DIFFRACTION METHODS IN MATERIAL SCIENCE. PD Dr. Nikolay Zotov Lecture 7

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1 DIFFRACTION METHODS IN MATERIAL SCIENCE PD Dr. Nikolay Zotov Lecture 7

2 OUTLINE OF THE COURSE 0. Introduction 1. Classification of Materials 2. Defects in Solids 3+4. Basics of X-ray and neutron scattering 5. Diffraction studies of Polycrystalline Materials 6. Measuring Powder Diffraction Patterns; 7. Microstructural Analysis by Diffraction 8. Diffraction studies of Thin Films 9. Diffraction studies of Nanomaterials 10. Diffraction studies of Amorphous and Composite Materials 2

3 OUTLINE OF TODAY S LECTURE Microstructure of polycrystalline materials Size Broadening Microstrain Broadening Residual Stresses 3

4 Polycrystalline Materials Grain Distributions Sand LaSrFeCoO 3 (LSFC) powder Metals; alloys 4

5 Polycrystalline Materias Microstructure Grains; Grain-Size Distribution Deformation of Grains (Microstrains) Orientation of grains Random orientation Preffered orientation (Texture) Grain Boundaries; Voids; Microcracks Residual Stresses (Marcostrains) 5

6 I(Q) ~ F 2 L(Q) Effects of Grain Size L(Q) Laue Function = [sin 2 (2pKQ.a/2)/sin 2 (2pQ.a/2)] [sin 2 (2pMQ.b/2)/sin 2 (2pQ.b/2)] [sin 2 (2pPQ.c/2)/sin 2 (2pQ.c/2)] It is the product of three sinc(x) = sin 2 (Nx)/sin 2 (x) functions (N = K, M or P) where K, M and P are the number of unit cells along x,y,z directions. N (single crystal) sinc(x) d(x) N Small: Broadening of the diffraction profile and decrease of maximum intensity N = 10 N = 20 N = N = 100 Oscilations Noise Grain-size distribution: 2000 The real profile is a superposition 0 of the lines of grains with different small sizes further broadening sinc(x) x 6

7 l = 2dsin(Q) Bragg Law Derivation of Scherrer Equation ml = 2mdsin(Q) = 2Lsin(Q) 0 = 2DLsin(Q) + 2Lcos(Q)D(Q) finite difference L = DLsin(Q)/cos(Q)D(Q) L hkl DL = d L = dsin(q)/cos(q)d(q) = l /cos(q)2d(q) = l /cos(q)d(2q) L hkl = md hkl FWHM in radians L = l/cos(q)ß; L is the average length of the crystallites perpendicular to the surface!!!! Qualitatively: Small grain size leads to structural line-broadening 7

8 Example - Ag Nanoparticles XRD, Scherrer Equation L ~ nm fcc Anneal. Time, h < Size>(TEM), nm Shameli et al. (2012) 8

9 Crystallite-size broadening Example Lampater (MPI) 9

10 Microstrain Broadening I Deformations within individual grains Extension Compression Ds Ds = (2/l) cos(q) DQ (1) Distribution of deformations Distribution (overlap) of Diffraction profiles s s = 1/d Ds = 1/d o (1-e) - 1/d o (1+e) ~ 2e/d o D = 4esin(Q)/l (2) DQ = Ds/2cos(Q) = 2 e tg(q) ß ~ 2DQ = 4e tg(q) (3) 10

11 Correction for Instrumental Broadening (Single Line) 1/ Fit of sample and standard lines with pv functions Get 2Q B, FWHM and h for specimen and standard; 2/ Interpolate, if necessary, FWHMs and h of the standard at the position 2Q B of the peak of the sample; 3/ Calculate the Gaussian and Lorentzian components of the corresponding profiles: ß GS, ß LS, ß GI, ß L I INPUT: FWHM, h ß = ½ FWHM { p/2 h + ½ (p/ln2) ½ (1-h)} ß G = ß f G (h) ß L = ß f L (h) f G (h) = g 1 + g 2 h + g 3 h 2 + g 4 (1 + g 5 h) ½ ; f L (h) = c 1 + c 2 h + c 3 h 2 ; Delhez, Keijezer, Mittemeijer, Fres. Z. Anal. Chem. 312 (1982) 1 11

12 Correction for Instrumental Broadening (Single Line) 4/ Correct for instrumental broadening, assuming Lorentian distribution of crystallite sizes G = G 1 * G 2 s 2 (G) = s 2 (G 1 ) + s 2 (G 2 ) and Gaussian distributions of microstrains L = L 1 * L 2 (ß LR ) = (ß LS ) - (ß LI ) ; s(l) = s(l 1 ) + s(l 2 ) (ß GR ) 2 = (ß GS ) 2 - (ß GI ) 2 ;

13 Single-line Size/Strain Anaylsis Calculate L by the Scherrer equation L = l/cos(q B )ß L R Calculate the microstrain e = ¼ ß GR ctg(q)

14 Williamson Hall Plot Expression for the total structural FWHM: Loretzian shape: ß Total ~ ß Strain + ß Size = cos(q)ß = 4etg(Q) + l/cos(q)l = 4esin(Q)/cos(Q) + l/cos(q)l cos(q)ß Total = l/<l> + 4<e>sin(Q) sin(q) 14

15 Willamson - Hall Plots Example I NiTi Shape Memory Alloys ß hkl cos(q) Williamson-Hall Plot Thermally Cycled NiTi SMA <L>= 39 nm <e> = 0.16% sin(q) Zotov (2016) 15

16 Willamson - Hall Plots Example II NiTi shape memory alloy Mechanically cycled (100 cycles) ßcos(Q) Anisotropic line-broadening sin( ) Hooke s law Anisotropy of the Young s modulus: e = s/e; cos(q)ß Total = l/l + 4s/E hkl sin(q) Presence of dislocations: cos(q)ß Total = l/l + 4(p/2) 1/2 br 1/2 /C hkl sin(q); (b Burgers vector; C - dislocation contrast factors (hkl); r dislocation density) 16

17 Willamson - Hall Plots Example IV Nanocrystalline Ni (fcc) Csiszar et al., J. Appl. Cryst. 45 (2012) 61 17

18 Residual Stresses # Macroscopically deformed state # Compression (extension) of the d-spacings d hkl in each grain # Shifts of the positions of the diffraction lines # Bragg s law: Measurement of the d-spacings parallel to the surface d ; e Z = (d - d o )/d o

19 Residual Stresses Elastically-deformed grains (Hooke s law) e Z = e 3 = - n (e 1 + e 2 ) = - n/e (s 1 + s 2 ) (d - d o )/d o = - n/e (s 1 + s 2 ) s 1, s 2 and s 3 stresses along the principle axes of the stress tensor n Poisson s ratio Measurements of the residual stresses in the Laboratory coordinate system Presence of grains, which have different orientation (inclination) with respect to the surface Determination of all stress components s ij ; Koker & Zotov (2013) e fy = (d fy - d o )/d o

20 Residual Stresses e fy = s 1 [ s 11 + s 22 + s 33 (1 + ½ s 2 /s 1 )] + ½ s 2 t f sin(2y) + ½ s 2 (s f s 33 ) sin 2 (Y) s 11 s 12 s 13 Stress Tensor s 21 s 22 s 23 s 13 s 23 s 33 Tr(s) = s 11 + s 22 + s 33 = s 1 + s 2 +s 3 = I 1 (1st Invariant of the stress tensor) s 1 = - n/e hkl ; s 2 = 2 (1 + n)/e hkl ; X-ray compliences s f = s 11 cos 2 (f) + s 12 sin(2f) + s 22 sin 2 (f) t f = s 13 cos(f) + s 23 sin(f) Normal stress in the f direction Shear stress in the f direction

21 In-plane stress state: s 33 = 0 Residual Stresses No shear stresses: s 13 = s 23 = 0 e fy = s 1 [ s 11 + s 22 ] + ½ s 2 s f sin 2 (Y) = s 1 I 1 + ½ s 2 s f sin 2 (Y) s f = s 11 cos 2 (f) + s 12 sin(2f) + s 22 sin 2 (f) e f = 0 o s f = s 11 ; f = 90 o s f = s 22 ; f = 45 o s f = s 11 3/4 + s 12 + s 22 ¼; If s 12 = 0; Rotationally-symmetric in-plane stress state sin 2 (Y) If s 11 = s 22 = s and s 12 = 0; Rotationally-symmetric bi-axial in-plane stress state e fy = 2s 1 s + ½ s 2 s sin 2 (Y)

22 Residual Stresses Fig. 5.5 Fitzpatriclk et al., NIST (2005)

23 Residual Stresses YSZ Measurements of the (220) t (400) c Cu Ka; Ni Kß filter, D2Q = 0.02 o, 30 s/step Zotov (2010)

24 Residual Stresses 500 nm Ag 10 Å/s 500 nm Sn 10 Å/s Si + 50 nm SiO 2 Rossi & Zotov (2014)

25 Residual Stresses s = MPa # Co-K a -radiation, 40 kv 40 ma, # Graphite Monochromator in diffracted beam path # Parallel beam geometry with polycapillary collimator Ag (311) peak measured 2Q ~ 92 o ; f = 0 o ; Rossi & Zotov (2014)

26 Residual Stresses Rotationally-symmetric in-plane state Rossi & Zotov (2014)