DIY Stands Template and Calculator

sizzle,
I posted back in post 1891 I forgot to add that those are 8th inches. The answer was no.

BUT

C RAD made some different assumptions and say you are right on the edge. So if you understand how he ran the formulas and can meet those requirements then I guess go for it. Otherwise I would look into some other options. Maybe a 1x10 or a center support.
 
I wanted to build a all plywood stand. I was planning on using 3/4" birch plywood and using pocket holes and glue method of installing i attached a pic of some what how i wanted it also there would be a 3/4" piece of top to hold the tank which is rimless with the dimensions of 36x20x18 can anyone tell me if this would hold? Should I add some support on the corners or add something in the middle to help distribute the weight or is it good as is? Also to note if i did add something to support the corners the back left would be left out since the tank is drilled very close the the corner. Sorry im a noob to all this stand building and this will be my first time building one. Would appreciate all the advice and opinions.

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I wanted to build a all plywood stand. I was planning on using 3/4" birch plywood and using pocket holes and glue method of installing i attached a pic of some what how i wanted it also there would be a 3/4" piece of top to hold the tank which is rimless with the dimensions of 36x20x18 can anyone tell me if this would hold? Should I add some support on the corners or add something in the middle to help distribute the weight or is it good as is? Also to note if i did add something to support the corners the back left would be left out since the tank is drilled very close the the corner. Sorry im a noob to all this stand building and this will be my first time building one. Would appreciate all the advice and opinions.

Since you have a raw plywood edge everywhere that you will have to cover with decorative wood trim, you might as well glue, clamp, and screw from the outside. The pocket screws will mean working from inside. Pre assemble by tacking with nails and lining everything up and then drilling for the screws whether they are pocket or conventional. I don't see the top, but it is probably stacked on top as the bottom is. Seal the wood everywhere. You can use paint inside and on the bottom even if you want the outside stained to show the wood. Even if it takes more plywood, I'd use one piece on the front instead of the three shown.
 
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dpx, I see nothing wrong with your design as long as the tank frame rests on the plywood. Make sure the front top brace is at least 3.5 inches tall.
 
sizzle, I had a mistake in my calculation I had a 1x8 as being 0.5 x 7.5 it is really 0.75 x 7.5. The extra 1/4 makes a difference. You should be good with 1x8s if you place the legs as C-Rad and RE designate.
 
New 80G setup

New 80G setup

Hi all,

Im a green horn to marine aquariums but recently go an 80Gallon tank real cheap so couldnt resist :P

Tanks measurements are 67" x 12 1/2" x 27 1/2" thats L,W,H would appreciate any help with design of a stand or equipment you all may suggest.
 
First, let me say that I'm very grateful to RocketEngineer (and TheFishGuy, and others) for all the time and effort they've put into this thread, and I don't mean to critiize, but I found a significant error in the he's been using the deflection formula to determine what size beams are safe to use, and I thought I should say something.

I'll try to make this my last post on the topic of this deflection equation, except for posting a final corrected version of the Generic Stand Version of the Formula, to keep the thread focus from being too narrow.

Beam bending for a distributed load:

(5*P*L^4)/(384*Ixx*10^6) where P is weight per inch (P=W/L)

If P=W/L the foruma simplifies into the form I gave which is

(5*W*L^3)/(384*Ixx*10^6) where W is the weight of the tank.

Rocket,
The formula from you book is correct, but if W is the weight of the tank, you break the formula when you substiture W/L for P.

P and L already have precise meanings in the book formula, but W is a variable that you introduce, and if you define W as the tank weight, then it is not true that P=W/L. When you make that substitution, you are replacing P with a value that is at least two to four times greater than P, and so the value you get for deflection will be wrong (too large by the same factor).

The book formula is concerned only with forces acting along the length of a single beam, not a group of beams. The weight of the tank is always spread out over a group of two or more beams, so the formula doesn't care what the total weight of the tank is, it only cares about the amount of weight pushing down on the single beam with length = L.

The good news is that if you define W correctly, then it will be true that P=W/L, and the you will get correct deflection values from:
(5*W*L^3)/(384*Ixx*10^6)​

I found the same formula, and a very useful drawing to go with it (see attachment). If you study the picture it becomes clear that W is the weight pushing down along the length L of a single beam. Since we don't balance our tanks on a single beam when we build tank stands, "the weight of the tank" will always be divided between two or more beams, so when using the formula on any single beam in a tank stand, W will always be less than the weight of the tank (usually about 1/2 to 1/4 of the tank weight)

Also notice, in the drawing, that L is the distance between the two support points. The beam actually extends a little beyond each of the support points. The beam is longer than L, so it isn't correct to use the length of the beam for L in the formula, we should use the distance between the support points. For a tank stand with 2x4's for legs, the support points are at the inside corner edges of the 2x4's. So when using the formula, we should not use the actual length of the beam for L, we should use the distance between the legs. It might not seem like a big deal, but L gets cubed, and so the error is magnified if L is off.

If you want to use this formula to decide what size your beams should be there are two ways to go:

1) If your stand design is generic enough and meets certain assumptions, you can just plug the beam height, and the tank's length, width, and height into the "Generic Stand" version of the equation that I'll post soon, and it will deduce the value of W and L, and calculate the beam deflection.

2) Or, you'll need to figure out what W should be for the beam you are interested in. I went into that at length in a previous post, but the short version is that you need to figure out how many pounds you'll have per linear inch of beam, and multiply that by L (the number of inches between the legs along the beam we are interested in) If P is the number of pounds per inch along your beam, then W=P*L (look familiar?).
Great, so how do you get P so that you can get W? In most cases we design our stands so that the beams are arranged symetrically around the center point of the tank bottom, in which case the tank weight will be evenly distributed along every inch of every beam, and P will be the same for every beam. Any arrangement of beams that is "balanced" around the center point of the tank bottom will work, for example:
a) One beam under each of the two long sides.
b) A beam under each of the four sides.
c) Either of the above, plus one or more beams equally spaced between two parallel beams.​

Note: Cross members are not beams. For a beam to be a beam, it must be supported at each end by a leg (or part of a leg). If it is supported at each end by a beam instead of a leg, then it is a cross member, not a beam.

So for a stand with symetrically arranged beams, P (pounds per inch) will be the same for each beam, so it is legal to use values that span all beams to find P:
P = (tank weight) / (Sum of the lengths of all beams)​

Notice that for any single beam, W=P*L, so W will be greater for a longer beams than for shorter one.

I think that TheFishGuy and RocketEngineer have been using the tank weight for W (maybe 1/2 the tank weight), and the full length of the beam for L, instead of the distance between the legs, i.e., they've been using values for W and L that are larger than what the formula expects. Since L gets cubed in the formula, and since W and L are in the numerator of the formula, I think the deflection values they've been using to decide what size lumber to use have been too high, maybe two to four (or more) times too high.

I don't consider the [beams along the] short sides [of the tank stand] because in my origianl post I mention that the outside 2X4 legs are option which means the ends of the top frame don't always have full support. It depends on how the legs are arranged whether the ends have support or not.
Okay, so if the "beams" along the short side don't have a leg at each end, they aren't really beams, they're cross members, and you are right to only count the two long beams. In such a case, when the stand only has two beams, you should set W = (tank weight)/2, but are you setting W that way for all beams, even if the stand has more than two beams?

A few things to keep in mind:
1) The 1/8" [0.125" - maximum safe distance of] deflection is a feel good number. I don't know what the professionals use but its what I feel should be about right.

Sure, understood, and since nobody has ever posted on this thread that their tank broke because you told them to use a beam that was too small, it's been safe to use. But what if, as I suspect, you've been plugging the wrong values into the formula for W and L, and getting deflection values that are double, tripple, or more what is accurate? If you fix your spreadsheet, and start getting accurate deflection numbers that are a lot lower than you've been using, is it still safe to use beams that the formula (when used correctly) says have a max deflection of .125"? The way I think you've been using the formula, a beam with a correct deflection estimate of 0.1" would have been rated by you as having a deflection of maybe 0.25", and would have been considered too weak. Should we now consider such a beam strong enough because it rates less than .125, or should we lower the cut off point to something like 0.05"? It's not fair of me to expect you to be able to answer that, but we need to pick a number and go with it. Any thoughts on what it should be? Maybe we can take some measurements from commercially sold stands, calculate the deflection they have, and use that as the limit?
 

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C-RAD long post, but I think you have 3 issues with RE's formula. This what I have gathered through reading this thread.

Length of Bending Beam
As you noted RE threw out the end boards, because they may not be supported. So that only leaves the error caused by the leg support (length should be 9-10 inches less). I point to this statement BY RE.
NOW for L there are two ways of considering it. The way I normally do L is to use the length of the tank and call it a day. This is more conservative than necessary but it gives me a value to start from. If the bending is marginal, then I reduce L to represent the actual supported span on the long side.
This is also what I try do so we usually cover that issue. Remember only the three of us use the formula (right now) and few have asked for it. So if we cover the special cases all is good.

Second Issue Definition of W
I always assumed W was half the tank weight. Since RE checked my math (once) and got the same number he must also. IIRC one time he forgot to divide by 2 which is more evidence of his intended process when using this formula.

Distribution of W
Ok I guess you get this one. The weight actually over a leg should be tossed out. So we could almost 25% off ((10) / 48 = 21%). Or for a 6 foot 10/72 = 14%.

NOTE: The only way sizzle got the 1x8 (with my and I think RE's formula) was by reducing the beam length to 65 inches, and you will see I commented that the legs had to be as specified. :) The fact that I counted too much weight makes me happy since it was very close to 1/8th.
 
The tank is a 180, and i planned on using rockets original design to do the build, the layout in the first post. There will be 120lbs of sand and 180lbs of rock in the tank. If you need more info let me know.
A 180 gallons of salt waters weighs 1540 lbs (180 * 8.556). Sizzle then adds 120 lbs of sand and 180 lbs of rock for a grand total of (don't you love word problems) 1840 lbs. Now I realize the sand and rock will displace some water, but I also neglected to add the weight of the tank. So if sizzle gave real numbers (and I have no reason to think they didn't), then perhaps our estimate of 10 lbs per gallon is on the light side. Thoughts?

NOTE: I think a 180 weighs about 30 gallons of salt water.
Full 180 found online is 2100 lbs (found online) - 1840 = 260 / 8.556 = 30
Another site say 338 empty / 8.556 about 40 gallons.
So do we displace a fifth to a sixth of our water with rock and sand?
 
Since you have a raw plywood edge everywhere that you will have to cover with decorative wood trim, you might as well glue, clamp, and screw from the outside. The pocket screws will mean working from inside. Pre assemble by tacking with nails and lining everything up and then drilling for the screws whether they are pocket or conventional. I don't see the top, but it is probably stacked on top as the bottom is. Seal the wood everywhere. You can use paint inside and on the bottom even if you want the outside stained to show the wood. Even if it takes more plywood, I'd use one piece on the front instead of the three shown.

Thanks for the advice and yes i didnt draw the top which will also be a 3/4" piece.

dpx, I see nothing wrong with your design as long as the tank frame rests on the plywood. Make sure the front top brace is at least 3.5 inches tall.

Are you talkin about the edge of the tank because the tank i have is rimless with no bottom frame just a single piece of 1/2" glass will this still be okay or would i need to put some supports going along the middle?
 
A 180 gallons of salt waters weighs 1540 lbs (180 * 8.556). Sizzle then adds 120 lbs of sand and 180 lbs of rock for a grand total of (don't you love word problems) 1840 lbs. Now I realize the sand and rock will displace some water, but I also neglected to add the weight of the tank. So if sizzle gave real numbers (and I have no reason to think they didn't), then perhaps our estimate of 10 lbs per gallon is on the light side. Thoughts?

NOTE: I think a 180 weighs about 30 gallons of salt water.
Full 180 found online is 2100 lbs (found online) - 1840 = 260 / 8.556 = 30
Another site say 338 empty / 8.556 about 40 gallons.
So do we displace a fifth to a sixth of our water with rock and sand?

Lets assume a 180g tank is made from 1/2" thick glass. That means that the actual space for the water is 71" long and 23" wide. Also consider that the water isn't fully up to the rim so the height is reduced by 1/2" for the bottom glass and another 2" below the top rim.

71*23*21.5/231 = 152.0 Gallons. So yah, a tank that size weighing about the same as 30g of water is about right.

RocketEngineer
 
A 180 gallons of salt waters weighs 1540 lbs (180 * 8.556). Sizzle then adds 120 lbs of sand and 180 lbs of rock for a grand total of (don't you love word problems) 1840 lbs. Now I realize the sand and rock will displace some water, but I also neglected to add the weight of the tank. So if sizzle gave real numbers (and I have no reason to think they didn't), then perhaps our estimate of 10 lbs per gallon is on the light side. Thoughts?

NOTE: I think a 180 weighs about 30 gallons of salt water.
Full 180 found online is 2100 lbs (found online) - 1840 = 260 / 8.556 = 30
Another site say 338 empty / 8.556 about 40 gallons.
So do we displace a fifth to a sixth of our water with rock and sand?

I agree that 10 lbs/gal, while in the realistic range, is on the low side, and 12 lbs per gallon is just as realistic an estimate on the high side. Here is my reasoning:

Tank water is kept at a Specific Gravity of 1.026 g/cm^3, which converts to a weight of 8.5624 lbs/gal.

Glass, live rock, and sand are all made of very similar stuff, and all have a density of roughly 2.4 - 2.8 g/cm^3. To play it safe lets assume that they are all 2.8 g/cm^3, which is about 23.37 pounds per gallon. So a pound of glass/rock/sand displaces about .0428 gallons of water, which is 0.3665 pounds of water. That means that a pound of glass/rock/sand only adds about .6335 pounds of weight to the tank, above what the same volume of water would add.

That means that we can use the outside dimension of a tank, calculate what that volume of water would weigh, and then for every pound of glass/rock/sand we have, add .6335 pounds. We would end up with a very good estimate of total tank weight.

So how many pounds of glass/rock/sand should we expect a tank have per gallon? Using a 180 gallon tank (72x24x24) as an example, with 1/2" glass, 240 lbs of live rock, and a 6" deep sand bed (it could happen):

313 lbs of glass (13.4 gallons @ 23.37 lbs/gal)
425 lbs of sand (5.67 cu ft of @ 75 lbs/cu ft)
240 lbs of live rock

978 lbs total (weight in air)

978 * .6335 = 620 lbs added to tank weight (above what just water would weigh)

So @ 8.5624 lbs/gallon, 72x24x24 = 179.5 gallons would weigh 1537 lbs. Add to that 620 lbs for glass/rock/sand, and we get 2157 lbs. There will be a minimum of at least .5" of air at the top, so subtract out 0.5*71*23/231*8.5624 = about 30 lbs. Add 30 pounds back for a heavy wooden canopy and the grand total for a 180 gallon tank could realistically be as high as 2157 lbs.

2157 lbs / 179.5 gallons = 12.01 lbs per gallon.

On the low side, if we had .75" of sand, 100 lbs of live rock, and no canopy, we have a total around 1800 lbs, or 10.02 lbs per gallon.

So I'd say that the realistic range for the weight/gal of a tank is 10-12 lbs/gal. I guess that means go with 12 lbs/gal most of the time, to be on the safe side.
 
forgot to ask is there a rule or anything about how many pocket holes i should drill per inch or feet apart to join my plywood stand together?
 
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I have a 100 gallon tank that is 60" by 18" deep. Is it ok to do the design mentioned here but make it out of 1x4 instead of 2x4?
 
Are you planning to have a center leg?

Neither a 1x4 or 2x4 is going to have deflection small enough for a 100 gallon tank. You need a 2x6 for the original plan.
 
I have a 100 gallon tank that is 60" by 18" deep. Is it ok to do the design mentioned here but make it out of 1x4 instead of 2x4?
You can safely use 2x4's for the legs no matter what.
Even if you estimate the total tank weight on the high end, at 12 lbs per gallon, you can safely use 2x4's for the top four horizontal pieces IF AND ONLY IF all of the following are true:
1) The stand outside dimensions are no more than 1/4" more than the tank outside dimensions.
2) Each leg is made of two 2x4's connected at 90 degrees, forming a leg that is 5" along one side, and 3.5" along the other.
3) Each of the four top pieces has at least part of a leg under each end (This is done if #2 above is done)
4) The 5" side of each leg is along the 60" side of the stand, and the 3.5" side of each leg is along the 18" side of the stand. (The important thing here is that the unsupported span between the legs is 50.25" or less, NOT 53")​

You can't use 1x4's, but a 1x6 is even stronger than a 2x4, so 1x6's would be fine. If you use 1x6's, and 1x4's for legs, be sure to use the tacking strips along the inside of each leg as shown in the picture in the first post of this thread.

Note: The reason I'm giving you a different answer than TheFishman65 is that he's playing it extra safe by not making the assumptions I make above. Since you may not really follow the "original plan", he's assuming that only the two 60" beams have legs under them, not the 15" long beams, and he's calling the unsupported span between the front/rear legs 60", where I require that it be no more than 50.25". We're both using the same formula to calculate the amount of deflection under the load, but we're making different assumptions about how you will put the pieces together.
 
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