examination paper for tmr4195 design of offshore · pdf filenorsok code. department of marine...

Department of Marine Technology

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Examination paper for TMR4195 Design of offshore structures Academic contact during examination: Jørgen Amdahl Phone: 73595544 / 95745663 Examination date: 16.05.2013 Examination time (from-to): 09.00-13.00 Permitted examination support material: (D) – Neither printed nor handwritten

notes are permitted. Approved, simple calculator is permitted.

Other information: Language: English Number of pages: 19 Number of pages enclosed: 0

Checked by:

____________________________

Date Signature


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Problem 1 The operator of Polar Bear Oil Field, Student Oil, has made a new discovery close to Polar Bear Field. The new discovery is named Polar Fox Oil Field. The water depth at Polar Fox Field is 160m. Metocean conditions for Polar Fox are the same as those provided for Polar Bear Oil Field given in the end of the exam text. In order to learn more about the size of the new field a semi-submersible drilling rig is hired for drilling a couple of test wells. The drilling process has to be stopped if the heave motion of the platform exceeds the capacity of the heave compensator. One of the concerns of Student Oil is that the expected percentage down time due to the capacity of the heave compensator may limit the efficiency of the drilling program. In the contract with the owner of the drilling rig, Gold Fish Drilling, it is specified that that the heave compensator can account for platform heave of +/- d [m]. As a member of the Polar Fox project Team in Student Oil, your job is to consider the expected down time of the rig when operating at Polar Fox Oil Field.

a) Denote the wave spectrum by SHH(f). The transfer function between surface elevation process and heave motion is denoted HHX(f). These quantities can be considered known. Explain what you understand by the heave transfer function. Show how you can determine the distribution function for the 3-hour maximum heave motion of the platform. No numerical calculations are to be done, but you shall select distribution function for heave maxima, explain why you select this probability distribution and show how you can determine possible distribution parameters.

b) Student Oil criteria for accepting that drilling can take place in a given sea state is that the probability of exceeding the heave compensator capacity in a 3-hour sea state is less than p. Show how you can calculate the expected annual down time. You can assume that an all year scatter diagram of hs and tp is available.

An additional concern of Student Oil is the annual probability of mooring line failure in case of total loss of thrust. The owner of the rig, Gold Fish Drilling, has shown that the mooring lines of semi-submersible fulfill all rule requirements for drilling at Polar Fox Oil Field. To investigate robustness of the mooring system, Student Oil has introduced an additional company specific requirement. The requirement is that the probability of exceeding mooring line capacity in the unfavourable 10-2 annual probability storm of 3 hours duration after total loss of thrust shall not exceed 0.1.

c) The critical part of the most loaded mooring line is the top part of the mooring line, often referred to as the platform chain. A typical length of the active load bearing part


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of the platform chain is about 150m. For the semi-submersible of consideration this corresponds to 400 individual chain links. Based on testing of a large number of individual chain links from a platform chain with similar history of use as those used on the drilling platform, the distribution function of breaking load for an arbitrary individual chain is estimated and denoted FX(x). This is considered known in the following. Establish the expression for the breaking load of the platform chain which is exceeded by 95 out of 100 similar platform chains.

d) Based on a request of Student Oil, Gold Fish Drilling has carried out a model test experiment in order to estimate the distribution function of the 3-hour maximum mooring line load, Y, in the unfavourable 10-2 annual probability storm. No thrusters are applied in order to represent a case with complete loss of thrust. 6 repeats of this storm were done. Based on these six 3-hour tests, the expected 3-hour maximum is found to be ρ and the standard deviation is estimated to be 0.2ρ. Select a reasonable probabilistic model for Y. Explain why it is selected and estimate the parameters (or just the equations if the system of equations cannot easily be solved). Let us assume that the breaking load determined in c) is 1.3ρ and calculate the failure probability. If distribution parameters are not estimated, establish the expression for the target exceedance probability. Do you have any comments to the estimated exceedance probability?

Problem 2 In parallel with the planning of the test wells, Student Oil is also working with the development of the field plans for Polar Fox Field. It is decided that the field will be developed by two jacket platforms. You are Student Oils expert on environmental conditions, hydrodynamic loading and structural dynamics and one day you are on short notice called to a meeting with the engineering manager for the field development. There you are presented for the following questions from the manager:

a) The long term distribution of 3-hour maximum wave height is given by a 2-parameter Weibull model :

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=λ

γhhF

hH exp1)(3

, γ and λ are known.

For large waves, the corresponding base shear, B, can be approximated by B = a(H + bU)2.2 , H is individual wave height and U is depth averaged current. Can you establish the distribution function for B3h when U is set to zero? Assuming the long term distribution for B3h is determined, FB3h(b), how can we estimate long term extremes corresponding to an annual exceedance probability of q?

b) N-003 suggests that we shall use cD = 0.65 above +2m above still water level and cD = 1.05 below. If we use a Gaussian sea in the time domain simulations, N-003 suggests


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that we should use cD = 1.15 over the whole water column. Our consultant has used cD = 1.05 over the whole range both for the design wave calculations and the time domain simulations. He finds that there is some inconsistency between the quasi-static extremes obtained by the Stokes 5th design wave and the time domain simulations. The Stokes 5th value he obtains seem to correspond to a very large percentile in the distribution function of the 3-hour extreme value based on the time domain simulations. He claims that scaling the values by 1.15/1.05 is not enough. This is of some concern to us! Can you help us? Can you see any reason for this mismatch assuming that it is correct to use 1.05 for all depths. We don’t expect you to answer this precisely, but we are interested in hearing your views? We think you have some experience with this. I am on my way to a meeting in a couple of hours where I need your views so you should restrict your discussion to 3- 4 brief bullet points.


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Problem 3

Figure 1 Truss-work jack-up platform Figure 1 shows a jack-up platform that consists of a deck and a truss tower. The tower is made up of four parallels columns at a distance of 30 m, connected by braces. At the bottom the tower is connected to a foundation base. All columns and braces are circular, with yield strength of 420 MPa. The columns have diameter D = 1.8 m and thickness t = 65 mm. The functional topside loads are 200 MN. The 100 year environmental load consists of horizontal wind load of 11 MN acting z = 150 m above the foundation, and horizontal wave and current load resultant of 24 MN acting z = 90 m above the base. The tower weight is supposed to be balanced by buoyancy. The structure is considered to be a truss. The structure and loading are assumed to be symmetric.

a) Determine the global force resultants (i.e. global overturning moment, global shear and global axial force) for the tower at the foundation base, and the associated design axial forces in the columns and the braces for the lower storey according to the NORSOK code.


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b) Check the dimensions of the columns according to NORSOK N-004.

c) Discuss the buckling length factor that should be used for the diagonal braces. Propose a simple modification of the truss work to reduce the buckling length factor.

Assume that you have run a pushover analysis of the platform using USFOS, and you are asked: “What is the buckling strength of a (given) diagonal brace?” Explain in a few words how you will determine the buckling strength of the brace using the results of the USFOS analysis?

d) Assume that one of the diagonal braces in the lower storey has fractured due to impact

from a dropped object with an annual probability of occurrence of 10-4. Determine the axial forces in the braces for the lower storey when the tower is subjected to the 100-year wave load with one brace fractured.

Compare these axial forces with those found in pt. a). Will the ULS or ALS criterion be governing the design of the braces for this case? Discuss whether other truss-work layouts could improve the strength in damaged condition for the given impact damage.

e) A fatigue check shall be carried out for a K-joint 78 m above the foundation for 20

year design life. The nominal stress amplitude in the brace corresponding to 100 year response (N0=108.7 cycles) is σnom(0) = 50 MPa and the shape parameter in the Weibull distribution for the stress cycles is β = 1.0. The stress concentration factor is SCF =σ weld /σ nom = 2.0 . The SN-curve for the weld has the constants K=1012.15 and m=3.0. Is the fatigue design requirement complied with?

Problem 4 The platform in Problem 3 is subjected to impact from a supply vessel of 7500 tons displacement drifting sideways with a speed of 2.2 m/s. It hits the leg midway between two joints.

a) Use a simple calculation model to estimate the damage to the platform leg. The ship has a resistance to indentation which starts at 10 MN for zero indentation, increasing linearly to 40 MN for 1.0 m indentation. Introduce the assumptions you otherwise consider necessary. The plastic section modulus of circular sections may be calculated as Z=D2t

b) Discuss what capacity checks that need to be conducted to ensure that the calculation model assumed in pt. a) shall be valid.


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c) Discuss how the duration of the impact relative to the fundamental period of vibration of the platform will influence how the collision force is resisted as global shear force for the truss tower.


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Provided text book information Waves:

Wave propagating in positive x-direction: )sin(),( 0 kxttx −= ωξξ

Corresponding velocity potential, deep water: )cos(),,( 0 kxteg

tzx kz −= ωωξφ

Dynamic pressure, deep water: )sin(),,( 0 kxtegtzxp kz

d −= ωξρ

Horizontal particle velocity: x

tzxu∂∂= φ),,(

Circular frequency: Tπω 2= , T = wave period

Wave number: ,2λπ=k λ = wave length

Dispersion equation, deep water: gk=2ω Dynamics: Equation of motion: )(tfkxxcxm =++ )(tmkcI =++ θθθ θθ

Steady state solution for harmonic load, f(t) = f0 sinωt: )sin()( 0 φω −= tDAFkf

tx p

where

2/1

2

0

22

0

21

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟

⎠

⎞⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

ωωλ

ωω

DAF


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⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

= 2

0

0

1

2arctan

ωω

ωωλ

φ

Natural circular frequency: mk=0ω

Unit impulse response: { })(sin

exp)( 0 t

Mt

th dd

ωω

λω−=

Convolution integral: ∫ −=t

dthFtx0

)()()( τττ

Dynamic amplification for impulse type loading:

Wind loads:

Wind pressure: 2

21

cap uCq ρ= , aρ is mass density of air (= 1.226kg/m3 for dry

air at 15oC) , uc is the characteristic wind speed, Cp is a pressure coefficient which can be taken equal to 1 for horizontal and


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vertical surfaces.

Wave loads:

Slamming loads: AucF s2

21 ρ=

Morison load: uDcuuDczf MD 4

||21)(

2

πρρ +=

Wave loading regimes:

Difraction analysis Morisons equation

Velocity potential:

∑=

Φ+Φ+Φ=Φ6

1kkdi

Force per unit length

udcuudcf MD 24

||21 ρπρ +=

Drag termdominates

Mass termdominates

LinearizingDrag term ?

Response statistics: Response spectrum: )(|)(|)( 2 fsfHfsXX ΞΞΞΞ=

Variance: ∫=

fXXX dffs )(2σ


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Zero-up-crossing frequency: 0

20

X

X

mm

=+ν ; ∫=f

XXk

Xk dffsfm )(

Rayleigh distribution: ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

2

21exp1)(

XX

xxFσ

Most probable largest maximum out of n maxima from a Rayleigh distribution:

n

xF nX1)~(1 =− è nx Xn ln2~ σ=

Gumbel distribution: ⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧ −−−=

βαxxFX expexp)(

[ ] βα 57722.0+=XE [ ] β28255.1=XSTD

Weibull distribution: ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=λ

ρxxFX exp1)(

Log-normal distribution: fX (x) =12πκ x

exp − 12ln x−θκ

⎡⎣⎢

⎤⎦⎥2⎧

⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Where: [ ]pTE ln=θ and [ ]pTVar ln2 =κ

Exponential distribution: ⎭⎬⎫

⎩⎨⎧

⎟⎠⎞⎜

⎝⎛−−=λxxFX exp1)( , [ ] λ=XE


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Selected tables from NORSOK N-001 and N-003 N-001

In the accidental damage limit state for progressive collapse, the load coefficients shall be 1.0 for all loads. The material coefficient shall be 1.15 in the ULS check and 1.0 in the ALS check.

Fatigue damage in a period with N cycles when the stress range has a Weibull distribution and

the SN-curve is given by N= KS−m : D = NS0m

K(lnN0 )m/β Γ(m / β +1) ,

K and m are constants of the SN-curve S0 corresponds to P S ≥ So[ ] =1/ N0 (if S0 is the 100 year value, N0=108.7) β is the shape factor of the Weibull distribution Gamma function Γ(a): a 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Γ(a) 1.00 1.33 2.00 3.32 6.00 11.6 24.0


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From N-003


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Summary from Metocean Design Basis for Polar Bear Oil Field

Table 1 Extreme wind speeds (m/s) with 1, 10, 100 and 10000 year return period for different averaging time intervals 10 m above MSL.

Return period [years] All year 1h 10min 1min

1 32.0 35.4 39.7 10 36.0 40.0 45.2 100 40.0 44.8 51.0

10000 46.0 52.0 59.8 Table 2 The normalized logarithmic design wind profile, u(z,t)/U0 for Uo = 40 m/s.

Height (m) Averaging time 1h 10min 1min 15s 3s

10 1.00 1.12 1.27 1.37 1.47 20 1.11 1.22 1.37 1.45 1.56 40 1.21 1.32 1.45 1.54 1.63 80 1.32 1.42 1.55 1.63 1.72

150 1.41 1.50 1.62 1.70 1.78 Table 3 Marginal omni directional extremes for the significant wave height, Hs, and corresponding values for the spectral peak period, Tp. Sea state duration: 3 hours.

Annual probability of exceedance

Extreme sea states Hs (m) Tp (s) 90% range of Tp (s)

0.63 11.0 14.2 12.1 - 16.5 10-1 13.0 15.1 13.1 - 17.3 10-2 14.9 16.0 14.0 - 18.2 10-4 18.2 17.5 15.5 - 19.7

Table 4 Design waves, Stokes' 5th order profile.

Annual probability of exceedance)

Wave height

(m)

Height above

MSL (m)

Wave period (s) Mean value

90% interval

0.63 22.0 12.3 12.7 10.9 14.8 10-1 25.5 14.6 13.6 11.8 15.6 10-2 29.0 16.5 14.4 12.6 16.3 10-4 36.5 21.1 15.8 13.9 17.7

Table 5 Omni-directional extremes for the 10 minutes mean current speed (cm/s) versus depth for

1-, 10- and 100-years return period.

Depth (m) 1-year return period 10-year return period 100-year return period Surface 106 117 126

30 92 102 110


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75 85 98 110 3 m above seabed 66 79 85

Table 6 Water levels with return periods of 100 and 10000 years.

Return period (year) 100 10 000

Tidal amplitude (m) 1.0 0.0 Storm surge (m) 0.9 1.1 Height of wave crest (m) 17.6 22.4 Extreme water level above MSL (m) 19.5 23.5

Figure 1 Contour lines in the Hs – Tp plane for annual probabilities of 0.63, 10-1, 10-2 and 10-4

(Return periods of 1, 10, 100 and 10 000 years). Duration of sea state is 3 hours.

From NORSOK N-004


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NORSOK standard N-004 Rev. 2, October 2004

NORSOK standard Page 19 of 287

6.3.2 Axial tension

Tubular members subjected to axial tensile loads should be designed to satisfy the following condition:

M

yRdt,Sd

fANN

J d

(6.1)

where NSd = design axial force (tension positive) fy = characteristic yield strength A = cross sectional area JM = 1.15.

6.3.3 Axial compression

Tubular members subjected to axial compressive loads should be designed to satisfy the following condition:

M

cRdc,Sd

AfNN

J d

(6.2)

where

NSd = design axial force (compression positive) fc = characteristic axial compressive strength JM = see 6.3.7

In the absence of hydrostatic pressure the characteristic axial compressive strength for tubular members shall be the smaller of the in-plane or out-of-plane buckling strength determined from the following equations:

1.34for f]28.00.1[f y2

c d� OO (6.3)

1.34for f9.0f y2c ! OO

(6.4)

Ef

iʌk

ff cl

E

cl l O

(6.5)

where

fcl = characteristic local buckling strength CO = column slenderness parameter

fE = smaller Euler buckling strength in y or z direction E = Young’s modulus of elasticity, 2.1�105 MPa k = effective length factor, see 6.3.8.2 l = longer unbraced length in y or z direction i = radius of gyration



6.3.2 Axial tension

Tubular members subjected to axial tensile loads should be designed to satisfy the following condition:

M

yRdt,Sd

fANN

J d

(6.1)

where NSd = design axial force (tension positive) fy = characteristic yield strength A = cross sectional area JM = 1.15.

6.3.3 Axial compression

Tubular members subjected to axial compressive loads should be designed to satisfy the following condition:

M

cRdc,Sd

AfNN

J d

(6.2)

where

NSd = design axial force (compression positive) fc = characteristic axial compressive strength JM = see 6.3.7

In the absence of hydrostatic pressure the characteristic axial compressive strength for tubular members shall be the smaller of the in-plane or out-of-plane buckling strength determined from the following equations:

1.34for f]28.00.1[f y2

c d� OO (6.3)

1.34for f9.0f y2c ! OO

(6.4)

Ef

iʌk

ff cl

E

cl l O

(6.5)

where

fcl = characteristic local buckling strength CO = column slenderness parameter

fE = smaller Euler buckling strength in y or z direction E = Young’s modulus of elasticity, 2.1�105 MPa k = effective length factor, see 6.3.8.2 l = longer unbraced length in y or z direction i = radius of gyration

Assume: γ =1.15 and fcl=fy


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2

z

2

Ez

ikEAʌN

»¼º

«¬ª

l

(6.30)

k in Equation (6.29) and Equation (6.30) relate to buckling in the y and z directions, respectively. These factors can be determined using a rational analysis that includes joint flexibility and side-sway. In lieu of such a rational analysis, values of effective length factors, k, and moment reduction factors, Cm, may be taken from Table 6-2. All lengths are measured centreline to centreline.

Table 6-2 Effective length and moment reduction factors for member strength checking Structural element k Cm

(1)

Superstructure legs - Braced 1.0 (a) - Portal (unbraced) k(2) (a) Jacket legs and piling - Grouted composite section 1.0 (c) - Ungrouted jacket legs 1.0 (c) - Ungrouted piling between shim points 1.0 (b) Jacket braces - Primary diagonals and horizontals 0.7 (b) or (c) - K-braces(3) 0.7 (c) - Longer segment length of X-braces(3) 0.8 (c) Secondary horizontals 0.7 (c) Notes: 1. Cm values for the cases defined in Table 6-2 are as follows:

(a) 0.85 (b) for members with no transverse loading, Cm = 0.6 - 0.4 M1,Sd/M2,Sd, where M1,Sd/M2,Sd is the ratio of smaller to larger moments at the ends of that portion of the member unbraced in the plane of bending under consideration. M1,Sd/M2,Sd is positive when the number is bent in reverse curvature, negative when bent in single curvature. (c) for members with transverse loading, Cm = 1.0 - 0.4 NSd/NE, or 0.85, whichever is less, and NE = NEy or NEz as appropriate.

2. Use effective length alignment chart in Clause 12. 3. At least one pair of members framing into the a K- or X-joint shall be in tension if the joint is not braced out-of-

plane. For X-braces, when all members are in compression, the k-factor should be determined using the procedures given in Clause 12.

4. The effective length and Cm factors given in Table 6-2 do not apply to cantilever members and the member ends

are assumed to be rotationally restrained in both planes of bending.

6.3.8.3 Interaction shear and bending moment Tubular members subjected to beam shear force and bending moment should be designed to satisfy the following condition provided that the direction of the shear force and the moment vectors are orthogonal within r 20q:

Rd

Sd

Rd

Sd

VV

4.1MM

�d for 4.0VV

Rd

Sd t (6.31)

Fundamental period of vibration: KMk2πT ml=

M = total mass


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K = characteristic stiffness

Impulse in impulsive domain: ( )∫= maxy

0dyyR2I Mklm

Dynamic response for a SDOF system

0.1

1

10

100

0.1 1 10

td/T

y max

/yel

= 1.1

Rel/Fmax= 0.8

= 1.0=

= 0.9

= 1.2= 1.5

=0.1= 0.7

= 0.6= 0.5Rel/Fmax=0.05 = 0.3

yel y

R

Rel

FFmax

td0.50td

k1

k3 = 0.5k1 =0.2k1 =0.1k1k3 = 0k3 = 0.1k1

k3 = 0.2k1

k3 = 0.5k1


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Transformation factors for beams with various boundary - and load conditions.

Load case

Resistance domain

Load Factor

kl

Mass factor km

Load-mass factor

klm

Maximum resistance

Rel

Characteristic stiffness

K

Dynamic reaction V Concen-

trated mass

Uniform mass

Concen-trated mass

Uniform mass

Elastic

Plastic bending

Plastic

membrane

0.64

0.50

0.50

0.50

0.33

0.33

0.78

0.66

0.66

8MLp

8MLp

3845 3EIL

0

4NLP

0 39 011. .R F+

038 012. .R Fel +

LyN maxP2

Elastic

Plastic bending

Plastic

membrane

1.0

1.0

1.0

1.0

1.0

1.0

0.49

0.33

0.33

1.0

1.0

1.0

0.49

0.33

0.33

4MLp

4MLp

48

3EIL

0

4NLP

0 78 0 28. .R F−

075 025. .R Fel −

LyN maxP2

Elastic

Plastic bending

Plastic

membrane

0.87

1.0

1.0

0.76

1.0

1.0

0.52

0.56

0.56

0.87

1.0

1.0

0.60

0.56

0.56

6MLp

6MLp

56 4

3. EIL

0

6NLP

0525 0 025. .R F−

052 002. .R Fel −

LyN maxP3

Elastic

Elasto-plastic

bending

Plastic bending

Plastic

membrane

0.53

0.64

0.50

0.50

0.41

0.50

0.33

0.33

0.77

0.78

0.66

0.66

12MLps

( )8 M M

Lps Pm+

( )8 M M

Lps Pm+

384

3EIL

3845 3EIL

3073EIL

⎛⎝⎜

⎞⎠⎟

0

4NLP

0 36 014. .R F+

039 011. .R Fel +

038 012. .R Fel +

LyN maxp2

Elastic

Plastic bending

Plastic

membrane

1.0

1.0

1.0

1.0

1.0

1.0

0.37

0.33

0.33

1.0

1.0

1.0

0.37

0.33

0.33

( )4 M M

Lps Pm+

( )4 M M

Lps Pm+

192

3EIL

0

4NLP

0 71 0 21. .R F−

075 025. .R Fel −

LyN maxP2

Mps= Plastic bending moment at support, Mpm = Plastic bending moment at mid-section

F=pL

L

L/2

F

L/2

L/3 L/3 L/3

F/2 F/2

F=pL

L

F

L/2 L/2

examination paper for tmr4195 design of offshore · pdf filenorsok code. department of marine...

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