calculation of log. plastic strain 1

[feaplastic]

Hello users,

From Nominal stress-strain curve(compressive), I calculated logarithmic plastic strain using following formula.

Log plastic stain = Nominal strain - (true comp stress/E)

Is it right? I get ascending values till 75% strain then my log plastic strain descends. Abaqus asks for a value in ascending order. What is wrong with my calculation?

My calculation:
Nominal Stress Nominal Strain Nom.strn-(true strs/E)
0,24 0,05 0,013636364
0,27 0,1 0,059090909
0,3 0,15 0,104545455
0,32 0,2 0,151515152
0,34 0,25 0,198484848
0,35 0,3 0,246969697
0,38 0,35 0,292424242
0,4 0,4 0,339393939
0,43 0,45 0,384848485
0,47 0,5 0,428787879
0,51 0,55 0,472727273
0,58 0,6 0,512121212
0,67 0,65 0,548484848
0,82 0,7 0,575757576
1,07 0,75 0,587878788
1,54 0,8 0,566666667


True stress is equal to nominal stress since my poisson ration is zero.

Thanks for your time. Any help would be nice.

 

[feaplastic]

Hello,

in above post, I have given the wrong formula for log plasitc strain.

Log plastic strain = True strain - (true stress/E)

= Ln(1+Nom strain) - (Truestress/E)

New log plastic strain as follows,

0,012426528
0,054401089
0,094307397
0,133836708
0,1716284
0,209333961
0,242528835
0,275866176
0,306412041
0,334252987
0,360982204
0,382124841
0,399260136
0,406385827
0,397494576
0,354453332

Here, again i have the same problem with last 2 values.

Some one can clarify me.

Thanks.

 

[prost]

Since you didn't give us the E, I tried to back calculate the E using the following. You defined Log Plastic Strain as
Log plastic strain = Ln(1+Nom strain) - (Truestress/E)

My strength book says Truestress==nom. stress*(1+nom. strain). Is this consistent with your definition of Truestress?

Back calculating using the numbers you gave:
E=Truestress/(Ln(1+nom. strain)-Log plastic strain)
Is there agreement there? If so, I get a different "E" for every strain/stress combination. Perhaps you can tell me what I've done wrong?

engr engr Your True True My calc.
stress strain plastic strain Stress Strain E
0.24 0.05 0.01243 0.252 0.04879 6.93
0.27 0.1 0.05440 0.297 0.09531 7.26
0.3 0.15 0.09431 0.345 0.13976 7.59
0.32 0.2 0.13384 0.384 0.18232 7.92
0.34 0.25 0.17163 0.425 0.22314 8.25
0.35 0.3 0.20933 0.455 0.26236 8.58
0.38 0.35 0.24253 0.513 0.30010 8.91
0.4 0.4 0.27587 0.560 0.33647 9.24
0.43 0.45 0.30641 0.624 0.37156 9.57
0.47 0.5 0.33425 0.705 0.40547 9.90
0.51 0.55 0.36098 0.791 0.43825 10.23
0.58 0.6 0.38212 0.928 0.47000 10.56
0.67 0.65 0.39926 1.106 0.50078 10.89
0.82 0.7 0.40639 1.394 0.53063 11.22
1.07 0.75 0.39749 1.873 0.55962 11.55
1.54 0.8 0.35445 2.772 0.58779 11.88

The input data, nom. stress and nom. strain, looks interesting, as the data doesn't appear to cross the (0,0) point (that is, extrapolating back to zero nom. strain from the smallest values of nom. strain, the nom. stress at nom. strain=0 is not equal to zero). Is there an offset problem with the original data measurements?

 

[feaplastic]

Hello Prost,

Thanks for your reply.

column 1 - Nom stress
Column 2 - Nom strain
Column 3 - True compressive stress
Column 4 - True compressive strain
Column 5 - True plastic comp. stain

True comp stress =N om.str/(1-pois rat*nom.strn)^2
True Strain = ln(1-nom.strn)
True comp. plastic strain = True.strn-(true strs/E)

In my earlier post and in your calculation, basic mistake was negative sign was not added(compressive strain) in true strain calculation.

E = 6.6 MPa
Poisson = 0

Moreover, in your forrmula of True stress, i cannot include zero poisson ratio. I have given mine also. Pls, check and correct me if i am wrong. For my case, true stress and nominal stress are same.

0,24 0,05 0,24 -0,051293294 -0,087656931
0,27 0,1 0,27 -0,105360516 -0,146269607
0,3 0,15 0,3 -0,162518929 -0,207973475
0,32 0,2 0,32 -0,223143551 -0,2716284
0,34 0,25 0,34 -0,287682072 -0,339197224
0,35 0,3 0,35 -0,356674944 -0,409705247
0,38 0,35 0,38 -0,430782916 -0,488358674
0,4 0,4 0,4 -0,510825624 -0,571431684
0,43 0,45 0,43 -0,597837001 -0,662988516
0,47 0,5 0,47 -0,693147181 -0,764359302
0,51 0,55 0,51 -0,798507696 -0,875780423
0,58 0,6 0,58 -0,916290732 -1,00416952
0,67 0,65 0,67 -1,049822124 -1,151337276
0,82 0,7 0,82 -1,203972804 -1,328215229
1,07 0,75 1,07 -1,386294361 -1,548415573
1,54 0,8 1,54 -1,609437912 -1,842771246


For your last question, I dont have any offset problem in my experimental data. If you tell me, how did you find this problem, then i could look further.

Thanks

feaplastic

 

[prost]

I guessed that there was an offset because when strain equal to zero, stress was not equal to zero. Say you used a line to extrapolate from the two smallest strain value to strain=0. The slope of that line is 0.6=(0.27-0.24)/(0.1-0.05). Using that slope, extrapolate from the last point (0.05,0.24) back to the point (0,stress0). The value of stress0 is 0.21. Normally I would expect a test which measures stress and strain to give you a curve that has the value (0,0) as the first point (zero strain gives you zero stress, that is).

Why doesn't zero strain give you zero stress?

 

[prost]

I like your formula for 'true stress' better than the one that commonly appears in many books, including my mechanics of materials book, and the ASM handbook Vol. 19. In these two references, the derivation of the 'true stress' includes an assumption that the material is undergoing 'large plastic strains' similar to what you see in a test coupon that begins to 'neck down'. Because the strains are plastic, they assume that the material is essentially incompressible (that is, Poisson ratio=0.5). With that assumption,

truestress=nom. stress*(1+nom. strain)

But if you don't make that assumption, just calculate the deformed area of the coupon gage section using the Poisson's ratios, then you derive the equation you have written:

truestress=N om.str/(1-pois rat*nom.strn)^2

It is puzzling that these two equations give you similar results for SMALL STRAINS! That seems counterintuitive to me because you are assuming the strains are very large, which will give you incompressible deformation. But your approximation of large strains gives you
truestress=nom. stress*(1+nom. strain)

which doesn't work very well for large strains.

 

[prost]

I am not an expert here, wish an expert would answer this one. However, I think I might have located the problem. You calculated True Strain = ln(1-nom.strn)--did you use

(1 - nom. strn)

because your nom. strn is really compressive, or negative, rather than positive as you show in Column 2? If you put that negative sign there, don't you have to use a positive sign in the formula

True comp. plastic strain = True.strn-(true strs/E)

instead of the negative?

You show your 'true compressive stress' as column of positive values. If you use that sign convention, then don't you have to use the following formula?

True comp. plastic strain = True.strn+(true strs/E)

When I make those assumptions, the True comp. plastic strain monotonically increases (that is, always increases).

 

[prost]

My mistake--the values for True comp. plastic strain are "monotonically decreasing" since they start at a small negative number (-0.149) and gets more negative, the last point is -1.376.

 

[40818]

Prost, it seems this post has certainly got your attention with all your replies!!

 

[prost]

my one unacceptable vice--obsessiveness about this kind of stuff. Doesn't that drive you crazy? Seems like there is something simple that is wrong, not a conceptual problem.

 

[feaplastic]

For your slope calculation, you have taken 2 values (0.24, 0.27),which are in non linear range according to my strs-strn curve. I have linear curve till 0.03 strain. So, when you calculate your slope from plastic region and then if you use it for linear range extrapolation, it wont work.

If you see my strs-strn curve, you could understand. Both have different slope. Imagine, for steel matl, you calculate slope from plastic region and use the same slope to calculate elastic region stress. It wont work, right?...

For your next question, i did not put negative sign in nominal strain cólumn and instead i changed the formula in my post.

I agree, your approach is professional.

For your last question, comp plastic strain should increase numerically with negative sign(compression). I dont see any confusion here. Could you clarify further. Hope, some experts will put an end to this problem.

Thanks a lot for your curiosity. I reviewed all my formulas again and hope i wont forget in future.

Thanks

feaplastic

 

[prost]

You didn't give us data for strain<0.05, so how were we to know? I used the best information I had available. I assumed you gave me the entire curve. With the changes I recommended in the data and the formulae, I get the following (I think if you copy this, paste into a TXT file, then read into Excel, everything should line up:

True True True True
Nom Nom Compressive True Compressive Compressive True Compressive
stress strain Stress Strain Plastic Strain Stress Strain Plastic Strain
-0.2 -0.05 0.24 -0.05129 -0.08766 -0.24 -0.05129 -0.01493
-0.3 -0.1 0.27 -0.10536 -0.14627 -0.27 -0.10536 -0.06445
-0.3 -0.15 0.3 -0.16252 -0.20797 -0.3 -0.16252 -0.11706
-0.3 -0.2 0.32 -0.22314 -0.27163 -0.32 -0.22314 -0.17466
-0.3 -0.25 0.34 -0.28768 -0.33920 -0.34 -0.28768 -0.23617
-0.4 -0.3 0.35 -0.35667 -0.40971 -0.35 -0.35667 -0.30364
-0.4 -0.35 0.38 -0.43078 -0.48836 -0.38 -0.43078 -0.37321
-0.4 -0.4 0.4 -0.51083 -0.57143 -0.4 -0.51083 -0.45022
-0.4 -0.45 0.43 -0.59784 -0.66299 -0.43 -0.59784 -0.53269
-0.5 -0.5 0.47 -0.69315 -0.76436 -0.47 -0.69315 -0.62194
-0.5 -0.55 0.51 -0.79851 -0.87578 -0.51 -0.79851 -0.72123
-0.6 -0.6 0.58 -0.91629 -1.00417 -0.58 -0.91629 -0.82841
-0.7 -0.65 0.67 -1.04982 -1.15134 -0.67 -1.04982 -0.94831
-0.8 -0.7 0.82 -1.20397 -1.32822 -0.82 -1.20397 -1.07973
-1.1 -0.75 1.07 -1.38629 -1.54842 -1.07 -1.38629 -1.22417
-1.5 -0.8 1.54 -1.60944 -1.84277 -1.54 -1.60944 -1.37610

 

[feaplastic]

Just i gave the tabular values, i got frm manufacturer. I did not realise this problem. Excuse me.

After changing the sign, i get the same values.

Thanks a lot for your help, especially the change in sign.

Regards,

feaplastic