Elongation in mechhanical property 3

[jfymsam86]

Hi
I have a simple question about Elongation in mechanical property.
Is that meam Ultimate Strain for example for A36 is 0.21?
I’m going to use 3-line Stress – strain diagram.
But according to table (attached), by E= 205000 and Yield stress = 250, so yield strain is 0.001219.
But 120 ε is 0.1463, not 0.21.
What is my mistake? What’s the difference between them?

Thanks in advance

 

[JStephen]

Is there some reason to assume that it fractures at 120ε?

 

[jfymsam86]

Thanks for your reply
I'm not sure, but most of reliable articles used that.(attached pictures)
did I understand the meaning of elongation right or not?

 

[jfymsam86]

Anyone knows that what's the difference between them?

 

[Ron]

Elongation is ultimate strain

 

[BAretired]

If yield strain is 250/205,000 = 0.0012195, then ultimate strain should be 474/205,000 = 0.002312 assuming that E remains constant. 120ε is the total yield strain of a 120" (10'-0") long sample in inches.

BA

 

[rb1957]

"assuming that E remains constant" ... that's a big assumption ... i'd've thought elongation would be closer to 10%, as the material behaves plastically ...
yield = 250MPa, ey = 250/205000 = 0.00122
uts = 474Mpa, eu = 0.10 (approx)

Quando Omni Flunkus Moritati

 

[BAretired]

Modulus of Elasticity falls off dramatically between first yield and ultimate stress but ultimate strain cannot be precisely related to yield strain by a constant factor such as 120ε. That is why I believe the term 120ε relates to something other than ultimate strain. It could be the total strain in mm of a bar 120mm long or the total strain in inches of a bar 120" long.

BA

 

[jfymsam86]

Thanks all
I anylized the materials in Abaqus, by both ways, but there are too many differrences between the result.
However wich one is better to use: 120 ε and 14.63% or 21% ???

 

[rb1957]

"However wich one is better to use: 120 ε and 14.63% or 21% ???" ?

what "one" are you talking about ?


Quando Omni Flunkus Moritati

 

[jfymsam86]

I need to introduce the stress related to plastic strain to Abaqus. Before I defined it as:
250 0
251 0.0134146 (11ε)
474 0.146341 (120ε)

it was according the diagram of 3-Line stress - strain.
but because of its differnce till 0.21, i thought it's wrong.
and I didn't find 0.1 that you bid in any refernces.

 

[rb1957]

i'd verify the definitio of "plastic strain" as Abaqus uses it.

it could be the strain added onto the elastic strain ...
total strain = elastic strain+plastic strain = stress/E+x

your points look "odd" to me ...

(250,0) ... ok that's the yield point
(251, 0.013) ... boy that's a large change in slope !? considering the yield strain is something like 0.001 !??

you should be able to final the elongation for your material (and heat treat) in your specs.
it looks like you set the yield point (250,0), the failure point (474,0.10-474/205000), and some intermediate point that gives Abaqus a smooth transition.

Quando Omni Flunkus Moritati

 

[jfymsam86]

Thanks again
I've heared that for defining plastic behavior to Abaqus, you shouldn't have 2 point with same stress, even you have the horizontal line like attached file.
yes, that's large change in slope, because of plastic behavior.
but we can't say: 474/205000, because in hardening region E[sub]sh[/sub]=E/33.
E=205000 is only for Elastic region
Am I right?

 

[rb1957]

it depends on what Abaqus means by plastic strain. if you input 0 for the yield point then Maybe Abaqus is saying ... total strain = Elastic strain + plastic strain in which case you extrapolate E to UTS ... yes, i know it's not valid as a final strain but i think it is valid if that's how Abaqus have defined their terms.

check the Abaqus manual and examples.

Quando Omni Flunkus Moritati