Question 4.5.1: 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}=2m,\,m_{22}={\frac{2}{3}}m L^{2},\,k_{11}=2k,\,k_{12}=k_{21}=-{\frac{k L}{2}},\,k_{22}={\frac{25k L^{2}}{8}},\,x_{1}=x is the translational displacement, and x_{2} = θ is the angular or rotational displacement.
a. Determine the natural frequencies and mode shapes of the 2-dof system.
b. For the given system with 2m = 2000 kg, 2k = 40,000 N/m, and L = 1.0 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}}}s{\mathrm{in}}\omega t={\binom{X}{\Theta}}\sin\omega t.
In this equation, X and Θ are the amplitudes of instantaneous displacements x_{i}. and symbolically not to be confused with the Laplace transform of x_{i}.
Substituting the above solution and its derivatives into Equation (i) and simplifying, 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
[Z(\omega)]{\binom{X_{1}}{X_{2}}}={\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}
Of course, one can determine the natural frequencies of the system by making use of Equations (A2a) and (A2b) in Appendix 4A.
\omega_{1}^{2}={\frac{-b+{\sqrt{b^{2}-4a c}}}{2a}} (A2a)
\omega_{2}^{2}={\frac{-b-{\sqrt{b^{2}-4a c}}}{2a}} (A2b)
However, one can simply apply the frequency equation from Equation (ii); that is, equating the determinant of the impedance matrix [Z(ω)] to zero,
\left|{\begin{array}{ll}{k_{11}-m_{11}\omega^{2})}&{k_{12}}\\ {k_{21}}&{{}(k_{22}-m_{22}\omega^{2})}\end{array}}\right|=0.
Substituting the system parameters given above and writing λ = ω²
\left|{\begin{array}{ll}(2k− 2mλ) & -{\frac{k L}{2}} \\ -{\frac{k L}{2}} & \left({\frac{25k L^{2}}{8}}-{\frac{2}{3}}m L^{2}\lambda\right) \end{array}}\right| = 0
Operating on this equation, one has
(2k-2m\lambda)\biggl(\frac{25k L^{2}}{8}-\frac{2}{3}m L^{2}\lambda\biggr)-\biggl(\frac{k L}{2}\biggr)^{2}=0.
This quadratic equation in λ gives
\lambda_{1}=0.9498\,\frac{k}{m},~~~\lambda_{2}=4.7377\frac{k}{m}
which gives the two natural frequencies as
\omega_{1}={\sqrt{0.9498{\frac{k}{m}}}}=0.9746{\sqrt{\frac{k}{m}}}, (iiia)
\omega_{2}={\sqrt{4.7377{\frac{k}{m}}}}=2.1766{\sqrt{\frac{k}{m}}}. (iiib)
In order to obtain the mode shapes, one substitutes {\boldsymbol{\omega}}_{1} into Equation (ii), resulting in
\left({\frac{X}{\Theta}}\right)_{(1)}={\frac{k_{12}}{m_{11}\omega_{1}^{2}-k_{11}}}={\frac{-{\frac{kL}{2}}}{2m\big(0.9498\,{\frac{k}{m}}\big)-2k}}=4.98L. (iva)
where the subscript (1) on the lhs of Equation (iva) denotes the amplitude ratio associated with the first natural frequency. Similarly, substituting {\boldsymbol{\omega}}_{2} into Equation (ii), one has
\left({\frac{X}{\Theta}}\right)_{(2)}={\frac{k_{12}}{m_{11}\omega_{1}^{2}-k_{11}}}={\frac{-{\frac{k L}{2}}}{2m(4.7377{\frac{k}{m}})-2k}}= − 0.067L (ivb)
Equations (iva) and (ivb) give the two mode shapes of the 2-dof system.
b. For the given system with 2m = 2000 kg, 2k = 40,000 N/m, L = 1.0 m, Equations (iiia) and (iiib) give the two natural frequencies as
\omega_{1}=4.3584\,\mathrm{rad/s}, (va)
\omega_{2}=9.7342\,\mathrm{rad/s}. (vb)
If one assumes ΘL = 1 so that Equations (iva) and (ivb) become
\left({\frac{X}{\Theta L}}\right)_{(1)}=4.98 (via)
\left({\frac{X}{\Theta L}}\right)_{(2)}=-\,0.067. (vib)
Thus, the required two normal modes can be written as
\Phi_{1}={\binom{4.980}{1.000}} (viia)
\Phi_{2}={\binom{-0.067}{1.000}} (viib)