SECOND APPROACH TO COMPUTE THE RESTORING COEFFICIENTS

“This draft is written in rough manner”

This approach is closer to reality because takes in account more motion parameters

and has no simplification for what concerns tethers motion.

Two tethers are used in this analysis

Surge restoring coefficients

Cij = forces in i-degrees of freedom due to unit displacement of j-degree of freedom

keeping all other degrees restrained.

C11 = give a displacement in the first degree and find out the force in the first degree.

C12 = give a displacement in the second degree and find out the force in the first

degree.

I apply a displacement along surge direction

Initially the tethers tension is T0

When I give a force in x1 I have a k11 and of course I have an effect in heave direction

so I will have a force k31 and of course along y axis I will have a rotation because

tethers create have now a vertical component and they create a rotation about y axis

∆T 1=AEl0

[(√x12−l2 )−l ]

Equilibrium in the surge direction

C11x1=2 (T 0+∆T 1 ) sin γ x

sin γ x=x1

√ x12+l2

It is gonna be restored by the horizontal component force of the tensions by an angle

ys.

C11=2 (T0+∆T1 ) sin γ x

x1=2 (T 0+∆T 1 )

√x12+l02

C12=0 because there is no motion or displacement along y axis. It is because since

there is no movement about y axis the rotation along this axis is pitch which is zero.

C41=0 no moment about x axis (so no roll about x axis)

C61=0there is no rotation about z axis due to symmetry

k11 has an influence in the heave direction so let’s find k31

C31(∆)=2 [T0 cos γ x+∆T1 cos γ x−T 0 ]

cos γ x=l0

√x12+l2

C31=2∆

[T 0 cosγ x+∆T 1cos γ x−T 0 ]

C51=−(C11h )

taking in consideration the vertical components of the tensions and compute the

moment about the point C. the moment is counteract k51. It is easy to note that the

horizontal component of the tensions is nothing but k11. Therefore, the moment about

the point C is k11 by h.

Sway restoring coefficients

C12=0 in the surge degree because of the heave displacement given in the sway

degree is zero

C52=0

C62=0 due to symmetry the platform will have no yaw rotation

C22=2(T 0+∆T 2)

√x22+l02

it is the same as C11 but the change in tension is different

C32=2∆

[T 0 cosγ y+∆T 2cos γ y−T0 ]

C42=−(C32h )

It is identical to the surge motion

Heave restoring coefficients

C13=0

When I give a displacement in surge, heave is generated, but when I give heave

displacement, why surge is not generated? When the tlp is floating because of the

buoyancy in excess, when I push my tlp right or to my left, there is a change in

buoyancy and the draft changes in tension and changing tension, the system restores

it back. Therefore surge sway are coupled with heave. If only I give a heave

displacement there is only a change in buoyancy T0 will not be there. So,

C23=0

C43=0 it is the force in the roll axis because of the displacement in heave direction.

Since there is no force along y axis (k23 = 0)

C53=0 since there is no force along x axis (k13 = 0)

C63=0 since the rotation about z axis is zero. By giving a displacement along heave

direction. It is not because of symmetry in this case. Since k13 and k23 are zero, the

moments about these forces are unbalanced moments about these forces about CoG

will remain zero. There is no activity.

The only thing we have here is

C33 (x3 )=2 [ A El + ρw g (D 2π4 )]

K33 cannot be zero because it is one of the leading elements of the stiffness matrix

diagonal. If it is zero the whole row is zero and this cannot be and you cannot invert

the matrix.

Second k33 is nothing but the change in buoyancy (change in draft) and the buoyancy

will be transferred only to T0. It means the restoration cannot happen.

It is given by the change in tension in the cable (AE/l)

C33=2[ A El +ρw g( D2π4 )] (x3 )

The stiffness matrix is not symmetric because for example k14 is never equal to k41. K41

is the displacement force in the surge degree because of rotation given in roll which

can never be equal to force displacement given in surge. One is a rotation and one is

the displacement so they cannot be combined.

Yaw restoring coefficients

It is a rotation about z axis

The position of the tethers will be twisted, but there is not buoyancy difference in my

FB value. So yaw influences only the tethers and so I will have a new l of tethers, l

which is a combination of l and θ6

L1=√ l2+θ62 (a2+b2)

Once known l1,

∆T 6=EA(l1−l)

l

C16=0Force in surge degree because of yaw rotation, it is identically rotating in the

horizontal plane, nothing is happening in the surge axis.

C26=0 Because for a unidirectional wave sway is of course zero

C46=0 Because sway C26 is zero

C56=0

C36 (θ6 )=2T 0( ll1−1)+∆T6 ll1 In the heave degree the force will be given by the change in tension

C66 (θ6 )=2 (T 0+∆T 6 ) (a2+b2 ) θ6l1

Roll restoring coefficients

I give a rotation θ4, initially there is a tension

The tension remains vertical. It is possible to note that the right part of the platform is

more immerged compared to the left one which is protruding up, there is a shift of

buoyancy from left to right. Therefore, the centre of buoyancy changes. Let us call it

as e4

s1=Pl2

+e4

s2=Pl2

−e4

Because of θ4, will be generated a moment k44

C14 is the force in the surge degree of freedom because of the rotation given.

C14=0 because surge is not affected at all.

C24=0 for a unidirectional wave k24 is zero because there is no effect of sway because

of roll angle this also remains zero.

∆T 4=EAl

P l2cosθ4=∆T 4' Remains the same because of the symmetry.

P l2cosθ4 is the component in direction of ΔT4

ΔT4 is caused by the initial tension AE/l

ΔT4 is the change in tension or pre-tension in the nearer leg

ΔT4’ is the change in tension or pre-tension in the farther leg

C34=2 (∆T 4+∆T 4' )

θ4

It is the force in heave degree because of the roll angle. The force in the heave degree

will be affected only by the change in tension. That is not causing disturbance. It is

divided by θ4 because of units

C44 (θ4 )=FBe4+2 (T0+∆T4 ) ( s1−e4 )−(1unodovrebbeessere )2 (T 0+∆T 4' ) ( s2+e4 )=ρw g πD2

4Plsin θ4+4

(T 0h sin θ4 )θ4

+4 AEl

Pl cosθ42

This part is to be reviewed

It is the force in the roll degree because of the displacement given in the roll degree.

The negative values come from the legs 1 and 2.

Knowing h, it is possible to know θ4

∆T 4'=Pl cosθ42

It is possible to plot a graph with θ4 varying

Pitch restoring coefficients

It is the same as the roll but

C15=0 because there are no forces in surge activated because of the rotation given in

pitch degree

C25=0 because there are no forces in sway activated because of the rotation given in

pitch degree

C35=2∆T 5+∆T 5'

θ5

It is like C34

C55=4 ρw gπD 2

4Pbsin θ5+4

T0h sin θ5θ5

+4AEPb cosθ5

2l (To be reviewed)

Properties of [C]

[C] is square, non-symmetric matrix of size 6x6

Heave has strong coupling with all degrees-of-freedom

Presence of off-diagonal terms reflect coupling between various degrees-of-freedom

Coefficients depend on change in tether tension which affects the buoyancy of the

system

Hence [C] is response dependent

Coefficients of [Cij] are non-linear (for example C32, C11)

Hence [C] is not constant but changes with instant of time

Mass MatrixThe mass matrix [M] is still 6x6 matrix, we assume that the structural mass is lumped

at each degree. When we deal with lumped mass we are talking about diagonal

matrix, but in this problem will not remain diagonal.

[M ]=(M 0 0 0 0 −M y g0 M 0 0 0 00 0 M M y g 0 00 0 M yg I 11 I 12 I 230 0 0 I 21 I 22 I 23

−M y g 0 0 I 31 I 32 I 33)

M 11=M 22=M 33=M Total mass of the body, which is added in all degrees of freedom

respectively in surge, sway and heave degree.

M 44=M 11r x2 M44 is the mass moment of inertia about x axis (r2

x is the gyration radius)

Similarly

M 55=M 22r y2

M 66=M 33r z2

You can compute easily r2x because we know the mass of inertia of the whole system

respect to the centre of gravity and we know the cross sectional area of all members

and you can easily find rx

Once you know the diagonal look at M11, you get Ma11 because it is added mass

M a11= ρw πD 2

4(Cm−1 ) xsurge

Look at the plane and i have an additional mass in surge degree and this mass will

create a moment about y axis, that is my Ma51. So it is the moment of ma11 about cog

Ma51 =

I have also Ma33 it is the added mass in heave direction. It is coming from set-down

behaviour

M a33=ρw πD2

4(Cm−1 ) xheave

I have also Ma53, because of the different heave there is a probability that some of the

columns may get immersed lesser and some of them immersed more. There is a

difference in buoyancy and so it