# KINEMATICAL AND DYNAMICAL ANALYSIS OF ... - SID

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Archive of SID
Iranian Journal of Science & Technology, Transaction B, Engineering, Vol. 32, No. B4, pp 325-339
Printed in The Islamic Republic of Iran, 2008

KINEMATICAL AND DYNAMICAL ANALYSIS OF MACPHERSON
SUSPENSION USING DISPLACEMENT MATRIX METHOD
*

M. S. FALLAH, M. MAHZOON
**

Dept. of Mechanical Engineering, Shiraz University, Shiraz, I. R. of Iran
Email: mahzoon@shirazu.ac.ir

Abstract In this paper, kinematics and dynamics of the MacPherson suspension are studied.
Displacement Matrix Method is utilized for this purpose. Camber and toe angle alterations are
derived by means of the rotation matrix. Kinematical characteristics of the MacPherson suspension
are displayed as functions of time and the wheel vertical displacement. Relations for velocities and
accelerations of key points are also obtained. Since internal forces and external forces are
important in stress analysis and for comfortability, these forces are calculated as functions of time
as well. Subsequently, a general analysis of the me chanism under the wheel sinusoidal
displacement is presented together with relevant figures and results.

Keywords Kinematics, dynamics, MacPherson suspension mechanism, camber angle, toe angle, kingpin angle,
caster angle, track alteration
1. INTRODUCTION

The MacPherson suspension was made in the Ford Company by Earl S. MacPherson for the first time in
1949. This type of suspension is widely employed in many modern vehicles because of its light weight
and compact size [1]. It can be used for both front and rear suspensions, but is usually found at the front,
where it provides a steering pivot (kingpin) as wel l as a suspension mounting for the wheel. A
MacPherson suspension, as shown in Fig. 1, consists of: 1) a control arm, 2) a tie rod, 3) a spindle and
piston rod, and 4) a strut; the control arm is connected to the chassis with a rotational joint. A spherical
joint connects the control arm and the strut. Wheels are mounted on the struts spindle. A cylindrical joint
connects the strut to the piston rod which is connected to the chassis with a spherical joint. A spring and a
damper are put in between the strut and the chassis along the piston rod to absorb vibration and shocks
caused by a bumpy road. The strut is also connected to the tie rod with a spherical joint [2].
The main functions of suspension systems are to adequately support the vehicle weight, to provide
effective isolation for the chassis from excitations due to rough roads, to maintain the wheels in the
appropriate steer and camber attitudes with respect to the road surface and to keep tire contact with the
ground. The wheel suspension characteristics are of vital importance for the road holding ability of the
tires, riding quality of the car, minimizing the transient forces to the body and the severity of the
environmental noise.
In the design process for choosing ride characteristics a mathematical model is essential. This can
lead to a better choice for system parameters and improvement of its behavior [3]. In order to study the
behavior of a MacPherson strut wheel suspension, various mathematical approaches and models have been
presented. The models are more or less complex depending on the requirements set by the authors. For

Received by the editors January 28, 2006; Accepted April 27, 2008.

Corresponding author

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Iranian Journal of Science & Technology, Volume 32, Number B4 August 2008
326

example, a spatial model of the Macpherson suspension to conduct kinematical analysis of this type of
structure was formulated by Cronin [1], and Stensson et al., [3]. Mantaras et al., [4] proposed three 2D
nonlinear models for Macpherson suspension to analyze the dynamical behavior of this mechanism.
Jonsson [5] carried out a finite element analysis for assessing deformation of the components of this
structure. In [2] and [6] two similar 3D models of MacPherson suspension were used to estimate its
dynamical parameters. Habibi et al. [7] implemented genetic algorithm method on a three dimensional
model of MacPherson suspension to optimize its design characteristics. Sohn et al. [8]-[11] proposed a 2D
model of Macpherson suspension for control applications. Lee and Han considered a simple model of
Macpherson suspension and designed a controller to control varying roll center during cornering [12]. A
comprehensive analysis of MacPherson suspension was performed by Reimpell et al. [13]. They reviewed
the main parameters of several suspensions and compared them with each other. In [14], another approach
to study variations of handling parameters of MacPherson suspension was proposed by Raghavan.

Fig. 1. MacPherson suspension mechanism [2]

In this paper, the displacement matrix method is employed to analyze the system behavior and a
comprehensive kinematical and dynamical analysis of Macpherson suspension is presented. As a main
advantage, implementation of this method is truly convenient for computer coding and also formulation of
kinematical and dynamical relations.
The main kinematical parameters which are investigated in this paper are: 1) camber angle, 2) toe
angle, 3) caster angle, 4) kingpin angle, and 5) track alterations. As a result of functional factors when the
wheels travel in bump and rebound in travel direction, the track changes. However, alteration of the track
size causes the rolling tire to slip and, on flat crossings in particular, causes lateral forces and may even
influence the steering. Caster and kingpin angles alterations affect the selfaligning torques and
consequently affect the stability and handling of the vehicle when wheels bounce or rebounce. Camber
angle alterations are due to the rubbing of tire and produce lateral force acting on the wheel. In order to
keep the vehicle stable, any change in the gradient of the toe angle characteristics as functions of wheel
travel should be avoided. In addition, to avoid increased tire wear and rolling resistance or impeding
directional stability, no toe angle changes should occur when the wheels compress or rebound [13]. For
B
0
D
E
A
1
P
B
1
C
1
j
1
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August 2008 Iranian Journal of Science & Technology, Volume 32, Number B4
327
z
0
Yaw

y
0
Pitch

Roll

dynamical analysis of the mechanism, the determination of velocities and accelerations of the key points
are necessary; therefore, relations for velocities and accelerations of these points should also be derived.
Internal forces are important in stress analysis and are crucial factors in the failure of the components
and joints. Therefore, in this paper all internal forces and moments acting on the mechanism are calculated
as functions of time. The equations for external forces and moments such as: 1) spring force; 2) damper
force; 3) road force, and 4) resisting torque are derived as they affect comfortability and also magnitude of
the internal forces.
2. DISPLACEMENT MATRIX METHOD AND CONSTRAINT EQUATIONS

One of the most effective methods to analyze mechanisms is through the use of displacement matrix
method. In this method, the general three-dimensional displacement matrix is given in terms of a
translation from a point ),,(
1111
zyxp to another point ),,(
2222
zyxp, and a rotation about a fixed
coordinate frame [15].





++
++
++
=





++
++
++
=
1000
1000
1331321312
1231221212
1131121112
1331321312333231
1231221212232221
1131121112131211
12
)zayaxa(z
)zayaxa(y
)zayaxa(x
R
)zayaxa(zaaa
)zayaxa(yaaa
)zayaxa(xaaa
]D[
D
(1)

where
ij
a are components of the rotation matrix. The rotation matrix can be described as the product of
successive rotations about the principal coordinate axes x
0
, y
0
and z
0
taken in a specific order. These
rotations define the roll, pitch, and yaw angles, which are denoted as

and,,
, and are shown in Fig. 2.
The resulting rotation matrix, R
D
, is then given by

R
D
=





++
++
=








coscossincossin
cossinsinsincossinsinsincoscoscossin
cossincossinsinsinsincoscossincoscos
333231
232221
131211
aaa
aaa
aaa
(2)

Another task in the displacement matrix method is the formulation of the holonomic constraint equations
which are specifically related to the joints between adjacent bodies. These are mathematical restrictions on
the mobility of the model so as to take away degrees of freedom of the multi-body system. The constraint
equations for the spherical-spherical (SS) link, the revolute-spherical (RS) link and spherical-cylindrical
(SC) link of the mechanism are listed below.

Fig. 2. Roll, pitch and yaw angles
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Iranian Journal of Science & Technology, Volume 32, Number B4 August 2008
328

a) Spherical-spherical (SS); link constraint equations

SS link is defined by a spherical joint at a point ),,(
0000
zyxp on a fixed body and another spherical
joint at a different point ),,(
1111
zyxp on a moving body. The constraint equation that specifies the
constancy of the distance between the two spherical joints is

=++
2
01
2
01
2
01
)zz()yy()xx( constant

Therefore, the displacement constraint for tie rod is:

222222
)()()()()()(
01
0
101000
BB
B
BBBB
j
BBBBB
zzyyxxzzyyxx
jj
++=++ (3)

where j subscript indicates the displaced point, i.e. B
j
, j = 2, 3, 4, .

b) Revolute-spherical (RS) link constraint equations

An RS link connects a revolute joint at a point ),,(
0000
zyxp with rotation axis ),,(
0000 zyx
uuuU on
a fixed body and a spherical joint at another point ),,(
1111
zyxp on a moving body. The constraint
equations which specify the constancy of distance between the revolute and spherical joints and the
perpendicularity between the revolute axis and the axis defined as the link are

=++
2
01
2
01
2
01
)zz()yy()xx(
constant,
0
010101
000
=

+

+

)zz(u)yy(u)xx(u
zyx

Consequently, the displacement constraints for control arm are:

222222
)()()()()()(
0
00010101
AAAAAAAAAAAA
zzyyxxzzyyxx
jjj
++=++ (4)

0)()()(
000000
=

+

+

AAzAAyAAx
zzuyyuxxu
jjj
(5)

),,(
0000 zyx
uuuU is unit vector of the DE line in the principal axis. In the subsequent analysis, the
revolute joints at D and E will be replaced by one revolute joint at point A
0
, found from the orthogonal
projection of vector DA
1
on the revolute axis.

c) Spherical-cylindrical (SC) link constraint equations

The SC link connects a spherical joint at a point ),,(
0000
zyxp on a fixed body and a cylindrical joint
at another point ),,(
1111
zyxp on a moving body with an axis of translation/rotat ion
),,(
1111 zyx
uuuU along the link axis. The constraint equations that specify that the straight line defined by
the link, or the cylindrical joint axis, remains a straight line during any displacement are

0
0101
11
=

)zz(u)xx(u
xz
, 0
0101
11
=

)zz(u)yy(u
yz
and
1
222
111
=++
zyx
uuu

So, the displacement constraints for the strut are (j
1
is an arbitrary point on strut):
For line C
1
C
0
:

0)()(
00
=

ccxccz
zzuxxu
jjjj
, 0)()(
00
=

ccyccz
zzuyyu
jjjj
and 1
222
=++
jjj
zyx
uuu (6)

For line j
1
C
0
:

0)()(
00
=

CjxCjz
zzuxxu
jjjj
, 0)()(
00
=

CjyCjz
zzuyyu
jjjj
(7)

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329
d) Kinematical equations
The assumptions adopted in Fig. 1 are summarized as follows:
1) The chassis is fixed; 2) All bodies are rigid; 3) Control arm is modeled by a rod.
The input to the system is the road profile which is applied at point P (Fig. 1). In this paper, the road
profile is assumed be a sinusoidal curve with the characteristics ,
)tsin(z
r
5
80

=
.
If displacement of point P in the vertical direction is equal to z
r
subject to road disturbance, point P
1

)z,y,x(
ppp
111
will move to point P
2
)z,y,x(
ppp
222
and the magnitude of its z coordinate will be equal
to:
12
PrP
zzz +=.
Consequently, the displacement matrix of spindle can be written as:

[ ]





++
++
++
=
1000
)(
)(
)(
1112
1112
1112
333231333231
232221232221
131211131211
AAAA
AAAA
AAAA
Spindle
zayaxazaaa
zayaxayaaa
zayaxaxaaa
D (8)

where
2
A is the new position of
1
A (Fig. 1) under the wheels vertical displacement. This matrix is used
to determine the new positions of points B
1
, C
1
and j
1
by the following relation:

[ ]





=





111111
111
111
111
222
222
222
jCB
jCB
jCB
Spindle
jCB
jCB
jCB
zzz
yyy
xxx
D
zzz
yyy
xxx
(9)

To solve the unknown parameters, the following constraint equations should be added to Eq. (9).





+++=
=++
++=++
=++
=
=
=
=
++=++
34333231
222222
222
222222
0
1112
020020020
0
10101020202
222
022022
022022
022022
022022
01
0
10102022
9
08
7
16
05
04
03
02
1
azayaxaz:
)zz(u)yy(u)xx(u:
)zz()yy()xx()zz()yy()xx(:
uuu:
)zz(u)yy(u:
)zz(u)xx(u:
)zz(u)yy(u:
)zz(u)xx(u:
)zz()yy()xx()zz()yy()xx(:
PPPP
AAzAAyAAx
AAAAAAAAAAAA
zyx
Cjyjjz
Cjxjjz
ccyccz
ccxccz
BB
B
BBBBBBBBB

As for the numerical scheme, the Newton-Raphson method is employed [16] and all unknown parameters
of the displacement matrix of the spindle are determined. Using this matrix, the spring force, which is vital
for dynamical analysis of MacPherson suspension mechanism, can be calculated.

3. MATHEMATICAL DETERMINATION OF THE KINEMATICAL PARAMETERS

In this section, kinematical parameters of the MacPherson suspension mechanism are described. As
mentioned before, θ and ψ correspond to camber and toe angles. Caster is the angle between the projection
of the steering axis (line A
1
C
0
) on xz plane and a line perpendicular to the road. Using this definition:
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Iranian Journal of Science & Technology, Volume 32, Number B4 August 2008
330

Caster angle
)(tan
0
01
j
j
AC
AC
zz
xx

=

. Kingpin is the angle between the projection of steering axis on the yz
plane and a line perpendicular to the road; therefo re, it can be stated as: Kingpin
angle
)(tan
0
0
1
j
j
AC
AC
zz
yy

=

, and track alteration is equal to: Track alteration
j
AA
yy

=
1
.

a) Velocity and acceleration matrices
The velocity and acceleration matrices are:
[ ]





++
++
++
=
0000
)(
)(
)(
1112
1112
1112
333231333231
232221232221
131211131211
AAAA
AAAA
AAAA
spindle
zayaxazaaa
zayaxayaaa
zayaxaxaaa
D




(10)

where
D
xy
xz
yz
D
R
aaa
aaa
aaa
R





=





=
0
0
0
333231
232221
131211










, and

[ ]





++
++
++
=
0000
)(
)(
)(
1112
1112
1112
333231332331
232221232221
131211131211
AAAA
AAAA
AAAA
spindle
zayaxazaaa
zayaxayaaa
zayaxaxaaa
D




(11)

where
D
zxzyyzx
xzyyzyx
yzxzyxx
D
R
aaa
aaa
aaa
R





+
+
+
=





=
22
22
22
333231
232221
131211










,
2222
zyx


++=.

Using these matrices, velocity and acceleration of any point can be determined.

[
]
[
]
[
]
TT
zyxDzyx 10


=
;
[
]
[
]
[
]
TT
zyxDzyx 10


=
(12)

Differentiating Eqs. (3) to (7), the velocity and acceleration constraints are derived.
Set of Velocity Equations:





++=
++=
++=
+=
+=
+=
++=
++=
++=
34
222
34
222
34
222
24
222
24
222
24
222
14
222
14
222
14
222
9
8
7
6
5
4
3
2
1
ayxz:
ayxz:
ayxz:
azxy:
azxy:
azxy:
azyx:
azyx:
azyx:
jxjyj
CxCyC
BxByB
jxjzj
CxCzC
BxBzB
jyjzj
CyCzC
ByBzB
























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331





++=
=++
=++
=++
=+
=+
=+
=+
=++
34
222
202020
022022022
222222
0222202
2
22
0222202222
0222202
2
22
0222202
2
22
022022022
18
017
016
015
014
013
012
011
010
ayxz:
zuyuxu:
)zz(z)yy(y)xx(x:
uuuuuu:
)yy(uyu)zz(uzu:
)xx(uxu)zz(uzu:
)yy(uyu)zz(uzu:
)xx(uxu)zz(uzu:
)zz(z)yy(y)xx(x:
PxPyP
AzAyAx
AAAAAAAAA
zzyyxx
CjzjzCj
y
jy
CjzjzCj
x
jx
CCzCzCC
y
Cy
CCzCzCC
x
Cx
BBBBBBBBB












Set of Acceleration Equations:





++++=
=++
=+++++
=+++++
=++
=++
=++
=++
=+++++
++++=
++++=
++++=
++++=
++++=
++++=
++++=
++++=
++++=
34
2
22
222
202020
2
2
2
2
2
2022022022
2
2
2
2
2
2222222
02222220222222
02222220222222
02222220222222
02222220222222
2
2
2
2
2
2022022022
34
2
22
222
34
2
22
222
34
2
22
222
24
22
22
22
24
22
22
22
24
22
22
22
14
222
22
2
14
222
22
2
14
222
22
2
18
017
016
015
2214
02213
02212
02211
010
9
8
7
6
5
4
3
2
1
az)(y)(x)(z:
zuyuxu:
zyx)zz(z)yy(y)xx(x:
uuuuuuuuu:
)yy(uyuyu)zz(uzuzu:
)xx(uxuxu)zz(uzuzu:
)yy(uyuyu)zz(uzuzu:
)xx(uxuxu)zz(uzuzu:
zyx)zz(z)yy(y)xx(x:
az)(y)(x)(z:
az)(y)(x)(z:
az)(y)(x)(z:
az)(y)(x)(y:
az)(y)(x)(y:
az)(y)(x)(y:
az)(y)(x)(x:
az)(y)(x)(x:
az)(y)(x)(x:
PzPxzyPyzxP
AzAyAx
AAAAAAAAAAAA
zyxzzyyxx
CjzjzjzCjyjyjy
CjzjzjzCjxjxjx
CCzCzCzCCyCyCy
CCzCzCzCCxCxCx
BBBBBBBBBBBB
jzjxzyjyzxj
CzCxzyCyzxC
BzBxzyByzxB
jx'zyjyjzyxj
Cx'zyCyCzyxC
Bx'zyByBzyxB
jyzxjzyxjxj
CyzxCzyxCxC
ByzxBzyxBxB





































Solving these linear sets of equations, velocity and acceleration of every key point can be calculated.

4. DYNAMICAL ANALYSIS OF MACPHERSON SUSPENSION

Figure 3 represents the MacPherson suspension model used in this paper. The suspension system is
attached to the vehicle body at points B
0
, C
0
and A
0
. The strut/shock absorber is composed of a cylindrical
joint with the axis formed by the collinear points C
0
, j
1
, C
1
, a viscoelastic damping element along the strut
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axis, an offset elastic spring between points H
1
and G
1
, and finally, a resisting torque element at point C
0
.
The MacPherson suspension has three degrees of freedom for motions of the wheel, however, two of them
are passive and only one degree of freedom is active. In as much as passive angular velocities of the tie
rod and strut do not directly contribute to the motion of the mechanism as a whole, they are ignored.

Fig. 3. Representation of MacPherson suspension

By means of sets of displacement, velocity and acceleration equations, positions, velocities and
accelerations of points B
1
, j
1
, and C
1
can be determined. Also, using kinematical relations, positions,
velocities and accelerations of mass centers of all links will be found. Subsequently, the forces exerted by
the spring and damper, resisting torque and the road force at point C
0
can be found as follows:
1-Spring force:
Direction: GH, Magnitude:)(
0
SGHk
s

,
0
S: Original length of spring,
s
k: Spring constant
2- Damper force:
Direction:
C
x
u (Along strut axis), Magnitude: )(
2
C
xk
uCd ×

where
2
C is on the wheel assembly
(spindle) and
k
d is damping constant.
3- Resisting torque )(
0
C
R:
Direction:)(
z
C
x
uu , Magnitude:
[
]
).(
zxkC
uurR
co
=, :
k
r Proportionality constant
R
F:
Using the work-energy method the road force will be found. First, kinetic and potential energies of the

system are calculated:
22222
2
1
2
1
2
1
2
1
2
1
)(I)(I)(I)(I)(I)energyKinetic(E
C
z
C
z
C
y
C
y
B
z
B
z
B
y
B
y
A
y
A
y
CCCCBBBBAA
++++= 

PmIII
E
E
z
E
z
E
y
E
y
E
x
E
x
EEEEEE


2
1
)(
2
1
)(
2
1
)(
2
1
222
++++

(13)

where
p is velocity of the wheel center,
j
x
j

and
j
y
j

are x and y components of angular velocity of
i
x
i
I,
iy
i
I and
i
z
i
I are moments of inertia of link i in its body frame x, y and z. In this
equation, links identified as A, B, C and E are the control arm, the tie rod, the strut and spindle, and xyz
C
0
j
1
B
0
G
1
H
1
C
1
P

A
1
B
1
A
0
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represents the local coordinates of each link, while XYZ shows the global coordinate frame.
p

in Eq. (13)
can be found using matrices,
[
]
Spindle
D and
[
]
Spindle
D

, as:

[
]
[
]
TT
SpindleSpindle
ppDD
21

= (14)
Also, potential energy is
)(
2
1
2
0
2
SGHkgzmU
srE
+= (15)

where m
E
is spindle mass. On the other hand, the work done on the system is equal to

ZCZCYCYCXCXCrR
C
CrR
)()R()()R()()R(zFRzFW
oooO


+++=×+×= (16)

where F
R
and
o
C
R are vertical force from the road and internal force acting on joint C
0
, respectively. So,

r
CC
R
z
R)energyPotential(U)energyKinetic(E
F
O
×+
=

(17)

In Eq. (13) the angular velocities of the bodies are expressed in the body local coordinate systems, in
contrast to the velocity matrices which are in terms of angular velocities expressed in the global coordinate
frame. The transformation tensor can be used to make the transformation between a vector expressed in
the global coordinate system V and the same vector expressed in the local coordinate system V

as
[
]
TT
VVL

=, where [L] is the second order transformation tensor for the body, [17, 18].

For example, the transformation tensor for body A (Control arm) is[ ]





=
AAA
AAA
AAA
z
Z
z
Y
z
X
y
Z
y
Y
y
X
x
Z
x
Y
x
X
A
uuu
uuu
uuu
L

where ),,(
AAAA
x
Z
x
Y
x
X
x
uuuu is a unit vector along the direction of
A
x (in the global coordinate system). For
the displaced positions of the mechanism, first the new direction of the unit vector should be calculated
from the rotational part of the displacement matrix as
TT
D
uuR
2
1
=, where
D
R is found from Eq. (2).

5. ANALYSIS OF JOINT REACTION FORCES AND MOMENTS

Figure 4 shows the joint reaction forces and moments for the MacPherson suspension.
Unknown reaction forces and moments are:
Control Arm, Body A

At revolute joint, A
0
:
AAAAA
z
A
x
A
z
A
y
A
x
A
MMFFF
00000
,,,,; at spherical joint, A:
AAA
z
A
y
A
x
A
FFF,,
where F and M indicate force and moment, respectively. Subscripts refer to the joint location and
superscripts refer to the vector direction.
Tie-rod, Body B

At spherical joint, B
0
:
BBB
z
B
y
B
x
B
FFF
000
,,; at spherical joint, B:
BBB
z
B
y
B
x
B
FFF,,
Strut, Body C

At spherical joint, C
0
:
CCC
z
C
y
C
x
C
FFF
000
,,; at cylindrical joint, C:
CC
z
C
y
C
FF, and J:
CC
z
j
y
j
FF,
There are a total of 21 unknown joint forces and moments. Since the motion of each body is already
known, all equations are linear in terms of unknowns. The method chosen here is to formulate Newtons
equations in the global coordinate system; the inverse of transformation tensor
1
][

L may be used to
transform a vector from a local coordinate system to the global coordinate system. Eulers equations are
formulated in the principal axes of the local coordinate system of the body. The 16 unknown forces and
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moments at joints, A, B
0
, B, C
0
, and C are determined, first by formulating the Newton-Euler equations for
bodies B, C and E in terms of these 16 unknowns.

Fig. 4. Joint reaction forces and moments for
MacPherson suspension

Strut

Spindle
Control Arm

Tie Rod
B
B
0
B
z
A
F
0

B
y
B
F
0

B
z
B
F
0

B
x
B
F

B
y
B
F

B
z
B
F

A
z
A
F
0

A
0
A
A
y
A
F
0

A
x
A
F
0

A
x
A
M
0
A
z
A
M
0
A
z
A
F

A
x
A
F

A
y
A
F

C
z
C
F
0

C
y
C
F

C
x
C
F
0

S
F

S
F

C
z
j
F
1

C
y
j
F
1

C
z
C
F
1

D
F

C
y
C
F
1

D
F

Rc
0
R
F

A
y
A
F

A
x
A
F

A
z
A
F

B
x
B
F

B
z
B
F

B
y
B
F

C
z
C
F
1

C
y
j
F
1

C
z
j
F
1

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Body B:

[ ]






=






+






Bm
Bm
Bm
BB
z
y
x
z
y
x
Z
Y
X
Lm
F
F
F
F
F
F
B
B
B
B
B
B



0
0
0
and
BBBBB
HFrFr

=+
00
(y, z components) (18)

where m
B
is tie rod mass,
B
j
F and
0
B
j
F are j components of internal forces acting at joints B and B
0
,
respectively. zyx ,, are acceleration components of center of mass of link BB
0
.
j
r and
i
H

are the
position vector of the corresponding point and derivative of angular momentum of the body around the
corresponding point.
Next, Newton-Euler equations for body A is used to determine the remaining five unknowns at joint
A
0

[
]
mAAAA
ALmFF

=+
0
(x, y, z components) and
AAAA
HMTT

=++
00
(x, z components) (19)

where M
A0
is the joint reaction torque for the revolute joint at point A
0
. F and T stand for acting forces and
torques, respectively.
6. SIMULATION OF THE MACPHERSON SUSPENSION MECHANISM

Simulation results of kinematical and dynamical analysis are presented under a sinusoidal vertical
displacement of the wheel and an amplitude of 80 mm (a typical road bump) is assumed for this
displacement. The initial positions for the key points of the MacPherson mechanism (all coordinates are in
mm) and system parameters are presented as follows (these characteristics are obtained from the R&D
office of Mehvar sazan -e- Iran Khodrow for Peugout 405 GLX):
A
1
= (503.46, 687.25, 191.5), C
1
= (525.478, 583.66, 403.968), D= (789.02, 379.5, 259.37),
j
1
= (527.327, 580.3961, 500), E= (501.17, 379, 243.67), p
1
= (509.23, 748.94, 279.66)
B
0
= (670.85, 312.5, 341), G
1
= (529.217, 609.579, 605.701), B
1
= (639.178, 664.471, 292.165),
H
1
= (530, 575, 850), C
0
= (534.42, 567.875, 868.4)
System parameters:
Spring constant:
mmN.k
s
615
=
, Original length of spring, mmS 245
0
=

Damping constant:
mmsNd
k
.12
=
along strut axis, Initial velocity:
smmp 0.0
1
=

o
C
R is a resisting torque acting on the strut and is proportional to the angular displacement of the strut
from the z axis with proportionality constant
k
.90
=

)
5
sin(80 tz
r

= mm acting in the vertical Z direction
All local coordinate systems are taken along the principal axes. Also, local coordinates and dynamical

Mass, m
A
=1.25 kg, Center of mass location: )0,0,150(),,(
1111
=
AAAm
zyxA
Local coordinate system (x
A
, y
A
, z
A
):
1010
1
AA)AA(x
A
=,
=
1
A
y along vector DE,
1
11
AAA
yxz

=

Moments of inertia at A
m1
:
2
500 mmkgI xA ×=,
2
2000 mmkgI yA ×=,
2
2500 mmkgI zA ×=

Mass, m
B
=1.4 kg, Center of mass location: )0,0,180(),,(
1111
=
BBBm
zyxB
Local coordinate system (x
B
, y
B
, z
B
) :
1010
1
BB)BB(x
B
=, 6276)0956903,0.0.0957787,(
1
=
B
y,
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1
11
BBB
yxz

=
. Moments of inertia at B
m1
:
2
500 mmkgI xB ×=,
2
2000 mmkgI yB ×=,
2
2500 mmkgI zB ×=

Mass, m
C
=1.30 kg, Center of mass location: )0,0,230(),,(
1111
=
CCCm
zyxC
Local coordinate system (x
C
, y
C
, z
C
):
1010
1
CC)CC(x
C
=,
=
1
C
y through point H,
1
11
CCC
yxz

=

Moments of inertia at C
m1
:
2
500 mmkgI xC ×=,
2
3000 mmkgI yC ×=,
2
3500 mmkgI zC ×=
Spindle

Mass, m
E
=17.00 kg; Center of mass location: )0,0,0(),,(
1111
=
EEEm
zyxE
Local coordinate system (x
E
, y
E
, z
E
):
=
1
E
x global X,
=
1
E
y global Y ,
=
1
E
z global Z
Moments of inertia at E
m1
:
2
12500 mmkgI
xE
×=,
2
21000 mmkgI
yE
×=,
2
21000 mmkgI
zE
×=
Alterations of camber angle, caster angle, toe angle and track are indicated in Figs. 5 to 8,
respectively. Damping and spring forces are calculated and their time variations and magnitudes are
shown in Figs. 9 and 10. Internal forces at joints A
1
and B
1
are shown in Figs. 11 to 16.

Fig. 5. Camber angle alteration versus
vertical displacement of wheel

Fig. 6. Caster angle alteration versus vertical
displacement of wheel

Fig. 7. Toe angle alteration versus vertical
displacement of wheel

Fig. 8. Track alteration versus vertical
displacement of wheel
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Fig. 9. Damping force versus time

Fig. 10. Spring force versus time

Fig. 11. x component of joint A
1
force versus time

Fig. 12. y component of joint A
1
force versus time

Fig. 13. z component of joint A
1
force versus time

Fig. 14. x component of joint B
1
force versus time

Fig. 15. y component force of joint B
1
versus time
Fig. 16. z component force of joint B
1
versus time

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Reimpell et al. have performed a comprehensive kinematical analysis for different types of
MacPherson suspension [13]. For example, they studied the track behavior of MacPherson strut
suspension for Opel and Audi, camber angle alterations of this suspension for a 3-series BMW and a
Mercedes, and caster angle alterations for Fiat Uno, VW Polo and a Mercedes. In addition, toe angle
alterations have been investigated for three separate cases which are different in their control arm length.
The range of alterations of each parameter for bounce and rebounce of one wheel is shown in Tables 1-4.
Comparing results presented in the tables, it is seen that there is a reasonable agreement between the
results presented in this paper and those of reference [13], and the trend of alteration of each kinematical
parameter is in accordance with the corresponding figures of the reference [13]. Also, for every
kinematical parameter, the range of alterations is acceptable. On the other hand, spring force and damper
force are proportional to displacement and velocity, respectively; as shown in Figs. 9 and 10. Because tie
rod is a spherical-spherical link, it can merely withstand force along x local direction and this is shown in
Figs. 14 to 16.

Table 1. Maximum track alteration (mm) for different types of MacPherson suspension

Opel Audi Studied Case
Bounce 3 2 3
Rebounce 11 14 24

Table 2. Maximum camber angle alteration (degree) for different types of MacPherson suspension

BMW Mercedes Studied Case
Bounce 1.3 1.2 0.5
Rebounce 1.7 2 2

Table 3. Maximum caster angle alteration (degree) for different types of MacPherson suspension

Fiat Uno Mercedes Studied Case
Bounce 0.1 2.2 0.5
Rebounce 0.2 1.2 0.8

Table 4. Maximum toe angle alteration (degree) for different types of MacPherson suspension

Case 1 Case 2 Studied Case
Bounce 1 0.6 0.18
Rebounce 1 1 0.7

7. CONCLUSION

All kinematical and dynamical relations for MacPherson suspension are derived. The displacement matrix
method has been employed as a general tool for simulating MacPherson mechanism. It requires no
commercial software but simply a non-linear algebraic equation solver. Kinematical parameters such as
camber angle, caster angle, toe angle, King-pin angle and, track alteration are investigated and their
variations are obtained. External forces such as damping and spring forces are calculated and their time
variations and their magnitudes are indicated. Internal forces, which are very important in the stress
analysis of the mechanism, are derived by the Newton-Euler method. Specifically, internal forces at joints
A
1
and B
1
are found.

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