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D2. Anelasticity of Solids

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D2. Anelasticity of Solids D2. Anelasticity of Solids I. OBJECTIVE OF THE EXPERIMENT Measurement of the elastic and anelastic moduli of different metals, observing the relaxation phenomenon in steel as well as Snoek relaxation. II. PHENOMENOLOGY OF ANELASTICITY When a solid that can not move is subject to a set of external stresses, it reacts by deforming. For small stresses, the deformation is instantaneous, proportional to the applied stress, and vanishes when the stress is removed. This is called elastic deformation. The relationship between stress and strain is given by Hooke’s law: It can be written E where E is Young modulus or J where J is elastic complaisance. Anelasticity in solids can be shown by this simple experiment (fig.1a). A stress , of low intensity is applied to a sample at the time t 0, and is held constant, while the strain , is recorded. We observe an instantaneous strain, eu J , where Ju is called the unrelaxed complaisance, and a so-called anelastic strain, a , that increases over time from zero to a limiting value, a When in equilibrium: e a J r where Jr is the relaxed complaisance t a J - J r u • a a Ju e e t a) b) Fig. 1: Anelasticity of solids The process through which a solid under a stress goes from one equilibrium state at , to another one at t = ∞, is called anelastic relaxation, and is defined by two main parameters. J J - relaxation intensity a r u et (1) e Ju - relaxation time EPFL-TRAVAUX PRATIQUES DE PHYSIQUE D2-2 If after a certain amount of time, the stress is removed, we will see the elastic strain vanish instantly, and the anelastic strain decrease with a relaxation time (fig 1). In this experiment, the deformation is completely reversible. The difference between the elastic and anelastic process, is that the latter isn’t instantaneous. In metals, anelastic strain is much smaller than elastic strain, i.e (Jr J u ) J u . We can describe the evolution of the anelastic strain a with an exponential: t aa(1 e ) (2) a) Elastic strain e and anelastic strain of a solid under a stress . b) Rheological model corresponding to te anelastic process in fig. 1 a, Jr Ju From equation (2) and in the case where (Jr J u ) J u , we can determine the equation of an ideal linear solid, characterizing the process described in fig. 1 Jr Ju (3) From a rheological standpoint, the ideal linear solid can be represented by the model in figure 1b (to be shown by the student) The measurements of the system in a static system are very delicate, since the average anelastic strain in metals is generally of the order of 109 to 108 , which is why we use dynamic methods to determine anelastic strain. A harmonic stress of frequency , exp(i t) , is applied to the solid. The linearity of “stress-strain” 0 relationships results in a strain that is also periodic, and of same frequency , exp(i( t )) 0 however, it will be phase-shifted with respect to the strain by an angle (due to anelasticity). Introducing the expressions of and in equation (3), yields: J *() (J()12 iJ()) (4) where J1 and J2 are respectively the real and J1 Jn J imaginary parts of the complex complaisance J * , J 2 J J J J r u 1 u 2 2 1 Diagram of “rotating vectors” Jr Ju J2 . (5) 12 2 Measuring the phase-shift between stress and strain, or measuring the tangent of the mechanical loss angle ( tan ), is actually a measure of the internal friction of the material Q1 J Q1 tan 2 (6) J1 By introducing (5) in (6) and supposing that (Jr J u ) J u , we get: 1 J J1 Ju 1 Q 22 et 2 2 (7) 1 J Ju 1 J where is the variation of the complaisance due to anelastic relaxation. J EPFL-TRAVAUX PRATIQUES DE PHYSIQUE D2-3 Equations (7) show that the internal friction, , has a maximum that depends on , centered in 1. This maximum, or internal friction peak gives us the intensity of the relaxation, as well as the relaxation time , (position of the peak on ). Fig. 2: Internal friction peak (Debye peak) and default associated elastic modulus. From a microscopic point of view, anelastic deformation can be interpreted as the motion of defects of the crystalline structure (elastic dipole associated to point defects, dislocations, grain boundaries,…) that go from one equilibrium state at 0 to a new equilibrium state for . The intensity of the relaxation , is proportional to the defect concentration, and the relaxation time , is a measure of their mobility. When the mechanisms that control the mobility of the defects are thermally engaged, we have: H kTb : 0 e ( H = activation enthalpy) And for an imposed frequency , we can see a peak that depends on the temperature TT P such that: H kTb 0e 1 (8) 1 from (7) and (8): Q1 (9) H 11 2ch ( ) k TP T The height of the peak is 2 (fig 2). Measuring the activation enthalpy H : st 1 method: Perform measurements at different frequencies. For the frequency i , the peak is located in 1 TPi . We then plot ln versus , whose slope is () Hk. TPi nd 1 2 method : From equation (9) we know the value of Q at mid-height (i.e. at a temperature T1 ) 1 1 1 Q Q H 11 4 2ch ( ) k TP T1 (10) H 11 H 11 Comparing (9) and (10) yields: ch ( ) 2 thus ( ) 1.317 k TP T1 k TP T1 EPFL-TRAVAUX PRATIQUES DE PHYSIQUE D2-4 2.633k 1 1 1 or H with () width of peak mid-height (11) 1 TTT () 21 T Anelastic Snoek relaxation x Snoek relaxation is due to interstitial point defects in the 1 form of solid solutions of C, O, N, H, in centered cubic (c.c.) crystals, such as carbon of iron. These impurities are in the octahedral sites of the c.c. structure. The generate elastic dipoles, of tetragonal 1 2 symmetry. The axis is parallel to the 4-symmetry axis of 3 an octahedron. For instance in figure 3, the interstitial atom located in position 1 can create a dipole of axis parallel to Ox1. It will push back the neighboring atoms (up and down in x the figure). 3 Interstitial atoms can be divided into three categories, since the axis of the dipole can be oriented according to Ox1, Ox2 or Ox3. x 2 Fig. 3: Possible octahedral site for interstitial atoms in a c.c. structure. Let N0 be the total number of defects or elastic dipole per unit volume. Then: 3 N0 N1 N2 N3 Ni (12) i1 where Ni is the concentration of dipoles whose axis is parallel to xi. When in equilibrium, and for no external stress ( 0 ), the 1,2,3 sites are equivalent. N Then: NNN 0 for (13) 1 2 3 3 and the extra elastic deformation due to the interstitial defects is the same in all three directions. When we apply stress in a given direction, one type of site can be preferred, and the defect densities will vary. N NNN 0 for 0 (14) 1 2 3 3 In order to simplify, consider a stress parallel to Ox1 ( 11 ) and observe the corresponding strain 11 . We have: C J (C 0 ) (15) e a n 1 3 where (16) c1 c0 const defines the strain variation due to migration of defects from 2 and 3 towards1. c0 is the atomic concentration of interstitials, c1 represents those oriented along Ox1. The anelastic deformation is due to the extra strain generated by this atomic movement. EPFL-TRAVAUX PRATIQUES DE PHYSIQUE D2-5 c (c 0 ) (17) a 1 3 We can calculate the relaxation parameters and : 2 cv a2 00 2 and (18) 9 Jn k b T P 36D where is the atomic concentration of interstitials, v0 their atomic volume, D their diffusion coefficient, and a the lattice parameter: H kTb D D0 e (19) where H = diffusion energy (enthalpy) of the interstitials, k the Boltzmann constant. The relaxation intensity is proportional to the concentration of relaxing atomic interstitials. The relaxation time is a measure of the activation energy of their diffusion. III. EXPERIMENTAL PROCEDURE tg can be obtained directly by measuring the phase-shift between stress and strain. However, for weak damping ( tan 103 ), it can be easier to use a resonant system, vibrating at its eigenfrequency. In this experiment, we will use a short vibrating rod (fig. 4). The vibration frequency f of an unattached rod is: eE e f 1.03 (20) l 2 0.224 l where E is the Young modulus, the density, e l and l the thickness and the length of the sample. Inverting equation (20) yields : l 4 Fig. 4: Flexion of a rod of length l. The vibration knots Ef 0.943 2 (21) are located at 0.224 from the end of the rod 2 l e Ai The internal friction, Q1 , is determined by the decrement rate of the oscillation amplitude (fig. 5) Ai+n 1 A tan Q1 ln i (22) nA in th th where Ai and Ain are the i and (i+1) vibration amplitudes respectively (fig. 5) c0 Fig. 5: Free decrement of an oscillating system. EPFL-TRAVAUX PRATIQUES DE PHYSIQUE D2-6 Measurement setup (fig. 6) The sample to measure is a small rod of length l and thickness e that is fixed with four thin steel wires that go through the knots of the fundamental vibration (cf.
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