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Plastic properties of Mild steel for Analysis in ABAQUS for a steel bumper design

Plastic properties of Mild steel for Analysis in ABAQUS for a steel bumper design

Plastic properties of Mild steel for Analysis in ABAQUS for a steel bumper design

(OP)
Hello all,

I have been working on a steel bumper design and have been analyzing it on ABAQUS.
I need plastic properties for mild steel (yield stress and plastic strain) which i need to input to get the
correct results.
Input is non temperature dependent and isotropic (not kinematic).
i have been trying to find the data on internet but cannot find it.
it will be great even if you can provide a formula to calculate these values.
I am stuck at the start of the analysis due to unavailability of these values.

I have attached a screenshot of the input window of ABAQUS where these values are to be put.

so anybody with either these values or a formula to calculate them, please help!

RE: Plastic properties of Mild steel for Analysis in ABAQUS for a steel bumper design

What do you really need to know?

Mild steel isn't a very precise material description.

Yield stress and elongation will only be part of the issue depending on how the bumper is formed.

You will probably need to know R and n values along with data from a Formimg limit diagram if you want a good simulation.

RE: Plastic properties of Mild steel for Analysis in ABAQUS for a steel bumper design

(OP)
Hey,


The steel is mild steel grade 1018
I have attached a screenshot of window where i need to input those values.
Is there a formula to generate that data with the R and n values and the forming limit diagram?

RE: Plastic properties of Mild steel for Analysis in ABAQUS for a steel bumper design

You can estimate the work hardening exponent from a true stress/true strain curve. If you have an engineering stress strain curve from a simple tensile test then you could convert the engineering stress strain values to treu stress and strain and then plot them on a log log basis to give a straight line.

The slope of this line would be 'n' the work hardening exponent.

This calculation would be based on the assumption that the material obeys a simple parabolic work hardening law. σ=kε^n

n is the work hardening exponent and k is the strength coefficient

If you sue stress/strain values between yield and maximum tensile stress (uniform strain) the estimate will not be to bad.

R value is concerned with anisotropy and a 1018 is likley to to be relatively anisotropic depending on fabrication history - it is difficult to estimate these values and I think you have to consider the material to be an isotropic solid.


Forming limit diagrams are difficult to estimate and rely on knowleged of the anisotropic yield behaviour of the material.


There are some suggested numerical methods to try to estimate this behaviour but they are relatively complex.

http://www.mech.northwestern.edu/ampl/papers/hy_jc...

This paper gives a good review.


RE: Plastic properties of Mild steel for Analysis in ABAQUS for a steel bumper design

Looking again at your screen input you seem to only need the Yield Stress and Plastic Strain.

Assuming (dangerous) that this is the plastic strain at yield there are several questions that could be raised.

Yield point is said the be analagous to the limit of proportionality or the Elastic limit.

In theory this means when 1 discocation has moved by 1 Burgers Vector the material has yielded.

This is a bit silly as we cannot measure strain at these levels. If we tried to estimate the onset of yield by measuring non linearity then yield points would simply become a function on extensometer accuracy and the better the extensometer the lower the yield point.

For materials which exhibit continuous yield the convention has been to use a 0.2% offest proof stress.

There are a few tant reasons why this value has beem used.

Firstly it is measurable using commercially available extensometers that maintaing their calibration for reasonable periods of time.

Secondly it is a characteristic that is relatively repeatable.


Thirdly it is roughly equivalent to the Luders Extensions commonly associated with steels that exhibit discontinuous yield.

I would use a platic strain of 0.2% for this value. At this low a strain there is no practical differnece between engineering and true strain.

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