Calculate Water Displacement of 295ft Barge

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SUMMARY

The discussion centers on calculating the water displacement of a barge measuring 295 feet in length, 52 feet in width, and a draft depth of 9 feet. The formula for displacement is established as volume (length x width x depth), resulting in a total displacement of 140,715 cubic feet. The conversation also explores the buoyancy principles, emphasizing that the weight of the water displaced equals the weight of the barge. Comparisons between aluminum, steel, and concrete barges highlight that aluminum may carry more weight due to its lower density, provided the dimensions are consistent.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Familiarity with basic geometry for volume calculations
  • Knowledge of material properties, specifically density of aluminum, steel, and concrete
  • Basic grasp of structural engineering concepts related to load distribution
NEXT STEPS
  • Research the properties of aluminum vs. steel in marine applications
  • Learn about the effects of hull shape on water displacement
  • Investigate the structural integrity of composite materials in barge construction
  • Explore advanced buoyancy calculations for irregularly shaped vessels
USEFUL FOR

Marine engineers, naval architects, and anyone involved in the design and construction of barges or similar vessels will benefit from this discussion.

djreidt
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what is the formula for calculating water displacement. I used to know it. it was something simple like length times width times .85 I think. Can anyone help me calculate the total displacement for a barge 295 feet long 52 feet wide with a designed dwaterline depth of 9 feet...
 
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Uh, would we be able to treat this barge like a big rectangular prism, 295' long and 52' wide, with 9' underwater? If so, the amount of water displaced is:

295' x 52' x 9' = 140715 cu. ft.
 
great now do I multiply the weight of the water to determine the weight of the loaded barge? if so what is it? I am trying to determine if an aluminum barge would carry more weight than a steel barge...or how about a "concrete " barge that weighs about 1.2 million pounds.
 
Well, if the boat has displaced so much water, and it is not sinking or being pushed up out of the water, then the density of the water, times the volume of water displaced gives you the mass of the water displaced. Multiply this by the gravitational constant and you get the weight of the water displaced. This is equal to the weight of the barge. In general, the buoyant force on an object is equal to the weight of the liquid displaced, and if an object is floating, the buoyant force and gravitational force (which is the exact same thing as the weight) are equal, but in opposite directions.

To solve your problem of whether an aluminum barge can carry more than a steel barge or a concrete barge, I'd have to know the basic shape of these barges, their dimensions, and their densities or weights/masses. Assuming that all the barges are the same shape and dimensions, then the barge made of the least dense material will fare best, that is, the barge with the least density, or equivalently, the least weight, or equivalently the least mass.
 
djreidt said:
I am trying to determine if an aluminum barge would carry more weight than a steel barge...or how about a "concrete " barge that weighs about 1.2 million pounds.
In the context of engineering, I believe it is appropriate to bring up some points here. Firstly, I would not recommend a concrete barge over a metal barge, since you would want to reinforce the concrete with metal anyway, to maximize the tolerable stress.

As far as aluminum vs. steel, it is an issue of volume vs. surface area. The load itself, and its distribution across the floor of the barge are big factors in determining the thickness of the metal. For that, you should consult some tables, probably (of which I know nothing about, but I'm sure they exist). Then, according to the outer volume of the barges (which should be the same for the aluminum and steel for comparison, I suppose), you find the mass of metal, which will give you the weight of metal. Add to that the weight of the load that you can fit inside the inner volume (which is less than the outer volume due to the thickness of the metal), and you will find the weight of the barge.

To find the depth (h) that the bottom of the barge will sink below the surface, you use:

V = lwh
V = m/ρ
W = mg

Just as a hunch, I don't think that the thickness is so critical, and I suspect that the lighter weight aluminum would be better. But, then again, I always see barges made out of (what I assume to be) steel. Perhaps it is cheeper to make a steel barge?
 
Structural aluminum generally has a higher strength than regular old ASTM A-36 carbon steel, and about the same as high strength steel. The trade-off is fatigue and sometimes product compatibility. Aluminum does not really have a fatigue stress limit that if you stay under that stress, it won't fatigue (break due to repetetive bending). Steel, on the other hand, does have that limit. Aluminum won't hold up at all to highly caustic product.

Concrete was wisely ruled out by a previous post.

The bottom line is that given the proper comparison between steel and aluminum, it COULD end up being the aluminum could carry more, since part of your buoyant force wouldn't also have to hold up as much "barge" structure as with the steel one and the remaining buoyant force could be used for additional product.

Most barges, however are of composite structure consisting of both steel and wood at the minimum.

So, given an aluminum barge and a steel barge whereas the material strength properties are similar and the dimensions are the same, you will likely be able to hold a bit more product with the aluminum barge at the same draft.
 
turin said:
In the context of engineering, I believe it is appropriate to bring up some points here. Firstly, I would not recommend a concrete barge over a ...

djreidt,

Pay no attention to turin and enginator. They are obviously basing their opinions on a bad experience at the National Concrete Canoe Competition. Not everyone's canoe wound up at the bottom of the Potomac. :biggrin:

http://www.asce.org/inside/nccc2004/rules.cfm
 
Last edited by a moderator:
BobG,
I have no idea what you're talking about, but concrete is brittle. It doesn't take nearly as much stress on a lateral member to make it snap. Metal is superior to concrete in that sense. I don't see what this has to do with canoes and potomacs; we're talking about barges.
 
turin said:
BobG,
I have no idea what you're talking about, but concrete is brittle. It doesn't take nearly as much stress on a lateral member to make it snap. Metal is superior to concrete in that sense. I don't see what this has to do with canoes and potomacs; we're talking about barges.

BobG has an emotional affinity towards concrete :smile:

Ok, even if we considered concrete, the members would be so big that your payload would need to be reduced to accommodate the structure. There you go, BobG... :wink:
 
  • #10
Loads of barges are made of concrete! And have you ever seen one which has sunk?
 
  • #11
turin said:
BobG,
I have no idea what you're talking about, but concrete is brittle. It doesn't take nearly as much stress on a lateral member to make it snap. Metal is superior to concrete in that sense. I don't see what this has to do with canoes and potomacs; we're talking about barges.

Sorry, I couldn't help myself. :redface:

I heard about this a couple years ago, because one of the local colleges competed in this. I just thought the idea of building concrete canoes was pretty funny.
 
  • #12
Hello...

I am wondering if you all could help me with a similar question. How do I determin the actual water draft of a barge. I have the dimensions (lenght/width/hull height). I know the gross tonnage. Is there a simple formula that will alow me to figure out just how far in the water she will draft?

Thanks.
 
  • #13
L = length, feet
W = width, feet
DH = draft height, feet
R = 62.4 lb/ft3 (density of water)
GW = gross weight, lb

L x W x DH x R = GW

Therefore:

DH = GW / (L x W x R)

This assumes a rectangular "squared off" bottom, with no slope to the hull bottom at all. If there is slope, then the equation is somewhat more complicated. The draft will be MORE than what this equation gives you. The basic idea is volume x water density = weight of water displaced, which equals gross weight of the object in the water.
 
Last edited:
  • #14
This isn't a reply, this is another question but I can't find the link to ask a question.
I am building a small pontoon boat. Two pontoons, 16" wide 12 ft. long. The platform is
8' x 12'. How deep should I make the pontoons in order to carry approximatley 500lbs.?
 

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