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Study and development of FEM-models used in expansion analyses of pipelines

Cristina Lindholm

Master of Scince Thesis Stockholm 2007

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Study and development of FEM-models used in expansion analyses of pipelines

by

Cristina Lindholm

Master of Science Thesis MMK 2007:17 MME 794 KTH Machine element

SE-100 44 STOCKHOLM

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Master of science thesis MMK 2007:17 MME 794

Study and development of FEM‐models used in expansion analyses of pipelines

Cristina Lindholm

Approved

2007‐03‐01 Examiner

Sören Andersson Supervisor

Ulf Sellgren Commissioner

Tore Søreide,REINERTSEN AS Contact person

Sigurd Trier

Abstract REINERTSEN AS performs expansion and buckling analyses of pipelines using ANSYS, a finite element modelling and analysis tool. In an expansion analysis the pipeline is modelled with thin‐walled pipe elements called PIPE20 which allows plastic deformation. However, in a recent analysis the results retrieved from the PIPE20 element were incorrect for strains about 1 percent. One main issue was that the PIPE20 element overestimated the strains.

The purpose with this Master Thesis was to become familiar with the theories and analysis methods used in pipeline design. By using the knowledge gained a short examination of the weakness in the PIPE20 element was carried out and a first development of a new model for expansion analysis in ANSYS was made.

First, a detailed initial study was made. The analytically derived functions were examined and a model was created in ANSYS using PIPE20 elements with linear‐elastic material model. This model was, when compared to the analytically expected results, very accurate and could thereby be used in further analysis.

A short study of the PIPE20 element was carried out for a linear‐elastic perfectly‐plastic material model. The results from PIPE20 were compared to analytically derived results and to results from Pipeline Analysis System (PAS) a 2D‐analysis tool developed at REINERTSEN. It was seen that results from calculations involving the PIPE20 element differs from the expected results. When the rotation is small and the cross‐section is almost completely elastic the expected value and PIPE20 element output is practically identical but with increasing rotation and plasticity the accuracy is lessen this due to the few number of integration points. When more than 23 percent of the cross section is plastic the element output is utterly incorrect. The conclusion is that the PIPE20 element is not suitable in applications where plastic deformation of a major part of the cross‐section is expected.

One solution is to create a model with a beam‐shell assembly. The PIPE20 elements in the model where inaccuracies have been found can be replaced by shell elements. A first development of such a model was made. The beam‐shell assembly was exposed to the same loads as a model with only PIPE20 elements. The strains where compared and was found to be smaller in the beam‐shell assembly than in the PIPE20 model, indicating that a model with a beam‐shell assembly can be used. Even though more development and testing are needed the first results are satisfying.

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Examensarbete MMK 2007:17 MME 794

FEM‐modeller för expansionsanalys av piplines

Cristina Lindholm Approved

2007‐03‐01 Examinator

Sören Andersson Handledare

Ulf Sellgren Uppdragsgivare

Tore Søreide,REINERTSEN AS Kontaktperson

Sigurd Trier

Sammanfattning REINERTSEN AS utför expansions‐ och bucklingsanalyser på pipelines genom att använda ANSYS. I dessa analyser modelleras rören med hjälp av ett tunnväggigt rörelement kallat PIPE20. Detta element tillåter plastisk deformation. I en analys nyligen utförd på REINERTSEN med töjningar på ca 1 till 1.5 procent var resultaten från PIPE20elementet felaktiga i jämförelse med resultat från andra applikationer. Ett huvudproblem var att PIPE20elemntet överskattar töjningarna.

Syftet med detta examensarbete är att sätta sig in i teorier och analysmetoder som ligger bakom konstruktionsarbetet för pipelines. Genom utnyttjande av den kunskap som erhållits gjordes först en kort undersökning av PIPE20elementets brister. Efter detta gjordes en första utveckling en ny expansionsmodell. En noggrann förstudie utfördes, de analytiskt härledda formlerna studerades och en modell med användande av PIPE20 element gjordes där en linjär‐elastisk material modell nyttjades. PIPE20modellen var, i jämförelse med de teoretiskt förväntade värdena, mycket noggrann och kunde därför användas i fortsatt analys.

En liten modell bestående av ett PIPE20element med plastiska egenskaper undersöktes. Förväntat moment vid olika rotationsförskjutningar jämfördes med utdata från PIPE20. Det fastslogs att utdata från PIPE20elementet skilde sig från det förväntade värdet av två anledningar. När rotationen är så liten att nästan hela tvärsnittarean är elastisk är det förväntade värdet och PIPE20modellens utdata så gott som identiska men med ökad rotationslast så minskar noggrannheten. Detta på grund av de få integrationspunkterna definierade i PIPE20elementet. Ökas rotationen ytterligare så mer än 23 procent av tvärsnittsarean är plastiskt deformerat är utdata från PIPE20elementet helt felaktiga. Slutsatsen är att PIPE20 inte är att rekommendera i analyser där en stor del av tvärsnittsarean förväntas bli plastisk deformerat.

Ett förslag till lösning är att skapa en ny modell där delar av modellen består av skalelement. Genom att byta ut de delar av modellen där problem med noggrannhet uppstår och ersätta dessa med skalelement kan en så kallad ”beam‐shell assembly” vara en förnuftigare modell. Ett första försök till utveckling av en sådan modell har gjorts. Den nya modellen och en modell bestående av enbart PIPE20element utsattes för samma analys och töjningar jämfördes. Det visade sig att töjningar i beam‐shell modellen var lägre än de funna i modellen med endast PIPE20element. Detta är en indikation på att en modell med både PIPE20element och skalelement kan användas i expansionsanalyser. Det är nödvändigt med vidare utveckling och testning av modellen, men de första resultaten är tillfredsställande.

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Table of Contents Abstract.................................................................................................................i Sammanfattning ............................................................................................... ii Table of Contents ............................................................................................ iii Table of Symbols and Abbreviations ...........................................................v 1 Introduction ................................................................................................. 1 2 Theoretical models ..................................................................................... 2 2.1 Effective axial force and submerged weight..................................................... 4 2.2 How the pipeline will buckle .............................................................................. 6 2.2.1 Vertical buckling effects ................................................................................. 6 2.2.2 Lateral buckling effects................................................................................... 8 2.2.3 Buckling behaviour on trigger berm‐Horizontally straight pipeline .... 10 2.2.4 Buckling behaviour on trigger‐berm ‐Horizontally curved pipeline .... 11 2.2.5 Conclusion...................................................................................................... 13

2.3 Feed‐in‐length and maximum allowable moment......................................... 14 2.3.1 Feed in length and strain.............................................................................. 14 2.3.2 Feed in length and deflection ...................................................................... 14 2.3.3 Combining feed in length in the case of lateral buckling, even seabed 15

2.4 How the axial effective force is built up over time........................................ 16 3 Verification of analytical model using ANSYS ................................. 18 3.1 The model .............................................................................................................. 18 3.2 The ANSYS analysis............................................................................................ 20 3.2.1 Vertical buckling............................................................................................ 21 3.2.2 Lateral buckling on vertical trigger‐berm.................................................. 22 3.2.3 Lateral buckling on even seabed................................................................. 22 3.2.4 The parameters and variables in the cases ................................................ 24

3.3 Results .................................................................................................................... 24 3.3.1 Vertical buckling............................................................................................ 24 3.3.2 Lateral buckling on vertical trigger‐berm.................................................. 27 3.3.3 Perfect straight pipe on perfect even seabed............................................. 29 3.3.4 Lateral buckling on even seabed................................................................. 30

3.4 Conclusion ............................................................................................................. 32 4 Plasticity, Ansys and PAS....................................................................... 33 4.1 Model of short plastic beam............................................................................... 33 4.1.1 Theory and analytical model ....................................................................... 34 4.1.2 PIPE20 and PAS model................................................................................. 36 4.1.3 Loading PIPE20 with moment..................................................................... 39

4.2 The importance of integration points‐a short study...................................... 39

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4.2.1 Analytical vs. numerical............................................................................... 39 4.2.2 Analytical vs. PAS ......................................................................................... 40

4.3 Conslusion ............................................................................................................. 41 5 Development and verification of a new model.................................. 42 5.1 Building a shell..................................................................................................... 43 5.2 Modelling a beam‐shell assembly .................................................................... 43 5.2.1 MPC................................................................................................................. 44 5.2.2 BEAM4 ............................................................................................................ 44 5.2.3 Study of the connections .............................................................................. 45

5.3 Modelling the Seabed and the contact ............................................................. 46 5.3.1 Forces due to pressure‐pipeline initially created on flat seabed ............ 46

5.4 Modelling Link elements ................................................................................... 48 5.4.1 Lowering the pipeline................................................................................... 48 5.4.2 Establishing contact with an un‐even seabed ........................................... 49 5.4.3 Forces due to pressure‐pipeline laid down flat seabed ........................... 49

5.5 Buckling of pipeline lying on trigger‐berm.................................................... 50 5.5.1 Vertical buckling............................................................................................ 50 5.5.2 Lateral buckling ............................................................................................. 52

5.6 Conclusion of the beam‐shell assembly and further development ........... 52 6 References .................................................................................................. 54 7 Appendix.......................................................................................................i 7.1 Appendix. Vertical buckling on even seabed....................................................i 7.2 Appendix. Lateral buckling on even seabed................................................... iii 7.3 Appendix. Combined buckling..........................................................................vi 7.4 Appendix. Parameters and variables............................................................... vii 7.5 Appendix. ANSYS input files for elastic PIPE20.........................................viii 7.6 Appendix. Graphs of vertical and lateral deflection..................................... xx 7.7 Appendix. Moments in partly plastic cross‐section. ..................................xxiv 7.8 Appendix. Input files for short beam in ANSYS and PAS ........................xxv 7.9 Appendix. Matlab input file for study of integration points ................xxviii 7.10 Appendix. Input file for beam‐shell assembly. MPC‐BEAM4.................xxix

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Table of Symbols and Abbreviations Latin symbols Index A Area 0 Initial value C Length between supports 1 Part one/principle 1 D Diameter 2 Part two/principle 2 E Young allow Allowable F Force ansys Value from ANSYS output g Gravity As‐laid Value as‐laid I Moment of inertia buckle Of buckle k Soil stiffness c Coating L Length calc Calculated value M Moment e Elastic N Axial force e External n Force per meter length eff Effective p Pressure feed‐in Feed‐in q Weigh per meter i Internal R Radius lat Lateral r Mean radius of pipeline lay From lay phase T Temperature lift When lift‐off t Thickness oper In operation u Elongation p Plastic V Shear force press Value when exposed to pressure v Lateral displacement p+T Pressure and temperature w Vertical displacement s Steal soil Soil Greek symbols sub Submerged α Heat coefficient true True δ Deflection vert Vertical ε Strain w Water θ Angel x Axial µ Coefficient of friction y Yield υ Poisons ratio ρ Density σ Stress

DNV Det norske veritas MPC Multi points constraints PAS Pipeline Analysis System

Design and development of FEM‐models used in expansion analyses Master Thesis MMK 2007 Cristina Lindholm

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1 Introduction REINERTSEN AS is a Norwegian main contractor supplying multidiscipline process facilities to the oil and gas industry offshore and onshore. They take on full scale analysis and design of pipelines based on the offshore standard developed by Det Norske Veritas (DNV) [2]. Submerged steel pipelines are used for transporting oil and gas from oil fields at the sea bottom to land or a platform. The difference in temperature and pressure in the hot fluids inside the pipe and the surrounding sea water will lead to axial expansion, rendering an elongation of the pipeline. If the expansion is restrained an axial compression force will develop in the pipeline. A large amount of inner forces in a pipeline is unwanted and numerous analyses have been made on how to lower the large forces in the pipeline. Due to the friction between the pipeline and the seabed, the large axial forces and moments that are built up in the pipe will cause deflection. This is called global buckling. It has been observed that when the pipeline buckles the effective axial force in the pipeline is lowered. This has lead to a discipline called designing for buckling. In order to do this accurate expansion and buckling analyses has to be performed. REINERTSEN AS performs expansion and buckling analyses on pipelines using ANSYS, a finite element modelling and analysis tool. In an expansion analysis the pipeline is modelled with thin‐walled pipe elements called PIPE20 which allows plastic deformation. However in a recent analysis the results received from PIPE20 has been proven incorrect for strains about 1 percent when compared to results from other applications. One issue is that PIPE20 overestimates the strains. This master thesis is divided into four chapters

• First the theoretical models used are derived. • Second a model using linear‐elastic material model for PIPE20 elements is developed

in ANSYS and verified against the theories. • Third the weakness found when using PIPE20 is investigate • Forth and last a new model is developed. By using a beam‐shell assembly, the PIPE20

elements where errors have been found can be replaced by shell elements. And by this create a higher accuracy in the model.


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2 Theoretical models Figure 1 demonstrates a pipeline at seabed with the coordinate system used in this thesis.

wz,

ux,

vy,

wz,

ux,

vy,

wz,

ux,

vy,

Figure 1. Demonstrating a pipeline at seabed with the coordinate system used in the thesis.

To be able to predict the behaviour of the pipeline during operation an analytical tool is needed. The construction of this tool is divided into three steps as described below.

• 1) Description of the forces acting on a small section of a pipeline

o This will give the submerged vertical weight subq and the effective axial force

effN . Both necessary tools to describe and understand the actions of pipelines.

o Due to pipe/soil interaction a pipeline is restrained. This will cause an unpredicted expansion and buckling of the pipeline. By designing for buckling [3]

The expansion problem is solved The effective axial force can be limited One can prevent buckling in unfavourable regions

• 2) Description of how a pipeline has a tendency to buckle. This is divided into studies

of vertical, lateral and combined buckling behaviour. o Study of vertical buckling on even seabed

This gives expressions of lift off length, buckleL⋅2 , maximum deflection 0vertδ and the needed axial force in the pipeline for lift‐off from

seabed, lifteffN .

o Study of lateral buckling on even seabed with uniform friction. There are several lateral buckling modes depending upon a number of

factors for example imperfections of the pipeline and the pipe/soil interaction. Here the first and second mode will be described.

Expressions for describing the length of the deformed pipeline, deflection at centre 0

vertδ and the needed force to initiate lateral sliding lateffN are presented.


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It is found that lifteff

lateff NN < on even seabed. Indicating that if there are

no imperfections on the pipeline lateral sliding will occur before vertical buckling.

o Buckling on trigger‐berm It is observed in the vertical case that the necessary effective lift off

force is inversely proportional to the initial deflection at the centre. To be able to control the vertical buckling imperfections has to be built into the system. For example using trigger‐berms.

Buckling behaviour on trigger‐berm/ un‐even seabed with a horizontally straight pipeline.

• This shows that the needed effective lateral buckling force is about 50% of the effective lift‐off force. Leading to the conclusion that lateral buckling will occur as soon as the pipeline lifts off the ground/trigger‐berm. This effect causes the pipeline to snap‐through. By building in lateral as well as vertical imperfections this un‐wanted behaviour can be controlled.

Buckling behaviour on trigger‐berm/un‐even seabed of pipeline with initial lateral imperfections.

• A pipeline with both vertical and lateral imperfections lowers the necessary effective axial force to initiate buckling.

• This gives a more controlled deflection development which is desired. Where the elongation of the pipeline is a slow process.

• 3) Description of how much a pipeline can be allowed to buckle.

o The buckling can be controlled by building in imperfections. But with continuant buckling the pipeline can be damaged. This can be controlled by limiting the elongation.

o The limit is the maximum allowable bending moment allowM in the pipeline which is a second order function of the deflection.

o The total elongation is the feed‐in length feedu

i) The feed in length is equal to the axial expansion ( buckleL⋅∆ε ) of the pipe.

• The change in strain ε∆ can be found by using Hooke’s law. ii) The feed‐in length can also be described as a second order function

of the deflection. o Using the second connection for feedu it is found that the allowable feed‐in

displacement is highly dependent on the allowable moment limit allowM . o By combining i) and ii) the allowable pipe distance between axially fixed

points allowC can be determined and the elongation is controlled.


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2.1 Effective axial force and submerged weight The understanding of the effective axial force effN and its effects is fundamental since effN

dominates how steel pipelines responses to loading. To illustrate the concept first consider a small section of a pipeline, Figure 2. The forces acting on the pipeline being

• Internal pressure, ip

• External pressure, ep

• Weight of steel pipe ss Aρ

ipep

trueN

gA ssρ

ipep

trueN

gA ssρ Figure 2. Forces acting on a submerged pipeline section in equilibrium

This introduces trueN , the true axial force acting on the steel pipe found by integrating the steel stress over the steel cross‐section area [4]. effN and trueN can cause but the difference is

fundamental, trueN is the force acting only on the steel pipe. effN depends of all axial forces,

this is further explained in the following. If the section is assumed to be a closed surface the effects of the external pressure can be understood using the law of Archimedes [4]

“The effect of the water pressure on a submerged body is an upward directed force equal in size of the weight of the water displaced by the body”

This means that the external pressure ep can be replaced by a vertical weight and axial force as displayed in Figure 3.

ep

ep

gAewρ

ee Ap

ep

ep

gAewρ

ee Ap

Figure 3. External forces acting on a pipeline section in equilibrium. The top pipe section is the real forces.

This is equal to the two pipes in the middle. The external pressure acting on the closed pipe section (midright) can be described as the bottom pipe section. This means that the only parts needed to describe

external pressure of a pipe section is the forces at the pipe ends and the buoyancy

A similar assumption can be made for the internal pressure ip replacing it with the weight of the content acting downward integrated over the surface and an axial force, Figure 4.


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ip

ip

gAiiρ

ii Ap

ip

ip

gAiiρ

ii Ap

Figure 4. Internal forces acting on a pipeline section in equilibrium. The only parts needed to describe

internal pressure of a pipe section are the forces at the pipe ends and the weight of the contents

These assumptions give an equivalent system to Figure 2.

trueN

ee Ap

gAewρ

gAiiρ

ii ApeffN

gAssρ

subq

trueN

ee Ap

gAewρ

gAiiρ

ii ApeffN

gAssρ

subq Figure 5. Resulting forces acting on a small section of a pipeline

The resultant of the vertical weights is the submerged weight: gAgAgAq ewiisssub ρρρ −+= [N/m] (1)

In case of coating an extra parameter gAq cccoat ρ= adds to the weight. The axial forces resultant is the effective axial force

iieetrueeff ApApNN −+= [N] (2)

When 0<effN a pipeline locked axially is in compression which could cause buckling. By

observing (2) it is understood that external pressure stabilizes the buckling effect while internal pressure destabilizes. An unrestrained pipeline is free to move axially. In that case the effective axial force is zero. This can be understood if the unrestrained pipe is seen with an end cap as in Figure 6. The forces acting on the end cap being internal and external pressure. This gives the true axial force as


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eeiitrue ApApN −= and the effective axial force

0=−+−= iieeeeiieff ApApApApN

ip

ep

trueN

End cap

ip

ep

trueN

End cap

Figure 6. The forces acting on an unrestrained pipeline can be found if the pipeline is seen with

end caps. The true axial force is given only by the internal and external pressure.

Consider a pipeline lying on the seabed free to move axially. If there were no soil/pipe‐interaction buckling would not be a problem. However, this is not the case since friction act as virtual anchors building up the effective axial force in the pipeline, Figure 7. If a pipeline is allowed to buckle, the development of effective force is modified as pipe feeds in to the buckle. The force in the buckle drops as the buckle develops [5], Figure 7.

x

effN

Restrained pipeline

Unrestrainedpipeline

Virtual anchorStraight pipeline

x

effNVirtual anchor

Buckle

x

effN

Restrained pipeline


Virtual anchor

x

effN

Restrained pipeline


Virtual anchorStraight pipeline

x

effNVirtual anchor

Buckle

Straight pipeline

x

effNVirtual anchor

Buckle

x

effNVirtual anchor

Buckle

Figure 7. The soil/pipe interaction creates virtual anchors which build up the effective axial force. This is the force that will make the pipeline deflect. With increasing buckling the effective axial force decreases.

This means that designing the pipeline to buckle solves the expansion problem and is a way to limiting the effective axial force. And since buckling will appear due to soil/pipe interaction built‐in buckling will prevent buckling in unfavourable areas.

2.2 How the pipeline will buckle It is concluded that one way to lower the effective axial force in the pipeline during operation is allowing it to deflect. To be able to control the buckling of the pipeline it is essential to understand how the pipeline has a tendency to move.

2.2.1 Vertical buckling effects The most basic case of buckling, vertical buckling of straight pipeline on idealized even hard seabed is displayed in Figure 8. The submerged weight is acting downwards on every part of the pipeline. The effective axial force is negative in compression.


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LL

]/[ mNqsub

x

w

0

0

0

=

=

=

Mdxdww

LL

]/[ mNqsub

x

w

0

0

0

=

=

=

Mdxdww

Figure 8. Vertical buckling behaviour on even, hard seabed

An infinite small section of the pipeline can be displayed as in Figure 9.

Figure 9. A small section of a pipeline.

Equilibrium from Figure 9 gives the differential equation for vertical buckling on even seabed

subeff qdx

wdNdx

wdEI −=+ 2

2

4

4

With the general solution [6] (symmetry gives even function) 2

2cos)( x

Nq

kxBAxweff

sub ⋅−+=

EIN

k eff=

The boundary condition 0)(,0)( =′= LwLw gives

⎟⎟⎠

⎞⎜⎜⎝

⎛−−+=

2sincos

sincos

2)(

22 xkLkx

kL

kLkL

kLL

Nq

xweff

sub

The third boundary condition 0)()( =′′= LwEILM gives 5.4)tan( ≈⇒= kLkLkL

The expression now reading

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⋅+=

2cos6.41.11)(

2xxEI

NNEI

NEI

Nq

xw eff

effeffeff

sub (3)

The buckling length is inversely proportional to the effective force

effNEIL 5.4= (4)

M

dMM +

effN

effeff dNN +

VdVV +

w′

dx

subq

dxww ′′+′


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Maximum deflection vert,0δ at 0=x , and lift‐off effective force offlifteffN − can now be expressed

by subq , effN and EI

020 96.3,7.15)0(

vert

subofflifteff

eff

subvert

EIqN

NEIq

wδ

δ === − (5) (6)

More details are given in appendix 7.1. lifteffN is negative when the pipe buckles. The true axial force when the pipe deflects then is

⎟⎟⎠

⎞⎜⎜⎝

⎛+−= 096.3

vert

subeeiitrue

EIqApApN

δ

For a normal pipe scenario the internal pressure will be dominating giving

096.3 0 >⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+> true

vert

subeeii N

EIqApAp

δ

This means that the steel pipe will be in tension when buckling occurs. This is an essential observation since it is against traditional engineering assumptions.

2.2.2 Lateral buckling effects When describing lateral buckling the effects of pipe/soil‐interactions must be considered. If the pipeline is considered ideally straight and lying on ideally even seabed the lateral restraint of the seabed will give the buckling mode. The lateral restraint is given by the lateral soil stiffness soilk which can be seen as a spring resisting the pipeline to move. If the spring is too weak, as for very soft soil, there is hardly any resistance when the pipeline starts to move.

The lateral (horizontal) dynamic stiffness soilk is defined as LLsoil Fk δ∆∆= / [N/m/m], where

LF∆ is the incremental horizontal force between pipe and soil per unit length of pipe, and

Lδ∆ is the associated incremental horizontal displacement of the pipe. Free spanning pipelines paragraph 7.3.6 [7]

If the seabed is considered stiff to hard the buckling mode will be similar to the vertical buckling case. In the same way as the ground prevents the pipeline to dig down in the vertical case the soil will be able to resist the shear forces and act as a fixed support, Figure 10. The friction is considered uniform.


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LL

]/[ mNqsub⋅µ

x

v

0

0

0

=

=

=

Mdxdvv

LL

]/[ mNqsub⋅µ

x

v

0

0

0

=

=

=

Mdxdvv

LL

]/[ mNqsub⋅µ

x

v

0

0

0

=

=

=

Mdxdvv

Figure 10. Lateral buckling behaviour on hard seabed. First buckling mode

effN will be similar to the vertical case with the difference of the friction coefficientµ .

096.3lat

sublateff

EIqN

δµ

= (7)

If the friction coefficient is less than one, lateral sliding will occur before vertical on hard even seabed. The shear force that the ground needs to be able to withstand is found by

effsubsub N

EIqLqLV 5.4)( ⋅=⋅= (8)

In the case of second mode buckling, Figure 11, the soil is not strong enough to withstand the shear forces at 1L giving a three‐buckle shape.

1L 2L

subq⋅µ

subq⋅µsubq⋅µ

v

x

1

2

21

21

21

21 0

vvvvvvvv

′′′=′′′′′=′′′=′==

000

2

2

==′=

Mvv

1L 2L

subq⋅µ

subq⋅µsubq⋅µ

v

x

1

2

21

21

21

21 0

vvvvvvvv

′′′=′′′′′=′′′=′==

000

2

2

==′=

Mvv

subq⋅µ

subq⋅µsubq⋅µ

v

x

1

2

21

21

21

21 0

vvvvvvvv

′′′=′′′′′=′′′=′==

000

2

2

==′=

Mvv

Figure 11. Lateral buckling behaviour on seabed. Second buckling mode, the seabed cannot withhold the

shear force at L1

Equilibrium as in Figure 9 gives the characteristic equation

)(2

2

4

4

vsignqdx

vdNdx

vdEI subeff ⋅−=+ µ

With the general solutions for section 1 and 2

,2

cossin)(,2

cos)( 232102

2101 x

Nq

kxBkxBxBBxvxNq

kxAAxveff

sub

eff

sub ⋅++++=⋅−+=

The five first boundary conditions at 1L given by continuity 0)0()(),0()(),0()(),0()( 211211211211 ==′=′′′=′′′′′=′′′ vLvvLvvLvvLv


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The three boundary conditions at 2L are caused by the soils lateral stiffness soilk . The elastic length, el , is the length for the moment at 2L to be damped out and highly dependent on

soilk as

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soile k

EIl =

Here the soil stiffness is assumed large enough to withhold the shear forces at 2L which will give a short elastic length. And hereby can the moment and the deflection at 2L can be considered zero.

0)(,0)(,0)( 222222 =′′=′= LvLvLv This is an adequate assumption for most soil types [8]. The boundary conditions give 1L and as 2L

kL

kL 63.4,92.2

21 == (9)

And v1 and v2

( )21 5.0cos885.2047.8)( xkx

NEIq

xveff

sub −+⋅

=µ

( )22 5.0cos789.1sin86.0918.2789.1)( xkxkxx

NEIq

xveff

sub +−−−⋅

=µ

(10)

See appendix 7.2 for more detailes The deflection in the centre is

20 93.11

eff

sublat N

EIq ⋅=

µδ (11)

Needed effective axial force to initiate sliding

045.3lat

sublateff

EIqN

δµ ⋅

= (12)

lateffN is lower than lift

effN indicating that a pipeline lying on stiff even seabed will slide lateral

before buckling vertical.

2.2.3 Buckling behaviour on trigger berm-Horizontally straight pipeline As seen in eq.(6) the lift‐off force is inversely proportional to the initial maximum deflection. On an ideally even seabed the force needed for lift‐off is infinitive. Since there always are imperfections on the seabed or the pipeline, an unknown small initial deflection will cause a large unpredictable lift‐off force. One solution to predict and control lift

effN is placing trigger‐

berms, Figure 12, at the sea bottom to create initial imperfections. A trigger‐berm is normally created by dumping stones at the seabed, this form a small hill for the pipeline to lie on.


11

]/[ mNqsub

LL

Post buckled

Trigger berm vert,0δ

vertδ

]/[ mNqsub

LL

Post buckled

Trigger berm vert,0δ

vertδ

Figure 12. Vertical buckling behaviour of pipeline lying on a trigger-berm

When lying on the trigger‐berm the pipeline is locked laterally preventing lateral buckling. When lift‐off occurs the contact pressure is reduced and the lateral friction is equal to zero. This can be seen as a fixed‐fixed system giving the lateral buckling force according to Euler.

2

2

2

2

)2(4

LEI

LEIN lat

effππ

==

Half the buckling length is given by eq.(4) giving the effective lateral buckling force as a function of effective lift‐off force

lifteff

lateff N

kEI

LEIN

25.20)/5.4(

2

2

2

2

2 πππ===

lifteff

lateff NN ⋅=⇒ 49.0 (13)

The conclusion is that lateral buckling appears as soon as lift‐off occurs. This means that when the pipeline is lifted from the trigger‐berm it will rapidly deflect lateral causing snap‐through Figure 13. To avoid critical snap‐through effects lateral imperfection can be build in.

latvert δδ ,

effN

lifteff

lateff NN ⋅= 49.0

vert,0δ

(Compression)

Lift off

Lateral bucklingVertical buckling Combined lat/vert buckling

Snap trough

latvert δδ ,

effN

lifteff

lateff NN ⋅= 49.0

vert,0δ

(Compression)

Lift off

Lateral bucklingVertical buckling Combined lat/vert buckling

Snap trough

Figure 13. The needed force for lift-off is depended of the vertical initial deflection. When lifted the

friction that is keeping the pipeline at its lateral position is reduced to zero which makes the pipeline snap lateral causing so-called snap-through effect.

2.2.4 Buckling behaviour on trigger-berm -Horizontally curved pipeline As seen in Figure 13 a trigger‐berm lower the effective lift‐off force but since the needed force for lateral buckling is lower than the needed force for continued vertical buckling the pipeline snaps lateral when it is lifted from the seabed


12

Built in lateral imperfections, Figure 14, can lower the snap‐through effect wz,

vy,

ux,vert,0δ

lat,0δ

wz,

vy,

ux,vert,0δ

lat,0δ

Figure 14. Curved pipeline lying on trigger‐berm gives a pipeline initially defected in both vertical

as lateral direction

To find the needed force to initiate lateral sliding for the combined case it is seen that the lateral friction resistance is zero when lift‐off occurs ( lift

effeff NN = ) [8]. The modified lateral

friction resistance is then

⎟⎟⎠

⎞⎜⎜⎝

⎛−= lift

eff

effsubsub N

Nqq 1*

Where lifteffN is given by eq. (7)

96.3,0 == bEIq

bNvert

sublifteff δ

The criterion to initiate lateral sliding is from eq.(12) and with the modified lateral friction resistance

45.3,0

*

== aEIq

aNlat

sublateff δ

µ

This gives the effective compression at start lateral sliding as (appendix 7.3)

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛++−⋅⎟

⎠⎞

⎜⎝⎛⋅⋅= 0

022

0

0

41

21

vert

lat

lat

vertlifteff

lateff a

bbaNN

µδδ

δδ

µ (14)

Figure 15 demonstrates how the needed force for lateral sliding decreases with increasing lateral imperfection.


13

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21Initial lateral imperfection [m]

Effective axial force at start slid

ing [M

N 1 meter triggheight

Needed axial effective for lift‐off 5.7 MN

Coefficent of friction, 0.6

Figure 15. A pipeline lying on a one meter trigger-berm, with increasing lateral imperfection the needed

force to initiate sliding is decreased.

Lateral imperfections on un‐even seabed give a lower effective axial force to initiate lateral buckling. This means that the snap‐through effect is eliminated and the deflection of the pipeline is more controlled. This is demonstrated in Figure 16

effN

time

Small initial lateral imperfection

Large initial lateral imperfection

effN

time

Small initial lateral imperfection

Large initial lateral imperfection

Figure 16. With a large lateral imperfection the needed axial force to initiate sliding is lowered and the pipeline slowly deflects instead of having snap‐through. This is caused since the effective axial

force is slowly built up over time.

2.2.5 Conclusion When snap‐through occurs it indicates that the pipeline contains large amount of inner forces. To avoid the snap‐through effects it is possible to create pipelines with initial defections. This would lower the effective axial when loaded with internal pressure and temperature rise. It is proven that lateral buckling will occur before vertical. By combining vertical and lateral initial imperfection the buckling behaviour can be considered satisfying. If the imperfections are large enough the pipeline will slowly expand and the snap‐through effect is eliminated.


14

2.3 Feed‐in‐length and maximum allowable moment The limiting parameter for a pipeline subjected to loads due to installation, seabed contours and high‐pressure/high‐temperature operating conditions is often found to be the bending moment capacity [9]. The maximum allowable bending moment allowM can be found as a second order effect of the deflection. When a pipeline is designed to buckle it is of interest to find how large the elongation can get, how much pipe that can be fed in. The elongation is called the feed in length infeedu − . By using infeedu − a connection between allowM and the

allowable pipe distance between axially fixed points allowC can be found. infeedu − can be

described in two different ways, either by using Hooke’s law or by using geometry.

2.3.1 Feed in length and strain Basic solid mechanics says that the change in strain in a linear‐elastic material is the change in length divided with the original length. This means that the feed‐in length can be described as

buckleinfeed Lu ε∆=−

buckleL is the total length of the deformed parts in the pipeline. The axial strain in a pipeline is given by Hooke’s law. If the conditions are such that the internal pressure is much larger than the external pressure the pipeline will be in plane stress [10]. This is the case in most design scenarios. But in deep water environment the external pressure can be large enough to affect the results. Therefore a three dimension stress state model is used.

[ ] TE radialhoopaxialaxial ∆⋅++−= ασσνσε )(1

(15)

where the stresses based on thin‐walled theory are

s

trueaxial A

xN )(=σ

tDpDp eeii

hoop 2−

=σ

2ei

radialpp +

−≈σ

radialσ is the stress at mid‐surface of the pipe wall when a linear stress distribution is assumed. By using the definition of effective axial stress eq.(2) in eq.(15) the total axial strain change laidasoperation −−=∆ εεε can be expressed as

TAt

DApApN

E si

siiieff ∆⋅+⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −∆−∆+∆=∆ ανε

21

21

(16)

The external pressure is considered constant. Note that ip∆ is the internal pressure difference relative to as‐laid.

2.3.2 Feed in length and deflection As seen in Figure 17 below a small section of the deformed pipeline is infeedu −∆ longer than in

the un‐deformed state. If the section is small enough the deformed pipeline section can be


15

considered straight and the deformed length can be expressed by the original length and the angle as in eq. (17).

0L

θ

w

0L

θ

w

Straight pipeline

Deformed pipeline

0L

θ

w

0L

θ

w

Straight pipeline

Deformed pipeline

Figure 17. The correlation between a straight section length and a deformed pipe

feeduLL

∆+= 00

cosθ

(17)

By using eq.(17) the elongation of the section can be described

θθ

cos)cos1(

0−

=∆ Lu feed

Taylor expansion gives

...!4!2

1cos42

−+−=θθθ

Since the angle is considered small infeedu −∆ can be simplified to

21/

2

2

0

2

0θθ⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅=∆ LLu feed

If the small section is infinitesimal feedfeed duudxLdxdwdx =∆==⇒→ ,,0 0θ

⎟⎠⎞

⎜⎝⎛=⇒

dxdwdxdu feed 2

1

Concluding that the total elongation of the entire deformed pipeline is

[ ] ∫∫ ⎟⎠⎞

⎜⎝⎛⇒⇒⎟

⎠⎞

⎜⎝⎛=

−

LL

Lfeed dx

dxdwsymmetrydx

dxdwu

0

22

21

(18)

2.3.3 Combining feed in length in the case of lateral buckling, even seabed Since lateral deflection will occur before vertical buckling, only lateral sliding will be described. For second mode lateral buckling the expression describing the deflection is eq. (10). By using it with eq. (18) the feed‐in length can be found as (appendix 7.2)

∫∫ =′+′=21

02/7

2/322

20

21

)()(96.82))(())((

L

eff

subL

feed NEIq

dxxvdxxvuµ

(19)

And the maximum bending moment, (at 0=x ) Figure 18, is

eff

sublat N

EIqvEIM

µ89.4)0( =′′=

(20)

(20)


16

subq⋅µ

subq⋅µsubq⋅µ

1L 2L

v

xfeedu

21

feedu21

latM

subq⋅µ

subq⋅µsubq⋅µ

1L 2L

v

xfeedu

21

feedu21

latM

Figure 18. Feed in length and moment at lateral buckled pipeline

By using eq. (19) the bending moment can be expressed as 7/27/47/3 )()(38.1 feedsublat uEIqM µ⋅=

The limiting parameter is allowM and it give the allowable feed‐in displacement as

22/3

2/7

)()(322.0

EIqM

usub

allowallowfeed µ

= (21)

The assumptions made in the beginning of this chapter that the bending moment is a relevant parameter to use as a limit for the pipeline is proven true. As seen in eq. (21) the allowed feed‐in length is highly dependent on the allowable bending moment. Another important factor is the friction coefficientµ . If the friction of the soil is high less elongation is allowed. Eq. (21) with eq. (16) gives the allowable pipe distance between two supports.

TAt

DApApN

E

uLC

si

siiieff

allowfeed

allowbuckle

allow

∆⋅+⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −∆−∆+∆

=∆

=αν

ε21

21

Or inserted for effN

TEAt

DApAp

MEIq

uC

si

siiiallow

sub

allowfeed

allow

∆⋅+⎟⎠⎞

⎜⎝⎛ −∆−∆+⋅−

=αν

µ21

245.3

(22)

When the pipeline is designed for buckling it is necessary that the buckling behaviour is controlled. With the increasing elongation of the pipeline in operation the bending moment keeps rising. Since the maximum allowed moment in the pipeline is a known parameter it is used to determine maximum allowed feed in. To limit the feed‐in length the maximum distance between two supports for free spanning pipelines is used.

2.4 How the axial effective force is built up over time The effective axial force is increasing with rising internal pressure and temperature until it reaches the needed force for deflection. If a pipeline is ideally straight the effective axial force will keep increasing with increasing pressure and temperature. Consider a fully restrained pipeline. The built up effective axial force at a certain internal pressure and temperature can be found by using eq. (16) where the elongation is zero

0),,,(),,,( )1()1()2()2( =−=∆ − elayeffilaidaseeffioper pNTppNTp εεε .


17

layeffN is the effective residual lay tension, the effective axial force the pipeline is subjected to

when laid down. This is a known parameter caused by the real lay down conditions. The equation now reads

021

2)(1

=∆⋅+⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −∆−∆+−=∆ TA

tD

ApApNNE s

isiii

layeffeff ανε

which gives

TEAt

DApApNN s

isiii

layeffeff ∆⋅−⎟

⎠⎞

⎜⎝⎛ −∆+∆−= αν

21

2 (23)

This is known as the fully constrained axial force [5] If the pipeline can be idealised as thin‐walled, eq. (23) can be approximated to

( ) TEAApNN siilayeffeff ∆⋅−−∆−≈ αν21 (24)

The error of the simplification is less than 1% for tDe / larger than 15 [4].


18

3 Verification of analytical model using ANSYS A model was built in ANSYS and exposed for the scenarios examined in the analytically derived functions in chapter 2. The results from ANSYS were compared with the analytical results. The purpose is to create an accurate model with linear‐elastic material model that can be converted into a model with a pipe with plastic abilities used in later analysis. The model uses the same theories and procedure as the model used for analysis at REINERTSEN. However the model developed in this thesis is a simplified version where three specified analysis are to be carried out.

Three different scenarios are analysed with an ANSYS generated model

• 1) Vertical buckling on even seabed. Using a trigger‐berm to get the pipeline to respond. The pipeline is restrained lateral and a rising temperature and internal pressure will build up effN to make it buckle vertical.

o The height of the trigger‐berm will be varied to get the berm height/ effN

diagram. • 2) Lateral buckling on trigger. The pipeline is laid‐down as in case one but a small

lateral force before the internal pressure and temperature is put on will make it deflect lateral.

• 3) Lateral buckling on even seabed. An initial imperfection created by forcing part of the pipeline to move will be used as a lateral trigger. As in the vertical case the internal pressure and the temperature rise will make it buckle.

o The lateral friction is varied to understand how this affects the needed buckle force.

3.1 The model The ANSYS model is build up by four element types. Each one explained below.

• PIPE20 • LINK10 • TARGE170 • CONTA175

TARGE170

PIPE20CONTA175

LINK10

Normal

TARGE170

PIPE20CONTA175

LINK10

Normal

Figure 19. The different element types used in the ANSYS model to create the buckling cases

For the pipeline the PIPE20 element is used.


19

PIPE20 is a uniaxial element with tension‐compression, bending, and torsion capabilities. The element has six degrees of freedom at each node: translations in the nodal, x, y, and z directions, and rotations about the nodal x, y, and z axes. The element has plastic capabilities.

• The element input data include two nodes, the pipe outer diameter and wall thickness, optional stress factors, and the isotropic material properties.

• Internal pressure and external pressure are input as positive values. • Only constant pressures are supported for this element. • Temperatures may be input as element body loads at the nodes.

Ansys element library. PIPE20 [1]

When the simulation starts the pipeline is above the seabed hold up by LINK10 elements. This is to simulate the most probably as‐laid mode and to create the initial stresses in the pipeline.LINK10 elements are tension‐only element making it suitable simulating lowering.

LINK10 is a 3‐D spar element having the unique feature of a bilinear stiffness matrix resulting in a uniaxial tension‐only (or compression‐only) element. With the tension‐only option, the stiffness is removed if the element goes into compression (simulating a slack cable or slack chain condition). This feature is useful for static guy‐wire applications where the entire guy wire is modelled with one element.

• The element is defined by two nodes, the cross‐sectional area, an initial strain or gap, and the isotropic material properties. The element x‐axis is oriented along the length of the element from node I toward node J.

• LINK10 has three degrees of freedom at each node: translations in the nodal x, y, and z directions. No bending stiffness is included in the tension‐only (cable).

Ansys element library. LINK10 [1]

The seabed and the trigger‐berm are built up by quadrilateral target elements TARGE170.

TARGE170 is used to represent various 3‐D “target” surfaces for the associated contact elements. This target surface is discredited by a set of target segment elements (TARGE170) and is paired with its associated contact surface via a shared real constant set.

• For any target surface definition, the node ordering of the target segment element is critical for proper detection of contact. The nodes must be ordered so that the outward normal to the target surface is defined by the right hand rule

• Each target segment of a rigid surface is a single element with a specific shape, or segment type. The segment types are defined by several nodes and a target shape code, TSHAP.

• QUAD is a 4‐node quadrilateral. Where 1st ‐ 4th nodes are corner points (UX, UY, UZ)

Ansys element library. TARGE170 [1]

To establish contact between pipeline and seabed contact elements CONTA175 are created on the pipeline nodes. The contact are initiated by using same real constant set for contact and target elements. The node at the middle of the pipeline share the same real constant set


20

as the trigger‐berm. The rest of the pipeline share same the real constant set as the seabed. In the lateral case where there is no triggerberm there is naturally only one real constant set.

CONTA175 may be used to represent contact and sliding between two surfaces (or between a node and a surface, or between a line and a surface) in 2‐D or 3‐D. [Here node‐to‐surface]. The element is applicable to 2‐D or 3‐D structural contact analyses. This element is located on the surfaces of solid, beam, and shell elements. Contact occurs when the element surface penetrates one of the target segment elements (TARGE169, TARGE170) on a specified target surface.

• The element is defined by one node. • The contact algorithm used has to be specified. • The penalty method uses a contact “spring” to establish a relationship between the

two contact surfaces. The spring stiffness is called the contact stiffness. • For the penalty method, normal and tangential contact stiffness are required. • ANSYS automatically defines default tangential contact stiffness that is proportional

to µ and the normal stiffness. • The maximum allowable elastic slip parameter is required. It is used to control

maximum sliding distance when the tangential contact stiffness is updated each iteration.

Ansys element library. CONTA175, 11.4.Performing a node‐to‐surface analysis [1]

The contact algorithm is the penalty method. This means that a spring in the target will keep the contact element up. The spring is represented by the vertical soil stiffness and given to ANSYS as the normal contact stiffness. This is similar to the lateral soil stiffness described in previous chapter. In the analytically solution the soil was considered very stiff and penetration was not allowed. The values given to the spring in the ANSYS model is the values of a very stiff soil [7] but penetration of some centimetres is allowed. This is more accurate to a real case. Since the trigger has to be able to withstand the entire pipeline at as‐laid a higher value of soilk is used for the trigger target‐contact set. The lateral soil stiffness is generated by the vertical soil stiffness and the mobilisation length, the maximum allowable elastic slip. This value is usually a couple of centimetres.

3.2 The ANSYS analysis

An analysis of a non‐conservative system is path dependent: the actual load‐response history of the system must be followed closely to obtain accurate results. An analysis can also be path dependent if more than one solution could be valid for a given load level (as in a snap‐through analysis). Path dependent problems usually require that loads be applied slowly (that is, using many substeps) to the final load value. Ansys 8.2. Basic Information About Nonlinear Analyses [1]

To be able to rely on the results it is necessary that the order of the different steps in the analysis corresponds to the actions of the real pipeline. A pipeline is first laid down which induces stresses in the pipeline due to the residual lay force and the bending of the pipeline caused by the shape of the seabed. This is the as‐laid condition where the pipeline is exposed


21

to only external pressure and the submerged weight. When the operation starts the pipeline is filled with hot medium giving internal pressure and this will rendering in rising temperature of the steel. To simulate this in ANSYS the solution is devided into several “timestep”. Every timestep is then divided into substeps where the loads are applies in small parts. The order in which the timesteps are given are called the “timeorder”.

3.2.1 Vertical buckling The timeorder of the vertical buckling case is described in four steps, Figure 20‐23. At the start of the analyses the link elements are locked at there upper node and the pipeline is locked axially at one end and having a residual tension force in the other, this to create the tensions caused by the real lay down conditions. The pipeline is simulated to “hang” above the seabed. (1)

1)Links locked in all directions

Pipe locked axially layeffN

1)Links locked in all directions

Pipe locked axially layeffN

Figure 20. Start condition for ANSYS analysis. The axial force in the last node is the residual

tension force. This is caused by the real circumstances when a pipeline is laid down.

The pipeline is loaded with external pressure as surface load and submerged weight eq.(1) as a nodal load. The link elements are in the next step lowered until the entire pipeline is resting on the seabed. (2)

2)

Lowering

ep

subq

Node forces due to buoyancy and weight of concrete, steel and gas

2)

Lowering

ep

subq

Node forces due to buoyancy and weight of concrete, steel and gas

Figure 21. When still hanging the pipeline is loaded with external pressure and submerged weight. To thereafter be lowered down.


22

Since the link elements are tension‐only they will go slack if they are compressed. When the pipeline is at the seabed the link elements are deactivated as they are no longer necessary. The pipeline is restrained laterally and the ends are fixed at there positions. (3)

3)

Pipe locked laterally. Axially at ends

Deactivate links3)

Pipe locked laterally. Axially at ends

Deactivate links

Figure 22. When the pipeline is locked at the seabed the lowering link elements are killed

In the last step, “the operation phase”, the pipeline is loaded with first internal pressure and then temperature. This starts building up the effective axial force which will make the pipeline deflect.

4)

T∆

Increasing temperature and pressure makesthe pipe expand

ip∆

4)

T∆

Increasing temperature and pressure makesthe pipe expand

ip∆

Figure 23. Deflection of the pipeline is caused by operation conditions of internal pressure and

rising temperature.

3.2.2 Lateral buckling on vertical trigger‐berm In the vertical case with lateral buckling the timeorder is almost similar with the difference of a small lateral force instead of lateral locked. This is to make it deflect lateral. Since the pipeline should not be able to buckle lateral without that small force the lateral restraining in case one should not be necessary but tests showed that without this restriction convergence problems occurred.

3.2.3 Lateral buckling on even seabed In the case with lateral buckling the pipeline is laid down as in the vertical case but without the trigger‐berm. It will therefore be completely straight when as‐laid. It is confirmed that


23

without an initial imperfection the needed buckling force is infinite. effN keeps rising with

increasing pressure and temperature according to eq. (24). To make the pipeline buckle laterally a force, preF , is applied at the nodes in the midsection around the “midnode”, as

seen in Figure 24.

L preF

Pipenodes

Midnode

L preF

Pipenodes

Midnode

Figure 24. To make the pipeline laterally imperfect a force is put on the nodes around the midnode

at the pipeline.

The timeorder for the lateral case is as in Figure 25.

effN latδ

""timelayeffN

605.5 e

lateffN

Preforce up Preforce downInternal pressure

Temperature

Stable postbuckling

Stable prebuckling

effN latδ

""timelayeffN

605.5 e

lateffN

Preforce up Preforce downInternal pressure

Temperature

Stable postbuckling

Stable prebuckling

Figure 25. Lateral loading case. Lateral deflection and effective axial force as functions of “time”.

First a force, preF , is applied to create the imperfections on the pipeline but since the model

is elastic the internal pressure needs to be applied as preF is lowered down to zero. This is

possible since the pipeline is not expected to buckle until the temperature is put on. The needed effective axial force for buckling is higher than the effective axial force that is reached with full internal pressure. At the last timestep, as the temperature is applied, effN will build

up to make the pipeline deflect. As ANSYS uses the Newton‐Raphson method for solving the problem the postbuckling behaviour cannot be followed. The pipeline will in infinite short time move from a stable prebuckling position to a stable postbuckling position. This means that the expected curvature is not possible to be find but the expected value of the effective axial force can be compared with the ANSYS values before and after buckling.


24

3.2.4 The parameters and variables in the cases The scenario used for the analyses is an steel pipeline with a concrete coating transporting hot gas at 130 meters water depth. When the pipeline is as‐laid the temperature is equal to the sea temperature 5°C and the inner pressure is zero. When the pipeline is in operation the temperature is 100°C and the internal pressure is 300 bar. In ANSYS the diameter used is for the steel pipe. The main purpose of the coating is to add weight and this factor is included in the submerged weight. The material model used is linear‐elastic.

eD sD iDeD sD iD

Figure 26. The pipeline model used has a concrete coating and transports gas.

Table 1. Parameters and variables used, for full input see appendix 7.4

Parameter Value Unit Description

eD 0.9 m External diameter

sD 0.8 m Steel diameter

iD 0.75 M Internal diameter

ip 30 MPa Internal pressure in operation

ep 3.7 MPa External pressure

T∆ 95 °C Temperature rise EI 946.5 MNm2 Bending stiffness

subq 2.197 kN/m Submerged weight

The variables 0vertδ 1,2,3 m Trigger‐berm height µ 0.4, 0.6, 0.8, 1 Friction coeff For ANSYS input files see appendix 7.5.

3.3 Results

3.3.1 Vertical buckling First the response of a laterally restrained pipeline lying on a trigger‐berm is examined. shows the as‐laid and Figure 28 the deflected pipeline.


25

Figure 27. Pipeline as‐laid on one meter trigger‐berm

Figure 28. Pipeline vertical buckled caused by internal pressure and temperature.

The analytical results are found using eq. (6) where the effective axial force is a function of trigger height. The analytical and ANSYS results for needed lift‐off force at different height of trigger is found in Table 2.

Table 2. The effective axial force at lift‐off. Comparison between analytical and ANSYS model. It is seen that the analytical result and the result from ANSYS correspond satisfyingly. It is also seen that with more initial deflection the error is lessen.

Trigger‐berm height Analytical lifteffN [MN] ANSYS lift

effN [MN] Error

1m 5.710 5.466 4.5%2m 4.038 4.096 1.4%3m 3.297 3.321 0.7% Figure 29 shows the buckling of the pipeline at a one meter trigger‐berm where the effective axial force is a function of the deflection.


26

Figure 29. The effective axial force as a function of deflection at midnode. The deflection has the trigger

height as reference. Figures of buckling at two and three meter trigger-berm in appendix 7.5. The effective axial force is built until it reaches the needed force for lift-off. With increasing deflection the

effective axial force is lowered until it reaches a stabilised value.

Table 3 shows the effective axial force before and after lift‐off. Observe that the pipeline is not entirely lifted from the trigger berm in the prebuckling substep.

Table 3. The effective axial force and vertical deflection of pipeline lying on one meter trigger-berm. Pre- and postbuckling values. When the needed effective axial force is reaches the pipeline buckles vertical. It can be seen that the pipeline is not entirely of the trigger-berm when deflected.

“Time” effN [MN] vertδ [m] 7.04 5.4650 -0.0016 vertδ still<0 7.08 2.9671 2.78 Vertical buckling

Studies of moment, shear force and true axial force were made at the one meter trigger scenario. The expected moment at lift‐off is

eff

sub

NEIq

MwEI 6.5)0()0( =−=′′


27

Expected shear force at touchdown at lift‐off is LqLwEI sub=′′′ )(

L is according to analytical formula eq. (4) 58 meters from the midpoint. By observing the nodal vertical displacement in ANSYS at the time just before lift‐off it is seen that touchdown is at 55.6 meter from midpoint and most penetration into the seabed is at 60.6 meter from midpoint.

Table 4. Moment and shear force at lift-off of pipeline lying on one meter trigger. The analytically derived moment and the moment from the ANSYS model agree.

Analytical ANSYS Moment 2.04 MNm 2.009MNmShear force == )58( mLV 0.127 MN == )6.60( mLV 0.139 MN == )6.55( mLV 0.128 MNTrue axial force and effective axial force are connected through the pressures eq. (2).

effsteeleeiitrue NApApN +−= )(

effN is negative since the pipeline is in compression. trueN cannot be found directly from

PIPE20 output but the stresses in the steel pipe can be compared with

zI

MA

N

s

truetrue +=σ

With effN and M taken from PIPE20 output the computed true stress should be equal to the

calculated. Comparison has been made when the pipeline is loaded with only pressure at mid of pipe cross section ( 0=z ), and with pressure and temperature at mid and top of cross section ( 4.0=z ). The connection between true axial force and effective axial force is verified with ANSYS.

Table 5. Effective axial force and moment from output are used to calculate the true stresses in the pipeline. This in then compared with the stresses from PIPE20 output.

ansyseffN [MN] ansys

trueN [MN] z [m] ansysM [MNm] calctrueσ [MPa] ansys

trueσ [MPa]

Press ‐4.9 6.5 0 ‐‐ 106 106P+T mid ‐1.8 9.6 0 ‐‐ 157 158P+T top ‐1.8 9.6 0.4 6.5 725 709

3.3.2 Lateral buckling on vertical trigger‐berm The second case is a pipeline lying on a vertical trigger. A small force at the midnode of 100N before the operation phase will make it deflect. The prediction is that the effective axial force will build up to the necessary lift‐off force calculated in Table 2. When the pipeline lifts the lateral friction, keeping the pipeline from buckling, is reduced to zero and the pipeline snaps laterally. A study is done at the one meter trigger scenario. By observing the actions of effN ,

vertδ and latδ the timestep before and after buckling (Table 6) it is seen that when effN has

built enough force to lift the pipeline off the trigger it deflects laterally as predicted. This is displayed graphically in Figure 30. The final configuration of the pipeline is shown in Figure 31.


28

Table 6. Effective axial force, vertical displacement and lateral displacement of an initially defect pipeline lying on a one-meter trigger-berm. When lifted from the trigger-berm the pipeline snaps lateral.

“Time” effN [MN] vertδ [m] (from trigger) latδ [m] Remark

7.0600 5.747 ‐0.000147 0.00135 vertδ <0, 7.0700 5.859 0.0408 0.00609 vertδ >0 7.0800 1.774 ‐0.0175 4.294 Lateral buckling

No verticaldeflection

Whenreaches theneeded lift‐off force thepipeline deflectslaterally

effN

No verticaldeflection

Whenreaches theneeded lift‐off force thepipeline deflectslaterally

effN

Figure 30. When the effective axial force has reached the value for needed lift‐off force the pipeline rises from the trigger‐berm and snaps laterally. Between times 6‐7 internal pressure is applied and

between times 7‐8 the temperature is raised.


29

Figure 31. Movement of pipeline lying on trigger‐berm with a small lateral imperfection loaded

with internal pressure and temperature

It is observed that a higher value of effN is needed to buckle the pipeline than in the clean

vertical deflection case. Here it is more accurate to the analytically calculated value. Possibly because the pipeline has to be lifted up before it can deflect lateral. In the vertical case the value of vertδ just before buckling is minus indicating that the pipeline is not above the trigger at the stable prebuckling time. It is possible that the effective axial force builds up some more but it is missed in the buckling analysis.

3.3.3 Perfect straight pipe on perfect even seabed As described before a perfect straight pipeline will not deflect during loading. Instead it keeps building up the effective axial force. Eq. (24) gives the connection between pressure temperature and effN . Figure 32 show how the analysis is done over “time”. A first

comparison is made when only internal pressure is applied and the second comparison when both internal pressure and temperature are fully applied.

( ) =−∆−≈ ν21iilayeff

presseff ApNN 5.054 MN

By adding the temperature the effective axial force is fond to be ( ) TEAApNN sii

layeff

Tpeff ∆⋅−−∆−≈+ αν21 =19.41 MN

The values from PIPE20 output =press

effN 5.056 MN

=+TpeffN 19.42 MN


30

MPapi 300 →= CT °→=∆ 950

No deflection

MPapi 300 →= CT °→=∆ 950

No deflection

Figure 32. An ideally straight pipeline does not deflect during pressure and temperature rise. The

effective axial force keeps building up.

When the full internal pressure is built up the effective axial force is ‐5.05 MN The expected true axial force in the pipe is then

=trueN 6.35 MN The stress in the pipeline is then

=trueσ 104.1 MN/m The stresses in the pipeline according to ANSYS is

=trueσ 104.1MN/m The straight pipeline reacts as predicted with very accurate results.

3.3.4 Lateral buckling on even seabed Lateral buckling is examined with four different coefficients of friction for the seabed. According to the analytical results eq. (12) the coefficient of frictions affects the results. For every lateral case the value of the lateral deflection just before and after buckling is used in the formula for determine the buckling force. This value is then compared with the value from PIPE20 output, Table 7. Graph of effN and latδ with friction coefficient of 0.6 is

displayed in Figure 33


31

Table 7. Effective axial force at initiation of lateral sliding (stable prebuckling position) and as deflected (stable postbuckling position). The errors are small in the analysis except for the case with a friction coefficient of 0.4. One reason can be that the initial lateral deflection is smaller than in the other cases. But this cannot explain the difference of 10 percent.

µ ansyslatδ [m] calc

effN [MN] ansyseffN [MN] error

0.4 Pre 0.0617 12.667 11.516 10% Post 4.841 1.430 1.408 1.5%0.6 Pre 0.0857 13.164 12.949 1.7% Post 4.796 1.760 1.723 2.1%0.8 Pre 0.1037 13.818 13.665 1.1% Post 4.668 2.060 2.007 2.6%1.0 Pre 0.1232 14.174 14.380 1.5% Post 4.598 2.320 2.253 3%

Figure 33. Effective axial force and lateral deflection as functions of “time”. This figure can be compared with figure 25. The preforce is applied in time 5‐6 and internal pressure time 6‐7.


32

The expected curvature of the pipeline is multiple buckle modes as seen in Figure 34.

Figure 34. Lateral deflection of pipeline gives a multiple buckle mode

According to analytical analysis the moment can be found as a function of friction and effN .

Eq.(20) says

eff

sublat N

EIqM

µ89.4=

In the case with µ 0.4 the postbuckling value of effN is 1.408. This give

=latM 2.889 MNm According to ANSYS output the value of the moment at postbuckling is

=ansyslatM 2.831 MNm

The derived moment correspond satisfyingly with the moment received in tha ANSYS analysis.

3.4 Conclusion The vertical case is satisfying especially when the deflection is high initially. The connection between big initial imperfection and accurate result can be seen in the lateral buckling scenario as well. When the lateral initial imperfection is small, as seen when the friction coefficient is 0.4, the difference between PIPE20 output and calculated result are bigger. A possible explanation is that the effective axial force has a large slope at the beginning of the effN ‐ vertlat δδ / curve figure 13 (eq. (7,12)). The exactly right value is hard to catch. When

the buckling occurs at the ANSYS analysis it is not necessarily because the full buckling force is reached if the timesteps are to large the real value can be missed. The connections between true axial force and effective axial force are found to be accurate, so are the equations describing moment and shear force. The effective axial force is built up as predicted and the buckling direction and modes are as expected. The conclusion is that the analytical model and the model used in ANSYS correspond satisfyingly.


33

4 Plasticity, Ansys and PAS The analyses presented in the previous chapters do not consider plasticity. In a real analysis that kind of a simplification is not acceptable. When a recent analysis involving plasticity was done at REINERTSEN the output from ANSYS PIPE20 element was of questionable accuracy. The reason for this is that another approach was used in the pipeline design. Normally a load controlled condition is necessary to be used where the limiting parameter is the bending moment. This is explained in chapter two. However, in this project a displacement controlled condition was used where the limiting parameter was the allowable strain. The allowable strains was about 1 to 1.5 percent but when this value was reached the PIPE20 element obtained incorrect result where the main issue was that PIPE20 overestimated the strains. This was a new problem that occurred since the strains when using a load controlled condition do not reach these levels. To exemplify the weakness of the PIPE20 element a simple one‐element model is used. The element is restrained in one end and has a structural displacement (rotation) on the other end. Studies of corresponding moment and stresses are done. The calculated bending moment and axial strain are compared with results from an analytical model and from a model build in PAS (Pipeline Analysis System). PAS is a program developed by REINERTSEN which uses the finite element method to calculate stresses and strains in pipelines.

• The study of the ANSYS model comprises the following

o The connection between rotation and moment is derived. o The results from the PIPE20‐ and the PAS model are compared with the

analytically derived results. o The stresses and strains from the PIPE20‐ and the PAS models are compared

• The results are found o It is seen that PIPE20 differs from the expected results due to two reasons.

When the rotation is small and the cross section is almost elastic the expected value and PIPE20 element output is practically identical but with increasing rotation and plasticity the accuracy is lessen. And when more than 23 percent of the cross section is plastic the element output is utterly incorrect.

• The errors when the rotation is small are probably caused of the few number of integration points in PIPE20. The impact of number of integrations points are studied

o Numerical o With PAS where number of integration points can be varied

4.1 Model of short plastic beam The modelled and analysed beam has the same cross section as the steel pipe used in the previous analyses and a length of two meter. The material is in the analytical study considered to be linear‐elastic ideal‐plastic with yieldσ at 448 MPa. However, to ensure that

the stiffness matrix is positive definite and non‐singular a bilinear curvature with a small inclining was used in the ANSYS and PAS analysis, Figure 35. This since the model in ANSYS and PAS will not converge if the stiffness matrix is singular.


34

ε

σ

⎩⎨⎧

==

%100450

εσ MPa

Perfect‐Plastic

ε

σ

⎩⎨⎧

==

%100450

εσ MPa

Perfect‐Plastic

Figure 35. Stress-Strain curve. The solid line is the stress-strain curve used in the analytical model. To ensure that the stiffness matrix in ANSYS and PAS is non-singular a stress-strain curve with a small

incline is used, this is shown as the dotted line.

4.1.1 Theory and analytical model Consider a thin‐walled beam fixed in one end and bent in the other with w′ radians, as shown in Figure 36.

w′ θ=′w

θ

R

L

w′w′ θ=′w

θ

R

Lθ=′w

θ

R

L

Figure 36. A beam subjected to a prescribed rotation w’.

The radius R of the curvature can be used to describe the rotation

RLLw

Rxw

Rxxw ==′=′′⇒=′− θ)(,1)()(

The pipeline is loaded until it is partly plastic. Figure 37 shows how the strain corresponds to the stresses and the rotation.

ϕ ab

r⋅2

eε pε

abw ′′ ϕ a

bϕ a

b

r⋅2

eε pε

abw ′′

r⋅2

eε pε

abw ′′

Figure 37. The plastic and elastic part of the cross‐section of a pipeline exposed to rotation w’ as

described in figure 36.

θσσσ

ε LE

aRE

awaE

yyye ⋅=⇒=⇒′′⋅==


35

The corresponding moment causing the stresses in the pipeline can be expressed by [11]

∫ ⋅⋅=π

ϕϕσ2

0)sin()( rdAM (25)

where ϕdrtdA ⋅⋅= .

dAyσ

a

b

0ϕ

ϕd

t

0=ϕ

2/πϕ =

r

dAyσ

a

b

0ϕ

ϕd

t

0=ϕ

2/πϕ =

dAyσ

a

b

0ϕ

ϕd

t

0=ϕ

2/πϕ =

r

Figure 38. Stress‐distribution over a quarter of the pipe cross section. a/r=sin(φ) percent of the

pipeline is elastic.

θσ

ϕϕEr

Lra y=⇒⋅= )sin()sin( 00

⎪⎩

⎪⎨

⎧≤

=plasticelse

elasticifra

y

y

,

,)sin()sin()sin()( 0

σ

ϕϕϕσ

ϕσ (26)

The moment can be expressed with one plastic and one elastic part. Due to symmetry there is only need to integrate over a quarter of the cross‐section Figure 38.

partPlastic

y

partElastic

y drtdrtM ϕϕσϕϕϕσ

π

ϕ

ϕ

)sin(4)sin()(sin4 2

2/

0

22

0 0

0

⋅+⋅= ∫∫

This gives the analytical function describing how the moment depends on the rotation. (Appendix 7.7)

02

,1

)cos()sin(

2 00

0

02 >>⎥⎦

⎤⎢⎣

⎡+= ϕπϕ

ϕϕ

σ trM y (27)

Since this model only is valid when the model is both elastic and plastic the first calculated value in Table 8 is from basic solid mechanics [11]. The model behaves like a plastic joint when pMM > where pM is the plastic moment

=⋅= trM yp σ2)2( 6.727 MNm (28)

The analytically expected values of the moment can be compared with the PIPE20 element and PAS output.


36

Table 8. The analytically expected value of the moment with different rotation loads. The remaining elastic part of the cross‐section is given in percent.

Rotation (rad) Percent part elastic ϕsin/ =ra

0ϕ [rad] Moment [MNm]

0.011 100 π/2 5.2058 0.012 93 1.1967 5.5532 0.013 86 1.0338 5.7674 0.014 80 0.9238 5.9218 0.015 74 0.8401 6.0391 0.016 70 0.7728 6.1313 0.017 65 0.7169 6.2054 0.018 62 0.6695 6.2660 0.020 56 0.5926 6.3587 0.022 51 0.5326 6.4256 0.025 45 0.4632 6.4960 0.05 22 0.2253 6.6706

4.1.2 PIPE20 and PAS model To be able to compare PAS and PIPE20 it is necessary that the input for both models corresponds. The geometry and material data are already presented and used similarly in the both programs. However the PIPE20 element has strict eight integration points around the circumference, Figure 39, where results are reported. This means that the results never can be more precise than the data found at the integration points. It is possible to change the number of integration points in PAS. Since PAS is a 2D‐program the number of integrations points is given for half plane. When comparing PAS with PIPE20 five integrations points are used since this should give the same resulting moment.

z

y

°90°45

°0 y

zz

y

°90°45

°0

z

y

°90°45

°0 y

z

y

z

Figure 39. PIPE20 has eight integration points around the circumference, to the left. In the comparison the number of integration points in PAS are set to five to make them correspond


37

Table 9. Moment taken from PIPE20 and PAS when at different rotation displacements

Rotation [rad] Moment PIPE20 [MNm] Moment PAS [MNm] 0.011 5.2028 5.2027 0.012 5.4800 5.4795 0.013 5.7166 5.7160 0.014 5.9531 5.9525 0015 6.1896 6.1890 0.016 6.3753 6.3776 0.017 6.3772 6.3776 0.018 6.2461 6.3776 0.020 6.0784 6.3776 0.022 6.1680 6.3776 0.025 5.8915 6.3776 0.05 5.7715 6.3778

When the rotation of the pipe section is over 0.0157 rad both the integration point at 90° and 45° are in the plastic area. Since the values at those location are the only ones delivering data to be read and calculated the moment should not be able to rise with increasing rotation displacement, Figure 40. Input files in appendix 7.8.

Figure 40. When the rotation is small the values taken from the integration points should

correspond to the figure to the left. With increasing rotation both the top integration point and the integration point at 45° should give a plastic value. The problem with few integration points are found when the rotation is further increased. The moment should keep rising but since no new

data can be found the value of the moment stays constant.

This behaviour is observed in PAS, after a 0.016 rad rotation the moment stabilises. However, the moment from PIPE20 is decreasing with increased rotation. This can be seen in Figure 41 below to the left. In the figure to the right three cases are tested in PAS to illustrate how insufficient number of integration points affects the results.


38

Figure 41. Moment as function of rotation for different models

With seven number of integration points the value of the moment is stabiled when more than 50 percent of the cross‐section is plastic. However for eight integrations points the moment wonʹt stabilise until almost 78 percent of the cross section area is plastic. Therefore the moment keeps rising with increasing rotation. As seen in the figure the deviation at 0.05 rad rotation is larger for the PAS model with eight integration points than for the PAS model with seven integration points. This is further investigated in next the chapter. As seen above the results from the PIPE20 output are unreliable when too large part of the cross‐section is plastic. This indicates that the stresses in PIPE20 behave oddly. In Table 10 the first principal stresses and strains at the integrations points are listed for three different rotation displacements. Before the 45° integration point is plastic, when the rotation is 0.015 rad, the stresses are similar. With increasing rotation the stresses in PIPE20 are falling and hence the values are no longer reliable.

Table 10. Stresses at integration points at different rotation displacements

PIPE20 PAS Int.point 1σ [MPa] pe εε + [%] 1σ [MPa] pe εε + [%] 90° 448.03 0.2906 448.00 0.2906 45° 425.39 0.2055 425.38 0.2055 0° ~0 ‐0.670∙10‐14 0 1.110∙10‐14 Compared at 0.015 rad rotation

Int.point 1σ [MPa] pe εε + [%] 1σ [MPa] pe εε + [%] 90° 432.80 0.3504 448.00 0.3487 45° 437.25 0.2478 448.00 0.2466 0° ~0 ‐0.6132∙10‐5 0 1.110∙10‐14 Compared at 0.018 rad rotation

Int.point 1σ [MPa] pe εε + [%] 1σ [MPa] pe εε + [%] 90° 378.62 MN 0.9761 448.02 0.9687 45° 398.98 MN 0.6902 448.01 0.6850 0° ~0 ‐0.6325∙10‐4 0 0 Compared at 0.05 rad rotation


39

4.1.3 Loading PIPE20 with moment The loading used at REINERTSEN are more common to be force based than displacement based. Therefore the PIPE20 model was subjected to a known moment instead of a known rotation. When the load exceeded the moment for which the 45° integration point should be plastic the analysis did not converge and hence no solution was able to get. The conclusion is that the PIPE20 element is not suitable in applications where plastic deformation of a major part of the cross‐section is expected.

4.2 The importance of integration points‐a short study Even though the previous analyses indicate that the problem with PIPE20 is not only caused by the insufficient number of integration points it was of interest to see how accurate a numeric model with finite number of points is to an analytical model. As already seen in Figure 41 using eight integration points instead of seven over half the cross section won’t necessary give a more accurate result.

4.2.1 Analytical vs. numerical With only a finite number of points where the stresses can be found the moment can be described as

)sin()(1

ii

N

ii rAM ϕϕσ ⋅⋅= ∑ (29)

Where N

rtAiπ2

⋅⋅= and N is the number of integrations point around the circumference. As

before the moment can be departed into one plastic and one elastic part according to eq.(26)

∑ ∑ ⋅+⋅=elastic

iplastic

yi

y Ntr

NtrM ϕπσ

ϕϕπσ sin2

sinsin2 2

0

22 (30)

By increasing the numbers of integrations points it can be seen how it affects the result compared with the integrated expected value of the moment. The graphs below shows the moment as a function of number of integrations points at half of the cross‐section. This is to be able to compare with PAS where the integrations points are specified for a half plane. Five integration points are as described in Figure 39. The connection is found at two different rotation loads. With a rotation of 0.014 rad the cross‐section is 80 percent elastic and with a rotation of 0.05 rad it has decreased to 22 percent. See appendix 7.9 for MATLAB file.


40

5 10 15 20 25 30 35 40 45 505.84

5.86

5.88

5.9

5.92

5.94

5.96

5.98x 106

Mom

ent

No. of integration points

0.014 rad rotation

5 10 15 20 25 30 35 40 45 56.35

6.4

6.45

6.5

6.55

6.6

6.65

6.7x 106


0.05 rad rotation0.014 rad rotation

0.05 rad rotation

Mom

ent

N N5 10 15 20 25 30 35 40 45 50

5.84

5.86

5.88

5.9

5.92

5.94

5.96

5.98x 106

Mom

ent


0.014 rad rotation

5 10 15 20 25 30 35 40 45 56.35

6.4

6.45

6.5

6.55

6.6

6.65

6.7x 106


0.05 rad rotation0.014 rad rotation

0.05 rad rotation

Mom

ent

N N Figure 42. The straight lines are the analytically expected value of the moment at different rotation

displacements, eq. (27). The oscillating lines are according to eq. (30) where the moment is a function of N, the number of integrations points. N is specified at half of the cross‐section.

4.2.2 Analytical vs. PAS The same analysis was made in PAS. The number of integrations points of half the cross‐section was changed while keeping the rotation load constant. The result is found in Figure 43 below.

Rotation 0.014 rad

5880

5900

5920

5940

5960

5980

6000

6020

6040

6060

5 10 15 20 25

Number of integration points

Mom

ent [kN

m]

Rotation 0.05 rad

6300

6400

6500

6600

6700

6800

6900

5 10 15 20 25


Mom

ent [kN

m]

Rotation 0.014 rad

5880

5900

5920

5940

5960

5980

6000

6020

6040

6060

5 10 15 20 25


Mom

ent [kN

m]

Rotation 0.05 rad

6300

6400

6500

6600

6700

6800

6900

5 10 15 20 25


Mom

ent [kN

m]

Figure 43. Expected analytically moments and moments received from PAS as a function of number of integration points. The graph to the left is a pipeline that is 80 percent elastic and the graph to

the right is a pipeline 22 percent elastic.

The same oscillating pattern as in the numeric analysis is observed in PAS. This explains why eight integrations points gave a more inaccurate result compared to the analytically expected than when seven integration points were used. One can by coincidence happen to use the “right” number of integration points and thereby get the correct result. Even though an increased number of integration points will give a less accurate result. In both analyses it is seen that to be sure to obtain an accurate value of the moment it is required to use more than 15 integrations points on half the cross‐section. It is obvious that the need for


41

integration points is bigger, the bigger the plastic part of the cross‐section is. The reason for this is already explained above, the numerical resolution of the cross‐section is too coarse. If it is of interest to know the exact stress distribution in the cross‐section naturally more integration points are needed.

4.3 Conslusion The first thing to regard is PIPE20’s unsuitability to be used when the cross‐section of the pipeline is expected to be more than 20 percent plastic. It is clear that until a better element formulation is available in ANSYS another approach to large‐plastic problems is needed. The problem of receiving imprecise result due to lack of integrations points is described above.


42

5 Development and verification of a new model The current ANSYS model used at REINERTSEN is a more general and advanced version of the model developed in this thesis. The REINERTSEN model is used to evaluate pipelines exposed to high pressure and temperature during periodic loading/unloading. As explained in chapter four when the model is allowed to reach strains up to 1.5 percent the results from ANSYS is incompatible with results from other applications. The analysis showed that the PIPE20 element overestimated the strains. Therefore it is of interest to find a new more accurate model when problems with strains over one percent occur. One suggestion of solutions is to replace the PIPE20 elements, where the results are questionable, with shell elements. A first study of a beam‐shell assembly is going to be carried out.

• The first step is creating a shell part that can interact satisfyingly with PIPE20

element. • Secondly the beam part and the shell part are connected. This creates a beam‐shell

assembly. o Two different ways of combining PIPE20 elements with shell elements are

examined MPC‐approach BEAM4 elements

o A study of deflection and axial force when the models are exposed to submerged weight are made. The results are compared to a “PIPE20only” model.

• Third step is creating a seabed and contact for the beam‐shell assembly o A straight pipeline created at the seabed is exposed to internal and external

pressure. The two different beam‐shell assemblies are compared with a PIPE20only model.

• For the shell part to work in the ANSYS model used at REINERTSEN, it is compulsory for it to be able to be lowered down at position. LINK10 elements are connected to the shell part using either BEAM4 elements or multi point constraints (MPC).

o Tests with lowering are done. By some reason using full weight of the pipeline is leading to convergence problems. This was temporarily solved by allowing the weight to be ten percent when lowering and set to 100 percent when the pipeline is at the seabed, before axially locked.

o Tests with pressure are made at the as‐laid pipeline on even seabed. The outcome of the test is used to choose model for continue testing.

• Using a linear‐elastic perfectly plastic pipeline with PIPE20only model and a similar beam‐shell assembly the strains as a function of the temperature is observed. The complaints being that PIPE20 overestimates the strains in the pipeline.

o Vertical deflection of the pipeline is examined. The results are that the strains in the shell‐beam assembly are smaller than in the PIPE20 model.

o Lateral deflection was to be examined as well but since there was problems with contact this part is not carried out entirely due to lack of time.


43

5.1 Building a shell

SHELL181 is suitable for analyzing thin to moderately‐thick shell structures. It is a 4‐node element with six degrees of freedom at each node: translations in the x, y, and z directions, and rotations about the x, y, and z‐axes. It is well‐suited for linear, large rotation, and/or large strain nonlinear applications. It has plasticity capabilities.

• Pressures may be input as surface loads on the element faces • Temperatures may be input as element body loads at the corners of the outside faces. • The element formulation is based on logarithmic strain and true stress measures

Ansys element library. SHELL181, [1]

The shell part is created with quadrilateral SHELL181 elements. 16 shell elements are used around the circumference. The shell part has to act as a PIPE20 element in order to interact with it. Since “the pipe element is assumed to have “closed ends” so that the axial pressure effect is included”[1] it is compulsory for the shell part to be created with end caps as well. The end caps on the shell part were first created with triangular SHELL63 element. These are similar to the SHELL181 except for there incapability of plasticity. The triangular SHELL181 elements are not recommended and since there is no need for plastic effect on the end caps SHELL63 element seemed appropriate.

Figure 44. The shell part is created with quadratic SHELL181 elements. Shells on the ends of the shell

part are a must so the end effects from PIPE20 elements are included.

These elements are the end caps used in the test with a pipeline created at seabed. However it turned out that when the pipeline is created above the seabed and exposed to external pressure the analysis did not converge. The exact reason for this is not examined. The solution here is to use triangular SHELL181 element which work where SHELL63 did not. Tests were made with SHELL181 elements and the same results were received for the forces in the shell part and the beam part as for usage of SHELL63. To create the submerged weight on the model developed in chapter three, nodal forces were applied. That approach is not suitable here instead a fixed density is used and gravity is simulated with ANSYS inertia command ACEL.

5.2 Modelling a beam‐shell assembly Two methods for combining the beam‐shell model are used. The first method is the multipoint constraints (MPC) option that ANSYS provides. It uses contact and target


44

elements to connect the nodes at the beam and to the nodes at the shell. The second method is using BEAM4 elements to attach the beam and the shell together.

5.2.1 MPC By this method, ANSYS generates MPC equations internally based on the contact kinematics. When using this method for building beam‐shell assemblies the end‐node at the beam must be a pilot‐node (TARGE170) and the shell nodes are the contact nodes (CONTA175), Figure 45.

Figure 45. Connection between PIPE20 beam elements and SHELL181 with MP-constraints. MPC uses

target and contact elements to establish connection of the beam-shell assembly

A surface‐based constraint is used to couple the motions of the nodes on the contact surface to the pilot‐node. This constraint can either be rigid or force‐distributed. It is recommended [1] that when having a beam‐shell assembly the force‐distributed constraint is used. “In this type of constraint, forces or displacements applied on the pilot node are distributed to contact nodes (in an average sense) through shape functions”, [1]. This means that if the pipeline is bent the shape of the shell part do not follow the rigid motion at the pilot node and can therefore change shape to the more accurate oval form.

5.2.2 BEAM4 The second method is building the beam‐shell assembly with BEAM4 as coupling elements, Figure 46.


45

Figure 46. Connection between PIPE20 beam elements and SHELL181 with BEAM4 elements

These elements have specified material constants. The density given is small so it will not affect the weight of the pipeline when exposed to gravity.

5.2.3 Study of the connections A short PIPE20only model and the combined models with the same length are locked at ends and exposes to submerged weight. The resulting deflection and axial force are seen in Figure 47. This analysis is made to verify the accuracy of the connection types. As argued later the stresses at the interface can be discussed in their precision. However the connection should be stiff enough to be able to give the same result at deflection and at forces at ends of the beam‐part for the combined models as for the PIPE20only model. No shell end caps are used in this analysis. It is observed that the MPC constraint and the PIPE20only are more similar to once another.

MN55.1

MN55.1

MN67.1

mµ4.72

mµ2.86

MPC

PIPE20only

BEAM4

mµ8.71MN55.1

MN55.1

MN67.1

mµ4.72

mµ2.86

MPC

PIPE20only

BEAM4

mµ8.71

Figure 47. Deflection and axial force are compared for models exposed to submerged weight. The models are locked at the ends. PIPE20only model at top and beam-shell models below. Connections are done with

MPC (middle) and with BEAM4 (at bottom).


46

5.3 Modelling the Seabed and the contact The contact element used for the shell part is CONTA173. It is similar to CONTA175 used for the PIPE20 elements with the difference of being surface‐to‐surface. The contact element is created on the shell elements that are situated on the underside of the shell part. On PIPE20 the contact elements are situated on the nodes. The nodes on the pipeline are on the same height as the centre of the shell. Here this was solved by creating two sea beds at two different levels. Another option is to use key option 11 for the CONTA175 element where the shell thickness of the pipeline can be set to be included.

5.3.1 Forces due to pressure-pipeline initially created on flat seabed The first bigger verification is creating each model directly on the seabed and let them be exposed to submerged weight and internal and external pressure. The resulting forces are compared. This is similar to the no‐buckling scenario presented in chapter three where the effective axial force are allowed to be built up without causing the pipeline to buckle. But since the pipeline is created already laying on the seabed the same prediction of effN ,eq.(24) ,

cannot be used. The assumptions for eq.(24) are that the external pressure is the same when as‐laid and during operation. Here this is incorrect since the external pressure is put on after it is on the ground. Figure 48 shows the positions of the elements in Table 11. The model used in the analysis is 500 meter with 20 meter shell part. 2128 shell elements were used at the shell part.

Figure 48. The tabulated elements are found at the demonstrated positions. The actual model used is 500

meter with 20 meter shell part. The shell part contained 2128 elements.


47

Table 11. The resulting forces and stresses of pipeline exposed to submerged weight and external and internal pressure. The direction of the shell forces are found in the figure above, n11 and n22 are forces per meter length

PIPE20only effN [MN] xσ [MPa]

‐4.313 116.3

MPC effN [MN] xσ [MPa]

Pipewest ‐4.326 116.1 Pipeeast ‐4.326 116.1 Pipeend ‐4.324 116.1 11n [MN/m] 22n [MN/m] Shellwest 7.032 3.198 Shellmid 10.01 3.312 Shelleast 7.032 3.198

BEAM4 effN [MN] xσ [MPa]

Pipewest ‐4.323 116.1 Pipeeast ‐4.323 116.1 Pipeend ‐4.322 116.1 11n [MN/m] 22n [MN/m] Shellwest 5.119 3.201 Shellmid 10.01 3.315 Shelleast 5.119 3.201 The first observation made is that the beam‐part is not affected of being in a beam‐shell assembly. The outputs from PIPE20 elements are almost the same in the PIPE20only model as in the combined. The second validations that need to be made are the compatibility between the beam part and the shell part. By using the effective axial force given by the PIPE20 output the true axial force can be calculated

eeiiefftrue ApApNN −+= =‐4.3 MN+11.4 MN=7.1 MN

The axial force per meter length, 22n , is the true axial force divided with the circumference of the pipeline. r is the mean radius of the pipeline.

rN

n truecalc

π222 = =7.1 MN/2.42 m=2.9e6 N/m

The calculated value is lower than the results from the both beam‐shell assemblies. This analysis can be performed in the opposite direction and use the axial force per meter from the shell element output as input. The expected true axial force is then

s

shellcakcx A

nr 222 ⋅=

πσ =2.42 m * 3.3 (MN/m)/0.061 m2=133 MPa

The radial force per meter length, 11n , is the difference in pressure divided with the mean radius. This value can be compared with the values from shellmid


48

rpp

n eicalc −=11 =26.3e6/0.3875=10.2 MN/m

The values from the shell part at ends (shellwest and shelleast) of the beam‐shell assembly are different for the MPC combined assembly and the BEAM4 combined assembly. However their accuracy can be discussed. St. Venantʹs principle states that “Statically equivalent systems of forces produce the same stresses and strains within a body except in the immediate region where the loads are applied. Thus the stresses calculated in the middle of a beam are not influenced by the way the ends are supported as long as the supporting forces and moments are statically equivalent to those in the mathematical model.” [12]

Here this means that since the effects at the interface of the beam‐shell assembly is not known the values there are not suitable to use in further analysis.

5.4 Modelling Link elements It is required that the beam‐shell assembly can be laid down by the same method as the PIPE20only model is. The LINK10 elements cannot be directly connected to the shell part, instead the same types of connection as for the beam‐shell assembly is used. A node at the centre of the shell part is used to create the lower part of the link element. This node is then connected to the shell part via MP‐constraints or BEAM4 elements. This gives three possible models to use in further analysis.

MPC‐MPC

BEAM4‐BEAM4

MPC‐BEAM4

MPC‐MPC

BEAM4‐BEAM4

MPC‐BEAM4

Figure 49. Three different connection types are examined. The first uses MP-constraints in both with the beam-shell connection and with the link shell-connection. The next two uses BEAM4 elements with the

shell-link connection. And the last one has BEAM4 elements to connect the beam-shell assembly.

5.4.1 Lowering the pipeline As described for the PIPE20 model in chapter three the pipeline is initially above the seabed and exposed to external pressure, submerged weight and the residual tension force, lay

effN .

The model is then lowered down to the seabed. By some reason applying full gravity ended in convergence problems when lowering the pipeline. The reasons for this are not further


49

investigated. Here letting the initial conditions for the weight to be ten percent solves this. When the pipeline is as laid 100 percent of the weight is applied. This approach is not acceptable in a general model but the argument here is that it is more important to validate the capability of the beam‐shell assembly when as‐laid.

5.4.2 Establishing contact with an un-even seabed Tests of laying down the pipeline on a trigger have been done. The trigger used is wider than the trigger used in the PIPE20only model. The problem with a too narrow trigger is that the pipeline falls through even with a high value of the normal contact stiffness. Raising this value to much will lead to convergence problems instead the solution was to broaden the trigger. Another suggestion for solution is to change the contact algorithm to the augmented Lagrangian method. This model works when the trigger is half a meter high but with a one meter trigger the pipeline falls through. A further study of this phenomenon is not done.

5.4.3 Forces due to pressure-pipeline laid down flat seabed To investigate which model best suited for further analysis each is lowered down on a before. The differences between the three beam‐shell assemblies are so small that only results from the MPC‐BEAM4 are represented in Table 12. The main differences between the beam‐shell assemblies are at the shell ends. As already concluded incomprehensibilities are to be expected at the interface of the shell‐beam assembly. These values are therefore not included in the table below.

Table 12. The resulting forces and stresses of pipeline exposed to submerged weight and external and internal pressure after lay down. The direction of the shell forces are found in the figure 47 above, n11 and n22 are forces per meter length

PIPE20only effN [MN] xσ [MPa]

‐5.056 104.1

MPC‐BEAM4 effN [MN] xσ [MPa]

Pipewest ‐5.068 103.9 Pipeeast ‐5.068 103.9 Pipeend ‐5.067 104.0 11n [MN/m] 22n [MN/m] Shellmid 10.02 2.913 As before, it is seen that the values for PIPE20 element in the PIPE20only model are consistent with the values for the beam‐shell assemblies. By using the effective axial force given by the PIPE20 element output the true axial force can be calculated and used to validate the correspondence between the shell part and the beam part

eeiiefftrue ApApNN −+= =‐5.07 MN+11.4 MN=6.33 MN

rN

n truecalc

π222 = =2.62 MN/m

And the expected true axial force based on the forces in the shell is

s

shellcalcx A

nr 222 ⋅=

πσ =116 MPa


50

The calculations here show a lower value of 22n than the shell element output value. In a further study it is necessary to create an analytical tool to verify the precision in the beam‐shell assembly. The main goal here is to decide which model to continue the analysis with. From the values in the tests with pressure no model seems more accurate than any other. However the MPC‐MPC model appears inappropriate to use due to need of more time to solve the problem than the other two models. During solving the message “a small pivot term” was more frequent for the MPC‐MPC model than for the other two assemblies. In the first small analysis where the beam‐shell connections ware investigated the BEAM4 elements gave a higher value of the deflection than the PIPE20only model and the MPC model. One possibility is that the BEAM4 connection is less stiff than the MPC connection. Based on this the model chosen for continue analysis is the MPC‐BEAM4 model.

5.5 Buckling of pipeline lying on trigger‐berm In the last analysis a pipeline lying on a one‐meter trigger‐berm and exposed to rising pressure and temperature is considered. The expansion of the pipeline due to the temperature rise will make it buckle. For the beam‐shell assembly the temperature load has to be applied in very small steps. The model used for the last analysis is 500 meter with a shell part of 40 meter. It is built using 4256 shell elements in the shell part. As the temperature rises part of the pipeline will be plastic. The strains in the beam‐shell assembly and the PIPE20only model are studied as a function of the temperature. To analysis are made. The first has a temperature rise of 95°C and the second analyse has a temperature rise of 111°C.

5.5.1 Vertical buckling As for the analysis of vertical buckling in chapter three the pipeline is locked lateral to force it to buckle vertical. When the temperature is about 5°C the pipeline deflects in both models. This indicates that they have similar stresses and effective axial force at that stage. The final configuration at 95°C of the vertical buckle beam‐shell assembly is found in Figure 50 See appendix 7.10 for input file.


51

Figure 50. Vertical buckling of a beam-shell assembly.

The elastic, plastic and total strains are found at the mid element on both models. For the beam‐shell assembly the top node at the top element of the cross‐section is used and for the PIPE20only model the integration point at the top is used. The strains at mid thickness are found as function of temperature in Figure 51 (95°C) and Figure 52 (111°C).

Temp °C

pe εε +plastε

elastε

ε

PIPE20only

Temp °C

pe εε +

plastε

elastε

ε

Beam‐shell

Temp °C

pe εε +plastε

elastε

ε

PIPE20only

Temp °C

pe εε +

plastε

elastε

ε

Beam‐shell

Figure 51. The models are exposed to a 95°C rising temperature. The corresponding strains as function of

the temperatures are showed.


52

Temp °C

pe εε +plastε

elastε

ε

PIPE20only

Temp °C

pe εε +plastε

elastε

Beam‐shell

ε

Temp °C

pe εε +plastε

elastε

ε

PIPE20only

Temp °C

pe εε +plastε

elastε

Beam‐shell

ε

Figure 52. The models are exposed to a 111°C rising temperature. The corresponding strains as function

of the temperatures are showed.

From the figures above it is seen that the strains are larger in the PIPE20 element than in the shell element, when the strains are over 1 percent as predicted.

Temp PIPE20only totε Beam‐shell assembly totε 95°C 0.96% 0.95% 111°C 1.7% 1.3%

This is satisfying as the beam‐shell assembly is supposed to deliver more accurate results. What is further interesting is how the plastic strains for the PIPE20 element are different after 40°C when the two analyses are compared. After this temperature the strains are found larger in the case where the total temperature rise is 111°C.

5.5.2 Lateral buckling Attempts with lateral buckling on trigger‐berm have been made using the beam‐shell assembly. However when the pipeline deflects lateral it loses the contact with the trigger. In the postbuckled stage it is lying on the seabed. Tests were made to increase the pinball radius, a function used in the contact algorithm to determine the contact status. But this only leads to convergence problems. The problem is probably due to the options made for the contact. In a development of the beam‐shell assembly the problems with lateral buckling should be studied further.

5.6 Conclusion of the beam‐shell assembly and further development The beam‐shell assembly is far from ready to be considered in a real analysis. It needs development and testing. The most critical points are listed below

• No analytical study of the shell elements that complement the comparison is done • Triangular SHELL181 elements were used even though they are not recommended • Gravity was not fully applied when the pipeline was initially laid down • A thin trigger is not possible to use with the penalty method as contact algorithm • In the lateral buckling case the pipeline was not able to stay on the trigger when

deflected. This is probably due to the options made for the contact.


53

However, the points indicating that a compliment model using beam‐shell assembly can be used are

• The elastic models seem to correspond • The beam‐shell assembly buckled as predicted in the vertical case. • When the strains exceed one percent the calculated strains are lower for the

SHELL181 element than for the PIPE20 element It is essential to bear in mind that the shell model represents a more exact simulation at the apex than the pure beam model. Factors that come in are ovalization and wall thickness changes that are included in SHELL181. And the restrictions on linear strain distribution do not come into the shell model. However, in various analyses the accuracy gained with a combined model is not needed and using a simplified model with only PIPE20 elements save time and resources. The combined model is therefore to be seen as a compliment to the PIPE20 model.


54

6 References [1] Inc. Ansys. Ansys 8.0 Documentation 2003. And Ansys 10.0 Documentation. 2005 [2] Det Norske Veritas. “Submarine Pipeline Systems”, DNV‐OS‐F101. 2000. [3 ] REINERTSEN Engineering AS. “Ormen Lange Project pipeline engineering”, doc.no.

11007600‐NH‐REE‐00003. (page 5) 2003 [4] Fyrileiv, O. and Collberg, L. “Influence of pressure in pipeline design‐effective axial

force”, Report OMAE2005‐67502.. 2005 [5] Carr, M., Bruton, D. and Leslie, D. “Lateral buckling and pipeline walking, a challenge for

hot pipelines”, Boreas paper no. OPT. 2003 [6] Soreide, T., Kvarme, S. O. and Paulsen G. Pipeline Expansion on Uneven Seabed. ISOPE

paper no. 2005‐HM‐03 [7] Det Norske Veritas. “Free Spanning Pipelines”, DNV‐RP‐F105. 2002 [8] Soreide, T., Kvarme, S.O. and Paulsen G. Nielsen, F.G. “Technical Challenges in Deep

Water Pipelines Design” [9] Hauch, S.R., Bai, Y. “Bending Moment Capacity of Pipes”. From Journal of Offshore

Mechanics and Arctic Engineering, ASME. 2000. [10] REINERTSEN Engineering AS. “ANSYS, Expansion and Global Buckling Analyses”,

doc.no. RE‐VER‐ANSYS. Page 57‐60. 2004 [11] Sundström, B. et al. ”Handbok och formelsamling i Hållfasthetslära”. Institutionen för

hållfasthetslära KTH. 1998. [12] D.L. DuQuesnay. “Mech 422 ‐ Stress and Strain Analysis”. 2002 [http://me.queensu.ca/courses/mech422/Notes422.pdf} [13] Nygård, T. Aglen, I. ”Pipe buckling and vortex induced vibration” Master thesis. 2004


i

7 Appendix

7.1 Appendix. Vertical buckling on even seabed

eff

sub

Nq

C = , EI

Nk eff=

2

21cos)( CxkxBAxw −+=

CxkxBkxw −−=′ sin)( CkxBkxw −−=′′ cos)( 2

kxBkxw sin)( 3=′′′ Boundary condition 1,2) 0)( =Lw , 0)( =′ Lw

1) 021cos 2 =−+ CLkLBA

2) 0sin =−− CLkLBk

kLk

LCBsin

−= , kLkL

kLCCLA

sincos

21 2 +=

22

21

sincos

sincos

21)( x

Nq

kLkx

kL

Nq

kLkL

kL

Nq

LNq

xweff

sub

eff

sub

eff

sub

eff

sub −−+=

Boundary condition 3) 0)()( =′′= LwEILM 0)cos()( 2 =−−=′′ CkLBkEILwEI

01sin

cos=−

kLkLkL

, kLkL =tan

By using Mathcad function “Given find”

0 2 4 6 80

2

4

6

8

tan x( )

x

x

Given

tan u( ) u

u 4:=

u Find u( ):=

u 4.493=

0 2 4 6 80

2

4

6

8

tan x( )

x

x

Given

tan u( ) u

u 4:=

u Find u( ):=

u 4.493=

effNEIL 5.4=


ii

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⋅+=

2cos6.41.11)(

2xxEI

NNEI

NEI

Nq

xw eff

effeffeff

sub

Maximum deflection at 0=x

verteff

sub

effeffeff

sub

NEIq

NEI

NEI

Nq

w δ==⎟⎟⎠

⎞⎜⎜⎝

⎛+= 27.156.41.11)0(

The needed axial effective force to initiate vertical buckling is then

096.3vert

sublifteff

EIqN

δ=


iii

7.2 Appendix. Lateral buckling on even seabed

,2

cossin)(,2

cos)( 232102

2101 x

Nq

kxBkxBxBBxvxNq

kxAAxveff

sub

eff

sub ⋅++++=⋅−+=

CxkxkAxv −−=′ sin)( 11 CkxkAxv −−=′′ cos)( 2

11 kxkAxv sin)( 3

11 =′′′ CxkxkBkxkBBxv +−+=′ sincos)( 3212

CkxkBkxkBxv +−−=′′ cossin)( 23

222

kxkBkxkBxv sincos)( 33

322 +−=′′′

The five first boundary conditions given by continuity 1‐5) 0)0()(),0()(),0()(),0()( 211211211211 ==′=′′′=′′′′′=′′′ vLvvLvvLvvLv 1) 3

213

1 sin kBkLkA −= 2) CkBCkLkA +−=−− 2

312

1 cos 3) kBBCLkLkA 21111 sin +=−−

4) 02

cos 21110 =⋅−+ LCkLAA

5) 030 =+ BB The last three boundary conditions at end, 2L 6‐8) 0)(,0)(,0)( 222222 =′′=′= LvLvLv

6) 02

cossin 222322210 =⋅++++ LCkLBkLBLBB

7) 0sincos 223221 =+−+ CLkLkBkLkBB

8) 0cossin 22

322

2 =+−− CkLkBkLkB BC 1‐6 This is calculations of the unknown value A0‐B3. They are to be multiplied

witheff

sub

NEIq ⋅µ

21 , kLbkLa ==

)sin(cos)sin2(

21 2

0 baabbaaA+

+−+=

)sin(sin2

1 babbaA

++−

−=

)sin(cos)sin2(20 ba

abbaB+

+−+−=

aB −=1

)sin(sin)sin2(2 ba

abbaB+

+−=

)sin(cos)sin2(23 ba

abbaB+

+−−=

BC 7‐8


iv

0)sin()cos()sin2(cos21 =

++

+−+−bababbab

0)sin()cos()sin2(

21cos2

)sin(cos)sin(2(2 2 =

++

+−++−−+

+−−bababbaabbb

baabba

By using Mathcad function “Given find”

Given

1 2 cos b( )⋅− a b− 2 sin b( )⋅+( )cos a b+( )sin a b+( )⋅+ 0

2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅− 2 cos b( )⋅− 0.5 b

2⋅− a b⋅+ a b− 2 sin b( )⋅+( )

cos a b+( )sin a b+( )⋅+ 0

a 3:= b 4.2:= start values

a

b⎛⎜⎝⎞⎟⎠

Find a b,( ):= a 2.9182862455= b 4.6327067738=

noggrannhet

1 2 cos b( )⋅− a b− 2 sin b( )⋅+( )cos a b+( )sin a b+( )⋅+ 2.2210661421 10

7−×=

2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅− 2 cos b( )⋅− 0.5 b

2⋅− a b⋅+ a b− 2 sin b( )⋅+( )

cos a b+( )sin a b+( )⋅+ 3.2642446524− 10

9−×=

A0 0.5a2

a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅+:=

A1 a b− 2 sin b( )⋅+( )−1

sin a b+( )⋅:=

B0 2− a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅+:=

B1 a−:=

B2 a b− 2 sin b( )⋅+( )sin a( )

sin a b+( )⋅:=

B3 2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅−:=

A0 8.0467759623=

A1 3.8850420758=

B0 1.7885786569=

B1 2.9182862455−=

B2 0.860362535−=

B3 1.7885786569−=

a 2.9182862455=

b 4.6327067738=

Given


2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅− 2 cos b( )⋅− 0.5 b

2⋅− a b⋅+ a b− 2 sin b( )⋅+( )

cos a b+( )sin a b+( )⋅+ 0


a

b⎛⎜⎝⎞⎟⎠

Find a b,( ):= a 2.9182862455= b 4.6327067738=

noggrannhet


7−×=

2 a b− 2 sin b( )⋅+( )cos a

Given


2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅− 2 cos b( )⋅− 0.5 b

2⋅− a b⋅+ a b− 2 sin b( )⋅+( )

cos a b+( )sin a b+( )⋅+ 0


a

b⎛⎜⎝⎞⎟⎠

Find a b,( ):= a 2.9182862455= b 4.6327067738=

noggrannhet


7−×=

2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅− 2 cos b( )⋅− 0.5 b

2⋅− a b⋅+ a b− 2 sin b( )⋅+( )

cos a b+( )sin a b+( )⋅+ 3.2642446524− 10

9−×=

A0 0.5a2

a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅+:=

A1 a b− 2 sin b( )⋅+( )−1

sin a b+( )⋅:=

B0 2− a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅+:=

B1 a−:=

B2 a b− 2 sin b( )⋅+( )sin a( )

sin a b+( )⋅:=

B3 2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅

( )sin a b+( )⋅− 2 cos b( )⋅− 0.5 b

2⋅− a b⋅+ a b− 2 sin b( )⋅+( )

cos a b+( )sin a b+( )⋅+ 3.2642446524− 10

9−×=

A0 0.5a2

a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅+:=

A1 a b− 2 sin b( )⋅+( )−1

sin a b+( )⋅:=

B0 2− a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅+:=

B1 a−:=

B2 a b− 2 sin b( )⋅+( )sin a( )

sin a b+( )⋅:=

B3 2 a b− 2 sin b( )⋅+( )cos a( )

sin a b+( )⋅−:=

A0 8.0467759623=

A1 3.8850420758=

B0 1.7885786569=

B1 2.9182862455−=

B2 0.860362535−=

B3 1.7885786569−=

a 2.9182862455=

b 4.6327067738=

( )21 5.0cos885.2047.8)( xkx

NEIq

xveff

sub −+⋅

=µ

( )22 5.0cos789.1sin86.0918.2789.1)( xkxkxx

NEIq

xveff

sub +−−−⋅

=µ


v

ufeedin1 A12 a

2⋅⎛⎜⎝⎞⎟⎠

A12 sin 2 a⋅( )

4⋅⎛⎜⎝

⎞⎟⎠− 2 A1⋅ sin a( ) a cos a( )⋅−( )⋅+ a

3

3

⎛⎜⎝

⎞⎟⎠

+:=

ufeedin2 B12

b a−( )⋅ B22 sin 2 b⋅( )

4⎛⎜⎝

⎞⎟⎠b2⎛⎜⎝⎞⎟⎠

− sin 2 a⋅( )4

⎛⎜⎝⎞⎟⎠

− a2⎛⎜⎝⎞⎟⎠

+⎡⎢⎣⎤⎥⎦

⋅+ B32 sin 2 b⋅( )

4− b

2+ sin 2 a⋅( )

4+ a

2⎛⎜⎝⎞⎟⎠

−⎡⎢⎣⎤⎥⎦

⋅+

b3

3

⎛⎜⎝

⎞⎟⎠

a3

3

⎛⎜⎝

⎞⎟⎠

−⎡⎢⎣

⎤⎥⎦

+

2 B1⋅ B2⋅ sin b( ) sin a( )−( )⋅+ 2 B1⋅ B3⋅ cos b( ) cos a( )−( )⋅+ B1 b2

a2

−( )⋅+ B2 B3⋅ sin b( )2

sin a( )2

−( )⋅−

2 B2⋅ cos b( ) b sin b( )⋅+ cos a( )− a sin a( )⋅−( )⋅+ 2 B3⋅ sin b( ) b sin b( )⋅− sin a( )− a sin a( )⋅+( )⋅−

:=

ufeedin1 55.7708928572= ufeedin2 27.192590769= ufeedin ufeedin1 ufeedin2+:=

dlat A0 A1+:=dlat 11.9318180381= som funktion av neff

Neff dlat:=Neff 3.4542463777=

Mlat A1− 1−( )−:=Mlat 4.8850420758= som funktion av neff

MlatdMlatNeff

:=Mlatd 1.4142135626= som funktion av deltalat

deltalatudlat

ufeedin

4

7

:= ufeedin 82.9634836262=

deltalatu 0.9554319851= som funktion av feedin

Mlatu Mlatd deltalatu⋅:= Mlatu 1.3823400344= som funktion av feedin

uallow1

Mlatu

7

2

:= uallow 0.321994089= som funktion av Mallow

...

...++

ufeedin1 A12 a

2⋅⎛⎜⎝⎞⎟⎠

A12 sin 2 a⋅( )

4⋅⎛⎜⎝

⎞⎟⎠− 2 A1⋅ sin a( ) a cos a( )⋅−( )⋅+ a

3

3

⎛⎜⎝

⎞⎟⎠

+:=

ufeedin2 B12

b a−( )⋅ B22 sin 2 b⋅( )

4⎛⎜⎝

⎞⎟⎠b2⎛⎜⎝⎞⎟⎠

− sin 2 a⋅( )4

⎛⎜⎝⎞⎟⎠

− a2⎛⎜⎝⎞⎟⎠

+⎡⎢⎣⎤⎥⎦

⋅+ B32 sin 2 b⋅( )

4− b

2+ sin 2 a⋅( )

4+ a

2⎛⎜⎝⎞⎟⎠

−⎡⎢⎣⎤⎥⎦

⋅+

b3

3

⎛⎜⎝

⎞⎟⎠

a3

3

⎛⎜⎝

⎞⎟⎠

−⎡⎢⎣

⎤⎥⎦

+

2 B1⋅ B2⋅ sin b( ) sin a( )−( )⋅+ 2 B1⋅ B3⋅ cos b( ) cos a( )−( )⋅+ B1 b2

a2

−( )⋅+ B2 B3⋅ sin b( )2

sin a( )2

−( )⋅−

2 B2⋅ cos b( ) b sin b( )⋅+ cos a( )− a sin a( )⋅−( )⋅+ 2 B3⋅ sin b( ) b sin b( )⋅− sin a( )− a sin a( )⋅+( )⋅−

:=

ufeedin1 55.7708928572= ufeedin2 27.192590769= ufeedin ufeedin1 ufeedin2+:=

dlat A0 A1+:=dlat 11.9318180381= som funktion av neff

Neff dlat:=Neff 3.4542463777=

Mlat A1− 1−( )−:=Mlat 4.8850420758= som funktion av neff

MlatdMlatNeff

:=Mlatd 1.4142135626= som funktion av deltalat

deltalatudlat

ufeedin

4

7

:= ufeedin 82.9634836262=

deltalatu 0.9554319851= som funktion av feedin

Mlatu Mlatd deltalatu⋅:= Mlatu 1.3823400344= som funktion av feedin

uallow1

Mlatu

7

2

:= uallow 0.321994089= som funktion av Mallow

...

...++

)0(1vlat =δ

)( 0lat

lateff fN δ=


vi

7.3 Appendix. Combined buckling

96.3,0 == bEIq

bNvert

sublifteff δ

45.3,0

*

== aEIq

aNlat

sublateff δ

µ

⎟⎟⎠

⎞⎜⎜⎝

⎛−= lift

eff

effsubsub N

Nqq 1*

When the needed force to initiate lateral sliding is reached lateffeff NN =

lifteff

lifteff

sub

sub

NN

qq

−= 1*

sub

sub

lat

vert

vert

sub

lat

sub

lifteff

lateff

qq

ba

EIqb

EIqa

NN *

0

0

0

0

*

δµδ

δ

δµ

==

{

01 0

02

0

02

2

0

022

=⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛⋅+

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⇒⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛

lat

vert

k

lat

vertlifteff

lifteff

x

lifteff

lateff

lifteff

lifteff

lat

vertlifteff

lateff

ba

ba

NN

NN

NN

ba

NN

δµδ

δµδ

δµδ

43421

⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+±=⇒=−⋅+

>44 344 21

0

2 141

210

kkxkkxx

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛++−⋅⎟

⎠⎞

⎜⎝⎛⋅⋅= 0

022

0

0

41

21

vert

lat

lat

vertlifteff

lateff a

bbaNN

µδδ

δδ

µ


vii

7.4 Appendix. Parameters and variables Parameter Value Unit Description

eD 0.9 m External diameter

sD 0.8 m Steel diameter

iD 0.75 M Internal diameter

steelE 207 GN/m Young’s modulus

υ 0.3 Poisson’s ratio alfa 12 10‐6/°C Heat coefficient

wρ dens 1027 kg/m3 Density water

cρ 2000 kg/m3 Density concrete

sρ 7850 kg/m3 Density steel

gρ 300 kg/m3 Density gas

ip 30 MPa Internal pressure in operation

ep 3.7 MPa External pressure

T∆ 95 °C Temperature rise

The variables 0vertδ 1,2,3 m Trigger‐berm height µ 0.4, 0.6, 0.8, 1 Friction coeff

Values used for PIPE20 in ANSYS Slot 0.02 m Maximum allowable elastic slip,

mobilisation length Fnk soil 1200 kN/m/m Normal stiffness/meter (for soil) Fnk trigger 1800 kN/m/m Normal stiffness/meter (for trigger)

layeffN 245 kN Residual lay tension

Trigger length 0.25 m Length of trigger Pipe length 1000 (vertical)

500 (lateral) m Length of pipeline


viii

7.5 Appendix. ANSYS input files for elastic PIPE20 The input files used in ANSYS are almost the same for the cases examined in chapter 3. Therefore the parts that are similar are only given once.

Input parameters and the modelling of the pipeline are similar

!----PREPROCESSOR------------------------------------------------------------- /prep7 /nopr !----PARAMETERS--------------------------------------------------------------- !-----pipeline parameters od= 0.8 ![m] outer diameter of pipe(steel) wt= 0.025 ![m] thickness of pipe(steel) length= 1000 ![m] length of entire pipeline overbed=4 ![m] pipe is initially 4 meter over the seabed !-----Steel parameters Es= 207e9 ![N/m2] Youngs modulus ny= 0.3 ![] poisson denss= 7850 ![kg/m3]density alfas= 12e-6 ![/*C] thermal !----link parameters lengthlink= 10 ![m] the linknodes is lengthlink long distancelinks= 5 !distance between pipenodes with links downlength=overbed ![m] the length the links will be downloaded !-----seabed parameters width= 10 ![m] +-y-value of seabed mulat= 0.6 ![-] lateral friction coefficient fknpm= -1200e3 ![N/m/m] vertical soil stiffness/meter slto= -0.02 ![m] allowable elastic slip, mob-length fknpmt= -1800e3 ![N/m/m] trigger stiffness triggheight= 1 ![m] height of trigger trigglength= 0.25 ![m] length of trigger weststart= -1 !start x value of seabed eastend= length+1 !end x value of sebed triggwest= (length/2)-trigglength/2 !start x value of triggerberm triggeast= (length/2)+trigglength/2 !end x value of triggerberm !----load parameters Nlay= 245e3 ![N] residual lay tension extpress= 3.7e6 ![Pa] external pressure intpress= 30e6 ![Pa] delta_pi temperature= 95 ![*C] delta_temp grav= 9.81 ![m/s2]


ix

densg= 300 ![kg/m3] density of gas densc= 2000 ![kg/m3] density of concrete densw= 1027 ![kg/m3] density of water Ag= 0.44178647 ![m2] internal area As= 0.06086836 ![m2] steel area Ac= 0.13351769 ![m2] area of concrete Ae= 0.63617251 ![m2] external area gasweight=densg*Ag*grav ![N/m] weight of gas steweight=denss*As*grav ![N/m] weight of steel conweight=densc*Ac*grav ![N/m] weight of concrete watweight=densw*Ae*grav ![N/m] buoyancy subweight=gasweight+steweight+conweight-watweight

!-----CREATING PIPE------------------------------------------ !-----pipeline elementtype, real constants and materialproperty- et,1,PIPE20 keyopt,1,6,1 r,1,od,wt mp,ex,1,Es mp,prxy,1,ny mp,dens,1,denss mp,alpx,1,alfas !-----nodes and elements in pipeline- nodp1= 1 !first nodenumber in pipe nodpn= 199 !last nodenumber in pipe (odd) nelemp= nodpn-1 !number of elements in pipe midnode=(nodpn+1)/2 elength=length/nelemp !lenght of an element n,nodp1,0,0,overbed !position of first pipenode n,nodpn,length,0,overbed !position of last pipenode fill,nodp1,nodpn !fill a row of nodes between first and…

!last pipenode numstr,elem,1 !element numbering from 1 e,1,2 !create a pipeelement between…

!node 1 and node 2 *repeat,nelemp,1,1 !create element between laststart+1…

!node,laststop+1 node. Number of… !in pipe elements times

nsel,all nsel,s,node, ,1,nodpn !select the pipenodes cm,pipenodes,node !make the pipenodes named pipenodes nsel,all nsel,s,node, ,2,nodpn-1 !select the midpipenodes cm,pipemidnodes,node !name the midpipenodes pipemidnodes


x

nsel,all esel,all esel,s,type, ,1 !select the pipe elements by type cm,pipeelem,elem !name pipeelement pipeelem esel,all !-----linkelement, real constants and materialproperty- et,2,LINK10 keyopt,2,2,0 !keypot(2)=0 tension-only the link…

!work as a wire r,2,0.01 !the links area mp,ex,2,2e11 !and the young modulus k=EA=high type,2 real,2 mat,2 !-----nodes and elements in links- linki= 201 !start link nodes with n,201 numstr,elem,201 !start element numbering from 201 (top 75 elem) *do,pipei,1,nodpn-1,distancelinks !for

!pipei=1:distancelinks:nodpn-1 *get,xpos,node,pipei,loc,x !get the x-location-position *get,ypos,node,pipei,loc,y !get the y-location-position n,linki,xpos,ypos,(overbed+lengthlink) !create linknode e,linki,pipei !create link element linki=linki+1 *enddo n,linki,length,0,(overbed+lengthlink) !last linknode above last…

!pipenode e,linki,nodpn !create last link element ! z^ ! | ! | . . . . . ('linknodes') ! | | | | | | ! | z z z z z (links) ! | | | | | | ! |=================== (pipeline) ! |------------------->x esel,all esel,s,type, ,2 !select the link elements by type cm,linkelem,elem !name linkelement linkelem esel,all nsel,all


xi

nsel,s,node, ,201,linki !select the link nodes cm,linknodes,node !name linknodes linknodes nsel,all

For the cases with a vertical trigger more seabed elements are used

!----CREATING SEABED----------------------------------------------------------- !-----seabed elementtypes, real constants and materialproperty- fkn=fknpm*elength ![N/m/'e'] vertical soil stiffness…

!per pipeelement fknt=fknpmt*elength !normal soil stiffness for trigger !----contact et,3,CONTA175 !node to surface contact keyopt,3,2,1 !contact keyopt(2)=1, penalty method keyopt,3,10,5 !contact keyopt(10)=5, fkt is updated at each iteration r,3, , ,fkn !real constant nr 3 is fkn rmodif,3,23,slto !real constant nr 23 is slto mp,mu,3,mulat !material property friction r,4, , ,fknt rmodif,3,23,slto !-----target------ et,4,TARGE170 !3D-surface tshap,quad !target shape 4-node quadrilateral !-----nodes in west seabed- n,301,weststart,-width,0 !first nodenumber 'south' in…

!'west'seabed n,302,triggwest,-width,0 !last nodenumber 'south' in 'west' seabed n,304,weststart,width,0 !first nodenumber 'north' in…

!'west' seabed n,303,triggwest,width,0 !last nodenumber 'north' in 'west' seabed ! 304---------303 ! |west | ! start|=======|===(pipeline) ! | | ! 301---------302 !-----nodes in east seabed- n,311,triggeast,-width,0 !first nodenumber 'south' in…

!'east' seabed n,312,eastend,-width,0 !last nodenumber 'south' in 'east' seabed n,314,triggeast,width,0 !first nodenumber 'north' in…


xii

!'east' seabed n,313,eastend,width,0 !last nodenumber 'north' in 'east' seabed ! 314---------313 ! |east | ! ==|=======| end(pipeline) ! | | ! 311---------312 !-----nodes in trigger-berm- n,321,triggwest,-width,triggheight !'west' nodenumber 'south'…

! in trigger n,322,triggeast,-width,triggheight !'east' nodenumber 'south'…

!in trigger n,324,triggwest,width,triggheight !'west' nodenumber 'north'…

!in trigger n,323,triggeast,width,triggheight !'east' nodenumber 'north'…

!in trigger ! 324,323 ! ---------- ! |w || e| ! |========|(pipeline) ! | || | ! ---------- ! 321,322 !-----create target element numstr,elem,290 !element numbering from 290 type,4 !element targe170 is used to make up the seabed real,3 !with real constants set 3 e,301,302,303,304 !create three targetelement counterclockwise…

!making them point towards the pipenodes e,311,312,313,314 !normal point towards the pipenodes type,4 ! real,4 !real constant set 4 for trigger e,321,322,323,324 !create trigger esel,all esel,s,type, ,4 !select the targe elements by type cm,targeelem,elem !name targeelement targeelem esel,all !-----create contact element- numstr,elem,301 !element numbering from 301 type,3 !element conta175 is used real,3 !with real constants mat,3 !and material property e,1 !create a contactelement at pipenode 1


xiii

*repeat,midnode-1,1 !creating all contactelements one for… !each pipenode

type,3 ! real,4 !real constant set 4 for midnode mat,3 ! e,midnode ! type,3 !element conta175 is used real,3 !with real constants mat,3 !and material property e,midnode+1 ! *repeat,midnode-1,1 !

For lateral and no buckling scenarios the seabed is even

!----CREATING SEABED----------------------------------------------------------- !-----seabed elementtypes, real constants and materialproperty- fkn=fknpm*elength ![N/m/'e'] vertical soil stiffness per pipeelement !----contact et,3,CONTA175 !node to surface contact keyopt,3,2,1 !contact keyopt(2)=1, penalty method keyopt,3,10,5 !contact keyopt(10)=5, fkt is updated at each iteration r,3, , ,fkn !real constant nr 3 is fkn rmodif,3,23,slto !real constant nr 23 is slto mp,mu,3,mulat !material property friction !-----target et,4,TARGE170 !3D-surface tshap,quad !target shape 4-node quadrilateral !-----nodes in west seabed- n,301,weststart,-width,0 !first nodenumber 'south' in…

!'west' seabed n,302,eastend,-width,0 !last nodenumber 'south' in 'west' seabed n,304,weststart,width,0 !first nodenumber 'north' in…

!'west' seabed n,303,eastend,width,0 !last nodenumber 'north' in 'west' seabed ! 304---------303 ! w| |e ! start|=======|end(pipeline) ! | | ! 301---------302 !-----create target element


xiv

numstr,elem,290 !element numbering from the 290 type,4 !element targe170 is used to make up the seabed real,3 !with real constants set 2 e,301,302,303,304 !create three targetelement counterclockwise…

!making the point towards the pipenodes !-----create contact element- numstr,elem,301 !element numering from 301 type,3 !element conta175 is used real,3 !with real constants mat,3 !and material property e,1 !create a contactelement at pipenode 1 *repeat,nodpn,1 !creating all contactelements one for…

!each pipenode esel,all esel,s,type, ,4 !select the targe elements by type cm,targeelem,elem !make the targeelement named targeelem esel,all

The solving is similar from step 1 to 5 for the cases

!----SOLVING------------------------------------------------------------- /solu antype,static,new !New static analyse solcontrol,on !solution control on nlgeom,on !includes large-deflection effects autots,on !automatic timestepping on nropt,unsym, ,off sstif,off !strength stiffening off cnvtol,f, ,0.01, ,1 !force tolerance cnvtol,u, ,0.05 !displacement tolerance outres,all,all !save all result data for all substeps /nopr !----BOUNDARY CONDITIONS-------------------------------------------------- time,1 !-Pipeline- d,linknodes,all,0 !all linknodes are restrained d,nodp1,ux,0 !the first pipenode cant move axially f,nodpn,fx,Nlay !the last pipenode has a axial force nsubst,1,1,1 solve


xv

!----TIME/LOAD STEP 2----------------------------------------------------- !-Initial loads- time,2 sfe,pipeelem,5,pres, ,extpress !apply external pressure…

!on pipe f,pipenodes,fz,-(subweight)*elength !submerged weight nsubst,5,10,4 ! solve !----TIME/LOAD STEP 3------------------------------------------------------- !-lower the pipeline down- time,3 d,linknodes,uz,-downlength !lowering the links nsubst,10,15,8, solve !----TIME/LOAD STEP 4------------------------------------------------------- !-lock the pipeline at ends and take away Nlay- time,4 fdele,nodpn,fx !remove force at endnode d,nodp1,all,%_fix% !lock first pipenode d,nodpn,all,%_fix% !lock last pipenode nsubst,1,1,1 ! solve !----TIME/LOAD STEP 5-------------------------------------------------------- !-kill the links- time,5 ekill,linkelem !the link elements arent needed nsubst,1,1,1, ! solve

Step 6 for Vertical buckling on trigger berm

!----TIME/LOAD STEP 6-------------------------------------------------------- !-apply constraines for pipe in lateral (y) direction- time,6 d,pipemidnodes,uy,%_fix% !dof-restraines for mid-pipenodes


xvi

nsubst,1,1,1 !for the vertical-only case solve

Step 6 for Lateral buckling on trigger‐berm

!----TIME/LOAD STEP 6-------------------------------------------------------- !-apply constraines for pipe in lateral (y) direction- time,6 f,midnode,fy,100 !lateral push nsubst,5,10,5 ! solve

Step 7‐8 and postprocessor for Vertical and lateral buckling on trigger‐berm

!----TIME/LOAD STEP 7-------------------------------------------------------- !-internal pressure rise- time,7 sfe,pipeelem,1,pres, ,intpress !apply internal pressure on pipe nsubst,25,100,25 solve !----TIME/LOAD STEP 8-------------------------------------------------------- !-temperature rise- time,8 bf,pipenodes,temp,temperature !bodyforce, temperature on pipenodes nsubst,25,100,25 ! solve !-----POSTPROCESSOR------------------------------------------------------------- /post1 cmsel,s,pipeelem etable,deflect,u,z !deflection etable,Neff,smisc,1 !Neff etable,shearfor,smisc,2 !Shear force etable,moment,smisc,5 !Moment etable,xstress0,ls,1 !axial stress 0* etable,xstres45,ls,5 !axial stress 45* etable,xstres90,ls,9 !axial stress 90* etable,xstre135,ls,13 !axial stress 135* etable,xstre180,ls,17 !axial stress 180* etable,xstre225,ls,21 !axial stress 225*


xvii

etable,xstre270,ls,25 !axial stress 270* etable,xstre315,ls,29 !axial stress 315* cmsel,all !-----TIME/HISTORY------------------------------------------------------------- /post26 numvar,200 timerange,3 theelement=nelemp/2 !the interesting pipe element thenode=theelement+1 !the equally interesting pipe node nsol,2,thenode,u,z,dVert0 !deflection for highest point esol,3,theelement,thenode,f,x,Neff !and the axial force esol,4,theelement,thenode,m,y,momenty ! esol,5,84,85,f,z,shear75 esol,8,86,87,f,z,shear65 esol,9,88,89,f,z,shear58 /grtyp,2 !xvar,5 !plvar,3 plvar,5,6,3 /axlab,y,- /axlab,x,time /devisep,font,1,menu /devisep,font,3,menu /replot prvar,3,5,6

Step 6‐9 for Lateral buckling on even seabed. For the “no buckling” scenario tryckbuck=0

!----TIME/LOAD STEP 6-------------------------------------------------------- !-make it imperfect- time,6 halvbuck=40 elembuck=nint(halvbuck/elength) tryckbuck=1500 ![N/m] kraftbuck=tryckbuck*elength f,midnode-elembuck,fy,kraftbuck, ,midnode+elembuck ! nsubst,10,10,4 !nsubst,1,1,1 solve !----TIME/LOAD STEP 7--------------------------------------- !-remove lateral force and add internal pressure- time,7


xviii

sfe,pipeelem,1,pres, ,intpress !apply internal pressure…

!on pipe f,midnode-elembuck,fy,0, ,midnode+elembuck !remove preforce nsubst,10,100,8 ! solve !----TIME/LOAD STEP 8-------------------------------------------------------- !-temperature rise- time,8 sfe,pipeelem,1,pres, ,intpress !apply internal pressure on pipe bf,pipenodes,temp,temperature !apply temp on pipe nsubst,20,100,20 ! solve !-----POSTPROCESSOR-------------------------------------------- /post1 cmsel,s,pipeelem etable,deflect,u,y !deflection etable,Neff,smisc,1 !Neff etable,shearfor,smisc,2 !Shear force etable,moment,smisc,5 !Moment etable,xstress0,ls,1 !axial stress 0* etable,xstres45,ls,5 !axial stress 45* etable,xstres90,ls,9 !axial stress 90* etable,xstre135,ls,13 !axial stress 135* etable,xstre180,ls,17 !axial stress 180* etable,xstre225,ls,21 !axial stress 225* etable,xstre270,ls,25 !axial stress 270* etable,xstre315,ls,29 !axial stress 315* etable,hstress0,ls,3 !hoop stree 0* etable,hstre180,ls,19 !hoop stree 180* pretab,xstress0,xstres45,xstres90,xstre135 pretab,xstre180,xstre225,xstre270,xstre315 cmsel,all !-----TIME/HISTORY------------------------------------------------------------- /post26 numvar,200 timerange,4 theelement=midnode-1 !the interesting pipe element


xix

nsol,2,midnode,u,y,dLat0 !deflection for highest point esol,3,theelement,midnode,f,x,Neff !and the axial force esol,4,theelement,midnode,m,y,Moment !and the axial force /grtyp,2 plvar,2,3 /axlab,y,Neff, dLat /axlab,x,time /devisep,font,1,menu /devisep,font,3,menu /replot prvar,2,3


xx

7.6 Appendix. Graphs of vertical and lateral deflection Vertical case. Vertical buckling on two meter trigger

Vertical buckling on three meter trigger


xxi

Moment at one meter

Shear force on one meter trigger


xxii

Lateral case Deflection and effective axial force. 4.0=µ

8.0=µ


xxiii

0.1=µ


xxiv

7.7 Appendix. Moments in partly plastic cross‐section.

partplastic

y

partelastic

y drtdrta

rdAM ϕϕσϕϕσ

ϕϕσπ

ϕ

ϕπ

)sin(4)(sin4)sin()( 22/

23

0

2

00

0

⋅+⋅=⋅⋅= ∫∫∫

Elastic part )sin( 0ϕra =

∫∫∫ −==⋅000

00

2

0

2

0

2

0

23

0 2)2cos(

21

)sin(4

sin)sin(

4)sin()(sin4

ϕϕϕ ϕϕ

σϕϕ

ϕσ

ϕϕϕσ

trd

trdrt yy

y

⎥⎦⎤

⎢⎣⎡ −

4)2sin(

2)sin(4 00

0

2 ϕϕϕ

σ try

Plastic part

)cos(4)sin(4 022

2/

0

ϕσϕϕσπ

ϕ

rtdrt yy ⋅=⋅∫

And together

⎥⎦

⎤⎢⎣

⎡+−⋅=⋅+⎥⎦

⎤⎢⎣⎡ −= )cos(

)sin(4)cos()sin(2

)sin(24)cos(4

4)2sin(

2)sin(4

00

00

0

020

200

0

2

ϕϕ

ϕϕϕ

ϕσϕσ

ϕϕϕ

σrtrt

trM yy

y

⎥⎦

⎤⎢⎣

⎡+⋅=⎥

⎦

⎤⎢⎣

⎡+−⋅

1)cos(

)sin(2)cos(

2)cos(

)sin(24 0

0

020

0

0

02 ϕϕ

ϕσϕ

ϕϕ

ϕσ rtrt yy


xxv

7.8 Appendix. Input files for short beam in ANSYS and PAS ANSYS input file

/clear /filname,plastichinge,on !----PREPROCESSOR----------------------------------------- /prep7 /nopr !----PARAMETERS------------------------------------------------- !-----pipeline parameters od= 0.8 ![m] outer diameter of pipe (steel) wt= 0.025 ![m] thickness of pipe (steel) length= 2 ![m] length of entire pipeline !-----Steel parameters Es= 207e9 ![N/m2] Youngs modulus ny= 0.3 ![] poisson denss= 7850 ![kg/m3]density alfas= 12e-6 ![/*C] thermal smys= 448e6 ![N/m] yield stress epsys= smys/Es sist= 450e6 !bilinear epsist= 1000e-3 !bilinear krok=0.018 !helt elast krok<0.01117rad /title, pipe utsatt for krokning %krok% rad !-----CREATING PIPE----------------------------------- !-----pipeline elementtype, real constants and materialproperty- et,1,PIPE20 r,1,od,wt mp,ex,1,Es mp,prxy,1,ny mp,dens,1,denss mp,alpx,1,alfas tb,kinh,1,1,2 !kinematic hardening with 2 points…

!one temp tbtemp,0.0 tbpt, ,epsys,smys !first point tbpt, ,epsist,sist !second point !-----nodes and elements in pipeline- nodp1= 1 !first nodenumber in pipe nodpn= 2 !last nodenumber in pipe nelemp= nodpn-1 !number of elements in pipe n,nodp1,0,0,0 !position of first pipenode


xxvi

n,nodpn,length,0,0 !position of last pipenode e,nodp1,nodpn !create a pipeelement between…

!node 1 and node 2 !----SOLVING------------------------------------------------------------- /solu antype,static,new !New static analyse solcontrol,on !solution control on nlgeom,on !includes large-deflection effects autots,on !automatic timestepping on cnvtol,f, ,0.01, ,1 !force tolerance cnvtol,u, ,0.05 !displacement tolerance nropt,unsym, ,off pred,off sstif,off outres,all,all time,1 d,nodp1,all,0 nsubst,1,1,1 solve time,2 d,nodpn,rotz,-krok nsubst,15,100,15 solve

Input files for PAS

File 1

EMPTY pasep Study of plastic behaviour 21 integrationspoints on half cross section deflection 0.011 START 1 3 2 1 2 3 3 0 0 1 5 1 2 15 1 100 0 100 0 1 1 0 0.0001 2 21 1 ONE 1 INITIAL X-COORDINATES 1 0 2 2 1 0 2 0 BOTTOM DATA MATERIAL 1027.0 1.2E-5 0.3 2.1643E-3 448.0


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15.000E-3 449.0 CROSS-SECT DATA 1 0.75000 0.0250 1 0.37500 0.0000 1 0.75000 7850.0 LOAD DATA 1 27.75 10.0 250E3 CONCENTRATED LOADS HISTORIES 1 0.0 0.0 0.00 0.0 0.0 0.0 15 0.0 0.0 0.00 0.0 0.0 1.0 BOUNDARY COND 1 1 1 2 1 3 POINT SPRINGS PRESCRIBED DEFLECTIONS 2 3 0.011 SPRING STIFFNESS -1000.0 5000000.0 0.60 0.90 0.005 1.0 100.0 5.0 0.0 0.0 0.005 1.0 100.0 5.0 -100.0 -1.0E7 0.00 0.00 10.0 0.0 PRINT DATA 15 1 3 ONE 1

File 2

0.0 -100 3.0 -100


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7.9 Appendix. Matlab input file for study of integration points clc; clear all; close all; De=0.8; Di=0.75; E=207e9; I=pi/64*(De^4-Di^4); EI=E*I; t=(De-Di)/2; Dm=De-t; r=Dm/2; sigy=448e6; L=2; wp=0.03; a=(sigy*L)/(E*wp); fi=asin(a/r); moment=[]; intpunkt=[]; for ip=5:1:50 iphel=(ip-2)*2+2; dfi=2*pi/iphel; elast=[]; plasti=[]; for i=0:dfi:(2*pi-dfi) if abs(sin(i))<=abs(sin(fi)) %elast elast=[elast;abs((sin(i)^2)/sin(fi))]; else plasti=[plasti;abs(sin(i))]; %plast end end M=dfi*r^2*t*sigy*(sum(elast)+sum(plasti)); moment=[moment;M]; intpunkt=[intpunkt;ip]; end Manal=(2*r)^2*sigy*t*((fi/(2*sin(fi)))+(cos(fi)/2)); Maxm=max(moment); Minm=min(moment); Uskillnad=max(moment)-Manal; Nskillnad=Manal-min(moment); elastpro=a/r; Manal=Manal*ones(length(intpunkt)); figure(1) plot(intpunkt,moment,intpunkt,moment,'*',intpunkt,Manal) ylabel('Moment') xlabel('No. of integration points') title([num2str(wp),' rad rotation'])


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elastdel=[]; procent=[]; rot=[]; for wprim=0.011:0.001:0.04 elastdel=[elastdel;(sigy*L)/(E*wprim)]; procent=[procent;elastdel/r;]; rot=[rot;wprim]; end R=r*ones(length(rot)); figure(2) plot(rot,elastdel,rot,elastdel,'*',rot,R) ylabel('elastisk del (0-r m)') xlabel('rotation rad')

7.10 Appendix. Input file for beam‐shell assembly. MPC‐BEAM4 Full input file for mpc‐beam4 shell‐beam assembly.

/clear /filname,95*C vertical buckling penalty,on !----PREPROCESSOR------------------------------------------------------------- /prep7 !----PARAMETERS--------------------------------------------------------------- !-----pipeline parameters od= 0.8 ![m] outer diameter of pipe (steel) wt= 0.025 ![m] thickness of pipe (steel) length= 500 ![m] length of entire pipeline overbed=4 ![m] pipe is initially 4 meter over the seabed !-----Steel parameters Es= 207e9 ![N/m2] Youngs modulus ny= 0.3 ![] poisson denss= 7850 ![kg/m3] density alfas= 12e-6 ![/*C] thermal smys= 448e6 ![N/m] yield stress epsys= smys/Es ![] yield strain sist= 449e6 ![N/m] bilinear stressstrain epsist= 25e-3 !-----Shell parameters shele= 40 ![m] about length of shellpart sides= 16 ![-] number of sides on shellpipe (/4!) ra=(od-wt)/2 ![m] radi smyss= 448e6 ![N/m] yield stress for shell


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epsyss= smys/Es ![] yield strain for shell sists= 459e6 ![N/m] bilinear stressstrain epsists=24.7e-3 !shell !----link parameters lengthlink= 10 ![m] the linknodes is lengthlink long distancelinks= 3 !distance between pipenodes with links downlength=overbed ![m] the length the links will be downloaded !-----seabed parameters width= 20 ![m] +-y-value of seabed mulat= 0.6 ![-] lateral friction coefficient fknpm= -1200e3 ![N/m/m] vertical soil stiffness…

!per meter slto= -0.02 ![m] allowable elastic slip, mob-length fknpmt= -3100e3 ![N/m/m] trigger stiffness. (high!) triggheight= 1 ![m] height of trigger trigglength= 2 ![m] length of trigger weststart= -1 !start x value of seabed eastend= length+1 !end x value of sebed triggwest= (length/2)-trigglength/2 !start x value of triggerberm triggeast= (length/2)+trigglength/2 !end x value of triggerberm !----load parameters Nlay= 245e3 ![N] residual lay tension extpress= 3.7e6 ![Pa] external pressure intpress= 30e6 ![Pa] delta_pi temperature= 95 ![*C] delta_temp grav= 1!9.81 ![m/s2] densg= 300 ![kg/m3]density of gas densc= 2000 ![kg/m3]density of concrete densw= 1027 ![kg/m3]density of water Ag= 0.44178647 ![m2] internal area As= 0.06086836 ![m2] steel area Ac= 0.13351769 ![m2] area of concrete Ae= 0.63617251 ![m2] external area gasweight=densg*Ag ![kg/m] weight of gas steweight=denss*As ![kg/m] weight of steel conweight=densc*Ac ![kg/m] weight of concrete watweight=densw*Ae ![kg/m] bouyancy subweight=gasweight+steweight+conweight-watweight densfix=subweight/As !fixed density /title, Pipes %length% m and shells %shele% m with %temperature% *C !-----CREATING PIPE---------------------------------------------- !-----pipeline elementtype, real constants and materialproperty-


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!-----beampart--------------------------------------------------- et,1,PIPE20 r,1,od,wt mp,ex,1,Es mp,prxy,1,ny mp,dens,1,densfix mp,alpx,1,alfas tb,kinh,1,1,2 !kinematic hardening with 2 points tbtemp,0.0 !temp 0.0 tbpt, ,epsys,smys !first point tbpt, ,epsist,sist !second point !-----nodes and elements in pipeline- nodp1= 1 !first nodenumber in pipe nodpn= 301 !last nodenumber in pipe (odd) nelemp= nodpn-1 !number of elements in pipe midnode=(nodpn+1)/2 !nodenumber at midnode elength=length/nelemp !lenght of an element n,nodp1,0,0,0 !position of first pipenode n,nodpn,length,0,0 !position of last pipenode fill,nodp1,nodpn !fill a row of nodes between first…

!and last pipenode numstr,elem,1 !element numbering from 1 e,1,2 !create a pipeelement between node 1…

!and node 2 *repeat,nelemp,1,1 !create element between laststart+1…

!node, laststop+1 node. Number of… !elements in pipe times

nsel,all nsel,s,node, ,1,nodpn !select the pipenodes cm,pipenodes,node !name the pipenodes pipenodes esel,all esel,s,type, ,1 !select the pipe elements by type cm,pipeelem,elem !name the pipeelement pipeelem esel,all !-----shellpart----------------------------------------------- et,2,shell181 r,2,wt rmodif,2,6,0 !added mass 0 mp,ex,2,Es mp,prxy,2,ny mp,dens,2,densfix mp,alpx,2,alfas tb,kinh,2,1,2 !kinematic hardening with 2 points


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tbtemp,0.0 !temp 0.0 tbpt, ,epsyss,smyss !first point tbpt, ,epsists,sists !second point type,2 $ real,2 $mat,2 !-----nodes and elements in shell l1=length/2-shele/2 !about where shell start l2=length/2+shele/2 !about where shell end pipew=node(l1,0,0) !node nearest l1 pipee=node(l2,0,0) !node nearest l2 startshell=nx(pipew) !where shell start endshell=nx(pipee) !where shell end shele=endshell-startshell !length of shellpart st_el=pipew !start deleting element nr en_el=pipee-1 !end deleting element nr *afun,rad pi=4*atan(1) !make pi be pi nods1=nodpn+1 !first nodenumber in shell nodsr=sides !number of nodes in one shellring dfi=2*pi/nodsr !raddistance between nodes sh_er=ra*dfi !length of element around ring sh_el=sh_er !sh_er~=sh_el nr_el=nint(elength/sh_el) !number of element along element…

!length sh_el=elength/nr_el !length of shell element…

!along pipe element length nodsl=nr_el*(en_el-st_el+1)+1 !number of nodes in shell *do,j,1,nodsr,1 !create first node ring fi=(j-1)*dfi nodnr=j+nodpn n,nodnr,startshell,ra*sin(fi),ra*cos(fi) *enddo xlast=startshell nodendlastring=nodnr *do,element,st_el,en_el,1 !create elements and nodes in shell edele,element !delete pipeelement *do,i,1,nr_el,1 !create so many rings per deleted… !pipeelement xp=i*sh_el+xlast !x-position of nodering i *do,j,1,nodsr,1 !create 16 nodes/ring fi=(j-1)*dfi


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nodnr=j+(i-1)*nodsr+nodendlastring n,nodnr,xp,ra*sin(fi),ra*cos(fi) *enddo *enddo *do,i,1,nr_el,1 !create shell elements *do,j,1,nodsr-1,1 ep2=j+(i-2)*nodsr+nodendlastring ep1=(j+1)+(i-2)*nodsr+nodendlastring ep4=(j+1)+(i-1)*nodsr+nodendlastring ep3=j+(i-1)*nodsr+nodendlastring e,ep1,ep2,ep3,ep4 *enddo ep2=nodsr+(i-2)*nodsr+nodendlastring !last element ep1=1+(i-2)*nodsr+nodendlastring !in ring ep4=1+(i-1)*nodsr+nodendlastring ep3=nodsr+(i-1)*nodsr+nodendlastring e,ep1,ep2,ep3,ep4 *enddo nodendlastring=nodnr xlast=xp *enddo nodsn=nodendlastring nsel,all nsel,s,node, ,nods1+2*nodsr,nods1+2*nodsr ! nsel,a,node, ,nods1+2*nodsr+7,nods1+2*nodsr+7 *do,i,3,132 nsel,a,node, ,nods1+i*nodsr,nods1+i*nodsr nsel,a,node, ,nods1+i*nodsr+7,nods1+i*nodsr+7 *enddo cm,shelltopnodes,node !select all nodes at y=0 (both top and

!bottom and name them shelltopnodes nsel,all !----shellends---------------------------------------------------- et,30,shell181 r,30,wt/3 mp,ex,30,Es mp,prxy,30,ny mp,dens,30,0 mp,alpx,30,alfas type,30 $ real,30 $mat,30 *do,i,nods1,nodsr+nodpn-1,1 e,pipew,i,i+1,i+1 !create shell elements at west end *enddo e,pipew,nodsr+nodpn,nods1,nods1


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e,pipee,nodsn-nodsr+1,nodsn,nodsn *do,i,nodsn-nodsr+1,nodsn-1,1 e,pipee,i+1,i,i !create shell elements at west end *enddo nsel,all nsel,s,node, ,nods1,nodsn !select the shellnodes cm,shellnodes,node !name shellnodes shellnodes nsel,all esel,all esel,s,real, ,2 !select the shell elements by type cm,shellelem,elem !name shellelement shellelem esel,all esel,all esel,s,real, ,30 !select the pipe elements by type cm,shellendelem,elem !name shellendelement shellendelem esel,all !----mpc----------------------------------------------------------------------- et,3,TARGE170 tshap,pilo keyopt,3,4,111111 !target keyopt(4)= all dof r,3 !two real constant set r,4 type,3 $ real,3 e,pipew !create pilot node 1 at west type,3 $ real,4 e,pipee !create pilot node 2 at east et,4,CONTA175 keyopt,4,2,2 !contact keyopt(2)=2, mpc keyopt,4,4,1 !force-distributed surface keyopt,4,12,5 !bonded always type,4 $ real,3 *do,i,nods1,nodsr+nodpn,1 !create contact nodes at west e,i *enddo type,4 $ real,4 *do,i,nodsn-nodsr+1,nodsn,1 ! create contact nodes at east e,i *enddo !-----linkelement, real constants and materialproperty------------- et,5,LINK10 keyopt,5,2,0 !keypot(2)=0 tension-only the link work…

!as a wire r,5,0.01 !the links area


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mp,ex,5,2e11 !and the young modulus k=EA=high type,5 $ real,5 $ mat,5 nodl1=nodsn+1 linki=nodl1 *do,pipei,1,pipew-1,distancelinks !for…

!pipei=1:distancelinks:nodpn-1 xpos=nx(pipei) n,linki,xpos,0,lengthlink !create linknode e,linki,pipei !create link element linki=linki+1 *enddo n,linki,startshell,0,lengthlink !last linknode above last pipenode e,linki,pipew linki=linki+1 !----beam4---- -create the conncetion between shell and links-------------------- kvadratsida=0.1 area=kvadratsida**2 Iet=(kvadratsida**3)/12 et,6,beam4 keyopt,6,2,2 r,6,area,Iet,Iet,kvadratsida,kvadratsida mp,ex,6,Es mp,prxy,6,ny mp,dens,6,0 mp,alpx,6,alfas !same distance between the links. The pipeelements were deleted but !the nodes are used here *do,pipei,pipew+1,pipee-1,distancelinks nods1pipei=(pipei-pipew)*nodsr*nr_el+nods1 nodsnpipei=nods1pipei+nodsr-1 type,6 $ real,6 $ mat,6 *do,i,nods1pipei,nodsnpipei,1 !create 16 beams e,pipei,I !for each link *enddo type,5 $ real,5 $ mat,5 xpos=nx(pipei) n,linki,xpos,0,lengthlink !create linknode e,linki,pipei !create link element linki=linki+1 *enddo linki=linki+1 *do,pipei,pipee,nodpn-1,distancelinks !


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xpos=nx(pipei) n,linki,xpos,0,lengthlink !create linknode e,linki,pipei !create link element linki=linki+1 *enddo n,linki,length,0,lengthlink !last linknode above last

!pipenode e,linki,nodpn nodln=linki nsel,all nsel,s,node, ,nodl1,nodln cm,linknodes,node nsel,all esel,all esel,s,type, ,5 !select the linkelements by type cm,linkelem,elem !NAME the linkelement linkelem esel,all esel,all esel,s,type, ,6 !select the beamelements by type cm,beamelem,elem !name the beamelement beamelem esel,all !-----SEABED----------------------------------------- fknp=fknpm*elength ![N/m/'e'] vertical soil stiffness

!per pipeelement fkns=fknpm !vertical soil stiffness for shell fknt=fknpmt !and for “triggershell” !----contact------ et,7,CONTA175 keyopt,7,2,1 !contact keyopt(2)=1, penalty method keyopt,7,10,5 !contact keyopt(10)=5, fkt is updated…

!at each iteration !----seabedcontact pipe20 r,7, , ,fknp !real constant nr 3 is fkn rmodif,7,23,slto !real constant nr 23 is slto mp,mu,7,mulat !material property friction type,7 $ real,7 $ mat,7 e,1 ! *repeat,pipew-1,1 !create contact elements on pipe e,pipee+1 ! *repeat,pipew-1,1 ! !----seabedcontact shell181


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et,8,CONTA173 keyopt,8,2,1 !contact keyopt(2)=1, penalty method keyopt,8,10,5 !contact keyopt(10)=5, fkt is updated…

!at each iteration mp,mu,8,0.4 !friction coeff for shell<my for pipe r,8, , ,fkns rmodif,8,23,slto r,9, , ,fknt !real constant nr 3 is fkn rmodif,9,23,slto !real constant nr 23 is slto type,8 $ real,8 $ mat,8 i=nodsl/2 j=nint(nodsl/2) *if,j,eq,i,then !if the shellrings are even an_elt=nint(trigglength/sh_el) i=an_elt/2 j=nint(an_elt/2) *if,j,eq,i,then an_elt=an_elt-1 *else an_elt=an_elt *endif triggis=nodsl/2-(an_elt-1)/2 !start shellring for trigger triggie=nodsl/2+(an_elt-1)/2 !end shellring for trigger *else !if the shellrings are odd an_elt=nint(trigglength/sh_el) i=an_elt/2 j=nint(an_elt/2) *if,j,eq,i,then an_elt=an_elt *else an_elt=an_elt+1 *endif triggis=(nodsl+1)/2-an_elt/2 triggie=(nodsl+1)/2+an_elt/2 *endif *do,i,1,triggis-1,1 !create contact elements n1=(i-1)*nodsr+nods1 !on westshell nodsrh=n1+(nodsr/2) *do,j,-1,2,1 e1=nodsrh-j+nodsr e2=nodsrh-j+1+nodsr e3=nodsrh-j+1 e4=nodsrh-j e,e1,e2,e3,e4 *enddo *enddo


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*do,i,triggie,nodsl-1,1 !create contact elements n1=(i-1)*nodsr+nods1 !on eastshell nodsrh=n1+(nodsr/2) *do,j,-1,2,1 e1=nodsrh-j+nodsr e2=nodsrh-j+1+nodsr e3=nodsrh-j+1 e4=nodsrh-j e,e1,e2,e3,e4 *enddo *enddo type,8 $ real,9 $ mat,8 *do,i,triggis,triggie-1,1 !create contact elements n1=(i-1)*nodsr+nods1 !on triggershell nodsrh=n1+(nodsr/2) *do,j,-1,2,1 e1=nodsrh-j+nodsr e2=nodsrh-j+1+nodsr e3=nodsrh-j+1 e4=nodsrh-j e,e1,e2,e3,e4 *enddo *enddo !----seabednodes and elem pipe20 et,9,TARGE170 tshap,quad nodb1=nodln+1 nodb2=nodb1+4 nodb3=nodb2+4 n,nodb1,-1,-width,-2 n,nodb1+1,length+1,-width,-2 n,nodb1+2,length+1,width,-2 n,nodb1+3,-1,width,-2 type,9 $ real,7 e,nodb1,nodb1+1,nodb1+2,nodb1+3 !----seabednodes and elem shell181 n,nodb2,startshell-1,-width,-2-ra n,nodb2+1,endshell+1,-width,-2-ra n,nodb2+2,endshell+1,width,-2-ra n,nodb2+3,startshell-1,width,-2-ra n,nodb3,length/2-trigglength/2,-width,-2-ra+triggheight n,nodb3+1,length/2+trigglength/2,-width,-2-ra+triggheight n,nodb3+2,length/2+trigglength/2,width,-2-ra+triggheight n,nodb3+3,length/2-trigglength/2,width,-2-ra+triggheight


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type,9 $ real,8 e,nodb2,nodb2+1,nodb2+2,nodb2+3 type,9 $ real,9 e,nodb3,nodb3+1,nodb3+2,nodb3+3 nsel,all nsel,s,node, ,nodb3,nodb3+3 cm,triggnodes,node nsel,all esel,all esel,s,type, ,9 !select the targeelements by type cm,targeelem,elem !name targeelement targeelem esel,all !----SOLVING------------------------------------------------------------- /solu antype,static,new !New static analyse solcontrol,on !solution control on nlgeom,on !includes large-deflection effects autots,on !automatic timestepping on cnvtol,f, ,0.01, ,1 !force tolerance cnvtol,u, ,0.05 !displacement tolerance nropt,unsym, ,off pred,off sstif,off lnsrch,on outres,all,all time,1 !residual tension force d,linknode,all,0 d,nodp1,all,0 d,nodpn,all,0 ddele,nodpn,ux f,nodpn,fx,Nlay nsubst,10,100,10 solve time,2 !external pressure and submerged weight acel, , ,grav !here gravity is set to 1 sfe,pipeelem,5,pres, ,extpress sfe,shellelem,2,pres, ,extpress sfe,shellendelem,2,pres, ,extpress nsubst,10,100,10 solve time,3 !lowering the pipe ddele,nodp1,uz d,nodpn,all,%_fix% ddele,nodpn,uz


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d,linknodes,uz,-2 nsubst,10,100,10 solve time,4 !lock the pipe and kill the “lowering device” fdele,nodpn,fx d,nodp1,all,%_fix% d,nodpn,all,%_fix% ekill,linkelem ekill,beamelem nsubst,1,1,1 solve time,5 !full gravity ddele,nodpn,all grav=9.81 acel, , ,grav nsubst,5,10,5 solve time,6 d,shelltopnodes,uy,%_fix% d,shelltopnodes,rotx,%_fix% d,shelltopnodes,rotz,%_fix% nsubst,1,1,1 solve time,7 !internal pressure d,nodpn,all,%_fix% sfe,pipeelem,1,pres, ,intpress sfe,shellelem,1,pres, ,intpress sfe,shellendelem,1,pres, ,intpress nsubst,10,100,10 solve time,8 !temperture (devided into two step so… !it is small step when snapping bf,pipenodes,temp,15 bf,shellnodes,temp,15 nsubst,30,100,20 solve time,9 !continue temperature rise bf,pipenodes,temp,temperature bf,shellnodes,temp,temperature nsubst,30,100,20 solve !-----POSTPROCESSOR

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