physical metallurgy 18 th lecture ms&e 410 d.ast [email protected] 255 4140
TRANSCRIPT
Martensite II
By now, this should be familiar !! So we will take time for a side tour into practical metallurgy: How to make a Japanese sword :-)
Watch this video
http://www.youtube.com/watch?v=fM9zhwdIRgY
called Bainite Katana. What Mr. Clark tries to makes is bainite in the center (for toughness) and martensite at the surface (for hardness).
Edgar Bain was a metallurgist at the US Steel Corporation (when they still had a research lab !)
The 1939 book
The Alloying Elements in Steel,
by Edgar C. Bain
is classic.
You can download it free from the WEB at
http://www.msm.cam.ac.uk/phase-trans/2004/Bain.Alloying/ecbain.html
Bainite is a form of Fe3C
Martensite is not a compound but Fe + C (of varying %)
Finer point:
The c/a ratio of the bcc phase is not 1. It is between 1 and 1.08 depending on the C content. That is martensite is tetragonal
The lighter blue part of the blue (111) plane in fcc is also (one half) of a (101) plane in the tetragonal cell which we could alternatively choose to describe the fcc lattice
The Kurdkjumov Sachs relationship applies to plain carbon steels and is as expected from the Bain construction
In Ni-Fe-C steels the orientation between austenite and ferrite is more like Nishiyama-Wassermann
(111) || (110))’ [211]||[110]’
The relationship, experimentally, has some scatter as steel by definition is alloyed (at least with C but in reality even in low C steels with many elements such as Mn….)
The two only differ by about 5%.
Think of the fcc phase as a ball. To get it into the Bain cell you squash it down (view from top is now a bigger circle). View from the side is an ellipse (“lying on its side”).
The Bain Transformation fcc => bcc. You can find an invariant line in the first step, but it won’t be invariant in the second…..
The problem
The fix
After the first compression you rotate the ellipsoid such that your original plane is again an invariant plane in the stretching!
HW 18-1
a) Find the angle that the invariant plane makes in the starting sphere with the x axis (horizontal axis) if the sphere is to be extended into an ellipsoid by a uniform compression in the z direction to 75% of its original height.
B) To keep this plane invariant, in a subsequent compression, find the angle you have to rotate the ellipse before stretching it in the z direction to 120%. (Hint, you can just as well cut a sphere into the ellipse prior to stretching it, and see on how you need to rotate the sphere). Give answer in degree, and if rotation is clockwise or anticlockwise.
Hint
I do not care how you do this. The previous page was done by using the stretch and compress image in MS Paint, with the rotation done in an other program that permitted rotation by degree. A few trial and errors got me there. Math is more exact but more work
The same, in a condensed form
The rotation is the matrix R in the lecture notes of lecture 16. (A rotation does not introduce strain, which is why we need to take the curl out of the displacement field …remember ?)
We went over this before. You need to minimize the macroscopic distortion that you set up in the austenite. You can only do this “on average” as you do not want to make infinitely thin twin plates in the martensite ! Twin boundaries cost energy (even though they are low, about 1/30 of surface energy, they add up)
Yes, we went over this before. And remember, you need the volume fraction of the twins that form in the martensite as an adjustable parameter. It allows you to “rotate back” the shear introduced by the austenite martensite transition.
The transformation
An other diffusion-less transformation… very interesting !
Geometry
The bcc lattice has an interesting stacking of (111) planes
The (111) planes have the lighter blue unit cell that contains one blue atom floating above. The blue atoms make the next layer of (111) planes. The plane is identical to the red plane but shifted
The stacking sequence of (111) planes in bcc using the light blue unit cell (which has 60o angles)
Two bcc planes can “collapse” into one new plane - shown in darker blue.
We now have hexagonal stacking with c/a = 0.63.
In pure metal the transition can be introduced by pressure
Or, you can add electrons….Hume Rothery driving force !
The phase is of great interest in the transition in Ti, which is in the hcp ( form below 882 C and bcc ( above it.
The collapse of the two planes into one is preceded by very large oscillations (“mutual visits”) called a soft phonon mode, of great interest to physicists.
We discussed Ti alloys earlier, so if you forgot go back.
Work from our own department
Stiff, strong, and tough… well
• Metals are tough because the high cohesive energy of the electron gas wants to hold the compensating ion charges together at all cost. So opening up a crack in a true metal is tough work (machine Cu !)
• Strong - well yes, in the sense above
• Stiff - oh..oh - unless you have something that impedes the movement of dislocations metals are soft as butter. Fortunately most metals have enough dislocations that they impede each other. In some metals with directional, covalent like bonds, the “friction force” on dislocations is higher (Peierls force) but in true metals like Cu, a single dislocation moves extremely easily.
Yes…. In particular because the d electrons can do all sorts of interesting things ! See previous lectures on color of metal oxides.
As to metal plating: At the age of 16 I worked in a metal plating factory plating Cu/Ni/Cr onto VW bumpers. This is an rather interesting process, so the appendix has slides how this is done.
Note: True elastic behavior provided by bonds is difficult to separate from reversible bowing out of dislocation, see previous
homework problem.
Unfortunately the IEEE pubs on line are not in color.
But note that the scale goes from 132 GPa to 189 GPa
A bit of a follow up: Young’s Modulus for Silicon
This variation, in turn, is frustrating the attempts to make perfect Si spheres as new units of the kilogram
HW 17-2
Calculate Youngs modulus for Si bars in which the tensile axis is || to 100, 110, and 111 directions
In low T deformation you need shear. The tensile axis and the shear plane are ~ 450 degree oriented. The specimen extends || to the stress axis indirectly, by multiple shear throughout
In high temperature deformation (creep) the tensile stress is coupled directly via diffusional mass transport to elongation
Sir Rudolf Ernst Peierls. Born in Berlin Germany, died in Oxford, GB. Assembled the first nuclear bomb , by hand, in Los Alamos, NM.
Frank Nabbaro Born in London, died in S. Africa
Peierls Stress
• A dislocation moving through a crystal moves in a rippled potential
•The stress needed to drive the dislocation is called the Peierls-Nabarro stress, p. Calculation of p is difficult, and depends on the ratio of the dislocation width w to the Burgers vector b.
• Simple model
yields for w=b (narrowest dislocation) G/180.
The dislocation can move at lower stress levels if it were to spread out, with a wider core w. When, and to what extend this occurs has been the subject of much discussion :-)
HW 18-3
Find the dislocation and make a sketch of it, using the familiar symbol
TiAl
The most famous: Sn in Cu => Bronze
The most you use : O in Si => IC silicon
Stress to “unlock” dislocations varies factor 5 as O content varies factor 5.
Flow stress (plastic) varies about a factor 1.3 as O content varies a factor 3
Stress Strain of Si at 800 C (wafer processing temperature
Classical model of moving dislocation “cutting forest dislocations”
We talked about this before. Does not hold if grains smaller than few tens of nanometers.
Derivation• the number of dislocations in a pileup N is proportional to both the applied stress A and the pileup length L, N ~ AL. (1)• The total stress on the lead or pinned dislocation,lead, arising from A and the interaction stress from thetrailing dislocations is given by lead = NA . (2)
Equations (1) and (1) give
Llead
A
And if you assume that when slead reaches a critical value you have Hall Petch
Particles must be either bypassed, dragged, or they must be cut.
Bypass is usually by Orowan Taylor but can also go via cross slip
After break away small unslipped regions surround the particles. These unsheared regions are separated from the slipped regions by dislocation loops left behind the moving dislocation
Dragging and Orawan Mechanism
Bypass : Orowan equation
Cutting particles
• Requires that particle is coherent
• Moves one half of particle, adds surface of order RbThe derivative of this force relative to dx (i.e. b) gives the force the particle exerts on the dislocation.
• This is balanced by the force moving the dislocation, length L, Force per unit length ab and thus
= R/bL
Diffusional creep
• 3 Stages
• Stage I => transitional creep, acquiring new equilibrium stress distribution. More later.
• Stage II => Steady state creep, only stage for which deformation laws are available.
• Stage III => Geometric instability (akin reduction in cross section in tensile test) lead to run away creep
In general, steady state creep, alias stage II creep, proceeds as
n 0
Where n has a value of 4…5. This has dramatic influence on the stress distribution in loaded beams that - elastically loaded - begin very slowly to creep
The compatibility equations require that the strain rate increases linearly from the neutral axis. But because the strain rate increases with the fourth to six power of the stress, the stress necessary to satisfy the compatibility equation, at the surface of the beam is dramatically lower, once the creep constitutional law takes over from the elastic stress distribution
Stage I creep
Vacancy transport can occur either through the bulk or along grain boundaries.
The first tends to dominate in metals, the second one in ceramics - rough guideline only !
Narbarro Herring creep is vacancy transport through the bulk. Same Naborro as encountered earlier.
Coble creep is vacancy transport along grain boundaries. Dgb can only be measured as product of the true D times the GB thickness.
Robert Coble worked at GE were he perfected the quartz envelope for the Na high pressure lamp. He then became a famous professor of ceramics at MIT. He became a professor emeritus in 1988, at the age of 60, but continued research. 4 years later, after he lost his research funding, he walked off into the ocean of Hawai and drowned.
As an example, I will discuss Selenium and the discovery of photoconductivity and then of the photovoltaic cell !
From Semiconductor Research before 1900… how is all started
In1873, Willoughby Smith arrived at the discovery of photoconductivity of selenium. How he reached at this observation has an interesting story. He was initially working with submarine cables. He set into experiments with selenium for its high resistance, which appeared suitable for his submarine telegraphy. Various experimenters measured the resistance of selenium bars, but the resistance as measured by them under different conditions did not agree at all. Then Smith discovered that the resistance actually depended on the intensity of incident light. When the selenium bars were put inside a box with the sliding cover closed, the resistance was the highest. When glasses of various colors were placed in the way of light, the resistance varied according to the amount of light passing through the glass. But when the cover was removed, the conductivity increased. He also found that the effect was not due to temperature variation.The first observation of photovoltaic effects in a solid system was made in 1876 again on selenium. W. G. Adams, along with his student R. E. Day at Cambridge. discovered that illuminating a junction between selenium and platinum had a photovoltaic effect. In 1883, Charles Edger Fritts, a New York electrician, built a selenium solar cell It consisted of thin selenium wafers covered with very thin semi-transparent gold wires and a protective sheet of glass. This was the first large area metal semiconductor junction device. However, it was very inefficient (η<1%) in converting solar energy into electrical energy.
The END
Excursion into practical metallurgy: Plating car bumpers
This process is rather lengthy, see appendix.
The base metal has to be decreased thoroughly, the old but now illegal method practiced in my company was to dump the entire bumper into carbon tetrachloride. A somewhat safer method is to use tetrachloroethylene, “drycleaner’s fluid”
All plating solutions were cyanide based, but contain numerous organic modifiers to control nucleation and growth. I put a paper on what these do on the 410 website
If the bumpers arrive new from the factory, like in the company I worked, they are visually inspected, small rust spots polished out, and then thoroughly decreased by dipping the entire bumper into hot dicloroethylene.
You hang the bumper into the plating bath by hooks that grab it from the back, as those spots won’t be plated.
The idea is to eliminate pinhole defects where the Cu plating did not take, by smearing the copper over the substrate. Our quality control was good, we only did spot polish. The objective is to have on the front of the bumper a uniform Cu layer on which the Ni can seed. Ni will not seed uniformly on steel.
At this stage, we inspected both sides of the bumper. Concave areas in the back might have blisters in which plating solution has been trapped (this is related to the electric field line distribution). Blisters were popped manually and polished out. Then the entire backside of the bumper was painted with “silver paint” which is suspension of Al particles in an organic vehicle.