Question 8.4: Use a finite element model to find the first two natural fre......
Use a finite element model to find the first two natural frequencies and normal modes of the uniform cantilever beam shown in Fig. 8.9(a), and compare the result with the exact answer. Divide the beam into three bending elements each of length L = 1 meter. Take Young’s modulus = E; second moment of area of the cross-section = I; mass per unit length = μ. The axial displacement of the beam may be assumed to be negligible.
The equations of motion, in local axes, for each of the unconnected elements, A, B, and C, as shown in Fig. 8.9(b), are given by Eq. (8.159).
\frac{\mu L}{420} \left [ \begin{matrix} 156 & 22L & 54 & -13L \\ 22L & 4L^{2} & 13L & -3L^{2} \\ 54 & 13L & 156 & -22L \\ -13L & -3L^{2} & -22L & 4L^{2} \end{matrix} \right ] \left \{ \begin{matrix} \ddot{z}_{a} \\ \ddot{\phi}_{a} \\ \ddot{z}_{b} \\ \ddot{\phi}_{b} \end{matrix} \right \} \\+ \frac{EI}{L^{3}} \left [ \begin{matrix} 12 & 6L & -12 & 6L \\ 6L & 4L^{2} & -6L & 2L^{2} \\ -12 & -6L & 12 & -6L \\ 6L & 2L^{2} & -6L & 4L^{2} \end{matrix} \right ] \left \{ \begin{matrix} z_{a} \\ {\phi}_{a} \\ {z}_{b} \\ {\phi}_{b} \end{matrix} \right \} = \left \{ \begin{matrix} F_{a} \\ M_{a} \\ F_{b} \\ M_{b} \end{matrix} \right \} (8.159)Since the global axes are parallel to the local axes, a transformation matrix is not required, and the relationship between global and local displacement coordinates is
\overline{z}_{0} = z^{A}_{a}; \overline{z}_{1} = z^{A}_{b} = z^{B}_{a}; \overline{z}_{2} = z^{B}_{b} = z^{C}_{a}; \overline{z}_{3} = z^{C}_{b}; \\ \overline{\phi}_{0} = \phi^{A}_{a}; \overline{\phi}_{1} = \phi^{A}_{b} = \phi^{B}_{a}; \overline{\phi}_{2} = \phi^{B}_{b} = \phi^{C}_{a}; \overline{\phi}_{3} = \phi^{C}_{b} (A)where the bars indicate global coordinates. Superscripts A, B, C indicate the element, and subscripts a, b indicate the end ( i.e. left or right) in Fig. 8.9(b).
The equations for the three elements, joined at nodes 1 and 2, but still free at the left and right ends, nodes 0 and 3, are formed by superimposing the terms of the mass and stiffness matrices at the joints. With L = 1 for all three elements, the mass and stiffness matrices, \left[\overline{m} \right] \mathrm{and} \left[\overline{k} \right], are then given by:
Since we require the beam be a cantilever, fixed at node 0, then the displacements \overline{z} _{0} \mathrm{and} \overline{\phi } _{0}, and the corresponding accelerations, at that node, are zero. The first two columns of the mass and stiffness matrices above can therefore be eliminated, since they would always be multiplied by zero. If, as in this case, the force and moment at the fixed end are not required, the first two rows of the mass and stiffness matrices can also be eliminated. The equations of motion of the uniform, cantilevered beam, in global coordinates, are then:
\frac{\mu }{420} \left [ \begin{matrix} 312 & 0 & 54 & -13 & 0&0 \\ 0 & 8 & 13 & -3 & 0 & 0 \\ 54 & 13 & 312 & 0 & 54 & -13 \\ -13 & -3 & 0 &8 & 13 & -3 \\ 0&0 & 54 & 13 & 156 & -22 \\ 0&0 &-13 & -3 & -22 & 4 \end{matrix} \right ] \left \{ \begin{matrix} \overline{\ddot{z}}_{1} \\ \overline{\ddot{\phi}}_{1} \\ \overline{\ddot{z}}_{2} \\ \overline{\ddot{\phi}}_{2} \\ \overline{\ddot{z}}_{3} \\ \overline{\ddot{\phi}}_{3} \end{matrix} \right \} \\ + EI \left [ \begin{matrix} 24 & 0 & -12 & 6 & 0&0 \\ 0 & 8 & -6 & 2 & 0 & 0 \\ -12 & -6 & 24 & 0 & -12 & 6 \\ 6 & 2 & 0 &8 & -6 & 2 \\ 0&0 & -12 & -6 & 12 & -6 \\ 0&0 &6 & 2 & -6 & 4 \end{matrix} \right ] \left \{ \begin{matrix} \overline{{z}}_{1} \\ \overline{{\phi}}_{1} \\ \overline{{z}}_{2} \\ \overline{{\phi}}_{2} \\ \overline{{z}}_{3} \\ \overline{{\phi}}_{3} \end{matrix} \right \} = \left \{ \begin{matrix} \overline{F}_{1} \\ \overline{M}_{2} \\ \overline{F}_{2} \\ \overline{M}_{2} \\ \overline{F}_{3} \\ \overline{M}_{3} \end{matrix} \right \} (D)\\ \mathrm {or in shorter form:} \\ \left[\overline{m} \right] \left\{\overline{\ddot{z} } \right\} + \left[\overline{k} \right] \left\{\overline{{z}} \right\} = \left\{\overline{F} \right\} (E)In this case, only the natural frequencies and mode shapes of the undamped system
are required, so the external forces and moments are set to zero, and the usual
substitution \left\{\overline{z} \right\} = e^{\mathrm{i}\omega t} \left\{\underline{\overline{z} } \right\} is made, leading to:
Using Eq. (D), Eq. (F) can be written as:
Using standard software gives the first two eigenvalues and eigenvectors, as follows:
\lambda _{1} = 0.3694 \times 10^{-3} \lambda _{2}= 0.01437 \\ \left \{ \begin{matrix} \overline{\underline{z} }_{1} \\ \overline{\underline{{\phi}} }_{1} \\ \overline{\underline{{z}} }_{2} \\ \overline{\underline{{\phi}} }_{2} \\ \overline{\underline{{z}} }_{3} \\ \overline{\underline{{\phi}} }_{3} \end{matrix} \right \}_{1} = \left \{ \begin{matrix} 0.1664 \\ 0.3118 \\ 0.5472 \\ 0.4439 \\ 1.0000 \\ 0.4470 \end{matrix} \right \} \left \{ \begin{matrix} \overline{\underline{z} }_{1} \\ \overline{\underline{{\phi}} }_{1} \\ \overline{\underline{{z}} }_{2} \\ \overline{\underline{{\phi}} }_{2} \\ \overline{\underline{{z}} }_{3} \\ \overline{\underline{{\phi}} }_{3} \end{matrix} \right \}_{2} = \left \{ \begin{matrix} -0.5895 \\ -0.5960 \\ -0.4219 \\ 0.9785 \\ 1.0000 \\ 1.6031 \end{matrix} \right \} (I)where the eigenvectors have been scaled to make the tip displacement \overline{\underline{{z}} }_{3} equal to unity in each case.
The natural frequencies are given by Eq. (H):
\omega _{i} = \sqrt{420\lambda _{i}} \sqrt{\frac{EI}{\mu } } (J)For the first mode, \lambda _{1} = 0.3694 \times 10^{-3}, and
\omega _{1} = 0.3939 \sqrt{\frac{EI}{\mu } } (K_{1})For the second mode,
\lambda _{2} = 0.01437 \mathrm{and} \omega _{2} = 2.456 \sqrt{\frac{EI}{\mu } } (K_{2})For comparison, the exact natural frequencies of a uniform cantilever are given by Table 8.1, as follows. From the table:
\beta _{1}L = 1.87510 \beta _{2}L = 4.69404 \omega _{i} =\beta _{i}^{2}\sqrt{\frac{EI}{\mu } }Noting that L = 3 m in this example,
\omega _{1} = \beta _{1}^{2}\sqrt{\frac{EI}{\mu } } = \left(\frac{1.87510}{3} \right) ^{2}\sqrt{\frac{EI}{\mu } } = 0.39066 \sqrt{\frac{EI}{\mu } } (L_{1})\\ \omega _{2} = \beta _{2}^{2}\sqrt{\frac{EI}{\mu } } = \left(\frac{4.69404}{3} \right) ^{2}\sqrt{\frac{EI}{\mu } } = 2.4482 \sqrt{\frac{EI}{\mu } } (L_{2})It can be seen that the first and second natural frequencies from the FE model are about 0.8 and 0.4% higher, respectively, than the exact results, in this case.
The global displacements in each mode are compared with the exact mode shapes of the uniform beam in Fig. 8.10. In both cases the displacements are given as a fraction of the tip displacement, z/z_{\mathrm{TIP}}, and the spanwise positions are given as a fraction of the total span. The differences are too small to show at the scale used, being all less than 0.5%.
| 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} |
