DIY Stands Template and Calculator

[Newbie Aquarist]

Wait. Is this about my tank? So now I need 2"x10"s instead of 2"x12"s? I think he is using the 73" where I can roll my sump in and out.

 

[bereanracer]

The 82 comes from longest unsupported portion of the 2x12s (two 2x6 long sides at 5.5in each and two 2x6 short sides at 1.5 each for 14in total; then 96-14=82). C-Rad is correct that an unsupported span of 82in using 2x10s would give you just under .1in deflection. I originally ran all the numbers with a 96 span. 2x12s give you a comfortable safety factor, which appears to be built into RocketEngineer's original design. 2x10s is more aggressively desiging to the limit. The cost of using 2x12s vs 2x10s is only ~$1 per 8ft piece.

As far as using different sizes for the 48in sides, it can be done but complicates the build a little to make the top all even. I'd stick with using the same all around.

 

[C-Rad]

Wait. Is this about my tank? So now I need 2"x10"s instead of 2"x12"s? I think he is using the 73" where I can roll my sump in and out.

Yes, I'm talking about your tank, and just letting you know that you can, if you prefer, use 2x10's instead of 2x12's, and it will still easily support your tank. The cost difference isn't significant, but you might like having 2" more clearance inside your stand. In fact, you can use the same design shown in bereanracer's drawing, but use 2x10's instead of 2x12's, and use 2x4's for legs, instead fo 2x6's. I built a spreadsheet using the formula given by RocketEngineer, and verified that bereanracer's design/drawing, but using 2x10's instead of 2x12's, and 2x4's for legs, instead fo 2x6's, will have a deflection of <= 0.1, which is fine.

I think that limiting the deflection to only 0.1" or less, is already a safety margin, so I think it's overkill to add another safety margin on top of that by going with 2x12's. Isn't that correct?

Bereanracer is correct that using too many different sizes of lumber in the same build, complicates the design, and that extra complexity may not be worth the extra clearance and reduced weight from using the smallest lumber you can safely get away with. On the other hand, sometimes extra space inside the stand (or the ability to make a shorter stand) can make a big difference, so it might be worth some added complexity. Your call.

You can use 2x10's and 2x4's, instead of 2x12's and 2x6's, without complicating the design, or compromising safety, so I would recommend at least doing that.

If a little more space is worth some added complexity, here's some info that might be useful (based on actual calculations, not guesses like my first post):

If you add two legs in the front, with only a 73" span between them, as I suggested earlier, then you could use a 2x8 along the front, and save a full 4" of clearance over using a 2x12, and still only have a deflection of 0.0970". The center horizontal piece (93" long) will still need to be at least a 2x10, so you would only have that extra 2" of clearance in the front 23" of the stand, but maybe you want that. If you add a 2x4 leg in the middle of the rear beam, then you can use a 2x4 or larger for that beam. the center beam must still be at least 2x10 though, unless you treat it like the front beam and add two legs, leaving only a 73" span, in which case you can use a 2x8 for the center beam also. Since there are three legs along each 48" side, the spans will never be more than 18.75" there, so you can go as low as a 2x4 along the sides.

 

[rahjelli]

I'm about to purchase a marineland 265 gallon aquarium (new style) 84x24x31. Can you please give me the run down of what I would need for a worry free stand? It's replacing a 120 with steel stand. This is the 1st time I will be trying a diy stand, but I have a table saw and jig saw and all the tools I think I would need... also is it possible to have the stand at 34-36 inches high? My tank is in the basement on a poured concrete foundation so no worries on weight. I am 5'11 and want an almost no bend over view into the tank. I will also like to have a 55 gallon QT tank underneath and a 29 gallon (existing) sump.

 

[TheFishMan65]

If you went with the basic design shown in the first thread (corner supports only) the top would need to be 2x10s. 34-36 inches would not be a problem. I would make sure the concrete is level if not it may cause some issues.

Now if you made your widest opening 50 inches (just enough to get the 75 in) then you could get away with 2x6s.

Is this a glass framed tank?

 

[rahjelli]

It's framed on top and bottom, black plastic standard 265 reef ready from marineland, nothing custom, the QT underneath will be is 55 gallon with a 29 gallon long sump. You think I would need a center vertical beam, Also the legs would need to be 2x10 as well?Thanks alot for the help

 

[TheFishMan65]

Legs only need to be 2x4s. With 2x10 no center legs are needed. With 2x6 an opening can be at least 50 inches (if larger I would need to check).

 

[x19]

This thread has been immensely helpful, and thanks to RocketEngineer and everyone else that has had made it so useful.

I've ordered a custom 36x24x24 rimless glass tank and have designed a plain black stand for it go on. The drain is on the back panel instead of on the bottom. I want to included space for the plumbing to go into the stand which means the stand extends behind the tank by 6.5" and thus the corner supports are not directly under the corners of the tank. The stand is all 2x4s.

Is this design still adequate to properly support the weight?

Thanks in advance everyone.

 

[Spaced Cowboy]

Just a note in case anyone is interested - you can see the process I went through building one of these stands over in my build thread - around the third post is where the stand build starts

Thanks again to RocketEngineer. Many, many thanks

Simon

 

[Chipie]

Hi,
I want to build a stand and canopy for a 20g long(30X12X12) and i'm trying to decide what kind of material to use. Was thinking about plywood but i'm not sure how i would attach it. Pocket screw jigs nowhere to be found around here
Would a frame of 1"X2" covered with a 1/4" plywood work? I'm trying to make it so that i have enough room inside to put a 30"X12" tank without making the stand much larger than the DT. Any ideas? And anyone that has a plywood stand please post a picture of how it is braced if you can.

 

[C-Rad]

Originally Posted by TheFishMan65 View Post
Rocket, would you mind posting the formula [for caclulating the amount of deflection (bowing) of different sized horizontal frame pieces]

(5*W*L^3)/(384*Ixx*10^6)

Where W is the load, L is the length of the span, and Ixx is the area moment of inertia for the beam. The 10^6 is the minimum modulus of elasticity for the lumber typically used in these stands and comes from page 16 of Mechanical Properties of Wood Chapter 5 and page 13 of Mechanical Properties Of Wood Chapter 4. The first one has a short list of properties while the second one has many different species. The author of the first link is one of the authors in the second.

To calculate Ixx I use the formula for a rectagular member which is 1/12*b*h^3 where b is the beam width and h is the beam height.

RocketEngineer

People on this thread very often ask "what's the minimum safe size of lumber I can use for my tank?".

To help people answer this common question for most tanks, I've come up with a simplified version of the formula given by RocketEngineer. It's the same formula, and returns the same results, it just makes some standard assumptions (listed below) and is simplified and easier to use. 2x4's are more than sufficient for all but the most ridiculously huge tank on the most ridiculously tall stand. So for all reasonable tanks, the only question is what size the four beams (top horizontal pieces) should be. If the deflection calculated by this formula is less than or equal to 0.1" then the beams will be strong enough, but, for this simplified version of the formula, only when the following assumptions are true:

1) Tank is rectangular, and supported around the four sided perimeter only (cross-members, between beams, are optional)
2) The tank design is the one given at the start of this thread, two 2x4's are used for each leg, and the legs (purple) are arranged exactly as shown in the picture (i.e., using two 2x4's each, a leg is 5"x3", and I assume that the 5" side of each leg is on a long side of the stand, and the 3" side of each leg is on a short side of the stand.)
3) Beams are 1.5" wide (which is true for 2x2, 2x3, 2x4, 2x6, 2x8, 2x10, 2x12, and others)
4) Length and width of the stand are the same as the length and width of the tank
5) Total tank weight is estimated at 11 lbs per gallon of tank volume, using outside dimensions of tank. (sea water weighs 9.8 lbs/gal)​

Under those assumptions, we can decide what size lumber we need to use for the top frame pieces (beams (red and yellow)) by plugging the length, width, and height of out tank (X, Y, and Z respectively), and the beam "height" (h), into the following formula. As long as the formula calculates a deflection of less than or equal to 0.1", then the beam height is sufficient:

Deflection of the Long Beam(in inches) = (X * Y * Z * (X-10)^4) / ((X+Y) * 4.032 * 10^8 * h^3)
​

Where (all in inches):

X is Tank and Stand Length (long side)
Y is Tank and Stand Width (Short side)
Z is Tank Height (NOT STAND height (used to estimate tank weight))
h is The "height" of the beam​

"Height" of standard beams:

2x2 1.5"
2x3 2.5"
2x4 3.5"
2x6 5.5"
2x8 7.25"
2x10 9.25"
2x12 11.25"
2x14 13.25"​

For example:
A 90 gallon tank is 48" x 18" x 24". Assuming I use 2x4's for the legs, what is the minimum safe size lumber for each of the two long horizontal beams? Any beam for which the formula gives a deflection of 0.1 or less will be fine, so let's check three beam sizes (each is 1.5" thick):

(48 * 18 * 24 * (48-10)^4) / ((48+18) * 4.032 * 10^8 * 2.5^3) = 0.0945" of deflection using a 2x3 beam

(48 * 18 * 24 * (48-10)^4) / ((48+18) * 4.032 * 10^8 * 3.5^3) = 0.0345" of deflection using a 2x4 beam


That tells you that you can safely use 2x3's for the frame, because the deflection is less than 0.1". A 2x4 would deflect only about 40% as much as a 2x3, so while a 2x3 would be sufficient, for all but the weakest tank, a 2x4 will give you that much more support.

RocketEngineer,
Two Questions:
Correct me if I'm wrong, but I think there is already a safety margin built into this formula, so there's no need to be afraid that, for example, 0.0945 is "too close" to 0.1 to be safe. Tanks are designed to tolerate a stand that allows a lot more than 0.1" of deflection. How much? That varies depending on the tank. To play it safe, so that even the weakest tank won't break on our stands, we've decided to set the limit at 0.1" of deflection in this thread. 0.1 is just a nice round number that we feel certain is more than safe enough to use as a limit.

According to this formula, I could use 2x2's (1.5" x 1.5") beams on a standard 65 gallon tank (36x18x25 tall) (Deflection of 0.0916").
Is there any reason not to trust this formula for a beam smaller than a 2x3 or 2x4? Does the modulus of elasticity change for lumber in that range?

 

[Lyfsux]

I haven't read every post in this mega thread, so forgive me...

I just ordered a 300g deep dimensions tank (72x36x27), and I see there stands only use what looks like to be 1x4's, how can these stands hold up this amount of weight?

Has anyone done a 300g DD tank and have a build thread? I PM'ed RocketEngineer, but no reply, I can imagine he is getting tons of PM's about this, so I will wait till someone posts and can point me in the right direction, thanks...

 

[TheFishMan65]

C-Rad,

Do you have a reference for the weight of salt water. I think (i looked a couple places) it is closer to 8.5. RE used 10 lbs to account for sand/rock.

RE original design criteria was less than a defection of 1/8 of an inch. That IMHO that should be the standard for this thread. I believe it was stated in the first post.

While that equation may work. I sure have a hard time understanding it. I see nothing about 11lb for the weight of the water. I think I will stick with the original, but what ever works for you.

While 2x2 and 2x3 may work by the formula. Have you ever tried to find a straight one? IMHO 2x4 are easier to work with and easier to come by and you don't save much going with the smaller sizes.

 

[TheFishMan65]

1x4 would work assuming that the widest opening is 22 inches so three doors. Most of the bending strength comes from the 4 inch direction. Making it a 1x5 (if it existed) cuts the deflection down by half (25% more wood). Making it a 2x4 cuts by about a 1/3 (100% more wood).

I lot of folks have used plywood for stand (no 2x4s).

Hope this helps.

 

[RocketEngineer]

RocketEngineer,
Two Questions:
Correct me if I'm wrong, but I think there is already a safety margin built into this formula, so there's no need to be afraid that, for example, 0.0945 is "too close" to 0.1 to be safe.

There is definately a safety margin in this design. Most of that comes from using the values of green wood vs kiln dried which is stronger. The reason for this margin is to account for any defects in the lumber (anyone looking at box store lumber knows EXACTLY what I'm talking about). And for when I run the numbers, anything below .125 is my limit so working for .1 is even safer.

According to this formula, I could use 2x2's (1.5" x 1.5") beams on a standard 65 gallon tank (36x18x25 tall) (Deflection of 0.0916").
Is there any reason not to trust this formula for a beam smaller than a 2x3 or 2x4? Does the modulus of elasticity change for lumber in that range?

I will say that finding a clear 2X3 is almost impossible and most 2X2s I have found are for railings and are therefore pressure treated (not good). For this reason, the minimum size of 2X material that I recommend is a 2X4. For smaller tanks, a 1X4 which is .75 wide will be a better option as you can find clear 1X4s fairly easily.

While I have yet to back out the formula, its good to see someone simplify things for the average individual.

RocketEngineer

 

[jdpiii3]

Thanks guys for all of your help thought you might like to see the finished product.

Build thread- http://www.reefcentral.com/forums/showthread.php?t=2040499

 

[C-Rad]

C-Rad,
Do you have a reference for the weight of salt water. I think (i looked a couple places) it is closer to 8.5. RE used 10 lbs to account for sand/rock.

I should have looked it up (my memory stinks) so thanks for the correction. My estimate of 11 lbs per gallon instead of 10 makes my version of the formula a bit overly conservative (but still useful). At a SG of 1.026 (grams/liter) you can calculate that out tank water weighs about 8.5657 lbs/gallon. Figuring in the submerged weight of glass, sand, live rock (3 lbs/gal), and assuming a canopy, I think 9.6 lbs/gal is an accurate estimate of total tank weight, so RE's nice round estimate of 10 lbs/gal is fine. If we call it 9.856 lbs/gal, the constant in my equation becomes 4.5 (in stead of 4.032) so lets use 4.5 in my equation instead of 4.032 (I wish RC would let me edit/correct my earlier post!)

RE original design criteria was less than a defection of 1/8 of an inch. That IMHO that should be the standard for this thread. I believe it was stated in the first post.

Right again (nuts!). So the cutoff should be a deflection <= 0.125 (not 0.1 like my post said)

While that equation may work. I sure have a hard time understanding it. I see nothing about 11lb for the weight of the water. I think I will stick with the original, but what ever works for you.

I didn't want to clutter the thread with all the steps, but if you can use RE's version of the equation, just plug the same numbers into both versions and notice that "my" equation always gives the same results as RE's equation (using the assumptions I described). This is math - no faith is required - just verify.

There is definitely a safety margin in this design. Most of that comes from using the values of green wood vs kiln dried which is stronger. The reason for this margin is to account for any defects in the lumber (anyone looking at box store lumber knows EXACTLY what I'm talking about). And for when I run the numbers, anything below .125 is my limit so working for .1 is even safer.

I don't believe in increasing an already conservative safety margin, I just remembered wrong when I said the deflection should be <= 0.1, I should have said <= 0.125. Thanks for correcting me.

finding a clear 2X3 is almost impossible and most 2X2s I have found are for railings and are therefore pressure treated (not good). For this reason, the minimum size of 2X material that I recommend is a 2X4. For smaller tanks, a 1X4 which is .75 wide will be a better option as you can find clear 1X4s fairly easily.

I'm surprised to hear that; maybe it's a regional thing. In Southern California I often find nice kiln dried 2x3's at HD, and while the 2x2's are often warped, and contain knots, untreated ones are easy to find (here). If go to HD with a tape measure, and a list of exactly what pieces I'll need for my stand, I can sort through and find long 2x2's that contain knot-free sections that are straight and long enough for the pieces I need. Then I have the HD guy cut the good pieces I need out of the junk I bought, and I take home straight, knot-free, untreated 2x2's to work with. Personally, I hate moving unnecessarily heavy stands, and sometimes the extra inch (or two, or four) of clearance inside the stand is important, so for me smaller is better. That's why this equation is so valuable, because it lets me use smaller lumber, but still know for sure that my stand will be more than strong enough (assuming that I only use good straight lumber of course).

While I have yet to back out the formula, its good to see someone simplify things for the average individual.

I wanted to make what I think is a very useful formula, accessible to as many people as possible, but I suspect they'll be afraid to use it until you check it, and tell them it's kosher, so please do that when you can, and thanks a lot for posting the formula, and for putting time into this thread. I suspect there are fewer tanks falling off stands because of the time you've given here.

Note: I started with the equation you presented, using the ME you gave, and in addition to the assumptions I described in my post, I used 231 cu in/gal. Using a weight for water of 9.856 lbs/gal (instead of 11 lbs/gal as in my original post) the simplified formula is:

Deflection of the Long Beam(in inches) = (X * Y * Z * (X-10)^4) / ((X+Y) * 4.5 * 10^8 * h^3)
​

Where (all in inches):

X is Tank and Stand Length (long side)
Y is Tank and Stand Width (Short side)
Z is Tank Height (NOT STAND height (used to estimate tank weight))
h is The "height" of the beam​

"Height" of standard beams:

2x2 1.5"
2x3 2.5"
2x4 3.5"
2x6 5.5"
2x8 7.25"
2x10 9.25"
2x12 11.25"
2x14 13.25"​

 

[Chipie]

Hi,
I want to build a stand and canopy for a 20g long(30X12X12) and i'm trying to decide what kind of material to use. Was thinking about plywood but i'm not sure how i would attach it. Pocket screw jigs nowhere to be found around here
Would a frame of 1"X2" covered with a 1/4" plywood work? I'm trying to make it so that i have enough room inside to put a 30"X12" tank without making the stand much larger than the DT. Any ideas? And anyone that has a plywood stand please post a picture of how it is braced if you can.

​

Nobody wants to help?

 

[TheFishMan65]

A 1x2 should take the weight, but I would be leary of it being too flimsy even with the ply wood.

C-Rad
Now I know what is wrong with your formula You can't handle 1x lumber. I guess it is the engineer in me I like to see where each number goes and how it affect the equation. To me the equation/step/siplicity doesn't matter since I have it in a spreadsheet.

 

[Chipie]

Thanks fishman. Maybe 2"X3" would be best then. But that means unfortunatly that i'm going to have to build the stand wider and longer than the display tank since the sump is the same dimensions as the DT. I guess there's no way around it.