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CHE 333 Class 15

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Strengthening at COLD temperatures. Metals basically all work in same way ... In this case the crystal structure is one that has a very high critical resolved ... – PowerPoint PPT presentation

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Title: CHE 333 Class 15


1
CHE 333 Class 15
  • Strengthening of Metals.

2
Strengthening at COLD temperatures
  • Metals basically all work in same way
  • which is to block dislocations or retard
  • Them.
  • Remember COLD is less than 0.3 Tm in
  • Kelvin

3
Strengthening Mechanisms
  • To optimize properties of metals, greater
    strengths can be achieved by several
  • techniques-
  • Cold Working.
  • Grain Size Control
  • Solution Strengthening
  • Second Phases.
  • New Phases.
  • These are the engineering alloys that are used
    for structural applications. Pure materials
  • are used for electronic and electrical
    applications or chemical applications.

4
Cold Working
As the number of dislocations increase they
interact and block each other. The first
dislocations will be the ones nearest 450 to the
applied stress where the resolved shear stress is
greatest. Therefore when another slip system
needs to be activated, the applied stress must
be increased to reach the critical resolved
shear stress on a new slip plane. So to increase
strain, the stress must be increased. Plastic
deformation results and so work hardening occurs
and the yield stress effectively increased,
strengthening the metal.
5
Dislocation Interactions
1
2
3
2
After 3, another slip system needs to be
activated by increasing the applied stress and so
meeting the critical resolve shear stress on a
new slip system.
6
Grain Size Control.
  • Grain boundaries block dislocation motion
  • as they change the orientation of slip planes
  • with respect to the applied stress. As the first
  • slip system activated will be the one nearest 45o
  • then all others will need more applied stress
  • to reach the critical resolved shear stresses
  • on planes which are not near 450 to the
  • applied stress.
  • In the figure, a dislocation is blocked by the
    grain
  • boundary as the (111) planes in the next grain
  • are not at 450 to the stress applied. This will
    lead
  • to a dislocation pile up where many
    dislocations
  • get blocked on the slip plane. This pile up
  • creates a stress build up in the next grain,
    adding
  • to the applied stress and so initiating slip in
    the
  • next grain. The smaller the grain size, the fewer
  • dislocations in the pile up and the higher the
  • applied stress to cause further slip. So the
    smaller
  • the grain size the higher the mechanical strength

(111)
30o
45o
s
7
Hall Petch Equation
  • Empirical Equation relating yield
  • strength to grain size.
  • sy so kd -1/2
  • sy yield stress for polycrystaline
  • so yield stress single crystal
  • k constant
  • d grain size.
  • The smaller the grain size the higher
  • the yield stress. Grain size can be controlled
  • by recrystallization and other techniques.

8
Solution Strengthening.
  • Add a solute to the metal, such as zinc to copper
    to create brass. The zinc atoms are
  • a different size and so affect dislocations.
  • Around a dislocations stress and
  • strain fields exist, compressive
  • above the slip plans and tensile
  • below it. A small atom can reduce
  • the compressive stress field
  • while a large atom can reduce
  • the tensile stress field. The applied
  • stress to move a dislocation will
  • therefore increase if the internal
  • stress field is decreased. The limit for
  • this is the Hume Rothery rules which
  • limit the amount of solute before
  • second phases form.

Compressive
Tensile
9
Dislocation Locking
For steels, there is an upper and lower yield
point. This is due to carbon in interstitial
sites locking the dislocations in place. The
small atoms reduce the strain energy of
a dislocation. It then requires more external
energy in the form of stress to move the
dislocation. Once the dislocation is free from
the the local carbon atom, less stress is
required to move it.
10
Dislocation Friction
  • Solute atoms have strain
  • fields associated with
  • them. As a result, as
  • dislocations move past
  • solute atoms, the energy
  • of the dislocation is
  • lowered and more stress
  • is required to keep it in
  • motion. This increases
  • the UTS of a material but
  • not the yield stress.

Strain
11
Second Phases.
  • The presence of second phases will strengthen
  • a material by blocking dislocation motion, and
  • requiring increased applied stress to produce
  • strain.
  • Second phases all work in the same manner
  • by blocking dislocation, their effectiveness
  • depends on the second phase distribution.
  • The spheroidal structure will be much weaker
  • than the eutectoid structure. The strength of the
  • eutectoid is a function of cooling rate, faster
  • cooling the plates are narrower and the
  • strength is higher than slow cooling rates
  • with wider plate spacing.
  • Aluminum alloys age hardening produces
  • optimum properties small particles which
  • interact with dislocations very effectively.

12
Second Phases
Dislocation pinned by particle
t
R
Second Phase Particle
Dislocation mobile
  • (Gb)/R

t shear stress to keep dislocation moving R
radius of curvature of dislocations As R
decreases, t increases so strengthening the
material. R decreases as the particles are closer
together, so the distribution is important
13
Dislocation Motion Large Particle Spacing
overaged.
Small spacing R small t will be large - peak age
Underage GP zones are FCC, like matrix so dont
block dislocations- solution strengthening only.
t
Dislocation loop after dislocation passes.
Large spacing, R smaller, and dislocation can
bend around particles and rejoin.
14
New Phases.
  • Best example would be steel transformation to
    martensite. Other alloy systems are also
  • capable of this type of diffusionless transfer
    such as titanium alloys Ti-6Al-4V, Fe-Ni
  • alloys. In this case the crystal structure is one
    that has a very high critical resolved shear
  • stress such as body centered tetragonal.
  • Other structures can produce high strength such
    as amorphous or glassy metals.
  • These metal alloys systems are quenched very
    rapidly, at a rate of several thousand
  • degrees per second. In this case, the resulting
    structures are not crystalline and so
  • have few dislocations and behave elastically to
    higher yield stresses.
  • Ni-Ti systems are a good example of these. They
    are metastable, and cannot be used at
  • temperature otherwise they gradually revert to
    their equilibrium crystal structure.

15
Homework
  • It the temperature is increased from 0.5 to 0.8
    of Tm (K) what effect does this have on the
    recrystallization process?
  • For aluminum estimate the shear stress required
    to move a dislocation when R 20nm use half the
    elastic modulus as the shear modulus. What is the
    effect of decreasing the radius of curvature of
    the dislocation?
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