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Chapter 10

Q. 10.1

Consider a three-DOF spring mass system shown in Figure 10.5. Determine its frequencies and mode shapes. Given: m_{1} = 2 m, m_{2} = 3 m, m_{3} = 2 m; k_{1} = k, k_{2} = 2 k, k_{3} = 2 k.

Annotation 2022-10-08 165441

Step-by-Step

Verified Solution

The equations of motion can be written by considering the dynamic equilibrium of each mass:

m_{1}\ddot{x} _{1}+k_{1}(x_{1}-x_{2})=0

m_{2}\ddot{x} _{2}+k_{2}(x_{2}-x_{3})+k_{1}(x_{2}-x_{1})=0               (i)

m_{3}\ddot{x} _{3}+k_{3}x_{3}+k_{2}(x_{3}-x_{2})=0

It may be noted that the DOF may be numbered either from top or from bottom. In the present case, these have been numbered from top. The equations will change accordingly. The mass matrix can be written as follows:

It is given that: m_{1} = 2 m, m_{2} = 3 m, and m_{3} = 2 m

[M]=\left[\begin{matrix} 2m & 0 & 0 \\ 0 & 3m & 0 \\ 0 & 0 & 2m \end{matrix} \right]

The stiffness matrix can be written as follows by rearranging Equation (i) in matrix form:

[K]\left[\begin{matrix} k_{1} & -k_{1} & 0 \\ -k_{1} & k_{1}+k_{2} & -k_{2} \\ 0 & -k_{2} & k_{2}+k_{3} \end{matrix} \right]

Given k_{1} = k, k_{2} = 2 k, and k_{3} = 2 k.

On substitution, we get

[K]=k\left[\begin{matrix} 1 & -1 & 0 \\-1 & 3 & -2 \\ 0 & -2 & 4 \end{matrix} \right]

det\left|K-m\omega ^{2}\right| =\left|\begin{matrix} k-2m\omega ^{2} & -k & 0 \\ -k &3k-3m\omega^{2} & -2k \\ 0 & -2k & 4k-2m\omega ^{2} \end{matrix} \right|

Let k/m = β

The expression for determinant can be written as follows:

det\left|K-m\omega ^{2}\right| =\left|\begin{matrix} \beta -2\omega ^{2} & -\beta & 0\\ -\beta & 3\beta -3\omega ^{2}& -2\beta \\ 0 & -2\beta & 4\beta -2\omega ^{2} \end{matrix} \right|

Or, it can be expanded as follows:

6(\omega ^{2})^{3}-21\beta (\omega ^{2})^{2}+16\beta ^{2}(\omega ^{2})-2\beta ^{3}=0

Its three positive roots are given as follows:

\left\{\omega ^{2}\right\} =\left\{\begin{matrix} 2.477\beta \\ 0.866\beta \\ 0.155\beta \end{matrix} \right\} or \left\{\omega \right\} =\left\{\begin{matrix} 1.574 \\ 0.93 \\ 0.394 \end{matrix} \right\} \sqrt{\frac{k}{m} }

Now mode shapes can be determined for each frequency using the following expression:

[K-m\omega ^{2}]\left\{X\right\} =\left\{0\right\}

\left[\begin{matrix}\beta -2\omega ^{2} & -\beta & 0 \\ -\beta & 3\beta -3\omega ^{2} & -2\beta \\ 0 &-2\beta & 4\beta -2\omega ^{2} \end{matrix} \right] \left\{\begin{matrix} x_{1}\\ x_{2} \\ x_{3} \end{matrix} \right\} =\left\{\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right\}     (ii)

Let us obtain the eigenvector corresponding to \lambda _{3} = \omega ^{2}_{3} = 2.477 β. Its value is substituted in Equation (ii), we get

\left[\begin{matrix} -3.954 & -1 & 0 \\ -1 &-4.431 & -2 \\ -1 & a-2& -0.954\end{matrix} \right] \left\{\begin{matrix} x_{13} \\ x_{23} \\ x_{33} \end{matrix} \right\} =\left\{\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right\}

Upon simplification,

-3.954 x_{13}-x_{23}+0=0    (iii)

-x_{13}-4.431 x_{23}-2 x_{33}=0     (iv)

-2 x_{23}-0.954 x_{33}=0     (v)

Equation (iii) and Equation (v) give,

\frac{x_{13}}{x_{23}} =\frac{-1}{3.954} and \frac{x_{23}}{x_{33}} =\frac{-0.954}{2} =\frac{-0.477}{1}

If the value of x_{33} in terms of x_{23} is substituted in Equation (iv), it gives the same ratio of x_{13} and x_{23} as before.

∴ In such cases, only the ratio of the elements is determined. A numerical value for each element of the eigenvector {x_{3}} may be obtained by arbitrarily assigning any one element as unity. For example, if x_{13} is assigned a unity, then

x_{23} = −3.954 and x_{33} = 8.289

This procedure is repeated for other eigenvalues as well.
For first mode, \omega _{1}^{2} = 0.155 β, and x_{11} = 1

\left\{\begin{matrix} x_{11} \\ x_{21} \\ x_{31} \end{matrix} \right\} =\left\{\begin{matrix} 1.000 \\ 0.689 \\ 0.374 \end{matrix} \right\}

For second mode, \omega _{2}^{2} = 0.866 β , and x_{12} = 1

\left\{\begin{matrix} x_{12} \\ x_{22} \\ x_{32} \end{matrix} \right\} =\left\{\begin{matrix} 1.364\\ -1 \\ -0.882 \end{matrix} \right\}

For third mode \omega _{3}^{2} = 2.477 β, and x_{13} = 1

\left\{\begin{matrix} x_{13} \\ x_{23} \\ x_{33} \end{matrix} \right\} =\left\{\begin{matrix}1.0\\ -3.954 \\ 8.289 \end{matrix} \right\}

Alternatively, it can be normalized with respect to x_{33} = 1 instead of x_{13} = 1

or,    \left\{\begin{matrix} x_{13} \\ x_{23} \\ x_{33} \end{matrix} \right\} =\left\{\begin{matrix}0.1206\\ -0.477 \\ 1.0 \end{matrix} \right\}

The mode shape matrix can be written as follows by arranging them column-wise or row-wise. In this case they have been arranged column-wise.

[\Phi ]=\left[\begin{matrix} \phi _{11} &\phi _{12} & \phi _{13} \\ \phi _{21} & \phi _{22} & \phi _{23} \\ \phi _{31} & \phi _{32} & \phi _{33} \end{matrix} \right]

where in \phi _{ij} i represents DOF and j represents mode number

 [\Phi ]=\left[\begin{matrix} 1.000 & 1.364 & 0.121 \\ 0.689 & -1 & -0.477 \\ 0.374 & -0.882 & a1.0\end{matrix} \right]

The mode shapes are shown in Figure 10.6.

Let us verify the results using MATLAB.
The command eig(K, M) gives two matrices V and D. Matrix V is for eigenvectors
arranged column-wise and matrix D is a diagonal matrix consisting of eigenvalues on the diagonal. The frequencies are obtained as follows:

λ = ω²

>>[V, D] = eig(K, M)

V =

\begin{matrix} -0.5195 & -0.4741& 0.0734\\ a-0.&3582 0.3476 & −0.2902 \\ −0.1942 & 0.3066 & 0.6069 \end{matrix}

D =

\begin{matrix} 0.1552 & 0 & 0 \\ 0 & 0.8665 & 0 \\ 0 & 0 & 2.4782 \end{matrix}

∴                          \omega ^{2}=\left\{\begin{matrix} 0.1552 \\ 0.8665 \\2.4782 \end{matrix} \right\} \frac{k}{m} or \omega =\left\{\begin{matrix}0.394\\ 0.930 \\ 1.574 \end{matrix} \right\} \sqrt{\frac{k}{m} }

In this option, the eigenvalues are obtained in increasing order.
Alternatively, the following command can be given in MATLAB.

MATLAB Verified Solution

Script Files

>> [V, D] = eig(M, K)

It again generates two matrices V and D. Matrix V is for eigenvectors arranged column-wise and matrix D is a diagonal matrix consisting of eigenvalues on the diagonal. The frequencies are obtained as follows:

λ = 1 / ω²

>>[V, D] = eig(M, K)

V =

\begin{matrix} −0.0466 & 0.5093 & 1.3185 \\ 0.1844 & −0.3734 & 0.9092 \\ −0.3855 & −0.3294 & 0.4928 \end{matrix}

D =

\begin{matrix} 0.4035 & 0 & 0 \\ 0 & 1.1540 & 0 \\ 0 & 0 & 6.4425 \end{matrix}

\omega ^{2}=\left\{\begin{matrix} \frac{1}{0.4035} \\ \\ \frac{1}{1.1540} \\ \\ \frac{1}{6.4425} \end{matrix} \right\} \frac{k}{m} or \omega =\left\{\begin{matrix} 1.574 \\0.93 \\ 0.394 \end{matrix} \right\} \sqrt{\frac{k}{m} }

In this option, the eigenvalues are obtained in decreasing order.

Annotation 2022-10-08 185158