4569846498

220 Multibody Systems Approach to Vehicle Dynamics
{ A }
P 1
⎡APx
⎤
⎢
A
⎥
Py mm/s
⎢ ⎥
2
⎣⎢
00 . ⎦⎥
(4.206)
We can now apply the triangle law of vector addition to equate the expression
for {A PG } 1 in equation (4.205) with {A P } 1 in equation (4.206) and
{A G } 1 in expression (4.177):
{A PG } 1 {A P } 1 {A G } 1 (4.207)
⎡ 6.
755 ⎤ ⎡584z
1764y⎤
⎡APx
⎤ ⎡ 0 ⎤ ⎡ 0
⎢
0.
303
⎥ ⎢
⎥
7
⎢ ⎥
⎢
⎥
⎢ ⎥ ⎢ 4z
176 4x
⎥ A 41 375.058
⎢
⎢ Py ⎥ ⎢
⎥ ⎢
2070 f
⎣⎢
157. 326⎦⎥
⎢
⎣
7 58
⎥
4y
4x
⎦ ⎣⎢
00 . ⎦⎥
⎣⎢
1354.093 ⎦⎥
⎣⎢
63 250 f
⎤
⎥
⎥
mm/s 2
⎦⎥
(4.208)
Rearranging (4.208) yields the next set of three equations required to solve
the analysis:
Equation 7 176 4y 58 4z A Px 6.755 (4.209)
Equation 8 2070f 2 176 4x 7 4z A Py 41 375.361 (4.210)
Equation 9 63 250f 2 58 4x 7 4y 1511.419 (4.211)
As with the velocity analysis this leaves us with nine equations and 10
unknowns. The last equation is again obtained by constraining the rotation
of the tie rod (Body 5) to prevent spin about its own axis:
{ 5 } 1 • {R HJ } 1 0 (4.212)
⎡ 0 ⎤
[ ]
⎢
5x 5 y 5z
228
⎥
0 mm/s
(4.213)
⎢ ⎥
⎣⎢
8
⎦⎥
Equation 10 228 5y 8 5z 0 (4.214)
The 10 equations can now be set up in matrix form ready for solution:
⎡ 0 3346 0 216 31 0 0 0 0 0⎤
⎡ f
⎢
2070 1840 216 0 19 0 0 0 0 0
⎥ ⎢ f
⎢
⎥ ⎢
⎢ 63 250 54 970 31 19 0 0 0 0 0 0⎥
⎢
⎢
⎥ ⎢
0 3346 0 51 44 0 8 228 0 0
⎢
⎥ ⎢
⎢ 0 1840 51 0 144 8 0 0 0 0⎥
⎢
⎢
⎥ ⎢
⎢ 0 54 970 44 144 0 228 0 0 0 0⎥
⎢
⎢ 0 0 0 176 58 0 0 0 1 0⎥
⎢
⎢
⎥ ⎢
⎢ 2070 0 176 0 7 0 0 0 0 1⎥
⎢
⎢
63 250 0 58 7 0 0 0 0 0 0
⎥ ⎢
A
⎢
⎥ ⎢
⎣⎢
0 0 0 0 0 0 228 8 0 0⎦⎥
⎣
⎢A
2
3
4x
4y
4z
5x
5y
5z
Px
Py
2
2
⎤ ⎡ 112.
881 ⎤
⎥ ⎢
⎥
5906.
732
⎥
⎢
⎥
⎥ ⎢ 415.833
⎥
⎥ ⎢
⎥
⎥ ⎢
32.
229
⎥
⎥ ⎢ 1539.
678 ⎥
⎥ ⎢
⎥
⎥ ⎢ 73.
753 ⎥
⎥ ⎢ 6.
755 ⎥
⎥ ⎢
⎥
⎥ ⎢ 41 375.361⎥
⎥ ⎢
⎥
1511.419
⎥
⎢
⎥
⎦
⎥ ⎣⎢
0 ⎦⎥
(4.215)