Kuwait UniversityCollege of Engineering and PetroleumMechanical Engineering Department

ME-459

Engineering Design

Easy Parallel Parking System

By

Abdullaziz AL-Zanadi 205112533

Duaij Al-Bawi 203112966

Fayez Shaheen 203215199

April 15, 2023

Instructor: Prof. Ahmad EL-Kholy

Abstract:

Parallel parking concept has been a painful and frustrating process since ever. In this

study, a system is devised to make such process more convenient and skill less. The system is

based on lifting the back of the vehicle using two hydraulic cylinders that are powered by a

24 V hydraulic power unit. The hydraulic power unit is composed of a hydraulic derive

centrifugal pump, oil reservoir and two electrical solenoids. The hydraulic cylinders lift the

car on two auxiliary rear wheels, which are in contact with the main car rear wheels. The

design was made to give the driver full control on the parallel parking process with minimum

cost and maximum durability.

The designed system was tested on a 4x4 vehicle and worked probably even with

some problems occurred in moving the vehicle. Many of these problems were resolved in

later stages while some of them still in. In the other hand some of the project goals were

achieved successfully while other was not. Hence needs to be modified first in order to apply

it.

2

Acknowledgement

This project could not be accomplished without the great help that we got from many

people specially the project advisor and our educator Pro. Ahmad EL Kholy and the entire

mechanical engineering department stuff such as Dr. Abullazim Falah and Dr. Mohamed Al-

Ostath. Also we want to give thanks to our families particularly Mr. Ahmad AL-zanadi who

helped use so much in technical work. In the other hand we can forget to thank colleagues

that helped us in this project on the top of them Yousef AL- Shamrok. At the end we should

express gratitude to Kuwait University which we all learned and got the knowledge that

helped us in our work.

3

Dedication

To all the people who helped us not just in this project, but also in our entire learning

path and to all of human race we give this project because that is what engineering about

making life easier.

4

Nomenclature:

w Distributed load

v Shear force

P Load- Pressure

M Moment

σ b Bending stress

S.F Safety factor

τ Shear stress

At Throat area

I Moment of inertia

r Radius of gyration

S Slenderness ratio

Sc Critical slenderness ratio

Pcr Critical load

σ t Tangential stress

σ L Longitudinal stress

σ H Hoop stress

σ y Yield stress

E Modules of elasticity

μ Coefficient of friction

ContentsAcknowledgement..................................................................................................................................3

5

Dedication...............................................................................................................................................4

Nomenclature:........................................................................................................................................5

Table of Figures:....................................................................................................................................7

List of Tables:.........................................................................................................................................7

1. Introduction:..................................................................................................................................8

2. Methodology:................................................................................................................................11

2.1 The Design Process:.............................................................................................................11

2.2 Detailed Design:...................................................................................................................15

2.2.1 Hydraulic System:........................................................................................................15

2.2.2 Mechanical System:.....................................................................................................16

2.3 Calculations:.........................................................................................................................18

2.3.1 Shear Stress and Pending Stress on connecting Pins:..............................................18

2.3.2 Shear Stress on Pin Stand Welding:..........................................................................21

2.3.3 Buckling on The Wheel Beam:..................................................................................22

2.3.4 Wheels Torque and Friction:......................................................................................24

2.3.5 Stresses Due to Presser in Hydraulic Cylinder:........................................................26

3. Results:..........................................................................................................................................28

4. Conclusion:...................................................................................................................................29

5. Future Recommendations:..........................................................................................................30

6. References:...................................................................................................................................31

7. Appendices:..................................................................................................................................32

7.1 Appendix A (Calculations):.................................................................................................32

7.1.1 Buckling With Moment Considered:.........................................................................32

7.1.2 Buckling Without Moment Considered:....................................................................33

7.1.3 Hydraulic Forces:........................................................................................................34

7.1.4 Hydraulic Cylinder Stresses Due to Pressure:..........................................................34

7.1.5 Shear Stress on Welded Wheel Joint:........................................................................35

7.1.6 Torque and Friction Calculations:.............................................................................35

7.2 Appendix B (Plots-Figures):...............................................................................................36

7.2.1 Base:..............................................................................................................................36

7.2.2 Brackets:.......................................................................................................................38

7.2.3 Wheel Beam:................................................................................................................38

7.2.4 Solenoid Manual:.........................................................................................................40

Table of Figures:

6

Figure 1: Over view of the process........................................................................................................9Figure 2: Four Wheel Steering (Alternative 1)....................................................................................12Figure 3: Alternative two.....................................................................................................................12Figure 4: Schematics diagram of the hydraulic system........................................................................16Figure 5: Base......................................................................................................................................17Figure 6: Schematics of mechanism....................................................................................................17Figure 7: Stress on pin.........................................................................................................................18Figure 8: Schematics of loads on pin...................................................................................................19Figure 9: First Cut location to find loads on pin..................................................................................19Figure 10: Second Cut location to find loads on pin............................................................................20Figure 11: Pin stand.............................................................................................................................21Figure 12: Welded area on the pin stand..............................................................................................21Figure 13: Wheel under buckling........................................................................................................22Figure 14: Friction between auxiliary wheel and the ground...............................................................24Figure 15: The frictional contact between the two wheels...................................................................25Figure 16: Hydraulic force...................................................................................................................27

List of Tables:

Table 1: Alternatives Comparison.......................................................................................................13Table 2: Alternatives Comparison.......................................................................................................14

1. Introduction:

7

This report will include several sections each one of them will focus on a point view

of the project. The first point will discuss the background of the project and a literature

review of previous work. The second point will discuss the methodology and the since

beyond the project mainly the part that include calculations and decision making. The third

focus point will be the results and discussion in which the final design of the system will be

explained and discussed mathematically and logically. Next the final conclusion will be

stated and the main problems and solutions will be mentioned. The final point will be the

recommendations for future work which is related to what can be done to improve the system

in future.

The objective of this project is to design a device that makes the parallel parking

process easier, hence that device should allow the vehicle to move in the side direction or at

least have the ability to rotate. And since rotating the vehicle is easier than moving it in the

side direction, hence the rotating concept will be considered. In order to have a successful

design the design should be cost effective and reliable.

In order to meet the project objectives there is a several things that should be taking in

consideration in designing process. First of all the design should be removable; in another

word the system should have the ability to be installed and removed easily without damaging

any of the vehicle components. The second consideration is that the design should be simple

as possible. Also the system should give the driver a full control on the process in order to

avoid any unexpected problems.

8

Many studies have been mad about parallel parking and many designs has been mad

previously such as Lexus company that created a computerize system to perform that task.

Also some students on the internet build a system to give the vehicle the ability to move in

the side direction also Ford made their own parallel parking system. As mentioned the

concept of parallel parking has been studied hard by many people or institutions, but the

problem still there most of these designs is even expensive, complicated or unfeasible to

many customers.

The main goals for this project it to make such system available for everyone with

affordable cost. The design was made to be cheaper than other product and also to perform

the task that it was made for. Also many customers prefer to take control of their vehicles

which is not the cause in many other designs, while in this design the whole process is

controlled by the driver.

The concept of this system is based on lifting the back vehicle on two auxiliary rear

wheels as shown in Figure 1 below.

Figure 1: Over view of the process.

9

These wheels are driven by hydraulic cylinders one for each wheel. Once the care is

lifted the movement of the vehicle in side direction will results from the contact

between one of the wheels with the main care wheel were it will act as a gear. The

hydraulic cylinders is driven by a hydraulic power unite which consists of driving

pump, oil tank, electrical motor and a flow controlling solenoid which is controlled by

an electrical switch.

10

2. Methodology:

In this section the main decisions and calculations that were used in the project will be

discussed briefly. But first of all the design has to be made, so this section will be divided

into three main subsections. The first subsection will explain the designing process and how

the overall image was generated. Then the second subsection will discuss how the chosen

design was made in details, while the third subsection will be related to the critical

calculation that was required to accomplish the final design.

2.1 The Design Process:

Previously the task details were mentioned, so in this subsection the design that

should meet the project goals should be made. The main goal of this project is to give the

vehicle the ability to rotate, and since there is many ways to do that the alternatives should be

compared as shown in Table 1 and 2 below.

The table is showing the main advantages and disadvantages of each method based

on many factors. The factors that were chosen to be the standards of the comparison are the

sources of the power that will cause the motion, the availability of the components, the

complicity of the design, the manufacturing difficulties, the portability and finally the cost.

The alternative one is to rotate the back wheels using the initial steering pump of the

vehicle (Figure 2) and the main problems with this method are the permanent change that

should be made to the vehicle, manufacturing the components, such system cannot be

removable and it will not be available to any vehicle.

11

Figure 2: Four Wheel Steering (Alternative 1).

While alternative two was to install a single wheel in the middle of the vehicle back

(Figure 3) and the problems of this choice was that the power source could be either external

motor or the main driving shaft of the vehicle. Both of these choices are not acceptable since

the using the shaft required applying permanent changes on the vehicle components, while

using electrical motor is too hard since it required a high torque motor will low speed or

using a typical electrical motor with gearing system which will increase both cost and level of

complicity.

Figure 3: Alternative two.

12

The third alternative was to take the advantage of the vehicle driving wheel motion in

order avoid applying any permanent changes on the vehicle components. In this case the best

way to do it is to use frictional contact with the main vehicle rare wheel, but that will cause a

balancing problem so another wheel should be installed in the other side to balance the

vehicle. Also another problem rise which is that if the two extra wheels are in contact with

the rear vehicle wheels they will rotate against each other which will prevent the motion,

hence only one of these extra wheels should be the driver and the other should be idle and

only for balancing purpose.

Table 1: Alternatives Comparison.

Alternative Source of Power Availability Level of Complicity

1 Vehicle steering pump

Many of the part are

not available and

should be

manufactured

Very complicated

specially in

controlling the motion

of the wheels

2Main vehicle shaft- external

motoravailable

Complicity in the

gearing either with

external motor or

main shaft

3 Vehicle main wheels AvailableNot complicated and

easy to figure it out

13

Table 2: Alternatives Comparison.

Alternative ManufacturingPortability

(Removable)Cost

1Hard to manufacture the

components the fitsPermanent

High manufacturing

cost

2 Moderate manufacturing

Permanent in case of

using vehicle shaft-

removable with

external motor

Moderate cost of

manufacturing and

components

3 Easy to manufacture Removable Moderate to low

Based on the comparison above alternative three was considered, hence a detailed

design should be made.

14

2.2 Detailed Design:

After the main picture became clear the chosen design should be figured out in

details, and in order to make a proper design each part of the system the is to be made should

be mentioned and discussed alone. After explaining each component the way of fixing these

components together is to be discussed. In general the system can be divided into two major

systems hydraulic system and mechanical system.

2.2.1 Hydraulic System:

In order to lift 900 Kg which is the half of the vehicle weight it is necessary to use a

powerful component either electrical motor or something else. Using electrical motor was an

available option, but the problem was in fixing it in a way to perform the task that is required.

So the other option was to use a hydraulic system which is capable of doing the task and easy

to integrate within the mechanical system.

The hydraulic system basically composed of hydraulic power unit hydraulic cylinders

and electrical controlling valves (solenoids) as shown in Figure 4 below. The hydraulic

power unit consists of oil storage tank, DC electrical motor and oil pump. In the other hand

the controlling valve is used to reverse the oil flow to give the ability of opening and closing

the hydraulic cylinders which are driven by oil pressure.

15

Figure 4: Schematics diagram of the hydraulic system.

2.2.2 Mechanical System:

The second part of the system is the mechanical system, which is basically composed

of simple mechanical components. The mechanical system can be divided into two major

parts the base structure and the wheels. The base is actually made specially to fit the vehicle

and to be connected to it by 6 bolts Figure 5. The major job of this base is to carry out the

other components and distribute the load on the vehicle body. In the other hand the wheels

are connected to beams that are fixed to the base with a pin to give the rod the ability to

rotate. Also these rods are connected to the hydraulic cylinders with pins too; hence that

mechanism can be modeled as a slider contact mechanism Figure 6“reference 1”.

16

Figure 5: Base.

Figure 6: Schematics of mechanism.

Even the mechanics and the hydraulic systems were mentioned independently they

are associated; since the hydraulic system provides the motion and power to the mechanical

system in order to lift the vehicle. “For more details about the dimensions see Appendix A”

17

2.3 Calculations:

In this subsection the methods used to find the important calculation will be

discussed. In any project it is not necessary to calculate the loads and stress in all the

component and also hard to done. Hence only the critical stresses on the components will be

calculated.

2.3.1 Shear Stress and Pending Stress on connecting Pins:

As mentioned before that only the critical calculation will be checked and one of the

most critical components is the pins that connect the wheels beam to the base. That pin is

subjected to several types of stresses which are pending, shear and bearing. Since the bearing

stress is not that important compared to the other two, so it will be neglected. And the other

two stresses are checked as following.

Figure 7: Stress on pin.

18

Figure 8: Schematics of loads on pin.

Figure 9: First Cut location to find loads on pin.

For the first cut:-

w1=P

2x1

V=−(w1)( x ) N

M 1=w1 x

2

2

19

Figure 10: Second Cut location to find loads on pin.

For the second cut:

w2=P/ x2

V 2=−w1 x1+w2(x−x1)

M 2=w1 x1(x− x1

2 )−( w2 ( x−x1)2

2 ) Since the bending stress act mostly on edges while shear stress is mostly on the center

and the bending force is for sure higher, then only bending stress on the edge need to be

checked.

σ b=Mmax y

I

Hence the safety factor can be found:

S .F=S y /σb

Calculated using the yield stress for low carbon steel.

20

2.3.2 Shear Stress on Pin Stand Welding:

One of the most critical loaded parts is the pin stand that connects the wheel bar to the

hydraulic cylinder. The importance of this part appears due to the load subjected to it by the

hydraulic cylinder. Figure 11 shows the part and the weld that need to be checked on it.

Were F is the force applied by the hydraulic cylinder.

Figure 11: Pin stand.

Figure 12: Welded area on the pin stand.

21

The stress on this weld is calculated as following:

Throat Area:

At=0. 707h(2b+d ) (m2 )

The shear force is equal to the horizontal component of the hydraulic cylinder force:

F t=Fhydrolic×cosθ

Were θ is the angle between the hydraulic cylinder and the base.

Then the shear stress is calculated by:

τ=Ft

A t

(N /m2)

Hence safety factor will be:

SF=Syτ

2.3.3 Buckling on The Wheel Beam:

In order to make sure that the wheel beam will not fail it is

quite important to know the effect of buckling on it. Even that the

beam is not too long the importance of the buckling stand up due to

the high load coming from the vehicle weight. The wheel beam is

fixed with hinge from the top and a wheel from the bottom as

Figure 13 shows, so it can be considered as pined-pined Euler’s

column and the buckling effects can be found as following:

22

First determine either it is short column of long one by finding slenderness ratio and

the critical slenderness ratio as following:

S= Lr

Were r is the radius of gyration which can be found as:

r=√ IA

I= π64

(OD 4−I D 4 )

A=π4O D2

While

Sc=√ 2π2 Eσ y

And since S was less than the critical one, hence the column is considered as short

column and the critical load is:

Pcr=Aσ y (1−( σ y

4 π2E )( Lr )2)

Hence

SF=Pcr

P

23

2.3.4 Wheels Torque and Friction:

One of the most important characters in moving ad wheeled mass is the friction and

torque calculations. The importance of these calculations is concerned to it is relation to the

motion of the vehicle and avoiding the slipping problems. The slipping can occur in two

locations between the ground and the wheel or between the main vehicle wheel and the

auxiliary wheel and they both will be calculated.

Figure 14: Friction between auxiliary wheel and the ground.

Figure 14 shows the wheel in contact with the ground and the maximum torque can be applied without the wheel slipping is calculated as:

T max1=Mgμstatic r

The coefficient of static friction μstatic between the ground and the wheel was taken

to be 0.65 which is the average coefficient of friction between rubber and dry asphalt.

While the required torque to move the vehicle is:

24

T req=M×a×r

With a assumed 1 m/s as an acceptable value.

Since T req<T max 1 hence there will be no slipping between the wheel and the ground.

Figure 15: The frictional contact between the two wheels.

Although it is not accurate to calculate the frictional contact this since the contact is

occurring with some angle and there is some other parameters involved, but it could give

estimation. The average coefficient of friction between rubber and rubber is 2.5. Figure 15

shows the two wheels in contact were N is the horizontal component of the hydraulic cylinder

force. Hence the maximum torque can be calculated as following:

T max2=N μr

If T max2>T req then there will be no slipping between the wheels, in the other hand if

T max2<T reqthen the wheels will slip and the vehicle will not move.

25

2.3.5 Stresses Due to Presser in Hydraulic Cylinder:

Since the hydraulic cylinder is subjected to high oil pressure, hence it is quite

important to find either if it can take that pressure or not. The pressure of the power unite was

give by the manufacturer as 2500 psi which is equal to 17.23 Mpa and the dimensions were

measured. In order to check whether the cylinder can take that pressure or not it is important

to know the type of stresses involved which are tangential stress, radial stress and

longitudinal stress. Since the radial stress is much smaller than the tangential stress then only

the tangential stress will be considered in addition to the longitudinal stress the occur only if

the cylinder is fully opened or closed.

The maximum tangential stress for thick walled cylinder which occurs at the inner

surface can be calculated as:

σ t=ri

2 pi

ro2−ri

2 (1+rori

2 ) While the longitudinal stress is:

σ l=ri

2 p i

ro2−r i

2

Also one of the most important stresses in pressurized cylinder is the hoop stress

which can be found as:

σ H=Prt

These values can be compared with the yield strength of low carbon steel in order to

find the safety factor.

Another important force to calculate is the force exerted by the cylinder (Figure 16)

which can be calculated as:

F2=P×π4

(d2 )2

26

F1=Pπ4

(d22−d1

2)

Were F2 is the cylinder expanding force and F1is the cylinder contraction force.

Figure 16: Hydraulic force.

27

3. Results:

In this section the calculation results will be posted and discussed to clear up the

image. First it is important to start with the hydraulic cylinder force since it is involved in

many other calculations and was found to be about 21.6 KN in case of pushing and 7.781 KN

in case of closing for each cylinder. These values of force are quite enough to life the vehicle

since the weight to be lifted is less than 900 Kg with two hydraulic cylinders.

Moving forward to in checking the hydraulic cylinders the results of the stresses

inside due to pressure were 94.6 Mpa as maximum value and gives a safety factor of 3.

Moving to the wheels and friction relation the wheels torque required were found less

than the maximum friction torque, hence the wheel will not slip on the ground. In the case of

the torque transmitted between the main car wheel and the auxiliary wheel also it was found

that is the maximum friction torque will be less than the required torque to move the vehicle.

The pin calculations showed that the stress is too low to be considered so it will be

neglected. And by checking the buckling on the wheel beam the result was also too small and

the safety factor was huge number so the beam is considered safe from buckling.

Finally moving to the welded joint between the hydraulic cylinder and the wheel base

the resultant safety factor was found to be about 1.7 which is quite safe. There may be more

calculations and results that can be made but that is all what the report gone cover. “for

further details about the calculations see appendix A”

28

4. Conclusion:

At the onset of this design project, there were many possible design methods to

perform parallel parking process. The project began by brainstorming and developing several

different ideas. Each member of the design team focused on a specific method to perform the

task. After developing each design method, the design group implemented a set of criteria to

determine the feasibility of each design. The criteria consisted of cost, configuration, and

availability of components. By evaluating the criteria for each design consideration, the group

concluded the chosen design would best meet the criteria that were established.

The chosen design met almost all the criteria that the group incorporated. As was

mentioned earlier, the chosen design has the best flexibility and feasibility between all the

suggested alternatives. The power source to generate the motion required was one of the

major criteria. The less complicity keeps the cost of the project a minimum.

Even that the design did not worked as well as expected, it is still a great idea that

needs some improvements. Therefore the next section will discuss the improvements that can

be made to get a better design.

29

5. Future Recommendations:

Since the system is not performing well yet and there were many modifications

planned but not accomplished, hence this chapter will mention some of these

recommendations in case if trying to apply the system on the future. First of all the problem

of the contact friction between the two wheels has to be solved by modifying the contact

position to avoid slipping. Also another great modification can be done by canceling the

power unit and alternate it by integrated large capacity steering pump connected to the engine

pelt. Another modification can be done by integrating the oil pipes with the vehicle bode to

give it a better shape and maybe there is much more modifications can be done but these are

the most important ones in order to have a good final product.

30

6. References:

1 – Richard G.Budynas , J. Keith Nisbett . Shigley`s Mechanical Engineering Design.

Eighth edition in SI units.(2008).

2 - Ferdinand Beer, Jr.,E. Russell Johnston, John DeWolf, and David Mazurek.

Mechanics of Materials. (2008)

3 - by John J. Uicker, Gordon R. Pennock, and the late Joseph E. Shigley. Theory of

Machines and Mechanisms. (2003)

4 - http://www.roymech.co.uk/Useful_Tables/Screws/Bolted_Joint.html

5 - http://www.roymech.co.uk/Useful_Tables/ARM/Safety_Factors.html

6 - http://www.engineeringtoolbox.com/hydraulic-force-calculator-d_1369.html

31

7. Appendices:

7.1 Appendix A (Calculations):

7.1.1 Buckling With Moment Considered:

σ max=FA (1+ ec

r 2 sec(√( FAE ) L

2 r ))While r=√( IA )OD=60×10−3m , ID=52×10−3m

F=4 .5×103 N , L=28×1 0−2m

Assuming e=L2=14×10−2m∧c=OD

2=30×1 0−3m

A=π4OD2=7 .03716×1 0−4m2

I= π64

(OD 4−I D 4 )=2 .7726×1 0−7m4

r=√ IA

=0 .02

E=210×109Pa σ all=210Mpa

σ max=73 .6 Mpa

SF=σall

σmax

=2 .8537

32

7.1.2 Buckling Without Moment Considered:

δ=L¿r ¿

r=√ IA

=0. 02 ¿I=27726×10−7m4 ¿¿ A=7 . 07×10−4 m2 ¿¿¿OD=60×10−3m ¿¿¿ ID=52×10−3 m ¿¿¿E=210Gpa ¿¿¿¿¿

σ all=450MpaL=42cm

δ=0 .420 .02

=21

δ cr=√2π2 E450×106

=96

δ<δ cr

Short geometry

Pcr=310 .5344kNSF=70. 34

33

7.1.3 Hydraulic Forces:

D small=3 .2cmP=2500 psi=17 . 23589 Mpa

Fhy . push=P×π4D2

D=48×10−3mFhy . push=21614=21 . 614kN=2. 2 ton

Fhy . pull=P×π4

(D12−D

rod2 )=7781 N=7 . 781kN=793 kg

7.1.4 Hydraulic Cylinder Stresses Due to Pressure:

Pi=17 .2 Mpa

ri=2×10−2m

ro=2 .4×10−2 m

σ t=ri2 Pi

ro

2−ri2

(1+ro

2

ri2

)=86 Mpa

σ L=ri2P i

ro2−r

i2

=39 Mpa

t=40×10−3 mr=2. 2×10−2m

σH=Prt

=94 .6 Mpa

Low carbon steel tensile strength = 284.1 Mpa

SF=3

34

7.1.5 Shear Stress on Welded Wheel Joint:

Throat area:

At=0. 707h(2b+d )b=3 .2cmd=1.8cmh=0 .3cmAt=1 . 74cm2=1. 74×10−4 m2

F t=21. 6kN

τ=F t

A t

=21.6kN1 . 74×10−4 m2

=124 .14 MPa

Sy=345×0. 6=207 Mpa

SF=Syτ

=207 MPa124 . 14 MPa

=1.67

7.1.6 Torque and Friction Calculations:

Tmax 1=900×9. 81×0. 65×(15×10−2 )=860 . 8N .m860 . 82

=430 N .m

T req=(900 )(1 )(15×10−2)=134 N .m

Since the torque required is smaller than the maximum torque, the wheel

will not slip on the ground

Tmax 2=N .μ . r

where N=Fhyd . push×cos(16 .087 )¿20 kW→Tmax 2=6000 N .m

The wheels will not slip over each other.

35

7.2 Appendix B (Plots-Figures):

7.2.1 Base:

36

37

7.2.2 Brackets:

7.2.3 Wheel Beam:

38

39

7.2.4 Solenoid Manual:

40

41

42