Trying to calculate barge capacity and displacement

[Viking89]

I am trying to calculate the capacity of Gabro (3g/cm3) in a barge that is 86.1 meters long, 21.6 meters wide, and 5.2 meters in depth. Additionally, the draft is 3.65 meters without any material. My cal is 3,187.8 tons. Is that correct and if so, how do I calculate the adds depth contributed by all 3,187.8 tons of Gabro loads into the barge?

 

[Bystander]

Are you intending to use this barge as an artificial reef? Take us through your calculation one step at a time, please.

 

[SteamKing]

I am trying to calculate the capacity of Gabro (3g/cm3) in a barge that is 86.1 meters long, 21.6 meters wide, and 5.2 meters in depth. Additionally, the draft is 3.65 meters without any material. My cal is 3,187.8 tons. Is that correct and if so, how do I calculate the adds depth contributed by all 3,187.8 tons of Gabro loads into the barge?

The first question is, what is the shape of your barge's hull?

My quick calculation shows that it is not just a box with the dimensions that you have listed. Does this barge have a rake (sloping bottom) at one or both ends? Some other kind of shape?

It would be helpful to see your calculation of the displacement of this barge at the draft of 3.65 meters. Did you have access to hydrostatic curves?

If you have a picture or plans of this barge, that would be most helpful.

 

[Viking89]

Thanks for responding. I am getting the pictures. At this point I am assuming a box barge. My calculation is basically to get volume(LxWxD) and divide by the density of Gabro (3g/cm3). From earlier thread, this should be 3,187.8 tons. This is only a theoretical tonnage at this point because then you have to calculate the added draft caust by this weight. Best I can figure, this is called TPC(tonnage per centimeter displacement). TPC = (LxW)/100 x 1.025. Because as I previously gave the non loaded draft at 3.65 m, my calculator says if I loaded the full 3,187 tons, the barge would sink another 1.7m, which means it would sink cause the overall depth was given at 5.2. So now the excercise is ...how much of the material could be loaded such that the barge would be considered stable. Center of gravity must have something to do with it?

 

[SteamKing]

Thanks for responding. I am getting the pictures. At this point I am assuming a box barge. My calculation is basically to get volume(LxWxD) and divide by the density of Gabro (3g/cm3). From earlier thread, this should be 3,187.8 tons. This is only a theoretical tonnage at this point because then you have to calculate the added draft caust by this weight. Best I can figure, this is called TPC(tonnage per centimeter displacement). TPC = (LxW)/100 x 1.025. Because as I previously gave the non loaded draft at 3.65 m, my calculator says if I loaded the full 3,187 tons, the barge would sink another 1.7m, which means it would sink cause the overall depth was given at 5.2. So now the excercise is ...how much of the material could be loaded such that the barge would be considered stable. Center of gravity must have something to do with it?

Just looking at your figures, it seems that something is off.

A barge 5.2 meters deep which has a draft of 3.65 meters without load seems a little heavy in my experience. There is too little carrying capacity to make using such a vessel worthwhile, IMO.

A typical box barge should draw around 1 meter or less when not loaded.

Depending on the route over which you want to transport this cargo, you will only be able to load this safely barge to a certain draft. Stability of the barge in the loaded condition and the strength of the barge should be carefully investigated, given that the cargo is so dense (specific gravity of 3.0).

Given the dimensions of this barge, does it carry a load line by any chance? If it does, then the load line draft will be the maximum draft to which the barge can be loaded.

 

[Viking89]

Yes, I think you are right. I need to challenge the draft figure and rerun the numbers. Thanks so much for your input...I will follow up when I get done better numbers

 

[nasu]

Your calculation is wrong from the start. Dividing volume by density is meaningless. You need to multiply volume by density yo get the mass. And to use consistent units.
And if you fill your box barge completely with a denser than water material it will sink, of course. So the added depth will be all the way to the bottom.

 

[Murshid]

Sir,
Any one can help me to do calculate Barge draft. Barge Size ( 550 cm x 156 cm x 29.5 cm ) Load is 350 Ton length of the load 10m Width 3.0m, please can give calculator too.

Thanks Sir

 

[SteamKing]

Sir,
Any one can help me to do calculate Barge draft. Barge Size ( 550 cm x 156 cm x 29.5 cm ) Load is 350 Ton length of the load 10m Width 3.0m, please can give calculator too.

Thanks Sir

Hi, Murshid, Welcome to PF.!

We ask members not to hijack existing threads by adding unrelated questions to the previous posts. This keeps things less confusing.

As to your specific question, it appears your barge (550 cm x 156 cm x 29.5 cm) is grossly undersized for carrying a 350-ton load which measures 10m x 3m.

A rectangular box having the dimensions of your barge would displace only about 2.5 metric tons of fresh water when fully submerged.

Do you have the correct barge dimensions? These dimensions seem to be more appropriate for a large skiff than for a working barge.

 

[Murshid]

Barge size is 54.864 x 15.5448 x 2.95 m and Barge capacity 1500 ton
Sorry to misunderstand. I am learning. Not to disturb others.
So my query is " My load is 350 Ton size above shown."
What the draft after load the cargo on the barge?

 

[Murshid]

Hi, Murshid, Welcome to PF.!

We ask members not to hijack existing threads by adding unrelated questions to the previous posts. This keeps things less confusing.

As to your specific question, it appears your barge (550 cm x 156 cm x 29.5 cm) is grossly undersized for carrying a 350-ton load which measures 10m x 3m.

A rectangular box having the dimensions of your barge would displace only about 2.5 metric tons of fresh water when fully submerged.

Do you have the correct barge dimensions? These dimensions seem to be more appropriate for a large skiff than for a working barge.

Barge size is 54.864 x 15.5448 x 2.95 m and Barge capacity 1500 ton
Sorry to misunderstand. I am learning. Not to disturb others.
So my query is " My load is 350 Ton size above shown."
What the draft after load the cargo on the barge?

 

[Murshid]

Hi, Murshid, Welcome to PF.!

We ask members not to hijack existing threads by adding unrelated questions to the previous posts. This keeps things less confusing.

As to your specific question, it appears your barge (550 cm x 156 cm x 29.5 cm) is grossly undersized for carrying a 350-ton load which measures 10m x 3m.

A rectangular box having the dimensions of your barge would displace only about 2.5 metric tons of fresh water when fully submerged.

Do you have the correct barge dimensions? These dimensions seem to be more appropriate for a large skiff than for a working barge.

Barge size is 54.864 x 15.5448 x 2.95 m and Barge capacity 1500 ton
Sorry to misunderstand. I am learning. Not to disturb others.
So my query is " My load is 350 Ton size above shown."
What the draft after load the cargo on the barge?

 

[SteamKing]

Barge size is 54.864 x 15.5448 x 2.95 m and Barge capacity 1500 ton
Sorry to misunderstand. I am learning. Not to disturb others.
So my query is " My load is 350 Ton size above shown."
What the draft after load the cargo on the barge?

You're not disturbing us, Murshid. We're here to help you.

Here are some other questions I have about this barge, now that its overall dimensions have been established:

1. What's the draft of the barge when it is not carrying any cargo?

2. What's the shape of the barge hull? Is it a rectangular box? Does it have some other shape to one or both ends of the hull?

3. Is this barge floating in fresh water or sea water?

 

[Murshid]

1. barge Draft 470mm
2. Bow side after 5m has slope. I mean front side from the to till 5 m has slope.
3. Fresh Water
Please give me calculator too.

 

[Murshid]

1. barge Draft 470mm
2. Bow side after 5m has slope. I mean front side from the to till 5 m has slope.
3. Fresh Water
Please give me calculator too.

 

[SteamKing]

1. barge Draft 470mm
2. Bow side after 5m has slope. I mean front side from the to till 5 m has slope.
3. Fresh Water
Please give me calculator too.

The barge draft with cargo will be roughly 0.47 + 0.86 = 1.33 m

My calculations are:

Fresh Water

Barge

L = 54.864 m
B = 15.5448 m
D =  2.95 m

TLS = 470 mm

1500 ton capacity

5 m Bow Rake

Load:
350 T
L = 10 m
w =  3 m

====================================
Head log depth = 0.61 m

2.95 - 0.61 = 2.34 m

Rake slope = 2.34 / 5 = 0.468

Barge light weight

V = (49.864 + 50.868)*0.5*15.5448*0.470
V = 367.98 m^3

W = 367.98 M.T.

For 350 T Load:

V = (50.868 + dT / 0.468)*0.5*15.5448 * dT = 350

(50.868 + dT / 0.468) * dT = 45.03

50.868 dT + dt^2 / 0.468 - 45.03 = 0

2.1368 dT^2 + 50.868 dT - 45.03 = 0

dT = [-50.868 + SQRT (50.868^2 + 384.87)] / 4.2736

dT = 0.855 m

T = 0.47 + 0.86 = 1.33 m

================================
Check GMT:

IT -

LWL = 50.868 + 0.855 / 0.468 = 52.695 m

IT = 52.695^3 * 15.5448 / 12 = 189,544 m^4

BMT = 189,544 / (367.98 + 350) = 264 m

KB = 1.33 / 2 = 0.67

KMT = 264.67 m

KG = (367.98 * 0.6 * 2.95 + 350 *(2.95 + 3)) / (367.98 + 350)

KG = 3.81 m

GMT = KMT - KG = 260.86 m

Barge is stable.

I have not included the effects of trim in this calculation.

Obviously in an actual barge loading, you would want to have the loaded vessel with zero trim and zero heel.

Stability calculation assumes the internal compartments of the barge are dry.

 

[jjNavalArch]

Hello from Philippines. Newbie here!

Same with Murshid, I'd like to ask your help and suggestions. Below are the given barge particulars:

Length=12m; Breadth=8m; Depth=1.5m and 29 tons will be loaded at center

It is a rectangular shape barge and has 1.5m rake distance in both ends. I'd like to know its light draft and load designed draft.

Thanks a lot.

best regards.

 

[SteamKing]

Hello from Philippines. Newbie here!

Same with Murshid, I'd like to ask your help and suggestions. Below are the given barge particulars:

Length=12m; Breadth=8m; Depth=1.5m and 29 tons will be loaded at center

It is a rectangular shape barge and has 1.5m rake distance in both ends. I'd like to know its light draft and load designed draft.

Thanks a lot.

best regards.

Those two drafts cannot be known without having an actual barge to analyze. The light draft, I assume, means the draft of the barge when it is empty, not carrying any cargo or ballast.

The light weight of the barge depends on its construction and thus is impossible to estimate without more information.

The load designed draft is something the designer picks, consistent with the barge being operated in a safe manner. The load draft will vary depending on whether te barge is operated strictly on inland waters or will travel some coastwise or offshore.

 

[jjNavalArch]

Those two drafts cannot be known without having an actual barge to analyze. The light draft, I assume, means the draft of the barge when it is empty, not carrying any cargo or ballast.

The light weight of the barge depends on its construction and thus is impossible to estimate without more information.

The load designed draft is something the designer picks, consistent with the barge being operated in a safe manner. The load draft will vary depending on whether te barge is operated strictly on inland waters or will travel some coastwise or offshore.

Can we use the gross tonnage formula (GRT=.2+.02log10 * V) to get the estimated mass displacement ?
I mean the load draft not the load designed draft, just got confused sorry for that.

Thanks Steamking.

 

[SteamKing]

Can we use the gross tonnage formula (GRT=.2+.02log10 * V) to get the estimated mass displacement ?
I mean the load draft not the load designed draft, just got confused sorry for that.

Thanks Steamking.

No, the gross tonnage of a vessel is not a measure of its deadweight carrying capacity. The gross and net tonnages are based on the internal volume of the hull and superstructure, in cubic meters, less exempted or excluded spaces.

https://en.wikipedia.org/wiki/Tonnage

In older tonnage systems, 1 gross ton = 100 cubic feet of internal volume, and there were many different kinds of spaces excluded from the internal volume of the vessel. It was possible, especially with smaller vessels, to design them according to the national tonnage rules such that two vessels would have completely different gross tonnages, although the exterior dimensions of the vessels (length, breadth, and depth) were very similar.

Most barges have light ship weights which fall into a narrow range, given the arrangement of the hull. Hopper barges which ply the Mississippi River have light drafts of about 1.5 feet in fresh water. Barges which are not arranged as open hopper barges, like deck barges, for instance, will have similar, but maybe not the same, light drafts.

 

[jjNavalArch]

No, the gross tonnage of a vessel is not a measure of its deadweight carrying capacity. The gross and net tonnages are based on the internal volume of the hull and superstructure, in cubic meters, less exempted or excluded spaces.

https://en.wikipedia.org/wiki/Tonnage

In older tonnage systems, 1 gross ton = 100 cubic feet of internal volume, and there were many different kinds of spaces excluded from the internal volume of the vessel. It was possible, especially with smaller vessels, to design them according to the national tonnage rules such that two vessels would have completely different gross tonnages, although the exterior dimensions of the vessels (length, breadth, and depth) were very similar.

Most barges have light ship weights which fall into a narrow range, given the arrangement of the hull. Hopper barges which ply the Mississippi River have light drafts of about 1.5 feet in fresh water. Barges which are not arranged as open hopper barges, like deck barges, for instance, will have similar, but maybe not the same, light drafts.

This is much appreciated Steamking. Thanks to you.

Anyways, can i ask you any suggestions in how to calculate the deadweight of the barge with its dimensions mentioned above? or Are there any methods to do barge calculations.

Thank you so much again Sir SteamKing

Best Regards

 

[SteamKing]

This is much appreciated Steamking. Thanks to you.

Anyways, can i ask you any suggestions in how to calculate the deadweight of the barge with its dimensions mentioned above? or Are there any methods to do barge calculations.

At this stage, you can only estimate the deadweight of the barge by assuming a light draft and a loaded draft and taking the difference in displacement at those two drafts. Later on, if more information about the barge becomes available, you can refine this initial estimate of the DWT.

There are quite a few methods of doing barge calculations. That's what the study of naval architecture is about.

Thank you so much again Sir SteamKing

Best Regards

 

[jjNavalArch]

At this stage, you can only estimate the deadweight of the barge by assuming a light draft and a loaded draft and taking the difference in displacement at those two drafts. Later on, if more information about the barge becomes available, you can refine this initial estimate of the DWT.

There are quite a few methods of doing barge calculations. That's what the study of naval architecture is about.

thanks a lot SteamKing!

 

[ChemRocks13]

The barge draft with cargo will be roughly 0.47 + 0.86 = 1.33 m

My calculations are:

Fresh Water

Barge

L = 54.864 m
B = 15.5448 m
D =  2.95 m

TLS = 470 mm

1500 ton capacity

5 m Bow Rake

Load:
350 T
L = 10 m
w =  3 m

====================================
Head log depth = 0.61 m

2.95 - 0.61 = 2.34 m

Rake slope = 2.34 / 5 = 0.468

Barge light weight

V = (49.864 + 50.868)*0.5*15.5448*0.470
V = 367.98 m^3

W = 367.98 M.T.

For 350 T Load:

V = (50.868 + dT / 0.468)*0.5*15.5448 * dT = 350

(50.868 + dT / 0.468) * dT = 45.03

50.868 dT + dt^2 / 0.468 - 45.03 = 0

2.1368 dT^2 + 50.868 dT - 45.03 = 0

dT = [-50.868 + SQRT (50.868^2 + 384.87)] / 4.2736

dT = 0.855 m

T = 0.47 + 0.86 = 1.33 m

================================
Check GMT:

IT -

LWL = 50.868 + 0.855 / 0.468 = 52.695 m

IT = 52.695^3 * 15.5448 / 12 = 189,544 m^4

BMT = 189,544 / (367.98 + 350) = 264 m

KB = 1.33 / 2 = 0.67

KMT = 264.67 m

KG = (367.98 * 0.6 * 2.95 + 350 *(2.95 + 3)) / (367.98 + 350)

KG = 3.81 m

GMT = KMT - KG = 260.86 m

Barge is stable.

I have not included the effects of trim in this calculation.

Obviously in an actual barge loading, you would want to have the loaded vessel with zero trim and zero heel.

Stability calculation assumes the internal compartments of the barge are dry.

Hi, I am new to this feed and to these calculations. I have figured out some the equations you have inserted the numbers right away but have not figured them all out. But I have not figured out where the number 50.868 has come from. I can not figure out what I am missing. I am trying to use this example to gage my understanding and eventually apply it to my company's gage slips. We measure light and heavy barges (box, rake, double rake) to gage how much product was loaded into the barge. I have no experience doing this and was asked to look into it due to my educational background. Please help me try and figure this. I would greatly appreciate it.

 

[berkeman]

Hi, I am new to this feed and to these calculations. I have figured out some the equations you have inserted the numbers right away but have not figured them all out. But I have not figured out where the number 50.868 has come from. I can not figure out what I am missing. I am trying to use this example to gage my understanding and eventually apply it to my company's gage slips. We measure light and heavy barges (box, rake, double rake) to gage how much product was loaded into the barge. I have no experience doing this and was asked to look into it due to my educational background. Please help me try and figure this. I would greatly appreciate it.

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