Primary design of an offshore jacket structure in English Channe

OE4651-12 Bottom Founded Structure Exercise Phase 1&2 Hand In Group Number: 8

Group Member :

Ang Li Hao Tan Yang Liu

4415671 4408950 4384954

Preface This is the first project we’ve done as candidates for master degree in TU Delft. It is a primary design of the Jacket offshore structure. We throw lights on many aspects related with the design of the bottom founded structure in offshore engineering. For instance, the preliminary design of the dimensions of the whole structure, computation of weight and buoyancy, computation of environmental loads on the structure, check for the static strength of piles and members, check for the static strength of joints, etc. It is such a comprehensive work that the knowledge absorbed in bachelor and recent few months in master are almost performed, so as to accomplish the project and optimize the design. It is a good lesson showing the complexity and the diversity of the work in offshore area. When deal with this exercise, we come across many difficulties, especially some details we neglect in the lectures. Through this precious opportunity, we have filled the blanks of all the ignored parts. It is with the help of Frank Sliggers and Jeroen Hoving that we can come through so many difficulties. Therefore I would like to thank our supervisor for the helpful illustration not only in the lectures, but also in their office time when they give some feedbacks to our work. The exercise is made by three persons, Ang Li, Hao Tan and Yang Liu. Most of the parts are done together so it is not possible to separate our parts. It is really a nice experience to enjoy a good teamwork and cozy work atmosphere like this. Hope everyone can be benefited and get prepared for the incoming exams and courses.

Yang Liu, Ang Li and Hao Tan

OE4651-12 Exercise Bottom Founded Structures

Design of an Offshore Jacket Structure

Group 8

Table of Contents 1.

Introduction ................................................................................................................................................. 2

2.

Structural Configuration ............................................................................................................................. 2

2.1

Primary Data Analysis ................................................................................................................................ 2

2.2

Deck Elevation ............................................................................................................................................ 3

2.3

Frames ......................................................................................................................................................... 4

2.3.1

Number of bays ........................................................................................................................................... 4

2.3.2

Geometry configuration .............................................................................................................................. 4

2.3.3

Horizontal Frame Lengths, Bay Heights and Bracing Angles .................................................................... 4

3.

Diameter and wall thickness ......................................................................................................................11

3.1

Diameter calculation ..................................................................................................................................11

3.2

Wall thickness calculation ..........................................................................................................................11

4.

Weight and buoyancy ................................................................................................................................ 13

5.

Hydrodynamic loading.............................................................................................................................. 13

5.1

Transformation of Joint Coordinate .......................................................................................................... 13

5.2

Selection of the stick range ....................................................................................................................... 14

5.3

Determination of the equivalent diameters of the various ranges ............................................................. 15

5.4

Estimation of the overall hydrodynamic loading ...................................................................................... 17

6.

Structure Analysis ..................................................................................................................................... 21

6.1

Design Load .............................................................................................................................................. 21

6.2

Foundation Check ..................................................................................................................................... 23

6.3

Member check........................................................................................................................................... 30

7.

Conclusion ................................................................................................................................................ 40

Reference ............................................................................................................................................................... 40 List of tables .......................................................................................................................................................... 41 List of figures......................................................................................................................................................... 42

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1. Introduction

The subject of this bottom founded exercise presents the schematic design of a jacket platform, which covers most issues of a preliminary bottom founded structure design, ranging from the determination of structural dimension to the calculation of the environmental loads. Starting from the research for the basic design, we guarantee the specializations of jacket platform to meet the requirements listed in the design criteria. The design process can be divided into four major parts as followed in phase 1: Structure configuration. Do research on functional specifications, environmental conditions and soil conditions. Gather Data and calculate input data, such as the maximum and minimum water depth and deck elevation Calculate weight and buoyancy of the jacket. . Determine the preliminary members, legs and pile diameters and prepare drawings. The weight and buoyancy can be calculated after we know the diameter of members and their locations by setting a global coordinate at the right leg of the vertical end-on side. Equivalent stick model. Using the method provided in hand-out, the 3-D structural model will be transformed into a layer equivalent stick model, which will simplify the task. Calculation for environmental loads. Since the stick model has been built, the environmental loads on the stick can be calculated through integration by using the Simpson’s method.

2. Structural Configuration 2.1 Primary Data Analysis In the basic design period of the jacket structural, the specific data of the operate location and the elevation of the deck were required in order to locate the uppermost horizontal frame and the work point. The summary of the required data and specifications are shown in Table 1. Table 1 Required Meteorological Data and Load Specifications Location

Southern North Data

Water depth(MSL)

31.6m

Wave height(100years)

9.6m

Wave period

14.2s

Astronomical tidal range

3.0m

Positive storm surge

1.2m

Negative storm surge

0.2m

Marine growth layer thickness

36mm

Subsidence at the end of production life

0.4m 2



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Table 1 continued Minimum elevation of the top framing above MSL

6m above MSL

Elevation of the topmost submerged frame

must remain flooded at all times

Initial pile diameter

42”

Deck leg size

36”

Wind load end-on direction

0.5MN

Wind load broadside direction

0.8MN

Yield stress of steel

345MPa

Yield stress for piles(if increase required)

430MPa

Calculations of the maximum and minimum water depth can be done by using the data above and the formulas below. The maximum water depth:

Dmax  water depth (MSL)+ half the tidal range + storm surge + subsidence +settlement Dmax  31.6 [m] + 0.5×3.0 [m] + 1.2 [m] +0.4 [m] = 34.7 [m]

Dmin  water depth (MSL) – half the tidal range – storm surge Dmin  31.6 [m] - 0.5×3.0 [m] -0.2 [m] = 29.9 [m] Considering about the safety and convenience for calculation, Dmax  34.7m and Dmin  30m

2.2 Deck Elevation The elevation of the lower deck level should be high enough to clear the wave crest reaching the platform, which will create an extra overturning moment and cause the failure of the structure. For this exercise this elevation may be calculated as the maximum water depth Dmax

plus 55% of the wave height ( H max ) plus 1,5 meters air gap.

Deck Elevation = Dmax + 55% × H max + air gap = 34.7 [m] + 0.55×9.6 [m] +1.5 [m] =41.48 [m] For convenience this number is rounded up to 41.5m. This number represents the elevation of the lower deck level from the seabed which includes 1m clearance for vertical extension from the pile to the deck , 2.5m for welding of the shims(above the uppermost horizontal framing) , 37m from the uppermost horizontal frame to the lowermost horizontal frame and 1m(including the diameter of the pipe) elevation of the lowermost frame from the mud line. The details show in the Figure 1. Distance between the uppermost and lowermost frame = Deck Elevation – top vertical gap – 2.5 – mud line = 41.5 [m] -1 [m] -2.5 [m] – 1 [m]

= 37 [m]

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Underside of deck

Figure 1 Vertical end-on side

2.3 Frames 2.3.1 Number of bays In the design, two bays were chosen for the reasons that it can reduce the cost for material, fabrication, cathodic protection and further inspection requirements. Moreover, using two bays can make the angles between the diagonal bracing locating between the values of 30-60 degree more easily. At last, two bays can guarantee the uppermost frame keeps dry and the second highest frame always under the water considering the two worst case scenarios.

2.3.2 Geometry configuration At the first we assumed a batter angle of 1:13, and found it’s hard to fulfil the requirements in the criteria with a height of 39m (from uppermost to lowermost). At the last we choose a batter angle of 1:10 and the calculations were carried out below: reference was made to Handbook of Bottom Founded Offshore Structures.

2.3.3 Horizontal Frame Lengths, Bay Heights and Bracing Angles Take the vertical end-on side shows in the figure 1 as an example: Assumed batter angle, 1:10 True batter angle, 1: 10 /

2

Distance between uppermost to lowermost frame, h=37m 

Lengths of the horizontal frames, For finding the last frame length, b2 :

b2  37  0.1 2  10  17.4m For finding the value of M: 4


M


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b2 17.4  1.319 b0 10

Therefore, the length of the middle horizontal frame, b1 is:

b1  M  b0  1.319 10  13.19m 

Height of the bays:

h1  h (1  m)  37 / (1  1.319)  15.95m h2  h  h1  37  15.95  21.05m 

Calculations of the angles of the diagonals:

  tan 1 (h1 b  0.1 h )  53.991 0 1 The principles of the calculations for the rest members (frames, braces, heights and angles) were the same with above. Figure 2 shows the 3-D model of the jacket. In the following parts of this report the figures showing the angles and dimensions are also shown.

Figure 2 The labels of the structure

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The coordinate system was chosen at the Dmax surface ( Au Bu Cu Du )so that the X-axis falls on the end-on side, Y-axis falls on the broadside and Z-axis shows the positive height of the structure. The dimensions of the members are given in Table 2 and illustrate in Figure 3. Table2 Dimensions of members Section

Segment End-on side

Diagonal Broad side

End-on side

Horizontal

Broad side

A0D1,C0B1

19.72

D1A2, B1C2

26.02

A0B1,D0C1

26.85

B1A2,C1D2

30.15

A0D0,B0C0

10

A1D1,B1C1

13.19

A2D2, B2C2

17.4

A0B0,C0D0

20

A1B1,C1D1

21.60

A2B2,C2D2

23.7

Figure 3 The labels of the structure

6

Length(m)



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Figure 4 Angles between members Figure 4, above, shows the different angles (actually showed the angle in vertical face) of the frame, from which we can see all the angles fall into 30-60 degrees which reach the requirements consider about the welding operation. The situations comes the same with the horizontal frames. In these horizontal frames, there are ten pipe guides. Considering the welding operation, the angles between two members are larger than 30 degree and smaller than 60 degree. Figure 5 shows the underlying horizontal framings of the jacket structure (combination of K brace and X brace). Further details in the three horizontal frames show in the Figure 6.

VERTICAL END-ON SIDE

INCLINED END-ON SIDE Figure 5 Bottom plan 7



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Figure 6 The angles in three plans

The joints in different frames has the same labels with the Figure 5 and their coordinates are given in Table 3, Table 4 and Table 5. Table 3 Joints in the top plan x

y

z

A0

3.7

0

3.3

B0

3.7

20

3.3

C0

13.7

20

3.3

D0

13.7

0

3.3

E0

3.7

9

3.3

F0

13.7

9

3.3

G0

7.0

0

3.3

H0

10.4

0

3.3

I0

7.0

4.5

3.3

J0

10.4

9

3.3

K0

7.0

4.5

3.3

L0

10.4

4.5

3.3

8



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Table 4 Joints in the middle plan x

y

z

A1

2.105

0

-12.655

B1

2.105

21.596

-12.655

C1

15.296

21.596

-12.655

D1

15.296

0

-12.655

E1

2.105

9

-12.655

F1

15.296

9

-12.655

G1

7.0

0

-12.655

H1

10.4

0

-12.655

I1

7.0

9

-12.655

J1

10.4

9

-12.655

K1

7.0

4.5

-12.655

L1

10.4

4.5

-12.655

Table 5 Joints in the lowermost plan x

y

z

A2

0

0

-33.7

B2

0

23.7

-33.7

C2

17.4

23.7

-33.7

D2

17.4

0

-33.7

E2

0

9

-33.7

F2

17.4

9

-33.7

G2

7.0

0

-33.7

H2

10.4

0

-33.7

I2

7.0

9

-33.7

J2

10.4

9

-33.7

K2

7.0

4.5

-33.7

L2

10.4

4.5

-33.7

Figure 7, below, shows the jacket location which presents by the underlying frame respect to the jack-up structure.

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Group 8

16.00

R7.50

D2

C2

8.00

5.00

5.00

17.40

45.00

9.00

A2

B2

14.70

Pile center line 10.00

Figure 7 Interface of the jacket and the jack-up rig ( showing jacket plan dimension at seabed )

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3. Diameter and wall thickness 3.1 Diameter calculation Initially the different type member diameters will be determined using rules of thumb: For the diagonal members, the slenderness KL / r is set to 80 (K=0.8), so the diameter D should thus be approximately 0.03 times its length, i.e. D=0.03L. Diameters of horizontal members in the vertical faces of the structure are also sized using the above rule, but their slenderness is increased to about 100. This results in D=0.023L.

3.2 Wall thickness calculation The D/t-ratio, i.e. ratio of diameter over wall thickness, for members varies over the structure. Due to handout information, members in the top part of the structure are often relatively thick walled with a D/t-ratio less than 30, while members near the bottom of the structure have much thinner walls with a D/t-ratio of more than 40. For weight estimation purposes however, we assume a D/t-ratio of 40. The stresses in the legs are generally moderate; here D/t = 60 is assumed. In any case the minimum wall thickness is 0,5 inch. Note that the diameters of members should always be smaller than the diameter of the member to which they are connected, by an amount such that a sensible welding configuration is achieved. Furthermore, the diameters of commercially available tubular go up in steps of 2 inches and wall thicknesses similarly go up in steps of 0,125 inches, so after calculation we will rounded up the diameters and wall thicknesses. As the main design has selected piles of 42’’, and consider the piles’ spare room in the leg, we select the jacket legs 6’’ larger in diameter than the piles, which is 48’’. Table 6 Diameter and wall thickness of members

Section

Round up

Round up

Diameter(inch)

thickness(inch)

-

48

0.875

40.903

-

48

0.875

C0_C2

40.903

-

48

0.875

D0_D2

40.702

-

48

0.875

A0_D0

10

0.254

10

0.25

A0_B0

20

0.508

20

0.5

B0_C0

10

0.254

10

0.25

C0_D0

20

0.508

20

0.5

A1_D1

13.191

0.305

12

0.375

A1_B1

21.596

0.508

20

0.5

B1_C1

13.191

0.305

12

0.375

C1_D1

21.596

0.508

20

0.5

Segment

Length(m)

A0_A2

40.702

B0_B2

Diameter(meter)

Leg

Frame 1

Frame 2

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Table 6 continued A2_D2

17.4

0.406

16

0.5

A2_B2

23.7

0.559

22

0.625

B2_C2

17.4

0.406

16

0.5

C2_D2

23.7

0.559

22

0.625

A0_B1

26.863

0.813

32

0.875

B1_A2

30.196

0.914

36

1

C0_B1

19.788

0.610

24

0.625

B1_C2

26.102

0.813

32

0.875

C0_D1

25.634

0.813

32

0.875

D1_C2

31.765

0.965

38

1

A0_D1

19.724

0.610

24

0.625

D1_A2

26.016

0.813

32

0.875

A0_K0

5.701

0.152

6

0.25

K0_E0

5.701

0.152

6

0.25

D0_L0

5.701

0.152

6

0.25

F0_L0

5.701

0.152

6

0.25

C0_E0

14.866

0.356

14

0.375

F0_B0

14.866

0.356

14

0.375

G0_I0

9

0.203

8

0.25

H0_J0

9

0.203

8

0.25

E0_F0

10

0.254

10

0.25

A1_K1

6.798

0.2032

8

0.25

K1_E1

6.798

0.2032

8

0.25

D1_L1

6.798

0.2032

8

0.25

F1_L1

6.798

0.203

8

0.25

C1_E1

18.238

0.457

18

0.5

F1_B1

18.238

0.457

18

0.5

G1_I1

9

0.254

10

0.25

H1_J1

9

0.254

10

0.25

E1_F1

13.19

0.305

12

0.375

A2_K2

8.491

0.203

8

0.25

K2_E2

8.491

0.203

8

0.25

D2_L2

8.491

0.203

8

0.25

F2_L2

8.491

0.203

8

0.25

C2_E2

22.778

0.508

20

0.5

F2_B2

22.778

0.508

20

0.5

G2_I2

9

0.254

10

0.25

H2_J2

9

0.254

10

0.25

E2_F2

17.4

0.406

16

0.5

Frame 3

Diagonal 1

Diagonal 2

Diagonal 3

Diagonal 4

conductor guide b0

conductor guide b1

conductor guide b2

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Table 6 continued Riser

Cassion

1 in-coming

37

0.356

14

0.375

2 out-coming

37

0.406

16

0.5

1 fire water

17.4

0.610

24

0.625

0.610

24

0.625

2 sump

31.4

4. Weight and buoyancy The weight is easily calculated by assuming the members are thin-walled hollow circular members. We use the formula below to calculate the weight of each member, then summary all the weight to get the total weight of the substructure. Weight=ρ*g*[ (outsideDiameter/2)2-(𝑖𝑛𝑠𝑖𝑑𝑒𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2)2]*π*Length Here we assume the density of steel ρ=7850kg/𝑚3, gravity of acceleration g=9.81𝑚/𝑠 2,then the total weight found in this way is roughly 2.63MN. When calculate the buoyancy of the substructure, we just consider the members are closed and in this case the buoyancy is the total volume of the structure which is submerged. Here we assume the density of seawater is 1025kg/𝑚3. Then we find the total buoyancy is roughly 1.85MN. Dry weight[MN] 2.63

Buoyancy[MN] 1.85

In-place weight[MN] 0.78

5. Hydrodynamic loading 5.1 Transformation of Joint Coordinate In order to calculate the hydrodynamic loading on the structure with simplification, the three dimensional structure model will be represented by a so-called equivalent stick. For the sake of the application of spreadsheet, here we set the wave travels in the X-axis direction. Then we can acquire the equivalent stick model through the rotation of the three dimensional model, using the following coordinate transfer:

xi  xi cos   yi sin  yi  yi cos   xi sin  zi  zi  Dmax This method is illustrated in Figure 8

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Figure 8 Transformation of joint coordinates

5.2 Selection of the stick range The ranges of the stick model should be defined in such a way that any member can be allocated unambiguously to one of these ranges. Here we set the origin of the coordinate at the maximum water depth

1 H  5.28 m 2



The upper boundary is at the wave crest elevation:

Dmax 



The lower boundary is at the seabed:

34.7 m

The allocation of the stick range is shown in Figure 9

14

Dmax



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Figure 9 Allocation of the stick range Normally, because of the length of the caisson, we should divide the 2nd bay into three ranges. In order to make the task easy, here we separate the calculation of caisson with that of the stick model. In other words, we merely calculate the load on the remained parts of the stick model without caisson, and then add the load of the caisson to the resulting load ( actually this only counts in this bay, so it won’t affect the calculation in other parts.). This method can reduce the range divisions and makes the labeling of the nodes more clear. As a consequence, we obtain the allocation of the ranges through the method above. The result is shown in Table 7 (in which we define boundaries Z_u and Z_l for all ranges with respect to SWL.) Table 7 Allocation of the ranges Range

Description

Zu (m)

Zl(m)

Height (m)

1

crest to top horizontal frame

5.3

3.3

2.0

2

the top horizontal frame

3.6

3.0

0.5

3

the upper part of 1st bay

3.3

0.0

3.3

4

SWL to MSL plus subsidence

0.0

-2.7

2.7

5

the lower part of 1st bay

-2.7

-12.7

10.0

6

the medium horizontal frame

-12.4

-12.9

0.5

7

the 2nd bay

-12.7

-33.7

21.0

8

the bottom horizontal frame

-33.4

-34.0

0.6

9

bottom horizontal frame to sea bed

-33.7

-34.7

1.0

5.3 Determination of the equivalent diameters of the various ranges The equivalent diameters of the various ranges is obtained through the method described in “The OE4651 equivalent stick model reviewed” (Ir. R. A. W. Dubbers, 9 January, 2004). In this method, we should define the following parameters: 15



Group 8

p  (yi  y j )2  (zi  z j )2 Li  j 

q

p  ( xi  x j )2

pD Li  j ( zu  zl )

Then we can acquire the equivalent diameters as following:

DDE 

q p Li  j

DIE 2  qD This procedure will be applied to all members. The results will be accumulated for all members (M) and assigned to their associated ranges (R). M

D

2 IE ( R )

  D 2 IE ( m) 1

M

D 2 IE ( R )   D 2 IE ( m) 1

Due to the marine growth, the stick diameters beneath the Mean Sea Level plus Subsidence should consider marine growth. So, for Z_u < MSL + Subsidence:

D  D  2* rMG The result of the determination of equivalent diameters of different ranges is shown in following tables. Table 8 Equivalent diameters of broad side Range

Description

Height

DDE(m)

DIE2(m2)

1


2.0

10.5

9.5

2


0.5

57.2

26.1

3


3.3

17.4

16.0

4


2.7

15.2

13.7

5


10.0

15.7

11.2

6


0.5

77.7

43.7

7

the 2nd bay

21.0

15.5

12.1

8


0.6

79.2

49.9

9


1.0

16.5

14.5

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Table 9 Equivalent diameters of end-on side Range

Description

Height

DDE(m)

DIE2 (m2)

1


2.0

10.6

9.5

2


0.5

22.9

7.6

3


3.3

16.1

15.1

4


2.7

13.8

12.7

5


10.0

14.3

10.0

6


0.5

49.9

23.9

7

the 2nd bay

21.0

14.5

11.2

8


0.6

76.0

40.7

9


1.0

10.3

13.3

Table 10

Equivalent diameters of diagonal side

Range

Description

Height

DDE(m)

DIE^2(m^2)

1


2.0

10.5

9.5

2


0.5

30.7

16.4

3


3.3

16.5

15.5

4


2.7

14.2

13.2

5


10.0

14.7

10.6

6


0.5

51.1

32.2

7

the 2nd bay

21.0

14.2

11.3

8


0.6

60.1

41.8

9


1.0

16.5

14.5

5.4 Estimation of the overall hydrodynamic loading The environmental loading of the slender members can be obtained through Morison equation. It is defined in two parts:

1  vx vx CD DDE 2 1 FIE  ax  D 2 IE  CM 4 FDE 

Since there is a 90 degrees phase shift between the drag and inertia force, the two force vectors are perpendicular, and hence the combined resultant force should be:

Fres  FDE 2  FIE 2 Then we should consider the terms in these two equations: (1). Design waves Wave loads can be determined using Airy’s theory. Through wave period we can obtain the angular velocity of the wave:

17



a   2



Group 8

cosh k ( z  d ) sinh kd

2  0.44(rad / s) T

Through Dispersion Relationship we can obtain wave number k (Eckart Approximation is applied):

 2  gk tanh(kd )   k0 d 

 2d g

kd   (tanh  )



1 2

Finally we get wave number k=0.0258 Then we need compute wave length:

c

 k

L  cT We obtain wave length L=243.8m For water depth/wave length is 0.14, we should use the following equations to compute water particle velocity and acceleration:

cosh k ( z  d ) sinh kd cosh k(z  d ) a   2 sinh kd v  

While, in this exercise we just use the equations of water particle velocity in deep water to make life easy. So actually here we adopt the following equation to obtain the water particle velocity and acceleration:

v  ekz

a   2e z The wave induced water particle velocities and accelerations decay exponentially from still water level down to the seabed. The horizontal velocity above MSL remains constant and the value of which equals to the value at MSL. From empirical data it is found that in most cases, the directional wave spreading factor is constant and equal to 0.906. Here the directional wave spreading factor is to be applied to water particle accelerations and velocities (2). Design current Here it is assumed that the current velocity is constant over the full instantaneous water depth. The current blockage factors for different wave headings are shown followings: End-on side:

0.8

Diagonal side:

0.85

Broad side:

0.8

(3). Stick model hydrodynamic coefficients The presence of any attachments such as anodes may be accounted for by increasing the drag coefficient. So the 18



Group 8

drag coefficient and inertia coefficient is adopted like these: Drag coefficient above MSL: 0.7 Drag coefficient below MSL: 0.75 Inertia coefficient:

2.0

(4). Maximum force and crest position As instructed in the Exercise hand out, for this exercise it is not necessary to calculate the crest positon for which the total force is at its maximum. Here it is omitted. In order to acquire precise loading by hand, we will use the Simpson’s Rule to apply numerical integration. As a result, the final force and moment can be computed though the equations below:

fu  4 f c  fl )(zu  zl ) 6 f z  4 f zc c f z l l M ( u u ) (uz zl 6 F (

)

Perform the whole calculation process for all the required wave direction(loading directions are depicted in figure 10 ), we can obtain the environmental force and moment applied on the structure. The result is shown in the tables beneath: ( the moment is obtained with respect to seabed) Table 11

Environmental loads on End-on side

level

description

fu (N)

fc (N)

fl (N)

F (N)

M (Nm)

1


61,352

61,352

61,352

121,477

4,736,407

2


128,666

128,666

128,666

65,363

2,483,776

3


93,832

93,832

93,832

309,645

11,255,604

4


69,516

67,061

64,728

181,119

6,043,238

5


70,698

73,666

55,440

698,149

18,991,886

6


191,115

197,701

188,916

99,131

2,185,428

7

the 2nd bay

56,375

57,759

37,307

1,224,572

15,187,218

8


193,638

196,909

191,877

109,259

109,305

9


27,272

27,173

26,817

27,130

13,603

Table 12 Environmental loads on Broad side level

description

fu (N)

fc (N)

fl (N)

F (N)

M (Nm)

1


60,916

60,916

60,916

120,614

4,702,745

2


336,115

336,115

336,115

170,746

6,488,360

3


101,176

101,176

101,176

333,880

12,136,546

4


76,217

73,524

70,967

198,576

6,625,686

5


77,681

80,923

60,915

766,980

20,864,340

6


301,544

311,643

298,072

156,311

3,446,015

7

the 2nd bay

60,132

61,588

39,788

1,300,175

16,105,465

8


203,902

207,009

202,042

114,925

114,973

9


42,681

43,079

41,986

42,830

21,473

19



Group 8

Table 13 Environmental loads on Diagonal side level

description

fu (N)

fc (N)

fl (N)

F (N)

M (Nm)

1


64,132

64,132

64,132

126,982

4,951,037

2


187,489

187,489

187,489

95,244

3,619,285

3


101,250

101,250

101,250

334,126

12,145,465

4


75,663

73,063

70,591

197,327

6,583,945

5


77,195

66,216

60,992

668,700

18,203,899

6


211,548

184,595

209,192

98,139

2,163,570

7

the 2nd bay

58,947

42,428

39,606

1,026,306

12,923,843

8


167,915

133,287

166,444

80,794

80,832

9


46,276

36,549

45,553

39,670

19,895

Figure 10 Loading directions All in all, the resultant shear force and moment on seabed is shown in table 14 and table 15 Table 14 Resultant hydrodynamic shear force and moment on seabed Shear force (MN)

Moment (MNm)

broad side

3.2

70.5

end-on side

2.8

61.0

diagonal side

2.7

60.7

Table 15 Resultant hydrodynamic and wind induced shear force and moment on seabed Shear force (MN)

Moment (MNm)

broad side

4.0

103.7

end-on side

3.3

81.8

diagonal side

3.6

99.8

20



Group 8

6. Structure Analysis 6.1 Design Load Load to be considered for the structural analysis:

Fw : Resulting wind load on the topsides GT : Total of the permanent and variable topsides loads

FH : Resulting hydrodynamic loads on the substructure GS : Total of the permanent and variable substructure loads, i.e. Structure weight minus its buoyancy These loads are shown in the Figure 11 below.

Figure 11 Applied load on the structure From the design loads above, we can now determine the total applied global loading for ULS by using the following formulas: Total vertical load, FG   G (GT  GS ) 21



Group 8

Base shear: FB = E ( FW  FH ) Overturning moment due to environmental loads: M OE   E (M OW  M OH ) Overturning moment due to permanent and variable loads: M OG   G (M OGT  M OGS ) Total overturning moment: M  M OE  M OG In which,

M OW  FW hW

M OH   hi FH，i i

M OGT  GT eGT M OGS  GS eGS These support reactions have been calculated in the following three conditions and are shown in the table below respectively：  Environmental loading  Symmetric vertical loading  Anti-symmetric vertical loading The partial load factors used in the calculation are manifested in Table 16. Table 16 Load factors Partial load factors

G

Q

E

Permanent and variable actions only

1.3

1.5

0

1.1

1.1

1.35

0.9

0.8

1.35

Extreme conditions; Action effects due to permanent and variable actions are additive Extreme conditions; Action effects due to permanent and variable actions are oppose

The calculations of different overturning moments and forces shown in Figure 11 are indicated in Table 17 to 20. Table 17 Calculation moment due to permanent load gravity and distances

Broad side

end-on side

diagonal

gravity and distances

Broad side

end-on side

diagonal

eGT(m)

0.94

0.25

0.88

eGS(m)

0

0

0

Gt(MN) M𝑂𝐺𝑇 (MNm)

16.68 15.68

4.15

Gs(MN) M𝑂𝐺𝑆 (MNm)

14.66

22

2.63 0

0

0



Group 8

Table 18 Moment due to wind load forces and distances

broadside

end-on side

Hw(m)

diagonal

51.50

Fw(MN)

0.50

0.80

0.64

Mow(MNm)

25.75

41.20

33.21

Table 19 Moment due to hydromechanics forces and distances

broadside

end-on side

Hw(m)

diagonal

31.97

FH(MN)

3.21

2.84

2.67

Moh(MNm)

102.46

90.66

85.27

Table 20 Total load and moment on the jacket structure Load type

broadside

end-on side

diagonal

Fg (MN)

21.24

21.24

21.24

Fb (MN)

5.00

4.91

4.47

Moe (MNm)

173.08

178.00

159.94

Mog (MNm)

17.24

4.57

16.13

Mo (MNm)

190.32

182.57

176.07

6.2 Foundation Check 6.2.1

Pile reactions

Instead of complex calculating all forces at once, we determine the support reactions at the sea floor, by separately considering 3 different types of loading.  Horizontal loading due to the environmental loads on the frame (wind, waves and current)  Symmetric part of the vertical loading due to the permanent and vertical loads  Anti-symmetric part of the vertical loading due to the permanent and vertical loads The problem was solved in three dimensions: broadside, end-on side and diagonal side. The loads calculated above in Table are the total loads which should divided by 2 when calculate the support reactions in each frame. The reactions of the foundation piles are considered in the local pile coordinate system, which is Pa for the axial force, Pt for the transverse pile reaction and Pm for the moments. The forces and moments are shown in the Figure12.

23



Group 8

Figure 12 Support reactions For end-on side, the value of 1 and  2 are identical due to the symmetric. While on broadside as the pile has been battered in one direction only, 1 is zero and  2 is 0.1. Furthermore, some assumptions are used in analysis. Transverse loads of the two piles are identical: Pt1  Pt 2  Pt Use the foundation model accounting for the inflection point at an equivalent depth d e . Assume that

1 and  2 are small enough, 1,  2

1

In which,

2de  4.5D Therefore,

de  4.5  42  0.0254  0.5

 2.40m The factors used in the calculation are presented in Table 21. Table 21 Assumption and parameters required broadside

end-on side

diagonal

D(m)

42

42

42

de(m)

2.40

2.40

2.40

Height of point C(m)

238

88

253.75

a1(radians)

0

0.1

0.1

a2(radians)

0.1

0.1

0.1

bt(m)

20

10

22.36

be(m)

24.66

18.36

30.74

24

OE4651-12 Exercise Bottom Founded Structures 


Group 8

Forces and moments caused by environmental loading

PtE 

M  1 h  FB  OE   2 h  de  h 

PaE ,1  PaE ,2  PaE 

FB de  M OE be

PmE ,1  PmE ,2  PmE  PtE de In which, M OE , FB are obtained from the calculations showed in Table 

Forces and moments caused by symmetric loading The symmetric part of the vertical loading due to the permanent and variable loading FG which is shown in Table 20. For a Jacket, the vertical load of topsides acts at mid-point of top brace and the vertical load of the substructure also acts at the mid-point of the top brace.

PtG ,1  PtG ,2  PmG ,1  PmG ,2  0 PaG ,1  PaG ,2  PaG  

FG 2

Forces and moments caused by anti-symmetric loading The anti-symmetric part of the variable loading is manifested by the corresponding overturning moment M OG which is shown in Table 20.

PtOG  M OG

sin   M OG  be be

PaOG  M OG

cos  M OG  be be

PmOG ,1  PmOG ,2  PmOG  de PtOG Combining all the support reactions and moments calculated above results in the total loading which is showed in Figure 13.

25



Group 8

Figure 13 Support forces and moments Therefore:

Pt ,1  Pt ,2  PtE  PtOG 

M 1 h  FB  OE  2 h  de  h

  M OG  b  e

Pm,1  Pm,2  de  PtE  PtOG  Pa ,1  PaE  PaG  PaOG 

FB de  M O FG  be 2

Pa ,1  PaE  PaG  PaOG 

FB de  M O FG  be 2

The final results of support reactions are shown in tabulated format in the Table 22. Table 22 Combined support reaction for total loading

6.2.2

broadside

end-on

Pt1,2 (MN)

1.02

0.69

Pm1,2 (MNm)

2.46

1.66

Pa1 (MN)

-1.21

-0.02

Pa2 (MN)

9.41

10.60

Maximum pile loads

When it comes to the foundation pile check, the maximum loads should be considered. As we can see in Figure 14, the maximum pile loads occur in the diagonal direction due to the fact that the moment arm is longer in this case, at the same time, the loading is stood by only one pile. For the overall equilibrium of the structure the bending moments 26



Group 8

within the piles may be neglected. These are relatively small compared to the overturning moment. Therefore, only the axial and lateral forces in the piles need to be considered for the overall equilibrium of the structure. Above all, the pile check are done in diagonal direction.

Figure 14 Piles under influence of overturning moment in diagonal direction In order to check the pile, the formulas below have been used. Vertical pile reaction, V1,2 



FG 4

M OE  M OG BS

Pile net shear, H1,2  FB  V1  V2   



The vertical forces and pile net shear can be solved along the axis of the pile. The load calculation result are shown in the Table 23. Table 23 Maximus pile loads diagonal Vertical Pile Reaction V1 (MN)

-0.66

Vertical Pile Reaction V2 (MN)

11.28

Pile Net Shear, 𝐻1,2 (MN)

0.06

Axial Compressive Load 𝑃𝑐 (MN)

11.23

Lateral Pile Force 𝑃𝑡 (MN)

1.18

Thus, compare the value of axial compression force and lateral force with the ultimate compression force and lateral forced gave in the hand-out with pile diameterφ42” , thickness 1.75” . Meanwhile. A factor 0.8 should be considered for the ultimate limit state of the pile, which can make the design more conservative. Ultimate compressive force ofφ42” , thickness 1.75”  22.1 0.8  17.68MN Ultimate lateral force ofφ42” , thickness 1.75”  3.5MN It can be seen that both compressive and lateral force are below the required ultimate values. So the pile is designed well and have no need to change to a larger one.

27

OE4651-12 Exercise Bottom Founded Structures 6.2.3


Group 8

Pile penetration check

It can be found that the required pile penetration is between 50-60 meters. In this case, the value of 50 was chosen. Then compared this value with the axial load which will applied on the pile, and the penetration can be calculated: 𝑃𝑖𝑙𝑒 𝑝𝑒𝑛𝑒𝑡𝑟𝑎𝑡𝑖𝑜𝑛, 𝑋 2

(

𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑜𝑛𝑒 𝑝𝑖𝑙𝑒

) = 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒

50

Therefore, 𝑃𝑖𝑙𝑒 𝑝𝑒𝑛𝑒𝑡𝑟𝑎𝑡𝑖𝑜𝑛, 𝑋 2

( 6.2.4

) =

50

11.23  35.64m 22.1

Pile Stress Check

Effectively the piles are fully clamped at about 4.5 pile diameters below the mudline. Alternatively we may assume that the piles are hinged at 2.25 pile diameters below the mudline. The bending moment in the pile near the mudline can now easily be determined: The stress due to the axial load is:

Pc A

fc 

where: A    D0  t  t , with D0 being the pile diameter and t the thickness The stress due to bending is: where: S 

 4

 D0  t 

2

t

fb 

M2 S

1 A  D0  t  4

The moment at 2.25D below the mudline is found by: M 2  2.25DH , with D being the pile diameter and H the pile net shear. Finally check whether:

where:

fc f  b  b  0.85 c Fc b Fy

c  0.85 and b  0.95 and Fy  345MPa for:  

Note: Fc  (1  0.25 ) Fy 2

Fc 

1

2

Where:

for:  

Fy



 

Fy E



KL Fy r E

2 2

and: r 

D D t I  av  0 , K 1 A 2 2 2 2

According to the process above, the check results are shown in the Table 24.

28



Group 8

Table 24 Checking of the pile for bending and axial force parameters

Stress check below mudline

Stress check above mudline

0.143

0.236

Area under stress caused by bending moment, S (m )

0.036

0.078

λ

0.894

0.695

𝐹𝑐 ,(Mpa)

432.081

714.257

345

345

φ𝑐

0.85

0.85

φ𝑏

0.95

0.95

r

0.361

0.465

Stress due to axial load, 𝑓𝑐 (Mpa)

77.611

46.950

𝑓𝑏 (Mpa)

125.152

160.127

4.576

12.418

Area of pile under stress, A(m2 ) 2

𝐹𝑦 ,(Mpa）

Stress due to bending,

M2=2.25DH(MN.m) Checking

0.53

0.566

satisfied

satisfied

From the result, we can read that the stress below and above mudline both under the criterion.

6.2.5

6.2.4 Stress check in bottom bay

The stability of the leg in the bottom bay may now be checked in a similar way as the through‐the‐leg pile. Any lateral loading may be ignored, as both the self‐weight and buoyancy is approximately acting along the axis of the leg, while the hydrodynamic loading near the sea floor is rather small. In this case, we use the same method of calculations above; however, still a minor difference exists, the length of consideration and calculation of M1 which is 0.25M2.The results are presented in Table 25. Table 25 Parameters used in calculation and checking of the leg segment parameters Area of pile under

Stress check below mudline

stress, A(m2 )

0.143

Area under stress caused by bending moment, S

(m2 )

0.036

λ

0.75

𝐹𝑐 ,(Mpa)

611.38

𝐹𝑦 ,(Mpa）

345

φ𝑐

0.85

φ𝑏

0.95

r

0.361

Stress due to axial load, 𝑓𝑐 (Mpa)

77.611

𝑓𝑏 (Mpa)

66.175

Stress due to bending,

M2=2.25DH(MN.m)

9.66

M1=0.25M2

2.41

Checking

0.35 satisfied

29



Group 8

All the check of pile has done, from which we can see the pile satisfied the required standard, and can withstand the topsides weight and self-weight well.

6.3 Member check In this section, we transfer the global loads to section forces to analysis the members. And here two simplifications are used to calculate member forces easily. 1) Apply equivalent point loads at the frame nodes. 2) Calculate the section force from horizontal loads and vertical loads separately then superimpose to get the total member force. 6.3.1

Horizontal section force

The horizontal forces are caused by hydrodynamic loads and wind loads, so the first step is to transfer the environmental loads into point force acting at the nodes of the frame. a) The wind load is transferred to the level of the top horizontal framing. Consider the top frame joints are hinged, we got the point on which horizontal wind load applying on the top frame and the point vertical acting on the top of the foundation piles, which is due to the overturning moment caused by wind load. See Figure 15.

Figure 15 Transfer wind loads to the substructure

Horizontal wind load on the top frame:

F

Vertical substructure frame load:



F

Wv ,1

W h, 1

F

F

Wv ,2

 FW 2

W h, 2

 aF W

b

t

Where a = Deck elevation + 10 – op horizontal frame elevation = 13.5 m The transferred wind load on broadside and end-on side are shown in Table 26, because we analyze a single frame, the force on each side should be divided by two.

30



Group 8

Table 26 Transferred wind load on substructure Transferred wind load

Broadside

End-on side

unit

FW

0.5

0.8

MN

FWh,1 , FWh,2

0.4

0.25

MN

FWv,1 , FWv,2

0.54

0.17

MN

The hydrodynamic load should also be transferred from distributed forces into lumped one, which apply on each horizontal frame. Herein we use geometry method to transfer the distributed load into lumped load, which can be learned easily from the Figure 16.

Figure 16 Geometry method The detailed calculation of geometry method is to add two parts of distributed loads into one lumped load on each frame. For example, we get FH ,1 by adding the distributed loads applying on A0 A1 and A1 A1,lower together and also the new lumped load matches the shear and moment caused by the relevant distributed load.

A1,lower A1   F H ,1  F H ,i  F H ,i A0 A1 ;

A1,lower A1      F H ,1 h1 F H ,i h i  F H ,ih i ; A0 A1 By using numerical method we get distributed load on each length. The lumped hydrodynamic load on broadside and end-on side are shown in Table 27.

31



Group 8

Table 27 Lumped hydrodynamic load Lumped hydrodynamic load

Broadside

End-on side

From top to MSL

FH ,1 (MN)

0.85

0.70

MSL to cassion tip

FH ,2 (MN)

1.78

1.61

0.66

0.61

Cassion tip to bottom

FH ,3

Then use the formulas below we can get the equivalent point forces which are shown in Table 28.

F1  ( FH ,1  FW ) E Fi   E FH ,i , i  2 Table 28 Equivalent point forces causing by environmental load Equivalent point load

Broadside

End-on side

F1

0.91

1.01

F2

1.20

1.09

F3

0.44

0.41

Consider horizontal forces only, we treat jacket as equivalent beam model and can calculate sectional shear and moments caused by environmental load, results are shown in Table 29. Shear forces and moments in different ranges can be calculated by the environmental point forces, here we can easily calculate the first horizontal frame using the formulas below,

S  PaE (1   2 ), M  PaE bt Also we can get the shear forces and moments of the third horizontal frame,

S  2PaE , M   PtE bs  2PmE Table 29 Sectional shear and moments

Range

Z(m)

Top of leg

Broadside

End-on side

Shear(MN)

Moment(MN.m)

Shear(MN)

Moment(MN.m)

39.5

-0.75

-150.11

-1.03

-103.37

1st

horizontal frame

38

0.16

-115.48

-0.02

-64.81

2nd

horizontal frame

22.05

1.36

-88.95

1.07

-40.84

3rd horizontal frame

1

1.81

-88.50

1.48

-40.43

Sea bed

0

2.12

-7.60

0.69

-5.05

32



Group 8

Vertical section force

According to the vertical force equilibrium, we use the formulas below to calculate relevant vertical forces, which are shown in Table 30.

VT ,1   G (GT ,1  GS ,1 )   E FWv,1 VT ,2   G (GT ,2  GS ,2 )   E FWv,2 Table 30 Vertical section forces

6.3.3

Broadside

End-on side

unit

Gt,1

3.78

3.96

MN

Gs,1

0.66

0.66

MN

Fwv,1

0.25

0.4

MN

Vt,1

4.54

4.54

MN

Gt,2

4.56

4.38

MN

Gs,2

0.66

0.66

MN

Fwv,2

0.17

0.54

MN

Vt,2

5.97

6.27

MN

Internal member force

 Internal member forces due to horizontal loads Here we use the cross-section method to calculate member forces, detailed force analysis is shown in figure 17. According to force and moment equilibrium, we can get formulas below, which are apply for all members of the jacket.

NB  N1 

S M (1  (1   2 )) cos   1 sin  Sb

M S sin  M  (1  (1   2 )) b cos   1 sin  Sb N2 

33

M b



Group 8

Figure 17 Cross-section analysis Internal member forces of broadside and end-on side are shown in Table 31 and Table 32. Table 31 Internal forces of broadside due to environmental load

Cross section position

b

θ

α1

α2

S

M

NB

N1

N2

[m]

[deg]

[rad]

[rad]

[MN]

[MN.m]

[MN]

[MN]

[MN]

0.1

0

-0.75

-150.11

-7.51

-7.51

Above 1st frame

20

Above

2nd

frame

21.6

36.46

0.1

0

0.16

-115.49

0.93

-6.28

-5.35

Above

3rd

frame

23.7

44.26

0.1

0

1.36

-88.95

2.69

-6.44

-3.75

0.1

0

1.81

-88.50

-3.70

-3.70

Above the sea bed

23.9

Table 32 Internal forces of end-on side due to environmental load

Cross section

b

θ

α1

α2

S

M

NB

N1

N2

position

[m]

[deg]

[rad]

[rad]

[MN]

[MN.m]

[MN]

[MN]

[MN]

Above 1st frame

20

0.1

0

-0.75

-150.11

-7.51

-7.51

2nd

Above

frame

21.6

36.46

0.1

0

0.16

-115.49

0.93

-6.28

-5.35

Above 3rd frame

23.7

44.26

0.1

0

1.36

-88.95

2.69

-6.44

-3.75

Above the sea bed

23.9

0.1

0

1.81

-88.50

-3.70

-3.70

All the internal forces are summarized in Table 33.

34



Group 8

Table 33 Summary internal member forces due to environmental loads Diagonal brace



Horizontal brace

NB,1

NB,2

NBh,1

NBh,2

NBh,3

Broadside

0.93

2.69

0.46

0.60

0.22

End-on side

1.90

3.03

0.51

0.54

0.21

Internal member forces due to vertical loads

Figure 18 Vertical loads applied at substructure According to the force distribute in figure 18, we can find the applied vertical loading at the top of the foundation piles is given by VT,1 and VT,2.

VT ,1   G (GT ,1  GS ,1 )   E FWv,1 VT ,2   G (GT ,2  GS ,2 )   E FWv,2 From the vertical equilibrium at the shim‐plate connection between frame and the 2 foundation piles, we find:

F

 VT ,1  VT ,3  PaV ,1 cos 1  0

F

 VT ,2  VT ,4  PaV ,2 cos  2  0

pile Z ,1

pile Z ,2

Substitution yields:

VT ,3   PaV ,1 cos 1  VT ,1  VT ,4   PaV ,2 cos  2  VT ,2  35

VT ,2  VT ,1 2 VT ,1  VT ,2 2

(1 

bt ) be

(1 

bt ) be



Group 8

Ergo:

VT ,3  VT ,4 After calculation, we summary the results in table 34. Table 34 Vertical force on the substructure bt[m]

be[m]

VT,1[MN]

VT,2 [MN]

VT,3 [MN]

VT,4 [MN]

Broadside

20

24.66

4.54

5.97

0.13

-0.13

End-on side

10

18.36

3.73

6.27

0.58

-0.58

For a jacket, although the loads VT,3 and VT,3 exist at points A and B, we assume them to be applied at respectively points C and D (figure 19).

Figure 19 Force analysis on point C and point D Use the law of Sines, we find

N Bh,1  VT ,4 tan  2 N L ,2  N L ,1  VT ,1

N B ,2  VT ,1

cos 

VT ,2 cos  2



 VT ,2 tan  2



 VT ,2 tan  2

sin(    1 ) 2 sin 1 sin(    1 ) 2

sin 



sin(    1 ) 2 cos 1



sin(    1 ) 2

We can use the same method to calculate the other frame members, then get the final result in table 35. Table 35 Summary of internal member forces due to vertical loads NBh,1

NB1

NL,1

NL,2

NB2

NBh,2

NBh,3

NL,3

Broadside

0

0.02

0.13

-0.13

0.012

0

-0.0087

0.17

End-on side

-0.06

0.17

0.44

-0.58

0.184

0

-0.1084

0.80

 Summary of internal forces Due to the forces result of each member caused by horizontal or vertical loads, we combine them together and get the 36



Group 8

final internal forces of each member in table 36. Table 36 Summary of forces for substructure members Diagonal brace

6.3.4

Horizontal brace

NB,1

NB,2

NBh,1

NBh,2

NBh,3

Broadside

0.95

2.70

0.46

0.60

0.21

End-on side

2.07

3.22

0.45

0.54

0.10

Lateral forces

For lateral loads, there are three components to be considered: 1. Lateral hydrodynamic loading (q1) due to waves and current. Determine this lateral load (for the case causing axial compression) applying Morison’s equation and using the water particle velocity and acceleration perpendicular to the member. For simplicity, this distributed load may be determined using the water particle velocity and acceleration at the centre of the member. Use the diameter including the marine growth. 2. Lateral loading (q2) due to the self‐weight of the member in air. Note that for an inclined member the component perpendicular to that member is relevant. Use the diameter without marine growth for the weight calculation. 3. Lateral hydrostatic (buoyancy) loading, dry (q3a) and flooded (q3b). There are two possible cases, either the member is dry, or it is flooded. For the overall platform weight we normally assume that all members are dry. However, an individual member may have been accidentally flooded. We never can tell beforehand which one this will be. Hence, for the member check we will have to consider both cases. It is once more the component perpendicular to the member that counts. Use for buoyancy again the brace diameter without marine growth. To calculate q3b there are 2 options: q3b is equal to the buoyancy of the steel cross sectional area only, or it is equal to the full buoyancy of the member minus the weight of the water within the member. Both options produce the same result of course. Next, determine the maximum value of the lateral loading; which is: max[±q1+ q2 − q3a,±q1+ q2 − q3b]. Once

the maximum lateral load q is known, the bending moment can be calculated based on M  ql

2

12

Table 37 shows the lateral member loads. Table 37 Lateral member loads Broadside Lateral force

Diagonal brace

End-on side

Horizontal brace

Diagonal brace

Horizontal brace

Unit

B1

B2

Bh1

Bh2

Bh3

B1

B2

Bh1

Bh2

Bh3

Length

26.86

30.20

20.00

21.60

23.70

19.72

26.02

10.00

13.19

17.40

[m]

q1

6.66

4.24

6.09

3.76

2.52

3.61

3.19

6.06

3.74

2.51

[KN/m]

q2

3.42

3.91

1.52

1.52

2.09

1.34

2.50

0.38

0.68

1.21

[KN/m]

q3a

4.20

4.73

0.00

2.04

2.47

1.73

3.07

0.00

0.73

1.30

[KN/m]

q3b

0.45

0.51

0.00

0.20

0.27

0.18

0.33

0.00

0.09

0.16

[KN/m]

q

9.63

7.64

7.62

5.09

4.33

4.78

5.37

6.44

4.33

3.56

[MN/m]

M

0.58

0.58

0.25

0.20

0.20

0.15

0.30

0.05

0.06

0.09

[MN/m]

37



Group 8

Substructure member check

Herein we can check stability for each member using axial force and bending moment results.

Pc A

fc 

The stress due to the axial load is:

where: A    D0  t  t , with D0 being the member diameter and t the thickness

fb 

The stress due to bending is: where: S 



 D0  t 

4

2

t

M S

1 A  D0  t  4

Finally check whether: rule= rule 

fc f Cm  b  b  0.85 c Fc b Fbn 1  f c FE 2

where:

c  0.85 ; b  0.95 ; Fbn  Fy  345MPa ; E  210GPa ; FE  Fy /  Cm  0.85 for:  

Note: Fc  (1  0.25 ) Fy 2

Fc 

1

2

Where:

for:  

Fy



 

Fy E



2 2

KL Fy r E

and: r 

D D t I  av  0 , K 1 A 2 2 2 2

According to the process above, the check results are shown in the Table 38. Table 38 Substructure member check Broadside Parameter

Diagonal brace

End-on side

Horizontal brace

Diagonal brace

Horizontal brace

B1

B2

Bh1

Bh2

Bh3

B1

B2

Bh1

Bh2

Bh3

L[m]

26.86

30.20

20.00

21.60

23.70

19.72

26.02

10.00

13.19

17.40

D[m]

0.81

0.91

0.51

0.51

0.56

0.61

0.81

0.25

0.30

0.41

t[m]

0.02

0.03

0.01

0.01

0.02

0.02

0.02

0.01

0.01

0.01

A[m2]

0.06

0.07

0.02

0.02

0.03

0.03

0.06

0.00

0.01

0.02

S[m3]

0.01

0.02

0.00

0.00

0.00

0.00

0.01

0.00

0.00

0.00

r[m]

0.28

0.31

0.18

0.18

0.19

0.21

0.28

0.09

0.10

0.14

λ

0.99

0.99

1.18

1.27

1.27

0.97

0.96

1.18

1.30

1.29

FC[MPa]

260.13

260.19

225.15

205.26

204.93

263.88

265.40

225.15

198.30

201.42

FE[MPa]

350.60

350.87

248.27

212.94

212.44

366.82

373.81

248.27

202.84

207.24

PC[MN]

0.95

2.70

0.46

0.60

0.21

2.12

3.22

0.43

0.54

0.10

ƒC[MPa]

17.24

38.10

23.05

30.45

7.88

71.69

58.28

87.66

61.51

6.16

M[MN.m]

0.58

0.58

0.25

0.20

0.20

0.15

0.30

0.05

0.06

0.09

ƒb[MPa]

53.10

36.82

103.75

80.81

55.14

35.25

27.75

175.34

96.29

58.05

rule

0.22

0.28

0.42

0.42

0.19

0.43

0.34

1.16

0.72

0.19

38



Group 8

Due to the results we can find that only Bh1 does not meet the stability check and should be redesign again. 6.3.6

Punching shear check

In this section, we check the punching shear strength of the joints between leg and the main Diagonal at the top of the jacket and at the bottom of the jacket on end-on side and broadside. We choose the Y-Joint model to check the shear strength. The punching shear model is shown in Figure 20.

Figure 20 Punching shear model

When di  d 0  2t0 ; N i , Rd 

f y0 3

t0 di

1  sin i /  M5 2sin 2 i

The punching shear check results are shown in Table 39 and Table 40 Table 39 Punching shear check for top leg joint location

top leg joint with diagonal brace

parameter

end-on side

broadside

t0[M]

0.02

0.02

di[M]

0.60

0.81

θ[Deg]

41.72

53.54

member

A0_D1

A0_B1

N1[MN]

15.94

15.77

Nb[MN]

2.12

0.95

status

satisfied

satisfied

39



Group 8

Table 40 Punching shear check for bottom leg joint location

bottom leg joint with diagonal brace

parameter

end-on side

broadside

t0[M]

0.02

0.02

di[M]

0.81

0.91

θ[Deg]

30.30

45.74

member

D1_A2

B1_A2

N1[MN]

33.41

21.28

Nb[MN]

3.22

2.70

status

satisfied

satisfied

7. Conclusion A two bay jacket is designed in this project under specific condition. The parameter and calculation process are well illustrated in this report with drawings in different views. Moreover, safety checks regarding the whole structure as well as the components are also included. The jacket is designed under Ultimate Limit State. For most of the cases the structure satisfy the safety check requirements. However, it still needs optimization, for it fails some checks. All in all, the phase 1 and phase 2 work are all accomplished with good results.

Reference [1] J. H. Vugts. Handbook of Bottom Founded Offshore Structure, Volume I. Delft University of Technology. 2002. [2] J.H. Vugts. Handbook of Bottom Founded Offshore Structure, Volume II. Delft University of Technology. 2002.

40



Group 8

List of tables Table 1 Table 2 Table 3 Table 4 Table 5 Table 6 Table 7 Table 8 Table 9 Table 10 Table 11 Table 12 Table 13 Table 14 Table 15 Table 16 Table 17 Table 18 Table 19 Table 20 Table 21 Table 22 Table 23 Table 24 Table 25 Table 26 Table 27 Table 28 Table 29 Table 30 Table 31 Table 32 Table 33 Table 34 Table 35 Table 36 Table 37 Table 38 Table 39 Table 40

Required Meteorological Data and Load Specifications………………………………………………..2 Dimensions of members……………………………………………………………………………...…6 Joints in the top plan…………………………………………………………………………………….8 Joints in the middle plan……………………………………………………………………...…………9 Joints in the lowermost plan…………………………………………………………………………….9 Diameter and wall thickness of members……………………………….………………….…………..11 Allocation of the ranges…………………………………………………………………..…………….15 Equivalent diameters of broad side…..…………………………………………………...…………….16 Equivalent diameters of end-on side…………………………………………….….…………………..17 Equivalent diameters of diagonal side……………………………………………………………...…..17 Environmental loads on End-on side………………………………………………..………………….19 Environmental loads on Broad side……………………………………………………….…...……….19 Environmental loads on Diagonal side…………………………………………..…….……...………..20 Resultant hydrodynamic shear force and moment on seabed…………………….…………………….20 Resultant hydrodynamic and wind induced shear force and moment on seabed…….................……...20 Load Factors………………………………………………………………………………...…...……..22 Calculation moment due to permanent load…………………………………..………………………..22 Moment due to wind load………………………………………………………………....……………23 Moment due to hydromechanics………………………………………………………………..………23 Total load and moment on the jacket structure…………………………………...…………………….23 Assumption and parameters required……………………………………………….………………….24 Combined support reaction for total loading……………………………………………………...……26 Maximus pile loads……………………………………………………………………………………..27 Checking of the pile for bending and axial force……………………………………………………….29 Parameters used in calculation and checking of the leg segment………………………………………29 Transferred wind load on substructure…………………………………………………………...……..31 Lumped hydrodynamic load…………………………………………………………………………….32 Equivalent point forces causing by environmental load………………………………………………..32 Sectional shear and moments……………………………………………………………...……………32 Vertical section forces……………………………………………………………………………..……33 Internal forces of broadside due to environmental load……………………..………………………….34 Internal forces of end-on side due to environmental load………………………..……………………..34 Summary internal member forces due to environmental loads…………………...…………………….35 Vertical force on the substructure……………………………………………………………………….36 Summary of internal member forces due to vertical loads……………………….……………………..36 Summary of forces for substructure members…………………………………………………………..37 Lateral member loads……………………………………………………………………………………37 Substructure member check……………………………………………………………………………..38 Punching shear check for top leg joint……………………………………………………..……………39 Punching shear check for bottom leg joint……………………………………………..……………….40

41



Group 8

List of figures Figure 1 Figure 2 Figure 3 Figure 4 Figure 5 Figure 6 Figure 7 Figure 8 Figure 9 Figure 10 Figure 11 Figure 12 Figure 13 Figure 14 Figure 15 Figure 16 Figure 17 Figure 18 Figure 19 Figure 20

Vertical end-on side……………………………………………………………………………………..4 The labels of the structure………………………………………………………………………………5 The labels of the structure………………………………………………………………………………6 Angles between members…………………………………………………………..…………..………7 Bottom plan…………………………………………………………………………………..…………7 The angles in three plans………………………………………………………………………………..8 Interface of the jacket and the jack-up rig……………………………………………………...………10 Transformation of joint coordinates……………………………………..……………………………..14 Allocation of the stick range…………………………………………………………...………………15 Loading directions……………………………………………………………………………..……….20 Applied load on the structure…………………………………………………………………………..21 Support reactions……………………………………………………………………………………….24 Support forces and moments……………………………………………………………...……………26 Piles under influence of overturning moment in diagonal direction………….……………………..…27 Transfer wind loads to the substructure…………………………………………..……………………30 Geometry method…. ……………………………………………………………………………….….31 Cross-section analysis……………………………………………………………………….…………34 Vertical loads applied at substructure…………………………………………………………………..35 Force analysis on point C and point D…………………………………………………………………36 Punching shear model…………………………………………………………………………………..39

42

OE4651 Bottom Founded Structures

F. Phase 1 – Summary Sheet Geometry 1. Elevations:

Dmax: …………………34.7 m………………

Dmin: …………………30 m…………………

Ddeck: …………………41.5 m………………

1:

…………………38 m…………………

2:

…………………22 m…………………

3:

…………………1 m……………………

4:

……………………………………………

2. Elevation hor. frames:

3. Piles:

Diameter: …………42 inch………

Wall thickness: …………0.75 inch………

Legs:

Diameter: …………48 inch………

Wall thickness: …………0.875 inch.……

Dry Weight:

…………………2.63 MN……………………

Buoyancy:

…………………1.85 MN……………………

In‐place Weight:

…………………0.78 MN..…………………

4. Substructure:

Hydrodynamic Loading (unfactored) 1. Base Shear: End‐on:

…………………2.8 MN……………………

Broadside:

…………………3.2 MN……………………

Diagonal:

…………………2.7 MN……………………

End‐on:

…………………61.0 MNm………………

Broadside:

…………………70.5 MNm.………………

Diagonal:

…………………60.7 MNm.………………

2. Overturning Moment:

OE4651 Bottom Founded Structures

G. Phase 2 ‐ Summary Sheet Vertical Loads (factored) 1. Substructure:

Dry Weight:

……………2.89 MN……………………

Buoyancy:

……………1.85 MN……………………

In‐place Weight:

……………1.04 MN……………………

2. Topsides Weight per pile: Horizontal Loads (factored) 1. Base Shear:

1: …………4.59 MN………

2: ……………4.59 MN…...…

3: …………4.59 MN………

4: ……………4.59 MN......…

End‐on:

…………………4.9 MN………………

Broadside:

…………………5.0 MN.………………

Diagonal:

…………………4.5 MN.………………

End‐on:

………………178.0 MNm……………

Broadside:

………………173.1 MNm……………

Diagonal:

………………159.9 MNm……………

2. Overturning Moment:

Maximum Support Reactions

Compression, max(Pa,1): ………………11.28 MN………………

Tensile, max(Pa,2):

…………………0.66 MN………………

Lateral, max(Pt):

…………………1.18 MN………………

Moment, max(Pm):

…………………2.46 MNm……………

Punching Shear Check Top leg‐joint:

Bottom leg‐joint:

Capacity:

…………15.94 MN..… Load:

……………2.12 MN...

Capacity:

…………33.41 MN….

……………3.22 MN…

Load:

Primary design of an offshore jacket structure in English Channe

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