Question 18.6: The plane frame shown in Figure E18.6a is modeled as a finit...
The plane frame shown in Figure E18.6a is modeled as a finite element system consisting of three elements, BA, BC, and CD. Columns BA and CD each have a length L, sectional moment of inertia I, and mass per unit length m. Beam BC is of length 2 L, moment of inertia 2 I, and mass per unit length 2 m. The elements may be considered as axially rigid. Using a consistent mass formulation, obtain the mode shapes and frequencies of vibration.
As shown in Figure E18.6a, the frame has three degrees of freedom, rotations at Nodes B and \mathrm{C}, and lateral translation along BC. The stiffness and mass matrices for individual elements corresponding to their local degrees of freedom, shown in Figure E18.6b, are obtained from Equations 18.35 and 18.39.
\begin{aligned}\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) \\\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)\end{aligned}
In addition the mass of beam BC contributes inertia force along the lateral translation degree of freedom u_{3}. The correspondence between the local and global degrees of freedom is shown in Table E18.6.
The element stiffness matrices are assembled in accordance with the correspondence between the local and global degrees of freedom to obtain the following global stiffness matrix
\mathbf{K}=E I\left[\begin{array}{ccc}\frac{8}{L} & \frac{2}{L} & \frac{6}{L^{2}} \\\frac{2}{L} & \frac{8}{L} & \frac{6}{L^{2}} \\\frac{6}{L^{2}} & \frac{6}{L^{2}} & \frac{24}{L^{3}}\end{array}\right] \quad \text { (a) }
Table E18.6 Correspondence between the local and global degrees of freedom in the frame.
\begin{array}{llcll}\hline \text{Local dof} & 1 & 2 & 3 & 4 \\\hline & \text{Global dof} & \\\hline \text{Element }BA & 1 & 3 & – & – \\\text{Element} BC & 1 & – & 2 & – \\\text{Element }CD & 2 & 3 & – & – \\\hline\end{array}
The global mass matrix is obtained in a similar manner. However, the term 4 m L must be added in the position (3,3) to account for the inertia contributed by the vibration of beam BC as a rigid body along degree of freedom 3 . The resulting mass matrix is given by
\mathbf{M}=m L\left[\begin{array}{rrl}0.1619 L^{2} & -0.1143 L^{2} & 0.0524 L \\-0.1143 L^{2} & 0.1619 L^{2} & 0.0524 L \\0.0524 L & 0.0524 L & 4.7429\end{array}\right] \qquad (b)
Since the stiffness and mass matrices carry terms with differing powers of the variable L, a numerical solution of the eigenvalue problem is difficult. This can be resolved if all displacements along all of the coordinates are measured in the same units. We carry out a coordinate transformation in which the rotations \theta_{1} and \theta_{2} are expressed, respectively, as u_{1} / L and u_{2} / L, where coordinates u_{1} and u_{2} are measured in units of length. The transformation is expressed as
\left\{\begin{array}{l}\theta_{1} \\\theta_{2} \\u_{3}\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{L} & 0 & 0 \\0 & \frac{1}{L} & 0 \\0 & 0 & 1\end{array}\right]\left\{\begin{array}{l}u_{1} \\u_{2} \\u_{3}\end{array}\right\}=\mathrm{Tu} \quad \text { (c) }
The transformed matrices are obtained from Equations 3.80 and 3.81
\begin{aligned}& \widetilde{\mathbf{K}}=\frac{E I}{L^{3}}\left[\begin{array}{rrr}8 & 2 & 6 \\2 & 8 & 6 \\6 & 6 & 24\end{array}\right] \\& \tilde{\mathbf{M}}=m L\left[\begin{array}{rrr}0.1619 & -0.1142 & 0.0524 \\-0.1143 & 0.1619 & 0.0524 \\0.0524 & 0.0524 & 4.7429\end{array}\right]\end{aligned} \quad \text { (d) }
Standard eigenvalue solution using the transformed matrices gives the following frequencies and mode shapes. Frequencies as multiples of \sqrt{\frac{E I}{m L^{4}}} are: \omega_{1}=1.9004, \omega_{2}=4.6609, \omega_{3}=14.5293. Mode shapes in coordinates u_{1}, u_{2}, and u_{3} are
\begin{aligned}\phi_{1}^{T} &=\left[\begin{array}{lll}-0.2741 & -0.2741 & 0.4636\end{array}\right] \\\phi_{2}^{T} &=\left[\begin{array}{lll}-1.3455 & 1.3455 & 0\end{array}\right] \\\phi_{3}^{T} &=\left[\begin{array}{lll}-3.2690 & -3.2690 & 0.0338\end{array}\right]\end{aligned}
If a lumped mass matrix is used, the only non-zero term in the mass matrix will be along the translational degree of freedom; its value being 2 \times \frac{m L}{2}+2 m \times 2 L=5 m L. An eigenvalue solution with the lumped mass matrix gives
\omega=1.833 \sqrt{\frac{E I}{m L^{4}}} \quad \phi^{T}=\left[\begin{array}{lll}-0.6 & -0.6 & 1.0\end{array}\right]