• In my opinion you should define a plastic potential law without a hardening / softening transition. In the material definition include only the softening curve. On the other hand this behavior works at constitutive level and the macroscopic response would has some hardening behavior according to the stress state. 
• Sincerely 

You provide the yield criteria by putting ultimate or yield stress.

@javier and Pankaj - Thanks for sharing your thoughts. I have specified a higher yield stress in tensile and compressive damage. The problem that persists is a common issue discussed in many forums.

"DURING THE CONVERSION FROM CRUSHING TO PLASTIC STRAIN Abaqus FOUND NEGATIVE AND/OR DECREASING VALUES OF PLASTIC STRAIN. VERIFY THAT THE DEGRADATION DATA UNDER *CONCRETE COMPRESSION DAMAGE IS CORRECT?"

cracking strain =  (total strain) - (total stress)/E0

plastic strain =  (cracking strain) - (d*(total stress))/((1-d)*E0)

E0 - Elastic modulus 

total strain and stress values as given in the txt file attached

I have checked the plastic strain values prior to adding them in ABAQUS. If I use the same values in the txt file the plastic strain fluctuates around a very small value 10^-20 (expected). 

This value can sometimes be negative or positive. So I round the total strain and stress values to around 6 decimal places and input the values to ABAQUS.

Problem still persists.  

Any reason for this to happen ?

Regards

Some remarks.

1. 'plastic strain = (cracking strain) - (d*(total strain))/((1-d)*E0)' is not consistent in terms of units. Perhaps there is a typo.

2. (total stress)/E0 represents some sort of elastic strain. I have never heard of total stress, do you mean equivalent stress? If it is equivalent stress then it is not correct because the elastic strain is a tensor quantity determined from eps_e = E-1:sig and after that we determine the equivalent elastic strain from the elastic strain tensor.

3. 10^-20 is a number n machine accuracy. you cannot rely in this number. Actually you must consider it as zero.

@Muhammad -  thanks for your comments. 

1. yes, sorry for that. Its a typo. corrected below

plastic strain = (cracking strain) - (d*(total stress))/((1-d)*E0)

2. correct. I believe you are familiar with stress representation wrt plastic and elastic stress.

total stress = elastic stress + plastic stress ; here plastic is rather nonelastic stress

3. correct - I have thus used rounded values such that  

plastic strain = (cracking strain) - (d*(total stress))/((1-d)*E0)

yields zero.  

IMHO There is no such thing like plastic stress. Plastic strain (or inelastic strain) is based on a dissipation potential thus there is no stored energy.  When there is no stored energy there must be no stress.

The recoverable strain i.e. elastic strain is a state potential and has a stored elastic potential energy which is 0.5 sig*eps_e.

If you have viscoplasticity then your stress can go temporarily above the elastic stress (over stress) but then with time it returns back to the stress based on elastic strains.

Hi Muhammad, yeah perfect. So the misunderstanding was with sigma_t - which is tensile stress in the tensile damage formulation. As you explained the formulation is given in

http://abaqus.software.polimi.it/v6.14/books/usb/default.htm 

section 23.6.3

unfortunately though, I get the following error

DURING THE CONVERSION FROM CRACKING TO PLASTIC STRAIN Abaqus FOUND NEGATIVE AND/OR DECREASING VALUES OF PLASTIC STRAIN. VERIFY THAT THE DEGRADATION DATA UNDER *CONCRETE TENSION DAMAGE IS CORRECT

the cracking  strain (eps_ck) values I use are as follows 

0

3.10862446895044e-20

1.02039329573588e-05

2.04075688529240e-05

0.00103934269033213

0.00205570934285863

0.00306985592835679

0.00408208358950695

0.00509265262026160

0.00610178800680207

0.00710968421688805

0.00811650933955205

0.00912240866326198

0.0101275077687210

0.0111319152021423

0.0121357247859054

0.0131390176157819

0.0141418637872457

0.0151443238876169

0.0161464502858023

0.0171482882470889

0.0181498768967190

0.0191512500527619

0.0201524369460091

0.0211534628422192

0.0221543495799577

0.0231551160354813

0.0241557785245626

0.0251563511498086

0.0261568461008677

0.0271572739139139

0.0281576436959342

0.0291579633185911

0.0301582395857897

0.0311584783785140

damage evolution values (dt)

 0

2.22044604925031e-16

0.0680262197157254

0.127547305330775

0.895985077872522

0.951717288360479

0.971473395049617

0.981270093631479

0.986948182221240

0.990550001104232

0.992972655989950

0.994670262200006

0.995896142277509

0.996801945740256

0.997483441052178

0.998003683051434

0.998405593904402

0.998719194014529

0.998965955647551

0.999161527586779

0.999317496916601

0.999442560392014

0.999543322169200

0.999624848512356

0.999691060596370

0.999745017146105

0.999789120702993

0.999825270056398

0.999854974157735

0.999879438106562

0.999899628641897

0.999916324429481

0.999930154958543

0.999941630828569

0.999951167474774

Although I get all plastic strain values equal to zero, still i get the error message given above.

Thanks again for your thoughts.

Regards

'plastic strain = (cracking strain) - (d*(tensile stress)/((1-d)*E0)' is not correct because you are relating strains with displacement.

Kindly check your data with the following (as given in the manual)

plastic/cracking strain = (total tensile strain) - (tensile stress)/((1-d)*E0)