Code stress for sustained case b31.3

Good day.
I am now on study to verify CaesarII version 2014 build140416 and have a record of it in our company, for audit purpose and the like.
I have a question regarding CaesarII 2014.
from ASME B31.3 2012 edition,
SL=((|Sa| + Sb)2 + (2St)2)1/2
but from CS2 quick reference guide
SL = [(Ia (Slp + |Fax / Ap|) +Sb)2 + 4St2]1/2

If SIF is 1.0, Code Stress is correct..
but if SIF is not equal to 1.0, result from CS2 is lower than from I computed based from ASME B31.3 formula.

I also read from CS2 help command that Ia could be taken also equal to Io.
but still Different result from manual computation to result of code stress from caesarII.

Could anyone clarify.

Thanks in Advance

In B31.3 -2012:
Sa=Ia(Fa/Ap) and Fa is axial load due to pressure and weight
In CAESAR II Fa due to pressure is P*Ain and Fa due to weight is Fax, so:
Fa=P*Ain+Fax
Sa, then, is equal to Ia((P*Ain+Fax)/Ap)
with P*Ain/Ap= Slp,
Sa=Ia(Slp+Fax/Ap)

and CAESAR II uses absolute value on axial stress due to weight

Sir Dave Diehl,
thanks for the reply and I Agree with your statement.

My question is Code Stress in Caesar Result
At node 30-40, Code stress is consistent with the formula because maybe Ia is taken as 1.0
but at node 25-29, Code stress is different is I used the formula taken Ia as 1.0.
I read in CaesarII help function Ia can be taken as same as Io or 1.0 whichever is greater. but even I used that statement in the formula, value of what I computed is greater that from Caesar result of code stress.

I believe if SIF is not 1.0, Code stress should be greater not lesser, please confirm.




Node Axial Bending Torsion Hoop Max Stress SIF(i) SIF(o) COde Stress Allowable Ratio Piping Code

25 5338.3 5851.1 -554.6 12286.0 13353.1 1.267 1.056 9789.6 132116.7 7.4 B31.3
29 5640.1 2847.5 976.2 12286.0 13024.7 1.267 1.056 8017.1 132116.7 6.1 B31.3

29 5640.1 2847.5 -976.2 12286.0 13072.4 1.267 1.056 8017.1 132116.7 6.1 B31.3
30 6133.4 2508.7 886.8 12286.0 12996.0 1.267 1.056 8208.8 132116.7 6.2 B31.3

30 6133.4 1566.8 -886.8 12286.0 12959.1 1.000 1.000 7901.8 132116.7 6.0 B31.3
40 6133.4 1596.7 886.8 12286.0 12964.4 1.000 1.000 7930.9 132116.7 6.0 B31.3

Sir Dave,
I'm sorry, Im not yet verifying version2014.
im currently doing 2013 version.
here is the formula in the CII quick reference guide.
1. why Ia is multiplied twice to Fa/Ap?

SL = [(Ia |Slp + Sa| +Sb)2 + (2St)2]1/2
Where: Slp = PDi2/(Do2 - Di2)
Sa = IaFa/Ap
Where: Ia is the longitudinal force index

2. does Ia should not be less than 1.0?
because if I assumed Ia to be 1.0 my manual computation of code stress should be lesser than from result in Caesar, but its the opposite.

rileykyrra,

check this .

SL = [(|Sa|+Sb)2 + (2St)2]1/2

where

Sa = Ia(Fa/Ap)

=>Sa = Ia |Slp +(Fax/Ap)|

Ia is multiplied once only not twice.

I agree.
and I tried this formula also in the result of CaesarII.assuming Ia is 1.0, so I assumed that what will I computed manually will be less than code stress in the result of CII, but what I computed manually is greater.

sample

for node 40
Sa=6133.4
Sb=1596.8
St=886.8
Code stress from CII=7930.9

asumming Ia=1.0
SL= (Ia(Sa+Sb)2+(2St)2)1/2
SL=7930.9 = code stress from CII

but for node 25
Sa=5338.3
Sb=5851.1
St=554.6
Code Stress from CII=9789.6

SL= (Ia(Sa+Sb)2+(2St)2)1/2
assuming Ia=1.0 (or whichever is greater than 0.75SIFo, from CII help box)
SL=11443.3
thus SL computed manually is greater than from code stress from CII.
strange answer, because i think Ia=1.0 is the least value of COde stress of sustained stress..

Please clarify.thanks

I found how CaesarII computed for Code stress is SIF is not equal to 1.0 but I doubt the answer.
sample
for node 25
Sa=5338.3
Sb=5851.1
St=554.6
Code Stress from CII=9789.6

SL= ((Sa+0.75Sb)2+(2St)2)1/2
SL=9789.6, thus it is consistent with Code stress for sustained case in CII output report.

but I find it strange, because SIF should intensify the stress not reduced?

Is Bending stress in CII output report already accompanied the SIF?
and if Sustained Case only 0.75 of Bending stress is taken into account.?

Is this something to do with Set Sustained Multiplier function in the CaesarII configuration editor box.?
Because ASME B31.3 permit 0.75 Sustained Multiplier?

The CAESAR II stress report lists the several stress components (e.g., bending stress, axial stress). Except for the Code Stress, I treat these as generic calculations that may or may not correspond to the term as it is used by a particular piping code.
I suggest you go back to the local forces and moments at 25 along with the node's section properties and pressure and work from that point.
I agree that an SIF or SSI should not be less than 1.0.

Sir Dave,
My conclusion is Bending Stress in stress extended report(even in sustained case refers to
Sb = [(iiMi)2 + (ioMo)2]1/2 / Z (bending stress for expansion case)
so CS2 trying to do is multiply Sb by 0.75 becase Ii and Io is 0.75 of ii and io respectively for sustained case.

where do I can find data for Ia and It?
it is not indicated/stated in ASME B31.3 Appendix D.

B31.3 does not offer values for Ia & It but allows their use if you have them using "more rigorous analysis". Since Ia=It=1, I see them as placeholders for future development. We should expect updated SIFs and SSIs coming from an update to B31J but I do not think Ia and It are included at this time.

thanks sir dave for the reply and information.