A Two-Degree-of-Freedom Quarter-Car Model Consider a quarter-car model shown in Figure 5.34a, where m1 is the mass of one-fourth of the car body and m2 is the mass of the wheel–tire–axle assembly. The spring k1 represents the elasticity of the suspension and the spring k2 represents the elasticity

Question

A Two-Degree-of-Freedom Quarter-Car Model

Consider a quarter-car model shown in Figure 5.34a, where [latex]m_{1}[/latex] is the mass of one-fourth of the car body and [latex]m_{2}[/latex] is the mass of the wheel-tire-axle assembly. The spring [latex]k_{1}[/latex] represents the elasticity of the suspension and the spring [latex]k_{2}[/latex] represents the elasticity of the tire. [latex]z(t)[/latex] is the displacement input due to the surface of the road. The actuator force, [latex]f[/latex], applied between the car body and the wheel-tire-axle assembly, is controlled by feedback and represents the active components of the suspension system.

a. Draw the necessary free-body diagrams and derive the differential equations of motion.

b. Determine the state-space representation. Assume that the displacements of the two masses, [latex]x_{1}[/latex] and [latex]x_{2}[/latex], are the outputs and the state variables are [latex]x_{1}=x_{1}, x_{2}=x_{2}, x_{3}=\dot{x}_{1}[/latex], and [latex]x_{4}=\dot{x}_{2}[/latex].

c. The parameter values are [latex]m_{1}=290 \mathrm{~kg}, m_{2}=59 \mathrm{~kg}, b_{1}=1000 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}, k_{1}=16,182 \mathrm{~N} / \mathrm{m}[/latex], and [latex]k_{2}=19,000 \mathrm{~N} / \mathrm{m}[/latex]. Use MATLAB commands to define the system in the state-space form and then convert it to the transfer function form. Assume that all the initial conditions are zero.

Accepted Answer

a. We choose the displacements of the two masses [latex]x_{1}[/latex] and [latex]x_{2}[/latex] as the generalized coordinates.

The static equilibrium positions of [latex]m_{1}[/latex] and [latex]m_{2}[/latex] are set as the coordinate origins. Assume

[latex]x_{1}>x_{2}>z>0,[/latex]

which implies that the springs are in tension and

[latex]\dot{x}_{1}>\dot{x}_{2}>\dot{z}>0.[/latex]

The free-body diagrams of [latex]m_{1}[/latex] and [latex]m_{2}[/latex] are shown in Figure 5.34b. According to the assumption, the mass [latex]m_{1}[/latex] moves faster than the mass [latex]m_{2}[/latex], and the elongation of the spring [latex]k_{1}[/latex] is [latex]x_{1}-x_{2}[/latex]. The force exerted by the spring [latex]k_{1}[/latex] on the mass [latex]m_{1}[/latex] is downward as it tends to restore to the undeformed position. Because of Newton's third law, the force exerted by the spring [latex]k_{1}[/latex] on the mass [latex]m_{2}[/latex] has the same magnitude, but opposite in direction. Other spring forces and damping forces can be determined using the same logic. Note that the gravitational forces, [latex]m_{1} g[/latex] and [latex]m_{2} g[/latex], are not included in the free-body diagrams.

Applying Newton's second law to the masses [latex]m_{1}[/latex] and [latex]m_{2}[/latex], respectively, gives

[latex] \begin{gathered} +\uparrow x: \sum F_{x}=m a_{x \prime} \ f-k_{1}\left(x_{1}-x_{2} ight)-b_{1}\left(\dot{x}_{1}-\dot{x}_{2} ight)=m_{1} \ddot{x}_{1} \ -f+k_{1}\left(x_{1}-x_{2} ight)+b_{1}\left(\dot{x}_{1}-\dot{x}_{2} ight)-k_{2}\left(x_{2}-z ight)=m_{2} \ddot{x}_{2} . \end{gathered} [/latex]

Rearranging the equations into the standard input-output form,

[latex] \begin{gathered} m_{1} \ddot{x}_{1}+b_{1} \dot{x}_{1}-b_{1} \dot{x}_{2}+k_{1} x_{1}-k_{1} x_{2}=f \ m_{2} \ddot{x}_{2}-b_{1} \dot{x}_{1}+b_{1} \dot{x}_{2}-k_{1} x_{1}+\left(k_{1}+k_{2} ight) x_{2}=-f+k_{2} z_{1} \end{gathered} [/latex]

which can be expressed in second-order matrix form (Section 4.1) as

[latex] \left[\begin{array}{cc} m_{1} & 0 \ 0 & m_{2} \end{array} ight]\left\{\begin{array}{l} \ddot{x}_{1} \ \ddot{x}_{2} \end{array} ight\}+\left[\begin{array}{cc} b_{1} & -b_{1} \ -b_{1} & b_{1} \end{array} ight]\left\{\begin{array}{l} \dot{x}_{1} \ \dot{x}_{2} \end{array} ight\}+\left[\begin{array}{cc} k_{1} & -k_{1} \ -k_{1} & k_{1}+k_{2} \end{array} ight]\left\{\begin{array}{l} x_{1} \ x_{2} \end{array} ight\}=\left[\begin{array}{cc} 1 & 0 \ -1 & k_{2} \end{array} ight]\left\{\begin{array}{l} f \ z \end{array} ight\} . [/latex]

b. Note that the inputs to the system are the actuator force [latex]f[/latex] and the road surface irregularity [latex]z[/latex]. The state, the input, and the output vectors are

[latex] \mathbf{x}=\left\{\begin{array}{l} x_{1} \ x_{2} \ x_{3} \ x_{4} \end{array} ight\}=\left\{\begin{array}{l} x_{1} \ x_{2} \ \dot{x}_{1} \ \dot{x}_{2} \end{array} ight\}, \quad \mathbf{u}=\left\{\begin{array}{l} f \ z \end{array} ight\}, \quad \mathbf{y}=\left\{\begin{array}{l} x_{1} \ x_{2} \end{array} ight\} [/latex]

The state-variable equations are then obtained as

[latex] \begin{aligned} \dot{x}_{1} & =x_{3}, \ \dot{x}_{2} & =x_{4} \ \dot{x}_{3} & =\ddot{x}_{1}=-\frac{k_{1}}{m_{1}} x_{1}+\frac{k_{1}}{m_{1}} x_{2}-\frac{b_{1}}{m_{1}} \dot{x}_{1}+\frac{b_{1}}{m_{1}} \dot{x}_{2}+\frac{f}{m_{1}} \ & =-\frac{k_{1}}{m_{1}} x_{1}+\frac{k_{1}}{m_{1}} x_{2}-\frac{b_{1}}{m_{1}} x_{3}+\frac{b_{1}}{m_{1}} x_{4}+\frac{1}{m_{1}} u_{1} \ \dot{x}_{4} & =\ddot{x}_{2}=\frac{k_{1}}{m_{2}} x_{1}-\frac{k_{1}+k_{2}}{m_{2}} x_{2}+\frac{b_{1}}{m_{2}} \dot{x}_{1}-\frac{b_{1}}{m_{2}} \dot{x}_{2}-\frac{f}{m_{2}}+\frac{k_{2}}{m_{2}} z \ & =\frac{k_{1}}{m_{2}} x_{1}-\frac{k_{1}+k_{2}}{m_{2}} x_{2}+\frac{b_{1}}{m_{2}} x_{3}-\frac{b_{1}}{m_{2}} x_{4}-\frac{1}{m_{2}} u_{1}+\frac{k_{2}}{m_{2}} u_{2} . \end{aligned} [/latex]

The output equation is

[latex] \mathbf{y}=\left\{\begin{array}{l} x_{1} \ x_{2} \end{array} ight\} [/latex]

Thus, the state-space representation is

[latex] \begin{aligned} \left\{\begin{array}{l} \dot{x}_{1} \ \dot{x}_{2} \ \dot{x}_{3} \ \dot{x}_{4} \end{array} ight\} & =\left[\begin{array}{cccc} 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ -\frac{k_{1}}{m_{1}} & \frac{k_{1}}{m_{1}} & -\frac{b_{1}}{m_{1}} & \frac{b_{1}}{m_{1}} \ \frac{k_{1}}{m_{2}} & -\frac{k_{1}+k_{2}}{m_{2}} & \frac{b_{1}}{m_{2}} & -\frac{b_{1}}{m_{2}} \end{array} ight]\left\{\begin{array}{l} x_{1} \ x_{2} \ x_{3} \ x_{4} \end{array} ight\}+\left[\begin{array}{cc} 0 & 0 \ 0 & 0 \ \frac{1}{m_{1}} & 0 \ -\frac{1}{m_{2}} & \frac{k_{2}}{m_{2}} \end{array} ight]\left\{\begin{array}{l} u_{1} \ u_{2} \end{array} ight\}, \ \mathbf{y} & =\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \end{array} ight]\left\{\begin{array}{l} x_{1} \ x_{2} \ x_{3} \ x_{4} \end{array} ight\}+\left[\begin{array}{ll} 0 & 0 \ 0 & 0 \end{array} ight]\left\{\begin{array}{l} u_{1} \ u_{2} \end{array} ight\} . \end{aligned} [/latex]

c. The following is the MATLAB session:

Question 5.5: A Two-Degree-of-Freedom Quarter-Car Model Consider a quarter...

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