DIY Stands Template and Calculator

[TheFishMan65]

Tanks the same width or length always seem to be a pain to stack.

Here is the best design I have seen (go ahead guess whose design )

 

[Sizzle]

I was wondering if 1x8 equally as strong as 2x6?

The stand dimensions will 72lx24wx36h, local wood suupliers in the area rarely have good straight 2x anything. However I can find dead straight 1xs with no issues.

Thanks for the help

 

[TheFishMan65]

For a 300 gallon tank on a 72 inch stand the 2x6 is stronger. But it may depend on the setup. If you tell me the tank size I can figure it out.

PS - neither one works in this scenario

 

[C-Rad]

I'm afraid I disagree with FishMan on this (maybe one of us needs to fix our spreadsheet)

I was wondering if 1x8 equally as strong as 2x6
The stand dimensions will 72lx24wx36h

For a 300 gallon tank on a 72 inch stand the 2x6 is stronger. But it may depend on the setup. If you tell me the tank size I can figure it out.
PS - neither one works in this scenario

Using the RocketEngineer's formula for deflection, and the dimensions Sizzle gave, I get that it's a 270 gallon tank (call it 2700 lbs) and that the deflection over a 62" span (the length of the section of the 72" beam that is between 2x4 legs that are arranged correctly) is

For a 1x8 (.75 x 7.25) ---- 0.1136" which is less than 0.125, and so is okay.

For a 2x6 (1.5 x 5.5) ----- 0.1301" which is a little more than 0.125, and so is not necessarily strong enough.


It's important to note that if you don't arrange the 2x4 legs correctly,
the span will be 65" instead of 62", and for that span, the deflection is above 0.125 for both sizes of lumber (0.1372, 0.1572), and so neither would be considered "safe enough".

 

[C-Rad]

I'm afraid I disagree with FishMan on this (maybe one of us needs to fix our spreadsheet)

I was wondering if 1x8 equally as strong as 2x6
The stand dimensions will 72lx24wx36h

For a 300 gallon tank on a 72 inch stand the 2x6 is stronger. But it may depend on the setup. If you tell me the tank size I can figure it out.
PS - neither one works in this scenario

Using the RocketEngineer's formula for deflection, and the dimensions Sizzle gave, I get that it's a 270 gallon tank (call it 2700 lbs) and that the deflection over a 62" span (the length of the section of the 72" beam that is between 2x4 legs that are arranged correctly) is

For a 1x8 (.75 x 7.25) ---- 0.1136" which is less than 0.125, and so is okay.

For a 2x6 (1.5 x 5.5) ----- 0.1301" which is a little more than 0.125, and so is not necessarily strong enough.


It's important to note that if you don't arrange the 2x4 legs correctly,
the span will be 65" instead of 62", and for that span, the deflection is above 0.125 for both sizes of lumber (0.1372, 0.1572), and so neither would be considered "safe enough".

 

[TheFishMan65]

Ok a 270 gallon tank, with a 62 inch opening and salt water weighing 10lb/gallon (although this does not change it much).
1x8 deflection is .237
2x6 deflection is .200
your equation
2x6 deflection .127
Hmm - something is wacky
how do you do a 1x?

 

[C-Rad]

I was wondering if 1x8 equally as strong as 2x6?

The stand dimensions will 72lx24wx36h, local wood suupliers in the area rarely have good straight 2x anything. However I can find dead straight 1xs with no issues.

Thanks for the help

Ok a 270 gallon tank, with a 62 inch opening and salt water weighing 10lb/gallon (although this does not change it much).
1x8 deflection is .237
2x6 deflection is .200
your equation
2x6 deflection .127
Hmm - something is wacky
how do you do a 1x?

One of the stated assumptions of the simplified version of the deflection formula that I gave is that the beams are 1.5" wide (2xWhatever's), so I can't use the simplified formula for a 1x beam. I do a 1x using RocketEngineer's original formula:

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

FishMan,
I think you're using RE's formula wrong, which is really easy to do, and illustrates why I thought a simplified version of the equation would help a lot.

I suspect that the problem is that you are plugging the weight of the tank into the equation for "W", which is not what "W" is. The weight of the tank is being supported by four beams, not just one beam. The formula is measuring the deflection of only one beam, and so "W" is the portion of the total weight that is being supported by the span between the legs of the single beam we are measuring the deflection of.

Here's how to use the formula correctly for this example (you'll see why a simplified version is a good thing):

The formula needs us to plug in numbers for:

Ixx --- (1/12 * b * h^3)
L ----- The length of the span between the legs of the beam we are measuring the deflection of.
W ----- The weight being supported by the span of length L (Not the total weight of the tank)​

Ixx is easy to get:

For a 2x6, b = 1.5" and h = 5.5" so Ixx will be (1/12 * 1.5 * 5.5^3) = 20.796875

For a 1x8, b = .75" and h = 7.25" so Ixx = 23.8173828125​

It's tricky to get correct values to plug into the formula for L and W, and easy to get wrong ones. Here's how to do it right (I'll type slowly...):

To get L (span between the legs) we need to know about the legs. In this example, let's assume that each leg is made by connecting two 2x4's at 90 degrees, so that each leg is 3.5" along one side, and 5" along the other side. The shorter the span (L), the less the deflection will be, so let's always put the 5" side of each leg along the long side of the stand, leaving the span between the legs on the long sides 3" less than it would otherwise be. Each leg takes up 1.5" + 3.5" = 5" at each end of each 72" beam, leaving a span between the legs of 62", so in the formula, L = 62".

To get "W", we need to know the "Total Tank Weight", The "Total Beam Length" (for all four beams), and "L" - The length of the span (between the legs) of the beam we are measuring the deflection of. Since the weight is evenly distributed, if the span "L" is X% of the "Total beam Length" then "W" will be X% of the "Total Tank Weight". So in general:

W = (L / "Total Beam Length") * "Total Tank Weight"​

"Total Beam Length" = 72+72+24+24 = 192"

There are 231 cubic inches in a gallon, so:
"Tank volume in gallons" = (72 * 24 * 36) / 231 = 269.3 gallons (call it 270 gallons)

"Total Tank Weight" = "Tank volume in gallons" * 10 lbs/gallon = 270 * 10 = 2700 lbs

So now we can calculate "W", the amount of weight pushing down on the 62" span.

W = (62 / "Total Beam Length") * "Total Tank Weight"
W = (62 / 192) * 2700 = 871.875 lbs​

Okay, now we have all the correct values to feed into the formula to calculate deflection:

Deflection of the beam = (5*W*L^3)/(384*Ixx*10^6)

For a 1x8 that is:
(5 * 871.875 * 62^3) / (384 * 23.8173828125 * 10^6) = 0.1136"

For a 2x6 that is:
(5 * 871.875 * 62^3) / (384 * 20.796875 * 10^6) = 0.1301"​

If you can plug numbers into a formula, you can get Ixx, but to get L and W requires a good understanding of what the formula is really doing. The formula isn't just plug and play, and until you understand what W and L really are, or at least how to calculate them, you can't really use this formula. Plus you need to know the weight of a cubic inch of water to find W, or the weight of a gallon, and that there are 231 cubic inches in a gallon. I hope this walk through helps. If not, you can use the simplified version of this formula for any stand that meets the assumptions (most stands that use the design given at the top of this thread will meet the assumptions I described)

 

[TheFishMan65]

I actually gathered this from your email. While I agree with most of what you said. I don't believe this is the way RocketEnginner calculates the vaules. I say this since he agreed with one of my answers. I believe he made the following assumptions for safety:

1. All of the weight is dsitributed over the 2 longest beams
2. He did not subtract the legs from the length

Rocket are my assumptions correct?

The things I will point out that I disagree with (although none are big deals)

"Total Beam Length" = 72+72+24+24 = 192"

This counts the corners twice you should probably subtract 1.5 inches from each measurement so you are using the center lines of each board. Better still would be the centerline of the tank frame.

Since the weight is evenly distributed

A reasonable assumption, but an FYI this is not always true especially with framed tanks.

 

[Sizzle]

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.

 

[Sizzle]

Tank is glass and framed

 

[TheFishMan65]

Sizzle, neither of those will work (assuming I am doing the equation correctly - see discussion above)
a 1x8 has a deflection of 1.98
a 2x6 has a deflection of 1.67
adding a center support would allow either of these to work

 

[ludiNano]

Stand is done and in spot, still need to do A LOT of leveling since may basement floor angels away from the front of the stand. I did follow the rocket engineer build template. Stand is really a stand within a stand, outer dimension's are 72.5 x 36.5 x 37.5 which is the dimension's I will upgrade to years down the road but there are also supports within the 72.5 x 36.5 x 37.5 that are 60.5 x 18.5 x 37.5 to hold my current tank. I also will be adding 2- 3/4'' pieces of plywood on top on the stand bringing the total hight to 39''.



And in reality the stand now is only 33'' wide since I still have to build the front wall which will be 3.5'' wide up to what the leveled stand hight turns out... if that makes sense... I am doing this because we had already acored the base plate (2x4) into the cement ( due to frameing happy father in-law while we were framing) and i want the tank to go to the edge of the wall.

What would be the best way to frame this front wall to properly support the 3.5'' hanging over the stand?

I was planning on using double up 2x4's for sure on the ends and then every 12''.

But would it be ok to double the ends and then just go single 2x4's every 12'' to use less wood? ( there will also be a 2x4 on top of the studs laying flat 3.5'' across.)

<a href="http://s461.photobucket.com/albums/qq340/ludinano/fish%20room/?action=view&current=IMAG0664.jpg" target="_blank"><img src="http://i461.photobucket.com/albums/qq340/ludinano/fish%20room/IMAG0664.jpg" border="0" alt="Photobucket"></a>

 

[TheFishMan65]

Formula under discussion, but I get less than a 16th inch of deflection for a 500 gallon tank with the 2x4 flat and 12 inch centers. Single 2x4s should be fine.

You probably plan this, but have the plywood be one piece across the stand and the new wall.

 

[ludiNano]

Thank you for the fast reply. However I believe I will be double 500lb and far more than that when it is upgraded to the 300 DD in the future.

 

[TheFishMan65]

Whoops I meant 500 gallons.

 

[C-Rad]

While I agree with most of what you said. I don't believe this is the way RocketEnginner calculates the vaules [...] I believe he made the following assumptions for safety:

1. All of the weight is dsitributed over the 2 longest beams
2. He did not subtract the legs from the length

Rocket are my assumptions correct?

If RE makes those assumption, I don't think he does it "for safety", but rather to make the formula a lot easier to use, by avoiding the two issues I described in my last post about calculating values for "L" and "W". Those two assumptions are shortcuts that make the results a lot less accurate, but could be considered acceptable because the error always shows more deflection than would be accurate, never less. The problem is that the inaccuracy is hidden, and the amount of inaccuracy is moderate for some tanks, but large for others. An overt and consistent way to increase the safety margin would be to get accurate deflection values from the formula, and then simply lower the amount of deflection you're willing to accept.

There is definately a safety margin in this design [formula]. 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. [...] And for when I run the numbers, anything below .125 [inches of deflection] is my limit so working for .1 is even safer.

The two shortcuts you are endorsing cause your result for deflection of a 1x8 for this 72x24x36 tank stand, w/ 2x4 legs, to be 139% too high (.2755" instead of .1154"). The accurate deflection value of .1136 is less than .125, and so would allow the use of a 1x8 for this stand, but your two short cuts cause the formula to erroneously disallow a 1x8 and even a 1x10 (.1326" deflection).

I admit that in most cases, the inaccuracy caused by these two assumptions won't be a big deal, because the worst thing that could happen is that you end up using larger beams than you need, which will seldom be a big problem, and never a safety problem. On the other hand, if this is going into a spread sheet anyway, why not go the extra mile, avoid the short cuts, and get better results?

This [C-Rad's example of using the deflection formula] counts the corners twice you should probably subtract 1.5 inches from each measurement so you are using the center lines of each board. Better still would be the centerline of the tank frame.

good catch!
My interpretation of the formula is correct, but I didn't follow my own interpretation when I walked through the example. I wrote:

"Total Beam Length" = 72+72+24+24 = 192"​

That's wrong because it is 6" longer than the actual total beam length. If you literally measured the cut beams you would get a "Total Beam Length" of 72+72+21+21 = 186" (or possibly 69+69+24+24 = 186"). Using 186 instead of 192, gives the correct value of W = 900 lbs (instead of the wrong W = 871.875 lbs). Using the center line would also give 186", but I think it's easier to think of it as "Total Beam Length"

That mistake caused my deflection result for a 1x8 to be 1.6% lower than it should have been (.1136 instead of .1154 - which still allows the use of a 1x8 beam for that tank). That same error is incorporated into the simplified version of the formula that I posted earlier, so I'll post a fixed version soon.

 

[RocketEngineer]

TheFishMan65 and C-Rad,

Back to school (well at least the books):

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.

I think part of the confusion is that in the "text books" P is the point load on a beam and W is the distributed load. I reverse them because W=Weight makes more sense to most people.

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. I don't consider the short sides 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.

Back to C-Rad's equation:

(5*W*L^3)/(384*(1/12*b*h^3)*10^6)​

->Replace W with (X*Y*Z/23.1) by assuming 10#/gal for tank+sand+rock+water
->Replace L with X
->Solve for the constants

(Y*Z*X^4)/(1.4784*10^8*b*h^3)​

The formula does not remove the leg support from the long side length and assumes that all of the weight of the tank is supported by the two long boards.

One of the reasons I stick with the true formula instead of the simplified version is it allows me to play around with having three front beams instead of two (the stand Newbie Aquarist was after is an example).

A few things to keep in mind:
1) The 1/8" deflection is a feel good number. I don't know what the professionals use but its what I feel should be about right.
2) The 10^6 is the modulus of elasticity for wood. For most species that we use in our stands this value is closer to 1.5*10^6 so there is a certain safety margin built into the formula from the start.
3) There are a number of different ways to run the values, some are more conservative than others but in the end, we are all still working with roughly the same selection of materials. If something is borderline, going a size bigger is up to the builder.
4) This is for a FISH TANK FOLKS! :rollface: This isn't rocket science (and remember I'm just a lowly engineer, not a scientist so I honestly can't tell you how hard rocket science really is). At the end of the day, this is a hobby that we do because we enjoy it. Twisted individual that I am, I enjoy designing things. If someone else has the money to build it and is nice enough to post pictures online than that's a bonus.

Hope that has cleared up a few questions about what values go where.

And for those folks whose eyes glazed over when the math talk started, don't worry. I have the same thing happen when folks start talking about the latest celebrity gossip. Just ask your question and it should get answered eventually.

RocketEngineer

 

[TheFishMan65]

Thanks Rocket. I just wanted to check some of my assumptions on on how you were using the formula. Sound like I was pretty close, but I forgot the end might not be supported - thanks for the reminder.

C-Rad, on other thing I just thought of in your simplified equation you assume the tank goes from stand end to stand end. Most stands I have seen pictures of tend to have a little space left and right so you might want to add an inch to the opening. But as RE says there is enough safety put in that maybe it does not matter much.

thanks for the discussion.

 

[C-Rad]

Stand is really a stand within a stand, outer dimension's are 72.5 x 36.5 x 37.5 which is the dimension's I will upgrade to years down the road but there are also supports within the 72.5 x 36.5 x 37.5 that are 60.5 x 18.5 x 37.5 to hold my current tank. I also will be adding 2- 3/4'' pieces of plywood on top on the stand bringing the total hight to 39''.

And in reality the stand now is only 33'' wide since I still have to build the front wall which will be 3.5'' wide up to what the leveled stand hight turns out... if that makes sense... I am doing this because we had already acored the base plate (2x4) into the cement ( due to frameing happy father in-law while we were framing) and i want the tank to go to the edge of the wall.

What would be the best way to frame this front wall to properly support the 3.5'' hanging over the stand?

I was planning on using double up 2x4's for sure on the ends and then every 12''.

But would it be ok to double the ends and then just go single 2x4's every 12'' to use less wood? ( there will also be a 2x4 on top of the studs laying flat 3.5'' across.)

Looks very well done. Good Job.

Single 2x4's, at 12" (center to center) is much more than enough, and there's no reason to use two at the ends. You could even space them 24" (center to center) and still support a tank twice as heavy as your big tank will be.
With a 72.5x36.5 footprint, even if your future tank were 36" deep, it would weigh less than 4100 lbs. If we ignore the support given by the plywood, and the internal frame, the external frame is 212" long, supporting 4100 lbs, which works out to 19.34 lbs per linear inch of frame. That means that the 10.5" span between each of the vertical 2x4's will only be supporting about 203 lbs. The deflection formula (with W=203, L=10.5, b=3.5, h=1.5) gives a negligible 0.0031" of deflection over the 10.5" span. If spaced 24" apart, the span between the legs, L, is 22.5", and the deflection is still only .0656" (up to .125 is acceptable). Don't worry about it being strong enough, but make sure that all the 2x4 legs are the same length, and that the top of the frame is very close to flat, so that the weight really is distributed evenly over the 212 linear inches of frame. The internal frame and double plywood could compensate for a lot of unevenness if need be. You have nothing to worry about.

I'm fairly sure that the kind of framing you are proposing (12") is required by code for first story support walls of two story houses, which are intended to hold up the second story, the roof, a few feet of snow on the roof, all the fish tanks, and gun safes that somebody might decide to put upstairs, plus the 50 people from over-eaters-anonymous that might come over to do synchronized jumping jacks. Not to mention that 8' tall walls are much weaker than 39" walls. You're covered.

Good job on the stand!

 

[Sizzle]

So I am lost can I use 1x8s? Or do i need to use 2x6s?
Thank you there will be beems on the top frame to help prevent twisting/warping of the wood.