Tank engineering question

[EllieGreene]

Hi.

I want to construct a 8 foot * 4 foot * 4 foot windowless tank, aka a pond out of wood. I have seen others tanks/ponds with Landscape timbers through bolted at the corners and center, Lined with Plywood and pond liner.

Now the bad part.. I want to "do the math" before getting started. The math is for three purposes really.. First, for safety, Second, so I can say I really understand it, and Third so I don't blow a tank a year from now when I have 500 lbs of fish in it.

I was working with the assumption that the Lowest Beams in a tank take the most stress, equal to the weight of all of the water above them.

So the bottom beam in a 4x8x4 foot tank needs to support 4x8x4 cubic feet * 64 lbs/cubic foot = 8192 lbs.

(Question #1, is this correct?)

Next up, the 8 foot span should logically be weaker than the 4 foot span. So I need to find an 8 foot long beam that will support ~8,000 lbs.

(Question #2, is this correct, or is it a smaller number?)

The American Wood Council Publishes (free online pdf) a Beam safe load table. According to it, an 8 foot long 4x4 beam is only safe for 833 lbs. The smallest timber suitable for 8,000 lbs is an 8x8. (eeks!)

(For those of you following along in the PDF, the Fb of Southern Yellow pine is between 1400 and 1600).

So now I'm confused. I know that others have built similar tanks out of el-cheapo 3x5 landscape timbers, but those should literally blow apart as soon as they are filled. What am I missing?

Thanks,
Ellie

 

[EllieGreene]

This doesn't fit.. If that is right then my tall 30 gallon's sides are withstanding ~300 lbs of pressure. I just don't see that being possible.

In diving class we learned 33 feet H20 = 1 ATM = 14.2psi. With a little division I get 0.43 psi/foot .. times 4 feet of water depth = 1.72 pounds per square inch. A plywood lined tank beam will be 8 feet long (=96 inches) x 6 inches tall* = 576 square inches. 576 sq/in times 1.72 pounds/sq/in = 990.72 lbs.

990 lbs of weight evenly distributed across an 8 foot beam sounds much much much more realistic. According to that, using a 4x6 for the bottom rail(s) should give me a 2x safety margin over the safety margin built in to the AWC tables.

(It sounds great.. But is it right?)

Comments/questions/opinions absolutely welcome.
-ellie

* - Actually the beam isn't this tall, that is the center-to center distance of the beam if I use 3x5 landscape timbers.

 

[sp33dstix]

hymm a good matlab problem =D
I would have to use that to figure it out. This is very interesting and I would really like to find this out as well. Looks like I have to look into water pressure at certain depths. I will let you know what I come up with once I am done.
I will work with this to see what I can come up with.
http://theory.uwinnipeg.ca/physics/fluids/node8.html
I would like to make a program to take in various tank dimensions and spitting out graphs with pressure vs height of the aquarium. I will post results once I am done with that.

 

[sp33dstix]

So I got a little bored and with the dimensions given I created the a graph of forces acting on both panes of the aquarium. Mind you I changed all the units to SI units because they are much easier to work with. Here is a picture I hope it helps.

Now before you bash me for using Canadian units use these graphs only as a quick reference. They should not be used in your final design. I just want to see if they make sense. But yea as you can see the deeper you go the more forces are acting on the walls of the aquariums/pond. Thr larger pane is the 8X4 pane and the smaller one is the 4X4 pane. Since pressure deals with areas it makes more sense that there is more force acting the further down you go in a pool of water.

 

[CHOMPERS]

You could have just said that the entire weight of the tank isn't going to be supported by one beam.










...but I like your way better.

 

[CHOMPERS]

...so I don't blow a tank a year from now when I have 500 lbs of fish in it...

The fish are neutrally buoyant so they weigh exactly the same as the amount of water they are displacing. A tank with one fish or a hundred fish will weight the same.

 

[CHOMPERS]

This doesn't fit.. If that is right then my tall 30 gallon's sides are withstanding ~300 lbs of pressure. I just don't see that being possible.

The weight of the water isn't the pressure in the tank. The pressure is measured in psi which is pounds per square inch.

psi= pounds/(inch^2)

To find the pressure at depth, you have to find the weight of the water over one square inch of horizontal surface. This can be at the bottom of the tank or an imaginary square at any depth. Water weighs approximately 8.3 pounds per gallon, and there are 7.5 gallons per square foot.

To find the force exerted on a pane of glass by the water pressure, you would need to divide the tank into one inch horizontal rows, and then find the pressure for each depth. Then multiply the pressure by the length of each row. Lastly, you would add the values for all of the rows.

 

[sp33dstix]

The weight of the water isn't the pressure in the tank. The pressure is measured in psi which is pounds per square inch.

psi= pounds/(inch^2)

To find the pressure at depth, you have to find the weight of the water over one square inch of horizontal surface. This can be at the bottom of the tank or an imaginary square at any depth. Water weighs approximately 8.3 pounds per gallon, and there are 7.5 gallons per square foot.

To find the force exerted on a pane of glass by the water pressure, you would need to divide the tank into one inch horizontal rows, and then find the pressure for each depth. Then multiply the pressure by the length of each row. Lastly, you would add the values for all of the rows.

That is what I did =D
Graphed it all out afterwards, although my division of the tank was much less then one inch. I used 500 intervals to get a more accurate measurement. But yea my bad for the SI units, I am way more familiar with those units then the imperial ones.

 

[sp33dstix]

Now to make sense of my graph. You would want to use a beam that could support at > 675 N (Newtons) of force acting along it's face. Since from the graph you can see that the most force acting on the larger pane is about 675 N. This translates to about 3002.4 lbf (foot lbs). You could use this value in finding which wood to use.

 

[sp33dstix]

The weight of the water isn't the pressure in the tank. The pressure is measured in psi which is pounds per square inch.

psi= pounds/(inch^2)

To find the pressure at depth, you have to find the weight of the water over one square inch of horizontal surface. This can be at the bottom of the tank or an imaginary square at any depth. Water weighs approximately 8.3 pounds per gallon, and there are 7.5 gallons per square foot.

To find the force exerted on a pane of glass by the water pressure, you would need to divide the tank into one inch horizontal rows, and then find the pressure for each depth. Then multiply the pressure by the length of each row. Lastly, you would add the values for all of the rows.

There is another way of finding pressure regarding depth of water. The formula is the following.
P(h) = Po + g*h*p
Where:
Po is atmospheric pressure of air
g is acceleration due to gravity
h is the variable of how deep you are in the water
p is density of water

I used this formula to find the pressure at certain depths and then I split the height of the aquarium into 500 divisions let dh = height/divisions and is constant. For each pane of the aquarium I ended up with 500 incremental areas. The incremental areas for the larger pane (8x4) is dh*8. The incremental area for the smaller pane (4x4) is dh*4. Now to graph out the forces I went through every division of the height, found a pressure with the above pressure forumla and for each division I multiplied it by the incremental area.

I have the source code if you want to take a look at it. (Used Scilab, matlab is too expensive)