Tank engineering question

[johnptc]

why not take the simple approach for the bottom beams

total weight of tank W

suppose 4 beams underneath
it would appear the two outside beams each support 1/6 of W or W/6

the inner two beams each support W/3

size the beams for the heavier load W/3

you can suppose equally distrubuted load over 96 inches

so you need beams 96 inches long able to support W/(3*96) pounds per inch

for a 10000 pound tank thats about 35 pounds per in or about 425 pounds per foot load

 

[EllieGreene]

If I understand correctly, I think you are saying is I got it right on my 2nd attempt and that the load supported by a given beam is equal to the wetted surface area in square inches times the depth to the center of the beam in feet times (14.2psi/33feet)?

Is that right?

 

[EllieGreene]

Johnptc, thanks, but I'm trying to size the beams to reinforce the sides of a rectangular plywood tank. Your formula fits for supporting the whole assembly but I'm okay on that point.. I'm on a concrete pad.

-ellie

 

[johnptc]

ok its probably about 2.5 psi for near the bottom

48x96x2.5 is the load on the wall figuring the bottom pressure exists at the top

which is very conservative.........its really the integral of (density)*(gravity)*height * dh

or 0.5*density*g*h*h at any depth h

or total force on the front wall 48x96x.5xdensityxgravityx48x48

 

[CHOMPERS]

...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.

Wow, a math savant in the making. You are going to be really surprised when you get into Calculus. That's a principle behind the Second Fundemental Theorem of Calculus.

 

[sp33dstix]

Wow, a math savant in the making. You are going to be really surprised when you get into Calculus. That's a principle behind the Second Fundemental Theorem of Calculus.

Ohh I know...I have done lots of "integrals" using matlab and incremental elements before. I have finished my 3rd year in electrical engineering, and am still alive. So much math your head can explode =D
But yea, I was also curious about forces of water in an aquarium as well, this question sort of just pushed me over the edge to find a numerical answer to this. So I did it.

 

[sp33dstix]

ok its probably about 2.5 psi for near the bottom

48x96x2.5 is the load on the wall figuring the bottom pressure exists at the top

which is very conservative.........its really the integral of (density)*(gravity)*height * dh

or 0.5*density*g*h*h at any depth h

or total force on the front wall 48x96x.5xdensityxgravityx48x48

keep in mind I included the pressure due to atmospheric air in my calculations...

 

[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.

I would like to add that from my calculations 675 N is equivalent to a person with mass 68.88 Kg = 151.85 lbs spreading their mass over the area. Now that I think about it, I am just finding instantaneous values of all the forces acting on specific points of the walls, I need to find a way to find it acting on preset size of wood( like an 8 foot long 2x4). I smell an integral....Also wish I remembered my calculus and physics...

 

[johnptc]

keep in mind I included the pressure due to atmospheric air in my calculations...

there is none as it acts on both sides and cancels out

 

[johnptc]

I would like to add that from my calculations 675 N is equivalent to a person with mass 68.88 Kg = 151.85 lbs spreading their mass over the area. Now that I think about it, I am just finding instantaneous values of all the forces acting on specific points of the walls, I need to find a way to find it acting on preset size of wood( like an 8 foot long 2x4). I smell an integral....Also wish I remembered my calculus and physics...

f=P x A
df(h)= P(h) dA
df(h)=density*gravity*bdh where b is unit width
f= integral P(h)dA from 0 to height H eg integral of xdx
f= 0.5*density*g*b*H*H
f= 0.5* (density of water) x gravity *96*48*48


4000 pounds ???