2.2: Mechanical System Modeling

Learning Objectives

After completing this section, you will be able to:

1. Describe three key components of mechanical subsystems: mass (or moment of inertial), damper (rotational damper), and spring (rotational spring).
2. Derive differential equations based upon Newton’s law for linear and rotational dynamic motions.
3. Model continuous mechanical systems based on finite element analysis.

Mechanical systems are often modeled using differential equations first. The system transfer function can be obtained by applying the Laplace transform to these differential equations. In this section, we discuss how to a model point mass-damper-spring system based on Newton’s law. We then briefly discuss how to model continuous mechanical systems using the finite element method. Note that there is another class of mechanical systems—such as fluid dynamic systems—that is often modeled by partial differential equations.

For the point mass-damper-spring system with linear motion, there are three basic components: mass, damper, and spring. We will first model these components individually. Consider that a force \(f(t)\) is applied to a point mass \(m\), resulting in a displacement \(x(t)\). The left plot of Figure 2.2 illustrates the free-body diagram.

Applying Newton’s law results in the following equation of motion for mass \(m\).

\[m\ddot{x}\left(t\right)=f(t) \label{2.17}\]

Assuming that \(\dot{x}\left(0\right)=x\left(0\right)=0\), equation \ref{2.18} shows the Laplace transform of equation \ref{2.17}.

\[ms^2X\left(s\right)=F\left(s\right), \ \dfrac{X(s)}{F(s)}=\dfrac{1}{{ms}^2} \label{2.18}\]

For the damper (the middle plot of Figure 2.2), using Newton’s law yields the following equation of motion.

\[b\dot{x}\left(t\right)=f(t) \label{2.19}\]

In turn, equation \ref{2.20} shows the associated Laplace transform of equation \ref{2.19}, assuming \(x\left(0\right)=0\).

\[bsX\left(s\right)=F\left(s\right), \ \dfrac{X\left(s\right)}{F\left(s\right)}=\dfrac{1}{bs} \label{2.20}\]

For the spring (the right plot of Figure 2.2), utilizing Newton’s law leads to the following equation of motion.

\[kx\left(t\right)=f(t) \label{2.21}\]

Equation \ref{2.22} shows the associated Laplace transform of equation \ref{2.21}.

\[kX\left(s\right)=F\left(s\right), \ \dfrac{X(s)}{F(s)}=\dfrac{1}{k} \label{2.22}\]

Now we are ready to discuss the process of modeling a point mass-damper-spring system to obtain associated ordinary differential equations and transfer functions. The process can be divided into the following three steps: Using the force relationships discussed above, draw the free-body diagram for the target mass-damper-spring system. Using Newton’s law (\(m\ddot{x}=\mathrm{s}\mathrm{ummation\ of\ all\ applied\ forces}\)), obtain the associated ordinary differential equations. Compute the Laplace transform on the ordinary differential equations to obtain the associated system transfer functions. The following examples demonstrate linear motion mass-damper-spring systems.

Example 2.5

Consider the simple mass-damper-spring system shown in Figure 2.3. Find the equation of motion (EOM) in the form of the ordinary differential equation and the associated transfer function.

Draw the free-body diagram

To draw the free-body diagram for the system shown in Figure 2.3, we must find the damping and spring forces applied to the mass \(m\). Using equations \ref{2.19} and \ref{2.21} results in the following.

\[f_b\left(t\right)=b\dot{x}\left(t\right), f_k\left(t\right)=kx(t) \label{2.23}\]

With damping force \(f_b\left(t\right)\) and spring force \(f_k\left(t\right)\) calculated in equation \ref{2.23}, we can draw a free body diagram (see the right diagram in Figure 2.4).

Find the equation of motion (EOM)

Applying Newton’s law yields

\[m\ddot{x}\left(t\right)=f\left(t\right)-f_b\left(t\right)-f_k\left(t\right)=f\left(t\right)-b\dot{x}\left(t\right)-kx(t) \label{2.24}\]

where the force in the same direction of the displacement \(x(t)\) is positive and vice versa. The resulting EOM is shown below.

\[m\ddot{x}\left(t\right)+b\dot{x}\left(t\right)+kx(t)=f\left(t\right) \label{2.25}\]

Find the Laplace transform of EOM

Computing the Laplace transform for equation \ref{2.25}, assuming zero initial condition, leads to the following.

\[ms^2X\left(s\right)+bsX\left(s\right)+kX\left(s\right)=F(s) \label{2.26}\]

The associated system transfer function with \(F\left(s\right)\) as input and \(X(s)\) as output is shown below.

\[G\left(s\right)=\dfrac{X(s)}{F(s)}=\dfrac{1}{ms^2+bs+k} \label{2.27}\]

Example 2.6

Consider a quarter-car driving on an uneven road with varying vertical displacement \(h(t)\) (see Figure 2.5). Here, the quarter-car and wheel masses are \(m_b\) and \(m_w\), respectively. The suspension damping coefficient and spring stiffness are \(b_s\) and \(k_s\), respectively.

The active suspension applied force and tire spring stiffness are \(f(t)\) and \(k_w\), respectively. Find the equations of motion and associated transfer functions. Follow the three steps described in Example 2.5 to find the EOM and associated transfer function. First, draw the free-body diagrams for \(m_b\) and \(m_w\) (see Figure 2.6). There are two point-masses in the system, resulting in two free-body diagrams.

With two free-body diagrams developed for \(m_b\) and \(m_w\), respectively, applying Newton’s law results in the following two differential equations of motion.

\[\begin{aligned} m_b{\ddot{x}}_1\left(t\right)+b_s\left({\dot{x}}_1\left(t\right)-{\dot{x}}_2\left(t\right)\right)+k_s\left(x_1\left(t\right)-x_2\left(t\right)\right) &=0 \\[4pt] m_w{\ddot{x}}_2\left(t\right)+b_s\left({\dot{x}}_2\left(t\right)-{\dot{x}}_1\left(t\right)\right)+k_s\left(x_2\left(t\right)-x_1\left(t\right)\right)+k_wx_2\left(t\right) &=f\left(t\right)-k_wh\left(t\right) \end{aligned} \label{2.28}\]

In turn, conducting the Laplace transform for equation \ref{2.28} yields the following, assuming zero initial condition

\[ x_1\left(0\right)=\dot{x}_1\left(0\right)=x_2\left(0\right)= \dot{x}_2\left(0\right)=0.\]

\[\begin{aligned} m_bs^2X_1\left(s\right) + b_ssX_1\left(s\right) + k_sX_1\left(s\right) &=b_ssX_2\left(s\right)+k_sX_2\left(s\right) \\[4pt] m_ws^2X_2\left(s\right) + b_ssX_2\left(s\right)+k_sX_2\left(s\right)+k_wX_2\left(s\right) & =F\left(s\right)-k_wH\left(s\right) + b_ssX_1\left(s\right) + k_sX_1\left(s\right) \end{aligned} \label{2.29}\]

Now, reorganize equation \ref{2.29} to obtain the system transfer functions as below.

\[\begin{array}{c} G_1\left(s\right)=\dfrac{X_1(s)}{X_2(s)}=\dfrac{b_ss+k_s}{m_bs^2+b_ss+k_s} \\ \begin{array}{c} X_2\left(s\right)=\dfrac{F\left(s\right)}{m_ws^2+b_ss+\left(k_s+k_w\right)}-\dfrac{k_wH\left(s\right)}{m_ws^2+b_ss+\left(k_s+k_w\right)}+\dfrac{\left(b_ss+k_s\right)X_1\left(s\right)}{m_ws^2+b_ss+\left(k_s+k_w\right)} \\ \\ = G_2\left(s\right)F\left(s\right)-k_wG_2\left(s\right)H\left(s\right)+G_3\left(s\right)X_1\left(s\right) \ \end{array} \end{array} \label{2.30}\]

The transfer functions in equation \ref{2.30} can be represented by the block diagram in Figure 2.7. (Section 2.4 will provide a detailed technique for drawing block diagrams.)

There is another class of mechanical systems with rotational motion, where the moment of inertia \(J\) is substituted for mass \(m\), a rotational damper for the linear damper, and a rotational spring for the linear spring. Figure 2.8 shows the free-body diagrams of the moment of inertia, rotational damper, and rotational spring. Using a similar approach to linear motion, we can establish torque \(τ\) and rotational displacement \(θ\) relationships for the moment of inertial, rotational damper, and rotational spring.

For the moment of inertia (the left drawing of Figure 2.8), applying Newton’s law results in the following equation of motion:

\[J\ddot{θ}\left(t\right)=τ(t) \label{2.31}\]

Now assuming that \(\dot{θ}\left(0\right)=θ\left(0\right)=0\), the Laplace transform of equation \ref{2.31} is shown below.

\[Js^2Θ\left(s\right)=T\left(s\right), \dfrac{Θ(s)}{T(s)}=\dfrac{1}{{Js}^2} \label{2.32}\]

For the rotational damper (the middle drawing of Figure 2.8), using Newton’s law yields the following equation of motion:

\[b\dot{θ}\left(t\right)=τ(t) \label{2.33}\]

In turn, the associated Laplace transform of equation \ref{2.33}, assuming \(x\left(0\right)=0,\) is show below.

\[bsΘ\left(s\right)=T\left(s\right), \ \dfrac{Θ(s)}{T(s)}=\dfrac{1}{bs} \label{2.34}\]

For the rotational spring (the right drawing of Figure 2.8), utilizing Newton’s law leads to the following equation of motion:

\[kθ\left(t\right)=τ(t) \label{2.35}\]

The associated Laplace transform of equation \ref{2.35} is as follows.

\[kΘ\left(s\right)=T\left(s\right), \ \dfrac{Θ(s)}{T(s)}=\dfrac{1}{k} \label{2.36}\]

By replacing the linear mass \(m\), damping coefficient \(b\), and spring stiffness \(k\) with moment of inertia \(J\), rotational damping coefficient \(b\), and spring stiffness \(k\), respectively, we can see that these three EOMs are similar to those for linear motion.

Example 2.7

Consider a motor with motor torque \(τ_m(t)\), moment of inertia \(J_m\), rotational friction \(b_m\), and angular displacement \(θ_m(t)\) that drives the load moment of inertia \(J_l\) through a flexible shaft with its rotational stiffness \(k\); see Figure 2.9.

To find the equations of motion and associated transfer functions, we will follow the three steps described in Example 2.5. First, draw the free-body diagrams for \(J_m\) and \(J_l\) (see Figure 2.10). Because there are two subsystems of moment of inertia, we have two free-body diagrams.

With the two free-body diagrams developed for \(J_m\) and \(J_l\), applying Newton’s law results in the following two equations of motion.

\[\begin{aligned} J_m{\ddot{θ}}_m\left(t\right) + b_m{\dot{θ}}_m \left(t\right) + k\left(θ_m\left(t\right)-θ_l\left(t\right)\right) &= τ_m(t) \\[4pt] J_l{\ddot{θ}}_l\left(t\right) + k\left(θ_l\left(t\right) - θ_m\left(t\right)\right) &= 0 \end{aligned} \label{2.37}\]

Computing the Laplace transform for equation \ref{2.37} yields the following, assuming zero initial condition

\[(θ_m\left(0\right)={\dot{θ}}_m\left(0\right)=θ_l\left(0\right)={\dot{θ}}_l\left(0\right)=0.\]

\[\begin{aligned} J_ms^2Θ_m\left(s\right) + b_msΘ_m\left(s\right)+k\left(Θ_m\left(s\right)-Θ_l\left(s\right)\right) &=T_m\left(s\right) \\[4pt] J_ls^2Θ_l\left(s\right)+k\left(Θ_l\left(s\right)-Θ_m\left(s\right)\right) &=0 \end{aligned} \label{2.38}\]

Now, reorganize equation \ref{2.38} to obtain the system transfer functions, as in equation \ref{2.39}.

\[\begin{aligned} Θ_m\left(s\right) &= \dfrac{T_m\left(s\right)+kΘ_l\left(s\right)}{J_ms^2+b_ms+k} \\[4pt] Θ_l\left(s\right) &=\dfrac{kΘ_m\left(s\right)}{J_ls^2+k} \end{aligned} \label{2.39}\]

To obtain the system transfer function from \(T_m\left(s\right)\) to \(Θ_l\left(s\right)\), substitute the top equation of \ref{2.39} into the bottom equation of \ref{2.39}. Similarly, find the transfer function from \(T_m\left(s\right)\) to \(Θ_m\left(s\right)\) by substituting the bottom equation of \ref{2.39} into the top equation of \ref{2.39}.

Having discussed modeling of point mass-damper-spring mechanical systems, we are now going to briefly discuss modeling of the continuous mechanical system, using an example. In general, we can use the finite element analysis method to find a linearized model, as mentioned at the beginning of this section. To demonstrate how to use the finite-element method to model a continuous mechanical system, consider a flexible beam interacting with a uniform flow field (the left diagram of Figure 2.11). The finite-element dynamic modeling approach can be adopted in this case. The beam is divided into \(n\) rigid-body elements (see the right diagram in Figure 2.11). These elements have equal length. Each is linked with its neighbor elements through joints modeled by a rotational spring \(k_i\) and a linear rotational damper \(b_i\), where index \(i\) (\(i=1,\ 2,\ \dots ,\ n+1\)) denotes the beam node and the \(i^{th}\) element is defined by the nodes \(i\) and \(i+1\). These dampers on the node collectively model the internal damping of the beam material and the hydrodynamic damping due to the interaction with the surrounding fluid (see the right diagram of Figure 2.11). In this case, these dampers could be nonlinear; however, in this section, they are assumed to be linear (or \(b_i\) is constant). Applied force \(f^f_i\) to the \(i^{th}\) beam is generated by the uniform flow field, which is modeled as a lumped load force applied to the element center of mass.

Figure 2.12 shows the free-body diagram for the \(i^{th}\) element, where \((x,y)\) are the coordinates in the inertial frame, \(J_i\) is the moment of inertia for the \(i^{th}\) element, \(l\) is the beam element length, \(f^x_i\) and \(f^y_i\) are respectively the reaction forces at the \(i^{th}\) node in the \(x\) and \(y\) directions, and \(f^x_{i+1}\) and \(f^y_{i+1}\) are the reaction forces for the \({(i+1)}^{th}\) node in \(x\) and \(y\) directions. Also, the solid dot in the middle of the element represents the center of mass for the \(i^{th}\) element.

For the \(i^{th}\) element, the governing equation for the rotational motion can be written in the following form based upon Newton’s law.

\[\begin{array}{c} J_i{\ddot{θ}}_i+b_i\left({\dot{θ}}_i-{\dot{θ}}_{i-1}\right)+k_i\left(θ_i-θ_{i-1}\right)-b_{i+1}\left({\dot{θ}}_{i+1}-{\dot{θ}}_i\right)-k_{i+1}\left(θ_{i+1}-θ_i\right)= \\ -\dfrac{f^x_i}{2}l{\cos θ_i\ }-\dfrac{f^y_i}{2}l{\sin θ_i\ }-\dfrac{f^x_{i+1}}{2}l{\cos θ_i\ }-\dfrac{f^y_{i+1}}{2}l{\sin θ_i\ } \end{array} \label{2.40}\]

Note that equation \ref{2.40} is valid for \(i=1,\ 2,\dots ,n-1\). For \(i=n\) (the last element), we need to set \(f^x_{n+1}=f^y_{n+1}=0\). Using Newton’s law, the translational motion satisfies the following two equations.

\[\begin{array}{c} m_i{\ddot{x}}_i\left(t\right)=f^f_i\left(t\right)+f^x_i\left(t\right)-f^x_{i+1}\left(t\right) \\ m_i{\ddot{y}}_i\left(t\right)=f^y_i\left(t\right)-f^y_{i+1}\left(t\right) \end{array}\label{2.41}\]

where \(m_i\) is the effective mass of the \(i^{th}\) beam element defined in Figure 2.12, and \({\ddot{x}}_i\) and \({\ddot{y}}_i\) are the mass acceleration in \(x\) and \(y\) directions at the center of mass \(m_i\). Note that equation \ref{2.40} is nonlinear. To obtain a linear system model, we assume that the bending angle \(θ_i\) (\(i=1,\ 2,\ \dots ,\ n\)) is small so that

\[{\sin θ_i\approx θ_i, {\cos θ_i\approx 1, \ i=1,\ 2,\ \dots ,\ n.\ }\ } \label{2.42}\]

This approximation leads to a maximum error of 1.5% when the beam angle \(θ_i\) (\(i=1,\ 2,\ \dots ,\ n\)) is less than or equal to 10 degrees (or 0.17 rad). With this assumption, both \({\ddot{x}}_i\) and \({\ddot{y}}_i\) can be approximated by the following equations.

\[\begin{aligned} \ddot{x}_i \left(t\right) &=\sum^{i-1}_{j=1} {\ddot{θ}_j(t)l + \dfrac{\ddot{θ}_i(t)l}{2}}, \, i=1,\, 2, \dots ,\, n \\[4pt] \ddot{y}_i (t) &=0, \, i=1,\, 2,\, ..,\, n \end{aligned} \label{2.43}\]

Because \(f^y_{n+1}\left(t\right)\) is zero, we have \(f^y_i\left(t\right)=0\) (\(i=1,\ 2,\ \dots ,\ n\)) from equations \ref{2.41} and \ref{2.43}; see [3] for details. On the other hand, from equations \ref{2.41} and \ref{2.43}, \(f^x_i\left(t\right)\) in \ref{2.40} can be solved as a linear combination of \(f^f_i\left(t\right)\) and \({\ddot{θ}}_i(t)\) (\(i=1,\ 2,\ \dots ,\ n\)) with \(f^x_{n+1}\left(t\right)=0\). The flow field force \(f^f_i\left(t\right)\) applied to each element is exerted by the surrounding medium; it can be expressed as

\[f^f_i\left(t\right)=wlC_dρ\dfrac{{\tilde{V}}^2_i}{2} \label{2.44}\]

where \(w\ \)is the beam element width, \(l\) is the beam element length, \(C_d\) is the beam drag coefficient, \(ρ\) is the fluid density, and \({\tilde{V}}_i\) is the relative beam velocity with respect to the fluid flow field at the center of the element. Note that \({\tilde{V}}_i=V_o+{\dot{x}}_i+l{\dot{θ}}_i/2\), where \(V_o\) is the fluid flow field velocity, and \({\dot{x}}_i+l{\dot{θ}}_i/2\) is the x-direction component of the beam element linear velocity at its center of mass. Again, assuming that \({\dot{θ}}_i\) is relatively small, we can approximate \({\dot{x}}_i+l{\dot{θ}}_i/2\) by \({\dot{x}}_i\). For the \(i^{th}\) element, the moment of inertia \(J_i\), rotational stiffness \(k_i\), and damping coefficient \(b_i\) can be expressed in terms of the beam dimensions, the properties of the beam material, and the fluid property as follows, assuming that each beam element has the same mechanical properties.

\[\begin{align} J_i & = \dfrac{m_il^2}{12}\approx \dfrac{m_el^2}{12} = J_e \\[4pt] k_i & = \dfrac{γwc^3}{12l}=k_e  \\[4pt] b_i & = 2ξ\sqrt{J_ik_i} = 2ξ\sqrt{J_ek_e}=b_e \end{align} \label{2.45}\]

where \(m_i=m_e\) is the effective mass for element \(i\); \(w\) and \(c\) are the beam element width and thickness, respectively; \(γ\ \)is the (effective) Young’s modulus of the beam material, and \(ξ\) is the critical damping ratio accounting for material and hydrodynamic damping. In addition, note that effective mass \(m_e\) represents the sum of the actual mass of element \(i\ \)and the added mass due to beam-fluid interactions [14], [15]. It can be expressed as

\[m_e=m_i+ρ_ewlc \label{2.46}\]

where \(ρ_e\) represents the effective density of the beam, which depends on the material density, fluid density, and beam geometry. Because all the beam elements are identical, they have the same moment of inertia \(J_e\), damping coefficient \(b_e\), and stiffness \(k_e\). We can rewrite equation \ref{2.40} in a compact matrix form below.

\[J\ddot{\overline{θ}}\left(t\right)+B\dot{\overline{θ}}\left(t\right)+K\overline{θ}(t)=Γ \label{2.47}\]

where \(J\), \(B\), and \(K\) are matrices of moments of inertia, damping coefficients, and spring constants, respectively. For the n-element beam model, these matrices are expressed as below.

\[J=J_e\left[ \begin{array}{cc} \begin{array}{ccc} 12\left(n-1\right)+4 & 12\left(n-2\right)+6 & \cdots \\ 12\left(n-2\right)+6 & 12\left(n-2\right)+4 & \cdots \\ \vdots & \vdots & \ddots \end{array} & \begin{array}{ccc} 30 & 18 & 6 \\ 30 & 18 & 6 \\ \vdots & \vdots & \vdots \end{array} \\ \begin{array}{ccc} 30 & 30 & \cdots \\ 18 & 18 & \cdots \ \ 6 & 6 & \ \cdots \ \end{array} & \begin{array}{ccc} 28 & 18 & 6 \\ 18 & 16 & 6 \\ 6 & 6 & 4 \end{array} \end{array} \right] \label{2.48}\]

\[B=b_e\left[ \begin{array}{cc} \begin{array}{ccc} \ 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} & \begin{array}{ccc} \cdots \ & 0 & 0 \\ \cdots \ & 0 & 0 \\ \cdots \ & 0 & 0 \end{array} \\ \begin{array}{ccc} \vdots \ & \vdots & \vdots \\ 0 \ & 0 & 0 \\ 0 \ & 0 & 0 \end{array} & \begin{array}{ccc} \ddots & \ \vdots & \vdots \ \ \cdots & \ 2 & -1 \ \ \cdots & -1 & 2 \end{array} \end{array} \right] \label{2.49}\]

\[K=k_e\left[ \begin{array}{cc} \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} & \begin{array}{ccc} \cdots & \ 0 & \ 0 \\ \cdots & \ 0 & \ 0 \\ \cdots & \ 0 & \ 0 \end{array} \\ \begin{array}{ccc} \vdots \ & \vdots & \vdots \\ 0 \ & 0 & 0 \\ 0 \ & 0 & 0 \end{array} & \begin{array}{ccc} \ \ddots & \ \vdots & \vdots \ \cdots & \ 2 & -1\ \ \cdots & -1 & 2\ \end{array} \end{array} \right] \label{2.50}\]

The angular displacement vector \(\overline{θ}\) and the fluid flow-induced torque vector \(Γ\) are expressed as

\[\overline{θ}(t)=\left[ \begin{array}{c} \begin{array}{c} θ_1\left(t\right) \\ θ_2(t) \end{array} \\ \begin{array}{c} \vdots \\ θ_{n-1}(t) \\ θ_n(t) \end{array} \end{array} \right], \ Γ\mathrm{(}\mathrm{t)}=\dfrac{l}{2}\left[ \begin{array}{c} \begin{array}{c} f^f_1\left(t\right)+2\sum^n_{j=2}{f^f_j\left(t\right)} \\ f^f_2\left(t\right)+2\sum^n_{j=3}{f^f_j\left(t\right)} \end{array} \\ \begin{array}{c} \vdots \\ f^f_{n-1}\left(t\right)+2\sum^n_{j=n}{f^f_j\left(t\right)} \\ f^f_n\left(t\right) \end{array} \end{array} \right] \label{2.51}\]

It is a bit complex to model a continuous mechanical system such as a cantilever beam. The purpose of this example is to show that the model can be developed by using finite-element analysis. You can find a detailed application of this example in [3]. Another example for a continuous mechanical system model, a Euler-Bernoulli beam with a small pin-joint plate, can be found in [16].

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