X-ray Diffraction from Materials

Transcription

1 X-ray Diffraction from Materials 8 Spring Semester Lecturer; Yang Mo Koo Monay an Wenesay 4:45~6:

2 Plastically ll eforme metal -Macro strain: uniform elastic strain over relatively large istance iffraction peak shift ue to the lattice planar spacing changes -Micro strain: nonuniform o elastic eas strain at short range iffraction peak broaening ue to nonuniform spacing change in each grain For real cases, both kins of strain are usually superimpose, an iffraction lines are both shifte an broaene, because not only the plane spacings vary from grain to grain but their mean value iffers from that of the uneforme metal. We will be concerne with the line shift ue to uniform strain. Theory of micro strain analysis is still investigating several group in the worl incluing my laboratory.

3 Example of elastic macro strain Clapping (mechanical force) C: compressive stress Weling (thermal expansion) T: tensile stress Resiual stress inuce by plastic flow in bening: (a) loae below elastic limit; (b) loae beyon elastic limit; (c) unloae. shae regions have been plastically straine

4 x ray x ray N P surface σ σ ψ rotation ti

5 Pi Principle i of fresiual stress measurement ( φ ψ ) ' ε, O S, S G,G, S 3,G : specimen axes 3 : Diffractometer axes after rotation of φ anψ The strain ' ε where ( φ, ψ ) ε ' = ( φ, ψ ) is ( φ, ψ ) efine as ( φ, ψ ) is the planar spacing O irection, ( φ, ψ ) = a a ε 3k l kl ε ' 3 is unstraine planar spaceing. The tensor transformation between the axes of specimen an those of iffractometer

6 a i is the irectional cosinebetweenthethe i-axisof iffractometer an the -axisof specimen. a i cosφ cosψ sinφ cosψ sinψ = sinφ cosφ cosφ sinψ sinφ sinψ cosψ y combiningthe equations; ' ε ( φ, ψ ) = ε cos φ sin ψ + ε sinφ sin ψ + ε sin φ sin ψ + ε 3 cos ψ + ε cosφ sinψ + ε sinφ sinψ 3 The relationship between stress an strain in isotropic materials; + ν ν ε i = σ i δ i σ kk E E whereν is Poisson' s ratio an E Young moulus. δ ' ε + ν sin E + ν ν + σ ( σ + σ + σ ) E E + ν + ( σ3 cosφ + σ 3 sinφ) sin ψ E ( φ, ψ ) = ( σ cos φ + σ sin φ + σ sin φ σ ) ψ i is elta function which has value, when i =.

7 This equation tell us the stress state of the specimen which is measure strain from iffractometer at the position of Φ anψ. The resiual stresses are usually estimate from an sin ψ as shown in the figure. an sin ψ. There are three types of (a) Isotropic an no shear stresses (b) Isotropic but shear stresses (c) () General cases (non - isotropic)

8 iaxial i stress The stress states are ivie by uniaxial, biaxial, an triaxial. The stress at right angles to a free surface is always zero at that surface, so that two stress components are eal with stress analysis; a s; ( φ ψ ) ' ε, With a stresse boy, there exists three prepenicu lar stresses ( σ, σ, an σ 3; principal stresses) which are normal to planes on which no shear stress acts. Consiering the surface stress state, principal stresses σ an σ are parallel to the surface, an σ3 is zero. ' ε where σ ν E sin φ + σ sin ( φ, ψ ) = σ sin ψ ( σ + σ ) φ + ν E = σ cos the specimen surface. φ ( φ + σ φ) which is the stress along φ irection on

9 Measuring the spacing of a (hkl) for severalψ at given φ of the specimen, one can get the graph of an sin ψ. The resiual stress σ φ can be calculate using the slope of the graph. Peak shift ue to resiual stress is very small (~. ) so the resiual stress σ can be approximate E cotθ σ φ = Δθ sin ψ ( + ν ) φ Triaxial stress For general stress state, one can estimate the stresses from the following proceure; Let us use the stress an stain equation an efine two parameters a a + + φ, ψ φ, ψ = + ν ν σ φ σ φ σ φ σ ψ + = cos + sin + sin sin + σ E E an a ; ν ( ) ( σ + σ + σ ) E + φ, ψ φ, ψ + ν a = = 3 3 sin E ( σ cosφ + σ sinφ) ψ

10 σ, σ, σ the graph a, an σ v.s. sin are calculate from the intersect an slope of ψ at φ =, 45, 9. n σ from the slope of the graph a v.s. sin ψ at φ = 3 an σ, 9. 3 are calculate Elastic constants for x-ray measurement To convert stresses from the measure strains, the values of E an ν of the materials is neee. However, E an ν is constant for x-ray measurement because they are efine at isotropic continuum boy. There are a few metho to calculate x-ray elastic constant using elastic constants of the crystal. Voigt: constant strain Reuss: constant stress Hill: average of Hill an Voigt

11 Measurement of The exact value of Since the absolute value of for the measurement of. is very important to get acurate ata of the resiual stresses. can not be measure irectly, inirect metho is use iaxial stress ssume the stress state t is biaxial i an Φ=, then If ψ ψ = ψ *, = * ψ = + ν σ E. sin ν ψ E ( σ + σ ) sin where m ν σ + + m m * = = ψ + ν σ + ν an m may be calculate form the graph of φ = ν are the slope of sin ψ v.s. at φ =. an 9. * fter ψ is obtaine,

12 Triaxial stress Similarly proceure is performe for triaxial stress state, sin ψ * = It is quite ifficult to get ssumeν =.5, m sin ψ * ( + ν ) σ + ν ( σ + σ + σ ) ( + ν )( σ σ ) the sin the above equation equation can be approxiamte as ν σ ν + + σ ψ * is given by using this equation. Let us make a simple form of this equation. σ σ ν m = ν + + m an m is the slope of the graph v.s.sin ψ at φ = φ an 9, respectively. The error which occurs from the assumption v =.5 can be estimate inserting this equation to the initial equation of triaxial stress state., ψ * Δ = ν = σ E However, the resiual stress is measure on the surface of the specimen so that σ small. This assumption is OK for the triaxial stress state measurment. is usually ν=.5 for the perfect isotropic elastic materials ν=. for metal.

13 Measurement of the iffraction peak position The measurement of peak positionis very important to estimateaccurateresiualstresses. The stanarmethoof the finingthe center of a iffraction lineis to fit a parabolato the top of thelinean take the axisof the parabolaas the line center. withits axisparallelto the y axisan vertex at (h,k) is ( x h) = p( y k) The equationof a parabola If weput x = θ an y = I, this equationrepresentsthe shapeof the iffraction line near its peak.. (h, k) One substitutes tes several eral pairs of θ, I values into the equation an solve for h by the metho of least squares. Metho of locating the center Of a iffraction line

14 Error sources for resiual stress measurement. Since focusing circle oes not intersect at receiving slit by tilting ψ, the intensity maximum position is change.. Misalignment of iffractometer 3. Specimen preparation: Grining an polishing may have surface amage Resiual stress on the surface after grining : 5μm is amage

15 Measurement of resiual stresses in non-isotropic materials The graph which has non - isotropic material behabiors an oscillation for These may cause elastic costants. sin ψ. i) non -uniform efomation of materials, an/or ii) anisotropy of x -ray The stuy on this fiel is still going g on several place in the worl. Crystal anisotropy causes a texture structure, an x-ray elastic constants can be calculate from etaile information of texture. v.s. Oscillation for v.s. sin ψ can be calculate using the obtaine x -ray elestic constants, a resiual stress of the anistropic crystal cn be evaluate.

16 Example; Dr. Chang-Hwan Chang Preferre orientations of col rolle steel sheet are {}<>, {}<>, an {}<>, of which reuction is 9% using ultra low carbon steel. The relationship between strain of hkl peak an the principal stresses becomes ε hkl = ( α α + β β + γ γ ) + ( αα + ββ + γγ ) 3 S + S = where α, β, γ are the irectional cosine between [hkl] an crystal axes, an α, β, γ are the irectional cosine between pricipal axes an crystal axes. One can get the elastic constants x - ray elastic constants for S 6 S 3 S + 6 the crystal. for 6 { } : S = S + = 77. [ MPa ] 3 6 { } : S3 = S + =. 75 [ MPa ] 6 { } : S = S =. 79 [ MPa ] 3 σ each preferre orientation from

17 The volume fraction of the crystallite group: {}<>; 8%, {}<>; 4%, an {}<>: 48%. The average x-ray elastic constants can be calculate these volume fraction. The calculate x-ray elastic constants using Voigt an Reuss moel are Voigt Reuss Moel : Moel : S 3, V = S 3, R =. [ MPa ] [ MPa ] MPa This figure shows the experimental results of which crystal has preferre orientation uner uniaxial loaing. The measure 6 elestic constant, t S3 = 8 MPa 6 [ ] This value is close to Hill's moel

18 9.5 Plastic Deformation ehavior of a Single Crystal Laue camera λ λ One can measure the slip an rotation of the slip plane X-ray Stereographic proection of the uniaxial tension P: position of tensile axis Φ : the angle between the tensile axis an the normal of the slip plane λ : the angle between the tensile axis an the slip irection

19 9.5 Plastic Deformation ehavior of a Single Crystal FCC: silp system {}<> ctive slip system: largest Schmi factor τ = σ cosφ cos λ If the tensile axis is P at starting point, is [ ]() an When another the tensile axis meet the active slip system P is moving as shown in the figure. slip system is acting for to the line [ ] -[], the eformation so that the tensile axis is moving to [ ]

20 9.5 Plastic Deformation ehavior of a Single Crystal This is Dr. Ki-hong Kim s ata One can observe the behavior of the silp system y x-ray iffraction peak mesurement.

21 9.6 Structure nalysis of Multi-layer Diffraction from multilyer n n n f f T : number : number = n + n of layer of layer : atomic scattering factor of : atomic scattering factor : average istance of = ( n + n )/ n T bilayer of ; mplitue of the iffracte x - ray from a multilayer becomes ( h,h,h ) exp { πi ( h n + h n + h n ) } f exp πih ( n + ) = n n n n= + f 3 exp πih n n n exp ih3 π n ( n + ) 3 =

22 9.6 Structure nalysis of Multi-layer Rewrite this equation; sin{ πih3n ( )} sin{ πih ( )} 3n f + f ( ) sin { π ih ( ) } sin { π ih ( ) } ( πih3nt N ) ( π ih N ) sin ( h,h,h3 ) sin where N is the total number of bilayer. = Since the multilayer is mae artificially, there is eviation from the perfect orering. Taking into account the isorering of the multilayer, let us calculate the structure factor. Deviation n n = n = n - th + δn + δn bilayer can be expresse where n is the average istance of () layer, an δn ( ) ( ) average istance. F The structure of ( q ) f exp( iq l ) z = - exp th layer becomes ( iq n ) exp( iq δn ) n z z = f l= exp z z ( iq ) is the eviation from

23 9.6 Structure nalysis of Multi-layer The structure factor for all N bilayer becomes F tot ( q ) = exp iq Λ [ exp( iq n ) F ( q ) + F ( q )] z N z m= where Λ = Λ + δ n + δn, an Λ = n + n m z z. z If δn follow the Poisson istribution, the ensemble average of δn may be expresse by exp where σ ( inx) = exp[ σ sin ( x Poisson )] Poisson The structure factor of F tot where S an F F ivsegence of δn. the ensemble average becomes ( q, σ ) = exp( iq Λ) exp( σ S ) z Poisson z Poisson exp exp ( iqznλ) exp( σ PoissonNS ) p ( iq zλ ) exp ( σ PoissonS ) { σ sin q / } + F ( q,σ ) [ F( qz,σ Poisson ) exp( iqzn ) exp Poisson ( z ) z Poisson ] = sin ( qz / ) + in ( qz / ) exp ( ) ( iqzn ) exp{ σ Poisson sin ( iqz / ) } qz,σ Poisson = f exp( iq ) ( q σ ) is the measure structure factor having the thickness fluctionations. tot z, Poisson z

24 9.6 Structure nalysis of Multi-layer Thickness measurement of multilayer X-ray θ t t : ( m layers) : ( m layers) θ Λ ( ) = + / Diffraction equation: very ifficult to have analytical solution Reflection equation: this is enough to get information of muitilayer Properties of multilayer from iffraction peaks: one may solve the iffraction equations which inclue the thickness fluctuation.

25 9.6 Structure nalysis of Multi-layer Example: Dr. Sang-Kook Kim

26 9. pplications of X-ray Diffraction Homework Due ate: May 7, 8 Solve the problems; 9., 9., 9.5 If possible, please solve 9.4 (Extra points)