# Dynamic Analysis of a Planar Hydraulic Rock-Breaker Mechanism with Multiple Clearance Joints

Authors: Ke Chen; Guojun Zhang; Rui Wu; Li Wang; Hongmei Zheng; Shunhua Chen
Journal: Shock and Vibration
 (2019)

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## Abstract

Clearance exists in the joint of a mechanism because of the assemblage, manufacturing tolerances, wear, and other conditions, and it is a focus of research in the field of multibody dynamics. This study built a planar hydraulic rock-breaker model with multiple joint clearances by combining the hydraulic cylinder model, the clearance joints based on the Lankarani–Nikravesh contact force model, and the Lagrange multiplier method. Dynamic simulation results indicated that multiple clearance joints can degrade the dynamic responses of a rock-breaker model, which can be decomposed to rapid vibrations and slow movements. The rapid vibrations are excited by coupling the spring-mass system of hydraulic cylinder and clearances. The effects of the clearance size, input force, damping coefficient, and friction on the dynamic behaviour of the rock-breaker mechanism are also investigated. The friction could reduce the rapid vibration state significantly, which is feasible for practical engineering applications. As compared with the traditional models without clearances, the present model provides not only better predictions for the theoretical study of the hydraulic rock-breaker but also useful guidance for reducing the vibrations of the hydraulic rock-breaker in practical engineering applications.


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## Body

## 1. Introduction

A hydraulic rock-breaker is a multibody mechanism, which is a powerful percussion breaker fitted to an excavator for demolishing rocks or concrete structures. It is widely applied in mining and construction engineering. The asymmetric hydraulic cylinder is used as the activator for a hydraulic rock-breaker, which attracts many research interests in recent years. Zhao et al. used the wavelet packet analysis method to distinguish different levels of leakage in the hydraulic cylinder [1]. Chen et al. developed a hydraulic cylinder with adaptive variable clearance and studied the static and dynamic performance, which can be applied to the hydraulic system with high frequency response and high speed [2]. Nizhegorodov et al. studied the vibratory problem driven by a hydraulic cylinder [3]. Ylinen et al. presented a model for a linear hydraulic actuator for multibody simulations considering the state variables, such as the translational degrees of freedom and the cylinder chamber pressures [4]. Because of the various loading conditions and complex hydraulic and mechanical subsystems, the mechanisms of hydraulic breakers similar to drifters or drills have attracted attention from the researchers. Giuffrida et al. built a model of a hydraulic breaker and used experimental data to verify the model predictions [5, 6]. Ficarella et al. optimized the impact energy of the breaker and minimized the power requirement to the feeding system [7]. Then, they used an experiment to test a hydraulic breaker’s blow impact energy and formulated a theoretical model to analyze the energy release during the impact of the piston against the chisel [8]. Kim et al. redesigned a urethane-steel-urethane multilayer upper cushion to replace the monolithic piece of the urethane cushion to absorb the noises and vibration of the hydraulic breaker system [9]. Oh et al. studied the hydraulic circuits of the drifter and analyzed the impact capability with regard to the penetration rate and varying kinetic energy, which is dependent on rock stiffness [10]. Li et al. built a model of a hydraulic rock drill. Thereafter, the team designed the overlapped reversing valve and proposed a fault diagnosis method [11, 12]. Song et al. used the transfer path analysis method to estimate the striking forces delivered to the hydraulic breaker housing [13].With regard to rock-breaker and excavator dynamics, Arai et al. considered the base, boom, and arm of the excavator to linearize the model and study the coupling vibration between the operating handle and body [14]. Tremblay et al. used a simplified model of a rock-breaker for the purpose of real-time control applications, which reduced the complexity of the dynamic model while preserving its accuracy and simplicity [15]. They assumed the revolute joints in the excavator to be ideal to simplify the mathematical modeling, simulation, and analysis of the dynamics of the mechanical system. In reality, however, revolute joints always have clearances to permit relative motion between connected bodies, manufacturing tolerances, and assemblage. Journal-bearing joints can have a radical clearance. When fracturing rock, a rock-breaker is subjected to impact forces with a certain frequency. Under such serious and heavy-load conditions during work, the joint clearance increases in size. The expanding clearance makes the operator feel vibration and noise, which could deteriorate dynamics of the hydraulic rock-breaker mechanism. However, although Zhang et al. have studied the nonlinear dynamics of an excavator with one clearance [16], the little literature on the effect of clearance on the dynamic performance of rock-breaker has rarely been reported, especially under multiple clearance joints.Clearance has attracted a considerable amount of research interest since the 1970s [17]. Erkaya and Uzmay assumed that the journal is always in contact with the internal surface of the bearing and substituted the clearance with a virtual massless link that connects the journal center to the bearing center [18]. Seneviratne and Earles found that the clearance can dramatically change the dynamic behaviour of a mechanical system [19]. Flores calculated the accuracy and efficiency of the presented approach by evaluating the total computation time consumed in each simulation. The result found that the dynamic responses of multibody systems are sensitive to the clearance size and the operating conditions [20]. The contact force model, which is commonly referred to as the penalty or elastic approach [21], allows the relative indentation of bodies and evaluates the reaction contact force that changes the acceleration of bodies in the Lagrange multiplier method, which is a mainstream approach for easy programming of a general multibody system. A series of contact force models have been investigated to describe the clearance in joints. In the nonlinear Hertz model [22], the contact forces are evaluated depending on the indentation between contacting bodies pressed against each other. However, this model cannot depict energy dissipation, and a damping factor has been introduced to address this weakness. Contact force models by Lankarani and Nikravesh [23] and Hunt and Crossley [24] are the most prevailing models and accepted by researchers widely. Recently, Ma and Qian developed a hybrid contact force model based on the Lankarani–Nikravesh contact force model and an elastic foundation model. They applied the discrete element theory to calculate the penetration of each point individually and obtained the total contact force by integrating the discrete forces in the contact area [25]. The effect of friction on the dynamic characteristics of clearance joints has also been investigated. Marques et al. compared several friction force models for a dynamic analysis on multibody mechanics [26]. Ambrósio proposed a modified Coulomb’s friction law that has been widely applied to prevent the direction of the friction force from changing at a tiny relative velocity [27]. Machado et al. compared the contact force models systematically and provided the information on selection of an appropriate model for different contact scenarios [28]. Tian et al. surveyed the most frequently utilized models of planar and spatial multibody mechanical systems with clearance joints. Then, they discussed the phenomena commonly associated with clearance joint models, such as wear, nonsmooth behaviour, optimization, control, chaos, uncertainty, and links’ flexibility [29].The abovementioned literature is mainly focused on mechanisms with only one clearance joint. Since the actual multibody systems have multiple clearance joints, the analysis can become much more complex, and more than one clearance joint needs to be considered. Megahed and Haroun [30] compared a slider-crank mechanism with two clearance joints to an ideal articulated pair model and found that the clearance has a significant effect on the system dynamics curve. Li et al. studied the dynamic response of a planar slider-crank mechanism with two clearance joints considering harmonic drive and link flexibility [31]. Lu et al. analyzed the dynamics of the steering mechanism and vehicle shimmy system with multiple clearance joints [32, 33]. Wei et al. studied the 9-DOF dynamics of a vehicle shimmy model based on Lagrange’s equation [34]. A series of factors have been considered in the literature to decrease the vibration dynamics due to clearances, such as the effect of flexible linkages and fluid lubricant on the mechanisms with clearances [35, 36]. Chen et al. [37] investigated the dynamic behaviours of a 4-UPS-RPS parallel mechanism considering joint clearances and flexible links. The flexible linkage and fluid lubricant have been shown to introduce effective stiffness and damping that stabilize the performance of multibody mechanical systems. Erkaya and Şefkatlioğlu used a spatial slider-crank mechanism as a numerical example, in which a multiaxis small-length flexural pivot named “pseudo-joint” is used to ensure the necessary mobility. The pseudo-joint is a good solution to decrease undesired reflections of joints with clearance on the system outputs [38]. Thereafter, Erkaya used computational and experimental investigations and discovered that a similar flexible connection between the adjacent mechanism links could minimize the clearance-induced vibration [39]. This is because the flexible connection and flexible link have a clear suspension on the dynamics of mechanism. However, the hydraulic rock-breaker is designed for a low-velocity and heavy-loading situation because the flexible connection and flexible link are unrealistic in practice as the flexible parts could not afford to handle the heavy loading. The clearance joint in a hydraulic rock-breaker is lubricated by grease. The thin grease film could bring damping to dissipate undesirable mechanical vibrations [40]. The hydraulic rock-breaker is under heavy-load conditions. When the journal reached the bearing surfaces, the lubricant film thickness is close to zero. Therefore, the journal and bearing will cause elastic deformation, which will make the reaction forces between the journal and bearing increase suddenly. Flores et al. proposed a force transition model to smooth the reaction forces [41]. When the clearance size becomes bigger and bigger with the use of rock-breaker, it is difficult to solve the dynamic equation. Bannwart et al. developed a hydrodynamic lubrication model to deal with the problem [42]. Zhao et al. employed a simulation and revealed that the clearance could help to reduce the slap noise and wear [43]. The damping related to hydrodynamic lubrication is still too tiny. As a result, an innovative method to decrease the clearance-induced vibration should be found.Nevertheless, the effect of multiple joint clearances on the dynamic properties of a rock-breaker in practical engineering applications has rarely been reported. Under actual working conditions, the clearances existing at joints interact with the hydraulic cylinders, which can cause the dynamic characteristics of the rock-breaker to deteriorate. Therefore, the changes in the rock-breaker dynamic response with multiple radical clearances need to be studied in the scientific and industrial fields for understanding the effect of clearance. In this work, we focus on a planar hydraulic rock-breaker model with multiple joint clearances. First, the hydraulic cylinder model, Lankarani–Nikravesh contact force model, and Lagrange multiplier methods are introduced in Section2. Second, a planar hydraulic rock-breaker model with multiple joint clearances and its parameters are applied as a numerical example in Section 3. Subsequently, a series of simulations based on the different clearance size, inputting force, damping coefficient, and friction on the dynamic performance of the hydraulic rock-breaker were studied, respectively, in Section 4. Finally, we illustrate the conclusion of the study in Section 5.


## 2. Dynamic Model of a Rock-Breaker System with Clearance

A rock-breaker consists of a boom, arm, and breaker, which are driven by hydraulic cylinders, as shown in Figure1. A model of a rock-breaker is presented here to demonstrate how clearances affect its dynamic behaviour. The rock-breaker model includes ten revolute joints with clearance. The joints at the top and bottom of boom cylinder 1 are denoted as P1 and P2. Similarly, the joints on the arm and breaker cylinders are designated as P3, P4, P5, and P6, respectively. The revolute joints, O3, A, and B, connecting the arm to the breaker also have clearances. Note that point P6 has two joints P6 and P6′ that are homocentric. The bearing of joint P6 connects to the rod of breaker cylinder 3, and the bearing of P6′ connects to link 1. Meanwhile, the sharing journal and link 2 are bonded together. The other joints in the clearance model are ideal revolute joints. The clearance model is the rock-breaker model with ten clearance joints. In contrast, all revolute joints in the ideal model are at an ideal condition.Figure 1
Rock-breaker model with clearance joints.


### 2.1. Model of the Hydraulic Cylinder

Hydraulic cylinder 2, which drives the arm, is taken as an example of a submodel. As shown in Figure2, it includes cylinder 2, rod 2, a forward chamber, and a return chamber. Rod 2 moves relative to cylinder 2, which makes volumes V1 and V2 of the chambers change. The radii of the bearings are larger than those of the journals. Thus, the centers of the bearings have the clearances P3 and P4, so the revolute joints are imperfect.Figure 2
Model of a hydraulic cylinder with clearance on the top.When the piston is moving, the continuity equation for the piston forward chambers can be written as follows [44]:(1)q1=A1dx2dt+Cipp1−p2+Cepp1+V1+A1x2βedp1dt,where q1 is the load flow; A1 is the area of the piston in the forward chamber; βe is the effective bulk modulus; V1 is the initial volume of the forward chamber; p1 and p2 are the pressures of the forward and return chambers, respectively; x2 is the piston displacement; and Cip and Cep are the internal and external leakage coefficients.When the valve is off and leaking is neglected,q1 is equal to zero. Then, equation (1) can be integrated in the [0, t] interval:(2)0=A1x2−x20+V10+A1x1βep1−p10,where x20 is the initial piston displacement.The pressurep1 of the forward chamber can be calculated from equation (2):(3)p1=p10−βeA1x2−x20V1+A1x2−x20.The pressurep2 of the return chamber can be determined in the same manner:(4)p2=p20+βeA2x2−x20V2+A2x2−x20,where A2 is the return tank flow, V2 is the initial volume of the return chamber, and p20 is the initial pressure of the return chamber.The forceFp on the piston is described by Newton’s equation:(5)Fp=p1A1−p2A2=m2x¨2+cx˙2+k2x2+F2x2−m2gsinθ2,where m2, c2, and k2 are the total mass, viscous damping coefficient, and stiffness, respectively, of rod 2. F2(x2) is the decomposition of the impact force Fk on the journal along the axial direction of the hydraulic cylinder, as described in the following section. It is evaluated as(6)Fkxk=cosθksinθk−sinθkcosθkFk,k=1,2,3,…,10.The change in length ofx2 relative to the initial value x20 is defined as(7)Δx2=x2−x20.The change in volume caused byx2 is very slight compared with V1. Thus, the change in amplitude of the pressure p1 is described by(8)Δp1=p1−p10=−βeA1Δx1V1.The pressure fluctuation of the other chamber is given by(9)Δp2=+βeA2Δx2V2.The variation in the force on the piston can be denoted as(10)ΔF2=Δp1A1−Δp2A2=−βeA1Δx2V10A1−βeA2Δx2V20A2.Then, the total stiffness of hydraulic cylinder 2 is calculated by(11)Kh2=ΔF2Δx2=−βeA12V10−βeA22V20.The natural frequency of hydraulic cylinder 2 is obtained as follows:(12)f2=Kh2m2.


### 2.2. Clearance Joint Model

The center points of the journal and bearing areOi and Oj, respectively.In Figure3, the eccentricity vector e can be written in the global coordinate system XOY with respect to the bearing and journal as follows:(13)e=rjP−riP=rj+AjsjP−ri+AisiP,where Ai and Aj are the transformation matrices of the bearing and journal, respectively, from the local coordinate system to the global coordinate system XOY and siP and sjP are the position vectors of the contact points. The absolute value of the eccentricity vector is given by(14)e=eTe.Figure 3
Revolution joint with a clearance model.The magnitude of the penetration depth is given by(15)δ=e−c.The unit vectorn along the eccentricity direction is defined as(16)n=ee=exeyT.The angleφ between the vector n and global x-axis is defined as(17)φ=arctanexey.The angleφ transformation matrix is given by(18)Aφ=cosφ−sinφsinφcosφ.The contact pointQ vectors ri and rj can be denoted as(19)rkQ=rk+AkrkP+Rkn,k=i,j.The relative tangential velocityvT and relative normal velocity vN are needed to calculate the contact force, where the relative velocity is projected along the parallel and normal directions of a collision:(20)ν=νNνTT=Aφr˙jQ−r˙iQ.The contact force model can be decomposed into the normal and tangential contact forcesFN and FT, respectively. FN is evaluated with the Lankarani–Nikravesh contact force model, which considers the contact deformation, relative velocity, geometry, and material properties to evaluate the normal contact force:(21)FN=Kδn+Dδ˙.The stiffnessK can be evaluated with equation (21), and δ˙ is the velocity of the relative penetration δ:(22)K=43σi+σjRiRjRi−Rj1/2,where Ri and Rj are the radii of the journal and bearing, respectively. σi and σj can be calculated using the following equation:(23)σk=1−vk2πEk,k=i,j,where νi, Ei and νj, Ej indicate Poisson’s ratio and Young’s modulus of each collision body. The damping D can be obtained using the following equation:(24)D=χδn=3K1−ce24δ˙−,where δ˙− is the initial velocity when the impact occurs, ce is the restitution coefficient, and the exponent n is usually 1.5 for a collision between metal bodies.Finally, the contact force can be evaluated as follows:(25)FN=Kδn1+31−ce24δ˙δ˙−.When relative sliding occurs, the tangential friction forceFT affects the journal and bearing to avoid sliding. This is described by Ambrósio [27] as follows:(26)FT=0,vT≤v0,−cfvT−v0v1−v0FN,v0<vT<v1,−cfFN,vT≥v1,where cf is the friction coefficient and vT is the relative tangential velocity. v0 and v1 are the given tolerances for the tangential velocity. The tolerances for tangential velocity in equation (26) are selected as v0 = 0.01 mm/s and v1 = 0.001 mm/s.In summary, equations (13)–(26) can be used to describe the contact force Fk as(27)Fk=FNkFTk,k=1,2,3,…,10.Figure4 illustrates the forces on hydraulic cylinder 2, which include the force Fp on the piston and impact force Fk.Figure 4
Forces on hydraulic cylinder 2.


### 2.3. Dynamic Constraint Equations

The kinematical constraints of a multibody system are independent and holonomic and can be expressed as(28)Φqr,t=0,where qr is the position and orientation vectors of the bodies in global coordinates and t denotes the time. The virtual power principle can be used to obtain the motivation equation of the system in Cartesian coordinates:(29)Mq¨+ΦqTλ=g,where M is the system mass matrix. The components of M indicate the masses and moments of inertia of each body in the system. ΦqT represents the transformation matrix. λ is the vector of Lagrange multipliers that determines the reactive forces and moment vector at ideal joints. g represents generalized forces, which include the external driving torques or forces, the centrifugal force, the Coriolis force, and the impact forces of clearance joints. The matrix of differential algebraic equations that describe the motion of a multibody system can be written as(30)MΦqTΦq0q¨rλ=gγ.


## 2.1. Model of the Hydraulic Cylinder

Hydraulic cylinder 2, which drives the arm, is taken as an example of a submodel. As shown in Figure2, it includes cylinder 2, rod 2, a forward chamber, and a return chamber. Rod 2 moves relative to cylinder 2, which makes volumes V1 and V2 of the chambers change. The radii of the bearings are larger than those of the journals. Thus, the centers of the bearings have the clearances P3 and P4, so the revolute joints are imperfect.Figure 2
Model of a hydraulic cylinder with clearance on the top.When the piston is moving, the continuity equation for the piston forward chambers can be written as follows [44]:(1)q1=A1dx2dt+Cipp1−p2+Cepp1+V1+A1x2βedp1dt,where q1 is the load flow; A1 is the area of the piston in the forward chamber; βe is the effective bulk modulus; V1 is the initial volume of the forward chamber; p1 and p2 are the pressures of the forward and return chambers, respectively; x2 is the piston displacement; and Cip and Cep are the internal and external leakage coefficients.When the valve is off and leaking is neglected,q1 is equal to zero. Then, equation (1) can be integrated in the [0, t] interval:(2)0=A1x2−x20+V10+A1x1βep1−p10,where x20 is the initial piston displacement.The pressurep1 of the forward chamber can be calculated from equation (2):(3)p1=p10−βeA1x2−x20V1+A1x2−x20.The pressurep2 of the return chamber can be determined in the same manner:(4)p2=p20+βeA2x2−x20V2+A2x2−x20,where A2 is the return tank flow, V2 is the initial volume of the return chamber, and p20 is the initial pressure of the return chamber.The forceFp on the piston is described by Newton’s equation:(5)Fp=p1A1−p2A2=m2x¨2+cx˙2+k2x2+F2x2−m2gsinθ2,where m2, c2, and k2 are the total mass, viscous damping coefficient, and stiffness, respectively, of rod 2. F2(x2) is the decomposition of the impact force Fk on the journal along the axial direction of the hydraulic cylinder, as described in the following section. It is evaluated as(6)Fkxk=cosθksinθk−sinθkcosθkFk,k=1,2,3,…,10.The change in length ofx2 relative to the initial value x20 is defined as(7)Δx2=x2−x20.The change in volume caused byx2 is very slight compared with V1. Thus, the change in amplitude of the pressure p1 is described by(8)Δp1=p1−p10=−βeA1Δx1V1.The pressure fluctuation of the other chamber is given by(9)Δp2=+βeA2Δx2V2.The variation in the force on the piston can be denoted as(10)ΔF2=Δp1A1−Δp2A2=−βeA1Δx2V10A1−βeA2Δx2V20A2.Then, the total stiffness of hydraulic cylinder 2 is calculated by(11)Kh2=ΔF2Δx2=−βeA12V10−βeA22V20.The natural frequency of hydraulic cylinder 2 is obtained as follows:(12)f2=Kh2m2.


## 2.2. Clearance Joint Model

The center points of the journal and bearing areOi and Oj, respectively.In Figure3, the eccentricity vector e can be written in the global coordinate system XOY with respect to the bearing and journal as follows:(13)e=rjP−riP=rj+AjsjP−ri+AisiP,where Ai and Aj are the transformation matrices of the bearing and journal, respectively, from the local coordinate system to the global coordinate system XOY and siP and sjP are the position vectors of the contact points. The absolute value of the eccentricity vector is given by(14)e=eTe.Figure 3
Revolution joint with a clearance model.The magnitude of the penetration depth is given by(15)δ=e−c.The unit vectorn along the eccentricity direction is defined as(16)n=ee=exeyT.The angleφ between the vector n and global x-axis is defined as(17)φ=arctanexey.The angleφ transformation matrix is given by(18)Aφ=cosφ−sinφsinφcosφ.The contact pointQ vectors ri and rj can be denoted as(19)rkQ=rk+AkrkP+Rkn,k=i,j.The relative tangential velocityvT and relative normal velocity vN are needed to calculate the contact force, where the relative velocity is projected along the parallel and normal directions of a collision:(20)ν=νNνTT=Aφr˙jQ−r˙iQ.The contact force model can be decomposed into the normal and tangential contact forcesFN and FT, respectively. FN is evaluated with the Lankarani–Nikravesh contact force model, which considers the contact deformation, relative velocity, geometry, and material properties to evaluate the normal contact force:(21)FN=Kδn+Dδ˙.The stiffnessK can be evaluated with equation (21), and δ˙ is the velocity of the relative penetration δ:(22)K=43σi+σjRiRjRi−Rj1/2,where Ri and Rj are the radii of the journal and bearing, respectively. σi and σj can be calculated using the following equation:(23)σk=1−vk2πEk,k=i,j,where νi, Ei and νj, Ej indicate Poisson’s ratio and Young’s modulus of each collision body. The damping D can be obtained using the following equation:(24)D=χδn=3K1−ce24δ˙−,where δ˙− is the initial velocity when the impact occurs, ce is the restitution coefficient, and the exponent n is usually 1.5 for a collision between metal bodies.Finally, the contact force can be evaluated as follows:(25)FN=Kδn1+31−ce24δ˙δ˙−.When relative sliding occurs, the tangential friction forceFT affects the journal and bearing to avoid sliding. This is described by Ambrósio [27] as follows:(26)FT=0,vT≤v0,−cfvT−v0v1−v0FN,v0<vT<v1,−cfFN,vT≥v1,where cf is the friction coefficient and vT is the relative tangential velocity. v0 and v1 are the given tolerances for the tangential velocity. The tolerances for tangential velocity in equation (26) are selected as v0 = 0.01 mm/s and v1 = 0.001 mm/s.In summary, equations (13)–(26) can be used to describe the contact force Fk as(27)Fk=FNkFTk,k=1,2,3,…,10.Figure4 illustrates the forces on hydraulic cylinder 2, which include the force Fp on the piston and impact force Fk.Figure 4
Forces on hydraulic cylinder 2.


## 2.3. Dynamic Constraint Equations

The kinematical constraints of a multibody system are independent and holonomic and can be expressed as(28)Φqr,t=0,where qr is the position and orientation vectors of the bodies in global coordinates and t denotes the time. The virtual power principle can be used to obtain the motivation equation of the system in Cartesian coordinates:(29)Mq¨+ΦqTλ=g,where M is the system mass matrix. The components of M indicate the masses and moments of inertia of each body in the system. ΦqT represents the transformation matrix. λ is the vector of Lagrange multipliers that determines the reactive forces and moment vector at ideal joints. g represents generalized forces, which include the external driving torques or forces, the centrifugal force, the Coriolis force, and the impact forces of clearance joints. The matrix of differential algebraic equations that describe the motion of a multibody system can be written as(30)MΦqTΦq0q¨rλ=gγ.


## 3. Numerical Examples

The parameters and initial state of the planer rock-breaker model with multiple clearance joints are listed below.Table1 presents the mass parameters of the rock-breaker model according to its geometric dimensions. Table 2 presents the point coordinates of the rock-breaker in the initial state. Table 3 lists the parameters of the hydraulic cylinders. The rod radius of the boom cylinder was 65 mm, and the piston radius was 110 mm. The arm cylinder and breaker cylinder had the same rod and piston to simplify the number of components and for easy maintenance. The rod radius was 55 mm, and the piston radii of the two cylinders were both 85 mm. Thus, the piston areas of the forward and return chambers were equal. The planar rock-breaker mechanism was modeled with ten joint clearances.Table 1
Mass and moment of inertia parameters of the rock-breaker model.
Body
Mass (kg)
Moment of inertia (kg·m2)
Boom
595
486
Arm
386
136
Breaker
198
23
Link 1
22
0.53
Link 2
16
0.29Table 2
Points of the rock-breaker in global coordinates.
Point
Coordinates (m, m)
Point
Coordinates (m, m)
O
1
[0.57, 1.66]T
P
4
[3.37, 3.38]T
O
2
[3.26, 2.98]T
P
5
[3.64, 2.95]T
O
3
[3.64, 1.48]T
P
6
[3.87, 1.89]T
P
1
[0.79, 1.32]T
A
[3.93, 1.53]T
P
2
[1.58, 2.76]T
B
[1.58, 2.76]T
P
3
[1.80, 3.26]T
P
b
[4.24, 0.11]TTable 3
Parameters of the hydraulic cylinders.
Body
Boom
Arm
Breaker
Initial length (m)
0.25
0.45
0.21
Mass of the rod (kg)
29.7
26.1
20.7
Moment of inertia of the rod (kg·m2)
2.91
2.63
1.36
Piston area of the forward chamber (mm2)
9503
5675
5675
Piston area of the return chamber (mm2)
6185
3299
3299
Initial volume of the forward chamber (L)
2.376
2.554
1.192
Initial volume of the return chamber (L)
4.02
2.144
2.639Table4 presents the geometric and physical parameters of all joint clearances. Young’s modulus, Poisson’s ratio, and friction coefficient were the same for each joint because the joints were made of the same material. The clearance size c was adjusted by varying the radius of the journal of joints.Table 4
Geometric and physical parameters of the revolute clearance joints.
Joint no.
Joint position
Radius of the bearing (mm)
Young’s modulus (GPa)
Poisson’s ratio
Friction coefficient
1
P
1
30
207
0.3
0.01
2
P
2
30
207
0.3
0.01
3
P
3
25
207
0.3
0.01
4
P
4
25
207
0.3
0.01
5
P
5
25
207
0.3
0.01
6
P
6
25
207
0.3
0.01
7
P
6
′
25
207
0.3
0.01
8
A
25
207
0.3
0.01
9
B
25
207
0.3
0.01
10
O
3
30
207
0.3
0.01In the simulation, the breaker cyclically struck the ground to induce the periodic input forceFb in each cycle parallel to the y-axis, which is shown in Figure 5. The force Fb can be expressed as(31)Fbt=0,t0<t≤t1,Fmaxt2−t1t−t1,t1<t≤t2,Fmaxt3−t2t3−t,t2<t≤t3,where Fmax is the maximum impact force and was set to 4, 8, 12, and 16 kN in this example. t1, t2, and t3 denote the changing points of time in every period and were set to 0.03, 0.06, and 0.11 s, respectively.Figure 5
Input forceFb in each period when the breaker strikes the ground.


## 4. Calculation and Analysis of Results

A set of initial conditions, positions, and velocities was assigned to the dynamic model equations for simulation in Section3. The simulation time step was set to 0.0001 s for balanced computational efficiency and accuracy, and the Newmark method was utilized for integration in the software ADAMS. At the beginning of the simulation, the bearings of the clearance joints were concentric with the journals. The total simulation time was set to 10 s in order to reach a steady state. The simulation parameters of the rock-breaker model are illustrated in Table 5.Table 5
Simulation parameters of the rock-breaker model.
Description
Value
Restitution coefficient
0.9
Dynamic friction coefficient
0.01
Integration step size (s)
1 × 10−4
Integration tolerance
1 × 10−6The motion of hydraulic cylinder 2, which moves the arm, was taken as an example and is shown in Figure6. As shown in Figures 6(a) and 6(b), the displacement x2 of rod 2 clearly differed in the ideal and clearance models. All clearances c in the clearance model were 0.5 mm. Figures 6(c) and 6(d) show the hydraulic force on the piston Fp, which was very similar to the displacement x2 in Figures 6(a) and 6(b). The relationship between the hydraulic force on the piston Fp and rod 2 displacement x2 is described in equation (10) by a linear relationship. Figures 6(e) and 6(f) illustrate the amplitudes of the change in pressure with the movement of rod 2 in each hydraulic cylinder chamber. The initial pressure p1 was 0.5 MPa, and the other chamber’s pressure p2 was 1 MPa. The piston connected to rod 2 caused the volumes of the chambers to change with the motion of x2. The bulk modulus βe connects x2 and the pressures, as shown in equations (8) and (9). The results showed that the compression and expansion of the oil caused a linear change in pressure according to the displacement x2.Figure 6
Differences in loads and displacements for the ideal and clearance models: rod 2 displacementx2, hydraulic force Fp on the piston, and hydraulic chamber pressure p1/p2 of the hydraulic cylinder in the ideal model (a, c, e) and clearance model (b, d, f).
(a)
(b)
(c)
(d)
(e)
(f)Figure7 illustrates the frequency distribution of x2 according to the Fourier transform. The peak frequency was 9.1 Hz in the ideal model. In contrast, the frequency distribution of x2 for the clearance model had the same peak of 9.1 Hz but also another peak frequency of 175 Hz. The latter was the natural frequency for the spring-mass system of hydraulic cylinder 2, which consisted of compressible oil and the masses of the rod and piston, and can be calculated using equation (12).Figure 7
Frequency distribution ofx2 in the ideal model (a) and clearance model (b) (c = 0.5).
(a)
(b)Figure8 shows the velocity and acceleration of rod 2 for 4.5–5.75 s to give more detail on the movement. The velocity and acceleration of rod 2 were smoother in the ideal model than those in the clearance model. The velocity amplitude of the ideal model was less than 0.11 m/s, while the velocity amplitude of the clearance model was 3.9 m/s, which is 35.5 times greater. The maximum acceleration of rod 2 in the ideal model was 22.1 m/s2 and is plotted in Figure 8(b). The acceleration of the clearance model had more peaks, and the maximum peak of 29,260 m/s2 was 1324 times bigger than that of the ideal model.Figure 8
Movement of rod 2: velocity and acceleration in the ideal model (a, b) and clearance model (c, d) (c = 0.5 mm, 4.5 s < t < 5.75 s).
(a)
(b)
(c)
(d)In this study, the velocity analysis method was used to determine which period movement was the rapid vibration. The velocity was defined to be equal to 0 at a peak point, which means thatx2 was at a peak. The interval time of the peak points was less than 0.006 s for the rapid vibration period considering the natural frequency of 175 Hz. The other interval was the slow movement period. Figures 8(a) and 8(c) show the results of the velocity analysis. The velocity in the clearance model had two alternating modes of motion: rapid vibration (red line) and slow movement (blue line). The rapid vibration corresponded to the frequency of 175 Hz, which was the natural frequency of hydraulic cylinder 2 as noted above. The acceleration of the clearance model was also divided into rapid vibration (blue) and slow movement (red) based on the above velocity analysis. The two movements can clearly be distinguished, which demonstrates the feasibility of the velocity analysis method.When the hydraulic breaker was working, the input forceFb acted on the rock-breaker and caused the arm to rotate counterclockwise. In the ideal model, the arm and rod 2 connected by an ideal joint caused the velocity of rod 2 to be related to the angular velocity of the arm. The arm slowly rotated clockwise because of gravity, and thus, the acceleration of rod 2 was nearly 1 m/s2. The velocity of rod 2 was smaller than that in the clearance model because of the greater moment of inertia of the arm and lack of impact forces caused by nonideal joints.The clearance model contained clearances between bearings and journals. Therefore, the velocity and acceleration of rod 2 demonstrated more peaks produced by impact forces. Figure9 shows the four periods t1, t2, t3, and t4, which were extracted to demonstrate the movement of rod 2 in the clearance model. According to the literature, the movement x2 can be classified into three states: freedom, collision, and contact [22].Figure 9
Displacementx2 of rod 2 in the clearance model (c = 0.5 mm, 5.5 s < t < 5.75 s).In order to compare the distances from journalsP3 and P4 to x2 based on the same initial value, the length l20 in Figure 2 was subtracted from the distance so that the two curves would coincide at the same reference. The black dotted line in Figure 9 is the distance between the two journals of the hydraulic cylinder after adjustment. In the clearance model, the clearances in joints P3 and P4 were 0.5 mm each. Thus, the sum boundary of the two clearances was 1 mm. Figure 9 plots two boundary lines as red dashed lines. The time spans obtained from the above velocity analysis were used to segment the movement x2 into rapid vibration (blue) and slow movement (red).Figure9 illustrates the significant rapid vibration (blue line) when the displacement amplitude of rod 2 was less than the sum clearance of 1 mm. In contrast, when the displacement amplitude of rod 2 was more than 1 mm, the journal began to connect with the bearing (red line) and created the penetration depth δ. The penetration depth δ in Figures 9, 10, 11, and 12 is too tiny and near to the clearance boundary line. In the study, the penetration depth δ is exaggerated 500 times in order to illustrate the penetration clearly. The hydraulic cylinder began to connect with the arm and move together with it, which is similar to the mechanism in the ideal model.Figure 10
Trajectories and impact forces in clearance jointsP3 (a, b) and P4 (c, d) (c = 0.5 mm).
(a)
(b)
(c)
(d)Figure 11
Impact forces and trajectories of the piston for four periods: (a, f, k, p)P3 impact force, (b, g, l, q) clearance P3 trajectory, (c, h, m, r) force Fp on the piston, (d, i, n, s) clearance P4 trajectory, and (e, j, o, t) P4 impact force.Figure 12
Trajectories in clearance jointsP3 and P4 when Fmax is 4 kN (a, b) and 16 kN (c, d) (c = 0.5 mm, t = 5–10 s).
(a)
(b)
(c)
(d)The two clearancesP3 and P4 at the top and bottom of hydraulic cylinder 2 were examined. Figure 10 illustrates the trajectories and impact forces in the clearances, within which the penetration depth δ was also exaggerated. The impact forces were divided into the free/collision state (red line) and the contact state (blue line) in correspondence to the rapid vibration and slow movement based on the velocity analysis method described previously. As shown in Figures 10(b) and 10(d), the red lines represent relatively gentle contact forces with maximum contact forces of 26,771 and 27,788 N, which are very close to each other. The blue lines indicate free and impact forces. The maximum impact force was 280,162 N for clearance P3 and 212,041 N for clearance P4. The contact forces in the slow movement were smaller than the impact forces in the rapid vibration. The dynamics of hydraulic cylinder 2 became worse with input force as reported in the literature [30].Figure11 shows the details of the hydraulic cylinder. There are five columns and four rank pictures according to the four periods t1–t4 in Figure 9. Each rank illustrates the P3 impact force, clearance P3 trajectory, force Fp on the piston, clearance P4 trajectory, and P4 impact force. The clockwise motion of the arm in periods t1 and t3 is shown in Figures 11(a)–11(e) and 11(k)–11(o). Figures 11(f)–11(j) and 11(p)–11(t) indicate the counterclockwise movement in periods t2 and t4.As the arm moved clockwise, the distance between the journalsP3 and P4 increased. Then, the bearings moved to the middle of the hydraulic cylinder, as shown in Figures 11(b), 11(d), 11(g), 11(i), 11(l), 11(n), 11(q), and 11(s). The blue triangle indicates the starting point of the rapid vibration trajectories, and the red dot on the other side represents the end of the slow movement trajectories. The bearings contained clearances that supplied free space for the journal to move during the rapid vibration. The journal and bearing collided and caused an impact force depending on the penetration depth when the bearing reached the rim of the journal, which can be determined using equation (21). The impact force was greater than the force Fp on the piston when hydraulic cylinder 2 was in free motion, so the direction of rod 2’s motion changed. The hydraulic oil bulk modulus βe acted on rod 2 just like a spring-mass system with the fixed frequency of 175 Hz, which triggered the rapid vibration (blue) shown in Figure 11. Subsequently, the vibration amplitude of rod 2 decreased because of the internal damping and clearance constraints. In Figure 11, the red lines illustrate the slow movement stage when the bearing was in contact with the journal. This produced a continuous contact force that was nearly equal to the force Fp on the piston, so rod 2 moved smoothly as shown in Figures 11(a), 11(c), 11(e), 11(f), 11(h), 11(j), 11(k), 11(m), 11(r), 11(o), 11(p), and 11(t).Figures11(f)–11(j) and 11(p)–11(t) display the counterclockwise motion of the arm, which shortened the distance between the journal and bearing. The trajectories were separated from the middle of the hydraulic cylinder. The chaotic trajectories in blue were affected by the impact force, friction, and gravity. When the bearing arrived at the edge of the journal, a continuous contact force was generated that was equal to the force Fp on the piston.


### 4.1. Effect of the Breaker Input Force

The influence of the breaker input maximum force atFmax = 4, 8, 12, and 16 kN was investigated, as shown in Figure 13. At Fmax = 4 kN, hydraulic cylinder 2 stabilized after 7 s because the impact force of the breaker was too small to resist the influence of gravity on the arm. The extension of hydraulic cylinder 2 provided a tensile force that eventually balanced with gravity. When Fmax = 12 and 16 kN, the input maximum force was large enough to oppose gravity and cause the arm to rotate counterclockwise. Hydraulic cylinder 2 provided a propulsive force that prevented the arm from rotating counterclockwise. Note that less time was required to enter the stable state when Fmax was 16 kN than that when it was 12 kN. This is because the former provided a larger moment to the arm, which caused the arm to move faster.Figure 13
Displacementx2 of rod 2 with different Fmax.
(a)
(b)
(c)
(d)However,Fmax = 8 kN did not stop vibration because the impulse moment of the force of the breaker was approximate to the impulse moment of the effect of gravity on the arm. The random impact force of the clearance joint made it difficult for the simulation to converge to a stable state at 8 kN.Figure12 shows the trajectories of Fmax = 4 and 16 kN reached the stable state in 5–10 s. The bearing was in contact with the journal and penetrated the journal periodically. The trajectories of Fmax = 4 kN were separated from each other, which indicates the extension of the hydraulic cylinder, as shown in Figures 12(a) and 12(b). Figures 12(c) and 12(d) illustrate the hydraulic cylinder in the stroke state.Figure14 shows the maximum forces in the rapid vibration and slow movement stage at a clearance of 0.5 mm. The maximum contact force at P3 (red line) ranged from 27,788 to 31,542 N. The maximum contact force at the opposite side P4 was 26,771–37,028 N. The fluctuation ratios for the forces with the two clearances were 138.32% greater than the minimum force of 26,771 N. With the rapid vibration, the maximum P3 impact force was 140,088–173,012 N, which represents a fluctuation of 123.5%. The P4 impact force varied from 183,253 to 242,387 N, which has a fluctuation ratio of 132.3%. The contact forces were smaller than the impact forces in the rapid vibration stage.Figure 14
Maximum contact and impact force inP3 and P4 with different Fmax.Based on the above velocity analysis, the rapid vibration total time (VTT) for the whole simulation process should be decreased to diminish the impact force caused by the clearance, which will aggravate joint wear. The velocity was obtained by differentiating the displacementx2 with time because this is easily measured in actual experiments. In this work, the rapid vibration quit time (VQT) and VTT were defined and used as measurement parameters for collisions. As shown in Figure 15, the force vibration quit time (FVQT) and force vibration total time (FVTT) were obtained according to the free/collision and contact states of the P3 (diamond) and P4 (square) impact forces. The VQT and VTT of rod 2 were close to the FVQT and FVTT, which verifies that the VQT and VTT can be used as parameters for the impact force.Figure 15
Vibration quit time (VQT) and vibration total time (VTT) based on an analysis of the velocity and impact force with differentFmax.


### 4.2. Effect of the Clearance Size

The influence of the clearance size is needed to be analyzed because the clearance increases gradually during rock-breaker operation, which increases the free space between the bearing and journal. The velocity analysis method was used to investigate the VQT and VTT.AtFmax = 4 kN, the VQT and VTT remained small when each clearance was less than 2.5 mm. VQT was 6.3 s and VTT increased when the clearance was 3.0 mm. When the clearance was greater than 3.0 mm, rod 2 could not stop vibrating. The whole simulation process was not stable at any clearance when Fmax = 8 kN. The VTT increased with the clearance. As shown in Figures 16(c) and 16(d), the VQT and VTT increased with the clearance size when Fmax = 12 kN and 16 kN. Instability eventually occurred when the clearance was greater than 3.0 mm. When Fmax = 12 and 16 kN, the motions became chaotic when the clearance was 3.5 mm or greater. Figures 16(c) and 16(d) show that the VQT and VTT were shorter when Fmax was 16 kN than those when Fmax was 12 kN. The simulation outputs show that the dynamics of the system become worse. Meanwhile, VQT and VTT prolong with the clearance size [16, 20].Figure 16
Vibration quit time and vibration total time based on an analysis of the velocity of rod 2 with differentFmax: (a) Fmax = 4 kN; (b) Fmax = 8 kN; (c) Fmax = 12 kN; (d) Fmax = 16 kN.
(a)
(b)
(c)
(d)Thus, the clearance should be limited to 2.5 mm to prevent the rock-breaker from vibrating. This is because 2.5 mm was a key changing value for the clearance models VTT and VQT at different input forces. The following sections discuss how to reduce the VTT and VQT for the 2.5 mm clearance model whenFmax = 8 kN.


### 4.3. Effect of Damping Coefficient

Figures17 and 18 show that increased damping coefficient did not shorten the VQT until the damping was 4 kN/(m/s). In contrast, the VTT was reduced. Rod 2 demonstrated some rapid vibration until the stable state was reached, but it quickly decreased when the damping was 4 kN/(m/s). When D = 5 and 6 kN/(m/s), the VQT and VTT clearly decreased. However, this approach is difficult to use in engineering practice because of the large damping values required.Figure 17
Displacementx2 of rod 2 with different damping coefficients D with c = 2.5 and Fmax = 8 kN: (a) D = 1 kN; (b) D = 2 kN; (c) D = 3 kN; (d) D = 4 kN; (e) D = 5 kN; (f) D = 6 kN.
(a)
(b)
(c)
(d)
(e)
(f)Figure 18
Vibration quit time and vibration total time of rod 2 with different damping coefficients (c = 2.5, Fmax = 8 kN).


### 4.4. Effect of Friction

A friction force of less than 150 N in the hydraulic cylinder did not clearly reduce the VQT, as shown in Figures19 and 20. Only the VTT was reduced. When the friction reached 200 N, the movement of rod 2 reached a stable state, but some rapid vibration remained. The VQT and VTT rapidly decreased when the friction was greater than 200 N. This friction value is smaller than the internal friction of a traditional hydraulic cylinder. It can be feasibly realized in engineering practice by adjusting the O-ring parameters of the hydraulic cylinder.Figure 19
Displacementx2 of rod 2 with different frictions f with c = 2.5 and Fmax = 8 kN: (a) f = 50 N; (b) f = 100 N; (c) f = 150 N; (d) f = 200 N; (e) f = 250 N; (f) f = 300 N.
(a)
(b)
(c)
(d)
(e)
(f)Figure 20
Vibration quit time and vibration total time of rod 2 with different frictions (c = 2.5, Fmax = 8 kN).In the preceding part of the paper, Section2 introduced the hydraulic cylinder model, Lankarani–Nikravesh contact force model, and Lagrange multiplier method, and a planar hydraulic rock-breaker model with multiple joint clearances were investigated in Section 3. The effect of the different clearance size, inputting force, damping coefficient, and friction on the dynamic performance of the clearance joint was studied, and they illustrated the dynamics of the hydraulic rock-breaker are sensitive to the varying parameters. The present findings are of significance to the theoretical study of the nonlinear dynamic response of the hydraulic rock-breaker.


## 4.1. Effect of the Breaker Input Force

The influence of the breaker input maximum force atFmax = 4, 8, 12, and 16 kN was investigated, as shown in Figure 13. At Fmax = 4 kN, hydraulic cylinder 2 stabilized after 7 s because the impact force of the breaker was too small to resist the influence of gravity on the arm. The extension of hydraulic cylinder 2 provided a tensile force that eventually balanced with gravity. When Fmax = 12 and 16 kN, the input maximum force was large enough to oppose gravity and cause the arm to rotate counterclockwise. Hydraulic cylinder 2 provided a propulsive force that prevented the arm from rotating counterclockwise. Note that less time was required to enter the stable state when Fmax was 16 kN than that when it was 12 kN. This is because the former provided a larger moment to the arm, which caused the arm to move faster.Figure 13
Displacementx2 of rod 2 with different Fmax.
(a)
(b)
(c)
(d)However,Fmax = 8 kN did not stop vibration because the impulse moment of the force of the breaker was approximate to the impulse moment of the effect of gravity on the arm. The random impact force of the clearance joint made it difficult for the simulation to converge to a stable state at 8 kN.Figure12 shows the trajectories of Fmax = 4 and 16 kN reached the stable state in 5–10 s. The bearing was in contact with the journal and penetrated the journal periodically. The trajectories of Fmax = 4 kN were separated from each other, which indicates the extension of the hydraulic cylinder, as shown in Figures 12(a) and 12(b). Figures 12(c) and 12(d) illustrate the hydraulic cylinder in the stroke state.Figure14 shows the maximum forces in the rapid vibration and slow movement stage at a clearance of 0.5 mm. The maximum contact force at P3 (red line) ranged from 27,788 to 31,542 N. The maximum contact force at the opposite side P4 was 26,771–37,028 N. The fluctuation ratios for the forces with the two clearances were 138.32% greater than the minimum force of 26,771 N. With the rapid vibration, the maximum P3 impact force was 140,088–173,012 N, which represents a fluctuation of 123.5%. The P4 impact force varied from 183,253 to 242,387 N, which has a fluctuation ratio of 132.3%. The contact forces were smaller than the impact forces in the rapid vibration stage.Figure 14
Maximum contact and impact force inP3 and P4 with different Fmax.Based on the above velocity analysis, the rapid vibration total time (VTT) for the whole simulation process should be decreased to diminish the impact force caused by the clearance, which will aggravate joint wear. The velocity was obtained by differentiating the displacementx2 with time because this is easily measured in actual experiments. In this work, the rapid vibration quit time (VQT) and VTT were defined and used as measurement parameters for collisions. As shown in Figure 15, the force vibration quit time (FVQT) and force vibration total time (FVTT) were obtained according to the free/collision and contact states of the P3 (diamond) and P4 (square) impact forces. The VQT and VTT of rod 2 were close to the FVQT and FVTT, which verifies that the VQT and VTT can be used as parameters for the impact force.Figure 15
Vibration quit time (VQT) and vibration total time (VTT) based on an analysis of the velocity and impact force with differentFmax.


## 4.2. Effect of the Clearance Size

The influence of the clearance size is needed to be analyzed because the clearance increases gradually during rock-breaker operation, which increases the free space between the bearing and journal. The velocity analysis method was used to investigate the VQT and VTT.AtFmax = 4 kN, the VQT and VTT remained small when each clearance was less than 2.5 mm. VQT was 6.3 s and VTT increased when the clearance was 3.0 mm. When the clearance was greater than 3.0 mm, rod 2 could not stop vibrating. The whole simulation process was not stable at any clearance when Fmax = 8 kN. The VTT increased with the clearance. As shown in Figures 16(c) and 16(d), the VQT and VTT increased with the clearance size when Fmax = 12 kN and 16 kN. Instability eventually occurred when the clearance was greater than 3.0 mm. When Fmax = 12 and 16 kN, the motions became chaotic when the clearance was 3.5 mm or greater. Figures 16(c) and 16(d) show that the VQT and VTT were shorter when Fmax was 16 kN than those when Fmax was 12 kN. The simulation outputs show that the dynamics of the system become worse. Meanwhile, VQT and VTT prolong with the clearance size [16, 20].Figure 16
Vibration quit time and vibration total time based on an analysis of the velocity of rod 2 with differentFmax: (a) Fmax = 4 kN; (b) Fmax = 8 kN; (c) Fmax = 12 kN; (d) Fmax = 16 kN.
(a)
(b)
(c)
(d)Thus, the clearance should be limited to 2.5 mm to prevent the rock-breaker from vibrating. This is because 2.5 mm was a key changing value for the clearance models VTT and VQT at different input forces. The following sections discuss how to reduce the VTT and VQT for the 2.5 mm clearance model whenFmax = 8 kN.


## 4.3. Effect of Damping Coefficient

Figures17 and 18 show that increased damping coefficient did not shorten the VQT until the damping was 4 kN/(m/s). In contrast, the VTT was reduced. Rod 2 demonstrated some rapid vibration until the stable state was reached, but it quickly decreased when the damping was 4 kN/(m/s). When D = 5 and 6 kN/(m/s), the VQT and VTT clearly decreased. However, this approach is difficult to use in engineering practice because of the large damping values required.Figure 17
Displacementx2 of rod 2 with different damping coefficients D with c = 2.5 and Fmax = 8 kN: (a) D = 1 kN; (b) D = 2 kN; (c) D = 3 kN; (d) D = 4 kN; (e) D = 5 kN; (f) D = 6 kN.
(a)
(b)
(c)
(d)
(e)
(f)Figure 18
Vibration quit time and vibration total time of rod 2 with different damping coefficients (c = 2.5, Fmax = 8 kN).


## 4.4. Effect of Friction

A friction force of less than 150 N in the hydraulic cylinder did not clearly reduce the VQT, as shown in Figures19 and 20. Only the VTT was reduced. When the friction reached 200 N, the movement of rod 2 reached a stable state, but some rapid vibration remained. The VQT and VTT rapidly decreased when the friction was greater than 200 N. This friction value is smaller than the internal friction of a traditional hydraulic cylinder. It can be feasibly realized in engineering practice by adjusting the O-ring parameters of the hydraulic cylinder.Figure 19
Displacementx2 of rod 2 with different frictions f with c = 2.5 and Fmax = 8 kN: (a) f = 50 N; (b) f = 100 N; (c) f = 150 N; (d) f = 200 N; (e) f = 250 N; (f) f = 300 N.
(a)
(b)
(c)
(d)
(e)
(f)Figure 20
Vibration quit time and vibration total time of rod 2 with different frictions (c = 2.5, Fmax = 8 kN).In the preceding part of the paper, Section2 introduced the hydraulic cylinder model, Lankarani–Nikravesh contact force model, and Lagrange multiplier method, and a planar hydraulic rock-breaker model with multiple joint clearances were investigated in Section 3. The effect of the different clearance size, inputting force, damping coefficient, and friction on the dynamic performance of the clearance joint was studied, and they illustrated the dynamics of the hydraulic rock-breaker are sensitive to the varying parameters. The present findings are of significance to the theoretical study of the nonlinear dynamic response of the hydraulic rock-breaker.


## 5. Conclusions

This study investigated the dynamic responses of a planar rock-breaker mechanism with multiple joint clearances, and the stiffness of the hydraulic cylinder was caused by the effective bulk modulus. The model of the rock-breaker with multiple clearance joints was proposed based on the combination of the hydraulic cylinder model, the clearance joints based on the Lankarani–Nikravesh contact force model, and the Lagrange multiplier method. The main results of the numerical simulation and dynamic analysis are summarized as follows:(1)
The rod of hydraulic cylinder 2 was taken as an example, and the dynamic response of the rod clearly changed with clearances by alternating between rapid vibration and slow movement. This is in contrast to the ideal model without clearances. The stiffness of the hydraulic cylinder caused the rod to vibrate rapidly in correspondence to the natural frequency.(2)
When the rapid vibration coincided with impact forces, the dynamic response of the planar rock-breaker mechanism degraded and aggravated wear. A series of scenarios indicated that the rapid vibration total time and rapid vibration quit time of the rod movement increase with the clearance size. Meanwhile, some certain values of the impact force could maximize the total time and the quit time mentioned above.(3)
Two approaches to reduce the rapid vibration were considered: increasing the damping coefficient and increasing the friction. The simulation results showed that damping can reduce rapid vibration but is difficult to be used in practical engineering because of the large values involved. However, increasing the friction can significantly reduce the rapid vibration state of rod 2 and is feasible for practical engineering applications.The present work gives more insight into the changing dynamics of the planar hydraulic rock-breaker mechanism with multiple clearance joints and provides a theoretical support for the further study of the hydraulic rock-breaker.



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*Source: 4718456-2019-02-04.xml*