Exampe 1 Derive the equations of motion for system shown in Figure 1.
Figure 1 - Spring mass damper system
Solution To obtain differential equations for each system component the free-body diagrams of masses must be drawn and Newton's second law of motion must be applied. The coordinates describing the positions of the masses are measured from their static equilibrium positions as shown in Figure 2.
Figure 2 Free body diagram of body with mass \(m_i\).
The application of Newton's II law of motion to mass \(m_i\) gives:
\begin{eqnarray}
m_i\ddot{x}_i &=& k_{i+1}(x_{i+1}-x_i) + c_{i+1}(\dot{x}_{i+1} - \dot{x}_i) - k_i(x_i - x_{i-1}) - c_i(\dot{x}_i - \dot{x}_{i-1}) + F_i(t) \quad i = 2,3,...,n-1.\nonumber
\end{eqnarray}
The equations of motion for first body of mass \(m_1\) and the last body with mass \(m_4\) are derived from previous equation by setting to \(i=1\) and \(i=4\) along with \(x_0 = 0\) and \(x_5 = 0\) can be written as:
\begin{eqnarray}
m_1\ddot{x}_1 + (c_1+c_2)\dot{x}_1 - c_2\dot{x}_2 + (k_1 + k_2)x_1 - k_2x_2 &=& F_1\nonumber\\
m_4\ddot{x}_4 - c_4\dot{x}_3 + (c_4+c_5)\dot{x}_4 - k_4x_3 + (k_4+k_5)x_4 &=& F_4\nonumber
\end{eqnarray}
The previous equations of motion for all bodies in the system can be written in matrix form as:
\begin{eqnarray}
\mathrm{\mathbf{M}}\ddot{\vec{x}} + \mathrm{\mathbf{C}}\dot{\vec{x}} + \mathrm{\mathbf{K}}\vec{x} &=& \vec{F},\nonumber\\
\end{eqnarray}
where \(\mathrm{\mathbf{M}}\), \(\mathrm{\mathbf{C}}\), and \(\mathrm{\mathbf{K}}\) are mass, damping, and stiffness matrices respectively. The \(\ddot{\vec{x}}\), \(\dot{\vec{x}}\), \(\vec{x}\), and \(\vec{F}\) are acceleration, velocity, displacemetn and force vectors respectively. The full form of the \(\mathrm{\mathbf{M}}\), \(\mathrm{\mathbf{C}}\), and \(\mathrm{\mathbf{K}}\) matrices can be written as:
\begin{eqnarray}
\mathrm{\mathbf{M}} &=&
\begin{bmatrix}
m_1 & 0 & 0 & 0\\
0 & m_2 & 0 & 0\\
0 & 0 & m_3 & 0\\
0 & 0 & 0 & m_4
\end{bmatrix}\nonumber\\
\mathrm{\mathbf{C}} &=&
\begin{bmatrix}
(c_1 + c_2) & -c_2 & 0 & 0\\
-c_2 & (c_2+c_3) & -c_3 & 0\\
0 & -c_3 & (c_3 + c_4) & -c_4\\
0 & 0 & -c_4 & (c_4 + c_5)
\end{bmatrix}\nonumber\\
\mathrm{\mathbf{K}} &=&
\begin{bmatrix}
(k_1 + k_2) & -k_2 & 0 & 0\\
-k_2 & (k_2+k_3) & -k_3 & 0\\
0 & -k_3 & (k_3 + k_4) & -k_4\\
0 & 0 & -k_4 & (k_4 + k_5)
\end{bmatrix}\nonumber\\
\end{eqnarray}
Example 2 Determine the stiffness influence coefficients of the system shown in Figure 3.
Figure 3 - 3 Spring & 3 Mass System
Solution: As seen from Figure 3 the \(x_1, x_2\), and \(x_3\) are displacements of the masses \(m_1, m_2,\) and \(m_3\), respectively. First the displacement of body with mass \(m_1\) is set to 1 (\(x_1 = 1\)) while displacements of bodies with masses \(m_2\), and \(m_3\) are set to zero (\(x_2=x3 = 0\)). With summation of all horizontal froces the following system of equations are obtained: \begin{eqnarray} k_1 &=& -k_2 + k_{11}\nonumber\\ k_{21} &=& -k_2\nonumber\\ k_{31} &=& 0\nonumber \end{eqnarray} The solutions of the previous system of equations can be written as: \begin{eqnarray} k_{11} &=& k_1 + k_2\quad k_{21} = -k_2,\quad k_{31} = 0\nonumber \end{eqnarray} The next step is to set displacements to \(x_1 = 0, x_2 = 1, x_3 =0\). Then the system of equations are: \begin{eqnarray} k_{12} + k_2 &=&0\nonumber\\ k_{22} - k_3 &=& k_2 \nonumber\\ k_{32} &=& -k3\nonumber \end{eqnarray} The solutions of previous system of equations can be written as: \begin{eqnarray} k_{12} &=& -k_2, \quad k_{22} = k_2 + k_3 \quad k_{32} = -k_3 \nonumber \end{eqnarray} In third case the displacemetns of bodies with masses \(m_1, m_2\), and \(m_3\) are set to \(x_1 = x_2 = 0,\) and\(x_3 = 1\). Then summing up all horizontal forces the following system of equations is obtained: \begin{eqnarray} k_{13} &=& 0\nonumber\\ k_{23} + k_3 &=& 0\nonumber\\ k_{33} &=& k_3 \nonumber \end{eqnarray} The solutons of previous system of equations are: \begin{eqnarray} k_{13} &=& 0\quad k_{23} = -k_3\quad k_{33} = k_3\nonumber \end{eqnarray} The stiffness in matrix form can be written as: \begin{eqnarray} \mathrm{\mathbf{K}} &=& \begin{bmatrix} (k_1 + k_2) & -k_2 & 0\\ -k_2 & (k_2+k_3) & -k_3\\ 0 & -k_3 & k_3 \end{bmatrix}. \end{eqnarray} Example 3 Find the flexibility influence coefficient of the system shown in Figure 4.Figure 4 - 3 Spring & 3 Mass System
Solution: The flexibility influence coefficinet, which will be denoted with \(a_{ij}\), can be determined by applying unit force at each body and then perform equilibration of forces in horizontal direction. So for the first body of mass \(m_1\) the unit force \(F_1 = 1\) is applied. The forces to the bodies of mass \(m_2\) and \(m_3\) are set to zero \(F_2 = F_3 =0\). The resulting deflections of the masses \(m_1, m_2\), and \(m_3\) are, by definition,\(a_{11}, a_{21},\) and \(a_{31}\), respectively. The equilibrium of forces in the horizontal direction gives system of equations which can be written as: \begin{eqnarray} k_1a_{11} &=& k_2(a_{21} - a_{11}) + 1\nonumber\\ k_2(a_{21} - a_{11}) &=& k_3 (a_{31} - a_{21})\nonumber\\ k_3(a_{31} - a_{21}) &=& 0\nonumber \end{eqnarray} The solution can be written as: \begin{eqnarray} a_{11} &=& \frac{1}{k_1} \quad a_{21} = \frac{1}{k_1}, \quad a_{31} = \frac{1}{k_1}.\nonumber \end{eqnarray} In the second case the unit force \(F_2 = 1\) is set to body with mass \(m_2\) the forces to bodies with mass \(m_1\) and \(m_3\) are set to zero. The equilibrium of forces in the horizontal direction gives system of equations which can be written as: \begin{eqnarray} k_1(a_{12}) &=& k_2 (a_{22}-a_{12})\nonumber\\ k_2(a_{22} - a_{12}) &=& k_3 (a_{32} - a_{22}) + 1 \nonumber\\ k_3(a_{32} - a_{22}) &=& 1\nonumber \end{eqnarray} The solution can be written as: \begin{eqnarray} a_{12} &=& \frac{1}{k_1} \quad a_{22} = \frac{1}{k_1} + \frac{1}{k_2}\quad a_{32} = \frac{1}{k_1} + \frac{1}{k_2}.\nonumber \end{eqnarray} In the last case the unit force acts on body with mass \(m_3\) while forces on the first two bodies are set to zero. The equilibrium of forces in the horizontal direction gives system of equations which can be written as: \begin{eqnarray} k_1 a_{13} &=& k_2 (a_{23} - a_{13})\nonumber\\ k_2(a_{23} - a_{13}) &=& k_3(a_{33} - a_{23})\nonumber\\ k_3(a_{33} - a_{23}) &=& 1.\nonumber \end{eqnarray} The solutions can be written as: \begin{eqnarray} a_{13} &=& \frac{1}{k_1}\quad a_{23} = \frac{1}{k_1} +\frac{1}{k_2} \quad a_{33} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3}. \nonumber \end{eqnarray} Example 4 - Obtain the flexibility matrix of the beam (weightless beam) shown in Figure 5. The beam is simply supported at both ends and on the beam three masses are placed at equal intervals. The assumptions is that beam is uniform with stiffness EI.
Figure 5 - Simply supported beam with three bodies of masses \(m_1\), \(m_2\), and \(m_3\).
Solution: The influence coefficinets \(a_{1j}\) can be found by applying a unit load at the location of \(m_1\) and zero load on the \(m_2\) and \(m_3\). \begin{eqnarray} a_{11} &=& \frac{9}{768}\frac{l^3}{EI}\quad a_{12} = \frac{11}{768}\frac{l^3}{EI}, \quad a_{13} = \frac{7}{768}\frac{l^3}{EI}.\nonumber \end{eqnarray} The same approach is applied on body with mass \(m_2\) and \(m_3\) and the following solutions are obtained: \begin{eqnarray} a_{21} &=& a_{12} = \frac{11}{768}\frac{l^3}{EI}\quad a_{22} = \frac{1}{48}\frac{l^3}{EI}, \quad a_{23} = \frac{11}{768}\frac{l^3}{EI}\nonumber \end{eqnarray}, and \begin{eqnarray} a_{31} &=& a_{13} \frac{7}{768}\frac{l^3}{EI}\quad a_{32} = a_{23} \frac{11}{768}\frac{l^3}{EI}, \quad a_{33} = \frac{9}{768}\frac{l^3}{EI}.\nonumber \end{eqnarray} In matrix form the influence coefficient can be written as: \begin{eqnarray} \mathrm{\mathbf{A}} &=& \frac{l^3}{768EI} \begin{bmatrix} 9 & 11 & 7\\ 11 & 16&11\\ 7& 11& 9 \end{bmatrix}\nonumber \end{eqnarray} Example 5 The compressor, turbine and generator of thermal poweplant are shown in Figure 6. This is torsional system where \(J_i\) denote the mass moments of inertia of the three componenets. The \(M_{ti}\) indicate the external moments acting on the components, and \(k_{ti}\) is torsional spring constants of the shaft between components. Derivie the equation of motion using Lagrange's equation.Figure 6 - Torsion system consisting of compressor, turbine and generator.
Solution: In this case the \(q_1,q_2\), and \(q_3\) are \(\theta_1, \theta_2,\) and \(\theta_3\), respectively. The kinetic energy of the system can be written as:
\begin{eqnarray}
T &=& \frac{1}{2}J_1\dot{\theta}_1^2 + \frac{1}{2}J_2\dot{\theta}_2^2+\frac{1}{2}J_3\dot{\theta}_3^2\nonumber
\end{eqnarray}
The potential energy of the shaft can be written as:
\begin{eqnarray}
V = \int_0^\theta k_t\theta d\theta = \frac{1}{2}k_t\theta^2\nonumber
\end{eqnarray}
The total potential energy of the system can be expressed as:
\begin{eqnarray}
V &=& \frac{1}{2}k_{t1}\theta_1^2 + \frac{1}{2}k_{t2}(\theta_2 - \theta_1)^2 + \frac{1}{2}k_{t3}(\theta_3 - \theta_2)^2\nonumber
\end{eqnarray}
The external moments applied to the components can be written as:
\begin{eqnarray}
Q_j^{(n)} &=& \sum_{k=1}^3 M_{tk}\frac{\partial \theta_k}{\partial q_j} = \sum_{k=1}^3 M_{tk}\frac{\partial \theta _k}{\partial \theta_j} \nonumber
\end{eqnarray}
Using previous equation the following expressions are obtained:
\begin{eqnarray}
Q_1^{(n)} &=& M_{t1}\frac{\partial \theta_1}{\partial \theta_1} + M_{t2}\frac{\partial \theta_2}{\partial \theta_1} + M_{t3}\frac{\partial \theta_3}{\partial \theta_1} = M_{t1}\nonumber\\
Q_2^{(n) } &=& M_{t1}\frac{\partial \theta_1}{\partial \theta_2} + M_{t2}\frac{\partial \theta_2}{\partial \theta_2} + M_{t3}\frac{\partial \theta_3}{\partial \theta_2} = M_{t2}\nonumber\\
Q_3^{(n) } &=& M_{t1}\frac{\partial \theta_1}{\partial \theta_3} + M_{t2}\frac{\partial \theta_2}{\partial \theta_3} + M_{t3}\frac{\partial \theta_3}{\partial \theta_3} = M_{t2}\nonumber
\end{eqnarray}
The Lagrange's equation in general form can be written as:
\begin{eqnarray}
\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q_j}}\right) - \frac{\partial T}{\partial q_j} + \frac{\partial V}{\partial q_j} &=& Q_j^{(n)}\quad j = 1,2,...,n\nonumber\\
\end{eqnarray}
By substituting all previous components into the lagrange differential equation the following solution is obtained:
\begin{eqnarray}
J_1\ddot{\theta}_1 +(k_{t1} + k_{t2})\theta_1 - k_{t2}\theta_2 &=& M_{t1}\nonumber\\
J_2\ddot{\theta}_2 +(k_{t2} + k_{t3})\theta_2 - k_{t2}\theta_1 - k_{t3}\theta_3 &=& M_{t2}\nonumber\\
J_3 \ddot{\theta}_3 +k_{t3}\theta_3 - k_{t3}\theta_2 &=& M_{t3}\nonumber
\end{eqnarray}
The previous three equations can be written in matrix form as:
\begin{eqnarray}
\begin{bmatrix}
J_1 & 0 & 0\\
0 & J_2& 0\\
0 & 0 & J_3
\end{bmatrix}
\begin{Bmatrix}
\ddot{\theta}_1\\
\ddot{\theta}_2\\
\ddot{\theta}_3
\end{Bmatrix}
+\begin{bmatrix}
(k_{t1} + k_{t2}) & -k_{t2} & 0\\
-k_{t2} & (k_{t2} + k_{t3}) & -k_{t3}\\
0 & -k_{t3} & k_{t3}
\end{bmatrix}
\begin{Bmatrix}
\theta_1\\
\theta_2 \\
\theta_3
\end{Bmatrix}
&=&
\begin{Bmatrix}
M_{t1}\\
M_{t2}\\
M_{t3}
\end{Bmatrix}
\end{eqnarray}
Solution: As seen from Figure 3 the \(x_1, x_2\), and \(x_3\) are displacements of the masses \(m_1, m_2,\) and \(m_3\), respectively. First the displacement of body with mass \(m_1\) is set to 1 (\(x_1 = 1\)) while displacements of bodies with masses \(m_2\), and \(m_3\) are set to zero (\(x_2=x3 = 0\)). With summation of all horizontal froces the following system of equations are obtained: \begin{eqnarray} k_1 &=& -k_2 + k_{11}\nonumber\\ k_{21} &=& -k_2\nonumber\\ k_{31} &=& 0\nonumber \end{eqnarray} The solutions of the previous system of equations can be written as: \begin{eqnarray} k_{11} &=& k_1 + k_2\quad k_{21} = -k_2,\quad k_{31} = 0\nonumber \end{eqnarray} The next step is to set displacements to \(x_1 = 0, x_2 = 1, x_3 =0\). Then the system of equations are: \begin{eqnarray} k_{12} + k_2 &=&0\nonumber\\ k_{22} - k_3 &=& k_2 \nonumber\\ k_{32} &=& -k3\nonumber \end{eqnarray} The solutions of previous system of equations can be written as: \begin{eqnarray} k_{12} &=& -k_2, \quad k_{22} = k_2 + k_3 \quad k_{32} = -k_3 \nonumber \end{eqnarray} In third case the displacemetns of bodies with masses \(m_1, m_2\), and \(m_3\) are set to \(x_1 = x_2 = 0,\) and\(x_3 = 1\). Then summing up all horizontal forces the following system of equations is obtained: \begin{eqnarray} k_{13} &=& 0\nonumber\\ k_{23} + k_3 &=& 0\nonumber\\ k_{33} &=& k_3 \nonumber \end{eqnarray} The solutons of previous system of equations are: \begin{eqnarray} k_{13} &=& 0\quad k_{23} = -k_3\quad k_{33} = k_3\nonumber \end{eqnarray} The stiffness in matrix form can be written as: \begin{eqnarray} \mathrm{\mathbf{K}} &=& \begin{bmatrix} (k_1 + k_2) & -k_2 & 0\\ -k_2 & (k_2+k_3) & -k_3\\ 0 & -k_3 & k_3 \end{bmatrix}. \end{eqnarray} Example 3 Find the flexibility influence coefficient of the system shown in Figure 4.
Solution: The flexibility influence coefficinet, which will be denoted with \(a_{ij}\), can be determined by applying unit force at each body and then perform equilibration of forces in horizontal direction. So for the first body of mass \(m_1\) the unit force \(F_1 = 1\) is applied. The forces to the bodies of mass \(m_2\) and \(m_3\) are set to zero \(F_2 = F_3 =0\). The resulting deflections of the masses \(m_1, m_2\), and \(m_3\) are, by definition,\(a_{11}, a_{21},\) and \(a_{31}\), respectively. The equilibrium of forces in the horizontal direction gives system of equations which can be written as: \begin{eqnarray} k_1a_{11} &=& k_2(a_{21} - a_{11}) + 1\nonumber\\ k_2(a_{21} - a_{11}) &=& k_3 (a_{31} - a_{21})\nonumber\\ k_3(a_{31} - a_{21}) &=& 0\nonumber \end{eqnarray} The solution can be written as: \begin{eqnarray} a_{11} &=& \frac{1}{k_1} \quad a_{21} = \frac{1}{k_1}, \quad a_{31} = \frac{1}{k_1}.\nonumber \end{eqnarray} In the second case the unit force \(F_2 = 1\) is set to body with mass \(m_2\) the forces to bodies with mass \(m_1\) and \(m_3\) are set to zero. The equilibrium of forces in the horizontal direction gives system of equations which can be written as: \begin{eqnarray} k_1(a_{12}) &=& k_2 (a_{22}-a_{12})\nonumber\\ k_2(a_{22} - a_{12}) &=& k_3 (a_{32} - a_{22}) + 1 \nonumber\\ k_3(a_{32} - a_{22}) &=& 1\nonumber \end{eqnarray} The solution can be written as: \begin{eqnarray} a_{12} &=& \frac{1}{k_1} \quad a_{22} = \frac{1}{k_1} + \frac{1}{k_2}\quad a_{32} = \frac{1}{k_1} + \frac{1}{k_2}.\nonumber \end{eqnarray} In the last case the unit force acts on body with mass \(m_3\) while forces on the first two bodies are set to zero. The equilibrium of forces in the horizontal direction gives system of equations which can be written as: \begin{eqnarray} k_1 a_{13} &=& k_2 (a_{23} - a_{13})\nonumber\\ k_2(a_{23} - a_{13}) &=& k_3(a_{33} - a_{23})\nonumber\\ k_3(a_{33} - a_{23}) &=& 1.\nonumber \end{eqnarray} The solutions can be written as: \begin{eqnarray} a_{13} &=& \frac{1}{k_1}\quad a_{23} = \frac{1}{k_1} +\frac{1}{k_2} \quad a_{33} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3}. \nonumber \end{eqnarray} Example 4 - Obtain the flexibility matrix of the beam (weightless beam) shown in Figure 5. The beam is simply supported at both ends and on the beam three masses are placed at equal intervals. The assumptions is that beam is uniform with stiffness EI.
Solution: The influence coefficinets \(a_{1j}\) can be found by applying a unit load at the location of \(m_1\) and zero load on the \(m_2\) and \(m_3\). \begin{eqnarray} a_{11} &=& \frac{9}{768}\frac{l^3}{EI}\quad a_{12} = \frac{11}{768}\frac{l^3}{EI}, \quad a_{13} = \frac{7}{768}\frac{l^3}{EI}.\nonumber \end{eqnarray} The same approach is applied on body with mass \(m_2\) and \(m_3\) and the following solutions are obtained: \begin{eqnarray} a_{21} &=& a_{12} = \frac{11}{768}\frac{l^3}{EI}\quad a_{22} = \frac{1}{48}\frac{l^3}{EI}, \quad a_{23} = \frac{11}{768}\frac{l^3}{EI}\nonumber \end{eqnarray}, and \begin{eqnarray} a_{31} &=& a_{13} \frac{7}{768}\frac{l^3}{EI}\quad a_{32} = a_{23} \frac{11}{768}\frac{l^3}{EI}, \quad a_{33} = \frac{9}{768}\frac{l^3}{EI}.\nonumber \end{eqnarray} In matrix form the influence coefficient can be written as: \begin{eqnarray} \mathrm{\mathbf{A}} &=& \frac{l^3}{768EI} \begin{bmatrix} 9 & 11 & 7\\ 11 & 16&11\\ 7& 11& 9 \end{bmatrix}\nonumber \end{eqnarray} Example 5 The compressor, turbine and generator of thermal poweplant are shown in Figure 6. This is torsional system where \(J_i\) denote the mass moments of inertia of the three componenets. The \(M_{ti}\) indicate the external moments acting on the components, and \(k_{ti}\) is torsional spring constants of the shaft between components. Derivie the equation of motion using Lagrange's equation.
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