Exam 2006_SH_ proposed solution
Problem 1
a) Return period is the expected time between occurrences of the target phenomenon. It
should be pointed out that is an average measure, i.e. individual occurrences may occur closer
in time as well as much more separated in time.
There is a one – to –one occurrence between return period in years, r, and the annual
ecxceedance probabilitiy, q, for long return periods, i..e:
r = 1/q
Inverting this gives the annual exceedance probability of the actual accidental event:
q = 1/4000 = 2.5*10-4.
It is seen that the exceedance probability of the actual event is larger that 10-4, i.e. the
structure must be checked against this event.
It should also be noted that if this type of accidental load will increase with reduced annual
exceedance probability, a somewhat worse scenario of this accidental type is most probably to
be checked.
b)
The given distribution function is the distribution function of the peak significant wave height
of an arbitrary storm. We are going estimate the storm event corresponding to an annual
exceedance probability of 10-2. If the selected storms had represented the annual maximum
storm, then we could use this probability directly. Now we have to correct for the fact that
there typically are more of this type than 1 per year.
The expected number of storms in 100 years is given by: n100 = 33/24*100 = 137.5. This
means that the value corresponding to an annual exccedance probability of 10-2 are found by
solving:
1  FH s , p (h0.01 ) 
1
137.5
 h0.01 = 14m
1
Problem 2
a) The resulting mass when moving in heave is: M = (60 + 90)*106 = 150*106 kg
The stiffness is related to the waterline area:
K = gAWL = 1026*9.81*4*172 = 11.64*106 N/m
Natural period:
T0 = 2*sqrt(M/K) = 22.6s
b) If the wave period is very long, a wave amplitude is the same as a increase of water level
by 5m, i.e. the static heave motion is 5m. What we need to determine is the dynamic
amplification when the wave period is 21s. This can be calculated from the expression for
DAF given in the provided textbook information. The natural frequency is 0.278rad/s, while
the exitation frequency is 0.299rad/s. The relative damping is given to be 0.25. Introducing
these figures in the expression for DAF yields DAF=1.785. This means that the heave
amplitude may reach 8.9m, i.e. the double amplitude is 17.8m. This is more than the damaged
heave compensator can handle and it is recommended that drilling is stopped prior to the
arrival of this swell system.
c) The load history is shown in the figure below.
F(t)
F0
t1
t1 +10
t
(s)
The load history is given by:
F (t ) 
0;
F0
(t  t1 );
10
F0 ;
if t 10
if t1  t  t1  10
if t  t1  10
The heave motion, denoted x3(t), and defined such that x3(t) is positive downwards can now
be calculated by solving the Duhamel integral:
2
t
x3 (t ) 
1
F ( ) e  0 (t  ) sin  d (t  )d

M 0
where  0 is the un-damped natural frequency and  d is the damped natural frequency. M is
the mass associated platform motion. This is equal the sum of of the dry mass of the platform
and the added mass. Mass will of course increase slightly as the module gradually is installed,
but this increase is so small that it can be neglected.
Introducing the force function we obtain:
F0 t
(  t1 ) e  0 (t   ) sin  d (t   )d ;

10 M t1
if t1  t  t1  10
x3 (t ) 
F
x3 (t1  10)  0
M
t
e
  0 ( t   )
sin  d (t   )d ;
if t  t1  10
t1  10
The requested amplitude will be the maximum value of x3(t). Qualitatively, x3(t) could look
like (the motion should look like a damped sinusoidal wave). The requested value is the
maximum. :
F0
x3max
x30
t1
t1+ 10
10 6  9.81
 0.84m
11.64 10 6
For a conservative estimate we can assume that the load transfer is instantaneous. Then we
can find the dynamic amplification as the DAF corresponding to impulse type d) where the
duration is much larger than the natural period. (The module will stay in that position). It is
then seen that the DAF can not be larger than 2. Accordingly, a conservative heave amplitude
to the project is:
The static heave is given by: x30 
x3, max  2 x0 1.68m
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Problem 3
a) From the deep water dispersion relation we have:
2
 2 

 g

 T 
2

 1.56 T 2
b)
For very long waves, the loads will be more or less the same at all three legs. This will not
correspond to a representative and typical wave period.
As the wave length equals the distance between the neighbouring leg, i.e.  = 30m, the load
on all legs are equal. The corresponding wave period is:

30
T

 4.4s
1.56
1.56
c)
Regarding extreme loads we see that for the larges waves, the ratio between wave height and
leg diameter is well above 10. From the figure given under text book information, we then see
that the load is dominated by the drag term. The drag term is very non linear and the
frequency domain analysis is not adequate. This means that we should not accept the
consultants recommendations of doing an extreme value analysis in the frequency domain
using a transfer function obtained for a much lower sea state.
The consultant’s suggestion may well be a valid approach for fatigue assessments.
Considering a sea state of 7m as representative for a sea state within the range of sea states
important for fatigue, we may say that the transfer function could be valid for this sea states if
non-linearity is not too important. With the information that the typical wave height in a 7m
sea is about 3.5m, it is seen that the wave height to leg diameter ratio is so small that the drag
term is not very important. We may well work with a linearized analysis. A transfer function
obtained as indicated in the problem text could be one way of obtaining an equivalent linear
relation between wave process and response. We could therefore let the consultant go on
doing some fatigue assessments.
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