A Two-Mass System: Suspension Model Figure 2.4 shows an automobile suspension system. Write the equations of motion for the automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car mass above one wheel. A system consisting of one of the four wheel suspensions

Question

A Two-Mass System: Suspension Model

Figure 2.4 shows an automobile suspension system. Write the equations of motion for the automobile and wheel motion assuming one-dimensional vertical motion of one quarter of the car mass above one wheel. A system consisting of one of the four wheel suspensions is usually referred to as a quarter-car model. Assume that the model is for a car with a mass of [latex] 1580 \ kg, [/latex] including the four wheels, which have a mass of [latex] 20 \ kg [/latex] each. By placing a known weight (an author) directly over a wheel and measuring the car’s deflection, we find that [latex] k_s = 130,000 \ N/m. [/latex] Measuring the wheel’s deflection for the same applied weight, we find that [latex] k_w \simeq 1,000,000 \ N/m. [/latex] By using the results in Section 3.3, Fig. 3.18(b), and qualitatively observing that the car’s response as the author jumps off matches the [latex] ζ = 0.7 [/latex] curve, we conclude that [latex] b = 9800 \ N·sec/m.[/latex]

Accepted Answer

The system can be approximated by the simplified system shown in Fig. 2.5. The coordinates of the two masses, [latex] x [/latex] and [latex] y, [/latex] with the reference directions as shown, are the displacements of the masses from their equilibrium conditions. The equilibrium positions are offset from the springs’ unstretched positions because of the force of gravity. The shock absorber is represented in the schematic diagram by a dashpot symbol with friction constant [latex] b. [/latex] The magnitude of the force from the shock absorber is assumed to be proportional to the rate of change of the relative displacement of the two masses—that is, the force [latex] = b(\dot{y} − \dot{x}). [/latex] The force of gravity could be included in the free-body diagram; however, its effect is to produce a constant offset of [latex] x [/latex] and [latex] y. [/latex] By defining [latex] x [/latex] and [latex] y [/latex] to be the distance from the equilibrium position, the need to include the gravity forces is eliminated.

The force from the car suspension acts on both masses in proportion to their relative displacement with spring constant [latex] k_s. [/latex] Figure 2.6 shows the free-body diagram of each mass. Note that the forces from the spring on the two masses are equal in magnitude but act in opposite directions, which is also the case for the damper. A positive displacement [latex] y [/latex] of mass [latex] m_2 [/latex] will result in a force from the spring on [latex] m_2[/latex] in the direction shown and a force from the spring on [latex] m_1 [/latex] in the direction shown. However, a positive displacement [latex] x [/latex] of mass [latex] m_1 [/latex] will result in a force from the spring [latex] k_s [/latex] on [latex] m_1 [/latex] in the opposite direction to that drawn in Fig. 2.6, as indicated by the minus [latex] x [/latex] term for the spring force.

The lower spring [latex] k_w [/latex] represents the tire compressibility, for which there is insufficient damping (velocity-dependent force) to warrant including a dashpot in the model. The force from this spring is proportional to the distance the tire is compressed and the nominal equilibrium force would be that required to support [latex] m_1 [/latex] and [latex] m_2 [/latex] against gravity. By defining [latex] x [/latex] to be the distance from equilibrium, a force will result if either the road surface has a bump ([latex] r [/latex] changes from its equilibrium value of zero) or the wheel bounces ([latex] x [/latex] changes). The motion of the simplified car over a bumpy road will result in a value of [latex] r(t) [/latex] that is not constant.

As previously noted, there is a constant force of gravity acting on each mass; however, this force has been omitted, as have the equal and opposite forces from the springs. Gravitational forces can always be omitted from vertical-spring mass systems [latex] (1) [/latex] if the position coordinates are defined from the equilibrium position that results when gravity is acting, and [latex] (2) [/latex] if the spring forces used in the analysis are actually the perturbation in spring forces from those forces acting at equilibrium.

Applying Eq. (2.1) to each mass and noting that some forces on each mass are in the negative (down) direction yields the system of equations

[latex] \pmb F = m \pmb a [/latex] (2.1)

[latex] b(\dot{y} − \dot{x}) + k_s(y − x) − k_w (x − r) = m_1 \ddot{x} , [/latex] (2.8)

[latex] −k_s(y − x) − b(\dot{y} − \dot{x}) = m_2 \ddot{y}. [/latex] (2.9)

Some rearranging results in

[latex] \ddot{x} + \frac{b}{m_1} (\dot{x} - \dot{y}) + \frac{k_s}{m_1} (x-y) +\frac{k_w}{m_1} x = \frac{k_w}{m_1}r , [/latex] (2.10)

[latex] \ddot{y} + \frac{b}{m_2} (\dot{y} - \dot{x}) + \frac{k_s}{m_2} (y-x) =0. [/latex] (2.11)

The most common source of error in writing equations for systems like these are sign errors. The method for keeping the signs straight in the preceding development entailed mentally picturing the displacement of the masses and drawing the resulting force in the direction that the displacement would produce. Once you have obtained the equations for a system, a check on the signs for systems that are obviously stable from physical reasoning can be quickly carried out. As we will see when we study stability in Section 3.6, a stable system always has the same signs on similar variables. For this system, Eq. (2.10) shows that the signs on the [latex]\ddot{x} , \dot{x}[/latex], and x terms are all positive, as they must be for stability. Likewise, the signs on the[latex]\ddot{y} , \dot{y}[/latex], and y terms are all positive in Eq. (2.11).

The transfer function is obtained in a similar manner as before. Substituting [latex] s [/latex] for [latex] d/dt [/latex] in the differential equations yields

[latex] s^2 X (s) +s \frac{b}{m_1} ( X(s) -Y (s)) + \frac{k_s}{m_1}(X(s) -Y(s)) + \frac{k_w}{m_1} X(s) = \frac{k_w}{m_1} R(s) , \ s^2 Y(s) + s \frac{b}{m_2} (Y(s) -X(s)) + \frac{k_s}{m_2} (Y(s) -X(s)) =0 , [/latex]

which, after some algebra and rearranging, yields the transfer function

[latex] \frac{Y(s)}{R(s)} = \frac{\frac{k_wb}{m_1 m_2}\left(s + \frac{k_s}{b} ight) }{s^4 + \left(\frac{b}{m_1} + \frac{b}{m_2} ight) s^3 + \left(\frac{k_s}{m_1}+\frac{k_s}{m_2}+\frac{k_w}{m_1} ight) s^2 +\left(\frac{k_w b }{m_1 m_2} ight) s+ \frac{k_w k_s }{m_1 m_2} } . [/latex] (2.12)

To determine numerical values, we subtract the mass of the four wheels from the total car mass of [latex] 1580 \ kg [/latex] and divide by [latex] 4 [/latex] to find that [latex] m_2 = 375 \ kg. [/latex] The wheel mass was measured directly to be [latex] m_1 = 20 \ kg. [/latex] Therefore, the transfer function with the numerical values is

[latex] \frac{Y(s)}{R(s)} = \frac{1.31 e 06\left(s + 13.3 ight) }{s^4 + \left(516.1 ight) s^3 + \left( 5.685 e 04 ight) s^2 +\left( 1.307 e 06 ight) s+ 1.733 e 07 } . [/latex] (2.13)

Question 2.EX.2: A Two-Mass System: Suspension Model Figure 2.4 shows an aut...

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