TP Mammoet Trailers Stability&Lashing - Rev03-2

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Mammoet Employees Training & Development Programme TP 06 – Trailer Stabilit y & Lashing

Trailer Stability & Lashing TP 06 EN – Rev03

© Mammoet Holding Holding B.V., 2004.

All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically, by print, photo print, or recording or otherwise without prior written permission from Mammoet Holding B.V.

3 2 Rev.

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Revised for comments

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February 23 , 2006

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Issued for Training & Development

April 1 2005

Description

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Mammoet Field Employees

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Training & Development Programme

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Date February 23 rd , 2006

Subject

TP 06 – Trailer Stability & Lashing

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Table of Contents 1

2

3

4

Mammoet Training & Development Programme

4

1.1

Foreword

4

1.2

Safety is a recurring theme in this course.

4

1.3

Goal Training & Development Programme

5

1.4

Benefits

5

1.5

Legislation

5

1.6

Scope

5

1.7

QSE Requirements

6

1.8

Share knowledge

7

1.9

Target group

7

1.10 Symbols used in this manual

7

1.11 Subject specific terms and abbreviations

8

Trailer stability

9

2.1

Mechanical stability

10

2.2

Structural stability

10

2.3

Dynamic stability

10

Mechanical Stability

11

3.1

Mechanical stability principles

12

3.2

Determination of the tipping lines

18

3.3

Hydraulic Platform Trailer

23

3.4

Influence of the hydraulic setup

25

3.5

Using turn tables

31

3.6

Determining the system Center of Gravity

34

3.7

Mechanical stability by numbers

36

3.8

Disabling axles

39

3.9

Fixed fields

40

3.10 Calculating Mechanical stability

41

3.11 Summary

45

Structural Stability

46

4.1

General principle

46

4.2

Structural stability by numbers

48

4.3

Influences on the Structural stability

49

4.4

Calculation of the Axle loads

50

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6

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4.5

Calculation of the Structural stability

53

4.6

Summary

56

Dynamic influences

57

5.1 5.2

Wind pressure 62 Wind pressure calculation.Error! calculation. Error! Bookmark not defined.

5.3

Centrifugal force

58

5.4

Calculation of the centrifugal force

59

5.5

Acceleration / Deceleration forces

60

5.6


60

5.7

Summary

62

Lashing & Blocking

65

6.1

Transport forces

66

6.2

Acceleration forces

66

6.3 6.4

Centrifugal forces Driving on slopes

6.5

Lashing under angles

71

6.6

Lashing of cargo on saddles

74

6.7

Lashing of cargo on turn tables

75

6.8

Single file SPMT’s

77

6.9

Summary

78

3

68 Error! Bookmark not defined.

7

Conclusion

79

8

Calculation examples

80

8.1

Three-point suspension

81

8.2

Four-point suspension

84

8.3

Stability on strength

85

8.4

Dynamic axle loads

86


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1 Mammoet Training & Development Programme 1.1

Foreword

Mammoet is a company, which obliges itself to have all its employees properly trained for the duties they will be carrying out. Each employee is therefore obliged to follow the training, which is offered by his employer and corresponds with his job description. This book on "Trailer Stability & Lashing" provides the most important facets for operators in the horizontal transportation profession who would like to develop their skills towards operating trailers. Safe tr ansportation is a profession, and each day there are new things to be learned for those who want to progress in this f ield.

1.2

Safety is a recurring theme in this course.

The Mammoet Training & Development Programme has been created from the recognition of the requirement for all employees who operate and repair plant and machinery to be competent in their chosen trade disciplines. As market leaders in our industry of Hoisting and Transporting we aim to set standards that will improve safety and revitalize Health and Safety agenda by the adopting of a “Full Compliance Culture”. Too many people within our industry are injured or suffer from health problems. The unacceptable high levels of injuries and suffering can be reduced by simple precautions, such precautions are built into this Training & Development Programme and if we all work together to achieve the targets that have been set we can make our industry a healthier and safer place to work in.


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Goal Training & Development Programme

This manual explains most important i ssues regarding the requirements of operating a trailer. While regional responsibilities may deviate between different countries, Mammoet aims to raise health, safety, environmental and quality standards on a worldwide basis b y giving recognized levels of training and competence for all employees worldwide. This all is resulting in a certification and qualification system that will be accepted and recognized at all Mammoet offices and projects on a worldwide basis. 1.4

Benefits

The Training & Development Programme, used and followed correctly, will provide employees with recognized skills, competence and qualifications. This will lead to improved health and safety awareness, better employment promotion prospects, more flexibility within the workforce and a much higher qualified workforce because the training standards leads the industry. This all will help improve customer satisfaction and the industry’s image. 1.5

Legislation

It is appreciated that some Mammoet employees work in many countries and therefore encounter different legislation. Throughout this course delegates will be taught the "Mammoet Standard" which will in most cases satisfy all countries. However, upon commencement of any operations in a new country advise should be sought to ensure full compliance with local requirements. All members of the workforce must be made aware of any local legislation requirements which impacts on their work operations and ensure compliance. The certificates granted after completing the complete Trailer Stability & Lashing course is not a replacement for local required requirements! 1.6

Scope

The Training and development programme will pr ovide certification for plant and machinery operators on a worldwide basis, participation in the scheme will be mandatory for all employees who require various training and competence in the trade as known in our industry.


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QSE Requirements

All participants within the scheme will be required to have completed the Mammoet Induction test prior to commencement. Quality, Safety and Environment is fundamental within our work disciplines and of paramount importance to the integral part of business performance and shall receive foremost priority. At all levels we must be committed to achieving a high level of Health, Safety, W elfare and Environmental performance by means of compliance with all Legal Acts, Regulations, Codes of Practice, Enforcing Authorities Guidance Notes, Industry Best Pr actice Standards, Corporate, Client, Regional, Local and Provincial QSE requirements and of continual cost-effective improvement of risk awareness from our undertakings and of the importation of risk from others.

Throughout the training you will continuously be required to undertake tests related to safety matters, which are both general to everyday safety and specific to the discipline that you have elected to follow. Test will take the form of one or all of the following, a) Multiple choice question papers. b) Written question papers. c) Verbal questions. During the induction you are issued a Mammoet Safety Guide which gives general information on safety matters within all our trade disciplines, also contained within the guide are the Mammoet Safety Rules which apply at all times, regardless of whether you are working on Mammoet premises or on a clients site. It is a requirement of your employment to ensure that you familiarize yourself with t hese rules. Safety is a culture that is very much influenced by attitude and behavior, each individual should be aware that they are their own “Safety Officer” with the responsibility for their actions whilst at work, equally, each individual must “Look Out” for others and ensure that nothing they do will have a detrimental effect on issues of which they have no control.


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Share knowledge

This manual is made by and for Mammoet employees. Each Mammoet employee has the responsibility to share his or her knowledge. The information in this manual is subject to change, therefore the latest version can be found on our corporate intranet website and always succeeds any previous release. If you that feel any of this information you about to read is incomplete, incorrect or even misleading, please contact the T&DP coordinators in The Netherlands at the following e-mail address:

[email protected] Your co-operation is highly appreciated.

1.9

Target group

This course is meant for personnel with or without experience with operating trailers, who want to qualify as certified Mammoet trailer operator and possibly move up to other functions in the branch of horizontal and vertical moving. Employee must be in possession of fo llowing Mammoet T&DP certificate before commencing this course: 1. Mammoet Induction course 2. Rigging Course

1.10

Symbols used in this manual

CAUTION CAUTION IS USED WHEN PERFORMING AN OPERATION THAT MAY BE CRITICAL TO SA FETY AND MAY REQUIRE AGREED DEVIATION FROM THE GIVEN PLANS. CAUTION

NOTE NOTE IS USED FOR MARKING AN IMPORTANT OPERATION AS WELL AS FOR ADDITIONAL INFORMATION.


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Subject specific terms and abbreviations

•

SWL:

Safe working load: The load at point of failure or t ipping reduced with a safety factor.

•

CoG:

Center of gravity The application point of the weight of an object.

•

g: Gravitational acc. 9.81m/s². We use 10m/s²


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Trailer stability The intention of this course is to explain the trainee the concepts of stability, so he can identify stability critical issues. This manual contains many formulas and calculations, which can aid the trainee into understanding the stability concept. It is not the intention that the trainee learns these formulas or calculations by head, they merely support the provided explanations. In general we review the “stability” of a trailer as: “When will the trailer tip over or loose its cargo due to the fact that trailer is driving on a camber or gradient”. Even though the result being similar, there are many factors that have a direct or indirect influence on this type of “stabilit y”. • • • • • • • • • • • • •

The height of the center of gravity of the cargo. The weight of the cargo. The offset of the center of gravity of cargo from trailer centre. The height of the trailer-deck. The track width of the trailer (combination). The weight of the trailer (combination). Type of suspension of the trailer: mechanical, pneumatic or hydraulic. Securing of cargo onto the transport vehicle (lashing). Wind pressure on the cargo and trailer. Acceleration/Deceleration forces from braking, speeding up. Driving speed. Road gradient Supporting capabilities of the road.

To make an objective verdict on the “stability” of a transport, we need to separate this global classification into various definitions, as commonly used in the construction and engineering community. - Mechanical stability (stability on tipping). - Structural stability (stability on strength). - Dynamic stability.


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Mechanical stability

In simple terms, the mechanical stability is defined as: “When will either load and/or trailer tip over”. Within Mammoet, this term is often referred to as “Stability on Tipping” 2.2

Structural stability

The structural stability can be defined as “When will any of the transport components be overloaded and fail”. Within Mammoet, this term is often referred to as “Stability on Strength” 2.3

Dynamic stability

The dynamic stability can be defined as “For which external factors will the t ransport be overloaded or tip over”. Both the mechanical stability and structural stability are based on the STATIC situation, and are part of the “initial” stability and are unconcerned of any form of movement or external forces. The dynamic stability considers the forces induced by movement such as: •

Acceleration.

•

Deceleration.

•

Wind.

•

Driving through a corner.

As the influence of the dynamic forces on the static stabilities cannot be simply “deducted” from the static stabilities, we in Mammoet have agreed to maintain the static stability angles and that the dynamic forces will affect the axle loadings.


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Mechanical Stability In practice we find that a transport combination will tip over as per one of the below scenarios: A. B. C. D.

The load tips over on the trailer deck. The load and trailer tip over. The load skids over the trailer deck, causing the trailer to tip The load makes the trailer tip over a little, then slides of the trailer deck.

A

B

C

D

To determine the transports mechanical stability, we need to find out which of these scenarios (A, B, C or D) will occur. To do this we need to look at the stability of each individual part of the system, and determine the weakest link, so first we must find a wa y to establish the stability of an object.


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Mechanical stability principles

This chapter discusses the general terms and definitions concerning mechanical stability. Before explaining the theory applicable to the stability of trailers, it is essential that you understand these basic concepts. Stability: A system is Stable when it comes back to its original state aft er it has been submitted to an external disturbance. In an Unstable system, a disturbance will produce an ever-increasing change from the original state. A system is considered Neutral when, after being submitted to an external disturbance, it will eventually come to equilibrium but does not return to its original location. Equilibrium: In mechanics, equilibrium is a stable situation in w hich forces cancel one another. To define stability we need to differentiate three system states: • Stable • Neutral • Unstable These states are explained by the following images:

Stable: Original state

Disturbed state

Final state after some oscillations

Neutral Original

New rest position

Unstable

Original state

Disturbed state

The ball will start rolling


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In this context we can define mechanical stability as: “Stability is the ability of a system to return to its original position after a disturbance.” We can explain this definition in more detail through an example: If we take a book, or a square block, it will have a certain volume and weight, depending on its measurements. The total weight of the book is formed by the weight of the book jacket + weight of the pages + weight of the 1 ink. It is obvious that the center of gravity of the total weight of the book is somewhere within the dimensions of the book. With some objects the center of gravity is easier to determine then others, in the case of our book it is most likel y in the middle, i.e. on half the length, half t he width and half the height. When we place the book facedown on the table, it will have a big supporting area, which is the part of table where we have placed the book. The center of gravity G can be found at a distance of ½ d above the table.

Image 1 As the weight of the book presses on the table, the table pushes the book back with the same force. The force that is pushing back is the called reaction force . Each point in the supporting surface pushes a part of the weight of the book up so it stays in its place. The total reaction force R is concentrated in a single point. This is the application point of the resultant of al l reaction forces and is called the support point (S). When there are no external forces working on the book, it will remain on the table.

1

In mechanics, weight is considered to be a force, caused by the gravity of the earth. Gr avity is working on every object around us. When an object consists of a combination of multiple parts, their gravity forces can be combined. The sum of all forces is called the resultant of the combined forces. The application point of this resultant is named the centre of gravity of the object. All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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When we introduce an external force to the book by lifting the right side of the book from the table, point G describes an arch as left side is being raised as shown in image 2.

Image 2 When we remove the external force holding the book, it will drop back onto the table. The reason that the book drops back into its original position can be explained as follows: As soon as the right side of t he book is lifted free from the table, the support point S will move to the left side of the book S’. The weight of the book is still working from point Z. When the book is released (by removing the external force) it is under the influence of two equal forces: one force in point G vertically down and one force in point S’ vertically up. These two forces result in moment working in clockwise direction, bringing the book back into its origi nal position. This moment is named the stability moment. We describe the state of the book in its facedown position as “stable” as it has a “great degree of stabilit y”. We measure the degree of stability by the angle at which the object becomes “unstable” from its rest position. When we place the book in an upright position on its bottom side, point G is located on a distance of ½ h above the table. This distance is considerably larger than the distance ½ d. The point of support S finds itself straight under G in the top of the table (Image 3). The width of the support surface of the book wi ll be large enough to allow the book to stand upright, without falling.

Image 3 All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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When the book gets thinner, its tendency to tip over gets bigger. This will definitely be the case when we start to push the right topside of the book. When we gently push the topside of the book, point S will directly shift to point S’ and initially, point G will remain to the right of S’. When we release the book, it will ret urn back to its original position, meaning it still has stability. When we increase the force pushing on the topside of the book and point G is positioned directly over S’, we create a condition where the book can either tip over or return to its original position. At this point the state of the book has become Neutral. In this state even a small force on the topside will disturb the state of the book; it will either tip over, or fall back into its original position. Since the above-mentioned condition is reached quicker than when the book is placed face down on the table, we can conclude that the stability of the book in an upright position is less than when it is placed face down on the table. When we allow the book to lean over even further, point G will be positioned left of S’ and a new moment appears; G has reached its highest point in the described arch and will continue left and down. The book no longer returns to its original position. Stable and unstable conditions are called balance states. Stable

Stable

Critical

Unstable

Intermediate summary: From all this we can conclude that when a book is placed face down on a table and lifted onto one side, point G describes a circular arch. As long as point G can get higher in the described arch, there is stability. Initially, this stability is large, but will reduce as we lift t he book further from the table. As soon as point G has reached the highest point in the described arch and finds itself directly over the point of support, it is in an unstable state. Immediately after reaching the highest position the book will tip over. The force of your hand lifting the side of the book is an external force. The book itself remains unchanged (it cannot lift itself).


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We have seen that the stability moment caused the book to return to its original position. A heavier book will produce a larger stability moment. Similar and equally sized forces are present when the book is in a neutral balance, but the book will no longer returns to its original position nor w ill it tip over. Apparently the w eight of the book is not the only factor that influences the magnitude of the stability moment. We w ill try to work out this factor based on image 4.

Image 4 Imagine a thin steel bar placed face-down on a table, being held in point S’ and rotated so it is describing the arch depicted in image 4. The center semicircle is the described arch of the center of gravity G during the movement. Immediately after one side of the bar is lifted free fr om the table, the point of support finds itself in S’. Line “a” is a vertical line through the support point S’. In position “I” the distance between G and the line “a” is “X” cm. It is difficult to rotate the bar in S’. In position “II” the distance between G and the line “a” has become smaller: “Y” cm. It is getting easier to rotate the bar in S’. In position “III” the plate is vertical and the distance between G and line “a” has now reduced to 0cm. As the bar is now balancing, it takes no effort to hold the bar in S’. From this we can conclude that as the bar is reaching position III, it is gradually getting easier to rotate the bar. When we continue the rotation after reaching the vertical position, G will follow its downward path and the effort required to stop the bar will continuously increase. This demonstrates that the distance between the forces of the moment is an important factor. The distance between the forces of a moment is called: “moment-arm”.


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Summary: Weight is a force something that happens to any object. Weight is the influence of the gravity of the earth on an object. Mass measures inertia, which means how fast it will react on speed changes. Center of gravity is the application point of the weight of an object. When multiple objects are moving together, their CoG’s and weights can be joined to form the “combined centre of gravity”. Force acting on a body causes that body to accelerate, i.e. to change its velocity. An object is more or less stable by its t endency to return to its original, undisturbed position. The initial stability is always bigger than when the object is already moving / rotating, i.e. the tendency to return to its original position will decrease while rotating. We quantify the degree of stability by the stability angle. The stabilit y angle is the angle at which the object becomes “unstable” when rotation it from its original position. This stability angle depends only on the ratio between the Height of CoG and the Tip Arm (a).

 Tip arm(a)    Height CoG 

S tability Angle = arcTan 

If the height of the Center of Gravity i s high, then the stability angle will small. When the Tip arm is large, the stability moment will be large and so is the stability angle.


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Determination of the tipping lines

The most important factor issue to determine the stability of a s ystem is to define the tipping points (S’) or when reviewing a three dimensional system: the tipping lines. When considering complete stability systems, such as a trailer carrying a load it gets more complicated to determine the correct tipping lines. W hen an incorrect tipping point or line is assumed, the calculation based on this assumption will not be correct. 3.2.1

The book on the table

The book on the table from the previous chapter actually has four tipping l ines. The active tipping line will depend on where you pick-up the book (apply an external force). When lifting the book in A, the active tipping line will be TLx1. When lifting the book in B, the active tipping line will be TLy1. The tipping line for lifting in C is TLx2, and for D it is TLy2

3.2.2

A

B C

D

Mechanical and Pneumatic suspension systems

A mechanical suspension system uses coils or leaf springs to marginally vary the axle height to deal with the unevenness of the road. The more the coil is compressed (or bend) the larger the reaction force. With mechanical suspension the axle loads are very d ifficult to determine, as the axle load depends on the amount of compression in the springs which are not equalized. This means that each axle is a support point and for vehicles with more then 2 axles we get more than 4 support points in our s upport system. If we assume that the spring compression is equal, we can group these axles by their contribution to the stability. We can also calculate a virtual pressure point by the summation of the position and contribution of each axle to the group. A Pneumatic system uses compressed air to vary the axle height. The pressure of this air depends on the loading of the axle and is regulated using a valve. Smarter systems can lift axles when its pressure falls below a certain value. The air system can also be used to raise and lower the tru ck or trailer but the stroke (height difference) is limited.


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When the axle loads are equal, the combined pressure points are easily found as displayed below. The fifth wheel (the connection pivot between truck and trailer) on the truck provides 2 supports on its hinges.

When the axle loads are not equal you can calculate the pressure point per the following example. The hydraulics of the first two axles is combined with the cylinders in the “neck” of the trailer. Please note that to calculate the axle pressures by hand, you need to know exactly how the trailer distributes the loads of the axle to the hydraulic neck. In this case the axle and kingpin loads are calculated using the NOVAB trailer program provided by the Nooteboom trailer company, resulting in the following arrangement. As the loads of the rear 6 axles are equal, we can combine them to find a pressure point. The pressure point of the kingpin combined with the 2 first trailer a xles is found using the following calculation: P = (F1 x D1 + F2 x D2 + F3 x D3 + F4 x D4) / (Ftotal) P = (17.5 x 0 + 6 x 5.067 + 6 x 6.427 + 6 x 7.787) / (17.5 + 6 + 6 + 6) P = 3.259m Please note: The two points in the kingpin combined with the pressure point on the two front axles reduces the width of the stability lines.


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CAUTION FOR ALL ARTICULATED TRANSPORT: WHEN THE TRAILERS ORIENTATION IS ROTATED ON THE TRACTOR THE STABILITY IS REDUCED DUE TO THE SWIVEL IN THE FIFTH WHEEL. IT IS THEREFORE CAUTION

ADVISED TO REVIEW THE FIFTH WHEEL AS A SINGLE SUPPORT POINT


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Hydraulic suspension

Hydraulic or hydro-pneumatic suspension systems are used to equalize the axle loads. A hydraulic suspension system will provide a higher grade of stability because the axle pressure is has no related to any form of compression. Also the hydraulics allow the operator to level the trailer deck. Below is a simplified hydraulic scheme of a 6 axle hydraulic trailer ( one side only). Normally when driving, the trailer height is not changed so valves VA and VB are closed, this results in closed system A and B. When the trailer operator opens valve VA or VB and the pump delivers oil, the trailer will raise its front or rear. These closed systems are generally referred to as a hydraulic group or field. When axle 1 is raised by an obstacle on the pavement, the oil from cylinder C1 will be pushed into cylinder C2, pushing the axle A2 down. In this case the displaced oil by Axle A1 will cause an equal displacement of Axle A2. This displacement will result in a tilt angle of the trailer deck. When an obstacle in the pavement raises axle 3, the oil from cylinder C3 will be pushed into cylinders C4, C5 and C6 pushing down axles A4, A5 and A6. In this case the displaced oil by Axle A3 will move Axles A4, A5 and A6 by 1/3 times the displacement of Axle A3. Due to the self-equalizing effect of the hydraulic oil and the equal diameters of the cylinders in the system, we know that the resulting axle load in each field will be equal, causing a resulting force in the center of this field. In our case the center of field A is exactly between A1 and A2. The center or pressure point of field B is located exactly between A4 and A5.


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3.2.4

Ref.

Rev.

3

Hydraulic/Pneumatic suspension

When a hydraulic suspension system is required to make fast changes in the axle heights, due to driving at high speeds, the resistance of the fluid going through the valves in the system cause high peak pressures. This effect is caused by the fact that load and trailer have a tendency to remain at the same height in relation to the center of the earth (inertia). For this to happen, the distance between the trailer deck and the pavement temporarily needs to be reduced. In a hydro-pneumatic suspension system the compression of a gas medium is used to reduce the peak pressures in the system caused by high d ynamic forces. The most common used gas medium in most trailers is nitrogen.

Hydraulic OILoil Hydraulic

M

GAS Gas

When one of the axles in this hydraulic set-up is raised by a bump in the road, the displaced oil of the cylinder connected to the axle is pushed into the accumulator, resulting in a higher pressure of the oil and gas. If this is a continuous situation, the oil wil l start flowing from the axles with a higher pressure to the axles with a lower pressure until all pressures are equalized. When the axle moves back before the pressures are equalized, the pressure in oil and gas is already reduced so the oil will not start f lowing to the other cylinders.


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3.3

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Hydraulic Platform Trailer

The suspension system of a platform trailer is a hydraulic system as described in paragraph 3.6. By changing the hydraulic set-up of the platform trailer, you can determine and alter the tipping lines of the system. 3.3.1

4-point suspension

A trailer in a 4-point suspension configuration uses 4 hydraulic fields, where the pressure in each field will be equal for each axle in this field. The combination of axles in these fields will produce a combined pressure point as indicated by the black dots. These pressure points are the support points of the trailer. When we connect the support points, we will find the tipping lines indicated by the black lines. The Tip Arm (a) is the distance from the CoG to the nearest tipping-line.

3.3.2

3-point suspension

The tipping lines of a trailer w ith 3-point suspension are determined in similarly manner as described for the 4-points system. The result is a triangle shown in the image. The Tip Arm (a) is the distance from the CoG to the nearest tipping-line. It may be evident that this distance is smaller for a 3-point than for a 4-point suspension.


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Ref.

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This defines the stability lines in the horizontal plane, but as we have seen with the book, we also need the height distance between the tipping plane and the CoG of the object in the vertical direction. The suspension of a hydraulic platform trailer is formed by a mechanical axle stand and rocker arm with a hydraulic cylinder, allowing the trailer deck to be raised and lowered. On this rocker arm, a swivel allow s the axle to turn over its longitudinal axis to allow for road cambers and small level differences. The image below shows a side view of some axles in their highest and lowest position. The centerline of the swivel is not parallel to the road for every tr ailer height. To remove this variation and to simplify the calculation input we assume the worse case: Assume the pivot points on the wheel hub centers as indicated by the black dots in the rear-view drawing. Pivot point

Scheuerle ’94 series

Rear-view KAMAG

Axle pivot of Cometto MX trailer.


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Ref.

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The tipping lines we have found in this section can be used to calculate the initial stability of the trailer. However, the swiveling motion of the wheel hub is limited by t he clearance between the tire and the axle stand, which can vary for different heights of the trailer deck. When the trailer s tarts tipping and the clearance diminished, in theory the trailer will start to t ip over the tire edge. In reality the t ire will deform due to the extreme and unbalanced tire loads and the actual tipping point can no longer be determined. To reach this type of stability, we already gained some speed in the tipping motion, making the calculation very complex. As the trailer first needs to negotiate the initial stability stage, it will be sufficient t o calculate the stability by using the initial tipping lines as found in this chapter.

3.4

Influence of the hydraulic setup

From the previous chapter, it was made clear that the Tip Arm ( a) largely depends on the hydraulic setup of the trailer. Before we discus the influence of the hydraulic setup on the mechanical stability of t he trailer we first need to explain some conventions as used by Mammoet and trailer manufacturers. Choosing the most appropriate hydraulic setup depends on various circumstantial issues. T his paragraph provides some rules of thumb to make these decisions. These rules are numbered by their initial priority, however, depending on the situation, you may need t o deviate or change their priorit ies. When the stability of a transport becomes more critical, the practical issues become less important. When there are less stability issues, it is better to obey the practical rules as work may be executed in a more standard way, making it obvious to all what the situation is.


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3.4.1

Ref.

Rev.

3

Rules of thumb and Terminology

Rule 1: Always choose a hydraulic setup that provides sufficient stability. Rule 2: Try to match the number of cylinders in each field. During lift ing or lowering operations, the trailer will remain reasonably level, small corrections can be made after setting the general height. Also it makes the trailer arrangement more responsive upon leveling a particular field. A f ield with a large number of cylinders requires more oil to get the same stroke of the axles. If it is not possible to match the number of cylinders in the fields, put the excess axles in the fields providing the sideways stability. In general, especially for single file trailers, the sideways stability wi ll be the smaller then the stabilit y over the front or rear. The fields providing the sideways stability are the fi elds that split the trailer left and r ight. The floating field provides stability over the front or the rear. In a four point setup, all fi elds contribute to the stability in all directions. Rule 3: Always try to make the hydraulic setup so that the point in the tr iangle points in or from the driving direction. This makes negotiating ramps easier, as we do not need to compensate to keep the trailer level in the sideways direction. In stability critical situations, you might need to resort to the second image. In this case it becomes mandatory to level the trailer deck for every bump in the road surface. Preferred setup

Triangle point

Triangle base


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Rule 4: When considering self propelled trailers, position the base of the triangle toward the position of the powerpack. The center of gravity of the power pack is not exactly in the middle and this setup provides a higher degree of stability. Also the oil going to the f ields providing side-ways stability has to negotiate fewer hydraulic connections (resistance!) making the trailer more responsive.

The standard way to indicate the hydraulic setup is to mention the ratio of front : rear axle lines, where the first number is always the split fields (left/ri ght). Examples:

Split fields providing sideways stability

Floating Field


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3.4.2

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Rev.

3

Deciding on a 3 or 4-point system

We will try to explain why you should decide on a 3- or 4-point suspension system. Trailer manufacturers always presume the utilization of the 3-point suspension system as a 3-point system does not allow any diagonal load transfer induced by uneven road levels. 4-Point suspension systems can introduce internal forces in the trailer, such as torsion when it i s not properly leveled. These load transfers might overload the fields in the system. A 3-point suspension system with a centrally positioned load combined on a level underground result in equal axle loadings on the trailer. On a transverse gradient of the road, the shift of the CoG causes a shift of the load from one of the sideways split fields to the o ther side, while the front field pressure remains the same as on a level road. To remove the shift in CoG, t he trailer can be leveled by pumping oil in the heavy loaded field. On a transversal slope one of the three field has to absorb all the eccentric forces, while one field retains the same load as on a level surface. To level the t railer, more oil has to be pumped in the heavy loaded field. The disadvantage of the t hree-point suspension is the reduced transverse stability arm in the tipping triangle. This also reduces the allowable tilt angle of the trailer, i.e. the stability on tipping angle. The 4-point suspension system provides a larger tra nsverse stability arm and has a better degree of stability on tipping. However, due to diagonal load transfer on an uneven road surface, the operator needs to monitor and adjust the axle pressures continuously in order not to overload any of the fields. If this i s not properly done there is a chance either the trailer or load will be overloaded. Rule 5: Always start with a 3-Point system, when the stabilit y is critical, or when a stability critical exercise needs to be performed, always resort to a 4-point suspension. This means that we need to define “critical”. Mammoet uses a mechanical stability limit of 7 degrees. In practice this means: When a transport has a mechanical stability angle smaller 7°, switch to a 4-point system. In doubt, contact the engineering department to make a stability calculation. 3.4.3

Splitting hydraulic fields

In the previous paragraphs, the importance of the tip arm (a) was explained. This tip arm(a) may vary f or different setups of a single trailer. In this example the hydraulic front t o back ratio is varied to find the maximum possible tipping arm. The height of the CoG will be k ept constant. The resulting table on the next page shows why we should never reduce the number of cylinders in the fields that provide side-ways stability. All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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CAUTION

Ref.

Rev.

3

CAUTION THE TABLE AND DRAWINGS BELOW ARE FOR S TUDY PURPOSES ONLY AND SHOULD NOT BE PRACTISED. THESE NUMBERS ARE PROVIDED FOR REFERENCE, THE CALCULATION IS DETAILED IN THE LATER PARAGRAPHS.

9:1

8:2

7:3

6:4

5:5

4:6

3:7

2:8

1:9

A (triangle length)

7000

7000

7000

7000

7000

7000

7000

7000

7000

X (point-CoG)

6300

5600

4900

4200

3500

2800

2100

1400

700

C (leg length)

7037

7037

7037

7037

7037

7037

7037

7037

7037

Tip arm (a)

649

577

505

433

361

288

216

144

72

From the above results it would appear that for mechanical stability, the best option is to split the trailer 9:1. In a later paragraph we will se why this is not the overall best option. Though highly important, mechanical stability is not the only criteria to decide on the hydraulic setup.


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Ro-Ro operations

JETTY BARGE When driving off or onto a barge or ship, it is advised to orientate the stability triangle in such a direction that the floating field will roll onto the barge first. This ensures that the trailer and cargo can be leveled using the quay as a fixed base, while the barge can move under the axles of the floating field. After t he floating field has been rolled on, the barge already holds some weight and is less affected by the movement of the water. As a ro-ro operation is considered to be stability critical situation, i t might be required to momentarily switch to a four point suspension system. When using a four point suspension during the roll-on or roll-off, it is mandatory to properly and constantly monitor the axle pressures. When the transport is half way onto the barge, the trailer assists the barge to remain le vel due to the fact that the “rear” of the transport has two support points on the quay that keep the barge level. When the final axles roll onto the barge, this support is lost and the barge will assume its natural position which might be a bigger angle then the stability angle of the trailer.


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3.5

Ref.

Rev.

3

Using turn tables

A turntable or bolster allows the trailer to rot ate under the load, when this load is carried by multiple sets of trailers (dollies). This rotation makes it easier to go thr ough turns making the transport more flexible and maneuverable.

Deck

Shoe - slipper Skid-Sliding ring Without shoes:

The turn table provides various degrees of freedom: Due the ball-joint in the center, it can rotate i n all three directions, while maintaining its position. Some bolsters have a ball-joint with a bolt-nut connection to limit vertical movement, but usually it’s being kept in place by the weight on top. The shoes restrict the rotation to the sides. Some bolster allow the shoes to be left out. I n that case, the bolster provides only front-to-back pivot. A bolster with the shoes provides 3 points of support while only the two outside supports actually provide lateral stability so the turntable is “on two points”. W hen the shoes are removed, there is only one support, so it is “on one point”. In practice we can combine these two types of supports. If we use both turntables with shoes, the combination of load plus turntables will be “on 4 points”. When we remove the shoes from one of the turntables then the combination will be “on 3 points”. When all shoes are removed, the combination has only 2 support points and will be unstable by definition.


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When using turntables (bolsters) you have to consider the stability of the trailers and the load with the turntables separately and transpose their results to c ombine them: Stability triangles of trailers Stability lines of load on turntables

Combined stability lines The black rectangle is the combined stability of the entire system. What is made clear from the above picture is that when driving straight ahead, the combined stability is a litt le bit better than the stability of the single trailer underneath the bolsters, due to the angle on the triangle. We can therefore conclude that the stability of the combination is the same or better then the stabilit y of the single trailer when we assume the saddle load to be applied at the height of the CoG of the cargo. But even when the stability of the combination is bigger than the stability of the single trailer, the actual l imit when driving will be the smaller one of the two, in this case being the tipping angle of the single trailer, as the trailer wi ll tip over before reaching the combined stability angle. A specific situation would be when both trailers would be angled under the load as depicted in the image below.

The entire transport combination would tip over the line A-A, which is at the same distance as the trailer tipping line. As we already assumed the trailer tipping angle as the limit , this situation results in the exact same limiting angle. Also we need to review the influence of rotating the trailer w hile negotiation a turn, or when it is required for project specific reasons to maneuver the trailers into a certain position. This is to demonstrate that the stability of the combination is only increased when the trailer is being positioned under an angle.


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The image below shows various different types of set-ups for using bolsters. Each trailer can have a different hydraulic suspension setup.

1

2

3

Trailer 1 = 3p Trailer 2 = 3p Bolster = 4p



Option 1 shows a traditional 3 point suspension on both trailers combined with a four point suspension in the bolsters. This option ensures maximum stability, but increases the risk off load transfer as stability forces will travel through load and bolsters. Option 2 is only for theoretical reference should never be practiced, as the 2 point suspension will be kept upright by the trailer on 3-points when driving straight ahead, but when driving through corners, the stability will reduce to zero. Option 3 may only be practiced when you can ensure that the load is able to transfer the up righting moment required for the floating bolster. This would be the case when transporting square beams that are lashed properly to the bolster table. Another example is a reactor on saddles, but you need to ensure that the fixation between saddle and reactor is rigid enough to transfer the loads. The mayor advantage is that there is no chance of load transfer.


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3.6

Ref.

Rev.

3

Determining the system Center of Gravity

As discussed in the conclusion of mechanical stability, the next important factor to find is the height of the Center of Gravity. We already know that the tipping plane is at the height of the center of the axle hubs, so by adding the height of the trailer above the axle hub to the height of the CoG of the cargo to the trailer deck, we find the height of CoG for the system. However, when doing this, we are already reviewing the stability of a combined system, i.e.: Trail er + Cargo. As explained at the start of this chapter, if w e are to determine the stability of a combined system, we need to look carefully at which of the described scenarios will take place, or i n other words, determine the stability of each individual item when moving. If we consider cargo placed on top of a trailer, the axles of the trailer will experience the weight of t his object plus its own weight as a single force that is lead through its wheels onto the road. This is because the weight of the cargo is pressing down on the trailer deck and the weight of the cargo and the trailer deck together are pressing down onto the axles. We therefore need to combine the weight of the trailer and the weight of the cargo. Furthermore, when the system is placed on an incline and the transport is rotated, the centers of gravity will rotate together as long as the transport remains “stable”. This in turn means we can combine these CoG’s as long as the system is stable. This combined center of gravity will always be located between the CoG of the cargo and the CoG of the Trailer: The below example shows the stability triangles for a three point suspension.


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If we do the same for less wide load (narrow), and different ratios between the weight of the trailer and the weight of the cargo, we find that when the weight of the cargo is much lighter then the trailer, the combination has a higher degree of stability then the cargo on the trailer deck. In all other cases the stability of the combination will the limiting factor. Within Mammoet we have agreed to the following compromise to determine the height of the transport CoG: -

When cargo is not lashed, use the CoG of the Cargo to determine stability angle. When cargo is sufficient vertically lashed, use the combined CoG to determine stability angle.

Sufficient vertical lashing means: “The lashing should be strong enough to be able to lift the trailer, so cargo plus trailer will keep acting as a single system when the stabilit y angle of the cargo has been exceeded. In the chapter “Lashing and Blocking” is explained how to calculate the required amount of lashing.


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3.7

Ref.

Rev.

3

Mechanical stability by numbers

Here we have calculated the stability angles for various standard trailer arrangements. This can only be done for standard trailer configurations, as there are an unlimited amount of trailer combination possibilities.

CAUTION

CAUTION THE FOLOWING TABLES ARE FOR STUDY PURPOSES ONLY AND H AVE ONLY LIMITES USE AS THE ARE ONLY VALID FOR STANDARD HYDRAULIC SE TUPS AND WHILE COG IS LOADED IN THE CENTER OF THE TRAILER IN ALL DIRECTIONS.

Trailer type: Single KAMAG / Scheuerle'84/'94/2004

Limi t: Lines All

Hydraulic Setup

Tip Arm (a)

7

°

Longitudinal spacing:

1.4

Transversal spacing:

m

1.45

m

Deck height:

1.5

m

Ti pping plane:

0.3

m

Height CoG above Trailer deck [m] 2.00

2.25

2.50

2.75

3.00

3.25

3.50

3.75

4.65

5.00

4 Point 4:2 6:2

0.725 0.476 0.539

12.7 8.4 9.5

11.8 7.8 8.8

11.0 7.3 8.2

10.4 6.8 7.7

9.7 6.4 7.3

9.2 6.1 6.9

8.7 5.7 6.5

8.3 5.4 6.2

7.0 4.6 5.2

6.6 4.3 4.9

8 10

5:3* 7:3

0.449 0.505

7.9 8.9

7.4 8.3

6.9 7.7

6.4 7.2

6.1 6.8

5.7 6.4

5.4 6.1

5.1 5.8

4.3 4.9

4.1 4.6

10 12

6:4* 8:4

0.433 0.482

7.7 8.5

7.1 7.9

6.6 7.4

6.2 6.9

5.8 6.5

5.5 6.1

5.2 5.8

4.9 5.5

4.2 4.7

3.9 4.4

14

10:4

0.516

9.1

8.5

7.9

7.4

7.0

6.6

6.2

5.9

5.0

4.7

14 16 16

9:5* 11:5 10:6*

0.465 0.497 0.452

8.2 8.8 8.0

7.6 8.2 7.4

7.1 7.6 6.9

6.7 7.1 6.5

6.3 6.7 6.1

5.9 6.3 5.8

5.6 6.0 5.4

5.3 5.7 5.2

4.5 4.8 4.4

4.2 4.5 4.1

18 20 20

12:6 14:6 13:7*

0.483 0.507 0.471

8.5 8.9 8.3

7.9 8.3 7.7

7.4 7.7 7.2

6.9 7.3 6.7

6.5 6.8 6.3

6.1 6.4 6.0

5.8 6.1 5.7

5.5 5.8 5.4

4.7 4.9 4.5

4.4 4.6 4.3

22

15:7

0.494

8.7

8.1

7.6

7.1

6.7

6.3

5.9

5.6

4.8

4.5

22 24

14:8* 16:8

0.461 0.483

8.1 8.5

7.6 7.9

7.0 7.4

6.6 6.9

6.2 6.5

5.9 6.1

5.6 5.8

5.3 5.5

4.5 4.7

4.2 4.4

6 8

*For study purposes only. **This table is only valid when CoG is loaded in center of trailer


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Ref.

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Trailer type: Double KAMAG / Scheuerle'84/'94/2004

Limi t:t: Lines All

Hydraulic Setup 4 Point

Tip Arm (a)

7

3


°

1.4

m


2.9

m

Deck height:

1.5

m

Ti pp pping plane:

0.3

m

11.00


5.75

6.00

6.25

6.50

6.75

7.00

7.25

10.50

1.450

12.2

11.7

11.3

11.0 11.0

10.6 10.6

10.3

10.0

9.7

7.0

6.7

6

4:2

0.914

7.7

7.4

7.2

6.9

6 6.7 .7

6.5

6.3

6 6.1 .1

4.4

4.2

8

6:2

1.053

8.9

8.6

8.3

8.0

7 7.7 .7

7.5

7.3

7 7.1 .1

5.1

4.9

10 12

7:3 8:4

0.994 0.953

8.4 8.0

8.1 7.8

7.8 7.5

7.5 7.2

7 7.3 .3 7 7.0 .0

7.1 6.8

6.9 6.6

6 6.7 .7 6 6.4 .4

4.8 4.6

4.6 4.4

14 16

10:4 11:5

1.025 0.989

8.6 8.3

8.3 8.0

8.0 7.8

7.8 7.5

7.5 7.5 7.3 7.3

7.3 7.0

7.1 6.8

6.9 6.6

5.0 4.8

4.8 4.6

18 20 22

12:6 14:6 15:7

0.960 1.010 0.984

8.1 8.5 8.3

7.8 8.2 8.0

7.5 7.9 7.7

7.3 7.7 7.5

7.1 7.1 7.4 7.4 7.2 7.2

6.8 7.2 7.0

6.6 7.0 6.8

6.4 6.8 6.6

4.6 4.9 4.8

4.5 4.7 4.6

24

16:8

0.963

8.1

7.8

7.6

7.3

7.1 7.1

6.9

6.6

6.5

4.7

4.5

*For study purposes only. **This table is only valid when CoG is loaded in center of trailer Trailer type: Triple KAMAG / Scheuerle'84/'94/2004

Limi t:t: Lines All

Hydraulic Setup

Tip Arm (a)

4 Point

7

°


1.4

m


4.35

m

Deck height:

1.5

m

Ti pp pping plane:

0.3

m


9.00

9.50

10.00

10.50

11.00

11.50

12.00

16.50

17.00

2.175

12.6

12.0

11.4

10.9

10.5

10.1

9.7

9.3

7.0

6.8

6 8

4:2 6:2

1.288 1.521

7.5 8.9

7.1 8.4

6.8 8.0

6.5 7.7

6.2 7.4

6.0 7.1

5.7 6.8

5.5 6.5

4.1 4.9

4.0 4.7

10 12 14

7:3 8:4 10:4

1.454 1.404 1.517

8.5 8.2 8.8

8.1 7.8 8.4

7.7 7.4 8.0

7.3 7.1 7.7

7.0 6.8 7.3

6.7 6.5 7.0

6.5 6.3 6.8

6.2 6.0 6.5

4.6 4.5 4.8

4.5 4.4 4.7

16 18 20

11:5 12:6 14:6

1.468 1.429 1.504

8.6 8.3 8.8

8.1 7.9 8.3

7.8 7.6 8.0

7.4 7.2 7.6

7.1 6.9 7.3

6.8 6.6 7.0

6.5 6.4 6.7

6.3 6.1 6.5

4.7 4.6 4.8

4.6 4.4 4.7

22 24

15:7 16:8

1.468 1.438

8.6 8.4

8.1 8.0

7.8 7.6

7.4 7.3

7.1 7.0

6.8 6.7

6.5 6.4

6.3 6.2

4.7 4.6

4.6 4.5



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Trailer type: Single 3m wide Conventional Trailer

Limi t:t: Lines All

Hydraulic Setup 4 Point

Tip Arm (a)

7

3


°

1.51

m


1.8

m

Deck height:

1.2

m

Ti pp pping plane:

0.3

m

5.00


3.25

3.50

3.75

4.00

4.25

4.50

4.75

4.65

0.900

12.9

12.2

11.5

10.9

10.4

9.9

9.4

9.0

9.2

8.6

6

4:2

0.588

8.5

8.0

7.6

7.2

6 6.8 .8

6.5

6.2

5 5.9 .9

6.0

5.6

8

6:2

0.668

9.7

9.1

8.6

8.1

7 7.7 .7

7.3

7.0

6 6.7 .7

6.8

6.4

10 12

7:3 8:4

0.626 0.597

9.1 8.7

8.5 8.1

8.0 7.7

7.6 7.3

7 7.2 .2 6 6.9 .9

6.9 6.6

6.6 6.3

6 6.3 .3 6 6.0 .0

6.4 6.1

6.0 5.7

14 16

10:4 11:5

0.641 0.617

9.3 8.9

8.7 8.4

8.2 7.9

7.8 7.5

7.4 7.4 7.1 7.1

7.0 6.8

6.7 6.5

6.4 6.2

6.5 6.3

6.1 5.9

18 20 22

12:6 14:6 15:7

0.599 0.629 0.613

8.7 9.1 8.9

8.2 8.6 8.3

7.7 8.1 7.9

7.3 7.7 7.5

6.9 6.9 7.3 7.3 7.1 7.1

6.6 6.9 6.7

6.3 6.6 6.4

6.0 6.3 6.1

6.1 6.4 6.3

5.7 6.0 5.9

24

16:8

0.599

8.7

8.2

7.7

7.3

6.9 6.9

6.6

6.3

6.0

6.1

5.7


Trailer type: Double 3m wide Conventional Trailer

Limi t:t: Lines All

Hydraulic Setup

Tip Arm (a)

7


°

1.51

m


3.4

m

Deck height:

1.2

m

Ti pp pping plane:

0.3

m


7.75

8.00

8.25

8.50

8.75

9.00

9.25

12.75

13.00

6

4 Point 4:2

1.700 1.061

11.4 7.1

11.1 6.9

10.8 6.7

10.5 6.6

10.2 6 6.4 .4

9.9 6.2

9.7 6.1

9.5 5 5.9 .9

7.0 4.4

6.9 4.3

8

6:2

1.227

8.3

8.0

7.8

7.6

7 7.4 .4

7.2

7.0

6 6.8 .8

5.1

5.0

10 12 14 16 18 20

7:3 8:4 10:4 11:5 12:6 14:6

1.161 1.114 1.199 1.157 1.125 1.183

7.8 7.5 8.1 7.8 7.6 8.0

7.6 7.3 7.8 7.6 7.4 7.7

7.4 7.1 7.6 7.4 7.2 7.5

7.2 6.9 7.4 7.2 7.0 7.3

7 7.0 .0 6 6.7 .7 7.2 7.2 7.0 7.0 6.8 6.8 7.1 7.1

6.8 6.5 7.0 6.8 6.6 6.9

6.6 6.4 6.9 6.6 6.4 6.8

6 6.5 .5 6 6.2 .2 6.7 6.5 6.3 6.6

4.8 4.6 5.0 4.8 4.7 4.9

4.7 4.5 4.9 4.7 4.6 4.8

22

15:7

1.153

7.8

7.5

7.3

7.1

6.9 6.9

6.8

6.6

6.4

4.8

4.7

24

16:8

1.128

7.6

7.4

7.2

7.0

6.8 6.8

6.6

6.5

6.3

4.7

4.6



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3.8

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Disabling axles

In some cases it might be required to disable axles to reduce the bending moment in the spine beam of the trailer, depending on the load case. When disabling axles on a three-point suspension, you need t o be aware that there is a chance that by doing this you reduce the tipping stability. On a four point suspension you only lose stability on strength, but the mechanical stability remains intact. Depending on the resulting hydraulic setup, the mechanical stability angle can be bigger or smaller.

3.9

Using spacers

A similar effect can be expected when using spacer frames. Even though it would be practical to split the hydraulics frontback at the position of the spacer, it will reduce the mechanical stability drastically. Never increase mechanical stability by re-ordering the hydraulic fields when this results i n large differences in the number of axles in the fields. This w ill increase the risk of overloading the axles in the field(s) with fewer axles.


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Fixed fields

A fixed field is a h ydraulic field that is preset to a certain pressure. You can compare a fixed field with the fourth field in a 4-point suspension, where the difference is that a second operator will monitor and adjust the pressure to a constant value. Sometimes the construction of the cargo is unable to transfer loads, and when it is impossible to physically connect the trailer, you can use a fixed field. This field will form an extra support point for the construction. The use of fixed fields can result in overload or loss of stability when the pressures in the fixed fields are not properly monitored and adjusted. You should only resort to using fixed fields w hen the construction on the trailer requires this, or when it i s impossible to hydraulically connect the trailers.


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3.11

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3

Calculating Mechanical stability

So far, we have learned that we measure the mechanical stability is the angle at which the transport t ips over, and know how to determine the tipping lines and level of the tipping plane of a trailer. When the center of gravity of the cargo resides between these tipping lines, the state of the transport is “stable”. However, when the center of gravity of the cargo finds itself above one of these tipping lines, the t ransport combination becomes Neutral, but Critical. If the center of gravity resides beyond the tip-lines, the trailer w ill tip-over (unstable). When we consider a hydraulic platform trailer using a four-point suspension, the center lines of the axles form the tipping points of the trailer, and the level of the tipping plane is at the center of the wheel hubs. The images below show these various states. We are now able to determine the maximum gradient or road camber for this particular transport before the trailer starts tipping over. To do this we require the dimensions of the tip arm, CoG height of the reactor, height of the trailer, axle height and the saddles height. Stable

Neutral

Unstable

Transport CoG

Tip plane Tip arm The height of the center of gravity of the cargo is calculated as: Formula 1

Height CoG = Height trailer + Height saddle + height CoG cargo - Trailer axle height. The maximum gradient using the formula:

Formula 2

Tan α ( ) o

=

Tip arm(a) Height CoG


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3.11.1

3.11.2

Ref.

Rev.

3

Example: Transport details

SPMT trailer height SPMT axle spacing SPMT track width SPMT axle height SPMT CoG height Trailer weight

= 1500mm = 1400mm = 1450mm = 300mm = 750mm above ground level = 33t (26[6Liner] + 7[PPU])

Reactor weight Diameter reactor Saddle height

= 120t = 4000mm = 675mm

Example: 4 point suspension

Due to the 4-point suspension the smallest Tip arm in this example is half t he track width (a=725).

So entering these values into our formulas results in: Height CoG = 1500

Tan ( α ) =

725 3875



= 10.6°

=

+

675 + (4000/2) - 300

0 .187 = 18 .7 %

=

3875mm

(Formula 1)

(Formula 2)

a

This means that the trailer will start to tip over at an angle of 10.6°so the trailer deck angle should be kept below that value. If we couple additional trailers sideways, we effectively improve the Track Base so our stability angle will increase. If we use timber blocks underneath the saddle of the reactor (100mm), we effectively increase the height of the CoG, so the stability angle will get smaller.


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3.11.3

Ref.

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Example: 3-point suspension

In case of the three point suspension, we need to do so more work to find the Tip Arm due to the geometry of the hydraulic setup.

We need to determine the dimensions of the triangle and calculate the smallest tipping arm. A = 3 axles B = half track X = 2 axles

= 3 x 1400 = 1450 / 2 = 2 x 1400

Pythagoras: C=

(4200 2

= 4200mm = 725mm = 2800mm C² = A² + B²

+

725 2 )

a = (725 * 2800) / 4266

= 4262mm = 476mm

The height of CoG remains unchanged (3875) so we can complete the f ormula: Tan ( α ) =

476 = 0 .123 = 12 .3 % 3875



= 7.0°

(Formula 2)

a

This means that the trailer will start to tip over at an angle of 7.0°so the trailer deck angle should be kept below that value. In case of a three point suspension you can improve the Tip Arm (a) by changing to a different hydraulic setup, though the improvement will not be that large.


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3.11.4

Ref.

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3

Example: 3 point suspension with combined CoG

By combining the center of gravity of cargo and trailer, the height of the center of gravity used in the formula is reduced, as the trailer center of gravity will always below the load. Lowering the height of the CoG is always more favorable for stability on tipping, but is often not possible, due to the shape of the cargo. The required friction to prevent the load from shifting on the trailer deck can be achieved by using plywood between the trailer-deck and the cargo support area, see chapter “Lashing and Blocking” for more details.

You can calculate the height of the combined CoG as follows: Multiply all weights with their respective heights of CoG and summarize them. Divide this sum by the total weight to find the combined CoG.

Formula 3

Height CoG (h) =

∑ (weight x height) ∑ weight

The reactors height of CoG above ground level is: 1500(trailer) + 675(saddle) + 2000(reactor) = 4175mm

Height CoG =

(750 x 32 ) + (4175 x 120 ) (33 + 120 )

=

52500 = 3436 mm 153

(Formula 3)

This makes the Height CoG above tipping plane: Height CoG = 3436 - 300mm

= 3136mm

Trailer is still on 3-point suspension, so use 476 as tip arm! Tan ( α ) =

476 = 0 .152 = 15 .2 % 3136

(Formula 2)

= 8.6°

a


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Summary

When neglecting any lashing arrangement, the mechanical stability of a trailer is only dependant of: -

Hydraulic setup of the trailer (Tipping Arm) Height of Cargo CoG, above tipping plane.

Below picture shows a trailer on 4 point suspension Stable

Neutral

Unstable

Transport CoG

Tip plane Tip arm When sufficient lashing is provided so we can use the combined CoG of trailer and cargo, the mechanical stability of a trailer also depends on: -

Weight of Cargo Weight of Trailer

We need to look at the mechanical stability in every tipping direction of the trailer. When the trailer uses a 3point suspension, there are three tip directions. When it is on a four point suspension there are four tipping directions. Mammoet limits the mechanical stability angle to 7 degrees. This means that any transport with a mechanical stability angle below 7 degrees is considered as critical and should be reviewed by a senior engineer. This limit for the angle was decided on, because the actual position of the CoG and the actual angle of the trailer is difficult t o monitor in practice.


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4

Ref.

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3

Structural Stability Structural stability is more identifiable in the f ield then mechanical stability. After loading, you can check if the CoG is centrally placed by checking the hydraulic pressures of the fields. Independent of the selected hydraulic setup, the pressures should be equal when the trailer is level. When there are differences in the axle pressures, the CoG is off-centered. After checking the placement of the load on the trailer, it i s good practice to checkout the reaction of the hydraulic setup of the trailer by setting the trailer deck just a little off-level (0.5 - 1 degree) using a spirit level. By monitoring the hydraulic gauges during the rotation, you can make a prediction of the pressure differences at larger angles. You will also get a feeling for the response time of the hydraulics, i.e. how quick will the hydraulic system of the trailer react. Alw ays check this while the load is still above its supports, and always use “LOWERING” to make the pressures different, or when equalizing pressures. Always try to “RIASE” all fields in t he system at the same time while performing the check.

4.1

General principle

When we are checking the stability on strength, we actually check at which angle the axles of the platform trailers will start to get overloaded due to the displacement of the center of gravity by the trailer angle. Structural stability angle

G

G

Transport CoG

Ra

Rb

Ra

Rb = Maximum

Image 9 When the cargo is centered on the trailer, both axles (Ra and Rb) will each take an equal amount of the load. If the platform trailer i s standing on a gradient, (Rb) will get a larger amount of the load then (Ra). A greater gradient results in a bigger difference.


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This load transfer is cause by the movement of the center of gravity. The larger the tr ansverse distance between the center of gravity of the cargo and the center of the trailer, the more l oad will be transferred from axle Ra to axle Rb. When the center of gravity of the cargo is exactly above axle Rb, axle Rb will carry the entire load of the cargo whilst axle Ra carries nothing. Considering the loading conditions of our trailers, the maximum possible axle load is already exceeded, so there is a limit for MaxX.

The maximum structural stability angle will be found when the maximum axle load is achieved. From the image we can clearly see that the structural stability angle is also dependent of the Height of CoG. When the cargo is not loaded in the centre of the trailer, the actual distance actX is changed and depending on w hich side of the trailer you are reviewing will become bigger or smaller.

G

When actX grows bigger, the distance between MaxX and actX becomes smaller, reducing the structural stability angle. The weight and maximum axle load are the other factors that impact the structural stability angle. In fact it is the ratio between the maximum axle load and axle capacity that limits the structural stability angle. If an axle is loaded with 15 tons (gross, i.e. ow n weight + cargo weight), and its capacity is 17 tons, there is an additional 2 tons capacity to take a shift in the CoG. When the axle is fully loaded to its maximum capacity of 17 tons, there is 17 – 17 = 0 t ons left, so even a slight movement in the CoG will cause an overload, meaning: The structural stability i s 0 degrees. In general Mammoet has adopted a minimum structural stability angle of 2 degrees. This small angle was chosen as the pressures on the trailers are easily monitored and adjusted. All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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4.2

Ref.

Rev.

3

Structural stability by numbers

Below we have prepared a table for some standard SPMT configurations with the CoG loaded in the center of the trailer. These tables were prepared for study purposes and are therefore for reference only. Trailer type: Single KAMAG / Scheuerle'84/'94 Longitudinal spacing: 1.4 m Cap.: Cap.: Limit: Lines

Hydraulic Setup

Loading ratio

Total Weight*

17 ton/axle 34 ton/line 2 °

Transversal spacing: 1.45 Deck height: 1.5 Tipping plane: 0.3

m m m


3.00

4.00

5.00 8.00

14.00

20.00

30.00

50.00

60.00

6 4:2 6 4:2 6 4:2

25% 50% 75%

51.0 102.0 153.0

44.5 18.1 6.2

18.7 6.4 2.1

15.3 5.2 1.7

12.9 4.3 1.4

8.8 2.9 0.9

5.3 1.7 0.5

3.8 1.2 0.4

2.6 0.8 0.2

1.5 0.5 0.1

1.3 0.4 0.1

6 4:2

100%

204.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

6 4 Point

25%

51.0

56.3

27.3

22.6

19.3 13.3

8.1

5.8

3.9

2.4

2.0

6 4 Point 6 4 Point 6 4 Point

50% 75% 100%

102.0 153.0 204.0

26.5 9.4 0.0

9.7 3.2 0.0

7.9 2.6 0.0

6.6 2.2 0.0

4.5 1.5 0.0

2.7 0.9 0.0

1.9 0.6 0.0

1.3 0.4 0.0

0.8 0.2 0.0

0.6 0.2 0.0

8 6:2

25%

68.0

48.1

21.0

17.2

14.6

9.9

6.0

4.3

2.9

1.8

1.5

8 6:2 8 6:2 8 6:2

50% 75% 100%

136.0 204.0 272.0

20.4 7.0 0.0

7.3 2.4 0.0

5.9 1.9 0.0

4.9 1.6 0.0

3.3 1.1 0.0

2.0 0.6 0.0

1.4 0.4 0.0

0.9 0.3 0.0

0.6 0.2 0.0

0.5 0.1 0.0

8 4 Point 8 4 Point

25% 50%

68.0 136.0

56.3 26.5

27.3 9.7

22.6 7.9

19.3 13.3 6.6 4.5

8.1 2.7

5.8 1.9

3.9 1.3

2.4 0.8

2.0 0.6

8 4 Point 8 4 Point

75% 100%

204.0 272.0

9.4 0.0

3.2 0.0

2.6 0.0

2.2 0.0

1.5 0.0

0.9 0.0

0.6 0.0

0.4 0.0

0.2 0.0

0.2 0.0

25% 50%

85.0 170.0

46.2 19.1

19.8 6.8

16.2 5.5

13.7 4.6

9.3 3.1

5.6 1.9

4.0 1.3

2.7 0.9

1.6 0.5

1.4 0.4

10 7:3

75%

255.0

6.6

2.2

1.8

1.5

1.0

0.6

0.4

0.3

0.1

0.1

10 7:3

100%

340.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

10 6:4* 10 6:4*

25% 50%

85.0 170.0

41.8 16.6

17.1 5.8

14.0 4.7

11.8 3.9

8.0 2.6

4.8 1.6

3.5 1.1

2.3 0.7

1.4 0.4

1.2 0.4

10 6:4*

75%

255.0

5.6

1.9

1.5

1.3

0.8

0.5

0.3

0.2

0.1

0.1

10 6:4*

100%

340.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

10 7:3 10 7:3

10 4 Point

25%

85.0

56.3

27.3

22.6

19.3 13.3

8.1

5.8

3.9

2.4

2.0

10 4 Point 10 4 Point 10 4 Point

50% 75% 100%

170.0 255.0 340.0

26.5 9.4 0.0

9.7 3.2 0.0

7.9 2.6 0.0

6.6 2.2 0.0

2.7 0.9 0.0

1.9 0.6 0.0

1.3 0.4 0.0

0.8 0.2 0.0

0.6 0.2 0.0

4.5 1.5 0.0


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TP 06 – Trailer Stability & Lashing Trailer type:

Ref.

Single KAMAG / Scheuerle'84/'94 Cap.: Cap.: Limit:

Lines

Rev.

Hydraulic Setup

Loading ratio

Total Weight*

3

Longitudinal spacing: Transversal spacing: Deck height: Tipping plane:

17 ton/axle 34 ton/line 2 °

1.4 1.45 1.5 0.3

m m m m


3.00

4.00

5.00 8.00

14.00

20.00

30.00

50.00

60.00

12 8:4

25%

102.0

44.8

18.9

15.5

13.1

8.9

5.4

3.8

2.6

1.6

1.3

12 8:4 12 8:4

50% 75%

204.0 306.0

18.3 6.3

6.5 2.1

5.2 1.7

4.4 1.4

2.9 0.9

1.8 0.6

1.3 0.4

0.8 0.2

0.5 0.1

0.4 0.1

12 8:4

100%

408.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

12 4 Point 12 4 Point

25% 50%

102.0 204.0

56.3 26.5

27.3 9.7

22.6 7.9

19.3 13.3 6.6 4.5

8.1 2.7

5.8 1.9

3.9 1.3

2.4 0.8

2.0 0.6


75% 100%

306.0 408.0

9.4 0.0

3.2 0.0

2.6 0.0

2.2 0.0

1.5 0.0

0.9 0.0

0.6 0.0

0.4 0.0

0.2 0.0

0.2 0.0

18 12:6

25%

153.0

44.9

19.0

15.5

13.1

8.9

5.4

3.9

2.6

1.6

1.3

18 12:6 18 12:6

50% 75%

306.0 459.0

18.4 6.3

6.5 2.1

5.3 1.7

4.4 1.4

3.0 1.0

1.8 0.6

1.3 0.4

0.8 0.2

0.5 0.1

0.4 0.1

18 12:6

100%

612.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

18 4 Point

25%

153.0

56.3

27.3

22.6

19.3 13.3

8.1

5.8

3.9

2.4

2.0


50% 75%

306.0 459.0

26.5 9.4

9.7 3.2

7.9 2.6

6.6 2.2

4.5 1.5

2.7 0.9

1.9 0.6

1.3 0.4

0.8 0.2

0.6 0.2

18 4 Point

100%

612.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

*Total weight includes trailer weight and power pack. **This table is only valid when CoG is loaded in center of trailer

4.3

Influences on the Structural stability

In the previous paragraphs, we looked at the structural stability in the transverse direction only, but you should also consider the load transfer front-to-back when the trailer is driving down a slope. However, a height difference in the length results in a smaller angle then a height difference in the wi dth due to the large distance between pressure points(i.e. the width is generally smaller then the length). This changes when we change the hydraulic setup of the trailer so we need to make sure that we look at all possibilities. • • • • • •

The height of the center of gravity of the cargo. The weight of the cargo. The offset of the center of gravity of cargo from trailer centre. The height of the trailer-deck. The weight of the trailer (combination). The hydraulic setup of the trailer.


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Ref.

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3

Calculation of the Axle loads

The calculation of the maximum load on a field is largely related to t he calculation of the actual axle pressures in the static condition. When the cargo is loaded in the center of the trailer it is easy to calculate the axle pressures, by dividing the weight of cargo + trailer by the amount of axles, but we cannot use this method when the CoG is not in the center of the trail er. In any case we can use the previously determined support points of t he hydraulic fields, and the reference distance of the CoG to the tipping lines of the trailer. We then transpose the axle reactions in line with the tipping line. For the example we use the previously used transport combination.

The triangle height (La) is also the support distance between Ra and Rb+Rc. We are now able to calculate the field loads: Ra =

(G x CoGa ) La

And: Rb + Rc =

(G x (La - Coga) ) La

When the CoG is in the center of the trailer, t hen Ra equals Rb so we can easily divide by 2. Rb = Rc =

(G x (La - Coga) ) 2 x La


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When the CoG is not in the center and we wish to calculate Rb, we first need to find the tri angle height Lb and the CoGb. As we know the height (La) and width (Track w idth w) of the triangle, we can calculate the area of the triangle. Also we are able co calculate hypotheses “B” between Rc and Rb. If we take hypotheses “B” as the base of the triangle, we can find the height of the triangle (Lb) in the direction of Rb. Area Triangle

=

Area Triangle = B=

W x La 2 B x Lb 2

( La 2 + ( 0 .5 xW 2 )

Combining these makes: 2

( La W x La B x Lb = = 2 2 La x W Lb = ( La 2 + ( 0 .5 xW 2 )

+

( 0 .5 xW 2 ) x Lb 2

To find CoGb is even more elaborate: Zoom-in on so we review sub triangle G-Rc-Rb, with legs B, K and M, for which we can calculate their lengths. B=

( La 2 + ( 0 .5 × W ) 2 )

M=

( CoGa 2 + ( 0 .5 × W + Y ) 2 )

K=

(( La − CoGa ) 2 + ( Y ) 2 )

With these lengths we can calculate the area of triangle (G-Rc-Rb): Area =

Sx ( S − B )x ( S − M )x ( S − K )

Where S=

1 x ( B + M + K ) 2

From this we calculate the height of that triangle. CoGb = 2 × Area × B All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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Example: Calculation of the axle loads

Trailer on 3-point suspension! Diameter reactor = 4000mm Trailer height = 1500mm Track width = 1450mm Saddle height = 675mm Axle height = 300mm SPMT axle spacing = 1400mm Reactor weight = 120t Trailer weight = 32t (26[6Liner] + 6[PPU]) Trailer CoG height = 750mm above ground level Maximum axle-line load = 17t/axle (34t per line). In our example: W = 120tCargo + 33t Trailer La = 3 axles x 1400 CoGa = 1 axle x 1400 Ra =

(153 x 1400 ) 4200

=

= 153t = 4200mm = 1400mm 51.0 t

As the CoG is in the center of the trailer we can use Formula (X) to fin d Rb and Rc. Rb = Rc =

(153 x (4200 - 1400) ) 2 × 4200

=

51.0 t

This result in the following axle loads: Field A (4 cylinders) = 51.0 / 4 = 12.75t Field B (4 cylinders) = 51.0 / 4 = 12.75t Field C (4 cylinders) = 51.0 / 4 = 12.75t Fortunately for us, these results are the same as when we simply divide the total weight by the number of cylinders: 153.0t / 12cylindres = 12.75t. With these axle loadings there is 17.0t – 12.75t = 4.25t of capacity left to cope for any additional forces, such as a load transfer due to the gradient of the road. If the trailer would be fully loaded (17.0t), there would be 0t capacity left, and result in a structural stability of 0 degrees!


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Calculation of the Structural stability

The maximum allowable displacement of the center of gravity can be determined with the help of the maximum allowable load of the axle. The larger the transverse distance between the center of gravity of the cargo and the center of the trailer, the more load will be transferred from axle Ra to axle Rb. When the center of gravity of the cargo is exactly above axle Rb, axle Rb will carry the entire load of the cargo whilst axle Ra carries nothing. When calculating the stability on strength, we use a similar method, except this time we know t he reaction “Ra-Max” which equals the “Field capacity”. This enables us to calculate the maximum allowable CoG position MaxX. We need to do this in each field direction i.e. over each tipping line.

MaxX =

L * Field capacity total Weight(G)

This determines the Maximum possible distance between the pivot and the CoG. If we look the image below which indicates the location of these distances we can determine that we can calculate the Structural stability angle as:

Tan α =

MaxX - actX Height CoG Transport


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Example: Transport details

For the following example we use the same arrangement as in the chapter “Mechanical stability” SPMT trailer height SPMT axle spacing SPMT track width SPMT axle height SPMT CoG height Trailer weight

= 1500mm = 1400mm = 1450mm = 300mm = 750mm above ground level = 33t (26[6Liner] + 7[PPU])

Reactor weight Diameter reactor Saddle height

= 120t = 4000mm = 675mm

We already calculated the Total W eight (153.0t ) and the combined CoG height (3136mm ). We also need the field capacity: As we have 3 fields with each 4 cylinders of 17.0t so: Ra max = Rb max = Rc max = 4 x 17.0

= 68.0t

We determine length L from the hydraulic setup of the trailer La = 3 axles = 4200mm

MaxX =

4200 × 68.0 = 1867 mm 153

This makes the maximum possible angle: Tan α ( ) o

a

=

= 8.5°

1867 - 1400 = 0 .149 = 14 .9 % 3136 (When tilting toward Ra)


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To calculate MaxX in direction of Rb max (shown in red): As the hydraulic setup is symmetrical and the CoG is loaded in the center, so the calculation of RB and RB is exactly t he same. From the mechanical stability calculation we found that the tip arm (a) was 476mm. As th e CoG is loaded in the center (front-back AND left-right), we can assume the triangle height to be 3 x the tip arm, so Lb = 3 x 476 = 1428mm

This results in a MaxX of MaxX =

1428 * 68.0 = 635 mm 153

As the combined CoG has not changed (3136), this makes the maximum possible angle: Tan α ( ) o

a

=

= 2.9°.

635 - 476 = 0 .050 = 5 .0 % 3136 (When tilting to either side)

Note: It may be obvious that as long as the hydraulic setup provides a base that is w ider then the height of the triangle, the side-ways stability will be smaller then f ront-back. Also, when the CoG is not on the centre line of the trailer, the difference between MaxX and actX will be smaller, reducing the structural stability angle. This calculation also proves that, “side-ways stability fields” with a large number of axles wil l improve sideways stability. This in turn reduces the number of axles in the fl oating field, so we can only do this until the sideways stability equals the forward stability. All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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Summary

The structural stability angle mainly depends on ratio between the axle load and the axle capacity and the height of the Center of Gravity. An offset of the cargo’s Center of Gravity from the center of the trailer will reduce the structural stabilit y considerably. We need to look at the structural stability in every tipping direction of the trailer, if the trailer uses a 3-point suspension, there are three tip directions, when it is on a four point suspension there are four tipping directions. Mammoet uses a structural stability limit of 2°. This means that any transport with a structural stability angle below 2 degrees is considered as critical and should be reviewed by a senior engineer. This limit is smaller then the 7°for mechanical stability as the hydraulic pressures are an accurate readout for the actual camber of the trailer.


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5

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Dynamic influences Until now, we have only considered the static stabilities (mechanical and structural) of the platform trailers. In reality there are many external factors working on trailer and cargo which influence these stabilities. The main external factors are: • • •

Centrifugal force. Acceleration/Deceleration forces. Wind pressure.

An accurate prediction of the expected forces can be made through calculations, but it would be very impractical to use the actual formulas and expected factors as they may vary per situation. In stead, Mammoet has adopted impact factors to cover f or a large range of issues: Impact factor for centrifugal forces of transports at high speeds: Impact factor for centrifugal forces of transports at low speeds:

50% 2%

Impact factor for acceleration/deceleration for transports at high speeds: Impact factor for acceleration/deceleration for transports at low speeds:

80% 5%

As the wind area of the cargo has a large influence on the wind pressure, we need to be calculated for each different situation. The wind force depends on the maximum expected wind speed and t he projected area of the cargo. This wind force causes a moment on the trailer in both longitudinal and transverse direction. This moment affects the axle pressures and reduces the mechanical stability. The centrifugal force causes a moment in the lateral plane of the transport that affects the axle pressures and reduces the mechanical stability. In general the fields providing the side-ways stability absorb the centrifugal forces, depending on the hydraulic setup. The acceleration/deceleration force causes a moment in the driv ing direction of the transport that affects the axle pressures and reduces the mechanical stability. In general either the floating field or the split fields absorb the acceleration/deceleration forces, depending on the hydraulic setup. All combinations of the above mentioned forces add up to the dynamic axle loads. Within Mammoet we have agreed to maintain the static stability angles and that the dynamic forces will affect the axle loadings only.


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Centrifugal force

When you round a corner in your car, you seem to be pushed to the outside of the curve. This is referred to as the centrifugal force. This force is not due to something actually pushing you in that direction, but b y your body's inertia trying to keep you moving in a straight line. The car is actually curving around in front of you and intercepting you in your straight line path. The car door pushes you towards the center of the curve and makes you change direction and move in a circle with the car.

This centrifugal force depends on the mass of the object, the speed of rotation, and the distance from the center. The more massive the object, the greater the force; the greater the speed of the object, the greater the force; and the less distance from the center, the greater the force. Similar to what was explained in the paragraph regarding the wind pressure, this centrifugal force causes a moment working on the trailer, affecting the mechanical stability and the axle pressures. The difference here is that the trailer experiences this force only in one direction, laterally on the driv ing direction.

The effect of this centrifugal force is only small for SPMT transports due to the lower speeds, but conventional transporters can be seriously affected by this centrifugal force. It is therefore important to reduce the speed of the transport while negotiating a turn or corner.


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The formula to calculate the centrifugal force is defined as: Centrifugal Force = m x v² / r

Where “v” is the speed in m/sec and “r” the curve radius in m.

As working with this formula is very impractical due to the different speeds and cornering radii, Mammoet has adapted an Impact factor to cover the most likely occurring situations. This factor is different for Conventional Transport (General Cargo) and for Self Propelled transport. High speed transport: Average minimum Curve radius taken at speed : 60 m Average maximum speed through corner : 60 km/h (=16.7m/sec) Centrifugal Force = m x 16.7² / 60 Centrifugal Force = m x 4.62

[N] [N]

Centrifugal Force = 0.462 x m

[kg]*



rounded to

: 50%**

Slow speed transport: SPMT average minimum Curve radius : 10 m SPMT maximum speed : 5 km/h (= 1.389 m/sec) Centrifugal Force = m x 1.389² / 10 Centrifugal Force = m x 0.193

[N] [N]

Centrifugal Force = 0.0193 x m

[kg]*



rounded to

: 2%

* If we want to express force in kg or tons we divide by 10 (F=m*a (a=10m/s²)) ** This 50% is the factor advised by trailer manufacturer Nooteboom. When performing a site move with conventional or hydraulic platform trailers at low speed, use the above factor of 2%, as speed is the governing factor. The additional axle load can now be calculated as: Rturn =

TotalWeigh t × Im pactFactor × HCoGcombin ed Track width × Number of cylinders

Note: Track width = distance between hydraulic pressure points! All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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3

Acceleration / Deceleration forces

Acceleration forces occur in the driving direction of the transport and are caused by the objects inertia. Inertia is the resistance an object has to a change in its state of motion, that is, if an object has a certain speed, its want to maintain that speed, or when it is stationary, it wants to remain stationary. To start or stop moving or to change its velocity will require an external force. Inertia has cost the crash test dummy in the image dearly, the car was forced to change its speed by the wall in front of it and the force to accomplish this has wrecked his car. The crash test dummy’s body also has inertia, and when not restrained, the body would be flying through the windshield. Fortunately he was wearing his seatbelt and the airbag was deployed so the energy of the impact was absorbed. In general, acceleration and brake forces occur in longitudinal direction of the transport. By reducing the abruptness of a change of speed we considerably reduce the forces involved. 5.4

Calculation of the acceleration force

From the first two laws of Newton we know that: Acceleration force Speed difference

F=mxa v = a x t

Acceleration force

F = m x v / t

(Where m = mass and a = acceleration) (Where v = speed and t = time difference in sec.)

Thus: If speed difference is large, so will be the acceleration force. If the time to accomplish this speed difference is small, the acceleration force will be large. As working with this formula is very impractical due to the different speeds and times, Mammoet has adapted an Impact factor to cover the most likely occurring situations. This factor is different for Conventional Transport (General Cargo) and for Self Propelled transport.


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Conventional Transport: To determine the maximum impact factor for conventional transport we can look at the maximum possible grip that can be provided by the vehicle. In general, most vehicles are able to make their wheel spin or slide on the asphalt, either by putting the pedal to the metal, or by braking. This means that t he limit is the friction between the tyre and the road surface. From various references we can find the following table. From this table we find, the maximum coefficient for rubber on any material is 1.2 and this is probably a very special type of rubber as used in racing tyres and can only be achieved at particular temperatures of the tyre. From the specifications of the tyre manufactures, the globally provided coefficient is 0.8 for normal truck tyres.

Sliding friction coefficients Material

Dry

Wet

Greasy

Wood on wood

0.20 - 0.50

0.20 - 0.25

0.15 - 0.05

Metal on wood

0.20 - 0.50

0.20 - 0.25

0.10 - 0.02

Metal on metal

0.10 - 0.25

0.10 - 0.20

0.10 - 0.01

Rubber on concrete

0.50 - 1.20

0.45 - 0.80

-

Rubber on asphalt

0.35 - 1.20

0.25 - 0.80

-

Rubber on gravel

0.40 - 0.85

0.40 - 0.8

-

Rubber on rock

0.55 - 0.75

0.55 - 0.75

-

Rubber on ice

0.07 - 0.25

0.05 - 0.10

-

Rubber on snow

0.10 - 0.55

0.30 - 0.60

-

The maximum friction force (F friction = m x g x µ .) is 0.8 * Weight, or

: 80%

Self Propelled transport: The above mentioned 80% would be much too high for self propelled transport. The speeds of the transports are already lower, and the difference in the reaction time of all hydraulic or pneumatic systems already exceed the actual deceleration time of the transport at those speeds. To calculate a more realistic figure, an emergency stop time of 3 seconds while transporting 100t at a speed of 5km/h (=1.38m/s). Acceleration force = m x v / t Acceleration force = m x 1.38 / 3 Acceleration force = m x 0.462

[N] [N] [N]

Acceleration force = m x 0.0462

[kg]*



rounded to

: 5%

* If we want to express the Acceleration force in kg or tons we divide by 10 (F=m*a (a=10m/s²)) To convert the acceleration/brake forces into an additional axle load we have to divide the longitudinal moment caused by this force, by the distance of the centers of gravity of the representative fields and the number of axles in this field. All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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5.5

Ref.

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3

Wind pressure

When wind is blowing against an object it builds up pressure again the “wind catching” area of that object. Wind is free to blow from any direction, so w e need to look for this wind pressure in each direction. The total wind pressure represents a force that has its attachment point in the wind pressure point. When the resulting wind force is working on the side and/or the front/rear of the object it will create a moment on that object. If the wind is strong enough, the object will eventually tip over, so there must be an effect of this wind force on the mechanical stability of the object. In fact, the wind has no influence on the geometrical shape of the points of support, but because its combined with the weight of the object with the lateral wi nd forces resulting in a total force which is working under a specific angle. The effect caused by the wind is similar to placing the object on a gradient.

Besides affecting the mechanical stability of the object, it also influences the support reactions. At the side that is receiving the wind, the load is reduced while the opposing side will receive higher loadings. Besides building up pressure on the windward side of the object, the sheltered side of the object experiences a vacuum due to the faster moving air. Cross section

Plan view


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The amount of wind pressure depends on the actual wind speed and the shape of the object. If an object has large open spaces, the air building up the pressure can “escape” through these open spaces, r educing the pressure in front of the object. When an object has rounded corners, the air is “guided” over these corners, again reducing the total wind pressure. •

• • •

Rounded shape with nothing sticking out: Rounded shape with something sticking out: Rounded shape with many things sticking out: Block shape:

5.6

80% 100% 120% 120%

Wind pressure calculation.

Wind force

=

wind pressure (P wind ) x cargo wind area x shape factor

Depending on the wind direction you want to calculate, use the front-view area (width x height) or t he side view area (length x height) as the cargo wind area. The wind pressure on the surface of the load determines the resulting wind force. The wind pressure can be 2 approximated by: Pressure = ½ x (density of air) x (wind speed) x (shape factor) [N] The density of air is 3 about 1.25 kg/m . The shape factor (drag coefficient) depends on the shape of the body. The wind speed must be expressed in m/s. See below table: Beaufort scale

Description

Limit [m/s]

Limit [knots]

Pressure [kg/m²]

Description on Land

0

Calm

0.2

1

0

Smoke rises vertically

1

Light winds

1.5

3

0.14

2

Light breeze

3.3

7

0.68

3

Gentle breeze

5.4

12

1.82

4

Moderate breeze

7.9

18

3.9

Raises dust and loose paper; small branches are moved.

5

Fresh breeze

10.7

24

7.1

Small trees in leaf begin to sway; crested waves form on waters

6

Strong breeze

13.8

31

11.8

Large branches in motion; whistling heard in telephone wires

7

Near gale

17.1

38

18.2

Whole trees in motion; inconvenience felt when walking against wind.

8

Gale

20.7

46

26.7

Twigs break off trees; progress generally impeded.

9

Strong gale

24.4

54

37.1

Structural damage occurs -roofing dislodged; larger branches break

10

Storm

28.4

63

50.2

Trees uprooted; considerable structural damage.

11

Violent storm

32.5

72

66.2

12+

Hurricane

over 32.5

over 73

>66.2

Wind felt on face; leaves rustle; ordinary vanes moved by wind.

Very rarely experienced - widespread damage


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The additional vertical loads per hydraulic field are obtained by dividing the appropriate wind moment by the width of the track-width in both longitudinal and transverse direction. The additional axle load is then the division of this force by number of axles in the field.

Rturn =

F wind × H pressurepo int Support distance × Number of cylinders

This results in the following formulas for the axle loads due to longitudinal or transversal wind: Transversal:

RTturn =

Lc arg o × H c arg o × P wind × H pressurepo int × Shapefacto r Track width × Number of cylinders

Longitudinal:

RLturn =

W c arg o × H c arg o × P wind × H pressurepo int × Shapefacto r Support distance × Number of cylinders

NOTE: At Mammoet we always assume a maximum of 6 Beaufort. Above this w ind speed, more detailed engineering is required.


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6

Ref.

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3

Lashing & Blocking As explained in the mechanical stability chapter, the instability of the transport can resort into one of t he following scenarios: A. The load tips over on the trailer deck. B. The load makes the trailer tip over. C. The load skids over the trailer deck, causing the trailer to tip D. The load makes the trailer tip over a little, then slides of the trailer deck.

A

B

C

D

The function of lashing and blocking is to fixate the cargo on the deck of the trailer when under the influence of the transportation loads. Basically it is to prevent scenarios A, C and D. This must mean that lashing is very important, because if lashing is done properly there is only one scenario left, where the combination of trailer and cargo tip over. As is made clear from these scenarios, the cargo needs to be properly secured by using the lashing and blocking in the correct direction.


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3

Transport forces

The transport forces discussed in this section largely depend on the transport speed and applicable accelerations which in turn are determined by the type of transport: • •

Road transport. Site moves.

(High speed transports) (Slow speed transports)

The transport forces we need to examine are the following: • • •

Acceleration forces Centrifugal forces Driving on slopes

(speeding-up and braking) (turning in a corner)

The acceleration and centrifugal forces are d ynamic forces and we have already discussed them in the chapter “Dynamic stability”. The other subject handles driving on slopes. When driving up a longitudinal slope, there is some load transfer from front to back, but as the tra iler is relatively long, there is li ttle chance of instability or overloading. However when driving through a bend with camber, the risk of losing trailer stability as shown in scenario A, B and D is much higher due to the relative narrow width of the trailer. I t is therefore assumed that the transverse stability is governing. 6.2

Acceleration forces

When a transport combination is traveling on the road and its needs to slow down, the cargo has a tendency to keep moving in the direction it was moving. The same is true for acceleration, when the object is stationary it needs to be forced to start moving. This effect is called inertia and both forces are referred to as acceleration forces. The direction of t hese forces is longitudinally and horizontally and can be countered by forward and backward lashing. If the cargo is not secured to the trailer with additional lashing, the friction between the cargo and the trailer deck (in the drawing the transport beams are used) is the only thing keeping the cargo from sliding over the trailer deck.


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The friction force can be calculated by multiplying the friction coefficient between the two materials (i n this case wood on steel) with the Normal force (downward force) on the transport beams. The friction coefficient when using plywood in the table below, it ranges from 0.20 to 0.5, w e choose to use 0.25.

Sliding friction coefficients Material

Dry

Wet

Greasy

Wood on wood

0.20 - 0.50

0.20 - 0.25

0.15 - 0.05

Metal on wood

0.20 - 0.50

0.20 - 0.25

0.10 - 0.02

Metal on metal

0.10 - 0.25

0.10 - 0.20

0.10 - 0.01

The impact factors for acceleration can be found in the chapter “Dynamic stability”: For transports at high speeds this factor is 80% For transports at low speeds this factor is 5%

High speed transports: Facc = Weight Cargo x 0.80 Ffrict = Weight Cargo x 0.25

(plywood)

This means that the friction force alone cannot keep the load in position when the maximum brake force is applied. The required lashing in the forward direction is 0.80-0.25 = 0.55 x weight or: Required lashing high speed with plywood: Required lashing high speed without plywood:

55% x Weight Cargo [t] 80% x Weight Cargo [t]

Slow speed transports: Facc = Weight Cargo x 0.05 Ffrict = Weight Cargo x 0.25

(plywood)

The resulting friction is five times higher than what i s required when using plywood, even when we assume metal on metal with a friction coefficient of 0.1, the friction is two times higher. For slow speed transport NO additional lashing or blocking is required.


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3

Centrifugal forces

When a truck is driving through a corner, the cargo has a tendency to continue its motion in a straight li ne, in stead of following the trailer that i s going through the corner. The force causing this is the centrifugal force, which depends on the weight of the cargo, the distance to the center line and the speed. The direction of this force is laterally and horizontally and can be countered with sideways lashing. Again we can use the curve impact factors we found in the chapter “Dynamic stability”: For transports at high speeds this factor is 50% For transports at low speeds this factor is 2% For high speed transport this results in: Fturn = Weight x 0.50 Ffrict = Weight x 0.25

(plywood)

The amount of side-ways lashing and/or blocking w hen using plywood is = 0.50-0.25 = 0.25 x Weight: Required lashing high speed with plywood = 25% x Weight [t] Required lashing high speed without plywood = 50% x Weight [t] For slow speed transport this results in: Fturn = Weight x 0.02 Ffrict = Weight x 0.25

(plywood)

The resulting friction is more then ten times higher than what is r equired when using plywood, even when we assume metal on metal with a friction coefficient of 0.1, the friction is 2.5 x higher. For slow speed transport NO additional lashing or blocking is required.


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6.4

Ref.

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3

Trailer Camber

Driving on slopes will cause the load to want to slide off. In general a transverse slope results in a bigger angle then a longitudinal slope. It is therefore assumed that the transverse gradient will be the governing factor. In theory, the lashing required to be able to use the combined stability angle would need to be able to lift one side of the trailer, regardless of the weight of the load. This would mean that the trail er could be on its side and the load would still be fixed on the trailer. As this is not the most practical way to use a t railer, we use the maximum expected camber to calculate the horizontal and vertical lashing. We limit the maximum expected camber to 22 degrees which is the maximum stroke on one side of a single wide SPMT trailer. As this is the trailer type with the smallest track-width and trailer deck width, this is also the worst case and a reasonably conservative method to determine horizontal and vertical lashing f or driving on a road with camber.

In this situation, the lashing force required to lift t he left side of the trailer can be calculated as: Fv Lashing = (670 x W Trailer) / 1793 = W Trailer x 37.4% Note that the vertical lashing is required on two sides of the tr ailer. The force that makes the load slide of the trailer deck can be resolved by the horizontal lashings: Fh Lashing = W Cargo x sinus (22°) = W Cargo x 38.2% Please note that the horizontal blocking or lashing is required in both directions. All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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It might be clear that if the trailer is actually wider, or the stroke in an axle is less then the previously mentioned 600mm, the lashing requirements for driving on a slope will be reduced. The table below indicates some of the forces that can be expected for that type of trailer (arrangement).

Width Trailerdeck

Width of support [mm]

Max stroke [mm]

Max camber [º]

Vertical lashing

Horizontal lashing

Single SPMT Double SPMT

2430 5330

1450 2900

600 600

22.5 7.9

37.4% 35.2%

38.2% 13.7%

Triple SPMT

8230

4350

600

4.7

34.6%

8.2%

Single Cometto Double Cometto

3000 6300

1800 3600

600 600

18.4 9.5

37.5% 36.4%

31.6% 16.4%

Nooteboom 8axle MCO-00-MCA-28 Nooteboom 8axle MCO-121-08V-25-96 Nooteboom 4axle MCO-581-04V

2500 2720 2470

2000 2200 1967

200 200 200

5.7 5.2 5.8

44.4% 44.7% 44.3%

10.0% 9.1% 10.1%

Goldhofer 5 axle SKP-H5

2750

2250

200

5.1

45.0%

8.9%

Goldhofer 5 axle SKP-H5

2750

1850

450

13.7

40.2%

23.6%

Trailer type

As is shown from the table, the maximum possible camber is the main denominator for the lashing requirements. Please note that in most cases, plywood would suffice to make up the horizontal lashing as this results in 25% friction! Please note that when we enlarge the spacing between the trailer trains, the required amount of vertical and horizontal lashing will reduce, maximum possible camber is smaller.


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Lashing under angles

In the previous paragraphs we discussed the lashing in each direction of the trailer:

• •

•

Longitudinal (Horizontal) Transversal (Horizontal) Vertical

In reality it is not always possible or practical to attach the lashing in the exact direction where it is required so many times we have to compromise. When we place lashing under an angle, we need to ensure that the effective capacity of the lashing in the required direction is still sufficient to withstand the transport loads. To determine the resulting components from a given capacity, we can use the following figures. - cosine (45) = 0.707 - cosine (30) = 0.866 - cosine (60) = 0.5

~ 70% ~ 90% = 50%

To determine the required capacity from the components, use the following figures: - 1 / 0.707 = 1.414 - 1 / 0.866 = 1.154 - 1 / 0.5 = 2

~ 140% ~ 115% = 200%


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Example

In the example picture, the box is fixed to a SPMT using 2 (front and rear) crossways lashings. The angles of these lashings are approximate 45 degrees to the horizontal. Trailer weight: 32t + 6t (PPU) = 38t. Cargo weight: = 90t. Single trailer SPMT has maximum lope of 22.5º and therefore requires 38.2% horizontal lashing. Additionally it requires an impact factor for turning of 2% in the transverse direction, and a 5% impact factor for acceleration/deceleration in the longitudinal direction. Transversal Longitudinal

= (38.2 % + 2 %) x 90t = 5 % x 90t

= 36.8t = 4.5t

Use plywood so we can reduce the requirements in the horizontal directions by 25% Transversal Longitudinal

= (38.2 % + 2 % – 25 %) x 90 = (5 % – 25 %) x 90

= 13.7t = 0.0t

With the lashing angles of 45º makes the required lashing (140%): For transversal = 13.7 x 140% = 19.2t For longitudinal = 0 x 140% = 0.0t

(can’t be less then 0)

(left AND right)

Required Vertical Lashing: The image on the right shows that for this particular case, the stability of the cargo is bigger then the angle of the trailer deck when the trailer fails. It also shows that the combination already tips over at a smaller angle. This means the box is stable by itself and makes the required vertical lashing 0t. However, the transversal component, is governing and we need two ratchets with a capacity of 10t each to cover one direction. So in total we need 4 chains of 10t to prevent the load from sliding over the deck side-ways.


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Another example:

A box of 180t is placed on a double SPMT. The angles of these lashings are approximate 45 degrees to the horizontal. Trailer weighs 2 x (4 x 8 + 6) = 76t.

Transversal Longitudinal

= 13.7 % (slope) + 2 % (corner) x 180 = 5 % (acceleration) x 180

= 28.3[t] = 9.0[t]

Use plywood so we can reduce the requirements in the horizontal directions by 25% Transversal = 13.7 % (slope) + 2 % (corner) – 25 % (friction) x 180 = 0.0[t] (can’t be less then 0) Longitudinal = 5 % (acceleration) – 25 % (friction) x 0 = 0.0[t] (can’t be less then 0) Required lashing: For transversal and longitudinal

=0

x 140%

= 0.0t

Vertical lashing: In this case, both the combined stability and cargo stability are bigger then the maximum possible angle. Therefore, no vertical lashing is required.


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Lashing of cargo on saddles

When a column is loaded on the trailer using saddles, they may require lashing when the saddles are relatively high or when they are unfit to transfer the loads created by accelerations of the transport. The unsecured saddle might to flip over in between the trailer and column during the accelerations. The forces induced by acceleration are transferred through the saddles, to the trailer, causing a moment that needs to be resolved by the weight of the column in the saddles. Occasionally the column manufacturer welds clips on the saddle or lashes the saddle to the column using binding straps over the column. This reduces the risk of flipping the saddles considerably. When a column is b eing transported on a ship without any additional horizontal bracing in the ships hold, it is safe to assume that the saddle arrangement can cope with the acceleration forces of the trailer. When in doubt, it is recommended to reduce the moment over the saddles by adding to add 50% horizontal lashing for when traveling a high speeds or 5% when traveling at lower speeds. Below is an example of a possible solution to connect the lashing. When using this type of arrangement, one of the lashings will take the acceleration forces while the other will resolve the braking forces. To use this type of arrangement, the ratio between saddle height and the distance needs to be below 1:10.


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When the height of the saddles is insufficient and the saddle arrangement is still potentially unstable, you can invert the lashing as shown in the diagram below. A potential problem with thi s type of lashing is to find the appropriate lashing points on the column.

6.8

Lashing of cargo on turn tables

When the cargo is transported on saddles, these same rules apply to using turntables, where this effect is increased by the fact that the pivot point of the table cannot transfer the moment induced by the acceleration forces. When using turntables you also need to consider the pull-forces of the rear dolly, especially if the dolly has no propulsion system. In this situation, the cargo drags the rear dolly along and these dragging forces are transferred through the saddles. To avoid these forces going through the saddles, w e can use tirfor wires between the decks of the two turntables. The rolling resistance of normal truck tires is 2% of t he axle/wheel load or: the total rolling resistance of the transport arrangement is 2% of the total weight. In case of using turntables, the required capacity for the tirfor between the dolly’s is equal to the rolling resistance of the rear dolly, which is 2% of the sum of the portion of the weight carried by the rear dolly plus the w eight of the dolly itself.

Weight of Reactor Weight of Trailer Weight of Turntable Weight of Load-spreading Total Load on Dolly

= 200t with a center of gravity at equal distances to the saddles = 42t = 4.5t = 5t = 100 + 42 + 4.5 + 5 = 151.5t

Rolling resistance

= 151.5 x 2% = 3.03t


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When the transport needs to negotiate slopes lik e ramps and wedges, there is an added effect that the rear dolly is being pushed down the slope by the weight of the reactor. We need to add the effect of the ramp to the rolling resistance of the dolly.

The effect of the slope is calculated as follows: Slope (example) = 1:20 = 2.86° Total Load on Dolly = 100 + 42 + 4.5 + 5 = 151.5t Horizontal component on slope

= sin (2.86) * 151.6t = 7.57t additional

For acceleration and braking, we assume the impact factor which we have determined f or slow moving transports in the dynamics chapter, so that would be 5%. Total Load on Dolly Horizontal component

= 100 + 42 + 4.5 + 5 = 151.5t = 151.5t * 5% = 7.57t

CAUTION WHEN IT IS REQUIRED THAT THE TRACTOR PUSHES THE TRANSPORT INSTEAD OF PULLING, THE USE OF A TIRFOR WIRE IS FUTILE AND YOU NEED TO RESOLVE THE HORIZONTAL FORCES USING

CAUTION

LASHING BETWEEN THE COLUMN AND THE TURN-TABLE DECK.


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Single file SPMT’s

Special attention should be given for movements with a Single file SPMT (s ingle wide trailer, single train) which, due to the small track width, have a low transverse stability on tipping. The most practiced method to increase the sideways stability of a single file trailer is to use a four point suspension. Avoid driving single file trailer sideways! The acceleration and deceleration forces in the transverse direction combined with t he lower stability angle of a single train SPMT can cause the trailer to tip over. Use diagonal steering instead. To further worsen the above effect, the tipping width is reduced as the wheels will start acting as pivot points. Depending on the height of the trailer, the distance between the wheel centers provide a variable width of the tipping plane. The same goes for single file trailers positioned under large constructions. If the trailer i s insufficiently secured to the construction, the trailer has a tendency to move from underneath the cargo, especially when the train is in a single field. In that case the oil can flow freely from the right to the left cylinder. The resulting stability of the support to the construction will be less t hen when there is hydraulic separation. Use bracing between supports, or tie the load to the trailer, which basically restricts t he support from tipping.

The image on the far right shows a Kamag trailer. As the “knees” of this trailer point in one direction, this will create an eccentricity moment when the train is in a single field ( note that the axle pressures are equal due). This moment needs to be resolved by the support and the construction to be equilibrium, so the support needs to be able to do that. All rights reserved. No part of this publication may be reproduced, stored in a database or retrieval system, or published, in any form or by any means, electronically, mechanically by print, photo print, or re cording or otherwise without prior wr itten permission from Mammoet Holding B.V.. Office

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Summary

From this chapter we have learned that: • • •

•

The purpose of lashing and blocking is to secure the cargo on the trailer. Lashing should be provided in the required direction. When using lashing under an angle, it resolves multiple directions, but it decreases the horizontal and vertical components. Special care should be taken for single file SPMT’s


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Conclusion In practice the stability on strength and stability on tipping will always need to be calculated to see which of the stabilities is critical.

The smallest value is the governing value and should never be exceeded. Stability angle Limits: • Minimum stability on tipping angle: • Minimum stability on strength angle:

7° 2°

Dynamic impact factors: Wind forces need to be calculated by area and shape • Rolling resistance of tires • Road transport: • General acceleration/brake impact factor • General cornering/turn impact factor Site moves: General acceleration/brake impact factor • General cornering/turn impact factor • Cargo shape factors: • Tubular shape with almost no protrusions: • Tubular shape with almost no to some protrusions: • Tubular shape with a lot of protrusions: Block shape: •

2% W cargo. 80% W cargo. 50% W cargo. 5% W cargo. 2% W cargo.

0.8 1.0 1.2 1.2

At Mammoet engineering these forms of stability are calculated with the use of a calculation sheet in Excel and the acTrailer program. After entering the required data into the program or the calculation sheet it automatically calculates the axle pressures and the allow able stability angles and reports back if the situation is allowable or not.


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Calculation examples In the following example we look at transporting a vertical reactor of 70 ton on a single Scheuerle SPMT with the following details: Weight of reactor Trailer Weight trailer Trailer CoG Axle pivots

: 80t : Single train SPMT : 2 x 16t (4liner) + 6.7 (power pack) : 1500mm above road surface (When fully extended) : 285mm above road surface.

As this object already looks unstable, and we are using a single wide trailer, thi s example will look at all the possibilities to improve the stability of the transport.


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Three-point suspension

Stability on tipping

In this example we use 8 * 2 = 16 cylinders  group1 = 4 cylinders, group 2 & 3 = 6 cylinders The left image shows the stability triangle in more detail. From this we deduct the stability arm (a): 2

725 2 )

C

= (5600

a

= (725 * 4200) / 5646

+

= 5647mm = 539mm

When we assume that the cargo is not firmly connected to the transporter. This makes the height above tipping plane: 9000 – 285 = 8715mm.

Tan α = 539 / 8715 = 0.0618 = 6.18% = 3.53°< 7°, NOT OK!

α


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When we use chains to lash the cargo firmly to the trailer, w e need to ensure that the vertical component of the used lashing is sufficient to lift the trailer. In our case this is 38.7 Ton x 37.4% per side. As the angle of the lashings is at approximately 60 degrees we know we need at least: 115% x 37.4% x 38.7t / 2 = 8.3t (So use 10t ratchets).

If we assume the weight trailer to be (2 x 16t (4liner) + 6.7 (power pack)) = 38.7t and the height of the CoG on 1500mm above the road level, the combined CoG of Cargo + Trailer will be: X = (80 * 9000 + 38.7 * 1500) / (80 + 38.7) = 6555 mm This makes the height above tipping plane 6555 – 285 = 6270mm Tan α = 539 / 6270 = 0.086 = 8.6% = 4.9°< 7°, NOT OK!

α


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To reduce the combined CoG even more, we place 50Tons of additional counterweights on the trailer deck. The required amount of vertical lashing is for this situation is (38.7 + 50) x 37.4% per side. As the angle of the lashings is at approximately 60 degrees we know we need at least: 115% x 37.4% x (38.7t + 50) / 2 = 19.0t each side (So use 4 x 10t ratchets).

Combined CoG of Cargo + Counterweight + Trailer: X = (80 * 9000 + 38.7 * 1500 + 50 * 2100) / (80 + 38.7 + 50) = 5234mm This makes the height above tipping plane 5234 – 285 = 4949mm Tan α = 539 / 4949 = 0.109 = 10.9% = 6.2°< 7°, NOT OK!

α


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Four-point suspension

As the trailer configuration in the previous paragraph is still not sufficiently resolved, w e change its hydraulic setup to a 4-point suspension. As per the diagram below (4 x 4 x 4 x 4).

8.2.1

Stability on tipping

The previous calculated heights of the COG will be the same for the 4-point suspension so:

Cargo not fixed to trailer (8715mm):

tan α = 725 / 8715 = 0,083 = 8.3% α = 4.76°< 7°, NOT OK!.

Cargo fixed to trailer (6270mm):

tan α = 725 / 6270 = 0.116 = 11.6% α = 6.60 °< 7°, NOT OK!

Cargo fixed + 50t CWT (4949mm):

tan α = 725 / 4949 = 0.146 = 14.6% α = 8.30°> 7°, OK!!!!


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Stability on strength

Now that we know that the combination can be transported safely without the risk of tipping over, we need to determine the stability on strength: The total weight of the combination = 80 + 38.7 + 50 = 168.7t Maximum allowable axle load 17t

17 x 8 x 1450 = X x 168.7

 x

= 1169mm

Maximum allowed CoG movement: Maximum allowed α

1169 – (1450 / 2) = 444mm tan α = 444 / 4949 α = 5.1°> 2°, OK!!!! As the stability angle on strength is the smallest value (5.3°< 8.3°), this will be the governing stability angle for the transport.

STABILITY LIMIT 5.1°


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Dynamic axle loads

Axle loads in static condition (no wind or accelerations): (Please note that the eccentricity of the PPU has not been accounted for in this example) Axles in Field Axles in Field Axles in Field Axles in Field

A B C D

= 168.7 / 16 = 168.7 / 16 = 168.7 / 16 = 168.7 / 16

= 10.5t = 10.5t = 10.5t = 10.5t

Additional axle loads due to acceleration / braking Acceleration impact factor: 5% H acc.

= 168.7 x 5% = 8.435t

V acc. V acc. V acc.

= (8.435 x 4949) / (2 x 2800) = 7.45t (per 8 axles) = 0.93t/axle

Additional axle loads due to cornering Cornering/Turn factor = 2% of the combined transport weight H turn H turn V turn V turn V turn

= 0.02 x 168.7 = 3.374t = (3.374 x 4949) / (1450) = 11.52t (per 8 axles) = 1.44t/axle


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Additional axle loads due to transfers wind 2

Allowable wind force for transport = 8 Beaufort = 25 kg/m = 0,025 t/m

2

2

Shape area of Reactor Shape area of CWT (l=1.65, h=1.8)

= 42.5 m , shape factor = 0.8 2 = 3.0 m , shape factor = 1.2

F wind op Reactor F wind op CWT

= 0.8 * 42.5 * 0.025 = 1.2 * 3.0 * 0.025

= 0.85t = 0.09t

Z Reactor Z CWT

= ½ * 8500 + (1500 - 300) = ½ * 1800 + (1500 - 300)

= 5450 mm. = 2100 mm.

V t-wind V t-wind V t-wind

= (0.85 * 5450 + 0.09 * 2100) / (1450) = 3.325t (per 8 axles) = 0.42t/axle

Additional axle loads due to longitudinal wind 2

Allowable wind force for transport = 8 Beaufort = 25 kg/m = 0,025 t/m

2

2

Shape area of Reactor Shape area of CWT (l=2.4, h=1.8)

= 42.5 m , shape factor = 0.8 2 = 4.32m , shape factor = 1.2

F wind op Reactor F wind op CWT

= 0.8 * 42.5 * 0.025 = 1.2 * 4.32 * 0.025

= 0.85t = 0.13t

Z Reactor Z CWT

= ½ * 8500 + (1500 - 300) = ½ * 1800 + (1500 - 300)

= 5450 mm. = 2100 mm.

V l-wind V l-wind V l-wind

= (0.85 * 5450 + 0.13 * 2100) / (2 * 2800) = 0.88t (per 8 axles) = 0.11t/axle


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TP Mammoet Trailers Stability&Lashing - Rev03-2

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