Concrete Barge Design need Help for computation for my Undergrad Thesis

Hello everyone, I am a 5th year Architecture student and would like to ask for help in designing the concrete basin/barge foundation for my floating home design that would have a maximum of 2 floors, and hopefully use the concrete basin as a basement if possible.

Based on my research from the Netherlands, the average weight of a house with fixtures and furniture, plus the family that would occupy the home, would be roughly around 60,000kg to 100,000kg and would be able to accomodate either a bungalow style house or 2 to 3 floor level house.

The dimensions for the shape of the basin would be 8x15 meters. I am in need of help in computing the needed depth for the basin, and the computation for it to float, and the freeboard to be around 1.5m from the waterline, considering this structure would be built and placed along an enclosed lake connected to a river via water gates.

Attached below is a mockup design of the layout and plans. Hopefully, if it is possible, I would like to expand the walls to the very edge of the 8x15m barge instead of building within it, if the calculations offer a possibility of making it happen.

 

Welcome to the Forum Morileck.

You mention that you have attached a mock up design of the barge, but there is nothing attached to your post.

If you have a floating concrete box that is 15 metres long and 8 metres wide, then if the total weight (displacement) of it is 60,000 kg (60 tonnes) it will float at a draft of 0.5 metres.
If you want to have a freeboard of 1.5 metres, then the depth of the barge will be 2 metres.
If the displacement is 100 tonnes, then the draft will be approx 0.83 metres.
However if you are then looking at putting a two (or three even) floor structure on this barge, you might have problems with stability.

 

Hello Bajansailor, thank you so much for replying! it means a lot to get some help here in the forum from experienced experts in the field
Sorry, it seems that the mockup didn't

Thank you so much for replying! It means a lot getting replies and answers from seasoned experts in the field and line of work.
Upload properly. Here is the mockup listed as a floor plan.

Would it be possible to at least have a 3-meter depth for the box, and change the freeboard to be around at least 0.6 meters or more? How would the computation for that look, taking into account the weight being 100 tonnes as the baseline for the design, and just having the 1 floor and basement be?

I would be honored if it would be okay to see how you computed everything, along with the steps on how you came up with the solution of the draft, freeboard, and depth of the barge. It would greatly help me in my Thesis sir Bajansailor.

I am also planning to use mooring poles and fixed rolling rollers so that the structure just moves vertically along with the change of water level.

 

It's called the Archimedes principle and you should have heard of it by now.

In a thesis I would expect a detailed calculation for the entire weight of the structure. The only estimations I would permit would be for the furnishings, and I would welcome samples of furniture weight.

First question: how much does a reinforced concrete box 15x8x3m weigh, considering local building code for a basement in a high groundwater area (hydrostatic pressure is present)? You can make free use of additional load bearing internal walls as needed.
Second question: how much does your house weigh, considering the chosen construction method? (I can see columns on ~4m center, but I have no ideea what they're made of (reinforced concrete presumably), what the infill is, what the ceiling and roof is made from, what the finishings are, etc. To the result you add the estimated weight of people, furniture, etc.

 

Thank you for the insightful questions. I will be getting back to you on the questions, and regarding the construction method, it would be either timber framing or steel framing, and the house would be from 70 tonnes to 100 tonnes, including the live load, but not including the float itself.

I've heard and familiarized myself with the Archimedes principle, but I still lack the necessary knowledge and experience to give a proper, feasible computation.

 

Positive buoyancy = weight of structure minus the weight of the displaced water (submerged volume x density of water)

 

Archimedes is very simple really - the weight (displacement) of an object floating in water is the same as the weight of the water that is displaced.
If you could freeze the water around your barge, and then remove the barge from the water, and you then filled up the resulting hole with more water, the weight of the water to fill the hole is the same as the weight of the barge.

So for your barge, assuming it is a simple rectangular shape, if it is 15 metres long, and 8 metres wide, floating at a draft of 0.5 metres, then the displacement is = 15 x 8 x 0.5 = 60 tonnes.
If it was floating at the same draft in salt water, rather than fresh water, the displacement would be greater - 61.5 tonnes - because the density of salt water (1.025) is greater than the density of fresh water (1.000).

 

Archimedes was a very cunning old man who displaced as much as he lost his weight.

 

Thank youu for explaining the computation, I can understand it from the point of view of a person with a bit of knowledege of the field.

 

Although the term "positive buoyancy" is redundant, it is paradoxical that the subtraction you refer to with those terms must always be negative, because otherwise the floating object would cease to be buoyant because it would sink. Should a floating object have a negative value for its "positive buoyancy"? Is this because of Archimedes' principle? Is there something I'm not understanding? Thank you.

 

Look, from a practical perspective it's really simple, in fresh water one cubic meter of hole floats 1t. For a house weighing 70-100t you need a hole in the water with a volume of 70-100m3. To make that hole you need something to keep the water out, and that something has its own weight, wich must be added to the weight of the house. The result is you need a bigger hole. Since the footprint of the base plate is fixed at 8×15m the only thing that changes is how deep the hole gets, that's the height of the walls that keep the water out (called draft). 0.834m is for the 100t house, so the question becomes how much the concrete barge weighs, because that weight must fit in the remaining 2.166m.
Example: if we asume a hollow concrete box 15×8×3m with a uniform thickness of 0.25m, we get a total concrete volume of 88.125 cubic meters. At 2.5t/m3 that weighs 220.31t + 100t for the house = 320.31t. The needed hole in the water will be 320.31m3. 15x8x3m has a volume of 360m3, so the barge will sink until 320.31m3 are submerged, that means to a depth of 2.67m, leaving 0.33m as freeboard.

In an academic paper all of this must be explained properly by applying the appropriate equations. You can't just say "I know that 1t of structure needs 1m3 of hole in the water" just like you can't say "the house will weigh 100t" and "the concrete barge will weigh 220t". The professor will want proof for all of that, you must demonstrate that you can indeed calculate the total volume and weight of the structure according to the local law. Otherwise you will be one of those arhitects who are forever in conflict with either the structural engineer, or the client.

P.S. Do you have a 100mm dividing wall under the kitchen island?

 

The weight can't just be claimed by saying there is a house in the Neatherlands that weights the same. The weight of every item has to be accounted for, and its center of gravity calculated so the total center of gravity can be calculated from the addition. There are no shortcuts. There is software that can do some of the calculations, but the data still has to be entered by hand.

 

Hi, good day. I just got back into the forum and analyzed what you said, and I agree that I can't just put on weight just because there is a reference to a different structure. I would like to ask how you got the depth of 2.67 and the total concrete volume of 88.125? I would love to learn the breakdown of the computation and formula for computing the depth/draft and the freeboard. is the 2.5t/m3 constant for the weight of the concrete?

 

By any chance, is this how you computed it? I tried ChatGPT, but I don't want to rely on AI too much regarding such important details and computations. So it is best to consult experts in the field such as the people here in the forum who have lots of experience.

 

Chat gpt calculated the volume of the walls by subtracting the air volume from the total voume. The other way is to calculate the volume of the individual components and add them up. The calculation is simple, the volume of a rectangular object is L x W x H. In this case there are four walls and two plates (bottom and top). You do the same if the object has a more complex shape, like interior walls, pillars, different thickness walls, etc. you break the structure down into simple shapes.

I have chosen a uniform thickness of 0.25m for convenience and because it's a common basement wall and plate thickness. You can optimize concrete use by making thinner walls with a frame, pillars, structural dividing walls, etc.

2.5t/m3 is the usual assumed weight for steel reinforced concrete. For a finished design you would use the actual number composed from the weight of the steel reinforcement and the specified concrete type.

Draft = volume/area, it's a derivation of arhimedes buoyancy formula.