4569846498

Modelling and analysis of suspension systems 217
{ } { }
f R f
3 1 3 AB 1 3
⎡230⎤
⎢
0
⎥
⎢ ⎥
⎣⎢
14 ⎦⎥
rad/s
2
(4.179)
⎡A
⎢
⎢
A
⎣⎢
A
Dx
Dy
Dz
⎤ ⎡ 0 0.855 0 ⎤ ⎡203.345⎤
⎥ ⎢
⎥
⎥
0.855 0 14.039
⎢
112.260
⎥
f
⎢
⎥ ⎢ ⎥
⎦⎥
⎣⎢
0 14.039 0 ⎦⎥
⎣⎢
3355.321⎦⎥
⎡A
⎢
⎢
A
⎣⎢
A
Dx
Dy
Dz
⎤ ⎡ 96.030
⎤ ⎡3346
f
⎥ ⎢
⎥
47 281.651
⎥
⎢
f
⎢
⎥ ⎢
1840
⎦⎥
⎣⎢
1576.804 ⎦⎥
⎣⎢
54 970 f
3
⎤
⎥
⎥
mm/s
⎦⎥
(4.180)
(4.181)
Determining an expression for the relative acceleration {A DG } 1 of point D
relative to point G using values for { 4 } 1 and {V DG } 1 found from the earlier
velocity analysis gives
{A DG } 1 { 4 } 1 {V DG } 1 { 4 } 1 {R DG } 1 (4.182)
⎡A
⎢
⎢
A
⎣⎢
A
DGx
DGy
DGz
4z
4y
4z
4x
4y
4x
⎡ADGx
⎤ ⎡ 16.851 ⎤ ⎡ 31 4z
2164y
⎤
⎢
A
⎥ ⎢ ⎥ ⎢
⎥ 2
(4.184)
⎢ DGy ⎥
0.139
⎢ ⎥ ⎢19 4z
2164x⎥
mm/s
⎣⎢
ADGz
⎦⎥
⎣⎢
193.122⎦⎥
⎢
⎣
19 ⎥
4y
314x
⎦
We can now apply the triangle law of vector addition to equate the expression
for {A DG } 1 in equation (4.184) with {A D } 1 in equation (4.181) and
{A G } 1 in equation (4.177):
{A DG } 1 {A D } 1 {A G } 1 (4.185)
3
3
3
⎤ ⎡ 0 1.446 10 0.945
⎥ ⎢
⎥
⎢1.446 10 0 8.774 10
⎦⎥
⎢
3
⎣
0.945 8.774 10 0
⎡ 0
⎢
⎢
⎢
⎣
⎡ 16.
851 ⎤ ⎡ 31 4z
2164
⎢
0.
139
⎥ ⎢
19
⎢ ⎥ ⎢ z 216
⎣⎢
193.
122⎦⎥
⎢
⎣
194y
314
x
y
0
4 4x
3 3
0
3
⎡ 0 14 0 ⎤ ⎡115⎤
⎢
14 0 230
⎥ ⎢
239
⎥
mm/s
⎢
⎥ ⎢ ⎥
⎣⎢
0 230 0 ⎦⎥
⎣⎢
15
⎦⎥
2
⎤ ⎡19⎤
⎥ ⎢
⎥ 31
⎥
mm/s
⎢ ⎥
2
⎥
⎦ ⎣⎢
216 ⎦⎥
⎤ ⎡ 96.030
⎤ ⎡3346
f3
⎤
⎥
⎢
⎥ 47 281.651
⎥
⎢
1840
⎢
⎥ ⎢
f
⎥
3
⎥
⎥
⎦ ⎣⎢
1576.804 ⎦⎥
⎣⎢
54 970 f3
⎦⎥
⎡ 0 ⎤ ⎡ 0
⎢
41 375.058
⎥
⎢
2070
⎢
⎥ ⎢
f
⎣⎢
1354.093 ⎦⎥
⎣⎢
63 250 f
2
⎤ ⎡204.345⎤
⎥ ⎢
⎥ 1.866
⎥
⎢ ⎥
⎥
⎦ ⎣⎢
17.829
⎦⎥
2
2
(4.183)
⎤
⎥
⎥
mm/s 2
⎦⎥
(4.186)