NTNU Faculty of Engineering Science and Technology Department of Marine Technology TMR 4195 DESIGN OF OFFSHORE STRUCTURES

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1 NTNU Faculty of Engineering Science and Technology Department of Marine Technology EXERCISE 4 TMR 495 DESIGN OF OFFSHORE STRUCTURES Distr. Date: 9 th Feb 4 Sign: Q. Chen Mandatory Exercise This exercise is mandatory and will count 5% to the final mark. The solutions must be delivered to Marinne Kjølås s office E. before 3. p.m. Thursday 5 th March. 4. The exercise should be marked with the students registration number only. Please also visit the website of this course to download corresponding reference. It is encouraged to answer all the questions in English, but not mandatory. Questions regarding the problem text or printing errors will be answered in C.69, or by calling or qiaofeng@marin.ntnu.no PROBLEM Fig. illustrates a production plant consisting of a floating platform and an articulated tower, linked by a hinged connection. The floater is a converted drilling rig. The dimensions of pontoons and columns are determined by requirements to floatability, stability and motions. The braces include horizontal ones shown in Fig. b and vertical inclined braces between the columns. The braces have a diameter of m. Risers made of steel tubes are supported along the articulated tower and over the bridge to the floating body. At the hinge joints, the steel tubes are replaced by flexible tubes. In addition, the bridge is used as the support of pipelines as well as a foot-bridge. The bridge is designed as a truss-work, with hinges at the ends, as illustrated in Fig. c. a) Assume that the bridge is simultaneously subjected to the following characteristic loads: evenly distributed, vertically acting dead and live loads q v = kn/m evenly distributed, horizontally acting wind loads q h = 5 kn/m axial force with return period of years (see Fig. c) due to sealoads on the floating body and the tower, N = 5 MN Determine maximal forces in horizontal members of the bridge. Hint: Consider the truss-work bridge as a simply supported beam and determine the moment in the beam. Assume that this bending moment is carried by a force couple in the horizontal members and determined the axial force accordingly. Note that the axial force due to sealoads also cause a bending moment.

2 b) Determine the dimensions of a lower horizontal member for the following characteristic axial loads in the ULS condition: dead loads and live loads (due to q v ) 9 kn (tension) wind (due to q h ) 9 kn (compression) sealoads (due to N) 5.3 MN (compression) Use steel with yield strength diameter of 45 mm for the horizontal members. σ = 4N / mm and ultimate strength = u 37 N / mm σ and a Dimensioning should be carried out according to NORSOK Standard N-4 Steel Structures Section 6.3 (Download from the course website). See given data in the end of the problem text. State the reasons for the choice of buckling length. Just indicate new assumption of the plate thickness if the dimensions do not fulfil ULS criteria after first iteration, it s not necessary to check the criteria again. Hint: Assume tubular horizontal members with a diameter of 45mm and plate thickness of mm. Then calculate the ultimate strength and introduce the material factor γ m. Check whether the design strength R d = R C /γ m (where R d = f u.a) is grater than the design load effect S d = γ P/L S P/L(c) + γ E S E(c) where S P/L(c) = load effect (axial force) due to permanent and live loads S E(c) = load effect (axial force) due to wind and sealoads. (c) = refer to characteristic load. c) Assume that the nominal stress amplitude in the braces is determined based on the design wave approach. Assume that the stress amplitude which refers to a years return period (N = 8 stress cycles) is 7 MPa for a horizontal brace. Assume that anodes, gangway etc. are welded to the horizontal brace by means of fillet welds. The relevant SN-curve is N= KS -m, based on a nominal stress range S and logk =.455 and m = 3.. Is the fatigue criterion fulfilled? (See given information). Choose the values of other parameters needed. If the fatigue criterion is not satisfied, how can it possibly be fulfilled for an existing platform?

3 Figure Floating platform and articulated tower connected with bridge Seen from side view Seen from top view d) 3D drawing of bridge 3

4 Figure Converted drilling platform a) Perspective b) Arrangement of horizontal brace Given:. Inertia moment of circular cross section: R t I = π R t t + R 3 I. Radius of inertia: i = ( =.355 D for a thin - walled circular cylinder) A 3. Check of local buckling of axially loaded circular tube can be neglected when D/ t < E /(9σ ) where D is the diameter t is the thickness E is the elasticity modules, E =. 5 N/mm σ is the yield strength 4. Elastic buckling stress of simply-supported tube with a length of l : π σ = E E λ where λ = l/ i 4

5 5. The fatigue damage, D with a Weibull distributed stress range and SN-curve: N = KS -m m N s D = Γ( m / β + ) m / β K ( lnn ) where K, m are constants in the SN-curve s corresponds to P [ S s ] = / N β shape factor of the Weibull distribution Gamma function, Γ(a) a Γ(a)

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7 PROBLEM The probability of fatigue during a time period, T, is given by [ ] p f = P f D () The fatigue resistance may be described by the SN-curve. The stress range is assumed to be Weibull-distributed. The cumulative fatigue damage is predicted by the Miner-Palmgren hypothesis. Based on these assumptions, the cumulative damage during N stress cycles is given by D = N K m S m Γ + S = a m m ( ln N ) ξ K ξ Where: f - cumulative damage at failure, i.e. D at failure. N - the number of stress cycles during the time period T. S - the level of stress range for which the probability of exceedance is / N. K, m - parameters of the SN-curve. ξ - shape parameter in the Weibull-distribution Γ - the gamma function a) Show that Eq. is equivalent with Eq. 3 when f, S and K are stochastic variables and a is a constant. [ ln mln S + ln K ln ] Pf = P f a (3) Hint: See approach for = P [ R S ] P f when R and S are lognormal random variables. b) Assume that m = 3 and ξ =. We will consider fatigue during a period of years. The number of stress cycles, N, is assumed to be 8. Based on these values, we get a = The stochastic variables f, S and K are assumed to be independent. The distributions and parameters are given in Table. Calculate the probability of failure for the cases A and B in this table. In case B, an uncertainty of f is taken into account. What does this uncertainty represent? And how significant is the contribution of this uncertainty? Hints: The logarithm of a lognormal variable has a normal distribution, i.e. = lnx is Gaussian distributed if X is lognormal. The sum of Gaussian variables is also Gaussian. The mean value and the variance of = lnx when X is lognormal, is given by µ ln µ.5σ σ = ln + V. = and ( ) X X () 7

8 CASE VARIABLE MEAN VALUE C.O.V. DISTRIBUTION A B TABLE f S K f S K E E+3 Description of the stochastic variables c) According to design codes, the fatigue limit state requirement is given by Deterministic D = k c i= ni N i d Where D c means characteristic value of D. (4) The characteristic value of K is defined by ln KC = µ σ (5) lnk lnk Calculate the characteristic value of K based on the data given in the Table. Show that the characteristic value of D for the data in Table is.. The design code requires different values of d dependent on the consequences of failure and access for inspection and repair. Calculate the implied probability of failure if d =.33 and d =.. Use the data for case B in the above table. Assume that you refer to the same weld joint (i.e. K c is given). Comment briefly on the results in item b) and c). d) In the following item the fatigue design criterion of NPD is to be evaluated by a simplified assessment of the probability of failure of the platform as a whole system. Let the system failure following a fatigue failure be written as P FSS = P P (6) FSS FF FF Where P FSS - System failure probability P FF - Fatigue failure probability P FSS FF - System failure conditioned on fatigue failure The purpose is to assess the implication of the code requirements of NPD: P FF can then be determined by assuming that Eq. (4) is exactly fulfilled for the relevant given d (as already done in item c)). 8

9 P FSS FF can be determined based on the assumption made about consequence of failure when assuming d, i.e. when no account of inspection is made. d =. if the system is statically determinate, i.e. failure of the system occurs immediately after fatigue failure. d =.33 if the ALS criterion of NPD is fulfilled when one brace (joint) is assumed to have fatigue failure. To estimate P FSS FF the following approach can be used. Consider a given tubular joint in different jacket structures (A and B). Assume that fatigue failure of a joint in jacket A implies failure of the jacket, while after fatigue failure of a joint, jacket B will survive a -year wave, with safety factors γ S = γ R =.. α Let the load effect S ( S = k ψ H ) and strength R be lognormally distributed, and µ R = BR RC. Assume further that B., V =., V =. 5, and that the factor α is., µ =., R = R S H = V H =. H = V =.5, V, H. 75, H H. 9 (where H n is the expected ψ maximum wave height in n years). = Find the conditional probability of failure of jacket B for and years, respectively, after fatigue failure of a joint. Comment on the fatigue design criteria of NPD in view of the results in item c) and d). ψ Hints: To calculate P FSS FF you can consider the damaged jacket as a component with resistance R and load effect S and use the result presented in lecture notes / transparencies. 9