E4215 Lec10 Dislocations+and+Slip II
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Transcript of E4215 Lec10 Dislocations+and+Slip II
E4215Mechanical Behavior of Materials
Lecture 10 Dislocations and Slip ‐ II
Prof. K. Barmak
1
Outline
• Single crystals deformed in tension– Slip systems– Schmid factor– Critical resolved shear stress– Rules for identifying slip systems
• Peirls‐Nabarro stress• Dislocation motion
– Glide, cross slip of screw dislocations– Climb
• Jogs and kinks
2
Slip Systems ‐ I
• 5 independent slip systems are needed for an arbitrary shape change
• Dislocations glide on planes of highest atomic density (i.e., on close‐packed planes) in close‐packed directions– fcc Al, Cu, Au, Ag, Ni– bcc Fe, W, Nb
• Slip direction is in the slip plane, therefore, it is normal to the normal to the plane
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{111} 110 {110},{112}or{123} 111
Slip Systems ‐ II• hcp: depends on the c/a ratio
• Basal slip if c/a > ideal ratio 1.633: • Basal slip or prism slip if c/a <ideal ratio:
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{0001} 1120
{1010} 1120
More densely packed than basal plane when c/a < ideal ratio
Slip Systems in FCC Metals• Note that since {} and <> mean families of planes and directions, there are a number of distinct slip system for a given {hkl}<uvw>
• So for example for the face centered cubic crystal there are twelve distinct slip system
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{111} 110 4 slip planes 3 slip directions in each plane
111
111
111
111
110 101 011
Note that the slip directions must be in the slip planes. Therefore, the normal to the slip plane, here, [hkl] must be normal to the slip direction [uww]. The dot product must be zero, i.e., must have hu+kv+lw=0.
e.g., in (111)
Dislocations in FCC Metals
• Relation between slip direction and Burgers vector– slip direction
– Burgers vector
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1 1102
22
a
b
b
110
Vector joining the nearest neighbor atoms
a is the lattice parameter
Dislocations in BCC Metals – I • Close(st)‐packed planes: {110}• Close‐packed directions: <111>• Burgers vector:
– b = a/2<111>– Magnitude of Burgers vector
• Glide occurs on {110} planes, in <111> directions• Each unit cell contains 6 {110} type planes (not counting the planes with the negative of all indices, e.g., count (110) but not (‐1 ‐1 0) )– Each {110} plane contains two <111> directions
• Therefore there are 12 {110}<111> type slips system
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3b2
a
Dislocations in BCC Metals ‐ II
• In BCC, slip can also occur on {112} and {123} planes in <111> directions
• Each unit cell contains 12 {112} planes– Each {112} plane contains 1<111> direction
• Thus there are 12 {112}<111> slip systems
• Each unit cell contains 24 {123} planes– Each {123} plane contains 1<111>
• Thus there are 24 {123}<111> slip systems
• Thus there are total of 48 slip systems in BCC metals.
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Dislocations in BCC Metals ‐ III
• Slip on {110} is most prevalent• Three {110} planes intersect a <111> direction, therefore, unit screw dislocation can easily move from one {110} to another {110} or to {123} and {112} planes, resulting in wavy slip lines.
• Extended dislocations are uncommon in BCC.
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Dislocations in HCP Metals
• Relation between slip direction and Burgers vector– Close‐packed directions: – Burgers vector:– Magnitude of Burgers vector: a
10
11201120
3a
http://www.cmmp.ucl.ac.uk
Deformation of a Single Crystal in Tension
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Normal to slip plane
Applied force
Applied force, F
Slip direction
Slip plane
FA
cos cosFA
(Engineering) stress is force per unit (initial) area:
Resolved shear stress
Slip occurs in a particular crystal direction on a particular slip when the resolved shear stress reaches a critical value, C . Slip therefore occurs on the slip system with the highest Schmid factor.
Schmid factor
Area normal to the load, A
Normal stress in the oblique plane
2cosN
Example 1
• A face centered cubic single crystal is loaded in the [100] direction. What is the Schmidfactor for the slip system given below?
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[110](111)
0.408
Example 2
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Critical Resolved Shear Stress
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• The experimental evidence of Schmid’s Law is that there is a critical resolved shear stress. This is verified by measuring the yield stress of single crystals as a function of orientation. The example below is for Mg which is hexagonal and slips most readily on the basal plane (all other crss are much larger).
“Softorientation”,withslipplaneat45°totensileaxis
“Hardorientation”,withslipplaneat~90°totensileaxis
coscos
Simple Rules for Identifying Slip Systems
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OILS Rule
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OILS Rule ‐ Example
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Diehl’s Rule: FCC Crystals ‐ I
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Diehl’s Rule: FCC Crystals ‐ II
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Diehl’s Rule – BCC Crystals
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Slip and Rotation of Crystal
• Before and after deformation • Note the rotation of crystal when using fixed grips
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Dieter
Geometric Softening and Hardening
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Conjugate Slip System ‐ I
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Conjugate Slip System ‐ II
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Conjugate Slip System ‐ III
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Crystal Rotation During Plastic Flow
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Slip Traces ‐ I
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Slip Traces ‐ II
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Slip Traces
• Slip traces are observed where dislocation glide planes intersect the polished face of a crystal– Therefore, a slip trace represents a direction comment to the two crystal planes
• In cubic crystals, the slip trace can be obtained via the cross product of the normals to these two planes
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Before glide
After glide
Slip trace on crystal face
h1 k1 l1 h1 k1 l1h2 k2 l2 h2 k2 l2
1 2 1 2 1 2 1 2 1 2 1 2k l l k l h h l h k k h
Peirls‐Nabarro Stress ‐ I• The critical resolved shear stress needed to cause dislocation glide is the Peirls‐Nabarrostress (or Peirls stress) , which represents the inherent lattice resistance to the movement of dislocations
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2 2exp1
1
p
edge
screw
G wb
dw
w d
Poisson’s ratio
Interplanar spacing
Burgers vector
Peirls Stress ‐ II
• c (or p) exists• Highest for covalent solids• Lower for metals• Decreases as width of the dislocation core increases
• Decreases with increasing temperature
31
Outline
• Single crystals deformed in tension– Slip systems– Schmid factor– Critical resolved shear stress– Rules for identifying slip systems
• Peirls‐Nabarro stress• Dislocation motion
– Glide, cross slip of screw dislocations– Climb
• Jogs and kinks
32
Dislocation Motion
• Two types of dislocation motion– Glide: Dislocation moves in glide plane, i.e., the plane that contains both the dislocation line and the Burgers vector
– Climb: Dislocation moves out of glide plane normal to the Burgers vector
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Dislocation Glide ‐ I
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• Recall the effect of dislocation motion in a crystal: passage causes one half of the crystal to be displaced relative to the other. This is a sheardisplacement, giving rise to a shear strain.
[Dieter]
Stress‐Strain Curve• Deformation at intermediate temperatures mediated by dislocation glide:
35
Dislocation Glide ‐ II• Dislocation Glide under the application of a shear stress:
• Conservative motion (volume conserved)• Glide of many dislocations = slip
– Slip: two undeformed blocks of the crystal move relative to each other
36
Do not confuse how b was defined, with the direction in which the dislocation will move under an applied shear stress.
Dislocation Glide ‐ III• For edge dislocations, glide plane is uniquely defined since dislocation line and Burgers vector are perpendicular to each other
• Screw dislocations have their line and Burgers vector parallel, therefore, they can easily change their glide plane– This is known as cross slip
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Plastic Shear Strain and Dislocation Glide
• Plastic shear strain associated with dislocation glide
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tan mDshear strain b xh
Magnitude of Burgers vectorMobile dislocation density
Average distance moved by dislocation
Dislocation Climb• Climb is non‐conservative (volume not conserved)
– Vacancies and diffusion are required for climb, and, as a result higher temperatures are required for this type of motion
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Plastic Strain and Dislocation Climb• Plastic deformation of crystalline materials at high temperatures is termed creep
40
Plastic (engineering) strain parallel to b resulting from climb of edge dislocation
eng mh b x
h
Average distance climbed by a dislocation
h
h
d
Jogs and Kinks ‐ I• Jogs move the dislocation line from one plane to another plane– Jogs can form by dislocation climb since climb does not occur over the whole length of the line; it occurs only on certain segments of the line, resulting in the formation of jogs
– Jogs can also form by intersection with other dislocations
41
Jogs and Kinks ‐ II• The jog should be contrasted with kinks, which displace the dislocation line on the same plane
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Jogs and Kinks ‐ III• The jog and kink on an edge dislocation
43
Jogs and Kinks ‐ IV• Jogs and kinks have the same b as the rest of the dislocation• Kinks do not impede dislocation motion, because they are on the same
slip plane• Jog on edge does not impede dislocation glide• Jog on screw does impede glide because has edge character, and requires
climb for motion
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Shear stress