Question 19.3: The flexural beam shown in Figure E19.3a has negligible mass...
The flexural beam shown in Figure E19.3a has negligible mass, is rigidly supported against translation at its left hand end, and supported by a spring of stiffness \alpha k at its right-hand end. The beam supports a mass 2 m at mid-span and another mass m at the right-hand end. The flexural stiffness of the beam is represented by k=48 \mathrm{EI} / \mathrm{L}^{3}. Analytically derive the frequency response functions H_{f f}, H_{s f}, and H_{s s}, where f refers to the free d.o.f 1 , and s refers to the support d.o.f. 2, as shown in the figure. Now compute the function \tilde{H}_{f f} from Equation 19.69; compare the result with that obtained directly for rigid vertical supports at both ends of the beam. The damping is 5% in each mode. Given α = 4.
The flexibility matrix corresponding to the two d.o.f. is given by
\begin{aligned}\mathbf{A} &=\left[\begin{array}{cc}(1 / k)+(1 / 4 \alpha k) & (1 / 2 \alpha k) \\(1 / 2 \alpha k) & (1 / \alpha k)\end{array}\right] \\&=\frac{1}{k}\left[\begin{array}{cc}1.0625 & 0.1250 \\0.1250 & 0.2500\end{array}\right]\end{aligned} \text { (a) }
Hence, the stiffness matrix is
\mathbf{K}=\mathbf{A}^{-1}=k\left[\begin{array}{rr}1.00 & -0.50 \\-0.50 & 4.25\end{array}\right] \text { (b) }
The mass matrix is
\mathbf{M}=m\left[\begin{array}{ll}2 & 0 \\0 & 1\end{array}\right] \quad \text { (c) }
The two frequencies of the beam are
\omega_{1}=0.683 \sqrt{k / m} \quad \omega_{2}=2.07 \sqrt{k / m} \text { (d) }
A Rayleigh damping matrix is used to provide 5 \% damping in each of the two modes, so that
\mathbf{C}=\alpha_{0} \mathbf{M}+\alpha_{1} \mathbf{K} \quad \text { (e) }
where \mathrm{C} is the damping matrix and the coefficients \alpha_{0} and \alpha_{1} are obtained from the following with \xi=0.05
\begin{aligned}&\alpha_{0}=\frac{2 \xi \omega_{1} \omega_{2}}{\omega_{1}+\omega_{2}}=0.0514 \sqrt{\frac{k}{m}} \\&\alpha_{1}=\frac{2 \xi}{\omega_{1}+\omega_{2}}=0.0363 \sqrt{\frac{m}{k}}\end{aligned} \text { (f) }
Substituting the values of \alpha_{0} and \alpha_{1} in Equation (c) we get
\mathrm{C}=\sqrt{k m}\left[\begin{array}{rr}0.1391 & -0.0182 \\-0.0182 & 0.2058\end{array}\right](\mathrm{g})
Denoting the exciting frequency by \Omega we have
where \beta=\frac{\Omega}{\sqrt{k / m}}.
The frequency response functions are now obtained from
\begin{aligned}&\left\{\begin{array}{l}H_{f f} \\H_{s f}\end{array}\right\}=\mathbf{B}^{-1}\left\{\begin{array}{l}1 \\0\end{array}\right\} \\&\left\{\begin{array}{l}H_{f s} \\H_{s s}\end{array}\right\}=\mathbf{B}^{-1}\left\{\begin{array}{l}0 \\1\end{array}\right\}\end{aligned} \text { (i) }
The absolute values of the frequency response functions are plotted in Figure E19.3b as functions of \beta. Functions H_{s f} and H_{f s} are of course identical. It may be noted that in practice the frequency functions derived from Equation (i) will, in fact, be obtained through modal testing.
The frequency response function for rigid support condition is determined from Equation 19.69 using the response functions calculated from Equation (i). Its absolute value is plotted in Figure E19.3c. The response function for rigid support condition can also be directly evaluated from the first element of the matrix in Equation (h), so that
\tilde{H}=\frac{1}{k\left(-2 \beta^{2}+0.1391 i \beta+1\right)} \text { (j) }
The two sets of the results for \tilde{H} are identical.