1. Unit 6: WAVE FORCE ON SMALL
DIAMETER MEMBERS
The Morison’s equation, Total Wave Force On The
Entire Member Length, Wave Forces Using Stokes (V)
Theory, Calculation Of Wave Forces Using Dean’s
Theory, Wave Force On Inclined Members (introduction
only-rigorous mathematical treatment to be avoided),
Wave Slam, Limitations of the Morrison’s Equation
2. • A structural member is considered to be of ‘small
diameter’ when its diameter is less that about 0.15 times
the wave length; for example, members of Jacket
structures and piled jetties.
• When member diameter is small incident waves do not
get much scattered by the obstruction and in that case
the equation given by Morrison et al. (1950) becomes
applicable.
3. Morrison equation
• It states that the total force, FI, in-line with the wave direction
can be obtained by addition of the drag, FD, and the inertia, FI
components, i. e.,
• The force due to drag is proportional to kinetic head, i. e.,
Where ρ=mass density of fluid, A= area of object projected on
a plane held normal to flow direction, u = flow velocity
T D I
F F F
2
1
2
D
F Au
4. • CD is coefficient of drag. Its value depends on body shape,
roughness, flow viscosity and several other parameters.
• The direction of wave induced water particle velocity reverses
after every half cycle,
• The force of inertia is proportional to mass times the fluid
acceleration: V = volume of fluid displaces by the
object, = acceleration of fluid
• Cm =Coefficient of Inertia. It depends on shape
of the body, its surface roughness and other parameters.
u
V
FI
u
V
C
F m
I
.
u
u
u
A
C
F D
D
2
1
2
2
1
u
A
C
F D
D
Most of the structural members are circular in cross section
5. u
u
DL
C
F D
D '
2
1
u
d
C
F m
I
4
2
u and vary along L’ and considering unit pile length
.
u
u
d
C
u
Du
C
F m
D
D
4
2
1 2
FT = in-line (horizontal) force per meter length at member axis
at given time at given location.
u
Du
CD
2
1 = in-line (horizontal) water particle velocity
at the same time at the same location.
u
d
Cm
4
2
= in-line (horizontal) water particle acceleration
at the same time at the same location.
6. • u = f(sinθ) = f(cosθ) hence they are out of phase by 90°
and not maximum at the same time
• Basically CD and Cm are functions of size and shape of the
object.
• If that is fixed then they depend on Keulegan-Carpenter
number, Reynold’s number as well as roughness factor.
• Keulegan-Carpenter number: KC ratio of maximum drag to
maximum inertia
.
u
2
max
2
1
max
u
D
C
F D
D
2
2
2
2
2
2
2
max cos
sinh
)
(
cosh
kd
T
z
d
k
H
u
.
max
2
4
max
u
d
C
F m
I
sin
sinh
)
(
cosh
2
2
2
max
.
kd
T
z
d
k
H
u
7. D
T
u
C
C
kd
kd
D
H
C
C
F
F
m
D
m
D
I
D max
2
1
sinh
cosh
1
max
max
D
T
u
F
F
I
D max
max
max
umax= Maximum velocity in the wave cycle
T= wave period, D= Diameter
The above ratio also stands for (Total horizontal motion of the
particle / Diameter).
If KC < 5 then inertia is dominant,
If KC >15 then drag is dominant and regular eddies are shed
at downstream section at frequency of fe=Sν/D where S =
Strouhal No. ≈ 0.2. Alternate eddy shedding gives rise to
alternate lift forces due to pressure gradient across the wake.
8.
9. • Roughness factor: Encrustation around cylindrical members
• Structural members are in course of time covered by sea
weeds, barnacles, shell fish etc. Due to this, effective diameter
changes, effective mass increases, flow pattern, eddy structure
changes . Finally the wave force also changes. Lab studies
have shown that Cm does not change much. CD changes
appreciably and can become 2 to 3 times more than the initial
value.
• Many laboratory and field studies have been made to assess
the effects of all unaccounted factors like eddy shedding , past
flow history, initial turbulence , wave irregularity directionality,
local conditions, data reduction techniques. But experiments
are inconclusive.
10. Recommendations:
For Indian conditions CD=0.7 ; Cm=2 are generally used.
Dean : CD =0.7-1.2 ; Cm=2
A.P.I. : CD =0.6-1.0 ; Cm=1.5-2
Shore Protection Manual: CD -Refer Fig.
Cm=1.5 if Re>5x105
=2 if Re <5x105
= otherwise 5
Re
2.5
5 10
x
11. TOTAL WAVE FORCE ON THE
ENTIRE MEMBER LENGTH
SWL
0
Z=-d
Drag
X
Z
Inertia
z=-z
Variation of drag and inertia over a vertical
12. Consider a vertical located at x=0 as shown above.
Consider Linear Theory
2
cos
2
H t
T
as x=0
cosh 2
cos
sinh
H k d z t
u
T kd T
2
.
2
2 cosh 2
sin
sinh
H k d z
u t
u
t T kd T
t
t
hd
z
d
k
T
H
D
C
u
Du
C
F T
T
D
D
D
2
2
2
2
2
2
2
cos
cos
sinh
cosh
2
1
2
1
When t=0, max
D
D F
F
t
hd
z
d
k
T
H
d
C
u
d
C
F T
m
m
I
2
2
2
2
.
2
sin
sinh
cosh
2
4
4
When t=0, max
I
I F
F
13. But
max
I
F When
t
T
2
sin =1 or when t
T
2
=
2
Or when t=
4
T
At this time t
T
2
cos =
2
cos
=0 Hence D
F =0
Note: When max
D
F occurs I
F =0
When max
I
F occurs D
F =0
max
D
F occurs after time
4
T
when max
I
F occurs.
If is small, =0 , if is not small , =
T
t
H
2
cos
2
14.
d
I
d
D
T dz
F
dz
F
F
d
I
d
D dz
z
d
F
dz
z
d
F
M
Hence total horizontal force on entire member length at any time ‘t’ : TI
TD
T F
F
F
2 2 2
2 2
cosh
1 2 2
cos cos
2 sinh
TD D D
d d
H k d z t t
F F dz C D dz
T kd T T
2 2
2
2 2
cos cos
1
cosh
2 sinh
D
d
H t t
C D k d z dz
T kd
2 2
2 2
cos cos sinh 2
1
2 sinh 2 4
D
d
H t t k d z k d z
C D
T kd k k
15. [Using 2 sinh 2
cosh
2 4
x x
x
]
2 2
2 2
cos cos
1 1
2 sinh 2
2 sinh 4
D
d
H t t
C D k d z k d z
T kd k
2 2
2
2 sinh 2
1 1
cos cos
2 4 4 sinh
D
d
k d z k d z
H
C D t t
k kd
2
2
2 sinh 2
cos cos
32 sinh
D
TD
z
k d z k d z
C D
F H t t
k kd
2
2
2
2 cosh
sin
4 sinh
TI I m
d d
H k d z
d
F F dz C t dz
T kd
2 2
sin sinh
2
4 4 sinh
m
d
t k d z
d H
C
kd k
Hence
2
2 sinh
1
sin
4 2 sinh
TI m
z
k d z
d
F C H t
k kd
16. Similarly, T DT IT
M M M
2
2
2 2
2 sinh 2 cosh 2 2 1 cos cos
64 sinh
D
DT
H
C D
M k d z k d z k d z k d z t t
k kd
2
2
2
sinh cosh 1
sin
2 4 sinh sinh
m
IT
H k d z k d z k d z
C d
M t
k kd kd
17. WAVE FORCES USING STOKES (V)
THEORY
• Water particle kinematics are
calculated at every m length of
the vertical structural member (at
its center along the immersed
length of the member axis) using
the Stokes Fifth Order theory.
• Corresponding forces are
worked out using the Morrison’s
equation at every such segment
and then they are added up to
cover the full member length.
SWL
35’
375’
75’
4’
Force
20. Wave Slam
• When wave surface rises, it slams
underneath horizontal members near the
SWL and then passes by them
• Where Cs = π (theoretically for circular
cylinder)
• The American Petroleum Institute (API)
: Calculate total individual member loads
and not the global horizontal base shear
and overturning moments.
• Impulsive nature of this force however
can excite natural frequency of the
members creating resonant condition
and large dynamic stresses.
2
1
2
z s z
F C Du
21. Limitations of the Morrison’s Equation:
• Physics of wave phenomenon is not well represented in it.
• The drag force formula and the inertia force formula involve
opposite assumptions. The former assume that the flow is
steady while the latter implies that the flow is unsteady
• Real sea effects like ‘transverse forces’, ‘energy spreading
(directionality)’ are unaccounted for.
•There is a high amount of scattering in values of CD and Cm
• Inaccuracies in the wave theory based values of water
particle kinematics get reflected in the resulting force
estimates.
