Seeking a simple formula to find deflection of flat ends on tanks 1

[MickE]

Does anyone know a simple way to determine the deflection of flat suspended tank bottoms?
Or if the maximum deflection was set to a percentage of the diameter, say by visual acceptability, how should I go about finding the required thickness? For simplicity and if it is a part of the solution, could we assume the bottom strake thickness was the same as the bottom and that a knuckled radius transition was used?
This would only be applied to relatively small diameters, say up to 2 meters diam.
Many thanks for any help.

 

[ksuengrng]

An excellent source for flat plate calculations is Roark's Formulas for Stress and Strain by Warren C. Young (6th ed.)

Table 24 has many different loadings and variations for you to choose from and apply. We would run this calculation and then compare it to an acceptable deflection (either .5% of the tank diameter, or diam/300, depending on client, tank location, etc...). We would then increase or decrease the thickness accorndingly, or add bracing to the suspended bottom.

A quick way to calculate the thickness is to treat the top like a beam. Take a "section" of the roof and treat it like a beam with the loading scheme it will see during operation, (fluid height, pressure, vacuum, etc...). Now whether you treat the ends as fixed or simply supported will need to be determined by the connections you have to the sidewall, AND what you, or your client, are comfortable with.

ASME also has a flat plate calculation, but unless this is a code vessel, I would look at treating it like a beam. The ASME calc will is much more conservative.

From past experiences, flat suspended bottoms do not handle deflection well, unless you are real thick, or brace the heck out of it.

 

[MJCronin]

MickE....

Let me see if i understand this.... you have a flat "suspended" tank bottom ? Does this mean that you have a vertical axis tank with support of the tank shell ? ( By legs or some other means ?) Is the tank to be filled with a liquid ? Is the tank metallic, PVC or perhaps fiberglass ?

Is this a special "custom" design ? Or something purchased at a yard sale ?

Most respectable tank fabricators would steer a customer away from flat bottomed design of any substantial diameter....A "flanged and dished design" is typically used ( that's why fabricators make this type of tank head)...Vertical axis tanks with diameters greater than about 40 inches usually have these dished heads.

"ksuengrg" is giving you good advice and refering you to the right source, however, it has been my experience that by the time you have reinforced a flat bottomed tank to take any substantial load, you might as well have installed a dished head,

Contact Enerfab in Cincinatti, OH for more info.....

http://www.enerfab.com/heads3.htm

Hope that this helps....................................................MJC
hope that this helps.....

 

[MickE]

Thanks to both of you for taking the time to respond.
They are metal tanks MJCronin, with suspended bottoms, ie on legs or skirt. I have Roark and found the approach overwhelming in its complexity.
As you have both pointed out, a flat bottom on any support is almost theoreticaly not feasible but in practice, for small tanks seems to be quite OK. Maybe 2 mtrs is pushing it but 1 mtr would be expected to be quite ok if say the bottom was of 5mm (minimum API recommendation for corrosion reasons) mild steel.
What I was hoping for was either a rule of thumb or simplified formula that might determine the cut off between say the 1mtr and 2mtr diameters for instance but could be used on any size up to that cutoff with thinner eg stainless steel tanks.
Most fabricators in fact rarely use less than 2mm thickness, for ease of welding or tend to be influenced by experience to determine adequate "robustness" to resist the rigours of delivery and installation.

 

[butelja]

The following formulas for circular flat plates come from Machinery's Handbook. I would advise comparing the results with the formulas from Roark & Young.

For a simply supported edge with distributed load:

S = 0.39*W/t^2 d = 0.221*W*R^2/(E*t^3)

For a fixed edge support with distributed load:

S = 0.24*W/t^2 d = 0.0543*W*R^2/(E*t^3)

In formulas above,
S = stress
W = total weight supported (NOT pressure)
t = plate thickness
R = plate radius
E = elastic modulus of plate

I tend to agree with the previous comments that an unsupported flat plate would probably only be economically feasible on very small tanks.

 

[meca]

There is a simple program to calculate the deflection at this link

http://www.mecaconsulting.com/Archon/ARCHON-WinPlate.htm

 

[butelja]

Also keep in mind that all linear plate deflection equations are valid only when the maximum deflection is less than about 1/2 of the plate thickness. Larger deflections than this impose membrane stresses in addition to bending stresses, and cause nonlinearity in the deflection vs. load relationship.

If properly accounted for, this nonlinearity can be used to advantage to design a lighter weigh structure. It will require the addition of a compression ring around the shell to resist the radial inward tension caused by the membrane stress in the plate.

 

[MickE]

Thank you for that butelja.
You've helped me get my thinking sorted out.
When calculating deflection using the Roark formulae, I kept getting results that were clearly wrong...overstated. Like the deflection of a 2.0mm bottom on a 1000 litre, 1200mm diam tank is more than 5 mtrs. That simply doesn't happen, it's nearer 45mm in practice.
The reality is that there are many small flat bottomed tanks in use in industry that deflection theory says can't exist. I've been involved with their construction over the years, and this theoretical deflection has always bothered me.
What is happening you're saying is that the membrane stresses are governing, right?
I was going to think of that next
Do you happen to know where I can get info on the design approach?
Thank you.

 

[butelja]

In the 5th edition of Roark & Young, section 10.11 covers the effect of large deflection and diaphragm stresses. After you solve for the diaphragm stress, this stress multiplied by the plate thickness will give you a radial load in lbs/in. This load is used to size the compression ring.

Although some small portion of the bottom of the tank shell will "participate" in the compression area, it is simpler and more conservative to neglect the shell participation and assume only the compression ring resists the radial load.

The compression ring is checked for elastic stability using a relatively large safety factor (>3). The critical radial load on a uniformly loaded ring is given in chapter 14 or Roark as:

P_cr = 3*E*I / r^3

I hope this is helpful.

 

[MickE]

butelja
Many thanks for your assistance, it is truly appreciated.
Perhaps you could help me with one more question?
In the Machinery's Handbook formula, I presume that d is the deflection, correct? The units of the formula are
mass x length^2/E x length^3
If d is deflection, should I presume it to be deflection away from the plane of the plate as one would expect?
I ran the fixed edge support formula for a 1/8in. thick tank bottom, 48in. diam, holding 53ft3 (3,300lbs)and d came out at 1 3/4in. (1.766) or 3.7% of the span or diameter. This is of the order of what I'd expect B-)
The stress however was 50,800psi, suggesting permanent strain...not what I expected B-(
I note your advise re comparing with Roark. Thank you, I will.
By the way, my interest is with stainless steel tanks.

 

[butelja]

For deflections above 1/2 the plate thickness, you must consider the nonlinear effects. Therefore any answer that the "simple" formulas give above 1/16" in your case is incorrect. At a deflection of about 1 plate thickness, bending and membrane stresses are of similar magnitude. Above 3 plate thicknesses, the membrane stresses are much larger than the bending stresses. For steel tanks, a maximum deflection of about 1% of the diameter will usually cause stresses in the neighborhood of 20-30 KSI when both bending and membrane stresses are considered.

 

[butelja]

MickE,

I ran your particular case using the methodolgy in section 10.11 (Large deflection theory) of Roark & Young. Here are the results for fixed and held edge:

INPUTS:
P = 2.0 psi (W = 3,619 lbf)
D = 48 in
t = 0.125 in
E = 29*10^6 PSI
nu = 0.3

RESULTS:
Center Deflection = 0.373 in
Center Stress = 13,571 psi (6,716 psi bending; 6,855 psi membrane)
Edge Stress = 13,680 psi (10,330 psi bending; 3,433 psi membrane)
Radial Force = 418 lbf/in (for compression ring sizing)
I_ring > 0.266 in^4 (4:1 safety factor against buckling)
A_ring > 0.501 in^2 (20 KSI design stress in ring)

 

[MickE]

Many thanks for doing that butelja, those answers restore my confidence in Mr Roark

Looks like I'm going to have to track down a copy that I can borrow...certainly can't afford to buy one.

 

[BigAL]

Get a copy for yourself. Tell those close to you to get it for Christmas or birthday. It certainly beats getting ties or underwear.