## ASME MPC Curves (Annex 3.D)

## ASME MPC Curves (Annex 3.D)

(OP)

Hello guys,

ASME Annex 3.D has some equations that can be used to create True Stress X True Strain curves for several classes of materials.

No problem putting them on an Excel.

I'll start my doubt showing a piece of an example that I have generated with these equations (just what I believe to be the first point ---> yield stress), for SA-285C (Su=380 MPa; Sy=205 MPa; E=200000 MPa):

st [MPa] ets e_elástico ETS_PLÁSTICO

205 0.003143883 0.001014851 0.002129031

The first collumn is the stress, ets= total true strain; e_elastico=elastic strain and ETS_PLÁSTICO=plastic strain.

My doubt is: shouldn't the value of plastic strain be zero for the yield stress??

Other question: I've seen one ASME Examples book where it is found the proportional limit using these equations. And plastic strain was set to zero for the proportional limit. Ok! But the example doesn't explain how to obtain the proportional limit from the ASME Annex 3.D MPC equations.

If necessary, I can send my Excel spreadsheet.

Thanks a lot if someone can clarify me.

ASME Annex 3.D has some equations that can be used to create True Stress X True Strain curves for several classes of materials.

No problem putting them on an Excel.

I'll start my doubt showing a piece of an example that I have generated with these equations (just what I believe to be the first point ---> yield stress), for SA-285C (Su=380 MPa; Sy=205 MPa; E=200000 MPa):

st [MPa] ets e_elástico ETS_PLÁSTICO

205 0.003143883 0.001014851 0.002129031

The first collumn is the stress, ets= total true strain; e_elastico=elastic strain and ETS_PLÁSTICO=plastic strain.

My doubt is: shouldn't the value of plastic strain be zero for the yield stress??

Other question: I've seen one ASME Examples book where it is found the proportional limit using these equations. And plastic strain was set to zero for the proportional limit. Ok! But the example doesn't explain how to obtain the proportional limit from the ASME Annex 3.D MPC equations.

If necessary, I can send my Excel spreadsheet.

Thanks a lot if someone can clarify me.

_______________________________

Hervandil Sant'Anna

Petrobras

## RE: ASME MPC Curves (Annex 3.D)

I've read one post where it was being caught the proportional limit for a plastic strain of 1e-6.

Is this make sense (I mean, is it written in ASME or wherever else?) or is just an arbitrary small plastic strain value?

Thanks a lot!

_______________________________

Hervandil Sant'Anna

Petrobras

## RE: ASME MPC Curves (Annex 3.D)

## RE: ASME MPC Curves (Annex 3.D)

Hello TGS4,

First of all, when I meant "I've read one post..." I was talking about your post (coincidence!!!)

So, my doubt arisen after I read the ASME III Examples Manual.

In the example 4.6 of this manual, the stress-strain curve is calculated.

And in step 2 it is calculated the proportional limit, that I'll reproduce here:

"STEP 2 – Determination of the Proportional Limit

The proportional limit is determined by evaluating the point at which the true plastic strain is equal to

zero. This is determined through an iterative procedure. Note that the true stress (σts) is equal to the

true elastic stress (σes) plus the true plastic stress (σps), i.e.:

σts = σes + σps

Also, at the proportional limit, there is no plasticity, therefore,

e_ts = e_es = sigma_t/E

And:

g1 + g2 = 0

An iterative procedure is used to determine the value of σt = 80 ksi and εts = 0.002923."

I have coded this example, and obtained this (showing just three points):

true_stress H e1 e2 g1 g2 total_true_strain e_elástico ETS_PLÁSTICO

80000 -18.14749818 1.08067E-08 4.24862E-05 1.08067E-08 9.43384E-21 0.002922812 0.002922802 1.08067E-08

117000 -2 0.001875248 0.004919791 0.001841519 8.84884E-05 0.006204605 0.004274597 0.001930008

146244 10.76252612 2.228108526 0.08 9.99367E-10 0.08 0.085343019 0.005343018 0.080000001

As you can see, the "ASME rules" doesn't make many sense for me, since the example also seems to have choosen an arbitrary value for the plastic strain in the proportional limit: ~1e-8.

What do you think?

_______________________________

Hervandil Sant'Anna

Petrobras

## RE: ASME MPC Curves (Annex 3.D)

## RE: ASME MPC Curves (Annex 3.D)

This is the example "4.6 Example Problem E-KD-2.2.4 – Generate a Stress-Strain Curve for Use in Elastic-Plastic Finite Element Analysis" from the book ASME PTB-5 (2013): ASME Section VIII – Division 3 Example Problem Manual".

This is the data from this example:

4.6 Example Problem E-KD-2.2.4 – Generate a Stress-Strain Curve for Use inElastic-Plastic Finite Element Analysis

Generate a true stress – true strain curve for use in elastic-plastic finite element analysis. Generate

this curve for SA-723 Grade 2 Class 2 material at 150°F.

Material Data:

• Engineering Yield Strength (0.2% Offset)(Sy) = 117,000 psi @ 150°F per Section II-D

Table Y-1

• Engineering Tensile Strength (Su) = 135,000 psi @ 150°F per Section II-D

Table U

• Modulus of Elasticity (Ey) = 27,371 ksi per Section II Part D

• Material Parameter (εp) = 2 x 10-5

_______________________________

Hervandil Sant'Anna

Petrobras

## RE: ASME MPC Curves (Annex 3.D)

Yes, in the PTB-5 Example Problem 4.6 E-KD-2.2.4, they have used an arbitrary value of 1e-8 for the plastic strain in determining the proportional limit. In my opinion, such a choice is arbitrary. For ASME Section VIII, Division 2 problems, I typically use 1e-6 plastic strain as my arbitrary limit.

Your numbers appear correct. What don't you understand?

## RE: ASME MPC Curves (Annex 3.D)

First of all, I checked the equations used to generate the curves: both codes have the same equations to generate the MPC stress-strain curves.

And what I didn't understand is what I wrote some posts above (reproduced here again):

"STEP 2 – Determination of the Proportional Limit

The proportional limit is determined by evaluating the point at which the true plastic strain is equal to

zero. This is determined through an iterative procedure. Note that the true stress (σts) is equal to the

true elastic stress (σes) plus the true plastic stress (σps), i.e.:

σts = σes + σps

Also, at the proportional limit, there is no plasticity, therefore,

e_ts = e_es = sigma_t/E

And:

g1 + g2 = 0

An iterative procedure is used to determine the value of σt = 80 ksi and εts = 0.002923."

The choosen proportional limit in this example 4.6 from PTB-5 has nothing to do with the proposed rule, that g1+g2=0. It seems to be arbitrary, as we both noticed. I was just trying to confirm this with someone else that understand these subjects.

And you have just confirmed my thoughts: the choose for the proportional limit using this model is arbitrary!!

Thanks a lot!!

_______________________________

Hervandil Sant'Anna

Petrobras