4569846498
Modelling and analysis of suspension systems 239 Referring back to equation (4.329) and the free-body diagram in Figure 4.76 we can see that it is convenient to sum moments of forces acting on Body 2 about the mass centre G 2 to eliminate the inertial force m 2 {A 2 } 1 acting through the mass centre. In order to carry out the moment balance we will need to establish new relative position vectors {R EG2 } 1/2 , {R FG2 } 1/2 and {R GG2 } 1/2 . We will also need to define the vector components in metres for consistency. Working first in frame O 1 we have: {R EG2 } T 1 [0.115 0.155 0.005] m {R FG2 } T 1 [0.115 0.155 0.005] m {R GG2 } T 1 [0.0 0.120 0.004] m Applying a vector transformation for the vector {R EG2 } 1 from frame O 1 to O 2 gives ⎡1 0 0⎤ ⎡ 0 1 0⎤ ⎡ 0.115 ⎤ ⎡0.155⎤ { R EG2 } 1 2 ⎢ 0 0 1 ⎥ ⎢ 1 0 0 ⎥ ⎢ 0.155 ⎥ ⎢ 0.005 ⎥ m ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ 0 1 0⎦⎥ ⎣⎢ 0 0 1⎦⎥ ⎣⎢ 0.005 ⎦⎥ ⎣⎢ 0.115 ⎦⎥ (4.332) ∑ Applying the same vector transformation to {R FG2 } 1/2 and {R GG2 } 1/2 gives us the three relative position vectors, referenced to the correct frame O 2 and in consistent units, needed for the moment balance: {R EG2 } T 1/2 10 3 [155 5 115] m {R FG2 } T 1/2 10 3 [155 5 115] m {R GG2 } T 1/2 10 3 [120 4 0] m Before writing the rotational equations of motion we can first determine the moment balance of the constraint forces acting at E, F and G: { M } { R } { F } { R } { F } { R } { F } G2 12 EG2 12 E21 12 FG2 12 F21 12 GG2 12 G24 12 10 3 ⎡ 0 115 5 ⎤ ⎡F ⎢ 115 0 155 ⎥ ⎢ F ⎢ ⎥ ⎢ ⎣⎢ 5 155 0 ⎦⎥ ⎣⎢ F E21x2 E21y2 E21z2 ⎤ ⎥ ⎥ 10 ⎦⎥ 3 ⎡ 0 115 ⎢ 115 0 ⎢ ⎣⎢ 5 155 5 ⎤ ⎡F 155 ⎥ ⎢ ⎥ ⎢ F 0 ⎦⎥ ⎣⎢ F F21x2 F21y2 F21z2 ⎤ ⎥ ⎥ ⎦⎥ 10 3 ⎡0 4 0 ⎤ ⎡F ⎢ 4 0 120 ⎥ ⎢ ⎢ ⎥ ⎢ F ⎣⎢ 0 120 0 ⎦⎥ ⎣⎢ F G24x2 G24y2 G24z2 ⎤ ⎥ ⎥ Nm ⎦⎥ (4.333) Considering next the rotational inertial terms we have: ∑ { M } [ I ] { 2} [ 2] [ I ] { 2} G2 12 2 22 12 12 2 22 12 (4.334)
240 Multibody Systems Approach to Vehicle Dynamics ∑ { M } G2 1 2 ⎡I2xx 0 0 ⎤ ⎡2 x2 ⎤ ⎢ 0 I2yy 0 ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ 2y2 ⎥ ⎣⎢ 0 0 I2zz ⎦⎥ ⎣⎢ 2z2 ⎦⎥ ⎡ 0 ⎢ ⎢ 0 ⎢ ⎣ 2y2 2x2 0 2z2 2y2 2z2 2x2 3 ⎡1.5 10 0 0 ⎢ 3 ⎢ 0 38 10 0 ⎢ ⎣ 0 0 3810 ⎤ ⎡I2xx 0 0 ⎤ ⎡ ⎥ ⎢ 0 I2yy 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎦ ⎢⎣ 0 0 I2zz ⎦⎥ ⎣⎢ 3 ⎤ ⎡ 0 ⎤ ⎥ ⎢ ⎥ 0 ⎥ ⎢ ⎥ ⎥ ⎦ ⎣⎢ 10.642⎦⎥ ⎤ ⎥ ⎥ Nm ⎦⎥ 3 ⎡ 0 12.266 0⎤ ⎡1.5 10 0 0 ⎢ 12.266 0 0 ⎥ ⎢ 3 0 ⎢ ⎥ ⎢ 38 10 0 ⎣⎢ 0 0 0⎦⎥ ⎢ ⎣ 0 0 38 10 (4.335) Equating (4.333) with (4.335) yields the rotational equations of motion for Body 2: Equation 4 (115F E21y2 5F E21z2 115F F21y2 5F F21z2 4F G24y2 ) 10 3 0 (4.336) Equation 5 (115F E21x2 155F E21z2 115F F21x2 155F F21z2 F G24x2 0.120F G24z2 ) 10 3 0 (4.337) Equation 6 (5F E21x2 155F E21y2 5F F21x2 155F F21y2 120F G24y2 ) 10 3 404.396 10 3 (4.338) At this stage the observant reader will note the inertial terms in (4.335) have only yielded a numerical value for the moment balance about the principal Z 2 -axis. This makes sense as the Z 2 -axis has been chosen to be parallel to the fixed axis of body rotation through points E and F. In the absence of components of angular velocity or acceleration about X 2 and Y 2 equations (4.336) and (4.337) above simplify to a static moment balance. It should also be noted that when rotation is constrained about a single principal axis the right-hand part of (4.334), [ 2 ] 1/2 [I 2 ] 2/2 [ 2 ] 1/2 , is entirely zero to indicate a lack of gyroscopic terms in the absence of rotational coupling. Choosing a body centred axis system for the upper wishbone Body 3, with an axis parallel to an axis through points A and B, would yield a similar formulation. This would not, however, be the case, for example, if we continued to set up the equations for Body 4 where there is no single fixed axis of rotation. Before leaving the area of formulating equations of motion for dynamic analysis we should also ensure that the impression is not given that the use 2x2 2y2 2z2 3 ⎤ ⎡ 0 ⎤ ⎥ ⎢ ⎥ 0 ⎥ Nm ⎢ ⎥ ⎥ ⎦ ⎣⎢ 12.266⎦⎥
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240 Multibody Systems Approach to Vehicle Dynamics<br />
∑<br />
{ M }<br />
G2 1 2<br />
⎡I2xx<br />
0 0 ⎤ ⎡2<br />
x2<br />
⎤<br />
<br />
⎢<br />
0 I2yy<br />
0<br />
⎥ ⎢<br />
⎢<br />
⎥ ⎢<br />
<br />
⎥<br />
2y2<br />
⎥<br />
⎣⎢<br />
0 0 I2zz<br />
⎦⎥<br />
⎣⎢<br />
2z2<br />
⎦⎥<br />
⎡ 0 <br />
⎢<br />
⎢ 0 <br />
⎢<br />
⎣<br />
2y2 2x2<br />
0<br />
2z2 2y2<br />
2z2 2x2<br />
3<br />
⎡1.5 10 0 0<br />
⎢<br />
3<br />
⎢ 0 38<br />
10 0<br />
⎢<br />
⎣<br />
0 0 3810<br />
⎤ ⎡I2xx<br />
0 0 ⎤ ⎡<br />
⎥ ⎢<br />
0 I2yy<br />
0<br />
⎥ ⎢<br />
⎥ ⎢<br />
⎥ ⎢<br />
<br />
⎥<br />
⎦<br />
⎢⎣<br />
0 0 I2zz<br />
⎦⎥<br />
⎣⎢<br />
<br />
3<br />
⎤ ⎡ 0 ⎤<br />
⎥ ⎢<br />
⎥ 0<br />
⎥<br />
⎢ ⎥<br />
⎥<br />
⎦ ⎣⎢<br />
10.642⎦⎥<br />
⎤<br />
⎥<br />
⎥<br />
Nm<br />
⎦⎥<br />
3<br />
⎡ 0 12.266 0⎤<br />
⎡1.5 10<br />
0 0<br />
⎢<br />
12.266 0 0<br />
⎥ ⎢<br />
3<br />
<br />
0<br />
⎢<br />
⎥ ⎢ 38 10 0<br />
⎣⎢<br />
0 0 0⎦⎥<br />
⎢<br />
⎣<br />
0 0 38<br />
10<br />
(4.335)<br />
Equating (4.333) with (4.335) yields the rotational equations of motion for<br />
Body 2:<br />
Equation 4<br />
(115F E21y2 5F E21z2 115F F21y2 5F F21z2 4F G24y2 ) 10 3 0<br />
(4.336)<br />
Equation 5<br />
(115F E21x2 155F E21z2 115F F21x2 155F F21z2<br />
F G24x2 0.120F G24z2 ) 10 3 0 (4.337)<br />
Equation 6<br />
(5F E21x2 155F E21y2 5F F21x2 155F F21y2<br />
120F G24y2 ) 10 3 404.396 10 3 (4.338)<br />
At this stage the observant reader will note the inertial terms in (4.335) have<br />
only yielded a numerical value for the moment balance about the principal<br />
Z 2 -axis. This makes sense as the Z 2 -axis has been chosen to be parallel to the<br />
fixed axis of body rotation through points E and F. In the absence of components<br />
of angular velocity or acceleration about X 2 and Y 2 equations<br />
(4.336) and (4.337) above simplify to a static moment balance. It should<br />
also be noted that when rotation is constrained about a single principal axis<br />
the right-hand part of (4.334), [ 2 ] 1/2 [I 2 ] 2/2 [ 2 ] 1/2 , is entirely zero to indicate<br />
a lack of gyroscopic terms in the absence of rotational coupling. Choosing a<br />
body centred axis system for the upper wishbone Body 3, with an axis parallel<br />
to an axis through points A and B, would yield a similar formulation.<br />
This would not, however, be the case, for example, if we continued to set up<br />
the equations for Body 4 where there is no single fixed axis of rotation.<br />
Before leaving the area of formulating equations of motion for dynamic<br />
analysis we should also ensure that the impression is not given that the use<br />
2x2<br />
2y2<br />
2z2<br />
3<br />
⎤ ⎡ 0 ⎤<br />
⎥ ⎢<br />
⎥ 0<br />
⎥<br />
Nm<br />
⎢ ⎥<br />
⎥<br />
⎦ ⎣⎢<br />
12.266⎦⎥