Quadratic Equation of Vessel Thickness

Chapter 12 Solution Manual - Engineering Mechanics ...

Good answer. I was thinking E should be in the equation. I would warn OP to beware of equations of unknown origin - not only can you get thrown off by the units, but you don't know the limitations and range of use.

origin is in question for the layman like myself. codemaster seems to be familiar with it.

I am thankful to all experts who have tried to help me out.

equation : t^2 - (p*D/2*f)t - (1.5*U)p*D^2*(1-0.2*D/L)/[200*f] = 0

if we consider the said equation as a quadratic a*x^2+b*x+c=0

then a = 1; b = -(p*D/2*f); c=-(1.5*U)p*D^2*(1-0.2*D/L)/(200*f)

1. I do not know the origin of this equation but it has something to do with resistance against elastic failure and vessel under external pressure.

2. for the above formula, I have following data to offer

p = vacuum = 0.1 N/mm^2,

f = 70 MN, D = 4000 mm,

L = 7002mm,

out-of-roundness =1.5%,

modulus of elasticiy=1.67*10^5 MN/m^2.

I can derive t using ASME and putting that 't' in above equation back calculated U. But could not correlate U with available data?

3. Based on above, does U has something to do with out-of-roundness or ovality ?

3. As suggested by Codemaster (most sensible, simple and short answer) if I take U = 0.26 or 0.30, thickness comes out to be 19mm using above equation while p=KE(t/do)^m gives me t in the range of as 216 to 319 mm? Am I making some gross mistake ?

Thanks once again to all good souls

http://www.engineersedge.com/material_science/pressure_vessel_external_pressure_calculations_9857.htm

thanks for your kind input..horace40

I could found it out (actually I am quite steadfast at what I want).

U is out of roundness. This equation is as per German Code AD-Merkblaetter.

If U is given as 1.5% it is to be substituted as such i.e. as 1.5 and not as fraction.

Thanks everyone for taking interest.

Hope this forum has a brand new equation for checking safe external pressure against plastic deformation.

How is E accounted for? Is the equation for steel vessels only?

When you say in #5 "I can derive t using ASME" do you mean the formula for thickness for internal pressure t = P*D/(2*f)? That corresponds to U = 0, as it's given by putting U = 0 in your original formula. What pressure, 0.1 N/mm²?

I'm sceptical about the formula, as U appears to be arbitrary, and if you pick U = 0 the thickness comes out far too small (same thickness as for internal pressure). Does the German Code suggest figures for U? Does it recommend a safety factor on calculated thickness?

From your figures for f and E it sounds like aluminium, or perhaps steel at elevated temperature.

Using your formula with U = 1.5, I get t 18mm for external pressure 0.116 N/mm². Formula from Roark is buckling pressure

P = 1.6*E*t²/(L*D)*((1 - ν²)^-3*(2*t/D)²)^0.25

but I wouldn't say it was brand new! Make sure you get the brackets in the right place! I'd like to paste in the formula from Mathcad but never got the hang.

This yields P = 0.31 N/mm², but a safety factor needed. For structural steelwork BS 449 applies a SF ~ 3 to loads calculated from Euler buckling, so using a similar SF here gives pretty good agreement. But that might be just a coincidence, as other have agreed I'd expect to see E in the formula.

This is called peer's review...lol....Thanks for grilling me...Here are explanations

1. U = Out of roundness and German code specifies 1.5% for brand new vessel

2. The equation is a disguise. It's actual purpose is to determine safe pressure against plastic deformation which I came to know afterward. So if you rearrange the formula for getting p is would be

p = 2*f*(t/D)* 100(t/D) / [100(t/D)+1.5*U*(1-0.2*D/L)]

3. The equation could be used to verify the adequacy of thickness calculated from other equation which could be p=KE(t/D)^m which is nothing but allowable pressure considering elastic deformation.

4. p = 0.1 N/mm^2 means 1 Bar external pressure

5. if the "Quadratic disguised equation" mentioned in step 2 gives answer of p greater than 0.1, this means the vessel with the choosen/calculaated thickness will resist elastic as well as plastic deformation.

Hope I am thorough...Thanks to all gurus once again for your interest.

Regards,

Viren

Some points not entirely clear.

There is a formula in Roark for elastic deformation of a very long cylinder, p = 2*E*(t/D)³ (ignoring a factor (1 - ν²) which is fairly close to 1), same as your p = KE(t/D)^m if K = 2 and m = 3. You haven't given K and m, but using those figures gives p = 0.3bar, which is not enough for full vacuum. Your post implies to calculate it both ways and presumably choose the worst case for design thickness. If so, t = 27mm needed to give p = 1bar. If L is shorter t comes down (at least according to Roark) as already discussed, but you've only mentioned p = KE(t/D)^m.

Your formula p = 2*f*(t/D)* 100(t/D) / [100(t/D)+1.5*U*(1-0.2*D/L)] has aspects of plastic and elastic deformation. It's somewhat analogous to the Rankine-Gordon formula for struts in compression, which for short struts reduces to direct compression, limited by the material compressive strength, and for long ones to the buckling formula, limited by the elastic modulus.

Thanks dear Codemaster for your deep interest !

1. You are absolutely correct about plastic and elastic deformation aspects of the formula.

2. I have not provided value of K,E,m which are as follows : K = 0.4809, E = 1.67*10^5 N/mm^2, m = 2.48. This will give thickness of about 16.726 mm for p = 0.10132 N/mm^2 and d=4000 mm. Substituting t=16.726 mm in that "Quadratic formula" gives pressure as roughly 0.1016N/mm2 meaning that t is just matching the requirement and 18 mm should be considered for safer design.

3.Idea is to verify the both aspects of deformation - elastic as well as plastic and take on safest bet.

4. Rankinne-Gordon formula analogy is ok and quite similar to the above formula.

Thanks once again for your deep interest.