Heterogeneous Deformation in Bulk Metallic Glasses. A.R. Yavari. European Reseach & Training Network Ductile BMG Composites

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1 Heterogeneous Deformation in Bulk Metallic Glasses A.R. Yavari European Reseach & Training Network Ductile BMG Composites

2 Bulk metallic glasses (BMG) can have yield strengths greatly exceeding those for conventional high-strength alloys, with maximum reported at 5.3 GPa [Inoue et al Nature Materials (2003)] and have unique mechanical properties, such as perfect elastic behaviour over 2% elastic strains. However, they show little macroscopic plastic deformation before fracture due to heterogeneous deformation and shear softening.

3 Flow in metallic glasses: In Cohen and Turnbull s theory, the probability p(v)dv of finding an atom with free volume between v and v+dv is: p(v)dv δ δ.v = ( )exp( )dv v v f where v f is the average free volume per atom and δ is a geometric factor between 1 and ½. In order for an atom to be a potential jump site, its free volume must be larger than v, the effective hard-sphere size of the atom. Therefore the total probability that an atom is a potential jump site is: f (0).v p(v ) = δ p(v)dv= exp( ) = v v Therefore, fraction of potential jump sites = Δf. (2) f C f δ.v exp( v f ) (1)

4 fraction of potential jump sites = Δf. (2) Net number of forward jumps per second per potential jump site =.[exp( νd where is the Debye frequency. δ.v exp( v m m ΔG τω/2 ΔG + τω/2 ) exp( )] k T k T B B f (3) ) The deformation flow rate γ t is then given by multiplication of fraction of potential jump sites eq.(2), by the net number of forward jumps per site per second eq.(3) γ = νd Δf t δ v exp( v f m τω ΔG )sinh( )exp( ) 2k T k T B B» (4)

5 γ = ν d t Δf exp( δ v v f )sinh( τω )exp( 2k T For low stesses τω<<2kt, 2sinh(τΩ/2kBT) τω/kbt and homogeneous deformation (Δf = 1), this flow equation yields Newtonian flow: B ΔG k B m T ) η = τ = T ( γ t) νd Ω v k T 0 f B vf k BT (5) The random walk 3-d coefficient of atomic diffusion D = (1/6)Γλ 2 where Γ is jump frequency (probability eq.(1) δ.v exp( ) v times number of jumps on this site jump and λ the jump distance. k B δ v exp( D ΔG )exp( δ v exp( Leading to the Stokes-Einstein equation: D.η = kbt/6(ω/λ2) = 1 6 f m νd λ ) 2 = η νd exp( ΔG k δ v exp( v f B T ΔG )exp( m ) m ) m ΔG )exp( k T B )

6 Free volume creation and shear softening : Energy necessary to squeeze an atom with volume v into a smaller hole of volume v : ΔG e = (S/v) (v-v) 2 (8) where S= (2/3) μ (1+υ)/(1-υ) = B (1-2υ)/(1-υ), μ is the shear modulus, B the bulk modulus and υ the Poisson ratio. Then the driving free energy of mechanical origin available for free volume creation becomes : ΔG = τω - (S/v) (v-v) 2 (9)

7 Then net free volume creation per second can be calculated to be m 2kT δ.v τω v ΔG Δ + vf = νd cosh 1 exp( ) exp( ) S v f 2kBT δ vf k BT (10) And net annihilation by diffusive jumps (relaxation) : Δ v f = v n D δ v νd exp( v f m ΔG )exp( ) k T B (11) Setting the two rates equal gives the steady-state free-volume as a function of local applied stress: δ v v f = v S n D 2kT / cosh τ Ω 2k T B -1 (12)

8 SEM image of ribbon stacks after gentle milling showing a thickness reduction from initial 30 microns about 1 to 3 microns Reduced volume per atom showing relaxation of heavily deformed foils of the ZrCuAlNiTi glass near Tg. Yavari AR, Le Moulec A, Inoue A, Nishiyama N, Lupu N, Matsubara E, Botta WJ, Vaughan G, Di Michie M, Kvick Å. Acta Mater 2005; 53:

9 But increased free volume should drop viscosity, lead to further shear in the same location until fracture with a single shear band evolving into the final crack Does this happen? ) T k ΔG )exp( v v exp( 0 η ) T k ΔG )exp( v v exp( Ω T k t) ( τ η B m f B m f B = = = δ δ ν γ d

10 The binary CuZr glass was repeatedly found to undergo plastic deformation of more than 50% in uniaxial compression

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13 In CuZr metallic glass microscopic examination shows presence of a fine dispersion of nanocrystals with dimensions of the order of 5 nm. TEM examination after heavy deformation

14 Diffraction patterns obtained in transmission before and after heat treatment in the synchrotron beam

15 Typical TEM images of the heat treated samples. Clearly nanocrystals of an average size of about 5 nm have formed Zr 55 Ti 5 Al 10 Cu 20 Ni 10 BMG-nanocomposite Zr55Al10Cu20Ni10Pd5 BMG-nanocomposite

16 -stress versus strain curves for these BMGs obtained in compression before and after the heat treatment resulting in the weak changes in the diffraction patterns. - As-cast BMGs all break at or below the limit of the elastic deformation range near 2%, - Heat treated samples all show extensive plastic strain to fracture 2000 Zr 55 Ti 5 Al 10 Cu 20 Ni 10 :BMG-Nanocomposite Zr 55 Ti 5 Al 10 Cu 20 Ni 10 : BMG stress (MPa) 1500 Strain Rate : s (a) Strain (%)

17 - As-cast BMGs all break at or below the limit of the elastic deformation range near 2%, - Heat treated samples all show extensive plastic strain to fracture Zr 55 Al 10 Cu 20 Ni 10 Pd 5 :BMG-Nanocomposite Zr 55 Al 10 Cu 20 Ni 10 Pd 5 : BMG Zr 55 Ti 5 Al 7.5 Cu 22 Ni 8 Ga 2.5 : BMG-Nanocomposite Zr 55 Ti 5 Al 7.5 Cu 22 Ni 8 Ga 2.5 : BMG Stress (MPa) Strain rate: s -1 (b) Stress (MPa) Strain rate: s -1 (c) Strain (%) Strain (%)

18 The external surfaces of the deformed samples show a maze of slip bands and steps at near 45 degree angle to the compression axis with nearly perpendicular slip systems crossing and jogging each other s surface steps Zr 55 Ti 5 Al 10 Cu 20 Ni 10 -BMG nanocomposite Zr 55 Ti 5 Al 10 Cu 20 Ni 10 -BMG nanocomposite

19 Samples were then taken out of the light beam and ion milled to produce a central hole and mounted on the tensile deformation setup built for in-situ deformation in a model 300kV JEOL 3010 transmission electron microscope (TEM).

20 TEM images of a typical shear band formed at the onset of deformation (when the applied tensile stress first reaches the composite structure s yield strength of about 2 GPa) show band is seen to be only nm thick.

21 Here a long shear band is seen trailed by a crack. The crack itself forming zigzags.

22 another blocked crack that widens (blunting) as it tries to escape blockage in a shear band containg a large volume fraction of crystallites.

23 Thus, as nanoparticles grow during shear, the solid fraction increases and the viscosity in the bands is expected to increase sharply. Therefore, contrary to the case of heterogeneous deformation of a monolithic metallic glass, in the case of these metallic glassbased nanocomposites, nanocrystal formation/growth leads to hardening of the shearing zone and shear delocalisation

24 (Cu 0.5 Zr Ti ) 99 Sn 1 Stress, σ/mpa strain rate: 4X10-4 s Strain, ε (%) Experimental evidence for nanocrystal nucleation DURING deformation in shear bands T. Zhang et al, ISMANAM-2005 Paris July 2005

25 Experimental evidence for nanocrystal nucleation DURING deformation in shear bands Lee et al, Acta Mater 2005

26 -As shear generates free volume and reduces viscosity, any nanocrystals present in the shear zone before deformation can grow rapidly and new ones can nucleate as the zone becomes liquid-like. - Active shear band regions are then constituted of two phases: 1) the glassy phase that with reduced viscosity and increased temperature behaves liquid-like 2) the nanocrystals that are a growing solid phase component => the deformation behaviour is like that of a semi-solid slurry. - at a given high shear rate, the mechanical properties of semi-solid slurries of a variety of materials including metals depend exponentially on the varying fraction of the solid component. (as for example reproduced from Krieger IM, Dougherty TJ. Trans. Soc. Rheol 1959; 3: 137)

27 2000 Zr 55 Ti 5 Al 10 Cu 20 Ni 10 :BMG-Nanocomposite Zr 55 Ti 5 Al 10 Cu 20 Ni 10 : BMG stress (MPa) Strain Rate : s -1 (a) Strain (%) 1800 Zr 55 Ti 5 Al 10 Cu 20 Ni 10 :B M G -h eat treated Zr 55 Ti 5 Al 10 Cu 20 Ni 10 : BMG - as prepared stress (MPa) S tra in R ate : s -1 4,0 4,1 4,2 4,3 4,4 4,5 4,6 4,7 4,8 4,9 5,0 S tra in (% )

28 Specimens are bars L = 7 mm high, 3 mm wide and 2 mm thick. The strain rate dε /dt = (1/L )(dl/dt) being 8x 10-4 s -1 The imposed cross-head advancement speed during the compression is about 5.6 µm/s. Zr 55 Ti 5 Al 10 Cu 20 Ni 10 -BMG nanocomposite

29 A 1% drop in stress corresponds to a 1% drop in elastic strain Shear of γ 2%. Since sample length is 7 mm, 1% of = 2% corresponds to 1.4 mm which indeed is the typical size of a shear band step in compression. Thus 1% load drops (serration events) basically correspond to single shear band emission events. Bands are oriented at near 45 degrees to cross-head axis (of applied load) so multiply by (2) 1/2 to get shear steps of roughly 2 microns.

30 Thus it can be said with confidence that the serration events correspond to individual shear band activation events. Now consider the elastic energy released due to the formation of a shear band. Q e = τdγ = Eγ 2 /2 ΔQ e = τ e γ e /2 - (τ e - Δτ e ) (γ e - Δγ e )/2 τ e.δγ e + Δτ e.γ e 2%τ e.γ e

31 ΔQ e = 2%τ e.γ e = 2%. 2 x 10 9 Pa = 4 J/cm 3 with specific heat C p 3 J/K.cm 3, does the sample heat up by ΔT ΔQ e / C p 1.3 K? If heat release is contained in the shear band (adiabatic heating), then you disperse ΔQ e from the whole specimen volume V sample = 0.2 cm x0.3 cm x0.7cm into by the shear band volume V band V band 0.2 cm x0.3 cm x20 nm and ΔT (ΔQ e / C p ).(V sample / V band 1.3 x K

32 It has recently been shown by Lewandowki anf Greer (Nature Materials Jan. 2006) that shear band nucleation results in a surge in temperature by over 1000 K. Do we expect that this phenomenon to help nucleation/growth of nanocrystals in shear bands? H = 0.4 kj m -2 H = 2.2 kj m -2 Temperature Rise, Δ T (K) ΔT = 207 K Distance, x ( μm)

33 In reality, the heating and subsequent cooling of shear bands occurs much faster than in melt-spinning so by itself cannot explain deformation-enhanced nanocrystal formation. Thus, while this heating is a very important phenomenon in deformation of BMGs, it is the destruction of the glass's medium range order in shear bands that seems to allow nanocrystal formation and shear delocalisation leading to enhanced ductility unexpected in heterogeneous deformation.

34 Thank you for your attention

35 Formation of nanocrystal clusters during deformation also affects crack propagation as cracks propagate behind and into severely sheared zones. Crack tips are stress concentrators and for tip radius R tip, crack propagation occurs under applied stress: σ effective (tip) = σ applied (a/ R tip ) 1/2 where a is the crack length. As the crack is forced to widen to escape blockage, the tip radius increases and the effective stress σ effective at the tip drops and the crack is blunted, leading to increased toughness.