CHE 333 Class 16

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CHE 333 Class 16

• Strengthening of Metals.

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Second Exam Nov 12

• From Class 8 to class 16 will be covered.

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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

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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.

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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.
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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.
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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
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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.

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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
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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.
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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
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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.

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Second Phases
Dislocation pinned by particle
t
R
Second Phase Particle
Dislocation mobile

• (Gb)/R

• - shear stress to keep dislocation moving
• G shear modulus
• 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

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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.
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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.

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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?