2 019 D
2.019
Desiign off O
Ocean Systems
t
Lecture 13
Mooring
g Dynamics
y
(II))
March 28, 2011
Static Analysis: Catenary Cable
z
ϕW
x
TH
T
h
ϕ
Anchor
s
Seafloor
ϕO = 0
xB
x
Image by MIT OpenCourseWare.
Cable configuration:
Tension along the cable:
w
TH
s = sinh 
w
 TH
T = TH + w h + (w + r gA ) z
TH
z−h =
w

x


w
cosh 
 TH

 
x  −1
 
Tz = w s
Catenary Solution ⎯⎯ Key Results (without Elasticity)
• Minimum line length required (or suspended length for a given fairlead tension)
for gravity anchor:
lmin = h
2Tmax
wh
−1
1
2
• Horizontal force for a given fairlead tension:
TH = T − wh
• Horizontal scope (length in plan view from fairlead to touchdown point):
x=
TH
w
sinh
−1
• Vertical force at the fairlead:
Tz = wlmin
wlmin
TH
Simple Examples
• Given Tmax= Tbr =1510 KN, w=828 N/m, h=25m, then
l min = h
¡ 2T
max
wh
h
−1
¢ 12
= 25 ∗
q
2∗1510∗103
828∗2
25
− 1 = 300.93m
TH = T − wh = 1510 ∗ 103 − 828 ∗ 25 = 1489KN
Tz = wlmin = 828 ∗ 301 = 249KN
x=
TH
w
sinh
−1
³
wlmin
TH
´
=
1489∗103
828
sinh−1
828∗301
1489∗103
= 300m
• Given x=270m, w=828 N/m, h=25m, then
h=
TH
w
h
³ ´
i
cosh Twx
−1
H
T = TH + wh = 1221kN
lmin = h
¡ 2T
wh
−1
¢ 12
= 271m
Tz = wlmin = 828 ∗ 271 = 224KN
TH = 1200kN
Cable Load-Excursion Relation
Vessel Moored with One Anchor Line
TH (kN)
ϕW
ls
l
h
x
Anchor
X
Image by MIT OpenCourseWare.
X(m)
X = l − ls + x
¡
¢1
TH 2
X = l − h 1 + 2 wh +
Restoring
g Coefficient:
C11 =
dTH
dX
"
=w ¡
TH
w
−2
TH
1+2 wh
³
cosh−1 1 +
¢1/2 + cosh
−1
³
wh
TH
1+
´
wh
TH
Image by MIT OpenCourseWare.
´
#−11
Simple Example
Given: A ship experiences a total mean drift force (in surge) of 50KN, wave
frequency oscillation of amplitude ζ 1 = 3 m and frequency 2π/10 rad/s, what is the
total tension in the cable?
Steady tension:
T0H = 50kN
Mean position:
X0 = 92.5m
Restoring coefficient:
C11 (T0H ) ≈ 10kN/
10kN/m
m
TH = T0H
) = T0H
(ωt + α))] = 50 − 30 cos((ωt + α))
0 + TH (ω)
0 + [−C11 ζ1 cos(
|TH | = 80kN
T = TH + wh
h
Plus effect due to slowly varying motion
Catenary Solution ⎯⎯ Key Results (with Elasticity)
• Horizontal force for a given fairlead tension T:
TH = AE
q¡
T
AE
¢2
+1 −
2wh
AE
− AE
• Minimum line length required (or suspended length for a given fairlead tension)
for gravity anchor:
lmin =
• Vertical force at the fairlead:
p
1
w
2
T 2 − TH
Tz = wlmin
• Horizontal scope (length in plan view from fairlead to touchdown point):
x=
TH
w
sinh−1
³
wlmin
TH
´
AE: stiffness of the cable
+
TH lmin
AE
Analysis of Spread Mooring System
Total mooring line force/moment:
M
1
F
y
THi
ψi
(xi, yi)
x
n
= ∑ THi cos y i
i=1
n
F2 M = ∑ THi sin y i
i=1
n
F6 M = ∑ THi [ xi sin y i − yi cos y i ]
i=1
Total mooring line restoring coefficients:
n
C11 = ∑ ki cos 2 y i
i=1
n
Image by MIT OpenCourseWare.
• Mean position of the body is determined
by balancing force/moment between
those due to environments and mooring
lines
C22 = ∑ ki sin 2 y i
i=1
n
C66 = ∑ ki ( xi sin y i − yi cos y i )
i=1
n
C26 = C62 = ∑ ki ( xi sin y i − yi cos y i ) sin y i
i=1
• Iterative solver is usually applied
2
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2.019 Design of Ocean Systems
Spring 2011
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