Steady-Sliding Equilibrium Point

It is obvious that a = 0 is the trivial solution. To determine its stability, dal jda is derived and evaluated at a = 0. From (6.43) 

From (6.46), the condition for the stability of the trivial solution is bo 0. From (6.44) and by substituting the original system parameters, we have 

From (6.47), the condition for the stability of the steady-sliding state becomes 

It is interesting to note that, (6.48) is accurate to 0 ) when compared with what was found from linear eigenvalue analysis, i.e., (6.8)  

Unfortunately, the other possible solutions (i.e., stable or unstable limit cycles) can only be found numerically due to the complexity of the averaged equations. However, some important insights can be gained by examining (6.43). 

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Considering small perturbations around the steady-sliding equilibrium point where Vb — y > 0, the linearized equation of motion is found from (4.4) as... 

As expected, a = 0 is the trivial solution of (4.27). Similar to the case of the pure-slip motion, the stability of the steady-sliding equilibrium point (i.e., the origin) is evaluated from the sign of da /d L=o- From (4.27) one finds... 

Fig. 4.22 Unstable steady-sliding equilibrium point, C(0eq) —3 < 0 4.3.4.3 Painleve s Paradoxes...

For the parameter values that satisfy the above inequality and A Vt < 0, Painleve s paradoxes occur and the equation of motion given by (4.59) or the linearized equation given by (4.63) are no longer valid due to violation of existence and uniqueness of the solution. However, as we will see in the numerical examples below, the steady-sliding equilibrium point is indeed unstable for such values of system parameters. 

The introduction of the new variables, yi = u and y2 = u, converts (6.5) into a system of first-order differential equations. The Jacobian matrix of this system evaluated at the origin (i.e., steady-sliding equilibrium point) is found as... 

Assuming To >0, the steady-sliding equilibrium point becomes unstable if... 

The above instability threshold can be stated alternatively in terms of the applied axial force, R. The steady-sliding equilibrium point is unstable if... 

The stable/unstable regions in the space of parameters R and d are shown in Fig. 6.1. Expectedly, when negative friction damping is present (d < 0), there is a limiting value of axial force, beyond which the steady-sliding equilibrium point is unstable. This limit proportionally increases with the increase of the damping in the lead screw supports. 

Fig. 6.1 Region of stability of the steady-sliding equilibrium point in the space of applied axial forces, R, and gradient of friction/velocity curve...

Fig. 6.2 System trajectories for c = 2 x 10 < Qr unstable steady-sliding equilibrium point (0,0)...

In Sects. 5.5 and 5.6, we have introduced two 2-DOF models for the lead screw drives. In this section, the equations of motion of these models are transformed into matrix form and linearized with respect to their respective steady-sliding equilibrium point. These equations are then used in the next sections to study the local stability of the equilibrium point and the role of mode coupling instability mechanism. In this chapter, for simplicity, the coefficient of friction, n, is taken as a constant. 

The equations of motion of the 2-DOF lead screw drive with axially compliant lead screw support are given in Sect. 5.6 by (5.30) and (5.31). Assume i = 2t where Q is a constant. The following change of variables is used to transfer the steady-sliding equilibrium point to the origin ... 

Figure 7.1 shows the stability region of the undamped 2-DOF model in the kc — p parameter space. The hatched region corresponds to the parameter range where the two natural frequencies are complex, and the steady-sliding equilibrium point is unstable. The boundary of this region is the fiutter instability threshold where co = co2-... 

To study the stability of the steady-sliding equilibrium point, the eigenvalues of the Jacobian matrix (evaluated at y = 0) are calculated. The Jacobian matrix is given by (F 0, O 0)... 

Figure 8.3 shows the region of stability of the steady-sliding equilibrium point as r0 and Co are varied. Due to the presence of the kinematic constraint in the system and for a self-locking lead screw drive (i.e., fi > tan X), both of these parameters can attain negative values. The origin of the linearized system is unstable whenever To < 0 or Co < 0. 

Fig. 8.3 Stability/instability regions of the steady-sliding equilibrium point as Fq and Cq are varied...

Fig. 8.4 Unstable steady sliding equilibrium point due to damping kinematic constraint instability mechanism g > tan/l, Wo > 0, Q > 0, and tan/lcx > c...

By decreasing R (the axial force) the line = 0 in Fig. 8.4 moves towards the origin. Fig. 8.5 shows a situation where the steady-sliding equilibrium point is in the N < 0 half plane. 

From the above discussions, one can conclude that, if — R > 0 is large enough such that every trajectory starting from uo,—Q.), where mq < (r-/ /kmr tan 1)+ cil/k) reaches the origin asymptotically, then the steady-sliding equilibrium point is globally stable. Otherwise, the region of attraction is only a subset of state space. 

The linearized equation of motion in matrix form in the neighbourhood of the steady-sliding equilibrium point (i.e. y = y = 0) is... 

Because of the symmetry of the mass (Mq), stiffness (K), and damping (C) matrices, the steady-sliding equilibrium point cannot loose stability by flutter. Divergence is also ruled out since det(K) > 0. As for the possibility of paradoxes, we set... 

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