ASME 2007 protection against local failure elastic-plastic analysis

[esseq]

I am designing a 28"x24" integral lateral tee 25 degrees and I would like/need to use the ASME VIII div.2 PART 5 2007 with an elastic perfectly plastic material model.

I have only the design pressure as a load. I performed an elastic-plastic analysis.

I am not sure about the procedure reported in the section 5.3 and specifically in the subsection 5.3.3 "Elastic-Plastic Analysis".

STEP 1. Perform an elastic-plastic stress analysis based on the load combinations for the local criteria given in Table 5.5 (This table seems to indicate a load equal to 1.7*(P+PS+D)).

IS IT CORRECT TO RUN THE FEM ANALYSIS IN MY CASE WITH 1.7*DESIGN_PRESSURE?

STEP 2 Determine...the "total equivalent plastic strain, epsilon_peq.

WHAT IS THE DEFINITION OF TOTAL EQUIVALENT PLASTIC STRAIN AND WHICH IS THE ASME SECTION IN WHICH I CAN SEE THE DEFINITION?

STEP 3 The triaxial strain epsilon_L is:

epsilon_L = epsilon_LU*exp(-(alfa_SL/(1+m2)) * ((s1+s2+s3)/(3se))-(1/3) )

I have an A694 F52 material. For a FERRITIC STEEL the table 5.7 states:

m2= 0.6*(1-R)=0.12
alfa_SL=2.2
epsilon_LU=0.365

the FEA in the most loaded point states:

s1=426 MPa
s2=164 MPa
s3=44 MPa
se=338.4 MPa

thus epsilon_L=0.206

I HAVE CALCULATED FOR THE MOST LOADED POINT IN MY PIECE AN epsilon_L = 0.206. IT SEEMS THAT I CAN HAVE AN EQUIVALENT PLASTIC STRAIN IN THAT POINT UP TO 20.6%. IS IT RIGHT BECAUSE IT SEEMS VERY HIGH TO ME (BUT I DON'T KNOW EXACTLY WHAT IS THE EQUIVALENT PLASTIC STRAIN)?

STEP 4. OK I CAN CONSIDER epsilon_cf = 0. NO QUESTIONS

STEP 5. OK NO QUESTIONS

 

[TGS4]

STEP 1. Perform an elastic-plastic stress analysis based on the load combinations for the local criteria given in Table 5.5 (This table seems to indicate a load equal to 1.7*(P+PS+D)).

IS IT CORRECT TO RUN THE FEM ANALYSIS IN MY CASE WITH 1.7*DESIGN_PRESSURE?

No - you should use 1.7 times the design pressure plus the static head pressure plus the dead load.

STEP 2 Determine...the "total equivalent plastic strain, epsilon_peq.

WHAT IS THE DEFINITION OF TOTAL EQUIVALENT PLASTIC STRAIN AND WHICH IS THE ASME SECTION IN WHICH I CAN SEE THE DEFINITION?

The total equivalent plastic strain, epsilon_peq, is the total equivalent plastic strain that you will obtain from your finite element analysis program. It does not appear to be defined otherwise, and that would appear to be an oversight. Please write a letter to the ASME Section VIII Code Committee indicating this oversight.

I HAVE CALCULATED FOR THE MOST LOADED POINT IN MY PIECE AN epsilon_L = 0.206. IT SEEMS THAT I CAN HAVE AN EQUIVALENT PLASTIC STRAIN IN THAT POINT UP TO 20.6%. IS IT RIGHT BECAUSE IT SEEMS VERY HIGH TO ME (BUT I DON'T KNOW EXACTLY WHAT IS THE EQUIVALENT PLASTIC STRAIN)?

Why do you think that this value is high?

 

[esseq]

Regarding the total equivalent plastic strain I sent a letter to the ASME committee asking what is the exact definition.

I am preparing some documents to discuss about the reason why I suppose that 20.6% is high.

thank you for the moment.

 

[TERIO]

The usual definition of equivalent plastic strain is described in the attached document, although I don't know for certain that this is exactly what the code intends.

The allowable effective strain, epsilon_L, is based on the strain from uniaxial tension, epsilon_Lu, the stress state, and some fitting parameters. The term

[(s1+s2+s3)/(3se)-(1/3)]

becomes zero for uniaxial tension, s1=se=s, s2=s3=0, and the allowable strain becomes the strain to failure from uniaxial tension. The term (s1+s2+s3)/(3se) is typically called the stress triaxiality ratio. Failure tends to occur at lower strains for positive triaxialities (tensile mean stress) than it does for negative (compressive).

If interested in background on this try to dig up the paper,

J. Hancock and A. Mackenzie, “On the mechanisms of ductile failure in
high strength steels subjected to multi-axial stress-states,” Journal of
the Mechanics and Physics of Solids, vol. 24, pp. 147–169, 1976.

 

[esseq]

Dear Terio,

Regarding the definition of equivalent plastic strain you sent me it refers (if I am not wrong) to the strain rate, thus if there is a dynamic effect (even slow as the creep effect). But in my case I can consider that the strain rate is almost zero thus the equivalent strain (following your definition) is the integral over time of zero that is zero. But the paragraph 5.3.3 of the ASME states something about the local failure without taking into account the time effects. Thus I suppose that the definition of the total equivalent plastic strain (of ASME) is different from the one you gave me.

Anyway I think you haven't found the definition inside the ASME code so we have an additional clue that the ASME does not mention the definition. In this case I think that the best thing to do is waiting for the ASME Committe reply.

Thank you for your reference I am going to see if I can get the article.

 

[TGS4]

TERIO - please note that the failure mode discussed in Article 5.3 is not ductile failure. It's the special case whereby one has a state of high tri-axiality, which would render the von Mises (equivalent) stress near zero (because the equivalent stress is calculated by the differences between the principal stresses). It's a real, but rare failure mode.

 

[TERIO]

esseq,

You are correct, I did not see a definition in the code but would be interested to hear the response to your inquiry.

As you suggest, the definition I sent does integrate the strain rate over time, but it does not necessarily have any dynamic effect. It would be equivalent to integrate (add up) increments of strain (e.g., epsilon_p_dot*dt = depsilon_p).

Also, as TGS4 suggested whatever FE code you are using should provide equivalent plastic strain as a result, and most likely provides the definition it uses in a theory manual.


TGS4,

Maybe my explanation was a little unclear. I agree with you that the failure under high stress triaxiality has very limited ductility (e.g., a notched tensile bar). The reference that I had listed suggests the exponential dependence of the strain to failure on the stress triaxility. I also agree that Von Mises can not pick up this type of failure since it is independent of mean stress (based on differences as you point out).



General Comment/Question:

I am curious why the Code writes these equations in terms of principal stresses. With the older versions using Tresca criterion this was necessary. However, Von Mises can be written in terms of stress components in any coordinate system not just principal coordinates. The mean stress (s1+s2+s3)/3 is also independent of coordinate system and can be replaced by (sx+sy+sz)/3.



Tom