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Work hardening and Forest Hardening

by Chemist20
Tags: forest, hardening, work
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Bavid
#2
Mar29-12, 11:18 AM
P: 32
They are just different models. Depending on the length scale of interest and the application one of them will be more appropriate.
Work Hardening is a simple (i.e., it does not require one to know the material's internal structrue) mathematical model of the hardening phenomena observed in a material's plastic stress-strain behaviour. According to this model, the hardening of the material is described as some function of the plastic work done. Work hardening models do not care if dislocations exist or not.
Forest hardening assumes the existence of dislocations; and the density of these dislocations increases as plastic deformation progresses. Consequently, it becomes harder to move new dislocations across this "forest" of pre-existing dislocations, resulting in hardening of the material. You can now choose to describe hardening as the increase in shear stress required to move a dislocation through the 'forest' as a function of the dislocation density in the forest. Physically, the observation of "work hardening" at the scale of the specimen is the overall effect of "forest hardening" due to dislocations at the micron scale.

Therefore when hardening=func(plastic work done) --> work hardening
and when hardening=func(forest dislocation density) --> forest hardening
Chemist20
#3
Mar31-12, 04:02 AM
P: 87
Quote Quote by Bavid View Post
They are just different models. Depending on the length scale of interest and the application one of them will be more appropriate.
Work Hardening is a simple (i.e., it does not require one to know the material's internal structrue) mathematical model of the hardening phenomena observed in a material's plastic stress-strain behaviour. According to this model, the hardening of the material is described as some function of the plastic work done. Work hardening models do not care if dislocations exist or not.
Forest hardening assumes the existence of dislocations; and the density of these dislocations increases as plastic deformation progresses. Consequently, it becomes harder to move new dislocations across this "forest" of pre-existing dislocations, resulting in hardening of the material. You can now choose to describe hardening as the increase in shear stress required to move a dislocation through the 'forest' as a function of the dislocation density in the forest. Physically, the observation of "work hardening" at the scale of the specimen is the overall effect of "forest hardening" due to dislocations at the micron scale.

Therefore when hardening=func(plastic work done) --> work hardening
and when hardening=func(forest dislocation density) --> forest hardening

But everywhere it says that: "Work hardening, also known as strain hardening or cold working, is the strengthening of a metal by plastic deformation. This strengthening occurs because of dislocation movements within the crystal structure of the material". So Work hardening does depend on dislocations, its not like it just ignores them right?

Astronuc
#4
Mar31-12, 10:20 AM
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Work hardening and Forest Hardening

I think Bavid's explanation is good. The mechanism of work hardening depends on dislocations, and there are always dislocations in materials, but dislocation density will vary depending on cold work and annealing.

There are also stages in work (strain) hardening. I wonder if some folks use 'forest hardening' to refer to Stage II and beyond. Stage I is characterized by easy glide.

http://www.osti.gov/bridge/servlets/...3/10170133.pdf
STAGE II: FOREST THEORY
The consequence of this secondary slip for the flow stress is that the dislocations produced are mostly "forest" dislocations with respect to the primary slip system. The term "forest" refers to the concept that the flow stress on a given slip plane is determined by the short range interaction of mobile.
. . .
See also - http://aero.caltech.edu/~ortiz/talks/tms-04.pdf
Bavid
#5
Mar31-12, 10:39 AM
P: 32
@Chemist20: We came to know relatively recently (1950s) that dislocation interactions are responsible for hardening. The idea of 'forest hardening' came after this time, when people actually started to try and model dislocations interacting with a forest of other dislocations.

Work Hardening ignores dislocations in the sense that it does not REQUIRE that you know anything about dislocations and how they interact. Work hardening is therefore empirical, while forest hardening explains the actual mechanism of the hardening. Again, as I said before, it is the same observed hardening phenomenon, only interpreted differently. Work hardening occurs DUE TO [forest dislocation interactions] in the crystal structure.
Chemist20
#6
Mar31-12, 05:21 PM
P: 87
Ok, let me see if i got this right:

Forest hardening arises due to dislocation interactions, i.e. The higher density the easier it is for them to interact. Consequently, work hardening takes places, meaning its more difficult to plastic deform the material. Is this right??

Then... Work hardening is like the consequence of forest hardening.


Thanks so much for all your help!!!!!
Bavid
#7
Mar31-12, 06:35 PM
P: 32
Yeah, you got that all right.
Chemist20
#8
Apr1-12, 12:01 PM
P: 87
Quote Quote by Bavid View Post
Yeah, you got that all right.
perfect! thanks so much!!
Paulibus
#9
May23-13, 08:58 AM
P: 175
Dislocations are 'units' of plastic shear, as it were. Metals deform plastically by shearing; the movement of dislocations enables shear. Dislocations are line defects that spread over surfaces; like wrinkles in a carpet that is being pulled across a floor. Moving the wrinkle-line causes a bit of shear if it spreads across the entire carpet. Clever nature has found a bit-by-bit way of enabling shear, rather than an all-at-once mechanism which is much, much more difficult.

If other dislocation lines intersect the surface, like a forest of trees, they impede the movement of dislocations that are causing shear on that surface. Dislocations have to cut through the trees. This is the cause of work-hardening. Hence the name 'forest hardening'. You've got it exactly right now, Chemist 20.


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