hydraulic design report copy

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DEPARTMENT OF MECHANICAL ENGINEERING

ALL Sections must be completed using Block Capitals or typed

CODE AND TITLE OF COURSEWORK

Course code:

MECH3010

Title:

DESIGN OF SERVO HYDRAULIC CONTROL SYSTEM

FOR AN ARTICULATED INSPECTION AND CLEANING

WORK PLATFORM

STUDENT NAME: Leheng LI, Linfei WU, Wenguang CHENG, Fanfu WU,

Shangzhou DUAN

DEGREE AND YEAR: 3rd Year MEng Mechanical Engineering

LAB GROUP: N.A

DATE OF LAB. SESSION: N.A

DATE COURSEWORK DUE FOR SUBMISSION: 18th Nov 2014

ACTUAL DATE OF SUBMISSION: 18th Nov 2014

LECTURERS NAME: Dr. A.Ducci

PERSONAL TUTORS NAME: N.A

RECEIVED DATE AND INITIALS:

I confirm that this is all my own work (if submitted electronically, submission will be taken as confirmation that this is your own work, and will also act as student signature)

Signed:

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1. Component Summary

2. Introduction

3. Hydraulic System Design

3.1 Basic Assumptions

3.2 Rotary Actuator R1

3.3 Linear Actuator L3





3.8 Pumping System

4. Control System

5. References

Appendix I: Sample Calculations and Details on Actuator Selection

Appendix II: MATLAB code

Appendix III. Rotary Actuators Calculation

Appendix IV: Percentage Distribution of Marks

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1. COMPONENT SUMMARY

Model w/o SF w/ SF

Min

Strok

e

Max

Stroke

Bo

re

R

o

d

Press.

(bar)

Pull

Force

(each)

Push

Force

(each)

L

3

Parker MMA

MF4

32040

N

64080

N 0 5460

8

0

5

0 250 79000N 128000N

L

2

Parker MMA

MP3(x2)

60306

5N

12061

29N 3610 6750

2

5

0

1

6

0

250 747000

N

1250000

N

L

1

Parker MMA

MT4(x2)

12762

07N

19143

10N 1060 5710

3

2

0

2

0

0

250 1265000

N

2050000

N

R

3

Parker

4000M

94988

Nm

18997

6Nm

N.A

204 9525Nm

R

2

Parker

3000M

14068

3Nm

28136

7Nm 204 338955Nm

R

1

Parker MR

2400I

4199

Nm

8398

Nm 250 451939Nm

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2. INTRODUCTION

Hydraulic systems are very widely used in a variety of engineering applications. From

the brake system of an everyday family car to industrial machinery to space shuttles,

hydraulic control or power system can be easily found in numerous engineering

products and solutions. They possess many relative advantages over other form of

power systems (e.g. electric, pneumatic etc.), such as a high power-to-weight ratio,

precise low-velocity control, and better transmission of large forces and torques, hence

their favoured use, particularly in heavy engineering systems.

The purpose of this design project is to analyze an articulated crane system that is used

to inspect railway bridge systems and design a complete hydraulics system, including

actuators, valves and power systems that can power the crane’s movements. The crane’s

rough design and dimensions are ready-made and provided to the project group, yet the

material of the components are not available and will have to be assumed.

The crane component and hydraulic system are used to power the pannier, which

provide the operators in the pannier 6 degrees-of-freedom to assume the optimum

position to inspect the infrastructures. To achieve this 6 hydraulic actuators are used,

with 3 rotational actuators and 3 linear actuators, coded R1-R3 and L1-L3. The

mounting position of R1, R2, R3 and L1 are already designed into the crane structure,

and the mounting positions of L2 and L3 will need to be decided in the hydraulic system

design. For each actuator, the design process involves 3 major steps:

1. Analysis of the loads acting on the analyzed actuator, and iterations to find the crane

configuration that gives the highest load (worst-case scenario). For certain actuators

(namely L1 and L2), the kinematics constraints of the mechanisms need to be analyzed

before starting the load analysis.

2. Analysis of the packaging requirements and travel range of the actuator, with the

most important parameters being the maximum and minimum lengths of the actuator.

3. Selecting an appropriate actuator from the catalogues of hydraulic system solution

providers. The catalogues of Parker Hannifin and Rexroth Bosch are used, and the

smallest/lightest actuator that satisfies the parameters set in the first two steps is

selected.

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When all six actuators are selected, the required power and flow rate can be found, and

a suitable power system and ancillary equipment (control valves, pump, motor,

accumulator, filter) can be selected according to the actuator requirements. The control

interface available to the operators in the pannier is also discussed, as well as the

required safety measures and possible integration of a computerized control system,

although no detailed designs are produced.

Finally, all the design parameters are examined with respect to BS7171:1989

(Specification for Mobile Elevating Work Platforms). It is worth noting that this British

Standard document is already superseded by BS EN 280:2013, so certain aspects of the

hydraulics system might not conform to the new standard, and all results from this

design should be inspected against the new standard before being used.

3. HYDRAULIC SYSTEM DESIGN

3.1 SYSTEM OVERVIEW

The crane consists of five main control arms, as shown in the diagram below:

From left to right, the 5 arms are: Wagon Boom, Boom Turntable, Inner Boom, Outer

Boom and Telescopic Boom. The controlled pannier is connected to the telescopic boom

only and is in the bottom right hand corner. The actuator locations(including an

assumed location for L2) is shown in the diagram as well.

The rotation of the inner boom with respect to the boom turntable is controlled by a

mechanism(as shown below) which involves a tie link that is connected to the actuator

L3

L2

R2

R3 L1

R1

Fig 3.1.1 The crane system, including locations of the actuators

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L1, and a push link. This allows the inner boom to have a higher range of rotation than

that if L1 is directly connected with the inner boom(as L1 will clash with the turntable).

The complete crane system is mounted on a standard gauge railway wagon, which is not

modelled in the diagrams.

It is worth noting that due to the lack of certain dimensions from the designers, the

CATIA model made from the provided drawings and dimensions is not 100% accurate.

There will almost certainly be dimension and mass discrepancies, but these are covered

by the large safety factor.

In order to visually simplify the system, a separate system simulator is created in CATIA

V5 using the same set of dimensions, which simplified the system into a 2D ‘stick

drawing’, as shown below:

L1

Fig 3.1.2 The tie link/push link mechanism between L1 and the inner boom

Fig 3.1.3 The ‘stick-drawing’ simulator made in CATIA V5

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This simulator is used to examine the geometries of the inner, outer and telescopic

booms, therefore the wagon boom is not modelled.

3.2 REQUIREMENTS

Several requirements regarding the actuators and kinematics of the crane system are

already outlined by the designers of the crane:

- Actuator L1 is required to have an actuation length of no less than 4653mm.

- Actuator L1 is required to have a movement range of 1058mm to 5711mm.1

- Actuator L2 is required to allow a 0°-110°rotation of the outer boom with regards to

the inner boom.

- Actuator L3 is required to have an actuation length of no less than 5460mm.

- Actuator R2 and R3 are required to be able to rotate ±90° from their respective

stowed positions.

- Actuator R1 is required to be able to rotate ±180°from its horizontal position.

3.3 BASIC ASSUMPTIONS

Some assumptions are made about the crane system and the design process in order to

aid the design process, as outlined here:

- The safety factor (SF) used in all calculations will be 2, apart from actuator L1 which

will use a SF of 1.5.

- All joint bearings on the system are assumed to be frictionless.

- The materials for all components are assumed to be steel, with a density of 7860kg/m3.

Except for the pannier, which is assumed to be made of aluminium with a density of

2710kg/m3.

- It is assumed that the stresses on the components are already analyzed by the crane

designers, and the components will not deform or fail under the sustained loads.

- The inner boom and wagon booms are assumed to be hollow(partially hollow for the

inner boom). Although not specified by the designers, calculations of the masses of these

booms indicate that if these booms are to be solid, their masses become unreasonably

large, making it impossible to find suitable actuators.

1 By satisfying this requirement, the first requirement will be satisfied as well, so the first requirement is

ignored in the design process.

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- Only one actuator will be moving at any time, no two actuators will be operating

together(except for L1 where two actuators are used at the same location).

- When analyzing the system configurations, it is assumed that there is enough ballast on

the railway wagon carrying the crane system to prevent the system from overturning.

- The only external/environmental effect acting on the system, apart from the masses of

the components and inertial forces, is wind force. According to BS7171:1989, it is

assumed to be 100N/m2 for each square meter of component surface area encountering

the wind, which is roughly equivalent to a 12.5m/s wind.

- The maximum movement speed of the pannier is assumed to be 0.4m/s when moving

vertically, and 0.7m/s when slewing (moving horizontally), as required in BS7171:1989.

The acceleration time is assumed to be one second, hence the linear acceleration of the

pannier is assumed to be 0.4m/s2 for vertical movements and 0.7m/s2 for horizontal

movements.

- The hydraulic fluid used is incompressible and there are no losses in the pipelines.

- The masses of the pipelines, fluid in the pipelines and pipeline fixtures are neglected.

- The gravitational acceleration is assumed to be 9.81m/s2.

Any further assumptions made during the calculations of the individual actuators can be

found in the individual calculation sections.

3.4 ROTARY ACTUATOR R1

The rotary actuator R1 is used to keep the pannier horizontal at all times regardless of

the outer boom angle, ensuring the safety of the operators inside. The load on it consists

of wind load on the operators/pannier, and the mass of the operators and tools, with the

worst case scenario being all the mass torque located on one end of the pannier with the

wind torque acting in the same direction. The detailed calculations can be found in

Appendix II. The calculated load on R1 is 4199Nm, and with a safety factor of 2, the R1

torque is 8398Nm.

Chosen Actuator:

Parker MR series radial piston motor , Model 2400I

- Maximum torque: 9525Nm

- Starting torque: 8572.5Nm

Fig 3.4.1 Parker MR Series

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- Motion Range: 360 degrees

- Maximum Continuous Pressure: 250bar

- Mass: 325kg

- Flow Rate: 20l/min

The Parker MR 2400I is chosen for its lightness, making it suitable to be installed on the

rather small area on the tip of the telescopic boom. Details on the actuator choice can be

found in Appendix II.

3.5 LINEAR ACTUATOR L3

Actuator L3 provides linear motion to the telescopic boom

and hence the pannier. It pushes and pulls the telescopic

boom relative to the outer boom, and its angle to the

horizontal is determined by the outer boom angle. This

actuator bears the load of:

- The operators and equipment(including wind load and

load caused by movement of the operators)

- The pannier(including wind load on the pannier)

- Actuator R1

- The telescopic Boom(including wind load on the

boom)

- Load caused by acceleration

By creating a free-body diagram of the telescopic boom, we

can resolve the forces in the direction of the telescopic

boom2:

( )

Which can be rearranged into:

2 It is assumed that all the load acting perpendicular to the boom is absorbed by the bearings between the

telescopic boom and the outer boom.

Fwind, pannier

Fwind, telescopic

mpannier

g

mtelescopicg FL3

θOuter

Fig 3.5.1 Load acting on L3

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( ) (eq 3.5.1)

Where wind loads can be calculated in accordance with BS7171:1989 by:

where S is the surface area of the component against wind. For the

pannier, S is 2.182m2.3 For the telescopic boom, it is dependent on the extension of L3

and the angle between the boom and the horizontal, in the worst case scenario where

the L3 is at its maximum extension and vertically upwards, S is equal to 4.452m2.

The acceleration in eq3.1 is in accordance with BS7171:1989 to be 0.4m/s.

From the equation a plot of against can be generated in to find the angle at

which the load on L3 is the highest. It is not hard to see that the maximum angle is when

the telescopic boom is pointing vertically upwards, and the plot (which can be found in

Appendix) confirms this.

Therefore, from eq 3.5.1, the maximum load on L3 is calculated to be 32040N, and

multiplying it with a safety factor of 2 gives an L3 load of 64080N.

The actuation length is 5460mm, as required by section 3.2.

Chosen Actuator:

Parker MMA series ‘Mill Type’ actuator, mounting style MF4 (Rear Flange and foot

mounting) with 80mm diameter Bore and 50mm diameter Piston Rod and oversized

ports

- Maximum Pulling force: 79000N

- Maximum Pushing force: 128000N

- Actuation Length: 5460mm

- Diameter: 210mm

- Actuation speed: 0.28m/s

- Continuous Pressure: 250bar

- Recommended Flow Rate: 85L/min

The Parker MMA series actuators are available in ‘any reasonable actuation length’,

hence the actuation length is chosen so that there is no excessive actuation length to

3 Including 1.4m2 for the operators, as per BS7171:1989 5.3.3.3.

Fig 3.5.2 Parker MMA with

MF4 mounting

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keep the mass of the actuator to a minimum. The ‘oversized port’ option is taken to

increase the actuation speed from 0.14m/s to 0.28m/s in order to allow the pannier to

move faster, and the new movement speed is still well below the speed required by

BS7171:1989.

The detailed calculation(including MATLAB code) and selection process can be found in

the appendix.


Actuator L2 controls the angle between the inner boom and the outer boom, with a

range of 0°-110°as per section 3.2. The actual maximum and minimum actuation

length is determined by the mounting positions of L2 on the inner and outer boom.

The geometry of the booms and L2 is shown in the diagram below. The distance from

the inner/outer pivot to L2’s mounting point on the inner boom is named ‘b’, the

distance of L2’s mounting point on the outer boom from the projection of the

inner/outer pivot on the outer boom is named ‘a’, and the actual distance from the

inner/outer pivot to the L2 mount is named ‘a’’(a-dash). The distance from the line

representing L2 to the pivot point is ‘c’.

The moment of the environmental and mass forces about the inner/outer pivot can be

calculated in three parts, the mass component is:

( ) (

)

(eq.3.6.1)

a

b

a’

c

L2

Fig 3.6.1 L2 geometry

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The wind component is4:

( )

( )

(eq.3.6.2)

The inertial component is:

(eq.3.6.3)

Where is the angular acceleration (or angular

velocity/1, as per section 3.3), calculated by:

(eq.3.6.4).

Therefore the total moment acting on the inner/outer

pivot is:

(eq.3.6.5)

From which the value of L2 can be calculated. It is assumed that all of the forces caused

by the moment is taken up by L2, with none going to the inner/outer bearing.

From the equations, some observations can be made:

- From eq.3.6.1, 3.6.2 and 3.6.4, it can be seen that all moment components are

maximized when L3 is at its maximum of 5460mm.

- From eq.3.6.5, it can be seen that for a constant , the value is maximized

when c is at its smallest.

From the geometries, it is not hard to find that c is minimized when the angle between

the inner/outer booms are at their maximum, i.e 110 degrees. Hence the only variable in

the equations are c(which can be calculated using trigonometry from a and b values, as

shown in the appendix) and the outer boom angle with the horizontal, (which is

dependent on inner boom angle and the inner/outer boom angle, and the latter is fixed

at 110 degrees). Therefore a MATLAB script is written to calculate the L2 load for every

4 According to BS1989:7171, the wind force are assumed to act at the centroid of the component.

Fwind, pannier

Fwind, telescopic

mpannierg

mtelescopicg

FL2

θOuter

Mouterg

Fwind, outer

Fig 3.6.2 Load on L2

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possible outer boom angle(by varying the inner boom angle) using assumed a and b

values. In the same script, the maximum and minimum actuation lengths of L2 are also

calculated using trigonometry.

The obtained data is compared with the available actuators in the catalogue, and the

calculation process is repeated until a suitable actuator can be matched to the data.

In the end the maximum load for L2 is calculated to be 603065N, and with a safety factor

of 2, the load is 1206130N. This happens when the inner boom angle with the horizontal

is at 70 degrees (upwards) and the inner/outer angle is at 110 degrees. The maximum

calculated actuation length of L2 is 6750mm, and the minimum length is 3610mm,

giving a stroke length of 3140mm.

Chosen Actuator:

A pair of Parker MMA series ‘Mill Type’ actuators, mounting style MP3 (Cap Fixed

Eye)with 250mm diameter bore and 160mm diameter piston rod and oversized ports


(1494000N in pair)


(2500000N in pair)


- Diameter: 412mm


- Recommended Flow Rate: 340L/min

(680L/min in pair)


Once again, Parker MMA series actuators are chosen, and the actuation length is once

again selected for there to be no excessive length, and the oversized port option is taken

up for the same reason in section 3.5. As the maximum pulling force of a single actuator

is not enough to power L2, a pair of the same actuators is used to complete the task. The

mounting style is chosen to allow for the extra rotation degree-of-freedom to be added

to L2, and since there will not be any other degrees-of-freedom, the spherical bearing

option in the mounting is not taken up.


Actuator L1 controls the inner boom angle, and since the maximum and minimum

actuation lengths are already set in the requirements, the only parameter to be

calculated is the maximum load on L1. This is done by first resolving moment about the


MP3 mounting

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inner boom/boom turntable pivot to obtain the force on the push link, then resolving

moment about the tie link/boom turntable pivot to obtain the force on L1. Since it is

assumed in section 3.6 that no load goes into the inner/outer bearing, the calculation of

L1 will hold the same assumption, and all the forces from the outer boom, telescopic

boom and pannier will come in the form of the L2 force. Hence the moment about the

inner/turntable pivot can be calculated as:

(eq.3.7.1)

Where L2 to pivot and push link to pivot lengths can be calculated with trigonometry,

and the angular acceleration can be calculated using the same method as outlined in

section 3.6.

After is obtained, the moment about the tie link/turntable pivot is:

(eq.3.7.2)

From which L1 can be calculated.

The only unknowns in the equation are the inner boom angle, and the L2 force, both of

which depend on the inner boom angle (obviously) and the inner/outer boom angle5.

The inner boom angle can be related to L1 length using geometrical methods, therefore

a 3D plot of can be generated for the full range of L1 and inner/outer boom angle.

In the end, the maximum L1 load is found to be 1276206N, with a safety factor of 1.5, the

load will be 1914309N.

Chosen Actuator:

A pair of Parker MMA series ‘Mill Type’

actuators, mounting style MT4 (Intermediate

Trunnion)with 320mm diameter bore and

200mm diameter piston rod


(2530000N in pair)


(4100000N in pair)

5 To simplify the calculation, the L2 geometry obtained in section 3.6 will not be used, and only using the

inner/outer angle will achieve exactly the same effect.


MT4 mounting

15


- Diameter: 530mm


- Recommended Flow Rate: 340L/min (680L/min in pair)


As with L2 and L3, Parker MMA series actuators are chosen, as their top of the range

actuators can perfectly satisfy the system’s needs without the need to use more

heavy-duty actuators. The oversized port option is not available on this bore hence it is

not used. Much like L2, the maximum pulling force of a single actuator is not enough to

power L1, and a pair of the same actuators is used. The intermediate trunnion mounting

allows the actuator(s) to be mounted with the trunnion connected to the turntable,

satisfying the packaging requirements outlined in section 3.3.


Rotary actuator R2 enables the booms to rotate with regards to the wagon boom, adding

additional mobility for the pannier, enabling the operators to reach a wider range of

positions. The load on R2 consists of types of torques – inertial torque caused by the

Moment of inertia of the components when they are accelerating, and wind torque

caused by wind. Detailed calculations can be found in Appendix II and Appendix III. The

results from the calculations indicate a maximum torque of 94988.1Nm, and with a

safety factor of 2 this translates to an R2 torque of 189976.1Nm.

Chosen Actuator:

Parker M series ‘Mill Series’ Rack and Pinion Rotary Actuators, face mounted, model

3000M, double rack.

- Maximum torque: 3000000lb-in(338954.5Nm)

- Motion Range: 90 degrees in both directions

- Rotational Tolerances: -0, +2 degrees


- Bore: 247.65mm (9.75 inches)

- Displacement: 28.3l

- Flow Rate: 603l/min

- Mass: 2222.6kg

Fig 3.8.1 Parker 3000M

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The Parker 3000M satisfies our requirements, and will be face mounted to the bottom of

the wagon boom, with its shaft mounted in the boom turntable. Details on actuator

selection can be found in Appendix II.


The calculation for actuator R3 is almost identical to that of R2, except that the wagon

boom’s moment of inertia and wind forces are taken into account. Details of the

calculation can be found in Appendix II and III.

The results showed that the torque on R3 is 140683.3 Nm, with a safety factor of 2, the

required load for the actuator is 281366.6Nm.

Chosen Actuator:

Parker M series ‘Mill Series’ Rack and Pinion Rotary Actuators, face mounted, model

4000M, double rack.

- Maximum torque: 4000000lb-in(451939.3Nm)

- Motion Range: 90 degrees in both directions

- Rotational Tolerances: -0, +2 degrees


- Bore: 285.75mm (11.25 inches)

- Displacement: 39.3l

- Flow Rate: 281.4l/min

- Mass: N.A

Details on the selection can be found in Appendix II.

4. CONTROL SYSTEM

4.1 PUMP

The maximum continuous pressure that the pump has to sustain, from the actuator

selections in section 3.1 to 3.9, is 250bar, while the maximum flow rate that the pump

has to produce is 680L/min. By comparing this data with the catalogues, the pump that

satisfies this requirement is the Rexroth Bosch A4VSG. The A4VSG is an axial piston

swash plate pump with a high efficiency, and at size 500, it produces a maximum flow

rate of 900L/min at maximum efficiency, and 750L/min at 1500rpm, which is more than

enough to power the system providing that only one actuator is power at any given time.

Fig 3.9.1 Parker 4000M

17

The nominal pressure and maximum pressure allowed for this pump are 350bar and

400bar respectively, also satisfying the system’s requirements.

4.2 MOTOR

The A4VSG pump requires a maximum power of 525kW and a maximum torque of

2783Nm. To satisfy this, the CAT 3412C Engine is chosen to power the system. The

3412C is a V8, four stroke diesel engine that produces a minimum power of 476kW and

a maximum power of 551kW that satisfies the pump’s required power.

4.3 CONTROL VALVES

Control valves are required in the system to control the actuation of the actuators. One

control valve is required for each actuator. Since the actuators are required to move in

both directions(push and pull), a 4/3 valve is chose, which allows the system to move in

both directions, as well as blocking all ports to keep the system in an equilibrium

position.

Fig 4.1.1 Rexroth Bosch A4VSG

Fig 4.2.1 CAT 3412C

18

From the catalogues, a 4/3 valve that can sustain the system’s maximum pressure of

250bar needs to be found, and 6 Rexroth Bosch 4WEH 25NG 4/3 control valves will be

used to control the actuators. These valves have a maximum pressure of 350bar and a

maximum flow rate of 1100L/min, both values well above our requirements. These

valves are pilot operated, as the force required to activate the valve cannot be provided

manually. It includes two spool valves, with the first being operated by a solenoid, and

the second being operated by ‘pilot oil’ from the first spool valve, much like using a low

voltage signal to control a high voltage system in electric control systems.

4.4 FILTER

The actuators use mineral oil as the hydraulic fluid, which can be easily contaminated. In

addition to the sealing systems in the control system components, a filter is required to

maintain the purity of the oil to maintain the performance of the hydraulic system. A

Rexroth Bosch Duplex tank mounted return line filter of size 1000 is to be used on the

return line of the hydraulic system before the fluid returns to the reservoir. The selected

filter can sustain a flow rate of up to 1270L/min, well above our requirement.

Fig 4.3.1 Rexroth Bosch 4WEH Fig 4.3.2 The two spool valves of a

pilot operated control valve

Fig 4.4.1 Rexroth Bosch Duplex

filter

19

4.5 COOLING DEVICE

A cooling device is required to stop the mineral oil in the hydraulic system from

overheating, which could decrease the performance of the system and, in the worst-case

scenario, cause a complete system failure.

Very few cooling devices available on the market are able to sustain the high flow rate A

Bowman RK400-1698-6 with a maximum flow rate of 900L/min is almost the only

cooler available for the system.

4.6 ACCUMULATOR

An accumulator is not required in the hydraulic system, however it is an useful addition

to the system in making the system more efficient as it reduces the pumping

requirements without increasing the energy used.

A Rexroth Bosch HAB 5000/6000psi(344bar/413bar) accumulator of 2.5 gallon(11

liters) is chosen, which allows a maximum flow rate of 160 gallons of minutes(727 litres

per minute).

Fig 4.5.1 Bowman RK series filter

Fig 4.6.1 Rexroth Bosch HAB accumulator

20

5. SAFETY FEATURES

5.1 RECOMMENDED OPERATING PROCEDURES

- During the operating process, the magnitude and moving directions of the platform

should be identified on the screen of the control system using easily identifiable

numbers or symbols.

- Install strain gauges on different parts of the crane system. If the strain in exceeds a

certain threshold value, the alarm should warn the controller immediately,

allowing the operators to know when the crane might fail.

- Making sure the system does not move without control signal issued by the

operators by regularly checking the control valves.

- Set a lockable switch to prevent unauthorized use.-

- Set an emergency button for some emergency conditions. At that time, the

emergency button can stop the whole operation and cut off the current in the

control circuit.

- Install shields to cover some important buttons to prevent accidental use.

- The qualifications of the staffs should be checked to ensure that they know the

safety protocols in the event of failure.

- Install flow rate in the pipelines so that if the flow rate exceeds the designed value

of any of the actuators, sensors will activate the pressure relief system or flow relief

valves.

- All staff should wear special working uniform and helmet, and the floor of the

pannier should be covered by non-slip materials.

- Periodic maintenance and checking is necessary.

- The system only can start work under all regulations and preparations are ready.

5.2 THE COMPUTER SYSTEM

A computer system can be applied to the operation of the system. It just treat the

whole device in a 3D system. The size of all parts are defined, so if you type the input

data such as translation, elevation and rotation distance, these orders will be carry out

by all actuators in order to move the pannier to the designed terminal point.

In order to achieve automatic transversing, the computer system can be integrated to a

set of motion controllers (e.g. angular, speed, torque controller) attached on the

articulated structure. Therefore feedback from the structure’s motions will be measured

simultaneously by the controller, followed by analysing from computers. Also the

software which from the same pack with controllers is preferably used to prevent

inaccurate feedback.

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5.3 COMPARISON OF CONTROLLERS

According to various type of machines, Each controller has its unique advantages in

certain fields of application. Generally they can be combined easily, which is especially

important for complex components. The individual hardware platforms always have the

same system properties, i.e. functionality and engineering are always identical -

irrespective of which platform is used.

- Obstacle-detecting sensor: Some obstacle-detecting sensors that are appropriately

fixed on different parts of the whole structure can measure the distance between

itself and the obstacle. These sensors are linked with the computer as the input

data. If we want to move the pannier to one particular position, the software will

process and calculate the moving distance and rotating angle for different actuators.

In this case, we still have to make sure the hardware is reliable and durable. Finally,

the designed position can be achieved.

- Motion controller with software: from this combination of control system, not only

the input motion from staff can be processed, but also unexpected interferes (e.g.

strong wind) can be monitored by the close loop controllers and therefore make

motors act on a more precise way.

- Drive-based motion control: The drive-based motion control system sets

itself apart as it is a complete system comprising open-loop control and

drive. This makes the machine significantly more compact with an

especially fast response.

Fig 5.2.1 Example of a computer control software from DELTA

company

22

- PC-based motion control: The most PC based controllers are ready to

switch on, no complex installation and with real-time expansion from

the designer.

6. REFERENCES

DUCCI, A. (2014). DESIGN OF SERVO HYDRAULIC CONTROL SYSTEM FOR AN ARTICULATED

INSPECTION AND CLEANING WORK PLATFORM. 1ST ED. [PDF] AVAILABLE AT:

HTTPS://MOODLE.UCL.AC.UK/MOD/RESOURCE/VIEW.PHP?ID=1700255

W3.SIEMENS.COM, (2014). HARDWARE PLATFORMS - MOTION CONTROL SYSTEM SIMOTION - SIEMENS.

[ONLINE] AVAILABLE AT:

HTTP://W3.SIEMENS.COM/MCMS/MC-SYSTEMS/EN/AUTOMATION-SYSTEMS/MC-SYSTEM-SIMOTION/HARDWARE

-PLATFORMS/PAGES/HARDWARE-PLATFORMS.ASPX

BRITISH STANDARDS INSTITUTION, (2014). BS 7171:1989 - SPECIFICATION FOR MOBILE ELEVATING WORK

PLATFORMS.

BRITISH STANDARDS INSTITUTION, (2014). BS EN 280:2001+A2:2009 - MOBILE ELEVATING WORK

PLATFORMS. DESIGN CALCULATIONS. STABILITY CRITERIA. CONSTRUCTION. SAFETY. EXAMINATIONS AND TESTS.

BRITISH STANDARDS INSTITUTION, (2014). BS 1757:1986 - SPECIFICATION FOR POWER-DRIVEN MOBILE

CRANES

MERIAM, J. (2014). ENGINEERING MECHANICS: DYNAMICS. 7TH ED. WILEY.

23

APPENDIX I: SAMPLE CALCULATIONS AND DETAILS ON ACTUATOR SELECTION

I.I COMPONENT MASSES AND MOMENT OF INERTIAS

The masses of all components are obtained by modelling the components in CATIA V5

using the provided dimensions and calculating the masses by applying the built-in

materials to the components and recording the resulting mass from CATIA V5. The

masses obtained using this method can be found in the MATLAB code Appendix II.I. As

stated in the code, all masses are in kilograms. It is inevitable that these masses are

inaccurate due to the lack of some key pieces of data (e.g. radius of the curved section on

the tie link), but the large safety factor should prevent the discrepancies from causing

any failures.

The moment of inertias(MOIs) used in the inertial forces calculations are also obtained

by CATIA V5 using the built-in ‘measure inertia’ function. The MOI of the components

are taken about the axis about which the moments are calculated – Centerline for the

pannier shaft for R1, inner/outer pivot for L2, turntable/inner pivot and tie

link/turntable pivot for L1, turntable R2 axis for R2 and wagon boom base axis for R3.

The obtained values can also be found in Appendix II.I.

Finally, the center of gravity(CG) of all the components are obtained using CATIA’s

built-in function, from which all the mass forces will be assumed to act.

I.II ACTUATOR R1

I.II.1 Sample Calculation

Actuator R1 ensures that the pannier is horizontal to the ground regardless of the outer

boom and inner boom angles, making sure that the operators are safely protected and

won’t suffer from any catastrophes. There are two types of loads acting on R1, the first

being the wind torque, the second being the mass torque6. The worst case scenario for

this actuator is when all the mass loads are on the same side and all acting at the point

on the pannier furthest away from the actuator (that is, right at the end of the pannier),

and when all the wind load are acting in a direction that the wind torque is in the same

direction as the mass torque, as shown in the diagram below:

6 It is worth noting that the mass/inertial torque from the pannier itself will not be taken into account as

the pannier is expected to be horizontal all the time, and the line of action of the weight of the pannier goes

through the R1 axis.

24

The mass of the operators and equipments in total are 350kg, therefore the mass torque

is:

The wind acting on the operators(assumed to be 0.7m2 as per BS7171:1989), and the

area of the pannier that is subject to wind is: .

Therefore the wind torque is(assuming W_Operator acts at the top of the

pannier):

Therefore the total torque that R1 has to take is 4199Nm.

I.II.2 Actuator Selection

The packaging constraint on the tip of the telescopic boom meant that the actuator had

to be rather compact. It is suggested that the Parker MR series of so-called ‘radial piston

motors’ can be suitable for the task, and in the catalogue it is found that the MR2400I

model is very suitable to be used for actuator R1, with a starting torque of 8572.5Nm

and a maximum torque of 9525Nm. It is designed for ‘high torque low speed’ operations,

and the task of levelling the pannier fits this description rather well. The actuator can

also satisfy the requirement of rotating 180 degrees in both directions, and can in fact

rotate 360 degrees, which satisfies our requirement.

I.III ACTUATOR L3

I.III.1 Sample Calculation

As described in section 3.5, the force on L3 on purely dependent on the value of ,

and by plotting a diagram of against L3 load, the maximum load configuration

can be found. As the effect of the mass forces are much higher than that of the wind

forces, it is not surprising to find that the maximum load configuration is when all the

mass forces are acting directly on L3 – when is equal to 90 degrees, at which

point no wind forces are acting on L3.

M_operator*g

W_operator

W_pannier

Fig I.II.1

Forces on R1

25

Therefore the load is:

( )

There is no angular acceleration involved in the movement of L3, only linear

acceleration along the outer boom, and since the outer boom is pointing vertically

upwards, the acceleration is simply 0.4m/s / 1s = 0.4m/s2 as per BS7171:1989.

I.III.2 Actuator Selection

The project group has obtained catalogues of three series of actuators from the website

of Parker Hannifin – the 2H series, the 3H series and the MMA series. From the load

calculated, the 83.6mm bore actuators of the 2H series and the 80mm bore actuators of

the MMA series both satisfy the load requirements. In the end the MMA series is chosen

as it is advertised to be ‘fatigue-free’ at its operating pressure of 250bar, which is an

ideal feature.

Next the mounting style is chosen. MF4 rear flanged mounting is chosen as a rear flange

can be easily mounted to the end of the outer boom with a few bolt holes drilled out, and

there is no need for the extra degrees-of-freedom provided by other mounting styles.

Side mountings are also considered, however in the end due to the long actuating length

of the cylinder it is found that side mounting is impossible to be implemented inside the

booms, and the actuator will have to be placed outside the booms, with higher

possibilities of contamination and damage.

Fig I.III.1

Plot of outer

boom angle

verses L3

Load

26

Finally the loads are compared with the bore and piston selection charts in the catalogue

to find the bore size. Due to the existence of the piston rod, the pulling force of the

cylinder will always be smaller than the pushing force, so the main objective is to find a

cylinder with a pulling force higher than the 64080N required. The bore and piston

combination of 80mm and 50mm satisfies this criteria, and although both 200bar and

250bar can provide enough force to power the cylinder, 250bar is chosen to leave a

larger safety margin, even in the existence of a large safety factor.

I.IV ACTUATOR L2

I.IV.1 Sample Calculation

To calculate the load on actuator L2, the kinematics of the L2 mechanism must first be

analyzed. Below is the diagram shown first seen in section 3.6:

The a and b values are assumed, and therefore known. The value of ‘c’ can be calculated

as long as the value of the inner/outer boom angle(the angle between a and b) is known.

First we need to transfer the value of a into the value of a’, as the value of a cannot be

applied to the triangles: √ .7 For an ‘a’ value of 2m, a’ is equal

to√ . Since is known, we can calculated the

L2 value under this angle: √ , for a=2m, b=5m and

, √ . With all

7 E1+0.225 is the value from the inner/outer pivot to the upper surface of the boom, and 0.225 is half the

height of the boom, it is assumed that the L2 is mounted on the middle of the outer boom, hence the weird

expression.

a

b

a’

c

L2

Fig I.IV.1

L2 Geometry

27

three sides of the triangle known, the knowledge of either b/L2 or a’/L2 will give us the

value of c. For example:

and now we calculate the moment about the inner/outer pivot that L2 has to act against.

As evident from eq.3.6.1, 3.6.2, 3.6.3 and 3.6.5, this moment depends on the inner boom

angle and L3 length. The L3 length that maximum that maximizes the load can be easily

identified as the one that gives the highest moment arm, i.e when L3 is at its maximum

of 5.46m. The MATLAB code from Appendix II.III shows that this moment is maximum

when the inner boom angle is 70 degrees, which is when the outer boom is horizontal to

the ground( ). It is not hard to see that this

configuration maximizes the moment arm of the mass forces, which has overpowered

the decrease in wind load. Therefore this angle:

( )

( )

( ) ( )

( )

Fig I.IV.2

Plot of inner

boom angle

versus L2

load

28

( )

Therefore, L2 load is:

∑

I.IV.2 Actuator Selection

Like L3, the Parker MMA series catalogue is consulted for the selection of L2. L2 needs to

be able to exert a much higher force than L2. Quite a few actuators are able to exert 1.2

million newtons in push, but very few can exert this amount in pull due to the larger

force reductions in larger bore cylinders due to larger piston rods. Also, as the distance

between the inner and outer boom in the crane’s stowed position is 469mm, as shown

below, the 320mm bore cylinders cannot be used as they won’t fit into the gap. The

maximum cylinder bore is 250mm, and since this does not provide the required force

even under maximum pressure, two actuators are used, with one mounted on each side

of the outer boom. As stated in section 3.6, due to the extra degree-of-freedom needed,

the cap fixed eye mounting is used to allow the rotation.

I.V ACTUATOR L1

I.V.1 Sample Calculation

Actuator L1 has a rather complex kinematics system that requires solving before any

load calculations can take place. The length of L1 controls the angle of the inner boom

through the tie link, and the motion pattern of all the components are required for the

calculations.

The project group first attempted to solve the kinematics analytically using

trigonometry. First, the relationship between L1 and the tie link is examined.

D

B L1

A

O

Fig I.V.1

L1 Geometry

(1)

29

As above, the distance AO, BO and L1 are all known(with a give L1 value):

√ , . Hence we can

work out the angle AOB using cosine rule:

For an L1 value of 1.6m, the angle is equal to 24.17 degrees, therefore the angle between

the tie link and the horizontal

Now the relationship between the tie link and the inner boom is examined.

The task of solving the geometry of the quadrangle is made easier by connecting DE and

separating it into two triangles. The angle between EF and the horizontal is what is

required, so the analysis starts from the bottom triangle:

√

DO=B1=1.871m, EO=sqrt((h4-h5)2+(A3-A2)2)=0.945m. For L1 = 1.6,

(

) , hence, DE=1.857m.

Now we apply a similar procedure to the top triangle:

D F

B

D

O

E

Fig I.V.2

L1 Geometry

(2)

30

Where DF=B3, EF=sqrt(B42+C22), hence

To find the angle between EF and the horizontal, we must first find the angle between

DE and the horizontal:

29.36-74.52=76.12 degrees.

Therefore the inner boom angle is 180-76.12-33.01=70.87degrees.

This method is transformed into the MATLAB code in appendix II.V, and a plot of inner

boom angle against L1 is plotted to check its validity. The inner boom angle relationship

is also obtained in CATIA empirically by varying L1 with 0.1m intervals and recording

the resulting inner boom angle.

The resulting graph shows that this method somewhat erroneous, and as de-bugging

took much time and was stopping the group’s progress, it is decided that the L1/inner

boom angle relationship will be obtained by fitting a polynomial trend line to the

empirical curve and using the equation of the trend line as the relationship. The push

link angle, L1 angle, as well as the moment arms of L1 and push link are all obtained

using this method to ensure accuracy.

Fig I.V.3

L1 vs. inner boom

angle curve for

empirical model

and analytical

model

Fig I.V.4

Plotted empirical

curves from CATIA

data

31

Finally the L1 load calculation can start. The load is calculated in two parts – firstly

moment is taken about the turntable/inner boom pivot to resolve for the force on the

push link, then moment is taken about the tie link/turntable pivot to find the force on L1.

For the turntable/inner boom pivot:

Since it is assumed all of the forces acting on the outer boom/telescopic boom and

pannier are reacted by L2, and there are no forces at the pivot bearing, it is assumed

here that the L2 force is equal to the contribution of the masses and wind forces of all

the above components, and the only force acting on the inner boom are the push link

force, the inner boom’s mass and L2 force.

As per eq.3.7.1:

For L1=1.6m and the inner/outer boom angle at its maximum of 110degrees, the push

link force is:

Next this force is transferred to L1 through the tie link:

M_inner*g

F_L2

F_PL

Fig I.V.5

Load on Push Link

F_L1 F_PL

Fig I.V.6

Load on L1

32

Eq.3.7.2 describes the relationship between the push link force and L1 force:

From which L1 can be calculated, as the moment arm is already known through

empirical methods:

By varying L1 length(thus the inner boom angle) and the inner/outer boom angle, a 3D

plot of the force on L1 is produced, and the maximum is found to be when the inner

boom is at 36 degrees and inner/outer at its maximum of 110 degrees, with the

maximum push link force being 1467797.8867N, and maximum L1 force

1276206.6767N.

I.V.2 Actuator Selection

The final linear actuator choice is still made from the Parker MMA series catalogues, and

straight away the MT4 mounting style is chosen – any rear mounted cylinders with an

actuation length of 4653mm will not fit L1’s minimum length of 1058mm, hence it will

have to be either front flange mounted or trunnion mounted, and from the turntable

shape trunnion mounting seems to be the more suitable option.

L1 suffers from the same problem as L2, in that there are no single actuators that are

able to provide the level of force required, particularly in pull. The largest force that an

MMA series actuator can provide (320mm bore and 200mm piston rod) is 2050000N in

push and 1250000N in pull. Hence like L2, two actuators will act in tandem. This is a

rather compromised solution, and the turntable mounting point might require

redesigning to fit two actuators in.

Fig I.V.7

3D plot of L1 load

and inner and

inner/outer boom

angle

33

I.VI ACTUATOR R2

I.VI.1 Sample Calculation

Actuator R2 only needs to counter two main types torques – the inertial forces caused

by the angular movement of the components, and the wind forces on the components.

The inertial forces are calculated by: ∑ .

Since I is so much larger than , it is not hard to see that the configuration that causes

the load on L2 to be maximum is the configuration that gives the highest moment of

inertia of the components. That is, when the pannier is furthest apart from R2. A

moment arm calculator is created in MATLAB to find this configuration, as can be seen

in appendix II.VI, and the maximum position is found to be when the inner boom is at 36

degrees and when the inner/outer angle is at 110 degrees. This configuration is fed into

CATIA V5, and the corresponding moment of inertia values

are obtained. The pannier is 24.5m from R2 in this

configuration, hence the angular velocity that causes the

maximum allowed velocity in BS7171:1989 of 0.7m/s

is:

. With the assumption of

acceleration time of 1s, the angular acceleration is

0.0286m/s2. Therefore the inertial torque for this

configuration is: (

)

.

The wind torque can be calculated as: ∑ ,

and with the r values measured from CATIA, the wind

torque is:

(The full list of moment of inertias, radiuses and wind forces

can be found in Appendix III.)

Therefore the total torque on R2 is .

I.VI.2 Actuator Selection

With a safety factor of 2, the required actuation force of R2 has to be above 189976.1Nm.

From the Parker Hannifin M series catalogue (which uses imperial units of lb.in for

34

torque), all the actuators above 2000M satisfies this target. In the end the 3000M

actuator is chosen to provide a larger safety margin, as the 2000M produces only

220000Nm at 204bar, which leaves a safety margin that is slightly too small. As required

in section 3.2, it has to be able to rotate 90 degrees both ways, hence the 90 degree

version is chosen.

Finally, for mounting, the actuator is only available with face mounting and foot

mounting, and in its application face mounting is the better option as it can be directly

mounted to the wagon boom without the need of an extra mounting bracket.

I.VII ACTUATOR R3

I.VII.1 Sample Calculation

The calculation method of R3 is the same to that of R2, with an extra component

included(wagon boom) and with different radiuses and MOIs used due to the different

axis location. Due to the fact that the equation calculating them are the same, the

configuration that causes the maximum torque on R3 is exactly the same as that of R2.

The inertial torque is: (

)

The wind torque

is:

Hence the total torque that R3 has to exert is:

I.VII.2 Actuator Selection

Again, the Parker M series rotary actuators are chosen. With a safety factor of 2, R3

needs to exert 281366.6Nm of torque. Both 3000M and 4000M series actuators can be

used, and it is decided in the end that the 4000M actuator will be selected for the same

reason that the 3000M is used for R2 – there is a higher safety margin when 4000M is

used(about 150kN versus the 50kN of 3000M). Again, according to the requirements in

3.2, the motion range is 90 degrees.

The mounting option selected is face mounting, for the same reason that it is selected for

R2, that no extra brackets are needed to mount the actuator.

APPENDIX II: MATLAB CODE

35

II.I MAIN PROGRAM %Load Calculator For MECH3010 Coursework of Hydraulic Systems Design

%Written by: LeHeng LI, Linfei WU

%Last Updated: 12.Nov.2014

%All Lengths in m unless stated, all weights in kg unless stated.

%Data

%%

disp('Initializing...');

%Environmental Data

g = 9.81;

%Boom Turntable Data

A1 = 2.420; A2 = 0.715; A3 = 1.320;

h1 = 1.265; h2 = 0.605; h3 = 0.138; h4 = 0.960; h5 = 0.234;

%Tie Link Data

B1 = 1.871; B2 = 0.660; B3 = 1.133; B4 = 0.220;

%Inner Boom Data

C1 = 11.880; C2 = 1.024; C2_dash = sqrt(B4^2+C2^2);

%Outer Boom Data

D1 = 0.375; E1 = 0.688; E2 = 9.900; E3 = 0.248; E4 = 0.220;

%Telescopic Boom Data

F1 = 0.820; F2 = 9.130; F3 = 0.550; L3 = 5.460;

%Pannier Data

P1 = 0.809; P2 = 0.893; P3 = 0.283; P4 = 0.420; P5 = 0.221; P6 = 0.1859; P7 =

0.1943;

%Intrinsic Angles

Theta_1_Rad = atan((h4-h5)/(A3-A2)); %Angle between Inner Boom and Tie Link holes

on Turntable

Theta_1_Deg = toDegrees('radians', Theta_1_Rad);

Theta_2_Rad = atan(B4/C2);

Theta_2_Deg = toDegrees('radians', Theta_1_Rad); %Inner boom Push link connection

Angle

Theta_3_Rad = atan((h1-h2+h4)/(A1+A2));

Theta_3_Deg = toDegrees('radians', Theta_3_Rad); %Angle betwen L1 link and Tie

link holes on Turntable

%Component Mass

M_Pannier = 599.898;

M_Outer = 4101.725;

M_Tele = 2187.437;

M_Inner = 6262.841;

M_TieLink = 1049.413;

M_PushLink = 190.454;

36

M_Turntable = 7407.415;

M_Wagon = 3042.096;

M_Operators = 350;

%Component CG

L_CG_Tele = 5.978847; %From Flat End

L_CG_Outer = 2.383444;

h_CG_Outer = 0.611082; %From Inner Conn End

L_CG_Inner = 5.671185;

h_CG_Inner = 0; %From Turntable Conn End

%Component MOI

%MOI around Pannier Axis(R1)

I_Pannier_Axis = 246.301;

%MOI around Inner/Outer Bearing(L2)

I_Tele_IO = 335664.646;

I_Outer_IO = 61179.642;

I_Pannier_IO = 138386.775;

%MOI around Turntable/Inner Bearing(L1)

I_Pannier_TI = 433030.485;

I_Tele_TI = 1.249*10^6;

I_Outer_TI = 858487.046;

I_Inner_TI = 325052.768;

%MOI around R2

I_Pannier_R2 = 470008.886;

I_Tele_R2 = 1.372*10^006;

I_Outer_R2 = 1.017*10^6;

I_Inner_R2 = 441196.111;

I_Turntable_R2 = 5063.101;

I_TieLink_R2 = 2643.287;

I_PushLink_R2 = 904.314;

%MOI around R3

I_Pannier_R1 = 644801.439;

I_Tele_R1 = 1.944*10^6;

I_Outer_R1 = 1.72*10^6;

I_Inner_R1 = 1.048*10^6;

I_Turntable_R1 = 160242.858;

I_TieLink_R1 = 41885.897;

I_PushLink_R1 = 9259.547;

I_Wagon_R1 = 18256.412;

%Tie Link MOI around Turntable

I_TieLink_T = 2544.423;

%Push Link MOI around Tie Link

37

I_PushLink_T = 680.417;

%Wind Load

W_Inner = 100*1.1*0.6*11.88;

W_Tele = 100*1.1*4.452; %N

W_Pannier = 100*1.1*2.182; %N

W_Outer = E2*0.7*100*1.1; %N

Load_L3

Load_L2

Load_L1

disp('--------------------------------------------------------------------')

;

disp('All Calculations Completed.');

II.II L3 CALCULATOR %Load Calculator(L3) For MECH3010 Coursework of Hydraulic Systems Design

%Written by: LeHeng LI



disp('Calculating L3...')

disp('-------------------------------------------------------------------');

Theta_T_Rad = 0:pi/180:2*pi; %Boom Angle in rad

acc_p = 0.4; %Maximum acceleration of pannier in m/s

F_L3 = (M_Tele + M_Pannier + M_Operators) * acc_p ...

+ M_Tele * g * sin(Theta_T_Rad) ...

+ (M_Pannier + M_Operators) * g * sin(Theta_T_Rad) ...

- W_Tele * cos(Theta_T_Rad)...

- W_Pannier * cos(Theta_T_Rad);

% plot (Theta_T_Rad, F_L3) %Plot of Tele angle vs. F_L3

[max_F_L3, ang_max_F_L3] = max(F_L3);

disp([' Max F_L3: ', num2str(max_F_L3), ' N']);

disp([' At: ', num2str(ang_max_F_L3), ' deg'])

disp('-------------------------------------------------------------------');

disp('Calculation Completed');

38

II.III L2 CALCULATOR %Load Calculator(L2) For MECH3010 Coursework of Hydraulic Systems Design

%Written by: LeHeng LI, Linfei WU



disp('Calculating L2...');

disp('-------------------------------------------------------------------');

%Modify a & b value here

a = 2;

disp(['a: ', num2str(a), ' m']);

b = 5;

disp(['b: ', num2str(b), ' m']);

%%

Theta_I_Rad = -84 * pi/180 : pi/180 : 157 * pi / 180;

Theta_IO_Rad = 111 * pi/180;

Theta_O_Rad = Theta_I_Rad - (pi - Theta_IO_Rad);

Theta_O_Deg = toDegrees('radians', Theta_O_Rad);

% disp(toDegrees('radians',Theta_O_Rad));

a_dash = sqrt(a^2 + (E1 + 0.225)^2);

Theta_A_Rad = acos(a/a_dash);

L2_Max = sqrt(a_dash^2 + b^2 - 2 * a_dash * b...

* cos(Theta_IO_Rad + Theta_A_Rad));

disp(['L2 Actuation Length: ', num2str(L2_Max), ' m']);

L2_Min = sqrt(a_dash^2 + b^2 - 2 * a_dash * b...

*cos(0 + Theta_A_Rad));

disp(['L2 Minimum Length: ', num2str(L2_Min), ' m']);

L3 = 5.460;

% disp(L2_Max);

Theta_L2_Rad = Theta_I_Rad - asin((a_dash * sin(Theta_IO_Rad...

+ Theta_A_Rad))/(L2_Max));

% disp(toDegrees('radians',Theta_L2_Rad));

c = b * sin(acos((b^2+L2_Max^2-a_dash^2)/(2*b*L2_Max)));

AngAcc_Outer = 0.4./15.14;

%%

39

Mass_Moment_L2 = (M_Pannier + M_Operators) .* g .* 15.14 * cos(Theta_O_Rad) ...

+ M_Tele .* g .* 12.388 * cos(Theta_O_Rad)...

+ M_Outer .* g .* 3.862 * cos(Theta_O_Rad)...

-(M_Pannier+M_Operators+M_Tele+M_Outer) * (E1+0.225) *

sin(Theta_O_Rad);

Mass_Moment_L2_Max = max(Mass_Moment_L2, [], 2);

plot(toDegrees('radians',Theta_I_Rad), Mass_Moment_L2)

Wind_Moment_L2 = W_Pannier * 15.14 .* sin(Theta_O_Rad)...

+ W_Outer * 3.862 .* sin(Theta_O_Rad) + W_Tele * 12.388 * sin(Theta_O_Rad)...

+ (W_Pannier + W_Outer + W_Tele) * (E1+0.225) * cos(Theta_O_Rad);

Wind_Moment_L2_Max = max(Wind_Moment_L2, [], 2);

% plot(toDegrees('radians',Theta_O_Rad), Wind_Moment_L2)

Inertia_L2 = (I_Outer_IO + I_Tele_IO + I_Pannier_IO) * AngAcc_Outer;

Inertia_L2_Max = max(Inertia_L2, [], 2);

%Moment on Inner/Outer Pivot

M_IO = Mass_Moment_L2 + Wind_Moment_L2 + Inertia_L2;

[max_M_IO, ang_max_M_IO_Deg_Dash] = max(M_IO, [], 2);

disp(' ');

disp(['Maximum moment on Inner/Outer Pivot: ', num2str(max_M_IO), '

Nm']);

ang_max_M_IO_Deg = ang_max_M_IO_Deg_Dash - 84;

disp(['at: ', num2str(ang_max_M_IO_Deg), ' deg']);

%%

%L2 Load

F_L2 = M_IO./c;

[max_F_L2, ang_max_F_L2_Deg_Dash] = max(F_L2, [], 2);

disp(' ');

disp(['Maximum L2 Load: ', num2str(max_F_L2), ' N']);

ang_max_F_L2_Deg = ang_max_F_L2_Deg_Dash - 84;

disp(['at Inner Angle: ', num2str(ang_max_F_L2_Deg), ' deg']);

disp(['at L2 Angle: ', num2str(toDegrees('radians',...

Theta_L2_Rad(ang_max_F_L2_Deg_Dash))), ' deg']);

% plot(toDegrees('radians', Theta_I_Rad), F_L2);

% plot(toDegrees('radians', Theta_I_Rad), M_IO);

% plot(Theta_I_Rad, M_IO, Theta_I_Rad, F_L2);

40

disp('-------------------------------------------------------------------');

disp('Calculation Completed')

II.IV L1 CALCULATOR %Load Calculator(L1) For MECH3010 Coursework of Hydraulic Systems Design

%Written by: LeHeng LI, Wenguang CHENG



disp('Calculating L1...');

disp('-------------------------------------------------------------------');

L1 = 1.2 : 0.1 : 5.7; %L1 Range

AngAcc_Inner = 0.4./(F1 + L3 + E2);

AngAcc_TieLink = 0;

%Modify a & b value here

a = 2;

disp(['a: ', num2str(a), ' m']);

b = 5;

disp(['b: ', num2str(b), ' m']);

%%

Theta_IO_Rad = 0: pi/180: 111 * pi/180;

Theta_IO_Deg = 0:1:111;

disp('Pre-allocating Matrices...');

%Pre-Allocation

Theta_O_Deg = zeros(46, 111);

Theta_O_Rad = zeros(46, 111);

Theta_L2_Rad = zeros(46, 111);

c = zeros(46, 111);

L2 = zeros(1,111);

Mass_Moment_L2 = zeros(46,111);

Wind_Moment_L2 = zeros(46,111);

Theta_TieLink_Deg = zeros(46);

Theta_PushLink_Deg = zeros(46);

Theta_PushLink_Rad = zeros(46);

Theta_PLI_Rad = zeros(46);

Theta_L2I_Rad = zeros(46,111);

Theta_I_Deg = zeros(46);

Theta_I_Rad = zeros(46);

Theta_L1_Deg = zeros(46);

41

M_IO = zeros(46,111);

PushLink_Load = zeros(46, 111);

F_PushLink = zeros(46, 111);

F_L2 = zeros(46,111);

F_L1 = zeros(46,111);

L1_TL_MA = zeros(46);

PL_TL_MA = zeros(46);

PL_TI_MA = zeros(46);

for j = 1:1:46

for k = 1:1:111

%Kinematics Calc

Theta_TieLink_Deg(j) = -0.0603 .* L1(j).^4 - 0.0766 .* L1(j).^3 +...

3.6699 .* L1(j).^2 - 36.038 .* L1(j) + 177.98;

Theta_PushLink_Deg(j) = -0.1563 .* (L1(j).^4) + 0.5786 .* (L1(j).^3) -...

4.8515 .* (L1(j).^2) + 13.922 .* L1(j) - 7.4517;

Theta_PushLink_Rad(j) = toRadians('degrees', Theta_PushLink_Deg(j));

Theta_I_Deg(j) = 0.958 .* (L1(j).^4) -13.87 .* (L1(j).^3) +...

67.413 .* (L1(j).^2) - 173.58 .* L1(j) + 271.57;

Theta_I_Rad(j) = toRadians('degrees', Theta_I_Deg(j));

Theta_L1_Deg(j) = -0.6132 .* (L1(j).^4) + 9.3072 .* (L1(j).^3) -...

53.448 .* (L1(j).^2) + 130.32 .* L1(j) - 93.954;

Theta_O_Deg(j,k) = Theta_I_Deg(j) - (pi - Theta_IO_Deg(k));

Theta_O_Rad(j,k) = toRadians('degrees', Theta_O_Deg(j,k));

PL_TI_MA(j) = -0.0043 * L1(j)^6 + 0.0932 * L1(j)^5 - 0.8136 * L1(j)^4 ...

+ 3.6842 * L1(j)^3 - 9.2504 * L1(j)^2 + 12.35 * L1(j) - 5.8582;

PL_TL_MA(j) = -0.0215* L1(j)^4 + 0.2747* L1(j)^3 - 1.3626* L1(j)^2 ...

+ 3.1269* L1(j) - 0.9644;

L1_TL_MA(j) = -0.0076 * L1(j)^6 + 0.1672 * L1(j)^5 - 1.5014 * L1(j)^4 ...

+ 7.0312 * L1(j)^3 - 18.204 * L1(j)^2 + 24.78 * L1(j) - 11.337;

%Pre Calc

L2(k) = sqrt(a_dash^2 + b^2 - 2 * a_dash * b...

*cos(Theta_IO_Rad(k) + Theta_A_Rad));

c(j,k) = b * sin(acos((b^2+L2(k).^2-a_dash^2)/(2*b*L2(k))));

%L2 Calc

Mass_Moment_L2(j,k) = M_Pannier * g * 15.14 * cosd(Theta_O_Deg(j,k))...

+ M_Tele * g * 12.388 * cosd(Theta_O_Deg(j,k))...

+ M_Outer * g * 3.862 * cosd(Theta_O_Deg(j,k))...

-(M_Pannier+M_Operators+M_Tele+M_Outer) * (E1+0.225) * sind(Theta_O_Deg(j,k));

42

Wind_Moment_L2(j,k) = W_Pannier * 15.14 .* sind(Theta_O_Deg(j,k))...

+ W_Outer * 3.862 .* sind(Theta_O_Deg(j,k)) + W_Tele * 12.388 *

sind(Theta_O_Deg(j,k))...

+ (W_Pannier + W_Outer + W_Tele) * (E1+0.225) * cosd(Theta_O_Deg(j,k));

M_IO(j,k) = Mass_Moment_L2(j,k) + Wind_Moment_L2(j,k);

Theta_L2_Rad(j,k) = Theta_I_Rad(j) - asin((a_dash * sin(Theta_IO_Rad(k)...

+ Theta_A_Rad))/(L2(k)));

Theta_L2I_Rad(j,k) = Theta_I_Rad(j) - Theta_L2_Rad(j,k);

F_L2(j,k) = M_IO(j,k)./c(j,k);

%PL Calc

PushLink_Load(j,k) = F_L2(j,k)*(C1-b)*sin(Theta_L2I_Rad(j,k)) ...

+ (M_Inner * g * cos(Theta_I_Rad(j)) * L_CG_Inner)...

+ (I_Pannier_TI + I_Tele_TI + I_Outer_TI + I_Inner_TI) *

AngAcc_Inner;

F_PushLink(j,k) = PushLink_Load(j,k) / PL_TI_MA(j);

%L1 Calc

F_L1(j,k) = (((I_TieLink_T+I_PushLink_T)) * AngAcc_TieLink + ...

F_PushLink(j,k) * PL_TL_MA(j)) /L1_TL_MA(j);

end

end

F_PushLink_Max_Dash = transpose(max(F_PushLink, [], 2));

F_L1_Max_Dash = transpose(max(F_L1, [], 2));

% plot(toDegrees('radians', Theta_I_Rad), F_PushLink_Max);

% surf(F_L1);

plot(L1, L1_TL_MA, L1, PL_TL_MA);

% plot(sec(Theta_PLI_Rad(j)));

% surf(cos(Theta_O_Rad));

% plot(L1, Theta_I_Deg,L1, Theta_TieLink_Deg, L1, Theta_PushLink_Deg, L1,

Theta_L1_Deg);

43

[F_PushLink_Max, F_PushLink_Max_Ang] = max(F_PushLink_Max_Dash);

[F_L1_Max, F_L1_Max_Ang] = max(F_L1_Max_Dash);

disp([' Max Force on Push Link: ', num2str(F_PushLink_Max), ' N']);

disp([' At(Inner Angle): ', num2str(F_PushLink_Max_Ang), ' deg']);

disp([' Max Force on L1: ', num2str(F_L1_Max), ' N']);

disp([' At(Inner Angle): ', num2str(F_L1_Max_Ang), ' deg']);

disp('-------------------------------------------------------------------');

disp('Calculation Completed');

APPENDIX II.V KINEMATICS CALCULATOR(NOT FUNCTIONING) %Kinematics Calculator For MECH3010 Coursework of Hydraulic Systems Design



%All Lengths in mm unless stated, all weights in kg unless stated.

%Data

%%

%Environmental Data

g = 9.81;

%Boom Turntable Data

A1 = 2420; A2 = 715; A3 = 1320;

h1 = 1265; h2 = 605; h3 = 138; h4 = 960; h5 = 234;

%Tie Link Data

B1 = 1871; B2 = 660; B3 = 1133; B4 = 220;

%Inner Boom Data

C1 = 11880; C2 = 1024;

%Outer Boom Data

D1 = 375; E1 = 688; E2 = 9900; E3 = 248; E4 = 220;

%Telescopic Boom Data

F1 = 820; F2 = 9130; F3 = 550; L3 = 5460;

%Pannier Data

P1 = 809; P2 = 893; P3 = 283; P4 = 420; P5 = 221; P6 = 1859; P7 = 1943;

%Intrinsic Angles

Theta_1_Rad = atan((h4-h5)/(A3-A2)); %Angle between Inner Boom and Tie Link holes

on Turntable

Theta_1_Deg = toDegrees('radians', Theta_1_Rad);

Theta_2_Rad = atan(B4/C2);

Theta_2_Deg = toDegrees('radians', Theta_1_Rad); %Inner boom Push link connection

Angle

Theta_3_Rad = atan((h1-h2+h4)/(A1+A2));

Theta_3_Deg = toDegrees('radians', Theta_3_Rad); %Angle betwen L1 link and Tie

44

link holes on Turntable

%Component Mass

M_Pannier_Steel = 1739.927;

M_Pannier_Aluminium = 599.898;

M_Pannier = M_Pannier_Aluminium; %Switch between Steel & Al

M_Outer = 4101.725;

M_Tele = 2187.437;

M_Inner = 25015.319;

M_TieLink = 1049.413;

M_PushLink = 190.454;

M_Turntable = 0;

M_Wagon = 0;

%Component CG

L_CG_Tele = 5978.847; %From Flat End

L_CG_Outer = 2383.444;

h_CG_Outer = 611.082; %From Inner Conn End

L_CG_Inner = 5671.185;

h_CG_Inner = 0; %From Turntable Conn End

%Component MOI

%MOI around Inner/Outer Bearing

I_Tele_IO = 0;

I_Outer_IO = 0;

I_Pannier_IO = 0;

%MOI around Turntable/Inner Bearing

I_Pannier_TI = 0;

I_Tele_TI = 0;

I_Outer_TI = 0;

I_Inner_TI = 0;

%MOI around R2

I_Pannier_R2 = 0;

I_Tele_R2 = 0;

I_Outer_R2 = 0;

I_Inner_R2 = 0;

I_Turntable_R2 = 0;

I_TieLink_R2 = 0;

I_PushLink_R2 = 0;

%MOI around R1

I_Pannier_R1 = 0;

I_Tele_R1 = 0;

I_Outer_R1 = 0;

I_Inner_R1 = 0;

45

I_Turntable_R1 = 0;

I_TieLink_R1 = 0;

I_PushLink_R1 = 0;

I_Wagon_R1 = 0;

%Tie Link MOI around Turntable

I_TieLink_T = 0;

%Push Link MOI around Tie Link

I_PushLink_T = 0;

%Wind Load

W_Inner = 0;

W_Outer = 0;

%Position Calulation

%%

%L1 to Tie Link Position

L1=1058:1:4653; %Change L1 Range

AO=sqrt((A1+A2)^2+((h1-h2)+h4)^2);

BD=B1+B2;

Theta_AOB = acos((L1.^2-AO^2-BD^2)/(-2*AO*BD));

% Alpha = pi - Theta_3_Rad - Theta_AOB;

% Combined Equation

Theta_TieLink_Rad = pi - atan((h1 - h2 + h4)/(A1 + A2))- acos(...

(L1.^2 - sqrt((A1 + A2)^2+((h1 - h2) + h4)^2)^2 - (B1 + B2)^2)...

/(-2*sqrt((A1 + A2)^2 + ((h1 - h2) + h4)^2)*(B1 + B2)));

Theta_TieLink_Deg = toDegrees('radians', Theta_TieLink_Rad);

% Kinematic Graph

% plot(L1,Theta_TieLink_Rad)

disp('Tie Link Angle Calculation Complete')

%%

%Tie Link Position to Inner Boom Position

BD = C2; BF = sqrt((A3-A2)^2+(h4-h5)^2); EF = sqrt(C2^2+B4^2);

DF = sqrt(BD^2+BF^2-2*BD*BF*cos(Theta_TieLink_Rad-Theta_1_Rad));

DE = B3;

Omega = acos((DE^2-DF.^2-EF^2)./(-2*DF.*EF));

Phi = acos((BD^2-DF.^2-BF^2)./(-2*DF.*BF));

Theta_I_Rad = (pi-Phi-Omega-Theta_2_Rad)+Theta_1_Rad;

Theta_I_Deg = toDegrees('radians', Theta_I_Rad);

46

Theta_I_Emp = ((1*10^-13).*L1.^4)- ((1*10^-8).*L1.^3) +...

((7*10^-5).*L1.^2) - (0.1736.* L1) + 271.57;

% Kinematic Graph

plot(L1,Theta_I_Deg, L1, Theta_I_Emp);

disp('Inner Boom Angle Calculation Complete')

%%

%Inner Boom to Outer Boom Position'

L2 = 0:2000; %Change L2 Range

GI = 1000;

HJ = 1000;

IJ = E1+350;

HI = sqrt(IJ^2+HJ^2);

Theta_4_Rad = atan(IJ/HJ);

Theta_4_Deg = toDegrees('radians', Theta_4_Rad); %Angle between L2 mount on outer

boom and inner mount hole

Theta_IO_Rad = acos((L2.^2-GI^2-HI^2)/(-2*GI*HI));

Theta_IO_Deg = toDegrees('radians',Theta_IO_Rad);

disp('Outer Boom Angle Calculation Complete')

% Kinematic Graph

% plot(L2, Theta_IO_Deg);

%%

%Pannier Position

%TBC

disp(' All Calculations Complete')

Appendix II.VI Moment Arm Calculator %Load Calculator(L1) For MECH3010 Coursework of Hydraulic Systems Design




L_Pannier_R2 = 0;

L_Tele_R2 = 0;

L_Outer_R2 = 0;

L_Inner_R2 = 0;

L_Turntable_R2 = 0;

disp('Calculating Moment Arm...');

disp('---------------------------------------------------------------

47

----');

Theta_I_Deg = -82:1:157;

L_Inner_Proj_x = C1 .* cosd(Theta_I_Deg);

L_Outer_Proj_x = (E2-E4+5.460+F1).*sind(110-(90-Theta_I_Deg));

L_Inner_Proj_y = C1 .* sind(Theta_I_Deg);

L_Outer_Proj_y = (E2-E4+5.460+F1).* cosd(110-(90-Theta_I_Deg));

L_x = L_Inner_Proj_x + L_Outer_Proj_x;

L_y = L_Inner_Proj_y + L_Outer_Proj_y;

MA = sqrt((L_x.^2 ...

+ L_y .^2));

[MA_Max, MA_Max_Ang_dash] = max(MA);

MA_Max_Ang = MA_Max_Ang_dash - 82;

disp(['Max Moment arm:', num2str(MA_Max), 'm']);

disp(['At: ', num2str(MA_Max_Ang), ' deg']);

APPENDIX III. ROTARY ACTUATORS CALCULATION

R2 R3 Wind

R(m) I(kg. m2) R(m) I(kg. m2) Area(m2) Wind

Force(N)

Pannier 24.5 361414.9 29.3 516359.7 2.183 218.3

Tele 22.1 1064000.0 26.8 1574000.0 2.826 282.6

Outer 13.5 752006.4 18.3 1369000.0 6.361 636.1

Inner 6.8 308920.1 11.2 845024.0 5.346 534.6

Turntable 0.8 5063.1 4.7 160242.9 0.253 25.3

Tielink 1.0 1078.7 5.7 34408.1 0.018 1.8

Pushlink 1.4 380.1 6.2 7261.9 0.242 24.2

Wagon N.A 2.4 18256.4 3.409 340.9

Inertial Torque 71093.9 107953.4

Wind Torque 23894.2 32729.9

Actuator

Torque 94988.1 140683.3

With SF 189976.1 281366.6

APPENDIX IV. PERCENTAGE DISTRIBUTION

The project group (Leheng Li, Linfei Wu, Wenguang Cheng, Fanfu Wu, Shangzhou Duan)

has decided on an equal split of the marks based on a roughly equal split of work load,

with all members receiving the same marks that this report has obtained.

hydraulic design report copy

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