Mathematical model
For a cantilevered marine riser, tangential force can be incorporated into the motion equations for dynamic analysis of a pipe subject to top horizontal and heave excitations.
Rotatory motions can be taken into account for a drilling riser. The rotatory kinetic energy includes rotatory kinetic energy of the pipe, the fluid, and the end-mass. For the effect of nozzle-induced tensile force, more accurate results can be achieved when assuming the force works in the tangential direction of the bottom end (i.e. r0Ldirection as shown in Figure 3.3). The pressurization at the outlet of the nozzle may have to be taken into consideration referring to [39].
For a simply-supported flexible marine, the internal flow effect would be more essential.
The motion equations can be derived by the modified Hamilton principle of changing mass, which was proposed and successfully applied in the pipe conveying fluid by McIver [26]. It has been successfully utilized in VIV study by Benaroya and Wei [150]. 3-D motion equations of flexible risers have been derived by Patel et al. [55], but the model has not been open to the public because it was carried out for an oil company (by private communication with prof. Patel).
When a free-hanging pipe is very long, large tension due to gravity can render the upper part of the pipe very stiff. In this case, high enough modes with short enough wavelength must be considered in the modal analysis. A large number of Galerkin modes are needed and it may put the calculations beyond the current computational capability [24]. According to author’ knowledge, mode functions may be gained by FEM method, referring to [46, 51, 75]. When there is a large end-mass attached at the bottom end of a long pipe, Houbolt’s finite difference method is still effective referring to [34]. CFD (computational fluid dynamics) method can be employed to investigate the wake mode and assist the pipe’s response predictions.
Dynamic analysis
If a cantilevered pipe is subject to external excitations (e.g. waves; top horizontal exci-tations) at low frequency (compared to vortex shedding), the steady displacement ysL
would be time-varying and dynamics would be very rich referring to experiments in [4].
Referring to [127], the van der Pol oscillator model may be used to study the VIVs of cylinders in tandem arrangement and shear current conditions. Our derived mo-tion equamo-tions account for that the displacement vary along the cylinder span and thus simulations can be conducted at various current profiles.
Calibration of wake coefficients
A new set of empirical wake coefficients can be calibrated based on our derived mathe-matical model and proposed solution methods.
Control system design
Appendix A
Derivation of Motion Equations
The coordinate systems are illustrated in Figure 3.2. Notations have been defined in Chapter 3. Referring to two dimensional (2-D) and 3-D dynamic study of a cantilevered pipe discharging fluid in the air by Semiler, Pa¨ıdoussis et al. in [12, 20, 151], the lateral displacement y and z can be considered small compared to the length of the pipe and defined as being of the order , recorded as y ∼ ϑ(), z ∼ ϑ(). The final motion equations account for nonlinear terms up to the third order ϑ(3) employing the inextensiblity condition and Taylor theorem. Terms exact to ϑ(4) or higher in the expressions of the energy will be omitted.
Figure A.1
1. Derivation of x0
As shown in Figure A.1, when the material point P moves toP∗,x0 = ∂x∂s = 1 +∂u∂s and x0 is always positive. By use of the inextensibility condition, the following is obtained:
x0 = (1−y02−z02)
1
2 = 1− 1
2y02− 1
2z02+ϑ(4). (A.1)
2. Derivation of x00
By use of the inextensibility condition x02+y02+z02 = 1, we can get
x0x00+y0y00+z0z00= 0. (A.2) Then the expression for x00 can be derived as follows:
x00 = −(y0y00+z0z00)
x0 ≈ −(y0y00+z0z00)(1 + 1 2y02+1
2z02)
≈ −(y0y00+z0z00). (A.3)
3. Derivation of δx
By use of the inextensibility condition x02+y02+z02 = 1, we can get
x0δx0+y0δy0+z0δz0 = 0. (A.4) Then the expression for δx can be derived as follows:
δx0 = −(y0δy0+z0δz0)
x0 =−(1 + 1
2y02+ 1
2z02+ϑ(4))(y0δy0+z0δz0)
⇒δx = − Z s
0
(1 +1
2y02+1
2z02+ϑ(4))(y0δy0+z0δz0)ds
≈ − Z s
0
(y0+ 1
2y03+ 1
2y0z02)δy0ds− Z s
0
(z0+1
2z03+1
2z0y02)δz0ds
≈ h
−(y0+1
2y03+1
2y0z02)δy−(z0+1
2z03+1
2z0y02)δz i
s
0
+ Z s
0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δyds +
Z s 0
(z00+3
2z02z00+1
2z00y02+z0y0y00)δzds
≈ −(y0+1
2y03+1
2y0z02)δy−(z0+1
2z03+1
2z0y02)δz +
Z s 0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δyds +
Z s
0
(z00+3
2z02z00+1
2z00y02+z0y0y00)δzds. (A.5) 4. Derivation of x˙0
By use of the inextensibility condition x02+y02+z02 = 1, we can get
Then the expression for ˙x0 can be derived as follows:
x˙0 = −(y0y˙0+z0z˙0) x0
≈ −(1 +1
2y02+1
2z02)(y0y˙0+z0z˙0)
≈ −y0y˙0−z0z˙0− 1
2y03y˙0−1
2y02z0z˙0−1
2z03z˙0−1
2z02y0y˙0+ϑ(5). (A.7) 5. One property of integrals
Z t2 t1
Z L 0
g(s)Z s 0
f(s)δyds dsdt=
Z t2 t1
Z L 0
Z L s
g(s)ds
f(s)δydsdt (A.8)
It can be demonstrated as follows:
Z t2 t1
Z L 0
g(s) Z s
0
f(s)δyds
dsdt
= Z t2
t1
hZ s 0
g(s)dsZ s 0
f(s)δyds dsi
L 0dt−
Z t2 t1
Z L 0
f(s)δyZ s 0
g(s)ds dsdt
= Z t2
t1
Z L 0
g(s)ds Z L
0
f(s)δyds
dsdt− Z t2
t1
Z L 0
Z s 0
g(s)ds
f(s)δydsdt
= Z t2
t1
Z L 0
Z L 0
g(s)ds
f(s)δydsdsdt− Z t2
t1
Z L 0
Z s 0
g(s)ds
f(s)δydsdt
= Z t2
t1
Z L 0
Z L s
g(s)ds
f(s)δydsdsdt. (A.9)
6. Derivation of x¨
By use of the inextensibility condition x02+y02+z02 = 1, we can get x0x˙0+y0y˙0+z0z˙0 = 0
⇒ x˙02+x0x¨0+ ˙y02+y0y¨0+ ˙z02+z0z¨0= 0. (A.10) From equation (A.7), it is easy to get
x˙02 ≈ (−y0y˙0−z0z˙0−1
2y03y˙0−1
2y02z0z˙0− 1
2z03z˙0−1
2z02y0y˙0)2
≈ y02y˙02+z02z˙02+ 2y0y˙0z0z˙0+ϑ(4). (A.11) By use of equation (A.10) and equation (A.11), the expression for ¨x0 can be derived:
x¨0 = 1
x0(−y˙02−y0y¨0−z˙02−z0z¨0−x˙02)
= (1 +1
2y02+1
2z02+ϑ(4))(−y˙02−y0y¨0−z˙02−z0z¨0−y02y˙02−z02z˙02−2y0y˙0z0z˙0+ϑ(4)).
Then the expression for ¨x can be derived as follows:
¨ x =
Z s 0
x¨0ds
≈ − Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0 ds
− Z s
0
3
2y02y˙02+1
2y03y¨0+1
2y02z˙02+1 2y02z0z¨0
ds
− Z s
0
3
2z02z˙02+ 1
2z03z¨0+1
2z02y˙02+1
2z02y0y¨0 ds
− Z s
0
2y0y˙0z0z˙0ds. (A.12)
7. Derivation of −
t2
R
t1
L
R
0
¨
xδxdsdt
By use of equation (A.5) and equation (A.12),
−
t2
Z
t1
L
Z
0
¨ xδxdsdt
≈ − Z t2
t1
Z L 0
n
y0δy+z0δz− Z s
0
y00δyds− Z s
0
z00δzdsZ s 0
y˙02+ ˙z02+y0y¨0+z0z¨0 dso
dsdt
= −
Z t2
t1
Z L 0
n y0δy
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0
ds o
dsdt
− Z t2
t1
Z L 0
n z0δz
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0
ds o
dsdt +
Z t2
t1
Z L
0
nZ s
0
y00δyds Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0 dso
dsdt +
Z t2
t1
Z L 0
nZ s 0
z00δzds Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0 dso
dsdt
= −
Z t2
t1
Z L 0
n y0
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0 dso
δydsdt
− Z t2
t1
Z L 0
n z0
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0
ds o
δzdsdt +
Z t2
t1
Z L 0
n y00
Z L s
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0
dsds o
δydsdt +
Z t2
t1
Z L 0
n z00
Z L s
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0 dsdso
δzdsdt (A.13)
can be achieved.
8. Derivation of
t2
R
t1
L
R
0
x0δxdsdt
By use of equation (A.1) and equation (A.5), the following can be achieved:
t2
Z
t1
L
Z
0
x0δxdsdt
= Z t2
t1
Z L
0
nh−(y0+1
2y03+1
2y0z02)δy−(z0+1
2z03+ 1
2z0y02)δz +
Z s 0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δyds +
Z s 0
(z00+3
2z02z00+1
2z00y02+y0z0y00)δzds i
+1 2y02
y0δy+z0δz− Z s
0
y00δyds− Z s
0
z00δzds +1
2z02
y0δy+z0δz− Z s
0
y00δyds− Z s
0
z00δzds o
dsdt
= Z t2
t1
Z L 0
(−y0δy−z0δz)dsdt +
Z t2
t1
Z L 0
Z s 0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δydsdsdt +
Z t2
t1
Z L 0
Z s 0
(z00+ 3
2z02z00+1
2z00y02+z0y0y00)δzdsdsdt
−1 2
Z t2
t1
Z L 0
y02 Z s
0
y00δyds+ Z s
0
z00δzds
dsdt
−1 2
Z t2
t1
Z L 0
z02Z s 0
y00δyds+ Z s
0
z00δzds dsdt
= Z t2
t1
Z L 0
(−y0δy−z0δz)dsdt +
Z t2
t1
Z L 0
Z s 0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δydsdsdt +
Z t2
t1
Z L 0
Z s 0
(z00+3
2z02z00+ 1
2z00y02+z0y0y00)δzdsdsdt
−1 2
Z t2
t1
Z L 0
h y00
Z L s
(y02+z02)ds i
δydsdt−1 2
Z t2
t1
Z L 0
h z00
Z L s
(y02+z02)ds i
δzdsdt
= Z t2
t1
Z L 0
h
(L−s)(y00+3
2y02y00+1
2y00z02+y0z0z00)ds i
δydsdt +
Z t2
t1
Z L 0
h
(L−s)(z00+3
2z02z00+ 1
2z00y02+z0y0y00)dsi δzdsdt
−1 2
Z t2
t1
Z L 0
h y00
Z L s
(y02+z02)dsi
δydsdt−1 2
Z t2
t1
Z L 0
h z00
Z L s
(y02+z02)dsi δzdsdt
− Z t2
t1
Z L 0
(y0δy+z0δz)dsdt. (A.14)
9. Derivation of
t2
R
t1
L
R
0
˙
x0δxdsdt
By use of equation (A.5) and equation (A.7), the following equation can be derived:
t2
Z
t1
L
Z
0
˙
x0δxdsdt
= Z t2
t1
Z L 0
h
−(1 + 1
2y02+1
2z02)(y0y˙0+z0z˙0) i
×h
−(y0+ 1
2y03+ 1
2y0z02)δy−(z0+1 2z03+1
2z0y02)δz +
Z s 0
(y00+ 3
2y02y00+1
2y00z02+y0z0z00)δyds +
Z s 0
(z00+3
2z02z00+1
2z00y02+z0y0y00)δzdsi dsdt
≈ Z t2
t1
Z L 0
h
(y0y˙0+z0z˙0)
y0δy+z0δz− Z s
0
y00δyds− Z s
0
z00δzds i
dsdt
= Z t2
t1
Z L 0
h
(y0y˙0+z0z˙0) y0δy−
Z s 0
y00δydsi dsdt +
Z t2
t1
Z L 0
h
(y0y˙0+z0z˙0) z0δz−
Z s 0
z00zδzdsi dsdt
= Z t2
t1
Z L 0
(y02y˙0+y0z0z˙0)δydsdt− Z t2
t1
Z L 0
h y00
Z L s
(y0y˙0+z0z˙0)dsi δydsdt
+ Z t2
t1
Z L 0
(z02z˙0+z0y0y˙0)δzdsdt
− Z t2
t1
Z L 0
h z00
Z L s
(y0y˙0+z0z˙0)ds i
δzdsdt. (A.15)
10. Derivation of Rt2
t1
RL
0 x00δxdsdt
By use of equation (A.3) and equation (A.5), the following equation can be derived:
Z t2
t1
Z L 0
x00δxdsdt
≈ Z t2
t1
Z L 0
[−(y0y00+z0z00)]
−y0δy−z0δz+ Z s
0
y00δyds+ Z s
0
z00δzds dsdt
= Z t2
t1
Z L 0
(y0y00+z0z00)
y0δy+z0δz− Z s
0
y00δyds− Z s
0
z00δzds dsdt
= Z t2
t1
Z L 0
y0y00
y0δy+z0δz− Z s
0
y00δyds− Z s
0
z00δzds
dsdt +
Z t2
t1
Z L 0
z0z00
y0δy+z0δz− Z s
0
y00δyds− Z s
0
z00δzds
dsdt
= Z t2
t1
Z L 0
y02y00δy+y0y00z0δz−y0y00 Z s
0
y00δyds−y0y00 Z s
0
z00δzds
dsdt +
Z t2
t1
Z L 0
z02z00δz+z0z00y0δy−z0z00 Z s
0
z00δzds−z0z00 Z s
0
y00δyds dsdt
= Z t2
t1
Z L 0
y02y00+z0z00y0−y00 Z L
s
y0y00ds−y00 Z L
s
z0z00ds δydsdt +
Z t2
t1
Z L 0
z02z00+y0y00z0−z00 Z L
s
z0z00ds−z00 Z L
s
y0y00ds
δzdsdt. (A.16)
11. Kinetic energy of a plain pipe
Z t2
t1
δT dt = m Z t2
t1
Z L 0
( ˙xδx˙+ ˙yδy˙+ ˙zδz)dsdt˙ +Mi Z t2
t1
Z L 0
h
( ˙x+U x0)δ( ˙x+U x0) +( ˙y+U y0)δ( ˙y+U y0) + ( ˙z+U z0)δ( ˙z+U z0)
i dsdt
= Z t2
t1
Z L 0
(m+Mi)( ˙xδx˙+ ˙yδy˙+ ˙zδz)dsdt˙ + Z t2
t1
Z L 0
MiU2(x0δx0+y0δy0+z0δz0)dsdt +
Z t2
t1
Z L 0
MiU(x0δx˙+y0δy˙+z0δz)dsdt˙ + Z t2
t1
Z L 0
MiU( ˙xδx0+ ˙yδy0+ ˙zδz0)dsdt
= Z L
0
h
(m+Mi)( ˙xδx+ ˙yδy+ ˙zδz)i
t2
t1
ds− Z t2
t1
Z L 0
(m+Mi)(¨xδx+ ¨yδy+ ¨zδz)dsdt +
Z L 0
h
MiU(x0δx+y0δy+z0δz) i
t2
t1
ds− Z t2
t1
Z L 0
MiU˙(x0δx+y0δy+z0δz)dsdt
−2 Z t2
t1
Z L 0
MiU( ˙x0δx+ ˙y0δy+ ˙z0δz)dsdt+ Z t2
t1
MiU h
˙
xδx+ ˙yδy+ ˙zδz i
L 0dt
= −
Z t2
t1
Z L
0
(m+Mi)(¨xδx+ ¨yδy+ ¨zδz)dsdt− Z t2
t1
Z L
0
MiU˙(x0δx+y0δy+z0δz)dsdt
−2 Z t2
t1
Z L 0
MiU( ˙x0δx+ ˙y0δy+ ˙z0δz)dsdt+ Z t2
t1
MiU( ˙xLδxL+ ˙yLδyL+ ˙zLδzL)dt
= −(m+Mi) Z t2
t1
Z L 0
h
¨ y+y0
Z s 0
( ˙y02+ ˙z02+y0y¨0+z0z¨0)dsi dsδydt
+(m+Mi) Z t2
t1
Z L 0
h y00
Z L s
Z s 0
( ˙y02+ ˙z02+y0y¨0+z0z¨0)dsds i
dsδydt
−(m+Mi) Z t2
t1
Z L 0
h
¨ z+z0
Z s 0
( ˙y02+ ˙z02+y0y¨0+z0z¨0)ds i
dsδzdt +(m+Mi)
Z t2
t1
Z L 0
h z00
Z L s
Z s 0
( ˙y02+ ˙z02+y0y¨0+z0z¨0)dsdsi dsδzdt
−2 Z t2
t1
Z L 0
MiUh
y˙0(1 +y02) +y0z0z˙0−y00 Z L
s
(y0y˙0+z0z˙0)dsi δydsdt
−2 Z t2
t1
Z L 0
MiUh
z˙0(1 +z02) +y0z0y˙0−z00 Z L
s
(z0z˙0+y0y˙0)dsi δzdsdt +
Z t2
t1
MiU( ˙xLδxL+ ˙yLδyL+ ˙zLδzL)dt
− Z t2
t1
Z L 0
h
MiU˙(L−s)(y00+3
2y02y00+1
2y00z02+y0z0z00) i
δydsdt
− Z t2
t1
Z L 0
h
MiU˙(L−s)(z00+3
2z02z00+ 1
2z00y02+z0y0y00)i
δzdsdsdt +1
2 Z t2
t1
Z L 0
h MiU y˙ 00
Z L s
(y02+z02)dsi δydsdt +1
2
Z t2Z Lh MiU z˙ 00
Z L
(y02+z02)dsi
δzdsdt. (A.17)
12. Strain energy of a plain pipe
Z t2
t1
δVsdt = 1 2EI
Z t2
t1
Z L 0
δ(y02y002+z02z002+ 2y0y00z0z00+y002+z002)dsdt
= EI Z t2
t1
Z L 0
(y0y002δy0+y02y00δy00+y00δy00)dsdt
| {z }
Vs1
+EI Z t2
t1
Z L
0
(z0z002δz0+z02z00δz00+z00δz00)dsdt
| {z }
Vs2
+EI Z t2
t1
Z L
0
δ(y0y00z0z00)zdsdt
| {z }
Vs3
Vs1 = EI Z t2
t1
Z L 0
(y0y002δy0+y02y00δy00+y00δy00)dsdt
= EI Z t2
t1
h
y0y002δy+y02y00δy0+y00δy0 i
L 0dt
−EI Z t2
t1
Z L 0
h
(y003+ 2y0y00y000)δy+ (2y0y002+y02y000)δy0+y000δy0 i
dsdt
= −EI
Z t2
t1
h
(2y0y002+y02y000)δy+y000δyi
L 0dt +EI
Z t2
t1
Z L 0
h−(y003+ 2y0y00y00)δy+ (2y003+ 6y0y00y000+y02y0000)δy+y0000δyi dsdt
= EI Z t2
t1
Z L 0
y0000+y003+ 4y0y00y000+y02y0000
δydsdt Vs2 = EI
Z t2
t1
Z L 0
z0000+z003+ 4z0z00z000+z02z0000
δzdsdt Vs3 = EI
Z t2
t1
Z L 0
δ(y0y00z0z00)dsdt
= EI Z t2
t1
Z L 0
y00z0z00δy0
| {z }
Vs3.1
+y0z0z00δy00
| {z }
Vs3.2
+y0y00z00δz0
| {z }
Vs3.3
+y0y00z0δz00
| {z }
Vs3.4
dsdt
Vs3.1 = Z t2
t1
Z L 0
(y00z0z00δy0)dsdt
= Z t2
t1
y00z0z00δy
L 0dt−
Z t2
t1
Z L 0
(y000z0z00+y00z002+y00z0z000)δydsdt
= −
Z t2
t1
Z L 0
(y000z0z00+y00z002+y00z0z000)δydsdt
Vs3.2 = Z t2
t1
Z L
0
(y0z0z00δy00)dsdt
= Z t2
t1
h
y0z0z00δy0i
L 0dt−
Z t2
t1
Z L 0
y00z0z00+y0z002+y0z0z000
δy0dsdt
= −
Z t2
t1
(y00z0z00+y0z002+y0z0z000)δy
L 0dt+
Z t2
t1
Z L 0
y000z0z00+y00z002+y00z0z000 +y00z002+ 2y0z00z000+y00z0z000+y0z00z000+y0z0z0000
δydsdt
= Z t2
t1
Z L
0
y000z0z00+ 2y00z002+ 2y00z0z000+ 3y0z00z000+y0z0z0000 δydsdt Vs3.1+Vs3.2 =
Z t2
t1
Z L 0
y00z002+y00z0z000+ 3y0z00z000+y0z0z0000 δydsdt Vs3.3+Vs3.4 =
Z t2
t1
Z L 0
z00y002+z00y0y000+ 3z0y00y000+z0y0y0000 δzdsdt
Vs = EI Z t2
t1
Z L 0
y0000+y003+ 4y0y00y000+y02y0000
δydsdt +EI
Z t2
t1
Z L 0
y00z002+y00z0z000+ 3y0z00z000+y0z0z0000
δydsdt +EI
Z t2
t1
Z L 0
z0000+z003+ 4z0z00z000+z02z0000 δzdsdt +EI
Z t2
t1
Z L 0
z00y002+z00y0y000+ 3z0y00y000+z0y0y0000
δzdsdt. (A.18)
13. Gravitational energy of a plain pipe
δ Z t2
t1
Vgdt = −mwg Z t2
t1
Z L 0
δxds
= mwg Z t2
t1
Z L 0
h (y0+1
2y03+1
2y0z02)δy+ (z0+1
2z03+1
2z0y02)δz
− Z s
0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δyds
− Z s
0
(z00+3
2z02z00+1
2z00y02+z0y0y00)δzdsi dsdt
= mwg Z t2
t1
Z L 0
(y0+1
2y03+ 1
2y0z02)δydsdt
−mwg Z t2
t1
Z L 0
(L−s)(y00+3
2y02y00+ 1
2y00z02+y0z0z00)δydsdt +mwg
Z t2
t1
Z L
0
(z0+1
2z03+1
2z02y0)δzdsdt
−mwg Z t2
t1
Z L 0
(L−s)(z00+3
2z02z00+1
2z00y02+y0z0y00)δzdsdt.(A.19)
14. Virtual work due to Virtual momentum transport
R.h = Z t2
t1
h MiU
∂rL
∂t +UτL
δrL
i dt
= Z t2
t1
MiU
˙
xLδxL+ ˙yLδyL+ ˙zLδzL dt
| {z }
Wvm.1
+ Z t2
t1
MiU2
x0LδxL+yL0 δyL+zL0 δzL dt
| {z }
Wvm.2
Wvm.2 = Z t2
t1
MiU2
x0LδxL+y0LδyL+zL0δzL
dt
= Z t2
t1
Z L
0
MiU2
x00δx+x0δx0+y00δy+y0δy0+z00δz+z0δz0 dsdt
= Z t2
t1
Z L 0
MiU2
x00δx+y00δy+z00δz dsdt
= Z t2
t1
Z L 0
MiU2
y02y00+y00+z0z00y0−y00 Z L
s
y0y00ds−y00 Z L
s
z0z00ds δydsdt
+ Z t2
t1
Z L 0
MiU2
z02z00+z00+y0y00z0−z00 Z L
s
z0z00ds−z00 Z L
s
y0y00ds
δzdsdt
⇒R.h. = Z t2
t1
Z L 0
MiU2 h
(1 +y02)y00+z0z00y0−y00 Z L
s
y0y00ds−y00 Z L
s
z0z00ds i
δydsdt +
Z t2
t1
Z L 0
MiU2 h
(1 +z02)z00+y0y00z0−z00 Z L
s
z0z00ds−z00 Z L
s
y0y00ds i
δzdsdt +
Z t2
t1
MiU
˙
xLδxL+ ˙yLδyL+ ˙zLδzL
dt. (A.20)
15. Kinetic energy of a bottom end-mass
Te = 1
2me( ˙x2L+ ˙yL2 + ˙y2L) = 1 2me
Z L 0
δ(s−L)( ˙x2+ ˙y2+ ˙y2)ds Z t2
t1
Tedt = Z L
0
meδ(s−L)
˙
xδx+ ˙yδy+ ˙zδz
t2
t1ds
− Z t2
t1
Z L 0
meδ(s−L)
¨
xδx+ ¨yδy+ ¨zδz dsdt
= −
Z t2
t1
Z L 0
meδ(s−L)h
¨
yδy+y0δy Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0 dsi
dsdt
− Z t2
t1
Z L 0
meδ(s−L) h
¨
zδz+z0δz Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0
ds i
dsdt +
Z t2
t1
Z L 0
meδ(s−L) hZ s
0
y00δyds Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0
ds i
dsdt +
Z t2
t1
Z L 0
meδ(s−L) hZ s
0
z00δzds Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0
ds i
dsdt
= −
Z t2
t1
Z L 0
meδ(s−L)h
¨ y+y0
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0 dsi
δydsdt
− Z t2
t1
Z L 0
meδ(s−L)h
¨ z+z0
Z s 0
y˙02+ ˙z02+y0y¨0+z0z¨0 dsi
δzdsdt
+ Z t2
t1
y00 Z L
s
meδ(s−L) Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0
dsdsδzdsdt +
Z t2
t1
z00 Z L
s
meδ(s−L) Z s
0
y˙02+ ˙z02+y0y¨0+z0z¨0
dsdsδzdsdt. (A.21)
16. Gravitational energy of a bottom end-mass
δ Z t2
t1
Vedt = − Z t2
t1
megδxLdt=− Z t2
t1
Z L 0
megδ(s−L)δxdsdt
= Z t2
t1
Z L 0
megδ(s−L)(y0+1
2y03+1
2y0z02)δydsdt Z t2
t1
Z L 0
megδ(s−L)(z0+ 1 2z03+1
2z0y02)δzdsdt
− Z t2
t1
Z L
0
megδ(s−L)hZ s
0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δydsi dsdt
− Z t2
t1
Z L 0
megδ(s−L)hZ s 0
(z00+3
2z02z00+ 1
2z00y02+z0y0y00)δzdsi dsdt
= Z t2
t1
Z L 0
h
megδ(s−L)(y0+1
2y03+1 2y0z02)i
δydsdt
− Z t2
t1
Z L 0
h
meg(y00+ 3
2y02y00+1
2y00z02+y0z0z00) Z L
s
δ(s−L)ds i
δydsdt
− Z t2
t1
Z L 0
h
meg(z00+3
2z02y00+1
2z00y02+z0y0y00) Z L
s
δ(s−L)ds i
δzdsdt +
Z t2
t1
Z L 0
h
megδ(s−L)(y0+ 1 2z03+1
2z0y02)i
δzdsdt. (A.22)
17. Virtual work due to the nozzle
δ Z t2
t1
WTLdt = Z t2
t1
TLδxLdt
= Z t2
t1
Z L 0
TLδ(s−L)δxdsdt
= −
Z t2
t1
Z L 0
TLδ(s−L)(y0+1
2y03+1
2y0z02)δydsdt
− Z t2
t1
Z L 0
TLδ(s−L)(z0+1 2z03+1
2z0y02)δzdsdt +
Z t2
t1
Z L 0
TLδ(s−L)hZ s 0
(y00+3
2y02y00+1
2y00z02+y0z0z00)δydsi dsdt +
Z t2
t1
Z L 0
TLδ(s−L)hZ s 0
(z00+ 3
2z02z00+1
2z00y02+z0y0y00)δzdsi dsdt +
Z t2
t1
Z L 0
TLδ(s−L)δydsdt+ Z t2
t1
Z L 0
TLz0Lδ(s−L)δzdsdt
= −
Z t2
t1
Z L 0
h
TLδ(s−L)(y0+1
2y03+ 1 2y0z02)
i δydsdt +
Z t2
t1
Z L 0
h
TL(y00+3
2y02y00+1
2y00z02+y0z0z00) i
δydsdt
− Z t2
t1
Z L 0
h
TLδ(s−L)(y0+1 2z03+1
2z0y02) i
δzdsdt +
Z t2
t1
Z L 0
h
TL(z00+3
2z02y00+1
2z00y02+z0y0y00)i
δzdsdt (A.23)
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