# Research on Coupled Dynamic Model of Tracked Vehicles and Its Solving Method

Authors: Yuliang Li; Chong Tang
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

In order to conveniently analyze the dynamic performance of tracked vehicles, mathematic models are established based on the actual structure of vehicles and terrain mechanics when they are moving on the soft random terrain. A discrete method is adopted to solve the coupled equations to calculate the acceleration of the vehicle’s mass center and tractive force of driving sprocket. Computation results output by the model presented in this paper are compared with results given by the model, which has the same parameters, built in the multi-body dynamic software. It shows that the steady state calculation results are basically consistent, while the model presented in this paper is more convenient to be used in the optimization of structure parameters of tracked vehicles.


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## Body

## 1. Introduction

Tracked vehicles are widely used in the field of military, agriculture, and construction industries. They are applied on the ground that has a low carrying capacity or wheeled vehicles cannot go through (swamp, snowfield, etc.) or are used for obstacle-crossing and climbing, and so forth because the tracks can be treated as self-carried road to increase the vehicle’s passing ability. Nowadays, there are lots of modeling methods about tracked vehicles which primarily treat the tracks as flexible [1–5] or rigid ones [6–9].Ma and Perkins employed a nonlinear finite element method to represent the track segment contacting with the terrain [1, 2]; Park et al. divided the track system contacting with terrain into a finite number of small elements and used structural analysis and the numerical iterative method to calculate the drawbar pull force [3]; Ma treated the tracks as flexible tracks and exerted the excitation directly on the road wheels to reduce the complexity of the model [4]; Han et al. regarded tracks as flexible ones and used springs to replace tracks between road wheels in order to build the ride comfort model of the tracked vehicle [5].The assumption of flexible tracks cannot fully reflect the dynamic performance of tracks. Rigid models are adopted by more researchers. Merhof and Hackbarth illustrated the actual structure of the tracks and set up their dynamic equations [6]; Rubinstein and Hitron described each track’s positions, movements, and the interaction with terrain by coordinate transformation in three-dimensional space [7]; Ma et al. regarded the tracks and the road wheels as a whole and analyzed each part in the track system [8]; Hu took heavy tracked vehicle as a research object and divided the track system into 4 parts, namely, the tracks in contact with terrain, the tracks in contact with front idler, the tracks in contact with driving sprocket, and the tracks between front idler and driving sprocket [9].Although the assumption of rigid tracks is approaching the truth, the effects of structure parameters on the performance are difficult to analyze, and generally the model cannot be used in the structural optimization. Therefore, this paper presents a simplified dynamic model based on the assumption of rigid tracks. A discrete method is used to solve the coupled equations. The computation results, namely, the acceleration of the vehicle’s mass center and the tractive force of driving sprocket, are compared with those obtained from RecurDyn to verify the model’s validity.


## 2. Dynamic Model of Tracked Vehicles

### 2.1. Random Road Excitations

In this paper gradeE road surface is taken as the random excitations for the tracked vehicles. The road model is built by making use of power spectral density function in time frequency domain Gq(f) [10–12] and the iteration of trigonometric series method [13]; the result is illustrated in Figure 1. The tracked vehicle’s highest speed is 60 km/h, the length of a track is 0.153 m, and so the time that a wheel passes by a track is 0.009 s. Thus the time interval for the discrete sampling of terrain excitation y(j) is 0.009 s in order to simulate the vehicle run on the road surface.Figure 1
Grade E random road surface.


### 2.2. Track-Terrain Contact

Currently the most widely used model for analyzing soil sinkage is Bekker’s equation [14, 15]:(1)σ=kcb+kϕzn,where, σ is the pressure, kpa; n is the pressure-sinkage exponent; kc is the soil cohesive coefficient, kN/mn+1; kϕ is the soil internal friction coefficient, kN/mn+2; b is the width of a track, m; z is the sinkage, m.Actually the tracked vehicle will have repeated loads on the same terrain when different wheels of the vehicle pass by it. This phenomenon can be more precisely described by the loading-unloading curve of soil model introduced by Wong [16] than Bekker’s equation, as shown in Figure 2.Figure 2
Loading-unloading curve of soil [16].As it can be seen from Figure2, OA is the continuous loading curve which can be described by (1); BA is the unloading and repeated loading curve in which the slope can be seen as a constant value ku, and the curve can be described by [17](2)σ=(kcb+kφ)znOA(kcb+kφ)zun-ku(zu-z)BA,where zu is the maximum sinkage, m; z is the sinkage, m.In order to build and solve the dynamic models of tracked vehicles, some simplifications, which have little effects on the final calculation results, are needed:(1) every y(j) is sequentially exerted on the two pins of a track so that the terrain under the track would be complete and continuous; this will lay the foundation for the calculation of shear forces acting on tracks by terrain; (2) in reality the angle of track changes when a boogie passes by it. However, since the time is very short, it is reasonable to assume that the track keeps the same angle during this period so that the coupled relationships among wheel, track, and terrain can be decoupled; (3) the average sinkage of both pins of a track after the wheel leaves it is regarded as the sinkage of the track when the wheel is on the middle of it, as shown in Figure 3.Figure 3
Average sinkage, where,c is the midpoint of the track; x is the road excitation; s is the average sinkage under the pins, namely s=(s1+s2)/2.At the first sampling time, terrain excitations on both pins of the track under the first road wheel are original excitationsy(1) and y(0). At the second sampling time, the excitations on the track under the first wheel are y(2) and y(1) that has sinkage and springback after the first sampling time and so are the following excitations on the track. For the tracks under the second to the sixth road wheels, the terrain excitations on both pins of them are excitations that have sinkage and springback after the former sampling time. Besides, there are 2 tracks at the right and left sides of one terrain excitation and at least one road wheel has passed by before. Therefore, all of the former sinkage and springback should be taken into consideration; namely, the original excitation should subtract the maximum average sinkage among all the ones and then add the springback to form the new excitations acting on the later tracks.


### 2.3. Mechanical Analysis of Tracks

In this model the inertial force and inertial moment of a track are neglected due to the fact that the mass of a track is very small compared with the whole vehicle. Thus, the forces and moments acting on a track can be illustrated in Figure4.Figure 4
Forces and moments acting on a track contacting with a road wheel, where,n and subscript n are the nth track, the same as below; subscripts i, l, r, s, x, z, and v are the ith road wheel, left pin, right pin, shear force, the direction along the track, the direction vertical to the track, and road wheel’s vertical force on the track, the same as below; cn is the midpoint of the nth track; Fnlx, Fnrx is the force components acting on both pins in the direction along the track, kN; Fnlz, Fnrz is force components acting on both pins in the direction vertical to the track, kN; Mnl, Mnr is the moments acting on both pins of the track, kN·m; Fiv is the vertical force acting on the track from road wheel, which equals the sum of the weight of a road wheel, inertial force of a road wheel and the suspension’s reactive force, kN; Fis is the shear force acting on a track from terrain, kN; g is the acceleration of gravity, m/s2; ml is the mass of a track, kg; L is the length of a track, m; βn is the angle of the track, rad; pn(x) is the pressure distribution function under a track, kpa.In order to compute the shear force acting on a track, the first step is to calculateFnlz, Fnrz and Mnl, Mnr. Here, it is supposed that the time period that a road wheel passes a track is dt. During this period the angle of a track, vertical and angular acceleration of the vehicle’s mass center, the vertical force acting on a track from road wheel Fiv, and the sinkage of excitation are calculated every dt, other physical variables are calculated every dt/3. Therefore, there are 4 calculation points in dt except the midpoint. Figure 5 illustrates this concept.Figure 5
Geometric and mechanical relationships between tracks, wherePn is the concentrated force obtained by summing the forces distributed along the whole track, KN.Firstly, let us analyze the interaction between the first road wheel and the track under it, and the analysis result also applies to other wheels and tracks. It is supposed that the tracked vehicle is initially static on a level terrain. After adt the angle of the track contacting with the road wheel (i-1, i=2) changes to βn because of the excitation from terrain. Fnlx, Fnlz acting on the track can be obtained from Figure 6:(3)Fnlx=Fa′cos⁡βa-θc+βn,(4)Fnlz=Fa′sinβa-θc+βn.Figure 6
Forces acting on the left pin of the track under the first road wheel, whereFa is the pretensioning force of tracks, kN; θc is the rotation angle of the vehicle’s mass center, rad; βa is the angle between the tracks that link front idler and the level terrain when the vehicle is static, rad. superscript “′” represents the values obtained at the former calculation point, the same as below (see (3) and (4)).Two force components of the pretensioning force, namely,Fnrx and Fnrz, are acting on the right pin. Figure 7 illustrates all the forces acting on the track at the first calculation point in the time period dt.Figure 7
Forces acting on a track at the first calculation point in the first time perioddt.The pressure distributionPn acting on the track from terrain is obtained by Figure 7 and the shear force (mentioned in Section 2.4) at the first calculation point can be calculated.In order to solve the coupled equations,βn, F(i-1)v, Fnlx, Fnlz, Pn, and F(i-1)s are kept unchanged at the next calculation point. At the second calculation point, Fnrx, Fnrz, and Mnr can be obtained according to equilibrium conditions. For instance, in Figure 8, we can have(5)Fnrx=Fnlx′+F(i-1)s′+mlgsinβn′,Fnrz=Fnlz′+Pn′-mlgcos⁡βn′-Fi-1v′,Mnr=Fnlz′L-mlgcos⁡βn′L2+Pn′L-l.Figure 8
Calculation ofFnrx, Fnrz, and Mnr at the second point in the time period dt.In Figure9, F(i-1)v moves L/3 forward, the pressure distribution acting on the track from terrain and the shear force can be calculated by Fnrx, Fnrz, and Mnr (5) and the unchanged values of βn, F(i-1)v, Fnlx, Fnlz, Pn, and F(i-1)s. Based on this method, the similar values can be obtained when the road wheel is at the midpoint, 2L/3, and L of the track.Figure 9
Forces acting on the track before the calculation of the shear force at the second calculation point in the time perioddt.In the second time perioddt, road wheel (i-1) moves on to track (n-1). At this junction point the tractive force acting on tracks (n) and (n-1) keeps unchanged. According to Figure 10, after the pressure distribution and shear force are obtained at the fourth calculation point in the first time period dt, Fnrx, Fnrz, and Mnr can be calculated as follows:(6)Fnrx=Fnlx′+mlgsinβn′+Fi-1s′,(7)Fnrz=Fnlz′-Fi-1v′-mlgcos⁡βn′+Pn′,(8)Mnr=Fnlz′L-Fi-1v′L-mlgcos⁡βn′L2+Pn′L-l.Figure 10
The calculation ofFnrx, Fnrz, and Mnr at the end of the fourth calculation point in the first dt.As illustrated in Figure11, after the road wheel moves onto track (n-1), F(n-1)rx, F(n-1)rz can be obtained from Fall′, which is the sum of force components of Fnrx, Fnrz in the direction along with track n. M(n-1)r equals Mnr′ (obtained by (8)): (9)F(n-1)rx=Fall′cos⁡βn+βn-1,F(n-1)rz=-Fall′sinβn+βn-1,M(n-1)r=Mnr′.Figure 11
Forces acting on the track (n-1) at the first calculation point in the second dt.To obtain the values at the following calculation points, the method is the same as the calculation points in the firstdt. When the sixth time period dt comes, random terrain excitation begins to be exerted on the track under the second road wheel because there are 5 tracks between two wheels, as illustrated in Figure 12.Figure 12
Forces acting on the left pin of trackn contacting with the second road wheel.In Figure12, F(n-5)lx, F(n-5)lz and M(n-5)l act on the same pin as F(n-6)rx, F(n-6)rz, and M(n-6)r. So they are active and reactive forces (moments). Then Fnlx, Fnlz, and Mnl can be computed by the equilibrium condition of track (n-5); namely,(10)Fnlx=(F(n-5)lx+5mlgsin(βn5))cos⁡(βn5+βn)+Fn-5lz-5mlgcos⁡βn5sinβn5+βn,Fnlz=(F(n-5)lz-5mlgcos⁡(βn5))cos⁡(βn+βn5)-Fn-5lx+5mlgsinβn5sinβn5+βn,Mnl=M(n-5)l-mlgcos⁡βn425L2.The method to calculateFnrx, Fnrz, and Mnr is the same as the one used to calculate the force or moment values of the track under the first road wheel in the first dt. In this way, forces or moments acting on the tracks under and between the road wheels can be obtained.As illustrated in Figure13, suppose track n contacts with the sixth road wheel. After calculating Fnrx, Fnrz, and Mnr, Fb′ can be obtained by(11)Fb′=Fnrxcos⁡βb-βn+θc,where Fb′ is the driving force given by sprocket, kN; θc is the rotation angle of the vehicle’s mass center, rad; βb is the angle between the tracks connected with the driving sprocket and the level terrain when the vehicle is static, rad.Figure 13
Forces acting on a track when the driving sprocket pulls the right pin of it.


### 2.4. Shear Forces Acting on Tracks

When tracks are pulled from terrain by the driving sprocket, tractive forces will be exerted on them by terrain. On the soft terrain, the tractive force is mainly due to the shear force. Therefore, the tractive force given by the driving sprocket can be obtained by calculating the shear force acting on tracks from terrain [18]. When a road wheel passes by a track in a time period dt, the pressure distribution under the track changes according to the wheels’ positions, and so do forces and moments on both pins of the track. Figure 14 illustrates the simplification of forces and moments acting on a track at a specific calculation point.Figure 14
Resultant forces of a track.After the simplification, the forces and moments exerted on a track are equal to a forceF and a moment M. For a rigid and small track, the pressure distribution under the track can be regarded as a linear distribution [19]. There are 3 different situations according to the resultant force F and moment M, as illustrated in Figure 15:(12)pmax⁡pmin⁡=FLb1±6eL,where pmax⁡, pmin⁡ are the maximum and minimum values of the distribution, kpa; L is the length of a track, m; b is the width of a track, m; e is the eccentricity, e=M/F, m.Figure 15
Pressure distributions under a track.
(a)
(b)
(c)A local coordinate can be placed on a track so that the pressure distribution function can be described as(13)px=k0x+b0,where p(x) is the pressure distribution function under a track, kpa; k0 is the slope; b0 is the intercept, m.Substitute (13) into Janosi shear stress equation [20] (14), shear force acting on tracks from terrain according to different pressure distribution functions can be calculated:(14)Fis=b∫0L(C+p(x)tanφ)(1-e-isL1/K)dx.Equation (14) can be reduced to(15)Fis=b1-e-isL1/KCL+12k0L2tanφ+b0Ltanφ,where L1 is the distance that road wheels pass by a track, m; b is the width of a track, m; C is the soil cohesion coefficient, kN/m2; φ is the internal friction angle of soil, rad; K is the shear modulus deformation of soil, m; is is the sliding rate of vehicle.


### 2.5. Dynamic Equations of Tracked Vehicles

The following assumptions are made to set up dynamic equations of tracked vehicles:(1) the vehicle is bilaterally symmetrical; (2) excitations exerted on both left and right tracks are exactly the same; (3) road wheels and tracks are rigid. Based on these assumptions, a half vehicle model can be used to build the dynamic model of the vehicle, as illustrated in Figure 16.Figure 16
Half vehicle model.(
1
) Initially the vehicle is static and in equilibrium. So the dynamic equations in z direction are as follows:(16)mcz¨c=∑i=16Fpi+∑k=a,bFkz,(17)Faz=Fa×sinβa+θc,Fbz=Fb×sinβb-θc,where mc is the mass of half vehicle, kg; zc is the displacement of mass center in z direction, m; Fpi is the forces acting on vehicle’ s body from the suspension system, kN; Fk, (k=a,b) is the forces acting on front idler and driving sprocket of tracks, kN; Fkz, (k=a,b) is the force components in z direction acting on front idler and driving sprocket of tracks, kN; βk, (k=a,b) is the angle between the tracks connecting the driving sprocket and the level road when the vehicle is static, rad; θc is the rotation angle of the mass center of the vehicle, rad.(
2
)
Pitching movement equations for the vehicle body,(18)Jcθc¨=∑i=16Fpilicos⁡θc-hisinθc+∑k=a,bFkx(lksinθc+hkcos⁡θc)+∑k=a,bFkzlkcos⁡θc-hksinθc,where Jc is the inertial moment when the vehicle rotates around the mass center, kg·m2; Fkx(k=a,b) is the force components in x direction acting on front idler and driving sprocket, kN; li, hi(i=1~6) is the horizontal and vertical distances of road wheels’ mass center from vehicle’s mass center, li(i=1,2,3) are positive, li(i=4,5,6) are negative; lk, hk(k=a,b) is the horizontal and vertical distances of front idler’s and driving sprocket’s mass center from vehicle’s mass center, m; la is positive, lb is negative.(
3
)
Forces exerted by suspension system,(19)Fpi=cs˙i+ksi.Since the mass center of the tracked vehicle rotates slightly when it is moving on the grade E random road surface, so(20)si=xir-zc-lisinθc≈xir-zc-liθc,where c is the damping coefficient, kN/(m/s); k is the spring coefficient, kN/m; xir is the excitations exerted on a track’s right pin, m; si is the stroke of the suspension system, m.(
4
)
The vertical force acting on the track from road wheel, Fiv,(21)Fiv=Fpi+mc6+mzg-mzai×cos⁡βn,i=1~6,where mz is the mass of a road wheel, kg; g is the acceleration of gravity, m/s2; ai is the road wheel’s acceleration, m/s2; βn is the angle of track n, rad.


## 2.1. Random Road Excitations

In this paper gradeE road surface is taken as the random excitations for the tracked vehicles. The road model is built by making use of power spectral density function in time frequency domain Gq(f) [10–12] and the iteration of trigonometric series method [13]; the result is illustrated in Figure 1. The tracked vehicle’s highest speed is 60 km/h, the length of a track is 0.153 m, and so the time that a wheel passes by a track is 0.009 s. Thus the time interval for the discrete sampling of terrain excitation y(j) is 0.009 s in order to simulate the vehicle run on the road surface.Figure 1
Grade E random road surface.


## 2.2. Track-Terrain Contact

Currently the most widely used model for analyzing soil sinkage is Bekker’s equation [14, 15]:(1)σ=kcb+kϕzn,where, σ is the pressure, kpa; n is the pressure-sinkage exponent; kc is the soil cohesive coefficient, kN/mn+1; kϕ is the soil internal friction coefficient, kN/mn+2; b is the width of a track, m; z is the sinkage, m.Actually the tracked vehicle will have repeated loads on the same terrain when different wheels of the vehicle pass by it. This phenomenon can be more precisely described by the loading-unloading curve of soil model introduced by Wong [16] than Bekker’s equation, as shown in Figure 2.Figure 2
Loading-unloading curve of soil [16].As it can be seen from Figure2, OA is the continuous loading curve which can be described by (1); BA is the unloading and repeated loading curve in which the slope can be seen as a constant value ku, and the curve can be described by [17](2)σ=(kcb+kφ)znOA(kcb+kφ)zun-ku(zu-z)BA,where zu is the maximum sinkage, m; z is the sinkage, m.In order to build and solve the dynamic models of tracked vehicles, some simplifications, which have little effects on the final calculation results, are needed:(1) every y(j) is sequentially exerted on the two pins of a track so that the terrain under the track would be complete and continuous; this will lay the foundation for the calculation of shear forces acting on tracks by terrain; (2) in reality the angle of track changes when a boogie passes by it. However, since the time is very short, it is reasonable to assume that the track keeps the same angle during this period so that the coupled relationships among wheel, track, and terrain can be decoupled; (3) the average sinkage of both pins of a track after the wheel leaves it is regarded as the sinkage of the track when the wheel is on the middle of it, as shown in Figure 3.Figure 3
Average sinkage, where,c is the midpoint of the track; x is the road excitation; s is the average sinkage under the pins, namely s=(s1+s2)/2.At the first sampling time, terrain excitations on both pins of the track under the first road wheel are original excitationsy(1) and y(0). At the second sampling time, the excitations on the track under the first wheel are y(2) and y(1) that has sinkage and springback after the first sampling time and so are the following excitations on the track. For the tracks under the second to the sixth road wheels, the terrain excitations on both pins of them are excitations that have sinkage and springback after the former sampling time. Besides, there are 2 tracks at the right and left sides of one terrain excitation and at least one road wheel has passed by before. Therefore, all of the former sinkage and springback should be taken into consideration; namely, the original excitation should subtract the maximum average sinkage among all the ones and then add the springback to form the new excitations acting on the later tracks.


## 2.3. Mechanical Analysis of Tracks

In this model the inertial force and inertial moment of a track are neglected due to the fact that the mass of a track is very small compared with the whole vehicle. Thus, the forces and moments acting on a track can be illustrated in Figure4.Figure 4
Forces and moments acting on a track contacting with a road wheel, where,n and subscript n are the nth track, the same as below; subscripts i, l, r, s, x, z, and v are the ith road wheel, left pin, right pin, shear force, the direction along the track, the direction vertical to the track, and road wheel’s vertical force on the track, the same as below; cn is the midpoint of the nth track; Fnlx, Fnrx is the force components acting on both pins in the direction along the track, kN; Fnlz, Fnrz is force components acting on both pins in the direction vertical to the track, kN; Mnl, Mnr is the moments acting on both pins of the track, kN·m; Fiv is the vertical force acting on the track from road wheel, which equals the sum of the weight of a road wheel, inertial force of a road wheel and the suspension’s reactive force, kN; Fis is the shear force acting on a track from terrain, kN; g is the acceleration of gravity, m/s2; ml is the mass of a track, kg; L is the length of a track, m; βn is the angle of the track, rad; pn(x) is the pressure distribution function under a track, kpa.In order to compute the shear force acting on a track, the first step is to calculateFnlz, Fnrz and Mnl, Mnr. Here, it is supposed that the time period that a road wheel passes a track is dt. During this period the angle of a track, vertical and angular acceleration of the vehicle’s mass center, the vertical force acting on a track from road wheel Fiv, and the sinkage of excitation are calculated every dt, other physical variables are calculated every dt/3. Therefore, there are 4 calculation points in dt except the midpoint. Figure 5 illustrates this concept.Figure 5
Geometric and mechanical relationships between tracks, wherePn is the concentrated force obtained by summing the forces distributed along the whole track, KN.Firstly, let us analyze the interaction between the first road wheel and the track under it, and the analysis result also applies to other wheels and tracks. It is supposed that the tracked vehicle is initially static on a level terrain. After adt the angle of the track contacting with the road wheel (i-1, i=2) changes to βn because of the excitation from terrain. Fnlx, Fnlz acting on the track can be obtained from Figure 6:(3)Fnlx=Fa′cos⁡βa-θc+βn,(4)Fnlz=Fa′sinβa-θc+βn.Figure 6
Forces acting on the left pin of the track under the first road wheel, whereFa is the pretensioning force of tracks, kN; θc is the rotation angle of the vehicle’s mass center, rad; βa is the angle between the tracks that link front idler and the level terrain when the vehicle is static, rad. superscript “′” represents the values obtained at the former calculation point, the same as below (see (3) and (4)).Two force components of the pretensioning force, namely,Fnrx and Fnrz, are acting on the right pin. Figure 7 illustrates all the forces acting on the track at the first calculation point in the time period dt.Figure 7
Forces acting on a track at the first calculation point in the first time perioddt.The pressure distributionPn acting on the track from terrain is obtained by Figure 7 and the shear force (mentioned in Section 2.4) at the first calculation point can be calculated.In order to solve the coupled equations,βn, F(i-1)v, Fnlx, Fnlz, Pn, and F(i-1)s are kept unchanged at the next calculation point. At the second calculation point, Fnrx, Fnrz, and Mnr can be obtained according to equilibrium conditions. For instance, in Figure 8, we can have(5)Fnrx=Fnlx′+F(i-1)s′+mlgsinβn′,Fnrz=Fnlz′+Pn′-mlgcos⁡βn′-Fi-1v′,Mnr=Fnlz′L-mlgcos⁡βn′L2+Pn′L-l.Figure 8
Calculation ofFnrx, Fnrz, and Mnr at the second point in the time period dt.In Figure9, F(i-1)v moves L/3 forward, the pressure distribution acting on the track from terrain and the shear force can be calculated by Fnrx, Fnrz, and Mnr (5) and the unchanged values of βn, F(i-1)v, Fnlx, Fnlz, Pn, and F(i-1)s. Based on this method, the similar values can be obtained when the road wheel is at the midpoint, 2L/3, and L of the track.Figure 9
Forces acting on the track before the calculation of the shear force at the second calculation point in the time perioddt.In the second time perioddt, road wheel (i-1) moves on to track (n-1). At this junction point the tractive force acting on tracks (n) and (n-1) keeps unchanged. According to Figure 10, after the pressure distribution and shear force are obtained at the fourth calculation point in the first time period dt, Fnrx, Fnrz, and Mnr can be calculated as follows:(6)Fnrx=Fnlx′+mlgsinβn′+Fi-1s′,(7)Fnrz=Fnlz′-Fi-1v′-mlgcos⁡βn′+Pn′,(8)Mnr=Fnlz′L-Fi-1v′L-mlgcos⁡βn′L2+Pn′L-l.Figure 10
The calculation ofFnrx, Fnrz, and Mnr at the end of the fourth calculation point in the first dt.As illustrated in Figure11, after the road wheel moves onto track (n-1), F(n-1)rx, F(n-1)rz can be obtained from Fall′, which is the sum of force components of Fnrx, Fnrz in the direction along with track n. M(n-1)r equals Mnr′ (obtained by (8)): (9)F(n-1)rx=Fall′cos⁡βn+βn-1,F(n-1)rz=-Fall′sinβn+βn-1,M(n-1)r=Mnr′.Figure 11
Forces acting on the track (n-1) at the first calculation point in the second dt.To obtain the values at the following calculation points, the method is the same as the calculation points in the firstdt. When the sixth time period dt comes, random terrain excitation begins to be exerted on the track under the second road wheel because there are 5 tracks between two wheels, as illustrated in Figure 12.Figure 12
Forces acting on the left pin of trackn contacting with the second road wheel.In Figure12, F(n-5)lx, F(n-5)lz and M(n-5)l act on the same pin as F(n-6)rx, F(n-6)rz, and M(n-6)r. So they are active and reactive forces (moments). Then Fnlx, Fnlz, and Mnl can be computed by the equilibrium condition of track (n-5); namely,(10)Fnlx=(F(n-5)lx+5mlgsin(βn5))cos⁡(βn5+βn)+Fn-5lz-5mlgcos⁡βn5sinβn5+βn,Fnlz=(F(n-5)lz-5mlgcos⁡(βn5))cos⁡(βn+βn5)-Fn-5lx+5mlgsinβn5sinβn5+βn,Mnl=M(n-5)l-mlgcos⁡βn425L2.The method to calculateFnrx, Fnrz, and Mnr is the same as the one used to calculate the force or moment values of the track under the first road wheel in the first dt. In this way, forces or moments acting on the tracks under and between the road wheels can be obtained.As illustrated in Figure13, suppose track n contacts with the sixth road wheel. After calculating Fnrx, Fnrz, and Mnr, Fb′ can be obtained by(11)Fb′=Fnrxcos⁡βb-βn+θc,where Fb′ is the driving force given by sprocket, kN; θc is the rotation angle of the vehicle’s mass center, rad; βb is the angle between the tracks connected with the driving sprocket and the level terrain when the vehicle is static, rad.Figure 13
Forces acting on a track when the driving sprocket pulls the right pin of it.


## 2.4. Shear Forces Acting on Tracks

When tracks are pulled from terrain by the driving sprocket, tractive forces will be exerted on them by terrain. On the soft terrain, the tractive force is mainly due to the shear force. Therefore, the tractive force given by the driving sprocket can be obtained by calculating the shear force acting on tracks from terrain [18]. When a road wheel passes by a track in a time period dt, the pressure distribution under the track changes according to the wheels’ positions, and so do forces and moments on both pins of the track. Figure 14 illustrates the simplification of forces and moments acting on a track at a specific calculation point.Figure 14
Resultant forces of a track.After the simplification, the forces and moments exerted on a track are equal to a forceF and a moment M. For a rigid and small track, the pressure distribution under the track can be regarded as a linear distribution [19]. There are 3 different situations according to the resultant force F and moment M, as illustrated in Figure 15:(12)pmax⁡pmin⁡=FLb1±6eL,where pmax⁡, pmin⁡ are the maximum and minimum values of the distribution, kpa; L is the length of a track, m; b is the width of a track, m; e is the eccentricity, e=M/F, m.Figure 15
Pressure distributions under a track.
(a)
(b)
(c)A local coordinate can be placed on a track so that the pressure distribution function can be described as(13)px=k0x+b0,where p(x) is the pressure distribution function under a track, kpa; k0 is the slope; b0 is the intercept, m.Substitute (13) into Janosi shear stress equation [20] (14), shear force acting on tracks from terrain according to different pressure distribution functions can be calculated:(14)Fis=b∫0L(C+p(x)tanφ)(1-e-isL1/K)dx.Equation (14) can be reduced to(15)Fis=b1-e-isL1/KCL+12k0L2tanφ+b0Ltanφ,where L1 is the distance that road wheels pass by a track, m; b is the width of a track, m; C is the soil cohesion coefficient, kN/m2; φ is the internal friction angle of soil, rad; K is the shear modulus deformation of soil, m; is is the sliding rate of vehicle.


## 2.5. Dynamic Equations of Tracked Vehicles

The following assumptions are made to set up dynamic equations of tracked vehicles:(1) the vehicle is bilaterally symmetrical; (2) excitations exerted on both left and right tracks are exactly the same; (3) road wheels and tracks are rigid. Based on these assumptions, a half vehicle model can be used to build the dynamic model of the vehicle, as illustrated in Figure 16.Figure 16
Half vehicle model.(
1
) Initially the vehicle is static and in equilibrium. So the dynamic equations in z direction are as follows:(16)mcz¨c=∑i=16Fpi+∑k=a,bFkz,(17)Faz=Fa×sinβa+θc,Fbz=Fb×sinβb-θc,where mc is the mass of half vehicle, kg; zc is the displacement of mass center in z direction, m; Fpi is the forces acting on vehicle’ s body from the suspension system, kN; Fk, (k=a,b) is the forces acting on front idler and driving sprocket of tracks, kN; Fkz, (k=a,b) is the force components in z direction acting on front idler and driving sprocket of tracks, kN; βk, (k=a,b) is the angle between the tracks connecting the driving sprocket and the level road when the vehicle is static, rad; θc is the rotation angle of the mass center of the vehicle, rad.(
2
)
Pitching movement equations for the vehicle body,(18)Jcθc¨=∑i=16Fpilicos⁡θc-hisinθc+∑k=a,bFkx(lksinθc+hkcos⁡θc)+∑k=a,bFkzlkcos⁡θc-hksinθc,where Jc is the inertial moment when the vehicle rotates around the mass center, kg·m2; Fkx(k=a,b) is the force components in x direction acting on front idler and driving sprocket, kN; li, hi(i=1~6) is the horizontal and vertical distances of road wheels’ mass center from vehicle’s mass center, li(i=1,2,3) are positive, li(i=4,5,6) are negative; lk, hk(k=a,b) is the horizontal and vertical distances of front idler’s and driving sprocket’s mass center from vehicle’s mass center, m; la is positive, lb is negative.(
3
)
Forces exerted by suspension system,(19)Fpi=cs˙i+ksi.Since the mass center of the tracked vehicle rotates slightly when it is moving on the grade E random road surface, so(20)si=xir-zc-lisinθc≈xir-zc-liθc,where c is the damping coefficient, kN/(m/s); k is the spring coefficient, kN/m; xir is the excitations exerted on a track’s right pin, m; si is the stroke of the suspension system, m.(
4
)
The vertical force acting on the track from road wheel, Fiv,(21)Fiv=Fpi+mc6+mzg-mzai×cos⁡βn,i=1~6,where mz is the mass of a road wheel, kg; g is the acceleration of gravity, m/s2; ai is the road wheel’s acceleration, m/s2; βn is the angle of track n, rad.


## 3. Decoupling of Dynamic Model

### 3.1. Coupled Dynamic Model

There exist couplings in the dynamic model of tracked vehicles:(1) if the shear force under a track is to be calculated, it is necessary to know the forces and moments acting on both pins of the track and the vertical force acting on the track by road wheels. But in order to get these forces and moments, the shear force must be first calculated, as mentioned in Section 2.3;  (2) in order to solve (16)–(18), the shear force and displacements of springs must be known. But, as described in (1), the shear force should be obtained through the vertical force acting on the track from road wheel and so are the displacements of springs. Obviously, the vertical force is related to the vibration and pitching movement of the tracked vehicle’s mass center, which means (16)–(18) are required to calculate the force. The coupling relationships among them are illustrated in Figure 17.Figure 17
Coupling relationships between tracked vehicle and terrain.


### 3.2. Model Decoupling Process

The process is shown in Figure18. Discretization is adopted to solve the coupled dynamic model. A very short sampling time is taken to precisely calculate the derivation and integral of equations or to obtain all the parameters that are needed to calculate other ones, and then substitute the parameters obtained at the former calculation point into the equations to get the parameters for the next calculation point. In the decoupling process, the pressure distribution and the shear force are calculated every dt/3 when a road wheel passes by a track. The vertical and angular velocities of the mass center and so forth are calculated every dt, in which Fb, calculated at the fourth calculation point in a dt, must be known beforehand. When the road wheel is at the midpoint of a track, the method introduced in Section 2.3, namely, the calculation equations for the pressure distribution at every calculation point, can be adopted to get the pressure distribution values, and thus the average sinkage of the track can be obtained.Figure 18
Decoupling process.


## 3.1. Coupled Dynamic Model

There exist couplings in the dynamic model of tracked vehicles:(1) if the shear force under a track is to be calculated, it is necessary to know the forces and moments acting on both pins of the track and the vertical force acting on the track by road wheels. But in order to get these forces and moments, the shear force must be first calculated, as mentioned in Section 2.3;  (2) in order to solve (16)–(18), the shear force and displacements of springs must be known. But, as described in (1), the shear force should be obtained through the vertical force acting on the track from road wheel and so are the displacements of springs. Obviously, the vertical force is related to the vibration and pitching movement of the tracked vehicle’s mass center, which means (16)–(18) are required to calculate the force. The coupling relationships among them are illustrated in Figure 17.Figure 17
Coupling relationships between tracked vehicle and terrain.


## 3.2. Model Decoupling Process

The process is shown in Figure18. Discretization is adopted to solve the coupled dynamic model. A very short sampling time is taken to precisely calculate the derivation and integral of equations or to obtain all the parameters that are needed to calculate other ones, and then substitute the parameters obtained at the former calculation point into the equations to get the parameters for the next calculation point. In the decoupling process, the pressure distribution and the shear force are calculated every dt/3 when a road wheel passes by a track. The vertical and angular velocities of the mass center and so forth are calculated every dt, in which Fb, calculated at the fourth calculation point in a dt, must be known beforehand. When the road wheel is at the midpoint of a track, the method introduced in Section 2.3, namely, the calculation equations for the pressure distribution at every calculation point, can be adopted to get the pressure distribution values, and thus the average sinkage of the track can be obtained.Figure 18
Decoupling process.


## 4. Case Study

The main parameters of a tracked vehicle are shown in Table1. The slipping rate of the vehicle is=0.025, the sampling time period dt=0.009 s. Parameters of the soil, that is, sandy loam, are kc=52.53 kN/mn+1, kφ=1127.97 kN/mn+2, C=4.83 kN/m2, φ=0.349 rad, K=0.025 m, and n=0.9, and the maximum sinkage = 0.2 m.Table 1
Main parameters of a certain type of tracked vehicle.
Parameters
Values
Mass of half vehiclemc/t
20
Moment of inertiaJc/kg·m2
115680.6
Damping coefficient of suspensionc/kN/(m/s)
5
Spring coefficient of suspensionk/kN/m
170
Pretensioning forceF0/kN
40
Horizontal distance of road wheels’ mass centers from vehicle’s mass centerli/m
[2.3,1.38,0.46,-0.46,-1.38,-2.3]
Vertical distance of road wheels’ mass center from vehicle’s mass centerhi/m
0.3
Horizontal distance of front idler’s and driving sprocket’s mass center from vehicle’s mass centerlk/m
[
3
,
-
3
]
Vertical distance of front idler’s and driving sprocket’s mass center from vehicle’s mass centerhk/m
−0.1
Width of a trackb/m
0.464
Length of a trackL/m
0.153
Radius of a road wheelr/m
0.30
Pitch radius of the driving sprocketrp/m
0.25
Radius of a carrier wheelrt/m
0.1
Radius of a front idlerry/m
0.2
Distance between two road wheels/m
0.92A virtual prototype model of the tracked vehicle was built using the module “Track-HM” in RecurDyn to test the accuracy of calculation, as illustrated in Figure19. The parameters used to build the virtual prototype are the same as the ones used in this paper, as shown in Table 1. Considering it needs some time for the virtual prototype model to accelerate to the required speed, aStep function is used to drive the model: (22)Stept,t0,h0,x1,h1,where t is the time, s; t0 is the initial time, 0; h0 is the initial function value, 0; x1 is the end time, 6 s; h1 is the maximum angular velocity of the driving sprocket. The maximum speed of vehicle is 60 km/h, the pitch radius of driving sprocket is 0.25 m, so its value is 67 rad/s.Figure 19
Virtual prototype model of tracked vehicle.The vehicle model is initially static and the friction between components is neglected. When the vehicle is in a uniform rectilinear motion after 6 seconds, the vertical acceleration of the mass center of the vehicle and the tractive force of driving sprocket can be compared with the results of this paper.It can be seen from Figure20 that when the tracked vehicle reaches stable state, the vertical acceleration of the vehicle body’s mass center calculated by the model presented in this paper is basically consistent with the result from RecurDyn, and the acceleration values are in the range of −6 m/s2~6 m/s2, which meets the requirement that the maximum vertical acceleration of tracked vehicle’s mass center should be less than 1 g. The maximum values of the model given in this paper and RecurDyn are 5.808 m/s2 and 5.938 m/s2, respectively. The error between them is 2.19%.Figure 20
Comparison of the vertical accelerations of the vehicle body’s mass center output by the model presented in this paper and RecurDyn.It can be seen from Figure21 that when the tracked vehicle reaches stable state, the tractive force calculated by our model is basically consistent with the result from RecurDyn if the friction among components of the vehicle and centrifugal force of the tracks are not taken into consideration. The error between our model and RecurDyn is 8.77%.Figure 21
Comparison of the tractive forces output by our model and RecurDyn.The results of the model presented in this paper have some deviations from the results output by RecurDyn. This is because(1) our model is built without consideration of inertia forces of tracks and slippage of the vehicle. (2) When a road wheel passes by a track, the track will certainly rotate a little angle. Although the time period that a road wheel on a track is very short, neglecting this angular change may affect the result. (3) Discretization of the computation process is used to decouple the coupled problems, computation errors may be inevitably induced into the results. However, generally the computation errors are in an acceptable range, which indicates the model introduced in this paper is useful for the future optimization of design variables.


## 5. Conclusion

This paper introduces a dynamic model of tracked vehicle when the vehicle moves on the soft random terrain. The contributions of the paper include(1) the soft random terrain excitation is introduced into the model, which makes it much closer to the reality; (2) a detailed track modeling method is introduced; (3) a decoupling method is introduced to solve the coupled dynamic model. Future work would be directed towards improving the modeling of tracks by taking the inertia forces, frictions, and so forth into consideration, so that the computation errors can be further reduced.



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*Source: 293125-2015-03-08.xml*