-o...

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