Wave force

Chapter 12

WAVE FORCES ON SLENDER
CYLINDERS

12.1 Introduction
Chapters 6 through 11 have handled the hydromechanics of large (‡oating) bodies in the
sea. Attention now switches to the hydromechanics of slender cylinders. Examples of such
cylinders include the leg or brace of an o¤shore space truss structure, a pipeline or even
an umbilical cable extending down to some form of remotely controlled vehicle.

12.2 Basic Assumptions and De…nitions
A slender cylinder in this discussion implies that its diameter is small relative to the wave
length.The cylinder diameter, D, should be much less than the wave length, ¸; the methods
to be discussed here are often usable as long as D < about 0:1 to 0:2.
¸
Derivations are done for a unit length of cylinder. Force relationships will yield a force per
unit length. This relationship must then be integrated over the cylinder length to yield a
total force. The implications of this unit length approach combined with the restriction
to slender cylinders is that the ambient water motions in the immediate vicinity of the
cylinder are all about the same at any instant in time. This is (assumed to be) true both
vertically and horizontally; the spatial variation in the undisturbed ‡ow near a unit length
of cylinder is simply neglected. A similar assumption was made for the heaving cylinder in
chapter 6, but this is not usually the case with a ship or other large structure as discussed
in the previous chapters.
The absence of a spatial variation in the ambient ‡ow as one moves from place to place near
the cylinder, makes it possible to characterize the ‡ow in the entire region of the cylinder
by the ambient ‡ow at one characteristic location. The axis of the cylinder is chosen as
that location; this simpli…es the bookkeeping.
The ‡ow around this cylinder segment will be considered to be two-dimensional - quite
analogous to strip theory for ships except that the axis of the in…nitely long cylinder is not
generally horizontal as it was for a ship. Flow components and any resulting forces parallel
0
J.M.J. Journée and W.W. Massie, ”OFFSHORE HYDROMECHANICS”, First Edition, January 2001,
Delft University of Technology. For updates see web site: http://www.shipmotions.nl.

12-2 CHAPTER 12. WAVE FORCES ON SLENDER CYLINDERS

to the cylinder axis are neglected; all forces are caused by the ‡ow - and later cylinder
motion - components perpendicular to the cylinder axis.
The axis system used here is identical to that used for the waves in chapter 5, see …gure
5.2. The origin lies at the still water level with the positive z-axis directed upward. The
wave moves along the x-axis in the positive direction.
The resulting water motions come directly from chapter 5 as well:

@ ©w dx cosh k (h + z)
u= = = ³ a! ¢ ¢ cos (kx ¡ !t) (12.1)
@x dt sinh kh
@©w dz sinh k (h + z)
w= = = ³ a! ¢ ¢ sin (kx ¡ !t) (12.2)
@z dt sinh kh
These can be simpli…ed for the following discussions, however. Since the location, x; of the
cylinder element is more or less …xed, the kx term in the above equations can be dropped.
For the moment, it is simplest to consider a vertical cylinder so that equation 12.1 will
yield the desired ‡ow velocity. All of this yields an undisturbed horizontal ‡ow velocity
given by:
cosh k (h + z)
u(z; t) = ³ a ! ¢ ¢ cos (¡!t) (12.3)
sinh kh
or at any chosen elevation, z:

u(t) = ua cos(¡!t) (12.4)
and since cos(¡!t) = cos(!t) the sign is often dropped so that:
u(t) = ua cos(!t) (12.5)
in which:
ua = amplitude the wave-generated horizontal
water velocity at elevation z (m/s)
! = wave frequency (rad/s)
Note that the elevation dependence in equation 12.3 has been included in ua in 12.4; this
dependence is not included speci…cally in the most of the following discussion.
Since the ‡ow is time dependent, it will have a horizontal acceleration as well. This can
be worked out to be:
u(t) = ¡! ua sin (!t)
_ (12.6)
The acceleration amplitude is thus given by:
ua = ! ua
_ (12.7)
Since potential theory describes waves so well, the above relations are assumed to hold for
any undisturbed wave ‡ow - even when viscosity is involved.

12.3 Force Components in Oscillating Flows
It is convenient to derive the relationships in this section for a smooth-surfaced vertical
cylinder. This restriction will be relaxed later in this chapter, however. Since potential
‡ows are so convenient for computations, this discussion of forces in oscillating ‡ows starts
with this idealization. The unit length of cylinder being considered is thus vertical and
submerged at some convenient depth below the water surface.

12.3. FORCE COMPONENTS IN OSCILLATING FLOWS 12-3

12.3.1 Inertia Forces
Remember from chapter 3 that D’Alembert proved that there is no resultant drag force
when a time-independent potential ‡ow is present. Here, it is the e¤ect of the ‡ow
accelerations that is of concern.
Consider …rst the undisturbed ambient (surrounding) ‡ow without any cylinder in it. Ac-
cording to Newton’s second law of motion, accelerations result from forces; this is univer-
sally true. Thus, the horizontal acceleration of the ambient ‡ow must be driven by a force
in the water which, in turn, must come from a horizontal pressure gradient. This pressure
gradient is present, even when there is no cylinder in the ‡ow. By examining the pressure
gradient force on a di¤erential ’block’ of ‡uid, one discovers that:

dp du
= ½
dx dt
= ½¢u _ (12.8)

which is nothing more than Newton’s second law applied to a ‡uid.
Given this information, what happens when a cylinder is inserted into this pressure and ‡ow
…eld? This question is answered using an approach which has the advantage of physically
explaining the separate contributions of two separate inertia force components; a faster,
but less ’transparent’ derivation will be given later.

Pressure Gradient Force
One must ’drill a hole’ in the ambient pressure gradient …eld in order to ’insert’ the cylinder.
For now, the fact that the cylinder wall is impervious is neglected completely; the ‡ow is
still undisturbed. Any force which this undisturbed pressure …eld exerts on the cylinder
can be computed by integrating this pressure around the perimeter of the circular hole.
This integral yields, knowing that the cylinder is symmetrical with respect to the x-axis
and has a unit length:
Z ¼
Fx1(t) = 2 ¢ p(R; µ; t) R cos µ ¢ 1 ¢ dµ (12.9)
0

in which p(R; µ; t) is the undisturbed pressure (N/m2) on the perimeter of the circle and
R is the cylinder radius (m).
The resulting force is computed just as was done in chapter 3, using …gure 3.16.
Since the pressure di¤erence across the cylinder at any distance, y; away from the x-axis
is:

¢p = ½ ¢ u ¢ ¢x
_ (12.10)
where ¢x is the width of the cylinder at distance y from its axis. The integral can be
simpli…ed again so that:
Z ¼
2
F x1(t) = 2 ¢ ¢p(R; µ; t) ¢ R ¢ cos µ ¢ dµ (12.11)
0

After integrating one gets:
Fx1(t) = ½ ¼ R2 ¢ u(t)
_ (12.12)


In which one should recognize ¼R2 ½ as the mass, M1; of ‡uid displaced by the unit length
of cylinder - the mass of ‡uid one would have ’removed’ when ’drilling the hole’ in the
pressure gradient …eld.
This inertia force term stems from the pressure gradient already present in the accelerating
‡ow - even before the cylinder was installed. It is equal to the product of the mass of water
displaced by the cylinder and the acceleration already present in the undisturbed ‡ow.
This force component is fully equivalent to the Froude Krilov force mentioned in chapter
6.

Disturbance Force
The cylinder was not allowed to disturb the ‡ow when Fx1 was computed; this error is now
corrected. Obviously, the cylinder is impermeable; ‡uid cannot actually ‡ow through the
cylinder wall. The cylinder geometry forces the ‡uid to go around it modifying all the local
velocities and thus accelerations. This can only occur if a force is exerted on the ‡uid, and
this force can only come from the cylinder.
Figure 3.12, which can be found in chapter 3, shows how the streamlines diverge and
converge around a cylinder in a potential ‡ow. One way to evaluate the extra force causing
this total disturbance …eld is to examine the kinetic energy change caused by the cylinder
as was done by [Lamb, 1932]. He evaluated the kinetic energy represented by the entire
(disturbed) ‡ow …eld around the cylinder and subtracted from that value the kinetic energy
of the undisturbed ‡ow in the same - theoretically in…nite - region. This yields in equation
form:
ZZ
1 ZZ
1
1 2 1
E= ½ ¢ [u (x; y; t)] dx ¢ dy ¡ ½ ¢ u2 (t) ¢ dx ¢ dy
1 (12.13)
2 2
cyl: wall cyl: wall

It is convenient to associate this energy with some sort of equivalent mass, M2; moving
with the ambient (undisturbed ‡ow) velocity, u1; so that:

1
E = M2 u2
1 (12.14)
2
Lamb discovered that:
M2 = ¼ R2 ½ (12.15)
or that M2 is simply the mass of ‡uid displaced by the cylinder segment (just as was M1)
so that:

Fx2 = ¼ R2 ½ ¢ u(t)
_ (12.16)
Note that Fx2 has the same form as Fx1 and that they both have the same phase as well.
Fx2 is analogous to the part of the di¤raction force which was in phase with the ship
acceleration in chapter 6.
A thoughtful reader may wonder why this second force component, Fx2 , was not present
in a constant current; after all, that cylinder was then impervious to the ‡ow, too. The
answer to this question lies in the fact that Fx2 does not result from the pattern itself, but
rather from its continuous build-up and break-down which occurs only in a time-dependent
‡ow. In a constant current there is no time dependent change and thus no Fx2.


Resultant Inertia Force
Potential theory indicates that the resultant force on a …xed cylinder in an oscillating ‡ow
is the sum of two terms:

Fx1(t) = ½ ¼ R2 ¢ u1(t)
_ from the ’hole’ in the undisturbed pressure gradient in the
ambient ‡ow. This is also know as the Froude-Krilov force.

Fx2(t) = ½ ¼ R2 ¢ u1(t) from the ‡ow disturbance caused by the impervious cylinder.
_

The resultant force is then:

FI (t) = Fx1 (t) + Fx2(t)
= 2 ¢ ¼ R2 ½ ¢ u(t)
_ (12.17)

Note that because this is still a potential ‡ow, there is no drag force. Also, since there is
no circulation, there is no lift, either.

Alternate Direct Calculation Approach
Another, possibly faster way to calculate the ‡ow disturbance force coe¢cient starts with
the potential function for an oscillating cylinder in still water. This approach is completely
analogous to that used in chapter 6 to determine the added mass of a ‡oating body. One
starts directly with the potential function just as was done there, and uses the Bernoulli
equation to calculate the pressure on the cylinder surface and then integrate this as was
done in chapter 3 to determine the resultant force.

Experimental Inertia Coe¢cients
The theoretical value of 2 in equation 12.17, above, is usually replaced by an experimental
coe¢cient, CM - often called the inertia coe¢cient. Remember that the theoretical
value of 2 is made up of 1 from Fx1 (the ambient pressure …eld) and 1 from Fx2, the
‡ow disturbance caused by the cylinder. In practice the 1 from the ambient pressure
…eld is usually considered to be acceptable; potential theory predicts the water motion in
undisturbed waves well. The coe¢cient from Fx2 is much less certain; the vortices in the
wake (in a real, [not potential!] ‡ow) disturb the theoretical ‡ow pattern used to determine
Fx2. This is taken into account by using a value C a - a ’coe¢cient of added mass’ -
instead. Usually Ca < 1. Note that Ca is quite analogous to the hydrodynamic mass used
in chapter 6. This is all summarize in the table below.

Force Force Experimental Theoretical Experimental
Component Term Coe¢cient Value Value

Froude-Krylov Fx1 1 1 1
Disturbance Fx2 Ca 1 Usually < 1

Inertia FI CM 2 Usually 1 to 2


Remember that:
jCM = 1 + Ca j (12.18)

and that Ca is associated with ‡ow disturbance.
The phrase ’added mass’ has just been used above, much like it is often used in ship
hydromechanics as in chapters 6 through 9. Ca is often interpreted as ’hydrodynamic mass’
- some mysterious mass of surrounding ‡uid; this interpretation can be very misleading and
dangerous, however. Consider the following true situation taken from ship hydromechanics.
An investigator was carrying out tests to determine the hydrodynamic coe¢cients for a
‡at-bottomed barge in shallow water. Attention was focussed on its heave motion and the
in‡uence of the barge’s (relatively small) keel clearance on the hydrodynamic mass. Tests
were carried out with various (average) keel clearances so that Ca could be determined as
a function of (average) keel clearance value. Figure 12.1 shows a cross-section sketch of the
set-up.

Figure 12.1: Cross Section of Barge Showing Keel Clearance

Since the relatively deeply loaded barge had vertical sides, these caused no waves or other
disturbance as the barge oscillated vertically; the only water that is initially really disturbed
by the vertical motions is the layer of water directly under the barge. The researcher who
carried out these tests then reasoned somewhat as follows: ”The mass of water under the
ship is directly proportional to the average keel clearance. This mass becomes less and
less as the keel clearance becomes smaller; it is therefore logical to expect Ca to approach
zero as the keel clearance becomes less and less.” His experiments proved however that C a
values became larger and larger as the keel clearance decreased.
The error here is the interpretation of Ca as if it represents a physical mass. It is not
this! Instead, C a (or even CM for that matter) should be interpreted as force per unit
acceleration or Acceleration . Returning to the experiments and the researcher above, C a
F orce

only represents an ’extra’ (in comparison the situation in air!) force needed to given the
barge a unit acceleration in the vertical direction. Thinking in this way, one can easily
reason that as the layer of water under the barge became thinner, it became more di¢cult
for it to ’get out of the way’ by being accelerated horizontally as the barge accelerated
downward. Conversely, it therefore took a larger force to give the barge its unit of vertical
acceleration as the keel clearance became smaller. With this reasoning, one gets the correct
answer: In the limit, Ca ! 1 as the keel clearance ! 0:


Fixed Cylinder in Waves
For a …xed cylinder in waves, one is confronted with both Fx1 and Fx2 so that equation
12.17 becomes:

FI (t) = Fx1(t) + F x2(t)
¼
= ½ CM D2 ¢ u(t) _ (12.19)
4
in which:
FI (t) = inertia force per unit cylinder length (N/m)
½ = mass density of the ‡uid (kg/m3)
CM = dimensionless inertia coe¢cient (-)
u(t)
_ = time dependent undisturbed ‡ow acceleration (m/s2 )

CM has a theoretical value of 2 in a potential ‡ow.

Oscillating Cylinder in Still Water
One might reason that the ‡ow around an oscillating cylinder in still water would be
kinematically identical to that of an oscillating ‡ow past a …xed cylinder and that the
resulting forces would be identical. This is not the case, however.
There is no ambient dynamic pressure gradient present in still water so that the …rst inertia
force term above, Fx1 the Froude-Krilov force, is now identically equal to zero. Thus, if
the cylinder is oscillating such that its velocity is given by:
_
X(t) = a cos(!t) (12.20)

then the resultant hydrodynamic inertia force on the cylinder will be:
Ä
FI (t) = ¡F x2(t) = ¡C a ¢ ¼ R2 ½ ¢ X (t) (12.21)

The minus sign indicates that the hydrodynamic resisting force is opposite to the direction
of cylinder acceleration. The value of Ca will generally not be larger than its theoretical
value of 1. Note as well that if one is measuring forces within an instrumented pile on
a segment of this oscillating (accelerating) cylinder, one will usually also measure a force
component proportional to the mass times acceleration of the (solid) cylinder element itself.
Force measurements in the lab are often corrected for this by …rst measuring forces while
oscillating the cylinder in air before the basin is …lled with water. This force is usually
considered to be the inertia force of the measuring element itself. Only a slight error is
made here by neglecting the aerodynamic resistance caused by the accelerating ‡ow pattern
in the still air.

12.3.2 Drag Forces
Experiments have shown (see chapter 4) that a drag force proportional to U 2 and the
cylinder diameter, D; is caused by a constant current; it is only reasonable to expect a
similar force to be present in a time-dependent real ‡ow as well. Since the drag force is in
the same direction - has the same sign - as the velocity in an oscillating ‡ow, the constant


current, U 2 ; is commonly replaced by its time-dependent counterpart, u(t) ju(t)j in order
to maintain a proper sign. Substituting relationships for u(t) and working this out yields:
1
FD(t) = ½ CD D u2 ¢ cos(!t) jcos(!t)j
a (12.22)
2
in which:
FD (t) = drag force per unit length of cylinder (N/m)
CD = dimensionless drag coe¢cient (-)
D = cylinder diameter (m)
ua = water velocity amplitude (m/s)
! = circular water oscillation frequency (rad/s)
t = time (s)

One should not, however, expect the values of CD for an oscillating ‡ow to correspond
with those found in chapter 4 for a constant ‡ow. Instead, they will have to be determined
again in a time-dependent ‡ow.

12.4 Morison Equation
J.E. Morison, a graduate student at the University of California at the time, wanted to
predict wave forces on an exposed vertical pile; see [Morison et al., 1950]. He simply su-
perimposed the linear inertia force (from potential theory and oscillating ‡ows) and the
adapted quadratic drag force (from real ‡ows and constant currents) to get the following
resultant force (per unit length):

F (t) = Finertia (t) + Fdrag(t) (12.23)
or:
¯ ¯
¯ ¯
¯ F (t) = ¼ ½ CM D2 ¢ u(t) + 1 ½ CD D ¢ u(t) ju(t)j ¯
_ (12.24)
¯ 4 2 ¯

in which the …rst of these two terms is the inertia force and the second represents the drag
force.
Note that in equations 12.24 the drag and inertia force components are 90± out of phase
with each other when seen as functions of time. This is a direct consequence of the phase
shift between velocity and acceleration in an oscillatory motion; check equations 12.4 and
12.6 if necessary. Examples of this will be shown during the discussion of coe¢cients and
their determination below; see …gure 12.2 later in this chapter as well.

12.4.1 Experimental Discovery Path
Morison formulated his equation simply by hypothesizing that the superposition of two
separate and well know phenomena (drag in a current and hydrodynamic inertia in an
accelerating ‡ow) would yield a viable solution for a vertical pile in waves. This section
explains how one comes to the same equation via experiments much like those for ships.
Readers should know from earlier chapters that a common technique in marine hydrody-
namics is to oscillate a body with a chosen displacement amplitude in still water and to

12.4. MORISON EQUATION 12-9

record its displacement and the force, F (t) acting on it as functions of time. Further, the
force record is resolved into two components: one in phase with the acceleration and one
in phase with the velocity.
The …rst is determined by multiplying F (t) by ¡ cos(!t) and integrating the result to get
an inertia force; the second comes from the integral of the product of F (t) and sin(!t) to
yield a component in phase with velocity.
One single test might not tell too much, but if testing were done with di¤erent excitation
amplitudes, but at constant frequency (or period), then a plot of the amplitude of the
inertia force component versus the oscillation acceleleration amplitude would be linear; the
plot of the drag force amplitude as a function of velocity amplitude would be quadratic.
Similarly, comparison of test results carried out with cylinders of di¤erent diameter would
show that the inertia force component was proportional to D2 , while the drag force would
be linearly proportional to D.
Putting all this together would indicate that the force on a cylinder was of the form:

F (t) = A ¢ D2 ¢ u(t) + B ¢ D ¢ [u(t) ¢ ju(t)j]
_ (12.25)
in which A and B are constants. It is then simple enough to use dimensional analysis and
common sense to express the unknown coe¢cients, A and B as:

¼ 1
A= ½ ¢ Ca and B= ½ ¢ CD (12.26)
4 2

12.4.2 Morison Equation Coe¢cient Determination
In this section, a vertical cylinder is assumed to be …xed in a horizontal sinusoidal oscillatory
‡ow. The force per unit length acting on the cylinder can be predicted using the Morison
equation with two empirical coe¢cients:
¯ ¯
¯ ¯
¯F (t) = + ¼ ½ CM D 2 ¢ u(t) + 1 ½ CD D ¢ u(t) ju(t)j¯
_ (12.27)
¯ 4 2 ¯

The values of the dimensionless force coe¢cients CD and CM can be determined experi-
mentally in a variety of ways. The …rst step, however, is always to get a recording of the
force, F; as a function of time, t. A characteristic of the ‡ow - usually the velocity - will
form the second time function.

Experimental Setup
These measurements can be made in a variety of test set-ups.

1. Oscillating ‡ows can be generated in a large U-tube. Unfortunately the ‡ow can
only oscillate with a limited frequency range - the natural oscillation frequency for
the installation - unless an expensive driving system is installed. An advantage
of a U-tube, on the other hand, is that its oscillating ‡ow is relatively ’pure’ and
turbulence-free. A discussion continues about the applicability of results from such
idealized tests in …eld situations, however. This topic will come up again later in this
chapter.


2. A second method is to impose forced oscillations to a cylinder in still water. The
‡ow - when seen from the perspective of the cylinder - appears similar to that in a
U-tube but the inertia force is not the same. Review the material above about Ca to
understand why this is so.

3. A third possibility is to place a vertical cylinder in regular waves. The waves are
generated by a wave maker located at one end of the experimental tank; they are
absorbed on an arti…cial beach at the other end. In this case it is often the wave
height (actually the water surface elevation) which is measured as a function of time.
The horizontal water velocity and acceleration at the location of the cylinder are in
this latter case determined using linear wave theory - see chapter 5:

! H cosh [k (z + h)]
u(z; t) = ¢ ¢ cos (!t)
2 sinh (k ¢ h)
= ua (z) ¢ cos (!t) (12.28)
!2 H cosh [k (z + h)]
u(z; t) = ¡
_ ¢ ¢ sin (!t)
2 sinh (k ¢ h)
= ¡ ! ¢ ua (z) ¢ sin (!t) (12.29)

in which:

! = 2¼=T = wave frequency (rad/s)
k = 2¼=¸ = wave number (rad/m)
z = elevation (+ is upward) from the still water level (m)
H = wave height (m)
h = water depth (m)
T = wave period (s)
¸ = wave length (m)
ua(z) = amplitude of horizontal water velocity component (m/s)

Note that even though ua is now a function of z; this will not really complicate matters
when studying the forces on a short segment of a cylinder. The change in ua over such a
short distance can be neglected.
With any of these methods, the resultant force on a section of the cylinder is often measured
by mounting that section on a set of leaf springs which are equipped with strain gauges.
These - via a Wheatstone bridge circuit and a proper calibration - provide the force record,
F (t) to use in conjunction with the measured or computed u(t) and u(t)._

Data Processing
Once the necessary data time series have been obtained, one is still faced with the problem
of determining the appropriate CD and CM values. Here, again, one has several options
dependent upon the computer facilities available.
Several methods are presented here, primarily for reference purposes:

1. Morison’s Method
Morison, himself, suggested a simple method to determine the two unknown coe¢-
cients, see [Morison et al., 1950]. His method was elegant in that it was possible to


determine the coe¢cients without the use of computers. (Computers - if available at
all - were prohibitively expensive when he did his work.) His approach was suitable
for hand processing and depended upon the realization that when:
u is maximum, u is zero so that at that instant, t1; F (t1) = FD and
_
u is maximum, u is zero so that at that instant, t2; F (t2) = FI :
_
Figure 12.2 shows a sample of an idealized measurement record. Under each of the
above speci…c conditions, equation 12.27 can be re-arranged to yield:

2F
CD = at an instant t1 when u = 0
_
½D ¢ ua jua j
4F
CM = at an instant t2 when u = 0 (12.30)
¼ ½ D2 ¢ ! ua

The method is simple, but it lacks accuracy because:

Figure 12.2: Measured Force and Velocity Record

- A small error in the velocity record can cause a signi…cant phase error. Since the


curve of F (t) can be steep (especially when determining CD in …gure 12.2), this can
cause quite some error in this coe¢cient. The e¤ect on CM is usually smaller.
- Information from only two instants in the time record is used to determine the
coe¢cients; the rest is thrown away.
Morison reduced errors by averaging the coe¢cients over a large number of measure-
ments (wave periods).
One might try to use this same approach at other time instants in the record. The
only di¢culty, however, is that one is then confronted with a single equation (for F
at that instant) but with two unknown coe¢cients. This cannot be solved uniquely.
A second equation could be created by examining the situation at a second, indepen-
dent time instant. A generalization of this would be to use the data pairs at every
instant with a least squares …tting technique. This is discussed below, but only after
another approach using Fourier series has been presented.
2. Fourier Series Approach
An entirely di¤erent method for determining the drag and inertia coe¢cients is based
upon the comparison of similar terms in each of two Fourier series: One for the water
motion and one for the force. Appendix C summarizes the theory behind Fourier
series.
Since modern laboratory data records are stored at discrete time steps (instead of
as continuous signals), the integrals needed to evaluate the Fourier coe¢cients are
replaced by equivalent sums.
Looking at this in a bit more detail, the water velocity and acceleration is already in
a nice form as given in equation 12.28. A single Fourier series term is su¢cient to
schematize this quite exactly. Since the inertia force, FI , is also well behaved, it can
be ’captured’ with a single Fourier series term as well.
The only remaining problem is the series development of the drag term; this requires
the development of a function of form:
f (t) = A cos (!t) ¢ j cos (!t)j (12.31)
This has been worked out in Appendix C as well. The resulting coe¢cients (given
there as well) for the …rst harmonic development of the quadratic drag turn out to
be:
Fourier
Coe¢cient Value
Constant, a0 0
Cosine, a1 8
3¼
¢ A = 0:849 ¢ A
Sine, b1 0
As shown in Appendix C, the drag force, dependent upon u juj, develops into a series
of odd-numbered harmonics in a Fourier series; only the …rst harmonic terms are used
here. Since this has an amplitude of 3¼ times the original signal, one must multiply
8

the …rst order harmonic of the force in phase with the velocity by a factor 38¼ to get
the amplitude of the quadratic drag force.
Once this has been done, then the determination of C D and CM from the analysis of
the compete force signal, F (t) is completely straightforward. Since the inertia force
component shows up now in the b1 term of the series development, the results are:
3¼ 4 b
CD = ¢ a1 and CM = ¢ 1
2 ½ ! u
(12.32)
4½D ¼D a


in which the following amplitudes are found:

a1 = velocity-dependent Fourier amplitude (kg/m2)
b1 = acceleration-dependent Fourier amplitude (N m)

Notice that with this method one has used data from the entire time record in the
determination of the Fourier series components and thus for the determination of
CD and CM . This should be an improvement over the method used originally by
Morison, but on the other hand, it is still only as accurate as the linearization can
be.

3. Least Squares Method
A third approach treats the basic Morison equation, (12.27) as a computational
approximation, F (t; CD; C M )computed , for the measured force record, F (t)measured . One
is now faced with only the problem of determining the (linear) unknown coe¢cients,
CD and CM . This is done by minimizing some residual di¤erence (or …t criterion)
function. The method of least squares uses a residual function of the form:
Z T
R(CD ; CM ) = [F (t)measured ¡ F (t; CD; CM )computed] 2 dt (12.33)
0

in which T is now the length of the measurement record.
Now one only needs to iteratively evaluate equation 12.33 for various values of CD
and CM until the residual function, R(CD; C M ) is minimized. If one were to plot this
function in three dimensions - with C D and CM on the two orthogonal horizontal
axes and R(C D; CM ) on the vertical axis, then one would …nd a sort of ’bowl-shaped’
function. It doesn’t take too much thought to realize that if the shape of the bottom
of this ’bowl’ is rather ‡at, then there are many combinations of C D and CM which
give about the same R(CD ; CM ) function value. The consequence of this is that it
is quite di¢cult to determine the ’best’ C D and CM values exactly in a numerical
way. On the other hand, it is theoretically possible to determine the minimum of the
@R @R
function .R(CD; C M ) by setting both of its partial derivatives, @CD and @CM equal to
zero analytically.

4. Weighted Least Squares Method
The least squares method, above, uses the entire time record for the determination of
CD and CM ; it shares that advantage with the Fourier series approach. On the other
hand, one can reason that for o¤shore design purposes, it is more important that the
Morison equation predict the force peaks accurately than to be as precise at
moments when the force is nearly zero. One way to improve the …tting near the peak
forces is to weight the di¤erence found above in equation 12.33 with - for example -
the measured or computed force value. Equation 12.33 can then become something
like:
Z T
Rw (CD ; CM ) = [F (t)measured] 2 ¢ [F (t)measured ¡ F (t; C D; CM )computed]2 dt
0

Of course the shape of the residual function - the shape of the ’bowl’ - will now be
di¤erent and hopefully steeper and deeper (not so ‡at on its bottom). Even if this


is not the case, however, one can expect that a Morison equation …tted in this way
will give a more accurate prediction of the force peaks.
Note that most any researcher can dream up his own residual or criterion function to
use in this approach. Residual values are usually not dimensionless quantities;
the absolute (numerical) value of R or Rw (or any other criterion function for that
matter) is quite irrelevant; only relative values are of interest. There is certainly
no point in comparing them by, for example, comparing values of R with those of
Rw . The only important matter is that of …nding the C D and CM associated with
the minimum value of the criterion function chosen.

5. Alternative Approach
This method illustrates an entirely di¤erent approach to the problem. It was used
by Massie some years ago - in an age when digital computers were still slow enough
to make numerical integrations a cumbersome process. Instead, integrations were
carried out using an analog computer; this could carry out these nearly instantly and
painlessly. The analog computer was coupled to a digital computer which read the
results of the integration and adjusted the coe¢cients accordingly for the next try.
Such a computer was called a hybrid computer.
The solution was based upon the following approach: First the Morison equation,
12.27, was written in the following form:

F (t) = +P ¢ CM ¢ u(t) + Q ¢ C D ¢ u(t) ¢ ju(t)j
_ (12.34)

in which P and Q are simply known constants. Both u(t) and F (t) had been measured
and were known functions of time.
The special approach feature was to re-arrange equation 12.34 by solving it for u(t)_
yielding:
1 1 Q CD
u(t) = ¢
_ F (t) ¡ ¢ ¢ u(t) ¢ ju(t)j (12.35)
P CM P CM
Equation 12.35 is, thus, a …rst order nonlinear ordinary di¤erential equation in u(t)
which has a given solution - the measured u(t) - but two unknown coe¢cients: 1=C M
and CD =C M . Values for the unknown coe¢cients were set by the digital portion
of the computer; the analog portion integrated the di¤erential equation to generate
a computed uc(t) and simultaneously subtract it from the measured u(t) to give a
residual which was also integrated over a time period in the analog portion. This
integral value was the residual function to be minimized using a numerical routine in
the attached digital computer. Notice that the criterion function is now based upon
the velocity record instead of the force record! Of course, various weighting functions
were tried as well.

Five di¤erent methods of determining CD and CM (or C a) coe¢cient values from a single
time record of water motion and force have been presented here. The frustrating result of
all this is that if one time record were to be analyzed with each of these methods, each
method would yield a di¤erent pair of CD and C M coe¢cient values! One can conclude -
correctly! - from this that it is impossible to determine exact values for these coe¢cients;
a tolerance of several percent is the very best one can expect.
It can also happen that one …nds widely varying values for CD or CM when comparing
results from two di¤erent time series with very similar test conditions. This can happen


with the drag coe¢cient, for example, when F (t) is inertia dominated as it is called.
Inertia dominated implies that the drag force is relatively unimportant so that since the
rest of the information used to compute FD is relatively small, this small value times any
coe¢cient value is still small. The converse is obviously also true: The inertia coe¢cient
value is unimportant if the force is drag dominated. More about the conditions which
can lead to this will be presented below in the discussion of the relative amplitudes of the
drag and inertia forces.

Cylinder Roughness
All of the above discussion has been for a smooth-surfaced (vertical) cylinder. Since o¤shore
structures accumulate marine growth very easily in at least the warmer seas, this modi…es
the hydrodynamic force computation in two ways: First, the cylinder can become larger -
a marine growth layer of 10 centimeters thickness will increase the cylinder’s diameter by
0.2 meters. This can be accounted for quite easily in the Morison equation. The second
in‡uence is that the roughness will in‡uence the boundary layer and vortex separation near
the cylinder. The drag and inertia coe¢cient values are generally adjusted to account for
this as will be seen later in this chapter.

Presentation Parameters
Now that C D and CM values have been found for a given ‡ow condition, it is logical to
want to present these results via a graph in which CD and CM are the dependent variables
plotted along the vertical axis. One must still choose a proper independent variable for
the horizontal axis, however. This would (ideally) include information on the wave (H; T
or something related to these), the ‡uid (½; º for example) and the cylinder (D is most
obvious choice for this, but it might include the roughness, too). Several possibilities for
making dimensionless combinations are discussed in this section.

1. Reynolds number
The Reynolds number for a constant current was given in chapter 4. This is modi…ed
here for unsteady ‡ow by replacing the constant current by the amplitude of the
oscillation velocity yielding:
ua ¢ D
Rn = (12.36)
º
in which:
Rn = Reynolds number (-)
ua = ‡ow velocity amplitude (m/s)
º = kinematic viscosity (m2 /s)

2. Froude Number
The Froude number can now be expressed using the velocity amplitude as well as:
ua
Fn = p (12.37)
g¢D
in which:
F n = Froude number (-)
g = acceleration of gravity (m/s2 )


The Froude number is associated primarily with free surface e¤ects while wave forces
can be exerted on cylinder elements which are so far below the sea surface that no
surface disturbance is generated. The Froude number is not really suitable for the
present purpose, therefore.

3. Keulegan Carpenter Number
[Keulegan and Carpenter, 1958] determined C D and CM values for various cylinders
in an oscillating ‡ow. They discovered that their data could be plotted reasonably
as a function of the dimensionless Keulegan Carpenter number:
¯ ¯
¯ ¯
¯KC = ua ¢ T ¯ (12.38)
¯ D ¯
in which:
KC = Keulegan Carpenter number (-)
T = oscillating ‡ow period (s)
This number can be de…ned in alternate ways. In a sinusoidal wave, ua = ! ¢ xa ,
in which xa is the (horizontal) water displacement amplitude. A bit of substitution
then yields:
water displacement amplitude x
KC = 2¼ ¢ = 2¼ a (12.39)
cylinder diameter D
which is very likely an important characteristic for the wake formation in the ‡ow as
well.
In deep water, the water displacement amplitude xa at the sea surface is identical
to the wave amplitude. This allows still another form in this speci…c situation:
H ³
KC = ¼ ¢ = 2¼ ¢ a (deep water only) (12.40)
D D
4. Iversen Modulus
Even before Keulegan and Carpenter did their work, [Iversen and Balent, 1951] sug-
gested:

ua ¢ D
_
Iv = (12.41)
u2
a
in which:
Iv = Iversen modulus (-)
ua = ‡ow acceleration amplitude (m/s2)
_
Knowing that in a sinusoidal wave ua = 2¼ ua ; and by doing a bit of algebra, one can
_ T
discover that:
2¼
Iv = (12.42)
KC
KC is more convenient to use in practice, however.

5. Sarpkaya Beta
[Sarpkaya and Isaacson, 1981] carried out numerous experiments using a U-tube to
generate an oscillating ‡ow. He found that the ratio:

D2
¯= (12.43)
º¢T


was convenient for plotting his data.
Just as with the Iversen modulus, this can be ’processed’ a bit to reveal that:

Rn
¯= (12.44)
KC
so that this is not really anything new, either.

6. Dimensionless Roughness
Cylinder roughness is generally made dimensionless by dividing it by the diameter,
yielding:

" roughness height
= (12.45)
D cylinder diameter

The Keulegan Carpenter number has survived as the most realistic and useful primary
independent parameter for plotting CD and CM : This is sometimes augmented by using
Rn , ¯ or D to label speci…c curves, thus introducing additional independent information.
"

12.4.3 Typical Coe¢cient Values
Hundreds (at least) of researchers have conducted laboratory tests to determine CD and
CM coe¢cients in one way or another and often for very speci…c situations. In many cases,
their objective and/or experimental set-up limited their range of test conditions so that
their results are quite restricted, too. Typical results are listed in this section.
The results of Sarpkaya’s experiments with smooth cylinders in U-tubes are presented as
graphs of the coe¢cients CD and CM as functions of ¯ and K C: Note that in …gure 12.3
the horizontal (K C) axis is logarithmic. Individual curves on each graph are labeled with
appropriate values of ¯.
[Clauss, 1992] for example suggests drag and inertia coe¢cient values given in the following
table:

Rn < 105 Rn > 105

CD CM CD CM
KC
< 10 1:2 2:0 0:6 2:0
> 10 1:2 1:5 0:6 1:5
Morison Coe¢cients Suggested by [Clauss, 1992]

Various design codes or rules also specify (or suggest) appropriate values for CD and CM .
Those published by [DetNorskeVeritas, 1989] or the American Petroleum Institute (API)
are the most widely accepted; the DNV suggestions for design purposes are shown in …gure
12.4.
The API as well as the SNAME have the simplest approach as listed in the table below:


Figure 12.3: Typical Laboratory Measurement Results from Sarpkaya


Figure 12.4: Suggested Drag and Inertia Coe¢cient Values from DNV


Smooth Rough

CD CM CD CM
API 0:65 1:6 1:05 1:2
SNAME 0:65 2:0 1:0 1:8

The following observations can be made from the above information:
- For low values of K C, the inertia coe¢cient C M is almost equal to its theoretical value
of 2 - at least if the K C value is used as a selection parameter. Also, one can notice
that the drag coe¢cient CD generally increases or stays rather constant till a value
of KC near 10 is reached.

- One sees as well from …gure 12.3 that the CD value gets lower as ¯ increases. This is
just the opposite of the trend observed with CM :

- Comparison of Sarpkaya’s curves (…gure 12.3) with those from DNV (…gure 12.4) show
that there can be quite some discrepancy in the value of CD or CM to choose.

- The DNV curves (…gure 12.4) as well as the other design and assessment codes include
roughness - that can easily result from marine growth, especially near the sea surface.
The roughness in …gure 12.4 is an D ratio.
²

- The API and SNAME recommendations seem rather simple in that they neglect the
KC number; Clauss adds that e¤ect, but in a more simple way that suggested by
DNV.

Comparisons
Examination and comparison of the various drag and inertia coe¢cient values presented
above shows that there is little agreement on exact values. This is true for smooth cylinder
values and even more so when a rough cylinder is involved. Di¤erences of up to roughly
40% can be found when comparing the drag or inertia coe¢cients suggested by the various
sources for a speci…c ‡ow situation.
This direct comparison of coe¢cient values can be misleading, however. In some cases a low
drag coe¢cient value can be at least partially compensated by a larger inertia coe¢cient.
After choosing a typical cylinder diameter and wave conditions, one can select appropriate
coe¢cients from each of the sources and compute the actual maximum force per unit length
upon which to base a comparison. Such an exercise can still lead to di¤erences of up to
about 30%. Luckily for survival design of o¤shore structures, the di¤erences found for
extreme wave conditions are generally less than this!
Even this comparison need not be correct. The ’purest’ approach is to select a typical
structure, and place it (in one’s mind) at a given location in the sea. Given this, one
should follow the entire procedure (given in each particular design or analysis code) to
select wave and current conditions and to translate these into forces on that structure.
These resulting forces should be compared. Carrying out such a comparison operation is
beyond the scope of this text, however.
One additional discovery that one will make when computing forces under …eld conditions
is that Sarpkaya’s data is a bit restricted for this. Indeed, the Reynolds numbers - needed
for his ¯ parameter - are much too low in nearly all laboratory situations.


12.4.4 Inertia or Drag Dominance
Now that CD and CM values have been presented, one should further re‡ect upon their use
and importance. The Keulegan Carpenter number can be a very important parameter for
this. Indeed, it can be used as an indication of the relative importance of drag versus inertia
forces in a particular situation. To prove this, one must work out the ratio of the amplitudes
of the drag and inertia forces. The 90± phase di¤erence between the force components is
completely neglected now; only the force component amplitudes are compared.

1
Fdraga ½ CD D ua juaj
2
= ¼
Finertiaa ½ CM D2 ! ua
4
2 CD jua j
= (12.46)
¼ CM D !
Note that the maximum value of juaj is the same as that of ua .
Since ! = 2¼=T ; then this can be reduced a bit to:

Fdraga 1 CD ua ¢ T
= ¢ ¢
Finertiaa ¼2 CM D
1 CD
= ¢ ¢ KC (12.47)
¼2 CM

Since 1=¼ 2 ¼ 1=10 and the value of CD is often a bit more than half the CM value, the
two force component amplitudes are about equal when K C is in the range of roughly 15
to 20:
Remembering the earlier de…nition of KC from equation 12.39:
2¼ xa
KC = (12.48)
D
then this means that xa=D will be about 3; this is big enough to generate a very respectable
set of vortices.
The Morison equation includes a nonlinear (quadratic drag) term which is 90± out of phase
with the inertia force. Many o¤shore engineers want to avoid using the entire Morison
equation (and the quadratic drag computation especially) unless it is absolutely necessary.
It would be convenient to have a simple way to justify neglecting either the drag term or
the inertia term in that equation. The Keulegan Carpenter number is an excellent help
with this:
² For low values of K C (K C < 3), the inertia force is dominant. The ‡ow ’does not
travel far enough’ relative to the cylinder diameter to generate much of a boundary
layer not to mention vortices; potential ‡ow theory is still applicable. Drag can
simply be neglected.

² For the next range until drag becomes signi…cant (3 < KC < 15); one will often
linearize the drag as has been explained earlier in this chapter.

² There is a range of K C (15 < KC < 45) in which one cannot really avoid using the
full Morison equation with its nonlinear drag.


² For high values of KC (KC > 45), the drag force is dominant. The vortex
shedding frequency becomes high compared to the wave frequency so the ‡ow tends
to behave more and more like a uniform ‡ow. Inertia can be neglected. Indeed,
the limit K C ! 1 corresponds to a constant current.

12.5 Forces on A Fixed Cylinder in Various Flows
This section describes the forces acting on a …xed cylinder in currents and/or waves. While
parts of it may seem like repetition of earlier work, its objective is to clarify the underlying
principles.

12.5.1 Current Alone
A …xed cylinder in a current alone will experience only a quadratic drag force (per unit
length) as already indicated in chapter 4. This force is assumed to be caused by the ‡ow
component:
Up = U sin ·
acting perpendicular to the cylinder axis so that the force can be expressed as:
1
Fc = ½ U 2 D CD sin2 · (12.49)
2
In these equations:

U = Total velocity vector (m/s)
Up = Perpendicular velocity component (m/s)
CD = Drag coe¢cient for constant current (-)
· = Cone angle between the velocity vector, U;
and the cylinder axis.
Fc = Current force per unit cylinder length (N/m)

See …gure 12.5 for a sketch showing the cone angle. The force, Fc ; will act in the direction
of Up of course; this is perpendicular to the cylinder axis and in the plane de…ned by the
cylinder axis and the approaching velocity vector, U:
Note that only the so-called cone angle, ·; is important in this computation. This is
su¢cient to describe the orientation of the cylinder relative to the current vector. It makes
no di¤erence whether the cylinder is in a vertical, horizontal or any other plane; it is only
the angle between the cylinder axis and total velocity vector which is important. (This
will be generalized to include inertia forces in waves below.)

12.5.2 Waves Alone
The time dependent ‡ow associated with waves requires the inclusion of inertia force com-
ponents into a force computation. Indeed, the basic Morison equation - derived for a unit
length of a …xed, vertical cylinder in waves is re-stated here for reference:
¯ ¯
¯ 1 1 ¯
¯F = ¼ ½ D M u(t) + ½ C D D u(t) ju(t)j¯
2 C _ (12.50)
¯ 4 2 ¯

12.5. FORCES ON A FIXED CYLINDER IN VARIOUS FLOWS 12-23

Figure 12.5: Cone Angle De…nition

in which:
F = Force per unit length of cylinder (N/m)
D = Cylinder diameter (m)
u(t) = Horizontal velocity component (m/s)
u(t) =
_ Horizontal acceleration component (m/s2)
How can this be generalized in the light of the above information on currents for a cylinder
having a non-vertical orientation? The following steps are suggested in which the inertia
force and the drag force are considered to be separate entities until the end. The following
sequence of steps must be carried out sequentially and at each time step for which results
are desired::

1. Determine the instantaneous water kinematics: velocity and acceleration (magnitudes
as well as directions) in a …xed x; y; z axis system. Relate their phase to that of a
reference such as the wave pro…le.

2. Knowing the cylinder axis orientation (in that same x; y; z axis system), determine the
instantaneous cone angles, ·I and ·D for the acceleration and velocity respectively.

3. Determine the instantaneous perpendicular components of acceleration and velocity
- up and up - as well as their directions. Use the results from the two previous steps
_
to do this. These two vectors will not generally be co-linear; they are both in the
plane perpendicular to the cylinder axis, however.

4. Evaluate the inertia and drag force components at each instant. Don’t forget that
the drag is quadratic! The direction of each force component will correspond to that
of its associated kinematics found in the previous step.

5. If the force on an entire member is needed, then now is the time to integrate these
separate force components over the length of the member in order to determine each
resultant at that time. The member support forces - and thus the equivalent loads
to be applied at the structure nodes - can be found by treating each member as a
simple beam.


6. Since the inertia and drag force components are not generally colinear, they can
be combined (if desired) via vector addition to yield the resulting force magnitude
and its direction (still in a plane perpendicular to the cylinder axis). This step is
not absolutely necessary for the computation of resulting forces on a large structure,
however.

Note that these …ve or six steps must be repeated at each instant in time. These steps are
not di¢cult in principle, but very careful bookkeeping is essential!
In many simple cases, each of the quantities needed for this methodology will be express-
ible in terms of nicely behaved and convenient functions so that the resulting force can be
described as one or another continuous time function. On the other hand, if the wave is
irregular and thus composed of many frequency and direction components, then the neces-
sary bookkeeping becomes too cumbersome for a hand calculation. The only requirement
for the force computation is that the water acceleration and velocity be known at any time.

Special Orientations
One can check his or her understanding of the above by evaluating the forces acting on
two special cases of a horizontal cylinder in a regular wave. These are in addition to the
vertical cylinder used during the Morison equation derivation.
If the horizontal cylinder segment is oriented with its axis parallel to the direction
of wave propagation (and thus perpendicular to the wave crests), then it will experience
a vertical force which has a time trace which looks much like that for a vertical cylinder -
see …gure 12.2. This force record will be shifted 90± in phase relative to a similar record
for a vertical cylinder, however. The relative phases of the resulting drag and inertial force
components on consecutive segments of the cylinder will correspond - with some constant
shift - to that of the wave pro…le on the sea surface.
The second case has the horizontal cylinder turned parallel to the wave crests (and
thus perpendicular to the direction of wave propagation). If the cylinder is situated in a
deep water wave (in which the horizontal and vertical kinematic components have the same
magnitudes) then one will …nd a resultant force of constant magnitude which sweeps around
the cylinder once per wave period. It may seem strange, but the horizontal component of
this force will have a purely sinusoidal form (except for a possible phase shift) independent
of the fact that quadratic drag is involved. Force components on consecutive segments of
this cylinder will have identical phases in this case as compared to the previous one.

12.5.3 Currents plus Waves
It is generally accepted practice to vectorially superpose the current velocity on the velocity
resulting from the waves before calculating the drag force. In a general case the wave and
current directions will not be co-linear making a vector sum necessary. Once this has been
carried out, however, one simply has to use the sequential steps given above to determine
the resulting force at any instant.
Why is it not acceptable to compute the wave force and the current force separately? The
current has no e¤ect at all on the ‡ow accelerations so that the inertia force is unchanged
by the current. The di¢culty lies with the quadratic drag force. Since:

Up + u2 < (Up + up )2
2
p (12.51)

12.6. FORCES ON AN OSCILLATING CYLINDER IN VARIOUS FLOWS 12-25

then a segregated treatment of the current drag and wave drag - which are superposed only
at the end of the computation - will lead to an underestimation of the forces involved.

12.6 Forces on An Oscillating Cylinder in Various Flows
Now that the hydrodynamic interaction of a …xed cylinder in a variety of ‡ows has been
explained, it is appropriate to discuss the hydrodynamic interaction of a moving cylinder
- again in a variety of ‡ow conditions.
A distinction will now have to be made between the (external) force exerted by the cylinder
on the surrounding water and the (internal, structural) force needed to cause the cylinder
(segment) to oscillate. In general, the internal force will often be one that is measured -
especially in a laboratory setting. This force includes the external hydrodynamic force but
also includes a force needed to accelerate the cylinder itself. Also, one should remember that
the hydrodynamic interaction force components will generally be in a direction opposite to
the actual velocity and acceleration of the cylinder.

12.6.1 Still Water
As indicated much earlier in this chapter, the Froude-Krilov force will be absent since
there are no ambient pressure gradients in water which is at rest. The inertia force will
be associated with a Ca value and there will be a drag force - associated with CD as well.
Analogous to the assumption made for a …xed cylinder, these forces will be associated with
the cylinder kinematics components (velocity and acceleration) which are perpendicular to
the cylinder’s axis.
The …ve steps used for determining the forces on a …xed cylinder can used here too, albeit
that the kinematics now is that of the cylinder instead of the water.

12.6.2 Current Alone
This interaction situation has already been discussed to some extent in chapter 4. One
should remember that the direction of cylinder oscillation and the current direction may
be quite di¤erent. Indeed, a vortex-induced vibration usually has its largest component
more or less perpendicular to the current direction. This results from the lift force - the
most important dynamic force in this ‡ow situation - which was discussed in chapter 4.
The drag component of the hydrodynamic interaction was quite well described in section
6 of that chapter too; there is no need to repeat that here.
Inertia forces will - in principle - now be present, too. They will be associated with a
Ca value since there is still no ambient time-dependent pressure gradient. These forces
will be opposite to the acceleration component which results exclusively from the cylinder
oscillation in this case.
In many realistic cases the approaching ‡ow velocity will be considerably larger than the
cylinder’s oscillation velocity. Also, since most oscillating cylinders are rather slender (an
umbilical cable to a ROV is an excellent example) the K C number will be large so that
inertia forces will be small anyway. In many practical situations, then, one considers only
a drag force as if it were exerted on a …xed cylinder. The drag coe¢cient is sometimes a
bit larger to account for the wider wake resulting from the cylinder oscillation.


12.6.3 Waves Alone
The inertia and drag forces are treated entirely separately here for clarity.

Inertia Forces
Waves will contribute both a Froude-Krilov force (from the ambient, time-dependent pres-
sure gradient) as well as a disturbance force from the encounter with the solid cylinder.
The cylinder oscillation, on the other hand, plays no role in the Froude-Krilov force but
it does contribute to the disturbance term. (It is implicitly assumed here that the motion
of the cylinder is small relative to the wave length so that no phase changes result from
this displacement. It is hard to conceive of a practical situation for which this assumption
does not hold.) When one keeps in mind that the direction of cylinder oscillation need not
coincide with the wave direction, then careful bookkeeping is called for.
One …nds the following inertia terms in the equation of motion:

M X(t) ¯ CM MD up (t) ¡ Ca MD X(t)
Ä _ Ä (12.52)

in which:
M = Mass of the cylinder segment (kg/m)
MD = Displaced water mass = ¼ D2½ (kg/m)
4
up(t) =
_ Perpendicular acceleration component from the waves (m/s2)
Ä
X(t) = Cylinder acceleration (m/s2)
The symbol ¯ has been used to segregate terms from the left hand side of the full equation
of motion from those on the right. Since only selected terms are included, true equality
cannot be guaranteed. The above relationship can be re-arranged by splitting the wave
force term into its two components so that:

M X(t) ¯ 1 MD up(t) + Ca MD up (t) ¡ Ca MD X(t)
Ä _ _ Ä (12.53)
If the cylinder acceleration corresponds exactly - in both magnitude and direction - to that
of the waves, then the last two terms in this latter equation cancel. This is logical; there
is then no disturbance at all and only the Froude-Krilov force remains.
In the more general case, all three hydrodynamic force components in equation 12.53 will
be present. It is often convenient to move all the force terms involving the cylinder motion
to the left hand side of the equation so that it becomes:

(M + C a MD ) X(t) ¯ C M M D up(t)
Ä _ (12.54)

This isolates the (unknown) cylinder motion on the left hand side of the equation and
places the time-dependent external exciting force on the right. This right hand side can
be evaluated (without knowing the cylinder motion) as a pure time function before the
di¤erential equation of the cylinder motion is evaluated.

Drag Forces
Drag forces result from the ‡ow disturbance and wake near the cylinder. Two quite dif-
ferent approaches to the description of hydrodynamic drag are used. They are discussed
separately here.

12.6. FORCES ON AN OSCILLATING CYLINDER IN VARIOUS FLOWS 12-27

Relative Velocity Approach It is reasonably simple to postulate that this wake de-
pends upon the motion of the water relative to the (moving) cylinder - the relative velocity:
_
u ¡ X. It results in a drag force proportional to the square of this relative velocity so that
one …nds the following velocity-dependent terms in the equation of motion:
¯ ¯
1
c X(t) ¯ ½ CD D (up(t) ¡ X (t)) ¯ up (t) ¡ X(t)¯
_ _ ¯ _ ¯ (12.55)
2
in which:
c = Material damping coe¢cient (N ¢ s/m)
up(t) = Time-dependent perpendicular water velocity (m/s)
_
X(t) = Time-dependent cylinder velocity (m/s)
The linear damping term on the left hand side of relation 12.55 involves the internal material
damping of the cylinder itself. This has nothing to do with hydrodynamic interaction which
is concentrated on the right hand side of the relation.
Notice that the quadratic nature of the drag force makes it impossible to segregate the
(unknown) cylinder velocity from the ambient water velocity when computing this exciting
force in this way. It is not possible compute this time-dependent excitation force component
from the water motion independent of the (also time-dependent) dynamic response of the
cylinder. This requires simultaneous step-by-step solution of the di¤erential equation in
the time domain. Additionally, an extra iterative loop must be included within each
_
time step in order to successively approximate values of X until the entire di¤erential
equation is satis…ed at each time step. This makes such time-domain computations rather
time-consuming - even with modern computers.

Absolute Velocity Approach A strict interpretation of this approach uses the princi-
ples of superposition in much the same way as they are used in the hydrodynamics of larger
structures. The forces resulting from the combined motion of the cylinder in waves plus
currents is treated as if it were made up of two independent phenomena: A force caused
by the waves plus current on a stationary cylinder proportional to up jupj plus a separate
_
force exerted on a cylinder oscillating in still water which is proportional to X 2. This
_
approach inherently ’fails to see’ the cross product (¡2 up X) term included automatically
in the relative velocity approach. Further, since the motions of the cylinder will usually
be considerably less that those of the surrounding water, the largest contribution to the
(external) drag force will therefore come from the up jup j term. This can be left alone
on the right hand side of the equation; it can be evaluated quite easily. The relatively
_
small X 2 term can be linearized and moved to the left hand side of the equation of mo-
tion. This linearized drag force now behaves in the same way as a linear damping; it has
the same e¤ect as increasing the structural damping which one normally includes in such
computations.
Alternatively, one can use an even more pragmatic approach by noticing that the latter two
terms of the time-dependent product in the expanded version of relation 12.55 involving
u are both smaller than the …rst term and are often of opposite sign; their combined
e¤ect will be small. One now linearizes this small e¤ect and associates it entirely with the
_
cylinder velocity, X; and treats it as above.
Even if the structure’s damping coe¢cient is not modi…ed, the result of either of these
approaches can be that the full di¤erential equation of motion for the cylinder has been


linearized and that the wave-caused excitation force has been isolated on its right hand
side. Each motion is treated as if it is in a …xed axis system. It is for this reason that this
approach is often referred to as the absolute motion approach.

Comparison Designers tend to be conservative in practice. As a result of this, they tend
not to modify the linearized damping when using the absolute motion approach. Since a
dynamic system with a lower damping will often have a larger response, the absolute motion
approach most usually leads to higher predicted structural responses (internal stresses or
even displacements, for example) than does the relative velocity approach. This is indeed a
strong motivation to use the absolute velocity approach when evaluating the performance
of a proposed design. The additional advantage of having a straightforward and simple
computational procedure comes as an added bonus.

12.6.4 Currents Plus Waves
Just as with the …xed cylinder, all hydrodynamic velocity components are generally super-
posed before starting a force computation. Once this has been done, the further treatment
is identical to that for an oscillating cylinder in waves alone as discussed above.

12.7 Force Integration over A Structure
The discussion above has concentrated on the forces on a unit length of cylinder or at least
no more than a single member of a large truss structure. It is now time to integrate these
forces over the length of the cylinder in order to determine loads which are relevant for
structural analysis and design evaluation.
Since many space truss analysis computer programs work with externally applied joint
loads, the goal should be to transform the distributed hydrodynamic load on each member
of the structure to equivalent concentrated loads at the structure’s nodes. Generally, this
procedure will have to be followed during a whole series of discrete time steps in order to
generate time-dependent loadings which may be needed for a dynamic analysis.
Many computer programs for this purpose work somewhat as follows:
As preparation,

² The nodes of the structure are numbered sequentially.

² Each node is assigned a set of X; Y; Z coordinates corresponding to the intersection
point of the member axes at that joint. (Any member eccentricity at the joint is
neglected in the hydromechanics.)

² Each member is speci…ed by its diameter and the numbers of the two nodes which it
joins. Each member’s geometric position and length is now de…ned.

Once this has been completed the following steps will be carried out for the dynamic loads
at each desired time step:

1. The water kinematics - both combined velocities and accelerations - will be computed
at each node location. It makes no di¤erence now whether the waves are regular or
irregular; there can even be directional spreading or additional currents involved.

12.7. FORCE INTEGRATION OVER A STRUCTURE 12-29

The result in all cases is known water kinematics at each node and at every chosen
time.

2. If a dynamic calculation is included, and relative velocity drag is used, then the
structure’s velocity and acceleration at each node will have to be estimated as well.
This is usually done by incrementally integrating the structure’s equations of motion
by working from the previous time step.

3. The results of the above two steps along with the known geometry allow the com-
putation of the force per unit length (for inertia and drag forces separately) at each
end of each member. These load intensities will have to be computed separately for
each member and at each of its end joints.

4. Most programs now consider each member to be a simple beam carrying a distributed
load which is assumed to vary linearly from one end to the other. Note that both the
intensity of the loading as well as the vector direction of the loading (about the axis
of the cylinder) may vary along its length. (Direction variation can be present when
the waves and current are not co-linear. It is also present when wave directional
spreading is involved.)

5. Basic mechanics yields the two reaction force components (relative to each member’s
axis!) at each of its ends.

6. It is now only a matter of bookkeeping to transform these individual member reaction
forces into equivalent X; Y; Z force components at each joint.

7. The above X; Y; Z components at each joint are summed to determine the total
equivalent applied dynamic load at that joint at the chosen moment. This is the
desired result.

8. If relative velocity hydrodynamics is being used this result must be checked. The
equations of motion of the structure must be integrated now to determine its new
velocity and acceleration at the end of the time step. If these values do no correspond
(within a chosen tolerance) with those estimated in step 2, then the above steps
must be repeated for another iteration cycle - within this same time step - until the
estimated and computed values appropriately coincide.

It is the extra iteration loop discussed in the last of the above steps that makes the rel-
ative velocity approach to hydromechanics so computationally ine¢cient relative to other
approaches.
Constant static loads, such as member weight and buoyant forces, can be included in the
above computation, but can be computed more e¢ciently separately before starting on the
dynamic computations. The computation principle is analogous to that for the dynamic
loads, by the way.


.

Chapter 13

SURVIVAL LOADS ON TOWER
STRUCTURES

13.1 Introduction
Chapter 12 has discussed how to compute the hydrodynamic forces on an element (of
unit length) of a single (slender) cylinder; a simpli…ed means of estimating the largest
hydrodynamic forces on an o¤shore tower structure will be handled in this chapter.
Figure 13.1 shows an isometric drawing of one of the larger o¤shore structures. The
annotations in that …gure will become clear in the course of this chapter.
Some refer to such a structure as a space frame, others see it as a space truss, some just
call it a tower. The terms ’frame’ and ’truss’ have quite di¤erent structural engineering
connotations which are not at all relevant to the hydrodynamic discussion in this chapter.
Another structural distinction is that a jacket is supported from the top by piles driven
through its legs while a tower is generally supported from below by piles driven through
sleeves - usually at the sea bed. This distinction is also irrelevant for the hydrodynamics
being discussed here; the term tower will be used more generically in this chapter to refer
to any three-dimensional structure made up from slender elements.
Imagine the bookkeeping and computational e¤ort needed to compute the total hydro-
dynamic forces on the structure shown in …gure 13.1 if this were to be done using the
elemental methods of chapter 12. Such a rigorous computation for a ’real’ tower structure
(with all its members: chords, braces, risers, etc.) is a very cumbersome undertaking,
indeed.
Many design engineers have no need for the computational accuracy suggested by a detailed
schematization of such a complex o¤shore structure - at least not during the preliminary
design phase. Instead, one more often needs a fast method of making a rough and preferably
conservative estimate of the hydrodynamic forces on a complex o¤shore structure. Such
a method can serve two purposes: Give a rough estimate for preliminary design or assure
that a detailed model is not making a major error. The objective of this chapter is to
outline a ’quick and dirty’ method to estimate the horizontal loads on a tower structure.
These loads generally yield the largest overall bending moments in the structure and thus
axial leg forces as well as the largest horizontal shear forces. These forces and moments are
also important for the foundation design. Maximum in-service bracing loads result from
0

13-2 CHAPTER 13. SURVIVAL LOADS ON TOWER STRUCTURES

Figure 13.1: Typical Larger O¤shore Tower Structure

13.2. ENVIRONMENTAL CONDITIONS TO CHOOSE 13-3

the horizontal wave and current loads as well.

13.1.1 Method Requirements
The results obtained with any approximation need not be exact - especially if one can
reasonably predict whether the ’real’ loads will be smaller (or larger) than the estimates.
Generally - and for preliminary design in particular - it is handy if the hydrodynamic loads
resulting from the approximation are larger than those which would follow from a more
sophisticated analysis.
When a preliminary design is based upon loads which are overestimated, it is most likely
that the resulting structure will be ’over-designed’; it will be a bit too big, or heavy, or
strong and thus probably too costly. Given this fact, then one would not expect the costs
of a structure resulting from a more detailed design to come out too high. Said in another
way, if the preliminary design survives an economic analysis, then the …nal design has a
good chance of surviving this too - at least to the extent that its total cost is determined by
hydrodynamics. An additional factor in practice is that topsides tend to become heavier
(as more equipment is added) rather than lighter during the design process. (Whether one
likes it or not, topside structures usually tend to get larger and heavier in the course of
their detailed design. This can result from equipment or throughput changes as well as
from modi…ed environmental or safety requirements.) To the extent that the tower design
is dictated by topside weight (if the water is not all that deep depending upon the sea
conditions), a bit of initial tower overdesign can prove to be handy during the later, more
detailed design analysis phase.
The objective of this chapter, therefore, is to come up with a computational procedure to
conservatively predict the hydrodynamic forces on a complex o¤shore tower structure.

13.1.2 Analysis Steps
Any hydrodynamic analysis of an o¤shore structure involves the following steps:
1. Selection of environmental conditions (raw data).
2. Schematization of the ambient hydrodynamics.
3. Schematization of the structure.
4. Computation of resulting (survival) forces and overturning moments.
These steps will be followed in the remainder of this chapter.

13.2 Environmental Conditions to Choose
Since one is looking for a maximum external load condition, it is common that this will
be caused by a maximum environmental input condition. (A signi…cant dynamic response
- which via resonance can lead to enhanced internal loads - can lead to internal response
maxima (leg forces, for example) which are caused by more moderate external sea condi-
tions. This is not considered here; it is not that common with tower structures, anyway.)
All of this means that extreme wind, wave and current conditions should be chosen for
this …rst design. Each of these involves at least two independent variables: Speed (in some
form) and direction. All of environmental inputs can be represented as vectors, but their
scalar magnitudes and directions are discussed separately here.


Wind Speed

Wind loads on o¤shore structures often play a relatively minor role in comparison to the
hydrodynamic loads. For an o¤shore wind turbine, even - on which one would expect
to have a relatively high wind load - the wind load is seldom greater than the combined
wave and current load unless the structure is placed in water somewhat less than about 20
meters deep (in the North Sea). This implies that the selection of design wind conditions
is often not all that important. When one does want to estimate wind loads, a maximum
one-minute wind gust is often chosen. This wind speed is usually measured at a ’standard’
elevation of 10 meters above the sea surface.

Current Speed

Maximum current speeds are usually chosen for survival design purposes as well. One could
select a speed corresponding to a maximum spring tide current, for example. In some cases
a velocity pro…le giving the current as a function of depth will be available, too. If not, it
is of course conservative to assume that the maximum current acts over the entire depth.

Wave Height and Period

One should choose wave height and period values such that a maximum wave force or
overturning moment is obtained. A high wave is obviously needed. If one assumes that
the worst part of a storm will have a duration of about 3 to 6 hours and that an extreme
wave - if seen as a single wave - will have a period of in the order of 15 to 20 seconds,
then one can expect an exposure to something in the order of 1000 waves during the peak
of the storm. The highest wave in a series of 1000 would have a chance of exceedance of
1000 : Substituting this in a Rayleigh wave height distribution yield a design wave height of
1

1:86 times the design signi…cant wave height chosen. (If one works out these limits more
exactly, one should expect something more than 540 waves and less than 1440 of them;
1 1
one thousand is pretty close to the average - by chance. If 540 and 1440 are used instead,
the wave height ratio is in the range 1.77 to 1.91; this makes no more than 5% di¤erence.)
Selecting the shortest wave period consistent with the chosen wave height will yield max-
imum water velocities and accelerations - at least near the water surface. On the other
hand, the hydrodynamics of waves with shorter periods ’die out’ faster at deeper locations;
a proper balance must be found between these two requirements. Wave breaking will, of
course, put a lower limit on the wave period for a given wave height; a very high and very
short wave will break so that one should also check any chosen combination of these to be
sure that the wave is not broken. The necessary relationships (for deep water and a quick
estimate) include:

H 1
¸ = 1:56 ¢ T 2 and < (13.1)
¸ 7
in which ¸ is the wave length (m), H is the wave height (m) and T is the wave period (s).
The above relations - which come directly from regular wave theory of chapter 5 - are less
dependable for larger and irregular waves at sea, however.
Since H will generally be rather large, ¸ will not be small, either. The signi…cance of this
will show up below.

13.2. ENVIRONMENTAL CONDITIONS TO CHOOSE 13-5

Figure 13.2: Schematic Plan of Tower and Approaching Wave Crests

Wind, Current and Wave Directions
In general, each of these independent physical phenomena will have its own direction,
independent - at least to some extent - from that of others. This is most obvious for the
relation between the tidal current direction and the wave direction. These seldom have
much correlation. The wind direction and the wave direction - in a major storm at least -
is usually rather well (but not perfectly) correlated of course.
The conservative choice (which is easy to work with too!) is to simply assume that all
three of these phenomena are colinear; they all come from the same direction.
Be sure to keep the direction bookkeeping correct. The wind direction is speci…ed usually
as the direction from which it comes, while a current direction is most commonly stated as
the direction to which it goes. It sounds inconsistent, but a northwest wind and southeast
current go in the same vector direction.
An independent direction consideration involves the orientation of the structure (about its
vertical axis) relative to the environmental conditions. Consider a simpli…ed situation as
shown in plan in …gure 13.2. The solid circles represent the legs of the tower as seen from
above. Two possible wave and current approach directions (indicated by the wave crests)
are shown: One with crests parallel to a face of the structure (and its X axis) and one with
crests parallel to its diagonal - the D axis in the …gure. If one assumes for simplicity (for
now) that the total external horizontal force as well as the overturning moment about an
axis at the sea bed is independent of the approach direction, one will still …nd larger axial
pile forces when the wave crests are parallel to a diagonal. On the other hand, since the
horizontal shear forces within the structure are carried primarily by the bracings, waves
approaching parallel to one side of the tower will generally lead to maximum forces in these
members.This is because the entire shear load is carried by only half of the braces then;
braces parallel to the wave crests carry essentially no load.

Phasing
All of the independent environmental phenomena - wind, waves and tides - are time de-
pendent. Do all of the maxima selected above then occur at one and the same instant?
Formally, the answer to this lies in a comparison of the periods of the various phenomena.
Since the tidal period (12 h 24 min) is long relative to the wind gusts and waves, it is


almost certain that a high wave or strong wind gust can occur when the tidal current is
high.
Looking next at the wind gust relative to the wave, its duration (1 minute) would be a
few wave periods long. Here, again, there is at least some …nite chance that a major wave
peak will coincide with this in time. On the other hand, many more waves will occur at
times when the design wind gust is not present!
Of course, assuming that the maxima of wind, waves and current do occur simultaneously
will lead to a conservative result; this is chosen here.

Implications
The choices made above already have signi…cant implications for the computations to be
carried out. For example, the wave length, ¸ , chosen above will be rather large relative to
the horizontal dimensions of (most) o¤shore structures being considered. This means that
there will be relatively little phase di¤erence between the ’upstream’ and ’downstream’
sides of the entire structure at any instant in time.
In chapter 12 the phase shift from one side to the other side (of a single cylinder) was
neglected. Now, with an extreme wave, the same reasoning is being applied to an en-
tire tower structure. Since phase di¤erences only tend to reduce the horizontal loads on
the structure, neglecting these will be conservative. Additionally, neglecting this phase
di¤erence will simplify the formulation of the structural model enormously.
Since the wave is relatively high, the Keulegan Carpenter number KC = ¼¢H (at the sea
D
surface and in deep water) will tend to be high as well. This indicates that the wave forces
will tend to be drag dominated - at least at the sea surface where they are largest.
Note that if the structure is not in deep water, the actual KC value will be even higher
than that estimated above; this can be checked by using the complete equations - rather
than deep water approximations - to describe the water motion. When the current is added
to the water velocity caused by the wave, then this drag dominance becomes even more
pronounced, of course. In all cases, the water velocity will decrease with depth so that the
KC value will decrease as well. One can expect the inertia force to increase in importance
as one moves down along the structure. A signi…cant current can prevent it from ever
playing much role, however.
Remember too from wave kinematics that the maximum horizontal water velocity in a
wave occurs under that wave crest or trough. The total resulting force (integrated over
the structure) will be greatest when the wave crest passes by, simply because more of the
structure is then exposed to the wave. Wave kinematics will have to be predicted over the
entire height from the sea bed to the wave crest in order to carry out such a computation.

13.3 Ambient Flow Schematizations
This section discusses the numerical models needed to translate the raw data on environ-
mental conditions into input data for a force model.

Wind
Wind velocity distributions have been discussed in chapter 4. If - as is standard meteoro-
logical practice - the wind is given at a standard elevation of 10 meters above mean sea

13.3. AMBIENT FLOW SCHEMATIZATIONS 13-7

level, then wind speeds at other elevations are often predicted from this value by using:

Vtw (z) ³ z ´0:11
= (at sea) (13.2)
Vtw (10) 10
in which:

z = desired elevation (m)
Vtw (z) = true wind speed at elevation z (m/s)
Vtw (10) = true wind speed at 10 meters elevation (m/s)

The exponent 0:11 in equation 13.2 is for sea conditions only; see chapter 4.
Some designers often reason that since wind loads on many o¤shore structures are often
relatively unimportant and hydrodynamic loads tend to be over-estimated, they can just
neglect wind loads altogether. This is certainly the very simplest approach that can be
chosen. On the other hand, the wind load has the longest moment arm when one is studying
the overturning moments about the structure’s base. It is best not to neglect them in the
overturning moment computation at least.

Waves
A maximum wave at sea results from the superposition of a large number of wave spectrum
components. The objective here is to replace this multi-component wave with a single
regular wave for computation purposes. Linear wave theory is certainly convenient for
predicting the water kinematics within such a wave, but it has one important drawback:
It predicts water motions only in the zone below mean sea level. As has been indicated
above, the water motions right up to the wave crest will have to be predicted. Methods
for doing this have been given in chapter 5. Of these, the use of a constant velocity - the
same as the velocity at z = 0 - or Wheeler stretching are the most popular.
Remember from chapter 5 that the (extreme) wave crest will be higher than H=2 above
the still water level as well. A common rule of thumb is that:

2 1
³ max = + H and ³ min = ¡ H (13.3)
3 3
With this, one can use either constant extension or Wheeler stretching along with the
complete equations for the water motion in order to calculate the horizontal components
of water particle acceleration and velocity at all elevations between the sea bed and wave
crest. All details of this can be found in chapter 5.
One should note the following about the waves formulas to be used:
1. If only a maximum velocity is needed, the time function in the wave can be neglected.
2. Since the wave length is usually considerably larger than the horizontal dimensions
of the total structure being considered, the phase relation, kx can be dropped too.
3. If the wave crest is involved, then the maximum crest elevation will follow from
equation 13.3.
4. The full equations (and not deep or shallow water approximations) must be used
when evaluating the horizontal kinematics in the wave. This is then valid for any
water depth and at any point under a wave pro…le.


Use of deep water wave theory - with its simpli…cations - can lead to less than conserva-
tive results as the water depth decreases; this is the reason that use of the full theory is
recommended above.
The wave height used should be the correct one for the actual water depth, h. Include the
shoaling in‡uence, if appropriate; see chapter 5. Use this same actual water depth when
computing ¸ as well.

Current
Since the current is constant and in the same direction as the wave propagation, it can
simply be added to the velocity component amplitude, ua (z), computed for the wave. If
the current velocity is given by V (z) , then the total horizontal velocity will become:
Ua(z) = V (z) + ua (z) (13.4)
Of course, V (z) = 0 for z > 0:

Remark
Upon re‡ection, one can conclude that the computational e¤ort needed to describe the
environmental hydrodynamics and aerodynamics has been simpli…ed considerably in com-
parison to the most general case:
- Only conditions under the wave crest are considered.
- Only hydrodynamic drag is considered.
- All spatial phase di¤erences are neglected.
These all reduce the computational e¤ort. On the other hand, one is forced to add a limited
amount of complication in order to:
- extend hydrodynamics up to the wave crest and
- have a solution valid for all water depths.
Even so, the overall result of all this is that the hydrodynamics has been considerably
simpli…ed. A simpli…ed schematization of the o¤shore structure will be discussed in the
following section.

13.4 Structure Schematization
The objective of this section is to replace the actual truss-like marine structure - with all
of its members and nodes - with a much simpler and computationally e¢cient equivalent
one for the purpose of estimating its external hydrodynamic drag loads. The formulations
derived in this section are valid only when all velocity or acceleration components are
co-linear. This means that the all waves and all currents come from the same direction.
This is completely in agreement with the schematization of the environment made above,
but it will not be true in a more general case. The formulations in this section will be
incorrect, however, if the current comes from another direction than the waves or even if
only directional spreading of the waves must be included.
The drag term in the Morison equation (for a vertical cylinder) is of the form:
1 2
Fdraga = ½ C D D ¢ Ua (13.5)
2
in which:

13.4. STRUCTURE SCHEMATIZATION 13-9

Fdraga = drag force amplitude per unit length of vertical cylinder (N/m)
CD = drag coe¢cient, to be discussed later (-)
Ua = horizontal velocity amplitude at the chosen elevation (m/s)

Since Ua will be positive in this case, Ua ¢ jUa j has been replaced by Ua .
2

Consider now a horizontal ’slice’ of the entire structure having a unit height; see …gure 13.1.
What happens as one sums the drag forces across all the members found at that level?
1
2
½; CD; and Ua in equation 13.5 remain constant. The remaining quantity, D times a unit
height, is simply an area. Since attention focuses on the horizontal forces on the structure,
this is an area projected on to a vertical plane, perpendicular to the ’slicing’ planes; it is
the area one would be able to measure on a side view photograph of the structure (if no
members were hidden behind others in that photo!). The ’photo’ should be made looking
in the direction of wave propagation, of course.
Continuing for now with a ’horizontal slice’ (of unit height) of the structure at a given
elevation, then the equivalent diameter, De ; which must be used in the Morison equation
drag term (in order to get the total drag force) is simply that total area (as seen on the
projection or photo) divided by the unit height.
Upon re‡ection, one will discover that at each elevation, z; one …nds the following contri-
butions to De :
² Leg chords (nearly vertical) each contribute their actual diameter, D:
² Horizontal braces (if present at the chosen elevation) at an angle µ relative to the
plane of water motion contribute:

De = L sin µ (13.6)

in which:

L = brace length (to the centerline of its nodes) (m)
µ = brace azimuth relative to the water motion plane (rad)

but only over the limited height, D:
² Sloping braces in the plane of the picture contribute:

D¢L D
De = = = D csc ® (13.7)
HB sin ®

in which:

HB = height of the bracing bay (m); see …g. 13.1
® = slope of brace relative to the horizontal (rad)
² Sloping braces in a vertical plane perpendicular to the plane of the picture (in the
plane of the water motion) contribute D:
² Sloping members with other spatial orientations - think of sloping braces when waves
approach a tower in a diagonal direction - require a bit more geometry and book-
keeping. Letting:


µ ¶
1 L 1
`= ¡1 = (csc ® ¡ 1) (13.8)
2 HB 2
then:
De = D ¢ f1 + ` ¢ [1 ¡ cos (2µ)]g (13.9)
in which µ is the bracing azimuth relative to the wave direction (rad).
This angle µ could alternatively be referred to as the angle between the (vertical) plane
which includes the brace and the vertical plane in which the ‡ow takes place. One might
note that equation 13.9 is quite general; it even works (in a degenerate way) for a vertical
cylinder.
One can argue that measuring each brace length to the centerline of each of its end nodes,
includes too much length; more than one member is being counted within each joint’s
volume. This is true, but it is often seen as a compensation for the fact that the ‡ow will
actually be more complex in the joint vicinity. This will - in turn - likely lead to higher
forces than would be predicted for a single straight member.
Since the diameter is a linear factor in the drag force relation, one can simply sum the
above diameters at any given elevation to come to a total equivalent diameter, De (z), to
use at that elevation. This procedure reduces the ’forest’ of truss and other members at
each elevation, z, to a single vertical cylinder segment.
For most structures, the resulting vertical cylinder will look rather ’lumpy’ in that its
diameter will not be constant over its length. Indeed, whenever horizontal members are
encountered, De will abruptly bulge out and become larger. It can also be larger where
the leg chords become larger or near the sea bed where extra legs or pile sleeves are often
included in the structure. Locations A and B in …gure 13.1 are such elevations.
Some may wish to simplify this schematization even more by ’smoothing out’ these diameter
bulges. This can be a dangerous operation, because the hydrodynamic forces are quite
elevation-dependent (and structure overturning moments are even more so) - especially in
the zone just below the sea surface. This is at best an operation which must be based upon
broad experience.

13.5 Force Computation
Now that the environment as well as the structure have been schematized, one is well on
his or her way to computing the hydrodynamic forces and associated overturning moment.
One remaining preparatory task is to select an appropriate drag coe¢cient. Usually a
single value is chosen for the entire structure.
How should the drag coe¢cient be selected? One wrong approach is to use the diameter
of the schematized pile, De ; and ua to compute the Keulegan Carpenter number, K C, and
then to select CD based upon these values. This is wrong because the equivalent cylinder
diameter, De cannot be found in the sea at all.
The diameter selected for determining CD should be more representative of those found in
the real structure which is being schematized. If one also discovers that CD is then rather
independent of the exact value of KC, then one is extra fortunate; the precise choice of
diameter used in this determination is not critical then anyway.
Should ua or the total velocity, U; be used to compute KC? Usually only the wave-caused
velocity component, ua; is used, but this can be a very interesting discussion topic.

13.6. FORCE AND MOMENT INTEGRATION 13-11

Once a proper CD value has been selected, then the (peak value of) the drag force per unit
elevation can be computed directly:
1 2
Fdraga (z) = ½ CD De (z) ¢ Ua (z) (13.10)
2
for the waves plus current, and:
1
Fwinda (z) = ½air Cd Aw (z) ¢ Vtw (z) (13.11)
2
in which A(z) is the projected area exposed to the wind.

13.6 Force and Moment Integration
The drag forces caused by the wind as well as waves and currents are known as a function
of elevation. All the necessary information is now available to compute the resulting
horizontal force and overturning moment on the (schematized) structure.

13.6.1 Horizontal Force Integration
The resulting horizontal force can now be computed by integrating Fdraga and Fwinda over
the appropriate height segment of the structure. This integration can most e¢ciently be
done using a spreadsheet program. This integration usually proceeds by computing the
forces (per unit length) at chosen elevations and then linearly interpolating the loading
between these values. The elevations to choose for this evaluation should be chosen based
upon the following criteria:

² If De or Aw changes abruptly, then one should evaluate the loading for each value -
just above and just below the transition.

² Additional successive elevations should be chosen close enough together so that lin-
ear interpolation between elevations still provides a reasonable approximation of the
exponential curve of the actual elevation function associated with the drag force.

The linear interpolation procedure suggested here replaces some form of elevation depen-
dent exponential decay function by a straight line. This is generally conservative and quite
in accordance with the objective of overestimating - if anything - the results. In order to
prevent this overestimation from becoming too great, one must be sure that the linear func-
tion used for a segment of elevation does not diverge too much from the actual elevation
decay function. This implies that …ner integration steps - shorter (in height) tower slices -
should be selected where conditions change rapidly. Sensitive locations can be found near
the water surface and wherever the structure changes abruptly.

13.6.2 Overturning Moment Integration
Overall structure overturning moments are usually computed about a horizontal axis at the
sea bed (mudline). The computation proceeds quite analogously to that used to compute
the horizontal force, but now one must include the appropriate moment arm with each
integration step. This is simply the elevation of that segment relative to the sea bed.


Alert readers can (correctly, in theory) point out that when a real structure - with ’real’
horizontal dimensions - includes horizontal members (this is usually the case, by the way!),
then the vertical water motions also induce overturning moments about the mudline. Luck-
ily, one only has to sketch the water motion as a wave progresses through the structure
to conclude that the vertical velocity components near the wave crest are small and that
this additional (small) moment acts counter (in the opposite sense) to that just computed
above. Once again, the objective of predicting an upper bound for the overturning moment
is achieved by neglecting this small (and very time-consuming!) detail.

13.7 Comparative Example
The ’proof of any pudding is in the eating’; this section demonstrates the results of compu-
tations carried out using the various alternative computation procedures. This is illustrated
here by working with an arbitrarily chosen standard case and then by varying one a single
variable (while keeping all the others constant) in order to observe its in‡uence.
The standard case involves:
Input Item Value
Wave Ht. H 15 m
Wave Per. T 12 s
as a wave whose crest extends 10 meters above the still water level. This yields a q
parameter for the Wheeler Stretching of 0:80. Further, in order to focus the comparisons
on the hydromechanics and to avoid a discussion of the structure or the drag coe¢cient,
the following quantity is keep constant from the wave crest elevation (+10 m) to the sea
bed.
1
½CD D = 1000 (13.12)
2
Further, only the drag force is considered here.
Computations have been carried out for each of the four treatments of the splash zone
discussed in chapter 5:

² Linear Theory - nothing above the still water level

² Extrapolated Linear Theory - linear theory functions are continued to the wave crest.

² Constant Extrapolation - the linear theory value as z = 0 is used for all positive z
values

² Wheeler Stretching - the pro…le is stretched to the wave crest.

Since the water motion in waves is more or less concentrated near the sea surface, one
would expect that the total horizontal force on a structure would increase more and more
slowly as the water depth continues to increase; each additional increment of structure
height (added at the bottom) adds less and less total horizontal force.
As the water depth approaches zero - at the other end of the range - one might reason that
the total horizontal force there should also approach zero as the tower height exposed to
the waves becomes less and less.
Figure 13.3 shows the total static horizontal shear force at the base of the structure (the
integral of the drag force from the wave crest to the sea bed) versus water depth.

13.7. COMPARATIVE EXAMPLE 13-13

The behavior of the curves on the right-hand side of the …gure is as expected; the behavior
on the left is not - although the curves do not extend completely to a zero water depth.
As the water depth decreases, the water motion in the wave becomes more and more like
that of a shallow water wave; the horizontal water velocities increase. This increased water
velocity - especially when used in a quadratic drag force formula - causes the force per
unit height of the structure to increase so rapidly that it more than compensates for the
corresponding loss of total tower height. The curves are not extended to zero depth because
wave breaking would limit the wave height. Neither wave breaking nor wave height changes
resulting from shoaling outside the breaker zone are included in this analysis.
The four curves on the …gure are located more or less as one would expect. It is no surprise
that extrapolated linear theory yields the largest force and that ’plain’ linear theory the
lowest.
Figure 13.4 shows the resulting moments. As would be expected, the overturning moment
increases signi…cantly with water depth. The relative positions of the four curves is logical
as well in view of the results above.


Figure 13.3: Horizontal Force versus Water Depth

Figure 13.4: Base Overturning Moment versus Water Depth

Chapter 14

SEA BED BOUNDARY EFFECTS

14.1 Introduction
So far, attention has focussed on ‡ows near and forces on man-made objects in the sea.
Now, attention is shifted to the largest object in the sea: the sea bed, itself. Along the
way, a few topics concerning forces on small man-made structures placed on or near the
sea bed will be discussed as well.
These are all cases in which the ‡ow of sea water over the sea bed is markedly in‡uenced
by the sea bed boundary layer; an exposed pipeline is an excellent example as is an item
of ship’s cargo that has been lost overboard and has sunk.
The in‡uence of the ‡ow boundary layer on the sea bed itself becomes important when one
considers the erosion or deposition of sea bed material near a man-made object. Erosion
around the piles of an o¤shore platform can leave a segment of the piles without lateral
support. It is harder to …nd and recover a valuable piece of ship’s cargo that has become
covered by the natural action of the sea bed.
The approach used in this chapter is not, in principle, really any di¤erent from that used
already. First attention is paid to the ‡ow - in the vicinity of the sea bed, in this case.
This is followed by a discussion of forces on objects (including the sea bed itself!) and the
consequences which these forces can have.

Figure 14.1: Axis System (plan view)
0

14-2 CHAPTER 14. SEA BED BOUNDARY EFFECTS

The axis system shown in …gure 5.2 in chapter 5 and here in …gure 14.1 remains consistent
with that used in o¤shore engineering:
- A current (if present) will ‡ow in the +X direction.
- The waves propagate in the +x direction.
- The positive wave direction (+x) makes an angle ¹ with the X -axis.
-The +z-axis is upward from the still water level.
Note that this last convention, especially, can be in contrast to that used by coastal engi-
neers (who often place the vertical coordinate origin at the sea bed). Their approach leads
to numerical computational di¢culties in deeper water.
The notation for some variables used in this chapter may not agree with that often used
by coastal engineers; these changes have been made to make the notation within this book
more consistent.

14.2 Boundary Layer under Currents and Waves
One should remember the following facts about boundary layers from basic ‡uid mechanics
courses or the earlier chapters of this book:
- They result from a velocity di¤erence between the ambient ‡ow and an object.
- They need time - or equivalently distance - to develop.
- Surface roughness plays an important role in their development.
The second of these items is more obvious if one remembers that distance is an integration
of velocity with respect to time.
What currents are important for this analysis? Tidal currents are driven by the gravita-
tional attraction of the sun and the moon. These attraction forces act essentially uniformly
over the entire depth of the sea, irrespective of the water depth at the given location. (This
is in sharp contrast to the situation with large scale oceanographic currents - which are gen-
erally less than a kilometer deep - and the water motion caused by wind waves - which are
even more ’surface-bound’.) Since the tidal current driving force is uniformly distributed
over the depth, one would expect that this current would also be uniformly distributed,
too. This is not the case, however - at least not in the vicinity of the sea bed. Here, the
current is in‡uenced by a friction force resulting from the water motion over the sea bed.
The Prandtl-Von Kármán logarithmic velocity distribution shown in …gure 14.2 results in
this case. Such a velocity distribution has its maximum at the sea surface and the velocity
reduces very slowly at …rst, but more and more rapidly as one gets nearer - in the pro…le
- to the sea bed. The exact shape of the pro…le depends upon the bed roughness.
Since the logarithm of small numbers is negative, the logarithmic velocity distribution
yields - strictly speaking - negative velocities in the immediate vicinity of the sea bed;
this is obviously unrealistic. This shortcoming is ’patched’ by using a linear (straight line)
velocity pro…le in the area nearest to the sea bed. This line has a velocity of zero at the
sea bed, and is tangent to the curve of the logarithmic velocity pro…le. The velocity at
this elevation of tangency is often referred to as Vt . Obviously, this linear velocity pro…le
has a constant slope, dV ; if the elevation of the tangency, zt , is known and …xed, then dV
dz dz
is simply proportional to V t.
The sea bed roughness (a length), which determines the details of the velocity pro…le can
be de…ned in either of two ways:
² If the sea bed is essentially ‡at, the roughness is de…ned in terms of the grain size of
the sea bed material.

14.2. BOUNDARY LAYER UNDER CURRENTS AND WAVES 14-3

Figure 14.2: Logarithmic Velocity Distribution

² Sandy sea beds especially, are often covered with small ripples which are perhaps a
centimeter or so high - much larger than a sand grain, in any case. The height of
these ripples then determines the bed roughness. Such ripples are often found on the
sea bed where ocean waves are present.

In both cases, the height zt is usually of the same order of magnitude as this roughness,
by the way.

14.2.1 Bed Shear Stress With Currents Alone
Newton postulated a friction model for two plates separated by a ‡uid - see chapter 4. It
resulted in a shear stress - to use the notation of this chapter:
dV
¿ =´¢ (14.1)
dz
in which:
¿ = shear stress at bed (N/m2 )
´ = dynamic viscosity (kg/m/s)
dV =dz = velocity gradient near bed (1/s)

Since the time scale in which a tidal current varies is so long, it can be treated as a constant
current for the purposes of this chapter; it ‡ows long enough for a well-developed boundary
layer to develop. This means that the shear stress, ¿, (caused now by the current) will also
be essentially constant with respect to a time interval of several minutes or perhaps even
an hour.
Constant current shear stresses also occur in rivers. If one considers a unit length (and
width) of a river section, one …nds that the energy input or driving force for the ‡ow comes
from the decrease in elevation (potential energy loss) over that length; the ‡ow resistance
comes from the shear stress between the river bed and ‡ow. This yields, in an equation
form:

¿=½ghi (14.2)


in which:

¿ = bed shear stress (N/m2)
½ = mass density of water (kg/m3)
g = acceleration of gravity (m/s2 )
h = water depth (m)
i = river surface slope dz=dx in the direction of ‡ow (-)

In the eighteenth century, the French hydraulic engineer, Antoine Chézy, developed an
empirical relationship for the depth-averaged velocity in a river. The earliest know record
of its publication is 1775. It is quite certainly the …rst uniform ‡ow formula for open
channels; it is still used today. It is a formula:
p
V =C h i (14.3)
in which C is an empirical coe¢cient which is not dimensionless; it has units of m1=2=s. It
also has an inverse relation to the bed roughness: the rougher the bed, the lower the value
of C.
It can be convenient to combine equations 14.2 and 14.3. Since there is essentially no
constant water surface slope o¤shore, one simply eliminates the slope term, i, from the
above two equations. This yields:

V2
¿=½g (14.4)
C2
Notice that the water depth falls out of this equation, too, but C must still be estimated.
This can be done for a constant current (at least) using:

12 h
C = 18 log (14.5)
r
in which: 1
C = Chézy Coe¢cient (m 2 /s)
h = water depth (m)
r = bed roughness (m)
All of this brings up an interesting question: Since Vt - the current velocity at the elevation
of the point of tangency - is directly proportional to the average velocity, V , how does
one explain that Newton’s approach relates ¿ to V , while at the same time, Chézy (see
[Herschel, 1897]) relates ¿ to V 2 ? One of these must be wrong, or there may be another
explanation.
One is not sure that C (or even ´ for that matter) remains constant for a wide variety of
‡ow conditions. Indeed, every river engineer knows that C is not constant. In practice,
there is generally more faith in Chézy than in Newton in this case, however, so that ¿ is
usually associated with a higher power of V or dV =dz:
The shear stress, ¿ , discussed so far has been the shear stress which the sea bed exerts
on the ‡ow. Newton’s Third Law of motion indicates, however, that this is also the shear
stress which the ‡ow exerts on the sea bed.
Before proceeding with this development, the discussion backtracks to discuss the boundary
layer and shear stress under waves.


14.2.2 Boundary Layer Under Waves
Wind waves are generally present at sea, but our river engineering colleagues did not
need to consider them when they came up with their velocity pro…le and bed shear stress
expressions. Potential theory - which by de…nition considers no friction - predicts that
the horizontal water motion velocity caused by surface waves decreases in some negative
exponential way as one proceeds deeper in the sea; see chapter 5. It is sometimes a
misnomer to assume that the water motion had completely died out at the sea bed; a
storm wave of 30 meters height and a period of 20 seconds still has a horizontal water
motion velocity amplitude of 0; 23 m/s at the bottom - in this case in 300 meters of water!
Even with more modest waves in shallower water, one can expect to have a wave-caused
water motion near the bed which cannot be neglected.
Since there is a motion of the water relative to the sea bed, one might expect a boundary
layer to be present. This motion is only the …rst of the three necessary conditions for a
boundary layer stipulated above, however. One can reasonably expect bed roughness to be
present, too; the third requirement is satis…ed. Concern centers on the second requirement:
That there is enough time (or distance) for the boundary layer to develop. Indeed, the ‡ow
in the example wave reverses every 10 seconds. There is no hope that a well-developed
boundary layer can be built up. Instead, a boundary layer of very limited thickness develops
in the immediate vicinity of the sea bed; the ‡ow above this layer remains ’ignorant’ of
the fact that the bottom (with its roughness) is present. This neglects the di¤usion of
turbulence originating at the sea bed.
Analogous to the treatment of the lowest part of logarithmic velocity pro…le, it is convenient
to assume that this wave boundary layer also will have a linear velocity pro…le. Continuing
the analogy, this means as well that the linear velocity gradient can be characterized by a
velocity at some chosen, known elevation above the sea bed; it is convenient to choose the
elevation zt for this, too. (One assumes that the boundary layer under the waves will be
at least this thick; this is safe in practice.) Since the boundary layer retards the ‡ow, it is
logical that the characteristic velocity for this shear stress determination will be less than
that predicted from wave theory. One generally assumes that:

ut = p ¢ ub (14.6)
in which:
ub = sea bed water velocity predicted by wave potential theory (m/s)
ut = characteristic water velocity for shear stress computations (m/s)
p = dimensionless constant with a value between zero and one (-)
The above relationship is true for all times during the wave period, but it is most often
used with velocity amplitudes.

14.2.3 Shear Stress Under Waves Alone
The characteristic velocity for waves, ut ; can be used in place of Vt in a shear stress relation
just as was done for constant currents. One should now, however, be aware that since ut is
a periodic function with zero mean (at least to a …rst order approximation), the resultant
shear stress - when averaged over at least a wave period - will now be zero; there is no
time-averaged resultant shear stress, even though it has non-zero instantaneous values.
The importance of this will become obvious later.


14.2.4 Shear Stress Under Waves Plus Currents
To combine the in‡uences of waves and current a way must be found to combine the
in‡uences of the two boundary layers considered separately above. This is done via a careful
superposition. Note that at an o¤shore location, the wave can propagate in any direction,
¹, relative to the (tidal) current. (This is in contrast to the Coastal Engineering situation
where the waves are usually nearly perpendicular to (¹ ¼ 90±) the current direction.)
Because the sea bed portions of each of the velocity pro…les - one caused by the current
and one caused by the waves - are linear, the resulting pro…le can be expected to be linear
in the near-bed zone as well. Since the current boundary layer was characterized by a
velocity Vt acting at the elevation of the tangency point, it can be convenient to choose
this same elevation for the wave boundary layer as well. Figure 14.3 shows a plan view
(one is looking down from above) of the velocity vectors at elevation zt above the bed.

Figure 14.3: Plan View of Flow Components Just Above the Sea Bed

In this …gure, V t has been normalized to 1 m/s. The wave velocity vector has an amplitude
of 0:5 m/s and makes an angle, ¹; of a bit over 60± with the X -axis which coincides with
the constant current velocity vector. The currents are added as vectors in this …gure. The
resultant velocity vector sweeps back and forth a bit in direction from one side of the axis
to the other. Its magnitude changes continually, too, from a maximum of 1:30 m/s to a
minimum of 0:90 m/s in this case.
The instantaneous shear stress will be proportional to the square of the instantaneous
velocity and its direction will correspond to that of the instantaneous velocity. (See …gure
14.4.)
This shear stress can be worked out a bit as follows.
The X and Y components of the wave current velocity will be of the form:

uX (t) = ua sin(!t) ¢ cos ¹
uY (t) = ua sin(!t) ¢ sin ¹ (14.7)

The constant current velocity, Vt (acting along the X-axis) will be added to uX (t).
The resulting bed shear stress magnitude at any instant is proportional to the square of
this resultant velocity so that:


Figure 14.4: Instantaneous Shear Stresses Under Waves Plus Currents

jVr2 j = (Vt + uX(t))2 + (uY (t))2
¡ ¢
= Vt2 + 2 Vt ¢ ua sin(!t) ¢ cos ¹ + u2 sin2(!t) ¢ cos2 ¹
a (14.8)
+u2 sin2 (!t) ¢ sin2 ¹
a (14.9)

The instantaneous magnitude of the resulting shear stress, ¿, is directly proportional to
this.

Time Averaged Shear Stress Magnitude
The magnitude of ¿ can be averaged over a wave period as well so that:
1 2 1
¿ cw = j¿ j ® V t2 + 0 + ua ¢ cos2 ¹ + u2 ¢ sin2 ¹ (14.10)
2 2 a
Noting that: cos2 ¹ + sin2 ¹ = 1, one comes to the …nal conclusion that:
1 2
¿ cw = j¿ j ® Vt2 +u (14.11)
2 a
which happens to be completely independent of ¹ ! In these equations (with all velocities
at height z = z t):

¿ cw = time averaged bed shear stress (N/m2 )
Vt = current velocity (m/s)
ua = amplitude of the wave motion (includes the factor, p) (m/s)
¹ = angle between the wave propagation and current directions (rad)

This average magnitude of the bed shear stress (independent of its direction) will be found
later to be important for the determination of sediment transport. Equation 14.11 makes
very clear that both the current and the waves contribute to the bed shear stress; waves,
if present, generally increase the average bed shear stress - in spite of the fact that the
average shear stress under the waves alone is identically zero as indicated earlier in this
chapter.


Time Averaged Shear Stress Components

Applying Newton’s third law again, the current reacts to the time averages of the X and
Y components of instantaneous shear stress. The corresponding ‡ow velocity components
responsible for each shear stress component are proportional to:

jVrX (t)j = (Vt + uX (t))2
2
(14.12)

jVrY (t)j = u2 sin2 (!t) ¢ sin2 ¹
2
a (14.13)

in which the subscript r denotes resultant.
It has been assumed in equation 14.12 that V t > uX to guarantee that VrX is never negative.
2

With this knowledge, then the time averaged shear stress in the X direction becomes:
µ ¶
1
¿x ® Vt2 + 0 + u2 cos 2 ¹ (14.14)
2 a
This indicates that the average bed shear stress in the direction of the current is increased
unless ¹ = ¼=2.
The average shear stress in the Y direction is even more interesting. It is proportional to
the time average of VrY jV rY j since ¿ is always in the direction of the current. The result
is, then:

1 2 2
¿y = u sin ¹ (14.15)
2 a
which is zero only when ¹ = ¼=2. This means that if ¹ 6= ¼=2, ¿ y 6= 0, and the resultant
bed shear stress is not co-linear with the constant current. This implies in turn that there
will be a resultant force acting on the water ‡ow which is perpendicular to the original
constant current direction. This force tends to divert the current so that ¹ does approach
¼=2; the current tries to turn to become parallel with the wave crests.
On the one hand, this shift in the current direction can actually take place more easily
o¤shore than it can near the coast where other boundary conditions such as the imper-
meability of the beach itself also contribute to the current’s behavior. On the other hand,
these boundary conditions have a much smaller in‡uence on the currents o¤shore; these
currents tend to be stronger in deeper water and the wave in‡uence near the bed is less
than it would be in shallower water, too.

This concludes our discussion of how the sea bed in‡uences the ‡ow - at least for now.
Results from above will be utilized in later steps, however. For now,the next step is to look
at the bed shear stress in the opposite sense - to examine how the bed shear stress a¤ects
the sea bed itself.

14.3 Bed Material Stability
This section discusses the forces on and stability of cohesionless grains of bed material.

14.3. BED MATERIAL STABILITY 14-9

14.3.1 Force Balance
The limit of stability of (cohesionless) bed material grains can be studied via a detailed
examination by an equilibrium of horizontal forces: A tiny wake - with low pressure - forms
just downstream of a soil grain on the bed surface; this yields a miniature drag force, FD;
as indicated in …gure 14.5.

Figure 14.5: Sea Bed Grain with Drag Force

This is resisted by a horizontal inter-granular friction force which is dependent in turn on
the vertical intergranular normal force. This latter force depends upon the net submerged
weight (weight minus buoyant force) of the grain and the vertical resultant of the hydro-
dynamic pressure force distribution around that grain; see …gure 14.6. (The working of
this latter under-pressure or lift has been described in chapters 3 and 4. It will come up
as well when pipelines are discussed in a later section of this chapter.)

Figure 14.6: Sea Bed Grain with Lift Force

The total force ’picture’ is shown in …gure 14.7.

Figure 14.7: Schematic of Forces on A Sea Bed Surface Grain


One can see from …gure 14.7 that a complete force balance - including the quite irregular
intergranular forces - would be cumbersome to carry out at best. An alternative more
global approach is therefore used instead.

14.3.2 Shields Shear Stress Approach
This alternative approach simply relates the time-average bed shear stress, ¿ or ¿ cw , to a
¹
stability parameter for the soil grains. This was …rst done by [Shields, 1936] for rivers. He
(as well as most others!) assumed that the river bed was so nearly horizontal that its slope
had no e¤ect on the stability of the bed grains.

Figure 14.8: Shields Grain Stability Curve

Figure 14.8 shows the Shields relationship. The dimensionless bed shear stress,

¿
(14.16)
¢½ ¢ g ¢ D

is plotted along the logarithmic vertical axis; a grain Reynolds number,

V¤ ¢ D
(14.17)
º

is plotted horizontally - again with a logarithmic scale.
In this …gure and these formulas:

14.4. SEDIMENT TRANSPORT PROCESS 14-11

g = acceleration of gravity (m/s2)
¿ = bed shear stress (N/m2)
¢½ = ½s ¡ ½w is density di¤erence (kg/m3)
½s = mass density of bed grains (kg/m3 )
½w = mass density of (sea) water (kg/m3)
D = diameter of the bed grains (m)
p
V¤ = V g=C is the so-called shear velocity (m/s)
C = Chézy coe¢cient (m1=2/s)
V = depth-averaged ‡ow velocity (m/s)
º = kinematic viscosity of (sea) water (m2/s)

The zone between the two curves in …gure 14.8 is the area of uncertainty between stability
below the band, and movement above. The fact that this boundary is a bit unclear, stems
from the fact that bed particles can interlock, etc., to some extent.

14.3.3 Link to Sediment Transport
One should be careful to note that bed material instability in a Shields sense is not su¢-
cient for material actually to be transported; instability simply indicates that the particle
’cannot sit still’. Two criteria must be met simultaneously for there to be net bed material
transport:
² Particles must be loosened from the sea bed; this is indicated by the Shields criterion,
and
² There must be a resultant current to provide a net transport of those particles.
Since a wave, alone, provides no net current or mass transport, it fails to satisfy this latter
criterion; it can only cause particles to ’cha-cha’ back and forth. On the other hand, if the
waves are intense enough to ’stir up’ the sea bed material, then only a very small resultant
current superposed on the waves can cause a very signi…cant bed material transport. Note
that this is true even when the current - if acting alone without the waves - would be too
weak to cause sea bed particle instability and thus transport.

14.4 Sediment Transport Process
Now that the stability of bed material grains has been discussed, attention switches to the
mechanisms by which such material is transported.

14.4.1 Time and Distance Scales
The work to be described here was …rst carried out for rivers. It was generally assumed
that the ‡ow conditions were not changing rapidly; a quasi-static or steady state solution
was found. This means then, that accelerations could be neglected and that conditions
remained essentially constant along a streamline. [Rijn, 1990] indicates that steady state
conditions - in terms of sediment transport - are restored within about 80 to 100 water
depths downstream from a major disturbance - such as a dam - in a river. With a typ-
ical river depth of 5 meters, this means that conditions become stable after about half a
kilometer. In other words, there is still a lot of river left (usually) in which a sediment
transport predicted by a steady state solution can be utilized.


For o¤shore conditions - in water ten times as deep, for example - this distance to regain
sediment transport equilibrium becomes 5 km - a distance which far exceeds the dimensions
of most o¤shore structures! This means that a completely stable sediment transport condi-
tion will never be achieved near an o¤shore structure as a result of its (local!) disturbance.
O¤shore engineers are continually confronted by the transient situation; this is in contrast
to the situation for coastal or river engineers. Even so, it is convenient for the explanation
to start with the stable or steady state situation - at least for now; the transient will be
picked up in a later section.

14.4.2 Mechanisms
How is bed material transported in a river? In principle there are three ways in which it
can be moved:
- solution,
- suspension and
- moving along the bed - sometimes called saltation.
Most minerals which make up the earth dissolve slowly in water and come out of solution
slowly, too. Transport via solution is of a molecular nature throughout the water; it is
not at all important for cases being considered here.
Particles in suspension tend to be relatively …ne; they move along with the water which
surrounds them. This transport can take place at any elevation in the ‡ow. It is suspended
transport which often makes water look turbid or ’hard to see through’.
Saltation, or bed load transport ’never really gets o¤ the ground’ - to put it popularly;
it rolls and bounces along the bed with a speed which is less than that of the adjacent ‡ow
in the sea bed boundary layer.
These latter two transports are discussed a bit more below.

Suspended Transport
It will later become obvious that suspended transport is only occasionally important for
o¤shore engineering applications. The following discussion is given for completeness and
to provide a basis of understanding for some other phenomena.
What mechanism keeps bed materials in suspension? Suspended sediment particles fall
back toward the sea bed with their fall velocity. (This was discussed in chapter 4.) Ma-
terial is moved back upward as a result of turbulent di¤usion and the fact that the water
exchanged upward has a higher sediment concentration than the water swapped downward
at the same time. This is illustrated in …gure 14.9.
Further, there is usually a free exchange of material between the ‡ow and the sea bed.
Generally, no suspended material is lost at the sea surface. This can all be put together to
yield a classical ordinary di¤erential equation for an equilibrium situation:

dc(z)
Vf c(z) + ²(z) =0 (14.18)
dz
in which:
Vf = particle fall velocity in water (m/s)
c(z) = sediment concentration at elevation z
²(z) = turbulent eddy viscosity at elevation z

14.4. SEDIMENT TRANSPORT PROCESS 14-13

Figure 14.9: Vertical Suspended Sediment Transport Balance

² is a measure of the scale of turbulence in the ‡ow. By assuming a distribution for ²s (z),
one can work out the solution to equation 14.18. After a bit of mathematical manipulation
(which is not really important for our insight here) the coastal engineers get a solution
which looks like:
µ ¶z¤
¡z a
c(z) = ca ¢ ¢ (14.19)
z+h h¡a
in which:
c(z) = sediment concentration at elevation z
ca = sediment concentration at a chosen elevation a > 0 above the bed
h = water depth (m)
z = vertical coordinate, + upward from the water surface (m)
z¤ = dimensionless parameter (-), dependent upon ¿ ; ½ and V f
(The exact background of z¤ is not important here)
Vf = the particle fall velocity (m/s)

The sediment concentration at some chosen elevation, a; must be known in order to deter-
mine the quantitative solution of equation 14.19. One will discover below that this comes
from the bed transport to be discussed in the next section.
Once c(z) is known, the total rate of suspended material transport follows directly from
the following integral:
Z sea surf ace
Ss = U (z; t) ¢ c(z) ¢ dz (14.20)
bed
in which:
Ss = suspended sediment transport (m3/s per meter width)
U(z; t) = velocity (from any cause) at elevation z and time t (m/s)

This is usually simpli…ed a bit if waves are involved; one is not interested in a truly
instantaneous sediment transport. Its time average (over a wave period) is much more
¹
relevant. This allows U(z; t) to be replaced in 14.20 by its time averaged value, U (z):


Bed Load Transport
Bed load transport ’never gets o¤ the ground’; it stays near the bed in a thin layer below
the suspended sediment transport. The same forces which determine particle stability also
govern bed load transport. In contrast to the situation with suspended transport, the bed
load moves more slowly than the water near the bed; most formulas for predicting the rate
of bed load transport, Sb, express this rate directly instead of via a concentration times
a velocity as was done with suspended transport. Indeed, the water velocity is changing
rapidly as a function of elevation here and the sediment concentration is hard to de…ne in
this region, too.
On the other hand, a sediment concentration - at some elevation - is needed in order to
determine the actual suspended sediment concentration pro…le as explained above. It is
convenient to perform this coupling near the sea bed - more speci…cally at an elevation
ht above the bed - the height at which the linear near-bed velocity pro…le is tangent to
the logarithmic Prandtl-Von Kármán pro…le. It is being assumed (quite arbitrarily) that
transport above this level takes place in suspension and that only bed load transport is
found below.
Hydraulic engineers have taken the very pragmatic step of converting the bed load transport
into a (form of) concentration by assuming that Sb takes place in a layer of thickness ht
and with a velocity equal to Vt . This equivalent ’concentration’ is then:

Sb
ca = (14.21)
ht ¢ Vt
Using this link, one can relate the entire steady state sediment transport, S = Sb + Ss ,
to one quantity: The bed transport, Sb. Attention can be concentrated on determining
this value. Before doing so, however, the relative importance of Sb and Ss for o¤shore
applications will be examined in this next section.

14.4.3 Relative Importance of Bed versus Suspended Load
Many comparisons can be included under this heading. River engineers are often interested
in the ratio of suspended load transport to bed load transport, Ss=Sb. This value can vary
widely in rivers, by the way; it can be high (thousands) for a muddy river such as the
Amazon or Mississippi, and very low (¿ 1) for a sparkling clear mountain stream tumbling
over rocks.
In o¤shore engineering on the other hand, one is wise to …rst consider the relative time
(or distance) scales within which Sb or Ss change. Consider what happens to each of these
quantities when - for example - the near-bed ‡ow is locally disturbed by a partially exposed
submarine pipeline, for example. Such a pipeline will typically be no more than a meter in
diameter and it may protrude 50 centimeters or so above the sea bed. At the same time,
the water depth can easily be in the order of 100 meters.
Using potential theory from chapter 3 to estimate the order of magnitude of the disturbance
caused by the half-buried pipe, one …nds that at a distance of at little as 1 meter above
the sea bed (above the pipe) the velocity has only been increased by 25%. At 2 meters
height this increase is only a bit more than 6%. The conclusion must be that the pipe only
disturbs the velocity …eld in its immediate vicinity. Consequently, the velocity gradients
and thus the local bed shear stresses and vertical di¤usion of sediment will be disturbed
only locally as well.

14.5. SEA BED CHANGES 14-15

One can expect the bed transport, Sb, to react quickly to these changes. Indeed, each
bed transport particle can stop any time it hits the bottom and this happens almost
continuously. The bed transport can adjust or adapt itself several times within a distance
of a few meters.
The material making up the suspended transport, Ss , on the other hand, does not fre-
quently come in contact with the bottom; it gets little opportunity to stop. Except that
‡ow in the immediate vicinity of the pipe - especially in its wake - may be a bit more tur-
bulent, the rest of the (tidal current) ‡ow does not even ’notice’ that the pipe is there; the
major part of the ‡ow as well as the suspended transport it carries is essentially undisturbed
by such small scale changes.

Some may argue that whenever Sb changes, Ss will change as well. Their reasoning follows
from the link established above between suspended and bed load transports. What they
fail to realize is that this theoretical ’link’ was established for an equilibrium situation
and that a change in Ss must take place via changes in its sediment concentration pro…le.
The driving forces for determining that pro…le - the turbulent di¤usion and the particle fall
velocity - are not (or only very locally) changed. It is indeed because of Ss that [Rijn, 1990]
concluded that a distance of 80 to 100 water depths is needed for an equilibrium sediment
transport to be reached in a river.

The conclusion to all this is that bed load transport reacts very quickly to ‡ow changes
which occur on a scale typical of o¤shore engineering objects, but that suspended transport
does not. O¤shore engineers seldom have to worry about the transport of suspended
material. Sb is almost always the most important transport component in an o¤shore
situation. Conversely, Ss is seldom important in an o¤shore situation!

14.5 Sea Bed Changes
14.5.1 Sediment Transport Not Su¢cient for Bed Changes
Having bed material instability (in the Shields sense) and having a resulting current to
transport that material is not usually su¢cient to cause a real morphological problem
(erosion or deposition). The presence of sediment transport past a point only indicates
that the bed material grains now present at that location will (probably) be replaced by
others within a very short time; a dynamic equilibrium can exist.
In order to have (or reveal) morphological changes as a result of sediment transport, one
must examine dS=dX, the change in sediment transport along a (resulting) streamline. If S
increases as the ‡ow proceeds from point A to point B, as shown in schematic cross section
in …gure 14.10, then the extra material transported past point B (if dS=dX is positive)
can only have come from the bed segment between A and B; a positive value of dS=dX
leads to erosion. Conversely, a negative value must lead to sedimentation or deposition of
material in the segment between A and B.
How does a time dependence a¤ect the situation? When the current changes slowly as a
function of time - such as with a tidal current - this makes no di¤erence. This variation
only introduces a dS=dt which is pretty much the same in the entire vicinity; this does not
- of itself - yield either a signi…cant erosion or deposition.


Figure 14.10: Longitudinal Bed Section Along Flow Line

14.5.2 Bed Change Time Scale
Coastal engineers work on a large scale. They are typically concerned with a beach which
can be hundreds of meters wide and many kilometers long. Hundreds or even thousands of
cubic meters of bed material must be removed or deposited in order to make a signi…cant
impact. This can take so long that the beach never really ’comes to rest’; it keeps on
changing continuously. Because of this, coastal engineers invest a major portion of their
e¤orts in determining the speed with which (coastal) conditions change. They need to
predict how much a given coastline will change in - for example - the coming decade as a
result of natural accretion and erosion resulting from waves and currents along the coast.
Typical objects for which sea bed morphology is important to o¤shore engineers include
a pipeline, the base of a jack-up platform leg or even an entire steel tower structure.
Smaller objects can include equipment lost overboard, an anchor or a communications
cable. O¤shore engineering morphological phenomena take place within a distance of
no more than some tens of meters. As a consequence of this, only a relatively very small
amount - very often less than a hundred cubic meters - of bed material needs to be removed
(or deposited) in order to reach a new equilibrium. One can intuitively feel (correctly)
that this can occur much faster. Indeed, morphological changes signi…cant for o¤shore
engineering often occur during a single tide period - usually when there is a storm going
on. This enhances the wave action and thus the magnitude of the bed shear stress. This in
turn stirs additional bed material loose from the sea bottom so that the resulting current
- locally in‡uenced by the pipeline or other object present - can transport it.

14.6 Laboratory Modeling
Before discussing applications, our attention will be temporarily diverted to the topic of
physical modeling of local morphological changes.

14.6.1 Theoretical Background and Scaling
Imagine the physical problem of modeling the morphology of a the area around a pipeline
in say 100 meters of water. Since waves are involved, one uses Froude scaling; see chapter 4
or appendix B. If the wave and current ‡ume has a maximum depth of 1 meter, one would
be forced to use a geometric scale of 1 : 100. At this scale, a cylinder with a diameter
comparable to that of a pencil would be a reasonable pipeline model. This is too small;
The Reynolds number, etc. become too distorted.

14.6. LABORATORY MODELING 14-17

A more inspired physical model is possible, however. Since o¤shore morphological problems
are dominated by bed load transport, there is no real need to model the suspended transport
correctly - or even at all for that matter. Since the bed load moves along in a thin layer
near the bed, why, then, is it necessary to model the entire ocean depth? Indeed, this is
no longer necessary at all!
With this knowledge, one can modify the laboratory model so that only some meters (in
height) of the near-bed ‡ow is reproduced as is shown in …gure 14.11.

Figure 14.11: Schematic Longitudinal Section of Field and Model Conditions

The height chosen must be such that the pipeline - which is now larger, too - does not
signi…cantly ’block’ the entire ‡ow; this would introduce extra in‡uences not found in
nature. A good rule of thumb is that the object being studied should not block more than
1=10 (or 1=6 at the very worst!) of the ‡ow cross section. With this, an exposed model
pipeline about 80 mm in diameter would require a water depth of about 800 mm, leaving
a freeboard in the ‡ume of 200 mm for waves or any other disturbances.
It should be realized as well, that this model no longer represents the entire ‡ow depth;
it only includes the lower meters of the prototype situation. What is the consequence of
this? What current velocity should now be used in the (new) model? It is no longer correct
just to scale the depth-averaged velocity from the prototype for use in the model. Instead,
the depth average of the current in the (lower) portion of the velocity pro…le - the part
actually being modeled - should be reproduced (to an appropriate scale, of course); this
has already been suggested in …gure 14.11.
What about waves, then, one can ask? The surface waves from the …eld will be much too
near the sea bed in the ’cut down’ model! Here again, it is the velocities near the bed
caused by the waves that must be reproduced to scale.
Since it will be generally necessary to reproduce the spatial e¤ects - in the horizontal
direction - it will be necessary to scale the wave length according to the geometric scale;
this - with the water depth - sets the wave period relationship in accordance with Froude
scaling. One obtains the proper bed velocity amplitude by adjusting the wave height; this
compensates for the fact that the water depth is not scaled in the same way as the rest of
the model.
One would probably not choose the actual current velocity to be used until this step was
completed by the way - in spite of the discussion above! One (alternate?) approach is to
scale the current velocity using the same ratio as was found for the wave motion velocities.
The key to successful physical modeling of sea bed sediment transport is to properly scale
the bed shear stresses in relation to the stability (in the Shields sense) of the bed material.
This fact may require that both water motion components be adjusted to achieve the


proper match of the sea bed ‡ow conditions to the stability of the bed material used in
the model. Indeed - to complicate matters even more - one may be using a di¤erent bed
material in the model than one …nds in the …eld.

14.6.2 A Modeling Experience
This relates to a modeling situation actually encountered by a thesis student in the mid
nineties. He was conducting lab experiments to check the stability of stone berms occa-
sionally used to cover otherwise exposed subsea pipelines. The tests were being carried out
in a wave and current ‡ume with wave propagation in the same direction as the current.
On the fateful day, a berm of loose stone had been built across the bottom of the ‡ume; it
had a slope of 1 : 5 on both the upstream and downstream sides; the crest height was such
that the water depth on top of the berm was about 90% of that in the rest of the ‡ume.
This was in accordance with all the criteria stated above. Figure 14.12 shows a (distorted)
cross-section of the berm and thus also a longitudinal pro…le of part of the ‡ume. The
water surface is well above the top of the …gure and only the surface layer of berm gravel
is represented in the sketch.

Figure 14.12: Longitudinal Section of Berm Model

As a check, the …rst tests of the day were carried out with just a current. After the current
had been increased in a few steps, the bed shear stress on the crest of the berm became
high enough to cause instability of the stones; quite a rapid erosion started - with the
stones being deposited downstream at the toe of the berm.
Remembering from the theory that waves, if present, increase the bed shear stress and thus
the sediment transport rate, the student hastened to start the wave generator in order to
be able to observe a really spectacular erosion process. Can you imagine his dismay when
- after the waves had been added - the stones in the berm remained stable! What was
wrong with the experiment? Is the theory wrong?
After recovering from the initial shock, he investigated the matter systematically as follows:
The distorted berm pro…le was …xed in place (by sprinkling …ne cement into its pores and
letting it set) and a series of local velocity pro…les were measured at various locations around
the berm using a laser anemometer. Since bed load sediment transport is governed by the
bed shear stress, the laser beams were set close to the berm surface to make measurements
in the boundary layer in order to determine the velocity gradients and thus infer the bed
shear stresses from the local ‡ow parameters.
Just as in the seemingly ill-fated experiments, tests were carried out both with a current
alone and with waves superposed on the current. To everyone’s surprise, the velocity pro…le

14.7. VERTICAL PILE IN CURRENT 14-19

measurements revealed that the velocity gradients (and thus the shear stress on the stones)
above the berm crest were greatest when just the current was present!
Apparently, when the waves were added to the current, they interacted with that current
and with the berm, and possibly other parts of the ‡ume itself, in such a way that the
gradients in the velocity pro…les above the berm crest were reduced instead of enhanced
by the waves. This con…rmed that the experiments still agreed with the basic theory; the
experiment was just a bit di¤erent that had been expected!
The morals of this whole experience are that the bed shear stress is important for deter-
mining bed load sediment transport and that such experiments must be carried out very
carefully. Persons trying to predict stability or erosion or deposition of bed material can
best do this by evaluating - only in a qualitative way if necessary - the local bed shear
stress and the changes which it undergoes as the ‡ow progresses along the streamline.

14.7 Vertical Pile in Current
The …rst application involves a somewhat isolated vertical pile which penetrates well into
the sea bed and protrudes above the bed to a height of at least several diameters. The
discussion which follows includes both the hydrodynamics and the resulting morphology -
both now in three dimensions.

14.7.1 Two Dimensional Approach
Remembering the 2-dimensional hydrodynamics (in a plan view, now) of the ‡ow around
a circular cylinder, one knows that there is a stagnation point at the most upstream side
of the cylinder and a wake behind the cylinder. Also, theoretically at least, the velocity
on each side immediately adjacent to the cylinder is twice that of the undisturbed ambient
‡ow.
Since there is no velocity at a stagnation point, one would expect little to happen to the
bed on the most upstream or ’leading’ part of the pile; at the sides where the velocity is
locally doubled, one might very reasonably expect an increased shear stress, too, yielding
a positive dS=dX and thus a local erosion. The extra turbulence caused by the wake could
enhance erosion on the lee side of the cylinder, too. This would leave the cylinder standing
in a ’pit’ (except on the upstream side).
One of the world’s greatest tragedies is the murder of a beautiful theory by a brutal set
of facts: One observes - in nature or even in a lab experiment - that there is a signi…cant
erosion hole on the entire upstream side of the cylinder. This extends along the sides (where
it is already expected) and ’fades out’ in the wake area. Apparently, the two-dimensional
approach is insu¢cient to explain what happens.

14.7.2 Three Dimensional Flow
The three-dimensional ‡ow pattern includes the velocity pro…le (in the vertical) caused
by bed friction in the ambient current. Figure 14.13 shows a longitudinal vertical section
with the approaching velocity pro…le shown on the left. The current shown here is quasi-
constant such as a tidal current. Waves are not needed for this explanation; they just
complicate the discussion in this case.


Consider now horizontal cross sections at two di¤erent elevations, A and B, in that …gure.
Because of its higher location, the approaching current at elevation A will be faster than
at location B.

V (A) > V (B) (14.22)

Figure 14.13: Pile with Approaching Velocity Pro…le

This implies, in turn, that the stagnation pressure at A, 1 ½ V 2 (A), will be greater than
2
the stagnation pressure farther down along the pile at point B. Along the vertical line of
stagnation points on the upstream side of the pile one will …nd a dynamic pressure gradient
which is steeper than the hydrostatic gradient, ½ g; there will be a residual quasi-static
downward pressure gradient along the upstream side of the pile! This pressure gradient
will result in a downward ‡ow along these stagnation points as indicated by the dashed
arrow in …gure 14.13. What happens to this ‡ow when it hits the sea bed?

Horseshoe Vortex
After colliding with the bed, the ‡ow turns upstream, along the bed against the approach-
ing ‡ow (where the approaching velocity is lowest) as shown very schematically in …gure
14.14. This ‡ow can turn upstream here along the bed because the kinetic energy of the

Figure 14.14: Schematic Detail of Horseshoe Vortex just Upstream from the Pile

approaching ‡ow is low here, anyway. After progressing a short distance - think in terms


of a pile diameter - it bends up and is swept back to the cylinder to form a vortex. This
vortex grows in length on each side; its ends (or tails) get swept around the cylinder by
the ‡ow so that this vortex takes on a U-shape resembling a horseshoe when viewed from
above.
A common feature of vortices is that they are local and have higher ‡ow speeds than
one would otherwise expect in the vicinity. A vortex has a relatively thin boundary layer
adjacent to an object or the sea bed so that relatively high velocity gradients and turbulence
(in their radial direction) result. Given this - and remembering that shear stresses are
dependent upon velocity gradients - it is not surprising at all that an erosion pit develops
on the entire front and sides of the pile.

What prevents this erosion from going on ’forever’ and making a pit of unlimited depth?
After all, the horseshoe vortex is not time limited. As the erosion pit gets deeper, the
slopes at its sides become steeper. This upward slope - as seen in the direction of local
‡ow - makes it more di¢cult for bed material to be transported. (Remember that slopes
have been neglected in sediment transport discussions.) In practice with a nice sandy bed,
one can expect this erosion pit to develop to an ultimate depth of in the order of 1:5 pile
diameters. All of these phenomena are illustrated in …gure 14.15.

Figure 14.15: Final Overall Bed Situation Near Pile

What happens to the material eroded from the pit, and what happens on the downstream
side of the pile? Here the vortices in the pile wake cause increased turbulence. Also, the
story about stagnation pressures can be repeated here, but it now works in the opposite
sense. A small secondary ‡ow of water coming from the sea bed - and even from the
horseshoe vortex - will be drawn upward into the wake. Is it surprising that this ‡ow also
carries sediment in suspension? Its (temporarily) increased turbulence makes this possible.
However, once the ‡ow is swept downstream past the pile, no new turbulence is added and
the vortices in the wake gradually die out. The sediment that had been picked up in the
immediate vicinity of the pile now falls out on a relatively extensive area downstream.


Flow Reversal
What happens when the tidal current reverses direction and - to make this discussion clearer
- ‡ows from north to south instead of from south to north? It has just been explained that
the ’original’ erosion pit generated by the south to north current was formed on the east,
south and west sides of the pile. Now, after the current reverses, the pit will form on the
east, north and west sides. The existing pit on the south side which is now in the wake
will most likely …ll up a bit with very loose material.

Final Erosion Pit
As a consequence of tidal action, therefore, one can expect - for design purposes - an erosion
pit to be formed completely around any slender vertical pile. The depth and width of this
pit will be in the order of 1:5 pile diameters. This, in turn, means that the lateral support
of the pile begins only 1:5 pile diameters below the sea bed; this can have consequences for
the design of the foundation (in a geotechnical sense) as well as the pile (in a structural
engineering sense).
Using the above rule of thumb for a pile 2 meters in diameter as an example, one …nds
that roughly 50 m3 of bed material will be removed to form the stable state erosion pit. In
onshore terms, this is not even two truckloads of earth - relatively little, indeed. One can
indeed expect this erosion pit to form quite rapidly - usually within a single tide period.

14.7.3 Drag Force Changes
Even though hydrodynamic forces on a pile are not of primary interest in this chapter, it
is convenient to discuss drag forces on the cylinder here, too - at least to the extent that
they are in‡uenced by the velocity gradient caused by bed friction.
One will already know that the drag force acting on a vertical cylinder in a constant current
of velocity, V , will be given by:
1
FD(z) = ½ C D D V 2 (z) (14.23)
2
in which:

FD(z) = drag force per unit length at elevation z (N/m)
½ = mass density of the ‡uid (kg/m3)
V (z) = approaching velocity at elevation z (m/s)
D = pile diameter (m)
CD = cylinder drag coe¢cient (m)

Does FD(z) change because the velocity pro…le is present or because of other factors not
accounted for by the strictly two-dimensional ‡ow pattern assumed in the earlier discussion
of forces in chapter 4? There are two additional e¤ects discussed so far: One at the sea
bed and one at the water surface.

Secondary Flow E¤ect at Sea Bed
The secondary downward ‡ow on the upstream side of the pile leads to the horseshoe vortex
formation at the bed, but it also supplies an extra volume of water which ‡ows around


the cylinder at the bed elevation. Looked at in another way, the ‡ow past the cylinder
near the bed is actually greater than one would expect to …nd based upon the undisturbed
approach velocity at that level.
Since the (measured) force is generally related to a velocity which in fact is smaller than
one would associate with that actually present, one will …nd a slightly larger value for the
drag coe¢cient, CD. A plot of CD versus depth will increase a few percent near the sea
bed.

Water Surface E¤ect

The pile will cause what looks like a standing wave at the water surface. It is in fact a
travelling wave which propagates upstream at a speed identical to that of the ‡ow at that
level. In this way, the wave stays synchronized with the pile location. This small wave
generates a dynamic pressure …eld which dies out exponentially with depth (just like that
of any other short wave). Since the wave crest (with higher pressure) is near the upstream
part of the pile and the wave trough (with lower pressure) is on the downstream side, this
wave will cause a net (additional) force component in the direction of the ‡ow.
When the total drag force at this level (including this additional component) is related to
the undisturbed near-surface ‡ow velocity, it is only logical that the associated CD value
will be a few percent higher at this elevation.

Free End E¤ect

A third e¤ect is more often encountered in the lab. One simple way to measure drag forces
in a towing tank is to extend a cylinder vertically downward into a towing tank and to tow
it while measuring its total resistance as a function of towing velocity. The force is then
assumed to be uniformly distributed over the submerged length of the cylinder in order to
ultimately arrive at a CD value. Is this correct?
This will not be precisely correct for the following two reasons: The force will be disturbed
(slightly) by the surface wave already discussed above, and there will be a three-dimensional
‡ow pattern near the free, submerged end of the cylinder. This end e¤ect will reduce the
force there.

None of these force disturbances are signi…cant enough for one to have to worry about
them when predicting loads on an o¤shore structure. The situation can be quite di¤erent
if one is determining a drag coe¢cient value from experiments in the lab, however.
These end e¤ects - as all three of these phenomena are sometimes collectively called - can
be eliminated for tests with currents only by adding thin but rigid horizontal plates to the
cylinder: One just below the water surface and one near its free end. The upper plate
should be larger than the wave length of the surface disturbance wave; the top of this plate
will absorb the wave’s dynamic pressure, thus preventing it from disturbing the pressure
…eld lower down in the water. The lower plate will force the ‡ow around the cylinder above
it to remain two-dimensional.
Obviously, it would be best if the drag force was measured now only on some segment
- located between the two guiding plates - of the total cylinder length. If this is done,
however, and the measuring section is far enough from the water surface and the free end
anyway, then it is not necessary to install the ‡ow-guiding plates in the …rst place.


If measurements are to be made in waves, it is impossible from the start to use plates
to guide the ‡ow around anything except a cylinder submerged horizontally with its axis
parallel to the wave crests.

14.8 Small Objects on The Sea Bed
There is a whole variety of small objects that can be found on or in the sea bed. In
some cases the objects are intentionally deployed; anchors, subsea positioning beacons, or
even mines and military listening devices fall into this category. Cargo items which fall
overboard are generally not intentionally deployed on the sea bed - think of an automobile
from a ferry or a small container being lifted to an o¤shore platform. The total list of
objects one can …nd on the sea bed is nearly endless.
In some cases it is necessary that an object remain exposed on top of the bed; the func-
tionality of a subsea beacon can depend upon this, for example. In other cases prompt
self-burial is desired as a means of reducing the chance of detection of certain military
devices. Here, again, the range of possibilities is broad.

14.8.1 Burial Mechanisms
How can an object which falls overboard become buried? There are several mechanisms
conceivable:
² It can hit hard enough to create its own crater which is then re-…lled by ’conventional’
sediment transport. The chance that this occurs is small, however. Most objects
don’t fall fast enough to have enough impact energy. Indeed, as indicated in chapter
4, the fall velocity of most objects in the sea is modest. The pile dropped vertically
was a striking (no pun intended) exception to this, however; see chapter 4.
² The object sinks into the soil under its own weight. This implies that there will be a
soil bearing failure under the object. For this to happen, either the object will have
to generate a high normal stress on the sea bed, or the bed material will have to be
relatively weak - think of a very soft mud in this latter case. Such self-burial will
often require a heavy and specially shaped object.
² An object can become buried as a result of local erosion and deposition. This is ’our’
type of problem which will be discussed more below.
² The object may be covered or exposed as a result of large scale bed form mobil-
ity. Many of the large sand banks or shoals along the Dutch coast migrate slowly
northward as a result of material being eroded on one side and re-deposited on the
other.
² Sudden large scale sea bed movements can take place, triggered by very high storm
waves or even by an earthquake. These can expose, cover, or even sweep away an
object in their path. This has happened from time on the continental slopes. This
was detected when transatlantic communication cables were suddenly broken and
swept away. The time lapse between the failures of successive cables even yielded
insight into the propagation speed of such slides.
² A last mechanism results from the possible slow build-up of excess pore pressure in
the soil - usually an initially loosely-packed …ne sand. This will be discussed below,
too.

14.8. SMALL OBJECTS ON THE SEA BED 14-25

Local Morphology
The morphology near the (small) object on the sea bed has much in common with the pile.
The object will generally stick up a bit but not have a nicely streamlined form. Since it is
sticking up into an ambient velocity pro…le, it is logical that a downward secondary ‡ow
develops on the upstream side; this will result in some form of horseshoe vortex on the
upstream side at the sea bed, just as near the pile. This vortex will again start creating
an erosion pit immediately upstream of and beside the object as shown (as a longitudinal
section) in …gure 14.16.

Figure 14.16: Small Object with Approaching Velocity Pro…le and Upstream Erosion Pit

Now, however, since the object does not penetrate signi…cantly into the sea bed, material
from under the object will fall into the pit; the object’s support is eroded on the ’upstream’
side, too. The result of this is often that the object ultimately tips forward into its ’own’
erosion pit. Depending on the exact shape of the object, this can then change the local
‡ow geometry signi…cantly as well.
Such a fate is common for small irregular objects such as odd cargo items no bigger than
a meter or so in maximum dimension. Short, stubby concrete cylinders - often used as
inexpensive moorings for navigation buoys, etc. are another excellent example of this.

Pore Pressure Buildup and Bed Instability
Pressure changes on the surface of the sea bed which result from surface (storm) wave action
can cause minute cyclic soil deformations. Loosely packed soil will ’try’ to consolidate, thus
reducing its void volume. Since the soil is saturated, water will have to escape during this
consolidation process. Fine soils - even …ne sand - can have a low permeability; this in
combination with the oversupply of pore water, leads to an increase in pore pressure. Since
Terzaghi’s rule that:

Total Stress = Grain (or E¤ective) Stress + Pore Pressure (14.24)
is valid, the increased pore pressure results in a reduction of e¤ective grain stress. In the
limit, a quicksand condition occurs in which the e¤ective stress has become too small to
withstand the applied loads. Such a soil then behaves more like a high density ‡uid instead
of a solid.
If the density of an object is less than that of quicksand (in the order of 1800 kg/m3 ), it will
’‡oat’ in the quicksand and move slowly and incrementally to the bed surface; a heavier
item sinks, instead. Pipelines - when …lled with air (just after installation) or even with gas


often have an overall density of about 1300 kg/m3 ; they have been known to ’‡oat’ upward
through a beach and become re-exposed. At the opposite end of the scale, electric and
some communications cables buried in the ground or sea bed will have an overall density
of more like 4000 kg/m3; they will indeed sink.
Note that it is not absolutely necessary for the quicksand condition to become ’fully devel-
oped’. The soil’s e¤ective intergranular stress and thus its shear strength is reduced as soon
as the pore pressure increases. Since the net vertical force exerted by the buried object - it
does not matter if it is positive or negative - will also cause shear stresses in the surround-
ing soil, it is only necessary that the soil’s shear strength be reduced below the imposed
stress level for failure to occur. Actual failure usually occurs slowly, the cyclic wave action
stimulates cyclic variations in the pore pressure so that the failure is intermittent rather
than continuous.
Relatively large pressure cycles are needed in order to build up su¢cient pore pressure for
this whole process to take place; this makes it essentially a shallow water phenomenon in
the marine environment. At leas one newly-installed pipeline has ‡oated up after being
buried across a beach in The Netherlands. Luckily the problem became apparent before
the pipeline was put into service.
In an onshore situation, electric cables laid in poor soil have been ’lost’ in that they have
sunk deeper - except where they are held in place in a junction house! In this case, the
vibration source was the tra¢c on the adjacent highway.
What can be done to prevent this problem - at least in the marine situation? There are
two possible solution approaches:
1. Consolidate the sand arti…cially as the pipeline or cable trench is being back…lled.
While this is theoretically possible, it is probably a pretty expensive solution for a
pipeline - even when one can work from the beach! Hydraulic engineers have used this
approach however during the construction of the dam in the mouth of the Eastern
Schelde. There, they feared that vibrations of the barrier support structures could
lead to pore pressure build-up and subsequent failure of the deeper sand layers.

2. Back…ll the pipeline trench with coarser material, providing su¢cient soil permeabil-
ity to prevent pore pressure build-up. This is the most commonly chosen preventative
measure - at least for o¤shore situations.

14.9 Pipelines
This section discusses forces on exposed pipelines as well as the sea bed morphology in
their vicinity. The discussion of hydrodynamic forces on pipelines has been delayed till
this point because the presence of the current velocity pro…le caused by bed friction plays
such an important role in the hydrodynamics.
A pipeline seems like a small object when seen in cross-section; in the third dimension it is
very much a one-dimensional or line-like object. Much of what was discussed about small
objects on the sea bed turns out to be applicable to pipeline cross-sections as well. The
reader must be careful, however; since di¤erences do exist. The following discussion will be
for a pipeline which is initially laying on the sea bed with only a minimum of penetration
into that bed (caused by its own weight). The current will ‡ow more or less perpendicular
to the pipeline route. The sea bed material is sand.

14.9. PIPELINES 14-27

14.9.1 Flow and Forces
Because the pipeline is bedded slightly in the sand, it will not allow ‡ow to pass under it; the
entire approaching ‡ow will have to pass over the pipeline. This ‡ow pattern is obviously
quite distorted relative to that for an isolated cylinder far away from any ‡ow-constraining
walls.

Drag Force
The drag coe¢cient for the pipeline will be somewhat higher than that for the cylinder
in the unrestricted ‡ow. This higher drag coe¢cient is then used in conjunction with the
undisturbed current at the elevation of the pipeline center line. Remember that there
is a strong velocity pro…le gradient here! Continuing the discussion of the horizontal
equilibrium, …rst, the drag force is resisted by a friction force which, in turn, depends upon
the contact force between the pipeline and the sea bed; horizontal stability depends upon
the vertical force balance, too. These forces are all illustrated schematically in …gure 14.17.
There will be something similar to a stagnation area - with an associated relatively high
quasi-static pressure - on the upstream side of the pipe near the bed.

Figure 14.17: Flow Situation and Forces on a Cross Section of Exposed Pipe

Lift Force
Because of the approaching velocity pro…le and the fact that all of the ‡ow must pass over
the pipe, the velocities on top of the pipe will be even higher than those predicted by
potential theory for an isolated cylinder. The concept of re‡ection was used in chapter 3
to model a potential ‡ow around a cylinder near (or in contact with) a ‡at bed, but this
still neglects the in‡uence of the velocity pro…le in the approaching ‡ow.
A high velocity along the top of the pipe implies low pressure there, while the water on
the underside of the pipe - near point P in …gure 14.17 - is nearly stagnant.
Conceptually, this begins to resemble the net ‡ow e¤ect of an isolated cylinder in a uni-
form current and surrounded by a circulation; the net force e¤ect will be a lift directed
perpendicular to the current - upward in this case. This force will have to be counteracted
by pipe weight.
Looking for a moment at the total vertical equilibrium, this lift force reduces the soil
contact force and thus, indirectly, the sliding friction resistance from the bed. Unless the
pipe is constrained somewhere else along its length, it will slide sideways before it lifts o¤
the bottom.


It is usually most economical to provide su¢cient pipeline weight (for larger lines) by adding
a high-density concrete coating; for smaller lines it can be more economical just to increase
the steel wall thickness a bit. In either case, however, the added weight increases the pipe’s
outside diameter. One can quickly realize that as the outside diameter is increased, the
drag and lift forces are increased, too! One is ’chasing one’s tail’ so to speak. Even so, an
iterative solution can be found so that a stable design can be achieved.

Morphology
What about the morphology? The approaching velocity pro…le will collide with the pipe
and cause the generation of a vortex on the upstream (or lu¤) side just as has been the
case for piles and small objects. Since the horizontal pipeline presents a less streamlined
shape to the ‡ow (compared to the vertical pile), the upstream vortex may not be as nicely
de…ned as the horseshoe vortex near a vertical pile. On the other hand, this vortex will be
(theoretically) as long as the segment of exposed pipe.
Because the ‡ow past the pipe cross-section is very unsymmetrical now, a larger and
stronger (in comparison to that for a free-standing cylinder) vortex will be formed on the
downstream (or lee) side of the pipe. It will be ’one-sided’, too, in that it will rotate in
only one direction: Clockwise if the ‡ow is from left to right as shown in …gure 14.17. For
those who are not sailors, the terms lu¤ and lee originally referred to the windward and
leeward side (or edge) of a sailboat sail.

Lu¤ and Lee Erosion The upstream vortex, with its high turbulence and sharp velocity
gradients, will cause what is called lu¤ erosion - a bit of a trench, often wider and shallower
than that near a pile - on the upstream side of the pipe.
The downstream vortex, too, will cause erosion - now called lee erosion - resulting in
another trench, now (obviously) on the downstream side. The resulting trenches are shown
somewhat schematically in image 2 of …gure 14.18. The series of images in this …gure depict
a whole series of pipeline self-burial steps.
As these two trenches develop, one can imagine that sand under the pipe becomes unstable
and falls gradually into the trenches. This loose sand is easily washed away.

Tunnel Erosion Ultimately the remaining ridge of sand under the pipe can no longer
support the pipe’s weight and resist the hydrodynamic pressure di¤erential between the
upstream and downstream sides of the pipe as well; it fails, letting water ‡ow under the
pipe. This ‡ow is initially squeezed between the pipe and the sea bed, restricting the
formation of the boundary layer; see …gure 14.18 part 3. Velocity gradients are then very
high on the bed under the pipe so that there is also a very large bed shear stress. Tunnel
erosion of material directly below the pipe can go quite fast! The high velocity ‡ow now
present under the pipe reduces the original upward lift force for two reasons: Less water
‡ows over the pipe now, and the pressure on the bottom side of the pipe is reduced as a
result of the high velocity now present there.

Pipeline Sag Looking at the pipeline as a structural element for a moment, the pipe
segment loses its vertical support as soon as the soil under it fails and tunnel erosion
starts. Shear forces in the pipe convey the weight of the suspended pipeline segment to
adjacent segments increasing their load on the intact sea bed and stimulating their failure


Figure 14.18: Pipeline Cross-Sections showing Progressive Self-Burial Steps

as well. The erosion tunnel can extend thus itself along the pipe axis crosswise to the ‡ow.
Obviously the pipe will start acting as beam, and as the erosion tunnel extends, the pipe
will sag under its own weight into its own trench formed by the tunnel erosion.
In some cases the pipeline axis is nearly down to the original sea bed level (but still sticking
up above the lu¤ and lee erosion trenches) while a narrow tunnel still exists under the pipe.
Ultimately, as the pipe continues to sink, it blocks less of the original ‡ow thus reducing
the driving force for the ‡ow in the tunnel under the pipe. At the same time, the tunnel
and streamlines under the pipe are getting longer, thus increasing the frictional resistance.
At some point in this development, the current under the pipe will become too weak to
transport (enough) sediment and the tunnel will become plugged with sand.

Repeated Cycle What happens next depends upon the extent to which the pipeline
still projects above the sea bed. If the pipeline is high enough, lu¤ and lee erosion will
start again so that the entire cycle including a new tunnel erosion phase is repeated. Of
course each repetition of this cycle leaves the pipeline a little lower relative to the original
sea bed. If the pipe gets deep enough - often its crown is then even below the original
sea bed level. Its disturbance to the ‡ow will be so slight then that local erosion will stop
and any remaining trenches will be re-…lled by the ambient sediment transport. The …nal
situation can be one in which the pipeline is completely buried and is even covered by a
few decimeters of sand!

Tunnel Erosion Stimulation It is of course …nancially lucrative for a pipeline owner
if this natural erosion process takes place; no costs are involved in this burial method!
Since the costs of trenching and covering a pipeline can easily amount to several hundred
guilders per meter of pipeline length, owners can be thankful for nature in this case! Dutch


Figure 14.19: Pipeline with Spoiler

regulatory authorities are cooperative, too: They require that pipelines with diameters of
16 inches (406 mm) or less be buried only within a year after their installation. Since one
good storm is often su¢cient to completely bury a pipe line in Dutch waters, there is a
good chance that this will occur - or, seen the other way, if it has not occurred within a
year it probably won’t occur at all. Then the owner will have to take action. The action
that will have to be taken is discussed in a later section.
Why are owners only required to bury smaller lines? The reasoning is that smaller lines
are weaker and can be more easily damaged by whatever may hit them - such as …shing
gear, an anchor or something which has fallen overboard. Larger lines are considered to
be strong enough. Also, since larger lines have a greater bending sti¤ness, they don’t sag
into their own erosion trenches as easily, either.
Do not get the impression that burial of a pipeline will actually provide that much pro-
tection against dragged anchors; anchors - especially those used in the o¤shore industry -
usually dig in a bit too deep to pass over a pipeline. Expressed another way, most pipelines
are not deep enough to escape a dragged anchor. On the other hand, pipelines are often
buried more deeply - using arti…cial means - when their exposure to anchors is abnormally
high - such as can be the case when a pipeline crosses a designated shipping channel.
Returning now to the main topic, how can the self-burial of a pipeline be stimulated? One
obvious way would be to stimulate tunnel erosion by increasing the natural ‡ow of water
under the pipe. This can be done by blocking the ‡ow of water over the pipe using what
is often called a spoiler.
A pipeline spoiler is a vane, which typically sticks upright from the pipeline crown or top
see …gure 14.19. It usually projects about a quarter of the pipe diameter above the crown.
The spoiler can be made of sti¤ but ‡exible plastic and it is held in place by plastic bands
placed at intervals around the pipe. It is installed just before the pipeline leaves the laying
ship during the installation process.
If the exposed pipeline and spoiler is hit by a towed object, it is designed to fold down
temporarily and then spring back into its upright position.
Obviously the presence of the spoiler increases the drag force on the pipeline, but - once
tunnel erosion has started - it makes the lift force even more negative so that the pipe
is ’pulled’ down toward the sea bed by the lift forces. This enhances its overall stability
against sliding by increasing its normal on the soil elsewhere along the pipe.

14.9.2 Cover Layers
It has just been shown that natural processes can cause a pipeline or other small object to
become buried in the sea bed. There are other situations in which it is desirable or even
required that arti…cial means be used to cover or otherwise protect an object on the sea
bed. Applications can be quite diverse:
² Cover an exposed pipeline or back-…ll its trench,


² Locally cover a pipeline so that a new pipeline can cross it,
² Cover a long distance power or communications cable to protect it,
² Build up intermittent supports to prevent a pipeline from sagging too much into a
deep ’valley’ in the sea bed.
Most of these applications should be obvious. The last one has been used when crossing
areas with hard and rough sea beds. The pipeline then tends to span from sea bed peak
to sea bed peak. If this results in too long a span, internal pipeline forces can exceed
allowable limits; an extra support is then needed. Such a support can be provided by a
mound of coarse (stable) material installed at an intermediate location along the span. Of
course the base of the support - up to the desired pipeline level - must be in place before
the pipe is installed. Its top will be 10 meters or so square to allow some tolerance for the
pipeline laying. After pipeline installation it is common practice to cover the pipe on top
of the support mound in order to guarantee that it remains …xed at that location.
Two separate problems should become obvious to the reader from this discussion:
² How to guarantee the stability of a cover layer or intermediate support?
² How to install the necessary materials e¢ciently - especially in water depths of a few
hundred meters?
These items will be discussed separately below.

Stability
In principle, the stability of any placed stone or gravel can be evaluated using the Shields
criterion. A detailed review of the o¤shore situation, however, shows that conditions are
not quite the same as in a river: The roughness of the dumped material will generally
be di¤erent from that of the natural sea bed. Figure 14.20 shows such a situation rather
schematically. This means that - at least at the upstream side - the approaching ‡ow

Figure 14.20: Cover Layer with Adjacent Sea Bed

velocity pro…le will be the one associated with the original sea bed roughness instead of
the one that could be expected to develop above a bed of the dumped gravel or crushed
stone. This velocity pro…le will probably generate a di¤erent shear stress as well. It should
be obvious that at least the …rst part of the cover layer will have to remain stable under
the in‡uence of this latter (ambient) ‡ow pro…le and resulting shear stress.
If the material placed on the sea bed distorts the general ‡ow pattern as well - think of
a berm or ridge of material covering an exposed pipeline - then the local in‡uence of this
‡ow distortion will have to be included in the analysis as well. Theory is often considered
to be a bit too crude - still - for this sort of prediction; model tests are still popular for
this. This has been (and can still be) an interesting experimental research area.
It has been pointed out that the stability of the …rst cover stones on the lu¤ side can be
critical. Is this the only concern? What about the start of the natural bed on the lee
side? This bed is exposed to a velocity pro…le which has adapted (at least for the lower


few meters) to the roughness of the cover material. Assuming that the cover material is
rougher than the sea bed - as will usually be the case - this ‡ow may at least be a bit more
turbulent than it would otherwise be. Local erosion of sea bed material on the lee side can
often be expected. This means that the ’trailing edge’ of the cover layer can be lost; it falls
into the downstream erosion pit. This can be compensated very pragmatically by making
the cover layer a bit wider. Then this loss will not be detrimental to the functionality of
the cover layer.

Installation
It is one thing to design a local cover layer; one must still install it e¢ciently in a water
depth of sometimes a several hundred meters. The hydraulic engineer’s approach of just
pushing gravel or stone overboard from a ship is …ne for building a breakwater in shallow
(from an o¤shore engineering point of view) water; it is not at all e¤ective in deep water!
In water deeper than a few tens of meters, cover layer material is often ’guided’ toward the
sea bed by a fallpipe. This is a long, more or less vertical pipe which extends from the
work ship to a point just a few meters above where the material is to be deposited. The
gravel or stone is then dumped into the top of the pipe; it will then come out the bottom
a while later.
The fall pipe, itself, only really needs to contain the stone or gravel being dumped. It can
be made up in at least three ways using either:
² ’Conventional’ pipe sections - often made from plastic to save weight. This makes a
closed pipe.
² A series of loosely coupled ’funnels’ such as those used onshore to guide building
renovation waste down into a container. Such a fallpipe allows water to enter at each
joint.
² A ’loosely braided hose’ made of chain links which contains the ‡ow of solids. This
is porous (to water) over its entire length.
Whatever type of fallpipe is used, it should be obvious (from chapter 12) that it will not
simply hang straight down from the workship. Indeed, the combined action of the ship’s
forward speed plus any currents will exert quasi-static drag forces on the pipe, causing
it to swing from the vertical. It may even respond to excitation coming from the ship’s
motion in waves as well. In order to be more certain that the lower end of the fall pipe is
exactly in the desired position, that end is often equipped with a remote controlled vehicle
or spider. This spider will have thrusters to compensate for small positioning errors.
The computation of the external hydrodynamic forces on a fallpipe is relatively straight-
forward. Even so, the prediction of its more complete static and dynamic behavior is not
a trivial task. Attention, here, however, focuses on its internal hydraulics.

Internal Fallpipe Hydraulics Since the pipe is open at the bottom, it will …ll with
water as it is initially deployed from the work ship. It is convenient at …rst to consider an
impermeable pipe and to keep its top end above the sea surface. The pipe will be …lled to
sea level with still water when material dumping starts.
The hydrostatic pressure in the surrounding sea water will match the static pressure re-
sulting from the column of clear water or later even the mixture of water and gravel in
the pipe. This is shown in …gure 14.21. Since the stone or gravel dumped into the pipe
will increase the overall density of the mixture in the pipe, this dumping will cause the


Figure 14.21: Pressure Distribution in and Around a Fallpipe (not to scale!)

liquid (mixture) level in the pipe to drop. The faster material is dumped into the pipe,
the higher the concentration of solids in the water in the pipe. Thus, the overall density of
the mixture will become higher, too. To compensate for this, the mixture level in the pipe
will drop even more.
The upper segment of the pipe - at least - will be subjected to a substantial net external
pressure as shown schematically in …gure 14.21. This pressure di¤erence forms a structural
engineering problem for the pipe, but this may not be the biggest problem, however. With
a constant rate of material dumping, the density of the mixture in the pipe will become
constant and the water in the pipe will come to rest. This means that the stone or gravel
dumped in the pipe will move downward in the pipe through essentially still water; it will
move with its fall velocity - which is not especially high! The table below summarizes some
typical soil grain fall velocity values (in water). This data can be found in chapter 4 as
well.

Particle Type Sand Grain Gravel Stone
Diameter (mm) 0; 2 20 100
Fall Velocity (m/s) 0; 02 1; 0 2; 35

The particles will fall relatively freely (and much faster) through the air in the pipe above
the water; they will be abruptly decelerated by their impact with the water (lower) surface
in the pipe. The only way the total mass transport of stone can be maintained with the
sudden lower velocity is for there to be many more particles per unit length of pipe. If this
’concentration’ becomes too high, the stone or gravel can bridge across the pipe and block

-34 CHAPTER 14. SEA BED BOUNDARY EFFECTS

it so that nothing more gets through; this is a disaster for the productivity of the entire
operation and embarrassing for the supervisor as well!
Even when this is working properly, the particles move relatively slowly - only with their
fall velocity - through the still water in the fallpipe. It would be attractive to increase the
productivity in terms of tons or cubic meters of solid material placed per hour using the
same fallpipe.
One way to increase the productivity of the entire system is to let water ‡ow down the
fallpipe along with the stones. Since at any elevation the external hydrostatic pressure is
greater than that in the pipe - see …gure 14.21 - one has only to provide an opening to
allow surrounding water to enter. This can be done at chosen intermediate elevations, at
the top only, or even continuously along the pipe. One way to let the water enter is to use
a porous pipe - such as the chain links hose - or simply to lower the top end of the pipe
below sea level so that water over‡ows into it at the top.
Now, the particles sink through the moving water and a ‡ow of water plus particles is
discharged from the bottom end of the pipe. For the same rate of material supply as was
used with the closed pipe, the concentration of particles discharged will be relatively lower
and the discharge velocity will be higher when water is also moving downward through the
fallpipe. Adding water is certainly a simple way to increase the productivity of the system,
but how far can one go with this?

Discharge Morphology What can happen when too much water is allowed to enter
the fall pipe? The discharge velocity gets so high that a vertical jet (of a mixture of water
and solids) collides with the sea bed and spreads out. It has even happened that this local
spreading current - with its higher density and thin boundary layer! - generates too much
local bed shear stress for the material being deposited.. Discharged stone will be swept
away from the place where it is wanted taking the reputation of the contractor with it!
Another interesting question involves the e¤ect of the grain size distribution of the material.
If all of the particles are essentially of the same size and density, then their fall velocities
will be more or less the same, too. For some applications, such as insulating a hot ‡ow line
between a subsea well and a production platform, it is desirable to use a well graded gravel
to cover the pipe. This reduces the permeability of the cover, thus reducing the heat loss
from the pipe via convected sea water in its vicinity. (Some crude oils become semi-solid
if they are cooled below their so-called pour point temperature; it can be very important
for the pipeline operator to keep things warm!)
When a well graded mixture of gravel sizes is dumped into a fallpipe, then the coarse
material will fall faster than the …nes. This means that when a discharging operation is
started, …rst only coarse material will be discharged; there will be a ’tail’ of …nes at the
end of the run as well. This need not be important - of itself - for a long pipeline, but it
can make a mess of an attempt to form a neat supporting mound of material by passing
slowly back and forth to build it up.
A second complication when discharging a well graded mixture of gravel is that the …nest
material must remain stable on the sea bed in the discharge jet. Segregation of the particles
can take place so that coarse material is left near the pipeline with the …ner fractions
deposited more to the sides. This can mean disaster for the insulation function of the
cover layer. It has led to litigation between the pipeline owner and covering contractor in
the past.

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