HSE Health & Safety

Executive

Ship collision and capacity of brace members of fixed steel offshore platforms

Prepared by Visser Consultancy for the Health and Safety Executive 2004

RESEARCH REPORT 220

HSE Health & Safety

Executive

Ship collision and capacity of brace members of fixed steel offshore platforms

Dr W (Pim) Visser VISSER CONSULTANCY

Zomervlinderberm 66 3994 WR Houten The Netherlands

Supply boat collision is one of the most critical accidents that can affect structural safety and integrity of fixed installations Pim Visser, an experienced engineer has disentangled much of the research complexity and provided a practical andaccessible resource for assessing impact resistance.

The capacity of members damaged due to vessel impact has been researched extensively because of the significant safety implications. However different approaches have been taken and, whilst many of the findings are in the public domain, it has not been an easy task for designers or assessment engineers to locate the information or apply it consistently.

The study was therefore commissioned by HSE to collate and assess the available research and methodologies, focusing particularly on bow or stern impacts on splash zone braces. Practical demonstrations draw on insight from safety cases, with the recommended approach being delivered in the form of a spreadsheet with worked examples for wider application.

Having compared the implications of four different norms for brace assessment, viz:

• Maximum joint rotation

• Ductility ratio

• Maximum brace deflection

• Maximum weld strain and strain hardening

results from sample calculations were in sufficiently close agreement for an averaged norm to be adopted in the final recommendation. Denting as well as member deformations are considered. Concerns for risers within the jacket can best be checked against a maximum deflection criterion of one metre.

Material and tubular joint considerations relevant to brace impact assessments are also addressed. It was concluded, for example, that:

• CTOD tests are performed at far too slow a rate to represent vessel impact conditions and use of Charpy test results isto be preferred.

• Although prior to brace failure tension may be significant, its contribution in impact energy absorption can generally bedisregarded simplifying the calculations significantly.

• Based on dynamics considerations, brace impact may be assessed as a purely local problem.

• Even if a node fails before the member, further failure at the joint in shear may not follow and subsequent bracebehaviour can be calculated.

Engineers designing for or assessing the capacity of vessel impact of jacket structures will find this report to be invaluable.

This report and the work it describes were funded by the Health and Safety Executive (HSE). Its contents, including any opinions and/or conclusions expressed, are those of the authors alone and do not necessarily reflect HSE policy.

HSE BOOKS

© Crown copyright 2004

First published 2004

ISBN 0 7176 2838 8

All rights reserved. No part of this publication may bereproduced, stored in a retrieval system, or transmitted inany form or by any means (electronic, mechanical,photocopying, recording or otherwise) without the priorwritten permission of the copyright owner.

Applications for reproduction should be made in writing to: Licensing Division, Her Majesty's Stationery Office, St Clements House, 2-16 Colegate, Norwich NR3 1BQ or by e-mail to hmsolicensing@cabinet-office.x.gsi.gov.uk

ii

CONTENTS

1 INTRODUCTION 1

2 SUMMARY 3

2.1 Major impact and platform findings 3

2.2 Norms for brace assessment under impact 4

2.3 On methodologies 4

2.4 Other findings 4

3 VESSEL AND SERIOUS INCIDENTS 7

3.1 Ship impact energy absorption capacity 7

3.2 Serious incidents and ship impact zone 7

3.3 Guidance on impact energy 8

3.4 Other considerations 9

4 PLATFORM GEOMETRY 11

4.1 Platform aspects 11

4.2 Other platform features 13

4.3 Risers and riser deflection effects 15

4.4 Risers, riser caissons and riser protection structures 16

4.5 Conductors 16

5 IMPACT DETAILS 17

5.1 Denting 17

5.2 Miscellaneous aspects of denting 19

6 BRACES - BEAM BENDING ANALYSES 21

6.1 Introduction 21

6.2 Properties of thin walled tubulars 21

6.3 Principle of Virtual Work 21

6.4 An elastic beam in three point bending 23

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6.5 Elasto-plastic three-point bending problem 24

6.6 Three-point bending with tension - elastic 25

6.7 Three-point bending with tension - tubular, plastic to API 26

6.8 Three-point bending with tension - joints, plastic to HSE guidance notes 29

6.9 Cross-bracing capacity 31

6.10 Simplified energy expression 32

7 MATERIAL AND JOINTS 35

7.1 Introduction 35

7.2 Material and fracture toughness 35

7.3 Geometry and local modelling 35

7.4 Strain rate effect 36

7.5 CTOD tests 36

7.6 Charpy test 37

7.7 Maximum strain criterion 37

7.8 Strain hardening 37

7.9 Joints 38

7.10 Impact close to a node 40

8 SAMPLE PROBLEMS 43

8.1 Description of the test cases 43

8.2 Vessel impact analysis 43

8.3 Variations in the sample problems 47

9 REFERENCES 51

APPENDIX A BEAM BENDING ANALYSIS

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I

ABBREVIATIONS

API American Petroleum Institute LAT low astronomical tide CTOD crack tip opening displacement LRFD load and resistance factor design DP DSV HSE

E

A DE Ed

u

&K

MA, ME

MC

MMMMMe

el

max

p

u

WTT

PPPPPP

1

2

d

e

u

Q R S T

p

u

E

WI

dynamic positioning MSL mean sea level diving support vessel SCF stress concentration factor Health and Safety Executive

LIST OF SYMBOLS

tube cross sectional area tube diameter

Young’s modulus dent energy

brace energy associated with plastic deflection (u) tube moment of inertia rate of stress intensity factor (for CTOD assessment)

plastic moment at the tube end points A and E plastic moment at the centre of the brace maximum elastic moment maximum elastic moment

maximum moment tube full plastic moment joint ultimate out-of-plane moment capacity lateral force due to impact lateral force before joint failure lateral force after joint failure force to cause a dent dd

maximum elastic lateral force ultimate lateral force force equal to P/2 shear resistance of tubular elastic section modulus tensile force plastic tensile force of tubular ultimate joint tensile force external work (in Principle of Virtual Work) internal work (in Principle of Virtual Work)

ZZ plastic section modulus of tubular d plastic section modulus of dented

tubular

wwwu

m

a quarter brace length (l /4) b a characteristic length l brace length

p plastic moment of the tube wall t wall thickness timpact impact duration u central plastic deflection

mean mean value of u centre central elastic deflection B elastic deflection at point B max maximum elastic deflection

dd dent depth ε& strain rate j angle over cross-section gA, gE ratio between joint strength and tube

strength at points A and E (gA, gE < 1.00)

l characteristic eigen value for beam

q

q

q

q

q

q

in tension / bending 1 end rotation in point A 2 end rotation in point E ave average end rotation crit. critical vale of end rotation max max. value of end rotation max maximum value of end rotation to

q

Marshall min minimum value of end rotation to

Marshall su ultimate stress sy yield stress

v

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1 INTRODUCTION

In the former HSE Guidance Notes1 and the Safety Case Regulations2 a number of accidental scenarios for fixed steel offshore structures have been identified. One of the potentially more critical accidental events is supply boat collision3. Common designs, when compliance with the HSE Guidance Notes is sought, address the supply boat collision through a combination on described forces, described impact energy4 together with elastic and plastic deformation of tubulars, the platform as a whole and the supply boat.

Figure 1.1 A typical jacket-vessel impact scenario

In the splash zone the most likely impact scenarios are:

· broadside impact of one of the legs of the platform · bow or stern impact of one of the braces in the splash zone.

This report deals with aspects related to the bow or stern vessel collision with one of the braces in the splash zone. In the industry there is a difference of opinion on how well these splash zone braces are able to resist vessel impact. Some engineers assume that under vessel collision the brace member will act as a string with a substantial impact resistance. But others are much more conservative and assume that the energy absorption capacity of the brace is limited because of limited rotational capacity at the joints. Hence the member could be severed by the impact, but the vessel would not be stopped by the braces. In the latter case consequential damage to risers and J-tubes near the outer faces of the jacket could occur as well. In any event the remaining structure is checked to ensure that brace forces during impact will not impair the integrity of the installation as a whole.

In order to obtain clarity on the behaviour of splash zone braces under vessel collision this report deals with the review on the forces and resistances of tubular braces and their end­connections under vessel collision. More specifically it concentrates on the following issues:

· check the loading considered for the assessment; · address the resistance side of the brace members,

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· determine the important resistance parameters (material, geometry, modelling, fracture mechanics properties).

In order to assess the impact resistance of splash zone braces in fixed steel platforms it is common practice to consider the following energy absorption mechanisms:

· local denting · elastic beam bending · plastic bending/hinge formation and rotation · plastic tensile strain · global deflection of the installation · local deformation of the vessel

Secondary impact resulting from ship ingress within the structure is, in general, excluded from the assessment. An exception should be made when these consequences are unacceptable, for example, if it could result in (gas) riser severance.

As a start for this report a review has been made of a joint industry project report on this subject5. Also a number of Safety Cases were reviewed with specific emphasis on impact to and resistance of splash zone braces in fixed steel platforms.

In the report the various items related to this subject matter are reviewed and some practical consequences, using sample problems, are highlighted. Specific mention should be made of the exercises in Appendix A on brace resistance. using both elastic and plastic methods, under bending and tension. Of special interest is the criterion for which tension becomes significant and it is found that, for many practical situations, the ignoring of tension is acceptable from an energy absorption point of view.

Besides the major conclusions in Section 2 the report is divided into a number of sections. These address vessel and incident aspects (Section 3), platform and riser information (Section 4), details on brace indentation (Section 5), a summary of Appendix A on brace resistance and of cross bracings (Section 6), observations on platform steels and joint strength (Section 7) and a final section on some sample problems (Section 8).

A major point is the identification of suitable norms for brace failure; these are of relevance both for hand calculations as well as for computer simulations. A summary of these norms is given in Section 2.2.

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2 SUMMARY

2.1 MAJOR IMPACT AND PLATFORM FINDINGS

The following results give a reflection of some findings on major impacts:

· on impact zone: ship collision will be at or close to the splash zone impact zone; for the study central impact was taken as the starting point;

· on impact energy: in the Guidance Notes 4.0 MJ is noted as a minimum impact resistance for offshore structures; this 4.0 MJ impact energy may well be a suitable norm for single components such as critical splash zone braces;

· on brace integrity: the norm for brace integrity will be based on ultimate brace deflection;

· on denting: denting leads to a reduction in member strength and should be taken into account;

· on large deformations: once excessive plastic deformation takes place both bending and tension strain concentrates in the hinges.

The major platform related findings are:

· on the platform design and construction date: difference should be made between first, second and third generation platforms;

· on stiffened nodes: for the collision resistance of splash zone braces it is essential to assess the node strength; it is noted that in the common platform design references little guidance is given on the ultimate capacity of stiffened nodes;

· on premature failing of a joint: if a node is weaker than the brace it may well fail before the full energy absorption capacity of the brace has been developed; however, failure in bending will not necessary imply failure in shear and this can be addressed during the assessment;

· on riser impairment: if a riser is connected to one of the outside braces its impairment during vessel collision will be site specific but may well be governed by Norm 3 (see below);

The effect of operator restrictions, e.g. leeward side offloading and operator requirements in the 500m zone, to reduce the probability of ship collision and thereby on the design impact energy, falls outside the scope of this study and will not be addressed in this report.

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2.2 NORMS FOR BRACE ASSESSMENT UNDER IMPACT

In the course of the development of this report the following norms have been identified for brace assessment:

· Norm 1: maximum joint rotation (Marshall criterion); · Norm 2: ductility ratio (elastic multiplication factor); · Norm 3: maximum brace deflection of 1.0 m; · Norm 4: maximum strain in weld combined with strain hardening.

For the sample problems given in Section 8 it appears that the minimum and maximum deflections found for these four norms are no more than a factor 2.0 apart. Therefore the average is taken as the final norm to calculate the energy absorption capacity.

2.3 ON METHODOLOGIES

As part of the review some fundamental aspects of tubular beams under three point bending, combined with tension both in the elastic and plastic regime have been studied. This led to some useful understanding of the plastic hinge mechanism, which can also be applied to tubular joints at the failure boundary.

It has been demonstrated that, prior to brace failure, tension could be significant but its contribution in impact energy absorption can, within the accuracy of the methods employed, be disregarded.

After joint failure in the chord saddle it is unlikely that the joints would fail under shear as well. This provides a suitable method to calculate brace behaviour after joint(s) failure.

Because of the many aspects involved it is doubtful if a few numerical tests can assist in developing sufficient understanding to nodes which, based on design equations, are of insufficient strength.

Similarly, full scale or model scale testing leads to little additional understanding because of the many local effects involved; it can only confirm effects and consequences of previously known behaviour. Therefore the benefits of physical testing are limited.

2.4 OTHER FINDINGS

Other significant observations made in the course of this work, some of which are addressed in some detail in this report:

· for stiffened joints the only simple assessment method is given in API-LRFD7;

· the use of brace stubs of thicknesses different from the brace thickness leads to additional complications;

· second and third generation platforms have better material properties than first generation platforms;

· the maximum strain criterion of 15% in the parent material and 10% in the weld is a useful norm for numerical analysis;

4

· based on dynamics considerations it is concluded that brace impact is a purely local problem;

· in some cases, particularly in older installations, risers may be connected in the splash zone to a primary brace and may therefore be subject to the consequences of vessel collision.

· the strain rate effect contributes to the brace strength;

· impact out of brace centre reduces the energy absorption capacity of the brace; this effect and the strain rate effect will compensate each other to some extent;

· CTOD (crack tip opening displacement) tests are performed at a rate in stress intensity which, for vessel impact, are too low by 2 orders of magnitude;

· strain hardening provides a useful contribution to smooth potential locations of high strain in numerical modelling.

Fenders or guard structures are used whenever risers are connected to the outside of the jacket structure; its design, however, falls outside the scope of this report.

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6

3 VESSEL AND SERIOUS INCIDENTS

3.1 SHIP IMPACT ENERGY ABSORPTION CAPACITY

Because of the size of the offshore supply vessels it is expected that broadside impact would be taken by the legs and not by a splash zone brace. For the bow and stern impact it is common practice to use the force indentation curves provided in the DnV Technical Note8 (see Figure 3.1). Some characteristics of the bow indentation curve are:

· the minimum force to show some bow indentation is 3.0 MN; · 5 MJ ship impact energy is taken at an impact force of 8.0 MN.

For stern impact the vessel can be considered infinitely stiff up to a force of 7.0 MN.

10

0

stern impact

bow impact

0.0 1.0 indentation (m)

Figure 3.1 Supply boat vessel indentation curve (from Ref. 8) (only part of the curve is shown)

The recommendation in DnV Note TN2028 is based on analytical methods together with model tests covering a range of typical vessel construction methods. In common with the UK approach as reflected in the Guidance Notes only vessels authorised to approach installations are considered.

For our review it is of relevance that forces due to bow impacts are much more gradually increasing with regards to vessel damage than for stern impacts.

With regards to the vessels and their deformation behaviour it is noted that vessel displacements have been increased since 1985 to 6000t displacement; for example Central North Sea and Southern North Sea structures have been subjected to a number of collisions involving vessels up to 10000t. Furthermore the DnV data are for 1.5 diameter (probably vertical) tubulars which is a suitable starting point for splash zone braces.

3.2 SERIOUS INCIDENTS AND SHIP IMPACT ZONE

The review of serious incidents in Ref. 5 is worthwhile. A table contains information on the nine worst collision incidents up to 1992 from Lloyd's Register and the Department of Energy Accident files; for example:

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· the damage sustained was in the range -5.5m to + 8.2m LAT;

In the assessment of historic impact events a simplified approach to back-calculate the total impact energy after a serious damage event is applied. The details of this approach are, most likely, based on the permanent deformation of the installation and the vessel velocity before and after the impact event.

A review of early impact incidents indicated that member severance can occur at low energy absorption and for such severance plastic deformation is limited. This is probably related to a first generation platform.

For ship impact it is necessary to consider the ship impact zone. Design considerations have reached an impact zone from -10 to +12m with respect to LAT. From the review of serious ship impact events it can be concluded that the ship impact zone can be taken from -5.5m to +8.2m with respect to the waterline with most impacts at the water level or above. For the current assessment the impact has been taken at:

· vessel impact at approximately LAT or at mid span.

Corrections for other points of impact are also identified in Sections 6.3.2.

Particularly an impact at a high level, close to a brace node, can be a concern, because in that case the validity of the design equations should be questioned. This scenario is addressed in some detail in Section 7.10.

3.3 GUIDANCE ON IMPACT ENERGY

There is quite a difference in approach in the various codes and guidances on vessel impact energy; for example, API-LRFD7 mentions the subject but there is no quantitative guidance.

The Norwegian codes on the other hand specify accidental impact by a 5000t vessel at 2.0m/s and require inclusion of added mass in longitudinal and broadside direction for bow, stern and side impact leading to an impact energy of 11 or 14 MJ. This impact energy can be distributed between the vessel and the structure; it is to be taken when the total platform integrity under vessel impact is to be ensured. In addition, an operational impact assumes a vessel impact energy of 0.5 MJ. The 2.0m/s impact velocity can be arrived at by taking a 4.0m wave (the maximum operating wave height for supply boat offloading) combined with a 6.0s wave period.

The Guidance Notes1 offer the same criteria but, in addition, the following clause can be found:

· for a fixed steel jacket structure it is recommended that the energy absorbed by the Installation should not be taken to be less than 4.0 MJ.

In addition it is stated that a lower value may be appropriate if supported by a study of the collision hazards and consequences. This value of 4.0 MJ is a suitable starting point when the integrity of a single component is assessed during vessel collision.

From the ship collision data base in Ref. 5 it follows that the main causes for ship impact are misjudgement and equipment failure. Therefore the chance of ship impact is

· related to the number of vessel visits;· related to the duration of vessels in the vicinity of the installation.

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The assessment of the consequential reduction in vessel impact energy falls outside the scope of this report.

3.4 OTHER CONSIDERATIONS

In the course of the review of information on accidental vessel impact some other, miscellaneous observations were found. In summary:

· the comment is made that the post-accidental impact strength of the installation should be equal to the loading corresponding to the environmental loads with a return period of at least one year; recommendations of this nature are particularly useful for design;

· the effect of dynamics is addressed in Section 4.2.1;

· proximity of risers/conductors (and J-tubes) and possible large gaps between the braces.

This report is related to brace impact and hence lower values of the impact force are to be expected than for leg impact.

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10

4 PLATFORM GEOMETRY

4.1 PLATFORM ASPECTS

In the North Sea a great variety of steel structures have been and are installed. Its impossible to identify all structural items of interest. Therefore only some of the features will be indicated in this report. For example, the following key-words provide some indication on the structures:

· not-normally manned installations · single column structures · various water depths ranging from 10m LAT up to 190m LAT · light-weight, lift-installed and heavy weight, launch or self-floating structures · first, second and third generation platforms.

From a structural assessment point of view the following features are of specific interest and will be addressed in some detail. A typical Central North Sea platform, showing some of the features of relevance for this assessment, has been sketched in Figure 4.1.

MSL

mudline

Figure 4.1 A sample of a North Sea substructure

4.1.1 Single diagonals and cross bracing with a central joint

This problem of cross-bracings is addressed in more detail in Section 6.9. For this case not only the strength of the end joints but the strength of the central X-joint should also be checked.

4.1.2 Relation between joint strength and brace strength

This is a critical issue and will be given further attention in Section 7.9 and in the evaluation of some sample cases in Section 8.

The interaction between bending and tension in the impacted brace is addressed in Sections 6.6, 6.7 and 6.8.

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4.1.3 Ultimate capacity of stiffened joint

The problem with a stiffened joint is that it creates hard spots in the circumference which initiate failure of the joint particularly under tension. This is recognised in API, which does not give any benefit to the stiffening of a joint (except for X-joints under compression).

The Guidance Notes’ equations1 have typically been developed for non-stiffened joints. They have a similar format to the API equations but no guidance is given on the application of these equations for stiffened joints. The following procedure for stiffened joints in tension can be adopted:

· for joints with a few heavy stiffeners the GN compressive equation (similar to API) apply;

· for heavily stiffened joints the lower of the GN tension capacity and the brace tensile strength is used.

4.1.4 Post-accident strength of a platform

The effect of the brace forces during impact should be assessed; of highest concern is the tension (T) in the impacting brace, which could result in an excessive compressive load (C) in another brace.

MSL

TC

C

Figure 4.2 Tension and compression in braces

In this context the brace has been looked at in isolation. This is in line with common engineering practice and acceptable when the brace can take the load in bending only. However, once tension becomes important it generates very high forces and the remainder of the structure, particularly the joints, has to be checked that it can accommodate these loads.

The amount of redundancy in the top part of the structure and the adequacy of joints and braces in this area as a result of tension in the braces falls outside the scope of this report. This

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problem can only be addressed using a full three-dimensional model of the top part of the structure.

4.1.5 Post-failure of a joint

It is important to pay some attention to the behaviour of a joint after it failed according to the code. This is particularly important if the node is weaker than the member is.

The question should then be asked if the joint, after bending and/or tension failure, is still able to carry relatively modest shear forces. In that case the node can be considered as a pinned connection without capacity to take any moments; however, failure of the node (provided it can take the shear load) does then not imply failure of the brace.

In the assessment the retention of shear capacity has been adopted to calculate a reduced energy absorption capacity of the brace (see Sections 7.9 & 8.3 and Table 8.1).

4.1.6 Hitting a stiff or a flexible member

It is stated that a reduction in impact energy absorption could be considered if a very stiff member, such as a grout filled leg, is hit. This follows from the comparison of the load­indentation curve of the vessel with the load (local) deflection curve of the platform. If, on the other hand, a soft member, such as a brace, is hit it can similarly be concluded that the vessel deformation energy is only a small proportion of the total vessel impact energy.

4.2 OTHER PLATFORM FEATURES

4.2.1 Platform dynamics

In an earlier report9 it was noted that vessel impact on one of the legs should be treated as a dynamic problem. This can be concluded by comparing the duration of vessel impact (timpact) with the first natural period of the platform (tplatform); if this ratio is less than 1.0 then dynamics cannot be disregarded. For brace impact these two parameters are:

timpact ≈ velocity

deflection. max =

0.2 0.1 0.1 -

= 0.5 - 1.0 s 4.1

tplatform ≈ 2.0 - 3.0 s 4.2

Hence the ratio between the two characteristic times is < 1.0 and platform dynamics are important.

In practical terms it means that, during brace impact, the platform as a whole has no time to move. Secondly, the mass of the brace is so small that the brace acts instantaneously. In conclusion all the impact energy has to be taken locally, by denting, brace/node deformation and by vessel indentation.

4.2.2 Hinge formation

During impact a hinge can be formed at the end of the brace and at the point of impact. In order to calculate the maximum energy absorption under plastic deformation it is necessary to have some understanding of the maximum hinge rotation that would be acceptable in a tubular.

13

l 3l 3l

A suitable paper on this topic is by Marshall et al6. It is used as a basis for plastic collapse of platforms when results for allowable end rotations were required for earthquake resistance of platforms. This 1977 paper is quite old but still in regular use failing any better expressions. The results have been used extensively for determining earthquake resistance of offshore platforms. The test set-up with which the following numbers were derived is given in Figure 4.3.

P/2

E

q

P/2

l ///3

l

3/l 3/l 3/l = 6.00 m

D = 0.27 m

Figure 4.3 Test set-up for establishing end rotation capacity (Ref. 6)

The paper presents a lower and an upper bound solution for the end rotation in a tubular:

3t ö2.5 t öæ æ×12800qmin = 122 × ÷

ø÷ qmax = ç

èçè DD ø

The material and welding have significantly improved since 1977. In addition the lower bound was considered to be particularly conservative.

Therefore it is proposed to take the log-mean average of these two expressions as a suitable norm for end rotation i.e.:

æ t ö 2.75

NORM 1: qave = 1250 × ç ÷ 4.4 è D ø

From a code point of view the tubular sections should have plastic properties, which, according to API-LRFD is for

D/t £ 13000/sy 4.5

which, for a yield stress of 340 MPa leads to D/t ≤ 38, which is satisfied for most North Sea platforms.

4.2.3 Ductility ratio

Another norm of n times the maximum elastic deformation, taken from the SCI guidance on blast and fire10, the so-called ductility ratio, can also be used. The value of the variable ‘n’ can be taken as 10 or 20. For large values of n the reduction in brace length will be substantial and other failure modes may then be introduced. Therefore the lower value of 10 will be used, or:

max.brace defl. NORM 2: ductility ratio = = 10 4.6

elastic brace defl.

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4.3

4.3 RISERS AND RISER DEFLECTION EFFECTS

In the splash zone risers are preferably located on the inside of the platform near one of the legs; in that way they receive structural protection from the strongest structural members in the platform. However, in the North Sea some risers are for the sake of convenience tied to the braces of the platform. In that case the impact on a brace has a direct effect on the riser and hence riser integrity may well be one of the governing criteria.

MSL

Figure 4.4 A typical riser connected to splash zone braces

The norm for the acceptable riser deflection should be riser integrity. Various norms have been encountered during the review of Safety Cases but the most appropriate one seems to be a maximum riser deflection of 1.0m.

For a 24" riser of 340 MPa steel at a span of 25m the elastic central deflection is given by:

Q × l3 s × l2

ywmax = = = m146.0 192 × E × I 12 × E × D

For large deflections the reduction in length may affect riser integrity. In the absence of tension a deflection of 1.0m and a support distance of 25m results in a length reduction of:

l2 2ó1 dw 1 0.1 ö öæ æ× dx » ×25= m 080.0 DL = ÷ø

÷ø ô

õ çè

çè2 dx 2 5.12

0

Therefore it is proposed to take this maximum deflection, but then as brace deflection, as another norm for brace assessment, i.e.:

NORM 3: umax = 1.0 m

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4.7

4.8

4.9

Other aspects on risers:

· risers could be connected to the inside of a cross brace particularly in the splash zone; at or below the first horizontal plan bracing below the water surface the risers may well be routed to the outside of the platform

· risers can be connected midspan or near a leg.

For the sake of this assessment the missing riser clamp between the riser and the diagonal does not affect the assessment for vessel impact except that the criterion for maximum riser deflection can be increased by the space between the riser and the brace.

4.4 RISERS, RISER CAISSONS AND RISER PROTECTION STRUCTURES

Brace impact and subsequent severance can have unacceptable consequences to nearby risers. In modern designs this effect is addressed by the use of riser caissons where the caisson itself provides extra riser protection; also risers are put close to a leg or behind a leg which make it less probable for an accidental boat impact to touch the riser.

An old design feature of having risers on the outside of the platform in the splash zone is no longer used in practice; the only reason for adopting this feature is that it enhances riser installation.

Also, for some North Sea structures risers are protected using special protection structures. These could be designed for operational impacts of 0.5 MJ. Due to the high level of redundancy in these structures their design and (re-)analysis falls outside the scope of this report.

4.5 CONDUCTORS

In many platforms conductors are inside the jacket structure and well protected by the platform braces and the group of conductors provide some protection for each individual conductor. Also conductors, with tubes-in-tubes provide some lateral resistance before a critical event on the platform would occur.

A single conductor as used in jack-ups for exploration and satellite development drilling may be more prone to an accidental impact scenario; this is addressed in some more detail in Ref. 11.

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5 IMPACT DETAILS

5.1 DENTING

The denting of a tubular is described by the equations by Amdahl12 or by Ellinas & Walker13. These equations for the impact force (Pd) and the impact energy (Ed), obtained from integration of the impact force as a function of the dent depth, are:

- Ellinas and Walker: Pd = 150 ×m p ×dd

D 5.1

5.1

Ed = 100 ×mp ×dd 5.2D

×- Amdahl: Pd = m 21 p ×dd

t 5.3

5.1 ×Ed = 14 ×m p dd 5.4t

where dd is the dent depth, D the diameter of the tubular, t its wall thickness and mp the plastic section property of the tube wall, given by:

t 2 s × y

mp = 5.54

For one specific value of D/t the two relations for Pd and Ed identical, namely for D/t = 50. For thicker braces, particularly compact braces, the Amdahl equation for Pd give a higher value for dd than the equation by Ellinas and Walker. For D/t = 30 the difference in dent depth is some 70%. The difference in force and energy absorption reflect the accuracy of this analysis.

The main difference is in the type of indentor: wedge-shaped or a rigid beam. The dent shape and profile are governed by the type of indentor and the angle (right angle to the tubular). In the Amdahl information the effect of the wedge width can also be estimated. For a width of the wedge equal to D/2 the effect as compared with values for a sharp wedge with the equations mentioned above, is no more than 20% different which is small in comparison with other approximations.

D

b

Figure 5.1 Dent geometry according to Amdahl (Ref. 12)

17

The ultimate or plastic moment and tension capacities are reduced due to the presence of the dent; this should be taken into account in the deformation capacity of the tubular.

The common method for change in plastic section properties due to denting is given: it assumes that the dented part does not contribute any more and only the non-dented part is considered (e.g. Ronalds3, see Figure 5.2).

Figure 5.2 Plastic stress distribution in pure bending for a dented tubular

For the above shape and stress distribution the plastic section modulus for a dented tubular, while ignoring the dented part (as defined by the dent depth dd), is:

sin jùZd = D2 × t × ê

é

ë cos (j / 2) -

2 ûú 5.6

while dd = D × (1- cos j) / 2 5.7

In graphical form the relation between Zd/Z0 and d/D is as shown in Figure 5.3.

0.00

0.20

0.40

0.60

0.80

1.00

0.00 0.10 0.20 0.30 0.40 0.50

relative dent depth (d/D)

Figure 5.3 Relation between plastic section modulus and dent depth

18

5.2 MISCELLANEOUS ASPECTS OF DENTING

The dented section is small in length; hence the relation describing the elastic deformation is virtually unaffected by the dent.

The effects of small dents on fatigue seems to be overemphasised: the highest SCF in the dent corresponds most likely with plain steel and not with a weld.

In principle, the large displacement, elasto-plastic analysis of the denting of a tubular could suitably be analysed using a suitable finite element program.

19

INTENTIONALLY BLANK

20

6 BRACES - BEAM BENDING ANALYSES

6.1 INTRODUCTION

Ship impact on a brace looks like a simple problem and yet there are some serious, fundamental complications. In Appendix A some of these complications are explained with the prime aim to develop further understanding.

In the subsequent analyses the effect of the denting of the tubular due to the applied force is ignored and the tubular will be assumed to be of adequate thickness to avoid local buckling.

6.2 PROPERTIES OF THIN WALLED TUBULARS

The braces in the splash zone can be considered as thin-walled tubulars (see Appendix A2). A thin walled tubular is defined by: t << D, where t is the wall thickness and D the diameter of the tubular.

This implies that simplified equations for the cross-sectional area (A), the elastic section modulus (S), the plastic section modulus (Z) and the moment of inertia (I) can be used. These are:

3 ×A = × p t D I = p

× D × t 6.1-1 8

2S = p

× D × t Z = D2 × t 6.1-2

4

Using the yield stress (sy) the maximum elastic bending moment and the ultimate, plastic moment of the tubular cross section are:

Mel,max = S s × y Mp = Z s × y 6.2

In addition, simplified equations can be developed for elastic-partially plastic sections.

6.3 PRINCIPLE OF VIRTUAL WORK

6.3.1 Centrally loaded beam

The principle of virtual work as applied to braces subject to vessel impact can best be demonstrated by a beam in bending loaded in the centre (see Appendix A3 and Figure 6.1).

The first step is to assume a reasonable displacement distribution for the structure under the applied loading. For a centrally loaded beam in bending it is the displacement distribution as shown in Figure 6.1. This displacement distribution is defined by the central deflection u or the end rotation q. As explained in Appendix A3 the principle of Virtual Work for this displacement distribution leads to the following relation between the ultimate force P and the internal moment distribution, with plastic moments at the end points A and E and the midpoint C, results in:

21

l

l

MpPu = 8 × 6.3 l

PROOF: The proof that this value for Pu is the correct failure load is found by assuming other deformation patterns and calculating the load at failure. The minimum value of P is the correct value for Pu. An example of the necessity of this check is found in the assessment of vessel impact at cross bracings where two failure modes are feasible for out-of-centre impact (see Section 6.9).

a a a

P

a

A B C D E

MA

MC

ME

P

a a a a

A B C D E

q u

l

l

Figure 6.1 Centrally loaded beam failing in bending

For boat impact the energy absorption capacity (E) is of interest, which follows from:

E = Pu q × max × l / 2 = 4 × M p q × max 6.4

where qmax is the maximum rotation at the joints before loss of integrity of the brace occurs.

6.3.2 Impact out of centre

If the impact force P is applied at point B rather than at point C the assumed deformation pattern will change as indicated in Figure 6.2. Using the same methodology as in the previous section the equations for Pu and E can be obtained (see Appendix A3.2).

At failure the following value for the failure load Pu is found:

×M 32 Mp pPu = = 7.10 × 6.5 3 × l l

22

6.6

For boat impact the energy absorption capacity (E) is of interest, which follows from:

E = Pu q × max × l / 4 = 7 . 2 × M p q × max

where qmax is the maximum rotation at joint A before loss of integrity of this joint occurs.

In summary: when the impact occurs out of centre, in this case at a quarter of the span, the impact force is somewhat higher by 33% and the impact energy absorption somewhat lower by 33% as compared to values found for impact in the centre of the beam.

P

a a a a

l

A B C D E

uq1 q2

l

Figure 6.2 Out-of-centre loaded beam failing in bending

Hence over quite some variation in the distance to the beam centre the impact energy is rather insensitive to the exact point of impact.

6.4 AN ELASTIC BEAM IN THREE POINT BENDING

An elastic beam in three point bending reflects the initial stage of ship impact on one of the splash zone braces (see Appendix A4). The simplified problem can be defined as follows:

· take a long slender tubular in three point bending, with fixed ends, loaded in the centre, while ignoring any axial restraint;

As shown in Appendix A4 the linearly elastic solution for the central deflection (wmax) of the tubular under three point bending and the maximum moment Mmax are:

P × l3 P × l wmax = and Mmax = 6.7

192 ×E ×I 8

where l is the length of the tubular. The maximum elastic moment is:

2Me = p

× D × t s × y 6.84

and from these two equations the maximum elastic force P can be obtained. As shown in Eqs. 6.2 & 6.3 the plastic limit load is:

23

pPu = 8 × M

where Mp = D2 × t s × y 6.9

l

Hence, the ratio between the elastic and plastic moment is p/4 = 0.785. This is also, for this particular case, the ratio between the maximum elastic load (Pe) and the ultimate load (Pu).

6.5 ELASTO-PLASTIC THREE-POINT BENDING PROBLEM

Once the maximum moment in the tubular exceeds the maximum elastic moment but is less than the plastic moment the tubular deforms elastically and partly plastically (see Appendix A5). But the bending moment at each point of the beam in the three point bending problem is exactly known. For loads in excess of Pe and less than Pu, or:

Pe < P < Pu

it can be calculated at which point in the circumference the yield stress is reached.

j

+sy

plastic

elastic

plastic

-sy

Figure 6.3 Stress distribution in partially plastic tubular in bending

By assuming elastic-perfectly plastic material behaviour and a Poisson’s ratio equal to zero (to avoid local end constraints) the relation between the actual moment M and the angle j of the plastic region (see Figure 6.3) can be found:

æ 2 j × - sin (2 j × ) + cos j÷÷

ö × D2 s × × t y 6.10M =

èçç sin 4 j× ø

The two limiting cases are:

for j = p / : 2 M = M p = D 2 × t s × 6.11-1

y

2for j = 0: M = Me = p

× D × t s × y 6.11-2 4

Subsequently, the radius of curvature at each point of the beam can be calculated (see Appendix A5) using the relation:

24

d2w = 1

× Me 6.12

dx2 cos j I E ×

Hence, even when the end moment is in excess of the elastic limit but less than the full plastic moment the radius of curvature over the length of the beam (or at each point of the beam) can be calculated; subsequently, by integration, the central deflection can be calculated.

It has been found that for an end moment of 98% of the plastic moment the central deflection is modestly more than the elastic limit deflection.

A first underlying thought was that, once the end reaches its plastic limit, then a hinge will be formed. This is true. However, at a very small distance away from the plastic limit moment the moment is somewhat less, say 98% of Mp. Hence the length of the plastic hinge is very small, and in the limit, as can be demonstrated, the length of the plastic hinge is infinitely small i.e. there is a singularity in the solution.

This calculation was found to be illustrative because it shows:

· in order to form a plastic hinge, an infinite strain in that section is required which would violate the condition on limiting strain.

Despite this flaw, analysis using this singularity are simple and form a powerful tool of adequate accuracy and therefore it will be employed in subsequent analyses.

6.6 THREE-POINT BENDING WITH TENSION - ELASTIC

An additional complication of the brace in elastic bending is to consider the elastic problem of combined bending and tension (see Appendix A6).

A three point bending of a uniform beam (AE) can be treated as a beam clamped on one end and with a point load on the other end (AB). The length (a) and the shear force (Q) are related to the length and the force on the beam as a whole by:

l P a = Q = 6.13

4 2

The general solution of this problem is:

w = cosh A lx + sinh A lx - Q × (a-x) + w B 6.141 2 T

Twhere: l2 = 6.15

E ×I

Using the two boundary conditions at x = 0 and the end condition M = 0 for x = a, the three parameters in the general solution, A1, A2 and wB can be determined (see Appendix A6). Two typical solutions, one for low tension and one for high tension, normalised for wC = -1.00 are given in Appendix A6 as well. The obvious conclusion is that for high tension the bending is confined to a much smaller part of the beam.

25

lll

P

a a a a

l

A B C D E

MA

MC

ME

P

a a a a

l

A B C D E

MA

MC

ME

P

a a a a

l

A B C D E

MA

MC

ME

l

a.a.a. CCCentententrararallllllyyy llloadoadoaded beed beed beamamam iiinnn bbbendiendiendingngng

Q = P/2

a = /4

A B

x

wx wB

Q = P/2

a = /4

A B

x

wx wB

Q = P/2

a = /4

A B

x

wx wB

4/l

b.b.b. AAA quaquaquartrtrtererer ofofof thththe ce ce centrentrentraaalllllly ly ly loadedoadedoaded beambeambeam

Figure 6.4 Elastic beam in bending and tension

6.7 THREE-POINT BENDING WITH TENSION - TUBULAR, PLASTIC TO API

Ship impact is, as a first approximation, considered as a 3-point bending problem. An additional complication is to consider the plastic problem of combined bending and tension (see Appendix A7). The concept of plastic hinges under combined bending and tension will be used.

For a plastic hinge in a tubular, under combined tension and bending, the following equation holds:

öM Tpæ ÷÷=0 6.16ç

çè

×- cos M 2 Tpp ø

where: Mp = D2 × t s× y and Tp = ×p D × t s× y

This equation, Eq. 6.16, can also be called the plastic yield surface for a tubular under combined tension and bending. It can be derived either from the plastic stress distribution in Figure A7.2 or by rearranging Eq. D.3.1-1 in API-LRFD7; therefore it will be called the plastic yield surface to API.

26

6.17

P

a a a a

A B C D E

MA

MC

ME

TT

P

a a a a

A B C D E

TT q

l

l

Figure 6.5 Centrally loaded beam failing in bending and tension

Once plastic hinges have formed the transverse resistance P as a function of the end rotation q

is:

8 × MP = 2 × T + q × 6.18 l

A combination of M and T will be sought which optimises the value of P. In other words P will be maximised under the constraint for a tubular under plastic deformation given above. The equations can be derived using the stress distribution as sketched in Figure 6.6.

The two extreme cases are:

(1) q ≈ 0 6.19-1

(2) q >> 8×Mp 6.19-2

2× Tp ×l

In case (1) the load is taken by bending only because the tension does not contribute. In the second case (2) the shear load P is taken by tension only; bending is only required to ensure that the beam can act as a string in its support and in the centre.

27

j

+sy

-sy plastic -

plastic ­tension

Figure 6.6 Stress distribution in fully plastic tubular in bending and tension

Using the above equations the force P can be expressed in a single variable, the parameter T. Subsequently P can be optimised for T. Or:

dP = 0 6.20

dT

The underlying assumption is that the plastic hinge, of infinitely small size, can redistribute the stresses from bending to tension and vice versa in accordance with requirements. Secondly it is assumed that the resistance P is a continuous function of a single variable, in this case T.

The final solution for T, using the expressions for Mp and Tp is (see Appendix A7):

T = Tp ×p 2 ×arcsin

ölæq × ÷÷çç

è 2 D× ø

and the corresponding value of M is found from the original expression for the yield surface.

A graph for P as a function of q is given in Figure 6.7. From this graph the two limiting conditions are obvious:

(1) if D l q × 0.1 < : the contribution by tension can be disregarded

(2) if D l q × 0.1 >> : there is no contribution by bending.

28

6.21

0.00

0.50

1.00

1.50

2.00

2.50

tension only

bending only

0.00 0.50 1.00 1.50 2.00 2.50 3.00

lq/D

Figure 6.7 Resistance of a centrally loaded beam in bending and tension

lq DThe total lateral resistance by bending combined with tension is larger than the resistance by

but this difference will only become significant for values of in excess of plastic bending; 2.0. As demonstrated later another failure mode will have overtaken namely the maximum joint rotation or the maximum strain.

Therefore, as a first approximation, the contribution by tension can be ignored, provided condition (1) that lq D = 1 is (approximately) satisfied.

Finally, the absorption energy is calculated by the equation:

q

lE = óôõ

P× ×dq2

0

A simplified expression for the energy is given in Eq. 6.36 of Section 6.10.

6.8 THREE-POINT BENDING WITH TENSION - JOINTS, PLASTIC TO HSE GUIDANCE NOTES

Joints under combined tension and out of plane bending follow a different yield surface as compared with tubulars as reflected by the equations given below (see Appendix A8). This is best described in the Guidance Notes.

As indicated in Section 6.7, for a plastic hinge in a tubular, under combined tension and bending, the modified API equations 6.16 and 6.17 hold for tubulars:

öM Tpæ ÷÷=0ç

çè

×- cosM 2 Tpp ø

where: Mp = D2 × t s × y and Tp = × p D × t s × y

are the plastic properties in bending and tension of the tubular.

For joints the Guidance Notes equation applies with a linear relation between M/Mu and T/Tu

or:

29

6.22

M T + = 1 6.23

M u Tu

where Mu and Tu are the ultimate capacity in bending and tension of the end joints.

Let’s assume that the tubular cross-section in the centre has the same property as the end joints and that the same failure surface can be used. Once plastic hinges have formed the transverse resistance P as a function of the end rotation q is:

M8 P = 2 × T + q × 6.24 l

A combination of M and T will be sought which optimises the value of P; in other words P will be maximised under the constraint given above for joints.

The two extreme cases are:

(1) q ≈ 0 6.25-1

4 ×M(2) q >> u 6.25-2

Tu ×l

In case (1) the load is taken by bending only because, for small deflections, the tension does not contribute. In the second case (2) the shear load P is taken by tension only; bending is only required to ensure that the beam can act as a string and that the boundary conditions are fulfilled.

Using the linear constraint equation the moment M can be eliminated so that P is expressed in one parameter T. Subsequently P can be optimised for T, or:

dP = 0 6.26

dT

4 ×MuThis leads to a unique solution for q (see Appendix A8): qcrit. = 6.27 l× Tu

For q < qcrit. the transverse load is taken by bending whereas for q > qcrit. the transverse load is taken by tension (see Figure 6.8).

Therefore, as a first approximation, the contribution by tension can be ignored, provided condition (1) that q q . crit = 0.1 is (approximately) satisfied.

30

i ly

q = 4 × M u

l×T u

0.00

0.50

1.00

1.50

2.00

2.50

tens on on

bending only

crit .

0.00 0.50 1.00 1.50 2.00 2.50

q q crit.

Figure 6.8 Plastic tubular in combined bending and tension (joint failure)

For a tubular the relations for Mp and Tp can be expressed in D, t and sy. If the joints are as strong as the tubular both under pure bending and pure tension, or:

Mu = Mp and Tu = T p 6.28

4 × Dit would then lead to: qcrit. = 6.29

× p l

and for l / D = 25: qcrit. = 0.05 6.30

6.9 CROSS-BRACING CAPACITY

As indicated in Figure 4.1 a ship can not only collide with a single brace but also a cross­bracing at or near the centre. Analysis using the principle of virtual work is given in Appendix A9.

P

u

b

P

u

b

deformation mode 1 deformation mode 2

Figure 6.9 Impact out of centre of the cross brace

The following situations are to be considered (see Figure A9.1):

31

· impact at the cross itself (b = 0, mode 1), resulting in:16 × Mpdouble capacity of a single brace P C = 6.31

l

· impact in the middle of a span (b = l / 4 , mode 2) resulting in: 4 × M 16 × Mp pimpact force P = = = PC 6.32 l / 4 l

· impact at a distance ‘b’ from the centre resulting in an impact force depending on the deformation mode (see Figure 6.9):

2 × M 2 × Mp pfor deformation mode 1 (local): Pu1 = + 6.33 b l / 2 - b

l / 2 16 × M pfor deformation mode 2 (global): Pu2 = × 6.34 l / 2 - b l

· for impact at a distance b = l / 8 the two modes result in the same impact force:

64 × MpP u1 = Pu2 = 6.35 3× l

More specifically: for b > l / 8 the centre will not deform and no plastic deformation of the non-impacted braces occurs.

When changing the value for ‘b’ there is a smooth change in impact force. However, due to the unchanging value of the maximum rotation at the joints, there is a rapid change in impact energy capacity when changing from the global deformation mode (mode 1) to the local deformation mode (mode 2):

· the allowable deflection (as governed by the maximum rotation) for b = l / 8 in the local mode is only 25% of the allowable deflection for b = 0.

Hence, the impact energy absorbing capacity for b = l / 8 shows a rapid jump:

Eu for b = l / 8 in the local mode = 44% of Eu for b = 0

Eu for b = l / 8 in the global mode = 100% of Eu for b = 0

Due to strain hardening this jump will be less abrupt in reality, but it should be recognised in any numerical analysis.

6.10 SIMPLIFIED ENERGY EXPRESSION

A number of solutions have been given in the previous paragraphs. It is noted that, once plastic deformation is serious, the deflections are much larger than the elastic deformation. Hence, as a first approximation, the elastic deformations can be disregarded and as a consequence the energy absorption due to bending can be calculated from:

E ≈ P × w max 6.36

32

Secondly, as shown in Sections 6.7 and 6.8, the contribution by tension becomes significant only for larger deflections and the change-over points can be calculated. Hence for deflections up to and somewhat in excess of this changeover point (see Figures 6.7 and 6.8) tension in the energy absorption expression can be disregarded. Hence the above equation is valid for many cases even if tension is significant, i.e.:

• if lq / D < 2.0 (see Fig. 6.7) or q/qcrit. < 1.5 (see Fig. 6.8) the above equation is a suitable equation to calculate the impact absorption capacity of a brace.

And P can be calculated using the expression:

MA + 2 × M C + MEP = 2 × 6.37 l

where MA and ME are the plastic moments at the two brace ends and MC is the plastic moment in the centre after incorporation of the effect of the dent depth.

Finally, a simplified expression and approximate relation for the reduction in length for a brace of length l , impacted at the centre with a central deflection u, can be obtained by considering Figure 6.1:

2 2u 2 æ ×lD

l =1-cos = q 1

2 ×ö÷ø

öuæ = 2× ÷

ø çè

çèl l

33

6.38

INTENTIONALLY BLANK

34

7 MATERIAL AND JOINTS

7.1 INTRODUCTION

In this section some of the aspects will be discussed which have a bearing on analysis or testing. The following topics will be addressed:

· material and fracture toughness · geometry and local modelling · strain rate effect · CTOD tests · Charpy test · strain hardening · the maximum strain criteria · joint strength.

The effects of these aspects on platform specific results will also be given.

7.2 MATERIAL AND FRACTURE TOUGHNESS

Platforms in the North Sea are often ranked in accordance to the year they were designed and built. Except for certain design issues, such as liftable jackets, minimum number of braces, etc. the difference between the various generations of structures from a material and welding point of view are:

· first generation: few requirements on material properties; standard Charpy tests · second generation: special requirements on joints: through thickness properties · third generation: steel and welding procedures with CTOD requirements

Finally, because of the very high local strains which will develop during vessel impact it is questionable if a PD6493 fracture assessment14 can lead to acceptable results (see Section 6.5). Also, for third generation platforms, a fracture assessment in accordance with PD6493 became a regular design feature. However, in view of the comments made below this may well be questionable for vessel impact assessments.

7.3 GEOMETRY AND LOCAL MODELLING

As described in Section 6.5 the geometric modelling of the critical region of the plastic hinge is not straightforward. Numerically, ductile failure of nodes can be computed reasonably well. For example VanderValk15 carried out extensive numerical analyses of joints under compression. In his case the load-deflection curve has a maximum and hence failure is caused by geometrical considerations, not by brittle or ductile fracture.

For joints under tension and bending this modelling is therefore much more complex. As for example stated in one of the papers on partially cracked joints16, the numerical modelling of a cracked node is not straightforward and should be executed with care.

35

7.4 STRAIN RATE EFFECT

The first approximation of the brace impact problem is the three point bending of a tube with clamped ends. For a velocity ( w& ) of 2.0m/s, a brace diameter (D) of 1.5m and a member length ( l ) of 25m, excluding the effects of an SCF, the maximum strain rate can be expressed by the following equation:

& D w 12 5.1 0.2 12 & = e

× × »

× ×

» s05.0 -1 7.12l 252

Using the Symonds’ relation17,18 this strain rate effect on the yield stress of the material can be calculated:

1ö÷ ø

/ p

1 ìïí

üïý

æ e & s = static , y ×{1+ 48.0 2.0 }( )e × &s dynamic , y s = static , y × +ç

èDïî ïþ

where: sy,dynamic = the dynamic yield stress at ε& sy,static = the static yield stress (at ε& = 10-4 s-1) D and p = material constants (for mild steel D = 40 s-1; p = 5)

These values for the material properties lead to the following change in (dynamic) yield stress:

s

s dyn , y = 1 + 48.0 × (e&) 2. 0

» 25.1 7.3 stat , y

Hence a 25% increase in the yield stress can be expected. It appears that this value is rather insensitive if the impact speed is reduced by a factor 2 or if an SCF of the order of 3.0 is introduced.

The Symonds’ relation is the most commonly used expression for strain rate effects and built in many non-linear finite element analysis packages. In a tubular this high strain rate is particularly observed in the outer fibres which contribute most to the moment capacity of the tubular. Hence, at least a 20% increase in plastic moment capacity can be expected during ship impact if the energy is taken by the tubular in bending.

However, if the joint provides most of the rotation then the stress distribution is much more complex and without further substantiation this increase should not be applied.

7.5 CTOD TESTS

The previously calculated strain rate can also be used to check the strain rate effects of the CTOD tests. According to BS744819 the rate in stress intensity during the CTOD test should be in the range:

& -1K BS » 5.0 - m MPa0.3 5.0 × s 7.4×

The strain rate in a brace, without stress concentration factor, can be determined by assuming that the yield stress of the material is reached at a deflection of 0.1m. Using a vessel speed of 1.0m/s the strain rate is:

36

7.2

ε& =sy

× vboat

= 345 0.1

× = 017.0 7.5E w .el 205000 1.0

For a defect of 2-4 mm at a strain rate of 0.05 s-1 , in combination with a stress concentration &(SCF) of 2.0 leads to the following value for K :

-1& &Kimpact » SCF × e × E × p a » 0.2 × 017.0 × 205000 × p 003.0 = m MPa700 5.0 × s 7.6×

&Hence the value for K during vessel impact is more than 2 orders of magnitude larger than during the CTOD test. (Comment: when the same arguments are applied to the extreme storm condition then the difference is 1 order of magnitude.) Therefore the validity of the CTOD test for vessel impact should be verified.

7.6 CHARPY TEST

The Charpy test leads, probably, to higher strain rates than those experienced during vessel impact. Therefore the condition that the material has good Charpy test values, even under temperatures much lower than the environmental temperatures, gives confidence in the good dynamic performance of the joints under vessel impact.

7.7 MAXIMUM STRAIN CRITERION

In some codes, for example for (chemical) plant design, criteria can be found on maximum strain under exceptional loading. This is a design feature and strains of 5-10% are quoted. In the review of Safety Cases a similar criterion was found which provide a suitable norm for vessel impact assessment:

· maximum strain criterion: 15% in the parent material and 10% in the weld

The maximum strain criterion could well be in contradiction of a pure fracture mechanics assessment of tubular joints, particularly for cracked tubular joints. Quite some work has been carried out over the past few years on this topic and it appears that small cracks have only a minimal effect on the ultimate capacity of tubular joints.

In some literature test data can be found where small defects lead to a large reduction in joint strength. However, as demonstrated in Ref. 20, the material and test details are not representative for offshore structures: i.e. the material properties are chosen such that defects in combination with very low parent material properties are tested.

If the joints are stronger than the brace the brace forms a hinge. In that case, as shown earlier, the hinge can form a singularity in the solution. Hence in a detailed FE analysis it can be expected that the maximum strain may well depend on the element size. This should be checked.

7.8 STRAIN HARDENING

In an ABAQUS type of analysis strain hardening can be incorporated. The anticipated effect of this strain hardening is that singularities in the elastic, ideally plastic solution, identified in Section 6.5, are smoothed. This will particularly be apparent when the plastic hinge would occur in the tubular. In this context it is useful to observe from the graphs in Ref. 17 that the strain hardening effects are virtually not affected by the strain rates expected during vessel impact

37

Under the 3-point bending test (without tension and combined with a linear model for the strain hardening) an indication of the rotation (q) due to plastic deformation has been obtained in Appendix A10. The final answer for this simplified problem is:

q = e all × l×

s u s - y 7.7 4 × D s y

where eall = maximum allowable strain in the tubular; l and D are the length and diameter of the brace; su and sy are the ultimate tensile stress at the maximum allowable strain and the yield stress. For example, for l / D = 25, eall = 10%, su/sy = 1.10 the maximum allowable hinge rotation (q) is found to be 0.07 rad; the associated central deflection is approximately 0.8 D.

By increasing the value for su/sy from 1.10 to 1.20 this allowable rotation is increased by a factor 2. But a value of su/sy lower than 1.10 for jack-up steels seems more realistic.

This leads to a norm for the brace end rotation as reflected by Eq. 7.7. This norm can be calculated using a combination of the maximum strain in the brace-end-weld together with strain hardening effects, or:

NORM 4: max. rotation using max. strain in weld combined with strain hardening

Note that the effects of strain hardening are incorporated (through the experiments) in the joint design equations and in the dent depth and dent energy equations.

7.9 JOINTS

For brace assessment under impact a set of four norms have been identified (see summary in Section 2.2). These norms, together with the resistance calculated in accordance with Section 6, assume that the joints are stronger than the brace. If the joints are stronger than the brace the brace forms a hinge. Next the following considerations should be considered:

· joint strength versus brace strength · one joint is weaker than the brace · both joints are weaker than the brace · the effects of brace stubs.

These topics will be addressed in some detail in the next sub-sections. Joint strength equation can be taken from Ref. 1 or 7.

7.9.1 Joint strength versus brace strength

For structures in earthquake sensitive areas it is good engineering practice to ensure that the joints are stronger than the braces. In that case it is ensured that the energy absorption capacity of the braces are fully utilised. And the same holds for braces in the splash zone which could be hit by boat impact.

For new designs such a recommendation can easily be incorporated but for existing structures the splash zone braces may not comply with this recommendation and the energy absorption capacity may well be affected as a consequence.

38

But even if the joints are modestly stronger than the braces there is no 100% guarantee that the brace will form a plastic hinge before the joint fails. The reason is that, for design, strengths are uniquely defined by the design equations whereas, in practice, strengths are always statistically distributed. And it is generally accepted that the standard deviation of the strength distribution for braces is smaller than for joints.

This implies that when:

Tp = Tu and Mp = Mu 7.8

where Tp and Mp are the plastic moment capacities in tension and bending of the brace and Tu

and Mu are the ultimate joint capacities in tension and bending, then, in the mean, the joints are stronger than the braces.

Therefore for this case the brace can fully develop its plastic energy absorption capacity. And, because bending is the prime energy absorption mechanism for the brace only a comparison will be made of the bending characteristics for joints and braces. The equation for the brace resistance, where MC incorporates the effect of denting, is:

M A + 2 × M C + MEP = 2 × 7.9 l

Other cases with joints weaker in bending than braces are addressed in the following paragraphs.

7.9.2 One joint is weaker than the brace

The three positions in a tubular joint under brace bending to be addressed are:

· chord saddle position in tension · chord saddle position in compression · chord crown position in shear

When one joint (Joint A) is weaker in bending than the connected brace, or:

Mp < Mu

then the bending energy absorption should be developed by the joint rather than by the brace. It is, however, rather unlikely that the joint can develop this absorption and it is expected to fail in the tension zone in the chord saddle position.

However, the chord saddle position in compression and the chord crown position in shear will retain their local integrity and hence the joint is still able to carry shear, but the bending resistance will substantially be reduced. Hence if Joint A has a lower joint strength than brace strength, reflected by a reduction factor gA, then the equation for the ultimate force capacity can be adjusted to reflect the failing of this joint in bending, while retaining its shear capacity or:

A × C · before joint A failure: P1 = 2 × g × MA + M 2 + ME 7.10

l

2 × M C + ME · after joint A failure: P2 = 2 × 7.11

l

39

The equation for P1 is used to calculated the dent depth: the reason is that the maximum dent depth is obtained for the highest lateral brace resistance; this force is highest just before failure of joint A. Subsequently, this dent depth can be taken to calculate the ultimate bending capacity in the centre (MC) for P2 and this value has to be used for the energy absorption capacity.

7.9.3 Both joints are weaker than the brace

Similarly if both Joint A and Joint E have lower joint strengths than brace strengths, reflected by reduction factors gA and gE, then the equation for the ultimate force capacity can be adjusted to reflect the failing of these joints in bending, while retaining their shear capacity or:

× CA E · before failure of joints A and E: P1 = 2 × g × MA + M 2 g + × ME 7.12

l

4 × M C · after failure of joints A and E: P2 = 7.13

l

The equation for P1 is used to calculated the dent depth: the reason is that the maximum dent depth is obtained for the highest lateral brace resistance; this force is highest just before failure of joints A and E. Subsequently, this dent depth can be taken to calculate the ultimate bending capacity in the centre (MC) for P2 and this value has to be used for the energy absorption capacity.

7.9.4 The effects of brace stubs and corrosion allowance

In many platforms brace stubs are used and the reason may well be that the fabrication process employs joint assembly prior to full jacket structure assembly. If the brace stub is of equal thickness as the remainder of the brace then no adjustment is necessary. However, if the brace stub is of heavier thickness then a comparison should be made between the lateral resistance of (a) the (shorter) brace between the closure welds of brace and stubs and (b) the (longer) brace between the joints.

If the brace stubs are stronger than the brace and also stronger than the joints the outcome of the use of brace stubs may well be negative from an impact energy absorption capacity point of view.

Finally, corrosion allowance is sometimes called ‘brace impact allowance’ and under proper corrosion protection may not be consumed in the operating life of the platform. Hence corrosion allowance can then be added to the brace thickness or the nominal brace thicknesses as given in ‘as-built’ drawings can be used for the assessment.

7.10 IMPACT CLOSE TO A NODE

In Section 6.3.2 it was indicated that the energy absorption in the central portion of half the length of the brace does not vary by more than 30%. For impact in the quarter section close to a node the following assumptions will be made:

· bow impact is the most likely scenario because the size of the stern will involve other braces or the leg as well.

· for bow impact 4.0 MJ impact corresponds approximately with a maximum impact force of 7 MN (see Figure 3.1) at 0.8m bow indentation.

40

· at impact close to the leg the joint may well fail in shear; little is known about the shear capacity of tubular joints and hence the following simplified expression will be used:

D × t s × / 3 < R < D 2 × t s × y / 3 7.14×y

The high value is valid for a non-damaged tubular. For a value after severe local damage Figure 7.1, showing an outline of the intersection between an inclining brace and the leg, will be applied; only the top and bottom parts A1A2 and B1B2 of lengths D/2 are assumed to contribute to the shear strength. The thickness is taken as t, which is the characteristic thickness of the tubular at A-A’ and B-B’ in Figure 7.1. Hence a conservative value is:

R = D × t s × y / 3 7.15

A1 A2

t

leg

B1 B2

D/2

D

D

A

A’

B

B’

brace

Figure 7.1 Simplified sketch of the brace-to-leg intersection

Using this expression 7.15 for the examples in Section 8 the condition that R > 7 MN is satisfied, but smaller braces could shear off during joint impact.

The impact force (Px) at a point x away from the node (for x see Fig. 6.4b) as compared with PC

(impact at the centre) is approximately given by:

Px = l 4

× æ

èç

1 x

+ 1

l - x ö

ø ÷ × PC »

l 4 × x

× PC 7.16

· for Px > 7 MN: all energy is taken by the impacting vessel

· for Px < 7 MN & P > 3 MN: the energy is shared between the vessel and the brace

· for Px < 3 MN: all energy has to be absorbed by the brace

41

This relation 7.16 also allows to calculate the length xP = 7 at which Px = 7 MN. If Figure 7.2 isrepresentative for 4.0 MJ impact energy two conditions can occur:

· a. the shear strength of the joint is greater than 7 MN, or

· b. the shear strength of the joint is less than 7 MN.

Then, depending on the node strength, it would follow (approximately) that:

· for x < l / 8: 4.0 MJ energy is taken by the vessel (Fig. 7.2a) or joint shear failure is likely at 4.0 MJ (Fig. 7.2b)

· for x > l / 8 & x < l / 4 energy absorption is shared between vessel and brace

· for x = l / 4 the brace energy absorption to Section 6.3.2 applies.

12.0

10.0

8.0

6.0

4.0

2.0

0.0

brace resistance joint shear strength bow resist. (7 MN)

A B C D E

12.0

10.0

8.0

6.0

4.0

2.0

0.0

i joi

( )

A B C D E

brace res stancent shear strength

bow resist. 7 MN

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75

a. Joint strength > 7 MN b. Joint strength < 7 MN

Figure 7.2 Resistance at impact near a joint

42

1.0

8 SAMPLE PROBLEMS

8.1 DESCRIPTION OF THE TEST CASES

In order to demonstrate the various items addressed in this report two sample-problems will be selected (Sample 1 and Sample 2). In Figure 4.1 a typical jacket structure is shown. Let’s assume, for the sake of this exercise, that the length of the splash zone brace is 25 or 35 m, or:

l = 25. 0 or 35. 0 m 8.1

In accordance with Ref. 21 on the structural design of offshore jackets, for main members in platforms designed for more severe environmental conditions a suitable range of l / D ratios is l / D = 30, but for Central and Northern North Sea platforms a ratio l / D = 25 seems more appropriate. This leads to:

D = 1.000 or 1.400 m 8.2

In accordance with API-LRFD for steel with a yield stress of 340 MPa the full plastic yield stress can be developed in bending for:

D / t <= 13000 / sy = 38 . 8.3

This leads to corresponding thicknesses for the two cases of:

t = 35 or 45 mm 8.4

Furthermore the following assumptions are made and subsequently will be varied:

· the joints are stronger than the brace · no brace stubs are used · impact in the centre of the beam.

The results are given in detail in Tables 8.2 and 8.3 at the end of this report while the mostrelevant data are summarised in Table 8.1.

8.2 VESSEL IMPACT ANALYSIS

For impact it will first be assumed that the effect of the denting can be ignored. In that case, for central impact, the plastic moments in the ends and at the centre can be determined. Without brace stubs these moments for the two cases are:

Mp = D2 t sy = 11.9 or 30.0 MJ 8.5

and the ultimate impact forces are:

Pu = 8 × Mp / l = 3.8 or 6.9 MN 8.6

These forces are quite substantial, but according to Figure 3.1 they are too small to cause any substantial vessel deformation under stern impact.

43

quantity Sample 1 Sample 2 units Ref. length l 25 35 m

diameter (D) 1.000 1.400 m

thickness 35 45 mm Mp = D2 t sy 11.9 30.0 MJ Eq. 6.2 initial Pu = 8 × Mp / l 3.8 6.9 MN Eq. 6.3

dent depth and dent energy

initial dent depth (dd ) 0.104 0.158 m Eq. 5.3

actual dent depth (dd ) 0.075 0.113 m Eq. 8.13

dent energy 0.2 0.4 MJ Eq. 5.4 brace central deflection

displ. from Marshall rotation 1.55 1.72 m Sect. 4.2.2

displ. from ductility ratio 0.86 1.21 m Sect. 4.2.3

max. brace deflection 1.00 1.00 m Sect. 4.3

displ. from strain hard. rot. 0.8 1.1 m Eq. 7.7

mean value displ. 1.05 1.25 m brace impact force and impact energy

final Pu 3.2 5.8 MN Table 8.2/3

brace energy absorption 3.4 7.2 MJ Table 8.2/3 vessel energy absorption

under bow impact 0.0 2.0 MJ Fig. 3.1

under stern impact 0.0 0.0 MJ Fig. 3.1 minimum energy absorption (brace + dent + vessel)

total energy 3.6 7.6 MJ one joint weaker than the brace

joint strength reduction (gA) 0.80 0.80 Sect. 7.9.2

dent depth 0.07 0.10 m Table 8.2/3

dent energy 0.1 0.4 MJ Table 8.2/3

brace energy abs. (one jt fail.) 2.3 4.8 MJ Table 8.2/3 two joints weaker than the brace

joint strength reduction (gA = gE) 0.80 0.80 Sect. 7.9.2

dent depth 0.06 0.09 m Table 8.2/3

dent energy 0.1 0.3 MJ Table 8.2/3

brace energy abs. (two jt fail.) 1.3 2.7 MJ Table 8.2/3

Table 8.1 Results for two sample cases (see Tables 8.2 and 8.3 for details)

44

However, the forces are large enough to cause some denting and using the Amdahl equations (see Section 5.1) the following, preliminary, results for the dent depths can be calculated:

dd = 0.104 or 0.158 m 8.7

The corresponding denting energy is:

Ed = 0.3 or 0.7 MJ 8.8

Earlier, the word ‘preliminary’ was used in calculating the dent depth. The reason is that the dent depth results in a substantial reduction in the plastic bending property (see Figure 5.3); and, as a consequence, result in a reduction in the dent force. Rather than solving this problem analytically, or by using the ‘Goal Seek’ option in Excel, a simple iterative scheme can be used which converges very rapidly (see Section 8.2.1).

In order to determine the brace plastic deformation energy the force Pu should be combined with a suitable displacement. Earlier the following criteria have been identified:

· Sect. 4.2.2: Marshall rotation · Sect. 4.2.3: ductility ratio of 10 on the elastic deformation uel

· Sect. 4.3: maximum displacement of 1.0m· Sect. 7.8: maximum strain in weld combined with strain hardening.

These values for the deflection are given in Table 8.1 (see Tables 8.2 and 8.3 for details). Thefollowing mean values for the two sample problems are obtained:

umean = 1.05 or 1.25 m 8.9

Hence the total energy which can be taken by the brace is found by combining the impact forceobtained at the end of the iterative scheme with this mean displacement:

Ep = 3.4 or 7.2 MJ 8.10

Hence, in both cases the energy criterion of 4 MJ is almost met or substantially exceeded.

Finally, the value of l q × / D for Fig. 6.7 and q/qcrit for Fig. 6.8 should be checked. Using Eq. 6.29 and the expression:

q = 2 u / l 8.11

the results are:

l q × / D = 2.1 and 1.6

and: q/qcrit = 1.6 and 1.4 8.12

Both results indicate (see Fig. 6.7 and 6.8) that the main contribution from the force resistance is developed by tension in the brace but that the energy absorption can suitably be approximated by the contribution using bending only. For example, for Figure 6.8, at q/qcrit = 1.5 the difference between the energy by bending and by bending/tension is no more than 8% (see Figure 8.1)

45

0.00

0.50

1.00

1.50

2.00

2.50

tension only

bending only

additional energy due to tension

qcrit . =

4 × M u

l×T u

0.00 0.50 1.00 1.50 2.00 2.50

q qcrit..

Figure 8.1 Additional energy due to tension at q/qcrit = 1.5

8.2.1 Details of the iterative scheme

The iterative scheme, mentioned in the analysis of the sample problems, calculates the impact force as a function of the dent depth. It can be described as follows:

initial condition: dd = 0

calculate subsequently:

- the plastic moment capacity in the centre for the previous dent depth

- the impact force, using the equation:

Pu = 2 × (MpA + M2 pC + MpE )/ l×

- the new dent depth

Here MpA, MpC and MpE are the plastic moments in the two end points (A and E) and in the centre (point C, see Figure 6.1) using the dent depth as determined earlier.

Or, in mathematical terms:

initial condition: dd = 0

calculate subsequently in step i:

- (MpC)i for the dent depth from the previous iteration (dd)i-1

- the impact force, using the equation:

(Pu) = 2 × [MpA + 2 × (MpC ) + MpE ]/ li i

46

- the new dent depth (dd)i

Using this iterative scheme the following results are obtained for the two sample problems (see Tables 8.2 and 8.3):

dd = 0.075 or 0.113 m Pu = 3.2 or 5.8 MN

Ed = 0.2 or 0.4 MJ 8.13

This iterative scheme is given at the bottom of the spreadsheets in Tables 8.2 and 8.3.

8.3 VARIATIONS IN THE SAMPLE PROBLEMS

8.3.1 One joint weaker than the brace

Using the joint strength equations of the Guidance Notes1 or API7 the ratio between the brace strength and the joint strength in out of plane bending can be established. This ratio can be input into the spreadsheet. For example a reduction of 20% results in a reduction of the maximum impact force for the two sample problems:

dd = 0.07 or 0.10 m initial Pu = 3.2 or 5.5 MN

Ed = 0.1 or 0.4 MJ 8.14

These results are some 10% lower than those established earlier for the case where the joints are stronger than the brace. Subsequently, it is assumed that the four rotations in Table 8.1 are not affected by a weaker joint but it should also be envisaged that the weaker joint could fail in bending being still able to a carry shear load. Hence, using the spreadsheet, the impact force after joint failure reduces to about 70% of the value before joint failure i.e. the force which determines the dent depth;

final Pu = 2.2 or 4.5 MN

Ep = 2.3 or 4.8 MJ 8.15

8.3.2 Two joints weaker than the brace

Similarly a reduction of 20% in joint strength as compared with brace strength results into a reduction of the impact force and impact energy, particularly since joint failure under bending (but not in shear) has to be assumed. The energies are (see Table 8.1):

Ep = 1.3 or 2.7 MJ 8.16

These values are only 40% of the energies that can be absorbed by the brace if the joints are stronger than the brace.

8.3.3 Brace stubs are used

As indicated in Section 7.9.4 brace stubs of different thickness can be addressed in the spreadsheet. Particularly the problem of brace stubs together with joints that are weaker than the brace or the brace stubs require some ad hoc adjustment to the input data in the spreadsheet.

47

8.3.4 Impact out of centre of the beam and effect of dynamics

Two corrections may well be required on the above results:

· impact out of centre e.g. an offset of l / 4 could lead to a reduction of impact energy of 33% (see Section 6.3);

· because of the short duration of the vessel impact the yield stress could well be increased by 25% (see Section 7.4).

It is within the accuracy of this assessment to let these two quantities cancel one another. Therefore the impact energies in Table 8.1 are valid for a range of points of impact.

48

L

brace properties Sample 1 units Sample 1 Sample 2 units D 1.000 m 1.000 1.400 mt at joint-1 (top) 0.035 m 0.035 0.045 mt in the centre 0.035 m 0.035 0.045 mt at joint-2 (bottom) 0.035 m 0.035 0.045 m

25.0 m 25.0 35.0 mFy 340 MPa 340 340 MPaE 205000 MPa 205000 205000 MPajoint-red. 1 (end-1) 1.00 (Sect. 7.9) 1.00 1.00joint-red. 2 (end-2) 1.00 (Sect. 7.9) 1.00 1.00Mpl-end-1 11.9 MNm (Eq.7.10)Mpl-centre 11.9 MNm (Eq.6.2)Mpl-end-2 11.9 MNm (Eq.7.12)

initial plastic bending properties Auxilliary properties P-plastic 3.808 MN (Eq. 6.3) D/t 28.6Energy 4.000 MJ (Sect. 3.3) norm 38.2 (Sect.4.2.3)u 1.050 m (= E/P-pl) T-pl 37.4 MN (Eq. 6.17)

brace denting (Sect. 5.1) Norm (S.=Sect.) centr. defl. (m) dent-Amdahl 0.104 m Norm 1 max. rot. at joint (S.4.2) 1.55 dent E&W 0.059 m Norm 2 ductility ratio (S.4.2) 0.86 energy-Amdahl 0.263 MJ Norm 3 max. brace defl. (S. 4.3) 1.00 energy-E&W 0.151 MJ Norm 4 max. str. of 10% (S. 7.7)

mean max. displ. 0.78 1.05

1. maximum rotation (Sect. 4.2.2)max. rot (Marshall) 0.124 approx. elongation (Eq. 6.38)assoc. deflection 1.549 m u 1.05 m

L 25.00 m 2. ductility ratio (Sect. 4.2.3) elongation 0.35 %

I (mom. of inertia) 0.014 m4 (Eq. 6.1-1) yield strain 0.17 %u-el at P-el. 0.086 m (Eq. 6.7) ratio 2.12ductility ratio 10.00

u-allowable 0.864 m Energy (final) mean displ. 1.05 m

3. max. brace deflection (Sect. 4.3) impact force 3.23 3.39

MNu-max 1.000 m E-beam MJ

4. max. all. strain of 10% (Sect. 7.7 & 7.8) dent-Amdahl 0.07 mmax strain 0.10 (Sect. 7.7)max. str. / yield str. 1.10corr. rotation 0.06 (Eq. 7.7)assoc. deflection 0.781 m

dent energy 0.16 MJ Premature joint failure (Eq. 7.11 & 7.13) E-one jt fail. 71 % E-two jt fail. 41 %

Reduced properties: impact at centre - iteration process (Sect. 8.2.1) iteration 0

imp. force P (Eq.6.37) 3.81dent Amdahl (Eq. 5.3) 0.10E-Amdahl (Eq. 5.4) 0.26phi (Fig. 5.2) 0.66Zpl (Eq. 5.6) 0.022Mpl (Eq. 6.2) 7.63

1 2 3 4 3.13 3.25 3.23 3.23 MN 0.07 0.08 0.07 0.07 m 0.15 0.16 0.16 0.16 MJ 0.54 0.56 0.55 0.55 rad

0.025 0.024 0.024 0.024 m3 8.44 8.29 8.32 8.31 MNm

Table 8.2: Excel results central impact Sample 1

49

L

brace properties Sample 2 units Sample 1 Sample 2 units D 1.400 m 1.000 1.400 mt at joint-1 (top) 0.045 m 0.035 0.045 mt in the centre 0.045 m 0.035 0.045 mt at joint-2 (bottom) 0.045 m 0.035 0.045 m

35.0 m 25.0 35.0 mFy 340 MPa 340 340 MPaE 205000 MPa 205000 205000 MPajoint-red. 1 (end-1) 1.00 (Sect. 7.9) 1.00 1.00joint-red. 2 (end-2) 1.00 (Sect. 7.9) 1.00 1.00Mpl-end-1 30.0 MNm (Eq.7.10)Mpl-centre 30.0 MNm (Eq.6.2)Mpl-end-2 30.0 MNm (Eq.7.12)

initial plastic bending properties Auxilliary properties P-plastic 6.854 MN (Eq. 6.3) D/t 31.1Energy 4.000 MJ (Sect. 3.3) norm 38.2 (Sect.4.2.3)u 0.584 m (= E/P-pl) T-pl 67.3 MN (Eq. 6.17)

brace denting (Sect. 5.1) Norm (S.=Sect.) centr. defl. (m) dent-Amdahl 0.158 m Norm 1 max. rot. at joint (S.4.2) 1.72 dent E&W 0.099 m Norm 2 ductility ratio (S.4.2) 1.21 energy-Amdahl 0.721 MJ Norm 3 max. brace defl. (S. 4.3) 1.00 energy-E&W 0.451 MJ Norm 4 max. str. of 10% (S. 7.7)

mean max. displ. 1.09 1.25

1. maximum rotation (Sect. 4.2.2)max. rot (Marshall) 0.098 approx. elongation (Eq. 6.38)assoc. deflection 1.716 m u 1.25 m

L 35.00 m 2. ductility ratio (Sect. 4.2.3) elongation 0.26 %

I (mom. of inertia) 0.048 m4 (Eq. 6.1-1) yield strain 0.17 %u-el at P-el. 0.121 m (Eq. 6.7) ratio 1.55ductility ratio 10.00

u-allowable 1.209 m Energy (final) mean displ. 1.25 m

3. max. brace deflection (Sect. 4.3) impact force 5.78 7.26

MNu-max 1.000 m E-beam MJ

4. max. all. strain of 10% (Sect. 7.7 & 7.8) dent-Amdahl 0.11 mmax strain 0.10 (Sect. 7.7)max. str. / yield str. 1.10corr. rotation 0.06 (Eq. 7.7)assoc. deflection 1.094 m

dent energy 0.43 MJ Premature joint failure (Eq. 7.11 & 7.13) E-one jt fail. 70 % E-two jt fail. 41 %

Reduced properties: impact at centre - iteration process (Sect. 8.2.1) iteration 0

imp. force P (Eq.6.37) 6.85dent Amdahl (Eq. 5.3) 0.16E-Amdahl (Eq. 5.4) 0.72phi (Fig. 5.2) 0.69Zpl (Eq. 5.6) 0.055Mpl (Eq. 6.2) 18.74

1 2 3 4 5.57 5.82 5.77 5.78 MN 0.10 0.11 0.11 0.11 m 0.39 0.44 0.43 0.43 MJ 0.55 0.58 0.57 0.58 rad

0.062 0.060 0.061 0.061 m3 20.96 20.53 20.61 20.59 MNm

Table 8.3: Excel results central impact Sample 2

50

9 REFERENCES

1. Offshore Installations: Guidance on Design, Construction and Certification, Fourth Edition, UK Health & Safety Executive, London, 1990 – withdrawn June 1998 (ref. HSE Operations Notice, ON27). Technical guidance republished in Offshore Technology Report Series. OTO 2001 013, ‘Loads’, available from www.hse.gov.uk.research/otopdf/2001/oto01013.pdf, includes vessel collision information.

2. A guide to the Offshore Installations (Safety Case) Regulations 1992, HSE publication, HMSO, London, 1992.

3. Ronalds, B.F., Vessel impact design for steel jackets, OTC 6384, 1990.

4. Kenny, J.P., Protection of offshore installations against impact, OTI 88 535, HMSO, London, 1988.

5. AME, Vessel impact on fixed platforms, Final Report Joint Industry Project, HSE number OTN-92-176, Feb. 1992.

6. Marshall, P.W., Gates, W.E. and Anagnostopoulos, S., Inelastic dynamic analysis of tubular structures, OTC 2908, 1977.

7. API-LRFD: Recommended practice for planning, designing and constructing fixed offshore platforms - Load and resistance factor design, 1st Ed., July 1993.

8. Det Norske Veritas, Technical Note Fixed Offshore Installations, TN A 202, Impact loads from boats, May 1981.

9. W. Visser, Dynamic effects resulting from ship impact on fixed steel offshore platforms, prepared for the Health and Safety Executive, Offshore Safety Division, under Warrant 6/3378, July 1996.

10. Interim Guidance Notes for the design and protection of topsides structures against explosion and fire, SCI report no SCI-P-112, Jan. 1992.

11. Visser, W. & Igbenabor, S.C., Resistance of jack-up conductors to boat impact, MaTSU report No MaTR 0479, March 1998.

12. Furnes, O. and Amdahl, J., Ship collision with offshore platforms, Intermaritec '80, 1980.

13. Ellinas, C.P. & Walker, A.C., Damage on offshore tubular bracing members, IABSE Colloquium, Copenhagen, 1983.

14. BSI, Guidance on methods for assessing the acceptability of flaws in fusion welded structures, PD6493, 1991.

15. VanderValk, C.A.C., Factors controlling the static strength of tubular T-joints, BOSS, Trondheim, 1988.

51

16. Stacey, A., Sharp, J.V. and Nichols, N.W., The influence of cracks on the static strength of tubular joints, Proc. 25th OMAE Conf., Vol. 3, p.435, Florence, 1996.

17. The effects of high strain rates on material properties, HSE-OTI 92-602, HMSO, 1992.

18 Symonds, P.S., Behaviour of materials under dynamic loading, Ed. J. Huffington, ASME, New York, p.106-124, 1965.

19. BSI, Fracture mechanics toughness tests, Part 1: Methods for determination of KIc, critical CTOD and critical J-values of metallic materials, BS7448, Pt.1:1991.

20 Visser, W., Potential contradiction in the fracture assessment of steel tubular joints, OMAE ’98 paper 2056, Lisbon 1998.

21 Visser, W., The structural design of offshore jackets, MTD Publ. 94/100, London 1994.

52

APPENDIX A BEAM BENDING ANALYSES

CONTENTS

A1 INTRODUCTION 55

A2 PROPERTIES OF THIN WALLED TUBULARS 55

A3 PRINCIPLE OF VIRTUAL WORK 56

A4 AN ELASTIC BEAM IN THREE POINT BENDING 59

A5 ELASTO-PLASTIC THREE-POINT BENDING PROBLEM 60

A6 THREE-POINT BENDING WITH TENSION - ELASTIC 61

A7 THREE-POINT BENDING WITH TENSION - TUBULAR, PLASTIC TO API 64

A8 THREE-POINT BENDING WITH TENSION - JOINTS, PLASTIC TO THE GN 69

A9 CROSS-BRACING CAPACITY 71

A10 PLASTIC HINGE WITH STRAIN HARDENING 72

53

INTENTIONALLY BLANK

54

APPENDIX A BEAM BENDING ANALYSES

A1 INTRODUCTION

Ship impact on a brace looks like a simple problem and yet there are some serious, fundamental complications. This Appendix will explain some of these complications and its aim is to develop further understanding.

In many vessel impact analyses it is found that a splash-zone brace has of the order of 1.0 MJ impact resistance. As a consequence:

· during the operational impact of 0.5 MJ the brace remains intact; · during the accidental impact of 4.0 MJ the brace may well fail.

In the course of this work four criteria were found to give an indication of the amount of plastic yielding in the plastic hinges in the brace:

· the mean value of the Marshall et al criteria for end rotation · a ductility ratio as used in blast resistance calculations · a maximum strain of 10% · a maximum displacement under impact of 1.0 m.

A2 PROPERTIES OF THIN WALLED TUBULARS

A thin walled tubular is defined by:

t << D A2.1

where t is the wall thickness and D the diameter of the tubular. The moment of inertia (I), the elastic section modulus (S) and the plastic section modulus (Z) are then given by:

I = p

× D3 × t

8 A2.2-1

S = p

× D2 × t Z = D2

× t A2.2-2 4

Using the yield stress (sy) the maximum elastic bending moment and the ultimate, plastic moment of the tubular cross section are:

Mel,max = S s × y Mp = Z s × y A2.3

55

Figure A2.1 Stress distribution in fully plastic tubular in bending

A3 PRINCIPLE OF VIRTUAL WORK

A3.1 Centrally loaded beam

The principle of virtual work as applied to braces subject to vessel impact can best be demonstrated by a beam in bending loaded in the centre (see Figure A3.1).

The first step is to assume a reasonable displacement distribution for the structure under the applied loading. For a centrally loaded beam in bending it is the displacement distribution as shown in Figure A3.1. This displacement distribution is defined by the central deflection u or the end rotation q. The relation between these two parameters is:

× q l u =2 A3.1

The External Work by the force P over the displacement u is:

WE = P × u A3.2

56

l

l

a a a

P

a

A B C D E

MA

MC

ME

P

a a a a

A B C D E

q u

l

l

Figure A3.1 Centrally loaded beam failing in bending

When the effect of the local dent in the centre is ignored, the Internal Work due to plastic moments (Mp) at the ends and at the centre in combination with the rotation q at the ends A and E and 2q in the centre of the beam is:

×WI = M 4 p q × A3.3

Equality of the Internal and External Work results in:

×P × u = P × × q l = M 4 p q × A3.4

2

At failure, by dividing this equation by q, the following failure load Pu is found:

Pu = 8 × Mp A3.5 l

The proof that this value for Pu is the correct failure load is found by assuming other deformation patterns and calculating the load at failure. The minimum value of P is the correct value for Pu. An example of this additional check is found in the assessment of vessel impact at cross bracings where two failure modes are feasible for out-of-centre impact.

For boat impact the energy absorption capacity (E) is of interest, which follows from:

×E = Pu q × max × l / 2 = M 4 p q × max A3.6

57

where qmax is the maximum rotation at the joints before loss of integrity of the joint occurs.

A3.2 Impact out of centre

If the impact force P is applied at point B rather than at point C the assumed deformation pattern will change as indicated in Figure A3.2.

P

a a a a

l

A B C D E

uq1 q2

l

Figure A3.2 Out-of-centre loaded beam failing in bending

Using the same methodology as in Appendix A3.1 the equations for Pu and E will change as follows:

q1 × l u = A3.7

4

The External Work by the force P over the displacement u is:

WE = P × u A3.8

The Internal Work due to plastic moments (Mp) at the ends A and E, and at the point of impact (B), in combination with the rotation q1 at the end A, q2 at end E and q1+q2 at the point of impact is:

WI = 2 × M p × (q1 + q2 ) A3.9

Since q2 = q1/3: WI = 8 × Mp × q1 / 3 A3.10

Equality of the Internal and External Work results in:

P × u = P × q1 × l / 4 = 8 × Mp × q1 / 3 A3.11

At failure, by dividing the above equation by q1, the following failure load Pu is found:

32 × Mp MpPu = = 10. 7 × A3.123× l l

58

/4

l

/4

l

/4

l

For boat impact the energy absorption capacity (E) is of interest, which follows from:

E = Pu × qmax × l / 4 = 2.7 × Mp × qmax A3.13

where qmax is the maximum rotation at joint A before loss of integrity of this joint occurs.

In summary: when the impact occurs out of centre, in this case at a quarter of the span, the impact force is somewhat higher by 33% and the impact energy absorption somewhat lower by 33% as compared to values found for impact in the centre of the beam.

Hence over quite some distance the impact energy is rather insensitive to the exact point of impact.

A4 AN ELASTIC BEAM IN THREE POINT BENDING

The simplified problem can be defined as follows:

· take a long slender tubular in three point bending, with fixed ends, loaded in the centre, while ignoring any axial restraint;

Furthermore the effect of the denting of the tubular due to the applied force is ignored and the tubular is of adequate thickness to avoid local buckling

P

a a a a

A B C D E

MA

MC

ME

P

a a a a

A B C D E

MA

MC

ME

P

a a a a

A B C D E

MA

MC

ME

l

a.a.a. CCCentententrararallllllyyy llloooaaaded beamded beamded beam iiinnn bebebendindindingngng

Q = P/2

a = l

A B

x

wx wB

Q = P/2

a = l

A B

x

wx wB

Q = P/2

a = l

A B

x

wx wB

4/l

bbb... AAA quaquaquartrtrtererer ofofof the cthe cthe ceeennntttrrraaalllllly ly ly loooaaadeddedded beabeabeammm

Figure A4.1 Centrally loaded elastic beam

59

The moment distribution in this model is linearly varying between the end points and the centre (see Figure A4.1a). Also the moment in the centre is identical with, but has opposite sign to, the moments at the end fixities. Hence the beam can be described by considering a quarter of the beam only. The linearly elastic solution for the central deflection (wcentral) of the tubular under three point bending and the maximum moment Mmax are:

P × l3 P × l wcentral = and Mmax = A4.1

192 ×E ×I 8

where l is the length of the tubular. As shown in Eqs. A3.5 & A2.2 the plastic limit load is:

Pu = 8×M p where Mp = D2

× t ×sy A4.2 l

The maximum elastic moment is:

Me = p

4 × D2

×t × sy A4.3

so that the ratio between the elastic and plastic moment is p/4 = 0.785.

A5 ELASTO-PLASTIC THREE-POINT BENDING PROBLEM

Once the maximum moment in the tubular exceeds the maximum elastic moment but is less than the plastic moment the tubular deforms elastically and partly plastically. But the moment at each point of the beam is exactly known. For loads in excess of Pe and less than Pu, or:

Pe < P < Pu A5.1

it can be calculated at which point in the circumference the yield stress is reached.

j

+sy

plastic

elastic

plastic

-sy

Figure A5.1 Stress distribution in partially plastic tubular in bending

By assuming elastic-perfectly plastic material behaviour and a Poisson’s ratio equal to zero, to avoid local end constraints, the relation between the actual moment M and the angle j of the plastic region (see Figure A5.1) can be found:

60

æ 2 ×j - sin(2 ×j) ö M = çç

è 4 ×sin j + cosj÷÷ ×D2

× t ×sy A5.2 ø

The two limiting cases are:

for j = p/2: M = D2 × t ×sy = Mp A5.3-1

for j = 0: M = p

4 × D2

×t × sy = Me A5.3-2

Subsequently, the radius of curvature of the beam can be calculated using the relation:

d2w = 1

× Me A5.4

dx2 cos j I E ×

Hence, as long as the end moment is less than the plastic moment the radius of curvature over the length of the beam can be calculated and subsequently the central deflection.

It has been found that for an end moment of 98% of the plastic moment the central deflection is modestly more than the elastic limit deflection.

A first underlying thought is that, once the end reaches its plastic limit, then a hinge will be formed. This is true. However, at a very small distance away from the plastic limit load the moment is somewhat less, say 98% of Mp. Hence the length of the plastic hinge is very small, and in the limit, as can be demonstrated the length of the plastic hinge is infinitely small and that the strains in the hinge are infinitely large: in mathematical terms this is called a singularity in the solution (see also the first paragraph of Appendix A10).

Thus, in order to form a plastic hinge, an infinite strain in that section is required and this would violate the third condition in Section A1 of limiting strain.

A6 THREE-POINT BENDING WITH TENSION - ELASTIC

Ship impact on a brace looks like a simple problem and yet there are some serious, fundamental complications. Some of the simple equations required for the further analysis will be put together in this note. Ship impact is, as a first approximation, considered as a 3-point bending problem, unless otherwise stated. The first additional complication is to consider the elastic problem of combined bending and tension.

61

P

a a a a

A B C D E

MA

MC

ME

TT

Q = P/2

a =

A B

T

x

wx wB

P

a a a a

A B C D E

MA

MC

ME

TT

Q = P/2

a =

A B

T

x

wx wB

l

4/l

Figure A6.1 Elastic beam in bending and tension

A three point bending of a uniform beam (AE) can be treated as a beam (AB) clamped on one end and with a point load on the other end. The length (a) and the shear force (Q) are related to the length and the force on the beam as a whole by:

l P a = Q = A6.1

4 2

The differential equation for the (elastic) deflections are:

2 wdE × I × dx2 - T × w = Q × (a-x) - T × w A6.2B

The general solution is:

Q×(a - x)w = A1 cosh lx + A2 sinh lx -

T + w B A6.3

T l2while: = A6.4

E ×I

The three boundary conditions are:

62

for x = 0: w = 0 A6.5-1

for x = 0: dw dx

= 0 A6.5-2

for x = a d2w dx 2

= 0 A6.5-3

The boundary conditions lead to the following equations:

for x = 0 and w = 0: A 1 - T aQ × 0w B =+ A6.6-1

for x = 0 and dx dw

= 0: 0T DA2 =+× l A6.6-2

for x = a and d 2w dx 2

= 0: A1 + A 2 tanh la = 0 A6.6-3

Addition of the first and second equation (the latter multiplied by a) results in:

A1 + (la)× A2 + wB = 0 A6.7

Subtraction of this equation from the third boundary condition (Eq. A6.6-3) leads to:

A 2 = -w B

la - tanh la A6.8

The boundary conditions translate into the following three relations for A1, A2 and wB:

A1 -Q × a

T + wB = 0 A6.9-1

l × A 2 = -Q T

A6.9-2

l 2 × cosh la + l 2

× sinh la = 0 A6.9-3

63

distance along the beam

-1.00

-0.80

-0.60

-0.40

-0.20

0.00 0.00 1.00 2.00 3.00 4.00

pure bending bending/tension

Figure A6.2 Deflection for a centrally loaded beam in bending and tension

The final solution is:

QA2 = - A1 = - A2 tanh la A6.10-1

l T

T wB =

Q ×a - A1 l

2 = A6.10-2 T E ×I

Two typical solutions, one for low tension and one for high tension, normalised for w C = -1 are given in Figure A6.2 at the previous page. The conclusion from this figure is that for high tension the beam remains (almost) straight over long sections while bending is confined to small end-sections to comply with the boundary conditions.

A7 THREE-POINT BENDING WITH TENSION - TUBULAR, PLASTIC TO API

Ship impact is, as a first approximation, considered as a 3-point bending problem, unless otherwise stated. An additional complication is to consider the plastic problem of combined bending and tension.

64

l

l

P

a a a a

A B C D E

MA

MC

ME

TT

P

a a a a

A B C D E

TT q

l

l

Figure A7.1 Centrally loaded beam failing in bending and tension

For a plastic hinge, under combined tension and bending, the following equation holds:

M æ p T ö

- cos ç × ÷ = 0 A7.1M ç

è 2 T ÷ p p ø

where: Mp = D2 × t ×sy and Tp = p × D× t ×sy A7.2

This equation, Eq. A7.1, can also be called the plastic yield surface for a tubular under combined tension and bending. It can be derived either from the plastic stress distribution in Figure A7.2 or by rearranging Eq. D.3.1-1 in API7; therefore it will be called the plastic yield surface to API.

Once plastic hinges have formed the transverse resistance P as a function of the end rotation q

is:

8×MP = 2× T×q + A7.3

l

A combination of M and T will be sought which optimises the value of P. In other words P will be maximised under the constraint for a tubular under plastic deformation given above.

65

j

+sy

-sy plastic -

plastic ­tension

Figure A7.2 Stress distribution in fully plastic tubular in bending and tension

The two extreme cases in Eq. A7.3 are:

(1) q ≈ 0 A7.4-1

(2) q >> 8×Mp A7.4-2

2× Tp ×l

In case (1), for small q, the load P in Eq. A7.3 is taken by bending only because the tension does not contribute. In the second case (2), for large values of q, the load P in Eq. A7.3 is taken by tension only; bending is only local and required to ensure that the beam complies with the boundary conditions.

Using the above equations the force P can be expressed in a single variable, the parameter T. Subsequently P can be optimised for T.

æ T öp çFrom A7.1: M = Mp ×cos ç 2

× T ÷

÷

è p ø

8×Mp æ p T ö

Substitution into A7.3: P = 2×T ×q + × cosç × ÷ A7.5 l ç

è 2 T ÷ p ø

Subsequently P can be optimised by calculating:

dP = 0

dT

8 × Mp p / 2 æ p T ö or: 2 -q × × × sin ç × ÷ = 0 A7.6 l Tp

çè 2 Tp

÷ø

The minimum value for T to carry the load P is:

66

2 æ Tp × l ×q÷

ö T = Tp × × arcsin

ç ç

2 ×p ×M ÷ ø

A7.7 p è p

Substitution of the equations for Mp and Tp in the yield surface equation results in:

æ l ö2T = Tp × × arcsinç ×q÷÷ A7.8ç

p è 2 ×D ø

and the corresponding value of M is found from the original expression for the yield surface (Eq. A7.1).

The absorption energy is calculated by the equation:

q

ó lE = ô P × × dq A7.9

õ 2 0

A graph for P as a function of q is given below. Figure A7.3a is normalised to the value of P for small values of q, i.e. normalised to P = 8Mp/l (see Eq. A7.3). The internal moment and tension in Figure A7.3b are normalised using the full plastic moment capacity (Mp) and tension capacity (Tp) in accordance with Eq. A7.2.

67

i ly

i ly

0.00

0.50

1.00

1.50

2.00

2.50

tens on on

bend ng on

0.00 0.50 1.00 1.50 2.00 2.50 3.00

lq/D a. Failure load in plastic tubular under combined bending and tension

1.40

1.20

1.00

0.80

0.60

0.40

0.20

0.00

pM/Mp T/T

0.00 0.50 1.00 1.50 2.00 2.50 3.00

lq/D b. T/Tp and M/Mp as functions of the deflection q

Figure A7.3 Resistance of a centrally loaded beam in bending and tension

From Fig. A7.3a the two limiting conditions in the impact force become apparent:

q× l(1) if < 1. 0 : the contribution by tension can be disregarded

D

q× l(2) if ³ 2.0 : there is no contribution by bending.

D

The total lateral resistance by bending combined with tension is larger than the resistance by plastic bending only, but this combination is only significantly different for large rotations. Therefore, as a first approximation, the contribution by tension is ignored, provided condition (1) above is (approximately) satisfied.

68

A8 THREE-POINT BENDING WITH TENSION - JOINTS, PLASTIC TO THE GN

Joints under combined tension and out of plane bending follow a different yield surface as compared with tubulars as reflected by the equations given below.

For a plastic hinge in a tubular, under combined tension and bending, the following equation holds for tubulars:

M æ p T ö

- cos ç × ÷ = 0 A8.1M ç

è 2 T ÷ p p ø

where: Mp = D2 × t ×sy and Tp = p ×D ×t ×sy A8.2

are the plastic properties in bending and tension of the tubular.

For joints the Guidance Notes (GN) equation (Ref. 1, Sect. A21.2.4.i) applies with a linear relation between M/Mu and T/Tu or:

M T + = 1.0 A8.3

M Tu u

where Mu and Tu are the ultimate capacity in bending and tension of the end joints.

Let’s assume that the tubular cross-section in the centre has the same property as the end joints and use the same failure surface. Once plastic hinges have formed, and in accordance with Figure A7.1 and Eq. A7.3, the transverse resistance P as a function of the end rotation q is:

8MP = 2× T×q + A8.4

l

A combination of M and T will be sought which optimises the value of P; in other words P will be maximised under the constraint A8.3 for joints.

The two extreme cases are:

(1) q ≈ 0 A8.5-1

4 ×M(2) q >> u A8.5-2

Tu ×l

In case (1) the load is taken by bending only because the tension does not contribute. In the second case (2) the shear load P is taken by tension only; bending is only local and required to ensure that the beam complies with the boundary conditions.

Using the constraint equation A8.3 the moment M can be eliminated from A8.4 so that P is expressed in one single parameter T:

æ T ö M = Mu ×çç1 - ÷ A8.6

è T u ø ÷

69

æ T ö8×Mu

Tand: P = 2×T ×q +

l ×èçç1 - ÷÷ A8.7

u ø

Subsequently P can be optimised for T by calculating:

dP = 0

dT

8×M uor: 2× q - = 0 A8.8 l ×T u

This leads to a unique solution for q:

4 ×Mu qcritical = A8.9

l× Tu

For q < qcritical the transverse load is taken by bending whereas for q > qcritical the transverse load is taken by tension. A graph for P as a function of q is given below. The force P in Figure A8.1 is normalised to the value of P for small values of q, i.e. normalised to P = 8Mp/l (see Eq. A8.4).

i ly

q = 4 × M u

l×T u

0.00

0.50

1.00

1.50

2.00

2.50

tens on on

bending only

crit .

0.00 0.50 1.00 1.50 2.00 2.50

q q crit.

Figure A8.1 Plastic tubular in combined bending and tension (joint failure)

For a tubular the relations for Mp and Tp can be expressed in D, t and sy. If the joints are as strong as the tubular both under pure bending and pure tension, or:

Mu = Mp and Tu = T p A8.10

4 ×Mp 4× D2 ×t × sy 4× D

it would then lead to: qcritical = = = A8.11 l× T l× p ×D ×t ×s p ×lp y

70

and for l / D = 25: qcritical = 0.05 A8.12

A9 CROSS-BRACING CAPACITY

Ship impact at a cross bracing can also be addressed using the principle of virtual work.

P

u

2/l

2/l

2/l

2/l

Figure A9.1 Impact in the centre of the cross brace

For impact in the centre (Figure A9.1) the capacity is double the capacity of a single brace:

16 × M por: PC = A9.1 l

For an impact at a distance ‘b’ from the centre, using the deformation modes as sketched in Figure A9.2, the impact force can be calculated using the principle of virtual work as explained in Appendix A3. From the equality of the Internal and External Work (see Eqs. A3.2 and A3.3) the values for Pb are found as:

2 × M 2 × Mp p1. for the local deformation mode (mode 1): Pb1 = + A9.2-1 b l / 2 - b

l / 2 16 × Mp2. for the global deformation mode (mode 2): Pb2 = × A9.2-2 l / 2 - b l

P

u

b

P

u

b

deformation mode 1 deformation mode 2

Figure A9.2 Impact out of centre of the cross brace

71

l

4 16 × M p 21 × MpAnd for b = l / 8: Pb1 = Pb2 = 3 × = A9.3

l l

This is sketched in the Figure A9.2. More specifically:

for b > l / 8 the centre will not deform and there is no plastic deformation in the other braces.

For impact in the middle of a span (Figure A9.2, b = l / 4 ) the impact force is:

4 × M 16 × Mp pP = = = PC A9.4 l / 4 l

When changing the value for ‘b’ there is a smooth change in impact force. However, due to the unchanging value of the maximum rotation at the joints, there is a rapid change in impact energy capacity when changing from the global deformation mode to the local deformationmode. More specifically:

· the allowable deflection (as governed by the maximum rotation) for b = l / 8 in thelocal mode is only 25% of the allowable deflection for b = 0.

· hence, the impact energy absorbing capacity for b = l / 8 shows a rapid jump:

Eb for b = l / 8 in the local mode = 44% of Eb for b = 0

Eb for b = l / 8 in the global mode = 100% of Eb for b = 0

Due to strain hardening this jump will be less abrupt in reality, but it should be recognised in any numerical analysis.

A10 PLASTIC HINGE WITH STRAIN HARDENING

It was noted elsewhere that the elastic-ideally-plastic material model leads to a singularity in the solution for the plastic hinge (see the last two paragraphs of Appendix A5). This appendix deals with the effect of strain hardening in alleviating this singularity.

P

a a a a

A B C D E

MA

MC

ME

l

Figure A10.1 Beam in three point bending

For the three point bending test the three moments MA, MC and ME are the same.

72

In order to simplify the process let us assume that all the material is in the outer fibre only. Therefore let’s replace the tubular by a square of height D, where the thickness of the vertical sides is taken equal to zero (t = 0) and the thickness of the top and bottom sides are taken equal to the thickness of the tubular (t). The plastic section modulus of the tubular of diameter D and this square of height D are the same: D 2

× t × sy .

The stress-strain curve reflecting strain hardening is shown in Figure A10.3.

D

t t

D

Figure A10.2 Equivalent plastic bending properties

sy

su

eall

Figure A10.3 Idealised stress-strain curve with strain hardening

In this model the elastic strains are disregarded. This seems acceptable, because eel. << eall. More specifically, eel. is approximately 0.15% and eall is approximately 10%.

73

P

a a a a

A B C D E

MA

MC

ME

Q = P

a =

A B b

wB = wC/2

plastic part elastic part

P

a a a a

A B C D E

MA

MC

ME

Q = P

a =

A Bb

wB = wC/2

plastic part elastic part

l

4/l

/2/2

Figure A10.4 Elasto-plastic beam with strain hardening

Over the length a in Figure A10.4 the major part acts elastically and a small part (b) plastically. If the end has a strain equal to eall corresponding with su in Figure A10.3 then the plastic part can be calculated from the linear moment distribution in the beam of length a. The length of this plastic part is:

b = su - sy

×a A10.1 s u

In this part the moment is known and also the stress and hence the strain in the outer fibre.

Or: for x = 0: s = su and e = eall A10.2-1

for x = b: s = sy and e = 0 A10.2-2

(b - x)×su + x× sy (b -and in between: s = e = x)×eall A10.2-3b b

The strain in the outer fibre is related to the change of curvature as follows:

D d2w e = A10.3

2 ×

dx2

Using these expressions the slope at x = b for the beam clamped at x = 0 can be calculated:

74

q = d2 w dx2

b ó õô dx =

2 D

× (b - x)×eall

b

b ó õ dx =

b ×eall D

A10.4 0 0

Substitution of the expression for b results into:

su - s yq = eall

× ×a A10.5D su

For: eall = 0.10; su = 1.20 sy; a/D = 6: q = 0.10

and the central deflection (governed in this case by the plastic hinge) is approximately:

wcentre ≈ 2 ×a × q = 0.20 a A10.6

For this example the deflection in the centre at first yield, using Eqs. A4.1 for a tubular of diameter D under three point bending is:

=

4 × s × a2

= 4 e × y × a × a = a 036.0 wcentre A10.7E × D D

Hence the plastic rotation, under limited plastic strain, is significantly larger (by a factor 6) than the deformation at first yield.

75

Printed and published by the Health and Safety ExecutiveC30 1/98

Printed and published by the Health and Safety Executive C0.06 05/04

ISBN 0-7176-2838-8

RR 220

78071 7 628384£15.00 9

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