Exam 2012: Solution SH problems
Problem 1
a) * Principles regarding structural design: N-001
Actions and action effects: N-003
Code check and utilization: N-004
* For 80m depth we do expect the natural period of jacket to be reasonably small, say less than 2s,
suggesting that dynamics can be skipped or at least estimated in a simplified way. The recommended
approach would therefore be the design wave method.
* Governing combination of environmental process is:
10-2 – annual probability wind, 10-2 annual probability wave, 10-1 – annual probability current and 10-2
– annual probability sea level (from N-003, Table 4. Page 16 in exam).
Wind:
Assuming wave induced response is governing, we could use the 1-minute mean speed
corresponding to 10-2 annual exceedance probability (3s gust wind will also be accepted):
51m/s at 10m above sea level (Table 1 and 2 in Metocean Report, page 17 of exam).
Waves:
Design wave method requires the 10-2 – annual probability individual wave height and a 90% band for
the associated wave period, i.e. Table 4, Metocean report, page 17:
h0.01 = 29m and <90%> band for period: 14 – 18.2s
Current:
10-1 annual surface current speed: 117cm/s (Table 5, metocean report, page 17/18 of exam)
Surface level:
Ttidal amplitude + 10-2 – annual probability storm surge: 1 + 0.9m = 1.9m. (Table 6, Metocean report,
page 18 of exam).
b) * We need the load history for the whole period, we need boundary conditions at the start of the
simulation, we have to introduce some assumption regarding the change of acceleration from one
time step to next. In order to make our assumptions accurate regarding acceleration, length of time
step should be small as compared the variation of the response time history .
* In order to obtain a 3 hour simulation of good quality simulation, simulation must in principle be
somewhat longer. One should skip the first part of simulation due to resonant effects that may occur
at start up. If a standard FFT with equidistant frequency components are used, the frequency
resolution must be 1/T if T is total required length of simulation.
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c) * The property of a probability paper for given distribution is that if this distribution is plotted in the
paper, it will appear as a straight line.
𝑥
𝛾
* Given distribution: 𝐹𝑋 (𝑥) = 1 − 𝑒𝑥𝑝 {− }
𝑥
𝛾
Reorganizing: 1 − 𝐹𝑋 (𝑥) = 𝑒𝑥𝑝 {− }
Taking the natural logarithm and change sign at both sides yields: − ln(1 − 𝐹𝑋 (𝑥)) =
1
𝛾
𝑥
Using the left hand side as the ordinate and x as the abscissa, the distribution will be straight line, i.e.
the probability paper would be an exponential probability paper (since the actual distribution is the
exponential distribution).
Problem 2
a) The minimum airgap that can be used in order to avoid deck impact for the 10-4 – annual probability
crest height is (see Table 6, Metocean report, page 18 of exam):
23.5m
Thus we have for this case: d = 80+23.5+10 = 113.5m
The natural period for this condition is given by: 𝑇113.5𝑚 = 2𝜋√
𝑚
𝑘
= 2𝜋√
𝑚
𝑐/𝑑 3
= 4.6𝑠
As deck is jacked to 50m above mean sea level, d is given by: d = 80+50+10 = 140m
The natural period for case reads: T140m = 7s.
It is seen that by increasing deck height above sea bed, the natural period increases considerably.
(The important distance is from sea bed to center of gravity of topside.)
a) Environmental contour method is used. Provided the worst sea state is correctly identified, one can
estimate the long term 10-2 – annual probability deck displacement by calculating the 90% value of
the extreme value distribution:
exp( − exp( −
𝑥0.01 −1.14
0.14
)) = 0.9  x0.01 = 1.46m
If we could neglect the short term variability we could use the most probable value or the median as a
proper estimate. However, short term variability can usually not be neglected and in order to account
for that we have to select a higher percentile (90%). The choice of 90% is an approximation.
Since we assume that an exceedance probability of 0.1 correspond to a return period of 100-year, an
exceedance probability of 0.01 should correspond to a return period of 1000 years and an
exceedance probability of 0.001 should correspond to 10000-year return period.
x0.001 = 2.11m
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This approach is not tuned for estimating 10-4 – annual probability values from the 10-2 annual
probability contour line. Since we are not accounting for the importance of the very severe sea states
regarding the exceedance of the 10-4 – annual probability response, it is expected that this estimate is
somewhat on the low side.
b) One could use Monte Carlo simulation. The latest distribution function is based on merely 12 data
points. By assuming this distribution to be correct, one can simulate many samples of size 12. Thus
we get large number of valid samples from our fitted model. If – say - we generate 200 samples of
size 12, we can establish a confidence band around our fitted model. If the old model (Eq. (2.2) falls
inside this interval, one can conclude that the observed difference may well be a result of statistical
uncertainties.
A possible realization from the Gumbel distribution can be obtained by generating a variable for a
uniform distribution between 0 and 1. Denoting this value by u and replacing F X(x) with this value, the
corresponding realization is found from Eq. (2.1). This step is repeated 11 times in order to have one
sample (of size 12) that could have been observed if our fitted Gumbel distribution was true.
Thereafter this procedure is repeated many times – say 200. Then you can estimate the distribution
function for each sample of size 12. Plotting all these in a Gumbel probability paper together with the
two distrbutions, Eqs. (2.1 and 2.2) we can deviation may well be due limited sample size if Eq. (2.2)
fall within “cloud” of simulated distribution functions.
The best approach for an accurate estimate is to do a full long term analysis. By such an approach
we don’t have to choose this approximate percentile. If contour method is to be applied, one should
increase the number of simulated observations underlying the extreme value distribution to at least 20
preferably more (say 40).
c) In order to use Eq. (2.3) for obtaining q-probability values for 𝑌̃, we have to determine the
corresponding exceedance probability per storm. The mean annual number of storms is:
4.83storm/year.
The target probability per storm thus reads: qstorm = q/4.83
The q-probability value for 𝑌̃ is then given by:
1 − 𝐹𝑌̃ (𝑦
̃)
𝑞 = 𝑞𝑠𝑡𝑜𝑟𝑚
If this value shall be correct, the short term variability of storm maximum response must be negligible.
In most cases that is not the case and the estimate is therefore on the low side .
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