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

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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:

𝐿𝑠 = 𝐿 − 𝐿1 − 𝐿2 = 100.75 𝑚

𝐿𝑙 = 𝐿𝑠 + 𝑇 ∗𝐿1

ℎ+ 𝑇 ∗

𝐿2

ℎ= 152.29 𝑚

The area of the trapezium can then be found:

𝐴𝑡𝑟𝑎𝑝 =(𝐿𝑠 + 𝐿𝑙) ∗ 𝑇

2= 2096.44 𝑚2

The displacement volume can be calculated by multiplying the area of the trapezium with the

width B of the pontoon:

∇𝑓= 𝐴𝑡𝑟𝑎𝑝 ∗ 𝐵 = 97484.46 𝑚3

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.

𝜌𝑓 = 1000 𝑘𝑔/𝑚3

𝜌𝑠 = 1025 𝑘𝑔/𝑚3

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:

∆𝑓= ∆𝑠

𝜌𝑓 ∗ 𝑔 ∗ ∇𝑓= 𝜌𝑠 ∗ 𝑔 ∗ ∇𝑠

The displacement volume in salt water can then be calculated:

∇𝑠=𝜌𝑓

𝜌𝑠∗ ∇𝑓= 95106.79 𝑚3

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

∇𝑠=1

2∗ (𝐿𝑠 + 𝐿𝑠 +

𝐿1

ℎ∗ 𝑇 +

𝐿2

ℎ∗ 𝑇) ∗ 𝑇 ∗ 𝐵

∇𝑠= 72.3075 ∗ 𝑇2 + 4684.875 ∗ 𝑇

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:

𝜕𝑇 = 𝑇𝑓 − 𝑇𝑠 = 16.57 − 16.23 = 0.34 𝑚

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:

𝜕∇ ≈ 𝐴𝑤 ∗ 𝜕𝑇 =0.025

1.025∗ ∇𝑓

𝜕𝑇 =0.025

1.025∗

∇𝑓

𝐴𝑤=

0.025

1.025∗ 𝐶𝑉𝑃 ∗ 𝑇

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:

∆𝑓 = 𝑚𝑝𝑜𝑛𝑡𝑜𝑜𝑛 ∗ 𝑔

𝑔 ∗ 𝜌𝑓 ∗ ∇𝑓 = 𝑚𝑝𝑜𝑛𝑡𝑜𝑜𝑛 ∗ 𝑔

𝑚𝑝𝑜𝑛𝑡𝑜𝑜𝑛 = 97484.46 𝑡𝑜𝑛𝑛𝑒𝑠

∆𝑠= 𝑚𝑝𝑜𝑛𝑡𝑜𝑜𝑛 ∗ 𝑔

𝑔 ∗ 𝜌𝑠 ∗ ∇𝑠= 𝑚𝑝𝑜𝑛𝑡𝑜𝑜𝑛 ∗ 𝑔

𝑚𝑝𝑜𝑛𝑡𝑜𝑜𝑛 = 97484.46 𝑡𝑜𝑛𝑛𝑒𝑠

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.

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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

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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 φ. 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 ∆NGZ and ∆NKP:

𝐺𝑍 = 𝑁𝐺 ∗ sin (𝜑)

𝐾𝑃 = 𝑁𝐾 ∗ sin (𝜑)

𝑀𝐺 = 𝑁𝐾 − ℎ

This gives the relationship between GZ and KP:

𝐺𝑍 = (𝑁𝐾 − ℎ) ∗ sin(𝜑)

𝐺𝑍 = 𝐾𝑃 − ℎ ∗ sin (𝜑)

Using this relationship, the lever arm can be calculated in Excel and plotted in function of the

heel angle φ. 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 [°]

Stability curve without load

Figure 5: Stability curve pontoon without load

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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 [°]

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 [°]

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 φ. 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 ∆ 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:

∆ = 𝜌𝑓 ∗ 9.81 ∗ ∇𝑓,𝑡𝑜𝑡= 1 051 954 847 𝑁

Figure 8: Forces acting on the pontoon with added mass

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∇𝑓,𝑡𝑜𝑡=𝑚𝑡𝑜𝑡

𝜌𝑓= 107 232.91 𝑚3

The weight of the added mass:

𝐹2 = 9.81 ∗ 𝑀 = 95 633 501 𝑁

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:

𝐺𝑈 =𝐺𝑊

cos(𝜑)

𝑈𝑉 = CU ∗ sin(𝜑)

From the geometric characteristics of the pontoon and the added mass, the distances GW

and CU can written as:

𝐺𝑊 =𝐵

2−

𝑇

2

𝐶𝑈 = 𝐷 +𝑇

2− ℎ − 𝐺𝑊 ∗ tan(𝜑)

After substituting these expression in the expressions for GU and UV, the expression for GV

can be found:

𝐺𝑉 = 𝐺𝑈 + 𝑈𝑉 =

𝐵2 −

𝑇2

cos(𝜑)+ sin(𝜑) ∗ (𝐷 +

𝑇

2− ℎ − tan(𝜑) ∗ (

𝐵

2−

𝑇

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:

𝐺𝑉 =14.965

cos(𝜑)+ sin(𝜑) ∗ (23.395 − 14.965 ∗ tan(𝜑))

Figure 9: Determination lever arm F2

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Both lever arms GZ and GV are functions of the heel angle φ. Now the moment equilibrium

equation can be written for these 2 forces:

∆ ∗ 𝐺𝑍 = 𝐹2 ∗ 𝐺𝑉

Solving this equation to φ 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 φ. 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 φ is smaller than the equilibrium angle φeq, 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 [°]

Moment equilibrium

Stability moment Overturning moment

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further than φeq 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 φmax 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 φmax 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:

𝐴 = 𝐵 ⟺ 𝐴 + 𝐶 = 𝐵 + 𝐶

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:

𝐴 ≈ ∆𝑥 ∗ (𝑦0

2+ 𝑦1 + 𝑦2 + ⋯ + 𝑦𝑁−1 +

𝑦𝑁

2)

with ∆x the interval between 2 discrete values, in this case ∆x is equal to 1, and N is the

number of intervals ∆x 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 [°]

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:

∆ ∗ 𝐺𝑍 = 𝐹2 ∗ 𝐺𝑉

𝐹2 ∗ 𝐺𝑉 = 𝐹2 ∗ (𝑥

cos(𝜑)+ sin(𝜑) ∗ (23.395 − 𝑥 ∗ tan(𝜑)))

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 ∗ 107 = 95 633 501 ∗ (𝑥 + 0.0209 ∗ (23.395 − 𝑥 ∗ 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.