4569846498

Modelling and analysis of suspension systems 221
Solving equation (4.215) yields the following answers for the 10
unknowns:
f 2 5.126 10 2 rad/mm.s 2
f 3 8.182 10 2 rad/mm.s 2
4x 29.386 rad/s 2
4y 3.737 rad/s 2
4z 20.847 rad/s 2
5x 23.361 rad/s 2
5y 6.466 10 2 rad/s 2
5z 1.843 rad/s 2
A Px 558.211 mm/s 2
A Py 36 163.671 mm/s 2
It is now possible to use the two scale factors, f 2 and f 3 , to calculate the
angular acceleration vectors { 2 } 1 and { 3 } 1 :
⎡230⎤
⎡11.
790⎤
2
{ 2} 1f2{ R } 1 5.
126 10
⎢
EF 0
⎥
⎢
0
⎥
rad/s
⎢ ⎥ ⎢ ⎥
⎣⎢
0 ⎦⎥
⎣⎢
0 ⎦⎥
2
(4.216)
⎡230⎤
⎡18.
819⎤
2
{ 3} 1f3{ R } 8.
182 10
⎢
0
⎥
1
⎢
AB 0
⎥
rad/s
⎢ ⎥ ⎢ ⎥
⎣⎢
14 ⎦⎥
⎣⎢
1.
145 ⎦⎥
(4.217)
In summary the angular acceleration vectors for the rigid bodies are as
follows:
{ 2 } T 1 [11.790 0 0] rad/s 2
{ 3 } T 1 [18.819 0 1.145] rad/s 2
{ 4 } T 1 [29.386 3.737 20.847] rad/s 2
{ 5 } T 1 [23.361 6.466 10 2 1.843] rad/s 2
We can now proceed to calculate the translational accelerations at all the
moving points, C, D, G, H and P, within this part of the model:
{A C } 1 {A CA } 1 { 3 } 1 {V C } 1 { 3 } 1 {R CA } 1 (4.218)
2
⎡A
⎢
⎢
A
⎣⎢
A
Cx
Cy
Cz
⎤ ⎡ 0 0.
855 0 ⎤ ⎡120.
555⎤
⎥ ⎢
⎥ ⎢
⎥
0. 855 0 14.
039 435.
157
⎥
⎢
⎥ ⎢ ⎥
⎦⎥
⎣⎢
0 14.
039 0 ⎦⎥
⎣⎢
1979.
499 ⎦⎥
⎡ 0 1.146 0 ⎤ ⎡115⎤
⎢
1. 146 0 18.
819
⎥ ⎢
141
⎥
mm/
⎢
⎥ ⎢ ⎥
⎣⎢
0 18.
819 0 ⎦⎥
⎣⎢
38
⎦⎥
s 2
(4.219)