Bus Suspension PPT

```Autar Kaw
Humberto Isaza
http://nm.MathForCollege.com
Transforming Numerical Methods Education for STEM Undergraduates
http://nm.MathForCollege.com
An Example to Show How to Reduce Coupled Differential Equations to a
Set of First Order Differential Equations
3
Figure 1: A school bus.
Figure 2: A model of 1/4th of suspension system of bus.
Problem Statement:
A suspension system of a bus can be modeled as above. Only
1/4th of the bus is modeled. The differential equations that
govern the above system can be derived (this is something
you will do in your vibrations course) as
2
M1
2
M
d x2
2
dt
2
d x1
dt
2
dx 2 
 dx
 B1  1 
  K 1  x1  x 2   0
dt 
 dt
(1)
dx 1 
 dx 2
dw 
 dx 2
  K 1  x 2  x1   B 2 
 B1 


  K 2 x 2  w   0
 dt

dt 
dt 
 dt

(2)
x1 ( 0 )  0 , v1 ( 0 )  0 , x 2 ( 0 )  0 , v 2 ( 0 )  0
Where
M1 
body
M  suspension mass
K  spring constant of suspension system
K  spring constant of wheel and tire
B  damping constant of suspension system
B  damping constant of wheel and tire
x  displacement of the body mass as a function of time
x  displacement of the suspension mass as a function of time
w  input profile of the road as a function of time
2
1
2
1
2
1
2
The constants are given as
m 1  2500 kg
m 2  320 kg
K 1  80 , 000 N/m
K 2  500 , 000 N/m
B1  350 N - s/m
B 2  15 , 020 N - s/m
Reduce the simultaneous differential equations (1) and (2) to simultaneous first order
differential equations and put them in the state variable form complete with
corresponding initial conditions.
Solution
Substituting the values of the constants in the two differential equations (1) and (2)
gives the differential equations (3) and (4), respectively.
2
2500
d x1
dt
2
320
d x1
dt
2
2
dx 2 
 dx
 350  1 
  80000  x1  x 2   0
dt
dt


dx 1 
dw 
 dx 2
 dx 2
 350 


  80000  x 2  x1   15020 
  500000 ( x 2  w )  0
dt
dt
dt
dt




(3)
(4)
,
,
Since w is an input, we take it to the right hand side to show it as a forcing function and
rewrite Equation (4) as
2
320
d x2
dt
2
dx 1 
dw
 dx 2
 dx 2 
 350 

 500000 w
  80000  x 2  x1   15020 
  500000 x 2  15020
dt 
dt
 dt
 dt 
(5)
Now let us start the process of reducing the 2 simultaneous differential equations
{Equations (3) and (5)} to 4 simultaneous first order differential equations.
Choose
dx 1
dt
dx 2
dt
 v1
(6)
 v2
(7)
then Equation (3)
2
d x1
2500
dt
2
dx 2 
 dx
 350  1 
  80000  x 1  x 2   0
dt 
 dt
can be written as
2500
dv 1
dt
2500
dv 1
dt
dv 1
dt
 350 v1  v 2   80000  x1  x 2   0
  350 ( v1  v 2 )  80000 ( x1  x 2 )
  0 . 14 v1  v 2   32  x1  x 2 
(8)
and Equation (5)
2
320
d x2
dt
2
dx 1 
dw
 dx 2
 dx 2 
 350 

 500000 w
  80000  x 2  x1   15020 
  500000 x 2  15020
dt 
dt
 dt
 dt 
can be written as
320
dv 2
dt
320
dv 2
dt
dv 2
dt
 350 v 2  v1   80000  x 2  x1   15020 v 2  500000 x 2  15020
dw
 500000 w
dt
  350 ( v 2  v1 )  80000 ( x 2  x1 )  15020 v 2  500000 x 2  15020
  1 . 09375 v 2  v1   250  x 2  x1   46 . 9375 v 2  1562 . 5 x 2  46 . 9375
dw
 500000 w
dt
dw
dt
 1562 . 5 w
(9)
The 4 simultaneous first order differential equations given by Equations 6 thru 9
complete with the corresponding initial condition then are
dx 1
dt
dx 2
dt
dv 1
dt
dv 2
dt
 v1  f 1 t , x1 , x 2 , v1 , v 2 , x1 0   0
(10)
 v 2  f 2 t , x1 , x 2 , v1 , v 2 , x 2 0   0
(11)
  0 . 14 v1  v 2   32  x1  x 2   f 3 t , x1 , x 2 , v1 , v 2 , v1 0   0
(12)
  1 . 09375 v 2  v1   250  x 2  x1   46 . 9375 v 2  1562 . 5 x 2  46 . 9375
 f 4 t , x1 , x 2 , v1 , v 2 ,
v 2 0   0
dw
dt
 1562 . 5 w
(13)
Assuming that the bus is going at 60 mph, that is, approximately 27 m/s, it takes
6m
 0 . 22 s
27 m / s
to go through one period. So the frequency
f 
1
0 . 22
 4.545 Hz
The angular frequency then is
  2    4 . 545
 28 . 6 rad / s
Giving
w  0 . 01 sin  t 
 0 . 01 sin  28 . 6 t 
and
dw
dt
 0 . 286 cos  28 . 6 t 
To put the differential equations given by Equations (10)-(13) in matrix form, we
rewrite them as
dx 1
dt
dx 2
dt
dv 1
dt
dv 2
dt
 v 1  1v 1  0 v 2  0 x 1  0 x 2 , x 1  0   0
(10)
 v 2  0 v 1  1v 2  0 x 1  0 x 2 , x 2  0   0
(11)
  0 . 14 v1  0 . 14 v 2  32 x1  32 x 2 , v1 0   0
(12)
 250 x1  1812 . 5 x 2  1 . 09375 v1  48 . 03125 v 2  1562 . 5 w  46 . 9375
 f 4 t , x1 , x 2 , v1 , v 2 ,
v 2 0   0
dw
dt
(13)
In state variable matrix form, the differential equations are given by
 dx 1
 dt
 dx
 2
 dt
 dv 1
 dt
 dv
2

 dt
where


  0
 
0

   32
 
  250


0
1
0
0
32
 0 . 14
 1812 . 5
1 . 09375
  x1  
  
x
1
 2  
  v1  
0 . 14
  
 48 . 03125   v 2  1562 . 5 w 

0
w  0 . 01 sin( 28 . 6 t )
and the corresponding initial conditions are
 x1 ( 0 )   0 

  
x 2 (0)
0

 
 v1 ( 0 )   0 

  
 v 2 (0)  0 


0


0
dw 
46 . 9375

dt 
0
See how MATLAB ode45 is used to solve such problems