PREX & TGS4 - Could you both please help with some Stress Linearization? 2

[AFG03082015]

Hello Prex, TGS4 and anyone else who may be able help!

From reading several other posts you both seem highly informed when it comes to stress linearization and was hoping you could help me out a little...

I am completing an FEA analysis to support an ASME VIII Div.1 vessel by completing the analysis to ASME VIII Div.2.

I have attached screenshots that show my 1/4 model and also the location of my SCL (The highlighted line in the centre of the three)

I have also attached the results file stating the six component stresses at each node's location along the SCL.

Would either of you be able to enlighten me as to how to proceed to calculate the Membrane Stress & Membrane + Bending Stress?

Any help would be greatly appreciated,

Thanks for your time...

 

[AFG03082015]

Results File Added

 

[TGS4]

In such a monoblock vessel, I would submit that linear elastic analysis is not appropriate.

If you are bound and determined to do so anyway, please be aware that I do not provide advice on SCLs or linearization - see 5.2.1.2, 5.2.1.3, and 5.2.1.4. And linearization is covered in Annex 5-A.

 

[NRP99]

As TGS4 suggested, thick vessels(monobloc vessel in your case) tend to behave differently than the thin vessels for which the stress linearization may lead to some wrong interpretation of results. I guess Monobloc vessels have nonuniform stress distribution.

Anyway you want to know stress liniearization in general please read this post-thread794-410119

Stress linearisation is very complicated subject and only experienced engineers could be able to do that.

Keep in mind the quote and suggestion from above thread from TGS4. He has very extensive experience in this field.

If you want my unvarnished opinion it's this: DO NOT use elastic stress analysis for demonstrating Protection Against Plastic Collapse and the subsequent considerations of stress linearization and classification.

 

[prex]

AFG03082015 is just asking how to do the linearization of the stress results, the use he will do of it is up to him.
Now, as you have the stresses at constant intervals along the SCL, the membrane stress compnents will be:
S[sub]mx[/sub]=(ΣS[sub]x[/sub]-(S[sub]x[/sub][0]+S[sub]x[/sub][36])/2)/36
and the same for the other 5 components. The only trick contained in the above formula is that the end values need be counted only in half. A more general formula, giving of course the same result and explaining that trick, is:
S[sub]mx[/sub]=(Σ[sub]0[/sub][sup]36[/sup](S[sub]x[/sub]+S[sub]x[/sub][i+1])/2)/36
or in a more general form (dx BTW is wrong in your table), valid also with non constant dx (with t=X[n]-X[0]):
S[sub]mx[/sub]=(Σ[sub]0[/sub][sup]n[/sup](S[sub]x[/sub]+S[sub]x[/sub][i+1])/2*dx)/t
Now for bending, according to App.5.A, the stress S[sub]x[/sub] is not considered, and the same for the shear stresses (as your model is axisymmetric, there can't be torsion of the SCL: do not ask me why it is done so, I can't understand, personally I would check whether considering all the stress components gives a more conservative result, as I would expect BTW).
So, using the fact that dx is constant through the thickness, for S[sub]y[/sub] (the same for S[sub]z[/sub]):
S[sub]by[/sub]=(Σ[sub]0[/sub][sup]36[/sup](S[sub]y[/sub]+S[sub]y[/sub][i+1])/2*(18-i))*6/36[sup]2[/sup]
This is for X=0, sign changed for X=t.
If dx was non constant:
S[sub]by[/sub]=(Σ[sub]0[/sub][sup]n[/sup](S[sub]y[/sub]+S[sub]y[/sub][i+1])/2*dx*(t/2-X))*6/t[sup]2[/sup]
Now you have all the membrane and (summing up at both ends of the SCL) the membrane+bending stress components, and you calculate the principal stresses and the equivalent ones at the ends of the SCL.

prex
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[AFG03082015]

Thanks All For Your Input, Advice & Help.

Prex Are you saying that dx would be 0.5mm at each node? and not a running addition at each node (0.5, 1.0, 1.5 etc...) as it currently is?

 

[prex]

Yes, correct. More exactly dx is 0.5 mm for each increment from one node to the following one: you have 37 nodes and only 36 dx increments.

prex
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[AFG03082015]

Thanks for your help Prex,

Once I have calculated Smx, Syx, Szx Etc... I use these as stress tensors to calculate the membrane stress?

I've got a litte confused because I found the following online that seems slightly different to the equation you stated before (Smx=(ΣSx-(Sx[0]+Sx[36])/2)/36)

The distance between each node along the SCL is the dimension dx. dx usually varies between nodes depending on how the model is meshed. Sum (integrate) dx*Sn over the length of the SCL from i to j. Repeat for each of the other 5 local stress components. This produces 6 stress components (Snm, Stm, Shm, Tntm, Tnhm, Thtm). These 6 components are either fed into a Tresca routine (ASME VIII-2 2006 edition) or von Mises (ASME VIII-2 2007 edition). The output is a single value for the membrane stress.

When the items are fed through a Von Mises routine, is the membrane stress the resultant Von Mises stress or the greatest difference between the principal stresses?

 

[prex]

You need to understand that the local value of stress to be used for the integration is not the nodal one, instead it is the value pertaining to each subsegment of the SCL. So the local value is (Sx+Sx[i+1])/2: with this is mind you'll be able to understand the formulae (BTW I made some mistakes with the indices, but if you understand the formulae, you'll be able to figure out what to do).
And I don't know what exactly is a Von Mises routine, but from a Von Mises routine you should get a Von Mises stress, and from a Tresca routine a Tresca stress?

prex
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[jsqwerty]

Hello Prex / AFG03082015,
I am working on something similar myself and have come unstuck in calculating the bending stress,
Did you manage to calculate it AFG03082015? If not could you talk me through it in more details Prex? In the ASME Div. 2 code (5-A.4.1.2) it is stated as...

Stress{ij,b} = 6/t^2 Integral Stress{ij} (t/2-x)

Is this the same as what you stated Prex?
Thanks

 

[prex]

Yes, nearly.
Assuming:
n as the number of subsegments in the SCL
X (i=0..n) the linear distance of the subsegment's ends from one end of the SCL
dx=X-X[i-1] (i=1..n) the length of each subsegment
t=X[n]-X[0] the length of the SCL
then
S[sub]by[/sub]=(Σ[sub]1[/sub][sup]n[/sup](S[sub]y[/sub]+S[sub]y[/sub][i-1])/2*dx*(t/2-(X+X[i-1])/2))*6/t[sup]2[/sup]

prex
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[TGS4]

To ensure compliance with 5-A.4.1.2, it should be noted that

Bending stresses are calculated only for the local hoop and meridional (normal) component stresses, and not for the local component stress parallel to the SCL or in-plane shear stress.

So, when calculating a von Mises membrane-plus-bending stress, so would use the membrane-plus-bending for the hoop and meridional components ONLY and then the membrane stresses for the other normal components (through-thickness) and the shears.

 

[AFG03082015]

TGS4,

are you saying the following would be my six stress tensors at end i of my SCL?

Snm+Snb, Stm, Shm+Shb, Tntm, Tnhm+Tnhb, Thtm

where
n= Normal
t= Tangential
h= Hoop
M= Membrane
b= Bending

 

[TGS4]

No.

You need to orient your SCL so that there is a local coordinate system aligned with a local hoop, meridional, and through-thickness direction. Then, your membrane-plus-bending tensor is:

t = through-thickness
h = hoop
m = meridional

σ_t-membrane   τ_th-membrane               τ_tm-membrane
τ_th-membrane  σ_h-membrane + σ_h-bending  τ_hm-membrane
τ_tm-membrane  τ_hm-membrane               σ_m-membrane + σ_m-bending

 

[Soave]

Dear TGS4, Prex and all

I'm studying for my graduate thesis about ASME Calculation Section VIII Div.2 and I appreciated and learned a lot with yours posts and helps.
Now I have elaborated my spreadsheet for linerarization of stress.
I have two questions:
1- How Can I obtain the bending stress for each node?
2- My calculations are correct?

I annexed my xls file and my test calculation with ansys.


Thanks for your attention and time

http://files.engineering.com/getfil...-18013a26c23f&file=linearization_engtips.rar

http://files.engineering.com/getfil...5-7d8fa5cdd92f&file=linearization_engtips.xls

 

[AFG03082015]

Hi Soave,

I'm a little confused by the values of L at nodes 1-4.

The nodes should be lined up sequentially along the stress classification line, therefore the value of L should increase in from node 0 up until node 48(just like that seen between node 5 and 48).

Between node 1-4 your value of L appears to increase by the value of 20.834 to 83.336 before dropping back down to 10.417

 

[Soave]

Hi, AFG0382015

Sorry for this wrong, the value at nodes 1-4 shall be divided by 10... So, Node: 1 = 2,0834; 2 = 4,1668; 3 = 6,2502; 4 = 8,336

Thanks

 

[TGS4]

Bending stress is not calculated at each node. It is the equivalent linearized stress distribution across a section that produces the same net forces and moments as the actual distribution.

 

[Soave]

TGS4,

About bending stress, How ANSYS Can calculated bending stress at each node?

And my bending stress calculation are correct?

http://files.engineering.com/getfil...c=212727627.1.1475512696363&__hsfp=3879665338

Thank you again.

 

[TGS4]

ANSYS doesn't calculate the bending at each node. It linearizes the component stresses along a path, and reports the average (membrane), bending, and peak of each component stresses along the path. Membrane and bending (and peak) are properties of the path, not a specific node.

In your calculation of membrane and bending of the component stresses, you need to consider the actual length fo each location, not the "n" of each location.