Exemple of Offshore Structures Calculation

8/4/2019 Exemple of Offshore Structures Calculation

1/25

- 1 -

Universit de Versailles et Saint Quentin-en-YvelinesLISV - Laboratoire dIngnierie des Systmes de VersaillesEquipe MMS : Mcanique des Matriaux et des StructuresBtiment Descartes RC27

45, Avenue des Etats-Unis78035 Versailles FRANCE

Master 2 SPI DSME

Responsable: Paolo VANNUCCI ([email protected])

U.E.: Mcanique pour lIndustrie du Ptrole

Cours: Offshore Engineering

Enseignant: P. VannucciA.U. 2006-07

Examples of offshore structures calculation

1. Response of a single-degree of freedom structure to a white noise.

A white noise is a completely random signal; this means that the energy is equally distributed

in the frequency domain, i.e. the spectrum is a constant, see the figures. We want to analyse

the response of a single-degree of freedom structure to a white noise.

The mean square of the response is

.)()()()(0

2

0

2 dffSifHdffSt FFxxx

==

The transfer function of such a system is

,)(

1

)( 2ff

f CiMKiH +=

and with the angular frequencyf=/2we get

.24

1)(

22 CfiMfKifH

+=

The spectrum of the force is of the type

,)( SfSFF =

so that

( ) ( ).

24

)(0 2222

2 df

CfMfK

Stx

+=

To perform integration, it is worth to transform the above expression as

( ) ( )

dM

K

K

Stx

+

=

2

1

21

)(0 2222

2

with


2/25

_______________________________________

- 2 -

P. Vannucci Examples of offshore structures calculation

.2

,2

MK

C

M

K

f==

In the case, relevant for real structures, of small , we get

.48

)(2

2

KC

S

M

K

K

Stx ==

The standard deviation of the response is hence

.2

1

22

1)(

KC

S

M

KS

Ktx ==

The distribution of the response being Gaussian with zero mean, the probability that the

response be less than x in absolute value is 68.27% (i.e. for the 68.27% of the time the

absolute value of the response is less than x), while it is of 95.45% for 2x. For the 50% ofthe time, in addition, the absolute value of the response is less than 0.6745 x.

To remark that x decreases with the damping and stiffness of the system in the same way,and that increases with the spectral density.

The mean square value is dominated by the peak in the transfer function, corresponding to the

natural frequency of the system,

.2

1

M

Kfs

=

The results found so far are substantially valid also when the spectral density is not constant,but sufficiently "flat", like the case in the figure.

A general remark: x is inversely proportional to the square root of . In the case of a

perfectly sinusoidal force, the magnification is 1/2. This means that damping has a greatereffect on a on a resonant response due to a sinusoidal force than on a response to a random

force.

F(t)

SFF(f)

S

SFF(f)

S

f

f

f

fs

fs

|H(if)|

t

Time history of a "white

noise" signal

White noise spectrum

Transfer function

Typical actual spectrum and

idealised white noise spectrum


3/25

_______________________________________

- 3 -


2. Study of a single-column structure.

Let us consider the simple case of a single-column structure, composed by a vertical circular

cylinder of diameterD, surmounted by a deck carrying some facilities. Such a structure issometimes used in offshore engineering to host some simple facilities (a loading terminal for

instance, or a light or radio instruments for navigation, or even a wind mill for power

generation).

The scheme is that in the figure, along with the mechanical model; the data are reported in the

table, and refer to a light tower for navigation aid. The level of the pile fixity, in the absence

of more detailed data, is assumed to be about 6D, i.e. 2 m.

Data Symbol Value

Pile external diameter D 324 mm

Wall thickness t 9.5 mm

Water depth d 8 mHeight of tip mass above SWL h 7 m

Pile mass m 73.8 kg/m

Tip mass M 300 kg

Inertial moment of the pile J 0.11610-3 m4

Young's modules of the pile E 205109N/m2

Depth of the pile fixity level s 2 m

Effective pile length l 17 m

Damping ratio 0.008

Water density 1015 kg/m3

a. To calculate the natural frequency of the structure.

Effective mass per unit length in water:

.kg/m1.2338.745.848.734

)2(

4

22

=++=

++=++=tDD

mmmm iw

Effective tip mass: it is calculated with the static scheme below

The result is MT= 1006 kg.

The bending stiffness of the pile is N/m.145203

3==

l

EJKT

The natural circular frequency is rad/s.8.31 ==T

T

M

K

The natural angular frequency is Hz.6.02

1

1 ==

N

M

l

Sea bed

SWL D

M

h

d

sPile fixity level

l

mw

m

d+s hMT

M


4/25

_______________________________________

- 4 -


b. To calculate the response to a monochromatic force: a horizontal sinusoidal force of

amplitudeFo= 330 N and frequencyNf= 0.33 Hz applied at the top.

The response is given by the general equation

.)()()()(ti

fffeiHtFiHtx

==

For a single-degree of freedom, the complex transfer function is

,)(

1)(

2ff

fCiMK

iH

+

=

and hence the amplitude of the response, in this case the horizontal displacement of the top of

the structure, is

( ).

)(

1)()()(

2220 o

ff

offof F

CMK

FiHiHFiHx

+

===

In this case:

.m032.0

,s/mN2.612

rad/s,07.22

=

==

==

o

ff

x

MKC

N

The maximum displacement can be calculated also as ,so xQx = where Q is the magnificationfactor andxs the static response:

( ),,

41

1

221

222K

F

xQo

s =+

=

with .030.02

,545.0 11

=====

M

Cf

We get hence .41.1andm023.0 == Qxs The dynamical response (in terms of deflection, but also of stresses and strains, as the system is considered as having a linear mechanical

behaviour) is hence 41% higher than the static case.

c. To calculate the wave force and the overturning moment spectra: the sea state is defined by

a significant wave heightHs= 1.5 m and a peak energy frequencyfm=0.33 Hz.

First of all, we calculate the wave spectrum (i.e. the spectral density of the surface elevation)

S(f); for this, we choose to use the Pierson-Moskowitz spectrum, which is given, as afunction ofHs and the zero-crossing period Tz, by:

( ).

4)(

4

316

4

2zT

z

s eT

HS

=

Tzis related tofm by


5/25

_______________________________________

- 5 -


.s16.24.1

124.14.1

2=====

mzmm

zz

fTf

T

Hence, in this case, the Pierson-Moskowitz spectrum is

( ) ( ).

10225.8

4)(

8)(

4

2

44

310462.1

5

3

1

45

2

16

45

24 fTf

z

sT

z

s ef

eTf

HfSe

T

HS zz

===

This wave spectrum is plotted in the next figure.

We take as cut-off frequency the valuef*=1.5 Hz; actually, beyond this value the waves havea such short period that their energy content is negligible.

Let us calculate now the mean square of the surface elevation :

( ) .m14.016)()( 2*

12

*

002

4

===

zTfs

feHdffSdffS

The standard deviation is hence .m374.0= Assuming a zero-mean Gaussian distribution

of, the surface elevation is hence in the interval 0.374 m for the 68.27% of the time.

The frequency at which most of the energy is concentrated is

.Hz45.0)*(erf11

)(

)(

)(

)(

44*

1

22

*

0

*

0

2

0

0

2

=

==

zTf

z

zf

f

e

Tf

TdffS

dffSf

dffS

dffSf

Using the linear approximation of Borgman, the wave force spectrum, concerning the linear

separation range, i.e. the case of the Morison's equation, is given by

+= 222222

222 4

8

sinh

cosh4)()( iudFF kfk

kd

kzffSfS

,

with

;4

,2

1 2DCkDCk MiDd

==

the mean square value of the particles horizontal velocity is given by

S(f)[m/Hz]

f [Hz]


6/25

_______________________________________

- 6 -


.)()()(0

2

0

2

== dffSfHdffS uuuu

To remark that the spectrum above depends not only upon f but also upon z, i.e. it is aspectrum of the sectional force at the heightzabove the sea bed.

For what concerns CD and CM we choose CD= 1.3 and CM=2; in this way, kd=213.76 kg/mand ki= 167.37 kg/m.

We need first to calculate u; the transfer function is

kd

kzffHu

sinh

cosh2)( = .

This function depends upon the wave numberk, which is related to the wave frequency by the

dispersion equation:

.2

tanh2

tanh 22

L

d

L

gfkdgk

==

The above wave spectrum shows that the wave with f


7/25

_______________________________________

- 7 -


The spectrum of the total force acting upon the column is obtained after integration overzandis given by:

.4

sinh

4

4tanh2

4cosh)(4

sinh

32)()(

22

2

22

22

222

02

222

22

2

22

+

+

+

=

dfg

dfg

dfg

kgf

dzzfg

z

dfg

kffSfS

i

du

dFFtot

The numerical integration overzgives the following diagram for the wave force spectrum:

The standard deviation of the total force spectrum can be calculated numerically, and gives:

.N441)()(*

00==

fFFFFF

dffSdffStottottot

The frequency at which the mean energy of the force occurs is

.Hz59.0

)(

)(

)(

)(

*

0

*

0

2

0

0

2

==

f

FF

fFF

FF

FF

F

dffS

dffSf

dffS

dffSf

tot

tot

tot

tot

f[Hz]

SFF(f)[N/Hz]

z= 8 m

z= 7.75 m

z= 7.5 mz= 7.25 m

z= 7 m

z= 6.75 m

SFFtot(f)[N/Hz]

f[Hz]


8/25

_______________________________________

- 8 -


It is interesting to notice that F>, because the energy increases with frequency.

The spectrum of the overturning moment can be calculated as well:

.4

cosh16

4sinh

4cosh8

16

4cosh)(32

4sinh

)()(

22

222442

22

22

22

22

2

0

22

2222

2

2

2

++

+

+

+

=

dfg

gdfg

dfg

dfg

dgff

kg

dzzfg

zzkf

dfg

fSfS

i

dudMM

We find also, with an analogous meaning, Nm,7.1289=M .Hz464.0=

d. To calculate the expected maximum value of the force and of the overturning moment in a

sea state of duration 5 hours.

Assumed the force to be a zero-mean Gaussian random process, we have

( )( )

( )( )

.ln2

ln2;ln2

ln2

+=

+=

TTM

TTF

MMMMmax

FFFFmax

Tis the duration of the sea state, in seconds: .s1800036005 ==T is the Euler's constant:=0.5772. We get henceFmax= 1956.8 N and Mmax= 5656.5 Nm.

e. To calculate the response, in terms of horizontal displacement at the SWL, of the structure

to the previous sea state.

The calculation of the response passes through the transfer function of the response. This

transfer function describes the way the structure responses dynamically to the exciting force

acting upon it. In this case, the exciting force, the wave force, is a force distributed on the

column, with a hyperbolic variation with water depth and maximum at the still water level.

SMM

(f)[Nm/Hz]

f[Hz]


9/25

_______________________________________

- 9 -


So, we need the response of the structure to such an exciting force. This can be done, but it is

very cumbersome. So, we prefer to follow another simpler way, which gives a sufficiently

approximated result.

In fact, as already said, the force has a hyperbolic distribution: this means that it varies very

quickly with the depth. A good approximation of it, and anyway a conservative

approximation, is to consider that the wave force is a concentrated load applied to the

structure in correspondence of the still water level, see the figure.

We know the transfer function for the case of the simple oscillator, that is, for the case of the

last model on the right in the figure above. In order to have a good approximation, the

equivalent mass Mo must be determined such that the natural frequency of the final model is

equal to that of the previous model (central model in the figure above).

Hence:

.kg4942

3

1 =

====

oT

T

oTo

T

T

o

oo

l

lM

K

KMM

M

K

M

K

The response of the structure is now given by the relationship

),()()(2

fSifHfStotFFxx

=

with SFF(f) the wave force spectrum found above and the transfer function

.

21

1)(

2222

2

+

=

ooo

N

f

N

fK

ifH

In this case, for what said aboveNo=N1= 0.6 Hz, and

N/m.713403

3==

o

ol

EJK Hence, see also the figure,

[ ].

000711.07778.215089395600

1)(

22

2

2

+

=

ff

ifH

|H(if)|[m/N]

f[Hz]

M

l

Concentratedwave load

lo=d+s

Mo

lo

Sea bed

SWL

Pile fixity level

Distributed

wave load

M

d

s

l


10/25

_______________________________________

- 10 -


The spectrum of the response is hence readily found and represented in the figure below.

To be remarked that the response at the natural frequency (0.6 Hz) is much more significant

than the response at the peak of the force spectrum ( 0.38 Hz). The frequency at which mostof the energy is concentrated is

.Hz595.0

)(

)(

)(

)(

*

0

*

0

2

0

0

2

==

fxx

fxx

xx

xx

x

dffS

dffSf

dffS

dffSf

to be remarked that this value is practically coincident with the natural frequency of the

structure, 0.6 Hz, because the response is narrow-banded.

The standard deviation of the response is

m051.0)()(*

00

== f

xxxxx dffSdffS

and the expected maximum horizontal displacement at the SWL in a 5 hours sea state is

( )( )

.m226.0ln2

ln2 =

+=

TTx

xxxxmax

For the 68.27% of time, the horizontal displacement is hence less than 5.1 cm.

f. To calculate the natural frequency of the structure by the Rayleigh method.

In this method, actually an energy method, the system is assumed to be undamped and thefrequency is found equating the maximum potential energy and the maximum kinetic energy

on a period. This leads to the expression of:

.

0

22

0

2

2

+

=

l

iii

l

yMdxAy

dxyEJ

In the equation above,y(x) is the deflection mode shape of the beam. In the Rayleigh method

y(x) must be assumed and the result depends upon its choice; nevertheless, usually simplechoices ofy(x), leading to simple calculations, give good results, and most part of times, the

choice ofy(x) is not critical. To show this point, we use two different choices fory(x). The

f[Hz]

Sxx(f)[m/Hz]


11/25

_______________________________________

- 11 -


fundamental restriction in the choice ofy(x) is that it must satisfy the boundary condition at

the point of fixity or support, in our case the clamped edge.

In the first choice we use a trigonometric approximation, ,2

cos1)( oyl

xxy

=

yo being the

end deflection. In this way it is

( )

( )

.Hz766.0

2

,rad/s81.4

3004.525.282

1047.1

kg;300

;kg9.334kg4.52kg5.282

N/m;1047.132

12

24

222

222

0

2

0

2

0

2

242

3

4

0

2

0

2

===

++

=

==

=+=+=

===

N

y

y

yyMyM

yyydxymmdxymdxAy

yyl

EJdxyEJdxyEJ

o

o

iooii

oool

wll

ooll

o

Let us now to take a polynomial fory(x): ( )22 4lxaxy = , with a a constant; we get:

( )

( ).Hz748.0

2,rad/s7.4

10695.14049.0767.1

1054.8

kg;10695.1

;kg10049.410767.1

N/m;1054.8

1214215

2142

213214

0

2

0

2

0

2

215

0

2

0

2

===++

=

=

+=+=

==

Na

a

ayM

aadxymmdxymdxAy

adxyEJdxyEJ

iii

lw

ll

ll

o

It can be easily checked that both the assumptions made fory(x) respect the conditions at theclamped edge, i.e.y(0)=y'(0)=0.

The difference between the first and the second approximation is 2.3% of the first; however,

the difference with respect to the natural frequency previously calculated with a rigid-like

model, 0.6 Hz, is 21.7%, which shows that this last was not a so good approximation.

g. To calculate the wave loads produced by aH= 1.5 m high wave having a period T= 3 s.

The frequency is =2/T= 2.094 Hz. The wave length can be calculated by the dispersionequation:

.2

tanh2

tanh2

2

L

d

L

g

Tkdgk

==

The numerical solution of this equation gives L= 14.03 m. The wave number is k=0.448 m-1and kd= 3.58>1: the deep waters condition can be applied. In fact, tanh kd= 0.998, so that

.m06.14

2

22 =

gT

Lgk


12/25

_______________________________________

- 12 -


To avoid the numerical solution of the dispersion equation, the deep waters approximation

can be used at first and once the value ofL found, its validity checked.

The ratio .2.0023.003.14

324.0===DH

kd(D/L)(H/L)

DTuKC

The highest value of the wave height for the given value of the wave length is

27

tanh7

max ==L

kdL

H ,

so that 5.075.0/ max >=HH .

The consequence of these evaluations is that for the case in object the wave action should be

calculated in the non-linear separation range that is, using the Lighthill correction to the

Morison's equation:

,54321 FFFFFF ++++=

with:

F1: the Froude-Krylov term, computed in deep waters and with the linear wave theory:;cos42.1231cos

4

21

1 ttCg aD

dst

F M

=== n

in this calculation we have get CM=2; in fact, the Reynolds number in deep waters is

56

109.3103 = L

H D

Re ; the diagrams usually employed give for a smooth

cylinder and for these values ofRe andKC, CM=1.8; conservatively, we take CM=2;

F2: the drag force, computed in the theoretical framework of the linear deep waterswave theory:

;sinsin77.589sinsin4

1 22 ttttDg aCF D ==

we have taken the value CD= 1.3, once again corresponding to the above values ofReandKCfor a smooth cylinder;

F3: due to the quadratic term of the velocity potential in a second order Stokes-likepotential decomposition:

;2sin13.102sin3280

ln162

342213 tt

k b

,kbg ads)(wF

dz

=

== =

F4: due to the kinetic pressure, relative to the only linear potential:;2sin7.512sin

4cos)(

2

1 22214 ttkbg adsF S

===

F5: present only if the cylinder cross the free surface; it is due to the fact that theboundary condition is written on the actual surface, and not onz=d:


13/25

_______________________________________

- 13 -


.2sin8.2064cos2

4

21

5 tFdltg

FC

==

=

The time variation of the five components is shown in the next figure.

The contribution of the first and second order terms is put in light in the next figure.

The second-order terms are smaller than the first-order ones, but in this case not completely

negligible. In fact, by differentiation and numerical solution, one gets the following values of

the maximum of the different components:

|F1+F2|max= 1231.42 N fort=0 (and t= 1.5 s); |F2+F3+F4|max= 268.63 N fort=0.375 s (and t= 1.125 s, 1.875 s, 2.625 s); (Ftot)max=1455.49 N fort=0.3 s.

The increase in the value of the maximum force considering the second-order terms is hence

of 224.07 N, i.e. the 18.2% of the maximum first order force; in addition, this maximum

produces with a delay equal to T/10.

For the purpose of the structural calculation, the bending moment at the clamped edge is perhaps more important. In this case, considering, as already done, the wave action as a

concentrated load applied at the SWL, we get .Nm9.14554maxmax

== otot lFM

h. To determine the minimum current velocity that produce flow-induced oscillations by

vortex shedding and the amplitude of these oscillations.

First of all we control the stability parameter .2

2D

mK es

= IfKs >1.8 in-line oscillations do

not occur, while ifKs>10 cross-flow oscillations do not occur.The effective mass per unit-length is given by

t

F1F2

F3F4

F5

F[N]

t

F1+ F2

F3+ F4+ F5

FtotF[N]


14/25

_______________________________________

- 14 -


,

)(

)()()(

)(

)(

0

2

0

222

0

2

0

2

++==

d

d l

dw

d

l

e

dss

lMdssmdssm

dss

dssmm

being the mode. If we take for the mode the polynomial approximation already used,

( )22 4lxax = , we get me=1526 kg/m. Hence, considering that the logarithmic decrement is= 2= 0.05, it isKs=1.4


15/25

_______________________________________

- 15 -


3. Wave slamming analysis.

The scope of this exercise is to calculate the dynamical response of a horizontal brace of a

jacket structure subjected to wave slamming. The brace can be considered as a beam clamped

at both the edges; it has a hollow circular section and it is empty. The wave can be described

by the linear wave theory and using the deep waters approximation.

The data are summarized below:

Beam length:L= 10 m; External diameter:D= 10"= 254 mm; Thickness: t=1/2"=12.7 mm; Young's modulus:E= 210 GPa; Steel density:s:=7850 kg/m3; Water density:w:=1015 kg/m3; Wave height:Hw=10 m; Wave period: Tw= 10 s; Damping ratio: = 0.03.

Preliminary calculations:

Cross area: ;m1063.94

])2([ 2322

=

=tDD

A

Mass per unit length: m=sA=75.57 kg/m;

Inertia moment: ;m1003.764

])2([ 4544

=

=tDD

J

Wave frequency: ;rad/s628.02

==w

wT

Wave number: ;m04.0 12

==g

k

Wave length: ;m13.1562 ==k

Lw

Wave steepness: ;128.0=L

H

Ratio to maximum wave steepness: ;897.0128.07)/(

/

max

==LH

LH

Wave slamming force per unit length: ;N/m4.159878

22 == HDf wws

Total wave slamming force: ;N159874== Lff wswstot


16/25

_______________________________________

- 16 -


Dynamical response calculation: the wave slamming is modelled as an impulsive distributed

force acting instantaneously on a system having only one degree of freedom (i.e., only the

first frequency is considered), determined by the mode shape. The response to such a force is

given by:

),()(),( thfxtxyws

=

where (x) is the mode shape and h(t) the impulse response function:

.1sin

1

1)( 2

2

= te

m

th t

The mode (x) and the frequency must hence be calculated. We use the Rayleigh approach,using as mode the static deformation of the beam under a unitary distributed load:

.2212

1)(

223

4

+=

xLLx

x

EJx

The deformation in the middle of the beam, under the wave slamming force considered as a

static load, is hence

.m028.0384

)()2

(4

====EJ

LfLxf wswss

The frequency is given by

,2

0

2

0

2

m

EJ

Ldxmy

dxyEJ

L

L

=

=

with

.45.22146

0

2

0

2

2 =

=

L

L

dxy

dxyL

Finally,

rad/s.2.99146

2==

m

EJ

L

The dynamic response can then be calculated:

).15.99sin(1033.1)()(

),( 976.24 tethxf

txy t

ws

==

This is also, for eachx, the dynamical magnification factor, i.e. the ratio between the dynamicand static response. This function is plotted below, and it is apparent that the dynamic effect is

very important but also rapidly vanishing. Actually, the fact that the wave period is much

higher than the natural period of the beam has as a result that the impulses given by two

successive wave slamming have not mutual influence. In other words, the dynamic response

is dominated by the response to only one slamming.


17/25

_______________________________________

- 17 -


4. Dynamics of a MINI-TLP.

The scope of this exercise is to analyze the dynamic behaviour of a MINI-TLP, i.e. of a TLP

having only one column. The scheme is that in the figure, and the data are summarized in the

table below. To a first approximation, used here, the tethers and the pontoons are considered

as rigid and the wave action on them is neglected. The tethers are linked to the pontoons andto the sea bed by pin joints; so, the system behaves articulated and has only one degree of

freedom (twisting is neglected here).

a. Wave force: the first thing to do is to calculate the wave force.

The frequency is =2/T= 1.01 Hz. The wave length can be calculated by the dispersionequation, written for deep waters:

.m602

m105.0 1-2

2 ===k

Lg

kgk

We have also kd= 52.35>1: the deep waters condition can actually be applied.

The ratio .2.025.060

15>==

L

DThe Keulegan-Carpenter number is

.4616.1tanh

max


18/25

_______________________________________

- 18 -


,m57.87

tanh7

max ==L

kdL

H

so that 5.09.0/ max >=HH .

The consequence of these evaluations is that for the case in object the wave action should be

calculated in the non-linear diffraction range. As the cylinder is truncated, the non-linearGarrett solution gives the wave forces. We use the simplified solution given by Vannucci.

Before calculating the wave loads, we compute some parameters used in the formulae:

column radius: b=D/2= 7.5 m; radius parameter: =kb= 52.36; depth parameter: =kd=0.78; clearance parameter: =kh=49.22; ratio /=0.94; wave amplitude: a=H/2=3.85 m; wave parameter: =ka=0.40; wave amplitude buoyancy:B= g a b =6.854106 N.

With the above parameters, by the following diagrams we can calculate the reduction

coefficientsx andz.

Considering the above diagrams and conservatively, we can take for both the coefficients the

value 1.

Finally, as


19/25

_______________________________________

- 19 -


,2sinh

2sinh2sinh22

2sinh

2sinh2sinh22)1()(

2

20

32

+++

++

+=

=

mmBX

mmD

with

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ),,,,,

,,,,,

,))((

,,2

12

122

11

mmmm

mmmm

mmmm

mmmmm

YYN

JJM

NMNM

MNNM

=

=

++

=

++

++

where

( ) ;I

I

1

sinh2sinh+2

8,,2

2

2212

2

+

=

=

n

n

n

n

m

m

nm

three terms are sufficient for the convergence of the above series: XD=3.33105 N;

a vertical drift:( )

+

+=

1=m

22

022

2

242,

1

2sinh

sinh8drr

NM

BZ mm

b

mm

D BA

with

,

1

2

12

22

1

+

+

=

=

h

bnI

h

rnI

n

n

hb

r

b

m

m

m

n

m

m

A

;

1

2

12

22

+

+

=

=h

bn

I

h

rnI

n

n

b

r

r

m

m

m

n

m

m

B

with six terms in the series above, we obtainZD= 1444.3 N;

a second order time depending horizontal force, approximated by the following term,due to only the first order velocity potential:

,2cos2sinh

222sinh2sinh

2sinh

222sinh2sinh32

22

22

21

21322

tQP

QPB

X

++

+

++

+=

with


20/25

_______________________________________

- 20 -


( ) ( )

( ) ( )

( )( );;;

;)1(1;)1(1

;1;1

21

21

22

1111

0

2

0

2

0

1

0

1

++

++++

=

=

=

=

++=

=+=

+=+=

==

mmmmm

mmmmmmmmmm

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

NMNMR

NNMMQMNNMP

R

QmmQ

R

PmmP

R

QQ

R

PP

with three terms in the series, we obtainX2=2.82106 cos 2tN;

a second order time depending vertical force, approximated by the following term, dueto only the first order velocity potential:

,2cos3

2sinh

12cosh

1

4 222

22

22

2322tVU

BZ

+

+

+=

with

( ) ( )

( ) ;2

;1;1

22222

00

mmmmmmmmm

m

m

m

m

m

m

m

m

NM; WNM; VNMU

W

VmV

W

UmU

+===

==

=

=

with three terms in the series, we obtainZ2=12.2 cos 2tN.

Finally, the results show that:

Z/X=0.045; XD/X=0.028; ZD/X=1.22104; X2/X=0.239; Z2/X=2.02108.

Hence, we will consider in the following only the forces X, Z, XD and X2. In addition, and

conservatively, we assume that Xand X2 have the same phase, while Zhas a phase of; infact, this is the condition which maximizes the displacements of the platform (when the

horizontal force is the highest, the vertical force is maximum downward). Hence, the waveactions are:

alongx:;Ncos10180.1 7 tX =

X2=2.82106 cos 2tN;

XD=3.33105 N;

alongz:.N)cos(1033.5

5

= tZ

h

M

A()

x

z

o

Z(t)

X tX2(t)

XD


21/25

_______________________________________

- 21 -


The mechanical model is hence that in the figure above, acted upon by the above forces; the

only degree of freedom is the angle ; the mass M is the total mass of the system plus theadded mass; considering the ratiop/D=2, the added mass coefficient is CA= 0.72 (obtained by

interpolation of the data in the figure below, from Sarpkaya & Isaacson).

Hence, the dynamical mass is

.kg10912.44

62

=+= TLPA MpD

CM

Conservatively, we neglect in the dynamical analysis the radiation damping, caused by thescattering of waves from the moving hull.

The stiffness of the system is given by the restoring force A, the buoyancy, which depends

upon :

[ ] .N)cos95.01(107864.8)cos1(4

)( 82

=+= gMhpgD

A TLP

The minimum restoring force is at rest, =0, where it is about 3 times the highest horizontal

force. The maximum possible angle is that for which the immersion of the hull leaves out ofwater the height ag+a=4.35 m. Hence, the maximum allowed additional immersion of the hull

is c(ag+a)=5.65 m. This value is obtained for an angle adm such that

.893,8988.01cos ==++

+= admh

cagap

The variation ofA with is shown in the figure.

We can now write the motion equation of the system:

.0=

LL

&t

The Lagrangian L is

,UK+=L

with:

K: kinetic energy of the system; we make the simplifying assumptions that the tethershave a negligible inertia with respect to that of the TLP and that the mass of the TLP is

concentrated at the top of the tethers (this last assumption is justified by the actual

lengths):

;2

1 2&

TLPIK=

[]

A()[N]


22/25

_______________________________________

- 22 -


ITLPis the inertial moment of the TLP with respect to the point o:

;2

1 222 &hMKhMI TLPTLPTLP ==

U: total potential of the forces acting upon the system:,

2 ZXXXAUUUUUU

D++++=

with

UA: potential of the restoring forceA(); ifp() is the draught of the hull andz() its vertical position, we have:

[ ] ;)cos1(2

1cos)(

2

1)(

2222 +== hpbghgMpbgzgMU TLPTLPA

UX: potential of the horizontal wave forceX; ifx() is the horizontal positionof the hull, we have:

;cossinsin)( thXhXxXUX ===

UX2: potential of the second-order horizontal wave forceX2:;2cossinsin)( 2222

thXhXxXUX ===

UXD: potential of the horizontal drift:;sin)( hXxXU DDXD

==

UZ: potential of the vertical wave force:).cos(coscos)( === thZZhzZUZ

To remark that the derivatives of the potentials with respect to give the moments of the

respective forces about the point o.

The equation of motion is hence:

[ ]

.0)cos(sincos2coscos

coscossin)cos1(sin

2

22

=+

++

thZhXthX

thXhhpgbghMhM

D

TLP&&

The equation is non-linear, and its numerical solution gives the following variation ofwith

the time:

t[s]

[rad]


23/25

_______________________________________

- 23 -


The maximum value of the angle is about 0.015 rad, i.e. 0.859 (which is


24/25

_______________________________________

- 24 -


having the wave frequency and the double of the wave frequency, superposed to a constantforce which gives the mean position (the last term). Introducing the numerical values gives:

.107.8)027.2cos(109.2)013.1cos(1006.5)( 343 += ttt lin

The variation of(t)lin is plotted in blue in the following figure, superposed to the non-linearresponse (in red).

The magnification factors for the two time depending forces are

.004.04

,016.022

2

22

2

2=

==

=

TLP

TLPX

TLP

TLPX QQ

Actually, the dynamical behaviour of the TLP diminishes the response to the exciting forces

with respect to the case where they act statically, essentially due to the fact that the wave

forces have a frequency much higher than the natural frequency of the TLP.

Finally, the linearized mean position is given by the third term in the above equation, which

corresponds to an angle of 0.498, the 13.7% more than the value found by the non-linearequilibrium equation. This fact is also visible on the diagrams of the figure above.

,lin [rad]

t[s]


25/25

_______________________________________P. Vannucci Examples of offshore structures calculation

References

O. C. Zienkiewicz, R. W. Lewis & K. G. Stagg (Eds.): Numerical methods in offshoreengineering. J. Wiley, 1978.

M. G. Hallam, N. J. Heaf & L. R. Wotton (Eds.): Dynamics of marine structures.Report UR8 CIRIA, 1978.

T. Sarpkaya & Isaacson: Mechanics of wave forces on offshore structures. VanNostrand, 1981.

B. Mc Clelland & M. D. Reifel (Eds.): Planning and design of fixed offshoreplatforms. Van Nostrand, 1986.

P. Vannucci: Cours de Mcanique pour l'Industrie du Ptrole Offshore Engineering.2006.

Some of the figures and subjects in this document have been partially taken from the above

authors; they are sincerely acknowledged.

__________________P.Vannucci

January 6, 2007

Date post:	07-Apr-2018
Category:	Documents
Upload:	wlueji
View:	238 times
Download:	1 times

Exemple of Offshore Structures Calculation

Documents