4569846498
Modelling and analysis of suspension systems 173 {} [C]{F} (4.62) Expanding equation (4.62) leads to a 12 12 matrix relating the motion of the left and right wheel centres to unit forces and torques applied to the wheel centres. From this perspective, matrix element C i,j is the displacement of system degree of freedom i due to a unit force at degree of freedom j where the degrees of freedom are the three displacements x, y and z and the three rotations Ax, Ay and Az at each of the left and right wheel centres. ⎡x ⎢ y ⎢ ⎢z ⎢ ⎢Ax ⎢Ay ⎢ ⎢ Az ⎢x ⎢ ⎢y ⎢ ⎢ z ⎢Ax ⎢ ⎢Ay ⎣ ⎢Az LW LW LW LW LW LW RW RW RW RW RW RW ⎤ ⎡C C C C C C C C C C C C ⎥ ⎢ C C C C C C C C C C C C ⎥ ⎢ ⎥ ⎢C C C C C C C C C C C C ⎥ ⎢ ⎥ ⎢ C C C C C C C C C C C C ⎥ ⎢C5,1 C5,2 C5,3 C5,4 C 5, C C C C C C C ⎥ ⎢ ⎥ ⎢C C C C C C C C C C C C ⎥ ⎢C C C C C C C C C C C C ⎥ ⎢ ⎥ ⎢C8,1 C8,2 C8,3 C8,4 C8,5 C8,6 C C C C C C ⎥ ⎢ C C C C C C C C C C C C ⎥ ⎢ ⎥ ⎢C C C C C C C C C C C C ⎥ ⎢ ⎥ ⎢C11,1 C11,2 C11,3 C11,4 C11,5 C11,6 C11,7 C11, C C C C ⎦ ⎥ ⎢ ⎣C C C C C C C C C C C C 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 1,10 1,11 1,12 2,1 2,2 2,3 2,4 2,5 2,6 2,7 2,8 2,9 2,10 2,11 2,12 3,1 3,2 3,3 3,4 3,5 3,6 3,7 3,8 3,9 3,10 3,11 3,12 4,1 4,2 4,3 4,4 4,5 4,6 4,7 4,8 4,9 4,10 4,11 4,12 5,5 5,6 5,7 5,8 5,9 5,10 5,11 5,12 6,1 6,2 6,3 6,4 6,5 6,6 6,7 6,8 6,9 6,10 6,11 6,12 7,1 7,2 7,3 7,4 7,5 7,6 7,7 7,8 7,9 7,10 7,11 7,12 8,7 8,8 8,9 8,10 8,11 8,12 9,1 9,2 9,3 9,4 9,5 9,6 9,7 9,8 9,9 9,10 9,11 9,12 10,1 10,2 10,3 10,4 10,5 10,6 10,7 10,8 10,9 10,10 10,11 10,12 11,8 11,9 11,10 11,11 11,12 12,1 12,2 12,3 12,4 12,5 12,6 12,7 12,8 12,9 12,10 12,11 12,12 ⎤ ⎡Fx ⎥ ⎢ Fy ⎥ ⎢ ⎥ ⎢Fz ⎥ ⎢ ⎥ ⎢ Tx ⎥ ⎢Ty ⎥ ⎢ ⎥ ⎢Tz ⎥ ⎢Fx ⎥ ⎢ ⎥ ⎢Fy ⎥ ⎢ ⎥ Fz ⎢ ⎥ ⎢Tx ⎥ ⎢ ⎥ ⎢Ty ⎥ ⎦ ⎣ ⎢Tz (4.63) From equation (4.63) it can be seen that the coefficients on the leading diagonal of matrix [C] directly relate the displacement or rotation to the associated force or torque applied at that degree of freedom. For example, in the absence of any other forces or torques, the vertical motion of the left wheel centre due to a unit vertical force applied at the left wheel centre is given by z LW C 3,3 Fz LW . Figure 4.32 illustrates this for another example where the vertical motion of the left wheel centre due to the application only of a unit vertical force applied at the right wheel centre given by z LW C 3,9 Fz RW . From Figure 4.32 it can be seen that for an independent LW LW LW LW LW LW RW RW RW RW RW RW ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎥ z LW C 3,9 Fz R Right wheel centre Left wheel centre Z X Fz RW Y Fig. 4.32 Application of compliance matrix to suspension system vehicle half model
174 Multibody Systems Approach to Vehicle Dynamics suspension without a roll bar C 3,9 would be zero in the absence of any mechanical coupling between the left and right suspension systems. The other elements of the compliance matrix are defined similarly. As stated the compliance matrix approach is well suited to investigate the effects of suspension movement due to compliance. By way of further example consider the definition used in ADAMS/Car for the calculation of Aligning torque Steer and camber compliance. The aligning torque steer compliance is the change in steer angle due to unit aligning torques applied through the wheel centres. Similarly the aligning torque camber compliance is the change in camber angle due to unit aligning torques acting through the wheel centres. Figure 4.33 illustrates the determination of steer angle resulting at the right wheel due to unit aligning torques acting through both the left and right wheel centres. Note that the usual symbol for steer angle is . For the Left wheel centre Tz LW Right wheel centre Z Tz RW Az RW X Y Fig. 4.33 Steer angle at right wheel due to aligning torques at left and right wheels Ax RW Left wheel centre Tz LW Right wheel centre Z Tz RW X Y Fig. 4.34 Camber angle at right wheel due to aligning torques at left and right wheels
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Modelling and analysis of suspension systems 173<br />
{} [C]{F} (4.62)<br />
Expanding equation (4.62) leads to a 12 12 matrix relating the motion of<br />
the left and right wheel centres to unit forces and torques applied to the<br />
wheel centres. From this perspective, matrix element C i,j is the displacement<br />
of system degree of freedom i due to a unit force at degree of freedom<br />
j where the degrees of freedom are the three displacements x, y and z<br />
and the three rotations Ax, Ay and Az at each of the left and right wheel centres.<br />
⎡x<br />
⎢<br />
y<br />
⎢<br />
⎢z<br />
⎢<br />
⎢Ax<br />
⎢Ay<br />
⎢<br />
⎢ Az<br />
⎢x<br />
⎢<br />
⎢y<br />
⎢<br />
⎢<br />
z<br />
⎢Ax<br />
⎢<br />
⎢Ay<br />
⎣<br />
⎢Az<br />
LW<br />
LW<br />
LW<br />
LW<br />
LW<br />
LW<br />
RW<br />
RW<br />
RW<br />
RW<br />
RW<br />
RW<br />
⎤ ⎡C C C C C C C C C C C C<br />
⎥ ⎢<br />
C C C C C C C C C C C C<br />
⎥ ⎢<br />
⎥ ⎢C C C C C C C C C C C C<br />
⎥ ⎢<br />
⎥ ⎢<br />
C C C C C C C C C C C C<br />
⎥ ⎢C5,1 C5,2 C5,3 C5,4<br />
C 5, C C C C C C C<br />
⎥ ⎢<br />
⎥ ⎢C C C C C C C C C C C C<br />
⎥<br />
⎢C C C C C C C C C C C C<br />
⎥ ⎢<br />
⎥ ⎢C8,1 C8,2 C8,3 C8,4 C8,5 C8,6<br />
C C C C C C<br />
⎥ ⎢<br />
C C C C C C C C C C C C<br />
⎥ ⎢<br />
⎥ ⎢C C C C C C C C C C C C<br />
⎥ ⎢<br />
⎥ ⎢C11,1 C11,2 C11,3 C11,4 C11,5 C11,6 C11,7 C11,<br />
C C C C<br />
⎦<br />
⎥ ⎢<br />
⎣C C C C C C C C C C C C<br />
1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 1,10 1,11 1,12<br />
2,1 2,2 2,3 2,4 2,5 2,6 2,7 2,8 2,9 2,10 2,11 2,12<br />
3,1 3,2 3,3 3,4 3,5 3,6 3,7 3,8 3,9 3,10 3,11 3,12<br />
4,1 4,2 4,3 4,4 4,5 4,6 4,7 4,8 4,9 4,10 4,11 4,12<br />
5,5 5,6 5,7 5,8 5,9 5,10 5,11 5,12<br />
6,1 6,2 6,3 6,4 6,5 6,6 6,7 6,8 6,9 6,10 6,11 6,12<br />
7,1 7,2 7,3 7,4 7,5 7,6 7,7 7,8 7,9 7,10 7,11 7,12<br />
8,7 8,8 8,9 8,10 8,11 8,12<br />
9,1 9,2 9,3 9,4 9,5 9,6 9,7 9,8 9,9 9,10 9,11 9,12<br />
10,1 10,2 10,3 10,4 10,5 10,6 10,7 10,8 10,9 10,10 10,11 10,12<br />
11,8 11,9 11,10 11,11 11,12<br />
12,1 12,2 12,3 12,4 12,5 12,6 12,7 12,8 12,9 12,10 12,11 12,12<br />
⎤ ⎡Fx<br />
⎥ ⎢<br />
Fy<br />
⎥ ⎢<br />
⎥ ⎢Fz<br />
⎥ ⎢<br />
⎥ ⎢<br />
Tx<br />
⎥ ⎢Ty<br />
⎥ ⎢<br />
⎥ ⎢Tz<br />
⎥ ⎢Fx<br />
⎥ ⎢<br />
⎥ ⎢Fy<br />
⎥ ⎢<br />
⎥<br />
Fz<br />
⎢<br />
⎥ ⎢Tx<br />
⎥ ⎢<br />
⎥ ⎢Ty<br />
⎥<br />
⎦ ⎣<br />
⎢Tz<br />
(4.63)<br />
From equation (4.63) it can be seen that the coefficients on the leading<br />
diagonal of matrix [C] directly relate the displacement or rotation to the<br />
associated force or torque applied at that degree of freedom. For example,<br />
in the absence of any other forces or torques, the vertical motion of the left<br />
wheel centre due to a unit vertical force applied at the left wheel centre is<br />
given by z LW C 3,3 Fz LW . Figure 4.32 illustrates this for another example<br />
where the vertical motion of the left wheel centre due to the application<br />
only of a unit vertical force applied at the right wheel centre given by<br />
z LW C 3,9 Fz RW . From Figure 4.32 it can be seen that for an independent<br />
LW<br />
LW<br />
LW<br />
LW<br />
LW<br />
LW<br />
RW<br />
RW<br />
RW<br />
RW<br />
RW<br />
RW<br />
⎤<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎦<br />
⎥<br />
z LW C 3,9 Fz R<br />
Right wheel<br />
centre<br />
Left wheel<br />
centre<br />
Z<br />
X<br />
Fz RW<br />
Y<br />
Fig. 4.32 Application of compliance matrix to suspension system vehicle<br />
half model