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Offline EngMath  
#1 Posted : 23 January 2021 19:19:33(UTC)
EngMath


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Poland

Hi,

in my calculations using SMath Studio I can't get MPa unit for the last formula to display properly. Instead of MPa I still get m MPa (even though the value is correct). I input the unit as described in tutorials - click on the right placeholder, click apostrophe button and type MPa.

Here's what I get:



What should I do to get rid of this unnecessary m before MPa ?

I also attached the Smath Studio file to this post.

Thanks in advance for your help.


MPa.sm (7kb) downloaded 15 time(s).

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Offline overlord  
#2 Posted : 23 January 2021 19:59:47(UTC)
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Originally Posted by: EngMath Go to Quoted Post
Hi,

in my calculations using SMath Studio I can't get MPa unit for the last formula to display properly. Instead of MPa I still get m MPa (even though the value is correct). I input the unit as described in tutorials - click on the right placeholder, click apostrophe button and type MPa.

Here's what I get:

What should I do to get rid of this unnecessary m before MPa ?

I also attached the Smath Studio file to this post.

Thanks in advance for your help.
MPa.sm (7kb) downloaded 15 time(s).

Units and result are correct, according to your calculation.
You multiply 'Pascal' with 'm' and 'm^2', which is M=p*L*A.
This results as 'Pa*m^3', then you multiply it with 6 and divide to 'm^2', which is sigma=6*M/t^2.
'Pa*m^3/m^2' results 'm*Pa' as unit, this is correct. Nothing wrong here.
If you can provide the tutorial page it would be more helpful.

I don't think this is right thing to do but if it will work for you,
you can get rid of 'm' by dividing calculation with 'm' unit.

2021-01-23_19-57.png

Regards
Offline fedeghi  
#3 Posted : 23 January 2021 20:02:48(UTC)
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I can't be sure since I don't know the topic of your calculation precisely, but it doesn't seem to be a SMath bug for sure.
SMath is displaying m*MPa because that's what your equation are computing for real.

Looks like you are solving something which is related to bending moment of a (circular?) plate, isn't it?

I'm assuming this since 6*M/t^2 is the solution for sigma in Roark tables for some plate geometries and various boundary conditions or load conditions...
If this is the case, beware of "M" meaning: it should be a "moment per unit length"... not a moment, but a distributed moment!

In other words, M in 6*M/t^2 should be considered [N*m / m], or in any case "force*length times unit length"
This would (correctly and consistently) result in stress measured in MPa.

Hope this helpes.
Bye!
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Offline EngMath  
#4 Posted : 23 January 2021 20:09:39(UTC)
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Thanks for reply. The formulas before sigma are from one book and the one for sigma is from another source. It seems that the last equation should use M [(N*m)/m] while previous equations results in M [N*m]. Interestingly, the value of sigma is correct (I verified it with numerical simulation).

By the way, these formulas are for stress in rectangular shell (pressure vessel) subjected to internal pressure.
Offline overlord  
#5 Posted : 24 January 2021 05:09:40(UTC)
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Originally Posted by: EngMath Go to Quoted Post
Thanks for reply. The formulas before sigma are from one book and the one for sigma is from another source.
By the way, these formulas are for stress in rectangular shell (pressure vessel) subjected to internal pressure.

As far as I remember, that sigma formula is for circular vessels.
Rectangular pressure vessels have different formulas and design should be taken very seriously.
Since they have corners unlike circular tanks or vessels, they tend to break from there.
Could you please check [ASME Sec VIII Div 1, Appendix 13] and [EN13445-3, Section 15].

Regards

Edited by user 24 January 2021 12:01:41(UTC)  | Reason: Not specified

Offline fedeghi  
#6 Posted : 24 January 2021 12:11:34(UTC)
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Take Overlord suggestion seriously, since a mistake in defining a pressure vessel geometry and thickness can have serious "consequences".

I have to underline that also the formula in your file, that determines area "A", seems quite odd: you get as a result a negative area, this should be carefully checked because it isn't definitely what one would expect from an area Wonder

I think that this thread is giving you the chance to dig deeper into the proper equations to be used and their meaning, since by mixing equations from two different references there is some risk of using them in an unproper way.

ASME VIII div 1 or EN13445-3 section 15 will guide you through a more complex but safe design, where both corners and shell midspan will be checked against membrane and bending overstresses.
Concepts as corrosion allowance and joint efficiency (to correlate between adopted thickness and Non Destructive Test after fabrication) are also included there.
And you get requirements for allowable stresses as well, which is an important topic in the pressure vessel industry.

Adopting a proven design and construction code will be required in case your vessel needs to comply to some pressure vessel safety regulation (based on the country where the equipment will be installed, or any other contractual agreement); you might be required to CE stamp (EU), or U Stamp (US and many other countries too..), or have compliance with EAC regulations (Russia, KZ, Belarus), etc...
If this is the case, again, you better stick to a coherent pressure vessel construction code (as the main ones mentioned above by Overlord).

In case you want to dig into any other source\book for rectangular vessel design, you may want to check:
- Dennis Moss "pressure vessel design manual" (check every equation again... remarkable book in terms of tons informations, methods, and logic behind, but there is some mistake here and there)
- Bednar (cant' remember the exact title), with more details about the derivation of a formula, while Moss is mainly a design handbook
... and many others

In the end, I found some time ago an old book which is titled "Tubular Structures", including a very clear and clever way to determine wall thickness and stiffener dimensions for rectangular powder silos, bins and bunkers. It was a very interesting reading (besides typos, some equations needed some adjustment..) and, even if powder is not going to behave as hydrostatic pressure at all, the book gave me a useful insight on how to model properly the structure with a "by formula" approach.. finally it was a good way to understand better how rectangle section silos are checked inside Eurocode (EN 1991 for structure, i.e silos).

Bye
Offline EngMath  
#7 Posted : 24 January 2021 12:56:47(UTC)
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Thanks for tips and references, I’ll try to find these books.

The formulas for A and M are from Polish book about the design of chemical processing equipment. In the book there’s no equation for stress in shell when bending moment is known so I’ve searched for it somewhere else. Indeed the equation sigma=6M/t^2 is for thin-walled pipes but I thought that it will work in this case too. Apparently the units don’t match but somehow the results are correct - I get the same value with finite element analysis. And I don’t think that it’s just a coincidence. So it seems that the formula for bending moment is correct and stress equation is also fine but needs to be rearranged somehow.
Offline fedeghi  
#8 Posted : 24 January 2021 13:08:59(UTC)
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Can't comment any further with the available input informations.
For sure equations for thin pipes should not be used on rectangular vessels.

Beware of FEM, you can get awesome results or awesome colored pictures, and that's not the same Smile
Boundary conditions, choice for modeling and meshing (brick elements, tetrahedral, surface elements) will make the difference too.
Take care about classifying stresses properly (I would suggest ASME VIII div.2 app.5) in case you need to produce any official calculation report that makes use of FEM.


Finally, sorry for misleading informations about the mentioned books: I just checked both MOSS and BEDNAR, apparently I remembered wrong info... those two doesn't cover rectangular vessels.
Stick to ASME VIII div.1 or EN-13445 for now.
Tomorrow I check other books and in case I will send you a PM with a temporary link to download.
I can't send you the ASME Code itself, it is assigned to a company user ID and I cannot share that.

BYe

Offline fedeghi  
#9 Posted : 24 January 2021 13:15:07(UTC)
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Oh look, I found something in my home pc, there's a chapter about non cylindrical vessel in Jawad/Farr book (another good book by the way!)
I will share a link to download the book via PM, it is a OneDrive link and it will expire in 24 hours.

The book resumes ASME VIII div.1 approach and well, you will find "something" for sure (it is also rich with other topics).

PM on its way in a couple of minutes.

Bye!
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Offline overlord  
#10 Posted : 24 January 2021 13:18:05(UTC)
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Originally Posted by: fedeghi Go to Quoted Post
Oh look, I found something in my home pc, there's a chapter about non cylindrical vessel in Jawad/Farr book (another good book by the way!)
I will share a link to download the book via PM, it is a OneDrive link and it will expire in 24 hours.

The book resumes ASME VIII div.1 approach and well, you will find "something" for sure (it is also rich with other topics).

PM on its way in a couple of minutes.

Bye!

I don't have that particular book, it would be great if you could PM to me too.
Crazy
Regards
Offline EngMath  
#11 Posted : 24 January 2021 17:19:10(UTC)
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Thank you very much for all your help. Especially for sharing this book. It seems that I've found the answer there. The book says that bending stress in the middle of the long side of rectangular shell is given by the following formula:

sigma = ((M*t)/(2*((t^3)/12)))*(1/E)

If we assume the weld joint efficiency E=1 we get:

sigma = (6*M)/(t^2)

That's the same equation I used before. This explains why my results were correct.

By the way, the book describes really interesting approach - treating the cross-section of rectangular shell as a structural frame.
Offline overlord  
#12 Posted : 24 January 2021 17:21:54(UTC)
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Originally Posted by: EngMath Go to Quoted Post
I get the same value with finite element analysis. And I don’t think that it’s just a coincidence.

Probably coincidence or wrong mesh, or FEM calculation is selected as circular.

Originally Posted by: EngMath Go to Quoted Post
That's the same equation I used before. This explains why my results were correct.

The equation you used doesn't consider the corners.
I won't disagree if you will continue this approach but it is nontraditional.

Here you can use this sheet if no stiffener shall be used.
Otherwise you have to alter formulas according to standard you will use.

Regards

rectangular.sm (59kb) downloaded 23 time(s).
rectangular.pdf (144kb) downloaded 22 time(s).


Originally Posted by: EngMath Go to Quoted Post
If we assume the weld joint efficiency E=1 we get;

PS: Never take weld factor as 1, 0.85-0.9 for perfect shop weld or 0.7 for field weld.

Take a look at this example of Ansys for 50mm thickness;
it will give you an idea how much 0.1MPa(1 bar) can effect on rectangular shell.
https://www.irjet.net/archives/V6/i6/IRJET-V6I6747.pdf

Edited by user 24 January 2021 17:53:27(UTC)  | Reason: Not specified

Offline fedeghi  
#13 Posted : 24 January 2021 18:53:32(UTC)
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Originally Posted by: overlord Go to Quoted Post

Originally Posted by: EngMath Go to Quoted Post
That's the same equation I used before. This explains why my results were correct.

The equation you used doesn't consider the corners.
I won't disagree if you will continue this approach but it is nontraditional.


... and moreover, this kind of equation will require as input a "distributed" moment, not a moment (this brings us back to the first post of this thread).
Corner stress can be the driver in this kind of design, so be sure to derive the complete set of distributed moments (in A, B, C points as ASME names them) and then complete every step of ASME guideline.

Originally Posted by: overlord Go to Quoted Post

Originally Posted by: EngMath Go to Quoted Post
If we assume the weld joint efficiency E=1 we get;

PS: Never take weld factor as 1, 0.85-0.9 for perfect shop weld or 0.7 for field weld.


ASME and EN will classify the weld based on welding detail (butt weld, fillet single side, fillet both side, etc...) and amount of NDT, and define appropriate J.E. for every case/combination.
Joint efficiency E = 1 is typical only for butt weld with 100% x-ray test (or equivalent volumetric test).
This implies very good welding results, with no defects (or, to say it more precisely, residual defects which comply with the selected code acceptance criteria for each kind of defect).
Otherwise, butt weld with "spot" x-ray check uses E=0.85 (each code defines what "spot" stands for, in terms of amount of xray)
Butt weld with no check is further de-rated to E=0.7

This from a general point of view: then, going back to rectangular vessels, it is quite unusual to xray 100% a rectangular vessel as far as I know, since application with high pressure (where 100% xray may be motivated by real concerns) wouldn't typically use rectangular geometries (not statically efficient).
So, I don't know your geometry or application, but E=1 may be debatable and, in many cases, not economical unless you make use of very special base material.
Also in those cases, it may be not applicable: I have a 6m*2m*12m rectangular electrostatic precipitator where, even if MOC is SAFF 2205, it didn't make sense to heavily xray to reduce wall thickness.
First reason: the outer longitudinal and axial stiffeners (MOC ASTM 304) are taking the higher loads, so xray on the shell wouldn't have led to thinning of the SAFF plates.
Second reason: xray do cost too Wink

Bye

PS: please remember about any corrosion allowance, if it is not zero..
Offline overlord  
#14 Posted : 24 January 2021 21:32:02(UTC)
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Originally Posted by: fedeghi Go to Quoted Post
Otherwise, butt weld with "spot" x-ray check uses E=0.85 (each code defines what "spot" stands for, in terms of amount of xray). Butt weld with no check is further de-rated to E=0.7

There is welding coefficient table in both standard if I remember correct.
But to ensure safety I usually stick with 0.85, 0.7 if it is field weld. Even with NDT.
Materials I had designed were usually huge (4-5m diameter) so full X-Ray was not possible.
There is no obligation to lower the safety stress except material cost. Increment should be not allowed however.

Regards
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