Mechanical properties of steel 11: texture, dislocations, defects
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00:00
Ship of the lineReference workPlane (tool)Mode of transportCommodore MAX MachineSheet metalFlatcarMaterialTexturizingPlane (tool)Reference workRolling (metalworking)SpantCartridge (firearms)TypesettingSteelShip of the lineKranbauFinger protocolSteeringOpticsPhotographic processingComputer animation
06:49
Roots-type superchargerSingle-cylinder engineSequinReplenishment oilerAngle of attackTexturizingCapital shipBoiler (power generation)Alcohol proofFinger protocolMaterialReference workTypesettingPlane (tool)SteelSheet metalComputer animation
13:38
Single-cylinder engineNut (hardware)Internal combustion engineGentlemanScoutingMcDonnell F-101 VoodooMixing (process engineering)VanCapital shipFiberInternational Space StationSheet metalPlane (tool)Single-cylinder engineTexturizingMaterialAngle of attackSchubvektorsteuerungShip of the lineScoutingCluster munitionAlcohol proofReference workShip classFirearmFighter aircraftRoots-type superchargerComputer animation
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Lawn mowerGentlemanVolkswagen GolfMechanicAmmunitionGlaze (painting technique)SchubvektorsteuerungTypesettingGamePhotographic plateFirearmScrewShip classFlightPlane (tool)ForgingSchubvektorsteuerungTypesettingToolFord TransitInternational Space StationSlip (ceramics)Narrow gauge railwayComputer animationEngineering drawing
28:51
GentlemanVolkswagen GolfSchneckengetriebeGemstoneAmmunitionSchubvektorsteuerungShip of the lineMixing (process engineering)Insect wingWeaponForgingToolShip of the lineFinger protocolPlane (tool)Ford TransitInternational Space StationTextileSchubvektorsteuerungComputer animationEngineering drawing
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Cut (gems)WeaponShip of the lineHot workingKümpelnAngle of attackMusical ensembleTypesettingSlip (ceramics)Slip (ceramics)SchubvektorsteuerungPlane (tool)SteelShip classMaterialBridge (nautical)ScoutingTypesettingShip of the lineRoll formingModel buildingFinger protocolCartridge (firearms)ForgingPressure vesselPenSizingStrappingMusical ensembleWork hardeningSpare partTire balanceScissorsGas turbineWireScrewGround effect vehicleFord FocusStock (firearms)Hang glidingAngle of attackTauComputer animation
Transcript: English(auto-generated)
00:02
talking about. So when you when you look at the like a piece of metal piece of steel right it consists of crystals right the little grains that you see under optical microscope are single crystals and and so if if you would be
00:23
able to do this and identify the crystal structure you'll be able to see identify the orientation of the unit cell. Yes, now when I say this it sounds very
00:41
you know simple orientation of the crystal. However, you know orientation doesn't mean anything if it's not with respect to reference frame right so so you have to you know orientation you know what's your reference. Obviously what
01:07
do I mean for instance if if our reference coordinate axis was exactly the same coordinate axis as for this unit cell this grain would be not you
01:24
know oriented and would be perfectly aligned to its own axis right so but obviously what what is important is the orientation vis-a-vis the what we call the the laboratory coordinate system that's us and and coordinate axis that
01:45
we define because it's you know it's an obvious one to choose for instance when you roll material yes the material you know you will choose a coordinate a laboratory coordinate axis you know along the rolling direction transfers to the
02:04
rolling direction perpendicular to the rolling direction right so this and so and so that's what we do we you know we define the orientation of all our crystals you know with respect that to that laboratory frame as you will
02:24
coordinate axis system and so what would if we do this we we find that some steels when you look at the orientation of the crystals you know
02:41
they're randomly oriented they don't have any particular preferred orientation other cases which happens very often when you deform the material heavily is that you see that the unit cells you know they're aligned not
03:01
maybe not very precisely but you know there is a preferred orientation right and this is what we call texture and it has a big impact on the mechanical properties this this this texture so what I just want to say so so there
03:23
are methods to describe this which are methods which which allow you to visually present this texture yes and so that that's what I want to talk about so one of the things you can do is use what we say a stereographic
03:46
projection right so what is stereographic projection relatively simple concept so just imagine here that this gray sheet here gray sheet
04:01
is a piece of rolled steel yes that I orient with the rolling direction in this direction the transverse direction in that direction so it's it's flat it's lying flat on the surface and then the normal direction is my third
04:22
axis and then and then I imagine a plane excuse me a sphere around this piece of metal right okay and now I focus on a small grain a single grain inside
04:41
this sheet of material yes enough so it and this is this grain this gray thing here that's this grain and I've chosen for as an example that this grain has a one-one-one axis one-one-one axis perpendicular to the
05:02
normal direction all right and so what I do in a in what we call a pole figure which is a a representation of orientation so for this little grain this little crystal here say the orientation is this so now I look at
05:25
this grain and I for this particular example here I also put the one-one-one direction parallel to the rolling direction this is actually a relatively
05:40
common orientation that the grains will have in a sheet of low carbon steel that's been rolled okay okay and now I focus on the one-oh-oh directions this so you can see them now they're they're well defined now right so they will have one orientation or direction that will be one specific
06:05
orientation for this particular grain with this particular structure yeah not structured but this particular orientation all right so now I do so this grain I is here yes and now what I do is I extend these axis you
06:21
very far out of this crystal till they hit my sphere that I have drawn imaginary sphere I've drawn around my sample okay and now I do what I do is that I make a stereographic projection right so stereographic projection is the
06:41
following I looked at the equator plane that's this light gray plane here of my sphere and I connect the points where the one-one one-oh-oh type axis intersect the sphere this this point I connect with the apex yes the lower
07:07
point here in the South Pole let's call it South Pole of my sphere here and I register the intersection points yes okay so I can basically do this for
07:25
many grains you know I just measured the grain measure the grain right and put all these orientations all these different o-one type poles on this plane yeah I know and I can just use this projection yes this stereographic
07:46
projection as a way to represent texture so let's let's see again how this works so this is what I just explained a for a single crystal my stereographic projection will be this this circular plane and then these
08:05
three what we call this point poles yes and you and so this is this circle point this circle stereographic projection sir and these are these three points this for a single grain say I do this for many grains you know
08:25
I look at the grain next to it and next to it and well if a material is textured yes then you know I'll have other yellow points which will be close to this original point this and I'll have many of them of course there's
08:45
many you know if I measure a lot of won't get many points and you can see here these are single measurements yes so I can now use this yes and then mathematically yes change this into a pole density yes it gives you an idea
09:07
of all right in this area yes I have four times the the pole density that I
09:22
should have if the orientation was totally random yes the orientation is totally random there will be you know there will be a you know definitely be a pole here right okay so so the pole density allows me to mathematically see how strong the the texture is so it's basically a mathematical step here
09:47
you you go from these discrete points to describing a pole density okay so you have this and and then of course it's not very practical this 3d
10:01
representation so you represented in 2d and then instead of having this pole densities these these peaks here you you just have density contours and so and this is a pole figure you know a very common way to represent texture
10:20
preferred orientations or absence of this preferred orientation yeah the problem is it with this approach you don't capture all the texture components you don't capture all detection there may be grains that have a specific
10:42
orientation and then grains which have another specific orientation yes when the texturing doesn't have to be all the grains of the same preferred orientation also half of the grains can prefer to be in one direction a large another large number of grains are have that specific orientation but
11:04
rotated by 30 degrees for instance yeah and so you wouldn't see this very clearly and you would need to use many pole figures now so what people did is
11:27
they came up with the idea of using orientation distribution function yeah and the approach is different you know that if you have a laboratory coordinate
11:46
axis a reference coordinate system and you have another orthogonal coordinate system other orientation I have two right you know that you can rotate one
12:02
of them into the other yes you can describe the relative orientation using Euler angles yes Euler angles and and to do this you need three Euler angles so it I'm not going to go into your which how it works you should know this from from math undergraduate math but anyway so to to
12:26
orthogonal axis you can relate their orientation using three Euler angles yes and that's the idea basically so you again have your
12:42
material for instance sheet material you define a laboratory coordinate axis rolling direction transfer normal directions it's one and then you have the cubic you know for steel you have cubic unit cells and so that means you
13:00
have another set of orthogonal axis and with the and you can describe their relative orientation by all your angles so you need three Euler angles for so for instance this grain here yes if I have these three Euler angles
13:22
capital Phi Phi 1 and Phi 2 yes I know the orientation of that particular grain relative to my laboratory coordinate system so so let's see how this works so we have here a 3d representation yes of Euler space so you have a
13:52
Phi 1 in this direction Phi 2 in this direction capital Phi in this direction do not ask me why capital Phi axis goes down right I have no idea why they
14:05
did that but you know there's no profound reason so anyway that's the way it's usually represented right what the z-axis down okay so any orient any
14:23
point in this what we call Euler space is an orientation right so what what you can do in in this approach to description of texture is every orientation of a grain you put point in the graph for instance this point
14:44
here corresponds to a crystal with a 001 plane parallel to the the sheet yes and a 1 bar 1 o direction parallel to the rolling direction so so this
15:06
particular orientation would look like this here so you can see a one of the the 001 axis parallel to the normal direction and in this direction here a 1 1 o direction parallel to the rolling so this would correspond to to this
15:23
and of course the point here 0 0 0 means that basically my unit cell is oriented perfectly parallel to the the unit vectors of my unit cell are parallel to the laboratory reference room so at any point here yes will
15:44
give me a specific orientation so that's for a single grain will be a single point yeah if I have many grains of course I will get many points if my material is randomly oriented these points will be distributed all over
16:05
Euler's space you know there will be no but what we what we see is that there is clustering of the points yes because of preferred orientation okay so again this is not very convenient to use the single measurements yes what we
16:24
do is we create a in three dimensions we create a density plot we make surfaces of equal density of orientation so and then 3d representation also not
16:47
very convenient yes and what we most of the time do is we make sections through this space so if we make a section through through this tube if
17:02
you want of orientations yes this is what we find yes so in in in steels and ferritic steels yeah the important texture components tend to lie on one plane in the Euler space yes so and that plane is defined by the fight to
17:28
angle equal to 45 degrees there in that section we see we can see most of the texture components yes and and so we can analyze very quickly whether or
17:41
not there is a strong texture so what's here on the on the left is a picture of this section where we have the specific texture components specific orientations yes and and here we have the actual what we call the
18:04
orientation distribution function and it's basically a section through the Euler space and these lines here are density contours of orientation so that means that here we have lots of orientation lots of grains are have are
18:22
close to this orientation lots of grains are close to this orientations and many grains have orientations parallel you know lying actually lying along this this line yes and and and this this line this line here and this line there they're they're they're called fibers because they
18:44
correspond to fiber textures this fiber here is called the gamma fiber and it corresponds to the grains with a 1-1-1 axis parallel to the sheet plane yes
19:04
this one here is the alpha fiber yes and it's corresponds to grains with a 1-1-0 direction parallel to the rolling direction yes that's how we
19:21
represent texture in in you know for polycrystalline materials and for for ferritic steels it's different for austenitic steels you know you have other preferred orientations and and so the the the texture the ODFs look
19:45
different of course yeah all right okay so let's just this was just like to make sure everybody was understanding what with these concepts was this is
20:03
also if I've also put it on the e-class so you can print it out and study the material if you want to right so let's now continue where we were so
20:30
from this last after class on Tuesday one of you it was you right no so was
20:41
it you one of you came to me and said as a problem with one of the formulas and that was right so I've in the meanwhile corrected this it's stacking fault energy right so so here in this formula here I had I'd written
21:15
here where you have f gamma I just written gamma right so it should be f
21:21
gamma it's a force yeah it's not a an energy term okay so if you could repair that all right all right so let's continue where we had left I think we
21:43
had shown yeah and so so I already told you that there were two fundamental differences between there was fundamental differences between gamma that equation is correct yeah yeah the equation with the distance
22:11
that is is not wrong right so we had said that fundamental difference between
22:23
gamma iron and austenitic steels and alpha iron and ferritic steels is that the stacking fault energy in austenite you can have depending on the composition and the temperature as we discussed you can have very low stacking fault energy yeah 10 milli joules per square meter to about
22:44
hundred and more yes so depending on the alloy you're studying you can have very narrow dislocation widths or very wide dislocation dissociation with ferrite is totally different the stacking fault image is very very high
23:04
yes and so it dislocation are never dissociated in ferritic steels or in alpha iron okay and we'll see today that this has an important impact right we discussed these two things here the the fact that when you're playing
23:29
around with dislocations dislocation reactions in the austenite you will be using this handy tool to be able to describe the dislocation reactions it's
23:47
it's a tetrahedron which consists of glide planes in the in the austenite and the edges are burgers vectors and the associated burgers vector and you have
24:00
different slip system in BCC right there 110 planes are prevalent will say something more about this today but 110 planes are prevalent glide system so this is a rhombic dodecahedron again all the planes here are 110
24:21
planes and the edges are burgers vectors and they are 111 so a of a of palm to 111 factors and these things are very convenient to much more convenient than the conventional unit cell yes because they allow us to
24:46
quickly see you know what what happens to dislocations right okay so this is where we were so you remember when I was talking about the Thompson
25:05
tetrahedral so you can all see this so you have you had we'll just use the note the letter notation rather than the so you have a B and C point this
25:23
this point here is this fourth point is D and the ABC plane is you know if you look at the lecture note that's the same as the 111 plane yes and we had here in the middle here we had this delta and so this vector here we
25:48
saw can dissociate into these two vectors so and we write the reaction a B is a Delta plus Delta B all right so this reaction now again there is a
26:17
convention to take into consideration is that when a dislocation
26:27
in FCC metals dissociates yes and it dissociates in two pieces where do you
26:43
put where do you put a Delta and where do you put Delta B right well it depends it depends on what kind of stacking fault you have you remember we
27:03
had to two options of stacking fault is it one option was to have a stacking fault that looks like a little sliver of HCP the other option was to have a stack of what it looks like a little sliver of twin and so it
27:21
basically depends however I told you that in most metal and alloys the stacking fault is intrinsic intrinsic so it looks like an HCP sliver
27:42
now why is why would that be well in general the argument is when you make an HCP sliver HCP stacking fault what you remember what we did is you
28:02
can look at it as if you had removed a plane removed a crystal of a lattice plane when you have the other type of stacking fault which looks like a little twin the equivalent the way you can make it yes is by inserting two
28:23
lattice planes put two lattice planes in so the extrinsic stacking fault have a higher energy yes then the intrinsic second calls so that is the reason why we see intrinsic stacking faults yes and that's the reason why we
28:42
can calculate intrinsic stack of faults using free energy difference of gamma and epsilon yes also if it if it was a twin you know would you know the twin has the same crystal structure so there the stacking fault energy would
29:01
basically be zero right so intrinsic and so when you when this happens when we have this yes the A Delta is on this side and Delta B is on that
29:27
side okay so so you remember this your right hand rule yes okay now if you
29:41
want to go one step further and you almost always have to do this in FCC because most of the time your dislocations are dissociated yes you do the same thing that gives you BA and then when it is dissociated you need to know you know which one is left which one is right the partial
30:03
with the Roman letter is always on the right yeah so I I just of course you know you forget this you know so I usually remember this right by thinking
30:23
Romans are always right just a memo memo technical way sentence to remember yeah so just because for me I of course this is convention you need
30:46
to know it and but it's it's it's it's very convenient and the reason is you know when you have dislocation reactions in FCC metals and alloys such
31:03
as austenitic steels when they meet yes the two partials may be reacting with each other forming another partial and if you don't have the the partials correct you know you may end up saying well you know they will repel there will be no reactions or whatever right so you in order to understand you
31:26
know more advanced concept and we'll see one today when we talk about lomer locks yeah you know you need to you need to know you need to know this but again this is you know the convention so important to remember the convention
31:43
and you'll be so that's for again don't start talking about partials in ferritic steels or alpha iron right you will really greatly upset me and don't talk about Thompson tetrahedra to to describe dislocations in alpha
32:01
iron right it's there's no dissociation in alpha iron and you certainly never use a Thompson tetrahedron to analyze dislocations in in fire okay right right so so so for instance remember the loop we had this location loop say for instance our so this would be for instance yes here for
32:30
say we have okay like this orientation and I have here a dislocation loop on this plane so let's let's let's make it a little bit geometrical now
32:47
like this and let's say this the burgers factor of this dislocation is in this direction it's B AB right that's the burgers factor so I'm going to use
33:08
my rule here so I've I've said so I know what the B is right so now I do have to to know where the extra half planes are I have to define a direction
33:28
so so what I do now is like okay this is my B vector yeah and here yes I'm going into the edge orientation right it's turning right look here it's edge
33:42
okay so I have B here you as is going around right you is going like this it follows the line right so I have you in this direction B in this direction so my extra half planes are up on this side on this side yes the B
34:02
is the same but you now points at me it points that the hero is pointing away from me now it points at me so so I do the same this is my line direction my burgers factor is still maybe so it's like this so my extra
34:20
half planes are down right okay and and so so here my extra half planes are down up here down of course when it's dissociated yes it means that the dislocation actually looks like this you know X two extra half planes pointing up and two extra half planes pointing down yes and so what you
34:47
need to have also so now if if I may let's oops have a look here so I said so a B is a Delta Delta B yes so Delta B is on this side a Delta is on
35:19
the right okay so it's it goes like this this this is a Delta this is this
35:28
one and Delta B is on the left it's it's like this all right and it's the same around here all around okay so that this factor so the inside here
35:48
this is on this side this is a Delta right it's a Delta everywhere yeah it's everywhere a Delta this partial dislocation okay so here the extra
36:01
half plane is up here the extra half plane is down yeah so the loop can still collapse so when the loop say becomes smaller and smaller yes this extra half plane will come together with this one and the dislocation will
36:21
disappear okay that's as it should be okay right okay and so now I need to jump ahead here because it was just some in something I had inserted right
36:43
and now we come to to to to to to this here and it's the we introduce a concept of a line tension that is similar but not the same as you know
37:08
surface tension of a liquid for instance when you have a liquid film like a soap film and you you you can extend it to extend the surface yeah the surface tension yes will make it such that you you need to apply a
37:28
force to make this surface larger okay so we have something similar in for dislocations and we call this the line tension and of course it's related
37:42
to the energy right so it's it's the derivative of the energy basically and what it basically means is that if you imagine a dislocation as a a wire
38:03
yes when you it will always want to be very straight yes and so if you imagine actually a this this dislocation all by itself yeah you you can look at
38:24
very very often in the future I instead of talking about dislocation yes I will talk about dislocation segments yes and basically that's the piece of dislocation we can be short or long whatever yes that is stuck at some
38:43
kind of pinning points yes so it's like it's like like the dislocation is like pinned in these two positions yeah the the reason why we use this image very much in practices is because that's usually what happens to
39:01
dislocations they they they not you know big dislocations which don't interact with anything yes in in your grains most of the time to just run into each other yes and they pin each other or you know if you're if you're
39:23
your steel contains precipitates you know they'll run into these precipitates that will prevent them from moving yeah or if you have alloying elements they will run into that particular alloying element and be more or less pinned by the presence of this island so that's why we you know
39:43
we usually think of dislocations not as long dislocations but as dislocation segments that are pinned yes okay so this here you have such a dislocation segment and we have you know we apply certain force to this
40:03
dislocation segment and say the forces are such that dislocation you know takes on a certain circular shape you have to imagine here I have some kind of shear force which are called tau yes and so I can you remember we we
40:26
can calculate what this force is on the dislocation using the peach color formula and so that that force will be tau times the the burgers factor of
40:41
dislocation right good now if I would cut a piece of this dislocation out of the dislocation it would in order to keep it in the same shape yes I would
41:12
I need to have something that balances the force I applied so so I cut this here so let's say this is a pair of scissors I cut this out and I look at
41:25
this piece and if I don't do anything yes it's going to run away right just because there's only this force so the the balancing force yes the thing that that keeps the the dislocation in equilibrium that's the line tension so
41:48
you imagine it as being a vector along the dislocation line yes and remember
42:07
it's it's a few moments later previously is that this this force is related to the energy of your dislocation so it's a derivative of
42:24
this energy and and so we have a based on that we know what the size typically
42:42
is of a the line tension it's GB square over two yes of course this equation is a very simplified equation it doesn't have it it just gives you a pretty good order of magnitude but it doesn't you know it's it's different
43:04
the this line tension will be different depending on the the type of dislocation you have yes depending on the dislocation direction etc so you there are many more advanced formulas but this is a pretty good one to work
43:22
with and certainly in in this course right all right so so this is the same thing here so we assume we will very often assume that the shape that the dislocation has is a it's a semicircle is a you know we can look
43:45
at it as being a segment of a circle yes again it does not necessarily have to be this way but it's a good approximation so if we have a segment DL the angle here is called d theta and and the the semicircle has a
44:08
certain radius radius of curvature so the line tension yes has a component
44:21
in the horizontal direction and component in the vertical direction the component in the horizontal direction they balance each other right these two balance each other so what is important here is the the vertical component of the line tension that you that you have to consider okay right so we can do some
44:54
really interesting calculations even with this very simple model so so first
45:04
of all we we look at what is the size of these the vertical component of the line tension so that is I have one here this is so the projection of T vertical projection you can easily calculate it because this
45:26
angle here yes is equal to half this angle that's from geometry okay so and the the force on the dislocation is tau yeah tau times B to be multiplied
45:48
with the length of the dislocation which is DL in this case so so tau B DL is balances T times d theta okay and well DL here is this angle d theta
46:11
times R right so DL can be replaced by R d theta yes and now I I take these two equations together and I just say I express the equilibrium of
46:22
forces yes this here is equal to that yes and I can now calculate the relation between the radius of curvature and the the the applied
46:42
external applied force yes so obviously if I have a very large shear force the R value R will go down and I'll have a more circular shape to the dislocation right so let's so this this equation that was down here is now
47:09
up here so there is a relation between the applied the shear force on the slip plane of the dislocation and the radius of the dislocation and there is
47:26
a an example here of how you apply this formula so for instance in when you fatigue materials when you fatigue steel you know what fatigue test is
47:41
right you have a oscillating force yes and you apply this for thousands of times yes well if you look at the microstructure of the the steel then what you find out is that you've got this incredibly nice three-dimensional
48:07
organization of dislocations you have very dense dislocation bands here and then channels with almost no dislocations that's one thing the
48:20
interesting thing is that these bands consists entirely of edge dislocations whereas the dis the few dislocations we see here are edge dislocations very interesting intriguing microstructure what you can what you also see is that rather regularly you you see that some dislocations are form
48:50
semicircular loops like like the one there on top so what I can do is I can now try to imagine okay could there be any stress in the material yes that
49:07
will cause the dislocation to assume this shape stress for instance due to the presence of this you know high density of edge dislocations and so what
49:20
I do is so I measured the radius of this this loop here and and I plug it into this formula yeah I don't need much more much to calculate because G is the shear modulus and B is my burgers factor so I know these things yes so
49:40
plug this in and I find 27 mega Pascal right and so you can actually use this the formula if you have in your microstructure managed to freeze in yes the the position of the dislocation before you make a sample or before you
50:06
remove the stress so what obviously in this case there was no external stress it was just stress inside the material internal stresses but in other cases you you know the dislocation will assume a certain shape because you apply
50:23
an external stress right in that case to freeze in the dislocation their position you will for instance irradiate the material yes so you create lots of point defects which will pin the dislocation in their stress position and then after that you make a sample and you can you know you
50:43
can measure the the shape of the dislocation the radius of the dislocations and then from there determine what was the shear stress on these dislocations just using this equation okay so now let's look at an
51:09
essential difference between ferrite ferritic steels austenite and austenitic steels and the big difference which is a direct consequence of the
51:23
stacking fault energy is the way cross-slip happens okay okay so let's just look at our slip planes here BCC slip planes okay and so let's just say
51:40
I have this you know the edges are burgers factors so I'm looking at the edges yes one two three four yeah so I have four burgers vector of course these two are the same yes so I have two two burgers vector actually I still
52:04
have four because I have I can have a burgers vector in this direction and in this direction so they are actually four right if I take into account the design of the so let's say we have the burgers factory and and I have a dislocation here so just to make my life our life easy yes we'll just make
52:25
it nice and rectangular yes and so this is the burgers vector yes and also for the sake of convenience I don't know if I will use it but let's say I've also defined a line direction all right so the can everybody see it
52:47
so say this I apply a stress and the dislocation loop the size of the dislocation loop increases yeah all right what happens here what what what
53:07
can happen when the dislocation encounters this plane well let's think this burgers factor is the same as this burgers factor so if I had instead
53:25
of having this situation if I had had a similar dislocation here like this that piece of dislocation would be exactly the same as that yes so for
53:42
these pieces of dislocations yes you can you can get what is called cross slip this piece of dislocation can just move move from here from here to
54:13
here when it does that it's on another glide plane yeah and it's cross slipped
54:26
now it cannot happen to all the segments of the dislocation it cannot happen to the edge parts yeah because the edge parts there so the
54:43
burgers like this but the line is in this direction right so it cannot go into the other slip plane another slip plane all right so only screw dislocations can cross slip only only screw dislocations can cross slip okay
55:06
only screw segments of dislocations now okay let's continue our story here so now I have I have this this plane this plane is the same as this
55:20
one orientation wise but it's it's not the same plane it's parallel to this one but at a certain distance now what happens if this dislocation that's now here which used to be there decides let's say because of the applied stresses yeah that he wants to go back to a plane that's parallel to this
55:44
there's no problem it can just go here so I can have this location move up down to another glide plane yes by cross slipping yes let's do the same
56:09
with our FCC crystal okay so so in this case I need to put it down on
56:35
this white board here okay so let's see what happens here if in this case
56:45
let's say we have a dislocation here no this and this would be the burgers vector this would be my line direction okay now this dislocation size
57:02
increases and the screw part arrives at this point yes so this plane that's that's this plane so it's also glide plane this burgers factor is also a
57:21
burgers vector in this plane so the dislocation can move the screw segment can move into this plane also it can also cross slip however yes and this is the big difference our dislocations here in FCC they are dissociated yeah
57:50
so the burgers vector here isn't parallel to this so let's see let's see
58:02
so let's say it has this vector yeah this vector and the other one has this vector you can see this vector here is not a vector of this plane yes so the dislocation cannot glide into the other glide the other glide plane because it's
58:27
dissociated and the dissociations the partial dislocation are not burgers vector of this cross slip plane so no no cross slip no cross slip unless we
58:41
manage to pinch the dislocation push the two partials back together yes and they and and remove the dissociation okay so this is what happens here this is what I just drew there so you have one dislocation here a screw segment here yes if it's dissociated yes first I will need to remove the
59:11
dissociation then I can get cross slip yes and then of course the dislocation will again dissociate yes making it again difficult for that dislocation to
59:27
cross slip again that happens when you have a low stacking fault energy steel if you have a very high stacking fault material steel then of course it's there is no dissociation and and the cross slip will be easy yeah
59:47
okay in ferritic steels we have huge stacking fault energy so there's never any problem for the screw dislocations to to hop from one glide plane to another one
01:00:00
Yes? And that is very, very fundamental difference, because in the case of ferrite and ferritic steels, your dislocations can easily change glide plane. They easily change glide plane. So if they run into an obstacle,
01:00:24
whatever, yes, a solute atom, another dislocation, a precipitate, yes, they can just go around it. They can just go change glide planes and circumvent the
01:00:43
obstacle. Not so in low stacking fault austenitic steels, yes. There, the dislocations will tend to stay on their original glide plane, whatever, yes. And so
01:01:04
as a consequence, these two crystal structures have very different strain hardening behavior. This material strain hardens much less than this one, yes, because
01:01:24
cross-slip makes the obstacles less efficient as pinning points. Okay, so I think it's a pretty good moment to stop, so I can let it sink in, as it were. I
01:01:54
tomorrow, at three o'clock, we have, we meet here again for the makeup
01:02:02
classes, yes. And that's the most important point I had to make. Thank you very much.
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