Physics question (structural)

her209

No Lifer
Oct 11, 2000
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I built a fish tank stand out of 2x4 wood that has six vertical legs, 4 in the back, 2 in the front. A 150 gallon fish tank sits on top though not directly any of the legs. That works out to about 1200 lbs. What's the weight distribution on each of the legs? Is it equal for all 6 or do the front 2 legs bear twice as much weight than the ones in the back?

Crude diagram using MS Paint

tankstand.jpg
 

SKORPI0

Lifer
Jan 18, 2000
18,405
2,309
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2 front legs - 25% each.
4 back legs - 12.5% each.

Assuming vertical load is equally distributed the whole area (weight of fish tank and water).
 

gaidensensei

Banned
May 31, 2003
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Yeah, I don't think it's in force equilibrium.

This is an interesting question as I've never had a theoretical question like this before, but I don't think it is in equilibrium (as in all tables carry the same acting force).

There are two main forces, don't forget not only do you have to calculate the mass of your fish tank, you also have to consider the mass of the table.
1gal of DI water is 8.32 lbs, 150 gallons translates to ~600kg of just DI water alone.

This is all assuming everything is equally distributed as mentioned, no angles or anything like that involved.
 
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CycloWizard

Lifer
Sep 10, 2001
12,348
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The design is statically indeterminate: there's no way to know how the loads will be distributed across all of the legs using a static, rigid body analysis. Are you just trying to determine a ballpark size for the legs based on the expected load? If so, you can be fairly conservative by assuming that each leg will carry 25% of the load (i.e. allow a factor of safety of 1.5).
 

Juked07

Golden Member
Jul 22, 2008
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The design is statically indeterminate: there's no way to know how the loads will be distributed across all of the legs using a static, rigid body analysis. Are you just trying to determine a ballpark size for the legs based on the expected load? If so, you can be fairly conservative by assuming that each leg will carry 25% of the load (i.e. allow a factor of safety of 1.5).

Couldn't we do better by adding reasonable assumptions to make the problem statically determinate? Assume symmetry and some sort of flex parameters for the load from above, and we should be able to get a unique solution..

And to address the OP, the legs on the 2 leg side will almost certainly bear more force (per leg) than the legs on the other side. In equilibrium (tank not falling over) we can't have any torque on the system, so the upward forces exerted by each face should be about equal, and there are more legs to spread the load over in the 4 leg case.
 

bonkers325

Lifer
Mar 9, 2000
13,077
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i built a fish tank stand out of 2x4 wood that has six vertical legs, 4 in the back, 2 in the front. A 150 gallon fish tank sits on top though not directly any of the legs. That works out to about 1200 lbs. What's the weight distribution on each of the legs? Is it equal for all 6 or do the front 2 legs bear twice as much weight than the ones in the back?

Crude diagram using ms paint

tankstand.jpg

from an equilibrium standpoint, you need 3 legs for it to be structurally stable. so, it'd be 33% per leg maximum.

if all 6 legs are active:

1__2___3__4

5_________6

1&4 = about 9% each
2&3 = about 16% each
5&6 = 25% each

the water weighs about 1300#, the tank itself has weight, and the wood frame itself also has weight. you're probably looking at 1450# altogether. so, just say 500# per leg in a worst-case scenario. less than 100psi on each leg, any 2x4 hardwood stud from home depot will work.
 

gaidensensei

Banned
May 31, 2003
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I don't think it's statistically indeterminate. I'm pretty certain, just did resource checking and this can theoretically be solved.

The load distribution is from stats, and the forces can be derived with newton's 2nd law.

I'm thinking it is like as Skorpio said, 25/25/12.5*4. Some quick calcs assuming his stand weights 25kg, fish tank 600kg.
F= ma = 0
F = 5886N + 245.3N = 6131.3N

front 2 legs @25%: 1533N downward force
back 4 legs @12.5%: 766N downward force
 

CLite

Golden Member
Dec 6, 2005
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The front 2 legs will carry 25% each, if they didn't the tank would flip over assuming the cg was in the middle.

The back 4 legs can't be determined, especially since you didn't provide geometry. Since I run FEA's as part of my job I spent 5 minutes making a stiff shell model to represent the tank bottom and fixed 6 of the nodes. I assumed a 24" deep by 60" long tank with a total load of 1200lb's

Corner legs , Middle legs
-38.784lb , 337.584lb

The vertical glass walls would provide additional rigidity so I doubt the back corner would actually have a lifting force. However the back two middle legs should be designed to carry more than 28% of the weight. If that is the reason for your question.
 
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PieIsAwesome

Diamond Member
Feb 11, 2007
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The 4 front legs will bear less weight. Add 2 more legs to the back or over-design.

Why do you really need to know? Its statically indeterminate to the 3rd degree, under a distributed load, and forces are in more than one plane. It can be solved, but it would be a pain and I don't think its necessary.
 

Greenman

Lifer
Oct 15, 1999
20,355
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Dump the two middle rear legs, add a cross brace or panel on the back for lateral support, do the same on the sides. Use bolts to assemble. This way load balance doesn't matter, so long as CG doesn't shift past any leg (earthquake).

2x4 material will be fine.
 

Bignate603

Lifer
Sep 5, 2000
13,897
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The loading on legs will be more dependent on which ones are slightly higher than the others. If one leg is slightly longer it will take more load.

To really make this safe you should do a rough estimate on the amount of weight on each leg (25% per leg is probably ok as an estimate) then making the legs able to take noticeably more than that, at least 1.5 times your estimate, maybe more. You'll also need to look at cross bracing and things to make sure it doesn't collapse.
 

Imp

Lifer
Feb 8, 2000
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Really? I'm the only civil engineer here?

It's technically indeterminate on one side, but like any engineering problem, there are so many assumptions/tips/tricks to solve it and get it close enough.

Please go talk to a carpenter. You're playing with 1200lb. Depending on how long your legs are, how well you reinforced everything, x6 2x4 pieces of wood may be pushing it.
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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www.slatebrookfarm.com
What matters more is where are the legs hitting the floor? Are some between joists and others directly above joists? You want that load distributed over joists; not just directly to the floor boards.
 

her209

No Lifer
Oct 11, 2000
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Had to reinstall computer... here are some 3D models of the stand

Isometric View

tankstand-3d-iso.jpg


Trimetric View

tankstand-3d-tri.jpg


Dimetric View

tankstand-3d-di.jpg


What I was planning on doing was reinforcing the front with two additional legs (see below) so in case one snaps, the second leg can still take up the load until I drain the tank and fix the leg.

tankstand2-3d-iso.jpg
 

MotF Bane

No Lifer
Dec 22, 2006
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Why are you so determined to have a gap in the front? Are you building a cabinet, storage for supplies, something like that underneath?

If this were my project, it would be done with 4x4's instead for the legs. You're also looking at a very top heavy structure if something hits it.
 

BoomerD

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Feb 26, 2006
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Even though the water itself works out to just a bit less than 8.5 lbs/gallon, by the time you add in the weight of the tank and any contents, (rocks, decorations, fish, etc.) most aquarists use 10 lbs/gallon to guesstimate the weight on the stand. (150 gallon glass tank can easily weigh 150 lbs by itself)

One thing the OP doesn't seem to be calculating is the bow of the top of the stand. A 2X4 placed so in the "tall config" MIGHT carry the weight without a problem...but it depends on the actual dimensions of the tank. (more than one type of 150 gallon tank)
Personally, I'd be leery of not having a center brace in the front. That's a lot of weight...and a lot of water spanning across the front of the stand.
Everyone who's been in the hobby very long has been indoctrinated into the "wet socks club," but I'd hate to "get my feet wet" with 150 gallons on the floor because the front of the tank sagged and caused the glass to crack...or a seam to fail.
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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from an equilibrium standpoint, you need 3 legs for it to be structurally stable. so, it'd be 33% per leg maximum.
You could easily have 90% of the weight on one leg, with two other legs barely supporting any weight, but keeping it balanced. (Stand up and lean to your left and to your right; your legs wouldn't be each supporting the same amount of weight.)
 

her209

No Lifer
Oct 11, 2000
56,352
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The tank sits like so on the stand.

tankstandwtank-3d-iso.jpg


Tank dimensions are 72" wide x 24" tall x 20" front to back
 

slayer202

Lifer
Nov 27, 2005
13,682
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You could easily have 90% of the weight on one leg, with two other legs barely supporting any weight, but keeping it balanced. (Stand up and lean to your left and to your right; your legs wouldn't be each supporting the same amount of weight.)

yeah, but you would be easier to tip over, no?


idk anything about this shit, but it seems like you guys are overcomplicating it
 

her209

No Lifer
Oct 11, 2000
56,352
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Why are you so determined to have a gap in the front? Are you building a cabinet, storage for supplies, something like that underneath?
I have another tank (135 G) that sits underneath which acts as my sump / secondary tank.
 

her209

No Lifer
Oct 11, 2000
56,352
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Even though the water itself works out to just a bit less than 8.5 lbs/gallon, by the time you add in the weight of the tank and any contents, (rocks, decorations, fish, etc.) most aquarists use 10 lbs/gallon to guesstimate the weight on the stand. (150 gallon glass tank can easily weigh 150 lbs by itself)
Tank is bare. No decorations, no rocks, no substrate. Just fish.

One thing the OP doesn't seem to be calculating is the bow of the top of the stand. A 2X4 placed so in the "tall config" MIGHT carry the weight without a problem...but it depends on the actual dimensions of the tank. (more than one type of 150 gallon tank)
The top has 2x4 in both "configs". There's 3 2x4 in the "flat" config and the box in the "tall" config. Don't know if that comes across on the model.

Personally, I'd be leery of not having a center brace in the front. That's a lot of weight...and a lot of water spanning across the front of the stand.
Everyone who's been in the hobby very long has been indoctrinated into the "wet socks club," but I'd hate to "get my feet wet" with 150 gallons on the floor because the front of the tank sagged and caused the glass to crack...or a seam to fail.
The tank is acrylic. While there is a big gap at the front, the 2x4 that runs across the front should take the weight and distribute it to the 2 vertical legs right? I was concerned with the ability of the two front legs to hold the weight over a long period.