physical metallurgy 22nd lecture ms&e 410 d.ast [email protected] 255 4140
TRANSCRIPT
A deeper look at solute solution strengthening
Important in nonferrous alloys (Bronze)
1) Interaction of solute with dislocation
2) Pinning
3) Unpinning
4) Precipitation
Single solute atom, straight edge dislocation
• Hydrostatic stress field
vr
Gbp zzyyxx
1
1sin
3)(
3
1
• Proportional to shear modulus of matrix
• Falls off with 1/r => long range
• Compressive (negative sign) above half plane
• Tensile (positive sign) below half plane
Interaction Energy, E
In the first approximation this is just
=> pv
where p is the pressure (from the dislocation) and v is the volume change due to the presence solute atom
This is the energy stored in the volume occupied in the impurity atom. This needs a correction for the long range stress field
=> E = pv .(3(1-)/1+) where is Poisson’s ratio,
being typically 1/3 the correction is about a factor of 1.5 which is insignificant given other effects (electronic etc….)
Force
Force is dE/dx|x (slope of the energy curve at x)
This gives
222 ))/(1(
)/(2
zx
zx
z
vGbFx
x/z is the angle as in 1/tg
The force is zero when the solute atom sits directly above and below the edge dislocation. It reaches a maximum at x/z = ….. (You may want to prove this to yourself)
The z can not be arbitrarily small. The smallest value is the distance to the next densest packed plane which is b|/-6 in fcc.
We see
• Force increases with decreasing z (distance between glide p.
• Fmax
22 z
vGbFx
22 z
vGbFx
Largest atom is about 14% bigger diameter (H.R. rules) about 1.5 atomic volume.
Lattice spacing z is about 1/2 b.
Fmax ~ 10-16 G
This is a very small force, so how can it do something ?
Because there are (depending on concentration) order of 1015 such interactions per cm of dislocation. It adds up but how ?
In alloy theory, one measures the change in lattice constant a as element B of concentration c is added to matrix A.
v = 3 where is misfit (in diameter)
The factor 3 comes from the 3 dimensions.
= d ln a/d cB
Inserting the glide plane spacing in fcc (which is 1/2 of the lattice constant !) in terms of b (d =0.408 b) than yields
Fmax ~ Gb2
Since b is on the order of 2 10-8 cm, and d is small this again, gives very small values for F as a fraction of G
Points to worry about
• The theory is continuum elastic, will break down at short distances
• We neglected the Friedel oscillations ( 2 kf see earlier lecs)
• We neglected difference in ionicity.
• We neglected diaelastic interactions.
• We neglected that solute atoms may settle on the SF between a dissociated dislocation at x =0. (Known as Suzuki pinning… has nothing to do with Suzuki motorcycles)
• We can not pin screw dislocation !!!!!
Screw dislocations
• have no hydrostatic stress field
• can not interact with solute atoms that only do volume expansion.
However there are impurities that can “pin” screw dislocations via impurity atom - screw dislocation interaction !
Can you guess which ones ?
Screw dislocations
• have shear field
• will interact with impurities that set up shear fields
Tetragonal stress fields
zz
yy
xx
00
00
00~
Of the C atom. This stress field does interact with the screw dislocation ikik is not zero
2 Remarkable facts about solute solution strengthening by C in the bcc lattice
• It has a tetragonal - not isotropic - stress field
• It does have a much higher misfit than the 14% for substitutional impurities a la H.R. (which above that will not dissolve) and still dissolve in the lattice - because it is an interstitial.
Di-elastic interactions of solute atoms with dislocations
• Impurity atoms changes elastic modulus locally
• In above example I made zero….. Elastically it’s a hole !
• Dislocation will try to move to impurity atom regardless if the impurity atoms is above or below the glide plane !
• Interaction will fall of with 1/r2
I will not do the details but
• d ln G/dxb I.e. the change is shear modulus with addition of B can be much larger than the change of lattice constrant (which is constraint by solubility). Up to 20 times larger
• Thus, the effect, although 2nd order, can be nearly as effective as the dilatory pinning discussed before
• It works for both edge and screws. (The BC conditions at a free surface for a screw dislocation are perfectly met with a single image screw, whereas the edge dislocation, in addition to the image, needs an additional superimposed field)
Suzuki pinning
• Ni is fcc
• Cr is bcc
A dissociated dislocation has 2 partials (a/6 112) separated by a stacking fault.
• Locally, the stacking is bcc !
• If it can diffuse* (!), Cr, will try to segregate to the S.F.
• Once there it lowers the energy
• The partials will separate wider !
• To move, the dislocation must provide this energy (“leaving a Cr rich, ribbon like region behind)
* Cr is not an interstitial diffuser !
Next step
We have pinned the dislocation - how many pinning points are there per unit length, and what is the minimum shear stress to move the dislocation along ?
Important parameters
• Area pinning density
• Distribution of pinning strength
The effective pinning density per area
• Assume pinning centers have all the same strength.
•Area density cb
• NO influence from solute atoms on glide planes above or below
We calculated this problem before, in the HW on the contribution of dislocation bowing to the measured elastic modulus
2/
1212
2
32
GbE
E
bl
R
lArea
ndislocatio
ndislocatio
Key assumption
Once the swept out area is equal to the area occupied - on average - by one solute atom, an other pinning center will be touched by the dislocation
his rather simple approach to a complex statistical problem is surpassingly good as computer simulations proofed. The dislocation is moving from I (initial) to F (final)
atomsoluteoneofAreaE
bl
R
lArea
ndislocatio
1212
32
T
At a given solute concentration, as increases, the dislocation are more and more curvedand touch more pinning points. That decreases l, the spacing between pinning points
How low can l go? If the obstacles are infinite strong, until you hit the geometric spacing, more or less.
Before that, so, the force on the anchor points (the pinning force is rather weak) will be such that the dislocation detaches.
Detachment from pinning points at
Maximum pinning force condition is t
c b l(c) = Fmac
which leads to
disloc EcFbArea
6/2/12/3max
B
dis
bc
El
6
The length decreases inversely with the sqrt of the stress, and solute concentration. A high line tension resists bowing and makes l (for a given stress level) larger
(1)
Ca is the area density of solute atoms
3/
3/)(
2/12/3
22/12/32
Area
Area
cGb
GbcGbb
c
c
Remember that Fmax went with Gb2 ?
Is this reasonable ?
To have a of 1/100 of G (a reasonable value for the yield stress), we need to get the spacing between the solute atoms to order 10 b (I am not worrying about ) That means a few percentages of volume concentration of solute which is reasonable value.
Better theories
• Take account of a distribution in Pinning Force provided by a variety of solute atoms
• Take in account solute atoms on planes other than the glide plane. I.e. Planes above and below
• Take into account summation with dielastic interaction
• Takes into account that at clusters of solute atoms (either by statistical fluctuation of by segregation) are present.
• Take into account that thermal activation permits dislocation to overcome solute pinning at a force below Fmax
(which is the T=0 case)
Temperature dependence
• At stress levels at which the dislocation applies a force F less than Fmax there is a potential hill in front of the dislocation
The simplest try is to represent the potential by a inverted parabola. This turns out to be quite good
2/33/2
max
2/3
maxmax
2/3
maxmax
2
11
1)(
o
FF
FzFFdxU
F
FFxconsxF
The dark area in the diagram, units force x distance, is a k energy required to overcome the
z is the “width” of the obstacle (Force times distance, z is effective distance)
substitution from equation (1)
(2)
Constant deformation critical shear stress, T dependence
Concept
• Dislocation density ~ constant
• Deformation velocity - hence dislocation velocity - constant
Then it must be true that for the U in eq (2) U/kT = constant2/33/2
1),(
oo T
TvT
o and To are adjustable constants that contain information
Constant deformation critical shear stress, T dependence
Concept
• Dislocation density ~ constant
• Deformation velocity - hence dislocation velocity - constant
Then it must be true that for the U in eq (2) U/kT = constant2/33/2
1),(
oo T
TvT
o is the critical shear stress at T=0.To is an adjustable constants that contain information on Fmax
Test
Single crystals of Ag with various additions of Al
The critical shear stress does not fall to zero, but to a plateau, which is believed, in this case, to be due to clusters of impurity atoms, rather than the inherent friction limit of the movement of dislocations, known as Peierls force shown for pure Ag
Various Al contents
Concentration dependence of critical shear stress (at constant T)
For a single dislocation, this went as carea1/2 see eq (1)
For many dislocations, there will be a pile up of dislocations in front of strong obstacles. The pile up increases the shear stress acting on the pinned dislocation above the applied stress. It is conceptually similar to the Hall Petch relation, except that grain size is now replaced by the average space between pinning points. An analysis gives
= Constant c2/3
The constant is a function of the pinning force
Cd is 5% bigger than Au, Zn is 5% smaller
Size Effect solute solution hardeners
Module Effect solute solution hardeners
Y axis : Slope of critical shear stress vs concentration
size effect
Module effect
X axis Combined size and module effect
Au single crystals
More solute atoms effects
• Cottrell atmosphere
• Snoek atmosphere
Cotrell atmosphere is condensation of C atoms in the stress field of a dislocation. Extend about 1 nm. There is no sharp boundary between a Cottrell atmosphere and carbide formation.
Snoek atmosphere is a “cloud” such that the C atom (that is centered on one out of 12 cube edge) around a dislocation are arranged such that the lower their energy in the stress field of the dislocation
In either case, “break away” required.
Basics
• Depinning results in a negative work hardening effect.
• Negative work hardening leads to localization of deformation into narrow bands, called Lueders bands, stretch marks, Portevin-Le-Chatelier effect.
• the localization is exactly analogue to the instabilities in Gunn diodes (that is for the EE types) or the oscillations in C arcs (for early radio buffs)
• Depinning can be as easy as a “fat atom” jumping from the tensile site to the compression side of an edge dislocation. This not only lets the dislocation go, it pushes it off.
•After 100 + years still an active area of research
The localization of strain into shear band - here computed for Al Mg Alloy
Contours in eV; positive energies indicate binding, that is lower total energy. Cores of the split partials are evident on the compression side of the dislocation. Arrows denote flux of solutes around the dislocation core in traditional continuum models. Right: magnification of the core region, with specific binding energies as indicated in eV, and arrows indicating the two cross-core transitions for which the activation enthalpy has been computed
The End
Appendix
Some interesting data for the military metallurgist ;-) … once the toughness is specified at -40 C (-40F) you knowit’s mil spec!
