4569846498
Modelling and analysis of suspension systems 205 Z 1 {V D } 1 O 1 Y1 X B 1 Body 3 A { 3 } 1 D Fig. 4.69 Angular and translational velocity vectors for the upper wishbone velocities at points within the system. An example of this is shown for the upper wishbone in Figure 4.69. Referring back to the earlier treatment in Chapter 2 we can remind ourselves that as the suspension arm is constrained to rotate about the axis AB, ignoring at this stage any possible deflection due to compliance in the suspension bushes, the vector { 3 } 1 for the angular velocity of Body 3 will act along the axis of rotation through AB. The components of this vector would adopt signs consistent with producing a positive rotation about this axis as shown in Figure 4.69. When setting up the equations to solve a velocity analysis it will be desirable to reduce the number of unknowns based on the knowledge that a particular body is constrained to rotate about a known axis as shown here. The velocity vector { 3 } 1 could, for example, be represented as follows: { 3 } 1 f 3 {R AB } 1 (4.103) Since { 3 } 1 is parallel to the relative position vector {R AB } 1 a scale factor f 3 can be introduced. This reduces the problem from the three unknown components, x 3 , y 3 and z 3 of the vector { 3 } 1 , to a single unknown f 3 . Once the angular velocities of Body 3 have been found it follows that the translational velocity of, for example, point D can be found from {V DA } 1 { 3 } 1 {R DA } 1 (4.104) It also follows that since point A is considered fixed with a velocity {V A } 1 equal to zero that the absolute velocity {V D } 1 of point D can be found from a consideration of the triangle law of vector addition giving {V D } 1 {V DA } 1 (4.105) A consideration of the complete problem indicates that the translational velocities throughout the suspension system can be found if the angular velocities of all the rigid bodies 2, 3, 4 and 5 are known. Clearly the same approach can be taken with the lower wishbone, Body 2, as with the upper wishbone using a single a scale factor f 2 to replace the three unknown components, x 2 , y 2 and z 2 , of the vector { 2 } 1 . Finally a consideration of the boundaries of this problem reveals that while points A, B, E, F and J are fixed the longitudinal velocity, V Px and the lateral velocity V Py at the contact point P remain as unknowns.
206 Multibody Systems Approach to Vehicle Dynamics Thus this analysis can proceed if we can develop 10 equations to solve the 10 unknowns: f 2 , f 3 , x 4 , y 4 , z 4 , x 5 , y 5 , z 5 , V Px , V Py Working through the problem it will be seen that a strategy can be developed using the triangle law of vector addition to generate sets of equations. As a starting point we will develop a set of three equations using {V DG } 1 {V D } 1 {V G } 1 (4.106) In this form the equation (4.106) does not introduce any of the 10 unknowns listed above. It will therefore be necessary to initially define {V DG } 1 , {V D } 1 and {V G } 1 in terms of the angular velocity vectors that contain unknowns requiring solution. The first step in the analysis can therefore proceed as follows. Determine an expression for the velocity {V G } 1 at point G: {V G } 1 {V GE } 1 { 2 } 1 {R GE } 1 (4.107) ⎡230⎤ { 2} 1 f2{ R } 1f ⎢ EF 2 0 ⎥ rad/s ⎢ ⎥ ⎣⎢ 0 ⎦⎥ (4.108) ⎡V ⎢ ⎢ V ⎣⎢ V Gx Gy Gz ⎤ ⎥ ⎥ f ⎦⎥ ⎡0 0 0 ⎤ ⎡115 ⎤ ⎡ 0 ⎤ ⎢ 0 0 230 ⎥ ⎢ 275 ⎥ ⎢ f ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2070 ⎥ mm/s ⎣⎢ 0 230 0 ⎦⎥ ⎣⎢ 9 ⎦⎥ ⎣⎢ 63 250 f2 ⎦⎥ 2 2 (4.109) Determine an expression for the velocity {V D } 1 at point D: {V D } 1 {V DA } 1 { 3 } 1 {R DA } 1 (4.110) { } { } f R f 3 1 3 AB 1 3 ⎡230⎤ ⎢ 0 ⎥ rad/s ⎢ ⎥ ⎣⎢ 14 ⎦⎥ (4.111) ⎡V ⎢ ⎢ V ⎣⎢ V Dx Dy Dz ⎤ ⎥ ⎥ f ⎦⎥ 3 ⎡ 0 14 0 ⎤ ⎡115 ⎤ ⎡3346 f3 ⎤ ⎢ 14 0 230 ⎥ ⎢ 239 ⎥ ⎢ f ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1840 3 ⎥ mm/s ⎣⎢ 0 230 0 ⎦⎥ ⎣⎢ 15 ⎦⎥ ⎣⎢ 54 970 f3 ⎦⎥ (4.112) Determine an expression for the relative velocity {V DG } 1 of point D relative to point G: {V DG } 1 { 4 } 1 {R DG } 1 (4.113) ⎡V ⎢ ⎢ V ⎣⎢ V DGx DGy DGz ⎤ ⎥ ⎥ ⎦⎥ ⎡ 0 ⎢ ⎢ ⎢ ⎣ 4z 4y 4z 4x 0 4y 4x 0 ⎤ ⎡19⎤ ⎡ 31 216 ⎥ ⎢ ⎥ ⎢ ⎥ 31 ⎢ ⎥ ⎢19 216 ⎥ ⎦ ⎣⎢ 216 ⎦⎥ ⎢ ⎣ 19 31 4z 4y 4z 4x 4y 4x ⎤ ⎥ ⎥ mm/s ⎥ ⎦ (4.114)
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206 Multibody Systems Approach to Vehicle Dynamics<br />
Thus this analysis can proceed if we can develop 10 equations to solve the<br />
10 unknowns:<br />
f 2 , f 3 , x 4 , y 4 , z 4 , x 5 , y 5 , z 5 , V Px , V Py<br />
Working through the problem it will be seen that a strategy can be<br />
developed using the triangle law of vector addition to generate sets of equations.<br />
As a starting point we will develop a set of three equations using<br />
{V DG } 1 {V D } 1 {V G } 1 (4.106)<br />
In this form the equation (4.106) does not introduce any of the 10<br />
unknowns listed above. It will therefore be necessary to initially define<br />
{V DG } 1 , {V D } 1 and {V G } 1 in terms of the angular velocity vectors that contain<br />
unknowns requiring solution. The first step in the analysis can therefore<br />
proceed as follows.<br />
Determine an expression for the velocity {V G } 1 at point G:<br />
{V G } 1 {V GE } 1 { 2 } 1 {R GE } 1 (4.107)<br />
⎡230⎤<br />
{ 2} 1 f2{ R } 1f<br />
⎢<br />
EF 2 0<br />
⎥<br />
rad/s<br />
⎢ ⎥<br />
⎣⎢<br />
0 ⎦⎥<br />
(4.108)<br />
⎡V<br />
⎢<br />
⎢<br />
V<br />
⎣⎢<br />
V<br />
Gx<br />
Gy<br />
Gz<br />
⎤<br />
⎥<br />
⎥<br />
f<br />
⎦⎥<br />
⎡0 0 0 ⎤ ⎡115<br />
⎤ ⎡ 0 ⎤<br />
⎢<br />
0 0 230<br />
⎥ ⎢<br />
275<br />
⎥<br />
<br />
⎢<br />
f<br />
⎥<br />
⎢<br />
⎥ ⎢ ⎥ ⎢<br />
2070 ⎥<br />
mm/s<br />
⎣⎢<br />
0 230 0 ⎦⎥<br />
⎣⎢<br />
9<br />
⎦⎥<br />
⎣⎢<br />
63 250 f2<br />
⎦⎥<br />
2 2<br />
(4.109)<br />
Determine an expression for the velocity {V D } 1 at point D:<br />
{V D } 1 {V DA } 1 { 3 } 1 {R DA } 1 (4.110)<br />
{ } { }<br />
f R f<br />
3 1 3 AB 1 3<br />
⎡230⎤<br />
⎢<br />
0<br />
⎥<br />
rad/s<br />
⎢ ⎥<br />
⎣⎢<br />
14 ⎦⎥<br />
(4.111)<br />
⎡V<br />
⎢<br />
⎢<br />
V<br />
⎣⎢<br />
V<br />
Dx<br />
Dy<br />
Dz<br />
⎤<br />
⎥<br />
⎥<br />
f<br />
⎦⎥<br />
3<br />
⎡ 0 14 0 ⎤ ⎡115<br />
⎤ ⎡3346<br />
f3<br />
⎤<br />
⎢<br />
14 0 230<br />
⎥ ⎢<br />
239<br />
⎥<br />
<br />
⎢<br />
f<br />
⎥<br />
⎢<br />
⎥ ⎢ ⎥ ⎢<br />
1840 3<br />
⎥<br />
mm/s<br />
⎣⎢<br />
0 230 0 ⎦⎥<br />
⎣⎢<br />
15<br />
⎦⎥<br />
⎣⎢<br />
54 970 f3<br />
⎦⎥<br />
(4.112)<br />
Determine an expression for the relative velocity {V DG } 1 of point D relative<br />
to point G:<br />
{V DG } 1 { 4 } 1 {R DG } 1 (4.113)<br />
⎡V<br />
⎢<br />
⎢<br />
V<br />
⎣⎢<br />
V<br />
DGx<br />
DGy<br />
DGz<br />
⎤<br />
⎥<br />
⎥<br />
<br />
⎦⎥<br />
⎡ 0<br />
⎢<br />
⎢ <br />
⎢<br />
⎣<br />
<br />
<br />
4z<br />
4y<br />
4z<br />
4x<br />
<br />
0<br />
4y<br />
4x<br />
<br />
<br />
0<br />
⎤ ⎡19⎤<br />
⎡ 31 216<br />
⎥ ⎢ ⎥ ⎢<br />
⎥ 31<br />
<br />
⎢ ⎥ ⎢19 216<br />
⎥<br />
⎦ ⎣⎢<br />
216 ⎦⎥<br />
⎢<br />
⎣<br />
19 31<br />
4z<br />
4y<br />
4z<br />
4x<br />
4y<br />
4x<br />
⎤<br />
⎥<br />
⎥ mm/s<br />
⎥<br />
⎦<br />
(4.114)