1. Mechanical Engineering Department Alabama A&M University Fall 2014, Lecture 16 Mechanical Behavior: Part II Dr. Aaron L. Adams, Assistant Professor

2. Learning Objectives…. What are the basic regions of plastic deformation in a stress- strain curve? What is meant by ‘work hardening’ coefficient? How do I calculate resolved shear stress? What is hardness and how is it measured? How does temperature influence deformation? Ductile to Brittle Transition Temperature? Creep? Ch. 7, pp. 201 – 222, 233 – 239; Ch. 8, pp. 262 – 279; Ch. 9, pp. 328 – 332, 342 – 347. Relevant Reading for this Lecture... 2

3. ELASTIC STRAIN Engineering Stress-Strain Curve Engineering Stress = F/A o Engineering strain Elastic Limit ELASTICDEFORMATION PLASTIC DEFORMATION PLASTICSTRAIN TOTAL STRAIN 5 3 Yield stress UTS (Tensile strength) Fracture stress ELASTICSTRAIN(recovered) Elastic Modulus E ΔσΔσ ΔεΔε

4. Elastic Strain Recovery Adapted from Fig. 7.17, Callister & Rethwisch 4e. Stress Strain 3. Reapply load 2. Unload D Elastic strain recovery 1. Load yoyo yiyi Some elastic deformation occurs before plastic deformation The elastic strain is recovered after the load is removed (even if the specimen fractures). Upon reloading the YS changes. 4

5. 5 Work Hardening Curve fit to the stress-strain response:  T  K  T  n “true” stress (F/A) “true” strain: ln(L/L o ) hardening exponent: n =0.15 (some steels) to n =0.5 (some coppers) An increase in  y occurs due to plastic deformation. Occurs between  y and TS on stress-strain curve.   large hardening small hardening  y 0  y 1 K = Stress at  = 1% 5

6. 6 (at lower temperatures, i.e. T < T melt /3) Plastic (Permanent) Deformation Simple tension test: engineering stress,  engineering strain,  Elastic+Plastic at larger stress (dislocation motion) pp plastic strain Elastic initially (bonds stretch) Adapted from Fig. 7.10 (a), Callister & Rethwisch 4e. permanent deformation after load is removed (result of dislocation motion) Plastic deformation occurs by dislocation motion (aka slip). Slip begins once the proportional limit is exceeded. Proportional limit 6

7. WHY DO METALS (AND SOME CERAMICS) WORK HARDEN? True Strain,  True Stress,  TS F  = K  n (describes work hardening), K = Stress at  = 1% Work hardening (aka, strain hardening) – increasing dislocation density Becomes harder. Need (more stress) to stretch (elongate) = work hardening Higher n hardening exponent: n = 0.15 (some steels) to n = 0.5 (some coppers) Dislocations tangle  TS = n 7

8. When the TS is exceeded, non-uniform deformation (i.e., “necking”) occurs Figure 6-27 Necking of a tensile test specimen within its gage length after extension beyond the ultimate tensile strength [Shackelford, 6 th Ed.] P YS Beyond the proportional limit the material accumulates strain. Rate of strain hardening can’t overcome the rate at which the cross- sectional area is decreasing Uniform, plastic deformation Non-uniform, plastic deformation Engineering stress-strain curve Why does this curve bend down? strain(UTS) = n 8 UTS Necking Work hardening

9. Consequences of Work Hardening High n → large value increases strength and hardness Engineering Consequences: (1)Poor machinability, cutting action strains material ahead of tool making it harder to cut. BUT (2) Higher n allow more uniform, plastic deformation (stretching). This can be good for sheet metal forming. 9

10. NOW LET’S DISCUSS DEFORMATION MECHANISMS AND STRENGTHENING

11. Why are Crystal Planes & Directions Important? ► Dislocations cause plastic deformation (i.e., slip). ► Slip occurs on specific crystal planes, in specific crystal directions on those planes. ► Plane + Direction = Slip System stress 11 1 RECALL

12. After Why are Crystal Planes & Directions Important? ► Dislocations cause plastic deformation (i.e., slip). ► Slip occurs on specific crystal planes, in specific crystal directions on those planes. ► Plane + Direction = Slip System stress 12 direction plane 1 RECALL

13. What stress is needed to cause slip? How do we calculate it?

14. How can I calculate what happened? F shear F normal F applied Force Decomposition F shear = F applied  cos( ) F normal = F applied  cos(  )   14 6 Applied forces can be decomposed into components on any coordinate system Slip plane Shear (in plane) Normal (perpendicular to plane) F shear F normal Force components can be converted into stress components Slip occurs on a slip plane in the slip direction

15. Determining shear and tensile stress Depending on the angle between the applied force and the planes (and directions) in the crystal, it takes different amounts of force to induce plastic deformation F shear F normal F applied  F = F applied  cos(  ) G = F applied  cos( ) F applied F normal F shear  A0A0 A 0 = A  cos(  ) A = A 0 / cos(  ) A A0A0 A   = F normal /A  = F shear /A  F shear /A = F applied  cos( ) / (A 0 / cos(  )) = (F applied / A 0 )  cos(  cos   applied  cos  cos  Review your dot product notes from earlier lecture! 15

16. Review: Critical Resolved Shear Stress Dislocations move in close packed directions on close packed planes because it is easier. (less work is required) 16

17. A relevant problem A single grain of nickel (Ni) in a metal plate is oriented so that it experiences a tensile load orientated along the [100] crystal direction. If the applied stress is 0.5MPa (72.5 psi), what will be the resolved shear stress, , along the [101] direction with in the (11-1) plane? The shear stress MUST be smaller than tensile stress 6

18. 18 Four Strategies for Strengthening: 1: Reduce Grain Size Grain boundaries are barriers to slip. Barrier "strength" increases with Increasing angle of misorientation. Smaller grain size: more barriers to slip. Hall-Petch Equation: Adapted from Fig. 8.14, Callister & Rethwisch 4e. (Fig. 8.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

19. 19 Four Strategies for Strengthening: 2: Form Solid Solutions Smaller substitutional impurity Impurity generates local stress at A and B that opposes dislocation motion to the right. A B Larger substitutional impurity Impurity generates local stress at C and D that opposes dislocation motion to the right. C D Impurity atoms distort the lattice & generate lattice strains. These strains can act as barriers to dislocation motion.

20. 20 Ex: Solid Solution Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. Empirical relation: Alloying increases  y and TS. Adapted from Fig. 8.16 (a) and (b), Callister & Rethwisch 4e. Tensile strength (MPa) wt.% Ni, (Concentration C) 200 300 400 01020304050 Yield strength (MPa) wt.%Ni, (Concentration C) 60 120 180 01020304050

21. 21 Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Result: Four Strategies for Strengthening: 3: Precipitation Strengthening Large shear stress needed to move dislocation toward precipitate and shear it. Dislocation “advances” but precipitates act as “pinning” sites with S.spacing Side View precipitate Top View Slipped part of slip plane Unslipped part of slip plane S spacing

22. 22 Internal wing structure on Boeing 767 Aluminum is strengthened with precipitates formed by alloying. Adapted from Fig. 11.45, Callister & Rethwisch 4e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) 1.5  m Application: Precipitation Strengthening Adapted from chapter- opening photograph, Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

23. 23 Four Strategies for Strengthening: 4: Cold Work (Strain Hardening) Deformation at room temperature (for most metals). Common forming operations reduce the cross-sectional area: Adapted from Fig. 14.2, Callister & Rethwisch 4e. -Forging A o A d force die blank force -Drawing tensile force A o A d die -Extrusion ram billet container force die holder die A o A d extrusion -Rolling roll A o A d

24. 24 Dislocation structure in Ti after cold working. Dislocations multiply and entangle with one another during cold work. Makes dislocation motion becomes more difficult. Increases strength. Adapted from Fig. 5.11, Callister & Rethwisch 4e. (Fig. 5.11 is courtesy of M.R. Plichta, Michigan Technological University.) Dislocation Structures Change During Cold Working

25. Temperature Effects on Deformation Adapted from Fig. 9.21, Callister & Rethwisch 4e. Ductile-to-Brittle Transition Temperature (DBTT)... BCC metals (e.g., iron at T < 914°C) Impact Energy Temperature High strength materials (σ y > E/150) Polymers More Ductile Brittle Ductile-to-brittle transition temperature FCC metals (e.g., Cu, Ni) 25 Can be assessed via IMPACT TESTING! final height initial height (Topic for next lecture) (Charpy)

26. Pre-WWII: The Titanic WWII: Liberty ships Problem: Used a type of steel with a DBTT ~ Room temp. Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(a), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Dr. Robert D. Ballard, The Discovery of the Titanic.) Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(b), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Earl R. Parker, "Behavior of Engineering Structures", Nat. Acad. Sci., Nat. Res. Council, John Wiley and Sons, Inc., NY, 1957.) Design Strategy: Stay Above The DBTT! 26

27. Creep – Time dependent deformation Permanent deformation at a constant stress (  ) vs. time Adapted from Fig. 8.28, Callister 7e. Primary Creep: slope (creep rate) decreases with time. Secondary Creep: steady-state i.e., constant slope. Tertiary Creep: slope (creep rate) increases with time, i.e. acceleration of rate.   0 t 27 (Elastic)

28. Occurs at high temperature, T > 0.4 T m (in degrees K) Adapted from Figs. 9.36, Callister & Rethwisch 4e.Creep elastic primary secondary tertiary What’s going on? Dislocations glide AND climb via diffusion … RECALL, diffusion is faster at higher T. Diffusion can contribute to deformation. 28 Look… another Arrhenius relationship General observation

29. HARDNESS ►The resistance of a material to deformation by indentation. ►Hardness can provide a qualitative assessment of strength or ductility. ►Hardness can not be used to quantitatively infer strength or ductility. ►Useful in quality control 29

30. Early hardness tests were based on scratching materials with minerals Figure 7.30 Comparison of several hardness scales. [from Callister & Rethwisch, 4 th Ed.] 30

31. HARDNESS Resistance to permanently indenting the surface. Large hardness means: -- resistance to plastic deformation or cracking in compression. -- better wear properties. e.g., 10 mm sphere apply known force measure size of indent after removing load d D Smaller indents mean larger hardness. increasing hardness most plastics brasses Al alloys easy to machine steelsfile hard cutting tools nitrided steelsdiamond 31

32. HARDNESS Hardness can be measured by different types of indenter tips/loads. Table 6.5 Hardness testing techniques 32

33. Hardness: Measurement ►Rockwell No major sample damage Each scale runs to 130 but only useful in range 20-100. Minor load 10 kg Major load 60 (A), 100 (B) & 150 (C) kg –A = diamond, B = 1/16 in. ball, C = diamond ►HB = Brinell Hardness TS (psi) = 500 x HB TS (MPa) = 3.45 x HB (These relationships ONLY apply for steel, cast iron, and brass.) 33

34. Summary The basic regions of deformation in a stress-strain curve: What is elasticity What is plasticity Necking Using the dot-product, you can solve for the resolved shear stress in a material….know how to do! Strategies for strengthening: Reduce grain size – Hall-Petch relationship Solid solution strengthening Precipitation strengthening Cold working Temperature dependent behavior DBTT Creep Hardness 34