4569846498

Modelling and analysis of suspension systems 239
Referring back to equation (4.329) and the free-body diagram in Figure 4.76
we can see that it is convenient to sum moments of forces acting on Body 2
about the mass centre G 2 to eliminate the inertial force m 2 {A 2 } 1 acting
through the mass centre. In order to carry out the moment balance we will
need to establish new relative position vectors {R EG2 } 1/2 , {R FG2 } 1/2 and
{R GG2 } 1/2 . We will also need to define the vector components in metres for
consistency. Working first in frame O 1 we have:
{R EG2 } T 1 [0.115 0.155 0.005] m
{R FG2 } T 1 [0.115 0.155 0.005] m
{R GG2 } T 1 [0.0 0.120 0.004] m
Applying a vector transformation for the vector {R EG2 } 1 from frame O 1 to
O 2 gives
⎡1 0 0⎤
⎡ 0 1 0⎤
⎡ 0.115 ⎤ ⎡0.155⎤
{ R EG2 } 1 2
⎢
0 0 1
⎥ ⎢
1 0 0
⎥ ⎢
0.155
⎥
⎢
0.005
⎥
m
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣⎢
0 1 0⎦⎥
⎣⎢
0 0 1⎦⎥
⎣⎢
0.005 ⎦⎥
⎣⎢
0.115 ⎦⎥
(4.332)
∑
Applying the same vector transformation to {R FG2 } 1/2 and {R GG2 } 1/2 gives
us the three relative position vectors, referenced to the correct frame O 2 and
in consistent units, needed for the moment balance:
{R EG2 } T 1/2 10 3 [155 5 115] m
{R FG2 } T 1/2 10 3 [155 5 115] m
{R GG2 } T 1/2 10 3 [120 4 0] m
Before writing the rotational equations of motion we can first determine
the moment balance of the constraint forces acting at E, F and G:
{ M } { R } { F } { R } { F } { R } {
F }
G2 12 EG2 12 E21 12 FG2 12 F21 12 GG2 12 G24 12
10
3
⎡ 0 115 5 ⎤ ⎡F
⎢
115 0 155
⎥ ⎢
F
⎢
⎥ ⎢
⎣⎢
5 155 0 ⎦⎥
⎣⎢
F
E21x2
E21y2
E21z2
⎤
⎥
⎥
10
⎦⎥
3
⎡ 0 115
⎢
115 0
⎢
⎣⎢
5 155
5 ⎤ ⎡F
155
⎥ ⎢
⎥ ⎢
F
0 ⎦⎥
⎣⎢
F
F21x2
F21y2
F21z2
⎤
⎥
⎥
⎦⎥
10
3
⎡0 4 0 ⎤ ⎡F
⎢
4 0 120
⎥ ⎢
⎢
⎥ ⎢
F
⎣⎢
0 120 0 ⎦⎥
⎣⎢
F
G24x2
G24y2
G24z2
⎤
⎥
⎥
Nm
⎦⎥
(4.333)
Considering next the rotational inertial terms we have:
∑ { M } [ I ] { 2} [ 2] [ I ] { 2}
G2 12 2 22 12 12 2 22 12
(4.334)