For a rigid rectangular block placed on a rigid horizontal ground [Fig. 2.2(a)], Shenton H. [95] obtained the relation between the horizontal ground acceleration and the coefficient of friction required to initiate the response of the block in rocking, sliding, or coupled rocking-sliding mode. Shenton H. [95] used force and moment balance equations to arrive at the separatrix between the different response modes. Here, we consider an analogous problem of a rigid rectangular block impacting a rigid horizontal ground. At the instant before the impact, the block is assumed to be translating with a horizontal velocity (u˙x >0) and a vertical velocity (u˙y < 0) in the presence of gravitational acceleration. Under the assumption that the impact between the block and ground is inelastic, i.e.,e = 0, the block
O1
2h 2b
O2
mg α
R
(a)
0 5 10
0 0.2 0.4 0.6 0.8 1
|u˙y/ ˙ux|
Coefficientoffriction(µ)
Rock
Equation 2.31
µ=hb |uu˙˙y
x|= µ1
|uu˙˙y
x|= hb
Rest
Slide Rock/Slide
(b)
Figure 2.2: (a) Rigid rectangular block and ground (b) motion of block after it impacts the ground as a function of initial velocity and coefficient of friction.
may come to rest, may start to rock, may slide, or may experience sliding coupled with rocking after the impact. Therefore, an approach similar to Shenton’s method, involving conservation of linear and angular momentum, may be used to arrive at the separatrix between the different modes as a function of the translational velocities before impact (u˙x andu˙y) and the coefficient of friction (µs =µd =µ) between the two surfaces [Fig.2.2(b)].
The details of this calculation are presented in Appendix B. For example, the separatrix between the rocking mode and the coupled rocking-sliding mode is given by:
˙ uy
˙ ux
= 4 + (hb)2−3µhb
µ[1 + 4(hb)2]−3hb (2.31) For this analysis, the rigid block [Fig.2.2(a)] considered has a half heighth= 0.6 m, half widthb= 0.2 m, and massm = 100 kg, and it may only come in contact with the ground at points O1 and O2. The motion of the block after impact is governed by Lagrangian mechanics with the reduced translational and rotational velocity of the center of mass after inelastic impact as initial conditions. These velocities are also obtained by conserving angular and linear momentum as shown in Appendix B. Here, we select two combinations of|u˙y/u˙x|andµcorresponding to the rocking and the coupled rocking-sliding modes and compare the solutions obtained from the rigid body dynamics algorithm with that obtained
by time-integrating the equations of motion for the corresponding response modes.
2.2.1 Rocking mode
The block will experience rocking coupled with sliding after impact foru˙y = −2.8 m/s,
|u˙y/u˙x| = 1.5, andµ = 0.9. The angular displacement of the block (θ) is positive when the block rotates aboutO2 and negative for rotations aboutO1. The non-linear equation of motion for the rocking mode can be obtained by moment balance about the rocking points O1 andO2:
Iθ¨=mgRsin(α+θ) θ <0 (2.32) Iθ¨=−mgRsin(α−θ) θ >0 (2.33) Here, α is a measure of the slenderness of the block [α = tan−1(b/h)], I is the moment of inertia of the block about corner O1 orO2, and R is the distance of the contact points from the center of mass of the block. In the current example, the block starts rocking about cornerO1after impact with the ground. Therefore, Eq.2.32is time integrated using the 4th order Runge Kutta method with the rotational velocity of the block after impact [Eq.B.10]
as initial condition. The reduction in the rotational velocity on inelastic impact during rocking, i.e., when point of rotation changes fromO1toO2or vice versa, is modeled using the expression presented by Kimura and Iida [64,65] and Housner [52]:
θ˙+
θ˙− = 1 + 3cos(2α)
4 (2.34)
Here,θ˙− andθ˙+ are the rotational velocities of the block before and after impact with the ground. The resulting angular displacement time history [θ(t)] of the block can be easily converted to arrive at the horizontal (x) displacement time history of the block’s center of mass, which is then compared against that obtained from rigid body dynamics algorithm Fig.2.3(a). The good agreement in the results indicate that the rigid body dynamics algo- rithm is capable of accurately predicting the response mode of the block and simulating the subsequent rocking response of the block.
0 1 2 3
−0.1
−0.05 0 0.05 0.1 0.15
time(sec)
x displacement of center of mass (m)
Rigid body algorithm Equation of motion
˙
uy=-2.8 m/s
˙
uy/u˙x =1.5 µ=0.9
(a)
0 0.5 1 1.5 2 2.5
−0.06
−0.04
−0.02 0 0.02 0.04 0.06
time(sec)
x displacement of center of mass (m)
Rigid body algorithm Equation of motion
˙
uy=-1 m/s
˙
uy/u˙x =1 µ=0.4
(b)
Figure 2.3: Comparison of x displacement time history of center of mass given by the rigid body dynamics algorithm against analytical solution for (a) rocking mode and (b) coupled rocking-sliding mode. For the rocking mode,u˙y =−2.8 m/s,|u˙y/u˙x|= 1.5andµ= 0.9. For the coupled rocking- sliding mode,u˙y =−1 m/s,|u˙y/u˙x|= 1andµ= 0.4.
2.2.2 Coupled rocking-sliding mode
The block will rock and slide after impact foru˙y = −1 m/s, |u˙y/u˙x| = 1and µ = 0.4.
The angular displacement of the block (θ) is positive when the block rotates aboutO2 and negative for rotations aboutO1. Ishiyama worked out the nonlinear equations of motion for the coupled rocking-sliding mode (Eqs. 1-3 in [58]). These equations of motion are time integrated using the 4th order Runge Kutta method to obtain horizontal (x) displacement time history of the block’s center of mass. When the block stops sliding, i.e, when the horizontal velocity ofO1 is zero (within a tolerance), the subsequent rocking response of the block is modeled using Eqs.2.32-2.34. For the combination of velocities and friction coefficient considered, the block stops sliding before the point of rotation changes fromO1 toO2.
The horizontal (x) displacement time history obtained from these differential equa- tions is compared with that obtained from the rigid body dynamics algorithm as shown in Fig.2.3(b). The results obtained using the rigid body dynamics algorithm agree well with the analytical results.