CHE 333 Class 17

Strengthening of Metals.

Strengthening at COLD temperatures

Metals – basically all work in same waywhich is to block dislocations or retardThem.Remember – COLD is less than 0.3 Tm in Kelvin

Strengthening MechanismsTo 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 bethe ones nearest 450 to the applied stress where theresolved shear stress is greatest. Therefore whenanother 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 deformationresults and so work hardening occurs and the yield stress effectively increased, strengtheningthe metal.

Dislocation Interactions1

2

2

3

After 3, another slip system needs to be activatedby increasing the applied stress and so meeting the critical resolve shear stress on a new slip system.

Grain Size Control.Grain boundaries block dislocation motionas they change the orientation of slip planeswith respect to the applied stress. As the firstslip system activated will be the one nearest 45o then all others will need more applied stressto reach the critical resolved shear stresseson planes which are not near 450 to the applied stress.In the figure, a dislocation is blocked by the grainboundary as the (111) planes in the next grainare not at 450 to the stress applied. This will leadto a “dislocation pile up” where many dislocationsget blocked on the slip plane. This “pile up” creates a stress build up in the next grain, addingto the applied stress and so initiating slip in thenext grain. The smaller the grain size, the fewerdislocations in the pile up and the higher theapplied stress to cause further slip. So the smallerthe grain size the higher the mechanical strength

(111)

s

45o

30o

Hall Petch EquationEmpirical Equation relating yield strength to grain size.

sy = so + kd -1/2

sy = yield stress for polycrystalineso = yield stress single crystalk = constantd = grain size.

The smaller the grain size the higherthe yield stress. Grain size can be controlledby recrystallization and other techniques.

Solution Strengthening.Add a solute to the metal, such as zinc to copper to create brass. The zinc atoms area different size and so affect dislocations.Around a dislocations stress andstrain fields exist, compressiveabove the slip plans and tensilebelow it. A small atom can reducethe compressive stress fieldwhile a large atom can reducethe tensile stress field. The appliedstress to move a dislocation willtherefore increase if the internalstress field is decreased. The limit for this is the Hume Rothery rules whichlimit the amount of solute beforesecond phases form.

Compressive

Tensile

Dislocation Locking

For steels, there is an upper andlower yield point. This is due to carbon in interstitial sites “locking”the dislocations in place. The smallatoms reduce the strain energy of adislocation. It then requires moreexternal energy in the form of stressto move the dislocation.Once the dislocation is free from thethe local carbon atom, less stressis required to move it.

Dislocation FrictionSolute atoms have strainfields associated withthem. As a result, asdislocations move past solute atoms, the energyof the dislocation islowered and more stressis required to keep it in motion. This increasesthe UTS of a material but not the yield stress.

Stress

Dislocation friction raises plastic deformation curve

Strain

Second Phases.The presence of second phases will strengthena material by blocking dislocation motion, and requiring increased applied stress to produce strain.Second phases all work in the same mannerby blocking dislocation, their effectivenessdepends on the second phase distribution.The spheroidal structure will be much weakerthan the eutectoid structure. The strength of theeutectoid is a function of cooling rate, faster cooling the plates are narrower and the strength is higher than slow cooling rateswith wider plate spacing.Aluminum alloys – age hardening producesoptimum properties – small particles whichinteract with dislocations very effectively.

Second Phases

Second Phase ParticleDislocation mobile

Dislocation pinned by particle

t

t = (Gb)/R

t = shear stress to keep dislocation movingR = radius of curvature of dislocationsAs R decreases, t increases so strengthening the material.R decreases as the particles are closer together, so the distribution is important

R

Dislocation Motion Large Particle Spacing – overaged.

t

Small spacing R smallt will be large - peak age

Large spacing, R smaller, and dislocation can bend around particles and rejoin.

Dislocation loop after dislocation passes.

Underage – GP zones areFCC, like matrix so don’t block dislocations- solutionstrengthening only.

New Phases.Best example would be steel transformation to martensite. Other alloy systems are alsocapable of this type of diffusionless transfer such as titanium alloys Ti-6Al-4V, Fe-Nialloys. 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 sohave 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?