Claas f1200 c1200 soja getreide lexion (type 442) service repair manual
Preliminary Design of a FOWT
1. Università degli Studi di Napoli Federico II
Facoltà di Ingegneria
Corso di Laurea in Ingegneria Navale
Elaborato di Laurea
Preliminary design of FOWT
English version
Relatore: Candidato:
Chiar.mo Pietro Rosiello
Prof. Antonio Campanile M66/099
Anno Accademico 2013-2014
9. Chapter 1
Stability and buoyancy
Hey guys, you are too troubled, I
see and this is not good; the
examination is important, of course,
but just three are really
fundamental things in life: the first
one is love, the second one is
studying, and finally, sex, drugs and
rock’n’roll.
G.G.
1
10. 1 – Statics and buoyancy
First we have to chose the structural plan. There are different turbines off-shore
structural configurations, but here we tried to limit the number of cylinders in order
to achieve a lightweight structure. Therefore, we tested both configurations with three
cylinders whose axes are arranged at the vertices of an equilateral triangle, and with four
cylinders whose axes are arranged at the vertices of a square.
Figura 1.1: Triangular Plan Figura 1.2: Square Plan
Fig.1.1shows how the rotation around the axis a leads to a higher immersion of the
individual cylinder (and therefore to an increased hydrostatic pressure); similarly, fig.1.21
shows how the rotation around b leads to the same hydrostatic pressure increase; the square
plan, unlike the triangular, offers more stability since it has the opposite cylinder which
increases the righting levers. Therefore we had to choose between three bigger cylinders and
four smaller. We chose the triangular plan because even if it is larger, we have in any ca-
se, a cylinder in less. Our aim therefore will be to achieve a configuration as shown in fig.??
The design of structure assumes as load a turbine 5-MW , whose features are known
from [1]We point out some reference data in the following tab. 1.1 .
Tabella 1.1: Reference Turbine
We will simply indicate with the abbreviation RNA (Rotor Nacelle Assembly ) this
Nacelle, Rotor and Tower assembly (or anything related to the payload).
Once shown the structure plan, we will clarify some sizes through a cross-sectional view of
fig.1.3:
2
11. 1 – Statics and buoyancy
Figura 1.3: Structure Front View
Where:
H stands for cylinder height
D stands for cylinder diameter
T stands for cylinder immersion
b stands for distance between the cylinders axes
1.1 Calculation procedure
The modus operandi is to set the structure geometric values, thus H as cylinder height
and D as cylinder diameter , and see what results brings the same configuration. Then,
we will make a configurations shortlist made of the most satisfactory results from which
we will chose that one used for our project.
Anyway, the metacentric hypothesis is the basic one. By splitting into various steps the
procedure, we will start by dealing with the subject of the buoyancy:
1. Once set H and D we assume that immersion T is half the height H. We decided that
by elimination : if T is less than H/2 the structure will present its own unnecessary
extension since, when the water reaches the downfloating angle φd
1
, it will not touch
the cylinders sides but the root, thus affecting the waterplane. From here onwards,
the righting moment will no longer be comforted by an isocarenic immersion thus
the steel above the limit point, shown in fig. 1.4. will be useless. This would increase
the barycentric height as well as the costs of the steel itself.
The assumption is similar in case where T > H/2. Then we assume so
T =
H
2
(1.1)
1
In pointing out this corner we have really committed an abuse of notation; it is right calling it
downfloatign angle if there were openings on the deck which justify such progressive flooding. In our case,
there are none, but we wanted to keep this notation to show better how at that corner the free surface
touches the deck.
3
12. 1 – Statics and buoyancy
Figura 1.4: Immerion less than H/2
2. Being the immersion available, we can calculate the hull volume without forgetting
that there are 3 cylinders; so we have
= 3πT
D2
4
3. 3. As a result we get displacement ∆
∆ = ρ
assumed ρ = 1.025t/m3
4. Let’s check now if the values of ∆ and just obtained are compatible with RNA
weights, plus the structure weight ; in effect we verify the buoyancy. At this step of
the project we had to set a parameter to assess its weight, ignoring the thicknesses :
therefore we have imposed a certain structure density ρcil such that
ρcil = 0.15t/m3
(1.2)
So
∆structure = ρcil (1.3)
Having
∆tot = ∆RNA + ∆structure
However, this similarity is just theoretical and it is necessary to forecast an additional
weight, namely that one of the ballast arranged inside the cylinders and that will
add up to the second member of the latter equation. So we have
∆tot = ∆RNA + ∆structure + ∆bal (1.4)
Where
∆tot = ∆RNA + 3πH
D2
4
ρcil + ∆bal
The check up will be performed by comparing this ∆tot with the ∆ found in Step 3.
We will have three possible cases
4
13. 1 – Statics and buoyancy
• ∆ = ∆tot
The hypothesis of T = H
2
is verified with ∆bal = 0.
• ∆ > ∆tot
The hypothesis of T = H
2
is verified with ∆bal > 0 to be valued.
• ∆ < ∆tot
Therefore the hypothesis of T = H
2
is not checked since we will need a ballast
weight negative to confirm exactly our calculations; for the physics laws that’s
clearly an absurd concept so we will discard immediately configurations of H
and D that lead us in this situation.
In order to achieve the step 4 check up, we have to calculate the suitable ballast
weight so that ∆ = ∆tot in case of ∆ > ∆tot which inter alia is the vast majority of
cases.
We get ∆balwith the following reasoning:
starting from
3πT
D2
4
=
We will replace with ∆tot/ρ
3πT
D2
4
= ∆tot/ρ
And then we explain ∆tot
3πT
D2
4
ρ = ∆RNA + ∆structure + ∆bal
Finally we have
∆bal = 3πT
D2
4
ρ − ∆structure − ∆RNA
We are now able to determine which are the values H, D and ∆balthat enforce the
buoyancy with the constraint T = H
2
. Sure of these arguments we are now able to
create an spreadsheet and set the calculations for a series of the values H and D
which meet ∆bal ≥ 0. All of these configurations will enforce the buoyancy; now let’s
focus on stability.
5. Static equilibrium angle φs
The off-shore design guidelines set this angle to 10 ° of which 3° are due to wave
oscillation.
So we have to set this angle to 7 ° expecting 3° due to wave oscillations.
The heeling moment exerted by the rotor thrust T; is assumed as a structure
operation condition; this thrust is calculated through the Betz coefficient denoted by
cT .
T =
1
2
ρcT Av2
5
14. 1 – Statics and buoyancy
Figura 1.5: Wave oscillations
Being available the diagram of fig.1.6 from [1], we can calculate the rotor thrust value
to the rated wind velocity of 11.4m/s, i.e. T = 645kN. Now we have to determine
the arm.
Figura 1.6: 5MW turbine features
We know that the thrust is applied to the rotor center, so the arm is the distance
between thrust center and the constraint where the anchorage is fixed. Therefore
we constrain the anchorages in the highest point of structure, or the cylinder top,
trying to get an arm as small as possible. Since we know all the RNA features we
can deduce from [1] that the arm is equal to 89.1 meters; hence the heeling moment
value M in operating conditions is:
M = T ∗ arm = 57470kNm (1.5)
So we have to fulfill the following checking
7°∆(r − a)g ≥ M
6
15. 1 – Statics and buoyancy
6. φd
calculation The downfloating angle is that angle which, once exceeded, doesn’t show
the same waterplane and so we will not have a righting levers linear diagram. It
therefore corresponds to the maximum value of the righting levers. In fig. 1.7 we can
see GZ, corresponding to φdand the mass HA (Heading Angle) that, as mentioned
previously, should correspond to the stress of (1.5) for the angle φs.
Figura 1.7: Righting arms
We can consider to be valid the following reports:
GZ = GMφd =
HA
φs
φd
AHA = HAφd
AGZ =
φd
0
GZdφ =
φd
0
HA
φs
φddφ =
1
2
HA
φs
φ2
d
The reference standard [2] requires that AGZ is at least 30% more than AHA.So we ap-
ply this condition multiplying AHAfor a value k that is equivalent to 1.3. Hence we get
AGZ = kAHA
1
2
HA
φs
φ2
d = kHAφd
φd = 2kφs
Once got φs then we have φd ≈ 18° .
There is also to be considered a nonlinear reserve stability obtained after we got φd
7
16. 1 – Statics and buoyancy
that presumably will be about 10% of the AGZarea which reduces the factor k from
1.3 to 1.2, thus obtaining φd = 17°:
φd = 2φsk0.9 ≈ 17
Anyway we choose not to consider (conservatively) this reserve stability so we set
the checking to 18°.
7. Wheelbase B calculation
Figura 1.8: Downfloating angle
The geometry tested shows that
b =
H
tanφd
− D
Hence we immediately get the wheelbase.
8. GM calculation: (r-a)
We get the value (r-a) from
GM = BM − (KG − KB)
as we can deduce from fig. 1.9
• We will calculate BM from
BM =
Ix
Where
Ix = 2π
D2
4
b2
4
that is, two times the buoyancy area multiplied by the arm to the square.
• KG calculation
We immediately get the height of the center of gravity from
G − 0 =
n
i=1 mi(Gi − 0)
mtot
8
17. 1 – Statics and buoyancy
Figura 1.9: (r-a) value
which in our case becomes
KG =
(cMRNA ∗ mRNA) + (cMstructure ∗ mstructure) + (cMbal ∗ mbal)
∆TOT
• KB calculation
We can calculate KB from
zB = M
since we are using a cylindrical geometry we can get KB from
KB =
3πD2
T2
8
Or more simply KB = T/2. At this point we have (r-a).
9. Static conditions verification
Once we have found (r-a) we can check if the righting moment meets the heeling
moment. If the H and D configuration tested here does not offer an adequate righting
moment, that is greater or equal to 57470 kN m about for an slant of 7°, then we
have to discard it. Therefore we check
M ≤ 7°∆(r − a)g
Where g is the gravitational acceleration and the 7° will be calculated in radians.
10. DNV Verification
As mentioned before, the other verification to be done according to [2] expects that
AGZ ≥ 1.3AHA.
1.2 Configuration H D b choice
Since the spreadsheet is available, it is certainly easier to perform some calculations for
different configurations and then check the results. Hence we have to make a choice between
9
18. 1 – Statics and buoyancy
the various pairs of H and D according to the weight of the structure ∆structureand the
wheelbase b length . For this purpose we have plotted the configurations values according
to masses standard of H/D (ranging from 0.5 to 3.75) thus obtaining two diagrams: one
according to ∆structure and another one according to b, respectively, as in fig. 1.10 e 1.11.
Figura 1.10: Diagram according to ∆structure
It is good of course to choose a configuration that presents a lower weight than the
other one, but as the diagrams show, as ∆structure decreases b increases. This could be
a matter not so much concerning the dislocation where the structure will be implanted,
but in relation to the creation of the tubular beams span, perhaps too wide, between the
cylinders; therefore if b is too large, it could lead to the a high tubular beams thicknesses
and paradoxically we might have more weight despite having chosen the lighter structure.
Then we considered appropriate not choose the lighter solution, i.e. H/D = 3.75 but an
H/D in a range from 2 to 2.5 as it offers the values b smaller and the weight differences
smaller than 3.75 . At this point we have to choose one among these and hence we get our
geometry.
Finally, we considered the ballast weight. Until now we thought our structure to be
as something abstract, without both tubular beams and anchorage but these ones have
their own weight, of course. In order to ensure that these weights do not affect the
KG position (and consequently the stability of the structure), we have to be able to
compensate decreasing the ballast. In effect we have to choose a configuration which allows
to compensate the beams and anchorage weights with a decrease of the ballast.
On the basis of all these considerations, we chose the following configuration H = 21m; D =
10m; b = 54.6m.
The chosen configuration leads to the following values:
We created the righting levers diagram with the spreadsheet. We checked up every
values obtained with the software, and more precisely we used the subroutine STABILITY
reports both in watertight integrity that in damage conditions.
10
19. 1 – Statics and buoyancy
Figura 1.11: Diagram according to b
H 21 m
D 10 m
T 10.5 m
∆RNA 697 t
cM 85 m
∆structure 742 t
cM 10.5
∆bal 1096 t
cM 2.3 m
∆ 2536 t
2474 m3
b 54.6 m
(r-a) 25.2 m
Ix 117204 m4
r 47.4 m
a 22.1 m
KB 5.3 m
KG 27.4 m
AGZ 1774948 m φ
AHA 1380515 m φ
Tabella 1.2: Chosen configuration
11
20. 1 – Statics and buoyancy
Figura 1.12: Bentley SACS
1.2.1 Report: intact condition
Medium precision, 57 sections, Trimming off, Skin thickness not applied.
Long. datum: MS; Vert. datum: Baseline. Analysis tolerance - ideal(worst case): Disp.%:
0.01000(0.100); Trim%(LCG-TCG): 0.01000(0.100); Heel%(LCG-TCG): 0.01000(0.100)
Loadcase - Loadcase 1
Damage Case - Intact
Fixed Trim = 0 m (+ve by stern)
Specific gravity = 1.025; (Density = 1.025t/m3
)
Fluid analysis method: Simulate fluid movement
Ballasting Item Name Quantity Total Total x y z
Mass [t] Vol m3
0 Lightship 1 1.439.700 0 14.015 46.500
1 Tank002 22.64% 365.400 365.400 -27.361 0 2.377
1 Tank003 22.64% 365.400 365.400 0 47.285 2.377
1 Tank001 22.64% 365.400 365.400 27.361 0 2.377
0 Total 2.535.900 1.096.200 0 14.810 27.396
Tabella 1.3: Intact condition
12
21. 1 – Statics and buoyancy
Figura1.13:Rightingarmsdiagraminwatertightintegrity
13
23. 1 – Statics and buoyancy
1.2.2 Report damage cndition
Damage conditions were analyzed according to two compartments: the first one consists
in imagine an half flooded cylinder and the second one concerning just a quarter of a
flooded cylinder.
The cylinders were subdivided using two bulkheads that divide it into four equal parts in
order to use the software STABILITY faster and linearly; then in order to analyze the
first damage condition we flooded just two of the four cylinder parties, and as regards the
second one we flooded just one of it. Here you are the first one.
Medium precision, 207 sections, Trimming off, Skin thickness not applied.
Long. datum: MS; Vert. datum: Baseline. Analysis tolerance - ideal(worst case): Disp.%:
0.01000(0.100); Trim%(LCG-TCG): 0.01000(0.100); Heel%(LCG-TCG): 0.01000(0.100)
Loadcase - Loadcase 1
Damage Case - DCase 1
Fixed Trim = 0 m (+ve by stern)
Specific gravity = 1.025; (Density = 1.025tonne/m3
)
Compartments Damaged - 2
Compartment or Tank Status Perm.% PartFlood.% PartFlood.WL
Tank009 Fully flooded 100
Tank012 Fully flooded 100
Fluid analysis method: Use corrected VCG
Tabella 1.5: First damage condition
15
30. 1 – Statics and buoyancy
Figura1.15:Rightingarmsdiagramintheseconddamagecondition
22
31. 1 – Statics and buoyancy
This last damage condition meets the regulation [2].
23
32. Chapter 2
Scantlings
Pietro, listen to me please, it is true
that I am more experienced than
you but I’m not so infallible, I can
mistake me too, so you mustn’t
believe all that I say. But above all
do not trust what you remember I
have said.
A.C.
24
33. 2 – Scantlings
Once determined the three cylinders geometric features, now we have to perform the
scantlings; we know that RNA is located in the center of the equilateral triangle formed
by the three cylinders axes whose base is the height of the cylinder top. Then we chose to
create a RNA support cylinder which we will call RNAS, almost a natural extension of
the tower, which will lead the connection with the cylinders under the deck limit, so that
we can already reduce conceptually the struts span. Anyway, the RNAS will not have a
height greater than H / 2 so as not to affect the statics arguments (since, of course, a
greater height would lead to have a new immersed volume).
So we state to have a structure consisting of tie beams and struts from the RNAS base to
the cylinders top and their base, in addition to the beams which ensure the connection
between the top cylinders and their bases. The beams, the tie beams and struts will be
spaced at least 200mmfor stability reasons.Fig. 2.1 shows what was said.
Figura 2.1: Structure Front view
Where
CYLINDERS BLUE
TIE BEAMS MAGENTA
STRUTS ORANGE
BEAMS GREEN
RNAS ROSSO
DECKS CYAN
Tabella 2.1: Colors Reference Structure
It ’s also helpful to have a feedback model in 3D
So the scantlings consists in the scantlings of the three cylinders, of the beams set,
of the tie beams and the struts aside from RNAS. For the scantlings of the cylinders
and the RNAS we used the reference standard Buckling and ultimate strength assestment
for off-shore structures from ABS [3], but as regards the set of the tie beams and struts,
we used the beam theory. The chosen steel has a yield stress of σ0 of 355 MPa, a value
commonly used in off-shore structures.
25
34. 2 – Scantlings
Figura 2.2: Structure 3D View
2.1 Cylinders and RNAS
We will calculate the RNAS load by the RNA weight plus a 50% increase due to the
inertial forces. Similarly, the cylinders load will be the sum of the RNA and RNAS weights
plus 50% due to the inertial forces. In order to make available RNAS weight included
within the loads of the cylinders, once known the RNA features, we will primarily perform
the RNAS scantlings.
The calculation ABS 1
method is based on tested methodologies with relative safety factors
which return a safety level suitable to the current practice. The internal structure given
by [3] ] for cylindrical support geometries shows rings and stringers which may or may not
be flanged.The following figure extrapolated from the legislation clarifies the rings layout
and their distances.
Then we have to distinguish the following directions: the longitudinal x, the radial r
and rotational θ while the internal rings are equidistant from each other, in particular the
stringers relating to s and the rings relating to l.
The rings sizes are explained in the following figures.
First we have to understand how the load acts on the structure then we will have four
applications concerning it:
• Compression evenly distributed in the longitudinal direction x: σa
• Cylinder full bending due to σb
• External pressure p
• Combinations of the previous one
The ABS shows two different characterizations of FPI (Floating Production Installation)
loading conditions according to the utilization factor η of the maximum allowable stress:
1
We will refer to [3] aalso mentioning some of its chapters, indicating the section (sec.) and the relevant
paragraph.
26
35. 2 – Scantlings
Figura 2.3: Shell Rings and Stringers
Figura 2.4: Vertical stringer section
Figura 2.5: Ring section
normal operation for which η = 0.60ψ and severe storm for which η = 0.80ψ; ψis the
adjustment factor which is different for the buckling of the shell, of the cylinder as a
kingpost and of the rings and stringers.
According to regulations there is a kind of hierarchy of the different buckling calculation
mechanisms (as regards the rings and the stringers); we have to perform scantlings in
order to ensure that our components thickness does not lead to a first local buckling and
then to a general buckling and finally to the bay buckling.
First, we have to calculate the longitudinal stress and the rotational one according to the
27
36. 2 – Scantlings
Sect.4 – 13 directives.
In particular, as shown in fig.2.3, we denote the longitudinal stress by σx,it is due to the
sum of a stress component given by the axial force σa plus a stress component given by
the bending moment σbas shown in section 4 - 13.1. As regards the rotational stress, we
follow the Sect.4 - 13.3 guidelines which locate the stress σθ at the middle of the shell
between two adjacent rings, and the σθR n the inner face of the ring flange ( when there is
one ) as shown by rF in 3.5.
These values will be considered in the calculation process of the various stability tests,
used by the legislation.
2.1.1 Local buckling
This may affect not only the shell but also each individual ring, that is, the shell may
show some buckling between two rings without deforming the latter as shown in following
figure:
Figura 2.6: Local shell buckling
So we have to test the local buckling both on the rings and on the shell.
• As regards the rings they have to be tested according to sect. 4-15 where we verify
that their moment of inertia iRiR (with their effective associated shell plating length)
is not less than that one proposed by the regulations; so we will have two different
tests for two different geometries: 15.1 as regards the rings and 15.3 as regards the
stringers.
28
37. 2 – Scantlings
• • It may happen that when the rings stiffness is low and a relatively high slenderness
of the curved panel is associated with them, the minimum buckling stress is flexural-
torsional. So we lose a large part of the efficiency of the shell initial shape. In sect.
4 - 9.1 verification hands in the following inequality
σx
ησCT
≤ 1
Where
σCT : flexural-torsional buckling stress compared to the ring axial compression in-
cluding the associated shell plating. It should be right to specify that as regards
the σCT calculation we had at first to get the σCL critical buckling stress on the
associated shell plating which corresponds to the half-waves number nfrom which
we get the Euler’s buckling stress σE. This factor nis assumed to be equal to 1 be-
cause the deck length (i.e. the size s) is so small to allow a single half-wave formation.
• • Let’s examine now sect. 4-5 concerning the shell plating local buckling. This kind
of buckling does not necessarily lead to a shell complete break as the stress can
be redistributed to the adjacent section. Anyway it is necessary to know this local
behavior because even if a structure can hold up despite a deformation, not sure
that it can also meet the alleged operational requirements. Verification proposed by
ABS is shown in the following inequality
(
σx
ησCxP
)2
− φP (
σx
ησCxP
)(
σθ
ησCθP
) + (
σθ
ησCθP
)2
≤ 1
Where
σCxP : critical buckling stress due to the axial compression or bending moment
σCθP : critical buckling stress due to the external pressure
• Finally in sect. 4 - 15.5 is shown a limitation to the rings geometries according to
the ratio of their height with their thickness dw/tw tw in order to limit precisely
the component slenderness. The legislation also allows not to complying with the
limitations explicated in this paragraph after an alternative study on the forces
involved with a resolution method. Anyway, we also tested this paragraph where
using supported beams, we will have the upper limit of the ratio equal to 9.6.
2.1.2 Bay buckling
This kind of buckling shows the shell that is deforming with the stringers but just
between two non-deformed rings:
So legislation distinguishes three possible cases to test this kind of buckling: cylinder
without any ring or stringer, cylinder with only rings and cylinder with rings and stringers.
The first two cases are discussed in sect. 4-3, while the last one in section. 4-7.
• The verification proposed by ABS concerning cylinders without rings or only with
rings shows the following inequality
(
σx
ησCxR
)2
− φR(
σx
ησCxR
)(
σθ
ησCθR
) + (
σθ
ησCθR
)2
≤ 1
29
38. 2 – Scantlings
Figura 2.7: Bay buckling
Where
σCxR: critical buckling stress due to the axial compression or bending moment
σCθR: critical buckling stress due to the external pressure
• The verification proposed by ABS concerning cylinders with rings and stringers
shows the following inequality
(
σx
ησCxBAe/A
)2
− φB(
σx
ησCxBAe/A
)(
σθ
ησCθB
) + (
σθ
ησCθB
)2
≤ 1
Where
σCxB: critical buckling stress due to the axial compression or bending moment
σCθR: critical buckling stress due to the external pressure
Ae: area of the effective section A: total area of the section
2.1.3 General buckling
This kind of buckling shows at the same time just one deformed ring or more than one
with adjacent shell plating and string:
General buckling meets verification if rings do not collapse; therefore, we have to refer
to the paragraphs already examined about local buckling: section 4 - 15.1 and sect. 4 -
15.3.
2.1.4 Shell buckling considered as a pillar
The shell is basically similar as a pillar. This can cause Euler buckling. To test this
kind of buckling we go back to sect. 4 - 11. We will have two tests to be performed:
• The first test it’s practically a condition shown in
λxE ≥ 0.50
30
39. 2 – Scantlings
Figura 2.8: General buckling
Where
λxE is the slenderness given by σ0/σE(C)
In particular
σE(C): Euler buckling stress.
So if the slenderness was greater than 0.5 we have to check as follows.
• All cylinder cross sections must meet the following inequality
σa
ησCa
+
σb
ησCx[1 − σa/(ησE(C))]
≤ 1
Where
σCa: critical buckling stress due to the compression
σCx: critical buckling stress due to the axial load or bending
In our case the first test is not exhaustive as λxE not exceed the value 0.08, it is therefore
unnecessary to proceed with the second test. This result is fairly predictable as if we have
a pillar with a diameter which is half of its height, then this means that we are having to
do with spandrel beams which admit the yield stress as a minimum crisis stress rather
than that Euler’s.
2.2 Deck and bottom
Deck and bottom structures are basically circular plates to which are applied rings, in
this case they are properly rings and radial rings having the same number and layout of
the shell cylinder stringers studied here , in order to respect a structural continuity.
We used the S.Timoshenko’s and S.Woinowsky-Krieger’s text Theory of plates and shells
[6] as reference for the scantlings. In particular we recalled the case of the load evenly
distributed on a supported plate. In fact, although the deck and the bottom will be welded
31
40. 2 – Scantlings
to the shell and the relative strings, given the scantlings in question, it’s unlikely you can
get a perfect fit.
Since at points equidistant from the center of the plate, the flexures will be the same, we
can consider the flexures in a single diametral section through the axis of symmetry, as in
figure
Figura 2.9: Supported plate layout
The deck loads were imagined as a water column of 75 cm stressing just the deck, but
multiplied by 2 in order to take into account the sea stresses. The bottom load is instead
the stress ρgH distributed on the bottom circular area; the bending moment is exactly the
loading action on the deck taken in account for the scantlings. The [6] give us the formulas
for the radial bending moment Mr and the tangential Mt, which are the following
Mr =
q
16
(3 + ν)(a2
+ r2
)
Mt =
q
16
[a2
(3 + ν) − r2
(1 + 3ν)]
Where
q is stress intensity evenly distributed
a is beam radius
r is the radius in which you are calculating the above moment
Now, once known the stress and its effect, the first pages of [6]consider the flexural
stiffness D for the beam:
DPlate = DP =
Et3
12(1 − ν2)
(2.2.1)
from which we get the thickness that the beam must have to withstand the stresses
referred to above. This D formulation is valid as mentioned for a supported beam, then
we can try to perform the scantlings of this thickness by analyzing the flexural stiffness
when we have a ringed beam in accordance with the
Dringedplate = DPR = DP +
EI
a1
Where E is the Young’s modulus and I is the following
I = hwtw[
h2
w
12
+ (
hw
2
− d)2
] + a1t[
t2
12
+ (d +
t
2
)2
] (2.2.2)
Where d standes for
d =
t
2
+
h2
wtw
2
− a1t2
2
a1t + hwtw
32
41. 2 – Scantlings
Dimensions are shown in fig. 2.10
Figura 2.10: Plate stringers
In the model of [6] the distance d starts at the beam axis rather than at the edge on
which stands the ring. Now we have to equate the DP with DPR and check whether the
values assumed are meeting the stress requirements.
2.3 Bulkheads
Studying statics we learned the bulkheads geometry number and layout. The pattern
[6] has been used also here for their scantlings, taking in account the case of the rectangular
supported beam, under the hydrostatic pressure stress. The reference diagram is the
following.
Plate under hydrostatic stress layout: β and β1 which are respectively included in the
bending moment formulations both according to x and y, which hold out the
Mx = βb2
q0
My = β1b2
q0
The results of the table are provided in according to two values: the first one identifies
the beam geometry and is a/b, the second one tells us what is the a percentage from which
the coefficient was calculated.
Since our a/b value is equal to 4.2, we obtain by interpolation the coefficients β and β1
from tab.2.2
β β1
a/b 0.25 0.5 0.6 0.75 0.25 0.5 0.6 0.75
5 0.0094 0.0187 0.0230 0.0309 0.0312 0.0623 0.0742 0.0877
4 0.0094 0.0192 0.0237 0.0326 0.0312 0.0617 0.0727 0.0820
4.2 0.0094 0.0191 0.0236 0.0323 0.0312 0.0618 0.0730 0.0831
Mx My
4.96E+07 1.01E+08 1.24E+08 1.70E+08 1.65E+08 3.26E+08 3.85E+08 4.39E+08
Tabella 2.2: β and β1 values
33
42. 2 – Scantlings
Figura 2.11: Plate under hydrostatic stress layout
So, then, we calculated the stresses to which the bulkheads were subjected.Once known
the moments values we got the section modulus W value from which then we calculated
the equivalent thickness t according to the fig.2.3.1.
W =
bt2
6
(2.3.1)
Now, similarly to the previous chapter, we compared the modulus just obtained with
the other one we should obtain in the case we have a ringed beam, i.e. with reference
in particular to fig.2.10where d does not start from the beam axis but from the edge on
which stands the ring.
W =
I
hw − d
(2.3.2)
Where we will have
d =
h2
wtw
2
− a1t2
2
a1t + hwtw
while the I value remains unchanged with respect to (2.2.2)
2.4 Crossbeams
The crossbeams scantlings, as for the tie beams and struts, follows the beam theory
using the safety factors steadied in the off-shore practice.
The components general layout, already clarified at the beginning of this chapter, is again
shown by treating in detail these structural components.
34
43. 2 – Scantlings
Figura 2.12: Crossbeams, ties beams and struts layout
2.4.1 Loads
The cylinders tubular beams are subjected to a scantlings load due to the roll stress
because of the waves. Worst situation, as well as the reference one, is when a wave has
its own crest on a cylinder and its trough on the other one. This situation generates
the maximum difference of immersed volume between the cylinders, so the hydrostatic
pressure.
Figura 2.13: Crest and trough on cylinders
The only wave that can generate this phenomenon is that one having the length λ
equal to twice the wheelbase b, because half of its length must be equal to the wheelbase
bby placing its own crest and trough on the axes of the cylinders. We have thus defined
wavelength:
λ = 2 ∗ b = 108.4m
The wave height ζ is accordingly defined supposing its ratio with the wavelength; a
valid value for which we have a real evidence is the following
λ
ζ
= 0.07
which returns ζ = 7.6m which returns a is equal to its half
35
44. 2 – Scantlings
ζ
2
= a = 3.8m
The cylinders geometry has already been defined, then we can understand how much
is this hydrostatic thrust difference that we define as ∆S.
So the first cylinder will show an increase in the immersed volume 1 that equal to the
decrease 2that the second cylinder shows. Then
1 = 2 = π
D2
4
a = 298m3
It is assumed, conservatively, that the increase (or decrease) in the volume is extended
from the waterline to the height athat remains constant from the cylinder axis up to its
ends as shown in fig. 2.14.Anyway it is an insignificant difference.
Figura 2.14: Increase in volume
Once known the increase (or decrease) in the volume, we can calculate the new immersed
volumes and the resulting buoyancy force:
+ 1 = π
D2
4
H
2
+ 1 = 1122m3
− 2 = π
D2
4
H
2
− 2 = 526m3
So we get buoyancy forces involved:
S1 = ρg( + 1) = 11270kN
S2 = ρg( − 2) = 5283kN
Which produce a Archimedean thrust difference ∆S = 5987kN. According to these
values we set the structural layout.
2.4.2 Structural layout
The structural layout which must be applied for the crossbeams scantlings is the
following:
Each crossbeam will then bear a constant shear and butterfly-shaped bending moment
that is maximum at the joints and zero at the center, in particular T = ∆S
2
and Mmax = Tl
2
.
36
45. 2 – Scantlings
Figura 2.15: Crossbeams structural layout
Figura 2.16: Crossbeams shear and bending moment
2.4.3 Crossbeam design
The section design will be made according to the section modulus W (multiplied by
the appropriate safety factor γ) which will be used to verify the ratio:
M
σamm
=
W
γ
(2.4.1)
Since the section modulus is a function of tubular beam diameters we will perform
the calculations for various diameters values up to meet the formulation (2.4.1). However,
given that the tubular section shows a stress distribution as evinced in fig.2.17,i.e. there
is a maximum shear where the moment is zero, we will perform the verification both for
shear and for the moment.
This checking will be performed according to the Jourawski’s approximate formula
assuming that in comparison to any one rope, the shear stresses components, in accordance
with the normal of the rope, may be confused with their average value according to the
following formula
τ =
T ∗ S
Is
where
• T is the shear designed on the bending axis (perpendicular to the axis of the tubular
beam)
37
46. 2 – Scantlings
Figura 2.17: Shear and Moment Distribution
• S is the static moment of the area neutral axis. We consider it just for the mid
section as it is zero for the whole annulus
• I is the section moment of inertia
• s is the thickness of the considered rope
The audit states that
√
σ2 + 3τ2 ≤ σamm
But in our case the maximum shear occurs exactly where the moment is zero, for which
we will just check
√
3τ ≤ σamm
The last check to be performed is concerning the critical load due to the external
pressure. The lower crossbeams, unlike those above, are immersed and therefore are subject
to this kind of stress that can cause buckling and deform the tubular beams forming some
lobes as shown in [4]. From the Mariotte’s formula we derive the compression stress shear
σh =
qD
2t
where q stands for the external pressure. The critical load in the elastic situation is
given by
σhE
σy
=
E
(1 − ν2)σy
(
t
D
)2
Whereas in the hinelastic situation
σhinelastic
σy
=
Et
E
σhE
σy
However, this formulation is not considered as the shear modulus Et is hard to
assess. Anyway, we can use the Perry-Robertson technique which assumes an initial beam
deformation in advance of the loading action of compression q, thus obtaining
σhu
σy
=
1
2
[1 +
σhE
σy
(1 + η
3D
2t
)] −
1
4
[1 +
σhE
σy
(1 + η
3D
2t
)]2 −
σhE
σy
Where η = 0.01 (maximum value detected over the whole length of pipes). This
formula assumes that the load which is at the elasticity limit is a critical stress, then the
38
47. 2 – Scantlings
material shows a highly resistant residual that makes very conservative this audit. We
could mitigate this overconfidence assuming that the collapse occurs when we reach the
full plasticity condition at the maximum moment section . However we decide to use this
formula as the chosen tubular beams sizes that will meet the yield strength, will meet also
this formula.
The checking is performed through the tables as the critical values both of the elastic and
inelastic field are available, depending on the ratio D/t.
D/t 10 20 30 40 50 60 70 80 90 100
σhE/σy 9.63 2.41 1.07 0.60 0.39 0.27 0.20 0.15 0.12 0.10
σhu/σy 0.86 0.70 0.53 0.38 0.28 0.21 0.16 0.12 0.10 0.08
Tabella 2.3: Compression buckling values
2.5 Tie beams and Struts
Layouts reference for these elements are shown in in fig. 2.18
Figura 2.18: Tie beams and struts layouts
Once known the values D, b, d, and H we can estimate immediately that α = 29° and
β = 61°. As for the loads, these are represented by the tower weight plus the RNAS weight
plus their 50% in order to take account to the inertial forces; moreover this load should be
distributed to the all six crossbeams that are holding it, then we will have to consider just
its sixth part for each crossbeam; we denote the load acting on each crossbeam with the
vector P.
2.5.1 Crossbeams design
The structural layouts are the same both for tie beams and for the struts, where the
classical cantilever beam at the ends will be loaded by the stress P . We will also break P
down into its components loading both the bending cantilever beam and the compression
one: so we have the Pf and the Pc.
Then the cantilever will be subject to a bending moment equal to M = Pf ∗ l, where
l is the span of the crossbeam . Similarly to crossbeams, we will use (2.4.1) in order to
39
48. 2 – Scantlings
Figura 2.19: Cantilever layouts
calculate our tubular beam sizes.
Further checks will be performed for buckling due to the component Pc. Therefore we have
to check the sizes meeting the yield stress in Euler’s buckling stress condition according to
the
σcr = σE =
π2
E
λ2
The risk we take when we use this formula consist in finding us in the Euler’s hyperbola
left area where there are spandrel beams and so the Euler values tend to infinite, thus
affecting a truthful verification; the figure is showing graphically this situation that Euler’s
hyperbola can cause:
σE
σy
=
1
λ2
Figura 2.20: Euler’s hyperbola
Therefore the Euler-Johnson formulation was applied as it solves this spandrel beams
problem without tending to infinite thanks to some small values of slenderness λ; for
σE ≥ σps, where σps is the structural proportionality stress , we will obtain
σcr = σy[1 −
σps
σy
(1 −
σps
σy
)
σy
σE
]
graphically we will have
40
49. 2 – Scantlings
Figura 2.21: Eulero-Johnson
2.6 Scantlings Report
In order to perform scantlings we have to start from the following values:
• Geometric parameters: H = 21 and D = 10 got from the study of the statics
• Off-shore steel parameters: yeld stress σ0 = 355MPa, modulus of elasticity E =
2.06x107
N/cm2
and Poisson’s ratio ν = 0.3 for steel
2.6.1 RNAS
The stress P to which is subject the RNAS is given by the RNA weight plus its 50%
due to the inertia forces . Hence we have P = 10256kN.Here we report tab.2.5that is
showing in tables the range of considered configurations as attested in fig. 2.4 and 2.5 for
the nomenclature of values. As well as for all the other sections, we get the chosen values
from considerations not purely economic (therefore we have a lightweight structure), but
also from the weldability requirements for the operator, as well as from the rings proximity
(in order to enable them to work well).
The chosen configuration is shown in tab. 2.4
SHELL RING STRINGER
θ t V l tw dw n V s tw dw n V Vtot PESO
DEG cm m3
cm cm cm m3
cm cm cm m3
m3
t
30 1.0 2.0 150 1.0 9.60 7 0.1 157 1.0 9.6 12 0.1 2.2 17.77
Tabella 2.4: RNAS chosen configuration
41
50. 2 – Scantlings
SHELL RING STRINGER
θ t V l tw dw n V s tw dw n V Vtot PESO
DEG cm m3
cm cm cm m3
cm cm cm m3
m3
t
360 6.8 13.3 1050 0.0 0.00 0 0.0 1885 0.0 0.0 0 0.0 13.3 106.45
45 1.0 2.0 262 1.0 9.60 4 0.1 236 1.0 9.6 4 0.1 2.1 17.02
45 1.0 2.0 262 0.9 8.64 4 0.1 236 0.9 8.6 4 0.1 2.1 16.79
45 1.0 2.0 262 0.8 7.68 4 0.0 236 0.8 7.7 4 0.1 2.1 16.59
45 1.0 2.0 262 0.7 6.72 4 0.0 236 0.7 6.7 4 0.0 2.1 16.40
45 1.0 2.0 210 1.0 9.60 5 0.1 236 1.0 9.6 4 0.1 2.1 17.16
45 1.0 2.0 210 0.7 6.72 5 0.0 236 0.7 6.7 4 0.0 2.1 16.47
45 1.0 2.0 210 1.0 9.60 5 0.1 236 1.0 9.6 4 0.1 2.1 17.16
45 1.0 2.0 210 0.9 8.64 5 0.1 236 0.9 8.6 4 0.1 2.1 16.91
45 1.0 2.0 175 1.0 9.60 6 0.1 236 1.0 9.6 4 0.1 2.2 17.31
45 1.0 2.0 175 0.9 8.64 6 0.1 236 0.9 8.6 4 0.1 2.1 17.02
45 1.0 2.0 175 0.8 7.68 6 0.1 236 0.8 7.7 4 0.1 2.1 16.77
45 1.0 2.0 175 0.7 6.72 6 0.1 236 0.7 6.7 4 0.0 2.1 16.54
30 1.0 2.0 175 1.0 9.60 6 0.1 157 1.0 9.6 12 0.1 2.2 17.63
30 1.0 2.0 175 0.9 8.64 6 0.1 157 0.9 8.6 12 0.1 2.2 17.28
30 1.0 2.0 175 0.8 7.68 6 0.1 157 0.8 7.7 12 0.1 2.1 16.98
30 1.0 2.0 175 0.7 6.72 6 0.1 157 0.7 6.7 12 0.1 2.1 16.70
30 1.0 2.0 150 1.0 9.60 7 0.1 157 1.0 9.6 12 0.1 2.2 17.77
30 1.0 2.0 150 0.9 8.64 7 0.1 157 0.9 8.6 12 0.1 2.2 17.40
30 1.0 2.0 150 0.8 7.68 7 0.1 157 0.8 7.7 12 0.1 2.1 17.07
30 1.0 2.0 150 0.7 6.72 7 0.1 157 0.7 6.7 12 0.1 2.1 16.77
Tabella 2.5: RNAS Report
2.6.2 Cylinders
The cylinders are subject to the stress P that is given by the RNA weight plus the
RNAS weight plus its 50% due to the inertia forces ; obviously just a third of this force
will act on each cylinder. Hence for each cylinder we will have P = 3507kN. Here we
report tab. 2.6 showing in tables the range of configurations considered as attested to fig.
2.4 and 2.5 for the nomenclature of values. The chosen configuration is shown in tab.2.7
2.6.3 Crossbeams, Tie Beams and Struts
As regards the Crossbeams, Tie beams and Struts we used a safety factor of 2 in order
to take account also of local stress.
As regards the Crossbeams, the load given by a bending moment is about 66000kNm,
whereas as regards the Tie beams and Struts, this moment is about 33000kNm.Tie
beams and Struts, as above, are also checked according to buckling by the Euler-Johnson
formulation for which the buckling is 1540kN.
The chosen configuration for the crossbeams is shown in tab. 2.8
2.6.4 Deck and Bottom
For each cylinder will have a deck and a bottom, whereas as regards the RNAS we will
have just the bottom since the other part of the cylinder will be welded directly to the
tower; the safety factor used in this section is 2. We examine the RNAS bottom for which
42
52. 2 – Scantlings
we has taken an hydrostatic pression (or neutral stress) load given by the RNAS height,
i.e. 10.5m; the resulting stress is of 105kN/m2
which should be met by the use of a not
ringed deck whose thickness isi 1.2cm.
Whereas as regards the cylinders bottom the reference height is 21m, then it’s equal to
the cylinder height, which returns a stress value of 211kN/m2
. The chosen configuration
is shown in tab. 2.9.
44
53. 2 – Scantlings
RING STRINGER
int tp hw tw d I int tp hw tw d I
cm cm cm cm cm cm4
cm cm cm cm cm cm4
143 1.00 9 1 0.30 219.79 37 1.00 9 1.00 0.97 173
143 1.00 9 1 0.30 219.79 112 1.00 9 1.00 0.37 214.19
143 1.00 9 1 0.30 220 187 1.00 9 1.00 0.23 224.89
143 1.00 262 1.5 96.44 2717415 262 1.00 9 1.00 0.17 229.80
Tabella 2.9: Cylinders bottom chosen configuration
Whereas as regards the deck the stress is 15kN/m2
and in order to contrast it we chose
a thickness value equal to 1cm.
2.6.5 Bulkheads
Final configuration is reported in tab. 2.10
45
55. 2 – Scantlings
2.6.6 Weights
Last Report concerns the total weights and is shown below
ELEMENT WEIGHT [t] n
CYLINDERS 137.9 3 413.8
DECK 6.3 3 18.8
BOTTOM 6.84 3 20.5
RNAS BOTTOM 2.26 1 2.3
BULKHEADS 23.1 6 138.6
CROSSBEAMS 116 6 696.0
TIE BEAMS 36 3 108.0
STRUTS 36 3 108.0
TOT 1506.0
Tabella 2.11: Total Weights
In order to understand better the structure we report some demonstrative image in 3D.
In particular in fig. 2.22 we hide its shell with its rings and stringers in order to show the
bulkheads and bottom rings geometry; moreover in fig. 2.23 we have again hidden its shell
in order to point out the full grid. We can see how the shell rings are connected to the
bulkheads transversal rings and also how the stringers (always as regards the shell) are
thrown on the bottom radial rings. All that in order to try to get a structural continuity
as much as possible, as the regulations recommend.
Figura 2.22: Cylinders bottom
47
57. Chapter 3
Mooring
What are you studying at the
University?
I’m a student in Naval Engineering
Wow ! Cool ! you will be a skipper,
then!
Many people
49
58. 3 – Mooring
3.1 Loads
As regards the mooring design we have to take in account the followings loads:
• Wind strength
Fw =
1
2
ρΣ[ηChCSA]v2
Where ρ is the air density increased by 20% in order to take account of water droplets
in the air during a storm.
η is a reduction factor for the possible presence of a screening surface to windward
A is the sail area
CS is the shape factor coefficient of the sail tested
Chis a factor coefficient taking into account the vertical distribution of the average
rated wind velocity at height zR. It is calculated by
Ch = (
z
zR
)2β
In particular zR is the reference size get at 10 meters and β = 0.15
• Flow strength
Fc =
1
2
ρΣ[CSA]v2
Where in this case, v is the flow velocity speed taken 1.5% of the wind velocity
• The rotor thrust obtained graphically in fig. 1.6
These three velocity masses contribute to get the "constant force" Fcost that will be
balanced by the pull of the mooring line. Moreover there will be another force that is
variable in the time, due to wave oscillations that make swinging just the body around its
equilibrium condition. Therefore we have to take into account these movements.
Studying the Pierson-Moskowitz’s wavy seas spectrum, we can get the significant wave
height HS from the wind velocity. Hence we get the maximum height from the ratio
HMAX = 1.86HSand finally the magnitude of the horizontal movement "a" such that
a =
1
8
HMAX
In fact this oscillation amplitude due to the wave, decreases the total shift limit that
the structure can withstand in accordance with the regulations.
The constant forces and oscillation amplitude calculation was performed for two wind
velocities (that’s the effective discriminant of this calculation); in particular cut-in velocity
11.4 m / s and cut-out one 25m / s. It might seem superfluous analyzing the rated wind
velocity stress, as this is about an half of the maximum one, but it is not so because, as
shown in fig. 1.6„ the rotor thrust maximum velocity is significantly less than the rated
one.. In fact this is due to the turbine that is always aiming to provide a constant power
even at the velocities higher than the rated one; it can mean that we are searching for a
lower blades velocity that generates a lower rotor thrust.
By this calculations we obtained a value of 899kN with a consequent shift due to the wave
of 3.1 meters for the rated wind velocity of 25m/s; whereas for a rated wind velocity of
11.4m/s we obtained a value of 703kN for the constant forces and a shift due to the wave
of 0.7 meters. vW = 25m/s condition was more conservative.
The limit imposed by the regulations as regards the structure shifts is 15 meters, hence the
new limit will be 11.9 meters where we have to develop a vertical shift of at least 899kN.
50
59. 3 – Mooring
3.2 Calculation case
The chosen configuration in order to moor the structure includes a mooring lines
provision at 60° that then will be anchored. This provision was chosen because the
cylinders are arranged according to the vertices of an equilateral triangle; thus forecasting
at least two mooring lines for each cylinder, they can easily get to an equiangular provision
; for symmetrical reasons is useful to have this provision in order to perform the calculation
just for only one mooring line as the other ones will be identical. Hence we will have
Figura 3.1: Mooring lines provision
The calculation of the vector forces shows that the most critical condition is when
very few lines are acting in contrast to the constant force and less efficiently. Given the
provision symmetry, we can get basically two cases: the first one when the constant force
acts perfectly parallel to a line, and the second one when the constant force acts in the
middle of two mooring lines. The second case is the most conservative because it shows
us that there are just four lines that are counteracting the constant force according to
their angular component of cos 30, in particular in this case we have two taut lines and
the other two loose. Whereas in the first case we have always four lines that are acting
and more effectively. The worst case is then shown here
3.3 Catenary
Mooring lines are one-dimensional components, their flexural stiffness is negligible in a
static analysis; hence the configuration that is constituted by its own weights , according
to a function called "catenary". In this case the mooring line is located almost entirely
in the water. In fact, the anchorages were arranged on the deck both in order to be in
sight (thus we make easier the inspections), and to reduce the moment due to the rotor
thrust ; therefore, the mooring line is just 10.5 meters clear of the water, so in a very
approximate way we consider it totally submerged. At the end of the mooring line there
is the anchor but this is not the point where the line touches exactly the seabed for the
first time. Whereas we have to foresee a line segment that rests on the seabed before the
line elevation beginning . This is very important because the anchor pulling capacity in
vertical direction is roughly equal to zero (in practice just its weight), so if there was not
51
60. 3 – Mooring
Figura 3.2: Azione forza costante
such line segment lying on the seabed without transferring the stresses according to the
horizontal direction, the anchor would still be forced to plow without offering its mooring
qualities. It’s for this reason that we’ll take in account the "immersed catenary" which
shows us in its own formulation that line segment lying on the seabed. The equations in
usual form are the following:
u = w
Tx
(Dt − D0)
H = Tx
w (cosh u − 1)
S = D0 + Tx
w sinh u
(3.3.1)
Where
S is the length of the line
H is the mooring line height on the seabed
Dt is the mooring line length
D0 is the length of the line segment resting on the seabed
Tx is the pulling component along x
Fig. 3.3 shows the catenary formulation
3.4 Mooring line design
We have to find such a catenary completely elevated after an horizontal displacement
almost of 15 meters of mooring on the deck; this has to have a pulling capacity able
to overcome the constant force after a stretching of no more than 11.9 meters from the
mooring initial position, such that
52
61. 3 – Mooring
Figura 3.3: Catenary
Fcost
2
≤ Txt cos α + Txl cos α (3.4.1)
Where we indicated with Txtthe taut lines pull and with Txl the loose lines pull
With these features the line will meet the standards imposed as, even in case of the worst
conditions of wind to 25m/s, the stretching obtained would be anyway lower or almost
equal to 11.9 meters, with an expected remaining spacing of 3.1 meters due to the action
of the waves; we head back so within the admitted spacing limits of 15 meters.
Furthermore, the values of the horizontal permitted displacements 11.9 meters and 3.1
meters, for the line change as follows:
• Constant force spacing limit : 11.9m
cos α = 13.7m
• Spacing limit due to the waves:3.1m
cos α = 3.6m
Greater clarification is offered in the following figure.
Our calculation are essentially based on two considerations: the height of the mooring
line H on the seabed was considered equal to the seabed assumed of 100 meters , and in
order to keep on a maintenance less effective in pulling, we planned using chains rather
than cables. Hence we show the procedure:
• We assumed the chain caliber (diameter) d from which we got the mass per unit of
length mwhich, in the usual application, is contained in the range
m = d2
(0.0215 ÷ 0.0232)
where d is in millimeters and m in kg/m
• Since the line runs almost entirely in the water, we have obtained the effective weight
(net buoyancy force) w by
w = 0.8694gm
• It is assumed the breaking stress, anyway contained in the range σa = 490 ÷
570N/mm2
53
62. 3 – Mooring
Figura 3.4: Catenary effective shifts
• The breaking load Q was calculated from the
Q =
σaπd2
2
• Assuming a safety factor s (in our case 3), we got the catenary value of u from the
formula
Q
swd
=
cosh u
coshu − 1
With this value we have practically defined a completely immersed catenary as we
see in the next steps
– Once found u and assumed the seabed, we can derive the ratio S/H from which
we get the S
S/H =
sinh u
coshu − 1
– - Hence we derive the ratio S − Dt/H from
S − Dt
H
=
sinh u − u
cosh u − 1
(3.4.2)
This ratio is very significant because the catenary values are calculated accor-
ding to it, among others appears exactly the mooring line length Tx value in
dimensionless form that affects this analysis according to the form Tx/wH
– Anyway we can derive Tx/wH from
1
cosh u − 1
Using spredsheet we could therefore obtain an immersed catenary series according to
the chain size and seabed . We have just to figure out which of these is able to perform a
stretching of at least 17.3 meters before the length D0 nullification, and so meet (3.4.1) a
54
63. 3 – Mooring
stretching of 13.7 meters.
As we got from [4]a good collection of values S − Dt/H that correspond to the values
Tx/wH,we imagined how would behavior a catenary from zero stress, and therefore move
it virtually until we come to our points of interest, i.e. Dt − 3.6m.
Figura 3.5: Optimal catenary search
Hence we start from the catenary vertically descending for which D0 = S − H and
consequently the S − Dt/H restituisce valore 1, returns a value of 1, and then we can
examine the catenary between S−Dt
H
= 1 and that one returned in (3.4.2).
In particular, we have to note down the value Tx/wHof each intermediate catenary and
finally we can obtain its corresponding value D from the taut lines progressing through the
D = S − H
S − Dt
H
.
Instead, as regards the loose lines, we derive the catenary corresponding to the value
Dl = D − 13.7m and we note also the value Tx/wH. If among these catenaries there is a
one that meets the pull required in (3.4.1) at the distance D = Dt − 3.6mthen the chosen
catenary in (3.4.2) will meet the regulations criteria. We can see the diagram in the figure.
Alternatively we can change catenary changing the caliber value.
The configurations limits are given in the tab. 3.1
ere there are the performed calculations relating to what has been said earlier about
the catenary; the starting values are given in tab. 3.2:
Then we get the values in tab. 3.3
Here we have the final values interpolated in bold that are showing a sufficient pull to
the shift D in tab. 3.4.
Il peso di ogni linea di ormeggio è di 29.7 tonnellate, con pretensione iniziale di 115
kN. In totale il peso delle sei linee è di 178.6 tonnellate.
In the initial condition, that is before the catenary stretching of 17.3 metres, the weight is
28 tonnes for each line, and 173 tonnes in total.
55
64. 3 – Mooring
Figura 3.6: Taut and loose lines
Configurations limits
d m w σ Q
mm kg/m N/m N/mm2
N
25 13 15 115 124 490 570 481056 559596
50 54 58 458 495 490 570 1924226 2238385
100 215 232 1834 1979 490 570 7696902 8953539
120 310 334 2641 2849 490 570 11083539 12893096
Tabella 3.1: Obtained catenary
d m w σ Q s Q/swd u θ S/d
mm Kg/m N/m N/mm2
N deg
53.0 60.4 570.0 490.0 2162059.8 3.0 12.6 0.412 22.9 4.9
Tabella 3.2: Assumed values
S D 0 S-Dt/H Dt Tx/wH Tx D
m m m kN m
492.8 393 0.1364 479.2 11.644 663.7 475.7
Tabella 3.3: Obtained catenary
S-D/H Tx/wH D Tx D-13.7 S-D/H Tx/wH Tx T tot
m kN m kN kN
0.180 6.561 474.82 374.0 461.1 0.317 2.1744 123.9 497.9
0.170 7.391 475.82 421.3 462.1 0.307 2.0181 115.0 536.3
0.160 8.382 476.82 477.8 463.1 0.297 2.5391 144.7 622.5
Tabella 3.4: Final values
56
66. 4 – Ballasting
Remembering what we said in Chapter about Statics, once known all the weights, we
have to understand how we can ballast the structure. We can use the tab. 1.2where we
are going to get the total movement value that has been suggested, i.e. ∆ = 2536t.
We report here the expression (4.0.1).
∆tot = ∆RNA + ∆structure + ∆bal (4.0.1)
This expression result should be subject to our new weight conditions in order to
calculate how is the ballast mass value. If this is negative we have to change design or try
to make a lighter structure, maybe breaking the beams spans. We will have therefore
∆tot = ∆RNA + ∆3cylinders + ∆RNAS + ∆Beams + ∆Chains − ∆Immersedbeams + ∆bal
In particular we note the term ∆immeserdbeamswhich is basically the hydrostatic thrust
from immersed beams. It would be wrong not to consider it because it was not aforesaid
in the Chapter about Statics (as it did not yet exist); moreover it is not a negligible mass.
Its value is equal to 888 tons returning a hydrostatic pressure of 870kN.From the inverse
formula we get ∆bal = 1048t, i.e. 350 ballast tons for each cylinder.
In order to ballast each cylinder of 350 tons, it have to be filled with water up to a height
of 4.3 meters, at least theoretically, for a diameter of 10 meters; we have to get to a
height of 4.5 meters for 350 ballast tons including the shell, the bulkheads, and the rings
thicknesses.
At this point it would be expected a deck as a “top closure” just on the ballast mass in
order to avoid sloshing stresses. However we choose to achieve just two full volumes until
to the top closure and two others less full in order to be used in a deeper seabed (as the
mooring weight would increase) or choosing maybe to use a lighter RNA. Then we achieve
compartments that reach the third ring height, that is 7.86 m from the bottom , and then
we will fill just two of them until to the top closure. The tween- deck at the closure of the
compartment will replace the ring of shell.
Figura 4.1: Ballasting
Taking always in account to [6] ] with a safety factor of 2, it is sufficient that these
decks are decks without strings with a thickness of 1 cm.
These components weight is therefore 5.7 tons then the new ballast weight according to
the decks is 1031 tons, and 344 per cylinder. Therefore filling just two compartments until
to the top closure the other two have to be filled up to 0.4 meters. Finally, we cannot not
appreciate a structure 3D vision including also the new component;
58
69. Bibliography
[1] J.Jonkman, S.Butterfield, W. Musial, G Scott Definition of a 5-MW reference wind
turbine for off-shore system development, Technical Report NREL/TP-500-38060
February 2009
[2] Det Norske Veritas Stability and watertight integrity, Offshore standard DNV-OS-
C301 April 2011
[3] ABS Buckling and ultimate strength assestment for off-shore structures, April 2004
(updated February 2014)
[4] A.Campanile Strutture off-shore 2012
[5] A.Campanile Tecnologia delle costruzioni navali, Marzo 2006
[6] S.Timoshenko, S.Woinowsky-Krieger Theory of plates and shells
[7] http://www.energoclub.it/
[8] http://www.nextville.it/
[9] http://www.greenstyle.it/
[10] Bollettino n.37 AIOM
61
