4569846498

Modelling and analysis of suspension systems 237
where
{V G2 } 1 {V G2E } 1 { 2 } 1 {R G2E } 1 (4.316)
⎡V
⎢
⎢
V
⎣⎢
V
therefore
⎡A
⎢
⎢
A
⎣⎢
A
G2x
G2y
G2z
G2x
G2y
G2z
⎤ ⎡0 0 0 ⎤ ⎡115⎤
⎡ 0 ⎤
⎥ ⎢
⎥ ⎢
⎥
0 0 12.266
155
⎥
⎢
61.33
⎥
mm/s
⎢
⎥ ⎢ ⎥ ⎢ ⎥
⎦⎥
⎣⎢
0 12.266 0 ⎦⎥
⎣⎢
5
⎦⎥
⎣⎢
1901.23⎦⎥
⎤ ⎡0
0 0 ⎤ ⎡ 0.0 ⎤
⎥
⎥
⎢
0 0 12.266
⎥ ⎢
61.33
⎥
⎢
⎥ ⎢ ⎥
⎦⎥
⎣⎢
0 12.
266 0 ⎦⎥
⎣⎢
1901.23⎦⎥
⎡0
0 0 ⎤ ⎡115⎤
⎢
0 0 10.
642
⎥ ⎢
155
⎥
mm/s
⎢
⎥ ⎢ ⎥
2
⎣⎢
0 10.642
0 ⎦⎥
⎣⎢
5
⎦⎥
(4.317)
(4.318)
⎡A
⎢
⎢
A
⎣⎢
A
(4.319)
Before progressing further it is important to refer back to Chapter 3 and
state that Newton’s second law is only applicable for a consistent set of
units. In effect this means either converting from millimetres to metres
before carrying out the dynamic analysis or incorporating a units consistency
factor UCF in the equations of motion:
∑
(4.320)
Since our current dimensions for length are in mm and we need to work in
SI our UCF value here is 1000. So converting {A G2 } 1 to m/s 2 gives
⎡AG2
x⎤
⎡ 0 ⎤
⎢
A
⎥
⎢ G2y⎥
⎢
23.373
⎥
m/s (4.321)
⎢ ⎥
2
⎣⎢
AG2
z⎦⎥
⎣⎢
0.869
⎦⎥
We also need to define a vector {g} 1 for gravitational acceleration which
for the reference frame O 1 used here would be
⎡gx
⎤ ⎡ 0 ⎤
⎢
⎢
g
⎥
y ⎥
⎢
0
⎥
m/s
⎢ ⎥
2
⎣⎢
gz
⎦⎥
⎣⎢
9.81⎦⎥
For Body 2 summing forces and applying Newton’s second law gives
∑
G2x
G2y
G2z
⎤ ⎡ 0 ⎤
⎥
⎥
⎢
23 372.797
⎥
mm/s
⎢
⎥
2
⎦⎥
⎣⎢
869.336
⎦⎥
{ F }
2 1
m { AG
}
UCF
2 2 1
{ F 2 } m 2 { A G }
1 2 1
(4.322)
(4.323)