Transcript CHE 333 Class 18

CHE 333 Class 17
Strengthening of Metals.
EXAM
• Exam on Nov 17.
• Classes 15 through 18.
• Same format as last ones.
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
Strengthening Mechanisms
To optimize properties of metals, greater strengths can be achieved by several
techniques:-
1.
Cold Working.
2.
Grain Size Control
3.
Solution Strengthening
4.
Second Phases.
5.
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.
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.
1
Dislocation Interactions
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.
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
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.
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
Compressive
the compressive stress field
while a large atom can reduce
the tensile stress field. The applied
stress to move a dislocation will
Tensile
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.
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.
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.
Str
ess
Dislocation friction
raises plastic
deformation curve
Strain
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.
Second Phases
Dislocation pinned by particle
R
t
Second Phase Particle
Dislocation mobile
t = (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
Dislocation Motion Large Particle
Spacing – overaged.
Underage – GP zones are
FCC, like matrix so don’t
block dislocations- solution
strengthening only.
Small spacing R small
t will be large - peak age
t
Dislocation loop after dislocation passes.
Large spacing, R smaller, and dislocation can bend around particles and rejoin.
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.
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?