CHE 333 Class 16

1. CHE 333 Class 16 Strengthening of Metals.

2. Second Exam – Nov 12 • From Class 8 to class 16 will be covered.

3. 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

4. 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.

5. 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.

6. 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.

7. 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

8. 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.

9. 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

10. 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.

11. Stress Dislocation friction raises plastic deformation curve 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

12. 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.

13. Second Phases Dislocation pinned by particle R t 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

14. Dislocation Motion Large Particle Spacing – overaged. Small spacing R small t will be large - peak age Underage – GP zones are FCC, like matrix so don’t block dislocations- solution strengthening only. t Dislocation loop after dislocation passes. Large spacing, R smaller, and dislocation can bend around particles and rejoin.

15. 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.

16. 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?