2.019 Design of Ocean Systems
Lecture 7
Seakeeping (III)
February 25, 2011 Motions and Wave Loads on a Barge z
y
B
D
x
L
Image by MIT OpenCourseWare.
A regular plane progressive incident wave in deep water travels along the x-direction:
ηI (x, t) = a cos(ωt − kx)
kz
ΦI (x, y, z, t) = − ga
e
sin(ωt − kx)
ω
To find the wave force and motion of the barge in the vertical direction using longwave and strip theory assumptions.
Heave Wave Excitation on a Barge (I)
FE3 = FI3 + FD3
Using the strip theory (which is valid for B/L << 1), we have:
FE3 =
R L/2
−L/2
fE3 (x)dx,
FI3 =
Froude Krylov force component:
fI3 (x) =
−
=
Z
=
=
=
In the limit ω → 0:
−L/2
fI3 (x)dx,
FD3 =
B/2
PI (x)nz dy
−B/2
B/2
−B/2
(−ρΦt (x, y, z = −D, t)dy
Bρgae−kD cos(ωt − kx)
=
FE3
Z
R L/2
Z
Z
L/2
fE3 (x)dx
−L/2
L/2
−L/2
Bρgae−kD cos(ωt − kx)dx
µ ¶
kL
2 −kD
e
cos ωt
sin
ρgaB
k
2
FE3 → ρgaBL cos ωt = ρgη(t)(BL)
R L/2
−L/2
fD3 (x)dx
Heave Wave Excitation on a Barge (II) Long-wave assumption: wave motion is a flow slowly varying in space and time.
The wave diffraction effect is approximated by the added mass effect.
fD3 (x, t)
˙
fD3 (x, t) = A2D
33 (x)V (x, t)
V (x, t) =
=
ΦIz (x, z = −D/2, t)
gak −kD/2
−
e
sin(ωt − kx)
ω
V̇ (x, t) = −gake−kD/2 cos(ωt − kx)
V̇ (t)
fD3 (x, t) = −gake−kD/2 A2D
33 cos(ωt − kx)
FD3 (t) =
R L/2
In the limit ω → 0:
kL
−kD/2 2D
f
(x,
t)dx
=
−2gae
A
sin
D3
33
2 cos ωt
−L/2
FD3 → 0
Heave Wave Excitation on a Barge (III) £
FE3 (t) = FI3 + FD3 = ρgaBe
−kD
−
Froude Krylov
Ca ∼ 1.0
100
Added mass
Fraude-Krylov
Total
50
0
-50
0
1
¤ ¡ 2
¢
Added mass effect
£ π
¤
2
= Ca ρ 2 (B/2) ,
Heave force/wave amplitude
A2D
33
−kD/2
gA2D
kae
33
2
3
4
Ship length/wavelength
Image by MIT OpenCourseWare.
k
sin kL
2 cos ωt
Radiation Force Added mass coefficient:
Wave damping coefficient:
Radiation force:
A33 =
R L/2
2D
2D
A
(x)dx
=
LA
33
33
−L/2
B33 → 0
with long-wave assumption
FR3 = −A33 ζ̈3 (t) = −LA2D
33 ζ̈3 (t)
Restoring Force
FS3 = −C33 ζ3 (t) = −ρgBLζ3 (t)
Equation of Motion
(M + A33 )ζ̈3 + Bv ζ̇3 + ρgBLζ3 (t) = FE3 (t)
If Bv =0, ζ3 (t) = ζ̄3 cos(ωt)
ζ¯3 (ω) =
−kD/2
kae
[ρgaBe−kD −gA2D
]( k2 ) sin kL
33
2
−ω 2 (M +A33 )+ρgBL
In the limit ω → 0: ζ¯3 =
ρgBLa
−ω 2 (M +A33 )+ρgBL
=a
ζ3 (t) = a cos ωt = η(x = 0, t)
Barge responds to move like a fluid particle in the limit of very long wave.
Natural Frequency
−ωn2 (M + A33 ) + ρgBL = 0
ωn =
³
ρgBL
M +A33
Natural period:
´1/2
=
Tn =
³
2π
ωn
gB
BD+A2D
33
´1/2
=
³
´1/2
g
D+Ca (π/8)B
Tn increases with D and B.
For example, for D=20m, B=60m, we have Tn = 13s.
Sample Results for Heave Motion Draft: D=12 m
Width: B=40 m
Natural period: Tn= 11.4 s
Bv = 8% critical damping
FE3
RAO
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Pitch Motion and Wave Loads on a Barge
Wave Excitation: FE5 (t) =
=
Z
h
L/2
−L/2
−xfE3 dx
ρgBae
−kD
Added mass and wave damping: A55 =
Radiation moment:
− gkae
R L/2
−L/2
−kD/2
A2D
33
x2 A2D
33 dx ,
iZ
L/2
−L/2
[−x cos(ωt − kx)]dx
B55 = 0
as ω → 0
FR5 = −A55 ζ̈5 (t) − B55 ζ̇5 (t)
Hydrostatic restoring moment: FS5 = −C55 ζ5 (t) = −[ρg∇(ZB − BG ) + ρg
Moment of inertia:
I55 = ρDB
R L/2
−L/2
R
Awp
x2 ds]ζ5 (t)
x2 dx
From the equation of motion for pitch, we can get pitch motion: ζ5 (t) = ζ̄5 sin ωt
ζ̄5
a
= .....
Sample Results for Pitch Motion Draft: D=12 m
Width: B=40 m
Bv = 8% critical damping
MN/m
Pitch moment/wave amplitude
6000
FE5
Added mass
Fraude-Krylov
Total
4000
2000
0
-7000
0
1
2
3
4
Ship length/wavelength
Degrees/m
RAO
Pitch angle/wave amplitude
0.9
0.6
0.3
0
0
1
2
3
4
Ship length/wavelength
Images by MIT OpenCourseWare.
Deck elevation at bow:
ZD = ζ3 (t) − (L/2)ζ5 (t) + H
where H is deck height
Bottom elevation at bow:
ZB = ζ3 (t) − (L/2)ζ5 (t) − D
where D is draft
Wave elevation at bow:
Zw = η(x = L/2, t) = a cos(ωt − kL/2)
If ZD < Zw , wave overtoping occurs;
If ZB > Zw , ship bottom is out of water
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2.019 Design of Ocean Systems
Spring 2011
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