Help on ASME-STS1 in regards to steel stack design. 1

[hoots711]

Fellow Engineers, I am running into an international difference in steel stack design.
Per the ASME, STS-1, Section 4.4, Eq 4-6, allowable stress cases are only valid when equation 4.6 is satisfied.

4-6 states that t/D <= 10Fy/E

basically, it limits your thickness based on diameter as Fy and E are going to be constants in my design.

The issue is in design of something like a 1.2M Diameter steel stack that is 40M high (free-standing). Per the above formula/condition, i cant use plate thicker than 15-16mm (after corrosion @ FY=34ksi & E=29Mill) and with that parameter i cannot design the stack to pass the allowable stress cases.

Our foreign counterparts design the above stack with most thickness around or over 20mm over the lower half of the stack.

Before i can debate which method is acceptable i have to understand why on Earth the ASME would be limiting thickness based on D in steel stacks.

The only thing i can possibly think of is that the rolling of thicker plate to smaller diameters could compromise the steel??? If this was the case i would think there would be thorough documentation on the subject. This also apears to not be the case as FY is on the right side of the formula (stronger steel would allow larger thicknesses... not likely a rolling issue)

Im striking out finding anything explaining equation 4-6. Can anyone offer any advice or interpretation?

I guess the basic question is: What do you do when equation 4-6 is not satisfied?

Am i mis-reading something here? I have brough this issue to 3 or 4 of our senior US structural engineers and they have all be puzzled by this (mostly by the fact that they havnt noticed this stipulation in previous designs..)

Why on earth would ASME limit MAXIMUM plate thickness by diameter??
Thanks and Best regards.

 

[JStephen]

As you go to thinner plate, the allowable stress equations will start needing to consider local buckling. Are you sure that equation is not just giving you a limitation on when one stress range or equation is applicable, rather than an absolute maximum thickness?

 

[hoots711]

Yes JStephen.

Section 4.4 of the ASME STS1 is for "Allowable stresses"
The first sentance in 4.4 is as follows: "The following formulas for determining allowable stresses are applicable for circular stacks and liners provided that eq (4-6) is satisfied.

They are as follows
Case 4.4.1 (Longitudinal Compression)
Case 4.4.2 (Longitudinal Compression and Bending Combination)
Case 4.4.3 (Circumferential Stress)
Case 4.4.4 (Combined Longitudinal and Circumferential Compressives Stress)

4.4.5 moves into stiffeners.

My company desings (and has designed) steel stacks by the above 4 cases for years.

In front of 4.4.1 is the following "For Stacks and liners meeting the requirements of eq. (4-6), the following four load cases must be satisfied."

Again, what do you do when 4.6 is not satisfied? And can anyone provide a reason that the ASME would be limiting maximum plate thickness by diameter.

The stack in question is outside of our "normal design" in being so tall and slender, but i have a hard time just going with our forign design while not understanding this ASME stipulation.

Thanks

 

[dhengr]

hoots711:
The t/D ratio is most certainly a plate/shape buckling criteria, or limit, probably somewhat subjectively (arbitrarily, the 10?) set, by ASME. D and t are intimately related in this type of buckling problem. Theory might indicate slightly different limiting values of t/D for different load combinations. What circumferential stress do you expect? Do they mean a torsional loading or a pressure one way or another, or dynamic loads? Do you need strakes on this stack?

I don’t have ASME, STS-1, so I don’t know exactly how they treat each of the load cases. Certainly, the design stresses for each case must be less than Fy of 34ksi, and are a function of t, D, rad. or dia., length and E. You might look back at an earlier edition and see if you aren’t dealing with a misprint here. The equation you show as (4-6) seems bass acwards to me, the ‘less than’ symbol should be pointing the other way. Given a ‘D’, I would expect ‘t’ to have to increase as the design stress increased to guard against buckling. I’d look at plate, plate arch, pipe, pipe column, shells and the like, for discussion on buckling stress limitations. My first look would be a number of Timoshenko books or Roark. I’ll try to look a little deeper later, if I don’t see any more activity on your thread.

 

[hoots711]

Thanks dhengr, I can assure you the formula of t/d <= 10FY/e is not miscopied. As far as it being a misprint, i have 2006, 2003, 2000 and an earlier version at my office and they all state the same exact formula.

Your back-asswards thoughts are exactly what the senior structural guys i talked to thought, but reversing the signs would make less sense as it would be setting a required minimum thickness of 14-16mm...

Keep in mind, this equation has no relation to loads, height, wind, eq, ect. It is purely D to t. (which is why it makes no sense to me!)

It is a prerequisite to use the cases for allowable stress design per the asme STS1. Those cases solve for a assumed t with trial and error to get your correct thicknesses as the results. (And here is where the above mentioned Loads, height, wind, eq, ice, etc all come into play)

We have the cases automated with excel and STAAD and I can see that the thickness provided by my European colleagues (of 20-22mm) are required at the base for this stack to pass design. The problem is... Equation 4-6 says i cant go above 16mm when using the design cases that result in a requirement of 22mm...

Confused yet? I and many other in my office are. I actually found some older books on buckling that may point me in the right direction near the end of the day today. Ill take a look at them tomorrow.

Thanks for your reply.

 

[paddingtongreen]

I have a couple of ideas but they are no more than speculation, however, do check with ASME. There should have been discussion of the standard in their proceedings before it was adopted, and there might well be a code case on this very question.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[dhengr]

hoots711:
I wasn’t saying you made a typo., but I still think your second para. is wrong, us old guy’s agree ASME has it bass acwords. I have never seen a situation where thickening a plate made a buckling problem worse, at a given stress level, unless it changes the buckling from one mode to another, and their cases are based on the former. Setting a max. thickness of 16mm, just makes no sense. You say your own calcs. confirm that you need 22mm from the stress standpoint. Although, Fy = 55ksi, and t = 22mm would apparently make everything just fine. Are they trying to force you to use a steel with an Fy way higher than the design stresses dictated by buckling with their eq. 4-6?

I think you should call ASME and find out who’s on the committee that works on that particular code. Get in touch with a couple of them, and find out what their take is on this conundrum, because you and all us old guys don’t understand the rationale, and we want to know. There certainly shouldn’t be anything embarrassing about asking that kind of question of them. Now days, you can’t trace three quarters of the stuff in the codes, their formulas, factors, limitations, etc., back to a simple, logical, apparent, or theoretical and tested basis.

 

[dhengr]

I meant to say or add, that in your second para. you have your thinking bass ackwards now too. Because, what you normally want to do is set a min. t, to prevent buckling. But, that normally has more to do with a working stress level, dia., E, etc. at which buckling may occur, than it does with Fy. Except, of course, that your allowable working stress is a percentage of Fy, and then throw in a few load factors or reduction factors, and you get a ten in the eq. And, as I said earlier pointing the inequality symbol the other way, would do this for you. A smaller dia. has a greater plate stiffening effect so you could get by with a thinner pl. as long as your stresses still checked.

paddington, speculate away, he will need these types of ideas as he goes to ASME. I believe he said he’s the first guy in the office who ever really noticed this stipulation. It certainly isn’t a rolling issue in the 1.2M dia. and 30-40mm thick range.

 

[hoots711]

Ill write a little more later (as it has been a hectic day), but perhaps these findings may spark some of your previous knowledge on the subject. In reviewing Troitsky (Tubular Steel Structures - Theory and Design) - Section 2.6 (Allowable Stresses) i find the following. (In Regards to local buckling)

"At Yield point for mild steel FY =36 KSI the limits are:"
92< D/t < 361

It also references that the AISI recommends eq 2-10 that states F_allow = [662/(D/t)] + .399FY

This is interesting to me bc it appears that when D/t reaches the lower limit (92) the math shows it is very close to .6FY

662/92 + .399(36) = 21.559 KSI
.6 (36) = 21.6 KSI

(The math shows that as D/t lowers past 92, the allowable grows past .6Fy)

For obtaining valid D/t ratios for other strengths they offer the bounds as
3,300/FY < D/t < 13,000/Fy

I know i had seen the 3,300/FY before and that lead me to the AISC (9th) Table B5.1 (limiting width thickness Ratios)

They match 3,300/FY as the "limiting width thickness Ratio" and for circular hollow sections, width.t.r is defined as D/t

I must run, and am still working through this with about 4 books open on my desk. I do not know if it means .6fy is acceptable as my allowable in this case... One issue with that is the ASME cases we use to check thickness.

Something in the bounds of 92 < D/t < 361 but close to the bottom calculates (im my particular case) out to an allowable of around 10 KSI (per the asme stress checks). Increasing plate 2mm to move beneath the lower bound would give me .6(34) = 20.4 KSi allow and that difference doesnt seem logical.

Must go, but perhaps this will spark your thoughts.

if nothing else, this is interesting and educational.

Thanks!
BR
Hoots

 

[dhengr]

Hoots:
Troitsky is saying .6Fy is your max. design stress in any case. AISC’s .67Fy assumes compact sections and all that good stuff, which you certainly don’t have here. Your allowable stress to prevent buckling will be less than .6Fy, and more so as t/D gets smaller, and you approach a flat plate. Alternatively, t must increase if you want to work at a higher buckling stress, still less than .6Fy. Again, t/D relates, in some fashion, to the stiffening effect of a smaller dia. wrt the thickness of the pl.

If you forget about the less-than symbol for the moment; ASME’s eq. 4.6 [ t/D = 10Fy/E ] looks an awful lot like Troitsky’s eq. (2.13) [sigma cr = (.6 gamma) E (t/R) ]. And, I suspect that your FY should really be F gamma. You see this very typo. on the abscissa of Troitsky’s fig. 2.5. Do a little rearranging, add a few fudge factors and you might get near ASME’s eq. I still think this would be a real interesting question to ask ASME.

Now I have to stop and read chap. 2, look at a few of my other ref’s. again, and see if what I said makes any sense.

 

[hokie66]

A lot of variance in opinion. I agree that the ASME equation makes little sense.

Found this in Gaylord and Gaylord, 1968 edition. Attributed to Ernest A Dockstader, retired Chief Structural Engineer, Stone & Webster...

"For A36 steel, the allowable stresses due to dead load plus wind or earthquake are 12000 psi for d/t <or= 200, and 2400000/(d/t) for 200 < d/t < 600".

 

[hoots711]

Im starting to wonder if this is an ASME issue as more sources are showing similar results/equations:

My head is spinning..

Gaylord & Gaylord, Struct. Eng. Handbook (3rd) -1990

page 30-8 (Chimneys)
Steel stacks
9. Allowable Stresses: The allowable longitudinal compressive stresses due to vertical load and and bending moment can be determined by:

(14): F =XY

X is broken into 3 ranges
0 < t/R < 8F_gamma/E
8F_gamma/E < t/R < 20Fy/E
t/r > 20Fy/E

at 1180mm D, 590mm R, 20mm t, im in the 3rd case

20/590 > 20(34)/29000
.033 > .023
This lists X as .5Fy

Y is the exact same calculation as used in the STS-1 cases.
Im my calcs it is .43.

This is tellig me that my allow is .43(.5)(fy) or .22Fy
7.3 KSI ???

This makes no sense to me. When you reach the boundry (in the original 4-6 ASME eq) your F_allow drops as your plate thickness increases???

In rough numbers at 1180mm D, the thickness / allow are:

6mm - 3746 PSI
8mm - 8369 PSI
10mm - 9407 PSI
12mm - 10180 PSI
15mm - 10843 PSI
20mm - 10623 PSI
24mm - 9254 PSI
30mm - 5213 PSI
(Per the ASME forumlas, allowables would be lower for 20-30mm plate using the above Gaylord calcs.)

the curve peaks right around 16mm which is exactly when eq 4-6 does not pass.

Does this make sense? at a constant diameter, how can increasing plate thickness reduce allowable long, compressive stress?


Dhengr, im still looking for some of your other references and the number of my coworkers who are puzzled or interested by this is growing daily...

We are starting to have some real concerns with our international designs that do not address this issue.

 

[dhengr]

Hoots:
Your losing me, because I don’t have the ASME spec. your keep referring too. Download a few pages of that so I can see what they are doing, and what their formulas and load cases look like. I do have the same Gaylord & Gaylord, Struct. Engr. Handbook that hokie cited, and I see that I’ve actually used that in design, about 25 years ago. Your 3rd. ed. will obviously have later research and testing as its basis, download a few pages of that too. My other ref’s. had nothing practical, they all had math that would blow your mind and got me to Troitshy’s figs. 2.2 & 2.4 in theory. I was hoping to see a chart or graph of R/t vs. bucking stress or some such.

I have seen the form of F = XY, which I think came out of the Univ. of Ill. in the early 1930s. But, I doubt that it’s the same one you are looking at, which I understand you see in Gaylord’s book. I don’t know what your gamma is, I can’t see it from here, but again we see [8F(gamma)/E ] as a limit; 8 instead of 10 now.

This particular buckling problem is so susceptible to any imperfections in the shape and loadings during testing, that no one has been able to do a test which approaches the theory very well. Thus, most of the design approaches are based on empirical eqs., with theory in mind, but to match the testing more closely; with fudge factors, etc. based on experience, added. I don’t think testing ever comes up to the theoretical buckling strength. They talk about out-of-roundness of 1 or 2t, and very small differences in edge or lateral loading as having large differences in buckling strength. Thus the large difference in empirical approaches, different guy’s best guesses.

The extremes of your buckling problem are: (A) flat pl., on edge and edge loaded, a very large D/t ratio; (B) a 12" round pipe with t = .5", acting as a canti. col.; and we think we have a fair handle on these two extremes. Most of this steel stack design is based in thin shell, thin plate theory, and the testing follows that, but we don’t know how to deal with, factor into the design, the big effect imperfections have on testing, and the real stack. There will be a blip in our design criteria, and in our design approach as we move from one theoretical regime, or buckling regime to another. But,t = 16mm and D = 1200mm, seems an arbitrary point to me. Again, armed with what you have I think you should talk with ASME, and/or the foreign agency.

 

[hoots711]

Sorry for the confusion dhengr, see attached scan of the asme-sts1. Section 4.4 is what we have been discussing (in regards to design when eq 4-6 is not satisfied)

Thanks again for the help and opinions. I think you are correct about contacting ASME about this, but im still hoping im missing something obvious here..

BR
Hoots

 

[hoots711]

Also, see attache Gaylord and Gaylord Steel Stacks.

Notice the reference and similarities to the ASME.

 

[hoots711]

2nd attempt at the gaylord scan. The first did not seem to post.

BR

 

[TERIO]

I am not familiar with this standard but the way I read section 4.4 it seems (although the wording is confusing) like if you have a "thin" shell by the t/D ratio being less than 10Fy/E you must meet those requirements. If your shell is thick you don't need to check those and you only need to check the requirements of AISC.

 

[hoots711]

Terio, i also had that thought... But what requirements do you check in the AISC? (its not like they have a steel stack design section)

Also, if you get a chance to review the Gaylord scan, it appears that they do have formulas for all cases of D to T.. and as i explained above, it would limit me to something like .22FY as my allowable when using thicker steel...How can thicker steel have a lower allowable at a constant D?

I feel like i have to be missing something here!

 

[hoots711]

Thank you all for the help and suggestions, I “believe” I have worked out a solution to the question of “what do you do when equation 4-6 is not satisfied?”

Please let me know if the following reasoning makes sense


When equation 4-6 is not satisfied design as follows.

1. Allowable Compression (per AISC 9th)
Table B5.1 states that the limiting width thickness ratio is D/t < 3300/FY for a compact section.
when EQ 4-6 is not satisfied, D/T is within the limit width thickness ratio of a compact section.

In checking for compression, eq (E2-2) AISC 9th is now valid

Fa = (12*PI*E) / 23*(Kl/r)^2

This value will be very close to the allowable yield stress (in regards to compression) from table C-36 which deals with slenderness (Kl/r)

These numbers are also close (but more conservative) than the above gaylord calculation of Fa =XY (The difference could be in safety factors as well as gaylord uses a factor of 2)

So now P/A < F_allow is satisfied for compression. (In reviewing the issue, compression was never the problem… it was the combined compression and moment)

2. Long. Compression and Bending

F.3 Allowable stress: bending of Box Members, Rect. Tubes and Circ. Tubes (AISC)
Eq (F3-1)
Fb=.66fy
This does make sense in reviewing the asme code. (attached a few posts up)
Ks becomes 0 (or very close to it) when eq4-6 is not satisfied because at the lower bound it becomes a negative number. In this case, at 1180mm D, a thickness of 13.8mm is the cutoff for eq 4-6.

Solving for Sbl in the asme (allowable combined long. Comp. and bending stress) we are directed back to eq 4-9. Sbl = Fy (1-.3Ks)Y / F.S. (Y becomes 1 as noted in case 4.4.2, F.S = 1.5 per table 4.4.6)

We are left with Fy*(1-0)*1 / 1.5 or 34/1.5 = 22.66 KSi

.66fy = 22.44 Ksi.

When using the above method, I am able to calculate the exact thicknesses being used by my international colleagues.

I have run this by my senior guys and in the local office, and while a few want to think about it for a while (I know… imagine that!), there has not been any alarms set off by this method.

Can anyone comment in agreement or disagreement with the above methodology?

Thanks and BR
-Hoots

P.S. – I kept having a hard time understanding lower allowables on thicker steel, but once I started thinking of the stack as a “slender column” it was like a bell went off in my head.

 

[hoots711]

I meant to add at the end of the bending calcs there is now no problem in satisfying the asme equation of P/A + MD/DI_section < Sbl when we use .66fy as the allow (or even .6 fy which we may do to be conservative_