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Stiffness of a rectilinear suspension with automatic length compensation branches
Mechanism and Machine Theory 56 (2012) 99–122
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Mechanism and Machine Theory journal homepage: www.elsevier.com/locate/mechmt
Stiffness of a rectilinear suspension with automatic length compensation branches Jing-Shan Zhao a,⁎, Xiang Liu a, Zhi-Jing Feng a, Jian S. Dai b, c a b c
State Key Laboratory of Tribology, Department of Precision Instruments and Mechanology, Tsinghua University, Beijing 100084, PR China Centre for Advanced Mechanisms and Robotics, Tianjin University, Tianjin 300072, PR China King's College London, University of London, London WC2R 2LS, United Kingdom
a r t i c l e
i n f o
Article history: Received 26 July 2011 Received in revised form 7 May 2012 Accepted 15 May 2012 Available online 21 June 2012 Keywords: Stiffness Suspension Length compensation branch Design
a b s t r a c t This paper focuses on the stiffness of a rectilinear independent suspension whose alignment parameters are invariable in mechanism theory during jounce and rebound. The suspension is a kind of multilink mechanism, the kinematic chains of which consist of 3-RRR distance compensation branches. However the capacity to keep a straight line motion depends on the resultant stiffness of the suspension. Therefore, this paper proposes a methodology to investigate the equivalent stiffness of the suspension. The fact that the resultant stiffness of the suspension is subjected to the stiffness of each 3-RRR distance compensation linkage raises a particular difficulty to build the model. So the deformation and the deflection of each branch are established first, and then the six-dimensional stiffness of the suspension is equivalently expressed in terms of the material properties of the four chains. © 2012 Elsevier Ltd. All rights reserved.
1. Introduction The main parameters such as caster, camber, and steer angles and so on determinate the handling of a vehicle [1]. The ride comfort is deteriorated by the vibration of suspension system excited by a road surface [2]. The roughness of the road and friction between road and wheels are the main reasons for the wheels' vibration [3]. The most important factors of vibration are the center height of gravity, the track width, the axle roll stiffness, suspension and tire compliance, and suspension roll stiffness [4]. The major vibration mode is the wheel hop of a suspension system. Hence the use of road-friendly vehicles is encouraged by road authorities because they can have higher static loads, but lower dynamic tire forces, consequently reducing road damage [5,6]. Conventional vehicle suspension design generally involves a trade-off between handling stability and ride comfort [7]. The rolling of the body, especially in independent suspensions, causes a change of kinematic parameters of the related mechanisms [8]. The dynamic characteristics of a vehicle are generally determined by the stiffness and damping characteristics of its suspension, and the sprung and unsprung masses [9]. A vehicle with a relatively stiff suspension is likely to possess good handling stability but poor ride comfort, while the reverse is true for a soft, compliant suspension [7]. To build a correct elastokinematic model of a given suspension, bushing stiffness must be determined [10]. But bush uncertainty and mechanical flexibility influence the movement error of suspension systems [11]. Therefore, it is difficult to precisely forecast the performance of a suspension system by quantified performance indices. The unavoidable vibration of a vehicle arises from combined bounce and pitch motions [12], while the longitudinal vibration at the driver location is predominated only by pitch motions of the vehicle [13]. It has also been stated that the ride comfort of heavy vehicles is generally dominated by the pitch plane motions, while the lateral and roll vibrations have been considered to be less important [12,14]. Therefore the synthesis goal is to fulfill a minimum variation of the wheel track, toe angle and camber angle during jounce and rebound of the wheels [15]. The requirements upon the motion of the rear wheels when synthesizing the ⁎ Corresponding author. Fax: + 86 10 62788308. E-mail address: [email protected] (J-S. Zhao). 0094-114X/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.mechmachtheory.2012.05.005
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Vehicle Body A1
A2
B1
B2
A4
A3
B3 B4
Knuckle
Fig. 1. Structure of a rectilinear suspension.
suspension mechanism are that the toe angle variation during compression and rebound should be minimum, excessive outward camber thrust on corners must be forbidden, excessive sideway thrust and consequent rear end steering impulses on single wheel bump or rebounds should also be excluded [16,17]. Consequently, the relationship between the properties of the lateral, vertical and longitudinal primary suspension stiffness should be investigated. Suspension compliance, compliant steer and wheel alignment variation are very important factors to improve a vehicle's stability, controllability and ride comfort [18]. Small differences on camber or toe angle or on orientation of steering axis may have a dramatically influence on the handling performance of a car either directly or through their rates of change with jounce-rebound [19]. Fergussona et al. [20] investigated the maintenance of wheels and the optimum relationship between the properties of the lateral and longitudinal primary suspension stiffness and the center plate friction of a self-steering three-piece bogie. Mun et al. [21] proposed an analytical method to calculate the torsional stiffness of a tubular beam with a closed cross-section in a torsional beam rear suspension system using linear beam theory. The accelerating and braking forces, the inertia forces resulting from turning are transmitted to the vehicle body via the suspension and its linkages; therefore, the design of suspension plays an important role in ensuring starting, turning, accelerating and braking stability [22]. Stiffness of the suspension is a primary index to represent the ride and handling stability of a vehicle. This paper proposes a rectilinear suspension which, in mechanism theory, could keep the alignment parameters of the wheel invariable. However, the capacity of the suspension to trace a straight line relies on the stiffness of the mechanism. Therefore, this paper will focus on this problem and investigate the equivalent stiffness of the rectilinear suspension.
2. Architecture of a rectilinear suspension with stretchable kinematic chains The architecture of a rectilinear suspension is shown in Fig. 1. The knuckle is connected with the vehicle body through four branches, AiBi (i = 1, 2, 3, 4), whose length along its axial direction can be changed automatically within a specified distance range. Each branch, as a matter of fact, is a 3-RRR parallel mechanism shown in Fig. 2.
A1 R11 R12 R13
π1
R21
R31
R32
R33 R22
R23
π2 B1
Fig. 2. A 3-RRR branch of the suspension.
π3
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Fig. 3. Possible motions of a planar RRR kinematic chain.
To investigate the mobility of the branch, the free motions of the RRR kinematic chain in plane π1 should be discussed from the viewpoint of geometry. This sub chain is simplified as Fig. 3 shows.
Fig. 4. The absolute coordinate frame on the suspension mechanism.
The free end point, B1, can translate left and right, up and down and rotate about any normal of the plane in a certain range determined by the lengths of the links. Therefore, point B1 could reach any point with any pose within such a region in π1-plane shown in Fig. 2. Similarly, the other two chains, R21R22R23 and R31R32R33, also have the capacity to reach any point with any poses within their corresponding regions, determined by the lengths of the links, in π2- and π3-planes, respectively. As a result, their common distal platform, R13R23R33, can only move along the intersection of π1-, π2- and π3-planes, which is line A1B1 in Fig. 2 or Fig. 3. This indicates that platforms R11R21R31 and R13R23R33 only have one relative degree of freedom of translation. So the branch shown in Fig. 2 can stretch and contract along its axial direction. The section of 3-RRR can be designed stiff enough to safeguard the strength requirement of the branch. Therefore, what follows will investigate the equivalent stiffness of the suspension by processing the branch as one stretchable link. 3. Equivalent stiffness of the suspension To investigate the stiffness of the suspension, the statics should be first analyzed. Generally, there are driving and braking forces along the x-direction, the lateral force along the y-direction, the vertical force from the ground and the moments and torques about the x-, y- and z-axes, which are shown in Fig. 4. Assume that onxnynzn is a local coordinate system which is attached to the knuckle, and the origin, on, is superimposed with the geometry center of the knuckle, xn directs backward and zn directs upward, so yn can be determined in accordance to the righthand rule. In addition, yn and zn are the symmetries of the knuckle. Initially, the local coordinate frame, onxnynzn, is completely superimposed with the absolute coordinate system oxyz. Therefore, the coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz as: 2
r B1
3 2 3 2 3 2 3 −b b −b b 4 5 4 5 4 5 4 ¼ 0 ; r B2 ¼ 0 ; r B3 ¼ 0 ; r B4 ¼ 0 5: a a −a −a
ð1Þ
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Fig. 5. Displacement of the knuckle in the x-direction.
The unit normal direction vectors of the chain planes which also represent the axial direction of the revolute joints of the chain, Bi (i = 1, 2, 3, 4), shown in Fig. 4 are: 8 > n1 > > < n2 > > n3 > : n4
¼ ½ − cosθu sinθu 0 T T ¼ ½ cosθu sinθu 0 : ¼ ½ − cosθd sinθd 0 T T ¼ ½ cosθd sinθd 0
ð2Þ
The unit normal direction vectors of the branches AiBi (i = 1, 2, 3, 4) shown in Fig. 4 are: 8 > k > > 1 < k2 > k3 > > : k4
¼ ½ cosφu sinθu cosφu cosθu − sinφu T ¼ ½ − cosφu sinθu cosφu cosθu − sinφu T ¼ ½ cosφd sinθd cosφd cosθd sinφd T ¼ ½ − cosφd sinθd cosφd cosθd sinφd T
ð3Þ
where θu denotes the subtended angle between the vertical plane passing through the upper chain axis and yoz-plane, and θd denotes the subtended angle between the vertical plane passing through the lower chain axis and yoz-plane, and φu denotes the subtended angle between the upper chain axis and xoy-plane, and φd denotes the subtended angle between the lower chain axis and xoy-plane, which are shown in Fig. 4. 3.1. Equivalent stiffness of the suspension in x-direction Suppose that there is a force, Fx, exerted to the center of the knuckle in the x-direction. The corresponding displacement of the knuckle, indicated by δx, is shown in Fig. 5. So the coordinate transformation from onxnynzn to oxyz is r B′ j ¼ Tr Bnj þ δron
ð4Þ
where rB′j denotes the vector coordinates in the oxyz frame, and T is the transformation matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, and rBnj represents the absolute vector coordinates in the onxnynzn frame, and δron indicates the vector of the origin of the local coordinate frame in the absolute coordinate frame, and 2
3 2 3 1 0 0 δx 4 5 T ¼ 0 1 0 ; δron ¼ 4 0 5: 0 0 1 0 Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (4) presents their absolute coordinates 2 r B′ 1
¼4
3 2 3 2 3 2 3 δx −b δx þ b δx −b δx þ b 5 4 5 4 5 4 ; r B′ 2 ¼ ; r B′ 3 ¼ ; r B′ 4 ¼ 0 0 0 0 5: a a −a −a
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
Therefore, one obtains 8 → ′ T > > > B1 B 1¼ ½ δx 0 0 > > → > < ′ T B2 B 2¼ ½ δx 0 0 → > > > B3 B′ 3¼ ½ δx 0 0 T > > > : →′ B4 B 4¼ ½ δx 0 0 T
103
ð5Þ
→
where Bi B′¼ i r B′ i −r Bi ; i ¼ 1; 2; 3; 4: The displacement of joint Bi (i = 1, 2, 3, 4) in its axial direction, ni, can be expressed as below: 8 → ′ > > > δ1 ¼B1 B 1⋅n1 ¼ −δx cosθu > > → > < δ2 ¼B2 B′ 2⋅n2 ¼ δx cosθu → > > > δ3 ¼B3 B′ 3⋅n3 ¼ −δx cosθd > > > → : δ4 ¼B4 B′ 4⋅n4 ¼ δx cosθd
ð6Þ
where the negative sign “−” indicates that the displacement direction of the joint is reverse to that of ni (i = 1, 2, 3, 4). Fig. 2 indicates that every branch can stretch or compress along its axis and rotate about the joint axes at both sides. As a result, each branch can only bear the force in the normal direction of its chain shown in Fig. 4, the torque about its axis and the bending moment around the direction perpendicular to both the chain axis and the normal of the chain plane. Because all the kinematic pairs are thrust bearing jointed, the torque and the bending moments are all small items compared with the normal force. For the sake of simplicity, the torque and the bending moments are not counted into the calculation of translational stiffness of the suspension. Suppose that the normal force exerted to joint Bi (i = 1, 2, 3, 4) by the knuckle is Fi (i = 1, 2, 3, 4) when the suspension only subjects to a force Fx the direction of which is along ni. Because the mobility of the parallel 3-RRR section shown in Fig. 2 is identical to a prismatic joint, each branch can be looked as a beam whose length is self adaptive. For the sake of simplicity, the flexural rigidity and the torsional rigidity of the branch are EI and GIp, respectively. Therefore, under the action of Fi (i = 1, 2, 3, 4) the deflection of the joint Bi can be expressed with 8 F l3 > ′ > > δ1¼ 1u > > 3EI > > > > ′ F l3 > <δ 2 ¼ 2 u 3EI ð7Þ 3 > > > δ′ ¼ F 3 ld > 3 > > 3EI > > 3 > > : δ′ ¼ F 4 ld 4 3EI where lu and ld stand for the lengths of the upper and lower chains at the instant, respectively. The compliant conditions require that 8 > δ > > < 1 δ2 > δ > > : 3 δ4
′
¼δ1 ′ ¼δ2 ′ : ¼δ3 ′ ¼δ4
ð8Þ
Substituting Eqs. (6) and (7) into Eq. (8) yields 8 3EIδx cosθu > > F1 ¼ − > > > l3u > > > 3EIδ cosθ > x u > > < F2 ¼ l3u 3EIδx cosθd : > > F3 ¼ − > > > l3d > > > > 3EIδ cosθ > x d > : F4 ¼ l3d
ð9Þ
The static equilibrium equations of the suspension are
−F 1 cosθu þ F 2 cosθu −F 3 cosθd þ F 4 cosθd −F x ¼ 0 : F 1 sinθu þ F 2 sinθu þ F 3 sinθd þ F 4 sinθd ¼ 0
ð10Þ
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Fig. 6. Deflection of the knuckle in the y-direction.
Substituting Eq. (9) into Eq. (10), one finds that the second equation is automatically satisfied because of the structural symmetry of the knuckle, and the first one can be rearranged to
Kx ¼
Fx cos2 θu cos2 θd ¼ 6EI þ δx l3u l3d
! ð11Þ
where Kx indicates the stiffness of the suspension in x-direction. 3.2. Equivalent stiffness of the suspension in y-direction Suppose that there is a force, Fy, exerted to the center of the knuckle in the y-direction. Considering that the structure and load Fy are both symmetrical about the plane of oyz, the forces and deformations of the branches are symmetrical about the plane of oyz. One can know that the knuckle might have a displacement in y-direction, δy, and a rotation about x-axis, γx, as Fig. 6 shows according to the force symmetry. So the coordinate transformation from onxnynzn to oxyz is r B′ i ¼ Tr Bni þ δron
ð12Þ
where rB′i represents the absolute vector coordinates in the oxyz frame, and T is the transformation matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, and rBni denotes the vector coordinates in the onxnynzn frame, and δron indicates the vector of the origin of the local coordinate frame in the absolute coordinate frame, and 2
1 T ¼ 40 0
0 cosγx sinγ x
2 3 3 0 0 − sinγ x 5; δr on ¼ 4 δy 5: cosγx 0
Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (12) presents their absolute coordinates 2
r B′ 1
3 2 3 2 3 2 3 −b b −b b ¼ 4 δy −a sinγ x 5; r B′ 2 ¼ 4 δy −a sinγx 5; r B′ 3 ¼ 4 δy þ a sinγ x 5; r B′ 4 ¼ 4 δy þ a sinγ x 5: a cosγx a cosγ x −a cosγx −a cosγx
The unit normal direction vectors of joints Bi (i = 1, 2, 3, 4) are changed to 8 ′ > n1 > > < ′ n2 > n′ > > : ′3 n4
¼ ½ − cosθu cosγx sinθu sinγ x sinθu T T ¼ ½ cosθu cosγ x sinθu sinγx sinθu : ¼ ½ − cosθd cosγx sinθd sinγx sinθd T T ¼ ½ cosθd cosγ x sinθd sinγx sinθd
ð13Þ
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105
Immediately, one obtains from Eqs. (1) and (13) 8 → > > B1 B′ 1¼ 0 > > > → > > < ′ B2 B 2¼ 0 → > > B B′ ¼ 0 > > > 3→ 3 > > : B4 B′ 4¼ 0
T
δy −a sinγx
að cosγx −1Þ
δy −a sinγx
að cosγx −1Þ
δy þ a sinγ x
−að cosγ x −1Þ
δy þ a sinγ x
−að cosγ x −1Þ
T T
:
T
The deflection of joint Bi (i = 1, 2, 3, 4) on its axial direction, ni (i = 1, 2, 3, 4), can be expressed as: 8 → > ′ > > > δ1 ¼B1 B 1⋅n1 ¼ δy −a sinγ x sinθu > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ δy −a sinγ x sinθu : → > ′ > > > δ3 ¼B3 B 3⋅n3 ¼ δy þ a sinγ x sinθd > > > → > > ′ : δ4 ¼B4 B 4⋅n4 ¼ δy þ a sinγ x sinθd
ð14Þ
Suppose that the normal force exerted to the joint Bi by the knuckle is Fi (i = 1, 2, 3, 4) the direction of which is along n′i when the suspension only subjects to a force Fy. Therefore, under the action of Fi the displacement of joint Bi can be expressed with 8 F l3 > ′ > > δ1¼ 1u > > 3EI > > > > F 2 l3u ′ > <δ 2 ¼ 3EI : > F 3 l3d > ′ > δ3¼ > > > 3EI > > 3 > > : δ′ ¼ F 4 ld 4 3EI
ð15Þ
The compliant conditions require that the expressions in Eq. (8) also hold. Substituting Eqs. (14) and (15) into Eq. (8) yields 8 > > 3EI δy −a sinγ x sinθu > > > F ¼ > > > 1 l3u > > > > > 3EI δy −a sinγ x sinθu > > > > < F2 ¼ l3u : > 3EI δy þ a sinγ x sinθd > > > > > > F3 ¼ > l3d > > > > > 3EI δ þ a sinγ x sinθd > y > > > : F4 ¼ l3
ð16Þ
d
The static equilibrium equations of the suspension are 8 < −F 1 cosθu þ F 2 cosθu −F 3 cosθd þ F 4 cosθd ¼ 0 F cosγ x sinθu þ F 2 cosγx sinθu þ F 3 cosγ x sinθd þ F 4 cosγx sinθd −F y ¼ 0 : : 1 F 1 sinγ x sinθu þ F 2 sinγx sinθu þ F 3 sinγ x sinθd þ F 4 sinγx sinθd ¼ 0
ð17Þ
Substituting Eq. (16) into Eq. (17), one then gets Fy = 0 if γx ≠ 0, which conflict with the reality. Therefore, there must be γx = 0. And one can find that the first equation is automatically satisfied and the second equation can be rearranged to ! 2 2 Fy sin θu sin θd ¼ 6EI þ Ky ¼ ð18Þ δy l3u l3d where Ky indicates the stiffness of the suspension in y-direction. 3.3. The equivalent torsional stiffness of the suspension about x-axis Suppose that there is a moment about the x-axis, Mx, exerted to the knuckle the corresponding rotational angle, shown in Fig. 7, is γx. Because the suspension is not rotational symmetrical about the x-axis, the knuckle might have a small displacement
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Fig. 7. Slope of the suspension under the moment of Mx.
within the yoz-plane accompanying a small rotational angle, γx. Assume that the center displacement of the knuckle under the T action of Mx is δron ¼ ð 0 δy δz Þ : The transformation matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, oxyz, can be expressed in term of the rotational angle of the knuckle, γx: 2
1 T ¼ 40 0
0 cosγx sinγ x
3 0 − sinγ x 5: cosγx
Therefore, the absolute coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz coordinate frame r B′ i ¼ Tr Bni þ δron :
ð19Þ
Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (19) presents their absolute coordinates 2
r B′ 1
3 2 3 2 3 2 3 −b b −b b ¼ 4 δy −a sinγ x 5; r B′ 2 ¼ 4 δy −a sinγx 5; r B′ 3 ¼ 4 δy þ a sinγx 5; r B′ 4 ¼ 4 δy þ a sinγx 5: δz þ a cosγ x δz þ a cosγx δz −a cosγx δz −a cosγ x
ð20Þ
With Eqs. (1) and (20), one immediately obtains 8 → > ′ > B1 B 1¼ 0 > > > → > > < B B′ ¼ 0
δy −a sinγx
δz þ að cosγ x −1Þ
δy −a sinγx
δz þ að cosγ x −1Þ
0
δy þ a sinγ x
δz þ að1− cosγx Þ
0
δy þ a sinγ x
δz þ að1− cosγx Þ
2 2 → > ′ > > B > 3 B 3¼ > > → > : B4 B′ 4¼
T T T
:
T
The deflection of joint Bi (i = 1, 2, 3, 4) on its axial direction, ni, can be expressed as: 8 → > ′ > δ1 ¼B1 B 1⋅n1 ¼ δy −a sinγ x sinθu > > > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ δy −a sinγ x sinθu : → > ′ > > sinθ δ ¼B B ⋅n ¼ δ þ a sinγ > 3 3 3 3 y d x > > > → > > : δ ¼B B′ ⋅n ¼ δ þ a sinγ sinθ 4 4 4 4 y d x
ð21Þ
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Fig. 8. Displacement of the end point on a branch.
Considering the small value of γx, the following approximations can be adopted:
sinγx ≈γx cosγx ≈1 nþ1
ð2n−1Þ
Þ where sinγx ¼ γ x − 3!1 γ3x þ 5!1 γ5x þ n⋯ þ ðð−1 þ ⋯ in which only the first item is retained for a first order approximation, and 2n−1Þ! γ x ð−1Þ 1 2 1 4 2n cosγ x ¼ 1− 2! γ x þ 4! γ x þ ⋯ þ 2n! γx þ ⋯ in which only the first item is retained for the approximation. Therefore Eq. (21) can be simplified as
8 > δ1 ¼ δy −aγx sinθu > > > > > < δ2 ¼ δy −aγx sinθu : > > > δ3 ¼ δy þ aγx sinθd > > > : δ ¼ δ þ aγ sinθ 4
y
d
x
Under the action of the moment, Mx, both the position and orientation of the joint axis of Bi (i = 1, 2, 3, 4) have been changed. The displacement of joint Bi consists of the bending and torsional components which are shown in Fig. 8. To illustrate the relationship of these angles, one can investigate the first branch. Suppose that the torsional angle of the branch, denoted by γ1, is shown in Figs. 8 and 9. The unit vector of B′1 in the oxyz-frame is 2
′
n1
1 ¼ 40 0
0 cosγ x sinγx
3 2 32 3 − cosθu 0 − cosθu − sinγx 54 sinθu 5 ¼ 4 sinθu cosγx 5: cosγ x sinθu sinγ x 0
So ′
2
2
n1 ⋅n 1 ¼ cos θu þ sin θu cosγ x that is 2
2
2
cosψ1 ¼ cos θu þ sin θu cosγx ¼ 1 þ sin θu ð cosγx −1Þ where ψ1 is the subtended angle between n1 and n′1. This can be rearranged to 2
1−cosψ1 ¼ sin θu ð1− cosγx Þ:
Fig. 9. Angular relationship in the deflection of a branch.
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Ultimately, one obtains 2
2 sin
ψ1 2 2γ ¼ 2 sin θu sin x : 2 2
Considering the small values of γx and ψ1 shown in Fig. 9, one finds that ψ1 ¼ γx sinθu :
ð22Þ
Because π ′ cos −ϕ1 ¼ n 1 ⋅k1 ¼ cosφu sinθu cosθu ð cosγx −1Þ− sinφu sinθu sinγ x ; 2 and taking into account of the small values of ϕ1 and γx, one gains ϕ1 ¼ −γx sinφu sinθu
ð23Þ
where ϕ1 is the subtended angle from n′1 to the branch plane. Substituting Eq. (22) into Eq. (23) yields ϕ1 ¼ −ψ1 sinφu : For the fact that ψ1 is a very small angle, ϕ1 is a tiny value, too. Therefore, cosϕ1 ≈1: From Fig. 9, one finds that cosψ1 ¼ cosϕ1 cosγ 1 ≈ cosγ 1 : Ultimately, one obtains γ1 ≈ψ1 ¼ γ x sinθu
ð24Þ
because ′ n1 n 1 ⋅k1 ¼ γ x cosφu sinθu > 0: The direction of the torsional angle, γ1, is same with the positive x-axis. Therefore, one can obtain 8 γ ¼ γ x sinθu > > < 1 γ 2 ¼ −γ x sinθu : γ ¼ γ x sinθd > > : 3 γ 4 ¼ −γ x sinθd
ð25Þ
Suppose that the torque about the axis of the branch is Ti (i = 1, 2, 3, 4) the direction of which is along ki (i = 1, 2, 3, 4) and the bending moment is Mi (i = 1, 2, 3, 4) the direction of which is defined by ki × ni shown in Fig. 10.
Fig. 10. Statics of the suspension.
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109
In vector form, the torque and moment loads can be rewritten as 8 > T > > 1 < T2 > T > > : 3 T4
8 > M1 ¼ ½ T 1 cosφu sinθu T 1 cosφu cosθu −T 1 sinφu T > > < ¼ ½ −T 2 cosφu sinθu T 2 cosφu cosθu −T 2 sinφu T M2 T > ¼ ½ T 3 cosφd sinθd T 3 cosφd cosθd T 3 sinφd M > > : 3 ¼ ½ −T 4 cosφd sinθd T 4 cosφd cosθd T 4 sinφd T M4
¼ ½ M1 sinφu sinθu M 1 sinφu cosθu M 1 cosφu T ¼ ½ M2 sinφu sinθu −M 2 sinφu cosθu −M2 cosφu T : ¼ ½ −M 3 sinφd sinθd −M3 sinφd cosθd M3 cosφd T T ¼ ½ −M 4 sinφd sinθd M 4 sinφd cosθd −M4 cosφd
Under these actions, the corresponding displacements and torsional angles of joint Bi (i = 1, 2, 3, 4) are 8 8 > > > > M1 l2u > > γ′1 ′ > > > > δ1 ¼ > > > > 2EI > > > > > > ′ > > M l2 > > < δ′2 ¼ 2 u < γ2 2EI and > > ′ M l2 > > > > γ3 δ′3 ¼ 3 d > > > > > > 2EI > > > > 2 > > > > ′ > δ′ ¼ M4 ld > > > : 4 : γ4 2EI
T 1 lu GIp T 2 lu ¼ GIp : T 3 ld ¼ GIp T l ¼ 4d GIp ¼
Considering the compliant requirements, one obtains 8 ′ > δ1 > > < ′ δ2 ′ > δ > > : ′3 δ4
¼ δ′ 1 ; γ′ 1 ¼ δ′ 2 ; γ′ 2 ′ ′ ¼ δ 3; γ 3 ′ ′ ¼ δ 4; γ 4
¼ γ1 ¼ γ2 : ¼ γ3 ¼ γ4
As a result, there are 8 > > 2EI δy −aγ x sinθu > 8 > > M ¼ > GI p γx sinθu > > > 1 > l2u T1 ¼ > > > > > > lu > > > > 2EI δy −aγ x sinθu > > GI p γx sinθu > > > > > > M ¼ T ¼ − < 2 < 2 l2u lu and : GI p γx sinθd > > 2EI δy þ aγ x sinθd > > > > T ¼ > > 3 > > ld > > > M3 ¼ > > > l2d > > GI p γx sinθd > > > > > : T4 ¼ − > 2EI δy þ aγ x sinθd > ld > > > : M4 ¼ 2 ld
ð26Þ
The static equilibrium conditions require that 8 < cosφu sinθu ðT 1 −T 2 Þ þ cosφd sinθd ðT 3 −T 4 Þ þ sinφu sinθu ðM 1 þ M 2 Þ− sinφd sinθd ðM 3 þ M 4 Þ−M x ¼ 0 cosφu cosθu ðT 1 þ T 2 Þ þ cosφd cosθd ðT 3 þ T 4 Þ þ sinφu cosθu ðM1 −M 2 Þ− sinφd cosθd ðM 3 −M 4 Þ ¼ 0 : : − sinφu ðT 1 þ T 2 Þ þ sinφd ðT 3 þ T 4 Þ þ cosφu ðM1 −M2 Þ þ cosφd ðM3 −M4 Þ ¼ 0
ð27Þ
Substituting Eq. (26) into Eq. (27), one can find that the last two equations hold automatically because of the structural symmetry of the suspension. Rearranging the first equation yields M x ¼ 2GIp γ x
! sin2 θu cosφu sin2 θd cosφd þ −4EIaγx lu ld
sin2 θu sinφu sin2 θd sinφd þ l2u l2d
! þ 4EIδy
sin2 θu sinφu sin2 θd sinφd − 2 lu l2d
!
where δy is an accompanying item under the action of Mx which can be neglected here because it is not explicitly expressed as any function of γx. Therefore, the equivalent torsional stiffness of the branch is K Mx
M ¼ x ¼ 2GIp γx
! ! sin2 θu cosφu sin2 θd cosφd sin2 θu sinφu sin2 θd sinφd þ þ −4EIa lu ld l2u l2d
ð28Þ
where KMx denotes the torsional stiffness of the suspension about the x-axis. 3.4. The equivalent torsional stiffness of the suspension about y-axis Suppose that there is a moment My about the y-axis exerting to the knuckle, the corresponding rotational angle of which is γy and is shown in Fig. 11. Because the suspension is not rotational symmetrical about the y-axis, the knuckle might have a small
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Fig. 11. Slope of the suspension under the moment of My.
displacement within the xoz-plane accompanying the small rotation γy. Assume that the accompanied displacement of the knuckle center under the action of My is δr on ¼ ð δx 0 δz ÞT : Therefore, the absolute coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz coordinate frame r B′ i ¼ Tr Bni þ δron
ð29Þ
where rB′i denotes the absolute coordinates of joint Bi, and 2 T¼4
cosγy 0 − sinγ y
0 1 0
3 sinγ y 0 5 cosγ y
where T denotes the rotational matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame. Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (29) produces their absolute coordinates 2
3 2 3 δx þ a sinγy −b cosγy δx þ a sinγ y þ b cosγy 5; r ′ ¼ 4 5; 0 0 r B′ 1 ¼ 4 B2 δz þ a cosγy þ b sinγy δz þ a cosγ y −b sinγy 2 3 2 3 δx −a sinγy −b cosγy δx −a sinγ y þ b cosγy 5; r ′ ¼ 4 5: 0 0 r B′ 3 ¼ 4 B4 δz −a cosγy þ b sinγy δz −a cosγ y −b sinγ y Therefore, one obtains 8 → > > B1 B′ 1¼ δx þ a sinγ y −b cosγ y þ b > > > → > > < B B′ ¼ δ þ a sinγ þ b cosγ −b 2
2
x
y
y
→ > > > B3 B′ 3¼ δx þ b−a sinγ y −b cosγ y > > > → > : ′ B4 B 4¼ δx −b−a sinγ y þ b cosγ y
δz þ a cosγy þ b sinγy −a T 0 δz þ a cosγy −b sinγy −a T : 0 δz þ a−a cosγ y þ b sinγy T 0 δz þ a−a cosγ y −b sinγ y T 0
The displacement of joint Bi (i = 1, 2, 3, 4) in its axial direction, denoted by ni, will be 8 → > ′ > > > δ1 ¼B1 B 1⋅n1 ¼ − cosθu δx þ a sinγ y −b cosγ y þ b > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ cosθu δx þ a sinγ y þ b cosγ y −b : → > ′ > > > δ3 ¼B3 B 3⋅n3 ¼ − cosθd δx −a sinγ y −b cosγ y þ b > > > → > > : δ4 ¼B4 B′ 4⋅n4 ¼ cosθd δx −a sinγy þ b cosγ y −b
ð30Þ
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Fig. 12. Rotation of a branch about the y-axis.
In a similar way to the above analysis, sin γy ≈ γy and cos γy ≈ 1 when the rotational angle is very small. Therefore, Eq. (30) can be further simplified to 8 > δ1 ¼ − δx þ aγ y cosθu > > > > > < δ2 ¼ δx þ aγy cosθu : > > δ3 ¼ − δx −aγy cosθd > > > > : δ ¼ δ −aγ cosθ 4
x
d
y
Under the action of My, the axis of joint Bi (i = 1, 2, 3, 4) changes its position and orientation as well, and the branch will have bending and torsional deflections which are shown in Fig. 12. To illustrate the relationship of these angles, one can investigate the first branch. Suppose that the torsional angle of the branch, denoted by γy, as shown in Figs. 12 and 9, one can obtain 2 ′
n1
¼4
cosγ y 0 − sinγy
0 1 0
3 2 32 3 − cosγ y cosθu sinγ y − cosθu 4 5 5 4 5: sinθu 0 ¼ sinθu sinγy cosθu cosγy 0
So ′
2
2
n1 ⋅n 1 ¼ sin θu þ cos θu cosγ y that is 2 2 2 cosψ1 ¼ sin θu þ cos θu cosγy ¼ 1 þ cos θu cosγy −1 where ψ1 is the subtended angle between n1 and n′1. It can be rearranged to 2 1−cosψ1 ¼ cos θu 1− cosγy and 2
2 sin
ψ1 2 2 γy ¼ 2 cos θu sin : 2 2
ð31Þ
Because γy and ψ1 are all small angles, one obtains from Eq. (31) ψ1 ¼ γy cosθu : From Fig. 9, one finds that π ′ cos −ϕ1 ¼ n 1 ⋅k1 ¼ cosφu sinθu cosθu 1− cosγ y − sinφu cosθu sinγy : 2
ð32Þ
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Because ϕ1 and γy are both small angles, Eq. (32) results in ϕ1 ¼ −γy sinφu cosθu
ð33Þ
where ϕ1 is the subtended angle from n′1 to the branch plane and γy is the twisting angle about the branch axis. Substituting Eq. (32) into Eq. (33) yields ϕ1 ¼ −ψ1 sinφu : For the tiny value of ψ1, it is safe to suppose that cosϕ1 ≈1: From Fig. 9, one obtains cosψ1 ¼ cosϕ1 cosγ 1 : Therefore, the following approximation holds cosψ1 ≈ cosγ 1 and then γ1 ≈ψ1 ¼ γ y cosθu
ð34Þ
because ′ s ¼ n1 n 1 ⋅k1 ¼ γ x cosφu cosθu > 0: The direction of the torsional angle, γ1, is the same with the positive x-axis. Similarly, one can obtain the torsional angle of the four branches 8 γ ¼ γ cosθ > > < γ 1 ¼ γ y cosθu u 2 y : γ ¼ γ cosθ > > : γ 3 ¼ γ y cosθd d 4 y
ð35Þ
Suppose that the torque about the axis of branch AiBi (i = 1, 2, 3, 4) is Ti the direction of which is along ki and the bending moment is Mi the direction of which is defined by ki × ni, which are shown in Fig. 10. In vector form, the torque and moment loads can be rewritten as 8 > T1 > > < T2 > T > > : 3 T4
8 ¼ ½ T 1 cosφu sinθu T 1 cosφu cosθu −T 1 sinφu T > M1 > > < ¼ ½ −T 2 cosφu sinθu T 2 cosφu cosθu −T 2 sinφu T M2 T > ¼ ½ T 3 cosφd sinθd T 3 cosφd cosθd T 3 sinφd M > > : 3 T ¼ ½ −T 4 cosφd sinθd T 4 cosφd cosθd T 4 sinφd M4
¼ ½ M1 sinφu sinθu M 1 sinφu cosθu M 1 cosφu T ¼ ½ M2 sinφu sinθu −M 2 sinφu cosθu −M2 cosφu T T : ¼ ½ −M 3 sinφd sinθd −M3 sinφd cosθd M3 cosφd T ¼ ½ −M 4 sinφd sinθd M 4 sinφd cosθd −M4 cosφd
Under these actions, the corresponding displacement in the axial direction and torsional angle of joint Bi (i = 1, 2, 3, 4) are 8 8 2 > > ′ > > M l > > γ1 ′ 1 u > > > > δ1 ¼ > > > > 2EI > > > > 2 > > ′ > > M l > > < δ′2 ¼ 2 u < γ2 2EI and 2 > > M l > ′ > > > γ ′3 δ3 ¼ 3 d > > > > > > 2EI > > > > 2 > > > > M 4 ld ′ ′ > > > > : δ4 ¼ : γ4 2EI
T 1 lu GIp T 2 lu ¼ GIp : T 3 ld ¼ GIp T 4 ld ¼ GIp ¼
Considering the compliant requirements, one obtains 8 ′ > > > δ′ 1 < δ2 > δ′ > > : ′3 δ4
′
′
¼ δ 1; γ 1 ′ ′ ¼ δ 2; γ 2 ′ ¼ δ 3 ; γ′ 3 ¼ δ′ 4 ; γ′ 4
¼ γ1 ¼ γ2 : ¼ γ3 ¼ γ4
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113
As a result, one gains that 8 > > 2EI δx þ aγy cosθu > 8 > > M1 ¼ − > > > > > l2u T1 > > > > > > > > > > cosθ 2EI δ þ aγ > > x u y > > > > > > < M2 ¼ < T2 2 lu and > > 2EI δx −aγy cosθd > > > > T3 > > > > M3 ¼ − > > 2 > > > > l > > d > > > > > : T4 > 2EI δx −aγy cosθd > > > > M4 ¼ : l2d
GI p γ y cosθu lu GI p γ y cosθu ¼ lu : GI p γ y cosθd ¼ ld GI p γ y cosθd ¼ ld ¼
ð36Þ
The static equilibrium conditions require that 8 <
cosφu sinθu ðT 1 −T 2 Þ þ cosφd sinθd ðT 3 −T 4 Þ þ sinφu sinθu ðM1 þ M2 Þ− sinφd sinθd ðM3 þ M4 Þ ¼ 0 cosφu cosθu ðT 1 þ T 2 Þ þ cosφd cosθd ðT 3 þ T 4 Þ þ sinφu cosθu ðM1 −M2 Þ− sinφd cosθd ðM 3 −M 4 Þ−M y ¼ 0 : : − sinφu ðT 1 þ T 2 Þ þ sinφd ðT 3 þ T 4 Þ þ cosφu ðM 1 −M 2 Þ þ cosφd ðM 3 −M 4 Þ ¼ 0
ð37Þ
Substituting Eq. (36) into Eq. (37) yields 8 > > > 4EIδx > > <
! sinφu cosθu sinφd cosθd cosφd cosθd cosφu cosθu − − þ 4EIaγ y ¼ −2GIp γy lu ld l2d l2u 1 0 ! : 2 2 2 2 sinφu cos θu δx þ aγ y sinφd cos θd δx −aγ y cosφu cos θu cosφd cos θd A ¼ My −4EI@ þ − lu ld l2u l2d
cosφu cosθu cosφd cosθd þ l2u l2d
> > > > > : 2GIp γy
!
So the torsional stiffness of the suspension about the y-axis is K My ¼
My ¼ 2GIp γy
! ! cosφu cos2 θu cosφd cos2 θd sinφu cos2 θu ðk1 þ aÞ sinφd cos2 θd ðk1 −aÞ þ − −4EI lu ld l2u l2d
ð38Þ
where KMy denotes the torsional stiffness of the suspension about the y-axis,
k1 ¼
−GIp
sinφu cosθu lu
− sinφdl cosθd þ 2EIa cosφdl2cosθd − cosφul2cosθu d u d : cosφu cosθu cosφd cosθd 2EI þ l2 l2 u
d
3.5. The equivalent torsional stiffness of the suspension about z-axis Suppose that there is a moment about the z-axis, Mz, exerted to the knuckle the corresponding rotational angle of which is γz shown in Fig. 13. Assume that the displacement of the knuckle center under the action of Mz is δr on ¼ ð δx δy 0 ÞT : Therefore, the absolute coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz r B′ i ¼ Tr Bni þ δron
ð39Þ
where T represents the rotational matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, oxyz, and 2
cosγz T ¼ 4 sinγ z 0
− sinγ z cosγz 0
3 0 0 5: 1
Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (39) produces their absolute coordinates 2
r B′ 1
3 2 3 2 3 2 3 δx −b cosγz δx þ b cosγz δx −b cosγz δx þ b cosγz ¼ 4 δy −b sinγz 5; r B′ 2 ¼ 4 δy þ b sinγz 5; r B′ 3 ¼ 4 δy −b sinγz 5; r B′ 4 ¼ 4 δy þ b sinγz 5: a a −a −a
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Fig. 13. Slope of the suspension under the moment of Mz.
Therefore, 8 → > > B1 B′ 1¼ δx −b cosγ z þ b > > > → > > < ′ B2 B 2¼ δx þ b cosγz −b → > > > B3 B′ 3¼ δx −b cosγ z þ b > > > → > : ′ B4 B 4¼ δx þ b cosγz −b
δy −b sinγ z δy þ b sinγz δy −b sinγ z δy þ b sinγz
0
T
0 T : 0 T 0 T
The displacement of joint Bi (i = 1, 2, 3, 4) in its axial direction, ni (i = 1, 2, 3, 4), can be expressed as: 8 → > ′ > δ1 ¼B1 B 1⋅n1 ¼ − cosθu ðδx −b cosγ z þ bÞ þ sinθu δy −b sinγz > > > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ cosθu ðδx þ b cosγ z −bÞ þ sinθu δy þ b sinγ z : → > ′ > > > δ3 ¼B3 B 3⋅n3 ¼ − cosθd ðδx −b cosγ z þ bÞ þ sinθd δy −b sinγ z > > > → > > : δ4 ¼B4 B′ 4⋅n4 ¼ cosθd ðδx þ b cosγ z −bÞ þ sinθd δy þ b sinγ z
ð40Þ
Considering sin γz ≈ γz and cos γz = 1 when γz is a tiny item, Eq. (40) can be further simplified as 8 > δ1 ¼ −δx cosθu þ δy −bγz sinθu > > > > > < δ2 ¼ δx cosθu þ δy þ bγz sinθu : > > δ3 ¼ −δx cosθd þ δy −bγz sinθd > > > > : δ ¼ δ cosθ þ δ þ bγ sinθ 4
x
d
y
z
d
Because the structure is symmetrical and the load is antisymmetrical, the deformations of the branches are antisymmetrical, there should be δy = 0. Then 8 δ > > < 1 δ2 > > δ3 : δ4
¼ −δx cosθu −bγ z sinθu ¼ δx cosθu þ bγz sinθu : ¼ −δx cosθd −bγ z sinθd ¼ δx cosθd þ bγz sinθd
ð41Þ
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115
Fig. 14. Deflection of a branch under the moment Mz.
Under the action of moment, Mz, both the position and the orientation of the joint axis of Bi (i = 1, 2, 3, 4) have been changed. The displacement of Bi on the suspension consists of the bending and torsional components which are shown in Fig. 14. The unit vector of B′1 in the oxyz-frame is 2
′
n1
cosγ z ¼ 4 sinγz 0
− sinγz cosγ z 0
3 2 32 3 − cosθu − cosγz cosθu − sinγz sinθu 0 0 54 sinθu 5 ¼ 4 − sinγz cosθu þ cosγz sinθu 5: 0 1 0
Because the vectors n1 and n′1 are coplanar, there is no torsional component of the branch 1, the same are true to the other branches. Suppose that the bending moment is Mi (i = 1, 2, 3, 4) the direction of which is defined by ki × ni. In vector form, the bending moment loads can be rewritten as: 8 > M1 > > < M2 > M > > : 3 M4
¼ ½ M 1 sinφu sinθu M1 sinφu cosθu M1 cosφu T ¼ ½ M 2 sinφu sinθu −M2 sinφu cosθu −M 2 cosφu T : ¼ ½ −M3 sinφd sinθd −M 3 sinφd cosθd M 3 cosφd T T ¼ ½ −M4 sinφd sinθd M4 sinφd cosθd −M 4 cosφd
Under these actions, the corresponding displacements in the axial direction of joint Bi (i = 1, 2, 3, 4) are 8 2 M l > > > δ′1 ¼ 1 u > > 2EI > > 2 > > M l > < δ′2 ¼ 2 u 2EI : 2 > M 3 ld > ′ > δ ¼ > 3 > > 2EI > > 2 > > : δ′ ¼ M4 ld 4 2EI
Considering the compliant requirements, one obtains 8 ′ > > > δ′ 1 < δ2 > δ′ > > : ′3 δ4
¼ δ1 ¼ δ2 : ¼ δ3 ¼ δ4
ð42Þ
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As a result, one finds that 8 2EI ðδx cosθu þ bγ z sinθu Þ > > > > M1 ¼ − > l2u > > > > 2EI ð δ cosθ þ bγz sinθu Þ > x u > > < M2 ¼ l2u : 2EI ðδx cosθd þ bγ z sinθd Þ > > M3 ¼ − > > > l2d > > > > 2EIðδx cosθd þ bγz sinθd Þ > > > : M4 ¼ l2d
ð43Þ
The static equilibrium conditions require that 8 < sinφu sinθu ðM1 þ M2 Þ− sinφd sinθd ðM 3 þ M 4 Þ ¼ 0 sinφu cosθu ðM1 −M2 Þ− sinφd cosθd ðM 3 −M 4 Þ ¼ 0 : : cosφu ðM1 −M2 Þ þ cosφd ðM3 −M4 Þ−M z ¼ 0
ð44Þ
Substituting Eq. (43) into Eq. (44) yields 8 ðδx cosθu þ bγz sinθu Þ sinφu cosθu ðδx cosθd þ bγz sinθd Þ sinφd cosθd > > > − ¼0 < l2u l2d : 4EIðδx cosθu þ bγ z sinθu Þ cosφu 4EIðδx cosθd þ bγ z sinθd Þ cosφd > > > − ¼ Mz :− 2 2 lu ld So the torsional stiffness of the suspension about the z-axis is
K Mz ¼
Mz 4EIbðk2 cosθu þ sinθu Þ cosφu 4EIbðk2 cosθd þ sinθd Þ cosφd ¼− − γz l2u l2d
ð45Þ
where KMz denotes the torsional stiffness of the suspension about the z-axis and k2 ¼
l2u sinθd cosθd sinφd −l2d sinθu cosθu sinφu : l2d cos2 θu sinφu −l2u cos2 θd sinφd
3.6. Equivalent stiffness of the suspension in z-direction The spring shock absorber connects the vehicle and the knuckle through two sphere joints shown in Fig. 15. The lower joint is denoted by C while the upper joint is denoted with D. Assume that the local coordinates of joint C in the coordinate frame onxnynzn
Fig. 15. The spring shock absorber and the suspension.
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117
are represented by Cn (xC, yC, zC). Also suppose that the absolute coordinates of joint D are denoted by Dn (xD, yD, zD). Suppose that the angle from the axis of the spring shock absorber to the xnonyn-plane is β when the suspension moves a distance of s under the weight of vehicle which is shown in Fig. 15, one can immediately obtain the following relations: 8 xD −xC > > < α ¼ arctan y −y D C : > β ¼ − arctan ðzD −zC −sÞ cosα > : yD −yC Assume that the knuckle is bearing a force Fz along the z-axis, under the action of which, the knuckle has small displacements along the x-, y- and z-axes, denoted by δx, δy and δz accompanying small rotations, γx and γz, about the x- and z-axes, individually. So the coordinate transformation from onxnynzn to oxyz is r ¼ Tr n þ δr on
ð46Þ
where rn denotes the vector coordinates in the onxnynzn frame, and r represents the absolute vector coordinates in the oxyz frame, and 2
sinγ x sinγy cosγ z − cosγ x sinγ z − sinγ x sinγy sinγz þ cosγx cosγ z sinγ x cosγy
cosγy cosγz T ¼ 4 cosγ y sinγ z − sinγy
3 cosγx sinγ y cosγz þ sinγ x sinγz cosγx sinγ y sinγ z − sinγx cosγz 5 cosγ x cosγy
2
δr on
3 δx 4 ¼ δy 5: δz
ð47Þ
ð48Þ
Because the values of the rotations, γx and γz, are very small, Eq. (47) can be simplified as 2
1 T ¼ 4 γz −γy
−γ z 1 γx
3 γy −γ x 5: 1
Assume the axial force exerted to the knuckle by the spring shock absorber is F which can be expressed as F ¼ ½ F sinα cosβ
F cosα cosβ
T
−F sinβ :
ð49Þ
In line with the equilibrium of the knuckle, one gains F z −F sinβ ¼ 0: The torques resulting from the action of F can be expressed by 2 3 3 Mx −F ðyC sinβ þ zC cosα cosβÞ 4 M y 5 ¼ r C F ¼ 4 F ðxC sinβ þ zC sinα cosβÞ 5: Mz F ðxC cosα cosβ−yC sinα cosβÞ 2
ð50Þ
Associating Eqs. (11), (18), (28), (38), and (45) with Eqs. (49) and (50) yields 8 F sinα cosβ > > > δx ¼ > Kx > > > > F cosα cosβ > > δ ¼ > y > > Ky > > < M γx ¼ x : K Mx > > > > My > > > > γy ¼ > K > My > > > M > > : γz ¼ z K Mz
ð51Þ
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Prior to the action of F, the coordinates of ball joint C in the coordinate system oxyz are r C ¼ ½ xC yC zC þ s T where s stands for the displacement of the knuckle under the self load of the vehicle. After the action of F, the coordinates of ball joint C in the coordinate system oxyz can be obtained by substituting rC into Eq. (46) 2
3 δx þ xC −γ z yC þ ðzC þ sÞγy r C ¼ Tr n þ δro ¼ 4 δy þ yC þ xC γz −ðzC þ sÞγx 5: δz þ zC þ s−xC γy þ yC γx Therefore, one obtains → ′ CC¼ δx −γ z yC þ ðzC þ sÞγy
δy þ xC γz −ðzC þ sÞγx
δz −xC γy þ yC γx
T
:
The unit axial vector of the spring shock absorber is nDC ¼ ½ sinα cosβ
cosα cosβ
T
− sinβ :
So, the deformation of the spring is →
Δl h¼ nDC ⋅ CC ′ i h i h i ¼ δx −γ z yC þ ðzC þ sÞγy sinα cosβ þ δy þ xC γz −ðzC þ sÞγx cosα cosβ− δz −xC γ y þ yC γ x sinβ:
ð52Þ
Suppose the stiffness of the spring is k, the force exerted to the knuckle is F ¼ −kΔl ð53Þ where “−” indicates that the direction of the displacement of the suspension is reverse to that of the elastic force of the spring. Substituting Eqs. (51) and (52) into Eq. (53) yields ! ! My F sinα cosβ M F cosα cosβ Mz M − z yC þ ðzC þ sÞ sinα cosβ−k þ xC − x ðzC þ sÞ cosα cosβ Kx Ky K Mz K K Mz K Mx ! My My M x þ x y sinβ: þ k δz − K My C K Mx C
F ¼ −k
ð54Þ
Substituting Eq. (50) into Eq. (54) yields ! F sinα cosβ F ðxC cosα cosβ−yC sinα cosβÞ F ðxC sinβ þ zC sinα cosβÞ − yC þ ðzC þ sÞ sinα cosβ Kx K Mz K My ! F cosα cosβ F ðxC cosα cosβ−yC sinα cosβÞ −F ðyC sinβ þ zC cosα cosβÞ þ xC − ðzC þ sÞ cosα cosβ −k Ky K Mz K Mx ! F ðxC sinβ þ zC sinα cosβÞ −F ðyC sinβ þ zC cosα cosβÞ xC þ yC sinβ: þk δz − K My K Mx F ¼ −k
ð55Þ
Eq. (55) can be simplified to Kz ¼
F sinβ 1 ¼ δz f k −f x −f y −f Mx −f My −f Mz
ð56Þ
where 1 sin2 α cot2 β cos2 α cot2 β ; fx ¼ ; fy ¼ ; 2 Kx Ky k sin β 2 2 2 y ð2z þ sÞ cosα cotβ þ zC ðzC þ sÞ cos α cot β þ yC ¼ C C ; K Mx 2 2 2 x þ xC ð2zC þ sÞ sinα cotβ þ zC ðzC þ sÞ sin α cot β ¼ C ; K My
fk ¼ f Mx f My
f Mz ¼
x2C cos2 α þ y2C sin2 α−2xC yC sinα cosα 2 cot β: K Mz
Suppose that the radius of the wheel is denoted by R, and L represents the distance from the central plane of the wheel to the knuckle, Lu and Ld stand for the initial lengths of the upper two links and the lower two links, respectively, Du stands for the
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119
Table 1 Parameters of the suspension with different branches. θu(°)
Initial C
a(mm)
R(mm)
E(GPa)
Du(mm)
r(mm)
43
ð0
50
320
206
95
15
Ld(mm)
θd(°)
Initial D
b(mm)
L(mm)
G(GPa)
Dd(mm)
m(kg)
234
48
ð0
75
60
79.38
55
441.5
−180
540 Þ
3800
3800
3600
3600
3400
3400
3200
3200 stiffness in x-direction stiffness in y-direction stiffness in z-direction
3000 2800
-50
0
50
3000
100
24.8
2800
24.7 24.6 24.5 24.4 24.3 24.2
Stiffness in z-direction (kN/m)
−80 −60 Þ
Stiffness in y-direction (kN/m)
Stiffness in x-direction (kN/m)
Lu(mm) 204
24.1
Travel of the Knuckle (mm) Fig. 16. Linear stiffness of the knuckle during jounce and rebound.
distance from the horizontal plane passing through the axis of the wheel at the equilibrium position to the fixed joints of upper branches, and Dd represents the initial distance from the horizontal plane passing through the axis of the knuckle at the equilibrium position to the fixed joints of lower branches. The natural frequency of a vehicle is an important parameter to measure the ride of the vehicle. Generally the natural frequency of the car is restricted to 1–1.3 Hz. Assume that the stiffness of the suspension is Kp, and the supported mass of spring is m, there is the following relationship 2
K p ¼ mð2πf Þ
ð57Þ
62
54
61
53
60
52
59
51
58
50
57
stiffness about x-axis stiffness about y-axis stiffness about z-axis
49 48
56 55
47
54
46
53
45
-50
0
50
100
52
Travel of the Knuckle (mm) Fig. 17. Rotational stiffness of the knuckle during jounce and rebound.
24 23 22 21 20 19 18 17
Stiffness about z-axis (kN.m/rad)
55
Stiffness about y-axis (kN.m/rad)
Stiffness about x-axis (kN.m/rad)
where f is the natural frequency of the vehicle and f = 1.2 Hz in general. Assume that the section of the branch is round and the radius is r. A set of parameters of the rectilinear independent suspension with length compensation branches listed in Table 1 is numerically simulated. The stiffness of the suspension during jounce and rebound is shown in Figs. 16 and 17. Fig. 16 indicates that the lowest stiffness in the x-direction is about 2900 kN/m, the lowest stiffness of the suspension in the ydirection is about 2950 kN/m which occurs at the upper limit or lower limit of the travel of the knuckle and the stiffnesses in the x-direction and the y-direction are symmetrical about the suspension initial point and the stiffness in the vertical direction keeps
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4 Nonlinear spring Linear spring
The axial load (kN)
3 2 1 0 -1 -2 -3 -4 -80
-60
-40
-20
0
20
40
60
80
100
120
Travel of the Knuckle (mm) Fig. 18. The characteristic curves of nonlinear spring and linear spring.
in a relatively horizontal line the largest change is within 3%. Fig. 17 indicates that the rotational stiffness of the suspension around x-axis changes within 45–55 kN/rad and firstly increases and then decreases, the rotational stiffness around the y-axis changes within 52–61 kN/rad and firstly decreases and then increases and the rotational stiffness of the suspension around z-axis changes within 17–24 kN/rad and firstly increases and then decreases. In order to keep a good frequency range and to guarantee good ride comfort and handling stability, the spring with variable stiffness is often used in applications. In engineering, the nonlinear relationship between force and deformation of the variable stiffness spring is usually expressed by
F ¼ Kx þ εKx
3
ð58Þ
where F is the axial load on the spring, x is the spring deformation, K represents the linear stiffness of the spring and ε is a parameter which represents the nonlinear coefficient. In the applications, the approximation of K = Kp is often adopted, so the nonlinear stiffness of the spring is
k¼
F 2 ¼ K p 1 þ εx : x
ð59Þ
Assume that ε = 501/m 2. The corresponding relationship of the axial load and the spring deformation is shown in Fig. 18. Substitute Eq. (59) into Eq. (56), with the parameters in Table 1 and Table 2, one can obtain the numerical simulation curves for the stiffness of the suspension shown in Figs. 19 and 20. Comparing Figs. 19 and 16, one can find that the linear stiffness in z-direction firstly decreases and then increases quickly with the application of nonlinear spring which is particularly useful to improve the comfort of a vehicle. Therefore, the vehicle has better ride comfort by using the nonlinear spring when the structure parameters are the same. From Table 2, one can find that the four branches of the knuckle are identical and the structure of the suspension is symmetrical. Therefore, the linear stiffnesses in x-direction and y-direction of the suspension are identical and the rotational stiffnesses about x-axis, y-axis and z-axis are symmetrical which verifies the theoretical results. Comparing Figs. 19 and 20, one can find that the stiffness of the suspension in the x-direction and the y-direction and the rotation stiffness of the suspension about x-axis and y-axis with parameters in Table 1 are larger than that in Table 2, the rotational stiffness of the suspension about z-axis with parameters in Table 1 is smaller than that of the suspension with parameters in Table 2. Therefore, different structure parameters result in different stiffnesses. Table 2 Parameters of the suspension with identical branches. Lu(mm)
θu(°)
Initial C
a(mm)
R(mm)
E(GPa)
Du(mm)
r(mm)
225
45
ð0
50
320
206
95
15
−80 −60 Þ
Ld(mm)
θd(°)
Initial D
225
45
ð0
−180
540 Þ
b(mm)
L(mm)
G(GPa)
Dd(mm)
m(kg)
75
60
79.38
55
441.5
3800
3600
3600
3400
3400 stiffness in x-direction stiffness in y-direction stiffness in z-direction
3200
3200
3000 2800
3000
38 36 34 32 30 28 26
2800
24
55
62
24
54
61
53
60
52
59
-50
0
50
100
Stiffness in z-direction (kN/m)
3800
121
Stiffness in y-direction (kN/m)
Stiffness in x-direction (kN/m)
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
58
50
57
stiffness about x-axis stiffness about y-axis stiffness about z-axis
49 48
56 55
47
54
46
53
45
-50
0
50
100
23 22 21 20 19 18
Stiffness about z-axis (kN.m/rad)
51
Stiffness about y-axis (kN.m/rad)
Stiffness about x-axis (kN.m/rad)
Travel of the Knuckle (mm)
17
52
Travel of the Knuckle (mm)
3300
3200
3200
3100
3100
3000
3000
2900
2900 stiffness in x-direction stiffness in y-direction stiffness in z-direction
2800 2700
2800 2700
2600 2500
2600
38 36 34 32 30 28 26
2500
24
52
60
28
50
58
48
56
-50
0
50
100
Stiffness in z-direction (kN/m)
3300
Stiffness in y-direction (kN/m)
Stiffness in x-direction (kN/m)
Fig. 19. Stiffness of the knuckle with parameters of Table 1.
54
44 42
52
-50
0
50
100
50
Travel of the Knuckle (mm) Fig. 20. Stiffness of the knuckle with parameters of Table 2.
26 24 22 20 18 16
Stiffness about z-axis (kN.m/rad)
stiffness about x-axis stiffness about y-axis stiffness about z-axis
46
Stiffness about y-axis (kN.m/rad)
Stiffness about x-axis (kN.m/rad)
Travel of the Knuckle (mm)
122
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
4. Conclusions This paper investigated the stiffness of a rectilinear independent suspension whose alignment parameters should be invariable in mechanism theory during jounce and rebound. The fact that the resultant stiffness of the suspension is subjected to the stiffness of each 3-RRR distance compensation linkage raises a particular difficulty to build the model. Therefore, this paper started from establishing the equivalent deformations and deflections of each branch of the suspension, and then the sixdimensional stiffness of the suspension is equivalently expressed in terms of the material properties of the four kinematic chains. This method well avoids the inverse problems of position equations which were generally used in the previous literature. Numerical simulations indicated that this methodology could provide an optimization basis for the design of the suspension. Acknowledgments This research was supported by the National Natural Science Foundation of China under grant 51175277, the Natural Science Foundation of Beijing under grant 3112014 and the Program for New Century Excellent Talents in University. The authors gratefully acknowledge these support agencies. References [1] Daniel A. Mántaras, Pablo Luque, Carlos Vera, Development and validation of a three-dimensional kinematic model for the McPherson steering and suspension mechanisms, Mechanism and Machine Theory 39 (6) (June 2004) 603–619. [2] J. Kang, Forced vibration analysis for dynamic coupling of finite-element spring-reduced suspension system, Proceedings of Institute of Mechanical Engineering, Part C: Journal of Mechanical Engineering Science 222 (3) (March 2008) 349–356. [3] Tadeusz Majewski, The properties of a dynamic eliminator for vehicle vibrations, Mechanism and Machine Theory 45 (10) (October 2010) 1449–1461. [4] T.G. Chondros, G. Michalos, P. Michaelides, E. Fainekos, An approximate method for the evaluation of the roll stiffness of road tankers, Proceedings of Institute of Mechanical Engineering, Part D: Journal of Automobile Engineering 221 (11) (November 2007) 1499–1512. [5] T.E.C. Potter, D. Cebon, D.J. Cole, Assessing road friendliness: a review, Proceedings of Institute of Mechanical Engineering, Part D: Journal of Automobile Engineering 211 (6) (June 1997) 455–475. [6] M. Costanzi, D. Cebon, An investigation of the effects of lorry suspension performance on road maintenance costs, Proceedings of Institute of Mechanical Engineering, Part C: Journal of Mechanical Engineering Science 221 (11) (November 2007) 1265–1277. [7] W.A. Smith, N. Zhang, Hydraulically interconnected vehicle suspension: optimization and sensitivity analysis, Proceedings of the Institution of Mechanical Engineers, Part D: Journal of Automobile Engineering 224 (11) (November 2010) 1335–1355. [8] Hosein Habibi, Kourosh H. Shirazi, Mohammad Shishesaz, Roll steer minimization of McPherson-strut suspension system using genetic algorithm method, Mechanism and Machine Theory 43 (1) (January 2008) 57–67. [9] J.B. Hoyle, Modeling the static stiffness and dynamic frequency response characteristics of a leaf spring truck suspension, Proceedings of the Institution of Mechanical Engineers, Part D: Journal of Automobile Engineering 218 (3) (March 2004) 259–278. [10] J. Meissonnier, P. Metz, J.-C. Fauroux, G. Gogu, C. Montezin, Iterative identification of stiffness parameters in a car suspension elastokinematic model, Proceedings of the Institution of Mechanical Engineers, Part D: Journal of Automobile Engineering 220 (11) (November 2006) 1477–1489. [11] D.-O. Kang, S.-J. Heo, M.-S. Kim, Robust design optimization of the McPherson suspension system with consideration of a bush compliance uncertainty, Proceedings of the Institution of Mechanical Engineers, Part D: Journal of Automobile Engineering 224 (6) (June 2010) 705–716. [12] D. Cao, S. Rakheja, C.-Y. Su, Pitch plane analysis of a twin-gas-chamber strut suspension, Proceedings of the Institution of Mechanical Engineers, Part D: Journal of Automobile Engineering 222 (8) (August 2008) 1313–1335. [13] T.D. Gillespie, Fundamentals of Vehicle Dynamics, SAE International, Warrendale, Pennsylvania, March 1992. [14] T.D. Gillespie, Heavy truck ride, SAE Paper 850001, 1985. [15] P.A. Simionescu, D. Beale, Synthesis and analysis of the five-link rear suspension system used in automobiles, Mechanism and Machine Theory 37 (9) (September 2002) 815–832. [16] J. Knapczyk, M. Maniowski, Stiffness synthesis of a five-rod suspension for given load–displacement characteristics, Proceedings of the Institution of Mechanical Engineers, Part D: Journal of Automobile Engineering 220 (7) (July 2006) 705–716. [17] Madhusudan Raghavan, Number and dimensional synthesis of independent suspension mechanisms, Mechanism and Machine Theory 31 (8) (November 1996) 999–1195. [18] Masataka Nasada, Effect on compliance and alignment variation with respect to stiffness of knuckle and shock absorber in a McPherson strut suspension, Vehicle System Dynamics 44 (2006) 171–180 (Supplement). [19] Yves A. Papegay, Jean-Pierre Merlet, David Daney, Exact kinematics analysis of car's suspension mechanisms using symbolic computation and interval analysis, Mechanism and Machine Theory 40 (4) (April 2005) 395–413. [20] S.N. Fergussona, R.D. Fröhlinga, H. Klopper, Minimising wheel wear by optimising the primary suspension stiffness and centre plate friction of self-steering bogies, Vehicle System Dynamics 46 (2008) 457–468 (Supplement). [21] K.-J. Mun, T.-J. Kim, Y.-S. Kim, Analysis of the roll properties of a tubular-type torsion beam suspension, Proceedings of Institute of Mechanical Engineering, Part D: Journal of Automobile Engineering 224 (1) (January 2010) 1–13. [22] N. Mirza, K. Hussain, A.J. Day, J. Klaps, Effect of component stiffness and deformation on vehicle lateral drift during braking, Proceedings of Institute of Mechanical Engineering, Part K: Journal of Multi-Body Dynamics 223 (1) (January 2009) 9–22.
Contents lists available at SciVerse ScienceDirect
Mechanism and Machine Theory journal homepage: www.elsevier.com/locate/mechmt
Stiffness of a rectilinear suspension with automatic length compensation branches Jing-Shan Zhao a,⁎, Xiang Liu a, Zhi-Jing Feng a, Jian S. Dai b, c a b c
State Key Laboratory of Tribology, Department of Precision Instruments and Mechanology, Tsinghua University, Beijing 100084, PR China Centre for Advanced Mechanisms and Robotics, Tianjin University, Tianjin 300072, PR China King's College London, University of London, London WC2R 2LS, United Kingdom
a r t i c l e
i n f o
Article history: Received 26 July 2011 Received in revised form 7 May 2012 Accepted 15 May 2012 Available online 21 June 2012 Keywords: Stiffness Suspension Length compensation branch Design
a b s t r a c t This paper focuses on the stiffness of a rectilinear independent suspension whose alignment parameters are invariable in mechanism theory during jounce and rebound. The suspension is a kind of multilink mechanism, the kinematic chains of which consist of 3-RRR distance compensation branches. However the capacity to keep a straight line motion depends on the resultant stiffness of the suspension. Therefore, this paper proposes a methodology to investigate the equivalent stiffness of the suspension. The fact that the resultant stiffness of the suspension is subjected to the stiffness of each 3-RRR distance compensation linkage raises a particular difficulty to build the model. So the deformation and the deflection of each branch are established first, and then the six-dimensional stiffness of the suspension is equivalently expressed in terms of the material properties of the four chains. © 2012 Elsevier Ltd. All rights reserved.
1. Introduction The main parameters such as caster, camber, and steer angles and so on determinate the handling of a vehicle [1]. The ride comfort is deteriorated by the vibration of suspension system excited by a road surface [2]. The roughness of the road and friction between road and wheels are the main reasons for the wheels' vibration [3]. The most important factors of vibration are the center height of gravity, the track width, the axle roll stiffness, suspension and tire compliance, and suspension roll stiffness [4]. The major vibration mode is the wheel hop of a suspension system. Hence the use of road-friendly vehicles is encouraged by road authorities because they can have higher static loads, but lower dynamic tire forces, consequently reducing road damage [5,6]. Conventional vehicle suspension design generally involves a trade-off between handling stability and ride comfort [7]. The rolling of the body, especially in independent suspensions, causes a change of kinematic parameters of the related mechanisms [8]. The dynamic characteristics of a vehicle are generally determined by the stiffness and damping characteristics of its suspension, and the sprung and unsprung masses [9]. A vehicle with a relatively stiff suspension is likely to possess good handling stability but poor ride comfort, while the reverse is true for a soft, compliant suspension [7]. To build a correct elastokinematic model of a given suspension, bushing stiffness must be determined [10]. But bush uncertainty and mechanical flexibility influence the movement error of suspension systems [11]. Therefore, it is difficult to precisely forecast the performance of a suspension system by quantified performance indices. The unavoidable vibration of a vehicle arises from combined bounce and pitch motions [12], while the longitudinal vibration at the driver location is predominated only by pitch motions of the vehicle [13]. It has also been stated that the ride comfort of heavy vehicles is generally dominated by the pitch plane motions, while the lateral and roll vibrations have been considered to be less important [12,14]. Therefore the synthesis goal is to fulfill a minimum variation of the wheel track, toe angle and camber angle during jounce and rebound of the wheels [15]. The requirements upon the motion of the rear wheels when synthesizing the ⁎ Corresponding author. Fax: + 86 10 62788308. E-mail address: [email protected] (J-S. Zhao). 0094-114X/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.mechmachtheory.2012.05.005
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Vehicle Body A1
A2
B1
B2
A4
A3
B3 B4
Knuckle
Fig. 1. Structure of a rectilinear suspension.
suspension mechanism are that the toe angle variation during compression and rebound should be minimum, excessive outward camber thrust on corners must be forbidden, excessive sideway thrust and consequent rear end steering impulses on single wheel bump or rebounds should also be excluded [16,17]. Consequently, the relationship between the properties of the lateral, vertical and longitudinal primary suspension stiffness should be investigated. Suspension compliance, compliant steer and wheel alignment variation are very important factors to improve a vehicle's stability, controllability and ride comfort [18]. Small differences on camber or toe angle or on orientation of steering axis may have a dramatically influence on the handling performance of a car either directly or through their rates of change with jounce-rebound [19]. Fergussona et al. [20] investigated the maintenance of wheels and the optimum relationship between the properties of the lateral and longitudinal primary suspension stiffness and the center plate friction of a self-steering three-piece bogie. Mun et al. [21] proposed an analytical method to calculate the torsional stiffness of a tubular beam with a closed cross-section in a torsional beam rear suspension system using linear beam theory. The accelerating and braking forces, the inertia forces resulting from turning are transmitted to the vehicle body via the suspension and its linkages; therefore, the design of suspension plays an important role in ensuring starting, turning, accelerating and braking stability [22]. Stiffness of the suspension is a primary index to represent the ride and handling stability of a vehicle. This paper proposes a rectilinear suspension which, in mechanism theory, could keep the alignment parameters of the wheel invariable. However, the capacity of the suspension to trace a straight line relies on the stiffness of the mechanism. Therefore, this paper will focus on this problem and investigate the equivalent stiffness of the rectilinear suspension.
2. Architecture of a rectilinear suspension with stretchable kinematic chains The architecture of a rectilinear suspension is shown in Fig. 1. The knuckle is connected with the vehicle body through four branches, AiBi (i = 1, 2, 3, 4), whose length along its axial direction can be changed automatically within a specified distance range. Each branch, as a matter of fact, is a 3-RRR parallel mechanism shown in Fig. 2.
A1 R11 R12 R13
π1
R21
R31
R32
R33 R22
R23
π2 B1
Fig. 2. A 3-RRR branch of the suspension.
π3
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
101
Fig. 3. Possible motions of a planar RRR kinematic chain.
To investigate the mobility of the branch, the free motions of the RRR kinematic chain in plane π1 should be discussed from the viewpoint of geometry. This sub chain is simplified as Fig. 3 shows.
Fig. 4. The absolute coordinate frame on the suspension mechanism.
The free end point, B1, can translate left and right, up and down and rotate about any normal of the plane in a certain range determined by the lengths of the links. Therefore, point B1 could reach any point with any pose within such a region in π1-plane shown in Fig. 2. Similarly, the other two chains, R21R22R23 and R31R32R33, also have the capacity to reach any point with any poses within their corresponding regions, determined by the lengths of the links, in π2- and π3-planes, respectively. As a result, their common distal platform, R13R23R33, can only move along the intersection of π1-, π2- and π3-planes, which is line A1B1 in Fig. 2 or Fig. 3. This indicates that platforms R11R21R31 and R13R23R33 only have one relative degree of freedom of translation. So the branch shown in Fig. 2 can stretch and contract along its axial direction. The section of 3-RRR can be designed stiff enough to safeguard the strength requirement of the branch. Therefore, what follows will investigate the equivalent stiffness of the suspension by processing the branch as one stretchable link. 3. Equivalent stiffness of the suspension To investigate the stiffness of the suspension, the statics should be first analyzed. Generally, there are driving and braking forces along the x-direction, the lateral force along the y-direction, the vertical force from the ground and the moments and torques about the x-, y- and z-axes, which are shown in Fig. 4. Assume that onxnynzn is a local coordinate system which is attached to the knuckle, and the origin, on, is superimposed with the geometry center of the knuckle, xn directs backward and zn directs upward, so yn can be determined in accordance to the righthand rule. In addition, yn and zn are the symmetries of the knuckle. Initially, the local coordinate frame, onxnynzn, is completely superimposed with the absolute coordinate system oxyz. Therefore, the coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz as: 2
r B1
3 2 3 2 3 2 3 −b b −b b 4 5 4 5 4 5 4 ¼ 0 ; r B2 ¼ 0 ; r B3 ¼ 0 ; r B4 ¼ 0 5: a a −a −a
ð1Þ
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J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
Fig. 5. Displacement of the knuckle in the x-direction.
The unit normal direction vectors of the chain planes which also represent the axial direction of the revolute joints of the chain, Bi (i = 1, 2, 3, 4), shown in Fig. 4 are: 8 > n1 > > < n2 > > n3 > : n4
¼ ½ − cosθu sinθu 0 T T ¼ ½ cosθu sinθu 0 : ¼ ½ − cosθd sinθd 0 T T ¼ ½ cosθd sinθd 0
ð2Þ
The unit normal direction vectors of the branches AiBi (i = 1, 2, 3, 4) shown in Fig. 4 are: 8 > k > > 1 < k2 > k3 > > : k4
¼ ½ cosφu sinθu cosφu cosθu − sinφu T ¼ ½ − cosφu sinθu cosφu cosθu − sinφu T ¼ ½ cosφd sinθd cosφd cosθd sinφd T ¼ ½ − cosφd sinθd cosφd cosθd sinφd T
ð3Þ
where θu denotes the subtended angle between the vertical plane passing through the upper chain axis and yoz-plane, and θd denotes the subtended angle between the vertical plane passing through the lower chain axis and yoz-plane, and φu denotes the subtended angle between the upper chain axis and xoy-plane, and φd denotes the subtended angle between the lower chain axis and xoy-plane, which are shown in Fig. 4. 3.1. Equivalent stiffness of the suspension in x-direction Suppose that there is a force, Fx, exerted to the center of the knuckle in the x-direction. The corresponding displacement of the knuckle, indicated by δx, is shown in Fig. 5. So the coordinate transformation from onxnynzn to oxyz is r B′ j ¼ Tr Bnj þ δron
ð4Þ
where rB′j denotes the vector coordinates in the oxyz frame, and T is the transformation matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, and rBnj represents the absolute vector coordinates in the onxnynzn frame, and δron indicates the vector of the origin of the local coordinate frame in the absolute coordinate frame, and 2
3 2 3 1 0 0 δx 4 5 T ¼ 0 1 0 ; δron ¼ 4 0 5: 0 0 1 0 Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (4) presents their absolute coordinates 2 r B′ 1
¼4
3 2 3 2 3 2 3 δx −b δx þ b δx −b δx þ b 5 4 5 4 5 4 ; r B′ 2 ¼ ; r B′ 3 ¼ ; r B′ 4 ¼ 0 0 0 0 5: a a −a −a
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
Therefore, one obtains 8 → ′ T > > > B1 B 1¼ ½ δx 0 0 > > → > < ′ T B2 B 2¼ ½ δx 0 0 → > > > B3 B′ 3¼ ½ δx 0 0 T > > > : →′ B4 B 4¼ ½ δx 0 0 T
103
ð5Þ
→
where Bi B′¼ i r B′ i −r Bi ; i ¼ 1; 2; 3; 4: The displacement of joint Bi (i = 1, 2, 3, 4) in its axial direction, ni, can be expressed as below: 8 → ′ > > > δ1 ¼B1 B 1⋅n1 ¼ −δx cosθu > > → > < δ2 ¼B2 B′ 2⋅n2 ¼ δx cosθu → > > > δ3 ¼B3 B′ 3⋅n3 ¼ −δx cosθd > > > → : δ4 ¼B4 B′ 4⋅n4 ¼ δx cosθd
ð6Þ
where the negative sign “−” indicates that the displacement direction of the joint is reverse to that of ni (i = 1, 2, 3, 4). Fig. 2 indicates that every branch can stretch or compress along its axis and rotate about the joint axes at both sides. As a result, each branch can only bear the force in the normal direction of its chain shown in Fig. 4, the torque about its axis and the bending moment around the direction perpendicular to both the chain axis and the normal of the chain plane. Because all the kinematic pairs are thrust bearing jointed, the torque and the bending moments are all small items compared with the normal force. For the sake of simplicity, the torque and the bending moments are not counted into the calculation of translational stiffness of the suspension. Suppose that the normal force exerted to joint Bi (i = 1, 2, 3, 4) by the knuckle is Fi (i = 1, 2, 3, 4) when the suspension only subjects to a force Fx the direction of which is along ni. Because the mobility of the parallel 3-RRR section shown in Fig. 2 is identical to a prismatic joint, each branch can be looked as a beam whose length is self adaptive. For the sake of simplicity, the flexural rigidity and the torsional rigidity of the branch are EI and GIp, respectively. Therefore, under the action of Fi (i = 1, 2, 3, 4) the deflection of the joint Bi can be expressed with 8 F l3 > ′ > > δ1¼ 1u > > 3EI > > > > ′ F l3 > <δ 2 ¼ 2 u 3EI ð7Þ 3 > > > δ′ ¼ F 3 ld > 3 > > 3EI > > 3 > > : δ′ ¼ F 4 ld 4 3EI where lu and ld stand for the lengths of the upper and lower chains at the instant, respectively. The compliant conditions require that 8 > δ > > < 1 δ2 > δ > > : 3 δ4
′
¼δ1 ′ ¼δ2 ′ : ¼δ3 ′ ¼δ4
ð8Þ
Substituting Eqs. (6) and (7) into Eq. (8) yields 8 3EIδx cosθu > > F1 ¼ − > > > l3u > > > 3EIδ cosθ > x u > > < F2 ¼ l3u 3EIδx cosθd : > > F3 ¼ − > > > l3d > > > > 3EIδ cosθ > x d > : F4 ¼ l3d
ð9Þ
The static equilibrium equations of the suspension are
−F 1 cosθu þ F 2 cosθu −F 3 cosθd þ F 4 cosθd −F x ¼ 0 : F 1 sinθu þ F 2 sinθu þ F 3 sinθd þ F 4 sinθd ¼ 0
ð10Þ
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Fig. 6. Deflection of the knuckle in the y-direction.
Substituting Eq. (9) into Eq. (10), one finds that the second equation is automatically satisfied because of the structural symmetry of the knuckle, and the first one can be rearranged to
Kx ¼
Fx cos2 θu cos2 θd ¼ 6EI þ δx l3u l3d
! ð11Þ
where Kx indicates the stiffness of the suspension in x-direction. 3.2. Equivalent stiffness of the suspension in y-direction Suppose that there is a force, Fy, exerted to the center of the knuckle in the y-direction. Considering that the structure and load Fy are both symmetrical about the plane of oyz, the forces and deformations of the branches are symmetrical about the plane of oyz. One can know that the knuckle might have a displacement in y-direction, δy, and a rotation about x-axis, γx, as Fig. 6 shows according to the force symmetry. So the coordinate transformation from onxnynzn to oxyz is r B′ i ¼ Tr Bni þ δron
ð12Þ
where rB′i represents the absolute vector coordinates in the oxyz frame, and T is the transformation matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, and rBni denotes the vector coordinates in the onxnynzn frame, and δron indicates the vector of the origin of the local coordinate frame in the absolute coordinate frame, and 2
1 T ¼ 40 0
0 cosγx sinγ x
2 3 3 0 0 − sinγ x 5; δr on ¼ 4 δy 5: cosγx 0
Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (12) presents their absolute coordinates 2
r B′ 1
3 2 3 2 3 2 3 −b b −b b ¼ 4 δy −a sinγ x 5; r B′ 2 ¼ 4 δy −a sinγx 5; r B′ 3 ¼ 4 δy þ a sinγ x 5; r B′ 4 ¼ 4 δy þ a sinγ x 5: a cosγx a cosγ x −a cosγx −a cosγx
The unit normal direction vectors of joints Bi (i = 1, 2, 3, 4) are changed to 8 ′ > n1 > > < ′ n2 > n′ > > : ′3 n4
¼ ½ − cosθu cosγx sinθu sinγ x sinθu T T ¼ ½ cosθu cosγ x sinθu sinγx sinθu : ¼ ½ − cosθd cosγx sinθd sinγx sinθd T T ¼ ½ cosθd cosγ x sinθd sinγx sinθd
ð13Þ
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
105
Immediately, one obtains from Eqs. (1) and (13) 8 → > > B1 B′ 1¼ 0 > > > → > > < ′ B2 B 2¼ 0 → > > B B′ ¼ 0 > > > 3→ 3 > > : B4 B′ 4¼ 0
T
δy −a sinγx
að cosγx −1Þ
δy −a sinγx
að cosγx −1Þ
δy þ a sinγ x
−að cosγ x −1Þ
δy þ a sinγ x
−að cosγ x −1Þ
T T
:
T
The deflection of joint Bi (i = 1, 2, 3, 4) on its axial direction, ni (i = 1, 2, 3, 4), can be expressed as: 8 → > ′ > > > δ1 ¼B1 B 1⋅n1 ¼ δy −a sinγ x sinθu > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ δy −a sinγ x sinθu : → > ′ > > > δ3 ¼B3 B 3⋅n3 ¼ δy þ a sinγ x sinθd > > > → > > ′ : δ4 ¼B4 B 4⋅n4 ¼ δy þ a sinγ x sinθd
ð14Þ
Suppose that the normal force exerted to the joint Bi by the knuckle is Fi (i = 1, 2, 3, 4) the direction of which is along n′i when the suspension only subjects to a force Fy. Therefore, under the action of Fi the displacement of joint Bi can be expressed with 8 F l3 > ′ > > δ1¼ 1u > > 3EI > > > > F 2 l3u ′ > <δ 2 ¼ 3EI : > F 3 l3d > ′ > δ3¼ > > > 3EI > > 3 > > : δ′ ¼ F 4 ld 4 3EI
ð15Þ
The compliant conditions require that the expressions in Eq. (8) also hold. Substituting Eqs. (14) and (15) into Eq. (8) yields 8 > > 3EI δy −a sinγ x sinθu > > > F ¼ > > > 1 l3u > > > > > 3EI δy −a sinγ x sinθu > > > > < F2 ¼ l3u : > 3EI δy þ a sinγ x sinθd > > > > > > F3 ¼ > l3d > > > > > 3EI δ þ a sinγ x sinθd > y > > > : F4 ¼ l3
ð16Þ
d
The static equilibrium equations of the suspension are 8 < −F 1 cosθu þ F 2 cosθu −F 3 cosθd þ F 4 cosθd ¼ 0 F cosγ x sinθu þ F 2 cosγx sinθu þ F 3 cosγ x sinθd þ F 4 cosγx sinθd −F y ¼ 0 : : 1 F 1 sinγ x sinθu þ F 2 sinγx sinθu þ F 3 sinγ x sinθd þ F 4 sinγx sinθd ¼ 0
ð17Þ
Substituting Eq. (16) into Eq. (17), one then gets Fy = 0 if γx ≠ 0, which conflict with the reality. Therefore, there must be γx = 0. And one can find that the first equation is automatically satisfied and the second equation can be rearranged to ! 2 2 Fy sin θu sin θd ¼ 6EI þ Ky ¼ ð18Þ δy l3u l3d where Ky indicates the stiffness of the suspension in y-direction. 3.3. The equivalent torsional stiffness of the suspension about x-axis Suppose that there is a moment about the x-axis, Mx, exerted to the knuckle the corresponding rotational angle, shown in Fig. 7, is γx. Because the suspension is not rotational symmetrical about the x-axis, the knuckle might have a small displacement
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Fig. 7. Slope of the suspension under the moment of Mx.
within the yoz-plane accompanying a small rotational angle, γx. Assume that the center displacement of the knuckle under the T action of Mx is δron ¼ ð 0 δy δz Þ : The transformation matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, oxyz, can be expressed in term of the rotational angle of the knuckle, γx: 2
1 T ¼ 40 0
0 cosγx sinγ x
3 0 − sinγ x 5: cosγx
Therefore, the absolute coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz coordinate frame r B′ i ¼ Tr Bni þ δron :
ð19Þ
Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (19) presents their absolute coordinates 2
r B′ 1
3 2 3 2 3 2 3 −b b −b b ¼ 4 δy −a sinγ x 5; r B′ 2 ¼ 4 δy −a sinγx 5; r B′ 3 ¼ 4 δy þ a sinγx 5; r B′ 4 ¼ 4 δy þ a sinγx 5: δz þ a cosγ x δz þ a cosγx δz −a cosγx δz −a cosγ x
ð20Þ
With Eqs. (1) and (20), one immediately obtains 8 → > ′ > B1 B 1¼ 0 > > > → > > < B B′ ¼ 0
δy −a sinγx
δz þ að cosγ x −1Þ
δy −a sinγx
δz þ að cosγ x −1Þ
0
δy þ a sinγ x
δz þ að1− cosγx Þ
0
δy þ a sinγ x
δz þ að1− cosγx Þ
2 2 → > ′ > > B > 3 B 3¼ > > → > : B4 B′ 4¼
T T T
:
T
The deflection of joint Bi (i = 1, 2, 3, 4) on its axial direction, ni, can be expressed as: 8 → > ′ > δ1 ¼B1 B 1⋅n1 ¼ δy −a sinγ x sinθu > > > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ δy −a sinγ x sinθu : → > ′ > > sinθ δ ¼B B ⋅n ¼ δ þ a sinγ > 3 3 3 3 y d x > > > → > > : δ ¼B B′ ⋅n ¼ δ þ a sinγ sinθ 4 4 4 4 y d x
ð21Þ
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107
Fig. 8. Displacement of the end point on a branch.
Considering the small value of γx, the following approximations can be adopted:
sinγx ≈γx cosγx ≈1 nþ1
ð2n−1Þ
Þ where sinγx ¼ γ x − 3!1 γ3x þ 5!1 γ5x þ n⋯ þ ðð−1 þ ⋯ in which only the first item is retained for a first order approximation, and 2n−1Þ! γ x ð−1Þ 1 2 1 4 2n cosγ x ¼ 1− 2! γ x þ 4! γ x þ ⋯ þ 2n! γx þ ⋯ in which only the first item is retained for the approximation. Therefore Eq. (21) can be simplified as
8 > δ1 ¼ δy −aγx sinθu > > > > > < δ2 ¼ δy −aγx sinθu : > > > δ3 ¼ δy þ aγx sinθd > > > : δ ¼ δ þ aγ sinθ 4
y
d
x
Under the action of the moment, Mx, both the position and orientation of the joint axis of Bi (i = 1, 2, 3, 4) have been changed. The displacement of joint Bi consists of the bending and torsional components which are shown in Fig. 8. To illustrate the relationship of these angles, one can investigate the first branch. Suppose that the torsional angle of the branch, denoted by γ1, is shown in Figs. 8 and 9. The unit vector of B′1 in the oxyz-frame is 2
′
n1
1 ¼ 40 0
0 cosγ x sinγx
3 2 32 3 − cosθu 0 − cosθu − sinγx 54 sinθu 5 ¼ 4 sinθu cosγx 5: cosγ x sinθu sinγ x 0
So ′
2
2
n1 ⋅n 1 ¼ cos θu þ sin θu cosγ x that is 2
2
2
cosψ1 ¼ cos θu þ sin θu cosγx ¼ 1 þ sin θu ð cosγx −1Þ where ψ1 is the subtended angle between n1 and n′1. This can be rearranged to 2
1−cosψ1 ¼ sin θu ð1− cosγx Þ:
Fig. 9. Angular relationship in the deflection of a branch.
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Ultimately, one obtains 2
2 sin
ψ1 2 2γ ¼ 2 sin θu sin x : 2 2
Considering the small values of γx and ψ1 shown in Fig. 9, one finds that ψ1 ¼ γx sinθu :
ð22Þ
Because π ′ cos −ϕ1 ¼ n 1 ⋅k1 ¼ cosφu sinθu cosθu ð cosγx −1Þ− sinφu sinθu sinγ x ; 2 and taking into account of the small values of ϕ1 and γx, one gains ϕ1 ¼ −γx sinφu sinθu
ð23Þ
where ϕ1 is the subtended angle from n′1 to the branch plane. Substituting Eq. (22) into Eq. (23) yields ϕ1 ¼ −ψ1 sinφu : For the fact that ψ1 is a very small angle, ϕ1 is a tiny value, too. Therefore, cosϕ1 ≈1: From Fig. 9, one finds that cosψ1 ¼ cosϕ1 cosγ 1 ≈ cosγ 1 : Ultimately, one obtains γ1 ≈ψ1 ¼ γ x sinθu
ð24Þ
because ′ n1 n 1 ⋅k1 ¼ γ x cosφu sinθu > 0: The direction of the torsional angle, γ1, is same with the positive x-axis. Therefore, one can obtain 8 γ ¼ γ x sinθu > > < 1 γ 2 ¼ −γ x sinθu : γ ¼ γ x sinθd > > : 3 γ 4 ¼ −γ x sinθd
ð25Þ
Suppose that the torque about the axis of the branch is Ti (i = 1, 2, 3, 4) the direction of which is along ki (i = 1, 2, 3, 4) and the bending moment is Mi (i = 1, 2, 3, 4) the direction of which is defined by ki × ni shown in Fig. 10.
Fig. 10. Statics of the suspension.
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109
In vector form, the torque and moment loads can be rewritten as 8 > T > > 1 < T2 > T > > : 3 T4
8 > M1 ¼ ½ T 1 cosφu sinθu T 1 cosφu cosθu −T 1 sinφu T > > < ¼ ½ −T 2 cosφu sinθu T 2 cosφu cosθu −T 2 sinφu T M2 T > ¼ ½ T 3 cosφd sinθd T 3 cosφd cosθd T 3 sinφd M > > : 3 ¼ ½ −T 4 cosφd sinθd T 4 cosφd cosθd T 4 sinφd T M4
¼ ½ M1 sinφu sinθu M 1 sinφu cosθu M 1 cosφu T ¼ ½ M2 sinφu sinθu −M 2 sinφu cosθu −M2 cosφu T : ¼ ½ −M 3 sinφd sinθd −M3 sinφd cosθd M3 cosφd T T ¼ ½ −M 4 sinφd sinθd M 4 sinφd cosθd −M4 cosφd
Under these actions, the corresponding displacements and torsional angles of joint Bi (i = 1, 2, 3, 4) are 8 8 > > > > M1 l2u > > γ′1 ′ > > > > δ1 ¼ > > > > 2EI > > > > > > ′ > > M l2 > > < δ′2 ¼ 2 u < γ2 2EI and > > ′ M l2 > > > > γ3 δ′3 ¼ 3 d > > > > > > 2EI > > > > 2 > > > > ′ > δ′ ¼ M4 ld > > > : 4 : γ4 2EI
T 1 lu GIp T 2 lu ¼ GIp : T 3 ld ¼ GIp T l ¼ 4d GIp ¼
Considering the compliant requirements, one obtains 8 ′ > δ1 > > < ′ δ2 ′ > δ > > : ′3 δ4
¼ δ′ 1 ; γ′ 1 ¼ δ′ 2 ; γ′ 2 ′ ′ ¼ δ 3; γ 3 ′ ′ ¼ δ 4; γ 4
¼ γ1 ¼ γ2 : ¼ γ3 ¼ γ4
As a result, there are 8 > > 2EI δy −aγ x sinθu > 8 > > M ¼ > GI p γx sinθu > > > 1 > l2u T1 ¼ > > > > > > lu > > > > 2EI δy −aγ x sinθu > > GI p γx sinθu > > > > > > M ¼ T ¼ − < 2 < 2 l2u lu and : GI p γx sinθd > > 2EI δy þ aγ x sinθd > > > > T ¼ > > 3 > > ld > > > M3 ¼ > > > l2d > > GI p γx sinθd > > > > > : T4 ¼ − > 2EI δy þ aγ x sinθd > ld > > > : M4 ¼ 2 ld
ð26Þ
The static equilibrium conditions require that 8 < cosφu sinθu ðT 1 −T 2 Þ þ cosφd sinθd ðT 3 −T 4 Þ þ sinφu sinθu ðM 1 þ M 2 Þ− sinφd sinθd ðM 3 þ M 4 Þ−M x ¼ 0 cosφu cosθu ðT 1 þ T 2 Þ þ cosφd cosθd ðT 3 þ T 4 Þ þ sinφu cosθu ðM1 −M 2 Þ− sinφd cosθd ðM 3 −M 4 Þ ¼ 0 : : − sinφu ðT 1 þ T 2 Þ þ sinφd ðT 3 þ T 4 Þ þ cosφu ðM1 −M2 Þ þ cosφd ðM3 −M4 Þ ¼ 0
ð27Þ
Substituting Eq. (26) into Eq. (27), one can find that the last two equations hold automatically because of the structural symmetry of the suspension. Rearranging the first equation yields M x ¼ 2GIp γ x
! sin2 θu cosφu sin2 θd cosφd þ −4EIaγx lu ld
sin2 θu sinφu sin2 θd sinφd þ l2u l2d
! þ 4EIδy
sin2 θu sinφu sin2 θd sinφd − 2 lu l2d
!
where δy is an accompanying item under the action of Mx which can be neglected here because it is not explicitly expressed as any function of γx. Therefore, the equivalent torsional stiffness of the branch is K Mx
M ¼ x ¼ 2GIp γx
! ! sin2 θu cosφu sin2 θd cosφd sin2 θu sinφu sin2 θd sinφd þ þ −4EIa lu ld l2u l2d
ð28Þ
where KMx denotes the torsional stiffness of the suspension about the x-axis. 3.4. The equivalent torsional stiffness of the suspension about y-axis Suppose that there is a moment My about the y-axis exerting to the knuckle, the corresponding rotational angle of which is γy and is shown in Fig. 11. Because the suspension is not rotational symmetrical about the y-axis, the knuckle might have a small
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Fig. 11. Slope of the suspension under the moment of My.
displacement within the xoz-plane accompanying the small rotation γy. Assume that the accompanied displacement of the knuckle center under the action of My is δr on ¼ ð δx 0 δz ÞT : Therefore, the absolute coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz coordinate frame r B′ i ¼ Tr Bni þ δron
ð29Þ
where rB′i denotes the absolute coordinates of joint Bi, and 2 T¼4
cosγy 0 − sinγ y
0 1 0
3 sinγ y 0 5 cosγ y
where T denotes the rotational matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame. Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (29) produces their absolute coordinates 2
3 2 3 δx þ a sinγy −b cosγy δx þ a sinγ y þ b cosγy 5; r ′ ¼ 4 5; 0 0 r B′ 1 ¼ 4 B2 δz þ a cosγy þ b sinγy δz þ a cosγ y −b sinγy 2 3 2 3 δx −a sinγy −b cosγy δx −a sinγ y þ b cosγy 5; r ′ ¼ 4 5: 0 0 r B′ 3 ¼ 4 B4 δz −a cosγy þ b sinγy δz −a cosγ y −b sinγ y Therefore, one obtains 8 → > > B1 B′ 1¼ δx þ a sinγ y −b cosγ y þ b > > > → > > < B B′ ¼ δ þ a sinγ þ b cosγ −b 2
2
x
y
y
→ > > > B3 B′ 3¼ δx þ b−a sinγ y −b cosγ y > > > → > : ′ B4 B 4¼ δx −b−a sinγ y þ b cosγ y
δz þ a cosγy þ b sinγy −a T 0 δz þ a cosγy −b sinγy −a T : 0 δz þ a−a cosγ y þ b sinγy T 0 δz þ a−a cosγ y −b sinγ y T 0
The displacement of joint Bi (i = 1, 2, 3, 4) in its axial direction, denoted by ni, will be 8 → > ′ > > > δ1 ¼B1 B 1⋅n1 ¼ − cosθu δx þ a sinγ y −b cosγ y þ b > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ cosθu δx þ a sinγ y þ b cosγ y −b : → > ′ > > > δ3 ¼B3 B 3⋅n3 ¼ − cosθd δx −a sinγ y −b cosγ y þ b > > > → > > : δ4 ¼B4 B′ 4⋅n4 ¼ cosθd δx −a sinγy þ b cosγ y −b
ð30Þ
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Fig. 12. Rotation of a branch about the y-axis.
In a similar way to the above analysis, sin γy ≈ γy and cos γy ≈ 1 when the rotational angle is very small. Therefore, Eq. (30) can be further simplified to 8 > δ1 ¼ − δx þ aγ y cosθu > > > > > < δ2 ¼ δx þ aγy cosθu : > > δ3 ¼ − δx −aγy cosθd > > > > : δ ¼ δ −aγ cosθ 4
x
d
y
Under the action of My, the axis of joint Bi (i = 1, 2, 3, 4) changes its position and orientation as well, and the branch will have bending and torsional deflections which are shown in Fig. 12. To illustrate the relationship of these angles, one can investigate the first branch. Suppose that the torsional angle of the branch, denoted by γy, as shown in Figs. 12 and 9, one can obtain 2 ′
n1
¼4
cosγ y 0 − sinγy
0 1 0
3 2 32 3 − cosγ y cosθu sinγ y − cosθu 4 5 5 4 5: sinθu 0 ¼ sinθu sinγy cosθu cosγy 0
So ′
2
2
n1 ⋅n 1 ¼ sin θu þ cos θu cosγ y that is 2 2 2 cosψ1 ¼ sin θu þ cos θu cosγy ¼ 1 þ cos θu cosγy −1 where ψ1 is the subtended angle between n1 and n′1. It can be rearranged to 2 1−cosψ1 ¼ cos θu 1− cosγy and 2
2 sin
ψ1 2 2 γy ¼ 2 cos θu sin : 2 2
ð31Þ
Because γy and ψ1 are all small angles, one obtains from Eq. (31) ψ1 ¼ γy cosθu : From Fig. 9, one finds that π ′ cos −ϕ1 ¼ n 1 ⋅k1 ¼ cosφu sinθu cosθu 1− cosγ y − sinφu cosθu sinγy : 2
ð32Þ
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Because ϕ1 and γy are both small angles, Eq. (32) results in ϕ1 ¼ −γy sinφu cosθu
ð33Þ
where ϕ1 is the subtended angle from n′1 to the branch plane and γy is the twisting angle about the branch axis. Substituting Eq. (32) into Eq. (33) yields ϕ1 ¼ −ψ1 sinφu : For the tiny value of ψ1, it is safe to suppose that cosϕ1 ≈1: From Fig. 9, one obtains cosψ1 ¼ cosϕ1 cosγ 1 : Therefore, the following approximation holds cosψ1 ≈ cosγ 1 and then γ1 ≈ψ1 ¼ γ y cosθu
ð34Þ
because ′ s ¼ n1 n 1 ⋅k1 ¼ γ x cosφu cosθu > 0: The direction of the torsional angle, γ1, is the same with the positive x-axis. Similarly, one can obtain the torsional angle of the four branches 8 γ ¼ γ cosθ > > < γ 1 ¼ γ y cosθu u 2 y : γ ¼ γ cosθ > > : γ 3 ¼ γ y cosθd d 4 y
ð35Þ
Suppose that the torque about the axis of branch AiBi (i = 1, 2, 3, 4) is Ti the direction of which is along ki and the bending moment is Mi the direction of which is defined by ki × ni, which are shown in Fig. 10. In vector form, the torque and moment loads can be rewritten as 8 > T1 > > < T2 > T > > : 3 T4
8 ¼ ½ T 1 cosφu sinθu T 1 cosφu cosθu −T 1 sinφu T > M1 > > < ¼ ½ −T 2 cosφu sinθu T 2 cosφu cosθu −T 2 sinφu T M2 T > ¼ ½ T 3 cosφd sinθd T 3 cosφd cosθd T 3 sinφd M > > : 3 T ¼ ½ −T 4 cosφd sinθd T 4 cosφd cosθd T 4 sinφd M4
¼ ½ M1 sinφu sinθu M 1 sinφu cosθu M 1 cosφu T ¼ ½ M2 sinφu sinθu −M 2 sinφu cosθu −M2 cosφu T T : ¼ ½ −M 3 sinφd sinθd −M3 sinφd cosθd M3 cosφd T ¼ ½ −M 4 sinφd sinθd M 4 sinφd cosθd −M4 cosφd
Under these actions, the corresponding displacement in the axial direction and torsional angle of joint Bi (i = 1, 2, 3, 4) are 8 8 2 > > ′ > > M l > > γ1 ′ 1 u > > > > δ1 ¼ > > > > 2EI > > > > 2 > > ′ > > M l > > < δ′2 ¼ 2 u < γ2 2EI and 2 > > M l > ′ > > > γ ′3 δ3 ¼ 3 d > > > > > > 2EI > > > > 2 > > > > M 4 ld ′ ′ > > > > : δ4 ¼ : γ4 2EI
T 1 lu GIp T 2 lu ¼ GIp : T 3 ld ¼ GIp T 4 ld ¼ GIp ¼
Considering the compliant requirements, one obtains 8 ′ > > > δ′ 1 < δ2 > δ′ > > : ′3 δ4
′
′
¼ δ 1; γ 1 ′ ′ ¼ δ 2; γ 2 ′ ¼ δ 3 ; γ′ 3 ¼ δ′ 4 ; γ′ 4
¼ γ1 ¼ γ2 : ¼ γ3 ¼ γ4
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113
As a result, one gains that 8 > > 2EI δx þ aγy cosθu > 8 > > M1 ¼ − > > > > > l2u T1 > > > > > > > > > > cosθ 2EI δ þ aγ > > x u y > > > > > > < M2 ¼ < T2 2 lu and > > 2EI δx −aγy cosθd > > > > T3 > > > > M3 ¼ − > > 2 > > > > l > > d > > > > > : T4 > 2EI δx −aγy cosθd > > > > M4 ¼ : l2d
GI p γ y cosθu lu GI p γ y cosθu ¼ lu : GI p γ y cosθd ¼ ld GI p γ y cosθd ¼ ld ¼
ð36Þ
The static equilibrium conditions require that 8 <
cosφu sinθu ðT 1 −T 2 Þ þ cosφd sinθd ðT 3 −T 4 Þ þ sinφu sinθu ðM1 þ M2 Þ− sinφd sinθd ðM3 þ M4 Þ ¼ 0 cosφu cosθu ðT 1 þ T 2 Þ þ cosφd cosθd ðT 3 þ T 4 Þ þ sinφu cosθu ðM1 −M2 Þ− sinφd cosθd ðM 3 −M 4 Þ−M y ¼ 0 : : − sinφu ðT 1 þ T 2 Þ þ sinφd ðT 3 þ T 4 Þ þ cosφu ðM 1 −M 2 Þ þ cosφd ðM 3 −M 4 Þ ¼ 0
ð37Þ
Substituting Eq. (36) into Eq. (37) yields 8 > > > 4EIδx > > <
! sinφu cosθu sinφd cosθd cosφd cosθd cosφu cosθu − − þ 4EIaγ y ¼ −2GIp γy lu ld l2d l2u 1 0 ! : 2 2 2 2 sinφu cos θu δx þ aγ y sinφd cos θd δx −aγ y cosφu cos θu cosφd cos θd A ¼ My −4EI@ þ − lu ld l2u l2d
cosφu cosθu cosφd cosθd þ l2u l2d
> > > > > : 2GIp γy
!
So the torsional stiffness of the suspension about the y-axis is K My ¼
My ¼ 2GIp γy
! ! cosφu cos2 θu cosφd cos2 θd sinφu cos2 θu ðk1 þ aÞ sinφd cos2 θd ðk1 −aÞ þ − −4EI lu ld l2u l2d
ð38Þ
where KMy denotes the torsional stiffness of the suspension about the y-axis,
k1 ¼
−GIp
sinφu cosθu lu
− sinφdl cosθd þ 2EIa cosφdl2cosθd − cosφul2cosθu d u d : cosφu cosθu cosφd cosθd 2EI þ l2 l2 u
d
3.5. The equivalent torsional stiffness of the suspension about z-axis Suppose that there is a moment about the z-axis, Mz, exerted to the knuckle the corresponding rotational angle of which is γz shown in Fig. 13. Assume that the displacement of the knuckle center under the action of Mz is δr on ¼ ð δx δy 0 ÞT : Therefore, the absolute coordinates of the revolute joints Bi (i = 1, 2, 3, 4) can be expressed in oxyz r B′ i ¼ Tr Bni þ δron
ð39Þ
where T represents the rotational matrix from the local coordinate frame, onxnynzn, to the absolute coordinate frame, oxyz, and 2
cosγz T ¼ 4 sinγ z 0
− sinγ z cosγz 0
3 0 0 5: 1
Substituting the local coordinates of the joints expressed by Eq. (1) into Eq. (39) produces their absolute coordinates 2
r B′ 1
3 2 3 2 3 2 3 δx −b cosγz δx þ b cosγz δx −b cosγz δx þ b cosγz ¼ 4 δy −b sinγz 5; r B′ 2 ¼ 4 δy þ b sinγz 5; r B′ 3 ¼ 4 δy −b sinγz 5; r B′ 4 ¼ 4 δy þ b sinγz 5: a a −a −a
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Fig. 13. Slope of the suspension under the moment of Mz.
Therefore, 8 → > > B1 B′ 1¼ δx −b cosγ z þ b > > > → > > < ′ B2 B 2¼ δx þ b cosγz −b → > > > B3 B′ 3¼ δx −b cosγ z þ b > > > → > : ′ B4 B 4¼ δx þ b cosγz −b
δy −b sinγ z δy þ b sinγz δy −b sinγ z δy þ b sinγz
0
T
0 T : 0 T 0 T
The displacement of joint Bi (i = 1, 2, 3, 4) in its axial direction, ni (i = 1, 2, 3, 4), can be expressed as: 8 → > ′ > δ1 ¼B1 B 1⋅n1 ¼ − cosθu ðδx −b cosγ z þ bÞ þ sinθu δy −b sinγz > > > > → > > > < δ2 ¼B2 B′ 2⋅n2 ¼ cosθu ðδx þ b cosγ z −bÞ þ sinθu δy þ b sinγ z : → > ′ > > > δ3 ¼B3 B 3⋅n3 ¼ − cosθd ðδx −b cosγ z þ bÞ þ sinθd δy −b sinγ z > > > → > > : δ4 ¼B4 B′ 4⋅n4 ¼ cosθd ðδx þ b cosγ z −bÞ þ sinθd δy þ b sinγ z
ð40Þ
Considering sin γz ≈ γz and cos γz = 1 when γz is a tiny item, Eq. (40) can be further simplified as 8 > δ1 ¼ −δx cosθu þ δy −bγz sinθu > > > > > < δ2 ¼ δx cosθu þ δy þ bγz sinθu : > > δ3 ¼ −δx cosθd þ δy −bγz sinθd > > > > : δ ¼ δ cosθ þ δ þ bγ sinθ 4
x
d
y
z
d
Because the structure is symmetrical and the load is antisymmetrical, the deformations of the branches are antisymmetrical, there should be δy = 0. Then 8 δ > > < 1 δ2 > > δ3 : δ4
¼ −δx cosθu −bγ z sinθu ¼ δx cosθu þ bγz sinθu : ¼ −δx cosθd −bγ z sinθd ¼ δx cosθd þ bγz sinθd
ð41Þ
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115
Fig. 14. Deflection of a branch under the moment Mz.
Under the action of moment, Mz, both the position and the orientation of the joint axis of Bi (i = 1, 2, 3, 4) have been changed. The displacement of Bi on the suspension consists of the bending and torsional components which are shown in Fig. 14. The unit vector of B′1 in the oxyz-frame is 2
′
n1
cosγ z ¼ 4 sinγz 0
− sinγz cosγ z 0
3 2 32 3 − cosθu − cosγz cosθu − sinγz sinθu 0 0 54 sinθu 5 ¼ 4 − sinγz cosθu þ cosγz sinθu 5: 0 1 0
Because the vectors n1 and n′1 are coplanar, there is no torsional component of the branch 1, the same are true to the other branches. Suppose that the bending moment is Mi (i = 1, 2, 3, 4) the direction of which is defined by ki × ni. In vector form, the bending moment loads can be rewritten as: 8 > M1 > > < M2 > M > > : 3 M4
¼ ½ M 1 sinφu sinθu M1 sinφu cosθu M1 cosφu T ¼ ½ M 2 sinφu sinθu −M2 sinφu cosθu −M 2 cosφu T : ¼ ½ −M3 sinφd sinθd −M 3 sinφd cosθd M 3 cosφd T T ¼ ½ −M4 sinφd sinθd M4 sinφd cosθd −M 4 cosφd
Under these actions, the corresponding displacements in the axial direction of joint Bi (i = 1, 2, 3, 4) are 8 2 M l > > > δ′1 ¼ 1 u > > 2EI > > 2 > > M l > < δ′2 ¼ 2 u 2EI : 2 > M 3 ld > ′ > δ ¼ > 3 > > 2EI > > 2 > > : δ′ ¼ M4 ld 4 2EI
Considering the compliant requirements, one obtains 8 ′ > > > δ′ 1 < δ2 > δ′ > > : ′3 δ4
¼ δ1 ¼ δ2 : ¼ δ3 ¼ δ4
ð42Þ
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As a result, one finds that 8 2EI ðδx cosθu þ bγ z sinθu Þ > > > > M1 ¼ − > l2u > > > > 2EI ð δ cosθ þ bγz sinθu Þ > x u > > < M2 ¼ l2u : 2EI ðδx cosθd þ bγ z sinθd Þ > > M3 ¼ − > > > l2d > > > > 2EIðδx cosθd þ bγz sinθd Þ > > > : M4 ¼ l2d
ð43Þ
The static equilibrium conditions require that 8 < sinφu sinθu ðM1 þ M2 Þ− sinφd sinθd ðM 3 þ M 4 Þ ¼ 0 sinφu cosθu ðM1 −M2 Þ− sinφd cosθd ðM 3 −M 4 Þ ¼ 0 : : cosφu ðM1 −M2 Þ þ cosφd ðM3 −M4 Þ−M z ¼ 0
ð44Þ
Substituting Eq. (43) into Eq. (44) yields 8 ðδx cosθu þ bγz sinθu Þ sinφu cosθu ðδx cosθd þ bγz sinθd Þ sinφd cosθd > > > − ¼0 < l2u l2d : 4EIðδx cosθu þ bγ z sinθu Þ cosφu 4EIðδx cosθd þ bγ z sinθd Þ cosφd > > > − ¼ Mz :− 2 2 lu ld So the torsional stiffness of the suspension about the z-axis is
K Mz ¼
Mz 4EIbðk2 cosθu þ sinθu Þ cosφu 4EIbðk2 cosθd þ sinθd Þ cosφd ¼− − γz l2u l2d
ð45Þ
where KMz denotes the torsional stiffness of the suspension about the z-axis and k2 ¼
l2u sinθd cosθd sinφd −l2d sinθu cosθu sinφu : l2d cos2 θu sinφu −l2u cos2 θd sinφd
3.6. Equivalent stiffness of the suspension in z-direction The spring shock absorber connects the vehicle and the knuckle through two sphere joints shown in Fig. 15. The lower joint is denoted by C while the upper joint is denoted with D. Assume that the local coordinates of joint C in the coordinate frame onxnynzn
Fig. 15. The spring shock absorber and the suspension.
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
117
are represented by Cn (xC, yC, zC). Also suppose that the absolute coordinates of joint D are denoted by Dn (xD, yD, zD). Suppose that the angle from the axis of the spring shock absorber to the xnonyn-plane is β when the suspension moves a distance of s under the weight of vehicle which is shown in Fig. 15, one can immediately obtain the following relations: 8 xD −xC > > < α ¼ arctan y −y D C : > β ¼ − arctan ðzD −zC −sÞ cosα > : yD −yC Assume that the knuckle is bearing a force Fz along the z-axis, under the action of which, the knuckle has small displacements along the x-, y- and z-axes, denoted by δx, δy and δz accompanying small rotations, γx and γz, about the x- and z-axes, individually. So the coordinate transformation from onxnynzn to oxyz is r ¼ Tr n þ δr on
ð46Þ
where rn denotes the vector coordinates in the onxnynzn frame, and r represents the absolute vector coordinates in the oxyz frame, and 2
sinγ x sinγy cosγ z − cosγ x sinγ z − sinγ x sinγy sinγz þ cosγx cosγ z sinγ x cosγy
cosγy cosγz T ¼ 4 cosγ y sinγ z − sinγy
3 cosγx sinγ y cosγz þ sinγ x sinγz cosγx sinγ y sinγ z − sinγx cosγz 5 cosγ x cosγy
2
δr on
3 δx 4 ¼ δy 5: δz
ð47Þ
ð48Þ
Because the values of the rotations, γx and γz, are very small, Eq. (47) can be simplified as 2
1 T ¼ 4 γz −γy
−γ z 1 γx
3 γy −γ x 5: 1
Assume the axial force exerted to the knuckle by the spring shock absorber is F which can be expressed as F ¼ ½ F sinα cosβ
F cosα cosβ
T
−F sinβ :
ð49Þ
In line with the equilibrium of the knuckle, one gains F z −F sinβ ¼ 0: The torques resulting from the action of F can be expressed by 2 3 3 Mx −F ðyC sinβ þ zC cosα cosβÞ 4 M y 5 ¼ r C F ¼ 4 F ðxC sinβ þ zC sinα cosβÞ 5: Mz F ðxC cosα cosβ−yC sinα cosβÞ 2
ð50Þ
Associating Eqs. (11), (18), (28), (38), and (45) with Eqs. (49) and (50) yields 8 F sinα cosβ > > > δx ¼ > Kx > > > > F cosα cosβ > > δ ¼ > y > > Ky > > < M γx ¼ x : K Mx > > > > My > > > > γy ¼ > K > My > > > M > > : γz ¼ z K Mz
ð51Þ
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Prior to the action of F, the coordinates of ball joint C in the coordinate system oxyz are r C ¼ ½ xC yC zC þ s T where s stands for the displacement of the knuckle under the self load of the vehicle. After the action of F, the coordinates of ball joint C in the coordinate system oxyz can be obtained by substituting rC into Eq. (46) 2
3 δx þ xC −γ z yC þ ðzC þ sÞγy r C ¼ Tr n þ δro ¼ 4 δy þ yC þ xC γz −ðzC þ sÞγx 5: δz þ zC þ s−xC γy þ yC γx Therefore, one obtains → ′ CC¼ δx −γ z yC þ ðzC þ sÞγy
δy þ xC γz −ðzC þ sÞγx
δz −xC γy þ yC γx
T
:
The unit axial vector of the spring shock absorber is nDC ¼ ½ sinα cosβ
cosα cosβ
T
− sinβ :
So, the deformation of the spring is →
Δl h¼ nDC ⋅ CC ′ i h i h i ¼ δx −γ z yC þ ðzC þ sÞγy sinα cosβ þ δy þ xC γz −ðzC þ sÞγx cosα cosβ− δz −xC γ y þ yC γ x sinβ:
ð52Þ
Suppose the stiffness of the spring is k, the force exerted to the knuckle is F ¼ −kΔl ð53Þ where “−” indicates that the direction of the displacement of the suspension is reverse to that of the elastic force of the spring. Substituting Eqs. (51) and (52) into Eq. (53) yields ! ! My F sinα cosβ M F cosα cosβ Mz M − z yC þ ðzC þ sÞ sinα cosβ−k þ xC − x ðzC þ sÞ cosα cosβ Kx Ky K Mz K K Mz K Mx ! My My M x þ x y sinβ: þ k δz − K My C K Mx C
F ¼ −k
ð54Þ
Substituting Eq. (50) into Eq. (54) yields ! F sinα cosβ F ðxC cosα cosβ−yC sinα cosβÞ F ðxC sinβ þ zC sinα cosβÞ − yC þ ðzC þ sÞ sinα cosβ Kx K Mz K My ! F cosα cosβ F ðxC cosα cosβ−yC sinα cosβÞ −F ðyC sinβ þ zC cosα cosβÞ þ xC − ðzC þ sÞ cosα cosβ −k Ky K Mz K Mx ! F ðxC sinβ þ zC sinα cosβÞ −F ðyC sinβ þ zC cosα cosβÞ xC þ yC sinβ: þk δz − K My K Mx F ¼ −k
ð55Þ
Eq. (55) can be simplified to Kz ¼
F sinβ 1 ¼ δz f k −f x −f y −f Mx −f My −f Mz
ð56Þ
where 1 sin2 α cot2 β cos2 α cot2 β ; fx ¼ ; fy ¼ ; 2 Kx Ky k sin β 2 2 2 y ð2z þ sÞ cosα cotβ þ zC ðzC þ sÞ cos α cot β þ yC ¼ C C ; K Mx 2 2 2 x þ xC ð2zC þ sÞ sinα cotβ þ zC ðzC þ sÞ sin α cot β ¼ C ; K My
fk ¼ f Mx f My
f Mz ¼
x2C cos2 α þ y2C sin2 α−2xC yC sinα cosα 2 cot β: K Mz
Suppose that the radius of the wheel is denoted by R, and L represents the distance from the central plane of the wheel to the knuckle, Lu and Ld stand for the initial lengths of the upper two links and the lower two links, respectively, Du stands for the
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119
Table 1 Parameters of the suspension with different branches. θu(°)
Initial C
a(mm)
R(mm)
E(GPa)
Du(mm)
r(mm)
43
ð0
50
320
206
95
15
Ld(mm)
θd(°)
Initial D
b(mm)
L(mm)
G(GPa)
Dd(mm)
m(kg)
234
48
ð0
75
60
79.38
55
441.5
−180
540 Þ
3800
3800
3600
3600
3400
3400
3200
3200 stiffness in x-direction stiffness in y-direction stiffness in z-direction
3000 2800
-50
0
50
3000
100
24.8
2800
24.7 24.6 24.5 24.4 24.3 24.2
Stiffness in z-direction (kN/m)
−80 −60 Þ
Stiffness in y-direction (kN/m)
Stiffness in x-direction (kN/m)
Lu(mm) 204
24.1
Travel of the Knuckle (mm) Fig. 16. Linear stiffness of the knuckle during jounce and rebound.
distance from the horizontal plane passing through the axis of the wheel at the equilibrium position to the fixed joints of upper branches, and Dd represents the initial distance from the horizontal plane passing through the axis of the knuckle at the equilibrium position to the fixed joints of lower branches. The natural frequency of a vehicle is an important parameter to measure the ride of the vehicle. Generally the natural frequency of the car is restricted to 1–1.3 Hz. Assume that the stiffness of the suspension is Kp, and the supported mass of spring is m, there is the following relationship 2
K p ¼ mð2πf Þ
ð57Þ
62
54
61
53
60
52
59
51
58
50
57
stiffness about x-axis stiffness about y-axis stiffness about z-axis
49 48
56 55
47
54
46
53
45
-50
0
50
100
52
Travel of the Knuckle (mm) Fig. 17. Rotational stiffness of the knuckle during jounce and rebound.
24 23 22 21 20 19 18 17
Stiffness about z-axis (kN.m/rad)
55
Stiffness about y-axis (kN.m/rad)
Stiffness about x-axis (kN.m/rad)
where f is the natural frequency of the vehicle and f = 1.2 Hz in general. Assume that the section of the branch is round and the radius is r. A set of parameters of the rectilinear independent suspension with length compensation branches listed in Table 1 is numerically simulated. The stiffness of the suspension during jounce and rebound is shown in Figs. 16 and 17. Fig. 16 indicates that the lowest stiffness in the x-direction is about 2900 kN/m, the lowest stiffness of the suspension in the ydirection is about 2950 kN/m which occurs at the upper limit or lower limit of the travel of the knuckle and the stiffnesses in the x-direction and the y-direction are symmetrical about the suspension initial point and the stiffness in the vertical direction keeps
120
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4 Nonlinear spring Linear spring
The axial load (kN)
3 2 1 0 -1 -2 -3 -4 -80
-60
-40
-20
0
20
40
60
80
100
120
Travel of the Knuckle (mm) Fig. 18. The characteristic curves of nonlinear spring and linear spring.
in a relatively horizontal line the largest change is within 3%. Fig. 17 indicates that the rotational stiffness of the suspension around x-axis changes within 45–55 kN/rad and firstly increases and then decreases, the rotational stiffness around the y-axis changes within 52–61 kN/rad and firstly decreases and then increases and the rotational stiffness of the suspension around z-axis changes within 17–24 kN/rad and firstly increases and then decreases. In order to keep a good frequency range and to guarantee good ride comfort and handling stability, the spring with variable stiffness is often used in applications. In engineering, the nonlinear relationship between force and deformation of the variable stiffness spring is usually expressed by
F ¼ Kx þ εKx
3
ð58Þ
where F is the axial load on the spring, x is the spring deformation, K represents the linear stiffness of the spring and ε is a parameter which represents the nonlinear coefficient. In the applications, the approximation of K = Kp is often adopted, so the nonlinear stiffness of the spring is
k¼
F 2 ¼ K p 1 þ εx : x
ð59Þ
Assume that ε = 501/m 2. The corresponding relationship of the axial load and the spring deformation is shown in Fig. 18. Substitute Eq. (59) into Eq. (56), with the parameters in Table 1 and Table 2, one can obtain the numerical simulation curves for the stiffness of the suspension shown in Figs. 19 and 20. Comparing Figs. 19 and 16, one can find that the linear stiffness in z-direction firstly decreases and then increases quickly with the application of nonlinear spring which is particularly useful to improve the comfort of a vehicle. Therefore, the vehicle has better ride comfort by using the nonlinear spring when the structure parameters are the same. From Table 2, one can find that the four branches of the knuckle are identical and the structure of the suspension is symmetrical. Therefore, the linear stiffnesses in x-direction and y-direction of the suspension are identical and the rotational stiffnesses about x-axis, y-axis and z-axis are symmetrical which verifies the theoretical results. Comparing Figs. 19 and 20, one can find that the stiffness of the suspension in the x-direction and the y-direction and the rotation stiffness of the suspension about x-axis and y-axis with parameters in Table 1 are larger than that in Table 2, the rotational stiffness of the suspension about z-axis with parameters in Table 1 is smaller than that of the suspension with parameters in Table 2. Therefore, different structure parameters result in different stiffnesses. Table 2 Parameters of the suspension with identical branches. Lu(mm)
θu(°)
Initial C
a(mm)
R(mm)
E(GPa)
Du(mm)
r(mm)
225
45
ð0
50
320
206
95
15
−80 −60 Þ
Ld(mm)
θd(°)
Initial D
225
45
ð0
−180
540 Þ
b(mm)
L(mm)
G(GPa)
Dd(mm)
m(kg)
75
60
79.38
55
441.5
3800
3600
3600
3400
3400 stiffness in x-direction stiffness in y-direction stiffness in z-direction
3200
3200
3000 2800
3000
38 36 34 32 30 28 26
2800
24
55
62
24
54
61
53
60
52
59
-50
0
50
100
Stiffness in z-direction (kN/m)
3800
121
Stiffness in y-direction (kN/m)
Stiffness in x-direction (kN/m)
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
58
50
57
stiffness about x-axis stiffness about y-axis stiffness about z-axis
49 48
56 55
47
54
46
53
45
-50
0
50
100
23 22 21 20 19 18
Stiffness about z-axis (kN.m/rad)
51
Stiffness about y-axis (kN.m/rad)
Stiffness about x-axis (kN.m/rad)
Travel of the Knuckle (mm)
17
52
Travel of the Knuckle (mm)
3300
3200
3200
3100
3100
3000
3000
2900
2900 stiffness in x-direction stiffness in y-direction stiffness in z-direction
2800 2700
2800 2700
2600 2500
2600
38 36 34 32 30 28 26
2500
24
52
60
28
50
58
48
56
-50
0
50
100
Stiffness in z-direction (kN/m)
3300
Stiffness in y-direction (kN/m)
Stiffness in x-direction (kN/m)
Fig. 19. Stiffness of the knuckle with parameters of Table 1.
54
44 42
52
-50
0
50
100
50
Travel of the Knuckle (mm) Fig. 20. Stiffness of the knuckle with parameters of Table 2.
26 24 22 20 18 16
Stiffness about z-axis (kN.m/rad)
stiffness about x-axis stiffness about y-axis stiffness about z-axis
46
Stiffness about y-axis (kN.m/rad)
Stiffness about x-axis (kN.m/rad)
Travel of the Knuckle (mm)
122
J-S. Zhao et al. / Mechanism and Machine Theory 56 (2012) 99–122
4. Conclusions This paper investigated the stiffness of a rectilinear independent suspension whose alignment parameters should be invariable in mechanism theory during jounce and rebound. The fact that the resultant stiffness of the suspension is subjected to the stiffness of each 3-RRR distance compensation linkage raises a particular difficulty to build the model. Therefore, this paper started from establishing the equivalent deformations and deflections of each branch of the suspension, and then the sixdimensional stiffness of the suspension is equivalently expressed in terms of the material properties of the four kinematic chains. This method well avoids the inverse problems of position equations which were generally used in the previous literature. Numerical simulations indicated that this methodology could provide an optimization basis for the design of the suspension. 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