Two DOF vehicle response analysis

8/2/2019 Two DOF vehicle response analysis

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DCU Faculty of Engineering & Computing.

School of Mechanical and Manufacturing Engineering

Two DOF vehicle

responseMM401 Mechanical Engineering System Simulation

Neville Lawless

10212298



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Contents1. Problem Constants .......................................................................................................................... 1

2. Free body diagram .......................................................................................................................... 2

3. Global problem equations .............................................................................................................. 4

4. Results ............................................................................................................................................. 5

1. Problem Constants

Vehicle mass (m) = 9.2tonnes = 9.2x103kg

Mass distribution = 37:63 (front: rear)

Total length of vehicle = 9.2 metres

Wheelbase = 5.6 metres

Vehicle velocity (V) = 45k.p.h = 12.5m/s

Wavelength 1 (λ1) = 4 metres

Wavelength 2 (λ2) = 6 metres

Stiffness of springs (k) = Front (Kf ) = 800 kN/m Rear (KR)=1050 kN/m



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Radius of gyration (r) = 1.95m

Amplitude = 0.04m

2. Free body diagram

Distance from front of vehicle to centre of gravity (a)

metresa

a

ondistributimassgthVehiclelena

733.5

63.0*1.9

%*

Distance from rear of vehicle to centre of gravity (b)

metresa

a

ondistributimassgthVehiclelena

367.3

37.0*1.9

%*

Values for lf and lr are then found to be:

lf = 4.233m lr = 1.367m

Moment of inertia (I)

2

2

2

34983

95.1*9200

)(

kgm I

I

mr I inertiaof moment

Natural frequency (f)

v f

f v

*



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There will be two different values for the natural frequency because there are two different

wavelengths to consider (f 1 for 4 metres and f 2 for 9 metres wavelengths)

hz f

hz f

083.26

5.12

125.34

5.12

2

1

Frequency (ω)

There will also be two different values for the frequency as there are two values to input as

the natural frequency (f 1 and f 2)

f 2

sec / 087.13**2

sec / 635.19**2

22

11

rad f

rad f

External forces

)(2

2

t kbSinFr

t kbSinFf

Where b is the wave road profiles amplitude = 0.04m

and is the phase lag for the rear wheel.

Phase lag (Φ)

As with the frequency there will also be two phase lags. This phase lag is how much the

cosine wave lags behind the sine wave and is only applied to the rear of the vehicle

)lrlf (2

rad

rad

86.56

)6.5(*2

79.84

)6.5(*2

2

1



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3. Global problem equations

r r t r r r f f f gg

r t r r f f

lF lF ll zk ll zk I θ

F F l zk l zk m z

f

..

..

Rearranging;

r r t r f f r r f f r r gg

r t lf r r r f

lF lF llk lk lk lk z I θ

F F kf lk k k zm z

f

22

..

..

In matrix format;

[ gg I ] ..

z..

θ r f k k lf r r kf lk

lf r r kf lk 22

f f r r lk lk { } r t F F

r r t lF lF f

letting; Z = zSin(wt) & = Sin(wt)

{ } = { } {

}

{ } { } { }

Yielding;

[ gg I ] { } r f k k lf r r kf lk

lf r r kf lk 22 f f r r lk lk

{ } r t F F

r r t lF lF f

{ } r f F F

r r t lF lF f

Applying Cramer’s rule:

| r t F F

r r t lF lF f

|

| |



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| r t F F

r r t lF lF f |

|

|

4. Results

On examining the mode shapes graph plot above we can see that the optimal ride position is

placed in front of the centre of gravity at a distance of just less than 0.5m

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

-2 -1 0 1 2 3 4 5 D i s p l a c e m e n t ( m )

Amplitude of vibration wrt centre of gravity (m)

Mode Shapes

4m Wavelenght

6m wavelenght

wavelenght @ 4m

x1 top = 2.72806E+11 X1= -0.03182 x2 top = -2.60613E+11 X2= 0.030395

x bottom = -8.5743E+12 x bottom = -8.5743E+12

wavelenght @ 6m

x1 top = -4.47493E+11 X1= 0.05219 x2 top = 1.80481E+11 X2= -0.02105

x bottom = -9.797E+11 x bottom = -9.797E+11

Two DOF vehicle response analysis

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