Simulation of Vehicle Ride Characteristics

Simulation of Vehicle Ride CharacteristicsMM401 Mechanical Engineering System Simulation

Neville Lawless

10212298

8/4/11

School of Mechanical and Manufacturing engineeringDUBLIN CITY UNIVERSITY

Glasnevin, Dublin 9,

Ireland

Contents

CONTENTS II

1 INTRODUCTION 1

1.1 Aims: 1

2 VEHICLE CHARACTERISTICS 2

2.1 Free body diagram 3

3 STATE EQUATIONS 5

4 CONSTRUCTION OF SIMULINK MODEL: 7

4.1 Responses at various locations 8

4.2 Force on the driver 9

4.3 Displacement and Force on the Rear Suspension 10

5 MODEL M-FILE CODE 11

6 RESULTS 13

6.1 Part 1 13

6.2 Part 2 14

6.3 Part 3 16

7 CONCLUSIONS 17

8 REFERENCES 18

Neville Lawless Page II

1 Introduction

The purpose of undertaking this assignment was to develop a Simulink model, which drew on a previous assignment, to determine the ride characteristic of a bus as it was subjected to various different external forces, due to the road profile it was to travel on. These were modelled by generating sine waves of wavelength 4m and 6m and also by an external road profile input. The vehicle was modelled as a 2 DOF system, having associated stiffness and damping properties on both front and rear suspension pairs and travelling with a constant velocity. It was also specified to determine the forces and displacement on the vehicle driver and also at another point along the length of the bus.

1.1 Aims : Construct state equations to represent the vehicle model.

Model these in Simulink using block diagrams and construct an M-file in Matlab to run the simulation.

Determine the response of the bus to sinusoidal road profiles of 4m and 6m wavelengths

Find the response of the four locations below as the speed of the vehicle increases in velocity.

A: near rear B: over rear wheels C: mid-point between B & D D: over front wheels

Determine the force felt by the driver (sitting at the front of the bus) if (s)he weighs 60 kg at 45 kph

Determine the force and displacement of the rear suspension in response to a road profile read from a matlab data file at 45 kph

Neville Lawless Page 1

2 Vehicle characteristics

Figure 1: Vehicle characteristics diagram [2]

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

Damping coefficients = Front (bf) = 350kNs/m Rear (br) = 350kNs/m

Radius of gyration (r) = 1.95m

Amplitude = 0.04m

2.1 Free body diagram


Figure 2 Free body diagram of the vehicle with lengths lf and lr

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

a=Vehiclelength∗mass distribution%a=9 .1∗0 . 63a=5 .733metres

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

a=Vehiclelength∗mass distribution%a=9 .1∗0 .37a=3 .367 metres

Values for lf and lr are then found to be:

lf = 4.233m lr = 1.367m

Distance for 5 locations along bus:

Setting the centre of gravity as zero and the front of the bus as the positive x- direction yields

A: near rear: Distance = -1mB: over rear wheels: Distance = lr = -1.367m C: mid-point between B & D: Distance = (span/2) – lr = (5.6/2)-1.367 = 1.433mD: over front wheels: Distance = lf = 4.233mE: Distance to driver: Distance = lf+overhang = 4.233+ 1.5 = 5.733m

Moment of inertia (I)


moment of inertia(I )=mr2

I=9200∗1 .952

I=34983kgm2

Natural frequency (f)

There will be two different values for the natural frequency because there are two different wavelengths to consider (f1 for 4 metres and f2 for 6 metres wavelengths)

f 1=12 . 54

=3 .125hz

f 2=12 . 56

=2 .083hz

Frequency (ω)

There will also be two different values for the frequency as there are two values to input as the natural frequency (f1 and f2)

ω=2πf

ω1=2∗π∗f 1=19. 635 rad /secω2=2∗π∗f 2=13 .087 rad /sec

External forces

F f=kbSin ( t )+c cos(t )

F r=kbSin ( t+Ø )+ccos (t+Ø)


v=λ∗f

∴ f=vλ

Where b is the wave road profiles amplitude = 0.04m; φ

is the phase lag for the rear wheel, c is the damping coefficient for the set of front or rear suspensions.

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

Φ=2π ( lf + lr )

λ

Φ1=2∗π (5.6 )4

=8.79 rad

Φ2=2∗π (5 .6 )6

=5 .86 rad

3 State equations

From the previous assignment the equations that describe the motion of the bus were derived. These were once again used and the terms corresponding to the damping coefficient now included.

Translation Equation:

m z+c (v−lf q )+c (v+lrq )+k f ( z−lf )+kr ( z+lr )=F f+F r


Rotation Equation:

❑ I ¿−c (v−l f q ) lf+c (v+lrq ) lr−k f ( z−lf θ ) lf+kr ( z+lrθ ) lr=−F f lf+F r lr

Rearranging the equations in state variable form:

z= 1m (−c (v−lf q )−c (v+ lrq )−k f ( z−lf )−kr ( z+lr )+F f +Fr )

❑ I ¿+v (c lr−c )+q (c (lf+lr2 ) )+z (kr lr−k f l f )+θ (k f lf2+kr lr

2 )=−F f lf+F r lr

State equations

Where q=θ & v= z

z= 1m {−v (2c )−q (c (lr−lf ) )−z (k f+kr )−θ (kr lr−k f lf )+F f+Fr }

❑= 1I¿

{−v (c lr−c)−q (c ( lf+lr2 ))−z (k rlr−k f lf )−θ (k f lf2+kr lr

2 )−F f lf+F r lr}


4 Construction of Simulink model:

Force Input

To model the 2 different types of inputs into the simulation model, two types of input blocks were used. One for the sinusoidal/cosine wave input and one for the provided road.mat profile file. The cosine wave input is also a sine wave source block that has a phase of pi/2. In figure 2 below it can be seen that gain blocks are used to multiply the corresponding inputs by the k, c or parameters and then added to obtain the F f=kSin (t )+c cos(t ) front force.

Figure 3: Simulink inputs

The same is done for the Fr input as well, with the addition of a phase lag applied to each. This is done as the force which is experienced by the font suspension at a time t will be the force experience by the rear suspension at a time equal to t+Ø. Where

the phase is calculated in section 2.1 Φ=

2π ( lf + lr )λ

A Global model switch to toggle between road profile and sine wave input was used also at the input stage. The switch was set to a threshold value of zero with a condition that u2 (user input value in Matlab)> threshold. I.e. to produce an output from either input, let control = 0 for road profile, or control =1 for sine wave.


Figure 4: Complete Simulink model block diagram

These input values were fed into the block diagram for both the translational equation Z (red flow of blocks) and the rotational equation θ (green flow of blocks).

The summation blocks were multiplied by the gain blocks of values 1/m and 1/Igg. This yields the two state equations. Both of these can be the integrated twice to obtain the 4 state variable z, v,θ, and q.

4.1 Responses at various locations

To obtain the displacement of the vehicle at any point along its length, the sum of the translational vertical displacement and the rotational displacement had needed to be found, this can be found from the simple equation:

Yy = z + xθ 1

Where:

Y = vertical displacement at a point (y) along the vehicles length

z = vertical displacement at the Centre of Gravity

x = Distance from y to the Centre of Gravity

θ = Rotation of the centre of Gravity


Part 1 of the assignment can be completed now with simulink by directly obtaining the values from the block diagram. The distance x was replace in the Matlab code with the variable “distance” which was varied for the respective sections of the m-file. The values for each can be found in section 2.1 previously.

4.2 Force on the driverUsing newton’s second law we can determine the force experienced by a driver of mass m.

F=MA. 2

Knowing the displacement at any point we can calculate the acceleration by differentiating twice. This is done for the translational component and added to the angular acceleration multiplied by the distance to the point.

Y y= z+x θ 3

However this value can be directly obtained again from the block diagram figure 4 as the input has already been integrated twice.

Figure 5: Translation equations showing state variables

Our term Y y now is the total acceleration of the driver. A gain block

of value M is used to give us the force F=MY yon the driver, as can

be seen in figure 6.


Figure 6: block diagram used to obtain force and displacements on the driver.

4.3 Displacement and Force on the Rear Suspension

Much in the same way as the responses were found for the

displacement and force felt by the driver, the responses over the

rear suspensions were modelled using the same blocks. This was

achieved by allowing a mass to be assigned to the rear of the

vehicle over the suspension. In figure 1 it is specified that the

weight distribution is 63:47 rear: front. Therefore a mass was

calculated by multiplying the bus mass by 0.63 and so allocated to

the rear suspension.

For this part of the assignment the road profile was needed to be

selected, so this is done by an input request from Matlab to the

user, further detailed in the next section.


5 Model M-file code

The input variables to the model that would remain constant

throughout the simulation were first declared, figure 7.

Figure 7: M-file declared variables

A variable called “control” was then defined and allocated a value which was to be read in from the main matlab menu. The code is as follows:

“control = input ('Enter 0 for road profile effects on rear suspension, or enter 1 for sine waves profile responses: ') % asks the user in the matlab main window which output they require ”

A conditional statement (if-else) was then used to select between the parts of the assignment which should be simulated. These depended on whether the road profile input was required or the sine wave input desired. The code is as follows:

if (control == 1) PART_1_RESPONSES_IN_RELATION_TO_VARIOUS_POSITIONS PART_2_FORCE_and_DISPLCAMENT_OF_DRIVER

else PART_3_FORCE_and_DISPLCAMENT_OVER_REAR_AXLE end;

The code correlates the input from the user against the if expression; if control = 1 output part 1 and 2 or else If the user


enters any other value it outputs part 3. No further error preventive coding was used here as the input of zero or one is self-explanatory for this assignment. However if a code such as this is to be used by users unfamiliar with its use, then other coding practices could be employed by means of while loops etc.The above else if expression calls m-files which contain the code to output the graphed responses of the vehicle. This is done solely for ease of use and clarity purposes. This code could have been included in the main matlab file, but was segregated to allow easier debugging.

Graphing

For all graphs, the same approach was used to plot the

graphs. This is specified below.

figure(1); % indication to matlab on which window to output a plot. subplot(4,1,1) % specifies a plotting window of 4 rows and 1 column, and then to plot in the first space. Ie top of graphing window

plot(Time.signals.values,z.signals.values,'r') % takes output values time and z from the Simulink model and plots them versus each other. The term ‘r’ plots the graph in red.

title('Displacement near rear wheels') % Graph title

ylabel('Displacement (m)') % Graph y label.

xlim ([0 10]) % Minimum and maximum x values.

ylim ([-.1 .1]) % Minimum and maximum y values.

hold on % Holds the data to be plotted and plots with the next provided data


6 Results

6.1 Part 1Responses in relation to various positions.

Figure 8: Displacements at different lengths along vehicle due to sine wave input.


Examining figure 8 shows that at nearly all points over the length

of the vehicle, a value of less than 0.07 m is the is achieved for the

displacement due to the forced vibration caused by the sine wave

input. In real life terms this value seems reasonable to what would

be expected/required by a passenger vehicle. At positions over,

and near the rear suspension, the displacements can be seen to be

nearly equal. Comparing this to the value at the midpoint which is a

further distance from the centre of gravity (COG), the value would

appear to decrease. Further comparing this to the values found at

the front suspension validates to the author that the model is

operating successfully.

From basic maths it can be shown that the arc length S (or vertical

displacement for small angles) is equal to the radius of the arc

times θ the angle subtended by the arc at the centre (centre of

gravity in this case). S = R θ so for a greater distance from the COG

the displacement is less.

At the rear and at the midpoint it can be seen that the displacement

due to the 6m sine wave dominates, however over the front wheels

these values seem to switch with the 6m wave causing significantly

less displacement and the 4m wave increasing significantly.

6.2 Part 2

Force and displacement of driver

Examining figure 9 gives a great indication to the effect that that

wavelength can hold on the response signal. The displacement of

the driver has a decrease of 100% when the sine wave input is

increased from 4 m to 6m. Again, this is the response which would

be expected. As the wavelength decreases the more frequent

interference is encountered due to the crest of each wave per unit

time. Closer inspection of figure 9 provides a good visual

indication to the different components of motion due to the


disturbances these being the transient and steady state response.

The initial transient stage dying off due to the damping present in

the system and the steady state forced vibration remaining

constant due to the sinusoidal input. These can be easier seen in

figure 10.

Figure 9: Displacement and force experienced by the driver


Figure 10: Transient, steady state and total response of a system [4]

The values of force experience by the driver hold a stark contrast

depending on the wavelenght. They are steadystate 5KN for 6m

wavelenght and 2KN for 4m wavelenght. Again these values seem a

reasonable approximation to real life situations. However, the true

value is not known to the author.

6.3 Part 3Force and displacment over rear axle due to road profile

input

Examining the force and displacement graphs generated by matlab

for a point over the rear suspension, it is seen that a maximum

positive displacement is

0.0175m and negative displacement is 0.0125m. The maximum

forces experienced hold a value of 75 and 100N. These values are

found at points where there is a dramatic jump in displacement.

Once again, referring to newton’s second law, F=MA, a large

change in displacement causes its second derivative, acceleration,

to also be large, thus leading to spikes occurring at these points.

Gradual increases/decreases in displacement cause no large

accelerations and so have only minor fluctuations on the matlab

output plots.


Figure 11: Force and displacement due to road profile.

7 Conclusions

From a previous assignment, the basic understanding and concept

of vehicle responses has been understood. Undertaking this report

has required a more detailed approach be taken. The first major

difference to this problem was that a damping system has been

incorporated into the vehicle suspension. This provides another

state variable to be derived. These being z, θ, v, & q. knowing

these, the state equations were derived for both the translational

motion of the system and also the rotational component of it. Once

these had been established a Simulink model was developed to

represent this. A matlab m-file was developed simultaneously to

control the model and provide us with the desired outputs. These

were to determine the response from the system at four locations

along the length of the bus, to determine the force and

displacement on the bus driver, and finally to plot the responses

over the rear suspension of the vehicle.


It was found that the displacements all lay under a value of 0.07m,

with 0.05m as an average value. Next a value of between 0.025m @

6m wavelength and 0.1m @4m wavelength were plotted for the

driver displacement, caused by a force of 5 & 20 KN respectively.

It was finally established from the road profile plot that the forces

transmitted through the vehicle are caused by sudden

displacements, which in turn transmits large accelerations through

the vehicle causing the spikes in force.

It is now felt by the author that having successfully completed the

task, that the use of matlab/Simulink, which have not been used

previously are a successful method of simulation vehicular

dynamics and many other engineering systems.

8 References

[1] Young, P., (2001), “Two Degree of Freedom (Forced)”, MM401

Lecture Notes, Dublin City University, Dublin.

[2] Young, P., (2011) “Assignment 2: Simulation of Vehicle Ride

Characteristics”, MM401, Dublin City University, Dublin.

[3] Young, P., (2001), “Assignment 1- 2011: Calculation of Response of 2-DOF System to Excitation” MM401, Dublin City University, Dublin.

[4] http://www.mfg.mtu.edu/cyberman/machtool/machtool/vibration/forced.html


http://www.mfg.mtu.edu/cyberman/machtool/machtool/vibration/forced.html

Simulation of Vehicle Ride Characteristics

Documents

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