Question 4.5.2: Consider a 2-dof system whose matrix equation of motion is g......
Consider a 2-dof system whose matrix equation of motion is given as
\begin{bmatrix} m_{11} & 0 \\ 0 & m_{22} \end{bmatrix}\begin{pmatrix} \ddot{x} _{1} \\ \ddot{x}_{2} \end{pmatrix} + \begin{bmatrix} k_{11} & k_{12} \\ k_{21} & k_{22} \end{bmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix} , (i)
where m_{11}=m,m_{22}=m r^{2},\,k_{11}=k_{1}+k_{2},\,k_{12}=k_{21}=k_{2}L_{2}-k_{1}L_{1},\;k_{22}=k_{2}L_{2}^{2}+k_{1}L_{1}^{2},\,x_{1}=x is the
translational displacement, and x_{2} = θ is the angular or rotational displacement.
If m = 4000 kg, k_{1}=k_{2}=20,000\,\mathrm{N/m},\,r=0.8\,\mathrm{m},\,m_{22}=m r^{2}=2560\,\mathrm{kg\,m^{2}},\,L_{1}=1.4\,\mathrm{m},\,\mathrm{and}\,L_{2}=0.9 m, find the natural frequencies and normal modes of the 2-dof system.
a. Since the system is undamped and therefore the responses have no delay or phase difference with respect to the applied force, one can write
{\binom{x_{1}}{x_{2}}}={\binom{X_{1}}{X_{2}}}{\cos{\omega{t}}}={\binom{X}{\Theta}}\cos\omega t.
In this equation, X and Θ are the amplitudes of instantaneous displacements x_{i}.
Substituting the above solution and its derivatives into Equation (i) and upon simplification, one has
\begin{bmatrix}(k_{11}-m_{11}\omega^{2}) & k_{12} \\ k_{21} & (k_{22}-m_{22}\omega^{2}) \end{bmatrix}{\binom{X}{\Theta}}={\binom{0}{0}} (ii)
or writing it in a more concise form
\left[Z(\omega)\right]\left(_{X_{2}}^{X_{1}}\right)={\binom{0}{0}}
where the so-called impedance matrix has been defined by Equation (4.18) as
\begin{bmatrix}(k_{11} – m_{1}\omega^{2}) & k_{12} \\ k_{21} & (k_{22} – m_{2} \omega^{2}) \end{bmatrix}{\binom{X_{1}}{X_{2}}}={\binom{F_{1}}{0}} (4.18)
[Z(\omega)]= \begin{bmatrix}(k_{11} – m_{11} \omega^{2}) & k_{12} \\ k_{21} & (k_{22} – m_{22} \omega^{2}) \end{bmatrix}
The frequency equation is
\left|{\begin{matrix}{(k_{11}-m_{11}\omega^{2})}&{k_{12}}\\ {k_{21}}&{{}(k_{22}-m_{22}\omega^{2})}\end{matrix}}\right|=0.
Substituting the system parameters given above and writing λ = ω²,
\left|{\begin{array}{ll}40,000 – 4000\lambda&10,000\\ 10,000&55,400−2560λ\end{array}}\right|=0.
Operating on this equation, one has
(40,000-4000\lambda)(55,400-2560\lambda)-(10,000)^{2}=0.
This quadratic equation in λ gives \lambda_{1}=9.2173,\,\lambda_{2}=59.8457, so that the two natural frequencies are
\omega_{1}={\sqrt{9.2173}}=3.036\,\mathrm{rad/s}, (iia)
\omega_{2}={\sqrt{59.8457}}=4.736{\frac{\mathrm{rad}}{s}} (iib)
In order to obtain the mode shapes, one first substitutes \omega_{1} into Equation (ii), resulting in
\left({\frac{X}{\Theta}}\right)_{(1)}=-3.18\,\mathrm{m/rad} (iva)
where the subscript (1) on the lhs of the equation denotes the amplitude ratio associated with the first natural frequency.
Similarly, substituting {{\omega}}_{2} into Equation (ii), one has
\left({\frac{X}{\Theta}}\right)_{(2)}=0.202\,\mathrm{m/rad}. (ivb)
Equations (iva) and (ivb) give the two mode shapes of the 2-dof system.