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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
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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
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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
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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)
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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)
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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
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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}
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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.
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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
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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.
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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.
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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
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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
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6 Results
6.1 Part 1Responses in relation to various positions.
Figure 8: Displacements at different lengths along vehicle due to sine wave input.
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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
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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
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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.
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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.
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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
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