If you are using the Hollomon Equation (sigma = K (epsilon)^n), then the short answer is no. If you are using the Ludwik Equation (sigma = sigma_0 + K (epsilon)^n), then the answer is yes.  However I strongly advise you to use the Voce equation which is linked to the Kocks-Mecking work hardening model. Good luck!

The dislocations can be either “strong” or “weak” obstacles to the movement of other dislocations, depending upon the types of interactions that occurs between moving dislocations. Work hardening or strain hardening can be described as the strengthening of the material by low temperature plastic deformation.

strain hardening exponent – measures the ability of a metal to harden is maximum in FCC structure (upto 0.5) and in BCC it is 0.2 and in HCP it is 0.05.  

If you are using the Hollomon Equation (sigma = K (epsilon)^n), then the short answer is no. If you are using the Ludwik Equation (sigma = sigma_0 + K (epsilon)^n), then the answer is yes.  However I strongly advise you to use the Voce equation which is linked to the Kocks-Mecking work hardening model. Good luck!

Dear All;

Thank you so much.

Dear Murat 

What is the best one for mesoscale metals? because this range still represents a problem!

Is it Kocks-Mecking work hardening model?

Thanks a lot, I appreciate your cooperation

Best regards

Sadeem

Dear Sadeem,

I suppose the decision to use which model will be driven by your experimental data.  Although I expect the Kocks-Mecking WH model to be still valid, I have not worked in this scale.  If I were you, I would plot work hardening rate versus true stress and determine whether the relationship becomes linear after the WH rate falls to about 3GPa.  If it does, then you can use Kock-Mecking and Voce.

Best,

Murat

Dear Murat

Thank you so much for your valuable answers.

Wish you the best

Sadeem

As a complement to the above, you should note that high work hardening coefficients of order 0.5 (e.g. the n in the Ludwik eqn) are only found for low stacking fault energy (SFE) alloys like austenitic stainless steels. High stacking fault energy fcc metals like most aluminium alloys usually have only relatively low strain hardening exponents of order 0.1-0.2. Al-5 Mg alloys however can have somewhat higher values.

in general the value of n depend on the type of the metal and so the strain hardening of the material  for most metals the value is below or equal 0.5 such as Al fro0.3 -.15 copper 0.5 steels from 0.4-0.25. please for be sure repeat your test and use tensile not compression as it is more accurate

Dear Julian and Mohamed

Thank you for your participation. I used tensile test and I got really high value of n if the used equation is Hollomon even more than 0.5 and its for 2024 Al, so I'm trying to look for better applied model to my data.

Basic question....have you used an extensometer for measuring strain?

No, It cannot be greater than 0.5

dear S. Sivaprasad

The test used Instron machine which by default contains extensometer for the direction of load but not for the transverse direction.

The strain-hardening exponent 'n' determined as slope in log(true stress) vs log (true plastic strain) plot is less than 0.5 (i.e., when Hollomon relation is valid) . For Cu, n ~ 0.5 and for Al, Al alloys it is in the range 0.1-0.22. The value of 'n' gives a valuable measure of stretch formability of a material, since true strain at necking is equal to n. Also, for materials (with low stacking fault energy) where dislocation cross slip is difficult, n is higher (n increases as cross slip becomes difficult).

A general comment is that It is better to use Voce equation (and apply Kocks-Mecking-Estrin model: - Y. Estrin, H. Mecking, Acta Metall., 32, 57, 1984) for describing the flow behavior (as suggested by Prof. Murat Tiryakioglu). Please see the following and may be useful:- (1) Yngve Bergström homepage: www.plastic-deformation.com:-link http://www.plastic-deformation.com/paper8.pdf. (2)  "Limitations of the Hollomon strain-hardening equation", A W. Bowen and P.G. Partridge, J. Phys. D: Appl. Phys., Vol. 7, (1974) 969. Please see the link if you have access:http://iopscience.iop.org/0022-3727/7/7/305/pdf/jdv7i7p969.pdf; and (3) Page 130 and Chapter 6 in book: Tensile Testing, 2nd Edition, edited by Joseph R. Davis, ASM International, 2004.

Dear Phaniraj C

Thank you so much for your nice advice.

Best regards

Sadeem

It is important to know at what level you are doing your modeling. If you are modeling individual slip systems, I would think a strain hardening exponent above 0,5 to be very exceptional (although maybe not impossible? I am looking at the theoretical guys here ;-) ). If you however are fitting a constitutive strength model to the output of a tensile test, without knowing the underlying deformation mechanisms, you can easily have strain hardening exponents higher than 1! For instance, if you take a magnesium material tested in a direction that will allow for easy twinning, you will get a concave upwards curve. Mathematically, most commonly used constitutive strength models can only accommodate this trend by introducing a strain hardening exponent greater than 1. This phenomenon can also be seen in strain hardening exponents mentioned in literature for titanium (alloys), where twinning can also be easily activated. It is essential to remember that most constitutive strength models implicitly assume deformation by a specific mechanism (e.g. dislocation movement). The moment you have other deformation mechanisms being activated, you can end up with very uncommon material parameters (although it may however be sufficient to mimic the real material behaviour).

You must be aware that, regardless of the contitutive law you use (Voce, Hollomon, Ludwik, Ramberg-Osgood (which is a sort of "inverse" Ludwik, so strain hardening exponents greater than 0.5 are the rule and not the exception), none can represent truly represent a experimental stress-strain relation. I read a work last month (can' t give the reference, I was reviewing it) in which the author modeled the stress-strain relation using piecewise functions (Voce, Hollomon and others). Their aim was to use this function in Finite Element simulations. Experimentally the concept of strain hardening exponent is also troublesome, if you do the standard log log plot from experimental data you will notice the graph is probably everything else than a straight line. This introduces the concept of a strain dependent strain hardening coefficient, which is, of course, absurd. 

Dear Frederik and Cláudio

I agree with you that the deformation mechanism play a major rule here, but I think the strain hardening coefficient need to be related to physical quantity because by its own it only well known for micro grains material but when the scale goes to meso or nano the problems arise with inordinary values of it.

Dear Murat and others who suggest using Voce

Is it worthy to use it if I have made the experiments using constant temperature with one value of it and with one value of strain rate

What experimental data I need to offer in order to fulfill the model?

Thanks a lot.

Dear Sadeem,

The answer is YES.  Here is how I suggest that you fit the model:

1. Plot work hardening rate versus true stress ( a smoothing algorithm will be helpful here)

2. Look for the the linear region that starts around 3 GPa. This represents Stage III work hardening. The region at higher work hardening rates is Stage II. Stage III WH rate is not affected by strain rate but Stage II is.  You may also look into the transition stress from Stage II to III.

3. Find the best fit line between WH rate and true stress.  The slope and intercept give you the Kocks-Mecking (KM) model.

4. Determine two Voce parameters from the KM parameters such that:

KM: WHR= WHRsub0 - k (true stress)

Voce: true stress = saturation stress - (saturation stress-stress_sub_0) exp (- true plastic strain/strain_sub_0)

The two are related by:

k = 1/strain_sub_0

WHRsub0 = saturation stress / strain_sub_0

5. Determine the third parameter of Voce (stress_sub_0) by the best fit to true stress-true plastic strain in the same region.

Good luck.

Thank you so much Prof. Murat

Best regards

Sadeem