Plastic deformation and creep in crystalline materials Chap. 11

Plastic deformation and creep incrystalline materials Chap. 11

Mechanical Properties of Materials Stiffness Resistance to elastic deformation Young’s modulus Strength Resistance to plastic deformation Yield stress Toughness Resistance to fracture Energy to fracture ductility Ability to deform plastically Strain to fracture

Uniaxial Tensile Test (Experiment 6) Gaugelength specimen

Result of a uniaxial tensile test STRENGTH  (Engineering stress) necking Ultimate tensile strength UTS Yield point plastic Yield strength y break elastic Area = Toughness STIFFNESS Slope = Young’s modulus (Y) DUCTILITY f(strain to fracture)  (engineering strain)

If there is a smooth transition from elastic to plastic region (no distinct yield point) then 0.2 % offset proof stress is used

During uniaxial tensile test the length of the specimen is continually increasing and the cross-sectional area is decreasing. True stress ≠ Engineering stress (=F/A0) True strain ≠ Engineering strain (=L/L0) True stress Ai = instantaneous area Eqn. 11.3 True incremental strain True strain Eqn. 11.4

Eqn. 11.5 K Strength coefficient n work hardening exponent

What happens during plastic deformation? • Externally, permanent shape change begins at sy • Internally, what happens?

? What happens to crystal structure after plastic deformation? Plastic Deformation

Some Possible answers Becomes random or amorphous Changes to another crystal structure Remains the same

How Do We Decide? X-ray diffraction No change in crystal structure! No change in internal crystal structure but change in external shape!!

How does the microstructure of polycrystal changes during plastic deformation? EXPERIMENT 5 Comparison of undeformed Cu and deformed Cu

SlipLines Before Deformation After Deformation

Callister Slip lines in the microstructure of plastically deformed Cu Experiment 5

Slip

Slip Planes, Slip Directions, Slip Systems Slip Plane: Crystallographic planes Slip Direction: Crystallographic direction Slip System: A combination of a slip plane and a slip direction

Slip Systems in Metallic Crystals Crystal Slip Slip Slip Plane Direction Systems FCC {111} <110> 4x3=12 (4 planes) (3 per plane) BCC{110} <111> 6x2=12 (6 planes) (2 per plane) HCP {001} <100> 3x1=3 (1 plane) (3 per plane)

Why slip planes are usually close packed planes? Why slip directions are close-packed directions?

Slip Systems in FCC Crystal (111) z y x

Tensile vs Shear Stress • Plastic deformation takes place by slip • Slip requires shear stress • Then, how does plastic deformation take place during a tensile test?

1 2 s s: Applied tensile stress N: Slip plane normal D: Slip direction N D F1: angle between s and N F2=angle between s and D Is there any shear stress on the slip plane in the slip direction due to the applied tensile stress? s

f1 2 Resolved Shear stress F  = F/ A Area=A FD = F cos2 N As = A cos1 D Area = As F

F F No resolved shear stress on planes parallel or perpendicular to the stress axis F F cos 2 = 0 cos 1 = 0

Plastic deformation recap No change in crystal structure: slip twinning Slip takes place on slip systems (plane + direction) Slip planes usually close-packed planes Slip directions usually close-packed direction Slip requires shear stress In uniaxial tension there is a shear component of tensile stress on the slip plane in the slip direction: RESOLVED SHEAR STRESS

s N D 1 2 s CRITICAL RESOVED SHEAR STRESS

If we change the direction of stress with respect to the slip plane and the slip direction cos 1 cos 2 will change. To maintain the equality which of the following changes takes place? 1. CRSS changes. 2. y changes Schmid’s Law:CRSS is a material constant.

Anisotropy of Yield Stress Yield stress of a single crystal depends upon the direction of application of load cos 1 cos 2 is called the Schmid factor

Active slip system Slip system with highest Schmid factor is the active slip system

Magnitude of Critical Resolved Shear Stress Theory (Frenkel 1926) Experiment

Potential energy Shear stress CRSS b b/2 b d

Critical Resolved Shear Stress Theory (GPa) 12 7 5 Experiment (MPa) 15 0.5 0.3 Ratio Theory/Exp 800 14,000 17,000 Fe (BCC) Cu (FCC) Zn (HCP)

Solution 1934 • E. Orowan • Michael Polanyi • Geoffrey Ingram Taylor

Solution • Not a rigid body slip • Part slip/ part unslipped

Slip Not-yet-slipped Boundary between slipped and unslipped parts on the slip plane Dislocation Line (One-Dimensional Defect)

Movement of an Edge Dislocation From W.D. Callister Materials Science and Engineering

Plastic Deformation Summary • Plastic deformation slip • Slip dislocations • Plastic deformation requires movement of dislocations on the slip plane

Recipe for strength? Remove the dislocation

Stress, MPa Fig. 11.6 700 50 strain Cu Whiskers tested in tension

Fe W Al2O3 Si 18-8 ss Ni Cu Effect of temperature on dislocation motion Higher temperature makes the dislocation motion easier Yield stress Eqn. 11.14 11.15 11.16 11.17 11.18 Fig. 11.8 T/Tm 0 0.7

Recipe for strength Remove the dislocation: Possible but Impractical Alternative: Make the dislocation motion DIFFICULT

Strengthening Mechanisms • Strain hardening • Grain refinement • Solid solution hardening • Precipitation hardening

Movement of an Edge Dislocation A unit slip takesplace only whenthe dislocationcomes out of thecrystal

During plastic deformation dislocation density of a crystal should go down Experimental Result Dislocation Density of a crystal actually goes up Well-annealed crystal: 1010 m-2 ? Lightly cold-worked: 1012 m-2 Heavily cold-worked: 1016 m-2

Plastic deformation and creep in crystalline materials Chap. 11