01.05.2017 Views

4569846498

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

136 Multibody Systems Approach to Vehicle Dynamics<br />

⎡ 1 1 1 1<br />

⎢<br />

AG BG CG DG<br />

⎢<br />

y y y y<br />

⎢AG BG CG DG<br />

⎢<br />

⎣ ? ? ? ?<br />

x x x x<br />

⎤ ⎡F<br />

⎥ ⎢<br />

F<br />

⎥ ⎢<br />

⎥ ⎢F<br />

⎥ ⎢<br />

⎦ ⎣F<br />

⎤ ⎡mg<br />

2 ⎤<br />

⎥ ⎢<br />

0<br />

⎥<br />

⎥ ⎢ ⎥<br />

⎥ ⎢ 0 ⎥<br />

⎥ ⎢ ⎥<br />

⎦ ⎣ ? ⎦<br />

(4.10)<br />

If a modified formulation is adopted, the problem can be posed more completely.<br />

If the rigid platform is presumed to be sprung in the manner of a<br />

normal road vehicle with an effective wheel rate, k A , k B , k C and k D , at A, B,<br />

C and D then a new formulation is possible. Writing A 2z , B 2z , C 2z and D 2z<br />

for the height of corners A, B, C and D on Body 2 and A 1z , B 1z , C 1z and D 1z<br />

for the ground height corners at A, B, C and D we can then define a preload<br />

Fp A , Fp B , Fp C and Fp D on each spring such that if the z co-ordinates of<br />

Body 2 at the corner are equal to the z co-ordinates of the ground, Body 1,<br />

at each corner then the spring load is equal to the preload. This leads to the<br />

following equations for the spring force at each corner:<br />

F Az k A (A 2z A 1z ) Fp A (4.11)<br />

F Bz k B (B 2z B 1z ) Fp B (4.12)<br />

F Cz k C (C 2z C 1z ) Fp C (4.13)<br />

F Dz k D (D 2z D 1z ) Fp D (4.14)<br />

At first sight it would appear this is simply a more elaborate formulation of<br />

the previous problem; there are still four unknown quantities (the corner<br />

heights of Body 2) and three equations (4.4), (4.8) and (4.9). However, consideration<br />

of the rigid platform yields a fourth relationship:<br />

D 2z C 2z (B 2z A 2z ) (4.15)<br />

This leads to, after some manipulation of (4.10) substituting in the spring<br />

equations (4.11–4.13) for F Az , F Bz , F Cz and relying on (4.15) to solve for<br />

F Dz , the equations in (4.16):<br />

⎡ kA kD kB kD kC kD<br />

⎤ ⎡A2<br />

z ⎤ ⎡1<br />

⎤<br />

⎢<br />

kAAGy kDDGy kBBGy kDDGy kCCGy k DG<br />

⎥ ⎢<br />

B<br />

⎥<br />

⎢<br />

D y⎥<br />

⎢ 2z<br />

⎥<br />

<br />

⎢<br />

⎢<br />

<br />

⎥<br />

2 ⎥<br />

⎣⎢<br />

kAAGx kDDGx kBBGx kDDGx kCCGx kDDGx⎦⎥<br />

⎣⎢<br />

C2<br />

z ⎦⎥<br />

⎣⎢<br />

3<br />

⎦⎥<br />

(4.16)<br />

where<br />

1 m 2 g k A A 1z Fp A k B B 1z Fp B k C C 1z Fp C k D D 1z Fp D<br />

(4.17)<br />

2 (k A A 1z Fp A )AG y (k B B 1z Fp B )BG y (k C C 1z Fp C )CG y<br />

(k D D 1z Fp D )DG y (4.18)<br />

3 (k A A 1z Fp A )AG x (k B B 1z Fp B )BG x (k C C 1z Fp C )CG x<br />

(k D D 1z Fp D )DG x (4.19)<br />

The equations in (4.16) may be solved by Gaussian elimination and substitution.<br />

It can be seen that preloads (Fp A , Fp B , Fp C and Fp D ), the ground<br />

Az<br />

Bz<br />

Cz<br />

Dz

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!