车辆系统动力学仿真大作业(带程序)

Assignment

Vehicle system dynamics simulation

学院：机电学院

专业：机械工程及自动化

姓名：

指导教师：

The model we are going to analys:

The FBD of the suspension system is shown as follow:

According to the New's second Law, we can get the equation:

2

)()(221211mg

z z c z z k z m --+-=∙∙∙∙

221212)()(z k mg z z c z z k z m w +-----=∙

∙∙∙

0)()()()(222111222111=-++--+-++--+∙

∙

∙

∙

∙

∙

∙

∙w w w w z L z k z L z k z L z c z L z c z m χχχχ

0)()()()(2222111122221111=-++----++---∙

∙

∙

∙

∙

∙

∙

∙w w w w z L z L k z L z L k z L z L c z L z L c J χχχχχ

d w w w w Q z L z k z L z c z m ,111111111)()(-=------∙

∙

∙

∙

∙χχ

d w w w w Q z L z k z L z c z m ,222222222)()(-=-+--+-∙

∙∙∙∙χχ

When there is no excitation we can get the equation:

2)()(221211mg z z c z z k z m --+-=∙∙∙∙

2

21212)()(z k mg z z c z z k z m w +-----=∙

∙∙∙

Then we substitude the data into the equation, we write a procedure to simulate the system:

Date:

⎪⎪⎪⎪

⎩⎪

⎪⎪⎪

⎨⎧==⋅==⋅===MN/m

0.10k

m 25.1s/m kN 0.20MN/m 0.1m kg 3020kg

2100kg 3250w 2l c k I m m by w b

Program :

1.For the no excitation situation.

Buid a file named rigid1.m

function dy=rigid1(t,y)

dy=zeros(4,1);

dy(1)=y(2);

dy(3)=y(4);

dy(2)=-(615.38*y(1)-615.38*y(3)+12.31*dy(1)-12.31*dy(3)-9.81);

dy(4)=-(5238.1*y(3)-476.19*y(1)+9.52*dy(3)-9.52*dy(1)-9.81);

Buid an other file named test1.m

[T,Y]=ode45('rigid1',[0 1.5],[0 0 0 0]);

figure (1)

plot(T,Y(:,1))

grid

on,xlabel('time(sec)'),ylabel('displayment(m)'),title('displayment of boge')

figure (2)

plot(T,Y(:,2))

grid on,xlabel('time(sec)'),ylabel('velocity(m/s)'),title('velocity of boge')

figure (3)

plot(T,Y(:,3))

grid

on,xlabel('time(sec)'),ylabel('displayment(m)'),title('displayment of wheel')

figure (4)

plot(T,Y(:,4))

grid on,xlabel('time(sec)'),ylabel('velocity(m/s)'),title('velocity of wheel')

We can get the figures as follows: The velocity of m1:

Dumping, then it goes to zero. The velocity of m2 and m3:

Dumping, then goes to zero.

The displacement of m1:

Dump to a constant.

The displacement of m2 and m3:

Dump to a constant.

When there are excitations, according to Newton's second Law and momentum equation we can get

0)()()()(32321=-++--+-++--+∙

∙∙∙∙∙∙∙z l z k z l z k z l z c z l z c z m χχχχ

0)()()()(3232=-++----++---∙

∙

∙

∙

∙

∙

∙

∙z l z kl z l z kl z l z cl z l z l c J χχχχχ

d Q z l z k z l z c z m ,222)()(22-=------∙

∙

∙

∙

∙χχ

d Q z l z k z l z c z m ,33333)()(-=-+--+-∙

∙∙∙∙χχ )(22,2x z k Q w d -= )

(33,3x z k Q w d -=

There are four excitations, each wheel set get one excitation. X1=0.005sin(4πt) X2=0.005sin(8πt) X1 and X2 act on boge 1 X3=0.005sin(4πt) X4=0.005sin(6πt) X1 and X2 act on boge 2