Question 19.2: For the uniform bar of Example 19.1 obtain the reduced stiff...
For the uniform bar of Example 19.1 obtain the reduced stiffness and mass matrices of the two components using free interface normal modes. Couple the matrices by enforcing compatibility of displacement along the interface between components 1 and 2.
The stiffness and mass matrices for components 1 and 2 are given in Example 1. Component 1 has 4 free interface normal modes. We retain the first 2 of these modes in the free interface CMS. These modes are given by
\boldsymbol{\Phi}_{k}^{(1)}=\left[\begin{array}{rr}-0.2741 & 0.7330 \\-0.5065 & 0.5610 \\-0.6617 & -0.3036 \\-0.7163 & -0.7934\end{array}\right] \qquad (a)
The residual flexibility matrix is obtained from Equation 19.41,
\mathrm{G}_d=\mathrm{G}-\mathrm{G}_{\mathbf{k}}=\mathrm{G}-\boldsymbol{\Phi}_k \boldsymbol{\Lambda}_k^{-1} \boldsymbol{\Phi}_k^T \quad(19.41)
which gives
The residual flexibility attachment mode is given by
The transformation matrix is
\mathrm{T}^{(1)}=\left[\begin{array}{ll}\boldsymbol{\Phi}_{k} & \boldsymbol{\Psi}_{a}\end{array}\right]^{(1)} \qquad (d)
Transformation yields the following mass and stiffness matrices
Component 2 has 1 rigid body mode and 4 flexible body modes. We retain the first 2 of the latter. The 3 modes are as follows
All of the modes have been mass normalized, so that M_{r r}=1. The inertia relief projection matrix is obtained from Equation 19.32,
\begin{aligned}\mathbf{f}_f &=\mathbf{f}-\mathbf{M} \boldsymbol{\Phi}_r \ddot{\mathbf{p}}_r=\mathbf{f}-\mathbf{M} \boldsymbol{\Phi}_r\left(\tilde{\mathbf{M}}_{r r}\right)^{-1} \boldsymbol{\Phi}_r^T \mathbf{f} \\&=\left[\mathbf{I}-\mathbf{M} \boldsymbol{\Phi}_r\left(\tilde{\mathbf{M}}_{r r}\right)^{-1} \boldsymbol{\Phi}_r^T\right] \mathbf{f}=\mathbf{P}_{\mathrm{I}} \mathbf{f}\end{aligned} (19.32)
which gives
The single attachment mode will be obtained by applying a unit load at the interface d.o.f. labeled 4. We therefore restrain d.o.f. 5 to obtain the augmented matrix \mathrm{G}_{r}. The submatrix of the stiffness matrix corresponding to the d.o.f. 1 to 4 and its inverse are as follows
\mathbf{K}_{i i}=\left[\begin{array}{rrrr}2 & -1 & 0 & -1 \\-1 & 2 & -1 & 0 \\0 & -1 & 2 & 0 \\-1 & 0 & 0 & 1\end{array}\right] \quad \mathbf{K}_{i i}^{-1}=\left[\begin{array}{llll}3 & 2 & 1 & 3 \\2 & 2 & 1 & 2 \\1 & 1 & 1 & 1 \\3 & 2 & 1 & 4\end{array}\right] \qquad (h)
The augmented flexibility matrix is thus given by
\mathrm{G}_{r}^{(1)}=\left[\begin{array}{ccccc}3 & 2 & 1 & 3 & 0 \\2 & 2 & 1 & 2 & 0 \\1 & 1 & 1 & 1 & 0 \\3 & 2 & 1 & 4 & 0 \\0 & 0 & 0 & 0 & 0\end{array}\right] \qquad (i)
The inertia relief flexibility matrix obtained from Equation 19.36
\mathrm{G}_f=\mathbf{P}_{\mathrm{I}}^T \mathbf{G}_r \mathbf{P}_{\mathbf{I}} \quad(19.36)
is
The residual flexibility matrix obtained from Equation 19.43
\mathrm{G}_d=\mathrm{G}_f-\mathrm{G}_k(19.43)
and 19.41 is
\mathrm{G}_{d}^{(2)}=\left[\begin{array}{rrrrr}0.1357 & -0.0625 & -0.0107 & -0.1661 & 0.0411 \\-0.0625 & 0.0625 & -0.0625 & 0.0625 & 0.0625 \\-0.0107 & -0.0625 & 0.1357 & 0.0411 & -0.1661 \\-0.1661 & 0.0625 & 0.0411 & 0.2089 & -0.0839 \\0.0411 & 0.0625 & -0.1661 & -0.0839 & 0.2089\end{array}\right]\qquad (k)
The residual flexibility attachment mode is obtained on multiplying the above residual flexibility matrix by the load vector \mathrm{f}_{a}^{T} = [0 0 0 1 0] .The resulting vector is simply the fourth column of the residual flexibility matrix. Thus we have
\left(\boldsymbol{\Psi}_{a}^{T}\right)^{(2)}=\left[\begin{array}{lllll}-0.1661 & 0.0625 & 0.0411 & 0.2089 & -0.0839\end{array}\right] \qquad (l)
The transformation matrix is that given in Equation 19.44
and the transformed mass and stiffness matrices are
The number of generalized coordinates for components 1 and 2 are, respectively, 3 and 4 for a total of 7 . There is one constraint related to the compatibility of displacement at the interface between the two components. Thus there are 6 independent generalized coordinates. The relationship between the two sets of coordinates is given by the transformation matrix \mathbf{S} obtained from Equations 19.54
\mathbf{u}_a^{(1)}=\mathbf{u}_a^{(2)} \quad(19.54)
and 19.56.
From Equation (1) \Psi_{a a}^{(2)}=0.2089 and hence \Gamma=1 / 0.2089=4.7859. We therefore have
Coupling transformation matrix is obtained from Equation 19.57 which gives
The transformed system mass and stiffness matrices are given by
As would be expected the mass and stiffness matrices are both fully populated.
An eigenvalue solution using the foregoing system matrices gives the following frequencies.
\omega^{T}=\left[\begin{array}{llllll}1.5733 & 4.7808 & 8.1727 & 11.877 & 16.300 & 22.462\end{array}\right] \sqrt{\frac{E A}{m L^{2}}} \qquad (r)These may be compared with the values given in Example 19.1, and are fairly close to the exact frequencies for a clamped free uniform bar undergoing axial vibration.