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Deformation of Single Crystals

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Deformation of Single Crystals Deformation of Single Crystals When a single crystal is deformed under a tensile stress, it is observed that plastic deformation occurs by slip on well‐defined parallel crystal planes. Sections of the crystal slide relative to one another, changing the geometry of the sample as shown in the diagram. Slip always occurs on a particular set of crystallographic planes, known as slip planes. Slip always takes place along a consistent set of directions within these planes –these are called slip directions. The combination of slip plane and slip direction together makes up a slip system. Slip systems are usually specified using the Miller index notation. For example, cubic close‐packed metals slip on 1 1 0{} 1 1 1 The slip direction must lie in the slip plane. Slip occurs by dislocation motion. To move dislocations, a certain stress must be applied to overcome the resistance to dislocation motion. Slip occurs when the shear stress acting in the slip direction on the slip plane reaches some critical value. This critical shear stress is related to the stress required to move dislocations across the slip plane. The tensile yield stress of a material is the applied stress required to start plastic deformation of the material under a tensile load. We want to relate the tensile stress applied to a sample to the shear stress that acts along the slip direction. resolved_____actingforce on slipF cos planeλ F τ = =cos =φ λ cos=cos σ φ cos λ R area of___ slip plane A A cosφ It is found that the value of τR at which slip occurs in a given material with specified dislocation density and purity is a constant, known as the critical resolved shear stress τC. This is Schmid's Law. The quantity CosφCosλ is known as the Schmid Factor (M) The tensile stress at which the material start to slip is the yield strength. τCY=cosσ φ cosλ In a given crystal, there may be many available slip systems. As the tensile load is increased, the resolved shear stress on each system increases until eventually τC is reached on one system. The crystal begins to plastically deform by slip on this system, known as the primary slip system. The stress required to cause slip on the primary slip system is the yield stress of the single crystal. As the load is increased further, τC may be reached on other slip systems; these then begin to operate. From Schmid's Law, it is apparent that the primary slip system will be the system with the greatest Schmid factor (M). τ = σ YC M BasicBasic ConsiderationsConsiderations Experimental technique Uniaxial Tension or Compression Experimental measurements showed that At RT the major source for plastic deformation is the dislocation motion through the crystal lattice. Dislocation motions occurs on fixed crystal planes (“slip planes”) in fixed crystallographic directions (corresponding to the Burgers vector of the dislocation that carries the slip) The crystal structure of metals is not altered by the plastic flow Volume changes during plastic flow are negligible Schmid’s Law • Initial yield stress varies from sample to sample depending on, among several factors, the position of the crystal lattice relative to the loading axis. • It is the shear stress resolved along the slip direction on the slip plane that initiates plastic deformation. • Yield will begin on a slip system when the shear stress on this system first reaches a critical value (critical resolved shear stress, crss), independent of the tensile stress or any other normal stress on the lattice plane. E. Schmid & W. Boas (1950), Plasticity of Crystals, Hughes & Co., London. Resolved Shear Stress τ = s⋅σ ⋅ n = σ cosφ cosλ τ σ = c c cosφ cosλ “Soft orientation”, with slip plane at 45°to tensile axis “Hard orientation”, with slip plane at ~90°to tensile axis Slip in single crystals Critical resolved shear stress (τcrss): minimum shear stress required to initiate slip. τ σ = crss This is when yielding begins (i.e. yield strength) y (cosφ cos λmax ) Condition for dislocation motion: τR > τcrss •Crystal orientation can make it easy or difficult to move dislocations. σ σ σ a) b) c) τR = 0 τR = σ/2 τR = 0 λ=90° λ=45° φ=90° φ=45° What happens in cases a and c (w.r.t. plastic deformation)? Example A tensile stress of 5kPa is applied parallel to the [432] direction in a cubic crystal. Find the shear stresses, τ, on the (11‐1) plane in the [011] direction. Solution Find the Schmidt’s factor for the slip system [ 432 ]⋅ [ 11 1 ] 5 Cosφ = = 0= . 536 42+ 3 2 + 2 2 ⋅ 1 2 + 2 1 + 2 29 1 ⋅ 3 [ 432 ]⋅ [ 011 ] 5 Cosλ = = 0 .= 6565 42+ 3 2 + 2 2 ⋅ 1 2 + 2 1 + 2 29 0 ⋅ 2 M= Cosφ ×0 λ Cos . = 352 τ0= .Mσ 352 = × 5kPa = 1kPa . 76 0⎡ 0 0⎤ The relationship between the applied stress σ =0⎢ 0 0⎥ [σapplied] and the shear stress τ is simply an ⎢ ⎥ example of a tensor rotation. For a tensile 0⎣⎢ 0 σ tensile ⎦⎥ stress: Employing the Euler [][]a = φ1Φφ 2 ⎡σ x1 0 τ ⎤ rotation transformation σ '= ⎢0σ 0⎥ ⎢ y1 ⎥ τ =a13 aσ 33tensile ⎣⎢τ0 σz1⎦⎥ sinτ= Φ sinφ2 Φ cos σtensile Because a13 is the cosine of the angle between x’ and z, we can now set cosλ= sin Φ φ2 sin θ= Φ τ =cosσ θ cosλ After Mechanical Behavior of Materials by K. wmanBo The relationship of the Schmid factor to the axial strain and the shear strain is given by: ε =cosγ θ cosλ Where ε is the strain along the direction of the applied stress and γ is the glide strain. The strains for an specific slip system can be expressed by SD=[uvw] and SPN=(hkl) ⎡ v h⋅ + u ⋅ k u ⋅ l +⎤ w ⋅ h u⋅ h ⎢ 2 2 ⎥ v⎢⋅ h + u ⋅ k w⋅ k + v ⋅⎥ l εij = γ⎢ v⋅ k ⎥ ⎢ 2 2 ⎥ u l⋅ + w ⋅ h w ⋅ + k ⋅ v l ⎢ w⋅ l ⎥ ⎣⎢ 2 2 ⎦⎥ Where γ is the magnitude of the simple shear in the slip system. Example 15⎡ 0⎤ 0 A cubic crystal is subjected to a stress state ⎢ ⎥ σ σ σ τ τ τ 0 0 0kPa x=15kPa, y=0 z=7.5kPa, yz= zx= xy=0, where ⎢ ⎥ x=[100], y=[010] and z=[001]. What is the shear 0⎢ 0 7⎥ . 5 stress on the (‐1‐11)[101] slip system? ⎣ ⎦ φ1 = −90 Φ =45 − φ2=54 − . 7 cos⎡ cosφ2− φ 1 coscos Φφ1 sinφ sin 2 sin2 φ cos+ 1 φ Φ cosφ1 φ 2 sinΦ φ2 ⎤ sin sin sina = ⎢− cosφ− φ cossin Φφ sin sinφ− cos φ cos+ φ Φ cosφ φ cosΦ φ ⎥ sin cos [] ⎢ 2 1 1 2 2 1 1 2 2 ⎥ ⎣⎢ sinΦ sinφ1 −sin Φ cosφ1 cosΦ ⎦⎥ cos 54 . 7 cos 90 cos 45 sin−⎡ 90 − sin − 54 − . 7 − cos − 54 . − 7 sin + − 90 − cos − 45− cos − ⎤ 90 sin 54 . 7 sin 45 sin 54 . 7 ⎢ ⎥ sin 54 .−− 7 cos[]a = ⎢ 90 −−−−− cos 45sin sin 90 90 cos cos −− 45 54 cos . 7 −+−−− 90 sin cos 54 . 54 7 − . 7 − sin⎥ 45 cos 54 . 7 ⎣⎢ sin− 45 sin − 90 sin− 45 − cos − 90 cos− 45 ⎦⎥ cos 45 sin⎡ 54− . 7 − cos − 54 − .− 70 sin. − 577⎤ −⎡ 45 0 sin . − 577 54 . 7⎤ 0 . 577 ⎢ ⎥ ⎢ ⎥ cos 45 cos[]a = ⎢ 54− . 7 − sin− 54 .− 70 . sin − 4086⎥ = ⎢ 45 cos− 0 . 816 54 − . 7⎥ 0 . 408 ⎣⎢ −sin − 45 0 cos− 0 45 .⎦⎥ 707⎣⎢ 0 0 .⎦⎥ 707 T=' ] a[ × T ][ ×T a 0 . 577−⎡ 0 . − 57715⎤ 0⎡0 . 0 577 577⎤⎡ 0− 0 . 408⎤ 0 . 707 T ⎢ ⎥⎢ ⎥⎢ ⎥ σ[][=0a ][ . σ ][ 408 a = ]⎢ 0− . 816 − ⎥0⎢ 0 . 0 408577⎥ 0−⎢ 0 . 0 − . 816⎥ 0 0 . 707⎣⎢ 0 00 .⎦⎥0⎣⎢ . 707 0 577 7⎦⎥ .⎣⎢ 5 0− . 408⎦⎥ 0 . 707 7 . 50⎡ − 5 . 3 − 3⎤ . 06 T ⎢ ⎥ σ[][=a ][ σ5 ][ a .= 3 ]⎢ − 3 . 75 2⎥ . 16 3 . 06⎣⎢− 2 . 16 11⎦⎥ . 25 The shear stress in the slip direction is in the x1 plane and z1 direction =‐3.06kPa Example: An FCC Cu is subjected to a uniaxial load along the [112] direction. What is most likely initial slip system? If CRSS is 50 MPa, what is the tensile stress at which Cu will start to deform plastically? Slip Slip Dir., •n • s M= σ cos φ = cos λ = Plane, n s cos φ cos λ MPa )( | || n | | || s | (111) 184 [0 1 1] + 22/3 3 /6 6 /9 184 [ 1 01] 3 /6 6 /9 u ndef [ 1 10] 0 0 367 ( 1 11) [0 1 1] 2 /3 3 /6 6 /18 smallest −122 [101] − 3 /2 − 6 /6 stress to 184 cause slip [110 ] 3 /3 6 /9 (yielding) [011] 122 (1 1 1) 2 /3 3 /2 6 /6 −367 [ 1 01] − 3 /6 − 6 /18 [110 ] 184 3 /3 6 /9 [011] 0 undef (11 1 ) 0 3 /2 0 undef [101] 3 /2 0 u ndef [ 1 10] 0 The initial Slip Systems (plane, direction) are then (1 11)[101], (11 1)[011] StrainStrain HardeningHardening ofof FCCFCC Crystals.Crystals. Shear Stress – Shear Strain Curves A typical shear stress‐ shear strain curve for a single crystal shows three stages of work hardening: Stage I = “easy glide” with low hardening rates; Stage II with high, constant hardening rate, nearly independent of temperature or strain rate; Stage III with decreasing hardening rate and very sensitive to temperature and strain rate. The extend of easy glide in a crystal depends on its orientation, the presence of dislocations (defects) and on the temperature. Stage I: •After yielding, the shear stress for plastic deformation is essentially constant.
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