Design of Offshore Wind Turbine

Ratnakar Gadi

March 7, 2016

1

Contents

1 Introduction 1

2 Environmental and Soil Conditions at North Sea 1

3 Assumptions used in the Design 3

4 Computation of Loads 44.1 Wind Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44.2 Wave and Current Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44.3 Structural Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.4 Added Mass Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.5 Top Mass Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.6 Buoyancy Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

5 Load Resistance Factor Design 9

6 Static Analysis 106.1 Concentrated loads for Bending Moment . . . . . . . . . . . . . . . . . . . . 10

6.1.1 Calculation of Diameter of Main Chord . . . . . . . . . . . . . . . . . 106.1.2 Calculation of Dimensions of Horizontal and Diagonal braces . . . . . 14

6.2 Bending Moment Diagram using Load distribution . . . . . . . . . . . . . . 146.3 Analysis in Ansys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.4 Critical Locations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

7 References 18

8 Appendix 198.1 Appendix-A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198.2 Appendix-B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288.3 AppendiX-C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1 Introduction

Wind Power is the use of air flow through wind turbines to mechanically power genera-tors for electricity. Wind Power, as an alternative to the burning fossil fuels, is plenti-ful,renewable,widely distributed,clean and produces no greenhouse gases during operation.The net land effects on the environment are far less than those of non-renewable powerresources.

Wind Power can be generated on-shore as well as offshore. But the main advantages ofusing offshore wind power is that we generally have a higher wind speed.

This report provides with some fundamental steps taken in the design of such an offshorewind turbine in the North Sea(54 deg N latitude and water depth of 40m).The first and mostimportant thing in the design of any structure is finding the loads the structure would faceduring its lifetime.In order to define loads on the structure, one needs to get a good thoroughknowledge of the environmental conditions and soil conditions over there. This is dealt issection 2.The assumptions used in the design are mentioned in Section 3.After finding theworst environmental conditions, these are used in determining the loads on the structure.Theprocedure associated with calculating the loads has been explained in the section 4.The Loadand Resistance Factors used are mentioned in Section 5.Once, the loads have been estab-lished, we need to perform a static analysis of the structure.The static analysis performedin three different ways is explained in section 6. Section 7 deals with the references,followedby Appendix in Section 8.

2 Environmental and Soil Conditions at North Sea

Wind, Waves and Current constitute the main environmental conditions for any offshorestructure. Determining them is very difficult because they vary spatially and temporally.Sincethe structure is designed by Ultimate Limit State method, the highest possible environmentalconditions are considered.

A wind speed of 50m/s has been chosen to the highest possible wind speed according toRef(1).It has been already mentioned in the Ref(1) that the highest significant wave heightoccurring is 12m. But Ref(1) suggested certain factor to be deducted to take into accountdepth induced breaking. This gives an significant wave height of 9m and mean wave period of11s.This is the not the maximum possible wave height that occurs.A calculation is performedto find the maximum wave height.

Using the corrected significant wave height, the variance of the Energy density spectraof the Waves is calculated.

Hs = 4(m0)1/2 (1)

m0 =(Hs)

2

16(2)

1

We assume that wave amplitude is guassian distribution.Then we have the followingprobability of finding a wave amplitude greater than ’a’:

P (a) =exp( −a2

2∗mo)

2 ∗ pi ∗ (m0)1/2(3)

To find the maximum wave amplitude,we take the probability of exceedance of 0.001.Then,weget an maximum wave amplitude of 8m.

We have a maximum current velocity of 115cm/s.Coming to the soil conditions,the soil is clay or dense Fine Sand as we can see from the

figure below:

2

This is taken from Ref(3).

3 Assumptions used in the Design

• The Wind profile assumed is a power law.The variations in the wind profile with respectto time are neglected.

• The wind Force is calculated by an approximate formula.

• The main chord of the Offshore Wind Turbine is considered to be fixed at the inter-section of main chord,horizontal braces and diagonal braces.

• All the Bending Moment and Shear Force Calculations are performed by assuming themain chord to be cantilever beam.

• the main chord,pair of diagonal braces and horizontal braces are taken into considera-tion in the various analysis.

• A direct consequence of the above assumption is that the wave and wind loads haveno directional spreading involved.This means that all the environmental loads act inthe same direction.

• The diagonal braces are dealt as pinned connections in the Truss Analysis

• The horizontal Braces just touch the ground,but are mechanically unconstrained.

• The material chosen is NV-60 with a yield stress of 390MPa.

• The wave forces are calculated without taking time dependence into considerationfor the analysis.Thus the total wave force that acts on the structure is low than thecomputed load.This would just make the design conservative.Also,if we use the exacttime load history of the wave forces,the maximum of the combination is 0.6-1 times ofthe computed load.More in section 3.2.

• The top mass load of 80,000 tonnes is considered as a lumped mass.

• The diagonal and horizontal braces are so small that they do not fail by bending,butfail from compression or tensile stresses.

3

4 Computation of Loads

4.1 Wind Loads

A parabolic wind profile is assumed.If the velocity at 10m high above the water is ’V’, thenthe velocity at any other ’z’ is given by:

Vz = V ∗ (z

10)1/8 (4)

z is the height at which velocity is calculated.After the velocity is computed, the drag force is calculated by:

FD = 0.6 ∗ Vz2 (5)

4.2 Wave and Current Loads

The deep water wave length is calculated using the deep water dispersion relation given by:

L =g(T )2

2 ∗ pi(6)

Based on depth to deep wavelength ratio,classification into deep water,intermediate waterand shallow water is made.

ds =d

(L0)(7)

Based on this ratio, if it is less than 0.05,it is shallow waters.If the ratio is greater than 0.5,itis deep waters.In between the ratios,it is intermediate waters. In this case:

ds = 0.1781 (8)

Thus,it is intermediate waters.To find the wave-number,we use the Eckhart approximateformula,given by:

k =alpha ∗ tanh(alpha)−1/2

d(9)

alpha =(w)2 ∗ d

g(10)

d is the depth of the water.k = 0.0311 (11)

4

L = 201m (12)

The horizontal water particle velocity and horizontal water particle acceleration are com-puted using the following formula:

ux =w ∗ a ∗ cosh(k(d + z)) ∗ sin(w ∗ t)

sin(h(k ∗ d))(13)

ud =−a ∗ cosh(k(d + z)) ∗ cos(w ∗ t) ∗ (w)2

sinh(k ∗ d)(14)

where d is the depth of the waterw is the angular frequencya is the amplitude of the waterz is the co-ordinate in the Z-directionux is the horizontal velocity of the water particleud is the horizontal acceleration of the water particle

Then Morrison equation is used to calculate the total wave forces on the structure.TheDrag force is given by:

Fd =(Cd) ∗ rho ∗ (ux + uc)

2 ∗D2

(15)

where Cd is the drag co-efficientrho is the density of the waterux is the horizontal water particle velocityuc is the velocity of waterD is the diameter of the monopileFd is the drag force per unit length on the monopile

The drag co-efficient used in the analysis is 1.05 as stated in Ref(2).

The inertia force on the structure is given by:

FI =(Cm) ∗ rho ∗ pi ∗ (D)2 ∗ (ud)

8(16)

where Cm is the inertia co-efficientrho is the density of waterD is the diameter of the monopileud is the horizontal acceleration of the water particleFI is the inertia force per unit length

5

The inertia co-efficient used in the analysis is 1.6 as stated in Ref(2).The total wave force on the structure is given by:

F = Fd + FI (17)

The Time Dependent Wave and Wind Drag Force and Wave Inertia Force are Shown inthe figure below for z=30m below Mean Sea Level:

The Time Dependent Total Force is shown in the Below Figure:

6

If we define:

v =TimeDependentWaveandCurrentForce

AmplitudeoftheWaveandCurrentForce(18)

Then a graph is plotted to show the variation of this parameter at z=0m and z=30m respec-tively from Mean Sea Level.

7

8

These graphs show that the minimum v of 0.6 and max v of 0.8.But in the analysis wetake v as 1.Thus the wave forces which are predicted are over conservative.So,we are on asafer side.

4.3 Structural Load

The structural mass is given by:

(Md) = rho2 ∗ pi ∗ ((R)2 − (R− t)2) ∗ L (19)

where L is the Length of the monopileR is the radius of the monopilerho2 is the density of the steel

4.4 Added Mass Loads

The added mass force is given by:

(Am) = rho2 ∗ pi ∗ (R)2 ∗ L (20)

(An) = rho2 ∗ pi ∗ (R)2 ∗ L (21)

Here Am refers to the Added Mass in the X-Direction and An refers to the Added Mass inY-Direction.

4.5 Top Mass Loads

The mass of the turbine is 80000 tonnes.This is the Top mass load.

4.6 Buoyancy Loads

The buoyancy loads are given by the volume of water displayed by the cylinder.

(Vb) = rho2 ∗ pi ∗ (R)2 ∗ L (22)

All the Values Calculated are displayed in Appendix-B

5 Load Resistance Factor Design

As per Ref(2),the following load factors are taken:

Type of Load Load factorDead Loads 1.1Live Loads 1.1environmental load 1.35

9

The dead loads include the mass of structure and any weight of the equipment on thestructure.The Live Loads include the weight of any component which changes during the lifetime(e.g. Water tanks).The environmental loads include the wave,wind and current loadson the structure.As per Ref(2),the Resistance Factors are taken:Loading Type Resistance FactorAxial Tension 0.95Axial Compression 0.85Bending 0.95Shear 0.95Hoop Buckling 0.8Connections 0.9-0.95

6 Static Analysis

A MATLAB program is coded which calculates the approximate Bending Moment as inAppendix-A.This code calculates the loads as concentrated loads acting at the respectivecentroid’s.There is second code which uses the load distribution calculated as a function oflength,integrates it to find the shear force and integrates it again to find Bending Moment.Ananalysis is done in ANSYS to find the maximum bending stress taking the 2D structure intoconsideration.

Here,a summary of each of the three is given:

6.1 Concentrated loads for Bending Moment

6.1.1 Calculation of Diameter of Main Chord

The forces are calculated per unit length of the structure.Then Simpsons Method is used tocalculate the total force,total moments and the respective centroid’s of the Loads.Using theseas an approximation,the bending moment and shear force are calculated.Then,the bendingstress and shear stress are calculated using the following formula:

sigb =Ma

S(23)

sigv =2Va

A(24)

Here sigb is the maximum bending stress on the structuresigv is the maximum shear stress on the structureMa is the maximum bending moment on the structureVa is the maximum shear force on the structureS is the elastic section modulus of the section,given by:

S =I

y(25)

10

where I refers to the moment of inertia given by:

I =pi((D)2 − (D − 2t)2)

64(26)

y refers to the maximum distance from the nuetral axis which in the Circular hollowsection is given by:

y =D

2(27)

Apart from these stress on the structure,we also get hoop stress on the structure due toexternal pressure on the cylinder,given by:

fh =pD

2t(28)

where fh is the hoop stress at any depth z

p is the pressure acting at that depth,given by:

p = rho ∗ g ∗ z + rho ∗ g(Hw

2)(cosh(k(d + z))

cosh(kd)) (29)

where rho,g have their respective meanings

Hw =a

2(30)

k refers to wave-numberd is the depth of the waterz is the vertical Z co-ordinate

Also,the member is held in compression due to the combination of lumped mass and selfweight of the structure.

sign =F

A(31)

where:

F = topmass + Md (32)

A refers to the area of the structure

The Von-Misses stress criteria is checked.The Von-Misses stresses are given by:

sigm =((sigb + sign)2) + (fh)2 + (sigb + sign − fh)2) + 6 ∗ (sigv)

2

2

1/2

(33)

11

where sigb is the bending stress,sign is the normal compression stress,fh is the hoop stress,sigvis the shear stress.The Factor of Safety is expressed as:

FS =sigmFy

(34)

where sigm is the Von-Misses Stress,Fy is the Yeild Stress.Assuming a Factor of Safety of 0.8,the diameter and thickness of the structure are calcu-lated.This Factor of Safety is suggested by standards for Ultimate Limit state.

All the calculations are performed by MATLAB.When checking for different criteria,correctionfor Bending Moment and shear force are provided,to take into account the Shear force andBending Moment calculated using Load distribution discussed in Section 6.2.Then,the dif-ferent standards as given by Ref(2),are checked for these diameters and thickness. TheCorrection Factors are defined as:

C1 =BM

Ma

(35)

where BM refers to maximum bending Moment obtained using load distributions andMa refers to the maximum Bending Moment obtained using Concentrated Loads.

C2 =SF

Va

(36)

where SF refers to maximum bending Moment obtained using load distributions and Va

refers to the maximum Bending Moment obtained using Concentrated Loads.The final diameter and thickness of the main chord obtained for this Factor of Safety are:

D = 5.7m (37)

t = 140mm (38)

For this simplified Method we have the following Shear Force,Bending Moment and NormalLoad distribution

12

13

6.1.2 Calculation of Dimensions of Horizontal and Diagonal braces

They fail by compression or tension.The same factor of safety is used as mentioned above.Aeffective truss analysis is performed by using Method of Joints to find the forces on thestructure as shown in Appendix-C.

We arrive at the following diameter and scantlings for the braces as shown in the table:

Member DiameterDiagonal Brace 1.2mHorizontal Brace 0.17m

Member thicknessDiagonal Brace 30mmHorizontal Brace 30mm

6.2 Bending Moment Diagram using Load distribution

The load distribution functions are fit using Excel and MATLAB.The load distribution curvesused are:

q = (0.004 ∗ x5− 0.223 ∗ x4 + 6.37 ∗ x3− 100.8 ∗ x2 + 1016 ∗ x− 840.6) ∗ 1.35forx < 40 (39)

q = (85.06 ∗ x2 − 12290 ∗ x + 522200) ∗ 1.35forx >= 40 (40)

where x refers to the distance from the free end of the cylinder.

14

Integration of the load curve gives the Shear Force.The equations for Shear Force Distri-bution are:

SF = (9x6)/10000−(6021x5)/100000+(17199x4)/8000−(1134x3)/25+(3429x2)/5−(113481x)/100(41)

SF = SF (39) + (9 ∗ x ∗ (4253 ∗ x2 − 921750 ∗ x + 78330000))/1000forx > 40 (42)

Here SF(39) refers to the shear force at x=39m.Integration of the Shear Force gives the Bending Moment.The equations for the Bending

Moment are given by:

BM = (9∗x7)/70000−(2007∗x6)/200000+(17199∗x5)/40000−(567∗x4)/50+(1143∗x3)/5−(113481∗x2)/200forx < 40(43)

BM = BM(39) + x2 ∗ ((38277 ∗ x2)/4000− (11061 ∗ x)/4 + 352485)forx >= 40 (44)

Below we can find graph for Normal Load distribution on the main chord.

Below we can find Shear Force distribution on the main chord.

15

Below we can find the Bending Moment Distribution on the Main chord.

16

6.3 Analysis in Ansys

An analysis is done to take the 3D effects of the structure and as expected gave low stresslevels when loaded with a uni-directional wave,wind and current loads due to the influenceof bending in one direction by the bending in the other direction.

An 2-d stress countour from Ansys is depicted in the figure below:

The maximum stress in the figure is without load factor,but if the load factor is takeninto account,the maximum stress is 230Mpa.The analysis is done with remote loads at every1m.It is done to compare whether the stress in the monopile to the calculated stress usingMATLAB.

6.4 Critical Locations

• The Bending Moment is maximum at the connection of the main chord with thediagonal and horizontal braces.This is critical location.

• The interface between the water and wind is critical because the region is subjectedto sudden changes in shear force.So there may be a possibility of local bending andbuckling.It is shown in deflection curve below:

17

The sudden change in deflection can be clearly noted at a length of 30m.

7 References

• Ref(1)-The Maximum Significant Wave Height in Southern North Sea,Final Report,ReportNo 14-94,February 1995.

• Ref(2)-Recommended Practices for Planing,Designing and Constructing Fixed Off-shore Platforms-Load and Resistance Factor Design,API Recommended practice 2A-LRFD,First Edition,July 1,1993.

• Ref(3)-Design of the Piles in the UK sector of North Sea Prepared by Geo-TechnicalConsulting Group for the Health and Safety Executive.

• Ref(4)-Design Methods for Offshore Wind Turbines at Exposed Sites,Final Report ofOWTES project,EU JOULE III project JOR3-CT98-0284.

18

8 Appendix

8.1 Appendix-A

The code in MATLAB which computes the approximate Bending Moments and checks thecriteria as specified by API is presented here.

D=5.7;t =0.137;%t =0.175z =1:40;v=D/ t ;i f ( ( v<=300)&&(t >0.0061))

Fw=0.6∗((50∗( z / 1 0 ) . ˆ ( 1 / 8 ) ) . ˆ ( 2 ) ) ∗D;f o r i =1:40

i f ( (mod( i ,2)==0)&&( i <40))c ( i )=4;

e l s e i f ( ( i ==1) | |( i ==40))c ( i )=1;

e l s ec ( i )=2;

endend

FW=sum(Fw.∗ c ) / 3 ;c1=sum(Fw.∗ z .∗ c )/(3∗FW) ;%c1 i s the c en t r o i d from the MSL% dont f o r g e t UX(31)= v e l o c i t y at MSLC1=c1 +30;%C1 i s the c en t r o i d from the z=−10 where we assume the s t r u c t u r e to be%clampedCd=1.05;Cm=1.6;uc =1.15;U=Ux+uc ;rho =1025;U=U’ ;Uxd=Uxd ’ ;mko=length (U) ;f o r i =1:mko

U1( i )=U(mko−i +1);Uxd1( i )=Uxd(mko−i +1);

endFd=0.5∗Cd∗ rho ∗(U1. ˆ2 )∗D;Fi =0.5∗Cm∗ rho ∗( p i ( )∗ (D/2 )ˆ2 ) .∗Uxd1 ;p=length (Fd ) ;

19

f o r i =1:31i f ( (mod( i ,2)==0)&&( i <31))

d( i )=4;e l s e i f ( ( i ==1) | |( i ==31))

d( i )=1;e l s e

d( i )=2;end

end

FD=sum(Fd .∗d ) / 3 ;FI=sum( Fi .∗d ) / 3 ;z1 =0:30;C2=sum(Fd .∗ z1 .∗d )/(3∗FD) ;C3=sum( Fi .∗ z1 .∗d )/(3∗FI ) ;

% s t r u c t u r a l loadMT=80000;g =9.81;rho1 =7850;m1=pi ( )∗ rho1 ∗ ( (D/2)ˆ2−((D−2∗t ) /2 )ˆ2 )∗70 ;vo l=pi ( ) ∗ ( (D/2)ˆ2−((D−2∗t ) /2 )ˆ2 )∗70 ;mass=(m1+MT)∗ g ;C4=(MT∗70)+m1∗(35)/ mass ;

%Added mass loads assuming almost cy l i nde r , but they may not be used%because the morr ison equat ion a l r eady has these c o n t r i b u t i o n s tak ing%roughness i n to accountAx=pi ( )∗ rho∗Uxd∗(D/2)ˆ2 ;Ay=pi ( )∗ rho∗Uxd∗(D/2)ˆ2 ;AX=sum(Ax.∗d ) / 3 ;AY=sum(Ay.∗d ) / 3 ;C5=sum(Ax.∗d .∗ z1 )/(3∗AX) ;C6=sum(Ay.∗d .∗ z1 )/(3∗AY) ;

% Buoyancy loadsV1=rho∗ pi ( ) ∗ ( (D/2)ˆ2)∗30 ;C7=15;

wd1=V1/mass ;d i sp (wd1)

% Hydrostat i c Pressurez2=max( z1 ) ;P=0.5∗ rho∗g∗( z2 ˆ2)∗ pi ( ) ∗ ( (D/ 2 ) ˆ 2 ) ;

20

C8=z2 /3 ;

%shear f o r c e at the supportVa=FW+FD+FI+P;%a x i a l f o r c e on the s t r u c t u r eHa=mass ;%bending moment at the supportMa=FW∗C1+FD∗C2+FI∗C3+P∗C8 ;

L=70;x=0:(L−1);M(1)=Ma;KL=[C1 C2 C3 C8 ] ;g=s o r t (KL) ;k1=g ( 1 ) ; k2=g ( 2 ) ; k3=g ( 3 ) ; k4=g ( 4 ) ;i f ( k1==C1)

F1=FW;e l s e i f ( k1==C2)

F1=FD;e l s e i f ( k1==C3)

F1=FI ;e l s e

F1=P;end

i f ( k2==C1)F2=FW;

e l s e i f ( k2==C2)F2=FD;

e l s e i f ( k2==C3)F2=FI ;

e l s eF2=P;

end

i f ( k3==C1)F3=FW;

e l s e i f ( k3==C2)F3=FD;

e l s e i f ( k3==C3)F3=FI ;

e l s eF3=P;

end

21

i f ( k4==C1)F4=FW;

e l s e i f ( k4==C2)F4=FD;

e l s e i f ( k4==C3)F4=FI ;

e l s eF4=P;

end

% Bending Moment Ca l cu la t i onf o r i =2:70

i f ( i<=k1 )M( i )=Ma−Va∗ i ;

e l s e i f ( ( i>k1)&&(i<=k2 ) )M( i )=Ma−Va∗ i+F1∗( i−k1 ) ;

e l s e i f ( ( i>k2)&&(i<=k3 ) )M( i )=Ma−Va∗ i+F1∗( i−k1)+F2∗( i−k2 ) ;

e l s e i f ( ( i>k3)&&(i<=k4 ) )M( i )=Ma−Va∗ i+F1∗( i−k1)+F2∗( i−k2)+F3∗( i−k3 ) ;

e l s eM( i )=Ma−Va∗ i+F1∗( i−k1)+F2∗( i−k2)+F3∗( i−k3)+F4∗( i−k4 ) ;

endendV(1)=Va ;% Shear Force Ca l cu l a t i onf o r i =2:70

i f ( i<=k1 )V( i )=Va ;

e l s e i f ( ( i>k1)&&(i<=k2 ) )V( i )=Va−F1 ;

e l s e i f ( ( i>k2)&&(i<=k3 ) )V( i )=Va−F1−F2 ;

e l s e i f ( ( i>k3)&&(i<=k4 ) )V( i )=Va−F1−F2−F3 ;

e l s eV( i )=Va−F1−F2−F3−F4 ;

endendH(1)=Ha ;k5=C4 ;% normal f o r c e l o c a t i o nf o r i =2:L

i f ( i<=k5 )H( i )=Ha ;

22

e l s eH( i )=0;

endendsubplot ( 1 , 3 , 1 )p l o t (x ,M) ;x l a b e l ( ’ length ’ ) ;y l a b e l ( ’ bending Moment ’ ) ;subp lot ( 1 , 3 , 2 )p l o t (x ,V) ;x l a b e l ( ’ length ’ ) ;y l a b e l ( ’ shear f o r c e ’ ) ;subp lot ( 1 , 3 , 3 )p l o t (x ,H) ;x l a b e l ( ’ length ’ ) ;y l a b e l ( ’ a x i a l f o r c e ’ ) ;

% choos ing NV−60 with a y i e l d s t r e s s o f 390MPa and Zeta =0.78%tmin=5.1mm%f o r t h i s t h i c k n e s s the maximum D=0.46m%D/t<=54.75

Fy=390∗10ˆ6;E=210∗1000∗10ˆ6;% c r i t e r i a f o r normal compress ion s t r engthlmd=2∗70∗(Fy/E) ˆ ( 0 . 5 ) / ( p i ( )∗D/ 4 ) ;

Fxe=2∗0.3∗E∗( t /D) ;

i f (v<=60)Fxc=Fy ;

e l s eFxc=(1.64−0.23∗v ˆ ( 0 . 2 5 ) )∗Fy ;

end

Fy1=min ( Fxe , Fxc ) ;i f ( lmd< (2ˆ0 .5))

Fcn=Fy1∗(1−0.25∗( lmd ˆ 2 ) ) ;e l s e

Fcn=Fy1/( lmd ˆ 2 ) ;endQ2=1.1∗mass ;A1=pi ( ) ∗ ( (D/2)ˆ2−((D−2∗t ) / 2 ) ˆ 2 ) ;f c =(Q2/A1 ) ;phi1 =0.85;

23

d1=abs ( f c )−phi1∗Fcn ;

i f ( d1<=0)d i sp l ay ( ’ nominal compress ion c r i t e r i a s a t i s f i e d ’ ) ;

e l s ed i sp l ay ( ’ nominal compress ion c r i t e r i a not s a t i s f i e d ’ ) ;

end

% bending c r i t e r i asd1=Dˆ4 ;sd2=(D−2∗t ) ˆ 4 ;I=(p i ( ) /64 )∗ ( sd1−sd2 ) ;y3=D/2 ;S=I /y3 ;M1=1.27∗Fy∗S ;i f (Ma>M1)

fb =(0.85∗1.3∗Ma/S ) ;e l s e

fb =0.85∗1.3∗Ma/S ;endm1=10340/Fy ;m2=20680/Fy ;i f (v<=m1)

Fbn=1.27∗Fy ;e l s e i f ( ( v>m1)&&(v<=m2) )

Fbn=(1.13−2.58∗(Fy∗D)/(E∗ t ) )∗1 . 27∗Fy ;e l s e

Fbn=(0.94−0.76∗(Fy∗D)/(E∗ t ) )∗1 . 27∗Fy ;end

phi2 =0.95;e2=fb−phi2∗Fbn ;i f ( e2<=0)

d i sp l ay ( ’ bending c r i t e r i a s a t i s f i e d ’ ) ;e l s e

d i sp l ay ( ’ bending c r i t e r i a not s a t i s f i e d ’ ) ;end

% shear f o r c e c r i t e r i aA=pi ( )∗ (D/2)ˆ2 ;fv =(2∗Va/A) ;Fvn=Fy / ( 3 ˆ ( 0 . 5 ) ) ;phi3 =0.95;

d3=fv−phi3∗Fvn ;

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i f ( d3<=0)d i sp l ay ( ’ shear c r i t e r i a s a t i s f i e d ’ ) ;

e l s ed i sp l ay ( ’ shear c r i t e r i a not s a t i s f i e d ’ ) ;

end

%pre s su r e check f o r Hoop s t r e s s buck l ingz4 =0:30;Hw=16;k =0.0311;d=30;Hz=z4+(Hw/2)∗ ( cosh ( k∗(d−z4 ) )/ cosh ( k∗d ) ) ;gh =1.1 ;g =9.8 ;p=gh∗ rho∗g∗Hz ;fh=(p∗D/(2∗ t ) ) ;phi4 =0.8 ;

G=(L/D)∗ (2∗v ) ˆ ( 0 . 5 ) ;ml1 =(1.6∗v ) ;ml2 =(0.825∗v ) ;i f (G>=ml1 )

cn =0.44/v ;e l s e i f ( (G<ml1)&&(G>=ml2 ) )

cn =0.44/v+0.21∗(v ˆ3)/(Gˆ 4 ) ;e l s e i f ( (G<(ml2))&&(G>=1.5))

cn =0.737/(G−0 .579) ;e l s e

cn =0.8;end

Fhe=2∗cn∗E∗ t /D;i f (Fhe<=(0.55∗Fy ) )

Fnc=Fhe ;e l s e

Fnc=0.7∗Fy∗(Fhe/Fy ) ˆ ( 0 . 4 ) ;endd4=fh−phi4∗Fnc ;i f ( d4<=0)

d i sp l ay ( ’ hoop s t r e s s buck l ing c r i t e r i a s a t i s f i e d ’ ) ;e l s e

d i sp l ay ( ’ hoop s t r e s s buck l ing c r i t e r i a not s a t i s f i e d ’ ) ;end

% combined bending and compress ion

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Fey=Fy/( lmd ˆ 2 ) ;lop =1−0.4∗( f c /( phi1∗Fcn ) ) ;Cmy=min( lop , 0 . 8 5 ) ;d5=0.8∗ f c /( phi1∗Fcn )+(0 .75/( phi2∗Fbn ) )∗ (Cmy∗ fb )/(1− f c /( phi1∗Fey ) ) ;d6=1−cos ( p i ( )∗ abs ( f c )/(2∗ phi1∗Fxc))+ fb /( phi2∗Fbn ) ;d7=fc−phi1∗Fxc ;i f ( ( d5<=1)&&(d6<=1)&&(d7<0))

d i sp l ay ( ’ the member i s good under combined bending and compression ’ ) ;e l s e

d i sp l ay ( ’ member f a i l s under bending and compression ’ ) ;end

%combined a x i a l compression , bending and h y d r o s t a t i c requirementfx=f c+fb +(0 .5 .∗ fh ) ;d8=fx −0.5∗ phi3∗Fhe ;d9=(( fx −0.5∗ phi3∗Fhe )/ ( phi1∗Fxe−0.5∗ phi3∗Fhe))+( fh / phi3∗Fhe ) . ˆ 2 ;%check d8>0 and d9<=1 manually at a l l z

cd1=0; cd2=0;mkl=length ( d8 ) ;mkl2=length ( d9 ) ;f o r i =1:mkl

i f ( d8 ( i )>0)cd1=cd1+1;

e l s efvb =0;

endend

f o r i =1:mkl2i f ( d9 ( i )>=1)

cd2=cd2+1;e l s e

fvb1 =0;end

end

i f ( ( ( cd1 ==0) | |( cd2==0))&&(d5<=1)&&(d6<=1)&&(d7<0))d i sp l ay ( ’ the member does not f a i l under a l l loads ’ ) ;

e l s ed i sp l ay ( ’ the member f a i l s ’ ) ;

ende l s e

d i sp l ay ( ’ choose d i f f e r e n t s e t o f parameters ’ ) ;

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endF=s q r t ( 0 . 5 ∗ ( ( fb+f c )ˆ2)+0.5∗(max( fh )ˆ2)+3∗( fv ˆ 2 ) ) ;FS=F/Fy ;

Z1=[−cos ( 0 . 8727 ) cos ( 0 . 8727 ) 0 ; s i n ( 0 . 8727 ) s i n (0 . 8727 ) 1 ;10∗ cos (0.8727)−10∗ cos (0 . 8727 )∗ s i n ( 0 . 8727 ) 10∗ cos (0 . 8727 )∗ s i n (0.8727)−10∗ cos ( 0 . 8727 ) 0 ] ;Q1=[Va ; Ha ;Ma ] ;F1=pinv (Z1)∗Q1;d i sp l ay (F1)%from the answer obtained , one can c l e a r l y des ign the d iagona l braces f o r%compress ion and not bending because the l ength o f the column i s smal l%the i n t e r e s t i n g f a c t i s that the middle h o r i z o n t a l member i s sub j ec t ed to%tens i on and should only comply with the t e n s i l e f o r c e r e s t r i c t i o nF11=F1 ( 1 ) ;F12=F1 ( 2 ) ;F3=F1 ( 3 ) ;

w2=max(BM) ;w3=max(SF ) ;we1=w2/S ;

we2=2∗w3/A;XCV1=(( abs (we1)+abs ( f c ) ) ˆ 2 ) ;XCV3=(abs (we1)+abs ( f c )−max( fh ) ) ˆ 2 ;F=(0.5∗XCV1+0.5∗((max( fh ))ˆ2)+0.5∗XCV3+3∗(we2 ˆ 2 ) ) ˆ ( 0 . 5 ) ;FS2=F/Fy ;MLJ=0.8;D21=(4∗abs (F1 ( 1 ) ) / ( p i ()∗390∗10ˆ6∗MLJ) ) ˆ ( 0 . 5 ) ;D31=(4∗abs (F1 ( 2 ) ) / ( p i ()∗390∗10ˆ6∗MLJ) ) ˆ ( 0 . 5 ) ;D4=(4∗abs (F1 ( 3 ) ) / ( p i ()∗390∗10ˆ6∗MLJ) ) ˆ ( 0 . 5 ) ;

DE=[D21 D31 D4 ] ;TE=DE/v ;

This code is executed until we get the diameter and thickness which satisfy the conditionthat the Factor of Safety equals 0.8.

Then a code is used to calculate the actual Bending Moment and Shear Force using theload distributions is expressed by the following MATLAB code:

syms x ;q1 =(0.004∗xˆ5−0.223∗xˆ4+6.37∗xˆ3−100.8∗xˆ2+1016∗x−840 .6 )∗1 .35 ;%t h i s i s only v a l i d f o r x<40q2 =(85.06∗xˆ2−12290∗x +522200)∗1.35;%t h i s i s v a l i d f o r x>=40

B1=i n t ( q1 , x ) ;B2=i n t ( q2 , x ) ;

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%B1 , B2 r e f e r s to Shear Force d i s t r i b u t i o n s from the r e s p e c t i v e Load D i s t r i b u t i o n sS1=i n t (B1 , x ) ;S2=i n t (B2 , x ) ;%S1 , S2 r e f e r s to the Bending Moment d i s t r i b u t i o n s from the r e s s p e c t i v e Shear f o r c e d i s t r i b u t i o n curve

f o r x=1:70i f (x<40)

SF( x)=(9∗xˆ6)/10000 − (6021∗xˆ5)/100000 + (17199∗xˆ4)/8000 − (1134∗xˆ3)/25 + (3429∗xˆ2)/5 − (113481∗x )/100 ;e l s e

SF( x)=SF(39)+(9∗x∗(4253∗xˆ2 − 921750∗x + 78330000))/1000 ;end

endf o r i =1:70

k ( i )=71− i ;k1 ( i )= i ;

end

p lo t (k , SF ) ;x l a b e l ( ’ length ’ ) ;y l a b e l ( ’ Shear Force ’ ) ;f o r x=1:70

i f (x<40)BM( x)=(9∗xˆ7)/70000 − (2007∗xˆ6)/200000 + (17199∗xˆ5)/40000 − (567∗xˆ4)/50 + (1143∗xˆ3)/5 − (113481∗x ˆ2)/200 ;

e l s eBM( x)=BM(39)+x ˆ2∗((38277∗xˆ2)/4000 − (11061∗x )/4 + 352485) ;

endendf o r i =1:70

k ( i )=71− i ;k1 ( i )= i ;

end

p lo t (k ,BM) ;x l a b e l ( ’ length ’ ) ;y l a b e l ( ’ Bending Moment ’ ) ;

8.2 Appendix-B

The wind Force is given by:Fw = 379.059KN (45)

The Wave and Current Drag Force is given by:

FD = 2110KN (46)

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The Wave Inertia load is given by:

FI = 1182.45KN (47)

The Structural Mass is given by:

Md = 1360.67373tons (48)

The Added Mass Loads on the Structure are:

Ax = 1480.03tons (49)

Ay = 1480.03tons (50)

The Buoyancy Load is given by:V = 804.06tons (51)

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8.3 AppendiX-C

The Free body diagram of the whole structure:

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The free body diagram of the lower horizontal chord with two diagonal braces is shownin the below figure:

The forces are calculated in the MATLAB code provided in Appendix-A.The value of the forces are mentioned here:

F21 = 332740KN

F4 = 6300KN

F31 = −323090KN

So the horizontal and the left diagonal brace are in tension and the right diagonal brace incompression.

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