Question 18.5: The simply supported beam of length L shown in Figure E18.5 ...
The simply supported beam of length L shown in Figure \mathrm{E} 18.5 is vibrating in the lateral direction. The beam is of uniform section, has flexural rigidity E I and mass per unit length m. It is divided into three finite element as shown in the figure, which also shows the global degrees of freedom. Obtain the first six vibration frequencies of the beam using (a) consistent mass formulation, (b) the mass matrix given by Equation 18.42,
\mathbf{M}^i=m L\left[\begin{array}{cccc}0 & 0 & 0 & 0 \\0 & \frac{1}{3} & 0 & \frac{1}{6} \\0 & 0 & 0 & 0 \\0 & \frac{1}{6} & 0 & \frac{1}{3}\end{array}\right] \qquad (18.42)
and (c) the lumped mass matrix given by Equation 18.44.
\mathbf{M}^i=m L\left[\begin{array}{llll}0 & 0 & 0 & 0 \\0 & \frac{1}{2} & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & \frac{1}{2}\end{array}\right] \qquad (18.44)
Compare your answers with the exact values given by Equation 15.59.
\omega_n=\beta_n^2 \sqrt{\frac{E I}{m}}=n^2 \pi^2 \sqrt{\frac{E I}{m L^4}} (15.59)
The stiffness matrix for element \mathrm{BC} is obtained from Equation 18.35
\mathbf{K}^i =\left[\begin{array}{cccc}\frac{4 E I}{L} & \frac{6 E I}{L^2} & \frac{2 E I}{L} & -\frac{6 E I}{L^2} \\\frac{6 E I}{L^2} & \frac{12 E I}{L^3} & \frac{6 E I}{L^2} & -\frac{12 E I}{L^3} \\\frac{2 E I}{L} & \frac{6 E I}{L^2} & \frac{4 E I}{L} & -\frac{6 E I}{L^2} \\-\frac{6 E I}{L^2} & -\frac{12 E I}{L^3} & -\frac{6 E I}{L^2} & \frac{12 E I}{L^3}\end{array}\right] \qquad (18.35)</p> </div> <p>
and is given by
\mathrm{K}_{BC} = \frac{EI}{l^3} \quad \begin{matrix} & \begin{array}{c} (3) & (4) & (5) & (6)\end{array}\\ \begin{array}{c} (3) \\ (4) \\ (5) \\ (6)\end{array} &\left(\begin{array}{cccc}4 l^2 & 6 l & 2 l^2 & -6 l \\6 l & 12 & 6 l & -12 \\2 l^2 & 6 l & 4 l^2 & -6 l \\-6 l & -12 & -6 l & 12\end{array}\right) \end{matrix} \qquad (a)
where l=L / 3.
In Equation a, the positions that individual elements of the given matrix will occupy after assembly in the global stiffness matrix have been indicated by specifying the global matrix row and column numbers that correspond to the rows and columns of the component matrix. This information is easily obtained by noting the correspondence between the element and global degrees of freedom. For example, in Equation a, element k_{23}=6 E I / l^{2} will be placed in row 4 and column 5 of the 8 by 8 global matrix. The stiffness matrices for elements \mathrm{AB} and \mathrm{CD} are identical to that for element BC. The correspondence numbers that relate the local degrees of freedom to the global degrees of freedom for all of the elements are shown in Table E18.5a.
On assembling the element matrices we get the following global stiffness matrix
Table EI8.5a Correspondence between the local and global degrees of freedom.
\begin{array}{lllll}\hline \text{Local dof} & 1 & 2 & 3 & 4 \\\hline & &{\text{ Global dof }} \\\hline \text{ Element} AB & 1 & 2 & 3 & 4 \\\text{Element} BC & 3 & 4 & 5 & 6 \\\text{Element} CD & 5 & 6 & 7 & 8 \\\hline\end{array}
We now apply the boundary conditions, which require that the lateral displacements along degrees of freedom 2 and 8 should be zero. This is achieved by deleting rows 2 and 8 and columns 2 and 8 from the matrix in Equation b. This gives
The consistent mass matrix for element \mathrm{BC}, as obtained from Equation 18.39,
\mathbf{M}^i =\frac{m L}{420}\left[\begin{array}{cccc}4 L^2 & 22 L & -3 L^2 & 13 L \\22 L & 156 & -13 L & 54 \\-3 L^2 & -13 L & 4 L^2 & -22 L \\13 L & 54 & -22 L & 156\end{array}\right] \qquad (18.39)
is given by
\mathbf{M}_{B C}=\frac{m l}{420}\begin{array}{c}& \begin{array}{c} (3)~ & (4)~ & ~(5) & ~(6) \end{array}\\ \begin{array}{c}(3) \\ (4) \\ (5) \\ (6) \end{array}&\left( \begin{array}{c} 4 l^{2} & 22 l & -3 l^{2} & 13 l \\22 l & 156 & -13 l & 54 \\-3 l^{2} & -13 l & 4 l^{2} & -22 l \\13 l & 54 & -22 l & 156 \end{array}\right) \end{array} \text { (d) }
The consistent mass matrices for elements \mathrm{AB} and \mathrm{CD} are identical, except for the correspondence numbers. On assembling the element mass matrices and deleting rows and columns 2 and 8 , we get the following global mass matrix
Using the stiffness and mass matrices from Equation c and e, respectively, the six frequencies are obtained from a Matlab subroutine and are shown in Table E18.4b. The tabulated value is in fact the coefficient \gamma_{n} in the frequency expression \omega_{n}=\gamma_{n} \sqrt{E I / m L^{4}}. The table also shows the frequencies obtained by dividing the beam into 1,2,3,4, and 5 elements of equal length. For the purpose of comparison the exact values of the frequency coefficient \gamma_{n}=n^{2} \pi^{2} are also shown.
The results presented in Table E18.5b show that for a given number of elements the frequency estimates deteriorate as the mode number increases. In general, the number of modes for which the estimate is reasonably close to the exact value is equal to the number of elements. The estimates improve with an increase in the number of elements. Also, in all cases the finite
Table E18.5b Frequencies of a uniform simply supported beam obtained from consistent mass finite element solution and the exact frequencies.
Table E18.5c Frequencies of a uniform simply supported beam obtained from concentrated mass finite element solution and the exact frequencies.
element estimates are larger than the exact values. This is a property of the Ritz method that uses a consistent mass formulation.
Part (b)
The element mass is in this case concentrated along the lateral degrees of freedom and the element mass matrices are given by Equation18.42. After assembling them and deleting rows and columns 2 and 8 of the assembled matrix we get the following global mass matrix
\mathbf{M}=m l\left[\begin{array}{cccccc}0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & \frac{2}{3} & 0 & \frac{1}{6} & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & \frac{1}{6} & 0 & \frac{2}{3} & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{array}\right] \qquad (f)
Frequency estimates obtained from the stiffness matrix given by Equation (c) and the mass matrix given by Equation (f) are shown in Table E18.5c. The estimates again converge from above to the exact solution, even though the formulation is not a true Ritz solution. However, for the same number of elements, the estimates are much less accurate than those obtained from a consistent mass formulation. Also, when only one element is used no estimate is obtained, because in this case all of the mass is assigned to the restrained degrees of freedom. Thus, even with 5 elements frequency estimates are obtained for only the first four frequency.
Table E18.5d Frequencies of a uniform simply supported beam obtained from lumped mass finite element solution and the exact frequencies.
Part (c)
The global lumped mass matrix is given by
The frequency estimates obtained from the stiffness matrix of Equation (c) and the mass matrix of Equation (\mathrm{g}) are shown in Table E18.5d. As in the concentrated mass solution of Part b, no estimate is obtained from just one element and 5 elements provide only 4 frequencies. However, unlike the case of concentrated mass solution, the frequencies converge to the true solution from below. Also, the estimates are much better than those obtained inn Part (b) although not as good as those obtained from the consistent mass formulation. It may, however, be noted that the computational effort required in finding the frequencies is significantly smaller than that in the consistent mass formulation. This is because the size of the problem can be substantially reduced by static condensation of the stiffness matrix to eliminate those degrees of freedom which have no mass assigned to them, and also because the mass matrix is diagonal.