GLOBAL RISER ANALYSIS METHODOLOGY

Wave scatter diagram analysis

The global riser analysis (from a fatigue assessment standpoint) comprise of the following activities

Collation of environment (global) data for the location Analysis of these environmental load – the method of analysis will depend on the level of

detail and conservatism desired Estimation of the deformation in the flexible pipe due to these environmental loads.

The first two steps above are mostly statistical. Fig.1 below shows a typical time trace of the water surface elevation of a particular location (of interest), from this time trace the parameters needed to characterize the wave are extracted using time domain e.g. significant wave height (one-third of the highest peaks), mean zero up-crossing period (average of the positive up-crossings) and root mean square period etc. Also, frequency domain (spectral) analysis could be used to obtain parameters which are related to the statistical parameters of a wave e.g. the mean spectra moment and the significant wave height are related as ***WD1

H s=4√mo 3-1

while the peak period is given as

Tm=√ mo 2

mo1 ** 3-2

mo = is the mean spectral moment

mo 1∧mo2 are the first and second spectral moment gotten from spectral analysis.

Armed with this information, it is possible to generate a wave scatter diagram of individual wave height and period using a probability distribution (e.g. Longuet-Higgins distribution) for a reference period of say one year.

In deterministic fatigue assessment, it is often convenient to split the generated scatter diagram of individual wave height and period into manageable bins (blocks) which is also associated with representative individual height and period. There are various different ways of achieving this –

The representative wave height is calculated from the average of the wave height of all bins making up the new bin e.g. assuming the scatter diagram of fig.2-4 above is to be blocked into five blocks, the first sea state will have a height

H1=

H1 ,low+H1 , high

2+H2 ,low+H2 , high

2+ .. .

Hn , low+H n,high

2n , 3-3

where n is the number of sea state blocked into one regular wave i.e.

H1=0.5+1 .5

2=1m

.

The representative period (a.k.a most probable period) is given as [Y. Bai].

T 1=0. 7+H10 . 4

3-4

This is often used when the wave classes are high in number. Another way suggested by [Sheehan] was to choose the representative height as the highest within the classes and the representative period as the weighted mean period of all classes. In this method, the number of wave classes that are blocked into one bin is chosen such that no single (new) bin has wave cycles greater the twice the number of cycles per block (bin) i.e.

nsingleblock≤2∗∑ ni

number of newblock 3-5

where possible. In this way blocks with higher wave cycles are smaller than those with less number of cycles.

To calculate the number of wave cycle within the new bin (block), the number of individual wave cycles calculated with Longuet-Higgins method is not used rather the stress range probability distribution is used to calculate the probability of the wave stresses being within that stress range. The stress range (encounter) probability is often assumed to be Weibull distributed [Sintef & ship design]. The Weibull two parameter stress range distribution is given as

P (σ lower ,i )=exp (−σ lower ,i

σ o)h

σ o=Δσ i

[ Ιn (no ) ]h 3-6

and

no=1

10−4

∆ σ i=stress rangedue ¿ seastate i ,

σ o=reference stress range **(look at again)

P (σ lower ,i )= is the probability of exceeding the lower stress in the range

σ lower ,i = lower stress value in the stress range due to the i

th sea state

h = the Wiebull shape parameter with values ranging *** (find the range)

no is the inverse of the reference probability level (taken as 104)

the probability of any stress range (cycle) is given by

PΔσi=P(σ lower ,i )−P (σupper ,i ) 3-7

this is a measure of the probability of the stress being within the lower and upper bound of the stress range [fatigue guide]. The number of stress cycles is simply

ni=PΔσ

i∗108

assuming a twenty years 3-8

(20∗365∗24∗3600=6 . 3×108 sec onds )

design life (often taken as a standard). The stress range is calculated from normal riser analysis (global and local) for each wave condition in new blocks.

Global deformation due to environmental loads

Now that the wave environment has been blocked into representative wave height and period, the next step is to calculate the pipes deformation due to this wave heights and periods. For design purposes, static analysis will be used to determine the configuration of the flexible pipe (riser) hence static analysis not will be necessary in fatigue assessment. But for the records, static analysis entails the solution of this differential equilibrium equation which results from fig.3-4 below.

−EId4 ydx 4

+ ddx (T dy

dx )+ f ( x )=0 3-9a

for equilibrium in the x-direction

dTdx

=w 3-9b

for equilibrium in the y-direction

The solution to these equations subject to the boundary conditions

x=0 , y=0⇒T=T0 will give the famous catenary cable equation

Y=T 0

w (cosh (wXT 0

)−1) 3-9c

This equation describes the catenary curve, hence the coordinates of the curve at any water depth (X, Y) are known from here. Also, this static analysis gives the tension at any location along the riser and the distance along the catenary curve once the water depth or the horizontal distance between the touchdown point and the vessel is known.

T k=T0+wY k 3-9d

EI is the bending stiffness of the riser

w is the riser submerged weight

T k tension in each element

SK=

T 0

wsinh (wxK

T 0) 3-9e

From the above equations, if the water depth, riser length, submerged weight are known, the

touch down tension T 0 can be calculated iteratively.

The angular inclination of any element to the global axis (X, Y) is very vital to the estimation of the nodal parameters relative the global axis.

θK=cos−1 (T 0

T k) or 3-10a

ΔX k=X k1−Xk 2 , ΔY k=Y k1−Y k2

lk=√ΔX k2+ΔY k

2 ,cos θk=ΔY k

lk,sinθk=

ΔXk

lk

3-10b

The angle are used to compute the transformation matrix between the fixed seabed and each element, the essence being to convert every force from the local coordinate system (x, y) to the global system of coordinate (X, Y). The transformation matrix for each node therefore will be (show the beam element)

T ( x , y→X ,Y )=(sinθk cosθk

−cosθk sin θk) 3-11a

and for the entire element we have

T ( x , y→X ,Y )k1,2=(T ( x , y→ X ,Y )1 02×2

02×2 T ( x , y→X ,Y )2 ) 3-12b

Note that the transformation matrix is same for both nodes as the angle does not vary within the element and the zeros here represent 2X2 null matrices.

The Mass Matrices

As noted above, the finite beam element could have its mass lumped at the nodes or evenly distributed using an interpolation polynomial.

Using the interpolation polynomial for a two dimensional system given in [A.M Roustad]

N=[1− xl,

xl ]

3-13

the mass of each element is

m|element=∫V

ρ⋅NT⋅N⋅dV≡¿ ρA∫0

l

NT⋅N⋅dx ¿

3-14a

⇒ [m ]|element= ρA∫0

l [1− xl

xl

]⋅[1− xl,

xl ]⋅dx⇒ ρ Al

6 [2 11 2 ]

3-14b

The structural mass of each element is computer by expanding the above taking into consideration the fact that the mass will always be unidirectional at each node,

3-15a

A∧A int = cross-sectional and internal area of the riser respectively

msk∧mint, fluid =structural mass of the riser and mass of internal fluid respectively

ρ s∧ ρfluid = density of steel and internal fluid respectively

l = length of each element

The internal fluid mass matrix will be similar with the structural mass matrix; the only difference will be in the cross-sectional area and density

3-15b

msk=ρ sAl

6 (2 0 1 00 2 0 11 0 2 00 1 0 2

)

mint, fluidk=ρfluid A int lk

6 (2 0 1 00 2 0 11 0 2 00 1 0 2

)

The matrix for the added mass is related to the external fluid and external area since the added mass effect is due to the interaction between the riser and the external fluid giving the riser additional mass, the added mass matrix is given as

mak=ρext , fluid Aext lk

6 (2 0 1 00 0 0 01 0 2 00 0 0 0

) 3-15c

[ Μ ]k= [msk ]+ [mint, fluidk ]+[mak ] 3-15d

mak , [Μ ]k =added mass and total mass matrix for each element

ρext , fluid = density of external fluid

Aext= external area

superscriptT = is the matrix transpose

V = volume of the element

This individual element mass matrix will first be transformed into the global reference frame and

then assembled using matrix concatenation method thus [A.M. Roustad]

[ Μ ]kXY =Τ ( x , y→X ,Y )1,2 [ Μ ]kΤ ( x , y→X ,Y )T

⇒ [Μ ]kXY=[Μ k

11 Μ k12

Μ k21 Μ k

22 ] 3-15e

Hence the global mass matrix on concatenation will be

[ Μ ]=[Μ 1

11 Μ 112

Μ 121 Μ 1

22+Μ 211 Μ 2

12

Μ 221 Μ 2

22+Μ 311 Μ 3

12

Μ 321 Μ 3

22+Μ 411

..

.

Μ n−122 +Μ n

11 Μ n12

Μ n21 Μ n

22

] 3-15f

Stiffness matrix

The stiffness matrices are derived by noting that the variation of the total potential energy at a boundary is equal to the work of external force on the virtual displacement. Mathematically we have

∫V

β⋅δu⋅dV +∫S

Ρ⋅du⋅dS=δU

3-16a

∫V

β⋅δu⋅dV= total body force in each element

∫S

Ρ⋅du⋅dS= total external force acting on the boundary of the element

δU = the change in the potential energy (virtual work) of the system

For a linear system δU=∫

V

τ⋅¿δE⋅δV ¿

**** 3-16b∴∫V

β⋅δu⋅dV +∫S

Ρ⋅du⋅dS=∫V

τ⋅¿ δE⋅δV ¿

This equation can be expressed in the form [Pav’s lecture]

{ f }=[ K ] {a } 3-16c

Where { f }∧ {a }= force and displacement vectors and

[ K ] = the global stiffness matrix. In this form

[ K ]=∫V

[B ]T [ D ] [B ]dVand **

{ f }=∫V

[ N ]T {β }dV +∫S

[ N ]T {Ρ }dS

[B ] = First derivative of the shape function of each element

[ D ] = matrix of the modulus of elasticity

{Ρ } = external load on the riser

{β } = the body force per unit volume

i.e. by substituting for δE ,δu∧τ in the above equation**** and noting that

δE→δε= ddx

{u }

δu= [N ] {a }

τ=[ D ] {ε }=[ D ] [L ] {u }=[ D ] [L ] [N ] {a }

[B ]=[ L ] [ N ]

dV =Adx

Using the same interpolation function as in the mass matrix above we have

B= ddx

[ N ]= ddx [1− x

lxl ]=[ 1

l1l ]

Substituting into ** above, we get

K= EAl [ 1 −1

−1 1 ]To reflect the number of degree of freedom of each element, the elastic stiffness matrix is written as

K EK=

EA k

l k [0 0 0 00 1 0 −10 0 0 00 −1 0 1

] 3-17a

Also, the geometric stiffness matrix which relates deformation in the lateral direction is given as

KGK=

T k

lk [ 1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

] 3-17b

K=K EK+KGK

=EA k

lk [0 0 0 00 1 0 −10 0 1 00 −1 0 1

]+ T k

lk [1 0 −1 00 0 0 01 0 1 00 0 0 0

] 3-17c

This is the stiffness matrix of each element in the local frame; it must be transformed to the

global reference frame before it can be concatenated to the global stiffness matrix. The procedure

is same as that for the mass matrix; the difference is that the stiffness matrix above is used

instead of the mass matrix.

DAMPING MATRIX

The structural acceleration and velocity of the riser pipe as a result of wave and current forces are continually attenuated (damped) as the motion progresses. Effectively, damping has to do with the dissipation of the energy of a vibrating system giving rise to decay in the amplitude of the vibration circle. The effect of damping is factored into the dynamic analysis of a system using (damping) ratios which are measure of the energy lost per cycle to the total energy of the un-damped system. In general, modelling the phenomenon of damping is quite involved but the current practice follows the Raleigh method which assumes that damping is proportional to the mass and stiffness of the entire system. Mathematically, the damping matrix is expressed as

[C ]=υ1 [ K ]+υ2 [M ] 3-18

The damping coefficients υ1∧υ2 are often expressed in terms of the damping ratio and the first and second mode natural frequency (from eigenvalue analysis) [Patel and Sarohia] .

υ1=2 (ξ1ωn2−ξ2ωn1 )

ωn 2

ωn 1

−ωn1

ωn2

3-19a

υ2=2 (ξ1 /ωn 2−ξ2 /ωn 1)

ωn2

ωn1

−ωn1

ωn2

3-19b

The damping ratio(ξ ) is often taken as 5% for the first two modes in analysis.

ξ1∧ξ2 = damping ratios

ωn1∧ωn2 = first and second mode natural frequencies

WAVE AND CURRENT (GLOBAL) LOADS ON RISER PIPE

At the moment, wave and current forces can only be quantified using the famous Morrison’s equation. The use of this equation means that the transformation matrix must be used to relate conditions at the node (local condition) to the global axis system since the current and wave kinematic properties are defined with respect to the global (fixed) coordinate system. Also, Morison’s equation has a projected area which implies that all wave and current loads act on the projected area normal to the direction of the fluid motion.

The Morison’s force is expressed as the vector sum of the drag and inertia forces at each node

FM k=Fk

D+FkI

3-20

FMk = total hydrodynamic loads calculated using Morison’s equation

FkI∧Fk

D= inertial and drag forces in each element respectively

Inertia forces at each node

FkI=

Cm ρw πD2 lk

4 [ v̇ k− q̈k ] 3-21

Cm=inertia coefficient

ρw= density of water

D =external diameter of riser

kw = wave number ( 2πLw

)Lw =wavelength

dn = depth of each node from the mean water level

Y = total water depth (negative downwards)

ω = wave frequency

H = wave height

q̈k =structural acceleration of the riser

σD = wave dispersion parameter (gkw tanh ( kY ) )

Note, it is not possible to evaluate v̇kunless it is transformed as shown below.

v̇k= [v̇kX , v̇k

Y ] , Wave acceleration in the global frame

v̇kX=ω2 H

2

cosh k (Y +dn )sinh kY

sin (kx−σD t ) 3-22a

v̇kY=−ω2 H

2

sinh k (Y +dn)sinh kY

cos (kx−σD t ) 3-22b

These accelerations are then transformed to the local frame ( x , y ) by multiplying the global water particles acceleration vector by the transformation matrix thus

[ v̇kx

v̇ky ]=[T ( x , y→X ,Y )1,2] [ v̇k

X

v̇kY ]⇒[ v̇k1

x

v̇k1y

v̇ k2x

v̇k 2y ]=[T ( x , y→X ,Y )1 0

0 T ( x , y→X ,Y ) 2] [ v̇k 1

X

v̇k 1Y

v̇k 2X

v̇k2Y ]

3-22c

Hence we substitute each of the acceleration component in ^^ into the inertial force formula stated above, to calculate the inertia force at each node in all degrees of freedom.

[Fk 1Ix

Fk 1Iy

Fk 2Ix

Fk 2Iy ]=Cm ρw πD

2 lk4 ([ v̇k 1

x

v̇k 1y

v̇k 2x

v̇k 2y ]−[ q̈k1

x

q̈k1y

q̈k2x

q̈k 2y ])

3-22d

Drag force at each node

FD=12ρwC DDvk

r|vkr|

3-23a

CD = drag coefficient

q̇k1x

= structural velocity of the riser

vkr|vk

r|= product of relative velocity

vkr= (vw,k+v c , k−q̇k ) - Superposition of the current, wave and structural velocities.

Wave velocity

vw ,k=[vw ,kX , vw ,k

Y ]

[vw ,kx

vw ,ky ]=[T ( x , y→X ,Y )1,2 ][vw,k

X

vw ,kY ]=[vw ,k1

x

vw ,k1y

vw ,k2x

vw ,k2y ]=[T ( x , y→X ,Y )1 0

0 T ( x , y→X ,Y )2] [ v̇k 1X

v̇k 1Y

v̇k 2X

v̇k 2Y ]

3-23b

vw ,kX =ωH

2

cosh k (Y +dn )sinh kY

sin (kx−σD t ) 3-23c

vw ,kY =ωH

2

sinh k (Y +dn)sinh kY

cos (kx−σD t ) 3-23d

Current velocity

vc , k=[vc , kX vc , k

Y ]

[vc , kx

vc , ky ]=[vc , k1

x

vc , k1y

vc , k2x

vc , k2y ]=[T ( x , y→X ,Y )1,2 ][vc , k1

X

vc , k1Y

vc , k2X

vc , k2Y ]=[T ( x , y→X ,Y )1,2 ][vc , k1

X

0vc , k2X

0] 3-24

Current velocity is zero in the vertical direction.

Y = the water depth and dn is the nodal depth**** other variable should be defined.

Note vc , k is gotten from environmental data measurement and depends on depth hence vc , k 1X

is the

current velocity measured at the depth of the node under consideration. While q̇k∧q̈k are calculated from numerical integration, the method of calculating them will be shown in the subsequent section. Also, the wave and current are assumed to be in phase to enable addition. A

good approximation of the wave phase angle φ=(kx−σD t ) can be evaluated by assuming different positions of the water surface elevation within the wave cycle i.e. zero crossing, crest, and trough. As an example

ηw=H2

cos (kx−σD t ), at the zero crossing ηw=0⇒φ=0 .5 π

The relative velocity for each element can then be represented in the matrix form as

[vr , k 1x

vr , k 1y

vr , k 2x

vr , k 2y ]=[vw ,k 1

x

vw ,k 1y

vw ,k 2x

vw ,k 2y ]+[vc , k 1

x

vc , k 1y

vc , k 2x

vc , k 2y ]−[ q̇k1

x

q̇k1y

q̇k 2x

q̇k 2y ]

3-25

Using this relative velocity vector, the drag force is calculated at each node of an element using eqn. ** above. The drag and inertial forces are summed vectorially (i.e. taking the phase into account) at each node.

[Fk 1Dx

Fk 1Dy

Fk 2Dx

Fk 2Dy ]=1

2ρwCD D [vr , k 1

x

vr , k 1y

vr , k 2x

vr , k 2y ]⋅[|vr , k1

x |

|vr , k1y |

|vr , k 2x |

|vr , k2y |

] 3-26

[Fk 1x

Fk 1y

Fk 2x

Fk 2y ]=[Fk1

Ix

Fk1Iy

Fk2Ix

Fk 2Iy ]+[Fk 1

Dx

Fk 1Dy

Fk 2Dx

Fk 2Dy ]

3-27

This total force is in the local reference frame and depends on the angular orientation of the element, in order to calculate the total force at the nodes; these forces will have to be transformed back to the global reference frame where all forces at each node will have the same sense. Again, using the local to global transformation matrix stated above, the nodal forces relative to the global frame is

[Fk 1X

Fk 1Y

Fk 2X

Fk 2Y ]=[T ( x , y→X ,Y )1,2 ][Fk 1

x

Fk 1y

Fk 2x

Fk 2y ]

3-28

In this form, the forces at each node of the entire element can now be summed provided they are in same direction i.e. each node will have contributions from two elements and the resultant force will be given by

FxkX =Fx , ki

X +F x , k (i−1 )X

i=1,2,3 . ..n 3-29a

F ykY =F y , ki

Y +F y , k ( i−1 )Y

3-29b

Here we are considering decoupled situation and the load term due to vessel motion is ignored, if it were to be the other way round then the load component due to vessel motion will be added accordingly.

RISER DEFORMATION

The essence of the above mathematical developments is to get a way of calculating the deformations due to wave and current forces using a finite element method. Knowledge of the deformation will give information on the stress-strain condition of that location in the riser pipe. The deformation is gotten from the solution of the dynamic equation

[ M ] q̈+ [C ] q̇+ [ K ]q=F (t )

Where [ M ] , [C ] , [K ] are the mass, damping and stiffness matrices and q ,q̇ , q̈are the structural

displacement, velocity and acceleration respectively. While F (t ) is the Morison’s force above i.e. the global force vector.

[F11

X

F11Y

F12X

F12X

.

.

.Fn 1

Y

Fn 2X

Fn 2X

]=[Μ ][q̈11x

q̈11y

q̈12x

q̈12y

.

.

.q̈n1

y

q̈n2x

q̈n2y

]+ [C ] [q̇11x

q̇11y

q̇12x

q̇12y

.

.

.q̇n1

y

q̇n2x

q̇n 2y

]+ [K ][q11x

q11y

q12x

q12y

.

.

.qn1

y

qn 2x

qn2y

] 3-30

The solution to this set of equation (in time domain) is an iterative process solved using numerical integration method e.g. fourth order Runge Kutta method, Wilson-theta algorithm, Newmark Beta method etc to get the structural acceleration, velocity and displacement.

Observe that the left hand side of this equation is wholly dependent on the wave and current condition and will not change during the iteration process. However for each time step, the right hand side will continually be compared with the left hand side to see when it equals the left hand side. Also, the left hand side consist of the summation of vectors i.e. the product of a square matrix and a vector is a vector.

To solve this massive set of equations, the following steps are followed here – Start the solution of the dynamic equation using the initial static condition of top tension and displacement thus:

Find the tension at each node and possibly at the middle of each element from the static tension equation given above – this will enable the computation of the stiffness matrix of each element. Starting from the estimated static effective top tension, the tension at each node is computed as

T top=T o+wd Where d is the water depth.From here, the tension in each element is calculated recursively as

T k=T top−wk−1

T k−1=T top−(wk−1+wk−2 ) k=(1,2 ,. .. , n ) 3-31

wk=(wair+wint . fluid−wext . fluid ) lk

Assume some static position ( t ) qk=

( t ) q (X k , Y k ) for each element in the global frame (hence the nodal coordinates are known). In this assumed static position, structural

velocity and acceleration ( t ) q̇k=

(t ) q̇ ( Xk ,Y k )∧(t ) q̈k=(t ) q̈ ( Xk ,Y k ) are zero. Using any of the

numerical integration methods stated above (Wilson-theta method used here) we get the structural acceleration velocity and displacement for the next time step as

( t+θΔt ) q=( t ) q+Δq( t+θΔt ) q̇=a1 Δq−a3 . ( t ) q̇−a4

( t ) q̈ 3-31( t+θΔt ) q̈=a0 Δq−a2 . (t ) q̇−a4

( t ) q̈

From these Wilson-theta integration expressions, knowing the initial displacement ( t ) qk as

assumed, we estimate the change in displacement when the wave and current loads are applied from the dynamic equation above ** this change in displacement is then substituted into the Wilson-theta velocity and acceleration equation for the next time step. This iteration process is continued until the convergence is achieved.

Also, observe that for each new displacement calculated, the nodal coordinate changes hence element length. Therefore, for each time step, the stiffness, mass and damping matrices will change; this change must be captured by continuously updating the matrices and the tension of each element since they both depend on the element length. the new tension is given by [Rumbod Ghadimi] as( t ) T k=

( t−1 ) T k+ΔT k

ΔT k≃EAt l k

2 [ ( t ) ( Xk 1−Xk 2) ( Δxk 1−Δxk 2)+ ( t ) (Y k 1−Y k2 ) (Δyk1−Δyk2 )] 3-32

( t−1 ) Tk =the static tension in the pipe initially. The new length will then be given by [A.M. Roustad]

( t+1 ) lk=( t ) l k+

( t ) T k( t ) lk

EA 3-33This new length is used to update the matrices at each time step. The change in global

and local displacement vector is gotten from( t+θΔt ) q=( t ) q+Δq .

Note also that the transformation matrix will be affected during the iteration.