Introduction to maritime technology
Stability of a pontoon
Prof. Dr. ir. Marc Vantorre
Dr. ir. Evert Lataire
Kevin Vivile
1st Master Faculty of Engineering and Architecture
Civil Engineering Maritime Technology
December 2018
2
Contents
1. Introduction ....................................................................................... 3
2. Geometric properties ......................................................................... 4
2.1 Displacement volume in fresh water............................................. 4
2.2 Draft change from fresh to salt water ........................................... 4
2.3 Mass of the pontoon .................................................................... 5
3. Stability of the pontoon ...................................................................... 6
3.1 Stability curve............................................................................... 6
3.1.1 Stability curve with no load .................................................. 6
3.1.2 Stability curve with mass M .................................................. 8
3.2 Heel angle when mass is placed at the side................................. 9
3.3 Maximum heel angle when mass is placed at once.................... 11
3.4 Transverse position of the mass ................................................ 13
4. Conclusion ...................................................................................... 14
3
1. Introduction
This project will analyse the stability of a pontoon. Firstly, some geometric properties of the
pontoon are calculated when the pontoon moves from fresh water to salt water, for example
the draft change. Next, the stability curve of the pontoon in fresh water will be calculated with
the help of the program Delftship. Then the heel angle will be calculated when a mass is placed
on the pontoon. Thereafter, the maximum heel angle will be determined when the mass is
positioned at once. Finally, the transverse position of the mass will be found so the side of the
pontoon is submerged until a distance h above the keel.
A drawing of the pontoon is given in Figures 1 and 2, and its dimensions are shown in Table 1.
L 155 m
L1 23.25 m
L2 31 m
D 32.55 m
h 17.44 m
B 46.50 m
T 16.57 m
Table 1: Dimensions pontoon
Figure 1: Drawing pontoon
Figure 2: Cross-section pontoon
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2. Geometric properties
2.1 Displacement volume in fresh water
The submerged volume of the pontoon is the displacement volume. This volume is the same
as the volume of water that is displaced by the pontoon. In Figure 1 this volume is represented
by the hatch.
The area of the submerged part is trapezoidal, so the small base Ls and the large base Ll must
be determined:
πΏπ = πΏ β πΏ1 β πΏ2 = 100.75 π
πΏπ = πΏπ + π βπΏ1
β+ π β
πΏ2
β= 152.29 π
The area of the trapezium can then be found:
π΄π‘πππ =(πΏπ + πΏπ) β π
2= 2096.44 π2
The displacement volume can be calculated by multiplying the area of the trapezium with the
width B of the pontoon:
βπ= π΄π‘πππ β π΅ = 97484.46 π3
Now that the displacement volume in fresh water is known, the draft change can be
calculated.
2.2 Draft change from fresh to salt water
Fresh water and salt water have slightly different densities. This causes a change in draft when
a floating body is moved from fresh to salt water or vice versa.
ππ = 1000 ππ/π3
ππ = 1025 ππ/π3
The subscripts f and s stand for fresh water and salt water respectively.
The buoyancy has to be equal in both fresh and salt water because the weight of the pontoon
remains constant. The buoyancy is equal to the weight of the displaced volume of water. This
can be written as:
βπ= βπ
ππ β π β βπ= ππ β π β βπ
The displacement volume in salt water can then be calculated:
βπ =ππ
ππ β βπ= 95106.79 π3
Now the displacement volumes in fresh and salt water are known, the change in draft can be
determined. First write an expression for the displacement volume in salt water in function of
the draft T:
5
βπ =1
2β (πΏπ + πΏπ +
πΏ1
ββ π +
πΏ2
ββ π) β π β π΅
βπ = 72.3075 β π2 + 4684.875 β π
This equation is only valid if the draft T is smaller than h otherwise the submerged volume will
not be a pure trapezoid anymore. But the expectation is that the draft T will decrease and still
be smaller than h when the pontoon is moved to salt water because the displacement volume
in salt water is smaller.
To get a value for the draft T in salt water, the last equation is solved and the draft Ts is equal
to 16.23 m. The draft in fresh water Tf is given in Table 1. To find the draft change, the
difference between the two drafts is calculated:
ππ = ππ β ππ = 16.57 β 16.23 = 0.34 π
As expected the draft decreases with a value of 34 cm when the pontoon is moved from fresh
to salt water.
Another method to calculate the change in draft is by using an approximating expression,
assuming that the waterline does not change significantly when the draft decreases:
πβ β π΄π€ β ππ =0.025
1.025β βπ
ππ =0.025
1.025β
βπ
π΄π€=
0.025
1.025β πΆππ β π
with Aw the horizontal area of the pontoon at the waterline, CVP the vertical prismatic
coefficient which has a typical value of 0.8 and T the draft in fresh water. This equation results
in an approximate value of 32 cm for the draft change. This value is approximately the same
as the result found with the analytic equation.
2.3 Mass of the pontoon
Another important geometric parameter besides the change in draft is the mass of the
pontoon. This is necessary to evaluate the stability of the pontoon. The mass can be easily
calculated by stating that the buoyancy force in fresh and in salt water is equal to the weight
of the pontoon:
βπ = πππππ‘πππ β π
π β ππ β βπ = πππππ‘πππ β π
πππππ‘πππ = 97484.46 π‘πππππ
βπ = πππππ‘πππ β π
π β ππ β βπ = πππππ‘πππ β π
πππππ‘πππ = 97484.46 π‘πππππ
As one can see, the mass of the pontoon is 97484.46 tonnes and does not change when the
pontoon moves from fresh to salt water.
6
3. Stability of the pontoon
3.1 Stability curve
The stability curve can be obtained by plotting the stability moment as a function of the heel
angle. With the help of Delftship the stability curves can be determined for the pontoon when
there is no load on it and when there is a mass M on it. But first the pontoon has to be
modelled in the program. In Figure 3 the model in Delftship is shown for the pontoon.
3.1.1 Stability curve with no load
In Figure 4 a sketch of the pontoon is shown when it is inclined. The stability curve is obtained
when the stability moment or the lever of the stability moment is plotted as a function of the
heel angle.
Figure 3: Model of the pontoon in Delftship
Figure 4: Sketch of the inclined pontoon without added mass
7
The notations used in Figure 4 are explained:
N = metacentre
G = centre of gravity of the pontoon
Z = projection of G onto the normal of the waterline
K = keelpoint
P = projection of K onto the normal of the waterline
In Delftship the cross curves can be calculated of the model and these curves give the distance
of KP with the corresponding heel angle Ο. But the distance GZ is wanted, so an expression
has to be found that relates the distance KP to GZ.
This relation can be found by using the similarity of the two triangles βNGZ and βNKP:
πΊπ = ππΊ β sin (π)
πΎπ = ππΎ β sin (π)
ππΊ = ππΎ β β
This gives the relationship between GZ and KP:
πΊπ = (ππΎ β β) β sin(π)
πΊπ = πΎπ β β β sin (π)
Using this relationship, the lever arm can be calculated in Excel and plotted in function of the
heel angle Ο. The stability curve is given in Figure 5.
-5
-4
-3
-2
-1
0
1
2
3
4
5
0 20 40 60 80 100 120 140 160 180GZ
[m]
Heel angle [Β°]
Stability curve without load
Figure 5: Stability curve pontoon without load
8
3.1.2 Stability curve with mass M
Now a different situation is considered. A cubic block with mass M is placed on the pontoon
and has dimensions T x T x T, with the value of T given in Table 1. The mass M is equal to
10% of the total mass of the pontoon. The mass of the pontoon was calculated in an earlier
paragraph and is equal to 97484,46 tonnes. This results in a mass of 9748.45 tonnes for the
added mass M. The new total mass is 107 232.91 tonnes. A higher mass causes a higher
water displacement. In Delftship the mass has to be changed and then the distance KP can
be calculated again. Using the same formulas as in the last paragraph, the lever arm GZ can
be calculated and plotted in function of the heel angle. The new stability curve is shown in
Figure 6.
Now the two obtained stability curves can be compared with each other. The difference
between the two curves can be easily seen when both curves are plotted in the same graph
(Figure 7).
-5
-4
-3
-2
-1
0
1
2
3
4
5
0 20 40 60 80 100 120 140 160 180GZ
[m]
Heel angle [Β°]
Stability curve with mass
Figure 6: Stability curve pontoon with added mass
-5
-4
-3
-2
-1
0
1
2
3
4
5
0 20 40 60 80 100 120 140 160 180GZ
[m]
Heel angle [Β°]
Stability curves
Without mass with mass
Figure 7: Stability curves for pontoon with and without added mass
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The biggest difference that can be noticed is a smaller peak value of the lever arm GZ when
the added mass is placed on the pontoon. A similarity is that the maximum heel angle is
equal to about 46 degrees for both situations.
3.2 Heel angle when mass is placed at the side
In this situation, the added mass is placed on the edge of the pontoon. This causes the
pontoon to heel with a heel angle Ο. The edge of the mass is aligned with the edge of the
pontoon and no trim occurs because the mass is longitudinally positioned. A sketch of this
situation is shown in Figure 8.
Three forces act on the pontoon. The first one, F1, is the weight of the pontoon itself without
the added mass which acts on the centre of gravity of the pontoon. The second force F2 is
the weight of the added mass which acts on the centre of gravity of the added mass C. The
final force is the buoyancy force β which acts on the centre of buoyancy B and is directed
upwards. The forces are shown in Figure 8.
To find the equilibrium heel angle when the mass is placed at the edge of the pontoon, the
moment equilibrium has to be written for the three acting forces around the centre of
gravity G of the pontoon. F1 goes through the centre of gravity so it will not create a moment
around G. The buoyancy force and the weight of the added mass however, will create a
moment around G. So the lever arms for these 2 forces have to be calculated.
The lever arm for the buoyancy is the distance GZ that has already been calculated in a
previous section and is represented by the stability curve. GZ is dependent on the angle of
inclination of the pontoon. The buoyancy force is equal to the weight of the displacement
volume and can be easily calculated as follows:
β = ππ β 9.81 β βπ,π‘ππ‘= 1 051 954 847 π
Figure 8: Forces acting on the pontoon with added mass
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βπ,π‘ππ‘=ππ‘ππ‘
ππ= 107 232.91 π3
The weight of the added mass:
πΉ2 = 9.81 β π = 95 633 501 π
The lever arm for this force can be calculated by using trigonometry. A figure with the
notations can be seen in Figure 9. The lever arm for F2 is the distance GV and is equal to the
sum of GU and UV. The following relations can be found:
πΊπ =πΊπ
cos(π)
ππ = CU β sin(π)
From the geometric characteristics of the pontoon and the added mass, the distances GW
and CU can written as:
πΊπ =π΅
2β
π
2
πΆπ = π· +π
2β β β πΊπ β tan(π)
After substituting these expression in the expressions for GU and UV, the expression for GV
can be found:
πΊπ = πΊπ + ππ =
π΅2 β
π2
cos(π)+ sin(π) β (π· +
π
2β β β tan(π) β (
π΅
2β
π
2))
Then the values found in Table 1 are substituted in this expression and the relationship of
the distance GV with the heel angle is found:
πΊπ =14.965
cos(π)+ sin(π) β (23.395 β 14.965 β tan(π))
Figure 9: Determination lever arm F2
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Both lever arms GZ and GV are functions of the heel angle Ο. Now the moment equilibrium
equation can be written for these 2 forces:
β β πΊπ = πΉ2 β πΊπ
Solving this equation to Ο will give the heel angle at the state of equilibrium. This equation
can be visualized in a graph by plotting both sides of the equation in function of the heel
angle Ο. The left side gives the influence of the buoyancy and has the same shape as the
stability curve. This is the stability moment. The right side represents the overturning
moment or the effect of the weight of the added mass. Both sides of the equation are
plotted in Figure 10.
As one can see on Figure 10, both sides of the equation intersect in 3 points. This means that
there are 3 equilibrium angles. But the most realistic equilibrium angle occurs at the first
intersection. The value of this intersection can be estimated and is equal to 24.8 degrees.
The irregularity in the curve of the right side of the equation exists because the cosine in the
denominator is equal to zero when the heel angle is 90 degrees.
The conclusion is that the heel angle of the pontoon is equal to 24.8 degrees when the
added mass is placed at the edge of the pontoon.
3.3 Maximum heel angle when mass is placed at once
In this situation the mass will be placed at once on the edge of the pontoon and the
maximum heel angle has to be calculated. A graphic representation of this situation is shown
in Figure 11. If the angle Ο is smaller than the equilibrium angle Οeq, then the overturning
moment is greater than the stability moment and the kinetic energy of the pontoon will
increase to a value equal to the area A. This kinetic energy causes the pontoon to heel
Figure 10: Moment equilibrium equation
-5E+09
-4E+09
-3E+09
-2E+09
-1E+09
0
1E+09
2E+09
3E+09
4E+09
5E+09
0 20 40 60 80 100 120 140 160 180
Mo
men
t [N
m]
Heel angle [Β°]
Moment equilibrium
Stability moment Overturning moment
12
further than Οeq and in this zone the stability moment is greater than the overturning
moment causing the kinetic energy to decrease. When the kinetic energy decreases until it is
equal to zero, then area A is equal to area B and the maximum heel angle Οmax is reached. At
this heel angle the kinetic energy is zero but the stability moment is greater than the
overturning moment so there is no equilibrium. This causes the pontoon to heel in the other
direction and the kinetic energy will rise again while the heel angle decreases. The pontoon
is oscillating about the equilibrium point and due to damping, the amplitude of the
oscillation will decrease and eventually become zero when the pontoon is finally at rest. So
the maximum heel angle Οmax is found when the area A is equal to area B. But because
calculating the area A and B is too complex, the area C is also considered when calculating
the areas:
π΄ = π΅ βΊ π΄ + πΆ = π΅ + πΆ
This makes it much easier because the areas under the curves are much easier to calculate
than just the areas A and B alone. Because the graph is made out of discrete values,
integration is not used to calculate the areas. Instead, the trapezium rule is used:
π΄ β βπ₯ β (π¦0
2+ π¦1 + π¦2 + β― + π¦πβ1 +
π¦π
2)
with βx the interval between 2 discrete values, in this case βx is equal to 1, and N is the
number of intervals βx that are considered.
The obtained value of the maximum heel angle is 43.5 degrees. This value is below the peak
value 46 degrees that was calculated in an earlier section. This means that the pontoon will
not capsize when the mass is placed at once on the side of the pontoon.
Figure 11: Determination maximum heel angle
0
500000000
1E+09
1,5E+09
2E+09
2,5E+09
3E+09
3,5E+09
4E+09
4,5E+09
5E+09
0 10 20 30 40 50
Mo
men
t [N
m]
Heel angle [Β°]
Moment equilibrium
Stability moment Overturning moment phi_max
A C
B
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3.4 Transverse position of the mass
In this section the transverse position of the added mass is determined so the side of the
pontoon is submerged until the distance h above the keel. This can be found with the help of
the software program Heelme. This program can calculate the heel angle when a
displacement volume is given. The program requires an input that describes the shape of the
pontoon. The y- and z-coordinates of 4 sections of the pontoon are given as an input. The
input file is shown in Figure 12.
Other parameters that have to be inputted in the program are the displacement volume of
the pontoon with the added mass on it, the accuracy of the calculations and the point where
the pontoon is rotated around (y critical value and z critical value). The displacement volume
has already been calculated and is equal to 107 232.91 m3. The chosen accuracy is in this
case 0.1% and the y critical value is equal to B/2 and z critical is equal to h. The result given
by the program is shown in Figure 13.
Figure 12: Input file Heelme
Figure 13: Result Heelme calculation
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As one can see, the calculated heel angle is 1.2 degrees. Now the expression of the moment
equilibrium is written in function of the unknown transverse displacement x:
β β πΊπ = πΉ2 β πΊπ
πΉ2 β πΊπ = πΉ2 β (π₯
cos(π)+ sin(π) β (23.395 β π₯ β tan(π)))
The value for the stability moment can be found by using interpolation or by using the graph
in Figure 10. In Figure 10 the value of the moment due to the buoyancy, the stability
moment, can be estimated for a heel angle equal to 1.2 degrees. This gives a value of
9.424*107 Nm for the stability moment. When using interpolation the same value can be
found. With this value and the value of the heel angle, the transverse position of the added
mass x can be easily calculated, assuming that the cosine of 1.2 degrees is approximately
equal to 1.
9.424 β 107 = 95 633 501 β (π₯ + 0.0209 β (23.395 β π₯ β 0.0209))
After solving this equation, the value of x is equal to 49.6 cm. So if the mass is placed at a
transverse distance of 49.6 cm from the centre of the pontoon, the side of the pontoon will
be submerged until a distance h above the keel point.
4. Conclusion
The first section mainly discussed the change in draft when the pontoon moves from fresh to
salt water. First the drafts were calculated in fresh and salt water and then the difference
between these two values was the draft change. The draft in fresh water was equal to 16.57
m and the draft in salt water was equal to 16.23 m. This resulted in a draft change of 34 cm.
The draft in salt water is lower than in fresh water because the density of salt water is
slightly higher than fresh water.
In the second section the stability of the pontoon was evaluated with and without an added
mass using the software program Delftship. The stability curves for the pontoon with and
without the extra mass were constructed and the conclusion was that the curve with the
mass had a lower peak value. Thereafter, the heel angle of the pontoon with the added mass
was calculated when the added mass was positioned at the edge of the pontoon. Using the
expression for moment equilibrium around the centre of gravity, the equilibrium heel angle
was found and was equal to 24.8 degrees. Then the maximum heel angle was calculated
when the mass was positioned at once on the edge of the pontoon. The sudden placing of
the mass causes the pontoon to oscillate and after a while the oscillation dampens until the
amplitude is zero. The maximum heel angle could be derived by equalizing the areas under
the curves of the stability and overturning moment. This resulted in a maximum angle of
43.5 degrees. For the last part the transverse displacement of the mass had to be calculated
so the side of the pontoon is submerged until a distance h above the keel. With the use of
the software package Heelme, the heel angle for this situation was found and after using the
moment equilibrium equation again the displacement x was found and was equal to 49.6
cm.