Question 8.3: Use the component mode synthesis, free interface, method to ......
Use the component mode synthesis, free interface, method to find the normal modes of the single system formed when the two uniform cantilever beams, R and S, of lengths L and 2L, respectively, as shown in Fig. 8.2, are subsequently rigidly joined at the free ends. The two cantilevers have the same mass per unit length, \mu, and the same values of Young’s modulus, E, and second moment of area of crosssection, I.
Represent the cantilever R by two assumed modes with modal displacements p_{1} \mathrm{and} p_{2}, defined as follows:
Represent the cantilever S by three assumed modes with modal displacements p_{3},p_{4} \mathrm{and} p_{5} defined by:
\mathrm{y}_{2} = \left(\frac{x_{2}}{2L}\right)^{2} p_{3} + \left(\frac{x_{2}}{2L}\right)^{3} p_{4} + \left(\frac{x_{2}}{2L}\right)^{4} p_{5} (B)where x_{1}, x_{2}, y_{1}, y_{2} \mathrm{and} L are defined by Fig. 8.2.
We shall need the following relationships:
From Eq. (A):
To derive the equations of motion for the two ‘substructures’ R and S, separately, we use Lagrange’s equations, which require expressions for the kinetic energies T_{R} \mathrm{and} T_{S}, and the potential energies, U_{R} \mathrm{and} U_{S}, of R and S, respectively.
T_{R} = \frac{1}{2} \mu \int_{0}^{L}{\dot{y}^{2}_{1} dx_{1}} = \frac{1}{2} \mu \int_{0}^{L}{\left[\left(\frac{x_{1}}{L}\right)^{2} \dot{p}_{1} + \left(\frac{x_{1}}{L}\right)^{3} \dot{p}_{2} \right]^{2} dx} = \frac{1}{2} \mu L \left(\frac{1}{5}\dot{p}^{2}_{1} + \frac{1}{3}\dot{p}_{1}\dot{p}_{2} + \frac{1}{7}\dot{p}^{2}_{2} \right) (G) \\ U_{R} = \frac{1}{2} EI \int_{0}^{L}{\left(\frac{d^{2}y_{1}}{dx_{1}}\right)^{2} }\;dx_{1} = \frac{1}{2} EI \int_{0}^{L}{\left(\frac{4}{L^{4}}p^{2}_{1} + \frac{12x_{1}}{L^{5}}p_{1}p_{2} + \frac{36x^{3}_{1}}{L^{6}}p^{2}_{2} \right) dx_{1}} \\ = \frac{EI}{2L^{3}} \left(4p^{2}_{1} + 12p_{1}p_{2} + 12p^{2}_{2}\right) (H)Applying Lagrange’s equations
\frac{d}{dt} \left(\frac{\partial T_{R}}{\partial \dot{p}_{i}} \right) + \frac{\partial U_{R}}{\partial p_{i}} = 0 (i = 1, 2)to substructure R gives the two equations:
\mu L \left(\frac{1}{5}\ddot{p}_{1} + \frac{1}{6}\ddot{p}_{2} \right) + \frac{EI}{L^{3}} (4p_{1} + 6p_{2}) = 0 (I) \\ \mu L \left(\frac{1}{6}\ddot{p}_{1} + \frac{1}{7}\ddot{p}_{2} \right) + \frac{EI}{L^{3}} (6p_{1} + 12p_{2}) = 0 (J) \\ \mathrm{or in matrix form:} \\ \mu L \begin{bmatrix} \frac{1}{5} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{7} \end{bmatrix} \begin{Bmatrix} \ddot{p}_{1} \\ \ddot{p}_{2} \end{Bmatrix} + \frac{EI}{L^{3}} \begin{bmatrix} 4 & 6 \\ 6 &12 \end{bmatrix} \begin{Bmatrix} p_{1} \\ p_{2} \end{Bmatrix} = 0 (K)Similarly, for substructure S:
Applying Lagrange’s equations:
\frac{d}{dt} \left(\frac{\partial T_{S}}{\partial \dot{p}_{i}} \right) + \frac{\partial U_{S}}{\partial p_{i}} = 0 (i = 3, 4, 5)to substructure S leads to:
\mu L \begin{bmatrix} \frac{2}{5} & \frac{1}{3} & \frac{2}{7} \\ \frac{1}{3} & \frac{2}{7} & \frac{1}{4} \\ \frac{2}{7} & \frac{1}{4} & \frac{2}{9} \end{bmatrix} \begin{Bmatrix} \ddot{p}_{3} \\ \ddot{p}_{4} \\ \ddot{p}_{5} \end{Bmatrix} + \frac{EI}{L^{3}} \begin{bmatrix} \frac{1}{2} & \frac{3}{4} & 1 \\ \frac{3}{4} & \frac{3}{2} & \frac{9}{4} \\ 1 & \frac{9}{4} & \frac{18}{5} \end{bmatrix} \begin{Bmatrix} p_{3} \\ p_{4} \\ p_{5} \end{Bmatrix} = 0 (N)Combining Eq. (K), representing substructure R, with Eq. (N) representing substructure S:
\mu L \left [ \begin{matrix} \frac{1}{5} & \frac{1}{6} & 0 & 0 & 0 \\ \frac{1}{6} & \frac{1}{7} & 0 & 0 & 0 \\ 0 & 0 & \frac{2}{5} & \frac{1}{3} & \frac{2}{7} \\ 0 & 0 & \frac{1}{3} & \frac{2}{7} & \frac{1}{4} \\ 0 & 0 & \frac{2}{7} & \frac{1}{4} & \frac{2}{9} \end{matrix} \right ] \left \{ \begin{matrix} \ddot{p}_{1} \\ \ddot{p}_{2} \\ \ddot{p}_{3} \\ \ddot{p}_{4} \\ \ddot{p}_{5} \end{matrix} \right \} + \frac{EI}{L^{3}} \left [ \begin{matrix} 4 & 6 & 0 & 0 & 0 \\ 6 & 12 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{3}{4} & 1 \\ 0 & 0 & \frac{3}{4} & \frac{3}{2} & \frac{9}{4} \\ 0 & 0 & 1 & \frac{9}{4} & \frac{18}{5} \end{matrix} \right ] \left \{ \begin{matrix} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \\ p_{5} \end{matrix} \right \} = 0 (O)These are still two separate sets of equations, and represent the two substructures not yet joined together.
The constraint equations are now derived by observing that at the junction, where x_{1} = L \mathrm{and} x_{2}=2L , when the substructures are joined, the displacements and slopes must be the same for the two beams, i.e.,
y_{1}(L) = y_{2}(2L) (P_{1})\\ \frac{dy_{1}}{dx_{1}}(L) = – \frac{dy_{2}}{dx_{2}}(2L) (P_{2})Note: In Eq. (P_{2}), the negative sign is due to the fact that x_{1} \mathrm{and} x_{2} are defined in opposite directions, affecting slopes but not displacements.
Substituting x_{1} = L \mathrm{and} x_{2}=2L into Eqs (A), (B), (C) and (E) gives
Combining Eqs (P_{1}),(P_{2}),(Q_{1}),(Q_{2}),(R_{1}) \mathrm{and} (R_{2}):
p_{1} + p_{2} – p_{3} – p_{4} – p_{5} = 0 (S_{1})\\ 2p_{1} + 3p_{2} + p_{3} + \frac{3}{2}p_{4} + 2p_{5} = 0 (S_{2})or in matrix form, corresponding to Eq. (8.60):
\left[A\right] \left\{p\right\} = 0 (8.60) \\ \left [ \begin{matrix} 1 & 1 & -1 & -1 & -1 \\ 2 & 3 & 1 & \frac{3}{2} & 2 \end{matrix} \right ] \left \{ \begin{matrix} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \\ p_{5} \end{matrix} \right \} = 0 (T)Equation (T) is now partitioned in the same way as Eq. (8.61). Since there are five modal coordinates, and two constraint equations, the number of independent coordinates, q_{i}, that will finally represent the complete system, is three. These can be any three of the five modes in the system, so choosing:
\left[\begin{array}{r :c}A_{1}& A_{2}\end{array}\right] \begin{Bmatrix}p_{d}\\ \hdashline p_{\mathrm{f}}\end{Bmatrix} = 0 (8.61)\\ q_{1} = p_{3}, q_{2}=p_{4}, q_{3}=p_{5},then:
\left[\begin{array}{r :c}A_{1}& A_{2}\end{array}\right] \left\{p\right\} = \left[\begin{array}{cc:ccc}1 & 1 & -1 & -1 & -1 \\2 & 3 & 1 & \frac{3}{2} & 2\end{array}\right] \left \{ \begin{matrix} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \\ p_{5} \end{matrix} \right \} = 0 (U)Now applying Eq. (8.64):
\begin{Bmatrix}p_{d}\\ \hdashline p_{\mathrm{f}}\end{Bmatrix} = \begin{bmatrix}-\left[A_{1}\right]^{-1} \left[A_{2}\right] \\ \hdashline I\end{bmatrix} \left\{p \right\}_{\mathrm{f}} (8.64) \\ \begin{Bmatrix}p_{d}\\ \hdashline p_{\mathrm{f}}\end{Bmatrix} = \begin{bmatrix}-\left[A_{1}\right]^{-1} \left[A_{2}\right] \\ \hdashline I\end{bmatrix} \left\{q \right\} (V)or numerically, noting that the order of the terms in {p} was not changed in this case,
The two separate subsystems in Eq. (O) are now joined together, by applying the transformation \left\{p\right\} = \left[\beta \right] \left\{q \right\}. Eq. (O) then becomes
\left[\hat{M} \right] \left\{\ddot{q} \right\} + \left[\hat{K} \right] \left\{q \right\} = 0 (Y)where, from Eq. (8.68):
and from Eq. (8.69):
Thus Eq. (Y), the equation of motion of the joined subsystems, expressed in numerical form is
\mu L \begin{bmatrix} 0.8857 & 0.8500 & 0.8333 \\ 0.8500 & 0.8357 & 0.8333 \\ 0.8333 & 0.8333 & 0.8413 \end{bmatrix} \left \{ \begin{matrix} \ddot{q} _{1} \\ \ddot{q}_{2} \\ \ddot{q}_{3} \end{matrix} \right \} + \frac{EI}{L^{3}} \begin{bmatrix} 28.50 & 33.75 & 39.00 \\ 33.75 & 40.50 & 47.25 \\ 39.00 & 47.25 & 55.60 \end{bmatrix} \left \{ \begin{matrix} q_{1} \\ q_{2} \\ q_{3} \end{matrix} \right \} = 0 (a)To find the natural frequencies and normal modes, since there is no damping, we may substitute \left\{q\right\} = \left\{\bar{q}\right\} e^{\mathrm{i}\omega t} \mathrm{and} \left\{\ddot{q} \right\} = -\omega^{2} \left\{\bar{q}\right\} e^{\mathrm{i}\omega t} in the usual way. Equation (a) can then be written as:
Using a standard eigenvalue/eigenvector program, solution of Eq. (b) gives the first two eigenvalues as \lambda_{1} = 6.206 \mathrm{and} \lambda_{2} = 49.44. The corresponding eigenvectors, arbitrarily scaled to make \bar{q}_{3} = 1, are
\left \{ \begin{matrix} \bar{q}_{1} \\ \bar{q}_{2} \\ \bar{q}_{3} \end{matrix} \right \}_{1} = \left \{ \begin{matrix} 2.371 \\ -3.103 \\ 1 \end{matrix} \right \} (d_{1}) \\ \mathrm{and} \\ \left \{ \begin{matrix} \bar{q}_{1} \\ \bar{q}_{2} \\ \bar{q}_{3} \end{matrix} \right \}_{2} = \left \{ \begin{matrix} 0.9267 \\ -1.978 \\ 1 \end{matrix} \right \} (d_{2})From Eq. (c), the two natural frequencies \omega_{1} \mathrm{and} \omega_{2} are
\omega_{1} = \frac{1}{L^{2}} \sqrt{\frac{EI}{\mu}} \sqrt{\lambda_{1}} = \frac{2.491}{L^{2}} \sqrt{\frac{EI}{\mu}} (e_{1}) \\ \mathrm{and} \\ \omega_{2} = \frac{1}{L^{2}} \sqrt{\frac{EI}{\mu}} \sqrt{\lambda_{2}} = \frac{7.031}{L^{2}} \sqrt{\frac{EI}{\mu}} (e_{2})The corresponding mode shapes can be plotted by substituting each eigenvector, from Eqs ((d_{1}) \mathrm{and} (d_{2}), in turn, into Eq. (X), to find the values of p_{1} \mathrm{to} p_{5}, for each mode. The local beam displacements, \mathrm{y}_{1} \mathrm{and} \mathrm{y}_{2}, are then given by Eqs (A) and (B). The two mode shapes, scaled to a maximum displacement of unity, are plotted in Fig. 8.3
Since the joined cantilevers form a fixed–fixed uniform beam of length 3L, it is possible to compare the results above with exact answers from Table 8.1. The first two normal mode frequencies are given by \beta_{1} (3L) = 4.7300 \mathrm{and} \beta_{2} (3L) = 7.8532.
Then the exact natural frequencies for a fixed–fixed uniform beam of length 3L are
The natural frequencies, \omega_{1} \mathrm{and} \omega_{2}, calculated by the component mode synthesis method, and given by Eqs (e_{1}) \mathrm{and} (e_{2}), are 0.24 and 2.6% higher, respectively, than the exact values given by Eqs (\mathrm{f}_{1}) \mathrm{and} (\mathrm{f}_{2}).
From Fig. 8.3, the mode shapes are seen to approximate to the first two modes of a uniform fixed–fixed beam.
| Table 8.1 Natural Frequencies and Mode Shapes for Uniform Beams in Bending | |||||
| End Conditions | Characteristic equation and roots \beta_{i}L | A | B | C | D |
| Simply-supported (Pinned-pinned) | \sin \beta_{i}L = 0 \\ \beta_{1}L = \pi \\ \beta_{2}L = 2\pi \\ \beta_{3}L = 3\pi \\ \beta_{4}L = 4\pi | 1 | 0 | 0 | 0 |
| Free-free | \cos \beta_{i}L \cdot \cosh \beta_{i}L = 1 \\ \beta_{1}L = 4.73004 \\ \beta_{2}L = 7.85321 \\ \beta_{3}L = 10.99561 \\ \beta_{4}L = 14.13717 | 1 | \frac{\sin \beta_{i}L – \sinh \beta_{i}L}{\cosh \beta_{i}L – \cos \beta_{i}L} | 1 | \frac{\sin \beta_{i}L – \sinh \beta_{i}L}{\cosh \beta_{i}L – \cos \beta_{i}L} |
| Fixed-fixed | \cos \beta_{i}L \cdot \cosh \beta_{i}L = 1 \\ \beta_{1}L = 4.73004 \\ \beta_{2}L = 7.85321 \\ \beta_{3}L = 10.99561 \\ \beta_{4}L = 14.13717 | -1 | -\frac{\sinh \beta_{i}L – \sin \beta_{i}L}{\cos \beta_{i}L – \cosh \beta_{i}L} | 1 | \frac{\sinh \beta_{i}L – \sin \beta_{i}L}{\cos \beta_{i}L – \cosh \beta_{i}L} |
| Cantilever (Fixed-free) (x is measured from the fixed end) | \cos \beta_{i}L \cdot \cosh \beta_{i}L = -1 \\ \beta_{1}L = 1.87510 \\ \beta_{2}L = 4.69409 \\ \beta_{3}L = 7.85475 \\ \beta_{4}L = 10.99554 | 1 | -\frac{\sin \beta_{i}L + \sinh \beta_{i}L}{\cos \beta_{i}L + \cosh \beta_{i}L} | -1 | \frac{\sin \beta_{i}L + \sinh \beta_{i}L}{\cos \beta_{i}L + \cosh \beta_{i}L} |
| Fixed-pinned (x is measured from the fixed end) | \tan \beta_{i}L – \tanh \beta_{i}L = 0 \\ \beta_{1}L = 3.92660 \\ \beta_{2}L = 7.06858 \\ \beta_{3}L = 10.21017 \\ \beta_{4}L = 13.35177 | 1 | -\frac{\sin \beta_{i}L – \sinh \beta_{i}L}{\cos \beta_{i}L – \cosh \beta_{i}L} | -1 | \frac{\sin \beta_{i}L – \sinh \beta_{i}L}{\cos \beta_{i}L – \cosh \beta_{i}L} |
