Mooring Lines and riser LWSCR Analysis

ANALYSIS & DESIGN OF LAZY-WAVE STEEL CATENARY RISERS

MSc. Subsea Engineering

Class Code NM958/NM521

Alexandros Venetsanos Reg.No 201459239 Scholar of Hellenic Petroleum S.A.

Scholar of Foundation of Education & European Culture

Angelos Papavasileiou Reg.No 201490748

Lecturer Dr. N.Srinil

Glasgow 2014

Analysis & Design of LWSCR & SCR

1 A.Venetsanos, A.Papavasileiou

INTRODUCTION

The main objective of this study is the investigation of the design parameters which affect the shape and the static loading of LWR. For this purpose, the catenary theory was applied. In addition, a Matlab code was developed for the evaluation of different scenarios. Furthermore, special cases where the LWR configuration degenerates into “Shaped” or “Steep Wave” SCR were examined. Finally, the influence of the hang-off point deflection in the geometry and the loads of the riser was taken into consideration.

Chapter 1 – Theory

The following analysis is based on two design options. Respective to figure 1, the

first option requires, the hang-off, the buoyancy and the touchdown catenary lengths, while

for the second option the total horizontal distance, the sag and the arch bend heights are

required. For both options, the effective mass ratio, and the total water depth must be

specified in advance. The whole approach to the determination of the geometry is

presented step by step in the paper of S.Li and C.Nguyen [1].

Figure 1

As for the axial load distribution, has been proved [2] that if the catenary theory is

applied and there are no external horizontal loads, then the horizontal component of axial

force at any point remains constant. In addition, the vertical component of the axial force at

the hang-off point is equal to the sum of the vertical forces applied throughout the riser

length (effective weight).By using these observations and taking into account the local slope

at each point, the axial force distribution can be specified.



The effective weight of the hang-off and touchdown segments is:

Qi,k = gMair +gπ

4(ριdi2 − ροdo2)

The effective mass ratio in the local coordinate system is:

EMratio =Qbuoyancy only − Qriser only

Qriser only

Thus, there are three areas of interest:

1 < EMratio The additional buoyancy is greater than the weight −1 < EMratio < 1 The additional buoyancy is lesser than weight

EMratio < −1 There is additional weight The effective weight of the buoyancy segment is: Qj = EMratio Qi,k

The vertical component of top tension is: TtopV = SiQi − SjQj + SkQk

If the geometry (x,y) is known, then the horizontal component of axial force at any location is:

TH =TtopV

dy

dx

The axial force at any location x is:

T(x) = |Tx

cos(tan−1(dy

dx))

|

The absolute value represents that although the slope can be either positive or negative, the axial force is always tensile. The axial stress at any location x is:

Sa(x) =T(x)

π4

(do2 − di2)

As for the bending stress, the catenary theory assumes that Sb=0 at any location.

However, it is possible to make an estimation of Sb by assuming that the geometry which

was calculated by applying catenary theory is very close to the one which would have been

calculated by using large deformation bending theory. In this way, the curvature at any

location is:

k(x) =|d2y

dx2|

(1+(dy

dx)

2)

3/2

The bending moment at any location x is:

M(x) = −EIk(x)

The maximum bending stress at any location x is:

MaxSb(x) = k(x)EI rmax

I



The above calculations require the first and the second difference (dy

dx ,

d2y

dx2). For this

purpose, there are two alternative approaches. The first is analytical and has to be applied

in each local coordinate system individually. The second is numerical and has to be applied

to the global coordinate system. On the one hand, the first method is most accurate.

However, it may not work correctly at points of discontinuity. On the other side, the

accuracy of the second method depends on the number of points and could ignore

discontinuities. In this study, the numerical approximation was adopted, and the total length

was divided into 10000 parts. The finite difference formulas are given below [3].

Forward difference approximation

dy

dx|

xi

≈−y(xi+2)+4y(xi+1)−3y(xi)

2h

d2y

dx2|

xi

≈y(xi+2) − 2y(xi+1) + y(xi)

h2

Backward difference approximation

dy

dx|

xi

≈3y(xi) − 4y(xi−1) + y(xi−2)

2h

d2y

dx2|

xi

≈y(xi) − 2y(xi−1) + y(xi−2)

h2

Central difference approximation

dy

dx|

xi

≈−y(xi+2) + 8y(xi+1) − 8y(xi−1) + y(xi−2)

12h

d2y

dx2|xi

≈y(xi+1)−2y(xi)+y(xi−1)

h2 (This formula was not used in this study for simplicity.)

As far as catenary theory does not take into account the shear stress, the total

maximum stress at any location is the sum of axial stress and maximum bending stress.

MaxStotal(x) = Sa(x) + MaxSb(x)



Figure 2

Figure 3



Figure 4

Figures 2,3,4 show the riser configuration together with the stress distribution. The

main observation is that the contribution of bending stress in total stress is imperceptible. In

addition, at the locations where axial stress has a local maximum, bending stress has a

local minimum. As a result, the maximum total stress which is one of the main design

parameters is not really affected by bending stress. Therefore, bending stress could be

neglected during a preliminary study.

Although bending stress is very small compared to axial stress, it is critical in case of

buckling. The bending radii which have been calculated by catenary theory are smaller than

the real ones [2]. This happens because in reality, bending stiffness prevents the

deformation of the riser. As a result, the calculated bending stress is below the real one and

the design is more conservative.

Finally, the bending stress diagramme appears to be discontinuous at the points

where the effective weight changes. In addition, bending stress is not equal to zero at the

hang–off location as it was expected due to the flexible joint. As far as bending theory is

based on a fourth-order differential equation, while catenary theory is based on a first order

differential equation and neglects bending moment and shear stress, the continuity and the

smoothness of the initial geometry (x,y) does not guarantee the continuity of the second

difference. This phenomenon does not have a physical explanation because, in reality, all

structures have even very small bending stiffness.



Chapter 2 – Matlab Code

The Matlab code “design.m” and the relative subroutines have been developed for

the parametric study and the design of different types of risers. More specifically, the above

programme is able to make geometrical and stress calculations for “Lazy Wave” and “Steep

Wave” Risers, while it can also combine the above types with a “Shaped” configuration. In

addition, this code can take into account the horizontal and vertical offset of the hang-off

point and update the geometry and the stress distribution due to these deflections (this

option is not available for “Steep Wave Risers”).

Figure 5



The above flow chart shows how the code works. Default input data like geometrical

dimensions fluid and material properties are specified in the lines (31-35) (43-47) (73-79) of

the file “design.m”. If the user wants to investigate different scenarios, he has to change the

default values in advance. After that, the code is ready to run. By starting the programme,

some messages appear on the screen in order to guide the user select the type of riser.

Then the effective weight of each segment is calculated and the nonlinear solver starts the

iterative procedure in order to determine the hang-off angle and the unknown lengths

(x1,x2,x3,x4,x5). The Matlab command fsolve has been used for this purpose. This

command is based on the “trust-region-dogleg” algorithm [4]. This method was chosen

because of the high rate of convergence. Once the iterations are completed, the

intermediate points of the riser configuration can be determined. Appropriate adjustments

have been done in order to express the local hyperbolic cosine functions in the global

coordinate system.

The calculation of the axial and bending stress distribution requires the first and

second difference respectively at any point of the riser. The numerical estimation of the first

difference uses forward, central and backward difference approximations for the first two

points, the intermediate and the last two points respectively. The numerical estimation of

the second difference uses only forward and backward difference approximations for the

first half and the second half number of points of each segment. If a “steep wave” riser

configuration has been selected initially, then the programme calculates the total stress and

gives a report together with the graphs of the geometry and the stresses. Otherwise, only

the graphs are printed, and the execution continues with the offset calculations.

The part of the code that is relative to the offset calculations is almost the same with

the previous one. The only difference is that not only the hang-off angle is unknown but also

the deflection of the TDP. As a result, the solver has to handle a system of two nonlinear

equations. For further explanations, please look at chapter 5.



Chapter 3 – Effect of Pipe Size & Fluids Properties

This chapter deals with the effect of pipe size and fluids properties (internal &

external) on the geometry and the maximum stress of the riser. Four different scenarios

were investigated. The lengths Si, Sj, Sk, the water depth and the effective mass ratio

remain steady in any case. As for the rest properties, in each scenario only one parameter

changes in order to highlight how this parameter influences the overall design. The only

exception is the first scenario where two different flexible pipes were examined.

Comparison between pipe 1 and pipe 2

Figure 6 Figure 7

EMratio V H Si Sj Sk Ya Ys θ max(Sa) max(Sb)

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 3,6269 1,1061

In general, the geometry is not affected by the size and the properties of the pipe.

The axial stress is greater in case of a small pipe, while the bending stress is higher in case

of a large pipe. The next examples highlight the influence of each property separately.

Po Pi do di EI Mair

1025 1025 150 105 4,47 37,86

1025 1025 341 259 50,95 141,62



Effect of internal fluid

Figure 8 Figure 9

The internal fluid does not affect the shape of the riser, but only the axial stress. As

far as in both cases the effective mass ratio remains the same, the increase of the total

weight because of the heavier internal fluid is equilibrated by the proportional increase of

the additional buoyancy. This observation could be useful only during the design stage

because after installation if the internal fluid properties change, it is not possible to add or

subtract buoyancy modules. Therefore, the effective mass ratio could not remain steady. As

it was expected the heavier fluid results in greater axial stress. As for the bending stress, it

remains the same because it is directionally corresponded to the geometry.

Po Pi do di EI Mair

1025 800 150 105 4,47 37,86

1025 1025 150 105 4,47 37,86


3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4,0661 1,0009

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009



Effect of weight in air

Po Pi do di EI Mair 1025 1025 150 105 4,47 37,86 1025 1025 150 105 4,47 50

Figure 10 Figure 11


3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 6,2137 1,0009

The pipe weight in air is another parameter that does not affect the geometry or the

bending stress when the effective mass ratio remains steady. The reason is the same as in

the previous case.

Effect of outer diameter

Po Pi do di EI Mair

1025 1025 150 105 4,47 37,86

1025 1025 170 105 4,47 37,86

Figure 12 Figure 13


3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 2,3032 0,6115



The increase of the external diameter leads to the rise of the cross-sectional area and the second moment of area. As a result, the axial and the bending stresses appear to be reduced. However, large pipes are prone to local buckling. This phenomenon has not been taken into account by catenary theory. One method that could investigate this kind of problems is the finite element analysis by using 3D solid elements that model the whole cross section. Effect of inner diameter

Po Pi do di EI Mair

1025 1025 150 105 4,47 37,86

1025 1025 150 120 4,47 37,86

Figure 14 Figure 15


3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 6,7869 1,2883

The increase of the internal diameter leads to the decrease of the cross-sectional

area and the second moment of area. As a result, the axial and the bending stresses go up.

Effect of bending stiffness

Po Pi do di EI Mair

1025 1025 150 105 4,47 37,86

1025 1025 150 105 6 37,86

Figure 16 Figure 17




3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,3437

Bending stiffness is a property that affects only the bending stress. As it was

anticipated, when the bending stress increase, the bending stress goes up proportionally.

As far as bending stiffness is the product of Young’s modulus and second moment of area,

in this study we assume that only the elasticity varies. Otherwise, the cross section area

also change, and many other parameters are affected, so it very difficult to express a

general result.



Chapter 4 – Effect of geometrical parameters

This chapter presents how large scale geometrical parameters like the lengths Si, Sj,

Sk, and the water depth together with the effective mass ratio influence the configuration of

the riser and the stress distribution. For this purpose, five different cases were examined.

The cross section, the material properties of the pipe and the internal fluid remain constant

in any case.

Effect of Effective mass ratio

Figure 18 Figure 19

The effective mass ratio is a parameter that shows the correlation between the

effective mass of the pipe and the effective mass of the additional buoyancy. When EM ratio is lesser than minus one, this means that instead of additional buoyancy there is extra weight. When EM ratio is equal to minus one, there is not any additional load and the riser has the configuration of SCR. As the EM ratio takes higher values, the riser configuration becomes wavier. As for the stresses, when EM ratio goes up, the axial stress goes down because the additional buoyancy equilibrates a greater part of the total weight. On the other hand, bending stress goes up because the geometry has locations with very small bending radii (sag bend, arch bend).


-5 150 287,5042 150 60 130 13,0867 -112,8862 41,235 24,0289 0,1639

-1 150 294,0040 150 60 130 0 0 42,3881 14,3418 0,0572

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009



Effect of buoyancy segment location

Figure 20 Figure 21


3 150 292,3111 100 60 180 117,801 117,801 54,3039 7,0834 0,3827

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009

3 150 49,0748 170 60 110 142,2672 53,2945 1,9190 3,4014 15,9147

In this scenario, although the total length of the riser and the length of buoyancy

segment remain constant, the location of buoyancy segment moves along the riser. While

the buoyancy modules move from the top end to the touch down point, the shape becomes

wavier, the total horizontal distance reduces, the axial stress reduces and the bending

stress increases. In the extreme case that the buoyancy modules are installed at the lower

end of the riser, the configuration degenerates to “Steep Wave”. (This case could be

investigated only by selecting the second option of the programme “design.m”).

Effect of the buoyancy segment length

Figure 22 Figure 23


3 150 296,2841 160 40 140 70,4853 68,8142 41,4462 7,4815 0,3297

3 150 289,8573 155 50 135 84,2617 82,9354 38,8081 5,5973 0,4730

3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009



In this case, the total length of the riser remains constant. In addition, the middle of

the buoyancy segment remains at the same location, while the buoyancy length extends

symmetrically along both sides. The main observations are that when the buoyancy length

increases, the axial stress goes down. This phenomenon could be explained because the

extra buoyancy counterpoises greater part of the total weight. On the other hand, while the

shape becomes curlier, the bending stress increases.

Effect of the riser length

Figure 24 Figure 25

In this scenario, the ratio of each segment length to the total length remains constant

(Si/S Sj/S Sk/S), while the total length varies. As the whole installation becomes larger, the

total weight increases and as a result the axial stress follows the some trend. On the other

hand, the shape becomes smoother leading to reduced bending stress. An interesting

observation is that even if the overall length becomes very large, the sag and the arch bend

heights change imperceptibly.

Effect of concentrated buoyancy

Figure 26 Figure 27


3 150 248,6873 150 60 130 116,3427 89,9356 28,0183 4.3754 1,0009

3 150 457,8163 225 90 195 115,2631 94,9199 49,6735 8,1899 0,3014

3 150 642,3752 300 120 260 115,3530 96,5461 60,0727 13,5117 0,1533




50 150 296,6991 179 2 159 53,0175 44,9894 42,3838 10,0386 3,7787

75 150 294,9779 179 2 159 65,0488 64,8000 41,2407 7,7894 8,1800

100 150 279,4537 179 2 159 88,1428 80,0736 36,2561 5,9148 18,5558

This example shows how a concentrated buoyant force could affect the configuration

of a flexible riser. The floater has been modelled as a buoyancy segment with very small

length and very large effective mass ratio. This assumption indicates that the existing code

(“design.m”) could be used for the study of hybrid risers.



Chapter 5 – Hang-off point offset

Figure 28 shows the initial geometry of a LWR and the new one when the top end

has a horizontal and a vertical displacement. The main assumption is that there is enough

laid down length so as at any case the slope at the touch down point remains zero.

Figure 28

Except for the offset of the top end, there is also a horizontal deflection of the TDP. As a

result the overall length of the riser changes by the TDP deflection. The new lengths of

each segment are given below:

Sinew = Si Sjnew = Sj Sknew = Sk + dxTDP

If DV , DH are the vertical and horizontal deflections of the top end, then the new water

depth and total horizontal distance of the riser are:

Vnew = V + DV

Hnew = H − DH + dxTDP

The minus sign in the later equation represents that the overall horizontal distance

increases when the top end moves towards the negative horizontal axis.

If dV , dH represent the vertical and horizontal deflections of the top end as percentages of

the initial water depth, then the last two equations could be modified.

Vnew = V (1 +dV

100)

Hnew = H − VdH

100+ dxTDP



The above modifications are useful during preliminary design because the majority of

regulations express the allowed deflections of the floating platforms as percentages of the

water depth.

If dV , dH are known, then the new shape of the riser could be determined iteratively in the

same way as the initial one was determined when the hang-off, the buoyancy and the

touchdown catenary lengths were known [1]. The only difference is that there is one more

unknown variable and one more equation for the check of the convergence. To conclude,

the unknown variables are:

New hang-off angle: θnew

TDP deflection: dxTDP

The converge criteria are:

|y1 − y2 − y3 + y4 + y5 − Vnew

Vnew| ≤ ε

|x1 − x2 − x3 + x4 + x5 − Hnew

Hnew| ≤ ε

The allowed error of the results could be specified by the selection of the very small number

(ε). The matlab non linear solver “fsolve” which was used, specifies the above tolerance by

default.

Effect of horizontal offset

Figure 29 Figure 30

EMratio V dV H dH Si Sj Sk Ya Ys θ max(Sa) max(Sb)

3 170 0 250 0 170,9427 60,5598 129,7893 120 90 22,2001 4,3165 1.0962

3 170 0 230,0295 10 170,9427 60,5598 126,8188 124,3598 86,9373 18,8288 4,127 1,3459

3 170 0 271,3463 -10 170,9427 60,5598 134,1356 115,1347 93,3504 26,3069 4,5797 0,8732

If the hang-off point has only horizontal offset and goes towards the TDP, then the axial stress decreases, while the bending stress increases. On the other hand when the top end moves to the opposite direction, the stresses follow also the opposite trend.



Effect of vertical offset

Figure 31 Figure 32


3 170 0 250 0 170,9427 60,5598 129,7893 120 90 22,2001 4,3165 1.0962

3 187 10 256,9291 0 170,9427 60,5598 136,7184 125,2016 103,0654 22,6535 4,5536 1,0131

3 153 -10 243,6571 0 170,9427 60,5598 123,4464 114,8125 76,9098 21,9536 4,0878 1,1725

If the hang-off point has only vertical offset and moves upwards , then the axial

stress goes up, while the bending stress goes down. On the other hand when the top end moves downwards, the stresses follow the opposite trend. Effect of combined top end deflection

Figure 33 Figure 34


3 170 0 250 0 170,9427 60,5598 129,7893 120 90 22,2001 4,3165 1.0962

3 187 10 279,7491 -10 170,9427 60,5598 142,5385 120,4 106,3364 27,0662 4,9003 0,7909

3 153 -10 224,2182 10 170,9427 60,5598 121,0075 119,1178 73,8793 18,6802 3,9304 1,4289

If the movement of the top end is a combination of both horizontal and vertical

deflections, it is not very easy to predict in advance the effect on the riser configuration and

the respective stresses. However, there are some special cases that the prediction is

possible. When the horizontal deflection is towards the TDP and the vertical deflection is

downwards, then the bending stress always goes up and the axial stress goes down. On

the other hand, when the horizontal deflection is away from TDP and the vertical deflection

is upwards, then the bending stress always goes down and the axial stress goes up.



Comparison between SCR and LWSCR

Figure 35 Figure 36

SCR


-1 150 0 294,00 0 150 60 130 0 0 42,39 14,34 0,06

-1 150 0 203,43 10 150 60 54,42 0 0 30,87 9,60 0,11

LWSCR


3 150 0 294,00 0 169,19 67,82 143,32 120 90 32,36 4,96 0,77

3 150 0 290,40 10 169,19 67,82 139,71 125,14 87,16 28,42 4,70 0,93

Figure 35 shows that if a floating unit is connected to the seabed either by a SCR or

by a LWSCR and there is also a possible offset, the movement of TDP in case of LWSCR

is very small compared to the one of SCR. In addition, figure 36 shows that the axial stress

and the respective variation due to the offset are smaller in case of LWSCR rather than in

case of SCR. These reasons indicate the increased ability of LWSCR to decuple the

vessel’s motion from TDP and the reduced fatigue damage compared to SCR.



Chapter 6 – Conclusions & Remarks

1) The catenary theory is simpler than large deformation bending theory and leads to

more conservative results from the aspect of stress calculations. As a result it could

be used safely for preliminary design.

2) The accuracy of the numerical calculations is proportional to the number of

subdivisions. As a result, different discretization may lead to slightly different results.

3) All parameters that affect the effective mass of a riser do not affect the riser’s

configuration, on condition that the effective mass ratio remains constant.

4) When the additional buoyancy increases either by installing bigger buoyancy

modules (greater EM ratio) or by increasing the number of buoyancy modules

(greater buoyancy segment length), the riser configuration becomes wavier

(“SCR”⟶”SHAPED” ⟶ “LWSCR”).

5) LWSCR has greater abrasion and wear resistance compared to SCR.



References

1. S. Li, C. Nguyen, Dec 2010, Dynamic Response of Deepwater Lazy-Wave Catenary

Riser, DOT International.

2. C.P. Sparks, 2007, Fundamentals of Marine Riser Mechanics, PennWell Corporation

3. M. Grasselli, D. Pelinovsky, 2008, Numerical Mathematics, Jones and Barlett

Publishers

4. J.E. Dennis, Jr. R. B. Schnabel, Numerical Methods for Unconstrained Optimization

and Nonlinear Equations, SIAM



Appendix

“design.m”

clear all clc global V H Si Sj Sk Qi Qj Qk x1 y1 x2 y2 x3 y3 x4 x5 ai aj ak Ys Ya T C do global di ni I J K dH dV V_new Si_new Sj_new Sk_new H_new Ya_new Ys_new u_new

ni=10001; % Number of points disp('This code determines the SCR configuration for two design options.') disp(' ') disp('For option 1 the hang-off catenary length, the buoyancy catenary') disp('length and the touchdown catenary length are required.') disp(' ') disp('For option 2 the sag and the arch bend heights are reguired.') disp(' ') disp('For both options the effective mass ratio and the total water depth') disp('must be specified.') disp(' ') disp('For option 1 please press 1 and then press enter.') disp('For option 2 please press 2 and then press enter.') Option=input('Option_');

if (Option~=1)&&(Option~=2) disp('ERROR Option must be either 1 or 2 please restart the code') break

elseif Option==1 disp('The design parameters are specified in the lines [31-35] & [73-79]')

% DESIGN PARAMETERS OPTION 1 % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% EMratio =3; %Effective mass ratio V = 150; %[m] Total water depth Si = 150; %[m] Length of the hang-off catenary Sj = 60; %[m] Length of the buoyancy catenary Sk = 130; %[m] Length of the touchdown catenary %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

elseif Option==2 disp('The design parameters are specified in the lines [43-47] & [73-79]')

% DESIGN PARAMETERS OPTION 2 % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% EMratio = 3; %Effective mass ratio V = 170; %[m] Total water depth H = 250; %[m] Total horizontal distance Ys = 90; %[m] Sag bend heigh Ya = 120; %[m] Arch bend height %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

disp('There are two alternative types of risers.') disp('For "Lasy Wave SCR" please press 1 and then press enter.') disp('For "Steep Wave SCR" please press 2 and then press enter.') T=input('Type of riser_');

if (T~=1)&&(T~=2) disp('ERROR Type of riser must be either 1 or 2 please restart the code')



break end

disp('If you want to combine the above type of riser with a "Shaped"') disp('configuration please press 1 and then press enter. Otherwise please') disp('press 0 and then enter.') C=input('Configuration_');

if (C~=1)&&(C~=0) disp('ERROR Configuration must be either 1 or 0 please restart the code') break end end

% RISER DIAMETERS & FLUIDS PROPRETIES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Po = 1025; %[kg/m^3] Density of external fluid Pi = 1025; %[kg/m^3] Density of internal fluid do = 150; %[mm] Outside diameter di = 105; %[mm] Inside diameter EI = 4.47; %[kN*m^2] Bending stiffness Mair=37.86; %[kg/m] Mass of empty pipe in air per unit length g = 9.81; %[kg*m/s^2] Acceleration of gravity %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%[kg/m] Effective mass of pipe with internal fluid per unit length m=Mair + pi/4 *(Pi*di^2 - Po*do^2) * 10^(-6);

% [N/m] Effective weight of the hang-off catenary per unit length Qi = m*g;

% [N/m] Effective weight of the buoyancy catenary per unit length Qj = EMratio * Qi;

% [N/m] Effective weight of the touchdown catenary per unit length Qk = Qi;

u0 = [45]; %[deg] Initial estimation of the hang-off angle details = optimset('display', 'iter');

if Option==1 [u,f1] = fsolve(@option_1,u0,details); %Call nonlinear equation solver H=x1+x2+x3+x4+x5;

elseif Option==2 [u,f2] = fsolve(@option_2,u0,details); %Call nonlinear equation solver end

j=0; for i=1:1:(ni+1);

x(i)=(x1+x2+x3+x4+x5)*(i-1)/ni;

if x(i)<=x1+x2 y(i) = (V-y1)+ ai*(cosh( (x(i)-x1)/ai ) -1); I=i; elseif x(i)<=x1+x2+x3+x4 y(i) = (V-y1+y2+y3) - aj*(cosh( -(x(i)-(x1+x2+x3))/aj )-1);



j=j+1; X(j) = x(i); Y(j) = y(i); J=i;

elseif x(i)<=x1+x2+x3+x4+x5 y(i)= ak*(cosh( (x(i)-(x1+x2+x3+x4+x5)) / ak )-1); K=i;

end end

% STRESS CALCULATION %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % [Mpa] Axial stress at any point of the riser Sa=Axial_Stress(y,Si,Sj,Sk,Qi,Qj,Qk);

% [MPa] Bending stress at any point of the riser

Sb=Bending_Stress(y,EI);

% [MPa] Total stress at any point of the riser Stotal=Sa+Sb; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% PLOTS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% figure(1) hold on plot(x,Stotal,'k'); xlabel('Horizontal Distance from Hangoff [m]') ylabel('Total Stress [MPa]') grid on

figure(2) hold on plot(x,Sb,'k'); xlabel('Horizontal Distance from Hangoff [m]') ylabel('Bending Stress [MPa]') grid on

figure(3) hold on plot(x,Sa,'k'); xlabel('Horizontal Distance from Hangoff [m]') ylabel('Axial Stress [MPa]') grid on

figure(4) hold on plot(x,y,'k'); if j>0 plot(X,Y,'r') end xlabel('Horizontal Distance from Hangoff [m]') ylabel('Elevation from Seabed [m]') grid on %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



% REPORT %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% if T==2 disp('REPORT') EMratio V H Si Sj Sk Ya Ys angle=u Maximum_Axial_Stress = max(Sa) Maximum_Bending_Stress=max(Sb) Maximum_Total_Stress=max(Stotal) break end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% OFFSET CALCULATION %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('Input the horizontal deflection of the hangoff point as') disp('a percentage of the water depth [%].') dH=input('dH = ');

disp('Input the vertical deflection of the hangoff point as a percentage') disp('of the water depth [%].') dV=input('dV = ');

u0_new = [45]; %[deg] Initial estimation of the new hang-off angle Dx0_TDP = [0] ; %[m] Initial estimation of the TDP deflection Initial_Values = [u0_new , Dx0_TDP];

% Call nonlinear equation solver [Angle__Dx_TDP , f3] = fsolve(@offset , Initial_Values);

j=0; for i=1:1:(ni+1);

x_new(i)=(x1+x2+x3+x4+x5)*(i-1)/ni;

if x_new(i)<=x1+x2 y_new(i) = (V_new-y1)+ ai*(cosh( (x_new(i)-x1)/ai ) -1); I=i; elseif x_new(i)<=x1+x2+x3+x4 y_new(i) = (V_new-y1+y2+y3)-aj*(cosh(-(x_new(i)-(x1+x2+x3))/aj)-1);

j=j+1; X_new(j) = x_new(i)+ V*dH/100; Y_new(j) = y_new(i); J=i;

elseif x_new(i)<=x1+x2+x3+x4+x5 y_new(i)= ak*(cosh( (x_new(i)-(x1+x2+x3+x4+x5)) / ak )-1); K=i; end x_new(i)=x_new(i)+ V*dH/100; end



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% NEW STRESS CALCULATION %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % [Mpa] New axial stress at any point of the riser Sa_new=Axial_Stress(y_new,Si_new,Sj_new,Sk_new,Qi,Qj,Qk);

% [MPa] New bending stress at any point of the riser

Sb_new=Bending_Stress(y_new,EI);

% [MPa] New total stress at any point of the riser Stotal_new=Sa_new+Sb_new; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% PLOTS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% figure(1) hold on plot(x_new,Stotal_new,'b'); xlabel('Horizontal Distance from Hangoff [m]') ylabel('Total Stress [MPa]') grid on

figure(2) hold on plot(x_new,Sb_new,'b'); xlabel('Horizontal Distance from Hangoff [m]') ylabel('Bending Stress [MPa]') grid on

figure(3) hold on plot(x_new,Sa_new,'b'); xlabel('Horizontal Distance from Hangoff [m]') ylabel('Axial Stress [MPa]') grid on

figure(4) hold on plot(x_new,y_new,'b'); if j>0 plot(X_new,Y_new,'r') end xlabel('Horizontal Distance from Hangoff [m]') ylabel('Elevation from Seabed [m]') grid on %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% REPORT %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('REPORT') disp('Main design parameters when there is no deflection of the platform.') EMratio V H Si Sj Sk



Ya Ys angle=u Maximum_Axial_Stress = max(Sa) Maximum_Bending_Stress=max(Sb) Maximum_Total_Stress=max(Stotal)

disp('Main design parameters when there is horizontal & vertical offset.') EMratio V_new dV H_new dH Si_new Sj_new Sk_new Ya_new Ys_new angle_new=u_new Maximum_Axial_Stress = max(Sa_new) Maximum_Bending_Stress=max(Sb_new) Maximum_Total_Stress=max(Stotal_new)

disp('If you want to see the authors of this code please press 1 and then') disp('press enter, else press any other number.') D=input('');

if D==1 load('Z','A') figure('units','normalized','outerposition',[0 0 1 1]) hold on image(A) end



“option_1.m” function f1 = option_1(u)

global V Si Sj Sk Qi Qj x1 y1 x2 y2 x3 y3 x4 y4 x5 y5 ai aj ak Ys Ya

if Sk==0 disp('This part of the code does not support Steep Wave Riser design.') disp('Please sellect a positive Sk value and restart the code.') return end

v = u*pi/180; %[rad] hang-off angle

S = Si+Sj+Sk; ai = (S -(1+Qj/Qi) *Sj)*tan(v); aj = Qi/Qj *ai; ak = ai;

S1 = S-(1+Qj/Qi)*Sj;

x1 = ai * asinh(cot(v)); y1 = ai * (cosh(x1/ai)-1);

x2 = ai * asinh((Si-S1)/ai); y2 = ai * (cosh(x2/ai)-1);

x3 = x2*Qi/Qj; y3 = y2*Qi/Qj;

S4 = Qi/Qj *(S-Si-Sj); x4 = aj * asinh(S4/aj); y4 = aj * (cosh(x4/aj)-1);

x5 = ai * asinh((S-Si-Sj)/ai); y5 = y4*Qj/Qi;

Ys = y4+y5-y2-y3; Ya = y4+y5;

f1=y1+y4+y5-y2-y3-V;



“option_2.m” function f2 = option_2(u)

global V H Si Sj Sk Qi Qj x1 y1 x2 y2 x3 y3 x4 y4 x5 y5 ai aj ak Ys Ya T C

v = u*pi/180; %[rad] hang-off angle y1=abs(V-Ys);

ai=y1/(cosh(asinh(cot(v)))-1); aj=ai*Qi/Qj; ak=ai;

x1=(ai*asinh(cot(v)));

y2=(Qj*(Ya-Ys)/(Qi+Qj)); y3=(Qi*(Ya-Ys)/(Qi+Qj));

if C==0 x2=(ai*acosh(y2/ai +1)); x3=(aj*acosh(y3/aj +1));

elseif C==1 x2=-(ai*acosh(y2/ai +1)); x3=-(aj*acosh(y3/aj +1)); end

if T==1 y4=(Qi*Ya/(Qi+Qj)); y5=(Qj*Ya/(Qi+Qj));

x4=aj*acosh(y4/aj +1); x5=ai*acosh(y5/ai +1);

elseif T==2 y4=Ya; y5=0;

x4=aj*acosh(y4/aj +1); x5=0; end

Si=ai*(sinh(x1/ai)+sinh(x2/ai)); Sj=aj*(sinh(x3/aj)+sinh(x4/aj)); Sk=ai*sinh(x5/ai);

f2=x1+x2+x3+x4+x5-H;



“Axial_Stress.m” function Sa=Axial_Stress(y,Si,Sj,Sk,Qi,Qj,Qk)

global do di Dy ni x1 x2 x3 x4 x5

% Numerical approximation of the first differece h=(x1+x2+x3+x4+x5)/ni; for i=1:1:(ni+1) if i==1 || i==2 % Forward diff. aprox. Dy(i)=(-y(i+2) + 4*y(i+1) - 3*y(i)) / (2*h);

elseif i==ni || i==ni+1 % Backward diff. aprox. Dy(i)=(3*y(i) - 4*y(i-1) + y(i-2)) / (2*h);

else % Central diff. aprox. Dy(i)=(-y(i+2) + 8*y(i+1) - 8*y(i-1) + y(i-2)) / (12*h); end

Dy(i)=abs(Dy(i)); end

% Vertical component of top tension Ttop_y = Si*Qi-Sj*Qj+Sk*Qk;

% Horizontal component of top tension Ttop_x = ( Ttop_y / Dy(1) );

for i=1:1:ni+1 % [N] Axial force at any point of the riser T(i) = Ttop_x / (cos( atan( Dy(i)) ));

% [Mpa] Axial stress at any point of the riser Sa(i) = abs( T(i)/( (do^2 - di^2)*pi/4 )); end



“Bending_Stress.m” function Sb=Bending_Stress(y,EI)

global x1 x2 x3 x4 x5 di do Dy ni I J K T

% Numerical approximation of the second difference

h=(x1+x2+x3+x4+x5)/ni; for i=1:1:(ni+1); if i<=I/2 % Forward diff. aprox. DDy(i)=(y(i+2) - 2*y(i+1) + y(i)) / h^2;

elseif i>I/2 && i<=I % Backward diff. aprox. DDy(i)=(y(i)-2*y(i-1)+y(i-2))/h^2;

elseif i>=I && i<=(I+J)/2 % Forward diff. aprox. DDy(i)=(y(i+2) - 2*y(i+1) + y(i)) / h^2;

elseif i>(I+J)/2 && i<=J % Backward diff. aprox. DDy(i)=(y(i)-2*y(i-1)+y(i-2))/h^2;

elseif i>=J && i<=(J+K)/2 % Forward diff. aprox. DDy(i)=(y(i+2) - 2*y(i+1) + y(i)) / h^2;

elseif i>(J+K)/2 && i<=K % Backward diff. aprox. DDy(i)=(y(i)-2*y(i-1)+y(i-2))/h^2; end end

for i=1:1:ni+1 KK(i)= abs(DDy(i) / (1+(Dy(i))^2)^(3/2) ); %[m^(-1)] Curvature end

% Curvature approximation at the points of discontinuity

if J~=[] KK(I) = 0.5*KK(I-1) + 0.5*KK(I+1); end if K~=[] KK(J) = 0.5*KK(J-1) + 0.5*KK(J+1); end

ro=do/2000; %[m] Outside radius ri=di/2000; %[m] Inside radius Ia=(ro^4 - ri^4)*pi/4; %[m^4] Second moment of area Sb=(KK*EI*ro/Ia)*10^(-3); %[MPa] Bending stress



“offset.m” function f3 = offset(Angle__Dx_TDP)

global V H Si Sj Sk Qi Qj x1 y1 x2 y2 x3 y3 x4 x5 ai aj ak global dH dV V_new Si_new Sj_new Sk_new H_new Ya_new Ys_new u_new

u_new = Angle__Dx_TDP(1); Dx_TDP = Angle__Dx_TDP(2);

Si_new=Si; Sj_new=Sj; Sk_new=Sk+Dx_TDP;

H_new=H - V*dH/100+Dx_TDP; V_new=V*(1+dV/100);

v = u_new * pi/180;

S = Si_new + Sj_new + Sk_new; ai = (S -(1+Qj/Qi) *Sj_new)*tan(v); aj = Qi/Qj *ai; ak = ai;

S1 = S-(1+Qj/Qi)*Sj_new;

x1 = ai * asinh(cot(v)); y1 = ai * (cosh(x1/ai)-1);

x2 = ai * asinh((Si_new-S1)/ai); y2 = ai * (cosh(x2/ai)-1);

x3 = x2*Qi/Qj; y3 = y2*Qi/Qj;

S4 = Qi/Qj *(S-Si_new-Sj_new); x4 = aj * asinh(S4/aj); y4 = aj * (cosh(x4/aj)-1);

x5 = ai * asinh((S-Si_new-Sj_new)/ai); y5 = y4*Qj/Qi;

Ys_new = y4+y5-y2-y3; Ya_new = y4+y5;

f3 = [y1+y4+y5-y2-y3-V_new ; x1+x2+x3+x4+x5-H_new];

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