4569846498

Modelling and analysis of suspension systems 213
Equation 2 9f vs 436 6x 3 6z 435.157 (4.160)
Equation 3 436f vs 9 6x 3 6y 1979.499 (4.161)
This leaves us with four unknowns, 6x , 6y , 6z and f vs , but only three
equations. We can use the same approach here as used with the tie rod in
the preceding analysis. Since the spin degree of freedom of Body 6 about
the axis C–I has no bearing on the overall solution we can again use the
vector dot product to enforce perpendicularity of { 6 } 1 to {R CI } 1 as shown
in Figure 4.71. This will yield the fourth equation as follows:
{ 6 } 1 • {R CI } 1 0 (4.162)
⎡ 3 ⎤
[ 6 6 6
]
⎢
9
⎥
x y z 0mm/s
⎢ ⎥
⎣⎢
436⎦⎥
(4.163)
Equation 4 3 6x 9 6y 436 6z 0 (4.164)
The four equations can now be set up in matrix form ready for solution.
The solution of a four by four matrix will require a lengthy calculation or
access as before to a program that offers the capability to invert the matrix:
⎡ 3 0 436 9
⎤ ⎡ f vs ⎤ ⎡ 120.555 ⎤
⎢
9 436 0 3
⎥ ⎢
⎥ ⎢
6 435.157
⎥
⎢
⎥ ⎢
x
⎥ ⎢ ⎥
(4.165)
⎢436 9 3 0 ⎥ ⎢6y
⎥ ⎢1979.499⎥
⎢
⎥ ⎢ ⎥ ⎢ ⎥
⎣ 0 3 9 436⎦
⎣6z
⎦ ⎣ 0 ⎦
Solving equation (4.165) yields the following answers for the four unknowns:
f vs 4.561 s 1
6x 0.904 rad/s
6y 0.245 rad/s
6z 1.159 rad/s
This gives us the last two angular velocity vectors upper and lower damper
bodies:
T
{ 6}
1
[ 0.904 0.245 1.159 ] rad/s
T
{ 7}
1
[ 0.904 0.245 1.159 ] rad/s
From equation (4.151) we now have
⎡ 3 ⎤ ⎡ 13.682 ⎤
VC 6C7 1
fvs RCI
4.561
⎢
9
⎥
⎢
41.045
⎥
mm/s
1 ⎢ ⎥ ⎢ ⎥
⎣⎢
436⎦⎥
⎣⎢
1988.378⎦⎥
{ } { }
(4.166)
Since the velocity vector {V C6C7 } 1 acts along the axis of the strut C–I the
magnitude of this vector will be equal to the sliding velocity V s :
V s | V C6C7 | 1988.641 mm/s (4.167)