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Subsea lifting operationsStavanger November 27-28 2007

Peter Chr. SandvikMARINTEK

Theory related to subsea lifting operations

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Content

Introduction, main critera for safe liftingLifting dynamics, simple equation of motion

Static and dynamic forces, wave forces on small objects Mass, stiffness and dampingResponse (motion) calculation, resonance

Large structures, 6 degrees of freedomWave forces on long structures (e.g. pipes)RAO calculation for ships, inherent limitations

Snatch and impact loadsStability

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The installation often gives the highest life-time forces on subsea equipment (1)

General

Templates, trawl protection

Suction anchor

Snatch at lift-off or after slackImpact after uncontrolled pendulous motionLocal loads from wave impact

Wave forces in the splash zone

Wave forces in the splash zoneSoil penetration forces

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The installation often gives the highest life-time forces on subsea equipment (2)

Jackups

ROV

Tools (ROT)

Spool pieces

Side forces at landing on seabed structure or the sea bed

Impact during launch and recoveryImpact at entry into the TMS

Landing impact

Forces during lift in airWave forces at /near the surface

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The installation often gives the highest life-time forces on subsea equipment (3)

Steel pipe

Cables

Flexible pipes

Bending stresses over stinger or at the sea bed, and during tie-in

"Kink" at pay-out after landing (due to rotation)

Curvature just above the sea bed

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Lift in general

Is the structure designed for the loads occurring during lifting and deployment?

Hydrodynamic forces

Limited lifting height may give large compressive forces from the slings

Measures:

Lifting frame

Spreader beam

Reinforcement(compression bar)

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Operation criterion:Ensure safe operation

Avoid:Excessive pendulum motion in air

Slack wire (when not intended)

Overload (in any lifting equipment)

Too hard landing

Do:Ensure acceptable stability

Have ability to handle unexpected changes

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Simple equation of motion

)()()()( tFxkxcxM =++ ωωω &&&

)()()()( 0 txkxkxcxM =++ ωωω &&&

Force excitation

Motion excitationF(t)

x0(t)

Force excitation Motion excitation

x

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Forces on the lifted object (Newtons, or kN)

Static forcesWeight (in air)BuoyancySubmerged weight

Dynamic forcesDampingInertia, moving objectInertia, wave force

Slamming force

cs = slamming coefficientx = body motionζ = wave particle motion

rrrd vvBvBF 21 +=

( ) ( ) xcVmxmmF aa &&&& ρ+=+=

( ) ( )ζρζρ &&&&aa cVmVF +=+= 1

2221

ra

rss vdhdcVvAcF ρρ ==

W = mgB = ρgVWs = mg - ρgV

Uncertaintyduring launchand recovery

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H-frame being taken on boardDrainage

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Wreck recovery

Unknown weight,weight distributionand stability

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Wave forces in the splash zoneExample: Template (1- 2)

1

218 x 18 x 7 m, 180 tonnes

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Wave forces in the splash zoneExample: Template (3- 8)

Large dynamic forces (± 150 T)

3 4 5

6 7 8

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Wire tension when lowering a body throughthe splash zoneExample

-15.00

-10.00

-5.00

0.00

5.00

10.00

15.00

0 500 1000 1500 2000 2500

Tension (kN)

Vert

ical

pos

ition

(m)

} Splash zone dynamics

Reducing

Weight in airWeight in water

!

!

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Wave kinematics (1)

Profile of regular waves propagating in x direction

Wave number (deep water)

Wave length

Propagation speed

Max. wave slope

kx)t( = −ωζζ sin0

gk

22 ωλπ==

256.12 Tk

==πλ

TT

vw 56.1==λ

0max

ζζ kdxd

=⎭⎬⎫

⎩⎨⎧

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Wave kinematics (2)

Wave particle velocity and acceleration

( ) ( )( ) ( )xktexkte

xktexktezk

zzk

x

zkz

zkx

−−=−=−=−=ωζωζωζωζ

ωζωζωζωζsincos

cossin0

20

200

&&&&

&&

Reduction factor with depth⎟⎠⎞

⎜⎝⎛

== λπ z

kz eeR2

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Wave kinematicsReduction with depth

Reduction of wave kinematics with depth

-100-90-80-70-60-50-40-30-20-10

0

0 0.2 0.4 0.6 0.8 1Depth reduction

Z (m

)

T = 4sT = 6sT = 10sT = 14sT = 20s

⎟⎠⎞

⎜⎝⎛

== λπ z

kz eeR2

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Wave kinematics in the VMO-rules

Parameter General expression(deep water)

Expression (VMO) Based on λ / Hs = 20

Extreme wave amplitude (m)

~Hs Hs

Wave particle velocity (m/s)Wave particle acceleration (m/s2)

3.1

Wave number (1/m)

Reduction with depth (-)

Hs⋅1.3

0.32 / Hs

ω2*Hs

k = ω2/g

)exp( kd− )/32.0exp( Hsd−

ω*Hs = 2π/T*Hs

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Wave force on a long member(heave force and pitch moment)

Wave force on an element dx

Total force, wave in worst position

dxcAf a ζωρ 2)1( +=

lengthwaveLLF

F == λλπ

πλ sin

0

02

0 )1( ζωρ acLAF +=

Harmonic wave:amplitude ζ

frequency ω

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3L / Wavelength

F / F

0,

M /

F0 L

ForceMoment

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Shielding-effect from the crane vesselExample of analysis results

1500

1750

2000

2250

2500

2750

3000

6 7 8 9 10 11 12Tp (s)

Max

liftw

ire fo

rce

(kN

)

Wavedir. 180 deg.Wavedir. 165 deg.Wavedir. 150 deg.

x

y21

3

45

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The mass: Structure mass (in kg or tonnes)

⎥⎦

⎤⎢⎣

⎡=

Im

M0

0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

mm

m

000000

m

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

II-I-

I-II-

I-I-I

= Iczz

cyz

cxz

cyz

cyy

cxy

cxz

cxy

cxx

c

Coupling terms = 0if symmetry

and origo in COG

Coupling terms = 0if origo in COG

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Hydrodynamic (added) mass, ma6 values - (for motion in 3 directions and rotation about 3 axis)

Plate

BoxSuction anchor

Added mass coefficient: Ca = ma / ρVρ = water density

V = reference volume

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Added mass - simple structures - 1

ca Vm ρα=

b4aV

2c

π=

a

b

Cylinder volume:

Geometry Formula b/a α

1.0 0.5791.2 0.6301.25 0.6421.33 0.6601.5 0.6912.0 0.7572.5 0.8013.0 0.8304.0 0.8715.0 0.8978.0 0.93410.0 0.947

Rectan-gular

a = shortest edge

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Geometry Formula b/a α β0 - 1.00

0.1 5.139 1.130.3 2.016 1.33

0.50 1.310 1.440.75 0.916 1.511.00 0.705 1.551.25 0.575 1.581.60 0.458 1.612.00 0.373 1.642.40 0.316 1.672.80 0.274 1.693.60 0.217 1.72

Rectangular block with rectangular base

α and Vc from rectangular plate, (1), Table 1

β from (1), this table

Rectangularblock withquadraticbase

a = base edge

Added mass - simple structures - 2

p

aVVm

ρβρα

==

4a579.0V

3p

π⋅=

ca Vm ρβα=

b4aV

2c

π=

aa

b

V = a2 b

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Added mass - simple structures - 3

p

saVVm

ρβρα

==

6a637.0V

a6

V

3p

3s

π

π

⋅=

=

ca Vm ρα=

b4aV

2c

π=

a

b

a

b

Geometry Formula b/a α β

0.8 to2,4 1.0 π/2 =

1.57

Same as for rectangular plate

Circular cylinder

Exclusive water inside the object.

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Added mass of ventilated structures

Added mass

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0 0.5 1 1.5 2

z(1-p)/(2Dp^2)

a/a0 Hatch 20, p=0.15

Hatch 18, p=0.25Roof #1, p=0.267Roof #2, p=0.47Roof #3, p=0.375

p = perforation ratio = open area / total area

Example

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Hydrodynamic mass for suction anchorsNumerical assessment compared to test results

D

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1 10 100 1000 10000

Amplitude / D * (1-p)/p^2

a / a

0

Calculated p=1%Measured p=1%Calculated p=3%Measured p=3%Calculated p=11%Measured p=11%

p = perforation ratio

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StiffnessGeneral

Axial wire stiffnessE = modulus of elasticityA = section areaL = length

Transverse stiffness

Hydrostatic stiffness

Rotation stiffness(spring k, distance a from

rotation center)Parallel springs

Springs in series

dxFkx∂

=

∑∑ =∆

= ii

tot kF

k

∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛==∆

itotitot kkkF

kF 11

LEAk =

LFF

kL

FFy

yy

yy =

∆=⇒

∆⋅=

WPAgk ρ=

2akK ⋅=

[Compression: k = 0]

[AWP = waterplane area]

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Guidewires - transverse force and stiffness

h

XFx

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Damping

Coulomb dampingFriction, hysteresis loss, ....

Linear dampingWave potential damping, material damping, oscillation damping in wind, ....

Quadratic damping

Hydrodynamic dragMorison's formulaNotice:cd for oscillating objects is larger

than cd for steady flow !

vvcF ⋅= 00

vcF ⋅= 11

vvcF ⋅= 22

rrdd vvAcF ρ21=

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Hydrodynamic damping of oscillations

The damping, expressed by the quadratic Morison formula, will have amplitude-dependent drag coefficientExample: cube

0.5

1

1.5

2

2.5

0 5 10 15 20

KC = 2 pi X / D

Dra

g co

effic

ient

, Cd

Only possible to calculate damping in harmonic oscillation

Solution:Adding a linear dragterm makes it possible to use constant coefficients

Cd for oscillating objects larger than in steady flow(factor ~2)

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Oscillation damping in wind

( )( )22

21

221

2 xvxvAcxvAcF

wwda

wdaw

&&

&

++=

+=

ρρ

Constant wind forceLinear wind damping

(Small)

x = body oscillation (inline with wind)vw = wind speed

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Damping linearization

Objective : To calculate dynamic behaviour by use of linear equations of motion. Method: Find linear damping that dissipates the same energy as the non-linear damping during one motion cycle

Linearized friction:

Linearized drag:

Total linearized damping:

0

00 x

c 4 = c*1 ωπ

c x 38 = c*

1 22 ωπ

vcccF e )( 1*12

*101 ++=

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Response of simple oscillating system

mc

mk2c =

kmT

mk

T =

4 + - 1

1 = F

k X = )H(

0

2

0

2

0

2 20

0

00

2

22

ωη

ππω

ηωω

ωω

ω

=

==

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

F(t)=x k + x c + x &&&m

ηωω 21

00

==X

XResponse at resonance:

F

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Response curve (RAO) of simple oscillating system

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Resonance periods - examples

Vertical oscillation in air:

∆≈∆=

∆

== 22220 gmgm

EAmLT πππ

∆ = elongation of wire due to weight mg

Vertical oscillation in water (long wire, wire mass mwincluded):

( )L

EAmam

T w31

330 2

++= π

LLgmg

mLF

mLT 22220 ≈===πππ

Pendulum oscillation in air:

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Snatch loads in the liftwire

MkVFF relstatic+=max

The snatch load is highest for:- Short lift wire (high stiffness)- Large hydrodynamic mass- Large relative velocity between

crane and load at snatch instant

The probably largest relative velocity at snatching should be used

Impact between objects or at landing can be assessed similarly.

k = wire stiffnessM = mass (incl. added mass)Vrel = relative velocity

Assumed:Short duration of impulsive load compared to motionperiod.

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Snatch-load: Lift-off from the sea bed

Assume a winch that can shift from constant tensionmode to lift mode when the crane passes its lowerpoint (i.e. when the winch stops taking in wire).

The lower end tension (F0) is below the object weight (W) at shift to lifting mode.

The lifting starts (the load moves) when F0 reaches W

The peak snatch load is reduced by selecting F0 closelybelow W

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Snatch load at lift-off from sea bedExample: Favourable lift-off time

1.00

1.251.50

1.75

2.00

2.252.50

2.75

3.00

100 200 300 400 500Water depth (m)

Fmax

/ W F0 / W = 0

F0 / W = 1 VMO formula

Module mass 12.6 tTotal dyn. mass 15.6 tSubmerged weight 107 kNWire diameter 38 mmCrane amplitude2.5 m

period 8.8 sspeed 1.8 m/s

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Calculation of RAO data for a ship

) F(=x k + x ))c(( + x ))(( 1 ωωω &&& ++ cmm a

6 coupled, linear equations of motion

Wave forces(only)

Frequency dependent added mass and damping

Mass matrix

Stiffness(heave, roll and pitch only)

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Frequency domain vs. time domain analysis ofcrane operations

Equations of motion can be solved directly if all coefficients areconstant (or function of frequency only)

However:Water entry/exit, slamming: M, c, k and F position dependentImpact, slack/snatch, winch operation: k variesDamping: quadratic drag, contact frictionNumerical analysis by time stepping is required (typical model):

Time domain analysis:Calculate and add all forces at t=tiEstimate accelerationIntegrate to find velocity and position at next time step, t=ti+dt

) F(=x k + x ))c(( + x ))(( 1 ωωω &&& ++ cmm a

(t)F + (t)F + (t)F = x k + |x|xB + xBxxB +

t)x(M

exlinesenvrr2r &&&&

&&10 +

∂∂

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Assume:Centre of buoyancy, CB, off centre of gravity, CG. The force centre, CF,

will be vertically below the hookheel angle

If center of vertical added mass is not inline with F:vertical excitation tilting oscillations

Lifting at points below CG should be analysed with care

Stability, tilt angle and angular oscillations

B

CF CG CB

F = mg-B

mg

BmgCGmgCBBCF

−⋅−⋅

=

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Underwater lifting operation(Not intended)

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Safe Job Analysis

Can anything go wrong ?

Shhh, Zog! ....Here come one now!