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Local criteria par 5.3.3 ASME VIII div2

Local criteria par 5.3.3 ASME VIII div2

Local criteria par 5.3.3 ASME VIII div2

(OP)
Hello!
Can anybody tell me how to evaluate local criteria (describing in formula 5.6 par.5.3.3) automatically for the whole object in ANSYS?

I found solution for this:
Use Element table. Create formula there and show in final Epeq/El.
But there is a problem - element table can`t involute. So I use a Taylor series to 5 summand. But I think I get a large error.

Here log for the criteria-

/POST1
SABS,0
ETABLE,s1,S,1
ETABLE,s2,S,2
ETABLE,s3,S,3
ETABLE,seqv,S,EQV
SADD,s12,S1,S2,1,1
SADD,s123,S12,S3,1,1
SEXP,seqv0,seqv,seqv,-0.5,-0.5
SMULT,se,S123,SEQV0,1,0.33333
SADD,se0,Se,,1,,-0.33333
SADD,se0k,Se0,,-1.774
SMULT,se0k2,Se0K,Se0K,0.5,1
SMULT,se0k3,Se0K,Se0K2,0.33333,1
SMULT,se0k4,Se0K,Se0K3,0.25,1
SADD,se0ke2,Se0K,Se0K2,1,1,1
SADD,se0ke3,se0ke2,se0k3,1,1
SADD,se0ke4,se0ke3,se0k4,1,1
SMULT,el,SE0KE4,,0.24,,
ETABLE,epeq,EPPL,EQV
SEXP,el0,EL,EL,-0.5,-0.5
SMULT,LocCrit,epeq,el0,1,1

Regards, Alex.

RE: Local criteria par 5.3.3 ASME VIII div2

Alex - one comment about using an element table - from the ANSYS help:

Quote:

...items are multivalued (varying over the element, such that there is a different value at each node). Because only one value is stored in the element table per element, an average value (based on the number of contributing nodes) is calculated for multivalued items.

When you have variation of stress and strain over elements, this approach may not provide you with the answers that you are looking for. Have you read my paper: http://proceedings.asmedigitalcollection.asme.org/... that talks about Local Failure? Your ETABLE approach may look efficient (if you can get it to work properly), but you are going to be missing much of the necessary nuance required. In order to have your results independent of the mesh you will need to have a prohibitively fine mesh.

One brief comment about your script - I don't think that the SEXP command, as you have written it, is doing what you think it is doing. I think that your problem lies there.

RE: Local criteria par 5.3.3 ASME VIII div2

(OP)
SEXP command - here I tried to explain my chain of actions
http://files.engineering.com/getfile.aspx?folder=4...
Unfortunately, I do not have a chance to read your paper.
So...you said that I should have fine mesh to get adequate results? I think mesh will be good. BUT!

I wanted to show another problem - replace e^x to 1 + 1/x + 1/(x^2)....large error...see the picture.
http://files.engineering.com/getfile.aspx?folder=4...

I tried to use arrays to work with these data...here example.

*set,sigma_t,245
*set,sigma_v,451
*set,alfa_sl,2.2

/post1
ESEL,S,MAT,,4
NSLE,S
*GET,MAX_UZEL,NODE,,num,max
*GET,Min_UZEL,NODE,,num,min
KOL_UZL=MAX_UZEL-Min_UZEL

*GET,KOL_UZL,NODE,,COUNT
*DIM,NAPR_YA_MASSIV,ARRAY,KOL_UZL,5,1
*DIM,EL,ARRAY,KOL_UZL,2,1

*DO,I,Min_UZEL,MAX_UZEL
*GET,NAPR_YA_MASSIV(I,1),node,i,s,1
*ENDDO
*DO,I,Min_UZEL,MAX_UZEL
*GET,NAPR_YA_MASSIV(I,2),node,i,s,2
*ENDDO
*DO,I,Min_UZEL,MAX_UZEL
*GET,NAPR_YA_MASSIV(I,3),node,i,s,3
*ENDDO
*DO,I,Min_UZEL,MAX_UZEL
*GET,NAPR_YA_MASSIV(I,4),node,i,s,eqv
*ENDDO
*DO,I,Min_UZEL,MAX_UZEL
*GET,NAPR_YA_MASSIV(I,5),node,i,EPPL,eqv
*ENDDO

*DO,I,Min_UZEL,MAX_UZEL
EL(I,1)=0.6*(1-sigma_t/sigma_v)*exp(-(alfa_sl/(1+0.6*(1-sigma_t/sigma_v)))*(((NAPR_YA_MASSIV(I,1)+NAPR_YA_MASSIV(I,2)+NAPR_YA_MASSIV(I,3))/(3*NAPR_YA_MASSIV(I,4)))-1/3))
*ENDDO

*DO,I,1,KOL_UZL
EL(I,2)=NAPR_YA_MASSIV(I,5)/EL(I,1)
*ENDDO
*DO,I,1,KOL_UZL
*VPUT,EL(I,2),NODE,I,U,x
*ENDDO


BUT! it takes a lot of time...and I have different types of elements and big model...so I should write different conditions for choosing them...I think it has perspective but I still don`t have any ideas how perform it. Have anybody universal (multipurpose) log-file to solve this problem?

Thanks!

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