.

Holy fuck.

Ill bump this, because it went clean over my fucking head.

No. They aren't larger; they're atomically denser. What alloy are you talking about? Is it a steel alloy, or are you working on the new transparent aluminum alloy for clear AR-15 recievers?

.

btt

Go here:  Ferium S53

You come here asking questions about this stuff?  I hope you are not building aircraft or something along those lines.  Geesh.

ETA:  Why don't you call the manufacturer?  They might know, or do you work for them?

.

btt

Materials Science and Engineering: an Introduction

By William Callister



the answer lies within.

I think that  Ashcroft and Mermin (Solid State Physics) would give you a decent explanation as well.

Bassically, you can have a slip fault along any of the preferential crystal directions, so if you slip a single plane over, the next plane can rebond, in this manner a linear dislocation can move through the material.  Any type of discontinuity in the crystal, be it an ampty lattice site, an inerstical attom, an impurity attom etc, causes the energy required for the slip to be greater, thus the material is stronger.

Sorry not a meturlagist, just a lowly Physicist.  (Which means I need drawing to explain anything well)

Quoted: I have a nanoscale M2C dispersion in my alloy.  This strengthens the metal through tempering. This is done to hinder dislocation movement, correct?  When an atom from a nearby plane breaks its bond, it can rebond with the edge of the dislocation.  This can cause the dislocation to move. My question is how does the M2C block this from happening.  It is because the atoms are larger than the alloy atoms? I also have a fine grain lath matinsite.  How does this help?   I know some of this stuff has to due wit the recovery after cold working. Scott

As long the dislocation movement is hindered, why does it matter if it was done by substitutional or interstitial atoms?  That question is more academic than practical in my mind (please correct me if I'm wrong).

How long was the material temepered and how many percent of lath martensite are still present?  The majority of the martensitic phase should be transformed to the globular form after tempering h
If I remember it correctly, the strengthening mechanism gained by the presence of lath martensite is primarily due to extensive lattice strain and cleavage.  This multiplies the presence of dislocations, which will hinder each other's movent (due to multiplications and pile ups).  The boundaries from the fine grained structure would make it harder for the dislocations to move because of the directional changes from grain to grain (I'm sure you already know that).

You think too much

Oh, by the way, M2C dispersion is still in effect an Iron - Carbide phase, so the strengthening mechanism is achieved through interstitial atoms (i.e., carbon).

Quoted: Materials Science and Engineering: an Introduction By William Callister the answer lies within.

I think the poster who asked this question is beyond an introduction to Materials Science and Engineering.  Good book, though.  Knew it by heart.  "Mechanical Metallugty" by Dieter is another good book, albeit a liitle too....what's that word, not detailed enough

.

btt

.

Quoted: I also need to find how when MC and M2C precipitates out and why it does and why/how the carbides get small. A fine grain has more boundary area and stops the movement of dislcations.   Why do these things happen in terms of energy, bonding, etc.

THey precipitate just like any other form of carbides, through nucleation and diffusion process, don't they?  The main ingredient for this transformation kinetic is low diffusion rate and high nucleation rate of the particles.  From the alloy's TTT diagram, would it be possible to figure out the heat treating schedule to see how the second phases precipitate out of the matrix?

As far as bonding and energy are concerned, fine grains do not only hinder the movement of dislocations due to drastic changes in prefered slip directions from grain to grain.  The grain boundary is also super rich with dislocations.  Dislocation densities along grain boundaries are significantly higer than that found at the center of the grain.  This causes dislocation multiplication and pile ups to occur, causing dislocation movements to stop due to entanglements.  Dislocations along grain boundaries must also "rearrange" themselves for metal to deform and the more grain boundaries you have, the more dislocations there are, the harder this process to take place, causing the fine grained material to be stronger.  Good for applications in ambient / slightly elevated temperature, but in elevated temperature, fine grain materials tend to do poorly.  HIgh dislocation densities also means higher stored energy, and with the introduction of additional energy from elevated temperature, grain boundaries starting to slide along each other, easing the deformation process.  That's why I understand that metals for turbine parts are directionally cooled to produce what is essentially a single grained material.

Is the above discussion what you're looking for?  OR did I just waste your time?

.

btt

.

Get it hot, and hit it hard.  Nothing else important to know about metals.

[Just kidding.  I would say a metallurgy text book from the university library would be a good start, or ask the prof if you can research his books.]

btt

Quoted: Also , would the MC and M2C be substitutional or intersitial precipitates?

Precipitates are clusters of atoms bonded together.  They are not substitutional or interstitial, but the precipitate's carbon atoms are interstitial atoms.  

Can you analyze the precipitate and the matrix composition using XRD / EDS?

.