Example 1 The reciprocating pump with mass of 200 kg is mounted at the middle of steal plate \(E = 200 \left[\mathrm{GPa}\right]\) with thickness of 20 mm, width 1000 mm and length of 5 m, clamped along two edges as shown in Figure 1 . During the pump's operation the plate is subjected to harmonic force of \(F(t) = 1000 \cos70t\). Find the amplitude of vibration of the plate.
Figure 1 - Plate supporting an unbalanced pump
Solution: The amplitude equaiton can be written in the following form
\begin{eqnarray} X &=& \frac{F_0}{k-m\omega^2}.\nonumber \end{eqnarray} To calculate the amplitude the stiffness, the natural frequency and the force magnitude must be determined \(F_0\) must be determined. The general form of harmonic force can be written as: \begin{eqnarray} F(t) = F_0\cos \omega t.\nonumber \end{eqnarray} By comparing the general expression with the harmonic force acting on a plate the \(F_0\) and the \(\omega\) can be determined. \begin{eqnarray} F_0 &=& 1000 \left[\mathrm{N}\right]\nonumber\\ \omega &=& 70\left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} To calculate the stiffnes the moment of inertia of the plate must be calcualted with following expression: \begin{eqnarray} I &=& \frac{1}{12}w\cdot t^3 = \frac{1}{12}\cdot 1 \cdot 0.02^3 = 6.66667\times 10^{-7} \left[\mathrm{m}^4\right]\nonumber \end{eqnarray} The stiffness can be calculated from the expression: \begin{eqnarray} k &=& \frac{192EI}{l^3} = \frac{192\cdot 200\times 10^9 \cdot 6.66667\times 10^{-7}}{5^3} = 204800 \left[\frac{\mathrm{N}}{\mathrm{m}}\right].\nonumber \end{eqnarray} Now the aplitude X can be easily calculated: \begin{eqnarray} X &=& \frac{F_0}{k-m\omega^2}\nonumber\\ X &=& \frac{1000}{204800-200\cdot 70^2}\nonumber\\ X &=& -0.00128999 \left[\mathrm{m}\right]\nonumber \end{eqnarray} Example 2 A spring mass system, with spring of stiffness of 2000 N/m, is subjected to a harmonic force of magnitude 100 N and frequency of 1 Hz. The mass is found to vibrate with an amplitude of 0.2m. The vibrations starts from rest \(x_0 = \dot{x}_0 = 0\) determine the mass of the system.
Figure 2 - Spring mass system
Solution:
\begin{eqnarray}
x(t) &=& \frac{F_0}{k-m\omega^2}(cos\omega t- \cos \omega_nt)\nonumber\\
x(t) &=& \frac{2F_0}{k-m\omega^2} \sin \frac{\omega_n+\omega}{2}t \sin \frac{\omega_n - \omega}{2}t\nonumber
\end{eqnarray}
\begin{eqnarray}
\frac{2F_0}{k-m\omega^2} &=& 0.2\nonumber \\
\frac{2\cdot 1000}{10000-m(125.664)^2} &=& 0.2\nonumber\\
m &=& 25.3303 \left[\mathrm{kg}\right]\nonumber
\end{eqnarray}
Example 3 Find the total response of SDOF (single-degree of freedom system) that consist of body with mass m = 20 kg, damper with damping coefficient of c=20 N-s/m, and spring with stiffness k = 20000 N/m. Initial conditions are \(x_0 = 0.01 \left[\mathrm{m}\right]\) and \(\dot{x}_0 = 0\). Determine the external force that acts on the system with \(F_0 = 1000 \left[\mathrm{N}\right]\), and \(\omega=100\left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\). Determine the free vibrations of SDOF system with \(F(t)=0\).
Figure 3 - Spring-mass-damper system
Solution In this example there are two cases that have to be investigated. The first case is the case when the force in the form \(F(t) = F_0\cos \omega t \) acts on the system. In second case the free vibration of SDOF system must be determined i.e. the force F(t) is equal to 0.
The natural frequency of SDOF system is equal to: \begin{eqnarray} \omega_n &=& \sqrt{\frac{k}{m}} = \sqrt{\frac{20000}{20}} = 31.6228 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right].\nonumber \end{eqnarray} The deflection under static force can be determined as: \begin{eqnarray} \delta_{st} &=& \frac{F_0}{k} = \frac{1000}{20000} = 0.05\left[\mathrm{m}\right].\nonumber \end{eqnarray} The damping ration is equal to: \begin{eqnarray} \zeta &=& \frac{c}{c_c} = \frac{c}{2\sqrt{km}} = \frac{20}{2\sqrt{20000\cdot 20}} = 0.0316228 \nonumber \end{eqnarray} The frequency \(\omega_d\) can be calculated as \begin{eqnarray} \omega_d &=& \sqrt{1-\zeta^2}\omega_n = \sqrt{1-(0.0316228)^2}\cdot 31.6228 = 31.607 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right].\nonumber \end{eqnarray} The frequency ration \(r\) is equal to: \begin{eqnarray} r &=& \frac{\omega}{\omega_n} = \frac{100}{31.6228} = 3.16228. \end{eqnarray} The parameter \(X\) and \(\phi\) are equal to: \begin{eqnarray} X &=& \frac{\delta_{st}}{\sqrt{(1-r^2)^2+(2\zeta r)^2}}\nonumber\\ X &=& \frac{0.05}{\sqrt{(1-3.16228^2)^2 +(2\cdot 0.0316228\cdot 3.16228)^2}}\nonumber\\ X &=& 0.00555417 \left[\mathrm{m}\right]\nonumber \end{eqnarray} \begin{eqnarray} \phi &=& \tan^{-1}\left(\frac{2\zeta r}{1-r^2}\right)\nonumber\\ \phi &=& \tan^{-1}\left(\frac{2\cdot 0.0316228 \cdot 3.16228}{1-3.16228^2}\right)\nonumber\\ \phi &=& -0.90502°. \end{eqnarray} The initial displacement \(X_0\) and phase angle \(\phi_0\) are determined by substituting initial conditions into following equations: \begin{eqnarray} x_0 &=& X_0 \cos\phi_0 + X\cos \phi\nonumber\\ \dot{x}_0 &=& -\zeta \omega_n X_0 \cos \phi_0 + \omega_d X_0 \sin \phi_0 + \omega X \sin \phi\nonumber. \end{eqnarray} Substituting initial condition into previous equation the displacement and phase angle are obtained: \begin{eqnarray} 0.01 &=& X_0 \cos \phi_0 + 0.00555417\cdot \cos (-0.90502°)\nonumber \\ 0.01 &=& X_0 \cos \phi_0 + 0.0055417\cdot 0.999875 \nonumber\\ X_0\cos \phi_0 &=& 0.00445899\nonumber \end{eqnarray} \begin{eqnarray} 0 &=& -0.0316228\cdot 31.6228\cdot X_0\cos\phi_0 + 31.607X_0\sin\phi_0 + 100\cdot 0.00555417\cdot (-0.0157949)\nonumber\\ 0 &=& -X_0\cos\phi_0 + 31.607X_0\sin\phi_0 -0.00877276 \end{eqnarray} By susbstituting \(X_0\cos\phi_0 = 0.00445899\) into previous equation the following expression is obtained \begin{eqnarray} -0.00445899 +31.607X_0\sin\phi_0 &=& 0.00877276\nonumber\\ 31.607X_0 \sin\phi_0 &=& 0.0132318\nonumber\\ X_0\sin\phi_0 &=& 0.000418634 \end{eqnarray} \begin{eqnarray} X_0 &=& \left[0.004458993^2 + 0.000418634^2\right]^{1/2}\nonumber\\ X_0 &=& 0.0044786\nonumber \end{eqnarray} \begin{eqnarray} \tan\phi_0 &=& \frac{X_0\sin\phi_0}{X_0\cos\phi_0} = 10.6513\nonumber\\ \phi_0 &=& 1.56916°.\nonumber \end{eqnarray} In case of free vibration the toal response is defined as: \begin{eqnarray} x(t) &=& X_0e^{-\zeta \omega_n t}\cos(\omega_d t - \phi_0)\nonumber \end{eqnarray} The \(X_0\) and \(\phi_0\) are determined from following expressions \begin{eqnarray} X_0 &=& \sqrt{x_0^2 + \left(\frac{\zeta \omega_n x_0}{\omega_d}\right)^2}\nonumber\\ X_0 &=& \sqrt{0.01^2 + \left(\frac{0.0316228\cdot 31.6228\cdot 0.01}{31.607}\right)^2}\nonumber\\ X_0 &=& 0.010005\nonumber \end{eqnarray} \begin{eqnarray} \phi_0 &=& \tan^{-1}\left(\frac{\dot{x}_0 + \zeta \omega_n x_0}{\omega_dx_0}\right)\nonumber\\ \phi_0 &=& \tan^{-1}\left(\frac{0.0316228\cdot 31.6228}{31.607}\right) \nonumber\\ \phi_0 &=& 1.06669° \end{eqnarray} Example 4 For system shown in Figure 4 under force vibration, a force will be transmitted to the base. Determine the force transmitted to the base of a viscously damped system subjected to a harmonic force in the steady state.
Figure 4 - Spring-mass-damper system
Solution The steady state response of the system can be expressed in the folowing form:
\begin{eqnarray}
x(t) &=& x_p(t) = X\cos(\omega_t-\phi) = X(\cos\omega t\cos\phi + \sin\omega t\sin\phi) = A\cos\omega t + B\sin \omega t\nonumber
\end{eqnarray}
where \(A = X\cos\phi\), and \(B = X\sin\phi\). Differentiation of previous equation gives the steady state velocity which can be written as
\begin{eqnarray}
\dot{x}(t) &=& -\omega X\sin(\omega t- \phi)\nonumber\\
\dot{x}(t) &=& -\omega X (\sin\omega t\cos\phi - \cos \omega t \sin \phi) \nonumber\\
\dot{x}(t) &=& \omega B \cos \omega t - \omega A \sin \omega t\nonumber
\end{eqnarray}
The force \(f_T (t)\) which istransmitted to the base can be written as the some of the forces due to the spring and the damper as:
\begin{eqnarray}
f_T (t) &=& kx(t) + c\dot{x}(t) = kX\cos(\omega t - \phi) - c\omega X \sin(\omega t - \phi)\nonumber
\end{eqnarray}
which can be written as:
\begin{eqnarray}
f_T (t) &=& kA\cos \omega t + kB \sin \omega t + c\omega B \cos \omega t -c \omega A \sin \omega t \nonumber\\
&=& (kA+c\omega B) \cos \omega t + (kB -c\omega A ) \sin \omega t\nonumber
\end{eqnarray}
If parts of the previous expression are denoted as\((kA+c\omega B) = F_t\cos\phi_T \) and \((kB-c\omega A) = F_t\sin\phi_T\) then the previous expression can be written as:
\begin{eqnarray}
f_T (t) &=& F_T \cos \omega t \cos \phi_T + F_T \sin \omega t \sin\phi_T \nonumber \\
f_T (t) &=& F_T\cos(\omega t - \phi_T)\nonumber
\end{eqnarray}
where
\begin{eqnarray}
F_T &=& \sqrt{(F_T\cos\phi_T)^2 + (F_T \sin\phi_T)^2}\nonumber\\
F_T &=& \sqrt{(kA + c\omega B)^2 + (kB - c\omega A)^2}\nonumber\\
F_T &=& \sqrt{(kX\cos\phi + c\omega X \sin \phi)^2 +(kX\sin\phi - c\omega X \cos \phi)^2}\nonumber
\end{eqnarray}
Performing the simplifications on the previous expression the previous expression can be written as:
\begin{eqnarray}
F_T &=& X\sqrt{k^2 + c^2\omega^2} \nonumber\\
F_T &=& F_0 \frac{\sqrt{k^2+x^2\omega^2}}{((k-m\omega^2)^2 + c^2\omega^2)}\nonumber\\
F_T &=& F_0 \frac{\sqrt{1+(2\zeta r)^2}}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\nonumber
\end{eqnarray}
\begin{eqnarray}
\phi_T &=& \tan^{-1}\left(\frac{2\zeta r^3}{1-r^2 +(2\zeta r)}\right)
\end{eqnarray}
Example 5 The model of a motor vehicle, shown in Figure 5 that can vibrate in the vertical direction while traveling over a rough road. The vehicle has a mass of 1500 kg. The suspension system has a spring constant of 800 kN/m and a damping ratio of \(\zeta = 0.5\) . If the vehicle speed is 20 km/h, determine the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude of Y = 0.05 m and a wevelength of 6m.
Figure 5 - Vehicle moving over a rough road
Solution The frequency \(\omega\) of the base excitation can be found by dividing the vehicle speed by the length of one cycle of road roughness.
\begin{eqnarray}
\omega &=& 2\pi f =\left(\frac{v\cdot 1000}{3600}\right)\frac{1}{6} = 0.290888 v\nonumber
\end{eqnarray}
In case where \(v = 20 \left[\frac{\mathrm{km}}{\mathrm{hr}}\right]\) the frequency of the base excitation:
\begin{eqnarray}
\omega &=& 0.29088\cdot 20 =5.81776 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber
\end{eqnarray}
The natural frequency of the vehicle is equal to:
\begin{eqnarray}
\omega_n &=& \sqrt{\frac{k}{m}} = \sqrt{\frac{800\cdot 10^3}{1500}} = 23.094 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber
\end{eqnarray}
The frequency ration is equal to:
\begin{eqnarray}
r &=& \frac{\omega}{\omega_n} = \frac{5.81776}{23.094} = 0.251917.
\end{eqnarray}
The amplitude ratio can be found from expressions
\begin{eqnarray}
\frac{X}{Y} &=& \sqrt{\frac{1+(2\zeta r)^2}{(1-r^2)^2+(2\zeta r)^2}} = \sqrt{\frac{1+(2\cdot 0.5 \cdot 0.251917)^2}{(1-0.251917)^2 + (2\cdot0.5\cdot 0.251917)^2}}\nonumber\\
\frac{X}{Y} &=& 1.30643\nonumber
\end{eqnarray}
The displacement amplitude of the vehicle can be calculated as
\begin{eqnarray}
\frac{X}{Y} &=& 1.30643\Big/\cdot 1.30643\nonumber\\
X &=& 1.30643\cdot 0.05 = 0.0653214 \left[\mathrm{m}\right]\nonumber
\end{eqnarray}
\begin{eqnarray} X &=& \frac{F_0}{k-m\omega^2}.\nonumber \end{eqnarray} To calculate the amplitude the stiffness, the natural frequency and the force magnitude must be determined \(F_0\) must be determined. The general form of harmonic force can be written as: \begin{eqnarray} F(t) = F_0\cos \omega t.\nonumber \end{eqnarray} By comparing the general expression with the harmonic force acting on a plate the \(F_0\) and the \(\omega\) can be determined. \begin{eqnarray} F_0 &=& 1000 \left[\mathrm{N}\right]\nonumber\\ \omega &=& 70\left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} To calculate the stiffnes the moment of inertia of the plate must be calcualted with following expression: \begin{eqnarray} I &=& \frac{1}{12}w\cdot t^3 = \frac{1}{12}\cdot 1 \cdot 0.02^3 = 6.66667\times 10^{-7} \left[\mathrm{m}^4\right]\nonumber \end{eqnarray} The stiffness can be calculated from the expression: \begin{eqnarray} k &=& \frac{192EI}{l^3} = \frac{192\cdot 200\times 10^9 \cdot 6.66667\times 10^{-7}}{5^3} = 204800 \left[\frac{\mathrm{N}}{\mathrm{m}}\right].\nonumber \end{eqnarray} Now the aplitude X can be easily calculated: \begin{eqnarray} X &=& \frac{F_0}{k-m\omega^2}\nonumber\\ X &=& \frac{1000}{204800-200\cdot 70^2}\nonumber\\ X &=& -0.00128999 \left[\mathrm{m}\right]\nonumber \end{eqnarray} Example 2 A spring mass system, with spring of stiffness of 2000 N/m, is subjected to a harmonic force of magnitude 100 N and frequency of 1 Hz. The mass is found to vibrate with an amplitude of 0.2m. The vibrations starts from rest \(x_0 = \dot{x}_0 = 0\) determine the mass of the system.
Solution In this example there are two cases that have to be investigated. The first case is the case when the force in the form \(F(t) = F_0\cos \omega t \) acts on the system. In second case the free vibration of SDOF system must be determined i.e. the force F(t) is equal to 0.
The natural frequency of SDOF system is equal to: \begin{eqnarray} \omega_n &=& \sqrt{\frac{k}{m}} = \sqrt{\frac{20000}{20}} = 31.6228 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right].\nonumber \end{eqnarray} The deflection under static force can be determined as: \begin{eqnarray} \delta_{st} &=& \frac{F_0}{k} = \frac{1000}{20000} = 0.05\left[\mathrm{m}\right].\nonumber \end{eqnarray} The damping ration is equal to: \begin{eqnarray} \zeta &=& \frac{c}{c_c} = \frac{c}{2\sqrt{km}} = \frac{20}{2\sqrt{20000\cdot 20}} = 0.0316228 \nonumber \end{eqnarray} The frequency \(\omega_d\) can be calculated as \begin{eqnarray} \omega_d &=& \sqrt{1-\zeta^2}\omega_n = \sqrt{1-(0.0316228)^2}\cdot 31.6228 = 31.607 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right].\nonumber \end{eqnarray} The frequency ration \(r\) is equal to: \begin{eqnarray} r &=& \frac{\omega}{\omega_n} = \frac{100}{31.6228} = 3.16228. \end{eqnarray} The parameter \(X\) and \(\phi\) are equal to: \begin{eqnarray} X &=& \frac{\delta_{st}}{\sqrt{(1-r^2)^2+(2\zeta r)^2}}\nonumber\\ X &=& \frac{0.05}{\sqrt{(1-3.16228^2)^2 +(2\cdot 0.0316228\cdot 3.16228)^2}}\nonumber\\ X &=& 0.00555417 \left[\mathrm{m}\right]\nonumber \end{eqnarray} \begin{eqnarray} \phi &=& \tan^{-1}\left(\frac{2\zeta r}{1-r^2}\right)\nonumber\\ \phi &=& \tan^{-1}\left(\frac{2\cdot 0.0316228 \cdot 3.16228}{1-3.16228^2}\right)\nonumber\\ \phi &=& -0.90502°. \end{eqnarray} The initial displacement \(X_0\) and phase angle \(\phi_0\) are determined by substituting initial conditions into following equations: \begin{eqnarray} x_0 &=& X_0 \cos\phi_0 + X\cos \phi\nonumber\\ \dot{x}_0 &=& -\zeta \omega_n X_0 \cos \phi_0 + \omega_d X_0 \sin \phi_0 + \omega X \sin \phi\nonumber. \end{eqnarray} Substituting initial condition into previous equation the displacement and phase angle are obtained: \begin{eqnarray} 0.01 &=& X_0 \cos \phi_0 + 0.00555417\cdot \cos (-0.90502°)\nonumber \\ 0.01 &=& X_0 \cos \phi_0 + 0.0055417\cdot 0.999875 \nonumber\\ X_0\cos \phi_0 &=& 0.00445899\nonumber \end{eqnarray} \begin{eqnarray} 0 &=& -0.0316228\cdot 31.6228\cdot X_0\cos\phi_0 + 31.607X_0\sin\phi_0 + 100\cdot 0.00555417\cdot (-0.0157949)\nonumber\\ 0 &=& -X_0\cos\phi_0 + 31.607X_0\sin\phi_0 -0.00877276 \end{eqnarray} By susbstituting \(X_0\cos\phi_0 = 0.00445899\) into previous equation the following expression is obtained \begin{eqnarray} -0.00445899 +31.607X_0\sin\phi_0 &=& 0.00877276\nonumber\\ 31.607X_0 \sin\phi_0 &=& 0.0132318\nonumber\\ X_0\sin\phi_0 &=& 0.000418634 \end{eqnarray} \begin{eqnarray} X_0 &=& \left[0.004458993^2 + 0.000418634^2\right]^{1/2}\nonumber\\ X_0 &=& 0.0044786\nonumber \end{eqnarray} \begin{eqnarray} \tan\phi_0 &=& \frac{X_0\sin\phi_0}{X_0\cos\phi_0} = 10.6513\nonumber\\ \phi_0 &=& 1.56916°.\nonumber \end{eqnarray} In case of free vibration the toal response is defined as: \begin{eqnarray} x(t) &=& X_0e^{-\zeta \omega_n t}\cos(\omega_d t - \phi_0)\nonumber \end{eqnarray} The \(X_0\) and \(\phi_0\) are determined from following expressions \begin{eqnarray} X_0 &=& \sqrt{x_0^2 + \left(\frac{\zeta \omega_n x_0}{\omega_d}\right)^2}\nonumber\\ X_0 &=& \sqrt{0.01^2 + \left(\frac{0.0316228\cdot 31.6228\cdot 0.01}{31.607}\right)^2}\nonumber\\ X_0 &=& 0.010005\nonumber \end{eqnarray} \begin{eqnarray} \phi_0 &=& \tan^{-1}\left(\frac{\dot{x}_0 + \zeta \omega_n x_0}{\omega_dx_0}\right)\nonumber\\ \phi_0 &=& \tan^{-1}\left(\frac{0.0316228\cdot 31.6228}{31.607}\right) \nonumber\\ \phi_0 &=& 1.06669° \end{eqnarray} Example 4 For system shown in Figure 4 under force vibration, a force will be transmitted to the base. Determine the force transmitted to the base of a viscously damped system subjected to a harmonic force in the steady state.