Exercise 3 SH_part 2006
Suggested solution
Problem 1
a) Mass when vibrating in water: M = (m + a33)
Stiffness: K = gAWL
Damping coefficient: C = 2 ξ 3 MK
b) For a linear system characterized by Eq. (1.1), the steady state heave response process will
be on the form:
x3(t) = x30 sin (t - )
The modulus part of the transfer function, the response amplitude operator, from force to
heave response is defined as the ratio between the heave amplitude and the force amplitude:
∣ H QX¨ 3 ω ∣ =
x 30
= RAO ω
Q30
The RAO is as the name suggests a scaling factor to be multiplied by the load amplitude for
obtaing the response amplitude. This quantity does not include any phase information.
The transfer function does also include information about the phase information shift, .
Generally speaking, the transfer function is a complex function given by:
H QX ω = ∣ H QX 3 ω ∣ e− iθ
3
In order to be a structure characteristic, a linear mechanical system is assumed, i.e as the load
is doubled so is also the response.
c) As mentioned above, it is the steady state response that can be written as suggested. This
solution assumes that the homogeneous solution has died out. This will take place if damping
is present.
If there is no damping or one is concerned with the response immediately after the force is
started, the resulting solution will look like:
x 3 t = x30 sin ωt − θ
− ξ 3 ω0 t
e
A sin ωd t
B cos ωd t
0 is the natural frequency (undamped) and d is the damped natural frequency. A and B
d) The transfer function characterises the steady state solution, i.e it is assumed that the
introductory noise caused by the homogeneous solution is assumed to have died out.
Introducing the steady state solution,
x 3 t = x 30 sin ωt − φ = x 30 cosφ sinωt − x 30 sin φcosωt
in the equation of motion yields after pooling terms with sint and cost:
[ K − ω2 M
ωC sin φ] x 30 sin ωt
cosφ
[−
K − ω2 M sin φ
ωC cosφ] x30 cosωt = Q30 sin ωt
If this equations shall be fulfilled at any time, we must require that the coefficient in front of
cos t is identical equal to zero. Solving that equation yields the an expression from which
the phase angle is found by inversion:
tg φ =
sin φ
ωc
=
cos φ
K − ω2 M
Similarly, equalling the coefficients in front of the sine terms, yields:
[ K − ω2 M
ωC tgφ] x30 =
Q30
cosφ
With tg given above and cos given by:
1
=
cosφ
1
tg2 φ
we find:
x 30=
Q30
2
K− ω M
2
ωC
2
=
Q30
K
1
M 2
1 − ω2
K
ω
C
K
2
Using that:
ω0 =
K
M
C = 2ξ 3 M K
β=
ω
ω0
x30 is shown to be equal to what is given by (1.2) in the problem text.
For a natural period of 23s and an exitation period of 16s, the frequency ratio, , becomes:
1.4375
For this frequency ratio and a relative damping of 15%, we DAF = 0.87, i.e. the inertial forces
helps to reduce the heave amplitude below the quasi-static level. As the period is increasing
towards 23s, DAF will of course increase above 1.
e) The RAO between force and response is given by:
RAOQX ω =
DAF
K
The RAO between wave and force, RAOQ(), can be found by exposing the platform to
harmonic waves with frequency  and amplitude 0. As the corresponding load amplitude is
found, we have as above:
RAOQ() = Q30/0
f) The RAO or absolute value of the transfer function from wave to heave response is denoted
∣ H ηX ω ∣ . Below we will for simplicity denote it by RAO().
As the significant wave height, hs and the spectral peak period, tp, are known, we can
determine the wave spectrum by e.g. select a JONSWAP spectrum parameterized in terms of
hs and tp, see e. g. the example metocean report. With the wave spectrum and the RAO
known, the respons spectrum is given by:
sX ω = RAO ω
2
Sη ω
If we assume the wave process to be Gaussian and both the load and the mechanical system
are linear, the response process is also Gaussian. The maxima of a Gaussian process
(reasonably narrow banded) follow a Rayleigh distribution. The most probable largest value,
x , during a time perod, T, is defined as the value which is only expected to be exceeded
once during the sea state, i.e:
P X max
{
x = exp −
1 x
2 σX
2
}
=
1
nT
The variance is found by:
σX =
∫
s X ω dω
The number of maxima is found as the time period T devided by the expected zero-upcrossing period, tX2. This period can be calculated as:
t X2 = 2π
mX2
mX0
mXn is spectral moment of order n:
∞
mXn =
∫
ωn SX ω dω
0
2
It is seen that mX0 = σX .
g) If we denote the maximum value in time T with XT, the distribution function of XT can be
found by:
F X x = [F Xmax x
T
nT
]
[
−
= 1− e
n
1 x 2 T
2 σX
]
Introducing x = x it is seen from above that the exponential term equals 1/nT, i.e:
P XT
x = 1− FX x = 1 −
T
1−
1
nT
nT
1−
∞
nT
1
= 0.63
e
This shows that the most probable value in time T will be exceeded in 63 out of 100
realizations of this sea state.
Problem 2
a) If w shall use the given target probability directly, we have to use the distribution
function of the annual extreme value distribution for the the target variable, i.e.
suggestion iii) in problem text.
b) F Y3h y is to be used for obtaining a value corresponding to an annual exceedance
probability of 10-2. In order to do so, we must find the exceedance probability per 3hour that corresponds to an annual exceedance probability of 10-2. The number of 3hour events per year is 2920. This means that the requested exceedance probability is:
10− 2
= 3.4⋅ 10− 6
2920
qULS =
3h
The ULS value of y is then found by solving:
1 − FY
3h
yULS = qULS
3h
Problem 3
a) The expected number of events with hs > 10m per year is: m1 = 33/24 = 1.375
With reference to the distribution function of the peak significant wave height of an
arbitrary event, the exceedance probability per event corresponding to an annual
exceedance probability of q is q/m1. This means that the value of significant wave
height expected to be exceeded by an annual probability of q is found by solving:
P H s , sp
hq = 1 − F H
s , sp
Solving this with respect to hq, yields:
hq − h0
q
=
θ
m1
{[
hq = exp −
]}
hq= h0 − θ ln
q
m1
Introducing given values for h0 and estimated 10-2 and 10-4 significant wave heights
are found to be:
h0.01 = 14.0m
h0.0001 = 17.7m
b) The parameters of the crest height distribution are found to be:
q= 0.01
c = 14.27m and c = 1.02m
q= 0.0001
c = 17.30m and c = 1.34m
The probability of exceeding the available freeboard, 22m, conditional on the
occurrence of these events then become:
{ {
{ {
P Cs
22m∣ H s, sp = h0.01 = 1 − exp − exp −
P Cs
22m∣ H s, sp = h0.0001 = 1 − exp − exp −
22 − 14.27
1.02
}}
22 − 17.30
1.34
= 5.1⋅ 10− 4
}}
= 2.95⋅ 10− 2
c) In order to find the unconditional probability of exceedance, the long term distribution
of crest height can be estimated. The long term (or marginal) distribution of storm
maximum crest height is given by:
∞
FC c =
s
∫
10
j max
F C s ∣ H s, sp c ∣ h f H s ,sp h dh ≈
FC∣ H
∑
j =1
s ,sp
c ∣ h j P H s, sp = h j
where
P H s , sp= h j = F H
s , sp
hj
Δh
Δh
− F H s, sp h j −
2
2
The summation above is easily carried out in spread sheet or Matlab etc.. Here a
spread sheet is used. The long term distribution for Cs, is shown in Table and figure
below.
Marginal probability of exceeding c
c (m)
P(C >c)
log10(P(C>c))
10 0.998060095
-0.000843308
12 0.461355735
-0.335964077
14 0.071024953
-1.148589042
16 0.009162384
-2.037991524
18 0.001233462
-2.908874287
20 0.000181661
-3.740739403
22 2.94496E-05
-4.530920156
24
5.2204E-06
-5.282296441
26 1.00202E-06
-5.999122375
Long term distribution of exceedance for C s
0
log10(P(Cs>c))
-1
-2
Deck level
-3
-4
-5
-6
-7
0
5
10
15
20
25
30
c (m)
It is seen from the table (and figure) that the probability of exceeding the available
freeboard by an arbitrary storm event of the type considered herein is:
P(Wave deck impact) = 2.94 * 10-5 per storm event. (Return period in terms of no of
storm events becomes 33898.)
The annual exceedance probability of the available freeboard becomes:
P(Wave deck impact per year) = 2.94*10-5*1.375 = 4*10-5
If we estimateted the annual probability of exceeding available freeboard by merely
considering the 10-2 probability storm, it is seen that the probability of exceeding this
level becomes: 5.1*10-4*10-2 = 5.1*10-6. Thus it is seen that this annual probability is
much smaller than the annual probability estimated when accounting for all storms.
 When estimating a marginal probability of exceedance one should include the
exceedance probabilities (weighted with the probability of the storm event) from all
important storm events, i.e. some sort of a long term distribution should be
established.
Problem 4
The static displacement, xst, is calculated to be 100mm. The acceptable deformation is
150mm. The question then is: What is the dynamic amplification?
There is not enough information in the problem text to do a proper detail calculation. But
can we say something from the available information? Let us do some simplifying
assumptions:
I. Let us assume that we can approximate the given load history with a history of the
shape referred to a c in figure below. This will be on the safe side, but since the
actual load history is rather skewed, this may be a good approximation to do at an
early stage where we are expected to express our concern without knowing too much
about the system.
Fig. 1 Horizontal axis is ratio of impulse duration to natural period.
II. It is seen from the figure that all we now need is to make a guess regarding the ratio
duration of impulse to natural period. This is abscissa axis in the figure below. From
the available information we can not calculate the natural period exactly. However,
it seems reasonable to assume that it is well below 1s. The duration of the impulse
could be guessed by looking at a 10-4 wave profile and estimate the time the tank
would be submerged, but again we the crest height of the 10-4 wave is not given. In
order to give a conservative assumption let as assume that the duration of the
impulse (1 in problem text) is 2s. This gives a ratio of 2.
From the figure we see that for an impulse of type c) and a duration ratio of 2, DAF
is roughly 1.75.
This means that the load bearing system will not take the loads if our assumptions
are correct.
The most critical assumption is possibly the shape, in particular if the load duration
ratio is larg. The shape does not have to change in front before the dynamic
amplification is considerably reduced for long duration ratios.
Regarding the duration ratio it is not likely that it will be lower than 1. This depends
very much on the solidity of the load bearing system for the tank.
This suggests that a simple consideration without doing any calculations suggests
that there may be problems with the tank if it is hit by a wave. The reasonable thing
to recommend is that more accurate calculations should be carried out. That will
involve a proper estimation of the natural period and finding the DAF for the given
shape of the load history.