# Dynamic Analysis of Slider-Crank Mechanism with a Cracked Rod

Authors: Shouguo Cheng; Shulin Liu
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

The dynamical equation of the slider-crank mechanism is established by using Lagrange equation and Newton’s second law. The slider-crank mechanism with an open crack rod is investigated and then establishes the equivalent mechanics model by a massless torsional spring to simulate the influence of the crack in the rod, and the mechanism of a cracked rod is divided into two subsystems. The dynamical equation of the slider-crank mechanism with a crack rod is established. Comparing the dynamic analysis results between with and without crack in the rod, the results show that the existence of the crack leads to a great change in the motion characteristics of the slider. The calculated maximum Lyapunov exponent is positive, which shows that the movement of the slider in the crank slider mechanism with a cracked rod is chaotic.


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## Body

## 1. Introduction

The precise relationship between the input and output is important for slider-crank mechanism [1]. For example in certain application, the rotation of crank is considered as the input and the displacement of the slider is considered as the output. Since the mechanism is not manufactured perfectly and cracks always exit which is known to be the source to reduce system reliability and precision [2, 3]. Moreover, it is difficult to calculate the influence by normal formula directly, if there is a crack in the rod. This kind of crack in the rod may cause the mechanism to exhibit nonlinear behaviour which should influence the mechanism dynamic when rod is activated according to crank rotating. So, this behaviour should be studied. Jin Zeng, HuiMa, Wensheng Zhang and RangchunWen mix elements by combining beam elements and solid elements to establish the finite element (FE) model for cracked cantilever beams and use the area damage factor to evaluate the crack levels [4]. Ugo Andresus, Paolo Casini use 2D quadrilateral to model the beam, then the natural frequencies (and the associated mode shapes) of the cracked cantilever beam were obtained [5], and they also use two-dimensional finite elements to consider the cantilever beam with an asymmetric edge crack as the plan problem and then conclude the behaviour of breathing crack which is simulated as a frictionless contact problem [6, 7]. Mihai Dupac and David G.Beale model the crack in rob by massless torsional spring, but the slider-crank mechanism is modeled by one equation of motion [8]. Andrea Carpinteri, AndreaSpagnoli, and SabrinaVantadori build up a general linear hardening rule for the fibers and a linear-elastic law for the matrix to assume an elastic-plastic crack bridging model [9]. O.Giannini, P.Casini, and F.Vestroni use finite element that has a bilinear element matrix with the discontinuity passing through the origin to model the cracked zone of the beam [10]. Pier Francesco Cacciola and Giuseppe Muscolino model the cracked beam by finite elements in which the closing crack model is used to describe the damaged element [11]. Ugo Andreaus, Paolo Baragatti, Paolo Casini, and Daniela Iacoviello present wavelet analysis for crack detection and quantification in beams based on static method, then the spatial wavelet transform is proven to be effective after comparing with the experimental study [12].Maryna V. Menshykova, Oleksandr V. Menshykov, and Igor A. Guz use the boundary integral equation method to solve the fracture mechanics problem, and they impose the constraints to the normal and tangential components of the contact force and the displacement discontinuity vectors to take the contact interaction of the crack faces into account[13].In this work, a new idea is created to divide the slider-crank mechanism into two subsystems by crack point, and the crack is modeled by massless torsional spring. Cracked rod is modeled as two successive equal rods connected with massless torsional spring. In this method, it is much easier to simulate the system with multiple cracks and easier to program by C language. Comparing with the analysis results between the slider-crank mechanism with crack and without crack, the conclusion shows that it is necessary to study crack influence when analyzing the mechanical system dynamic performance and vibration characteristic.


## 2. Motion Analysis of a Slider-Crank Mechanism

In order to study the difference of slider-crank mechanism dynamic motion with and without crack in the rod, the equation will be created and simulation will be done based on the calculation.


### 2.1. Equation of Motion on a Slider-Crank Mechanism without Crack in the Rod

A slider-crank mechanism without crack in the rod is modeled in Figure1 to study the effect of dynamic motion. A cyclic bending moment M is considered to crank the mechanism, and the rods OA and AB are considered to be rigid. The mechanism motion can be written as Lagrange equation and Newton’s second law [14–17].Figure 1
Slider-crank mechanism without crack in the rod.For this one degree of freedom mechanism, a designated valueφ is set as variable in the system, and M is the external moment. The crank angleφ is the angle between rod OA and the horizontal direction. The length of the rod OA isl1, and then the length of the rod AB isl2. The mass of rod OA ism1, and the mass of rod AB ism2. Rod OA inertial mass is J1=m1/l1∫0l1x2dx=1/3m1l12. Rod AB inertial mass is J2=m2/l2∫0l2x2dx=1/3m2l22.The center velocity of rod AB is vc0, and its center angular velocity is ωc0. vB represents the velocity of slider B. The total kinetic energy is T=∑n=13Tn, whereT1 is kinetic energy of rod OA,T2 is kinetic energy of rod AB, andT3 is kinetic energy of slider B. One can calculate the slider-crank mechanism kinetic energy by(1)T=∑n=13Tn=12J1φ˙2+12m2vc02+12J2ωc02+12m3vB2The generalized force can be written as(2)Qδ=∑Wδδφ=M·δφ+m1g·l1·cos⁡φ·δφ+m2g·l1·sin⁡φ·cos⁡φ·δφδφ=M+m1g·l1·cos⁡φ+m2g·l1·sin⁡φ·cos⁡φOne can write the Lagrange differential equation of motion for the slider-crank mechanism without crack as follows [18–21]:(3)ddt∂T∂φ˙-∂T∂φ=QφThe solution for the function is given as follows:(4)13m1l12φ¨+m2l12φ¨+1312m2l14φ¨l22+m3l22l12+l22φ¨l22=M+m1g·l1·cos⁡φ+m2g·l1·sin⁡φ·cos⁡φ


### 2.2. Equation of Motion on a Slider-Crank Mechanism with a Crack in the Rod

A slider-crank mechanism with a crack in the rod is modeled in Figure2 to study the effect of dynamic motion. Rod AB is considered to be composed of two rods AC and CB. Crack point C is at the middle of rod AB. A cyclic bending moment M is considered to crank the mechanism.Figure 2
Slider-crank mechanism with a crack in the rod.


#### 2.2.1. System 1 and System 2 Dynamic Equations

The slider-crank mechanism is considered as system 1 and system 2 which are divided by the crack. The equivalent mechanics model of the crack can be established by a massless torsional spring [22, 23].For system 1, there are two degrees of freedom; two designated valuesφ and θ1 are set as variable in the system. The crank angle φ is the angle between rod OA and the horizontal direction. θ1is the angle between rod AC and the horizontal direction. The center velocity of rod AC is vc1, and its center angular velocity is ωc1. The mass of rod AO is m1, and the mass of rod AB is m2. Because rod AO and rod AB are considered as material uniform distribution, the mass of rod AC mc1 is 1/2m2. The total kinetic energy for system 1 is T=∑n=12Tn, where T1 is kinetic energy of rod OA, T2 is kinetic energy of rod AC. One can calculate the system 1 kinetic energy by the following formula:(5)T=∑n=12Tn=12J1φ˙2+12mc1vc12+12Jc1ωc12The generalized force can be written as(6)Qφ=∑Wφδφ=Mδφ+m1g·cos⁡φ·1/2δφ·l1+1/2m2g·δφ·l1+1/4ωc1·l2·cos⁡θ1+Tc1·ωc01δφQθ1=∑Wθ1δθ1=Tco1δθ1+1/2m2g·cos⁡θ1·1/4δθ1·l2+M·δφ+m1g·cos⁡φ·1/2·δφ·l1δθ10One can write the Lagrange differential equation of motion as follows [24]:(7)ddt∂T∂φ˙-∂T∂φ=Qφφ˙=δφddt∂T∂θ˙1-∂T∂θ1=Qθ1The solution for the function is given as follows:(8)13m1l12φ¨+12m2l12φ¨+18m2θ¨1l1l2=M+12m1g·cos⁡φ·l1+12m2g·cos⁡θ1·l1+18m2g·l2·cos⁡θ1+Tc01·θ˙1φ˙132m2θ¨1l22+18m2φ¨l1l2+124m2l22θ1=Tc01+18m2gcos⁡θ1l2+M+m1g·cos⁡φ·12l1·φ˙δθ1For system 2, same as system 1, the generalized force can be written as(9)Qθ2=∑Wθ2δθ2=Tc02·δθ2+1/2·m2g·cos⁡θ2·1/4·l2δθ2QS=∑Wsδs=F·δsδs=FOne can write Lagrange differential equation as follows:(10)ddt∂T∂θ˙2-∂T∂θ2=Qθ2ddt∂T∂s˙-∂T∂s=QsThe solution for the function is given as follows:(11)724·m2·l22·θ¨2=Tc02+18·m2g·cos⁡θ2·l2δθ2m3·s¨=FThe mass of rod CB is 1/2m2, and m3 is the mass of slider B. Tco2 is the kinetic energy of rod CB. θ2 is the angle between rod CB and the horizontal direction. S is the distance for slider B. The center velocity of rod CB is vc2, and its center angular velocity is ωc2. F is the external force on slider B.


#### 2.2.2. Cracked Rod Calculation

The rod AB is considered as flexible, then, to simulate the crack by massless torsional spring [25–31], which is shown in Figure 3. Figure 3(a) is on crack dimension in the rod, and Figure 3(b) is on torsional spring.Figure 3
The geometry of cracked rod.
(a)
Crack dimension in the rod (b)
Torsional springThe equation of the deflection curve is(12)d4ydx4+k2·d2ydx2=0where k2=p/EI, p is the generalized force, and EI is the bending stiffness.The solution of (12) is(13)yx=D1+D2x+D3sin⁡kx+D4cos⁡kxwhere y(x) is the deflection, and the cross section bending angle is θ(x), bending moment is M(x), and one can calculate shearing force as follows [32]:(14)θx=dydxMx=-EI·d2ydx2Qx=dMdx-P·dydxThe flexible rod is considered as consistent on deflection, bending moment, and shearing force, so the relative angle of the torsional spring θc can be written as(15)θc=θ1xc-θ2x2=CEIM1xcwhere θ1xc is the relative angle between upper rod and the crack position, and θ2xc is the relative angle between lower rod and the crack position. The bending moment of the torsional spring is (16)M1xc=EIC·θc(17)C=5.346h1.8624ε2-3.95ε3+16.375ε4-37.226ε5+76.81ε6-126.9ε7+172ε8-143.97ε9+66.56ε10where C is the flexibility of the rotation spring which can be influenced by crack depth d and cross section height h [30, 33, 34].(18)ε=dhThen, the equation of slider-crank mechanism motion with a crack in the rod can be calculated as follows:(19)13m1l12φ¨+12m2l12φ¨+18m2θ¨1l1l2=M+12m1g·cos⁡φ·l1+12m2g·cos⁡θ1·l1+18m2g·l2·cos⁡θ1+Tc01·θ˙1φ˙132m2θ¨1l22+18m2φ¨l1l2+124m2l22θ¨1=Tc01+18m2gcos⁡θ1l2+M+m1g·cos⁡φ·12l1·φ˙δθ1724·m2·l22·θ¨2=Tc02+18·m2g·cos⁡θ2·l2δθ2m3·s¨=Fθc=θ1xc-θ2x2=CEIM1xcC=5.346h1.8624ε2-3.95ε3+16.375ε4-37.226ε5+76.81ε6-126.9ε7+172ε8-143.97ε9+66.56ε10


## 2.1. Equation of Motion on a Slider-Crank Mechanism without Crack in the Rod

A slider-crank mechanism without crack in the rod is modeled in Figure1 to study the effect of dynamic motion. A cyclic bending moment M is considered to crank the mechanism, and the rods OA and AB are considered to be rigid. The mechanism motion can be written as Lagrange equation and Newton’s second law [14–17].Figure 1
Slider-crank mechanism without crack in the rod.For this one degree of freedom mechanism, a designated valueφ is set as variable in the system, and M is the external moment. The crank angleφ is the angle between rod OA and the horizontal direction. The length of the rod OA isl1, and then the length of the rod AB isl2. The mass of rod OA ism1, and the mass of rod AB ism2. Rod OA inertial mass is J1=m1/l1∫0l1x2dx=1/3m1l12. Rod AB inertial mass is J2=m2/l2∫0l2x2dx=1/3m2l22.The center velocity of rod AB is vc0, and its center angular velocity is ωc0. vB represents the velocity of slider B. The total kinetic energy is T=∑n=13Tn, whereT1 is kinetic energy of rod OA,T2 is kinetic energy of rod AB, andT3 is kinetic energy of slider B. One can calculate the slider-crank mechanism kinetic energy by(1)T=∑n=13Tn=12J1φ˙2+12m2vc02+12J2ωc02+12m3vB2The generalized force can be written as(2)Qδ=∑Wδδφ=M·δφ+m1g·l1·cos⁡φ·δφ+m2g·l1·sin⁡φ·cos⁡φ·δφδφ=M+m1g·l1·cos⁡φ+m2g·l1·sin⁡φ·cos⁡φOne can write the Lagrange differential equation of motion for the slider-crank mechanism without crack as follows [18–21]:(3)ddt∂T∂φ˙-∂T∂φ=QφThe solution for the function is given as follows:(4)13m1l12φ¨+m2l12φ¨+1312m2l14φ¨l22+m3l22l12+l22φ¨l22=M+m1g·l1·cos⁡φ+m2g·l1·sin⁡φ·cos⁡φ


## 2.2. Equation of Motion on a Slider-Crank Mechanism with a Crack in the Rod

A slider-crank mechanism with a crack in the rod is modeled in Figure2 to study the effect of dynamic motion. Rod AB is considered to be composed of two rods AC and CB. Crack point C is at the middle of rod AB. A cyclic bending moment M is considered to crank the mechanism.Figure 2
Slider-crank mechanism with a crack in the rod.


### 2.2.1. System 1 and System 2 Dynamic Equations

The slider-crank mechanism is considered as system 1 and system 2 which are divided by the crack. The equivalent mechanics model of the crack can be established by a massless torsional spring [22, 23].For system 1, there are two degrees of freedom; two designated valuesφ and θ1 are set as variable in the system. The crank angle φ is the angle between rod OA and the horizontal direction. θ1is the angle between rod AC and the horizontal direction. The center velocity of rod AC is vc1, and its center angular velocity is ωc1. The mass of rod AO is m1, and the mass of rod AB is m2. Because rod AO and rod AB are considered as material uniform distribution, the mass of rod AC mc1 is 1/2m2. The total kinetic energy for system 1 is T=∑n=12Tn, where T1 is kinetic energy of rod OA, T2 is kinetic energy of rod AC. One can calculate the system 1 kinetic energy by the following formula:(5)T=∑n=12Tn=12J1φ˙2+12mc1vc12+12Jc1ωc12The generalized force can be written as(6)Qφ=∑Wφδφ=Mδφ+m1g·cos⁡φ·1/2δφ·l1+1/2m2g·δφ·l1+1/4ωc1·l2·cos⁡θ1+Tc1·ωc01δφQθ1=∑Wθ1δθ1=Tco1δθ1+1/2m2g·cos⁡θ1·1/4δθ1·l2+M·δφ+m1g·cos⁡φ·1/2·δφ·l1δθ10One can write the Lagrange differential equation of motion as follows [24]:(7)ddt∂T∂φ˙-∂T∂φ=Qφφ˙=δφddt∂T∂θ˙1-∂T∂θ1=Qθ1The solution for the function is given as follows:(8)13m1l12φ¨+12m2l12φ¨+18m2θ¨1l1l2=M+12m1g·cos⁡φ·l1+12m2g·cos⁡θ1·l1+18m2g·l2·cos⁡θ1+Tc01·θ˙1φ˙132m2θ¨1l22+18m2φ¨l1l2+124m2l22θ1=Tc01+18m2gcos⁡θ1l2+M+m1g·cos⁡φ·12l1·φ˙δθ1For system 2, same as system 1, the generalized force can be written as(9)Qθ2=∑Wθ2δθ2=Tc02·δθ2+1/2·m2g·cos⁡θ2·1/4·l2δθ2QS=∑Wsδs=F·δsδs=FOne can write Lagrange differential equation as follows:(10)ddt∂T∂θ˙2-∂T∂θ2=Qθ2ddt∂T∂s˙-∂T∂s=QsThe solution for the function is given as follows:(11)724·m2·l22·θ¨2=Tc02+18·m2g·cos⁡θ2·l2δθ2m3·s¨=FThe mass of rod CB is 1/2m2, and m3 is the mass of slider B. Tco2 is the kinetic energy of rod CB. θ2 is the angle between rod CB and the horizontal direction. S is the distance for slider B. The center velocity of rod CB is vc2, and its center angular velocity is ωc2. F is the external force on slider B.


### 2.2.2. Cracked Rod Calculation

The rod AB is considered as flexible, then, to simulate the crack by massless torsional spring [25–31], which is shown in Figure 3. Figure 3(a) is on crack dimension in the rod, and Figure 3(b) is on torsional spring.Figure 3
The geometry of cracked rod.
(a)
Crack dimension in the rod (b)
Torsional springThe equation of the deflection curve is(12)d4ydx4+k2·d2ydx2=0where k2=p/EI, p is the generalized force, and EI is the bending stiffness.The solution of (12) is(13)yx=D1+D2x+D3sin⁡kx+D4cos⁡kxwhere y(x) is the deflection, and the cross section bending angle is θ(x), bending moment is M(x), and one can calculate shearing force as follows [32]:(14)θx=dydxMx=-EI·d2ydx2Qx=dMdx-P·dydxThe flexible rod is considered as consistent on deflection, bending moment, and shearing force, so the relative angle of the torsional spring θc can be written as(15)θc=θ1xc-θ2x2=CEIM1xcwhere θ1xc is the relative angle between upper rod and the crack position, and θ2xc is the relative angle between lower rod and the crack position. The bending moment of the torsional spring is (16)M1xc=EIC·θc(17)C=5.346h1.8624ε2-3.95ε3+16.375ε4-37.226ε5+76.81ε6-126.9ε7+172ε8-143.97ε9+66.56ε10where C is the flexibility of the rotation spring which can be influenced by crack depth d and cross section height h [30, 33, 34].(18)ε=dhThen, the equation of slider-crank mechanism motion with a crack in the rod can be calculated as follows:(19)13m1l12φ¨+12m2l12φ¨+18m2θ¨1l1l2=M+12m1g·cos⁡φ·l1+12m2g·cos⁡θ1·l1+18m2g·l2·cos⁡θ1+Tc01·θ˙1φ˙132m2θ¨1l22+18m2φ¨l1l2+124m2l22θ¨1=Tc01+18m2gcos⁡θ1l2+M+m1g·cos⁡φ·12l1·φ˙δθ1724·m2·l22·θ¨2=Tc02+18·m2g·cos⁡θ2·l2δθ2m3·s¨=Fθc=θ1xc-θ2x2=CEIM1xcC=5.346h1.8624ε2-3.95ε3+16.375ε4-37.226ε5+76.81ε6-126.9ε7+172ε8-143.97ε9+66.56ε10


## 2.2.1. System 1 and System 2 Dynamic Equations

The slider-crank mechanism is considered as system 1 and system 2 which are divided by the crack. The equivalent mechanics model of the crack can be established by a massless torsional spring [22, 23].For system 1, there are two degrees of freedom; two designated valuesφ and θ1 are set as variable in the system. The crank angle φ is the angle between rod OA and the horizontal direction. θ1is the angle between rod AC and the horizontal direction. The center velocity of rod AC is vc1, and its center angular velocity is ωc1. The mass of rod AO is m1, and the mass of rod AB is m2. Because rod AO and rod AB are considered as material uniform distribution, the mass of rod AC mc1 is 1/2m2. The total kinetic energy for system 1 is T=∑n=12Tn, where T1 is kinetic energy of rod OA, T2 is kinetic energy of rod AC. One can calculate the system 1 kinetic energy by the following formula:(5)T=∑n=12Tn=12J1φ˙2+12mc1vc12+12Jc1ωc12The generalized force can be written as(6)Qφ=∑Wφδφ=Mδφ+m1g·cos⁡φ·1/2δφ·l1+1/2m2g·δφ·l1+1/4ωc1·l2·cos⁡θ1+Tc1·ωc01δφQθ1=∑Wθ1δθ1=Tco1δθ1+1/2m2g·cos⁡θ1·1/4δθ1·l2+M·δφ+m1g·cos⁡φ·1/2·δφ·l1δθ10One can write the Lagrange differential equation of motion as follows [24]:(7)ddt∂T∂φ˙-∂T∂φ=Qφφ˙=δφddt∂T∂θ˙1-∂T∂θ1=Qθ1The solution for the function is given as follows:(8)13m1l12φ¨+12m2l12φ¨+18m2θ¨1l1l2=M+12m1g·cos⁡φ·l1+12m2g·cos⁡θ1·l1+18m2g·l2·cos⁡θ1+Tc01·θ˙1φ˙132m2θ¨1l22+18m2φ¨l1l2+124m2l22θ1=Tc01+18m2gcos⁡θ1l2+M+m1g·cos⁡φ·12l1·φ˙δθ1For system 2, same as system 1, the generalized force can be written as(9)Qθ2=∑Wθ2δθ2=Tc02·δθ2+1/2·m2g·cos⁡θ2·1/4·l2δθ2QS=∑Wsδs=F·δsδs=FOne can write Lagrange differential equation as follows:(10)ddt∂T∂θ˙2-∂T∂θ2=Qθ2ddt∂T∂s˙-∂T∂s=QsThe solution for the function is given as follows:(11)724·m2·l22·θ¨2=Tc02+18·m2g·cos⁡θ2·l2δθ2m3·s¨=FThe mass of rod CB is 1/2m2, and m3 is the mass of slider B. Tco2 is the kinetic energy of rod CB. θ2 is the angle between rod CB and the horizontal direction. S is the distance for slider B. The center velocity of rod CB is vc2, and its center angular velocity is ωc2. F is the external force on slider B.


## 2.2.2. Cracked Rod Calculation

The rod AB is considered as flexible, then, to simulate the crack by massless torsional spring [25–31], which is shown in Figure 3. Figure 3(a) is on crack dimension in the rod, and Figure 3(b) is on torsional spring.Figure 3
The geometry of cracked rod.
(a)
Crack dimension in the rod (b)
Torsional springThe equation of the deflection curve is(12)d4ydx4+k2·d2ydx2=0where k2=p/EI, p is the generalized force, and EI is the bending stiffness.The solution of (12) is(13)yx=D1+D2x+D3sin⁡kx+D4cos⁡kxwhere y(x) is the deflection, and the cross section bending angle is θ(x), bending moment is M(x), and one can calculate shearing force as follows [32]:(14)θx=dydxMx=-EI·d2ydx2Qx=dMdx-P·dydxThe flexible rod is considered as consistent on deflection, bending moment, and shearing force, so the relative angle of the torsional spring θc can be written as(15)θc=θ1xc-θ2x2=CEIM1xcwhere θ1xc is the relative angle between upper rod and the crack position, and θ2xc is the relative angle between lower rod and the crack position. The bending moment of the torsional spring is (16)M1xc=EIC·θc(17)C=5.346h1.8624ε2-3.95ε3+16.375ε4-37.226ε5+76.81ε6-126.9ε7+172ε8-143.97ε9+66.56ε10where C is the flexibility of the rotation spring which can be influenced by crack depth d and cross section height h [30, 33, 34].(18)ε=dhThen, the equation of slider-crank mechanism motion with a crack in the rod can be calculated as follows:(19)13m1l12φ¨+12m2l12φ¨+18m2θ¨1l1l2=M+12m1g·cos⁡φ·l1+12m2g·cos⁡θ1·l1+18m2g·l2·cos⁡θ1+Tc01·θ˙1φ˙132m2θ¨1l22+18m2φ¨l1l2+124m2l22θ¨1=Tc01+18m2gcos⁡θ1l2+M+m1g·cos⁡φ·12l1·φ˙δθ1724·m2·l22·θ¨2=Tc02+18·m2g·cos⁡θ2·l2δθ2m3·s¨=Fθc=θ1xc-θ2x2=CEIM1xcC=5.346h1.8624ε2-3.95ε3+16.375ε4-37.226ε5+76.81ε6-126.9ε7+172ε8-143.97ε9+66.56ε10


## 3. Motion Comparison of a Slider-Crank Mechanism with and without Crack in the Rod

The simulation on Slider-crank mechanism with and without crack in the rod is done based on the upper equation calculation. The mechanism is cranked by a cyclic bending moment with regular angel velocity 300RPM (revolution per minute) which means the crank is driven one cycle every 0.2s. Crack parameters used in calculation are as follows: crack depth d=6mm and cross section height h=30mm. A summary of the properties of the experimental slider-crank model is given in Table1.Table 1
Properties of the experimental slider-crank mechanism.
Body Mass(kg) Length(m) Inertia moment(kg•m2) Crank 0.3 0.03 0.000072 Rod 0.036 0.12 0.0000624 Slider 1.25 - 0.00068For the slider-crank mechanism with a crack in the rod, the action response of the slider is plotted versus crank angle, and the obtained results are compared with those previously obtained slider actions on the slider-crank mechanism without crack in the rod. The slider-crank mechanism without crack in the rod can be abbreviated as SC1, and the slider-crank mechanism with a crack in the rod can be abbreviated as SC2.As shown in Figure4, there is the comparison on slider displacement between SC1 and SC2.Figure 4
Displacement of slider in three cycles.(1)There is cyclic fluctuation for slider displacement on SC1, and the displacement period 360deg can be easily found in the curve. There is no clear cyclic fluctuation for slider displacement on SC2. The main reason is that SC1 is the linear system, but SC2 is the nonlinear system. Nonlinear system is much complicated and has no regular period.(2)The maximum for slider displacement on SC1 is lower than the maximum for slider displacement on SC2. For SC2, the slider will move on further when the slider reaches the point which is the maximum displacement for SC1, because of the different inertia influenced by the crack. Furthermore, the maximum for slider displacement on SC2 is not a definite number in each fluctuation, due to the crack lead to a complicated nonlinear vibration system.(3)The slider displacement fluctuation tendency is the same between SC1 and SC2. That is because the crack just changes SC2 inertia and then changes the displacement. But the system motion tendency should be the same.(4)The slider trial on SC1 is exactly symmetrical. The slider trial on SC2 is not symmetrical; for example, at the beginning fluctuation, there is one sine wave in positive direction, and there are two sine waves (one bigger sine, one smaller sine) in negative direction. The reason is the two sine waves happened during the crank pull the crack rod change to push the rod, and then the crack depth d changed.As shown in Figure5, there is the comparison on slider velocity between SC1 and SC2.Figure 5
Velocity of slider in three cycles.(1)There is cyclic fluctuation for slider velocity on SC1, and the period is 360deg which can be easily found. There is no clear cyclic fluctuation for slider velocity on SC2. The main reason is the same with displacement analysis.(2)The maximum velocity on SC1 is much lower than the slider of SC2. The main reason is the same with displacement analysis.As shown in Figure6, there is the comparison on slider acceleration between SC1 and SC2.Figure 6
Acceleration of slider in three cycles.(1)There is cyclic fluctuation for slider acceleration on SC1, and the period is 360deg which can be easily found. There is no clear cyclic fluctuation for slider acceleration on SC2. The main reason is the same with displacement analysis.(2)The maximum of slider acceleration on SC1 is much lower than the value on SC2, which is the same with the displacement-velocity curve.From Figures4, 5, and 6, we can see that the displacement, velocity, and acceleration of the slider are all nonperiodic signals.


## 4. Nonlinear Dynamic Analysis

Nonlinear dynamics are often characterized by a chaotic behaviour of the system. Figures7 and 8 show that the phase trajectories of displacement-velocity and velocity-acceleration for the slider under the conditions of crank angular velocity 300 RPM.Figure 7
Phase trajectories of slider displacement-slider velocity.Figure 8
Phase trajectories of slider velocity-slider acceleration.From Figures7 and 8, it can be seen that the phase space curve fluctuates obviously, and there will be a deviation between different periods. The motion trajectory of the nonrepetition of the ring surface in the phase diagram shows that the system is in the quasiperiodic state.Lyapunov exponentλ is a good method to evaluate the system sensitivity based on initial condition, and it can be used to distinguish chaotic and nonchaotic. Negative and zero Lyapunov exponent means convergence to a predictable motion, and only one positive exponent will lead to a chaotic system. To estimate the Lyapunov exponent of time series, several approaches are suggested, like the method of Wolff, Kantz, or Rosenstein [35–39].The Lyapunov exponent is shown in Figure9 for the system of slider-crank mechanism with a crack in the rod. The displacement, speed, and acceleration of the slider Lyapunov exponent value are 0.0095, 0.0147, and 0.0301, respectively, which means it is chaotic system. That exactly explains why there is no regular period for the slider displacement, velocity, and acceleration on SC2.Figure 9
Lyapunov exponent of Slider on SC2.


## 5. Conclusions

The motion of slider-crank mechanism with a crack in the rod is analyzed by dividing the system into two systems linked by the crack. The crack of the mechanism is simulated by using massless torsional spring. After comparing with slider displacement, velocity, and acceleration between SC1 and SC2, we can conclude that the effect of impact with crack should not be ignored when analyzing the dynamic performance. The whole curve is periodic for the slider motion on SC1, whereas it is chaotic for SC2.



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*Source: 8540546-2018-09-02.xml*