Deflection due to water 3

[karthur]

I have an open top,rectangular water container. It is constructed with a bottom and four sides. The containter is 40" deep x 110" wide x 440" long. The walls are 1/4" thick.

I need to know how to figure what the deflection will be on the sides of the tank.

Thanks

 

[SperlingPE]

Is the container completely full of water?
You could do a static analysis of it to get a "rough" estimate. However, a finitie element analysis will yield far batter results.

 

[jhardy1]

As SperlingPE says, the depth of water fill is important. (Your view seems to show holes in the walls near the top of the side plates.) Also important is the dimensions of the flange around the top edge.

If you ignore the contribution of the top flange, and assuming the box is full of water to the brim, you will find a maximum deflection at the mid-point of the longer top edge of approximately 110 mm (approx. 4 1/2"). Bending stresses in the bottom of the wall are close to yield for mild steel, so this is probably an unacceptable solution.

If you have a suitably proportioned stiffener flange at the top edge, this will act as a ring beam, and will greatly reduce stresses and deflections in the wall plates.

 

[JStephen]

I believe "Formulas for Stress and Strain" has some rectangular-tank formulas; check there.

If not, you can use the rectangular plate equations. Assume simply supported on three (or four) sides and fixed on three (or four) sides and your case should be in between those two cases.

 

[karthur]

Julian Hardy,
Thanks for your answer.

As you assumed, the water level will be 40in deep. I felt the deflection would be high, so I added a 4inx4inx1/4 wall square tube around the top. How much will this "Rim Beam" reduce the stress and deflection in the side plates?

From what I calculated, the force on the side walls are given by F=wL*d^2 where w=62.4lb/ft^3, L=440in(36.7ft), d=depth of water 40in(3.3ft). I get this to be 12,697lb

I am still looking for a rectangular simply supported equation.

Thanks Again.

 

[JAE]

A finite element analysis may not be totally useful here...depending on your software. Many do not include in their formulae the aspect of tensile straightening.

If you have a cable, and is spans from A to B - and you hang a load on it at midspan, it will deflect into a "V" shape. If you add tension, it will draw upward, eventually almost straight.

In your box problem, the water will tend to bow the walls outward, but the orthogonal forces at each end will tend to straighten out the top edges of the tank....a typical finite element analysis may not include this effect and your stresses will be somewhat inaccurate.

 

[JStephen]

Another simplification that may help- treat the sides as vertical slats, design them as simply supported beams; then design the top stiffener to carry the reactions from the top. The flat-plate equations approach this approximation as the width becomes much larger than the height.

 

[karthur]

>Another simplification that may help- treat the sides as >vertical slats, design them as simply supported beams then >design the top stiffener to carry the reactions from the >top. The flat-plate equations approach this approximation >as the width becomes much larger than the height.

How do I determine what the reaction is at the top? Do I just design the top stiffener to carry this full load or would that be too much of a S.F.?

I calculated the loading on the side is around 13,000 lbs.

 

[aggman]

karthur,
If you don't already have it, get yourself a copy of Design of Welded Structures by Omer Blodgett. It can be found at the Lincoln arc welding society website and is relatively cheap. Section 6.5 is on the design of Tanks, Bins and Hoppers. It will have the information you need.

 

[karthur]

aggman,
Thanks for the suggestion. I read his column in Welding Design and Fabrication every month. I just ordered both the Design of Welded Structures and Design of Weldments.
Got both of them new for about $45.

Can't wait to get them now.

Thanks

 

[JStephen]

The load on the top stiffener would just be the beam reaction from the vertical strips of plate. Isolate a vertical strip of plate, sum moments about each end of it to find reactions.

 

[karthur]

aggman,
I got the books in over the holidays. I started by working throught my problem using problem 1, page 6.5-1 as a guide. The first thing that I ran into was that my a/b=40/444=.09. This is off the chart in Table 1,4D. From this chart, if the length is more than twice the height (a/b<.5), then the equations are not valid.

I wonder if Roark's "Formulas for Stress and Stain" lists an equation for beta and lamda? I don't have this book.

Any suggestions?

Thnaks

 

[aggman]

karthur,
Because of the large aspect ratio poissons effects are negligible on the deflection of the plate. Because of this you will be valid to design the plate as a one-way strip (simply supported) or beam from top to bottom. Design the top stiffner to carry the reaction. Be careful to limit deflections on the stiffner as it will increase the deflection in the plate. As far as the plate is concerned you should find that the results using (a/b < .5) will give you the same if not very close stresses and deflections as designing as a one-way strip. This is why they stopped at this aspect ratio. Generally you will not find direct formulas for beta and lambda, as they are generally generated using specific boundry conditions for each type of plate to solve series solutions. When you design the plate be sure to limit deflections to less than plate thickness / 2. This will ensure that membrane effects are negligible.

 

[IFRs]

It is critical to know how the tank is supported. If the tank is supported on the short ends, or hung from the holes in the sides, there will be significant changes to the analysis.

 

[karthur]

"It is critical to know how the tank is supported. If the tank is supported on the short ends, or hung from the holes in the sides, there will be significant changes to the analysis."

The tank rests on the floor. The holes are access and drain holes.

 

[Ron]

Compute the deflection of the top without the stiffener. The load in the stiffener is then the equivalent load to cause that deflection, neglecting the stiffness of the plate. Force the combination of deflections to zero and that will be your stiffener design.

 

[miecz]

Roarke's Formulas do contain solutions for flat plates with hydrostatic loading. These solutions are better than Blogett's for your case. However, they all have the same limitation as Blogett. They assume fixed or pinned edges. None of your edges is fixed or pinned. Even the bottom edge will rotate. The best solution is finite elements. Short of that, you want a conservative solution, so assume the bottom is pinned and use solution 3Da for the plate deflection at x=.5193h. Notice that this solition assumes the stiffener is infinitely stiff. With R1, calculate the deflection of the sifffener. Add .5193 times the stiffener deflection to the plate deflection to get the total. Note that the moment of inertia of the stiffener will nearly double if you include composite action with the tank plate. For the effective width of the plate, I would use the width of the stiffener plus 16 times the thickness of the plate. Also, since the stiffener is neither pinned nor fixed at its ends, the stiffener should be anlyzed as a 4 member frame. Deflections will be very sensitive to this assumption, so you can't assume a simple support.