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friends this is lecture five . in which we
will talk about new generation platforms .
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before doing this let us try to do some quick
numeric examples to understand the basic action
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stiffness frequency in period of a by a verity
of platforms for our basic understanding . let
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us take one example . of a spar platform are
spar buoy . let us say it operates . at water
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depth of . one forty meters it has a cylindrical
tank .
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one thirty seven meter height the diameter
of the tank .
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lets says twenty nine meters the corresponding
displacement . .
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sixty six into ten power three tons which
corresponds to your draft of hundred and seven
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meters what is asked is to compute . the hull
oscillation period . of the buoy .
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to calculate the heave oscillation . . we
need estimate the restoring force . in the
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vertical direction . this is generally produced
. by the vertical displacement of the spar
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.
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the restoring force .
is mainly . due to change of buoyancy effect
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. .
one may ask me a question what to be the contribution
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from the mooring lines so mooring lines do
not contribute . to the restoring force . even
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if they are very marginal .
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so let us now find . the vertical displacement
.
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the vertical displacement of . lets say x
meters . will actually produce the restoring
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force .
the restoring force is given by rho g pi r
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square into x . rho is taken as one zero two
five kg per cubic g is nine point eight one
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meter per second square radius in this example
is fourteen point five meter because the diameter
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is twenty nine meter so i am interested to
find stiffness because you know frequency
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is actually root of stiffness per mass so
to find stiffness . . one can say it is actually
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the restoring force for unit displacement
.
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so keeping x as unity . in equation one i
can now call the restoring force actually
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as stiffness which is actually equal to one
zero two five nine point eight one pi fourteen
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point five square which gives me . six point
six four into ten power six . newton per meter
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so now i have the value stiffness which is
given by . this value of six point six four
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ten power six newton per meter mass . in the
heave direction . d o f stands for degree
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of freedom . is given as sixty six ten power
three tons which is sixty six ten power six
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kg six point six ten power seven kg so now
i have mass in kg i have stiffness in newton
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per meter i can always find actual frequency
in s i units . which is k by m . which is
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going to be so many radians per second but
i am interested in finding the period i substitute
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we know periods omega is two pi by t so period
if you do substitution of k and m and get
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these equation back here the period what you
are call is going to be twenty point seven
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eight seconds so thats the typical period
of a spar platform in heave degree of freedom
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lets take another example . where i want to
find for a t l p . heave and surge oscillations
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. . we all know that in t l p buoyancy exceeds
a weight and t zero equalizes this weight
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to the top buoyancy so t zero values are generally
substantially high they actually act . as
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linear springs . whose stiffness is k n where
n is the total number of tendons generally
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tendons are not of a single piece they will
be in groups of four into three four into
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four that makes twelve that makes sixteen
etcetera so the stiffness of this spring . is
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given by n . a t e by l t we all know linear
stiffness is simply a e by l . so a t in this
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case is area of each tendon n is a total number
of tendons so that makes the total area . e
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modulus of elasticity . usually steel is used
as a material of course l is the length of
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the tendon . so this can be in meters this
can be in meters square this can be newton
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per meter square we will get stiffness in
. newton per meter
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if you want to find the period . in heave
degree of freedom for a t l p i can simply
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say two pi root of m that is mass in heave
degree length of the tendon by n a t e second
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when we know the mass of the t l p participating
in the heave degree of freedom i can always
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find the period in radiance per second or
in seconds usually the heave period varies
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from . two to four seconds of a t l p if you
want to do this in surge degree of freedom
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. .
then the restoring force . in surge degree
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of freedom is actually . f r we say it is
n t . t total x by l t where t t is initial
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pretension in each tendon so n t into t t
will give me the total initial pretension
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may be n newtons x is a surge displacement
. . l is the length of . each tendon . the
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horizontal restoring force . after we include
the tendon weight also .
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in that case f r x will be . n t . t t minus
w l t by two of x by l t .
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and period in surge is simply two pi root
of mass in surge degree . length of the tendon
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divided by n t t t .
lets take a typical value .
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lets say mass . is about two ten power seven
kg from the previous example lets say n t
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into t t is ten power seven newton again from
the previous example let the length of the
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tendon b two hundred meters therefore typical
surge period would be . two pi . two ten power
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seven two hundred meter . by ten power seven
which is . one twenty five . point six seven
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seconds where surge period you surely of a
t l p . . is anywhere from ninety five to
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about one twenty seconds .
we will take one more example . where we will
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talk about articulated tower . we all know
. articulated tower has a top size . supported
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by a tower . which is connected to the universal
joint to the seabed . the tower also has . buoyancy
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tank . .
and the ballast chamber . as we saw lets say
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this is my water level .
this is my mass of the deck .
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this is my mass of the buoyancy tank .
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essentially this tower acts as an inverted
pendulum .
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the degree of freedom . this tower has is
a rotation about the base . so theta is the
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degree of freedom .
so in that case the variable summer gens changes
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for the new position of the platform therefore
i should say . the righting moment .
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is given by . b rho minus m b . into g into
l b that gives my buoyancy tank value . minus
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. m d . g l d multiplied by theta where in
this case b stands for the buoyancy . which
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is provided . by the buoyancy tank l d . is
the length of the deck in sense distance of
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. c g of the deck from the articulated joint
. .
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l b distance . of center of buoyancy . from
the joint .
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the moment of inertia . for rotation . about
the articulated joint . is given by .
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m d l d square plus m b plus m b h . of l
b square where m b h . is the additional hydrodynamic
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mass . or what we otherwise call added mass
of the buoyancy tank . now the frequency . n
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pitch degree because we talking about rotation
is simply . square root i should say approximately
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g times of . b rho minus m b . into l g minus
. m d into l d divided by . m b plus m b h
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. of l b square [vocalized-noise] plus m d
l d square
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let us do one more example . i want to find
. the pitch and roll motion . of semisubmersible
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.
a typical semisubmersible has a deck . supported
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on columns .
rest thing on bottom pontoon . with all top
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side . with the risers connecting now . by
neglecting . the mooring cables influence
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.
righting moment .
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i should say m r is given by . .
rho g . v g m r theta r v g m r where g m
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r is defined as metacentric height of the
platform . lets say for roll degree of freedom
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similarly . for pitch degree of freedom m
p the righting moment . will be minus rho
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g v g m p theta . hence the natural periods
. in roll could be simply two pi . root of
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i r by rho g v g m r which is approximately
. thirty to fifty five seconds in terms of
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pitch this two pi i p by rho g v m p which
is approximately twenty to forty seconds . whereas
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in the rolls expression i r is mass moment
of inertia . in roll degree of freedom . and
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i p is the mass moment of inertia in pitch
degree of freedom .
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let us try to calculate . yaw motion of . platform
while calculating so the snapping effect . of
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tendons . is neglected what is snapping effect
. . snapping means alternate . slackening
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and tensioning . initial t zero is so high
slackening does not happen . we also agree
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in yaw motion . t l p is flexible . but the
platform is [debine/define] define symmetric
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. hence yaw motion is highly limited .
the restoring moment .
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in yaw motion . .
is given by . which otherwise . and period
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in yaw . is simply two pi by r y . i y l t
by n t l t where in this case i y is total
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mass moment of inertia in yaw degree of freedom
.
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so friends in this lecture we are trying to
understand through simple examples and equations
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how we can estimate the restoring force the
initial stiffness periods of vibration for
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different verities of offshore platforms under
simple normal free vibration conditions now
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the question comes what is the necessity for
the new generation platforms there are some
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lacunas . which the existing platforms had
one a very large hull displacement . very
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quick . restoration . which can damage the
connecting . risers . initiating fatigue failure
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. snapping effect .
corrosion . due to extended large . deep draft
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caissons in case of spar etcetera in all these
cases you will see that . the hull displacement
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. essentially in rotational degrees of freedom
. is challenging . actually it is undesirable
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. . so the new generation platforms . are
conceived in such a manner that this primary
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problem is eliminated .
so friends in the next lecture we will talk
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about those conceive geometry of new generation
platforms how are they conceived in idea and
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what would be the efficiency of this platform
in terms of eliminating rotation responses
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under normal environmental loads
thank you very much . .