Topic Outline Topic Outline Types Slip What is

Document Sample
Topic Outline Topic Outline Types Slip What is Powered By Docstoc
					    Topic Outline
     What is Slip?
     Where does Slip Occur?
     Slip Systems
     Burger’s Vector
     Schmid’s Law
     Slip Bands and Critical Stress
Your Responsibilities
Sample Calculations

                                      CHE 294   1
    We recognize two types of line defects.
                                     Looks as though we are
   Screw                             tearing apart the material.
                                       A dislocation line marks
                                      the boundary of the tear.
As we walk around
the line, we spiral
through the crystal
like a cork screw.                               Edge
  Looks like it could contain an additional
  plane of atoms ... this is not the case
                   because one plane is
Line defects are commonly called dislocations.
                                                          CHE 294   2
                     What is Slip?
          Slip is atom motion due to stress.
  unit cell under
  no stress
                                                apply a stress

bonds stretch to                                obtain a strain
their limit before                                apply more
breaking                                                stress

                                                  atoms slip

broken bond                                          release the
leads to a                                       stress, and the
defect inside                                        atoms relax
the material     the defect caused by slip remains         back

                                                           CHE 294   3
        Where does Slip Occur?
                 Compare these two planes.
                                              So slip occurs on
                             lattice planes       close packed
                            can easily slip     planes in close
                                   now apply          directions.
                                   a stress
                             lattice planes do not slip as easily

a (110) plane

BCC Crystal
                    a (100) plane

                                                          CHE 294   4
                  Slip Systems
       Slip Planes + Slip Directions gives Slip Systems.
 Lattice Slip Planes    Slip Directions # of Slip Systems
 FCC         {111}           <110>            4 x 3 = 12
 BCC         {110}           <111>            6 x 2 = 12
             {112}           <111>               12
             {123}           <111>               24
 HCP        {0001}          <1000>                3

You are responsible for learning
• placements of all slip systems in FCC
• placements of all {110}<111> slip systems in BCC
• the total number of slip systems in FCC, BCC, and HCP

                                                           CHE 294   5
 What are the Miller indices for each of the following directions?


                             [111]               z



   (1 1 0) plane of BCC crystal

                                                           CHE 294   6
    Defining the Burger’s Vector
       The Burger’s vector completes a loop around
                   the dislocation line.
                                            start here to
         Edge Dislocation                   make a loop
              +1 in x                  (this is arbitrary)

   +4 in y                           +3 in x    The length of the
                                             Burgers vector |b| is
                                           the distance between
                                              the two atoms that
Burger’s                                      complete the loop.
 Vector                                       You should learn
             -4 in x           -4 in y       how to determine
                                           |b| for cubic metals
                                                 and ceramics.
                                                             CHE 294   7
                      Slip in Ceramics
                          Slipping ions is difficult!

          +               +
  +            +
      +               +           Consider slip along
          +               +
              +                   <100> in NaCl
  +               +
                          We cannot move a cation into the slot
                          of an anion during this slip.
    We have to slip the cation past other cations into
    the slot of the next closest cation.

Slip in ceramics is limited by both ionic repulsion (like charges
repel each other) and the different sizes of the ions (cations
are larger, making the slip plane less smooth).

                                                                  CHE 294   8
       Line and Atom Motion
     How do the lines and atoms move during slip?

                               Atoms move parallel to b
                                  (in the slip direction
                                    on the slip plane)
                               Line moves parallel to b
                                  (on the slip plane)

How does this picture change for a screw dislocation?

                                                    CHE 294   9
       General Characteristics
  These are the general characteristics of dislocations.

           b Vector             Line            Atoms
Type        Points             Moves            Move
Edge          to line           || to b             || to b
Screw       || to line            to b              || to b

       means perpendicular      || means parallel

Atoms are always moving parallel to the Burgers vector in the
slip direction on the slip plane.
      Convince yourself that you understand this table.

                                                              CHE 294 10
                 Force Required for Slip
            How much applied force is needed to cause slip?
 Consider a
 single crystal       Fresolved         3. Resolve that force on
                                        to the slip plane ... in the slip direction.
 rod.                      slip
 1. Apply a
                           direction              τresolved = Fresolved / As
 tensile force
                                                                  4. The resolved
 along its
 axis.                                                            stress is a shear
        F                                    As                   stress.

                 Ao          φ            slip plane            5. Express the
                                                                resolved shear

σ = F /Ao
                                                                stress in terms of
                                                                the applied
                           normal to                            tensile stress.
2. Calculate the
                                             τrss = σ cos φ cos λ
                           slip plane
applied tensile stress.

     This formulation is known as Schmid’s law.
                                                                                CHE 294 11
                   Vector Notation
Schmid’s law can
be expressed in
vector notation
with Miller              cos λ = T • R / | T | | R|
You already know how           T
to calculate angles
between vectors using                             φ
Miller indices and dot
 An example using                               cos φ = T • n / | T | | n |
 Miller indices is ...
                                   (hT hR + kT kR + lT lR)
         cos λ =
                          (hT2 + kT2 + lT2 ) (hR2 + kR2 + lR2 )
                                                                    CHE 294 12
What is τrss in terms of σ for this example in a BCC crystal?

                        slip    τ along [1 1 1]

                                      σ along [1 1 0]

                               φ = 90o (by inspection)
                               λ = 35.3o (by calculation)
 (110) plane             τrss = 0!           Why?

                                                            CHE 294 13
  Slip Bands and Critical Stress
  Consider stress applied to a single crystal rod or grain.



Slip causes bands to
appear in a single crystal or             τ
within a grain.

The stress needed to start slip is the
critical resolved shear stress τcrss
                                                              CHE 294 14
        Your Responsibilities
             You should learn ....
• the slip systems given previously
• where to locate the Burgers vector in screw
 and edge dislocations
• how to calculate the length of the Burgers
 vector in metals and ceramics
• the direction of motion of dislocation lines
 and atoms relative to the Burgers vector in
 edge and screw dislocations
• how to use Schmid’s law backwards and
                                                 CHE 294 15

Shared By: