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Modelling and analysis of suspension systems 209
{ 5 } 1
J
J Body 5
UNIVERSAL
H
Vector approach
H
MBS approach
SPHERICAL
Fig. 4.70
Constraining the spin degree of freedom of the tie rod
about its own axis. This degree of freedom has no bearing on the overall
kinematics of the suspension linkage. If there is to be no spin component of
angular velocity parallel to the axis of the tie rod, spin, then the angular velocity
vector { 5 } 1 must be perpendicular to the line H–J as shown in Figure
4.70. Note that this is equivalent to a common practice in MBS modelling
where a universal or Hookes joint is used at one end of the link with a spherical
joint at the other end. The universal joint allows the tie rod to articulate
in a way that does not constrain overall suspension movement but constrains
the spin freedom that would exist if a spherical joint were used at each end.
Using the vector dot product to enforce perpendicularity, as described in
Chapter 2, yields the tenth and final equation required to solve this part of
the problem:
{ 5 } 1 • {R HJ } 1 0 (4.137)
0
[ 5x5y5
z] ⎡ ⎤
⎢
228
⎥
0 mm/s
⎢ ⎥
⎣⎢
8
⎦⎥
(4.138)
Equation 10 228 5y 8 5z 0 (4.139)
The 10 equations can now be set up in matrix form ready for solution. The
solution of a 10 by 10 matrix will require access to a mathematical or
spreadsheet program that offers the capability to invert the matrix.
⎡ 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
⎥ ⎢
V
⎢
⎥ ⎢
⎣⎢
0 0 0 0 0 0 228 8 0 0⎦⎥
⎣
⎢V
2
3
4x
4y
4z
5x
5y
5z
Px
Py
⎤ ⎡ 0 ⎤
⎥ ⎢
0
⎥
⎥ ⎢ ⎥
⎥ ⎢ 0 ⎥
⎥ ⎢ ⎥
⎥ ⎢
0
⎥
⎥ ⎢ 0 ⎥
⎥ ⎢ ⎥
⎥ ⎢ 0 ⎥
⎥ ⎢ 0 ⎥
⎥ ⎢ ⎥
⎥ ⎢ 0 ⎥
⎥ ⎢
⎥
3366
⎥
⎢ ⎥
⎦
⎥ ⎣⎢
0 ⎦⎥
(4.140)