OE4680 2013 - Lecture 16 - Discussion Exercise & Exam Questions

8/17/2019 OE4680 2013 - Lecture 16 - Discussion Exercise & Exam Questions

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MSc Offshore & Dredging Engineering

Faculty CEG, Department Hydraulic Engineering

Faculty 3mE, Department Maritime & Transport Technology

120 June, 2013


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Discussion Exercise & Exam Questions

OE4680 Arctic Engineering


Sea Ice conditions

Baidaratskaya Bay is part of the southwest Kara Sea…

The location of the structure is offshore…

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June,

2013 3


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Structural Configuration

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June,

2013 4


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Given Parameters

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June,

2013 5


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Overview Exercise Scoring

Answering all problems correctly, yields a 0,8 bonus to your exam grade.

This bonus is valid for the exam in Q4 (on Monday 24 June 2013, 14:00‐17:00)

and the retake (on Thursday 15 August 2013, 9:00‐12:00).

In total, there were 32 points to be earned, divided among the 4 problems as:

1a. [1] 2a. [2] 3a. [1] 4a. [1]

1b. [2] 2b. [2] 3b. [4] 4b. [8]

2c. [3] 3c. [1] 4c. [2]

2d. [1] 4d. [4]

1. [3] 2. [8] 3. [6] 4. [15]

Thus, every single point equals a (0,8/32 = ) 0,025 bonus.

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June,

2013 6


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Problem Statement 1

In the winter 2012‐2013, the mean daily air temperature at the considered location

in the Kara Sea was below the sea water freezing point from 16 September 2012

until 20 April 2013. The average mean daily air temperature during this period was ‐

16,7 °C.

For the calculation of ice thickness, the Kara and Chukchi Seas have the same site

specific constants; in the Chukchi Sea a total of 4096 freezing degree days yielded an

ice thickness of 2,24 m.

1. For the given weather conditions in the winter 2012‐2013,

a. calculate the number of accumulated freezing degree days in the Kara Sea.

b. determine the maximum undisturbed ice thickness in the Kara Sea

assuming linear

heat

conduction.

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Problem 1a

For the given weather conditions in the winter 2011‐2012:

a. calculate the number of accumulated freezing degree days in the Chukchi Sea.

The number of accumulated degree days is found as:

“Mean daily air temperature at the Chukchi Sea was below the sea water freezing

point from 16 September 2012 until 20 April 2013.”

The period from 16 September 2012 until 20 April 2013 yields a total of:

15+31+30+31+31+28+31+20 = 217 days.

“The average mean daily air temperature during this period was ‐16,7 °C.”

T a is therefore given as ‐16,7 °C.

The sea water salinity is 32, thus the freezing point of the sea water T b is ‐1,76 °C.

And thus:

20 June, 2013 8

a b a b FDD daysC T T avg T T n

16,7 1,76 217 3242 FDDC


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Instead of the undisturbed ice thickness, extreme ice floe

thicknesses should be used for the design loads. In other

words, from this point onwards use the extreme ice floe

thickness value specifically given to your group.

Assume that for extreme ice floe thicknesses in the Chukchi

Sea, the ice temperature at the floe surface is ‐15⁰ C.

2. For the depicted cylindrical structure, i.e. substructure A,

a. determine the ice action for an average sized isolated ice floe for limit force;

b. calculate the design action for ice crushing failure according to ISO19906;

c. give a conservative estimate of the penetration of the structure into an

average sized

isolated

ice

floe

for

a limit

energy

event

and

determine

the

corresponding limit energy ice action;

d. conclude which limiting mechanisms governs the ice action and explain why.

Problem Statement 2

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2 2

2 3 2 3

, 8 8

,

0,025 1,37 21 4,5 10 0,002 1029 3 4,5 10

120,1 147,3 267,4

LF floe

LF floe

F

F MN

Problem 2a

For the depicted cylindrical structure, i.e. substructure A,

a. determine the ice action for an average sized isolated ice floe for limit force;

For an isolated ice floe, where the ice floe is described as an equivalent circular floe

with a diameter Deq , the limit force action can be described as:

From the lecture on ice mechanics, we find that C d,a = 0,025 and C d,w = 0,002.

Furthermore, from table 1 we find that ρa = 1,37 kg/m3

, V a = 21 m/s, ρw = 1029 kg/m3. And V w = 3 m/s.

Additionally table 1 gives the equivalent diameter of a FY level ice floe as 3 to 6

km; an

average

sized

floe

then has an equivalent diameter of 4,5 km.

Thus:

20 June, 2013 11

2 2 2 2

, , ,8 8 LF floe d a a a eq d w w w eqC V D C V D


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Problem 2b

b. calculate the design action for ice crushing failure according to ISO19906;

According to ISO19906, we find the global crushing load through the global ice

pressure due to crushing (eqs. A.8‐20 and A.8‐21) as:

The ice strength coefficient for Arctic area’s is equal to: C R = 2.8 MPa.

m and n are empirical coefficients that depend on the ice thickness, but for the possible thicknesses are always found as: m=‐0.16, n=‐0.3.

The width of the structure is: w = 14 m and h1 is a unit variable: h1 = 1.

With h=2,1 m we thus find:

20 June, 2013 12

1

n m

G G G Rh w F p hw p C h h

1

48,6

n m

G R

h w F C hw MN

h h


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Problem 2c (1)

c. give a conservative

estimate

of

the

penetration

of

the

structure

into

an

average

sized isolated ice floe for a limit energy event and determine the corresponding

limit energy ice action;

The ice action for limit energy is determined from the work ‐energy principal:

Again, we assume an average sized ice floe with a diameter of 4,5 km.

Thus, with ρice = 910 kg/m3 and h = H2 = 2,1 m, the mass of the ice floe becomes:

We assume that F(x) = F G is constant.

The structure is considered to be offshore, so v beg = 0,3 m/s.

Penetration is then:

20 June, 2013 13

2 21 12 2 beg end F x dx p x w x h x dx mv mv

22 3910 4,5 10 2,1 30, 44 4 ice eqm D h Gkg

21 9

221

2 6

1,37 1028,2

48,6 10

beg

G beg

G

mv F x dx F x mv x m

F


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Problems 2c (2) and 2d

c. give a conservative

estimate

of

the

penetration

of

the

structure

into

an

average

sized isolated ice floe for a limit energy event and determine the corresponding

limit energy ice action;

So, under the assumptions made, we find that the penetration: .

As the diameter of the structure is only 14 m, this means that during a limit energy event, the structure gets fully enveloped .

Consequently, the corresponding limit energy ice action must be equal to the limit

stress load:

d. conclude which limiting mechanisms governs the ice action and explain why.

From question a. and b. we find: F LS


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Problem Statement 3

3. Using the ISO19906 provisions,

a. calculate the average ice salinity for design conditions;

b. calculate the corresponding brine volume and total porosity;

c. determine the

flexural

strength

of

the

ice

for

preliminary

design.

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Problem 3a

Using the ISO19906 provisions:

a. calculate the approximate ice salinity for design conditions;

The AVERAGE ice salinity of a growing first ‐year level ice sheet is found according to

the ISO19906 provisions by the following equation:

Clearly all given extreme ice thicknesses are > 0,34 m and thus, substituting the possible values gives a salinity (in ppt) as:

20 June, 2013 16

13,4 17,4 for 0,34

8,0 1,62 for 0,34

h h mS

h h m

1 1,8 : 5,084

8,0 1,62 2 2,1: 4,598

3 2,4 : 4,112

H S

S h H S

H S


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Problem 3b (1)



The corresponding brine volume follows from ISO19906 as:

For the calculation of the brine volume that corresponds to the average salinity S of

an ice sheet, we should also use the average temperature over the ice sheet.

“Assume that for extreme ice floe thicknesses in the Kara Sea, the ice temperature

at the floe surface is ‐15°C.”

In the lecture on ice mechanics, it was explained that the temperature in an ice floe

changes linearly over the height:

at the floe surface the temperature is given as ‐15 °C.

Looking at the heat flux through the ice, the temperature at the bottom of the ice sheet must be equal to the freezing point, i.e. ‐1.76 °C.

And thus we find that:

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49,180,53brineV S

T

15 1,76 2 8,38 avg T T C


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Problem 3b (2)



Substituting the average salinity S and the average temperature T yields:

According to the ice mechanics lecture, the air volume (in ppt) may be

approximated as:

And thus the porosity is found (in ppt) as:

20 June, 2013 18

1: 32,5449,180,53 6,4 2 : 29,43

8,383: 26,32

brine

brine brine

brine

H V V S S H V

H V

9101 1 1 8,49

917,8

bulk sea iceair

particles pure ice i

V T

1: 32, 54 8, 49 41, 03

2 : 29, 43 8, 49 37, 92

3: 26, 32 8, 49 34,81

brine air

H

V V H

H


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Problem 3c


c. determine the flexural strength of the ice for preliminary design; [2]

The flexural strength of the ice is defined in ISO19906 as:

Here, the brine volume should be substituted as the brine volume fraction , thus a

brine volume of 37,92 (in ppt) corresponds to a brine volume fraction 0,03792.

Thus, the flexural strengths corresponding to the different ice thicknesses are:

Note that these values are much higher than what ISO19906 notes reasonable!

20 June, 2013 19

5,881,76 bV f MP ea

5,88

1: 0,609

1,76 2 : 0,642

3: 0,678

b

f

V

f f

f

H MPa

e H MPa

H MPa


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To reduce the design actions, the structure has been redesigned and equipped with a cone.

3. For the depicted conical structure, i.e. substructure

B, and for the specific combination of parameters

specified for your group,

a. determine the diameter of the cone at the still water

level;

b. calculate the total horizontal design load for bending

failure according to ISO19906;

c. suppose that instead of an upward cone, a downward cone with the same

angle would be used; explain which components of the total design bending

load should be recalculated;

d. give an estimate of the total design load reduction by applying a downward

cone with the same waterline diameter as the upward cone.

Problem Statement 4

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Problem 4a

For the depicted conical structure, i.e. substructure B, and for the specific combination of parameters specified for your group:

a. determine the diameter of the cone at the still water level;

The diameter of the cone at the still water line follows from the radius of the top of

the cone, the distance of the top of the cone above water and the cone angle. From simple trigonometry we find the additional radius r add due to the cone as:

Consequently, the diameter of the cone at the still water level is found as:

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,

1: 6,21 4 : 5,03 7 : 4,056

2 : 5,79 5 : 4,69 8 : 3,75tan tan

3: 5,40 6 : 4,36 9 : 3,46

add add add

top c

add add add add

add add add

A r A r A r h

r A r A r A r A A

A r A r A r

,

1: 20,42 4 : 18,06 7 : 16,102 2 : 19,58 5 : 17,38 8 : 15,50

3: 18,80 6 : 16,72 9 : 14,92

c c c

c top c add c c c

c c c

A w A w A ww b r A w A w A w

A w A w A w


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Problem 4b (1)

b. calculate the total horizontal design load for bending failure according to ISO19906;

The total horizontal design load for bending failure is found according to ISO19906

as:

In general the breaking component HB is the main component, which is found as:

Here, the flexural strength f is the result of question 3c.

20 June, 2013 22

1

B P R L T H

B

f c

H H H H H F H

h

: Load required to break the ice blocks against the slope

: Load required to push the ice blocks up the slope: Load required to turn the ice block at the top of the slope

: Load required to push the

B

R

T

P

H

H H

H sheet ice through the rubble

: Load required to lift the ice rubble with the unbroken ice floe L H

0,253

0,255 2 212 1

0,684

sin coscos sin

C

w w B f C

Eh L

gh g v H w L

E


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Problem 4b (2)


Alternatively, the breaking component HB can (and should) be written as:

We find:

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0,252 3

0,255 2with:4 12 10,68

sin cos

cos sin

C C C

w w B f C

Ehw L L gh g v

H E

sin cos 0,8061 0.03, 5 52 : 1,362

cos sin 0,592

C A

0,25

9 3

2

1: 22,68 76,385 102 : 25,46 80,19

12 1029 9,81 1 0,33: 28,14 84,36

C C

C C C

C C

H L m m H L H L m m

H L m m


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Problem 4b (3)


The remaining load components can be rewritten as:

20 June, 2013 24

2

2 tan 11 1

tan 2 tan

sin cos tan sin cos0,5 1 cos 1

cos sin tan tan tan sin

tan 1 1 tan1 0,5 1 tan 1

tan tan tan t

P r i i

i r

R i r i

L r r i

H wh g e

w gh H e h h

H wh h g e

2

an

cos1,5

sin cosT i

c

H wh g


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Problem 4b (4)


The horizontal breaking load HB , as well as the other 4 components are now found

by “simply” substituting the calculated values into the ISO19906‐equations.

For example using H2, C1 & A5, we find:

(i.e. h = 2,1 m; μ = 0,03; α = 52°)

And the total force becomes:

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4,544 0,024 4,023 0,583 0,82110,4

4,5441 10,642 80,19 2,1

B P R L T

H B

f c

H H H H H F MN

H h

4,544

0,024

4,023

0,583

0,821

B

P

R

L

T

H MN

H MN

H MN

H MN

H MN


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Problem 4c (1)

c. suppose that instead of the upward cone, a downward cone with the same angle

would be used; explain which components of the total design bending load

should be recalculated;

If we would critically asses the physics of upward versus downward bending, we can state that:

The flexural strength of the ice is different for upward and downward bending.

The friction coefficient is different above and under water.

The weight of the ice on an upward slope should be replaced by its buoyancy

for a downward slope, i.e. the density should be replaced by the submerged

density . (See lectures on Ice Actions)

Note here that in ISO19906, the flexural strength is calculated as an average flexural

strength and

therefore

cannot

distinguish

between

up

‐/downward

bending!

Furthermore, we have only considered one ice‐structure friction coefficient.

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Problem 4c (2)

c. suppose that instead of the upward cone, a downward cone with the same angle

would be used; explain which components of the total design bending load

should be recalculated;

Thus the only change that remains is: The weight of the ice on an upward slope should be replaced by its buoyancy

for a downward slope, i.e. the ice density should be replaced by the submerged

density . ( ρi → ρw ‐ρi ; See lectures on Ice Actions)

So all components that are a function of the ice density should be recalculated, which

is all components except for the breaking component. (i.e. HP , HR , HL and HT )

Note here that, in principal, the component HT may be neglected completely, as an

ice block will likely turn before the end of the slope when its submerged.

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Problem 4d (1)


cone with the same waterline diameter as we have previously used for the

upward cone.

Looking at

the

equations

for

the

4 ice

components:

we find that we only need to recalculate HL! The rest can be directly factored.

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2

2 tan 11 1

tan 2 tan

sin cos tan sin cos0,5 1 cos 1

cos sin tan tan tan sin

tan 1 1 tan1 0,5 1 tan 1

tan tan tan t

P r i i

i r R i r i

L r r i

H wh g e

w gh H e h h

H wh h g e

2

an

cos1,5

sin cosT i

c

H wh g


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Problem 4d (1)


cone with the same waterline diameter as we have previously used for the

upward cone.

The total horizontal load than becomes:

For our parameter set (H2, C1, A5), we find:

And the total reduction becomes: F H,red = F H,up - F H,down = 5,0 MN

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,,

119

9101

B down P R L down w i H down down

B i

f c

H H H H F

H

h

, ,

4,544 0,003 0,526 0,14910,4 5,4

4,5441

0,642 80,29 2,1

H up H down F MN F MN


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Faculty CEG, Department Hydraulic Engineering

Faculty 3mE, Department Maritime & Transport Technology

3120 June, 2013


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20 June, 2013 32

Exam dates & Configuration

Exam datesExam: Monday 24 June, 14:00 – 17:00, DTC ‐TZ2

Re‐exam: Thursday 15 August, 09:00 – 12:00, tba

The exam

roughly

follows

the

setup

of

the

course

schedule:

General Arctic Engineering: 65‐75 %

Arctic regions, Arctic structures and ice features;

Ice loads and Ice actions;

Ice physics and/or ice mechanics; Ice management and/or Arctic escape, evacuation and rescue;

Capita Selecta.

Dynamics of ice‐structure interaction: 25‐35 %

Frequency lock‐in and ice‐induced vibrations;

Beam/Plate theory and Numerical modelling;

Industry experience;


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1

2

20 June, 2013 33

Arctic regions and ice features

Q: Name the seas these hydrocarbon

fields are situated in and describe the

ice features that you expect to

encounter at the given locations.

1. Barents Sea (Shtokman)

First‐year

level

ice,

First‐year ice ridges, and

(occasionally) Icebergs.

2. (Southern) Kara Sea

First‐year

level

ice,

First‐year ice ridges, and

Multi‐year ice ridges.

At the locations given in the figure below, offshore hydrocarbon fields are being developed or will be developed in the next few years.


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20 June, 2013 34

Arctic concepts

Q:

Discuss the

advantages

(pros)

and

limitations

(cons)

of

the

following

hydrocarbon production concepts for use in the Arctic:

Gravity Based Structure

Arctic Spar

Downward Conical Buoy

Let’s discuss the Downward Conical Buoy (pros & cons):

Good performance in level ice, as ice is forced to fail in bending.

Good rubble clearing due to buoyancy effects on broken ice.

Circular symmetry, therefore no ice‐vaning required.

Requires no offshore topside installation.

Optionally a disconnection system for (risers and) moorings.

Optionally variable draft, allowing use in both summer and winter.

Congested deck

area

Requires extremely heavy mooring system

Poor open water behaviour under extreme weather conditions

The risers and mooring systems may be exposed to broken ice.


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1

2

20 June, 2013 35

Arctic regions/concepts and ice features

Q: Which of the given concepts would you choose for each location and explain why.

1. Barents Sea (Shtokman) GBS: Too deep (320‐340 m)

Downward Conical Buoy: Assuming disconnecta‐bility and a variable draft makes this a viable option.

Arctic Spar: perfect for open water behaviour,

but it can not be disconnected for icebergs.

Least bad: DCB with extra options

Best option: FPSO (disconnection, open water)

2. (Southern) Kara Sea GBS: Viable option especially in the shallow parts

Downward Conical Buoy: Viable option as the

occurring downloads can be withstood, especially with ice management.

Arctic Spar: Too shallow.

Best option: DCB or GBS depending on water depthGBS, Arctic Spar or Downward Conical Buoy?


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20 June, 2013 37

Ice actionsQ: What are the mechanisms that limit the ice load on a structure during ice‐

structure interaction?

Limit Energy

Limit Stress

Limit Force

Q: Explain briefly what is meant by each of these limiting mechanisms.

Limit Energy:

The mechanism that occurs when the action is limited by the (relative) kinetic energy or

momentum of

the

ice

feature.

This

mechanism

is also

referred

to

as

Limit

Momentum.

Limit Stress:

The mechanism that occurs when an ice feature is driven against the structure and the driving forces are insufficient for the ice to fail and envelop the structure.

Limit Force:

The mechanism that occurs when the driving forces working on the ice feature are sufficient for the ice to fail as it interacts with the structure.


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20 June, 2013 38

Ice actions

Q: The ice loads exerted on a sloping structure are different from the ice loads exerted on a vertical structure. Explain on which of the structures the ice floe

exerts the lowest loads, explain why, and name the corresponding failure

modes.

Against sloping

structures

the

ice

fails

through

(ice) bending,

while on a vertical structure the ice fails through (ice) crushing.

The failure of ice through bending depends on the tensile strength of ice,

while crushing depends on the compressive strength of ice.

The tensile strength of ice is much lower then the compressive strength of ice

and therefore bending exerts lower loads on a structure than crushing.

Ergo, the loads will be lowest on sloping structures. Note however that rubble piling up

and/or adfreeze may diminish the advantages of sloping structures

To calculate static ice actions, we usually apply the ISO19906,

but we have extensively discussed this during the first part of this lecture.


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Bernal‐Fowler Rules

Q: The Bernal‐Fowler rules describe the arrangement of water molecules and

hydrogen atoms in the ideal crystalline structure of ice. Give the 4 Bernal‐

Fowler rules.

1. The water molecule is preserved in the ice lattice. Ergo, 1 O‐atom with 2

H‐atoms.

2. Each water molecule is tetrahedrically bonded to 4 neighbour water

molecules.

3. There is only 1 hydrogen atom per oxygen‐oxygen bond.

4. The hydrogen‐atoms are mobile so rules 1‐3 may be satisfied in any

configuration.

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Dynamics of Ice‐Structure Interaction

Q: What are the 3 main types of models that are available to model dynamic

interaction between sea ice and offshore structures? Give a short explanation

of each type of model.

Physically based models

This type of modelling tries to approach reality as much as possible by taking

into ac‐count the fundamental physical (micro‐ )properties of the

phenomenon to be modelled.

Empirical modelsModels based on data.

Phenomenological models

Models that try to mimic the behaviour of a certain phenomenon rather than

looking into the source of this behaviour.

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Dynamics of ISI ‐ exam 31‐08‐2012

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Q: The dynamic interaction between an ice floe and a sloping structure is

described by a loading cycle in 2 alternating phases. Describe this loading cycle

and identify its 2 phases.

1. Upon initial contact of the ice floe with the hull of the sloping structure, the tip of the ice floe is pushed downwards and the ice floe starts bending downwards

up to the point where the ice floe, here modelled as a beam, breaks in bending

at a certain distance from the interaction point at the tip of the ice floe. This is

the first phase commonly described as: “Bending up to failure” .

2. Once a piece of the ice floe (beam) breaks off from the ice floe, this piece (or

pieces of rubble) is pushed down the slope by the remaining ice floe, until the tip

of the remaining ice floe hits the hull of the sloping structure. This is known as

the second phase. Once the tip of the remaining ice floe hits the structure the ice floe is once again applying a direct load to the structure, and we thus

commonly describe this phase as: “Pushing rubble (down) until reloading” .

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Q: Which beam theory would you use to model the interaction between the ice

and the downward conical structure and why?

Euler ‐Bernoulli beam theory, also commonly known as the classical beam

theory.

When modelling the ice as a beam, the beam representing the ice can always

be considered to be long or slender; When beams are long/slender, shear

deformations and rotational inertia may be disregarded as is assumed for the

Euler ‐Bernoulli beam theory.

Additional Note:

When considering short beams, shear deformations and rotational inertia

should be taken into account according to Timoshenko‐Rayleigh beam theory.

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Q: Give the 4 boundary conditions for the interaction between the unfractured ice sheet

and the downward conical buoy, assuming that the downward conical buoy is fixed.

Assuming that the ice sheet is infinitely long, we can find a distance, say L, from the ice‐

structure interaction point where the deflection and the rotation of the ice are negligible.

This yields the first 2 boundary conditions as:

At the contact point between the ice sheet and the downward conical buoy, the boundary

conditions are due to the interaction with the rigid hull of the structure.

The bending moment in the beam follows

from the axial compression force as:

The last boundary condition follows from the

deflection of the beam tip, which is found as:

20 June, 2013 46

0

z x Lu

2

H2

0 2

z

x

u h EI F

x

0

z

x L

u

x

00 sin z xu t v t t


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For any floating structure that is used for drilling, it is important that its rocking motion

(roll/pitch) is minimal, because large rocking motions may damage the drillstring. As a first

check of the maximum roll/pitch motion, we assume that the horizontal and vertical loads

are constants, i.e. time‐independent, and that the damping of the structure in roll/pitch

may be neglected.Q: Give the equation of motion for the roll motion of the downward conical structure and

determine the corresponding natural frequency.

The equation of motion for the structure is found as:

Here, the natural frequency is found as:

20 June, 2013 47

2 V

V H V

r n

F a b J k F a F b F a b

J

9

9

12 100,195

315 10

r

n

k rad s

J


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During the

structures’

first

drilling

operations

in

the

Chukchi

Sea,

it is

observed

that

the

ice

floe velocity in the Chukchi Sea varies between 0,04 and 0,08 m/s and the length of the

pieces of ice that break off against the structure in bending ranges from 10 to 14 m.

Q: Determine the frequency range of ice failure against the conical structure and explain

whether frequency

lock

‐in

may

occur

while

operating

this

structure

in

the

Chukchi

Sea.

The lowest ice‐ failure‐ frequency is found for a combination of the biggest breaking

length and the lowest ice floe velocity:

Accordingly, the highest ice‐ failure‐ frequency is found for a combination of the smallest

breaking length and the highest ice floe velocity:

Previously, we found a natural frequency of 0,195 rad/s, which is 4 times bigger than the highest ice‐ failure‐ frequency. Due to the big difference, we may expect that the ice‐

failure‐ frequency and the structures’ natural rolling frequency will not synchronize.

Thus, we do not expect frequency lock ‐in to occur here.

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min

max

2 2 0,040,018

14

low

vrad s

max

min

2 2 0,080,050

10

high

vrad s


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Q: Assuming the

initial

conditions

, derive

the

expression

for

the

rotation .

The equation to solve is:

This yields:

Differentiation to time gives:Substitution into the initial conditions then yields:

Finally giving:

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0 0 0 t

2 V

n F a b

J

V V

2 Vsin cos

sin cos

P

n r n n

r

H n n

F a b F a bt F a b J k t A t B t

k t A t B t

cos sin n n n nt A t B t

V V V

0 cos 0 sin 0 0 0

0 sin 0 cos 0 0

n n n

r r r

A B A A

F a b F a b F a b A B B B

k k k

V 1 cos

nr

F a bt t

k


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For the design of this downward conical structure the maximum allowed roll/pitch

angle was found to be .

Q: Determine the vertical force that corresponds with the maximum allowed

roll/pitch angle.

Max rotation if:

The maximum force allowed is thus:

Note here that any angles should be rewritten to radians.

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V Vmax

max

2cos 1 1 cos

n n

r r

F a b F a bt t

k k

3

maxV,max

12 10 0,06118,82

2 2 36 1,39 4

r k F MN a b


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20 June, 2013 51

Good luck with the examon Monday 24 June

And enjoy the summer holiday!

Documents

OE4680 2013 - Lecture 16 - Discussion Exercise & Exam Questions