# damage stability calculation

Discussion in 'Stability' started by Dr34m3r, Dec 15, 2013.

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### Dr34m3rSenior Member

hi,

Can anyone tell me some reference source (book or any other) where i can find an example calculation for 1 compartment/tank damage calculation ( calculation new gz , hydrostatic parameters after damaged).

I actually want to know the basics steps /formula.

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### jehardimanSenior Member

Any of the basic Naval Architecture texts will have it. I know PNA and Basic Ship Theory do.

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### athvasSenior Member

Deterministic damage stability of Probabilistic Damage stability ?

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### Eric SponbergSenior Member

This is relatively easy to do on a computer with hydrostatic and stability calculation abilities. But the whole process can get quite involved. Here in the US, damaged stability calculations are required for a lot of commercial vessels, and generally for a number of loading conditions. For strictly passenger vessels, this involves departure condition (full fuel and water with all passengers) arrival condition (10% fuel and water with all passengers), and the same with just one person (or minimum crew) on board.

First, you have to have the hull and floodable compartments fully defined in the computer. You also need to know precisely (proved by stability test) where the center of gravity is of the vessel and in its various load conditions. For each load condition, you first calculate the intact stability. Then you tell the computer which compartments are flooded--you have to check a variety of compartment flooding conditions--and the computer will remove the buoyancy from that compartment and recalculate the equilibrium floating condition. The boat will settle in the water, and you have to make sure that the boat does not submerge too far. Generally, here in the US, the allowed sinkage limit is down to the Margin Line, which typically is 3" (76 mm) below the deck edge. The guidelines for the geometry of the margin line (and all the rest of intact and damage stability requirements) are stipulated in the US Federal Regulations, Title 46, Subchapter S.

For passenger vessels, in each load case, you have to assume that all of the passengers are going to make their way to the nearest point of egress in the flooded vessel. So if the boat is heeling to starboard, for example, and there is a point of egress off the vessel on the starboard side, you have to assume that all the passengers are congregating at that point of egress.
this is the worst possible scenario--personally, if I were on a flooded vessel, I would probably make my way to the high side. There are rules that stipulate how much deck area each person occuplies, so that you can determine how much area all those passengers are going to take up on the available deck. You have to calculate the new center of gravity (longitudinal, vertical, and transverse) of the boat in that condition, and use that center of gravity in the corresponding flooded condition stability calculation.

You do this analysis for every compartment that can be flooded, and check to see if the margin line submerges at any point along the length of the vessel on the low side. If it margin line does not submerge, then the flooded condition for the vessel is considered safe. If the margin line does submerge, then the vessel fails in that condition and something has to be done to correct it. Either you reduce the number of passengers, or reconfigure the bulkheads (or add more) to make the flooded spaces smaller, or some combination of both.

Doing all this work gets really involved. If water does get on deck in a flooding situation, such as through a scupper or freeing port, it might be able to get below into a floodable space through a hatch or a doorway. You have to take those factors into account.

I hope that helps put the problem of damaged stability into a little better perspective.

Eric

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### smartbightNaval Architect

Basic principles, stability intact-damage

Take a look at page 40. Basic damage.

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### jehardimanSenior Member

It is really easy to do by hand also, all you need is a set of Bonjeans, which you should have (along with a set of Draft & Other curves) before starting damage stability anyway. And while you need the exact KG to determine the exact GZ, damaged stability should be done in terms of KN looking for a curve of maximum KG to give the limiting GM required. See the latest IMO SOLAS damaged stability requirements.

Telling someone thay have to have a computer to do something is a disservice to both them and the community in general. The computer program should be a helpmate, not a subsititue to understanding. Getting off my soapbox now, have to go teach someone the Moment Distribution method.

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### Eric SponbergSenior Member

Well, yes, doing the calculations by hand using Bonjean curves is not difficult, just tedious, and computers are ubiquitous these days, so I felt my explanation was easier to give in this posting by referring to the computer as the appropriate tool for calculation. I was giving an overall explanation of tackling the task, not a detailed explanation of actually doing the hydrostatic and stability calculations.

A complete discussion of subdivision and damaged stability calculations, by hand, is given in Principles of Naval Architecture, Volume 1, published by the Society of Naval Architects and Marine Engineers.

Eric

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### Dr34m3rSenior Member

Thanks very much everyone for the discussion. trying to follow the advices i will post again if have some problem of understanding.

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### Dr34m3rSenior Member

Deterministic damage stability

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### TANSLSenior Member

I will write with all humility, without trying to be pedantic. It seems to me that the calculation of stability after damage is a complicated process and , if done by hand, impossible by the enormous amount of calculations required. You must have very clear concepts and know exactly what you are doing.
The floodable length calculation is much simpler but it is not easy. Probably it can be performed with a spreadsheet but doing enough simplifications. The first is that integration by Simpson's method , for example, or trapezoids , can not be sufficiently accurate.
Estimates of the flood of the various compartments of the ship , frankly , it seems difficult to make by hand. The option of adding a weight to simulate water shipped in each compartment , is simpler, but not exact . If we want to calculate the actual properties of the new waterplane, the thing is impossible to calculate by hand . How to calculate the actual area of ​​the flotation and its center of gravity ? . How to calculate the position of the center of buoyancy of the boat? And so , how to calculate the equilibrium position after damage, heel and trim?. Without that you can not make a correct calculation.
If many simplifications are introduced, one can obtain approximate results. But it seems not plausible that, with the calculation methods available today, no administration nor the Classification Society wants to accept those results.
And do not even want to talk about the probabilistic method.

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Not impossible at all.

As JEH and Eric note it is just a bit tedious. How do you think boats in the past before computers came along were calculated?? I like, I am sure Eric and JEH, were taught pre computer days, we did everything by hand. Tedious, yes, time consuming, yes....but it can be done.

Those were days were you had a person or dept, dedicated to doing just stability, as it took so long to produce the data required. Not impossible, just time consuming. (Certainly on big boats).

My first job after graduation...we had 1 computer in the whole company, for stability calc's. It had some 64kb of memory! The program used 11 floppy discs which had to be inserted in sequence and a single calculation took some 10mins (felt longer in those days), per angle of heel...after digitising the hull form into the program.

Today my computer program does this calculation in less than a second, for the whole range and conditions.

BUT...you can still do this by hand, if you understand how it is calculated. So, as Eric/JEH note, just read the text, if you haven't been taught or know how.

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### Dr34m3rSenior Member

some small question,

is the residual arm and the righting arm the same item ?

what is the heeling arm by definition ?

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### TANSLSenior Member

I agree with you Ad Hoc. I said "is imposible" and that is not true. I should have said unworkable in practice, rather than impossible. Any calculation to do with a computer, can be done by hand. There are other things which a computer does and by hand are impossible to do, but no calculations. So I have to learn to use words correctly.
When I did my first computations after graduation, even pocket computers existed.
That said, is it right to tell Dr34m3r "doing the calculations by hand using Bonjean curves is not difficult"?. I think not. With all due respect, I think should also be good clarify words like these. Because the important thing is not to use or not to use Bonjean, but knowing what to calculate and accurately calculate it and Bonjean, to calculate the equilibrium waterline, may not be very accurate.This is what I wanted to express in my previous post.

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Yes and no!

The right arm, or normally called the righting lever is the distance between two locations that is usually defined by the 2 letters, G and Z.

In the diagram below:

When a vessel heels, i.e. it rolls due to "some force", the vessel is no longer upright. As can be seen in the image, the G, the centre of gravity (VCG or KG) remains in the same location. The centre of buoyancy (KB or VCB) has moved from B to B1.

The righting lever is the distance from the KG, or "G" in the diagram to the vertical through the centre of bouncy. The angle is at 90 degrees. Where this intersects it is noted with the letter Z.

So when the vessel is heeled (or rolls), the righting lever required to bring the vessel back to up right is the distance GZ.

However, if the vessel is damaged, for whatever reason and this results in the centre of gravity going up, now located at G1, the righting lever is now G1Z1.

As you can see there is a difference in the length (size) of the righting lever (arm). The righting lever has reduced from GZ to G1Z1.

Both are called righting levers. However, when a vessels has changed from its original condition, as in this simple example, the final resulting righting lever is less, this being G1Z1 is called the residual righting lever. In other words, following damage or change to the vessel, how much righting lever is left when compared to the original undamaged/changed vessel.?

The reason for this is that once a vessel is damaged, you need to ensure that the vessel can still 'right itself'. Thus the residual righting lever is important. It can sometimes become negative..which means..yup..there is a serious problem!

A heeling arm, or lever, is created by any disturbance that causes a vessel to heel (roll) to one side. Since the word used is "lever" and in the diagram above it is obviously a distance, the heeling arm is simply a function of some "Moment" divided by the vessels displacement. The result is a lever, units in metre or feet which ever you use.

So, a wind heeling lever, is calculated from the exposed area to the wind, that area coupled with the wind speed is a force and its location is at the centroid of the area. Thus a force x distance = moment. If you divide that by the vessel's displacement you get a "heeling lever". What you can then do is draw that lever (as a curve/line) directly onto a GZ curve. Where that lever crosses the intact stability GZ curve, is the angle at which the vessel will heel, or roll, to, when exposed to a wind with the speed as calculated..

Does that makes sense now?

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### TANSLSenior Member

Estimated Ad Hoc, probably due to my ignorance on the subject, I was surprised by following statement:
I would not want to argue with you but to clarify my confused ideas. My question is: how can a flooding produce a rise of center of gravity?. No doubt it's possible but I do not see it. However, if the existing load in the flooded compartment is not lost , the c.of g. does not have to change. If the c. of g. does not change, which, I believe, occurs in many damages, what about the GZ and why?. Consider, for example, in a flood of the engine room, the weights do not change. Or are you referring to their apparent weight?. This would further complicate the calculations, especially if done by hand.
For me the problem is not knowing the position of G but the center of buoyancy of the ship flooded. Am I right or I'm wrong as a beginner?

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