Chapter 1 Exercises of Chapter I-1: Introduction 1.1 Selection of materials Exercise 1.1 Material for a flywheel This exercise illustrates rational selection of materials. It could have been given as an example in Sect. 1.1.7 of Volume I. We wish to design a flywheel to store as much energy as possible given the rotation speed and the thickness. There is no requirement regarding the radius. Which material should we choose? 1) Give the energy W stored in the flywheel. Solution Assume the flywheel to be a disk of constant thickness t 1 2 1M 2 2 Iω = Rω 2 2 2 M = π R2t ρ W= (1.1.1) I being the moment of inertia of the flywheel, M its mass, R its radius, ω the speed of rotation and ρ the density of the material. 2) What is the dimensioning factor of a flywheel? Solution The centrifugal force should be less than the one which could produce an explosion. This means that the maximum stress should be less than the tensile strength Rm (provided there is no microcrack present). However, the flywheel stops from time to time; the maximum stress varies in a cyclic manner between zero and its maximum value when the rotation velocity is reached. This produces fatigue of the material. The stress amplitude should be kept below the endurance limit of the 1 2 material (refer to Chapter 6 of Volume II). It can be considered that this endurance limit is equal to one third to one half of the tensile strength. All being considered, the dimensioning condition is for the maximum stress not to exceed some fraction k, including a safety factor, of the tensile strength. 3) Give the location and the value of the maximum stress in the flywheel. Solution For a disk, the maximum stresses are located on the axis (refer to mechanics textbooks). The hoop stress and the radial stress are equal there and given by: σ θ max = σ r max = 3+ν 2 2 R ρω 8 (1.1.2) 4) Write the dimensioning condition 3+ν 2 2 R ρω = kRm 8 (1.1.3) Being free to choose the dimension of the radius, what is the characteristic ratio of material properties to maximise the stored energy in the flywheel? Solution Eliminate R between (1.1.1) and (1.1.3). W = k2 π 16 ( 3 + ν ) 2 t Rm 2 ω2 ρ (1.1.4) To maximise W the performance ratio Rm2/ρ should be as high as possible. 5) Indicate the best selection of materials for the flywheel. Solution An Ashby map of tensile strength versus density (Fig. 1.1.1) shows that ceramic materials, composite materials and high strength steels are good candidates. Diamond has a performance ratio equal to about 3x107 in MPa and kg/dm3. That would be an expensive flywheel! For alumina it is about 2.5x106, 3x105 for high strength steel, 5x105 for carbon fibre reinforced polymer. For problems in applica- 3 tion, alumina cannot be considered and it turns out that carbon fibre reinforced polymer is the best choice. Fig. 1.1.1 Strength-density selection plot 1.2 Mechanical testing The following exercises aim at a better understanding of mechanical tests described in Sect. 1.3 of Volume I. Exercise 1.2 Tensile test, Considère's relation 4 We consider a tensile test performed with a cylindrical test piece and we wish to predict the load versus deformation curves assuming particular constitutive behaviours. The cross-section of the test piece as a function of time is denoted S(t) and its length L(t). Assuming constant volume: S(t) L(t) = S(0) L(0), express the relation between the true strain and the evolution of the cross-section. a) General case Assuming a constitutive uniaxial relation σ = σ ( ε , εɺ ) , write the equation relating the applied load P to the engineering strain denoted η (neglect the elastic deformation). Find the percentage elongation at maximum force Ag (Considère's relation). Solution Give the true strain as a function of the elongation. ε = log L L ( 0) (1.2.1) Calculate the evolution of the cross-section. S (t ) L (0) = = exp ( −ε ) S (0) L (t ) (1.2.2) Calculate the engineering strain . η (t ) = L (t ) − L (0) = exp ε − 1 L (0) (1.2.3) Deduce the evolution of the applied load. From (1.2.2) and (1.2.3) P ( t ) = S ( t )σ = S (0) σ ( ε , εɺ ) 1 + η (t ) (1.2.4) Calculate, in term of stress, the condition when the maximum force is reached. This is achieved when dP(η ) = 0 5 σ ( ε , εɺ ) 1 dP (η ) 1 dσ ( ε , εɺ ) =− + 2 S ( 0 ) dη (1 + η ) 1 + η (1 + η ) dε (1.2.5) And now from (1.2.5) the Considère relation for dP(η ) = 0 −σ ( Ag ) + dσ ( Ag ) = 0 dε (1.2.6) b) Strain rate independent plasticity Consider the case of a strain rate independent plastic material. Assume that its uniaxial behaviour follows the Hollomon relation: (1.2.7) σ = σ 0ε n Determine the percentage elongation at maximum load Ag. Solution Equation 1.2.6 yields: Ag = 100n (1.2.8) c) Strain-hardening independent viscous material Consider the case of a strain-hardening independent viscous material. Assume that its uniaxial behaviour follows the Norton law εɺ = (σ K ) M (1.2.9) Draw the load versus elongation curve of a material for which M = 4.2 and K = 100 MPa. The cylindrical test piece is 10 mm in diameter and 200 mm long. The cross-head velocity is 1.2 mm/min. Solution Equation 1.2.4 together with (1.2.1) yields: 1M Lɺ P (η ) = KA ( 0 ) L ( 0) 1 1 M +1 (1 + η ) (1.2.10) 6 Figure 1.2.1 gives the result in term of P(η)/P(0). Fig. 1.2.1 Load ratio vs engineering strain for a strain-hardening independent viscous material. d) Strain-hardening viscous material To evaluate the percentage elongation at maximum load for a strain-hardening viscous material, study the stability of a small local perturbation of the test piece radius. Solution (Hutchinson and Obrecht 1977) The test piece of cross-section S0(0) contains at the beginning of the test a local zone of cross-section S(0) a little smaller than S0(0). Use subscript 0 for the perfect test piece devoid of any perturbation, that is away from the local perturbation, and no subscript for the quantities within the perturbation. Denote ∆S (t ) = S (t ) − S0 (t ) and s (t ) = ∆S (t ) S0 (t ) . These quantities are negative and small. This allows linearisation of the equations. If their absolute values increase the perturbation is unstable. Write the quantities ∆σ(t), ∆εɺ and ∆ε in term of s(t). From (1.2.4) can be written to the first order: 7 ∆σ ( t ) S 0 ( t ) + σ 0 ∆S ( t ) = 0 ∆σ ( t ) = −σ 0 s ( t ) (1.2.11) From (1.2.2) can be obtained the strain rate difference: ∆εɺ = − ∆Sɺ ɺ ∆S + S0 2 = − sɺ ( t ) S0 S0 (1.2.12) Integration of the equation yields: ∆ε = s ( 0 ) − s ( t ) (1.2.13) Find the relation between ∆σ(t), ∆ε and ∆εɺ . From the uniaxial behaviour: ∂σ ∂σ ∆σ = ∆ε + ɺ ∆εɺ ∂ε 0 ∂ε 0 (1.2.14) Write the differential equation for s(t) and integrate, introducing the quantities ∂σ 1 ∂σ εɺ0 ( t ) and µ ( t ) = ν (t ) = ∂ε 0 σ 0 ( t ) ∂εɺ 0 σ 0 ( t ) Elimination of ∆ε, ∆σ and ∆εɺ yields the differential equation: sɺ + εɺ0 ( t ) ν (t ) ν ( t ) − 1 s = εɺ ( t ) s ( 0 ) µ (t ) µ (t ) 0 (1.2.15) Integration yields: t t t ν (t ) s ( t ) = s ( 0 ) exp − ∫ h ( t ) dt 1 + ∫ εɺ0 ( t ) exp ∫ h ( t ) dt 0 0 0 µ (t ) ( where h ( t ) = ) (1.2.16) εɺ0 ( t ) (ν ( t ) − 1) µ (t ) Discuss the stability of the perturbation. The parameter s being negative sɺ is negative when ν < 1; in that case s decreases so that its absolute value increases. This is the case of instability. Note that the rate of instability increases with the ratio εɺ0 µ . 8 For a strain rate independent material, µ = 0, and then the instability takes place when ν = 1, the Considère criterion. εɺ When ν = 1 , sɺ (ν = 1) = 0 s ( 0 ) , a very small quantity, the more so the higher µ the strain rate sensitivity. A high strain rate sensitivity has a stabilising effect. Necking becomes noticeable for a strain, which can be much higher than the one given by the Considère criterion. Hutchinson and Obrecht show results of numerical calculations, which illustrate this fact. Write the solution for a material whose uniaxial behaviour is given by σ = αε nεɺ m , with m = 1 M . With this constitutive equation, µ = m and ν = n ε 0 . Exercise 1.3 Torsion test Torsion tests are useful for measuring the shear modulus of elasticity and for the study of the high temperature behaviour of metallic alloys. This is particularly important for optimising hot working conditions. From torsion tests, finding relations between stress and strain is required. Thus, recording the torque as a function of the angular rotation must result in a determination of the corresponding stress, strain and strain rate. a) Elastic behaviour Which reasonable assumption can be made about the deformation of the test piece in a torsion test? Deduce the strain and the strain rate tensors in term of the angular rotation φ. Give the stress tensor for isotropic elastic behaviour. Solution Assume that the cross-sections remain flat and that diameters remain rectilinear. This is well verified experimentally. (i) Give the displacement field, in cylindrical coordinates. ur = 0 uθ = −rϕ ( z ) = −rz φ L uz = 0 (1.3.1) where φ is the angle of rotation measured at the extremity of the test piece and L its length. Hence the strain and strain rate tensors: 9 0 0 (ε ) = 0 0 φ r 0 − 2L 0 φ r − 2 L 0 0 0 (εɺ ) = 0 0 0 − φɺ r 2L 0 φɺ r − 2 L 0 (1.3.2) The maximum values are found for r = R , the radius of the test piece. (ii) Find the stress tensor. The strain is pure shear. The only component of the stress tensor is a shear stress: σ θ z = τ = 2µε zθ = µφ r L (1.3.3) µ being the shear modulus of elasticity. (iii) Give the maximum shear stress. τR = µ φ L R (1.3.4) (iv) Determine the corresponding torque. C= ∫ R 0 2πr 2τ d r = ∫ R 0 φ π φ π 2π µ r 3dr = R 4 µ = R3τ R L 2 L 2 (1.3.5) Thus recording φ and the torque as a function of time allows determining the strain, the strain rate, the maximum shear stress and the shear modulus of elasticity. b) Strain rate dependent plasticity Determine the critical torque for plastic deformation to begin. As it is the stressstrain relation which is to be found, determine the equivalent strain as a function of the angle of rotation φ and the relation between the torque C and the maximum shear stress τR. 10 Solution Use the von Mises criterion. From Eq. 1.3.5, the torque C0 when plasticity begins at the surface of the test piece is given by: C0 = π 2 3 R3 Rp (1.3.6) Rp being the yield strength. (i) Calculate the strain tensor for a large rotation. Keep the same assumptions than for small deformation. Write the coordinates of the displaced vector M ′ . M ′ = M − r (1 − cos ϕ ) e r − r sin ϕ eθ + u ( z ) e z (1.3.7) u(z) being the lagrangian displacement in the z direction, that is with respect to the position of a cross-section which was initially at z. Deduce the strain tensor. The strain tensor ∆ij is given by: 2∆ij = 1 1 ∂M ′ ∂M ′ (θ k , t ) (θ k , t ) − δij ∂θ j E i E j ∂θ i (1.3.8) where E i = ∂ M ∂θ i and δij is the Kronecker tensor. From (1.3.7): ∂M ′ = er − (1 − cos ϕ ) e r − sin ϕ eθ = cos ϕ e r − sin ϕ eθ ∂r 1 ∂M ′ = − 1 − (1 − cos ϕ ) eθ + sin ϕ e r = sin ϕ e r + cos ϕ eθ r ∂θ ∂M ′ r r = e z − φ sin ϕ e r − φ cos ϕ eθ + uz′ e z ∂z L L Using Eq 1.3.8, the lagrangian strain rate tensor can be calculated: (1.3.9) 11 0 0 (∆) = 0 0 φ r 0 − 2L 1 φ 2 0 φ r − 2L 2 r 2 + 2u ′ + u′ L (1.3.10) There is now an elongation of the test piece if the lagrangian displacement u is constant, that is a rigid body displacement. For the dilatation ∆zz to be zero, a lagrangian displacement must be introduced, such that: 2 1 r u ′ + 2u ′ + φ = 0 2 L 2 (1.3.11) that is: 12 1 r 2 u ′ z = −1 + 1 − φ 2 L () φ≤ 2 L R (1.3.12) Plane cross-sections do not remain plane. Denoting γ = φ r L , calculate the equivalent strain when the test piece length is kept constant. 2 ∆ = ∆ij ∆ij 3 12 = 1 3 ( γ 2+γ2 12 ) (1.3.13) (ii) Deduce the stresses, neglecting the elastic part. s33 = σ 33 − sθz = σ θz σ 33 =− 3 2γ σ =− 32∆ 2γ2 σ 3 2 ∆ (1.3.14) where σ is the equivalent stress, a function of the equivalent strain, which must be determined experimentally. 12 (iii) Deduce a relation between the maximum shear stress τR and the torque. Integration by parts of Eq. 1.3.5 yields: C= 2π 3 2π R τR − 3 3 ∫ τR r 3dτ (1.3.15) dτ ∂∆ φ d∆ dτ dr = d∆ ∂r L dγ d∆ (1.3.16) 0 with dτ = Differentiate C, given by Eq. 1.3.5, with respect to φ: R ∂τ R r ∂∆ dτ ∂C 2π = 2π ∫ r 2 dr = 2π ∫ r 2dr = 0 0 ∂φ ∂φ L ∂γ d∆ L ∫ R 0 ∂∆ dτ 3 r dr ∂γ d∆ (1.3.17) Now, from (1.3.15), (1.3.16) and (1.3.17): C= 2π 3 φ ∂C R τR − 3 3 ∂φ (1.3.18) (iv) Find now a relation between the torque and the maximum shear stress involving the rotation velocity. The same procedure as the one just used yields: C= 2π 3 φɺ ∂C R τR − 3 3 ∂φɺ (1.3.19) (v) Explain the procedure to determine, from the torsion test, the relation between the stress and the strain and the strain rate. From (1.3.18) and (1.3.19): τR = 1 ∂C ɺ ∂C +φ ɺ 3C + φ 2πR 3 ∂φ ∂φ (Note that Eq.1.15 in Volume I is mistaken.) (1.3.20) 13 Perform torsion tests at various rotation velocities or perform a torsion test changing suddenly the rotation velocity. In each test, record the torque as a function of the angle of rotation. Equation 1.3.20 gives the maximum shear stress. Equations 1.3.13 and 1.3.14 with γ = γ R = φ R L give the needed relation. Reference Hutchinson J W and Obrecht H (1977) Tensile instabilities in strain rate dependent materials, Proc 4th Int Conf on Fracture Waterloo Canada 1:101-116 14 Chapter 2 Exercises of Chapter I-2: Elastic Behaviour 2.1 Interatomic binding energy in crystals Exercise 2.1 Cohesive energy, bulk modulus An interaction potential u per unit mass has the form (cf. Sect. 2.2.3 in Volume I) u = − Ar − n + Br − m where m > n (2.1.1) By considering the compressibility of a sphere of radius equal to the interatomic distance at rest a, express the bulk modulus k = −Vdp / dV , where p is the pressure and V the volume, as a function of m, n and the cohesive energy u0 for the mass density ρ. Solution The interatomic distance at rest a is obtained from du / dr = 0 , namely (du / dr ) r = a = 0 ⇒ a = (mB / nA)1/( m− n ) (2.1.2) and hence u0 = u (a ) = (1 / n)(n − m) B(mB / nA) − m /( m− n ) (2.1.3) Let U = Mu where M is the mass of a sphere of radius a; at r = a we have 2 pdV = −(dU / dr )dr ⇒ p = − 1/ (4πa ) (dU / dr ) 2 2 2 2 dp / dV = − 1/ (4πa ) (d U / dr ) r = a (2.1.4) Since k = −Vdp / dV , we find k= ρa2 9 (d 2u / dr 2 )r = a (2.1.5) 15 Now (d 2u / dr 2 ) r = a = −n(n + 1) Aa − n− 2 + m(m + 1) Ba − m− 2 = −mn u0 / a 2 (2.1.6) and hence k = −(mn / 9) ρ u0 (2.1.7) showing that the numerical value of k increases with that of u0. 2.2 Anisotropic linear elasticity The constitutive equations for anisotropic linear elasticity need the use of fourth-order tensors of moduli or compliances. Due to the symmetry of the strain and stress tensors, the corresponding 81 components reduce to 21 in the general case. The number and the value of independent elastic constants may be still reduced by expressing the possible symmetries of the anisotropy, which can be investigated by use of the properties of sinusoidal wave propagation and of the necessary conditions of stability of the elastic equilibrium. Exercise 2.2 Wave propagation in cubic symmetry Investigate the propagation of plane sinusoidal waves in the [110] direction in a cubic crystal (cf. Sect. 2.3.5 in Volume I). Solution Entering the general form of a sinusoidal wave in the Lamé-Clapeyron equations (Eq. 2.86 in Volume I) with no body forces other than the inertia forces yields the classical condition for such waves to exist: det(Cijkl K j K l − ρω 2 δ ki ) = 0 (2.2.1) with C the elastic moduli, K the propagation vector, ρ the mass density and ω the pulsation. 16 In the [110] direction, K1 = K 2 = K / 2, K 3 = 0 ; substituting in Eq.(2.92) of Sect.2.3.5 of Volume I and using the Voigt notation(2.58) for the elastic moduli we have 1 2 (C11 + C44 ) K 2 − ρω 2 1 (C12 + C44 ) K 2 2 (C12 + C44 ) K 2 1 (C11 + C44 ) K 2 − ρω 2 2 0 0 1 2 0 0 = 0 (2.2.2) C44 K 2 − ρω 2 and then: 2 1 1 2 2 2 2 2 4 − ρω ( + ) − ρω C K C C K ( 44 ) 2 11 44 − (C12 + C44 ) K = 0 4 (2.2.3) a) the velocity vTz of transverse waves with displacement parallel to [001] derives from C44 K 2 − ρω 2 = 0 ; it is given by vTz = ωT z K = C44 (2.2.4) ρ 2 2 b) From 12 (C11 + C44 ) K 2 − ρω 2 = 12 (C12 + C44 ) K 2 it follows that ω 2 = (1 / 2 ρ ) (C11 + C44 ) K 2 ± (C12 + C44 ) K 2 (2.2.5) ω12 = ( K 2 / 2 ρ )(C11 + C12 + 2C44 ), ω22 = ( K 2 / 2 ρ )(C11 − C12 ) (2.2.6) giving the values The displacement (u1 , v1 ) associated with ω1 obeys −( K 2 / 2)(C12 + C44 )u1 + ( K 2 / 2)(C11 + C44 )v1 = 0 (2.2.7) i.e., u1 = v1 . So, these waves are longitudinal and their velocity vL is vL = (C11 + C12 + 2C44 ) / (2 ρ ) (2.2.8) 17 The displacement (u2 , v2 ) associated with ω2 obeys ( K 2 / 2)(C12 + C44 )(u2 + v2 ) = 0 (2.2.9) i.e., u2 = −v2 . These are transverse waves with displacement parallel to 110 and their velocity vT is vT = (C11 − C12 ) / (2 ρ ) (2.2.10) Thus if the value of ρ is known, the values of all the elastic moduli can be derived from measurements of vTz, vL and vT. Exercise 2.3 Wave propagation in transverse isotropy Investigate the propagation of plane sinusoidal waves in a plate normal to x3, constituted with a linear elastic material exhibiting transverse isotropy about the axis x3. Consider a wave vector K either normal or parallel to the plate. Is it sufficient for measuring all the elastic moduli CIJ? (cf. Sect. 2.3.5 in Volume I) Solution The propagation condition reads: det(Cijkl K j K l − ρω 2 δ ki ) = 0 , with Cijkl given in Voigt notation by C11 C12 C13 0 C12 C11 C13 0 C13 C13 C33 0 0 0 C44 0 0 0 0 0 0 0 0 0 that is 0 0 0 0 C44 0 0 0 0 0 0 (1 / 2)(C11 − C12 ) (2.3.1) 18 1 (C11 + C12 ) K1 K 2 2 A − ρω 2 1 (C11 + C12 ) K1 K 2 2 (C13 + C44 ) K1 K 3 with (C13 + C44 ) K1 K 3 B − ρω 2 (C13 + C44 ) K 2 K 3 =0 C − ρω 2 (C13 + C44 ) K 2 K 3 (2.3.2) A = C11 K12 + C66 K 22 + C44 K32 2 2 2 B = C11 K 2 + C66 K1 + C44 K3 C = C ( K 2 + K 2 ) + C K 2 44 1 2 33 3 1 with C66 = (C11 − C12 ) . 2 For a wave vector (0, 0, K) normal to the plate, the propagation condition reduces to C44 K 2 − ρω 2 0 2 0 0 C44 K − ρω 0 2 0 2 0 =0 C33 K − ρω (2.3.3) 2 which yields v1 = C33 /ρ and v2 = C44 /ρ . The displacement associated with v1 is (0, 0, u30 ), which corresponds to a longitudinal propagation (v1 = vL ) . For v2, we get u30 = 0 , which corresponds to a transverse propagation (v2 = vT ) . For a wave vector ( K cos α , K sin α ,0) parallel to the plate, we have K 2C1 − ρω 2 K 2 (C11 + C22 )sin α cos α K (C11 + C22 )sin α cos α K 2C2 − ρω 2 2 0 0 2 0 C44 K − ρω =0 2 (2.3.4) C1 = C11 cos α + C66 sin α 2 2 C2 = C11 sin α + C66 cos α 2 with 0 2 which yields v1 = C44 / ρ = vT1 (u10 = u20 = 0) and K4 K 2C1 − ρω 2 K 2C2 − ρω 2 = (C11 + C12 ) 2 sin 2 α cos 2 α 4 that is, after reduction (2.3.5) 19 2 ω 2 C11 + C66 ω 2 C11C66 + =0 2 − ρ ρ2 K2 K (2.3.6) and hence v2 = C66 ρ = The velocity v2 corresponds to C11 − C12 2ρ v3 = C11 ρ (2.3.7) u20 cos α =− , i.e., a transverse propagation in u10 sin α the plane of the plate (v2 = vT2 ) . The velocity v3 corresponds to u20 sin α = , i.e., a u10 cos α longitudinal propagation in the plane of the plate (v3 = vL ) . Note that, according to the property of transverse isotropy, the velocities vL , vT1 and vT2 do not depend on α. So, the knowledge of ρ and the measurement of these velocities give access to the moduli C11 , C12 (and then C66 ), C33 and C44 , but not to C13. This modulus can only be measured from an oblique propagation with respect to the plate plane. Exercise 2.4 Propagation of sinusoidal longitudinal waves Show that sinusoidal longitudinal waves cannot propagate in an arbitrary direction α in a linear elastic homogeneous body with arbitrary anisotropy. Consider then the special case of cubic symmetry. Solution The general equation of sinusoidal wave propagation is known to read (see Sect. 2.3.5 in Volume I) (C ijkl K j K l − ρω 2 δik ) uk0 = 0 (2.4.1) For longitudinal waves along α, we must have uk0 = λ K k with K i = Kα i and the velocity is vL = ω / K . Equation (2.4.1) then reads Cijklα jα lα k − ρ vL2α i = 0 (2.4.2) 20 These are three equations for two unknowns (αi), since α is a unit vector, and one parameter ( ρ vL2 ): there are no solutions in the general case. Nevertheless, let us suppose that such a propagation is possible in the special direction α0; the velocity vL0 would then be such that α i0 ( Cijklα 0j α l0α k0 − ρ (vL0 ) 2 α i0 ) = Cijklα i0α 0j α l0α k0 − ρ (vL0 )2 = 0 (2.4.3) and then vL0 would be given by vL0 = Cijklα i0α 0j α k0α l0 ρ (2.4.4) an expression which makes sense since C is definite positive and ρ > 0. For instance, for a cubic symmetry, we would have: C11 for < 100 >: C α α α α = C = C ⇒ v ijkl i j k l 1111 11 L <100> = ρ 1 4 for < 110 >: Cijklα iα jα kα l = (C1111 + 2C1122 + C2222 + 4C1212 )( ) 2 2C44 + C11 + C12 and vL<110> = 2ρ for < 111 >: C α α α α = (3C + 6C + 12C )( 1 ) 4 ijkl i j k l 1111 1122 1212 3 4C44 + C11 + 2C12 and vL<111> = 3ρ (2.4.5) We now focus on the special case of cubic symmetry. This can be done in a more convenient way by using the representation (see Exercise 2.11 below for a proof of these formulae): 21 C = (C11 − C12 )K ′ + 2C44K ′′ + (C11 + 2C12 )J 1 1 J ijkl = δij δ kl I ijkl = (δik δ jl + δil δ jk ) I = J + K 3 2 K′ = H - J K ′′ = I - H K = K ′ + K ′′ (2.4.6) where all the components of H, in the axes of the cubic symmetry, are zero but 0 0 0 H1111 = H 2222 = H 3333 = 1 . We find 0 H ijkl α jα lα k = H i0111α13 + H i0222α 23 + H i0333α 33 = α i3 1 1 J ijklα jα lα k = δij δklα jα lα k = α i 3 3 1 1 I ijklα jα lα k = (δik δ jl + δil δ jk )α jα lα k = (α i + α i ) = α i 2 2 (2.4.7) Hence: 1 ′ α jα lα k = α i3 − α i K ijkl 3 ′′ α jα lα k = α i − α i3 K ijkl (2.4.8) 1 1 Cijklα jα lα k = (C11 − C12 )(α i3 − α i ) + 2C44 (α i − α i3 ) + (C11 + 2C12 )α i 3 3 3 = (C11 − C12 − 2C44 )α i + (2C44 + C12 )α i = ρ vL2α i (2.4.9) and then namely (C11 − C12 − 2C44 )α13 + (2C44 + C12 )α1 = ρ vL2α1 3 2 (C11 − C12 − 2C44 )α 2 + (2C44 + C12 )α 2 = ρ vLα 2 (C − C − 2C )α 3 + (2C + C )α = ρ v 2α 12 44 3 44 12 3 L 3 11 If no αi vanishes, we must have α12 = α 22 = α 32 = velocity vL<111> = 4C44 + 2C12 + C11 . 3ρ (2.4.10) 1 (<111> direction) with the 3 22 If only one αi vanishes, the two others must be equal: this corresponds to the <110> direction with the velocity vL<110> = 2C44 + C12 + C11 . 2ρ If two αi vanish, the third one must be equal to 1: this corresponds to the <100> direction with the velocity vL<100> = C11 ρ . These results fit well with (2.4.5). They are valid only if C11 − C12 − 2C44 ≠ 0 . When C11 − C12 − 2C44 = 0 , the cubic symmetry is degenerated into isotropy: all the directions α are equivalent and the longitudinal propagation occurs with the velocity vLiso = C11 ρ = λ + 2µ 3k + 4 µ = , which is a classical result. ρ 3ρ Exercise 2.5 Anisotropy in cubic symmetry For the case of cubic symmetry find the bulk modulus as a function of SIJ; also Young's modulus in an arbitrary direction u. (cf. Sect. 2.3.2.6 in Volume I) Solution (i) Apply a hydrostatic pressure p. We get: σ ij = − pδij ⇒ ε11 = ε 22 = ε 33 = − p ( S11 + 2S12 ) (2.5.1) So, the relative volume change ∆V/V is ∆V / V = Tr(ε ) = −3 p( S11 + 2S12 ) (2.5.2) and then k =− pV 1 = ∆V 3( S11 + 2S12 ) (2.5.3) (ii) Perform a tensile test along the direction u: σ = σ u ⊗ u ⇒ σ ij = σ ui u j . Hence 23 ε11 2 2 2 σ = S11u1 + S12 (u2 + u3 ) etc. (circ. permutation) 2 ε12 = S u u etc. (circ. permutation) 44 1 2 σ (2.5.4) and then, with e = ε ij ui u j , we get for the Young modulus E: 1 e = = S11 + ( S44 − 2 S11 + 2 S12 )(u12u22 + u22u32 + u32u12 ) E σ (2.5.5) Exercise 2.6 Elastic anisotropy in hexagonal symmetry Show that, for linear elasticity, the property of hexagonal symmetry leads to the same relations between the elastic moduli CIJ as that of transverse isotropy (cf. Sect. 2.3.2.7 in Volume I). To do that, use the rotation matrices Aij(α) for the rotation angle α about the axis x3 of the hexagonal symmetry and express invariance with respect to such rotations for α = nπ / 3, n = 1 to 6 . Solution According to (2.64), we must have Cijkl = Aip (α ) Ajq (α ) Akm (α ) Aln (α )C pqmn cos α with [ A(α )] = − sin α 0 tem of Fig. 2.6.1. sin α cos α 0 (2.6.1) 0 0 α = nπ / 3 in the rectangular coordinate sys1 Fig. 2.6.1 Coordinate system (x3 normal to the plane). 24 In addition to the general symmetry relations Cijkl = C jikl = Cijlk = Cklij , we get, −1 0 0 for n = 3 , i.e., for [ A(π)] = 0 −1 0 : 0 0 1 C14 = C1123 = A1 p A1q A2 l A3 mC pqlm = −C1123 = 0 C15 = C1131 = A1 p A1q A3l A1mC pqlm = −C1131 = 0 C24 = C2223 = A2 p A2 q A2l A3mC pqlm = −C2223 = 0 C25 = C2231 = A2 p A2 q A3l A1mC pqlm = −C2231 = 0 C34 = C3323 = A3 p A3q A2l A3mC pqlm = −C3323 = 0 (2.6.2) C35 = C3331 = A3 p A3q A3l A1mC pqlm = −C3331 = 0 C46 = C2312 = A2 p A3 q A1l A2 mC pqlm = −C2312 = 0 C56 = C3112 = A3 p A1q A1l A2 mC pqlm = −C3112 = 0 For n = 1, 2, 4 and 5 , we have: ε /2 ε′ 3 / 2 0 [ A(nπ / 3)] = −ε ′ 3 / 2 ε / 2 0 with ε , ε ′ = ±1 0 0 1 (2.6.3) and then C13 = C1133 = A1 p A1q A3l A3mC pqlm = A1 p A1q C pq 33 = 1 / 4C1133 + 2(εε ′ 3 / 4)C1233 + 3 / 4C2233 C55 = C1313 = A1 p A3q A1l A3mC pqlm = A1 p A1l C p 3l 3 (2.6.4) = 1 / 4C1313 + 2(εε ′ 3 / 4)C1323 + 3 / 4C2323 With εε ′ = ±1 , this reduces to the four equations: C13 = 1 / 4C13 + ( C13 = 1 / 4C13 − ( C55 = 1 / 4C55 + ( C55 = 1 / 4C55 − ( 3 / 2)C63 + 3 / 4C23 3 / 2)C63 + 3 / 4C23 3 / 2)C54 + 3 / 4C44 3 / 2)C54 + 3 / 4C44 (2.6.5) 25 which leads to C36 = 0, C13 = C23 , C55 = C44 , C45 = 0. (2.6.6) We have also, taking account of the preceding results: C11 = C1111 = A1 p A1q A1l A1mC pqlm = = (1 / 16)C11 + ( 3 / 4)εε ′C16 + (3 / 4)C66 + (3 3 / 4)εε ′C26 + (3 / 8)C12 + (9 / 16)C22 C66 = C1212 = A1 p A2 q A1l A2 mC pqlm = (2.6.7) = (3 / 16)C11 − (3 / 8)C12 + (3 / 16)C22 + (1 / 4)C66 + ( 3 / 4)εε ′(C16 − C26 ) These 4 equations (since εε ′ = ±1 ) reduce to: C16 = C26 = 0 C11 = C22 C − C = 2C 12 66 11 (2.6.8) which are identical to the relations deriving from the property of transverse isotropy. Exercise 2.7 Shear modulus in anisotropic elasticity 1) Find the expression of the shear modulus for given shear plane and direction in anisotropic elasticity with arbitrary symmetry and specify this expression for various symmetries. Solution A uniform shear stress state with the amplitude τ is applied to an anisotropic linear elastic body along a plane with unit normal n and in the (unit) direction m: the resulting shear strain in the same directions (m, n) has the amplitude γ and the shear modulus µ is defined by τ = µγ . It depends on (m, n). We define the orientation symmetric second-order tensor R by S R = (m ⊗ n) 1 Rij = (mi n j + m j ni ) 2 (2.7.1) 26 We have σ = τ (m ⊗ n + n ⊗ m) = 2τ R , whereas γ is given by γ = 2m.ε .n = 2ε ij mi n j = ε ij (mi n j + m j ni ) = 2 R : ε (2.7.2) so that 1 µ = γ 2R : ε 2R : S : σ = = = 4R : S : R τ τ τ (2.7.3) Note that we have Tr( R) = 0 and R : R = 1 / 2 In the orthonormal coordinate system (m,n,t) with t = m × n , we have 0 τ 0 0 1/ 2 0 σ = τ 0 0 R = 1 / 2 0 0 0 0 0 0 0 0 (2.7.4) For an orthotropic symmetry, we have in the Voigt notation and in the axes of this symmetry S11 S12 S [ S ] = 013 0 0 S12 S 22 S23 0 0 0 S13 S 23 S33 0 0 0 0 0 0 S44 0 0 0 0 0 0 S55 0 0 0 0 0 0 S66 (2.7.5) and then we get 1 4µ = S11 R112 + S22 R222 + S33 R332 + 2S12 R11 R22 + 2S23 R22 R33 + 2S31 R33 R11 orth 2 23 2 31 (2.7.6) 2 12 + S44 R + S55 R + S66 R where the components Rij must be calculated in the axes of the orthotropic symmetry. For a quadratic symmetry with respect to x1 and x2, we would have 1 4µ = S11 ( R112 + R222 ) + S33 R332 + 2S12 R11 R22 + 2 S13 ( R11 + R22 ) R33 quad 2 23 2 31 2 12 + S44 ( R + R ) + S66 R (2.7.7) 27 For transverse isotropy about x3, we have in addition S66 = 2( S11 − S12 ) . We get 1 4µ = S11 ( R112 + R222 + 2 R122 ) + S33 R332 + 2S12 ( R11 R22 − R122 ) tr. isot 2 33 2 23 (2.7.8) 2 31 − 2S13 R + S44 ( R + R ) where the relation Tr( R) = 0 has been used. If the shear plane is normal to x3, we have 0 n = (0,0,1) m = (cos α ,sin α ,0) R = 0 cos α 2 0 0 sin α 2 cos α 2 sin α 2 0 (2.7.9) 1 cos 2 α sin 2 α = S44 ( + ) ⇒ µ = 1 / S44 , which, as expected, is inde4µ 4 4 pendent of α. If n is normal to x3, we can take it parallel to x1 and then and then 0 cos ϕ n = (1,0,0) m = (0,cos ϕ ,sin ϕ ) R = 2 sin ϕ 2 and then cos ϕ 2 0 0 sin ϕ 2 0 0 (2.7.10) 2( S11 − S12 ) 2 = S44 1 + − 1 cos ϕ . µ S44 1 For cubic symmetry, we find, with the “anisotropy parameter” a = 2( S11 − S12 ) S 44 which is greater than 1 1 µ = S44 {1 + 2(a − 1)( R112 + R222 + R332 )} cubic (2.7.11) 28 where the relations Tr( R) = 0 and R : R = 1 / 2 have been used. It is easy to check that 0 ≤ R112 + R222 + R332 ≤ 1 / 2 , so that we get, since a ≥ 1 : S44 ≤ 1 µ ≤ aS44 cubic 2) A very simple model for a polycrystal with cubic crystals could be constituted with two cubic crystals in series, subjected to the same shear stress τ, with the volume fractions c1 and c2 (with c1 + c2 = 1 ). The first crystal is sheared on the system (100 ) [ 011] and the second one on the system (110 ) 1 10 . Express the overall shear strain as the average of the two shear strains and derive the overall shear modulus. Comment this result. Solution We have τ 1 γ 1 = µ with µ = S44 1 1 1 τ γ = with = aS44 2 µ 2 µ2 (2.7.12) and hence γ = c1γ 1 + c2γ 2 = τ ( c1 µ1 + c2 µ2 )= τ µ (2.7.13) so that 1 µ = c1S44 + ac2 S 44 = ( a + c1 (1 − a) ) S44 (2.7.14) So, when the volume fraction c1 of the first orientation increases from 0 to 1, the overall shear modulus decreases (here, linearly) from aS44 to S44. This illustrates the fact that the anisotropy induced by a plastic flow and the resulting crystallographic texture can make the overall moduli decrease, even if no damage occurs. 29 Exercise 2.8 Elastic anisotropy in quadratic symmetry Justify the fact that the compliance matrix in Voigt notation for a linear elastic material with quadratic symmetry in the symmetry axes is frequently written in the form (“engineering notation”): ν ν ν 1 − 12 − 13 (or − 31 ) E1 E1 E3 E1 ν ν ν 1 − 12 − 13 (or − 31 ) E E E3 E1 1 1 ν 13 ν 31 ν 13 ν 31 1 − (or − ) − (or − ) E E E E E 3 1 3 1 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 µ13 0 0 0 1 µ13 0 0 0 0 0 0 1 µ12 (2.8.1) where the notations E, µ and ν refer to the isotropic Young modulus, shear modulus and Poisson ratio. Solution For an isotropic linear elastic material, the strain response to a tensile test is classically given by σ σ = 0 0 1 E 0 0 0 0 ⇒ ε = σ 0 0 0 0 0 − ν E 0 whereas the strain response to a simple shear test reads 0 0 ν − E (2.8.2) 30 1 2µ 0 0 τ 0 1 σ = τ 0 0 ⇒ ε = τ 2µ 0 0 0 0 0 0 0 0 0 (2.8.3) For a quadratic symmetry, the compliance matrix in Voigt notation in the symmetry axes reads (see Sect. 2.3.2.5 in Volume I) S11 S12 S [ S IJ ] = 013 0 0 S12 S11 S13 0 0 0 0 0 0 S44 0 0 S13 S13 S33 0 0 0 0 0 0 0 S44 0 0 0 0 0 0 S66 (2.8.4) For a tensile test along x1, we have σ 0 0 S11 0 σ = 0 0 0 ⇒ ε = σ 0 S12 0 0 0 0 0 0 0 S13 (2.8.5) Comparing with the isotropic case (7.2), we can put S11 = 1 E1 S12 = − ν 21 E1 (= − ν 12 E1 ) S13 = − ν 31 E1 (2.8.6) For a tensile test along x3, we get 0 0 0 S13 σ = 0 0 0 ⇒ ε =σ 0 0 0 σ 0 0 S13 0 0 0 S33 (2.8.7) so that we can write S33 = 1 E3 S13 = − ν 13 E3 (= − ν 31 E1 ) (2.8.8) 31 For a shear test on (x1, x2), we have 0 0 τ 0 S σ = τ 0 0 ⇒ ε = τ 66 2 0 0 0 0 and then we can put S66 = S44 = 1 µ13 1 µ12 S66 2 0 0 0 0 0 (2.8.9) . For a shear test on (x1, x3), we find similarly . Exercise 2.9 Stability of equilibrium in cubic symmetry Prove the inequalities (2.85) for the elastic moduli in a cubic system (cf. Sect. 2.3.3 in Volume I). Solution Given the definition of CIJ, it suffices to find the conditions for the associated quadratic form to be positive definite: all the eigenvalues must be positive; therefore all the diagonal sub-determinants must be positive, the conditions for which are the following: determinants of order 1: C11 > 0, C44 > 0 2: C112 − C122 > 0 3: (C11 − C12 )2 (C11 + 2C12 ) > 0 ⇒ C11 + 2C12 > 0 with no further conditions for orders 4,5,6. Exercise 2.10 Stability of equilibrium in transverse isotropy Find the conditions to be obeyed by the compliances S11 , S13 , S33 , S44 and S66 for the elastic equilibrium to be stable in transverse isotropy. For given values of S13 and S33 , draw the stability domain in the plane ( S11 , S66 ) . Solution 32 Like in Exercise 2.9, the quadratic form σ I S IJ σ J must be positive definite, i.e., σ I S IJ σ J > 0, ∀ [σ ] ≠ [ 0] : all the diagonal sub-determinants of SIJ must be positive, that is: order 1: S11 > 0, S33 > 0, S44 > 0, S66 > 0 S66 2 S 2 2 2 ) = S66 ( S11 − 66 ) > 0 S11 − S12 = S11 − ( S11 − order 2: ¨ 2 4 S S − S 2 > 0 11 33 13 order 3: S S S S11( S11S33 −S132 ) − ( S11− 66 ) S33 ( S11− 66 ) −S132 + S13 S13 ( S11− 66 ) −S11S13 > 0 2 2 2 S S This last condition reduces to S66 ( S11S33 − S132 − 33 66 ) > 0 . 4 Finally, the stability conditions read S33 > 0, S66 > 0, S44 > 0, S11 > S66 S132 + 4 S33 (2.10.1) For given values of S13 and S33, the stability domain is drawn in the (S11, S66) plane in Fig. 2.10.1. Fig. 2.10.1 The stability domain lies between the S11 axis and the oblique straight line. 33 2.3 Bounds and estimates for the elastic moduli of heterogeneous materials Material scientists and engineers are more and more concerned with the prediction of the overall behaviour and the local response of heterogeneous and composite materials. We investigate here the simpler case of linear elastic and thermoelastic heterogeneous materials. For random materials, the RVE (“representative volume element”) cannot be described completely so that the effective overall behaviour of the HEM (“homogeneous equivalent medium”) cannot be predicted unambiguously. One can then either make additional assumptions or approximations in order to derive estimates for the overall behaviour or look for bounds for the overall mechanical properties, by using variational approaches. Consideration of (either real or fictitious) eigenstrains or eigenstresses in the initial state is a useful tool for the derivation of both improved estimates and bounds, as illustrated by the various applications which have been made of Eshelby’s solution of the (ellipsoidal) inclusion problem. This approach can also be applied to heterogeneous thermoelasticity. Exercise 2.11: Voigt and Reuss bounds for the elastic moduli of an isotropic polycrystal with cubic crystals Find the Voigt ( µ V , k V ) and Reuss ( µ R , k R ) bounds for the shear and bulk moduli ( µ eff , k eff ) of a linear elastic isotropic polycrystal with cubic crystals. Solution From (2.135) in Volume I, namely < c−1 > −1 ≤ (S eff ) −1 = Ceff ≤< c > (2.11.1) we know that CV =< c >, S R =< s > . Note that the scalars cijij , ciijj , sijij and siijj are invariant with respect to any rotation, which means that they have the same value for all the grains of the polycrystal, so that we can derive V V < cijij >= cijij = Cijij , < ciijj >= ciijj = Ciijj R R < sijij >= sijij = Sijij , < siijj >= siijj = Siijj (2.11.2) 34 With the notations A = C11 − C12 , B = C11 + 2C12 , a = 2( S11 − S12 ) S 44 (2.11.3) we have cijij = 3C11 + 6C44 = B + 2 A + 3aA ciijj = 3C11 + 6C12 = 3B 3 1 2 3 sijij = 3S11 + S44 = + + 2 B A Aa 3 siijj = 3S11 + 6 S12 = B (2.11.4) Since the polycrystal is isotropic, CV and S R can be considered as isotropic too, and written as 1 1 K + V/R J 2 µ V/R 3k 1 1 = (δik δ jl + δil δ jk ) J ijkl = δij δ kl 2 3 CV/R = 2µ V/R K + 3k V/R J S V/R = with I = J + K I ijkl (2.11.5) so that we get finally (3a + 2) A 5 Aa eff 2(2a + 3) ≤ µ ≤ 10 B B ≤ k eff ≤ 3 3 (2.11.6) This last result shows that keff is determined unambiguously and has the value B/3. Note: This problem can be solved more straightforwardly by using the representation (2.4.6) given in Exercise 2.4 for cubic symmetry. It uses, in addition to the unit tensors I, J and K which are especially for isotropy, the new tensors H, K’ and K”, with 1 J ijkl = δij δ kl 3 K = K ′ + K ′′ 1 I ijkl = (δik δ jl + δil δ jk ) I = J + K 2 K′ = H - J K ′′ = I - H (2.11.7) 35 where all the components of H, in the axes of the cubic symmetry, are zero but 0 0 0 H1111 = H 2222 = H 3333 = 1 . It is easy to prove the following properties: K′ : K′ = K′ K ′′ : K ′′ = K ′′ H : H = H K ′ : K ′′ = K ′′ : K ′ = K ′ : J = J : K ′ = K ′′ : J = J : K ′′ = 0 1 1 1 −1 ( a′K ′ + a′′K ′′ + bJ ) = a′ K ′ + a′′ K ′′ + b J Q.F.( a′K ′ + a′′K ′′ + bJ ) > (<)0 ⇔ a′, a′′, b > (<)0 (2.11.8) where Q.F.(X) denotes the quadratic form associated with X. For instance, in any system of orthonormal axes turned by the orthogonal matrix [A] from the cubic axes, we get 0 H ijkl = Aim Ajn Akp Alq H mnpq = Ai1 Aj1 Ak 1 Al1 + Ai 2 Aj 2 Ak 2 Al 2 + Ai 3 Aj 3 Ak 3 Al 3 (2.11.9) and then 0 0 H ijmn H mnkl = Aip Ajq Amr Ans Amt Anu Akv Alw H pqrs H tuvw 0 = Aip Ajq Akv Alw H 0pqrs H tuvw = Aip Ajq Akv Alw H 0 pqvw (2.11.10) = H ijkl Similarly, we have 1 1 0 Aip Ajq Amr Ans δ mn δ kl H 0pqrs = Aip Ajq δ rs δ kl H pqrs 3 3 1 1 0 = Aip Ajq δkl H pqrr = δij δ kl = J ijkl 3 3 H ijmn J mnkl = (2.11.11) We have also H : J = J : H (= J). Consequently K ′ : K ′ = ( H − J ) : ( H − J ) = H − 2J + J = H − J = K ′ (2.11.12) and so on for the other equalities in (2.11.8). Using the Voigt notation in the cubic axes, we have 36 1 0 0 [ I] = 0 0 0 0 0 0 [ K′′] = 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 1 [H] = 1/ 2 0 0 1/ 2 0 0 1/ 2 0 0 1 / 3 1 / 3 1 / 3 0 0 0 1 / 3 1 / 3 1 / 3 0 1/ 3 1/ 3 1/ 3 [J] = 1/ 2 0 1/ 2 0 0 1/ 2 0 2 / 3 −1 / 3 −1 / 3 0 −1 / 3 2 / 3 −1 / 3 −1 / 3 −1 / 3 2 / 3 [K] = 1/ 2 0 1/ 2 1 / 2 2 / 3 −1 / 3 −1 / 3 0 −1 / 3 2 / 3 −1 / 3 (2.11.13) −1 / 3 −1 / 3 2 / 3 [ K ′] = 0 0 0 0 So, if we write c = a′K ′ + a′′K ′′ + bJ , we get from identification in the cubic axes 2 ′ b 3 a + 3 = C11 b 1 − a′ + = C12 3 3 1 2 a′′ = C44 which yields, as in (2.4.6) a′ = C11 − C12 ⇒ b = C11 + 2C12 a′′ = 2C 44 (2.11.14) 37 c = (C11 − C12 )K ′ + 2C44 K ′′ + (C11 + 2C12 )J (2.11.15) and, according to the third line of (2.11.8) s= 1 1 1 K′ + K ′′ + J C11 − C12 2C44 C11 + 2C12 = ( S11 − S12 )K ′ + S44 K ′′ + ( S11 + 2S12 )J 2 (2.11.16) For an isotropic elastic material, we have C11 − C12 = 2C44 and then (2.11.15) reads c = (C11 − C12 )(K ′ + K ′′) + (C11 + 2C12 )J = (C11 − C12 )K + (C11 + 2C12 )J (2.11.17) = 2µ K + 3kJ as we know already. From the definitions (2.11.7), we find easily the value of the invariants ′ = 2 Kijij = 5 K ijij ′′ = 3 J ijij = 1 K ijij ′ ′′ = 0 J iijj = 3 K iijj = 0 K iijj = 0 K iijj (2.11.18) which leads to < cijij >= 2a′ + 3a′′ + b < c >= 3b iijj 2 3 1 + + a′ a′′ b 3 < siijj > b < sijij >= (2.11.19) The overall isotropy implies the averages < K ′ > and < K ′′ > to be isotropic. So, < K ′ > can be put in the form ′ >= 2 = 5α + β < Kijij 2 < K ′ >= α K + β J with ⇒< K ′ >= K ′ 5 < Kiijj >= 0 = 3β (2.11.20) 3 and then < K ′′ >= K . The derivation of the Voigt and Reuss bounds for the 5 shear and bulk moduli follows straightforwardly. 38 Note that the last line of (2.11.8) would have also made obvious the resolution of Exercise 2.9. Exercise 2.12: Voigt and Reuss bounds for an elastic inhomogeneous material with cubic symmetry and isotropic constituents eff Derive the Voigt and Reuss bound for the elastic moduli (C11eff , C12eff , C44 ) of an inhomogeneous material with overall cubic symmetry constituted with elastic isotropic components. Solution For cubic symmetry, we use the same notations as in Exercise 2.11 A = C11 − C12 B = C11 + 2C12 (2.12.1) with C = C44 . Note that this leads to the relations S11 − S12 = A−1 S11 + 2S12 = B −1 S44 = C −1 (2.12.2) Remind also the results of Exercise 2.11.9 concerning the stability conditions for cubic symmetry, namely C11 > 0 C44 > 0 C112 − C122 > 0 C11 + 2C12 > 0 (2.12.3) which leads to: A > 0, B > 0, C > 0 . For an isotropic material, we would have (C11 − C12 )iso = 2 µ (C11 + 2C12 )iso = 3k (C44 )iso = µ (2.12.4) The Voigt bounding inequation reads E : (< c > −Ceff ) : E ≥ 0, ∀ E (2.12.5) According to what precedes, the positivity of (< c > −Ceff ) (Voigt bound) implies the inequalities 2 < µ > − Aeff ≥ 0 eff 3 < k > −B ≥ 0 < µ > −C eff ≥ 0 (2.12.6) 39 whereas the positivity of (< s > −S eff ) (Reuss bound), with help of (2.12.2), leads to the inequalities 1 −1 eff −1 2 < µ > −( A ) ≥ 0 1 −1 eff −1 < k > −( B ) ≥ 0 3 < µ −1 > −(C eff ) −1 ≥ 0 (2.12.7) Thus, we obtain the bounds 2 < µ −1 > −1 ≤ Aeff ≤ 2 < µ > −1 −1 eff 3< k > ≤ B ≤3< k > − 1 − 1 < µ > ≤ C eff ≤< µ > (2.12.8) eff Voigt and Reuss bounds for C11eff , C12eff , C44 can then be derived easily from their relations with A, B and C. Note: the representation (2.4.6) could have been used here again for a more direct resolution. Exercise 2.13: Voigt and Reuss bounds for the Young modulus and Poisson ratio of an isotropic elastic material with isotropic constituents Show that the laws of mixtures which yield in this case the right values of the Voigt and Reuss bounds for the shear and bulk moduli are no more valid for the Young modulus and the Poisson ratio. Solution For the Young modulus, we have E= 9k µ 3k + µ or 1 1 E −1 = µ −1 + k −1 3 9 (2.13.1) The Voigt and Reuss bounds for µ eff and k eff can be written as < µ > −1 ≤ µ eff −1 ≤< µ −1 > < k > −1 ≤ k eff −1 ≤< k −1 > (2.13.2) 40 so that we have from (2.13.1) 1 1 1 1 < µ > −1 + < k > −1 ≤ E eff −1 ≤ < µ −1 > + < k −1 > (=< E −1 >) 3 9 3 9 (2.13.3) and then −1 1 1 < E −1 > −1 ≤ E eff ≤ < µ > −1 + < k > −1 (≠< E >) 9 3 (2.13.4) For the Poisson ratio, we start from the general relations for an isotropic material (cf. Eqs (2.82) in Volume I) E = 2 µ (1 +ν ) k = E 3(1 − 2ν ) (2.13.5) and then k 2(1 +ν ) = 3 (1 − 2ν ) (2.13.6) µ which reads also 2(1 +ν )(3 + µ k )=9 (2.13.7) From the Voigt and Reuss bounds for the shear and bulk moduli, we can derive the inequalities 1 < k >< 1 µ ≤ > µ eff k eff ≤< µ >< 1 > k (2.13.8) According to (2.13.7), we then have the inequalities 9 9 ≤ 1 + ν eff ≤ −1 2 3+ < µ >< k > 2 3+ < k > −1 < µ −1 > −1 (2.13.9) which yield the searched Voigt and Reuss bounds for ν eff . It can be checked easily that they differ from the predictions of the laws of mixtures, namely < ν > and < ν-1 >-1, respectively. For example, for a two-phase material with uniform Pois- 41 son ratios (ν1 = ν2 = ν), the Voigt and Reuss bounds for ν eff do not coincide whereas both laws of mixtures yield the same estimate ν. Exercise 2.14: Elementary bounds and estimates for an elastic isotropic porous medium Derive the strain field in a spherical isotropic elastic inhomogeneity embedded in an infinite isotropic elastic matrix subjected to a uniform strain at infinity from the solution of the reference problem of a spherical inclusion by using Eshelby’s method of the “fictitious equivalent inclusion” (see Sect. 2.7.3 of Volume I). Use this result to derive the Voigt and Reuss bounds, the Hashin-Shtrikman bounds and the self-consistent estimate for the effective shear modulus of an elastic isotropic porous medium with an elastic isotropic incompressible matrix. Comment these results at high and low porosity. Solution F We have to choose the (uniform) stress-free strain ε in the inclusion so that the solutions of the two problems shown in Fig. 2.14.1 are identical. We know the solution of the classical inclusion problem (at right), namely (see Eq. 2.159 in VolI I ume I), with e , s the strain and stress (uniform) deviators in the inclusion, θ I = Tr(ε I ) and E 0 the strain deviator at infinity: F I 0 s = 2µ E + ( β − 1)e ⇒ I 0 F Tr(σ ) = 3k Θ + (α − 1)θ eI = E 0 + β e F 0 F θ = Θ + αθ with α = 1 +ν 2(4 − 5ν ) and β = . 3(1 −ν ) 15(1 −ν ) E 0 E µ, k µ, k µ *, k* 0 = µ,k F ε Fig. 2.14.1 The equivalent inclusion method (2.14.1) 42 The solutions of the inclusion and inhomogeneity problems are the same only if F F * 0 0 µ E + ( β − 1)e = µ E + β e ⇒ 0 F 0 F k Θ + (α − 1)θ = k Θ + αθ s I = 2µ * e I I * I Tr(σ ) = 3k θ (2.14.2) whence the known results (see Eq. 2.165 in Volume I): I e = µ 0 E * µ (1 − β ) + µ β θI = k Θ0 * k (1 − α ) + k α (2.14.3) Note that for a spherical cavity ( µ * = k * = 0 ), we get e Void = E 0 1− β θ Void = Θ0 1− α (2.14.4) Voigt and Reuss bounds From < µ −1 > −1 ≤ µ eff ≤< µ > , we have, for a porosity c and an incompressible matrix (ν = 1 / 2 ) < µ >= (1 − c) µ < µ −1 >→ ∞ (2.14.5) and then µ Reuss = 0 µ Voigt = (1 − c) µ (2.14.6) Hashin-Shtrikman bounds E0 µ, k E0 µ, k µ,k 0 µ,k F ε (1) (2) Fig. 2.14.2 Two auxiliary problems for the derivation of the HS+ bound 43 Since the softest phase is made of voids, the lower Hashin-Shtrikman bound is obviously zero: µ HS− = 0 . To derive the upper Hashin-Shtrikman bound, we solve the two inclusion problems of Fig. 2.14.2: e1 = E 0 E0 5E = 0 e2 = 1− β 3 (2.14.7) since, for ν = 1 / 2 , β = 2 / 5 . With E the macroscopic strain deviator, we must have E =< e >= (1 − c)E 0 + 5c 2c E 0 = (1 + )E 0 3 3 ⇒ E0 = 3 E 3 + 2c (2.14.8) so that e1 = 3 E 3 + 2c e2 = 5 E 3 + 2c (2.14.9) The stress tensors are given by s1 = 2µ e1 ⇒ < s >= s2 = 0 6(1 − c) µ E = 2µ HS+ < e > 3 + 2c (2.14.10) and then µ HS = + 3(1 − c) µ 3 + 2c (2.14.11) Self-consistent scheme E E µ SC , k SC µ SC , k SC µ,k 0 µ,k F ε (1) (2) Fig. 2.14.3 Two auxiliary problems for the derivation of the self-consistent estimate 44 We have to deal with two inclusion problems with the same matrix constituted with the effective medium ( µ SC , k SC ) and either the sound material ( µ , ∞) or the void ( 0,0 ) in the spherical inhomogeneity (see Fig. 2.14.3). We find for the strain deviators µ SC E e1 = SC µ (1 − β SC ) + µβ SC e = 1 E 2 1 − β SC (2.14.12) with E = (1 − c)e1 + ce 2 ⇒ µ SC (1 − c) c + =1 µ (1 − β SC ) + µβ SC 1 − β SC SC (2.14.13) i.e., c µ SC = µ 1 − 1 − β SC (2.14.14) where β SC , which depends on ν SC , has still to be determined (due to the voids, the effective medium is no longer incompressible). This can be obtained from the average equation for the volume dilatations: θ1 = 0 θ 2 = Θ 1 − α SC ⇒ cθ 2 = c Θ 1 − α SC =Θ ⇒ α SC = 1 − c (2.14.15) From α SC , we derive easily ν SC = (2 − 3c) / (4 − 3c) and then β SC = (2 + c) / 5 , so that we have, from (2.14.14) µ SC = 3(1 − 2c) µ 3−c (2.14.16) Actually, this result does not hold for c > ½ which would correspond to a negative effective shear modulus: beyond this value, according to the self-consistent model, the porous body has no more any mechanical resistance and the effective shear modulus vanishes. So, instead of (2.14.16), we have to write 45 µ SC = 3 < 1 − 2c > µ 3−c (2.14.17) where < A > means A as long as A ≥ 0 and 0 when A < 0. Comments For c << 1, we get µ HS+ 2c 5c ≈ 1− c − = 1− 3 3 µ SC µ ≈ 1 − 2c + c = 1 − 5c µ 3 3 (2.14.18) which shows that the upper Hashin-Shtrikman bound and the self-consistent estimate behave similarly at very low porosity, the latter being always smaller that the former (no bound violation): this corresponds to the fact that, at low porosity, the voids are far away from each other with a continuous matrix, which is the situation taken into account by the upper Hashin-Shtrikman bound treatment. At larger porosity, the voids are still included in a continuous matrix (closed porosity) for this treatment whereas the self-consistent scheme let both phases (sound material and voids) play an equivalent morphological role (open porosity), which yields a lower estimate for the effective shear modulus. This modulus even vanishes for c ≥ ½. Note that for a cubic lattice of spherical voids with the same radius a, these 4 3 πa π * voids can touch each other for c = 3 3 = (≃ 0.524) , which is close to ½. 8a 6 Exercise 2.15: Elementary bounds and estimates for an elastic isotropic twophase material with one rigid phase Derive the Voigt and Reuss bounds, the Hashin-Shtrikman bounds and the selfconsistent estimate for the effective shear modulus µ eff (or the corresponding shear compliance η eff = 1/ µ eff ) of a two-phase elastic isotropic material with isotropic and isotropically distributed phases. Phase (1) is supposed to be incompressible and phase (2) to be rigid. Comment these results at low and high volume fractions c of phase (2). Solution Voigt and Reuss bounds 46 From < µ −1 > −1 ≤ µ eff ≤< µ > , we get also < η −1 > −1 ≤ η eff ≤< η > , with η 2 = 0 . So, we have η eff ≤ (1 − c)η1 or µ eff ≥ µ1 1− c (= µ Reuss ) and 0 ≤ η eff or µ eff ≤ ∞ (= µ Voigt ) or η Voigt = 0 . Note that, from < k −1 > −1 ≤ k eff ≤< k > k eff → ∞ and then ν eff with k1 , k2 → ∞ , we have also = ½ ∀c , for any phase distribution. Hashin-Shtrikman bounds We use the same method as in Exercise 2.14 with adequate matrices and inclusions. For the upper bound, the matrix is rigid so that µ HS+ → ∞ (or η HS+ = 0). For the lower bound, the matrix is made of phase (1) and we get in a spherical inhomogeneity ( µ * , k * ), with the same notations as in Exercise 2.14: s= with β1 = 2 / 5 2µ * µ1 0 E µ1 (1 − β1 ) + µ * β1 since ν 1 = ½ . So, s= 0 (2.15.1) 10µ * µ1 0 E 3µ1 + 2µ * and, when µ * → ∞ , 0 s → 5µ1 E . The auxiliary strain deviator E is determined from the average strain equation: E =< e >= (1 − c)e1 + ce 2 0 with e1 = E , e 2 = 0 (2.15.2) 0 So, we have E = E / (1 − c) and then 2µ1 s1 = 1 − c E s = 5µ1 E 2 1 − c (2.15.3) From S =< s >= 2 µ HS− E , we get finally µ HS = − 2 + 3c µ1 2(1 − c) (or η HS− = 2(1 − c) η1 ) 2 + 3c (2.15.4) Self-consistent scheme Proceeding like in Exercise 2.14, we find, with β SC = 2 / 5 since ν eff = ½ ∀c : 47 e1 = 5µ SC E 3µ SC + 2µ1 e2 = 0 (2.15.5) and then E = (1 − c)e1 ⇒ 5(1 − c) µ SC =1 3µ SC + 2µ1 (2.15.6) which yields µ SC = 2 µ1 2 − 5c (or η SC = (1 − 5c )η1 ) 2 (2.15.7) Note that this expression is valid only as long as µ SC ≥ 0 , i.e., for c ≤ 0.4 ; this means that µ SC → ∞ (or η SC = 0) for c > 0.4 . This could be proved rigorously by making µ2 tend towards infinity in the second-order equation in µ SC and excluding the negative solution. Comments For c << 1, we get from (2.15.4) η HS 3c 5c ≈ 1− c − = 1− 2 2 η1 − (2.15.8) which shows, according to (2.15.7), that the lower Hashin-Shtrikman bound and the self-consistent estimate behave similarly at very low porosity (with η HS+ ≤ η SC ≤ η HS− , which ensures that there is no bound violation): this corresponds to the fact that, at low porosity, the rigid phase consists of inclusions embedded in a continuous matrix of phase (1), which is the situation taken into account by the lower Hashin-Shtrikman bound treatment. At larger porosity, the rigid phase is still embedded in a continuous deformable matrix according to this treatment whereas the self-consistent scheme let both phases play an equivalent morphological role, which yields a larger estimate for the effective shear modulus. From c = 0.4, this estimate is infinite, which corresponds to the formation of a rigid continuous “skeleton” which makes the whole material rigid itself. Exercise 2.16: The differential scheme for a two-phase elastic material The classical self-consistent scheme for a two-phase material is well adapted to the case of phases which play an equivalent morphological role. For a particle or 48 fibre reinforced composite, one phase is a continuous matrix and the second phase consists of particles (or fibres) embedded in this matrix and the classical selfconsistent scheme is no more adapted to this morphology. So, this scheme has been modified in different ways so as to be able to be used for modelling the overall behaviour of such materials. Among the simplest extensions of this scheme let us quote the “differential (self-consistent) scheme” and the “generalised selfconsistent scheme”. We consider here the first of these schemes; the reader interested in the second one is invited to consult Christensen and Lo (1979). For sake of simplicity, we restrict ourselves to the case of an isotropic two-phase material whose phases obey a linear isotropic incompressible elastic behaviour (shear moduli µ 1 and µ 2 and volume fractions c1 and c2). The basic idea consists in the incremental introduction of the second phase made of particles in a matrix which has been homogenised at each step, starting from the sole continuous phase, and to use the Eshelby-Einstein dilute approximation (see Volume I, Sect. 2.8.2, Eq. 2.191) for this incremental homogenisation procedure. 1) Derive the expression of µ EE from the tensorial expression of CEE given by (2.191) for the present case. Check that this result could have been obtained too from a first-order development of the self-consistent equation (as obtained, e.g., from Eq. 2.14.12 and the method used in the Exercises 2.14 and 2.15) for an infinitesimal volume fraction of second phase (this is the reason why the differential scheme is sometimes also named the “self-consistent” differential scheme). 2) Apply this result to the current situation where the second phase has still the volume fraction f (with 0 ≤ f ≤ c2 ) and the homogenised matrix has the shear modulus µ * ( f ) and where f is changed into f+df: what is the resulting variation of µ * ( f ) ? Derive the corresponding estimate µ DS of the differential scheme by integration of this differential equation. 3) What would be the prediction of the differential scheme for a rigid second phase? Compare this prediction with the elementary bounds and estimates of Exercise 2.15 and comment on the differences. Why, from a rigorous point of view, this result cannot be applied directly to a porous medium (where the second phase is constituted of voids)? What would be the prediction if it was applied nevertheless without modification and what is the rigorous prediction (it will be noticed that the difference is very small). Compare this prediction with the elementary bounds and estimates of Exercise 2.14 and comment on the differences. Solution 1) The tensorial expression of the Eshelby-Einstein estimate is (cf. Eq. 2.191 of Volume I) CEE = c mat + f (c2 − c mat ) I + P2mat : (c 2 − c mat ) −1 (2.16.1) 49 which would read with the notations of this exercise CEE = c1 + f (c 2 − c1 ) I + P21 : (c 2 − c1 ) −1 (2.16.2) with f infinitesimal. For isotropy, this yields µ EE β = µ1 + f ( µ 2 − µ1 ) 1 + 1 ( µ2 − µ1 ) µ1 −1 (2.16.3) Since phase (1) is incompressible, β1 = 2 / 5 and (2.16.3) becomes µ EE = µ1 1 + 5 f ( µ 2 − µ1 ) 3µ1 + 2µ2 (2.16.4) This means that the introduction of a very low volume fraction dc2 = f of phase (2) in phase (1) makes the shear modulus µ 1 of this phase be changed by dµ = µ EE − µ1 = 5µ1 ( µ 2 − µ1 ) dc2 3µ1 + 2µ2 (2.16.5) To derive the overall shear modulus estimate µ SC according to the classical selfconsistent scheme, we have, from (2.14.12) in Exercise 14 µ SC E e1 = SC µ (1 − β SC ) + µ1β SC µ SC e = E 2 µ SC (1 − β SC ) + µ 2 β SC (2.16.6) Since both phases are incompressible, the homogenised material is incompressible too so that β SC = 2 / 5 . So we have 5µ SC e = E 1 3µ SC + 2µ1 SC e = 5µ E 2 3µ SC + 2µ 2 and the strain average equation reads (2.16.7) 50 (1 − c) 5µ SC 5µ SC + =1 c 3µ SC + 2 µ1 3µ SC + 2µ 2 (2.16.8) or 5(1 −c) µ SC (3µ SC + 2µ 2 ) + 5cµ SC (3µ SC + 2µ1 ) = (3µ SC + 2µ1 )(3µ SC + 2µ2 ) (2.16.9) where µ SC is a function of c. For c infinitesimal, this equation can be differentiated and yields, with c ≈ dc and µ SC ≈ µ1 + dµ : −5dcµ1 (3µ1 + 2µ 2 ) + 5[ dµ (3µ1 + 2µ 2 ) + 3µ1dµ ] + ... ...5dc µ1 (5µ1 ) = 3dµ (8µ1 + 2µ 2 ) (2.16.10) or 10dcµ1 ( µ1 − µ 2 ) + 5(6 µ1 + 2µ2 )dµ = 3(8µ1 + 2 µ2 )dµ (2.16.11) 5µ1 ( µ1 − µ2 )dc + (3µ1 + 2µ 2 )dµ = 0 (2.16.12) and then which is, as expected, the same equation as (2.16.5). 2) Equation (2.16.12) can be used in the current state with µ1 = µ * ( f ) and dµ = dµ * . As for dc, it is obtained as follows: we start with a volume V containing the volume fV of phase (2); we add the volume αV of phase (2) and the new ( f + α )V volume fraction of phase (2) is f + df = . For α = dc (≪ 1) , we get (1 + α )V α = dc = df 1− f (2.16.13) so that we have to solve the equation 5µ * ( µ * − µ 2 ) which reads also df + (3µ * + 2µ 2 )dµ * = 0 1− f (2.16.14) 51 2 df 3µ * + 2 µ 1 * = − * * 2 dµ * = * − * dµ 1− f 5µ ( µ − µ 2 ) 5µ µ − µ 2 (2.16.15) This can be easily integrated (through the logarithms) into 1− f = λ µ2 − µ * ( µ * ) 2/5 (2.16.17) where the constant λ is determined from the initial condition ( µ * = µ1 for f = 0 ), i.e., λ= µ12/5 µ 2 − µ1 (2.16.18) The differential scheme estimate µ DS is finally obtained for f = c2 , that is 1 − c2 = µ12/5 ( µ2 − µ SD ) µ2 − µ SD µ1 = ( µ2 − µ1 )( µ SD ) 2/5 µ 2 − µ1 µ SD 2/5 (2.16.19) 3) If the second phase is rigid ( µ2 → ∞ ), we get µ1 1 − c2 → SD µ rig 2/5 or SD → µrig µ1 (1 − c2 )5/2 (2.16.20) which shows that, at variance with the classical self-consistent estimate, the differential scheme estimate remains finite and only tends to infinity when c2 → 1 : due to the differential homogenisation procedure, there is no more any possibility for a continuous rigid skeleton to be formed. On the contrary, when c2 ≪ 1 , these estimates as well as the Hashin-Shtrikman lower bound behave similarly 5c ( µ ≈ µ1 (1 + 2 ) . 2 When the second phase is constituted of voids, the foregoing calculations are, from a rigorous point of view, no more valid since neither the second phase nor the homogenised material is incompressible ( β ≠ 2 / 5 ). If, as a first approximation, this difficulty is omitted, we find from (2.16.19) 52 µ SD µ1 1 − c2 ≈ por SD µ1 µpor 2/5 or SD µpor ≈ µ1 (1 − c2 )5/3 (2.16.21) This would mean that, now again, the overall shear modulus only vanishes for c2 = 1 , as expected for a porous material with closed porosity. For c ≪ 1 , we find 5 SD µpor ≈ µ1 (1 − c2 ) , which conforms with (2.14.18) in Exercise 2.14. 3 An exact calculation would have given dµ µ = − (1 − dk = − k (1 − df 5(3k + 4µ )df dµ =− f )(1 − β ) µ (1 − f )(9k + 8µ ) ⇒ df dk = − (3k + 4 µ )df k f )(1 − α ) 4µ (1 − f ) (2.16.22) and then dµ 20µ 2 3k − 4µ = ⇒ µ = µ1 dk k (9k + 8µ ) 3k 5/3 (2.16.23) and finally 3 DS µ por µ1 (1 − c2 )6 = DS 3/5 µpor 2− µ 1 (2.16.24) It can be checked that this expression is very close to (2.16.21) and the qualitative comments given before are still valid. Exercise 2.17: Bounds and estimates for the elastic moduli of an isotropic polycrystal with cubic crystals This is an additional question to Exercise 2.11. Apply to the overall shear modulus of this polycrystal the method proposed in Volume I, Sect.2.8.2, aiming at constructing a whole set of estimates for the overall moduli Cest of heterogeneous materials. Due to the overall isotropy, spheres and an isotropic matrix (µ 0, k0) can be considered here. Give the expression of µ est as a function µ 0, β 0 (the Eshelby coefficient) and the crystal moduli. Derive the bounds of Voigt and Reuss and of Hashin and Shtrikman (assuming 2C44 > (C11 –C12)) for µ est and the equation for 53 the self-consistent estimate µ SC. Propose a graphic construction for the derivation of this estimate. Solution According to (2.188) in Volume I, the equation for Cest is < δcest : (I + P 0 : δc0 )−1 >= 0 (2.17.1) with P0 for a sphere and an isotropic matrix given by 0 PSph = α0 3k J+ 0 β0 3k 0 0 K α = 3µ 0 3k 0 + 4µ 0 β0 = 6(k 0 + 2µ 0 ) 5(3k 0 + 4µ 0 ) (2.17.2) and δc0 = c − C0 . Using the representation (2.4.6), we get δc est = (a′ − 2µ est )K ′ + (a′′ − 2µ est )K ′′ + (b − 3k est )J −1 β 0 ( a′ − 2 µ 0 ) (I + P : δc ) = 1 + K ′ + ... 2µ 0 0 0 −1 −1 (2.17.3) −1 β 0 (a′′ − 2µ 0 ) α 0 (b − 3k 0 ) ′′ ...1 + K + 1 + J 2µ 0 3k 0 and then the equation a′ − 2µ est a′′ − 2µ est b − 3k est ′ ′′ < K > + < K > + J = 0 (2.17.4) β 0 ( a′ − 2 µ 0 ) β 0 (a′′ − 2 µ 0 ) α 0 (b − 3k 0 ) + + 1+ 1 1 2µ 0 2µ 0 3k 0 From Exercise 2.11 (see Eq. 2.11.20), we know that, for an isotropic distribu2 3 tion of cubic crystals, < K ′ >= K and < K ′′ >= K . Consequently we get 5 5 b C11 + 2C12 est 0 est eff ) b − 3k = 0, ∀C ⇒ k = k = 3 (= 3 est 3(a′′ − 2µ est ) 2(a′ − 2µ ) =0 β 0 ( a′ − 2 µ 0 ) + β 0 (a′′ − 2µ 0 ) 1+ 1 + 2µ 0 2µ 0 (2.17.5) 54 The first result in (2.17.5) was already known from Exercise 2.11. The second equation yields the estimated shear modulus µ est as µ est = 2 µ 0 (1 − β 0 )(2a′ + 3a′′) + 5a′a′′β 0 20µ 0 (1 − β 0 ) + 2(3a′ + 2a′′) β 0 with β 0 = 2(b + 6 µ 0 ) 5(b + 4µ 0 ) (2.17.6) a′ = C11 − C12 , a′′ = 2C44 , b = C11 + 2C12 As we know from Sect.2.8.2 in Volume I, the bounds of Reuss and Voigt must be recovered from (2.17.6) when µ 0 tends to zero and infinity, respectively, that is 2 5a′a′′ 0 0 est Reuss µ → 0 ⇒ β → 5 , µ → 2(3a′ + 2a′′) = µ µ 0 → ∞ ⇒ β 0 → 3 , µ est → 2a′ + 3a′′ = µ Voigt 5 10 (2.17.7) It is easy to check that these bounds are the same as those found in Exercise 2.11, taking account of the changes in notations (note that a = a′ / a′′ ). To derive the lower (upper, resp.) Hashin and Shtrikman bound, we have to find the value of µ 0 which gives to the quadratic form associated to δc0 = c − C0 the smallest (highest, resp.) possible positive (negative, resp.) value. Since 3k0 = b, δc 0 = c − C0 = (a′ − 2µ 0 )K ′ + (a′′ − 2 µ 0 )K ′′ (2.17.8) Since 2C44 > (C11 –C12) (i.e., a” > a’) and taking account of the last property of (2.11.8) in Exercise 2.11, the lower (upper, resp.) Hashin and Shtrikman bound is reached when, in (2.17.6), 2 µ 0 = a′ ( 2 µ 0 = a′′ , resp.). Expressing first β 0 in (2.17.6) as a function of b and µ 0, we get µ est = f ( µ 0 ) = 8(2a′ + 3a′′)( µ 0 )2 + 3[10a′a′′ + b(2a′ + 3a′′) ] µ 0 + 5a′a′′b (2.17.9) 80( µ 0 ) 2 + 6 [ 5b + 2(3a′ + 2a′′)] µ 0 + 2b(3a′ + 2a′′) The Hashin and Shtrikman bounds are found by setting 2 µ 0 = a′ or 2 µ 0 = a′′ in (2.17.9). We find HS+ a′′(38a′a′′ + 12a′′2 + 16a′b + 9a′′b) µ = 36a′a′′ + 64a′′2 + 12a′b + 38a′′b 2 µ HS− = a′(42a′a′′ + 8a′ + 6a′b + 19a′′b) 2 24a′a′′ + 76a′ + 42a′b + 8a′′b (2.17.10) 55 The self-consistent estimate µ SC satisfies the equation µ est = f ( µ est ) where the function f(µ ) is defined in (2.17.9). This third-order equation reads µ SC = 8(2a′ + 3a′′)( µ SC ) 2 + 3[10a′a′′ + b(2a′ + 3a′′)] µ SC + 5a′a′′b 80( µ SC )2 + 6 [5b + 2(3a′ + 2a′′)] µ SC + 2b(3a′ + 2a′′) (2.17.11) Fig. 2.17.1 Graphical construction for the determination of the self-consistent estimate If µ est = f ( µ 0 ) is plotted as a function of µ 0 ≥ 0 , the self-consistent estimate is obtained as the ordinate of the point where the bisector of the first quadrant crosses this curve. This property suggests an iterative numerical resolution of the third-order equation (2.17.11) which, among several other equivalent methods with other initial values, would read (see Fig. 2.17.1) µ(0) = 0 µ(1) = f (0) µ(2) = f ( µ(1) ) ... µ( n+1) = f ( µ( n ) ) SC ( µ( n ) ) µ = lim n →∞ (2.17.12) Exercise 2.18: Elastic and thermoelastic behaviour of an elastic inhomogeneous material with cubic or isotropic symmetry and isotropic constituents This exercise is a continuation of Exercise 2.12 which is concerned with the investigation of the thermoelastic behaviour of the same material. 1) Consider the special case of identical shear moduli µ i in the phases (i). Show that the overall behaviour is then isotropic and derive the Hashin and Shtrikman bounds for the homogenised bulk modulus keff. Comment the obtained result. 2) This result can be corroborated by the construction, according to the procedure developed in Volume I, Sect.2.8.2, aiming at constructing a whole set of estimates for the overall moduli Cest of heterogeneous materials. Specify this method for es- 56 timating the sole homogenised bulk modulus keff by considering spherical inhomogeneities with moduli (ki , µi ) embedded in an infinite isotropic matrix with the moduli (k 0 , µ 0 ) , subjected to the uniform dilatation Θ0 δ at infinity and calculate 3 the corresponding estimate k . How can we derive from that the Voigt and Reuss and Hashin and Shtrikman bounds as well as the self-consistent estimate for the homogenised bulk modulus. What happens when all the shear moduli µ i are identical? Consider the special case of a two-phase material with µ1 ≥ µ 2 . 3) Use these results for proposing the construction of a whole set of estimates for eff the overall coefficient of thermal expansion α when these coefficients are isoest tropic in every phase. Focus on the case of a two-phase material and check that α eff obeys the relation given by Levin’s theorem. What happens when all the shear moduli µ i are identical? Solution 1) When µi = µ , ∀i , (2.12.1) and (2.12.8) give C11eff − C12eff =µ 2 (2.18.1) Ceff = 2 µ K + 3k eff J (2.18.2) eff = C44 and then so that the overall behaviour is then isotropic with µ eff = µ and, according to the Voigt and Reuss bounding, < k −1 > −1 ≤ k eff ≤< k > . Now, the property of local and global isotropy allows us to use the sharper Hashin and Shtrikman bounding (see (2.184) in Volume I), namely α − δk − − ) −1 > α + δk + ) −1 > + eff k k ≤k ≤ α − δk − −1 α + δk + −1 < (1 + ) > < (1 + ) > k− k+ with < k (1 + < k (1 + k − = min {ki } , k + = max {ki } , − − + δk = k − k , δk = k − k α− = + 3k − 3k + + = , α 3k − + 4µ 3k + + 4 µ (2.18.3) 57 So, we have (1 + α − δk − k − + (1 + α δk + k+ )−1 = (1 + 3 k − k − −1 3k − + 4 µ ) = 3k − + 4 µ 3k + 4µ 3k + + 4 µ ) = 3k + 4µ (2.18.4) −1 and then k HS− 3k − + 4µ k )> < > − µ 3 k + 4 3 k + 4µ k = = = = k HS+ (2.18.5) 1 α − δk − −1 3k − + 4µ < > < (1 + ) > < > 3k + 4µ 3k + 4µ k− < k (1 + α − δk − ) −1 > < k( since these expressions do not depend on k- (nor on k+) any more. Consequently, k eff is determined too and reads k eff = < k (3k + 4 µ ) −1 > < (3k + 4 µ ) −1 > (2.18.6) where µ is constant and k varies from one phase to another. 2) In the spherical inhomogeneity (i) (see Fig. 2.18.1) the volume dilatation θiest is given by 1 θiest = 1+ Θ0 = 0 α 0 δki k0 1 3k 0 + 4µ 0 0 0 Θ = Θ 3( k − k 0 ) 3ki + 4 µ 0 1+ 0i 3k + 4µ 0 Θ0 δ 3 µ 0 ,k 0 … µ i , ki …. (i) Fig. 2.18.1 Construction of a set of estimates for the overall bulk modulus (2.18.7) 58 and the mean stress σ mesti is σ mesti = kiθiest = ki (3k 0 + 4µ 0 ) 0 Θ 3ki + 4µ 0 (2.18.8) So, kest is given by k est = <σ <θ est m est k (3k 0 + 4µ 0 ) k > < > > 3k + 4µ 0 3k + 4 µ 0 = = 1 3k 0 + 4µ 0 > < > < > 0 3k + 4 µ 0 3k + 4µ < (2.18.9) since (3k 0 + 4µ 0 ) is constant. When µ 0 → 0, k est →< k −1 > −1 which is the Reuss bound. When µ 0 → ∞, k est →< k > which is the Voigt bound. The Hashin and Shtrikman bounds are derived with µ 0 = µ − = min {µi } for the lower bound and µ 0 = µ + = max {µi } for the upper bound. The self-consistent estimate k SC is obtained for µ 0 = µ SC , the self-consistent estimate for µ eff . When all the shear moduli µ i are identical, µ − = µ + = µ and the lower and upper Hashin and Shtrikman bounds for k eff (as well as the self-consistent estimate, which is intermediate) coincide to the exact value for k eff , namely k > 3k + 4µ = 1 < > 3k + 4µ < k eff (2.18.10) as we already know from (2.18.6). For a two-phase material, we get, with µ2 < µ1 k est c1k1 c2 k 2 + 0 0 3 k + 4 µ 3 k 3k1k2 + 4µ 0 < k > 2 + 4µ = f (µ 0 ) = 1 = c1 c2 4µ 0 + 3k1k2 < k −1 > + 0 0 3k1 + 4µ 3k2 + 4µ (2.18.11) The plot of k est = f ( µ 0 ) (see Fig. 2.18.2) yields especially the elementary bounds and estimates for k eff . 59 Fig. 2.18.2 A set of estimates for the effective bulk modulus of a two-phase material 3) We use the basic equation (see Eqs (2.119) and (2.125) in Volume I) connecting thermoelasticity with elasticity th th E =< B T : ε > ⇒ α eff =< BT : α > (2.18.12) where B T is the transposed stress concentration tensor of the purely elastic coneff centration problem. To derive an estimate of α , we use (2.18.12) with an estimate of B T deriving from Question 2. As a matter of fact, due to the isotropy of α and B T , we have simply α est =< bmestα > where bmesti = σ mesti and Σ m = k est < θ est > Σm (2.18.13) From (2.18.7) and (2.18.8) we get σ mesti = ki (3k 0 + 4 µ 0 ) 0 Θ 3ki + 4 µ 0 Σ m = k est < 3k 0 + 4µ 0 > Θ0 3ki + 4µ 0 (2.18.14) and then bmesti ki (3k 0 + 4µ 0 ) 0 ki ki Θ 0 0 3ki + 4µ 3ki + 4µ 3ki + 4µ 0 = = = (2.18.15) 0 0 1 k 3k + 4µ est est 0 k < < > > k < >Θ 3k + 4µ 0 3k + 4µ 0 3ki + 4 µ 0 So, we get 60 α est =< bmestα >= < kα (3k + 4 µ 0 ) −1 > < k (3k + 4µ 0 )−1 > (2.18.16) Levin’s theorem (see Eq. 2.127 in Volume I) for a two-phase material leads to the relation α eff =< α > +(S eff − < s >) : (s 2 − s1 ) −1 : (α 2 − α 1 ) (2.18.17) In case of local and global isotropy, this relation reads α eff =< α > +( 1 1 1 1 − < >)( − ) −1 (α 2 − α1 ) eff k k k2 k1 In order to get consistent estimates for α eff (2.18.18) , we must require this relation to hold for estimates too, namely α est =< α > +( 1 1 1 1 − < >)( − ) −1 (α 2 − α1 ) k est k k2 k1 (2.18.19) With k est given by (2.18.11), we get α est =< α > +( 4 µ 0 + 3k1k2 < k −1 > 1 1 − < k −1 >)( − ) −1 (α 2 − α1 ) (2.18.20) 0 3k1k 2 + 4µ < k > k2 k1 After some algebra, it can be checked that (2.18.20) and (2.18.16) coincide. Finally, when all the shear moduli are equal ( µi = µ , ∀i ), we take µ 0 = µ and obtain the exact value for α eff , namely α eff = < kα (3k + 4 µ ) −1 > < k (3k + 4µ ) −1 > (2.18.21) where µ is constant and k and α vary from phase to phase. Exercise 2.19: Thermal stresses in a two-phase material When the temperature is uniformly varied in an unloaded heterogeneous material, for instance during the processing route of a composite material, this gives rise to an internal thermal stress field which is important to be evaluated in view of a secure further use. As a first approach to this problem, show that these per phase averaged thermal stresses can be calculated in an elastic two-phase material as a 61 function of the elastic compliances s1 and s 2 , the coefficients of thermal expansion α 1 and α 2 , the effective coefficient of thermal expansion α eff , the volume fractions c1 and c2 and the temperature variation ∆T. Find a relation involving the efeff fective compliance S eff in place of α . Consider the special case when s1 and s 2 , S eff , α 1 and α 2 are isotropic and discuss the sign of the residual stresses when, as it is more usual in reinforced composite materials with phase (1) as a matrix and phase (2) made of particles or fibres, one has α 2 < α1 . Give the solution when S eff is estimated as its Voigt and Reuss bounds. Discuss the limits of such a first approach. Solution When the material is unloaded, we have < σ >= 0 (2.19.1) and, in addition, ε = ε e + ε th = s : σ + α ∆T ⇓ th (2.19.2) eff < ε >= E = α ∆T =< s : σ > + < α > ∆T that is eff < s : σ >= (α − < α >)∆T When α eff (2.19.3) is known (or estimated), the two equations (2.19.1) and (2.19.3) al- low us to calculate (or estimate) the per phase averaged residual stresses σ 1 and σ 2 in the case of a two-phase material. With c = c2 , these equations read (1 − c)σ 1 + cσ 2 = 0 eff (1 − c)s1 : σ 1 + cs 2 : σ 2 = α − (1 − c)α 1 − cα 2 ∆T When, instead of α eff (2.19.4) , S eff is known (or estimated), Levin’s theorem (see Eq. 2.127 in Volume I or (2.18.17) in Exercise 18) can be used in order to express the 62 relation between α eff and S eff since we are dealing with a two-phase material. This relation reads α eff =< α > +(S eff − < s >) : (s 2 − s1 ) −1 : (α 2 − α 1 ) (2.19.5) so that the system (2.19.4) can be transformed into (1 − c)σ 1 + cσ 2 = 0 (2.19.6) −1 eff (1 − c)s1 : σ 1 + cs 2 : σ 2 = S − (1 − c)s1 − cs 2 : (s 2 − s1 ) : (α 2 − α 1 )∆T or cσ 2 = −(1 − c)σ 1 = (s 2 − s1 ) −1 : S eff − (1 − c)s1− cs 2 : (s 2 − s1 ) −1 : (α 2 − α 1 )∆T (2.19.7) In case of isotropy for s1 , s 2 , S eff ,α 1 and α 2 , (2.19.7) reduces to 3( cσ 2 = −(1 − c)σ 1 = 1 1 − < >)(α 2 − α1 ) k eff k ∆T δ 1 1 ( − )2 k2 k1 (2.19.8) In this equation, the sign of the residual stresses does not depend on the sign of (k2 − k1 ) since < k −1 > −1 , which the Reuss bound for k eff , is always smaller than 1 1 k eff , so that eff − < >≤ 0 . So, for ∆T < 0 and α 2 < α1 , the average residual k k stresses are, as expected, compressive in phase (2) and tensile in phase (1). When 1 1 S eff is then estimated as its Reuss bound, Reuss =< > and the per phase average k k residual stresses vanish, which conforms with the underlying uniform stress assumption. When the Voigt bound is used, we get 3( cσ 2 = −(1 − c)σ 1 = 1 1 − < >)(α 2 − α1 ) <k > k ∆T δ 1 1 ( − )2 k2 k1 = −3c(1 − c)(α 2 − α1 ) k1k2 ∆T δ (1 − c)k1 + ck2 (2.19.9) 63 Roughly speaking, this estimate could be considered as a bound for the thermal stresses. Nevertheless, one has to keep in mind that this analysis yields estimates only for the per phase average residual stresses. If these averages are isotropic, there might exist deviatoric components whose average vanishes but which could play an important role in local plastification. In addition, the elastic moduli could depend on temperature, which has not been considered here. Actually, a whole thermo-elastoviscoplastic of the residual stress field would be necessary for a refined analysis. 2.4 Miscellaneous Exercise 2.20: Finite Element Method The principle of the Finite Element Method has been briefly exposed in Sect. 2.4.2 of Volume I. Without going into too much details of the practical implementation of this method, this exercise aims at studying the simplest way to proceed up to the complete resolution of an elementary problem. 1) For sake of simplicity, we restrict ourselves to 2D-problems in the (1,2) plane and triangular 3-node elements. For such a triangle, say 123, calculate the interpolation functions ϕ i = (ϕi1 ,ϕi 2 ) , i = 1,2,3 when a linear interpolation between the nodes is considered. Solution The trial displacement field in the considered triangle, u′( M ) = [u′( x, y ), v′( x, y )] , must be a linear function of both the coordinates ( x, y ) and the displacement components at the nodes of the triangle, that is u1′( x, y ) = u1′ f1 ( x, y ) + u2′ f 2 ( x, y ) + u3′ f 3 ( x, y ) v1′( x, y ) = v1′ f1 ( x, y ) + v2′ f 2 ( x, y ) + v3′ f 3 ( x, y ) (2.20.1) with the relations f1 ( x1 , y1 ) = f 2 ( x2 ,y2 ) = f 3 ( x3 ,y3 ) = 1 (2.20.2) f1 ( x2 , y2 ) = f1 ( x3 , y3 ) = f 2 ( x1 ,y1 ) = f 2 ( x3 ,y3 ) = f3 ( x1 ,y1 ) = f 3 ( x2 ,y2 ) = 0 So, the linear functions f i ( x, y ) read 64 f1 ( x, y ) = a1 x + b1 y + c1 f 2 ( x, y ) = a2 x + b2 y + c2 f ( x, y ) = a x + b y + c 3 3 3 3 (2.20.3) where the 9 constants ai , bi , ci are, according to (2.20.2), the solutions of the linear system of 9 equations a1 x1 + b1 y1 + c1 = 1 a2 x1 + b2 y1 + c2 = 0 a x + b y + c = 0 3 1 3 1 3 a1 x2 + b1 y2 + c1 = 0 a1 x3 + b1 y3 + c1 = 0 a2 x2 + b2 y2 + c2 = 1 a2 x3 + b2 y3 + c2 = 0 a3 x2 + b3 y2 + c3 = 0 a3 x3 + b3 y3 + c3 = 1 (2.20.4) From the definition u′( x, y ) = X i ϕ i ( x, y ) , we can write the transposed matrix T [ X ]123 as T [ X ]123 = [u1′ u3′ v3′ ] (2.20.5) 0 f3 f2 0 0 f 3 (2.20.6) v1′ u2′ v2′ f 0 f1 0 T and the matrix [ϕ ]123 as T [ϕ ]123 = 01 f2 with f1, f2 and f3 defined by (2.20.3) and (2.20.4), so that u ′( x, y ) T = [ X ]123 .[ϕ ( x, y ) ]123 ′ v ( x , y ) 123 [u′( x, y )]123 = (2.20.7) 2) We consider only plane strain situations in isotropic elasticity (Lamé coefficients λ and µ). The strain and stress matrices are written in the form T [ε ] = ε11 ε 22 2ε12 T [σ ] = σ 11 σ 22 2σ 12 (2.20.8) Calculate the 3x3 matrix [C ]PS of the elastic moduli. Calculate the stiffness matrix [ R ]123 for one triangle, say 123, with area A123 and thickness e. Solution In plane strain conditions, we have 65 σ 11= (λ +2µ )ε11+λε 22 σ 22 = λε11+(λ +2 µ )ε 22 σ 12 = 2µε12 σ 11 λ +2µ ⇒ σ 22 = λ 2σ 12 0 λ λ +2 µ 0 0 ε11 0 ε 22 (2.20.9) 2µ 2ε12 So, the matrix of the elastic moduli in plane strain conditions reads [C ]PS λ λ + 2µ = λ λ + 2µ 0 0 0 0 2µ (2.20.10) The elementary stiffness matrix [ R ]123 for the element 123 obeys the relation 1 e T T [ X ]123 .[ R ]123 .[ X ]123 = ∫123 [ε ′]123 .[C ]123 .[ε ′]123 dS 2 2 (2.20.11) T T T ∂ϕi1 ε11′ X i ϕ i = X i ε11′ ϕ i = X i ∂x T T T ∂ϕi 2 ′ X i ϕ i = X i ε 22 ′ ϕi = Xi with ε 22 ∂y 1 T ∂ϕi1 ∂ϕi 2 T T + ) ε12′ X i ϕ i = X i ε12′ ϕ i = X i ( 2 ∂y ∂x (2.20.12) ( ) ( ) ( ) ( ) ( ) ( ) Consequently, we can write [ε ′]123 = a1 0 a2 0 a3 0 b1 0 b2 0 b1 2 a1 2 b2 2 a2 2 b3 2 0 b3 .[ X ]123 a3 2 (2.20.13) and the elementary stiffness matrix for the element 123 reads T [ R ]123 = eA123 [ g ] .[C ]PS .[ g ] with A123 the area of the element 123 and the matrix [ g ] given by (2.20.14) 66 [g] = a1 0 a2 0 a3 0 b1 0 b2 0 b1 2 a1 2 b2 2 a2 2 b3 2 0 b3 a3 2 (2.20.15) 3) Find the equivalent nodal forces which are to be concentrated at the concerned nodes when a) a uniform density of forces F (q1 , q2 ) is applied inside the element 123 g b) a linear density of forces T [ (α s + β ) p1 ,(α s + β ) p2 ] is applied on the boundary edge 12 of length l, with 0 ≤ s ≤ l . Solution a) The energetic equivalence is ruled by the relation ∫ T 123 F .u ′ dV = X i ∫ F .ϕ i dV = X i ∫ (q1ϕi1 + q2ϕi 2 )dV = [ X ]123 .[ F ]123 (2.20.16) 123 123 Due to the properties of the interpolation functions f i ( x, y ) (linear variation in the triangles and 1 or 0 values at the nodes), it is easy to show that ∫ 123 f i ( x, y )dV = eA123 , ∀i 3 (2.20.17) Consequently, we find T [ F ]123 = eA123 [ q1 3 q2 q1 q2 q1 q2 ] (2.20.18) b) Here again, the equivalent nodal forces are defined from the energetic relation ∫ 12 g g l T T .u′ dS = X i ∫ T .ϕ i dS = eX i ∫ (α s+β )( p1ϕi1+ p2ϕi 2 )ds = [ X ]12.[ F ]12 (2.20.19) 12 0 On the boundary edge 12, f 3 = 0, f1 = l−s s , f 2 = . So, one has l l 67 l−s l (α l + 3β ) l ∫0 (α s + β ) l ds = 6 l s l (2 α l + 3β ) (α s + β ) ds = ∫0 6 l (2.20.20) and then [ F ]12 (α l + 3β ) p1 (α l + 3β ) p 2 α β (2 l + 3 ) p el 1 = 6 (2α l + 3β ) p2 0 0 (2.20.21) 4) Show how the foregoing results could be used in order to solve by the finite element method the problem of an isotropic elastic dam which is addressed in Exercise 8.5 (question 2) from a purely static point of view (see Fig. 2.20.1). y 0 x π/4 water dam A B Fig. 2.20.1 Sketch of the studied dam This dam can be discretised as indicated in Fig. 2.20.2. The basement 46 is fixed; the vertical edge 14 suffers the water pressure (linearly increasing with the depth), 16 is load-free and the dam itself is subjected to its own weight. Solution Interpolation functions: using a local coordinate systems, the four triangles are either in the position of triangle (1) (this is the case for (1), (2) and (4)) or in the position of triangle (3). In the first case, one has, with 1 (0, h), 2 (0, 0) and 3 (h, 0) 68 Fig. 2.20.2 Discretised structure: triangles (1) to (4); nodes 1 to 6. y f1 ( x, y ) = h h − ( x + y) f 2 ( x, y ) = h x f 3 ( x, y ) = h (2.20.22) and in the second case, with 1 (0, h), 2 (h, 0) and 3 (h, h) h− x f1 ( x, y ) = h h− y f 2 ( x, y ) = h x+ y−h f 3 ( x, y ) = h (2.20.23) Stiffness matrix: the [ g ] matrix reads differently for (1), (2), (4) and for (3). According to (2.20.15), we find [ g ](1),(2),(4) [ g ](3) 0 0 −1 h = 0 1h 0 1 h 2 0 −1 h 2 −1 h 0 = 0 0 0 −1 h 2 0 1h −1 h 0 −1 h 2 0 0 0 −1 h 2 0 1 h 2 0 −1 h 0 1h 0 1 h 2 1 h 2 0 1h 0 (2.20.24) 69 The same remark holds for the elementary stiffness matrices. From (2.20.14), (2.20.10) and (2.20.24) we find, with A(1) = A(2) = A(3) = A(4) = h 2 2 [ R ](1),(2),(4) [ R ](3) 0 −µ −µ 0 µ µ λ + 2µ λ −λ −(λ + 2 µ ) 0 λ + 3µ λ+µ − (λ + 2 µ ) − µ e = λ + 3µ −λ −µ 2 sym λ + 2µ 0 µ (2.20.25) λ 0 0 −(λ + 2 µ ) −λ λ + 2 µ µ µ 0 −µ − µ µ 0 −µ −µ e = λ + 2µ −λ − (λ + 2 µ ) 2 sym λ + 3µ λ+µ λ + 3µ The whole stiffness matrix for the structure is composed by transferring each component of an elementary matrix at its adequate place in the global matrix; when several components are transferred at the same place, they have to be added. Actually, the whole global matrix is not necessary for solving the problem with the finite element method. In the central final equation (Eq. 2.115 in Volume I), the submatrix [ R22 ] only is needed, if the first p values of Xi correspond to nodes where the displacement is given by the boundary conditions. In our case, this means that the lines and rows of the global matrix corresponding to nodes 4, 5 and 6 where the displacement is forced to vanish can be omitted. We are then left with the following global [ R22 ] stiffness matrix 0 µ λ + 2µ e [ R22 ] = 2 sym −µ −µ −λ −(λ +2µ ) 0 λ 2(λ +3µ ) λ +µ −2(λ +2µ ) 2(λ +3µ ) −(λ +µ ) 2(λ +3µ ) −(λ +µ ) (2.20.26) −2 µ λ +µ 2(λ +3µ ) µ 0 which, in the final equation, will be multiplied by [ X ] transposed of the matrix given in (2.20.5), since the displacements at 1, 2 and 3 are the sole unknowns. 70 Nodal forces: they are concerned with the weight of the dam and the water pressure. The force density associated with the weight has the form F (q1 , q2 ) studied in Question 3a, with q1 = 0, q2 = − ρ ′g , with ρ ′ the density of the constitutive material and g the acceleration of gravity. According to (2.20.18), we find T [ F ]123 = − eh 2 ρ ′g [ 0 1 0 1 0 1] 6 (2.20.27) As for the water pressure, the force density as also the form studied in Question g 3b, namely T [ (2h − y ) ρ g1 ,0] , with ρ the water density. In triangles (1) and (2), application of (2.20.21) yields [ F ]12 1 0 eρ gh 2 2 = 6 0 0 0 [ F ]24 4 0 eρ gh 2 5 = 6 0 0 0 (2.20.28) The global nodal force vector is obtained according to the same process as the one indicated above for the matrix stiffness. For the sole nodes 1, 2 and 3, we get ρ −ρ′ egh 2 6 ρ [ F2 ] = 6 −3 ρ ′ 0 −3ρ ′ (2.20.29) So, the final equation to be solved reads 0 µ λ + 2µ sym 0 u1′ ρ −ρ ′ −λ −(λ+2 µ ) λ 0 v1′ λ+µ −2(λ+2µ ) −(λ+µ ) u2′ gh 2 6ρ 2(λ+3µ ) . = (2.20.30) 2(λ+3µ ) −(λ+µ ) −2µ v2′ 3 −3ρ ′ 0 2(λ+3µ ) λ+µ u3′ 2(λ+3µ ) v3′ −3ρ ′ −µ −µ µ 71 From the nodal displacements we can then derive the displacement, strain and stress fields in the whole structure. Reference Christensen R.M., Lo K.H. (1979) Solution for effective shear properties in three phase sphere and cylinder models J. Mech. Phys. Solids, 27: 315-330. 72 Chapter 3 Exercises of Chapter I.3: Elastoplasticity 3.1 Deformation of single crystals Exercise 3.1 Active slip planes The active slip planes in a single crystal are dense planes of the crystallographic structure on which the shear stress is larger than the critical value τc. This is the Schmid and Boas law. Calculate the tensile stress σp for activation of slip on a plane, whose normal n makes an angle φ with the tensile axis, in a direction making an angle λ with this axis. Solution Use Cartesian coordinates one direction Ox1 of which coincides with the tensile axis. Write the stress tensor σij: σ (σ ) = 0 0 0 0 0 0 0 0 (3.1.1) Use the formula: ti = σ ij n j , ti being the stress components on the plane with unit normal ni: t1 = σ cos φ t2 = t3 = 0 (3.1.2) The resolved shear stress is obtained by the projection of the stress ti on the required direction m1 = cos λ τ = σ cos φ cos λ (3.1.3) 73 The critical tensile stress corresponds to τ = τ c in (3.1.3): σ p = τ c cos φ cos λ (3.1.4) Find the orientations of the slip plane and slip direction which minimise the critical stress for slip activation. Solution Choose the coordinate system such that Ox1 coincides with the tensile axis and that the unit vector n is perpendicular to the Ox3 axis. The coordinates of this vector are cosφ and sinφ. Let m1, m2, m3 be the coordinates of the unit vector in the slip direction with m1 = cos λ . The slip direction is perpendicular to n. Hence: m1 cos φ + m2 sin φ = 0 (3.1.5) m12 + m22 + m32 = 1 (3.1.6) Furthermore: From these two equations is deduced: cos 2 λ = 1− m32 sin 2 φ (3.1.7) so that the Schmid factor is: 12 mS = cos φ cos λ = cos φ sin φ (1 − m32 ) (3.1.8) It is maximised for m3 = 0 (thus mS = cos φ sin φ ) and φ = π 4 ( mS = 1 2 ). Find the relation which the spherical components sinϕ cosθ, sinϕ sinθ, cosϕ, of the tensile stress direction T must satisfy in order to activate a particular slip system ( hkl ) [ uvw ] in a F.C.C. single crystal, for example (111) 101 . Solution Express cosφ and cosλ as a function of h, k, l and u, v, w. Let N be the vector of components (h, k, l) normal to the slip plane and M the vector of components (u, v, w) along the slip direction: 74 12 T .N = h sin ϕ cos θ + k sin ϕ sin θ + l cos ϕ = ( h 2 + k 2 + l 2 ) cos φ (3.1.9) 12 T .M = u sin ϕ cos θ + v sin ϕ sin θ + w cos ϕ = ( u 2 + v 2 + w2 ) cos λ Equation 3.1.4 yields: 2 12 2 12 σp ( h + k + l ) (u + v + w ) = (3.1.10) τ c ( hsinϕ cosθ +ksinϕ sinθ +lcosϕ )( usinϕ cosθ + vsinϕ sinθ + wcosϕ ) 2 2 2 2 Note that h, k and l are equal to ±1 , while, one of the coordinates u, v and w being equal to 0, the other two have the appropriate values ±1 for the slip direction to be perpendicular to the normal to the slip plane. Table 3.1.1 lists the various slip systems in FCC crystals according to this rule. Inverting the + and – signs for the slip direction merely modifies the orientation. The absolute values of the cosines only enter the Schmid factor. Table 3.1.1. Slip systems in FCC crystals (111) (11 1 ) (1 11) (111) 110 101 011 1 10 10 1 01 1 - - - x x x - x x x - - x - x - x - x x - - - x ( ) In the particular case of the slip system 111 101 , Eq.3.1.10 yields: σp τc = 6 sin ϕ cos θ + sin ϕ sin θ + cos ϕ 1 − sin ϕ cos θ ( )( ) (3.1.11) Find the tensile direction which minimises the critical stress for slip activation on the 111 101 slip system. ( ) Solution ( ) The three vectors of components sin ϕ cos θ ,sin ϕ sin θ ,cos ϕ , (1,1,1) and (-1, 0, 1) must be in the same plane, so that: 75 sin ϕ cos θ sin ϕ sin θ cos ϕ 1 −1 1 0 1 1 = sin ϕ cos θ − 2 sin ϕ sin θ + cos ϕ = 0 (3.1.12) The second condition for maximising the Schmid factor is: cos φ = sin ϕ cos θ + sin ϕ sin θ + cos ϕ 3 = 1 2 (3.1.13) The solutions of Eqs 3.1.12 and 3.1.13 are: 1 1 cos ϕ = 6 + 2 = 0.908 1 = 0.976 sin θ = 12 7 2− 6 ( (3.1.14) ) The signs of cosϕ and of sinθ are such that the pole of the tensile axis is in the reference triangle on the stereographic projection. For this direction of the tensile axis σ p = 2τ c . Place the pole of this tensile axis on the stereographic projection. Solution On the stereographic projection the distance ρ of the pole to the centre of the base circle of radius R is equal to Rtan(ϕ/2), ϕ being the angle between the tensile axis and the direction 001 .This is demonstrated in the following way, x and y being the coordinates of a pole on the projection plane, r, θ and ϕ the spherical coordinates of a pole on the Ewald sphere of radius R/2: 2x 2y 2R = = R sin ϕ cos θ R sin ϕ sin θ R cos ϕ + 1 ( ) (3.1.15) Hence: 12 ρ = ( x2 + y2 ) = R sin ϕ ( cos ϕ + 1) = R tan ϕ 2 (3.1.16) 76 ( ) The values found for cosϕ and sinθ (Eq. 3.1.14) yield: ρ R = tg ϕ 2 = 0.219 Figure 3.1.1 shows the position of the tensile axis for easiest activation of the slip system 111 101 . ( ) Fig. 3.1.1. Stereographic projection showing the pole of the tensile axis for the easiest activation of the (111) plane in the [-101]direction. Which are the double slip directions? Solution ( )( )( ) Observing the stereographic projection, the reference triangle 001 , 011 , 111 , ( ) by two triangles: (001), (011), (111) and (001), (101), (111), within which the tensile vector activates respectively the (111)101 and the (1 11)011 slip syswithin which the tensile vector activates the 111 101 slip system, is bordered tems. When the pole of the tensile axis coincides with the lines belonging to two 77 adjacent triangles, the Schmid factors of the corresponding slip systems are equal. There is then double slip. Exercise 3.2 Geometrical hardening (Jaoul 2008) In the course of the plastic deformation by slip of a single crystal loaded by traction is observed a rotation of the slip planes. The angle of their normal with the tensile axis increases. The orientation becomes less and less favourable for slip. Hardening occurs, even if the critical shear stress remains constant. This is called geometrical hardening. The exercise is based on the book of Jaoul, which was reedited in 2008. In a single crystal, find the relation between the extension e in the tensile direction and the slip g in a slip direction making initially an angle λ0 with the tensile axis, along a slip plane whose normal makes initially an angle φ0 with the tensile axis. Assume free rotation of the specimen. Using Cartesian coordinates such that axis Ox3 coincides with the tensile axis and that the slip direction lies in the Ox1, Ox3 plane, find the strain tensor for the transformation corresponding to the slip g. Solution The coordinates of the unit vector along the slip direction are sinλ0, 0, cosλ0 . The coordinates of the unit vector normal to the slip plane are v1, v2, cosφ0. As it is perpendicular to the slip direction: cos φ0 cos λ0 ν 1 = − sin λ 0 1 sin 2 λ0 sin 2 φ0 − cos 2 λ0 cos2 φ0 ) 2 ( v2 = sin λ0 (3.2.1) The distance of point M, whose initial coordinates are a1, a2, a3, to the slip plane is aivi. Point M is displaced to point M’ whose coordinates are x1, x2, x3. The length MM’ is gaivi, so that: x1 = a1 + gai vi sin λ0 x2 = a2 x = a + ga v cos λ 3 3 i i 0 (3.2.2) 78 Hence the gradient tensor (see section A3.1.1 in Volume I) given by the partial derivatives of xi with respect to ai. The deformation tensor ∆ij can now be written. As the extension e is in the Ox3 direction, the ∆33 element only needs to be calculated: 2 ∆33 = g 2 cos 2 φ0 + 2 g cos φ0 sin λ0 (3.2.3) The extension e is related to this by: 2 2 2 2∆33da32 = dx − da = (1 + e ) − 1 da32 (3.2.4) Equating the two preceding relations, the slip g can be calculated as the positive root of a second degree equation, yielding: 2 g cos φ0 = (1 + e ) − sin 2 λ0 1 2 − cos λ0 (3.2.5) The relation between the tensile stress σ and the extension e is now sought. In order to do that, it is needed to introduce the current angles λ and φ in the course of deformation. Choosing a point M on the tensile axis with coordinates 0, 0, 1 the scalar product of the vector OM’ and the unit vector in the slip direction yields OM’cosλ. As the coordinates of OM’ are g cosφ0 sinλ0, 0, 1 + g cosφ0 cosλ0 it is found that: cos λ = g cos φ0 + cos λ0 (1 + 2 g cos λ cosφ 0 0 + g 2 cos 2 φ0 ) 1 (3.2.6) 2 and, using the preceding equation (3.2.5) relating g and e: (1 + e ) 2 − sin 2 λ0 cos λ = 1+ e 1 2 (3.2.7) The scalar product v.OM’ is equal to OM’cosφ so that: cos φ = It can also be written: cos φ0 (1 + 2 g cosφ 2 2 0 cos λ0 + g cos φ0 ) 1 (3.2.8) 2 79 cos φ = cos φ0 1+ e (3.2.9) The tensile flow stress σp is related to the critical shear stress τc by the Schmid relation found in Exercise 3.1 above: τ c = σ p cos φ cos λ (3.2.10) Equations (3.2.7), (3.2.8) and (3.2.9) yield: σp = (1 + e ) 2 2 cos φ0 (1 + e ) − sin 2 λ0 1 2 (3.2.11) τc It is an increasing function of the extension e, provided 2 sin λ0 ≤ 1 + e , even when there is no work-hardening (τc is a constant). Draw the trajectory of the tensile axis on the stereographic projection. Assume free rotation of the specimen. Solution It can be checked that the tensile axis remains in the plane containing the normal to the slip plane and the slip direction. This is achieved if cos φ = sin λ . Equations 3.2.7 and 3.2.9 allow checking that this is the case. The tensile axis rotates, the angle φ increasing, so that its pole on the stereographic projection moves away from the pole (111) towards the pole 101 . Starting from the position of the pole of the tensile axis determined in the preceding exercise, it can be seen (Fig. 3.2.1) that this pole, moving towards the pole 101 , meets the border line 001 , 111 (cosθ = sinθ) where double slip starts. If ( )( ) it goes beyond this line, the angle φ keeps increasing, corresponding to a decrease of the Schmid factor; in the meanwhile the homologous angle φ for the 1 11 011 slip system decreases; this system is then activated; the pole of the ( ) ( ) tensile axis moves towards the pole 011 ; the slip system 111 101 is activated again; the pole of the tensile axis moves back to the cosθ = sinθ line. Thus, the double slip is stabilised along this line; the pole of the tensile axis moves towards the pole 112 . 80 Fig. 3.2.1. Displacement of the pole of the tensile axis in the course of deformation of a FCC crystal. Single and double slip. The dotted lines show the double slip systems. 3.2 Twinning A twin is part of a crystal which is related to the parent crystal by an operation of symmetry. In FCC the twin is the mirror image of the parent crystal across the (111) plane, called the twin plane. 81 Exercise 3.3 Deformation of a twinned crystal A twinned FCC crystal, the (111) plane being the twin plane, is compressed in a direction 134 . Are the deformations of the crystal and the deformation of the twin compatible? Solution Both deformations will be compatible if the compression activates the same slip systems. The determination of the position of the slip systems in the twin is needed. The coordinates of a point P being p1, p2, p3, we want to determine the coordinates p'1, p'2, p'3 of its image P' in the twin. The equation of the plane (111) is: x1 + x2 + x3 − 1 = 0 (3.3.1) The distance d of the point P to the (111) plane is: d= p1 + p2 + p3 − 1 (3.3.2) 3 The direction cosines of the normal to (111) being 1 3, 1 3 ,1 3 , the co- ordinates of the vector PP' are (A, A, A) with A= −2 ( p1+ p2 + p3 − 1) 3. Therefore, the coordinates of P' are: p1 − 2 ( p2 + p3 ) + 2 p1′ = 3 p2 − 2 ( p1 + p3 ) + 2 p2′ = 3 p3 − 2 ( p1 + p2 ) + 2 p3′ = 3 (3.3.3) For a vector OP the image in the twin is O'P' = OP' – OO'. The coordinates of OO' are (2/3, 2/3, 2/3). 82 The location of the pole of the compression axis on the stereographic projection is found by noting that it is parallel to the plane (111) and that cos θ = −1 10 , so that tanθ = 3. Figure 3.3.1 shows this location. It lies in the triangle ( 001) , ( 0 11) , ( 1 11) . In this position it activates the 1 11 101 slip system. The Schmid factor is: ( ) mS = ( −1 + 3 + 4 )(1 + 4 ) = 26 6 15 13 6 (3.3.4) The compression axis being parallel to the (111) twin plane remains unchanged in the twin. It is the same for the vector of the slip direction. The image P' in the twin of the normal to the slip plane is such that: p1′ = p2′ = p3′ = 1 27 −5 27 1 27 (3.3.5) Its position is shown in Fig. 3.3.1.The Schmid factor in that case is: mS′ = ( −1 + 15 + 4 )(1 + 4 ) = 26 6 27 15 13 6 (3.3.6) It is the same as the one calculated for the parent crystal (Eq. 3.3.4). Compression in the 134 direction activates as well the 1 11 101 slip sys- ( ) ( ) tem in the parent crystal and its image, the 151 101 slip system, in the twin. The deformation is therefore compatible. 83 Fig. 3.3.1. Positions of the [-1,-3,4] compression axis, P in the parent crystal and P' in the twin on the stereographic projection. Exercise 3.4 Twinning and martensitic transformation The Bogers-Burgers model constitutes an explanation of the martensitic transformation of a FCC structure into a BCC one. Consider in a FCC structure a tetrahedron PVQS of base PVQ the faces of which are {111} planes. Calculate, in terms of aFCC<112>, the slip of the successive planes parallel to PVQ in the QT direction perpendicular to the PV one (QT is a <112> direction) to transform the {111} QPS and QVS planes in {110} planes of the BCC structure. What additional shear is needed to obtain a BCC structure? Solution Determine the angle between the 111 and the 11 1 directions in the BCC structure (arctan( 2 2 ), almost 70°). Determine the shear needed to achieve the transformation of a FCC (111) plane in a BCC (110) plane ( 2 2 − 3 2). Calcu- 84 late, in terms of aFCC<112> the slip of the successive compact {111} plane piled over the PVQ plane to obtain this shear (almost 1/3(1/6); the summit S moves to S’; a 1/6 aFCC<112> slip would create a twin). Observe the way the atoms are piled up in successive planes parallel to the QPS’ or QVS’ planes and determine the additional slip needed to create a BCC structure (1/8 aBCC<110>). 3.3 Dislocations Dislocations are linear defects in crystals. The displacement of dislocations produces plastic deformation. Their interactions with various obstacles, periodic potential in the crystal itself, other dislocations, grain boundaries and free surfaces, foreign atoms, increase the force needed to move them; this explains how yield strengths are modified. Exercise 3.5 Geometrical properties of dislocations This exercise illustrates Sect. 3.3.3 of Volume I. a) Show that at a node where three dislocations meet the sum of the Burgers vectors is zero. Solution Draw a Burgers circuit around one of the dislocations; then displace it to surround the other two. The closure displacement should remain unchanged. Be careful about the orientation of the lines. b) Determine the Burgers vector of the dislocation in the bubble raft shown in Fig. 3.13 (p. 175) of Volume I. Solution Draw a Burgers circuit as shown in Fig. 3.5.1. Carefully count the same number of bubbles along each side. The Burgers vector is the closure defect. 85 Fig. 3.5.1. Determination of the Burgers vector b by drawing a Burgers circuit on the bubble raft. Plastic irreversible deformation stores in crystals geometrically necessary dislocations. Several examples are given in Volume I (Sect 3.3.3.5). c) Estimate the density of geometrically necessary dislocations in a thin wall cylinder the radius of which has been increased from r0 to r0 + ∆r by internal pressure. Solution The increase of the radius induces a decrease of both the wall thickness and of the length of the cylinder. As the only component of the stress tensor is the hoop stress σ θ , the ratio of the longitudinal strain to the circumferential strain is: εz − σ θ 2 1 = =− εθ σθ 2 (3.5.1) The volume remaining unchanged, ε r = −1 2 . With e the thickness of the wall and l its length: e l r = = 0 e0 l0 r 12 (3.5.2) 86 Let a Burgers closed circuit follow the inner and outer circumferences of the cylinder, these two being joined by two segments across the wall. The variation of the lengths of these two segments is of second order. The increase of the length Lb of the circuit is thus: ( ) ∆Lb = 4π r0 + ∆r − 4πr0 = 4π∆r (3.5.3) This corresponds to the introduction of n edge dislocations of Burgers vector b such that nb = 4π∆r. Note that there will be about the same number of positive and negative edge dislocations because there is no bending of the wall. The corresponding density of geometrically necessary dislocations of length l is: 4π∆r ρG = l b ( 2πrel ) = 2∆r 12 be0 r0 ( r0 + ∆r ) (3.5.4) d) Find the density of dislocations resulting from the rapid cooling of an Al-SiC composite from 400°C to room temperature, given that the coefficients of expansion of Al and SiC are 23.10-6 and 5.10-6 respectively and the diameter of the SiC particles is 10 microns. Solution The difference of the coefficients of expansion ∆α results during cooling in the punching of edge dislocation loops in the aluminium matrix. The relative volume variation from the injection of a number N of such loops of diameter φ, that of the SiC particles, is equal to the relative volume variation due to the difference of the dilatation coefficients: Nbπφ 2 = 6∆α∆T 4V (3.5.5) ∆T being the temperature difference. The density ρG of these geometrically necessary dislocations is then: ρG = N πφ 24∆α∆T = V bφ In the aluminium matrix ρG = 5.7x1013 m-2. Exercise 3.6 Mobile dislocations (3.5.6) 87 The displacement of dislocations produces deformations. The average plasS tic strain rate is given by the equation εɺ p = ∑ γɺ g ( m g ⊗ n g ) , symbol S deg noting the symmetric part of the tensor and g relates to the various slip systems (Volume I, Sect. 3.5.1.2, p.310, Eq. 3.186). The average strain is given ( by δε = b ⊗ n S ) (δA V ), δA being the area swept by the dislocation and V the volume (Volume I,Sect. 3.3.6.4, p.196, Eq. 3.32b). a) Find the deformation induced by the climb of an edge dislocation over an area δA. Solution Choose a coordinate system such that Ox3 is along the dislocation line and Ox2 perpendicular to the glide plane, so that the components of the Burgers vector are (b1, 0, b3) and those of the vector l are (0, 0, 1). The average deformation is given ( )( ) by the above equation written: δε ij = bi n j + bj ni δA 2V , where the coordinates of the Burgers vector reduce to b1, the coordinates of the normal n to the swept area reduce to n1 and V is the volume of the body. Thus the deformation tensor reduces to ( δε11 = b1 δA V ) (3.6.1) Contrary to dislocation glide, this climb of the dislocation produces a volume change. b) Calculate the mean velocity of dislocations in copper to reach a deformation rate of 10-4 sec-1 in a tensile test, assuming a reasonable mobile dislocation density. Solution Remember that the shear rate dγ/dt is given by the relation γɺ = ρ bv , ρ being the mobile dislocation density, b the Burgers vector and v the average velocity of dislocations. The tensile strain rate Eɺ may be approximately related to the shear rate γɺ in the grains by the Taylor factor, about 3 for FCC materials (Volume I §3.5.3.6 p.348). For copper the interatomic distance (from Table D1, Appendix D in Volume I) is 2.56x10-10 m. Take 1012 m-2 as a reasonable mobile dislocation density. The solution is then v = 1.2 micron/s. 88 c) What are the nucleation rates of dislocations Nɺ on each activated slip system in a copper single crystal of diameter D = 1 cm and length L = 5 cm tested in tension along the 001 , 011 , 111 or 123 directions with a strain rate of 10-4s-1? Solution Find in the first place which are the active slip systems in each case. This is achieved by calculating the Schmid factor, which is equal to n3b3/b, b being in the active slip direction. They are listed in Table 3.6.1. Table 3.6.1. Schmid factors multiplied by √6 for the possible slip systems according to the orientation of the tensile axis. The highest Schmid factors of the active slip systems are in bold characters. Tensile axis 001 Slip plane 111 ( ) 110 101 011 1 10 101 01 1 - - - 0 1 1 1 1 0 2/3 2/3 0 16/14 - 011 111 123 001 (11 1 ) - 011 111 123 001 (1 11) 011 111 123 001 011 111 123 ( 111) 0 1 1 12/14 0 0 0 0 0 2/3 2/3 4/14 - 10/14 1 6/14 - 1 0 0 0 0 2/3 2/3 1 1 1 0 0 0 0 4/14 - 0 - - 0 0 4/14 0 - 4/14 - 1 4/14 89 The areas swept by the dislocations are A0/n3, if A0 is the cross-section of the test piece. Thus δ Aɺ V = Nɺ A0 n3 A0 L (3.6.2) Aɺ Nɺ = kn3b3 V n3 L (3.6.3) The strain rate is: εɺ33 = kn3b3 where k is the number of active slip systems. Hence: Lεɺ Nɺ = 33 kb3 (3.6.4) The results are in Table 3.6.2. Table 3.6.2. Nucleation rates of dislocations on each active slip system Tensile axis 001 Number of active slip systems k 8 Dislocation nucleation rate Nɺ (s-1) 3.45x103 011 4 9,7x103 111 6 4x103 123 1 25.8x103 Exercise 3.7 Force on a dislocation The Peach and Köhler formula (Volume I, Sect. 3.3.6.3, p.194, Eq. 3.29): f = b ⋅ σ ∧ l , relates the force f on the dislocation to the Burgers ( ) vector b, the stress tensor and the unit vector along the dislocation line. a) Show that the glide force is equal to τb and calculate the climb force. 90 Solution Take a Cartesian coordinate system such that Ox1 lies along the dislocation line and Ox2 is perpendicular to the glide plane. The components of the force on the dislocation are found by application of the Peach and Köhler formula with l1 = 1, l2 = l3 = 0 and, the glide plane of the dislocation being perpendicular to Ox3, b2 = 0. So: f1 = 0 f 2 = b1σ 12 + b2σ 23 = τ b f = − (b σ + b σ ) 1 13 2 33 3 (3.7.1) where f3 is the glide force, perpendicular to the dislocation line. The climb force is equal to – b3σ33 as the edge component is to be considered. b) Determine the force on a prismatic loop of dislocation, that is a dislocation the Burgers vector of which is perpendicular to the plane of the loop. How will the dislocation loop change if gliding is active? Solution Let the Ox2 axis be in a direction perpendicular to the plane of the loop; b2 is the only component of the Burgers vector. The coordinates of the unit vector along the dislocation line are: sinθ, cosθ, 0. The Peach and Khöler formula gives: f1 = −bσ 22 cosθ f 2 = bσ 22 sin θ f = bσ cos θ − bσ sin θ 3 12 23 (3.7.2) It is easy to check that the force is perpendicular to the dislocation line. The glide force f3 is positive if tanθ < σ12/σ23, negative otherwise. Part of the dislocation loop will glide in the positive direction of Ox2, the other part in the negative direction; the prismatic loop will become an elongated loop, the two long branches attracting one another to create a dipole. Exercise 3.8 Jogs A jog on a dislocation line is a small segment of length b, the Burgers vector length, perpendicular to the glide plane. Jogs are created when dislocations 91 cross (Volume I, Sect. 3.3.8.1, p.212). Their energy is estimated to be (1/10)µb3, µ being the shear modulus. Is there an equilibrium density of jogs along a dislocation line? Solution The equilibrium corresponds to a negative free enthalpy G = H − TS. Let n be the number of jogs per unit length of dislocation. They can occupy N = 1/b positions along the line. Then the enthalpy per unit length is given by: H=n µ b3 (3.8.1) 10 with µ = 45,000 MPa and b = 2.5x10-10, H = (n/N) 2.8x10-10 J. The entropy S is estimated from the number of combinations of N positions n to n: n n n n S = −kN log + 1 − log 1 − N N N N (3.8.2) where k is the Boltzman constant (1.38x10-23 J/K). S is maximum for n/N = ½; its value is kNlog2. With b = 2.5x10-10 this is equal to 3.8x10-12 J/K per unit line of dislocation. Now: n n n G=H −TS =N 2.8 × 10−10 + kT log N N N n n +1 − log 1 − N N (3.8.3) Stability is achieved if the quantity in bracket, which is negative, is smaller than −2.8x10-10 /kT = − (2x1013/T) (n/N). At 300 K this is − 6.7x1010(n/N). This condition is not achieved (except for extremely small values of the jog concentration). The conclusion is that jogs are unstable defects. Exercise 3.9 Local change of the volume near a dislocation Introduction of dislocations in a material produces but a very small increase of the overall volume. There are however local changes in the neighbourhood of the dislocation. This constitutes a sound foundation for the assump- 92 tion that plastic deformation takes place at constant volume (Volume I, Sect. 3.3.7.3, p.200). Find the change in volume in the neighbourhood of a dislocation. Draw the equal value contours. Solution Around a screw dislocation the only component of the strain tensor which is different from 0 is ε θz (Volume I, Sect. 3.3.7.1, p.197). There is no local volume variation in the neighbourhood of a screw dislocation. This is not so near an edge dislocation as: ε rr = ε θθ = b 1− 2ν sin θ 4π 1 − ν r (3.9.1) where b is the Burgers vector, ν the Poisson ratio, θ the polar angle and r the distance to the core. Then the volume variation is twice this value. The contours of constant volume variation obey equations such that r = Constant x sinθ ; they are circles in cylindrical coordinates, passing through the dislocation core and with their centre on the Ox2 coordinate axis. At each point where there is a positive volume variation corresponds an opposite point where it is negative with the same absolute value. Thus the overall volume does not change. Exercise 3.10 Interactions between dislocations A dislocation has a stress field which creates a force on neighbouring dislocations according to the Peach and Köhler equation (Volume I, Sect. 3.3.8, p.210). a) Find the equilibrium positions of two parallel edge dislocations on parallel slip planes. Solution Choose (Ox1, Ox3) as the slip plane of the first dislocation with Burgers vector b1, the dislocation line being oriented in the Ox3 direction. Use the equations of the stress field of an edge dislocation (Volume I, Sect. 3.3.7.2, p.200, Eq. 3.43a, noting that there is an exponent 2 missing in the denominator) and the Peach and Köhler equation, which yields: 93 f1 = − µ (b1 ⋅ b2 ) ( ) 8π 1 − ν h sin 4θ (3.10.1) where f1 is the glide force per unit length on the second dislocation with Burgers vector b2, µ is the shear modulus and ν the Poisson ratio, h is the distance between the slip planes and sinθ the ratio between h and the distance between the two dislocations. When the Burgers vectors of the dislocations have the same sign, they attract each other when -π/4 < θ < π/4, and repel each other otherwise; the stable position is found for θ = 0. The two dislocations, which must remain in their own slip plane, form the nucleus of a tilt wall. When the dislocations have opposite signs, the stable position is found for θ = π/4. The two dislocations form a dipole. b) What is the resolved shear stress τ needed to destroy a dipole of dislocations in copper assuming that the distance between the slip planes is equal to 4b. Solution To destroy the dipole the applied force τb on the second dislocation must exceed the maximum attraction force f1 when the distance between the two dislocations increases. This is found for an angle θ = 3π/8. Hence, from Eq. 10.1: τb = µ b2 8π (1− ν )4b (3.10.2) For copper, with µ = 45,000 MPa, ν = 0.33 and b = 2.56x10-10 m, τ = 670 MPa. When the two dislocations are in the same slip plane (h = 0) the stress of the first dislocation on this plane is (Volume I, Sect. 3.3.7, Eq. 3.35b and 3.43a): µ ( b1 ⋅ b2 ) for a screw σ 23 = 2πx1 σ = µ ( b1 ⋅ b2 ) for an edge 12 2π (1 −ν ) x1 c) Calculate the climb force on the second dislocation Solution The Peach and Köhler equation yields: (3.10.3) 94 f2 = µb2 sin 2 θ ( 2cos 2 θ + 1) 2π (1 −ν ) h (3.10.4) The maximum value is found for θ = π/4 and: f 2max = µb 2 2π (1 −ν ) h (3.10.5) Exercise 3.11 Interaction of a dislocation with an interface The stress field of a dislocation depends on the elastic moduli of the material in which it lies. On the other side of the interface, they are different; thus the stress field is also different there. The problem is to insure the continuity of the stress and strain fields at the interface. This is achieved by the introduction of image dislocations (Volume I, Sect. 3.3.8.4, p.216). Find the interaction of a screw dislocation in an isotropic material of shear modulus µ1 parallel to an interface with an isotropic material of shear modulus µ2. Solution Chose coordinate axes such that Ox1 is perpendicular to the interface, and Ox3 along the dislocation line. The displacements and the stresses, which the dislocation would create on the location of the interface, distant of d from the dislocation, in an infinite medium would be (Volume I, Sect. 3.3.7.1, p.197): x b arctan 2 2π d µ1b x2 σ 13 = − 2π d 2 + x22 u3 = (3.11.1) µ1 being the shear modulus of the material in which lies the dislocation. Notice that the displacement is independent of the shear moduli and, everywhere, is not influenced by the presence of the interface. The image dislocations must be located either at the same place than the real dislocation or at a symmetric place with respect to the interface. This is because, if there are n image dislocations with Burgers vectors ±b , the equality, whatever x2, 95 of the second equation (3.11.1) with the sum of the stresses due to the images, provides n equations; all the n distances di can be calculated. Now, the equality of the first equation (3.11.1) with the sum of the displacements due to the images is respected for one value only of x2. Thus, it is needed that di = ± d . Designate by βb the Burgers vector of the symmetrical image dislocation assumed to be in the material of modulus µ1, and by γb the Burgers vector of the image dislocation coinciding with the real dislocation and assumed to be in the material of modulus µ2. Matching displacement and stress on both sides of the interface µ − µ1 2µ1 shows that β = 2 and γ = . µ2 + µ1 µ2 + µ1 If the interface is a free surface, µ2 = 0 and β = −1, so that the corresponding image dislocation attracts the real dislocation. It feels an attractive force towards the free surface. In the general case, the force is attractive if µ2 < µ1, repulsive otherwise. The presence of an oxide layer can prevent the escape of dislocations and harden the crystal. Exercise 3.12 Dislocation loop The energy of a straight dislocation is of the order of 0.5µb2 per unit length, µ being the shear modulus and b the Burgers vector (Volume I, Sect. 3.3.7.5, p.202). For a dislocation loop, part of this energy disappears because each line segment along the line faces an opposite one the Burgers vector of which is of opposite sign. The energy of a dislocation loop was calculated 8Rc µ b2 Rc 1 1 (Nabarro 1967, p.75, formula 2.53 a). It is 1+ log 2 , 2 2 1 − ν e r0 where Rc is the radius of the loop, ν the Poisson ratio and r0 the core radius of the order of 10-9 m. a) Find the shear stress ratio τ/µ required to create ex nihilo a dislocation loop in copper. Solution The work W of the shear stress to expand the loop must be equal to the energy E of the loop: 96 2 2 R τ b 1 1 1 b Rc 8 Rc W = π c = 1 + log 2 a0 µ a0 2 2 1 −ν a0 a0 e a0 (3.12.1) With a0 ≃ b and 8 e 2 ≃ 1 , τ/µ reaches a maximum as a function of Rc/a0 for Rc∗ a0 ≃ 2.7 . Then τ/µ = 0.2. The corresponding energy is 3.4x10-19 J. It is too large with respect to the thermal energy kT for thermal fluctuations to generate the loop. b) Find the force F needed to punch a 10 µm in diameter prismatic dislocation loop in copper. Solution If the punch creates a prismatic loop, the displacement of the force is equal to the Burgers vector b and the work done is Fb. This must be equal to the energy of the loop for which the dislocation is of edge character all around the periphery. Hence: Fb = µb2 8R Rc log 2 c 4 (1 −ν ) e a0 (3.12.2) The result is F = 2.1x10-4 N. We neglected the energy of the area 2πRcb created by the punch. With γ s ≃ 0.1µ b , the extra force needed is 0.36x10-4 N. c) Find the critical velocity v for a particle of a specific mass ρ of 3,000 kg/m3 to punch a dislocation loop of 10 µm in diameter in copper. Solution ( )( ) Equating the kinetic energy of the particle 1 2 ρ 4π 3 Rc3v 2 to the energy of the prismatic loop of radius Rc, the critical velocity is found to be 0.26 m/s. Exercise 3.13 Peierls force Dislocations tend to lay along Peierls troughs of potential aligned with the most dense directions of the crystal structure. A force is needed to move a dislocation from one trough to the next, by the displacement of kinks. This is the lattice friction (Volume I, Sect. 3.4.2.1, p.225). 97 Calculate the Peierls force τP with respect to the shear modulus for a partial Shockley dislocation in a FCC structure with a Poisson ratio of 0.3, given that 2π a 2µ τP = exp − , a being the distance between the atomic slip planes. 1− ν 1− ν b Solution From the ratio a/b for a partial Shockley dislocation, 1 5x10-3. 2 , the result is τP/µ = Exercise 3.14 Strain-hardening Strain-hardening results from the multiplication of dislocations during plastic deformation. The Taylor formula, τ c = αµ b ρD , outlines the effect of the dislocation density ρD on the critical shear stress τc. (It is reminded that α is a parameter of the order of 1/3, µ is the shear modulus and b the Burgers vector). This formula results from the density of obstacles to the displacement of gliding dislocations offered by the forest of dislocations (Volume I, Sect. 3.4.2.2, p. 229). The dislocation density includes both the part due to the multiplication from Frank-Read sources and the part due to geometrically necessary dislocations. a) Estimate the amount of hardening of an extruded aluminium composite material due to 10 vol% of rigid inclusions 1 micron in diameter, the drawing ratio being 10%. Solution The extrusion introduces geometrically necessary dislocation loops punched from the rigid inclusions. We first need to determine the displacements to be compensated by dislocations. Let D0 be the initial diameter of the aluminium bar. After extrusion it becomes D and the volume remaining the same in plastic deformation the elongation is such that: LD 2 =1 L0 D02 (3.14.1) Let ∆D be the decrease of the diameter of the bar after extrusion and ∆L its increase in length. 98 D ∆D D = 1− D 0 0 2 ∆L = D0 − 1 L 0 D (3.14.2) If the volume fraction is fV, the mean distance d between the inclusions volume πφ /6 is: 3 13 π d = 6 φ (3.14.3) f V1 3 The elongation 2u in the extrusion direction of the edges of length d of a cube containing one inclusion is given by: D 2 L0 2u = ∆L = L0 0 − 1 d D (3.14.4) The area decrease of the cube containing one inclusion is such that: ( 2 πD02 2 2 = πD0 − π D0 − ∆D d − d − 2u r 4 4 4d 2 ( ) 2 ) (3.14.5) Let Nl be the number of geometrically necessary dislocation loops: D N l b = 2 u + ur = d D0 ( ) D 3 π 1 3 φ D D 3 0 − 1 = 0 − 1 (3.14.6) D 6 f V1 3 D0 D The density of geometrically necessary dislocations is: ρG = N l πφ d3 13 ( ) = 36π f 23 V 3 1 D D0 − 1 φ b D0 D (3.14.7) With the above values for the parameters ρG = 1.24x1015 m-2. The critical shear stress being estimated by the Taylor formula, it is found that τc ≈ 85 MPa. b) We wish to determine the efficiency of the work of deformation. 99 Assume that the strain-hardening curve for copper can be represented by σ = 120ε p0.25 in MPa, εp being the plastic deformation. Find the relation between the density of dislocations ρD and the strain, given that its value for εp = 0.4 is 1015 per m2. Solution Equating the ratio of stresses for a deformation of εp and for a deformation of 0.4 expressed in term of the corresponding dislocation densities, it is found that ρD = 1.58x1015 εp0.5 m-2. c) Compare the work done in plastic deformation and the stored energy. Solution The work done in plastic deformation is: Wp = ∫ εp 0 σ dε p = ∫ εp 0 120x106 ε p0.25dε p = 9.6x107 ε 1.25 J / m3 p (3.14.8) The stored energy Ws is that of the dislocations the total length of which is ρD m/m3. Hence: Ws = 0.5µ b2 ρD = 0.5µ b2 1.58x1015 ε p0.5 (3.14.9) For copper, with µ = 45,000 MPa and b = 2.56x10-10 m, ED = 2.3x106 εp0.5 J/m3. Hence the efficiency ratio: 0.5 Ws 2.3x106 ε p 0.024 = = 0.75 Wp 9.7x107 ε 1.25 εp p (3.14.10) This ratio is for instance equal to 0.027 for a deformation of 0.4. This small value comes from the heat dissipated in plastic deformation. Exercise 3.15 Depressed modulus of elasticity A small enough shear stress bends dislocation segments held by anchorage points without breaking them. As the segments sweep slip plane area, this produces a deformation. It disappears upon release of the stress as the dislocation segments, because of the line tension, jump back to their original po- 100 sition. The apparent modulus of elasticity, the depressed modulus, is thus smaller than the true one (Volume I, Sect. 3.4.2.2, p.229). a) Find the change in the shear modulus of elasticity µ resulting from the reversible movement of dislocations in a FCC material in which the dislocation density is 1012 m-2. Solution The relative change of the shear modulus is related to the relative increase of the shear strain δγ: δµ µ =− δγ (3.15.1) γ where δγ is related to the total area NδA swept by the N dislocation segments in volume V by δγ = NbδA V (3.15.2) If R is the radius of curvature of the segments of dislocation when the shear stress is τ, δA is given by: 12 2 L RL L δA = R arcsin − 1− 2 2R 2R 2 = 1 L3 16 R (3.15.3) where L is the average length of the dislocation segments. If the anchorage points of the dislocations are dislocation trees, L2 is equal to the dislocation density ρD: ρD = NL V (3.15.4) In addition, we have R= 0.5µ b2 τb (3.15.5) τ being the average shear stress on the various active slip planes (Volume I, Eq. 3.50 p. 204). From these various equations is deduced: 101 δµ µ =− 1 8 (3.15.6) b) Explain why the modulus depression should be greater after polygonisation. Solution Let d be the size of a subgrain, assumed to be cubic, and n the total number of dislocation segments of length l between anchoring points. There are V/d3 subgrains, 3V/d3 walls and (3V/d3)(d2/l2) anchoring points. Thus: ( ρ D )P = nl 3 = V dl (3.15.7) Hence δγ P = nb 3bl τ b 3lγ P 9γ δ AP = = = 2 P V 16d 0.5µ b 2 8d 8d ( ρ D ) P (3.15.8) index P referring to the polygonised material. ( δµ µ )P δµ µ = 9 d 2 ( ρD )P (3.15.9) Figure 3.15.1 shows an abacus giving this ratio as a function of the subgrain size and of the dislocation density. Exercise 3.16 Kinematic hardening Strain-hardening corresponds to an increase of the dislocation density. Dislocations form clusters and then cells within which the dislocation density is much smaller than in the walls. The behaviour then corresponds to kinematic hardening. The composite model of Mughrabi explains this (Volume I, Sect. 3.4.2.2, p.237). It considers the material to be a composite of two materials, the cells and the walls. The shear stress in each is given by the Taylor formula (Eq.3.100). In copper subjected to cyclic work-hardening the volume fraction of dislocation walls is estimated to be 5%, within which the dislocation density would be about 1016 m-2, against 1013 m-2 in the interior of the cells. Draw the cyclic hardening 102 curve, assuming absence of isotropic hardening, and estimate the amount of stored energy when the total strain amplitude is 0.01 after load removal (for copper µ = 45,000 MPa, ν = 0.33 and b = 2.56x10-10 m). Fig. 3.15.1. Dislocation density and subgrain size yielding a certain ratio of the depressed modulus of a polygonised material to that of the material with a uniform distribution of dislocations. Solution Apply the composite model of Mughrabi in which the shear stress is the average of the one of the cell interior τc and the one of the walls τw, each one given by the Taylor formula. 103 τ = (1 3) µb 1013 = 4 MPa c 16 τ w = (1 3) µb 10 = 128 MPa (3.16.1) The shear stress of the composite is the average: τ = 0.95τ c + 0.05τ w = 10 MPa (3.16.2) Figure 3.16.1 shows the behaviour. For the stored energy is the energy of the dislocations: 0.5µb2 J/m: WS = 0.95 ( ρ D )c + 0.05 ( ρD ) w µ b 2 ≃ 8.4x105 Jm −3 (3.16.3) This stored energy can be compared to the work W = τγ of the shear stress, which is about 1.4x105 J/m3. The difference is the heat produced. Exercise 3.17 Interaction of solute atoms with dislocations Solute atoms interact with dislocations owing to two effects: size effect and modulus effect. The first one comes from the local change of volume and distortion, which a solute atom creates, interacting with the stress field of the dislocations. The second one is of second order (Volume I, Sect. 3.4.3.1, p.249). a) Calculate the approximate interaction energy of an atom of zinc with an edge dislocation in copper (for copper µ = 45,000 MPa, ν = 0.33 and b = 2.56x10-10 m, and for zinc the interatomic distance is 2.67x10-10 m). Solution A zinc atom in substitution in the copper crystal has a larger volume than the solvent. It interacts with the hydrostatic component of the dislocation stress field. Neglect any modulus effect. The hydrostatic stress in the neighbourhood of an edge dislocation (Volume I, Eq. 3.110) is, r and θ being the polar coordinates: σm = − 1+ ν µ b sin θ 1− ν 3π r (3.17.1) 104 Fig. 3.16.1. Cyclic behaviour of copper containing dislocation cells according to the model of Mughrabi. The relative size increase due to a zinc atom is η = ∆a0/a0 = 0.043. (a0 = b/2 is the radius of the copper atom). The volume increase is: ∆V = 4πa03η = 1.13x10-30. The interaction energy, for θ = 0, is: U = −σ m ∆V = 2.74x10−30 1 J r (3.17.2) At a distance r = 2b, U = 5.3x10-21 J (0.03 eV) per solute atom. b) Find the approximate interaction energy of a carbon atom with a dislocation in α iron. 105 Solution A carbon atom occupies the middle of the edge of the cubic iron cell. The length of this edge is a = 2.866x10-10 m. The insertion of the carbon atom increases it to (2.482 + 1.544)x10-10 m. The resulting strain is log (4.026/2.866) = 0.339. The elongations in the two other perpendicular directions being zero, the relative volume increase is 0.339. The volume increase is 0.339 a3 = 6x10-30 m3. The interaction energy U = − σm∆V with an edge dislocation, calculated at a distance 2b, is deduced from Eq. 3.17.1 applied to iron. It yields U = 6x10-20 J (0.4 eV) per atom. There is no volume interaction with a screw dislocation the stress field of which being of zero hydrostatic stress. However, the distortion introduced by the carbon atom interacts with the shear components of the stress field. We calculate the coordinates of the displacement vector u in the coordinates of the screw dislocation the Burgers vector b of which lies on the Ox3 axis (Fig. 3.17.1). The vector u is collinear with A/A and u = 1.16x10-10 m. Hence: 2 1 u= e1 + e3 u 3 3 (3.17.3) The normal stress in the direction of u is: σn = 2 2 2 2 µ b x2 2 µ b sin θ σ 13 = − =− 3 3 2π x12 + x2 2 3 π r The coordinates of the carbon atom in position C1 are: (3.17.4) x1 = 0, x2 = − 3a 2 2 , x3 = 0 (Fig.3.17.1). In order to calculate the interaction energy U, we need to know the intensity of the force being displaced by u. It is equal to the average normal stress over a small area s2 multiplied by this same area. As σn varies as 1/x2, its average value over a small area remains equal to σn. It is therefore difficult to decide which value to choose for s. If we arbitrarily decide s2 to be equal to the cross-section of the carbon atom, we get: U≃ 4 u 2 µb rC = 2x10−20 J ( 0.13eV ) 9 a (3.17.5) This is a stronger interaction energy than for edge dislocations. Note that the interaction energy of the dislocation with the carbon atom situated in the centre of the cell face (position C2 in Fig. 3.17.1) is zero because in that case x2 = 0 so that σ13 = 0. This is also found to be the case for the carbon atom situated 106 in position C3 when the calculation of the stress in the direction of the displacement shows that it is zero. Fig. 3.17.1. Positions of carbon atoms inserted in a BCC cell and direction A of the displacement due to the insertion. 107 c) Find the carbon concentration needed to saturate all the Cottrell’s atmospheres in α Fe containing a dislocation density of 1014m-2, assuming that the saturation means that all the sites C1 closest to the dislocation line are occupied. Solution There are four carbon atoms in position C1 per cell along the dislocation line. The number of cells along the dislocation line per cubic meter is equal to ρ D (a 3) . 3 The number of cells per cubic metre is equal to 1/a and there are 2 Fe atoms per cell. For the saturation of the Cottrell atmosphere the atomic concentration of C atoms is equal to 2 3ρD a 2 = 2.8x10−5 d) What is the shear stress needed for an edge dislocation to escape from a saturated carbon atmosphere in α Fe? Solution Assume that the carbon atoms in the Cottrell atmosphere lie at a distance a below the dislocation line. The evolution of the interaction energy U per unit length when the dislocation glides away from the atmosphere is U =− 1 1+ ν a µ∆V 2 3π 1 − ν x1 + a 2 (3.17.6) where ∆V is the increase of the atomic volume in inserting a carbon atom and x1 is the distance away from the original position of the dislocation. The stress to pull 2 1 +ν ax1 away the dislocation is τ = µ∆V . It goes through a maximum 2 2 3π 1 −ν ( x + a2 ) 1 3 1 + ν µ∆V , that is 6,100 MPa! It is a very high value; the escape of a 8π 1 − ν ba 2 loop would be easier. Following Nabarro (Nabarro 1967) in the calculation of such a loop escape, the loop is represented by two “jogs” of length x1 and a free straight line of length w. τe = The various components of the energy UL are the interaction with the atmosphere, the increase of the line tension and the work done. The equation yielding the extremums as a function of the distance x1 is 108 ∂U L 2 Awax1 µb 2 = + − τ bw = 0 ∂x1 b ( x 2 + a 2 ) 2 π 1 (3.17.7) 1 1 +ν A= µ b∆V 3π 1 −ν ⇒ Setting x1 = atanφ, the energy goes through a minimum followed by a maxi- 3 3 A . (τe corresponds to φ = π/6 and 8 a 2 b2 τ c ≃ 10,000 MPa . For plastic deformation to be possible new unanchored dislocations must be created. Minimising with respect to w the activation energy, UL*, for the escape of the loop for values of φ close to π/6 yields: mum if w is large and τ ≤ τ e = w* = * L U = µb 2π τ e − τ ( 4 3π ) µ b a (1− τ τ ) (3.17.8) 12 2 e e) Determine the precise trajectory of carbon atoms diffusing towards a screw dislocation in α Fe (c0 = 0.001). Deduce the kinetics of the creation of the Cottrell atmosphere. Achieve the computation for temperatures of 25, 100 and 200°C, given the diffusion coefficient of carbon D = 2x10-6 exp (-29.93/RT) m2/s (activation energy in kcal/mole). Solution If U is the interaction energy of the carbon atoms, their velocity is v = D(−gradU)/kT . From (3.17.4): 2 3π 2 = 3π v= µb D πrC 2u ( e1 sin 2θ − e2 cos 2θ ) r2 kT µb 2 D πrC u ( er sin θ − eθ cosθ ) r2 kT (3.17.9) (It is reminded that rC is the radius of the carbon atom 0.772x10-10 m and u = 1.16x10-10 m. It follows that the trajectories are circles (Fig.3.17.2). They bring the carbon atoms to the dislocation atmosphere. 109 Fig. 3.17.2. Trajectories of carbon atoms towards the dislocation. From (3.17.9) the velocity along the circle of radius r0 is: v = −r0 ( )= d 2θ dt 2 µb 2 D 1 D πrC u = 4.15x10−29 2 2 3π r kT r kT (3.17.10) With r = 2r0cosθ, the integration of (3.17.10) yields the time t(r0) for a carbon atom to travel along the entire circle: () t r0 = 3 1 kT r0 kT = 2x1030 r03 2 µ b D D πr u 2 C 12π 2 (3.17.11) During this time an area S = πr02 has been emptied of carbon atoms. An area increment dS is swept during time dt: dS = πdr02 = 2π 0.5x10−30 D 3 kT 23 () t (r ) dt r0 13 (3.17.12) 0 This brings a number dN of carbon atoms to a unit length of dislocation equal to: 110 2c 2π 0.5x10−30 D dN = 30 kT a 3 23 () t (r ) dt r0 13 (3.17.13) 0 The number of carbon atoms needed to saturate the unit length of dislocation is: N= 4 3 a (3.17.14) Integration of (3.17.13) yields the time t needed to create a Cottrell atmosphere: t= 7x10−22 kT c03 2 D (3.17.15) The results are in Table 3.17.1. Table 3.17.1. Approximate times for the creation of a Cottrell atmosphere Temperature (K) 300 373 473 Time (s) (hours) (min) (days) 1.4x107 3830 160 2160 36 6 Exercise 3.18 Piobert-Lüders band When dislocations are anchored by Cottrell atmosphere, for plastic deformation to occur fresh dislocations must be created. This avalanche in a localised zone of concentrated stress results in the formation of bands of localised strain, Piobert-Lüders bands (Volume I, Sect. 3.4.3.1, p.256, Fig. 3.80) Find the angle made by a Piobert-Lüders band with the tensile axis in a thin sheet. Solution The band, which deforms plastically, is in plane stress and at the same time in plane strain in its own direction. If axis Ox'1, making an angle φ with the axis Ox1, 1 is along the band and Ox'2 perpendicular to it, Eε11′ = 0 = σ 11′ − σ 22 ′ . Figure 2 111 3.18.1 shows, on the Mohr circle, the positions which these stresses must therefore occupy. So, cos2φ = 1/3. Fig. 3.18.1. Piobert-Lüders band in a thin sheet and Mohr circle. 112 Exercise 3.19 GP zone hardening Guinier-Preston (GP) zones are formed during the first stage of precipitation of solute atoms. They are clusters of solutes which remain on the lattice sites of the solvent atoms. A classical example is that of aluminium copper alloys. (Volume I, Sect. 3.4.3.2, p.260) Figure 3.89 of Volume I (p. 264) shows a GP zone in a 1.7at.% Al-Cu alloy, which Karlik and Jouffrey (Karlik M and Jouffrey B (1996) Guinier-Preston platelets interaction with dislocations. Study at the atomic level. J Phys III 6:825-829) observed by high resolution transmission electron microscopy. They demonstrated that it was a 100% copper monolayer disk on {100} planes of the aluminium crystallographic lattice. The figure shows that a dislocation sheared the GP zone. Evaluate the hardening resulting from the precipitation of the GP zones in the 1.7at% Cu aluminium alloy. Solution The dislocations glide in {111} planes with [110] Burgers vector. Two kinds of interactions with the GP zones can be considered: elastic interaction and chemical interaction. The first one can be considered as the interaction of a gliding dislocation with the solute atoms disk represented by a dislocation loop of Burgers vector ηb, η being the misfit atomic parameter equal to 0.11 for the smaller Cu atoms in Al. A rough approximation consists in applying the equation giving the shear stress needed for crossing the forest of dislocations: τc = 1 µηb ρD 4 (3.19.1) The dislocation density is the total length of dislocations in a unit volume: ρ D = N disk 2πr (3.19.2) Ndisk being the number of disks of radius r in a unit volume. If the atomic concentration of precipitated copper atoms is c and a the parameter of FCC aluminium: c Hence: 4 2πr 2 = N disk 2 3 a a (3.19.3) 113 ρD = 4 c ar (3.19.4) Combining (3.19.1) and (3.19.4), with for aluminium µ = 2.6x1010, a = 4x10-10, b = a/21/2 = 2.8x10-10, η = 0.11 and r = 5x10-9, as measured on Fig. 3.89 of Volume I: τc = 1 µη b c = 2.8x108 c 2 ar (3.19.5) If c = 0.017, the critical shear stress is 36 MPa, a value of a factor 3 too low with respect to measurement of the yield strength. Moreover, the concentration of precipitated carbon is lower than 0.017. Elastic interaction is insufficient to account for the critical shear stress. If the slip plane of the dislocation is 1 11 , the Burgers vectors are 011 , ( ) ( ) which is parallel to the 100 plane of the GP zone, 10 1 and 110 ,which are inclined to this plane. When a 011 dislocation penetrates in a GP zone, it replaces a Cu−Cu bond by a Al−Al one. The difference of the bond energies is small. The GP zone offers little resistance to the crossing of such dislocations. When a 10 1 or a 110 dislocation crosses a GP zone, one Cu−Cu bond and one Al−Al bond are replaced by two Al−Cu bonds. This involves an energy ∆H = 2hAlCu − hCuCu − hAlAl . This is evaluated by Karlik and Jouffrey to be 1.72 eV = 2.7x10-19 J. Per unit area of the glide plane, the disks which cross it are in a volume 4r 2 . The number of disks per unit volume being given by (3.19.3) and as there are 2/3 effective disks, the number of effective disks crossing the glide plane is 16c 3 2 πar . A gliding dislocation zigzags between the crossing points with a small "amplitude" x for a "period" y, xy being the area per crossing point: xy = 3 2πar 16c (3.19.6) The average length of a crossed disk is πr/2. So, the average pinning energy is πr H= ∆H . 2b The gain of energy owing to the pinning action of the crossed GP zones is: 114 E1 = H πr = ∆H y 2by (3.19.7) The increase of the line energy is: 12 E2 = 2 2 µb 2 ( x + y ) − y 2 y = µb2 x 4 y 2 (3.19.8) Minimising E2 – E1 yields the zigzag shape: xc3 = π (xy )r∆H (3.19.9) 2µ b3 2 3 c y = ( ) 2 xy µ b3 (3.19.10) πr∆H It can be checked that xc/yc = 0.024 is indeed a small quantity. According to the hypothesis of Friedel (1964), escaping from a pinning point the dislocation sweeps the area (xy), so that: τ cb ( xy ) = yc ( E1 − 2 E2 ) = πr ∆H 4b (3.19.11) Using (3.19.6) the critical shear stress is: τc = 4c∆H 3 2ab2 = 140 MPa (3.19.12) Exercise 3.20 Irradiation hardening by cavities The vacancies produced in stainless steel by irradiation condense to create cavities. The result is swelling as well as hardening. Estimate the yield strength increase by a volume fraction fv of cavities of radius r. Solution 115 The hardening when a cavity is crossed by a dislocation comes from the elimination of a segment of this dislocation and hence of the corresponding energy. In reality the problem is more complicated as the stress concentration around the cavity plays also a role. We look only for the hardening coming from the annihilation. Take the origin of the coordinates at the centre of the cavity (Fig.3.20.1), y being in the direction of the displacement of the dislocation. The coordinates of a point on the surface of the cavity are rcosθ and rsinθ. Let L be the mean distance between cavities centres. With t the line tension, the radius of curvature of the dislocation is t/τb = (µb2/2)/τb, µ being the shear modulus, b the Burgers vector and τ the shear stress. Thus the angle ϕ of the dislocation with the x axis where it meets the cavity is such that: ( sin ϕ = L − 2r cos θ Fig.3.20.1. Dislocation crossing a cavity )µτb (3.20.1) 116 The work done by the force on the dislocation where it meets the surface of the cavity when it moves forward a distance d(rsinθ) is equal to the variation of the line energy (µb2/2)d(rcosθ). Hence: t cos ϕ sin θ − t sin ϕ cos θ = −t sin θ (3.20.2) This equation shows that the angle ϕ is equal to 2θ. The dislocation arc between the cavities becomes unstable (the Frank-Read condition is met) for θ = π/4. From (3.20.1), the evolution of the angle θ as a function of the shear stress is such that: τ µb = sin 2θ L − 2r cos θ (3.20.3) This shows that the dislocation finds no resistance until it crosses the cavity by the middle and that the shear stress must be increased until θ = π/4. At that point: τ c µb = 1 (3.20.4) L−r 2 It remains to evaluate L. According to the Friedel hypothesis the area swept by a dislocation until it meets another obstacle is equal to the average area per obstacle on the glide plane s. This is related to the volume fraction fv: f v = fs = πr 2 s (3.20.5) The Friedel hypothesis yields: s= 2 πr 2 πRc π µ b = = fv 2 2 2τ c 2 (3.20.6) Rearranging, it is found that: fv b τc = µ 2 2r (3.20.7) Note that the volume fraction has a limit, 0.9, when the cavities are close packed. 117 The cavities are not all crossed by the equatorial plane, so that the radius should be replaced by the average value 2r/π. Assuming then for instance r/b = 10 and fv = 0.4, the critical shear stress is equal to 0.035µ. 3.4 Composite materials Exercise 3.21 Composite wire A 2 mm diameter steel wire, coated with aluminium to give an external diameter of 2.5 mm, is subjected to traction. Which metal yields first? What is the yield strength of the composite wire? What is its modulus of elasticity? (The Young moduli E are 210,000 MPa and 70,000 MPa and the yield strength 200 MPa and 70 MPa respectively for the steel and for aluminium; they both have the same Poisson ratio). Solution Considering that the tensile strain ε of the two metals is the same, the stress in the steel part is equal to 210,000 ε MPa and to 70,000 ε MPa in the aluminium part. Thus the yield strength is reached for ε = 1/1,000 for the aluminium part and for ε = 1/1,050 for the steel one, which yields first. The cross-section of the steel part is π mm2 and that of the aluminium one is 0.562π mm2. The average stress in the composite wire for a strain ε is: Ecompε = 1 0.562 200,000ε + 70,000ε = 153, 200ε 2 2.5 4 2.52 4 (3.21.1) Ecomp being the Young modulus of the composite wire. When the yield strength is reached in the steel part, the stress in the composite is 153,200/1050 = 146 MPa. To reach the yield strength in the aluminium part, a further strain is needed: 1/1,000 – 1/1,050. It corresponds to a stress increase equal to: 1 0.562 1 ∆σ = − 2 70,000 = 1.5 MPa 1,000 1,050 2.5 4 (3.21.2) Assuming that there is no strain-hardening in the steel, the yield strength of the composite wire is 147.5 MPa. 118 Exercise 3.22 Fibre reinforced composites The fracture strength of a fibre reinforced composite is the larger of its strength at fracture of the reinforcing fibres or of the remaining strength of the matrix when the fibres are broken. If the volume fraction of reinforcing fibres is too low, the fracture strength of the composite is lower than the one of the matrix with no reinforcement (Volume I, Sect. 3.4.6.2, p.299). The shear lag model (Volume I, Sect. 3.4.6.3, p.300) shows that the tensile stress along the length of the fibres is maximum in their middle. The load transfer is the higher when the matrix is fully plastified around the fibres. The tensile stress in the fibres then increases linearly up to the maximum at the middle, where its ratio to the shear yield strength of the matrix is equal to the ratio of the fibres length to their radius. a) A composite material consists of 40% by volume of high strength steel wires (Rp = 1,050 MPa, A% = 0.15) aligned in an aluminium matrix for which the strain hardening curve is σ = 270 ε0.2 MPa. Find the fracture strength of the composite. Solution Neglecting the strain-hardening of the high resistance steel before fracture, the tensile stress σ in the composite is: σ = 0.4 × 1,050 + 0.6 × 270ε 0.2 MPa (3.22.1) When the steel wires fracture the tensile strain is 0.15 and the corresponding stress, according to Eq. 3.22.1, is 531 MPa. Increasing further the strain, the stress comes from the aluminium matrix only: σ = 0.6 × 270ε 0.2 MPa (3.22.2) The Considere condition gives the strain at maximum load: 0.2. According to Eq. 3.22.2 the corresponding stress is 117 MPa. The fracture strength of the composite corresponds to the breaking of the steel wires: 531 MPa. b) A boron fibres-aluminium composite has the following characteristics: unidirectional reinforcement; fibres: volume fraction 50%, length 100 mm, diameter 0.1 mm, breaking strength 3,000 MPa, Young modulus 415,000 MPa; matrix: yield strength 70 MPa, flow strength upon fibres breaking 70 MPa, breaking strength 200 MPa. Calculate the critical length for load transfer and the critical volume fraction. 119 Solution According to the shear lag model the critical length lcf of the reinforcing fibres of radius rf is such that the maximum tensile stress in the middle of their length is equal to their fracture resistance Rmf: Rmf = k m lcf rf = R mp lcf 2r f (3.22.3) where km and Rpm are respectively the yield strengths of the matrix in shear and in tensile loading. With the given values lcf = 0.1x3,000/70 = 4.3 mm. The critical volume fraction fvc is such that: σ R = f vc 3,000 + (1 − f vc ) 70 = 200 (3.22.4) Hence fvc = 4.4 %. c) How does the critical length for load transfer vary with temperature? Solution The yield strength of the matrix decreases when the temperature is increased. Hence, Eq.3.22.3 shows that the critical length increases. 3.5 Continuum crystalline plasticity Continuum crystalline plasticity combines the macroscopic approach of continuum plasticity with the crystalline structure of the considered materials. The plastic deformation is considered as resulting from glide occurring on easy glide systems when their resolved shear stress reaches a critical value. This critical shear stress evolves with strain-hardening which generally depends on the plastic activity of all the slip systems. Exercise 3.23 Latent hardening tests 120 Some aspects of the strain-hardening properties of single crystals can be investigated through “latent hardening tests” (see Sect. 3.5.1.2.d2 in Volume I): a primary specimen is pre-deformed by single slip (l) and from this are cut several secondary samples at different orientations. Each of these is then deformed again according to some system (g) for which the new critical resolved shear stress (CRSS) will depend on the short-range interactions between (l) and (g). The studied FCC crystal is supposed to obey the Schmid law with equal initial CRSSs τ 0 and the strain-hardening law τ cg = τ 0 + ∑ H gl γ l (γ l > 0) (3.23.1) l where τ cg is the CRSS on system (g), γ l the plastic glide on the active system (l) and H gl a strain-hardening matrix to be determined experimentally. The initial sample is a plate normal to 1 11 . 1) A tensile test is performed on this plate in the 123 direction. Find the resolved shear stress on the 12 easy glide systems {111} 110 , which are denoted according to the classical notation A : ( 111) B : (111) C : ( 1 11) D : (1 11) 1: [ 011] 2 : 0 11 3 : [101] 4 : 101 5 : 110 6 : [110] Which is the primary system? Write the yield stress σ 0 as a function of τ 0 . 2) This tensile test is performed up to the axial plastic strain ε1p and the axial stress σ 1 . Write, by use of (3.23.1), the self-hardening modulus on the primary system as a function of σ 0 ,σ 1 and ε1p . 3) A secondary sample is then cut from the initial plate in the direction 3 12 , at an angle of 60° with the 123 direction, and then submitted to a tensile test. The measured yield stress is σ 2 . Calculate the resolved shear on the easy glide systems. Which would be the active system in case of an isotropic strain-hardening matrix (i.e., H gl = H , ∀g , l )? What would be the H value? Actually, it is observed that, as long as ε1p < ε cp , the active system is the primary one but that, for ε1p > ε cp , the active system is the secondary one, namely C5. What are the consequences of this observation on the components of the H gl matrix which can be determined or bounded? Solution 1) Let k be the unit vector along the tensile axis. The stress tensor then reads 121 (3.23.2) σ =σk ⊗k where a ⊗ b denotes the tensorial product between the vectors a and b. The resolved shear stress τ on any slip system (n, m) is given by τ = m.σ .n = σ (k .m)(k .n) (3.23.3) With k along 123 , we get in the cubic axes k= 1 σ (−1, 2,3) ⇒ τ = (−m1 + 2m2 + 3m3 )(−n1 + 2n2 + 3n3 ) 14 14 (3.23.4) It follows that, on the 12 easy glide systems {111} 110 , the resolved shear stress is 14 14 14 14 6τ A 2 = 6σ 14 6τ A3 = 12σ 14 6τ A6 = 6σ 6τ B2 = 4σ 14 6τ B4 = 16σ 14 6τ B5 = 12σ 6τ C1 = 10σ 14 6τ C3 = 4σ 14 6τ C5 = 6σ 6τ D1 = 0 14 6τ D6 = 0 14 6τ D4 = 0 (3.23.5) Thus, the primary system is B4, with τ B4 = 8 7 6 σ ⇒ σ0 = 7 6 τ0 8 (3.23.6) 2) As long as the primary system B4 is the only active one and neglecting the change of orientation of the cubic lattice with respect to the tensile axis, we have on the system B4 τ= 8 σ 7 6 σ : ε p = τγ = σε p ⇒ γ = 7 6 p ε 8 (3.23.6) and then, according to (3.23.1) τ= 8 7 6 σ = τ 0 + H B4B4γ = 8 7 6 σ 0 + H B4B4 7 6 p ε 8 (3.23.7) 122 The self-hardening modulus H B4B4 can then be deduced from the primary tensile test by the relation H B4B4 = 32 σ 1 − σ 0 147 ε1p (3.23.8) 1 3) We still use (3.23.3) but now with k along 3 12 , that is k = (−3, −1, 2) . 14 We find τ= σ 14 (−3m1 − m2 + 2m3 )(−3n1 − n2 + 2n3 ) (3.23.9) The resulting resolved shear stresses on the 12 systems {111} 110 is then now 14 14 14 14 6τ A 2 = 12σ 14 6τ A3 = −4σ 14 6τ A 6 = −16σ 6τ B2 = −6σ 14 6τ B4 = −10σ 14 6τ B5 = −4σ 6τ C1 = 6σ 14 6τ C3 = −6σ 14 6τ C5 = 12σ 6τ D1 = 0 14 6τ D4 = 0 14 6τ D6 = 0 (3.23.10) In case of an isotropic strain-hardening matrix, the initial CRSS for this secondary test would be the same on the 12 systems and given by τ cg = τ 0 + H 7 6 p ε1 8 ∀g (3.23.11) and the active system would be the primary one, namely A6, (say ( 111) 1 10 to take the sign into account) with σ2 = 7 6 A6 7 6 7 6 p τc = ε1 τ 0 + H 8 8 8 (3.23.12) Thus, H would be equal to 8 σ −τ 8 7 6 2 0 32 σ 2 − σ 0 H= = p ε1p 7 6 147 ε1 (3.23.13) 123 Since H would also be equal to HB4B4, given by (3.23.8), this proves that σ2 would be equal to σ1, which is consistent with the assumption of isotropic strainhardening. If σ 2 ≠ σ 1 , which is supposed from now on, this means that the strainhardening is not isotropic. Let then Rg be the Schmid factor on system (g) (as deduced from (3.23.10), with a positive sign). Since A6 is observed to be the only active system in the secondary test as long as ε1p < ε cp , this means that, on any system (g), one has ε1p < ε cp ⇒ R gσ 2 ≤ τ 0 + H gB4 7 6 p ε1 8 (3.23.14) where the equality only holds for (g) = A6. This leads to the value of H A6B4 , namely H A6B4 8 σ −τ 8 7 6 2 0 32 σ 2 − σ 0 = = p ε1p 7 6 147 ε1 ( for ε p 1 < ε cp ) (3.23.15) and to lower bounds derived from (3.23.14) for H gB4 , g ≠ A6 . For (g) = C5, which is observed as active when ε1p ≥ ε cp , we have, with R C5 = ε1p ≥ ε cp ⇒ 6 7 6 p σ 2 = τ 0 + H C5B4 εc 7 8 6 7 (3.23.16) or H C5B4 = 8 3σ 2 − 4σ 0 ε cp 147 (3.23.17) Exercise 3.24 Pencil glide In most BCC crystals, plastic slip occurs along a <111> direction on planes including this direction ({110}, {112}, {123}... or even any plane in zone with <111>); this gives rise to what is called “pencil glide”: the plastic deformation resembles that of a stack of hexagonal shaped pencils (see Volume I, Sect. 3.3.2.1 and Sect 3.3.5.1). We aim here at expressing a plastic criterion according to 124 which, for a given stress state σ , plastic glide occurs in the <111> direction on the (non necessarily crystalline) plane in zone with it which suffers the maximum resolved shear stress when it reaches the critical value τ 0 . We then apply this criterion to the uniform stress state which has, in the cubic axes, the matrix 0 0 0 (σ ij ) = 0 p τ 0 τ p 2 p≥0 (3.24.1) and we compare the corresponding elastic domain in the ( τ , p ) half-plane with the one deriving with the Tresca criterion with the critical maximum shear stress τ 0 . Finally, we derive the form of the plastic strain rate tensor εɺ p , up to a multiplicative scalar factor λɺ > 0 . Solution We have to calculate max n (T (n).m ) over all the unit vectors n normal to m, for a given slip direction m, with T (n) = σ .n . Since we have T (n).m = n.σ .m = m.σ .n = T (m).n (3.24.2) max n (T (n).m ) = max n (T (m).n ) = τ (m) (3.24.3) it follows that where τ (m) is the shear stress on the plane normal to m . This shear stress is given by 2 2 τ (m) 2 = T (m) − (T (m).m ) = TT i i − ( Tk mk ) = σ ij m jσ ik mk − (σ ij mi m j ) 2 2 (3.24.4) In the cubic axes, for the <111> directions, we have mi = ± and then 1 ε = i 3 3 with ε i = ±1, ∀i (3.24.5) 125 1 3 τ (m) 2 = σ ijσ ik ε jε k − 2 1 (σ ijε jε j ) 9 (3.24.6) The considered plastic criterion, which is a special case of the Schmid law, implies that max <111> τ = ±τ 0 ⇒ max <111> τ 2 = τ 02 (3.24.7) that is 2 1 1 max <111> σ ijσ ik ε j ε k − (σ ij ε j ε j ) = τ 02 9 3 (3.24.8) The maximum has to be searched over the four <111> directions, namely (A) : [111] (B) : 11 1 (C) : 1 11 (D) : 1 1 1 ε1 = 1 ε 2 = 1 ε3 = 1 ε1 = 1 ε 2 = 1 ε 3 = −1 ε1 = 1 ε 2 = −1 ε 3 = 1 (3.24.9) ε1 = 1 ε 2 = −1 ε 3 = −1 For the stress state (3.24.1), we have 0 σ ijε j pε 2 + τε 3 1 τε 2 + pε 3 2 2 1 2 σ ij ε jσ ik ε k = ( pε 2 + τε 3 ) + τε 2 + pε 3 2 5 ⇒ (3.24.10) = p 2 + 2τ 2 + 3ε 2ε 3 pτ 4 3 σ ij ε i ε j = p + 2ε 2ε 3τ 2 and then the criterion (3.24.8) reads 2 1 5 2 1 3 2 max <111> p + 2τ + 3ε 2ε 3 pτ − p + 2ε 2ε 3τ = τ 02 9 2 3 4 (3.24.11) 1 2 1 max <111> p 2 + ε 2ε 3 pτ + τ 2 = τ 02 3 9 6 (3.24.12) or 126 For ε 2ε 3 > 0 ( ε 2ε 3 < 0 , resp.) i.e. for (A) and (D) (for (B) and (C), resp.), the maximum is reached for τ ≥ 0 ( τ ≤ 0 , resp.) and reads 2 2 2 (A) and (D), τ ≥ 0 ⇒ 3 p + 6 pτ + 4τ = 18τ 0 2 2 2 (B) and (C), τ ≤ 0 ⇒ 3 p − 6 pτ + 4τ = 18τ 0 (3.24.13) In the ( τ , p ) half-plane ( p ≥ 0 ), the yield locus consists of two portions of ellipses centred at the origin and symmetric with respect to the p half-axis, according to Fig. 3.24.1 Fig. 3.24.1 Yield locus for pencil glide in the (τ, p) half plane Since the Tresca criterion relies on a critical value τ 0 of the maximum shear stress, which is always larger than any resolved shear stress, the Tresca yield locus would necessarily be interior to the one in Fig. 3.24.1, which relies on the same critical value τ 0 of some resolved shear stress. In other words, the plastic flow would occur earlier according to Tresca’s criterion than according to the criterion studied here. We know (see Volume I, Sect. 3.5.2.3) that the Schmid law associated with a deformation mode by slip leads to the principle of maximum plastic work. This principle leads itself to the normality rule, which can thus be used here. The global load function f (σ ) reads f (σ ) = 3 p 2 ± 6 pτ + 4τ 2 − 18τ 02 or, for every system (3.24.14) 127 f (σ )(ABCD) = 6σ ijσ ik ε j ε k − 2(σ ij ε i ε j ) 2 − 18τ 02 (3.24.15) The normality rule reads ∂f εɺ = fij λɺ λɺ > 0 fij = ∂σ ij S p ij where α S (3.24.16) denotes the symmetric part of α . With (3.24.15), we get for every ac- tive system f ij (ABCD) = 6(σ ik ε j ε k + σ jk ε iε k ) − 4σ kl ε k ε l ε iε j (3.24.17) which leads in the cubic axes to the matrix −6 p − 8ε 2ε 3τ ( fij (ABCD) ) = −2ε1ε 3τ −3ε ε p − 2ε ε τ 1 3 1 2 −2ε1ε 3τ 6 p + 4ε 2ε 3τ 3ε 2ε 3 p + 4τ −3ε1ε 3 p − 2ε1ε 2τ 3ε 2ε 3 p + 4τ (3.24.18) 4ε 2ε 3τ and then to the deformation rates −3 p − 2τ −6 p − 8τ −2τ (εɺ A ) = λɺA −2τ 6 p + 4τ 3 p + 4τ −3 p − 2τ 3 p + 4τ 4τ 3 p − 2τ −6 p + 8τ 2τ p (εɺ B ) = λɺB 2τ 6 p − 4τ −3 p + 4τ 3 p − 2τ −3 p + 4τ −4τ p −6 p + 8τ p ɺ (εɺ C ) = λC −2τ −3 p + 2τ −6 p − 8τ p (εɺ D ) = λɺD 2τ 3 p + 2τ −2τ −3 p + 2τ 6 p − 4τ −3 p + 4τ −3 p + 4τ −4τ 2τ 3 p + 2τ 6 p + 4τ 3 p + 4τ 3 p + 4τ 4τ (3.24.19) p where it can be checked that the condition σ : εɺ > 0 is satisfied with positive val- ues for all the parameters λɺ . Assuming that two systems which suffer the same resolved shear stress have the same shear strain rate, we find finally 128 0 0 −12 p − 16τ p ɺ ɺ ɺ ɺ 0 12 p + 8τ 6 p + 8τ τ > 0 ⇒ λA = λD = λ ⇒ (εɺ ) = λ 0 6 p + 8τ 8τ (3.24.20) 0 0 −12 p +16τ τ < 0 ⇒ λɺ = λɺ = λɺ′ ⇒ (εɺ p ) =λɺ′ 0 12 p − 8τ −6 p +8τ B C 0 −6 p + 8τ −8τ Exercise 3.25 Card glide A hexagonal crystal is plastically deformed by single slip occurring on the sole basal plane, with unit normal c, along the direction corresponding to the maximum resolved shear stress τ on the basal plane, when it reaches the critical value τ0 (“card glide”). 1) Write the corresponding plastic criterion for a uniform stress state σ. Consider the special case of a simple shear stress T in the plane (x1, x2), along x2, with the c axis referred by the polar angles θ and ϕ in the rectangular axes (x1, x2, x3). 2) Find the yield stress Tc(θ) and the slip direction m(θ) when the c axis is such p that sin ϕ = 1 / 3, cos ϕ = 2 2 / 3. Find the plastic strain rate tensor εɺ for a given slip rate γɺ . Solution 1) The maximum resolved shear stress on a given plane is the shear stress. On the basal plane, it is given by τ = σ .c − (c.σ .c) c (3.25.1) The plastic criterion then reads f (σ ) = σ .c − (c.σ .c) c − τ 0 = 0 (3.25.2) The considered stress state has the matrix 0 0 (σ ) = 0 0 0 T 0 T 0 (3.25.3) 129 and the shear vector τ is sin θ cos ϕ c sin θ sin ϕ cos θ 0 σ .c T cosθ T sin θ sin ϕ c.σ .c = 2T sin θ cos θ sin ϕ (3.25.4) −2T sin 2 θ cos θ sin ϕ cos ϕ ⇒ τ T cosθ − 2T sin 2 θ cos θ sin 2 ϕ 2 T sin θ sin ϕ (1 − 2 cos θ ) The plastic criterion then reads 4sin 4 θ cos 2 θ sin 2 ϕ cos 2 ϕ + cos 2 θ (1 − 2sin 2 θ sin 2 ϕ ) 2 ... 2 τ =T = τ 0 (3.25.5) 2 2 2 2 ... + sin θ sin ϕ (1 − 2cos θ ) 2 2 2) With sin ϕ = 1 / 3, cos ϕ = 2 2 / 3, we get 2 32 4 2 2 2 1 2 2 2 τ = T sin θ cos θ + cos θ 1 − sin θ + sin θ (1 − 2cos 2θ ) 9 9 81 2 2 (3.25.6) 2 T2 = 1 + 2cos 2 θ ) = τ 02 ( 9 and then Tc (θ ) = 3τ 0 1 + 2cos 2 θ (3.25.7) So, Tc (θ ) increases from τ 0 to 3τ 0 when θ goes from 0 to π/2. The slip direction m can be calculated as 130 2 −4 2 sin θ cos θ 3 1 + 2cos 2 θ cos θ ( 9 − 2sin 2 θ ) τ 3τ m= = = τ T (1 + 2cos 2 θ ) 3 (1 + 2cos 2 θ ) 2 sin θ (1 − 2cos θ ) 2 3 (1 + 2cos θ ) (3.25.8) For a single slip on the system (c, m), we have S εɺ p = γɺ ( m ⊗ c ) (3.25.9) Using (3.25.4) and (3.25.9), we find (ε ) p 32sin 3 θ cos θ 2 2 sin θ cos θ ( 9 − 4sin 2 θ ) 2 2 sin 2 θ (1 − 4sin 2 θ ) − 9 (1 + 2cos 2 θ ) 3 (1 + 2cos 2 θ ) 9 (1 + 2cos 2 θ ) 2sin θ cos θ ( 9 − 2sin 2 θ ) 1 + 2cos 2 θ = Sym 2 3 9 (1 + 2cos θ ) 2 2sin θ cos θ (1 − 2cos θ ) Sym Sym 1 + 2cos 2 θ (3.25.10) Note that this result could have been obtained as well by use of the normality rule and of the plastic potential f (σ ) given by (3.25.2) through the relation ∂f εɺ = γɺ ∂σ p like in Exercise 3.24. S (3.25.11) 131 3.6 Macroscopic Formulation of Plastic Behaviour Such a formulation is based on a plastic criterion in the initial state and its evolution with the plastic flow (the yield function, including a law of workhardening) and on a law ruling the plastic flow. The initial plastic criterion may be isotropic, such as the Tresca and von Mises criteria, but it becomes generally anisotropic with an increasing plastic deformation, such as it occurs during the classical operations of metal forming (cold rolling, deepdrawing…) due to the formation of a crystallographic and morphologic texture in polycrystals. The work-hardening law may itself be isotropic or not, such as the kinematic hardening or any combination of isotropic and kinematic hardening. The description of the plastic flow is made easier when the material (then named a “standard material”) obeys Hill’s principle of maximum plastic work, which is the case for polycrystals whose grains deform only by plastic slip according to the Schmid law: thanks to the resulting normality and convexity rules, the yield function determines the plastic flow almost completely, up to a scalar function which can be identified from a simple test. Particular models, such as the Prandtl-Reuss and the Prager models, derived from these properties, are frequently used for a first approximation of the actual plastic behaviour under investigation. Exercise 3.26 Non-isochoric plastic flow We want to study the elastoplastic behaviour of a metallic material which has been processed according to a powder metallurgy route (sintering, hot isostatic pressing…), responsible for some residual porosity and then for a non-isochoric plastic flow. This porosity can be considered as distributed isotropically and the original powder as isotropic. Show why a plastic criterion described by the load function f (σ ) = A(Trσ ) 2 + BTr( s : s ) − C ( = Aσ 2 kk + Bsij sij − C ) A, B, C positive (3.26.1) where s is the stress deviator could be an adequate criterion for such a material. Derive the yield stress σ 0 for a tensile test and τ 0 for a simple shear test. What is the yield surface in the space of the principal stresses (σ 1 , σ 2 , σ 3 ) when A = B / 3 ?. What are the conditions for the plastic flow to occur? Assuming the 132 principle of the maximum plastic work to be valid, express the components of an increment dε ijp of plastic strain for a tensile test. Check that there is a plastic volume variation. To characterise the volume change during the plastic flow, derive the corresponding “plastic Poisson ratio” and check that it is different from ½. Solution The proposed criterion only uses symmetric invariants of the stress tensor σ . So, it is isotropic, as expected for an isotropic material. It depends on the stress deviator but also on Trσ and so on the hydrostatic pressure, as expected for a porous material. So, (3.26.1) can be an adequate yield function for the considered material. For a tensile test, we have σ 0 (σ ) = 0 0 2 3σ0 0 0 0 0 ⇒ (s) = 0 0 0 0 0 − σ0 3 0 0 0 σ − 0 3 (3.26.2) and then (3.26.1) yields 2B 2 3C A+ σ 0 = C ⇒ σ 0 = 3 3A + 2B (3.26.3) For a simple shear test, we have τ 0 (σ ) = 0 0 0 −τ 0 0 0 C 0 ⇒ 2 Bτ 02 = C ⇒ τ 0 = 2B 0 (3.26.4) 1 With sij = σ ij − σ kk δij , we have 3 1 1 sij sij = σ ij − σ kk δij σ ij − σ kk δij 3 3 2 2 1 2 1 = σ ijσ ij − σ kk + σ kk = σ ijσ ij − σ kk2 3 3 3 (3.26.5) 133 and then f (σ ) = Aσ kk2 + Bsij sij − C B = A − σ kk2 + Bσ ijσ ij − C 3 B 2 = A − (σ 1 + σ 2 + σ 3 ) + B (σ 12 + σ 22 + σ 32 ) − C 3 (3.26.6) When A = B / 3 , the equation of the yield surface in the space of the principal stresses reduces to σ 12 + σ 22 + σ 32 = C B (3.26.7) which is the equation of the sphere of radius C / B centred at the origin. The conditions for the plastic flow to occur read B 2 f (σ ) = A − 3 σ kk + Bσ ijσ ij − C = 0 ∂f : dσ = 2 Bσ + 2 A − B σ δ dσ > 0 ij kk ij ij ∂σ 3 (3.26.8) B 2 A − 3 σ kk + Bσ ijσ ij = C 2 Bσ dσ + 2 A − B σ dσ > 0 ij ij kk ll 3 (3.26.9) or If the material obeys the principle of maximum plastic work, the normality rule implies a plastic strain increment of the form dε ijp = ∂f dλ , dλ > 0 ∂σ ij (3.26.10) that is B dε ijp = 2 Bσ ij + A − σ kk δij dλ , dλ > 0 3 (3.26.11) 134 For a tensile test, this gives ( dε ) p 2B A+ 3 = 2σ 0 dλ 0 0 0 A− 0 B A− 3 0 B 3 0 (3.26.12) The associated relative volume variation is given by ( ) = dε Tr dε p p kk = 6 Aσ 0 dλ > 0 (3.26.13) since A > 0,dλ > 0 . The corresponding “plastic Poisson ratio” is νp =− p dε 22 α −3 B − 3A B = = , α = >0 p dε11 2 B + 3 A 2α + 3 A (3.26.14) It increases from −1 to 1/ 2 when α goes from 0 to + ∞ and vanishes for α = 3 (i.e., for A = B / 3 ). Exercise 3.27 Cold rolling induced plastic anisotropy 1) A metallic thin sheet has been cold-rolled in such a way that it has acquired an induced plastic anisotropy prior to further forming steps such as stamping or deep drawing. The corresponding yield function is expressed as f (σ , K ) = σ : B : σ − K 2 (3.27.1) where B is a positive fourth-order tensor (i.e., such that σ : B : σ ≥ 0, ∀σ ) and K is a hardening parameter. What are the minimal symmetries of B? Show that the Voigt notation can be used for it as well as it can be for elastic compliances S. In all what follows, we assume that the studied plastic anisotropy obeys a transverse isotropy around the axis x3 normal to the sheet. Using the results of Exercise 2.10 concerning the stability of elastic equilibrium in transverse isotropy, find the form of the BIJ matrix and the domain of the possible values for its components. 135 Solution Since we have σ ij = σ ji and σ ij Bijklσ kl = σ kl Bklijσ ij (3.27.2) Bijkl = B jikl = Bijlk = Bklij (3.27.3) it follows that So, we can use the Voigt notation as we do for elastic compliances S. The transverse isotropy obeyed by B leads in addition, as known for S in this case, to a matrix BIJ of the form B11 ( BIJ ) = B12 B11 Sym. B13 B13 B33 0 0 0 B44 0 0 0 0 B44 0 0 0 0 0 2( B11 − B12 ) (3.27.4) Furthermore, the positivity of B leads, as shown for S in Exercise 2.10, to the properties B33 ≥ 0 B66 ≥ 0 B44 ≥ 0 B11 ≥ B66 B132 + 4 B33 (3.27.5) 2) The yield function is now supposed not to depend on the hydrostatic pressure. Show that this implies that the yield function can be written as 2f (σ , K ) = F (σ 22 − σ 33 ) 2 + G (σ 33 − σ 11 ) 2 + H (σ 11 − σ 22 )2 + ... 2 ... + 2 Lσ 23 + 2 M σ 312 + 2 Nσ 122 − 2 K 2 (3.27.6) where G, M and N can be expressed as functions of F, H and L. Show what are these relations and, assuming that H > F, what is the variation domain of F, H and L. Solution Since the yield function is independent of the hydrostatic pressure p, we must have 136 (σ + pδ) : B : (σ + pδ) = σ + : B : σ , ∀σ , ∀p (3.27.7) 2 pB jjklσ kl + p 2 B jjll = 0, ∀σ , ∀p (3.27.8) B jjkl = 0 (3.27.9) B11 + B12 + B13 = 0 2 B13 + B33 = 0 (3.27.10) that is and then These 9 equations reduce to which implies also, since B66 = 2( B11 − B12 ) , the relation B11 = B66 B132 + 4 B33 (3.27.11) Referring to (3.27.5), this is associated to the fact that B is still positive, but not definite positive since we can have σ : B : σ = 0 with σ ≠ 0 , as soon as σ = pδ : ( p δ) : B : ( p δ) = 0, ∀p (3.27.12) So, the BIJ matrix only depends on three independent parameters such as B12, B33 and B44, since we have B11 = − B12 + B33 B , B13 = − 33 , B66 = B33 − 4 B12 2 2 Now, the yield function can be written as (3.27.13) 137 f (σ , K ) = (σ 11 σ 22 σ 33 σ 23 σ 31 σ 12 ) ... B11 ... B12 B11 B13 B13 0 0 0 0 B33 0 0 B44 0 Sym. 0 σ 11 0 σ 22 0 σ 33 2 −K 0 σ 23 0 σ 31 B66 σ 12 B44 (3.27.14) with the relations (3.27.13). Keeping B12, B33 and B44 as independent parameters, it reads B 2 f (σ , K ) = 33 − B12 (σ 112 + σ 22 ) + 2 B12σ 11σ 22 − B33σ11σ 33 − B33σ 22σ 33... 2 ... + B33σ 332 + B44 (σ 232 + σ 312 ) + ( B33 − 4 B12 )σ 122 − K 2 B B 2 = − B12 (σ 11 − σ 22 ) + 33 (σ 22 − σ 33 ) 2 + 33 (σ 33 − σ 11 )2 + B44σ 23 ... 2 2 ... + B44σ 312 + ( B33 − 4 B12 )σ 122 − K 2 (3.27.15) 2 This expression conforms with (3.27.6) if we put F = B33 G = B33 ⇒ G=F H = −2 B12 M =L L = B44 N = F + 2H M = B44 with N = B33 − 4 B12 H > F > 0, L > 0 (3.27.16) So, in what follows, we use the expression 2f (σ , K ) = F (σ 22 − σ 33 ) 2 + F (σ 33 − σ 11 ) 2 + H (σ 11 − σ 22 )2 + ... 2 ... + 2 Lσ 23 + 2 Lσ 312 + 2( F + 2 H )σ 122 − 2 K 2 (3.27.17) 3) Check that the tensile yield stress in any direction of the plane (x1, x2) making the angle α with x1 does not depend on α. We consider now plane stress states in the plane (x1, x2) of the form σ1 0 (σ ) = 0 σ 2 0 0 0 0 0 (3.27.18) Draw the corresponding yield surface in a plane (σ 1 , σ 2 ) after having shown that such a description makes sense. How does this surface evolve when the 138 strain-hardening is isotropic? How do the parameters H and F evolve in this case? Solution The stress matrix in the axes (x1, x2, x3) reads cos 2 α (σ ) = σ 0 sin α cos α 0 sin α cos α 2 sin α 0 0 0 0 (3.27.19) The tensile yield stress σ 0 is reached when f (σ , K ) vanishes, namely F (sin 4 α + cos 4 α ) + H ( cos 2 α − sin 2 α )2 ... 2 σ = 2K 2 2 ... + 2( F + 2 H )sin α cos α 2 0 (3.27.20) or σ 02 ( F + 2 H ) = 2 K 2 ⇒ σ 0 = K 2 , ∀α F+H (3.27.21) which, as expected from the property of transverse isotropy, is independent of α. For the considered plane stress state, the plastic domain is reached when 2 F (σ 12 + σ 22 ) + H (σ 1 − σ 2 ) = 2 K 2 ⇒ ( F + H ) (σ 12 + σ 22 ) − 2 H σ 1σ 2 = 2 K 2 (3.27.22) Due to the transverse isotropy, this criterion only depends on the principal stresses σ1 and σ2 and the elastic domain can be drawn in the plane (σ 1 , σ 2 ) . It is limited by an ellipse centred at the origin and with its axes making the angle π/4 with the axes x1 and x2. For isotropic strain-hardening, this ellipse is dilated by a homothetic transformation centred at the origin during the plastic flow, which implies that the ratio H/F remains constant. 4) The studied material is now supposed to be standard (i.e., to obey Hill’s principle of maximum plastic work). Derive from this property the value of the Lankford coefficient defined as the ratio r = εɺ2p / εɺ3p of the lateral and normal plastic strain rates during a tensile test along the x1 axis. Derive also the response to a balanced 139 biaxial tension ( σ 1 = σ 2 > 0 ). We then define an equivalent plane plastic strain rate by 1/2 p p p εɺequ = ( εɺαβ εɺβα ) α , β = 1, 2 (3.27.23) p Compare the values of the parameter εɺ3p εɺequ in the same hardening state for simple tension and balanced biaxial tension. Finally, discuss, from what precedes, the deep-drawing ability of such a thin sheet by comparing pure drawing ( σ 2 ≃ 0 ) and pure stretching ( σ 1 ≃ σ 2 > 0 ) deformation modes by taking both yield stress levels and risks of tearing into account. Solution From (3.27.17) written in the principal axes, the normality rule implies ∂f εɺ = λɺ ∂σ ij p ij S (λɺ > 0) with ∂f = H (σ 1 − σ 2 ) − F (σ 3 − σ 1 ) ∂σ 1 ∂f = F (σ 2 − σ 3 ) − H (σ 1 − σ 2 ) (3.27.24) ∂σ 2 ∂f = − F (σ 2 − σ 3 ) + F (σ 3 − σ 1 ) ∂σ 3 In simple tension, we then have εɺ1p = λɺ ( H + F )σ p εɺ2 = −λɺ H σ εɺ p = −λɺ Fσ 3 ⇒ r= εɺ2p H = >1 εɺ3p F (3.27.25) and then p ɺ H 2 + ( H + F )2 εɺequ = λσ εɺ3p p equ εɺ = ST F 2 H + (H + F ) 2 = 1 2 H H + 1 + F F (3.27.26) 2 140 In balanced biaxial tension ( σ 1 = σ 2 > 0 ), we get εɺ1p = λɺ Fσ p ɺ εɺ2 = λ Fσ εɺ p = −2λɺ Fσ 3 (3.27.27) and then p ɺ F 2 εɺequ = λσ εɺ3p p equ εɺ = 2 (3.27.28) BBT Since we have H F > 1 , it follows that εɺ3p p εɺequ < ST εɺ3p p εɺequ (3.27.29) BBT This proves that the thickness reduction, and then the risk of tearing, is larger in balanced biaxial tension than in simple tension. At the same time, the yield stress is larger too ( K 2 / F instead of K 2 / F + H ); the difference increases with H F . Consequently, the deep-drawing ability is better in simple tension and, more generally, in pure drawing than in pure stretching and this is favoured by larger values of H F , the Lankford coefficient, i.e., by a stronger coldrolling induced plastic anisotropy. 3.7 Plastic response of structures In the general case, the computation of the plastic response of structures needs the resolution of a difficult boundary value problem due to the multibranched and incremental nature of the constitutive equations (see Exercise 3.29 below). Thanks to adequate variational approaches, the problems are mainly solved numerically. Nevertheless, the problem is easier in some situations, which are mainly considered in the following: when the loading is proportional, a direct integration is possible, which transforms a “flow problem” into a “deformation problem”, so that plasticity is no more very far from nonlinear elasticity (Exercise 3.28); for rigid perfectly plastic 141 standard material, a limit yield analysis can be developed by use of variational principles and of rigorous or approximate powerful specific methods such as the slip lines technique and the method of rigid blocks, respectively (Exercise 3.30). Exercise 3.28 Cylindrical thin tube under proportional loading A circular cylindrical thin tube is constituted with an elastoplastic standard material obeying the von Mises criterion with isotropic hardening. It is subjected to the stress field 0 (σ )rθ z = 0 0 p 0 τ 0 τ p/2 p>0 (3.28.1) written in cylindrical polar coordinates (r, θ, z), with z along the tube axis. 1) What is the considered loading? Find, in the half-plane (τ, p), the initial yield surface (let σ 0 be the initial tensile yield stress). 2) Why can the plastic flow law be written as p dε = g( J 2 ) s d J 2 (3.28.2) where s is the stress deviator and J 2 its second symmetric invariant? Find the function g( J 2 ) by using the fact that, for this material, the tensile curve equation reads σ = σ0 + A ε p (3.28.3) 3) The tube is now subjected to a monotonic proportional loading. Give the corresponding relation between the equivalent stress and strain ( σ and ε ). Find the plastic strain tensor ε p for such a loading path. Consider the special case when p τ = p and find ε for the final value τ = p = 2σ 0 . Solution 1) The cylindrical tube is subjected to the internal pressure p and to the axial couple of torsion C = 2πRtτ where R is the mean radius of the cross-section and t is 142 the (small with respect to R) thickness of this section. The stress deviator in cylindrical polar coordinates is − p/2 ( s ) = 0 0 0 0 p/2 τ τ 0 (3.28.4) So we have J2 = 1 1 p2 p2 p2 + 2τ 2 ) = +τ 2 sij sij = ( + 2 2 4 4 4 In simple tension, J 2 = σ2 3 (3.28.5) . So, the von Mises criterion reads σ2 p2 +τ 2 = 0 4 3 or p 2 + 4τ 2 = 4σ 02 3 (3.28.6) In the (τ, p) half-plane (p > 0), this is the equation of a semi-ellipse symmetrical with respect to the axes (see Fig. 3.28.1). Fig. 3.28.1 Initial yield surface in the (τ, p) half-plane 2) The von Mises yield function with isotropic hardening reads 143 f (σ , Y ) = J 2 − K 2 (3.28.7) with Y the hardening parameters. For a standard material (i.e., obeying Hill’s maximum plastic work principle), the normality rule implies dε ijp = g(σ , Y ) f ij f kl dσ kl f ij = ∂f = sij ∂σ ij (3.28.8) So, with f kl dσ kl = skl dσ kl = skl ds kl = dJ 2 , we get dε ijp = g(σ , Y ) sij dJ 2 (3.28.9) where g(σ , Y ) must be a hydrostatic pressure independent isotropic function of σ and where the hardening parameters Y reduce to K2, i.e., the value of J2 during the plastic flow. Thus, we can write p dε = g( J 2 , J 3 ) s dJ 2 for J 2 = K 2 and dJ 2 > 0 (3.28.10) Like for the von Mises criterion, we can then neglect the influence of the third invariant J3 and we adopt (3.28.2) as the plastic flow law. The flow function g( J 2 ) can be deduced from a tensile test as indicated in Sect. 3.5.2.4 of Volume I (Eqs 3.239 and 3.240). The tensile test curve ε p = ϕ (σ ) is given here, from (3.28.3), by ϕ (σ ) = (σ − σ 0 ) 2 σ ≥ σ0 (3.28.11) 2(σ − σ 0 ) A2 (3.28.12) A2 and then we have g (σ ) = ϕ ′(σ ) = Since, in a tensile test, σ = 3J 2 , we get g(J2 ) = 2( 3J 2 − σ 0 ) A2 (3.28.13) 144 3) For a monotonic proportional loading path, we know (Sect. 3.5.3.2 of Volume I) that the relation between the equivalent stress and plastic strain is the same as the one between the axial stress and plastic strain in a tensile test, that is ε p = ϕ (σ ) = with ε p = (σ − σ 0 ) 2 A2 if σ > σ 0 (3.28.14) 2 p p ε ij ε ij . In addition, we have 3 ε ijp = 3sij 2σ ϕ (σ ) = 3sij (σ − σ 0 ) 2 2 A2σ if σ > σ 0 (3.28.15) When τ = p , the loading path is proportional (see Fig. 3.28.1) and the foregoing results may be used. We have − p/2 ( s ) = 0 0 0 p/2 p 0 3 p2 p 15 p ⇒ σ = + 3τ 2 = 4 2 0 (3.28.16) so that we get 2 3 p 15 2 ε = − σ 0 2 s 2 2 A p 15 p (3.28.17) or (ε ) p 0 0 2 −1/2 3 p 15 = 2 − σ 0 0 1/2 1 A 5 2 0 1 0 (3.28.18) For p = 2σ 0 , we get (ε ) p −1/2 0 0 2(8 15 − 15)σ 02 = 0 1/2 1 5 A2 0 1 0 (3.28.19) 145 Exercise 3.29 Deep-drawing analysis We consider a situation of deep-drawing of a thin metallic sheet similar to the one studied in Exercise 3.27 but here attention is focussed on macroscopic structural aspects whereas the material behaviour is somewhat simplified (no anisotropy, no strain-hardening). We restrict ourselves to the initial stage of the forming of a circular cylindrical flat-bottom can with the radius a, starting from a circular blank with the radius R and a thickness t much smaller than R, and to the mechanical analysis of the flange, that is this part of the blank which is located between the die and the blank-holder and is not yet in contact with the punch (see Fig. 3.29.1) Fig. 3.29.1 Scheme of the deep drawing forming of a flat-bottom thin sheet The flange is supposed to deform by isotropic pure drawing (the French notion of isotropic “rétreint”), a deformation mode according to which the exterior circular edge as well as any material circle centred on the symmetry axis z remain circular while its radius r, with a ≤ r ≤ R , is decreasing. Its initial state is supposed to be natural; the constitutive material obeys an isotropic elastic-perfectly plastic (no strain-hardening) behaviour, the von Mises criterion and the principle maximum plastic work. The friction action of the blank-holder and the associated pressure are neglected, so that both faces of the flange and its exterior edge (r = R) are free of forces. The action of the punch and of the rest of the can on the flange is schematised as a density of prescribed forces Tg on the section r = a of the form g T = − p er r =a ( p ≥ 0) (3.29.1) where p, which is progressively increasing from 0, can be considered as the loading parameter of the studied structure. The stress state in the flange is plane in 146 both the elastic and the plastic regimes (i.e., the only non-zero stress components are σ rr ,σ rθ and σ θθ and they depend on r and θ only). 1) Preliminary analysis of the stress and strain fields a) Write the equilibrium equations and the boundary conditions on r = a and on r = R . From now on, we suppose that σ rr ,σ rθ and σ θθ do not depend on θ. What does this assumption imply? b) What is the relation between σ rr , σ θθ and the tensile yield stress σ 0 inside a plastic zone? Draw the cross-section of the yield surface by the plane σ zz = 0 in the space of the principal stresses. c) In this plane, which part of the yield boundary corresponds to a “retreint” p mode of plastic flow implying εɺrrp ≥ 0 and εɺθθ ≤ 0 ? When does the flange thickness decrease and when does it increase by plastic flow? Solution a) The equilibrium equations read (see Volume I, Appendix F.2) ∂σ rr ∂σ rθ σ rr − σ θθ =0 ∂r + r ∂θ + r ∂σ rθ + ∂σ θθ + 2σ rθ = 0 r ∂θ r ∂r (3.29.2) The boundary conditions are σ rr ( R) = σ rθ ( R ) = 0 σ rr (a) = p σ rθ (a) = 0 (3.29.3) If σ rr ,σ rθ and σ θθ do not depend on θ, we get dσ rr σ rr − σ θθ =0 dr + r dσ rθ + 2σ rθ = 0 ⇒ σ = A rθ dr r r2 with A A = 2 =0 2 a R (3.29.4) and then σ rθ = 0 . The first equation (3.29.4) reads also σ θθ = with σ rr ( R) = 0 σ rr (a) = p . d(rσ rr ) dr (3.29.5) 147 b) The von Mises criterion reads J2 = 1 1 s : s = σ 02 2 3 (3.29.6) with 2σ rr − σ θθ 2σ − σ rr σ + σ θθ , sθθ = θθ , szz = − rr srr = 3 3 3 srz = szθ = srθ = 0 (3.29.7) and then J2 = 1 2 1 σ rr − σ rrσ θθ + σ θθ2 ) = σ 02 ( 3 3 (3.29.8) In the ( σ rr ,σ θθ ) plane, this is the equation of an ellipse centred at the origin with the principal axis inclined at 45° with the coordinate axes (Fig. 3.29.2). p c) The plastic flow obeys the normality rule εɺ = λɺ s (λɺ ≥ 0) : (εɺ ) p p λɺ εɺrr = ( 2σ rr − σ θθ ) 3 p λɺ = εɺθθ = ( 2σ θθ − σ rr ) 3 p λɺ εɺzz = (σ rr + σ θθ ) 3 (3.29.9) 148 Fig. 3.29.2 Cross-section of the yield surface by the plane ( σ rr , σ θθ ) p The conditions εɺrrp ≥ 0 and εɺθθ ≤ 0 imply 2σ rr ≥ σ θθ and σ rr ≥ 2σ θθ respectively, so that the admissible part of the ellipse is restricted to the arc ACB (see Fig. 3.29.2) with 2σ σ 2σ σ σ σ A − 0 , − 0 , C 0 , − 0 , B 0 , 0 3 3 3 3 3 3 (3.29.10) In addition, we have εɺzzp ≥ 0 (plastic thickness increase) on AC and εɺzzp ≤ 0 (plastic thickness decrease) on CB. 2) Elastic regime This regime is defined by an elastic response at any point of the flange. Let E be the Young modulus and ν the Poisson ratio. a) Show that the non-zero stress components must have the form σ rr = k +l r2 σ θθ = − k +l r2 (3.29.11) where the constants k and l have to be derived from the boundary conditions. Give the expression of the radial displacement ur(r). 149 ur ( R ) and find the maximum value of p, R pc say, for which this relation is valid. Find the location of the incipient plastic flow in the flange. b) Give the relation between p and δ = − Solution a) The strain compatibility implies dur σ rr −νσ θθ ε rr = dr = E − σ νσ u rr ε = r = θθ θθ r E ⇒ ε rr = d ( rε θθ ) dr (3.29.12) so that we have d ( r (σ θθ −νσ rr ) ) = σ rr −νσ θθ dr with σ θθ = d ( rσ rr ) dr (3.29.13) which leads to r2 d 2σ rr dσ k k + 3r rr = 0 ⇒ σ rr = 2 + l σ θθ = − 2 + l 2 dr dr r r (3.29.14) The boundary conditions are σ rr ( R) = 0 ⇒ σ (a) = p ⇒ rr k +l = 0 pa 2 R 2 R2 ⇒ k= 2 k R − a2 + l = p a2 l=− pa 2 R − a2 2 (3.29.15) So, we find σ rr = pa 2 ( R 2 − r 2 ) r 2 (R2 − a2 ) σ θθ = − pa 2 ( R 2 + r 2 ) r 2 (R2 − a2 ) (3.29.16) and then ur = rε θθ = r (σ θθ −νσ rr ) pa 2 =− ( (1 +ν ) R 2 + (1 −ν )r 2 ) (3.29.17) E E ( R 2 − a 2 )r 150 b) From (3.29.17) we find δ =− ur ( R ) 2a 2 = p or R E(R2 − a2 ) p= E(R2 − a2 ) δ 2a 2 (3.29.18) This relation is no more valid as soon as any plastic flow occurs in the flange. This happens when 2 max r∈[ a , R ] (σ rr2 − σ rrσ θθ + σ θθ ) = σ 02 (3.29.19) With help of (3.29.16) we find σ rr2 − σ rrσ θθ + σ θθ2 = p2a4 (R 2 −a 2 2 ) 3R 4 1 + 4 r (3.29.20) This expression has its maximum value when r ∈ [ a, R ] at r = a and this maximum equals σ 02 when p reaches the critical value pc given by pc = σ 0 (R2 − a2 ) 3R 4 + a 4 2σ 0 a 2 δc = 4 4 E 3R + a (3.29.21) 3) Elastoplastic regime In this regime, let us consider that the flange is divided into two zones: an elastoplastic one, defined by a ≤ r ≤ ρ , and a still elastic one, with ρ ≤ r ≤ R . We have to find the relation between p and ρ or between δ and ρ. a) elastic zone: the general form of the equations (3.29.11) is still valid but k and l take values different from (3.29.15). Find these values and the expression of δ as functions of ρ. b) plastic zone: - kinematics: how can the kinematic compatibility be assessed in the plastic zone and at its junction with the elastic zone? - statics: write the equations to be satisfied by σ rr and σ θθ in the plastic zone. Show that σ θθ can be expressed as a function of σ rr and then that σ rr is the solution of a first-order nonlinear differential equation. Solve this equation thanks to the variable change σ rr = 2σ 0 π sin(ϕ + ) 6 3 (3.29.22) 151 where ϕ (r , p ) has to be determined. Show that one can choose ϕ (r , p) = 0 when σ rr = −σ θθ = σ 0 / 3 and that, for the “rétreint” mode, ϕ (r , p ) may be supposed π π and + only. What are the conditions to be satisfied at 3 3 r = a and r = ρ ? Propose a computation method to derive ρ and δ . Show that to vary between − the existence of a solution needs the condition p ≤ 2σ 0 / 3 to be satisfied. Is this condition sufficient to ensure the existence of a solution? Comment on the variations of ρ and δ with p. Solution a) The boundary condition at r = R gives σ rr ( R) = k + l = 0 ⇒ k = −lR 2 R2 (3.29.23) so that we get σ rr = l 1 − R2 r2 σ θθ = l 1 + R2 r2 (3.29.24) The plastic criterion is satisfied at r = ρ , namely σ rr2 ( ρ ) − σ rr ( ρ )σ θθ ( ρ ) + σ θθ2 ( ρ ) = l 2 1 + 3R 4 2 = σ0 ρ4 (3.29.25) In order to conform with the “rétreint” deformation mode (see Question 1.c and the conditions 2σ rr ≥ σ θθ and σ rr ≥ 2σ θθ ) we have to choose 3R 4 l = −σ 0 1 + 4 ρ −1/2 =− σ0ρ 2 ρ 4 + 3R 4 (3.29.26) so that we get σ rr = σ0ρ 2 R2 2 − 1 ρ 4 + 3R 4 r Finally we have σ θθ = − σ 0ρ 2 R2 2 + 1 4 4 r ρ + 3R (3.29.27) 152 δ =− ur ( R ) σ ( R ) −νσ rr ( R) 2σ 0 ρ 2 2l = − θθ =− = R E E E ρ 4 + 3R 4 (3.29.28) b) The kinematic compatibility is a condition which has to be applied to the total strain rate field εɺ (r ) and not to the elastic and plastic parts separately. We have in the plastic zone ν e 1 +ν εɺ = E σɺ − E Tr(σɺ ) δ εɺ p = λɺ s λɺ ≥ 0 (3.29.29) where the time derivatives of ε , σ and λ may be understood as derivatives with respect to the loading parameter p as well. In addition, one has to prescribe the continuity of the associated displacement field at r = ρ . The stress field has to satisfy the equilibrium equations, the plastic criterion, the boundary condition at r = a and the continuity condition at r = ρ . d σ θθ = dr ( rσ rr ) σ rr (a) = p σ rr2 − σ rrσ θθ + σ θθ2 = σ 02 σ rr ( ρ ) = σ 0 (R2 − ρ 2 ) (3.29.30) ρ 4 + 3R 4 The first two equations yield, with the notation σ rr = y and y′ = dy/dr y 2 + ryy′ + r 2 y′2 = σ 02 (3.29.31) which is a first-order nonlinear differential equation for y (r ) . In order to solve this equation, it is more convenient to use a variable change suggested by a parametrisation of the elliptic yield surface of Fig. 3.29.2, namely y = σ rr = 2σ 0 π sin ϕ + 6 3 (3.29.32) It is easy to check that, for the three points A, B and C of Fig. 3.29.2, one has π 3 ϕ A = − ,ϕC = 0,ϕ B = π 3 (3.29.33) 153 so that the “rétreint” mode regime is fully described by the interval − π π ≤ϕ ≤ . 3 3 With this new variable ϕ (r , p ) , (3.29.32) becomes π π π π 3 sin 2 (ϕ + ) + r sin(ϕ + )cos(ϕ + )ϕ ′ + r 2 cos 2 (ϕ + )ϕ ′2 = 6 6 6 6 4 (3.29.34) or equivalently π π π r cos ϕ + 6 ϕ ′ + cos ϕ r cos ϕ + 6 ϕ ′ + sin ϕ − 6 = 0 (3.29.35) It is easy to check that forcing the second term of this equation to be zero leads to the same equation as that obtained by doing the same on the first term, but leads to values of ϕ larger than π/3. Actually, the “rétreint” mode corresponds to π π π r cos ϕ + ϕ ′ + cos ϕ = 0 − ≤ ϕ ≤ 6 3 3 (3.29.36) This equation may be rewritten as sin ϕ 2 2 3− ϕ ′ + = 0 ⇒ r cos ϕ exp ϕ 3 = cst cos r ϕ ( ) (3.29.37) The constant can be determined from the third equation of (3.29.30), i.e., ( ) (3.29.38) 2σ 0 π sin ϕ (a ) + = p 6 3 (3.29.39) cst = a 2 cos ϕ (a)exp ϕ (a ) 3 where ϕ (a) is given by σ rr (a ) = Note that this is possible only if p≤ 2σ 0 3 (3.29.40) Combining this condition with the yield equation (3.29.21), we must have 154 σ 0 (R2 − a2 ) 4 3R + a 4 ≤ p≤ 2σ 0 3 (3.29.41) Finally, the last equation (3.29.30) yields σ rr ( ρ ) = 2σ 0 π σ (R2 − ρ 2 ) sin ϕ ( ρ ) + = 0 6 3 ρ 4 + 3R 4 (3.29.42) This allows the derivation of ρ since ϕ ( ρ ) is given, from (3.29.37) and (3.29.38), by ( ) ( ρ 2 cos ϕ ( ρ )exp ϕ ( ρ ) 3 = a 2 cos ϕ (a)exp ϕ (a) 3 ) (3.29.43) where ϕ (a) is given by (3.29.39). From ρ ( p) , we get δ ( p ) by (3.29.28). Note that the conditions (3.29.41) are necessary but not sufficient for the solution to exist. We must have in addition the property ρ ≤ R , which yields a condition on ϕ ( ρ ) , and then on ϕ (a) , and then on p: this condition may be more restrictive than (3.29.41). The variations of ϕ (a, p) , ϕ ( ρ , p) and δ ( p ) with p are given by dϕ (a, p) = dp 3 π 2σ 0 cos ϕ (a, p) + 6 with ϕ (a, p) ≤ π dϕ ( a , p ) ⇒ ≥ 0 (3.29.44) 3 dp 2 4 4 3/2 dρ a (3R + ρ ) exp 3 (ϕ (a) − ϕ ( ρ ) ) = ≥0 dp 6σ 0 ρ R 4 ( R 2 − ρ 2 ) Since (3.29.45) ∂ϕ ( ρ , p) ∂ϕ ( ρ , p) ≤ 0 , this leads to ≤ 0 . Finally, we find, as expected ∂ρ ∂p 2 dδ 2a exp 3 (ϕ (a) − ϕ ( ρ ) ) = ≥0 dp E(R2 − ρ 2 ) (3.29.46) 4) Limit analysis Finally, we investigate briefly the plastic breakdown of the structure. Show that a complete (i.e., ρ = R ) plastic forming of the flange can only occur when the ratio R/a does not exceed a critical value; give a numerical approximation of this value 155 and draw an indicative plot of the relation p(δ ) when R/a is either lower or larger than this critical value. What is the maximum value of p in both cases? Comment qualitatively on the deformation mode in both cases. What practical recommendation can be given from what precedes in order to favour the deep-drawing of a can by diminishing the risk of an early tearing of the flange? Solution When ρ → R,ϕ ( ρ ) → − R2 π according to (3.29.42) and then (3.29.43) yields 6 π 3 3 2 exp − = a cos ϕ (a )exp ϕ (a ) 3 2 6 ( ) (3.29.47) The right member in this equation obeys, since ϕ (a) ≤ π / 3 ( ) a 2 cos ϕ (a) exp ϕ (a) 3 ≤ π 3 a2 exp 2 3 (3.29.48) So, (3.29.47) needs the condition R2 π 3 a2 π 3 3 exp − ≤ exp 2 6 2 3 (3.29.49) to be fulfilled, i.e., π 3 R2 1 exp ≤ ⇒ 2 a 3 2 When this condition is satisfied, ρ = R, δ = R ≤ 2.96 a σ0 and p = pmax ≤ (3.29.50) 2σ 0 . When it is 3 E 2σ 0 not, p reaches the value for ρ max < R . Then we have ϕ (a) = π / 3 and ρ max 3 is given by 156 2 ρ max cos ϕ ( ρ )exp with ( ) 3ϕ ( ρ ) = cos ϕ ( ρ ) = 3π a2 exp 2 3 R2 3 (3.29.51) 4 3R 4 + ρ max which would make it possible to derive ρ max (< R ) and then δ max (< σ 0 /E ) . Both cases are schematically illustrated in Fig. 3.29.3. Fig. 3.29.3 Schematic illustration of the (p, δ) relation according to the value of R/a In the first case (R/a < 2.96), the flange is fully converted into the can wall. In the second case (R/a > 2.96), there remains in the flange an elastic ring which cannot follow the plastic flow of the internal part of the flange: hence the risk of ductile tearing at the boundary of the plastic zone. So, in order to favour a deeper drawing, it can be recommended to limit the initial flange radius R at values smaller than three times the punch radius. Exercise 3.30 Method of rigid blocks compared with the slip lines technique As indicated in Sect 3.5.3.4 of Volume I, the method of rigid blocks is a convenient and simple method to build up a kinematic approach to limit analysis. The material is still considered as a standard, rigid-perfectly plastic one. The method consists in envisaging the flow as the blocks sliding relative to one another (it is an incipient flow): we then have a velocity field in a rigid solid with discontinuities at the interfaces between moving blocks, tangential to the interfaces. The rela- 157 tive movements of the blocks must be compatible one with another; this can be ensured by the hodograph method (Sect. 3.5.3.5c in Volume I): for a plane problem, we associate with the physical plane a velocity plane (the hodograph plane), in which m(u, v) is the image of a physical point M (x, y) at which the velocity components are (u, v); if O is the origin of the hodograph plane Om = v( M ) . The plastic power developed over an interface element dS with the velocity discontinuity ∆v is k ∆vdS (k is the flow limit in simple shear). At a metal-tool interface we assume a friction law of the Tresca, or boundary layer, type in which the frictional shear τi is independent of the normal stress: τ i = mk with 0 ≤ m ≤ 1 ( m = 0 corresponds to perfect lubrication and m = 1 to “stick friction”). If vf is the relative slip velocity at the interface element dS over which friction is acting, this is tangential to dS and the friction power over the element is mk vf dS . 1) Apply this method to the same problem of a flat punch acting on a semi-infinite body (plane strain, isotropic criterion, stick friction) treated in Sect 3.5.3.4 of Volume I from a static point of view. Use for that the 5 blocks scheme indicated in Fig. 3.30.1 (note that the triangles are right-angled isosceles ones) and then optimise this treatment with respect to the depth h, which is a/2 in Fig. 3.30.1, of the plastic zone. Compare the result with the one derived from the slip lines method. Fig. 3.30.1. Punch on a semi-infinite body: method of rigid blocks Solution In the hodograph plane (for example, 32 is parallel to AD, the seat of the velocity discontinuity 32 between blocks 3 and 2), we get the diagram of Fig. 3.30.2. 158 Fig. 3.30.2 Hodograph analysis (stick friction) The plastic power dissipated at the interfaces between the blocks per unit thickness, assuming no slip under the punch, is: over AD and A′D′ : 2( a 2 )V0 2 k = 2kV0 a over AC and A′C′ : 2(a 2 )(V0 / 2 )k = kV0 a ⇒ total : 6kV0 a over DC and D′C′ : 2kV0 a over BC and B′C′ : 2(a 2 )(V / 2 )k = kV a 0 0 (3.30.1) Hence, according to the upper bound theorem FV0 ≤ 6kV0 a, ∀V0 ⇒ F ≤ 6ka (3.30.2) Note that this upper bound is larger than the one derived from the slip lines method (see Sect. 3.5.3.5 in Volume I), namely ka (2 + π) . The estimate can be improved by optimising with respect to the depth of the plastic zone; thus if we assume that the same flow scheme applies over a deformed zone of variable depth h, where in the above we have fixed h = a/2 , we find for the plastic power P(h) P( h ) = 2kV0 ( a2 + 2h ) h (3.30.3) This upper bound for the limit force is minimum for h = a / 2 and it is equal to 4ka 2 ; it is still larger than the bound obtained by the method of slip lines. 159 2) We treat the same problem for conditions of perfect lubrication (no friction) between the punch and the body. Does the static approach developed in the former case and referred to in question 1) apply now without change? Use the method of rigid blocks according to the flow scheme of Fig. 3.30.3 (all the triangles are right-angled isosceles triangles) and find an upper bound for F. Use then the method of slip lines by considering two Prandtl fans with opening π/2 at A and A’ and homogeneous fields on both sides. Comment on these results. Fig. 3.30.3 Punch on a semi-infinite body (no friction): method of rigid blocks Solution The static approach developed for the case of stick friction under the punch treat the interface as a principal surface so that the shear stress vanishes, as it does when there is no friction; so, the associated lower bounds for F are still valid in the latter case and thus we have F ≥ ka (2 + π ) (3.30.4) The hodograph analysis of Fig. 3.30.3 leads to the diagram of Fig. 3.30.4 Fig. 3.30.4 Hodograph analysis (no friction) 160 So, the plastic power dissipated on the interfaces is over BA′ and BA : 0 (no friction) over BC and BC′ : 2(a / 2 2 )(V0 2 )k = kV0 a over AC and A′C′ : 2( a / 2 2 )(V 2 )k = kV a 0 0 ⇒ total : 6kV0 a (3.30.5) ′ ′ a V k kV a over DC and D C : 2( / 2)2 = 2 0 0 over AD and A′D′ : 2(a / 2 2 )V 2 k = kV a 0 0 over ED and E′D′ : 2(a / 2 2 )V0 2 k = kV0 a and we find FV0 ≤ 6kV0 a, ∀V0 ⇒ F ≤ 6ka (3.30.6) The slip lines under A (and A’) are suggested in Fig. 3.30.5 Fig. 3.30.5 Sketch of slip lines under A and A’ (no friction) On a line α, we have: p + 2kθ = cst On Ax, we get: θ = π / 2 σ 1 = 0 σ 2 = −2k p=k On A’A, we find: θ = 0 σ 2 = − pAA′ − k Consequently, we have pAA′ = k + k π σ 2 = − k (1 + π) − k = −k (2 + π) (3.30.7) and then FV0 ≤ −σ 2 aV0 ⇒ F ≤ ka (2 + π) (3.30.8) This upper bound is lower (and then better) than the one found in 3.30.6. In addition, it is equal to the lower bound (3.30.4). So, it gives the exact value of the limit punching force. 161 References Friedel J (1964) Dislocations. Pergamon Press Jaoul B (2008) Étude de la plasticité et application aux métaux. Mines Paris Tech Paris Nabarro F R N (1967) Theory of dislocations. Clarendon, Oxford. 162 Chapter 4 Exercises of Chapter I.4: Elastoviscoplasticity High temperature (T/TM > 0.4) deformation of crystalline materials involves diffusion flow as well as dislocation motion. Creep is a high temperature phenomenon in which a constant load results in continuing permanent deformation. Upon application of a fixed load, transient creep (or primary creep), characterised by a decreasing creep rate, is observed. After a certain time (strain), the strain rate decreases to a characteristic constant and minimum value called the steady-state creep rate. This stage is followed by a progressive increase of the creep rate corresponding to tertiary creep leading to failure. Constitutive equations have been developed to describe these three stages, in particular steady creep. All relate the creep rate to stress and diffusivity appropriate to the particular creep mechanism. Additionally, diffusion-controlled creep mechanisms explicitly relate creep rate to the material grain size. The creep rate increases with decreasing grain size for volume-diffusion-controlled (Nabarro-Herring) creep, and depends even more strongly on grain size when mass transport occurs along grain boundary (Coble creep). This creep mechanism is prevalent in small grain-sized materials, such as ceramics. 4.1 Mechanisms of creep deformation Exercise 4.1 Coble creep The basis of Coble creep is presented in Volume I, p. 402. This model was proposed for creep in polycrystals in which the strain rate is controlled by diffusion, not through the grains but along the grain boundaries, as it is assumed in the Herring-Nabarro creep (Volume I, p. 400). The original paper was published by Coble (1963). Let us consider a spherical grain of radius R submitted to a vertical uniaxial stress, σ. The driving force for diffusional creep was shown by Nabarro and Herring to result from a change in equilibrium vacancy concentration with stress. Their result can be reduced to a simple expression (see Volume I, Eq. 4.73) for the change in vacancy concentration. ∆C = C0σ Ω / kT (4.1.1) 163 where Ω ≈ b3 , b being the Burgers vector and C0 is the equilibrium vacancy concentration at temperature T in a stress-free crystal; σ is the component of the local stress normal to the boundary. For a spherical grain the stress direction defines the polar axis. The vacancy concentration at the equator is C0, and the vacancy concentration at the poles changes by the amount given by Eq.4.1.1. In this figure the applied stress is vertical. It is assumed that excess vacancies created at the pole diffuse to the equator exchanging atoms, which produces an elongation. Since each half of the sphere acts independently, we need only to consider one hemisphere. a) Indicate schematically the path of vacancies and atoms Solution Figure 4.1.1 is a sketch showing the vacancies and the atoms fluxes. A vacancies S1 S2 atoms α B Fig. 4.1.1. Sketch illustrating the motion of vacancies and atoms in the Coble creep mechanism b) The hemisphere is divided into two parts, one part corresponding to the source of vacancies (pole A), the other one corresponding to the sink (equator B). To simplify, the concentration of vacancies in each part is assumed to be uniform. The sphere keeps a constant volume; the sink surface must be equal to the source surface. Show that the corresponding parallel is obtained for α = π / 3 Solution The areas S1 and S2 can be easily calculated: 164 dS1 = 2πR sin α Rdα ( S1 = 2πR 2 1 − cos α ) 2 S2 = 2πR cos α (4.1.1) S1 = S2 ⇒ α = π / 3 ( ) c) The mean gradient of the concentration of vacancies is therefore ∆C / πR / 2 . The maximum value of this gradient occurs at α = π / 3 and can be written as: 1 dC ∆C =N R dθ α = π / 3 πR / 2 (4.1.2) where N is a proportionality factor equal to 2.15 (See Coble 1963). Calculate the flux of vacancies per second. Solution If δGB is the thickness of the grain boundary; the area, S, crossed by the vacancies at α = π / 3 is given by: ( ) S = δ GB 2πR sin π / 3 = π 3Rδ GB (4.1.3) The number of vacancies crossing this area is given by: J = DL N ∆C π 3Rδ GB = 7.45DLδ GB ∆C πR / 2 ( ) (4.1.4) d) Calculate the corresponding strain rate Solution The volume change associated with the flux of vacancies: J Ω = πR 2 dR = 7.45DLδ GB ∆CΩ dt (4.1.5) The mean strain rate is given by: εɺ = 1 dR J Ω 7.45DLδ GB ∆CΩ = = R dt πR3 πR 3 (4.1.6) 165 The vacancy diffusion coefficient, DL is related to the grain boundary diffusion coefficient, DGB, by DGB = DL C0 Ω. The strain rate can thus be expressed as (refer to Eq (4.60a and b) in Volume I): εɺ = 7.45σ DGBδ GBΩ πR3kT (4.1.7) When R is considered as one half of the grain diameter, one obtains εɺ = α DGBδ GBσΩ d 3 kT (4.1.8) where α is a numerical factor. This expression is similar to the Herring- Nabarro creep law except for the d-1/3 variation instead of d-1/2. Exercise 4.2 Enhanced role of diffusion creep in ceramics Diffusion creep with n = 1 (n =1/M in Eq. 4.5 in Volume I) is a common behaviour in ceramics. Give two reasons why the role of diffusion is enhanced in ceramics compared to metals (Cannon and Langdon, 1988) Solution (i) The grain boundary mobility in ceramics is lower than in metals. It is thus easier to stabilise, and maintain, a very fine grain structure. Whereas most metals (and especially pure metals) have grain sizes of 50-100 µm or larger, many ceramics have stable grain sizes of the order of 1-10 µm. As indicated in figures 4.33, 4.34, 4.35, and 4.36 (Volume I), diffusion creep with n = 1 extends to higher values of σ/µ when the grain size is reduced (see Eq. 4.76 in Volume I) and at the same time the creep rates also become faster. (ii) Diffusion creep becomes important in many ceramics because of the preferential enhancement of diffusion of one of the ionic species along the gain boundaries. The faster lattice diffusion coefficient for the cation usually dominates in creep diffusion creep regime. Exercise 4.3 Diffusion creep in alumina Using the creep deformation map for alumina (grain size =10 µm) given in Fig. 4.46 in Volume I, determine: 166 a) The shear stress (in MPa) that will give rise to a creep rate of 10-10/s at 1400°C b) The creep mechanism under these conditions. c) The stress at which power law creep becomes dominant in the temperature range between 1,400°C and 1,500°C. Hint: Melting point, TM = 2,320K Burgers vector, b = 4.76 x 10-10 m Shear modulus at 300K, µ0 = 1.55x105 MPa T dµ Temperature dependence of modulus, M = −0.35 µ 0 dT Solution a)T/TM = 1,673/2,320 = 0.72 1,373 = − 0.321x105 (MPa) ∆µ = − 0.35x1.55x105x 2,320 µ (1,400°C)= 1.23x105 (MPa) b) Figure 4.46a (Volume I) shows that σ is about 0.10 MPa. This corresponds to diffusion (Coble) creep. c) The stress is about 300 MPa 4.2 Mechanical models Exercise 4.4 Stress relaxation A bolt made of steel is used to assemble two plates which are maintained at 550°C. Creep tests at this temperature showed that the steady creep law can be written as εɺ = Bσ 3 , where B is a constant, such as for σ = 27.5 MPa, εɺ = 2.80x10-8/hour This bolt has been initially loaded to a stress σ0 of 70 MPa. Calculate the tightening stress remaining after one year of operation. Solution One neglects the part of the relaxed stress due to primary creep. This assumption is not necessarily justified (see Exercise 4.5) because of the low values of the creep strain involved in creep relaxation. So, B = 1.35 x 10-12 (h x MPa)-1. 167 During stress relaxation the elastic strain, ε e , is converted into a creep strain, ε vp , such as: dε vp dt =− dε e 1 dσ =− = Bσ n dt E dt (4.4.1) 1 1 − σ n−1 σ 0n−1 (4.4.2) After integrating one obtains: t= 1 BE n − 1 ( ) The numerical application leads to σ = 16.3 MPa Exercise 4.5 Creep under variable stress In many situations one has to deal with variable stresses (see Exercise 4.4). Creep tests were performed at 700°C on one heat of 316 stainless steel (see Calvet et al., 1972). The specimens were loaded continuously at constant stress rate such that σ = Kt. The strains which were obtained were limited so that the deformation behaviour could be described using only primary creep law. a) Two assumptions are usually made when the creep stress is variable, as shown in Fig. 4.10 (Volume I): (i) time hardening, (ii) strain hardening. Indicate the most appropriate assumption. Solution As in primary creep there is an increase in dislocation density, one is tempted to say that the strain-hardening assumption is better than the time-hardening rule. b) Tests at constant load showed that the primary creep law could be written as: εɺ = At − pσ n (4.5.1) Show that this law can also be expressed as: εɺ = Bε − qσ n′ Relate B to A and p; q to p and n' to n and p. Solution (4.5.2) 168 Integration of (4.5.1) leads to: 1 1 1 − p 1− p 1− p ⇒t = ε Aσ n 1 1− p ε = Aσ t 1− p n (4.5.3) The integration of (4.5.2) yields: 1 ε = (q + 1)Bσ n′ t q+1 (4.5.4) Eliminating t between (4.5.3) and (4.5.4) and using (4.5.2) one obtains: B= A 1 1− p p (1 − p )− 1− p ⇒q= p 1− p and n′ = n 1− p (4.5.5) c) Tests at constant increasing stress, σ = Kt were performed. The prediction of the strain with time can be made using either the time-hardening (Eq. 4.5.1) or the strain-hardening (Eq. 4.5.2) rules. Calculate as a function of time the evolution of the creep strains εTH and εSH corresponding to the two rules of time- and strain-hardening and the time to reach a given strain as a function of K. Solution As the strain hardening assumption predicts a higher strain rate the relative position of the curves calculated with the two extreme assumptions is the following (Fig.4.5.1): Time to reach a given strain b) Creep strain a) SH TH Time (a) K (b) Fig. 4.5.1. (a) Schematic evolution of creep strain with time in stress increasing creep tests. Comparison between strain hardening (SH) and time hardening (TH) rules; (b) Variation of time to reach a given strain as a function of K parameter 169 More quantitatively, εTH and εSH can be calculated as follows: Time-hardening: integration of Eq.4.5.1 gives: εTH = AK n n− p+1 t n − p +1 (4.5.6) Strain-hardening: integration of Eq.4.5.2 gives: 1 ε SH B q + 1 K n' ( ) = n' + 1 ( ) q +1 n' +1 1 A t q +1 = K nt n− p +1 1− p p ( n − p + 1) (1 − p ) (4.5.7) The ratio εTH/εSH is given by: p ε SH n − p + 1 = ≥1 ε TH 1 − p (4.5.8) because p < 1 and n > 0. Hence, the evolutions of the creep strain as a function of time and of the time to reach a given strain as a function of K are shown in Fig. 4.5.1. The results of the tests on 316 stainless steel corresponding to two values of K (Calvet et al., 1972) are shown below (Fig.4.5.2) where it is observed that the SH rule gives the best predictions. Fig. 4.5.2. Creep curves under increasing stress – Steel AISI 316 − θ = 700°C. Comparison between experiments and calculations; (a) K = 0.13 MPa/h; (b) K = 0.3 MPa/h (Adapted from Calvet et al., 1972) 170 Exercise 4.6 Relationship between creep ductility and time to failure accounting for tertiary creep This exercise is largely based on a publication by Phaniraj et al (2003). It deals not only with Chapter 4 in Volume I, but also Chapter 6 in Volume II. The shape of the creep curves in many materials is shown in Fig. 4.6.1. In many cases primary creep strain, εp, is relatively insignificant and is neglected in this sv exercise. The material deforms at constant rate, εɺ , in the stationary regime and then at an increasing rate in the tertiary regime. Fig. 4.6.1. Schematic creep curve showing negligible primary creep strain εp, tMGD as time to reach Monkman-Grant ductility, εt limiting tertiary creep strain and λ as damage tolerance factor (Phaniraj et al., 2003) According to continuum damage mechanics (CDM) (see Sect. 8.4.4, Volume II) damage is treated as an internal variable, D, and for uniaxial stress conditions it can be expressed by two coupled differential equations: m (4.6.1) k k Dɺ = Dɺ 0 (σ / σ 0 ) 1 / (1 − D ) (4.6.2) m εɺ = εɺ0 (σ / σ 0 ) 1 / (1 − D ) It is assumed that the exponents m and k are constant, independent of the applied stress, σ. a) Find the evolution of the damage parameter D and of the creep strain as a function of the reduced time: t/tr, tr being the end of life time. Show that, integrating (4.6.2), the time to fracture tr and the evolution of damage can be determined: Show that integration of (4.6.2) yields: 171 k tr = 1 / ( k + 1) Dɺ 0 (σ / σ 0 ) (4.6.3) corresponding to D = 1, and: 1 t k +1 D = 1 − 1 − tr (4.6.4) Solution Integration of (4.6.2) yields: 1 k +1 1 − D = 1 − ( k + 1) Dɺ 0 (σ / σ 0 ) t k with tr = 1 ɺ ( k + 1) D0 (σ / σ 0 )k 1 t k +1 = 1 − tr when D = 1 (4.6.5a) (4.6.5b) b) Show that, integrating (4.6.1) and using (4.6.5b), one obtains the evolution of the creep strain: t ε / εr = 1 − 1 − tr k +1− m k +1 (4.6.6) with εɺ 1 m−k ε r = ɺ0 (σ / σ 0 ) D0 k − m + 1 Solution Substituting Eq. 4.6.5a in Eq. 4.6.1, and integrating one obtains: (4.6.7) 172 ε = εɺ0 (σ / σ 0 ) m k − m +1 k +1 t k +1 tr 1 − 1 − k − m + 1 tr (4.6.8) The variation of strain with time is given by: ε = εɺ0 (σ / σ 0 ) m k −m +1 k +1 t k +1 tr 1 − 1 − k − m + 1 tr (4.6.9) The strain to failure is given by: m ε r = εɺ0 (σ / σ 0 ) tr εɺ k +1 1 m−k = ɺ0 (σ / σ 0 ) k − m + 1 D0 k − m + 1 (4.6.10) Thus one obtains: t ε = 1 − 1 − εr tr k − m+1 k +1 (4.6.11) g c) The Monkman-Grant law states that the product ( εɺ sv ) tr is a constant denoted MGD. The exponent g does not differ much from 1, and in this exercise it is assumed that g = 1. The ratio λ = ε r εɺ sv tr measures the relative importance of the tertiary creep strain. The time to reach the Monkman-Grant ductility is denoted tMGD. Calculate the ratio λ in terms of the reduced time tMGD/tr. Solution The ratio λ = ε r εɺ sv tr measures the relative importance of the tertiary creep strain. It can simply be expressed as: λ= εr εɺ svtr = k +1 k − m +1 Denoting the Monkman-Grant strain εMGD, Eq. 4.6.5 yields: (4.6.12) 173 t ε 1 − MGD = 1 − MGD εr tr 1λ (4.6.13) As εt = εr − εMGD, the tertiary ductility is such that: ε t tMGD = 1− ε r tr 1/ λ (4.6.14) This expression is only valid when D is negligible, in the stationary regime. Equation 4.6.11 can thus be written as: ε sv t = 1 − 1 − εr tr 1/ λ (4.6.15) where λ is assumed to be a constant independent of the applied stress. d) Show that the time to reach the Monkman-Grant ductility (MGD) can simply be related to the value of λ which measures the part of the ductility associated with tertiary creep. Comment the expression obtained between tMGD /tr and λ = ε r εɺsvtr . MGD is defined as MGD = εɺsvtr . (See Fig. 4.6.1) Solution Equation. 4.6.11 indicates that: ε t 1− = 1 − ε r tr 1/ λ (4.6.16) So, the ductility due to tertiary creep, εt, can simply be written as (see (4.6.14)) ε t tMGD = 1− ε r tr 1/ λ with ε t = ( λ − 1) εɺ sv tr and ε r = λεɺ sv tr Using the definition of λ, Eq. 4.6.17 can be rewritten as follows: (4.6.17) (4.6.18) 174 ( λ − 1) εɺsvtr λεɺsvtr 1/ λ t = 1 − MGD tr (4.6.19) One deduces that: tMGD tr λ λ − 1 = 1− = constant λ (4.6.20) The variation of tMGD / tr as a function of λ is shown in Fig. 4.6.2 where it is observed that this ratio reaches a value which is almost constant as soon as the damage tolerance factor λ is sufficiently large ≥ 5 . It can easily be shown that ( ) this constant value is equal to (1 − 1 / e ) ≈ 0.63 . Fig. 4.6.2. Variation of tMGD/tr as a function of λ in two steels (adapted from Phaniraj et al 2003) 4.3 Miscellaneous Exercise 4.7 Bending of creeping bars and application of deformationmechanisms maps The bending behaviour of a prismatic bar is investigated in the frame of classical assumptions used in the study of the strength of materials (small displacement, a planar section remains plane; the stress state is assumed to be uniaxial). The length of the bar is L; the bar is aligned along the x axis, the y axis being vertical. 175 This bar is submitted to a bending moment, M (Fig. 4.7.1). The viscoplastic law which is used is the simple Norton law: σ εɺv = K n (4.7.1) a) Rectangular beam (width b, height h) Show that at a given time, t, the profile of the stress σ xx = σ can be written as: 2y h σ = σM 1/ n (4.7.2) where σ M is the maximum stress in the beam at time t and σ is the stress along the x axis across the beam section. What are the simplifying assumptions made to obtain this expression? Remember that assuming that a section remains planar leads to: ε = εM 2y h or εɺ = εɺM 2y h (4.7.3) Calculate the maximum stress σ M as a function of the applied bending moment, M. Introducing the inertia moment, I, used in elasticity, show that the stress can be expressed as: σ = M h 2I ( ) f y, n (4.7.4) Check that the function, f, is equal to 1 for y = h/2 when n = 1. Indicate qualitatively the shape of this function f, as a function of y for different values of exponent n. An “equivalent” inertia moment (IC) can be defined for creep deformation. IC can be defined such that: σM = M h 2 IC (4.7.5) Derive the expression relating the curvature to the bending moment, M, and the time, t. Express IC as a function of the elastic inertia I. 176 Solution If the elastic strain is neglected, one can write: εɺ = εɺv = n σ K (4.7.6) This implies that: σ = σM 2y h 1/ n (4.7.7) The stress is maximum at the free surface and is null in the centre of the beam. The situation is similar to that observed in elasticity. The bending moment can be written as: M = 2b ∫ h/ 2 0 σ ydy (4.7.8) Using (4.7.1) and integrating, one obtains: σM = M h 2n + 1 2I 3n where I is the “elastic” inertia moment of the beam (I = bh3/12) Fig. 4.7.1. Sketch showing a bending bar (4.7.9) 177 Fig. 4.7.2. . Stress distribution in bending of a rectangular beam for different values of the exponent n. The elastic stress distribution is the same as that given for n = 1 One obtains: M h 2y σ = 2I h 1/ n 2n + 1 3n (4.7.10) The function f (y, n) is: f = 2n + 1 2 y 3n h 1/ n (4.7.11) The ratio between the bending stress and the maximum elastic strain, as a function of the distance from the centre line (2y/h) is shown in Fig. 4.7.2 for various values of the exponent n. The stress redistribution associated with creep deformation is relatively independent of the value of exponent n when it is larger than 1. It can be noted that when n tends to infinity, the maximum stress at the surface tends to 2/3 the stress calculated in elasticity. Equation 4.7.9 shows that IC = 3n I 2n + 1 (4.7.12) 178 b) Circular beam and deformation-mechanisms maps One considers the case where n =1. The corresponding creep mechanisms are diffusional ones (Coble or Herring-Nabarro creep: see Sect. 4.3.2.2 in Volume I). Creep of lead water pipes. Analysis of sagging pipes. Use of the deformation map for antimonial lead (Figures 4.7.3a and b. See Ashby and Jones 1980). Since pre-Roman times lead has been used for water conduits and roof coverings. It has now been largely eliminated for these usages One distinguishes “soft lead”, typically 99.5% Pb containing some impurities such as antimony, Sb, and “hard lead” containing 1 to 6% Sb. The earliest pipes were cast which produced large grain-sized (about 1mm) material. Pipes produced by plastic deformation had a smaller grain size (about 50µm) due to recrystallisation. Figures 4.7.3a and b indicate that the reduction in grain size from 1 mm down to 50 µm produces a large acceleration in creep strain rate at low stresses, as expected from Eqs. 4.71 and 4.76 in Volume I. The melting point of pure lead is 327°C (600 K). The addition of antimony lowers the melting point of lead to 525 K (eutectic temperature), such as at room temperature T/TM is between 0.50 and 0.57. Two examples of creeping lead pipes were examined by Ashby and Jones: one is an external drain pipe, 75 years old, the other one is an interior hot-water pipe of about the same age (Figures 4.7.4 and 4.7.5).The maximum strain in the rectangular section drain pipe occurs in the lower surface of the pipe. This strain was calculated from the radius of curvature, the depth of the sag, h, and the height, b, of the pipe. These measurements were made from Fig. 4.7.5 showing that the drain pipe has crept under its own weight. The shear strain rate for small sags (h<b) can be calculated as γɺ = 4 3bh / l 2t . Calculate the corresponding strain rate from this expression. The stress varies with position along the pipe; expressed as an equivalent shear stress it has a maximum value of 3x10-5µ. Is the observed strain consistent with that predicted from Fig. 4.7.3? What would have happened if the grain size had been 1mm? Solution Fig. 4.7.3. Deformation mechanisms of antimonial lead with a grain size of 50 µm ((a) at left) and 1 mm ((b), at right)(Ashby and Jones 1980) 179 Fig. 4.7.4. Creep of lead pipes (Ashby and Jones 1980). Fig. 4.7.5. Dimensions of drain and hot water pipe (Ashby and Jones 1980). One finds γɺ ≈ 6 x10−12 s −1 which corresponds to a very low strain rate. This strain rate is compatible with the results reported in Fig. 4.7.3a (d = 50 µm) for an applied stress of 3.5x10-5µ. An increase of the grain size by a factor of 1,000/50 = 20 would have produced a strain rate of about 10-15 due to Coble creep. However it is difficult to conclude unambiguously from the examination of Fig. 4.7.3a. This figure shows that, for a stress of 3.5x10-5 µ, and a grain size of 1 mm, diffusion flow and power-law creep are overlapping. The positions corresponding to the drain pipe and the hot water pipe are indicated in Fig. 4.7.3. 180 References Ashby MF, Jones DRH (1980) Engineering Materials 1. An introduction to their properties and Applications. Pergamon Press and applications of deformation-mechanism maps. See website: engineering.dartmouth.edu/defmech/chapter_19.htm Calvet JN, Le Bret P, Besnault R, Lehman D, Blanc B, Roulliay R, Castaing A (1972) Etude des déformations de fluage sous sollicitations complexes d’un acier inoxydable austénitique Z6CND18-12 (AISI 316) – Fluage sous charge progressive à haute température – Fluage sous contrainte thermique. Note technique RMA (72) 507 Cannon WR, Langdon TG (1988) Review-Creep of ceramics. Part II: an examination of flow mechanisms. J. Mater Sci 23:1-20 Coble RL (1963) A model for boundary diffusion controlled creep in polycrystalline materials. Journal of Applied Physics 34: 1679-1682 Phaniraj C, Choudary BK, Bhanu Sankara K, Raj B (2003) Relationship between time to reach 181 Chapter 5 Exercises of Chapter I.5: Viscoelasticity 5.1 One-dimensional responses in linear viscoelasticity The coupling of elasticity and viscosity, which is the characteristic feature of viscoelasticity, is responsible for the hereditary nature of this constitutive behaviour and for the specific delayed responses to various loadings: creep and retardation, relaxation, stress- and strain-recovery... One-dimensional linear viscoelasticity, which obeys the Boltzmann superposition principle, is fully characterised by only one scalar function (the creep or the relaxation function) of two time variables, and it can be described by an integral relation between the force and displacement variables. This relation reduces to a convolution product when non-ageing linear viscoelasticity is considered, so that the Laplace-Carson transform can be used for an easier resolution and for a conversion of the viscoelastic problem into an elastic one: the creep or the relaxation function only depends then on one time variable. The method of the spectral representation shows that all the constitutive behaviour characteristics are contained in the relaxation or retardation spectrum; when this spectrum reduces to discrete single lines, the constitutive equation reduces to a differential equation and can be represented by series or parallel assemblages of the Kelvin and Maxwell models. The Dynamic Mechanical Analysis (DMA) may be used to get the complex modulus from an experimental investigation and then to a complete description of the 1-D constitutive behaviour. Exercise 5.1 Elementary rheological models Calculate the constitutive differential equations for series and for parallel assemblages of one Kelvin model and one Maxwell model. For that, use both a direct method and the Laplace-Carson transform technique. Check that the short and long term creep and relaxation responses coincide with the intuitive predictions which can be made from the spring and dashpot representation. Solution 1) Burgers model (series assemblage): 182 For the Maxwell model, the differential equation is known to read (see Sect. 5.1.3.3 in Volume I), with obvious notations qɺM = Qɺ M QM + EM η M (5.1.1) For the Kelvin model, it is (5.1.2) QK = EK qK + η K qɺK For a series assemblage of these models, we have QM = QK = Q q = qM + qK qɺ = qɺM + qɺK (5.1.3) and then, from (5.1.2) Qɺ = EK (qɺ − qɺM ) + η K (qɺɺ − qɺɺM ) ɺɺ Qɺ Q Q Qɺ = EK qɺ − + + + ηK qɺɺ − EM η M EM η M (5.1.4) which may also be written as EK qɺ + η K qɺɺ = E η η ɺɺ Q + 1 + K + K Qɺ + K Q ηM η E EM M M EK (5.1.5) This result could have been derived straightforwardly by using the fact that, for a series assemblage, the resulting creep function is the sum of the creep functions of each element. According to Sect. 5.1.3.3, we get for the Laplace-Carson transformed resulting creep function f *( p ) = q *( p ) 1 1 1 ( E + ηM p )( EK + ηK p ) + EMηM p (5.1.6) = + + = M * Q ( p ) EM η M p EK + η K p EMη M p ( EK + η K p ) After reduction, it is easy to show that the second-order polynomial relation with respect to p which derives from (5.1.6) admits (5.1.5) as its original function of time. From the spring and dashpot representation of this model, it is obvious that it exhibits instantaneous elasticity (due to the unconstrained spring of the Maxwell model), unlimited creep (due to the unconstrained dashpot of the Maxwell model) and complete relaxation (due to the unconstrained dashpot of the Maxwell model). 183 This is corroborated by the behaviour of f*(p) and r*(p) given by (5.1.6) or derived from it for p → 0 (i.e., t → ∞) and p → ∞ (i.e., t → 0+ ) : 1 f * ( p) = lim f (t ) = (instantaneous elasticity) lim p →∞ t →0 + EM f * ( p) = limf (t ) = +∞ (unlimited creep) lim p →0 t →∞ r * ( p) = lim r(t ) = 0 (complete relaxation) lim t →∞ p →0 (5.1.7) 2) Parallel assemblage We now have qM = qK = q Q = QM + QK Qɺ = Qɺ M + Qɺ K (5.1.8) From (5.1.1), we have (Qɺ − Qɺ K ) (Q − QK ) + EM ηM ɺ Q − ( EK qɺ + η K qɺɺ) Q − ( EK q + η K qɺ ) = + ηM EM qɺ = (5.1.9) which may also be written as E η q + 1 + K + K ηM EM η M EK ηK Q Qɺ qɺɺ = + qɺ + η M EM EM (5.1.10) Like for the series assemblage, we can also use the Laplace-Carson transform technique with the fact that, for a parallel assemblage, the resulting relaxation function is the sum of the relaxation functions of each element. Thus, according to Sect. 5.1.3.3, we get for the Laplace-Carson transformed resulting relaxation function r *( p) = = Q*( p ) E η p = EK + η K p + M M * q ( p) EM + η M p ( EM + ηM p )( EK + ηK p ) + EMηM p ( EM + η M p ) which leads to the second-order polynomial relation with respect to p (5.1.11) 184 ( EM +ηM p ) Q*( p )= EM EK +( EMηK +EKηM +EMηM ) p+ηMηK p 2 q*( p) (5.1.12) It is easy to check that (5.1.12) admits the differential equation (5.1.10) as its original function of time. From the spring and dashpot representation, it is obvious that this model does not exhibit instantaneous elasticity (due to the dashpot of the Kelvin model), has limited creep (due to the spring of the Kelvin model) and cannot suffer a relaxation test (due to the dashpot of the Kelvin model, which cannot deform instantaneously). This is corroborated by the behaviour of r*(p) and f*(p) given by (5.1.11) or derived from it for p → 0 (i.e., t → ∞) and p → ∞ (i.e., t → 0+ ) : lim f * ( p) = lim f (t ) = 0 (no instantaneous elasticity) t →0 + p →∞ 1 f * ( p) = limf (t ) = (limited creep) lim p →0 t →∞ E K lim r* ( p) = lim r(t ) = ∞ (relaxation test impossible) t →0 p →∞ (5.1.13) Exercise 5.2 Creep recovery and relaxation 1) Preliminary creep recovery tests performed on a linear non-ageing viscoelastic material described by the parameters Q (load) and q (deformation) have shown that there remains a permanent strain qp proportional to the load Q0 and to the time interval τ during which this load has been imposed, say q p = AQ0τ , where A is a constant. Show from that property that the creep flow is unlimited with a finite limit creep rate and that the relaxation is complete. 2) A test at constant strain rate qɺ0 is performed on the same material. Show that Q tends toward a finite limit value Q∞ and derive the relation between Q∞ and qp (question 1), for fixed values of Q0, τ and qɺ0 .This test is performed up to t = θ and the load is then suppressed. Show that q then tends toward the permanent θ strain qp′ given by qp′ = A∫ Q (t )dt where A is the constant of question 1 0 Solution 1) A creep recovery test is described by the relations Q (t ) = Q0 [ Y(t ) − Y(t − τ ) ] q (t ) = Q0 [ f (t ) − f (t − τ ) ] ⇒ * * * * Q ( p ) = Q0 [1 − exp(− pτ )] q ( p ) = f ( p )Q ( p ) (5.2.1) 185 where Y(t) is the (Heaviside) unit step function and f(t) is the creep function. We have q p = lim q(t ) = lim q* ( p) = lim f * ( p)Q* ( p ) = lim f * ( p) .0 p →0 t →∞ p→0 (5.2.2) p →0 Since q p (= AQ0τ ) is finite and non zero, this means that f * ( p) → ∞ when p → 0 , i.e., that f (t ) → ∞ when t → ∞ : so, the creep flow is unlimited. Its time derivative for t → ∞ can be evaluated as follows * ( fɺ ) p q *( p ) = p f * ( p ) − f (0+ ) = * − pf (0+ ) Q ( p) (5.2.3) * () and, since f(0+) is necessarily finite, pf (0+ ) → 0 when p → 0 . So, fɺ like behaves p q *( p ) when p → 0 and we get Q* ( p ) p q *( p ) p q *( p ) lim fɺ (t ) = lim fɺ * ( p) = lim * = lim t →∞ p →0 p →0 p → 0 Q ( p) Q0 [1 − exp(− pτ ) ] (5.2.4) Now, we have lim p→0 1 − exp(− pτ ) =τ p lim q* ( p ) = lim q(t ) = qp = AQ0τ p →0 t →∞ (5.2.5) and then lim fɺ (t ) = t →∞ AQ0τ =A Q0τ (5.2.6) So, when p → 0 , we have pf * ( p ) ≈ A, f * ( p) ≈ A * 1 p , r ( p) = * ≈ , lim r(t ) = lim r* ( p ) = 0 p f ( p ) A t →∞ p →0 which means that the relaxation is complete. 2) We impose q (t ) = qɺ0tY(t ) . For the Laplace-Carson transforms, we get (5.2.7) 186 q* ( p) = qɺ0 p ⇒ Q* ( p ) = r * ( p ) q * ( p ) = qɺ0 r* ( p ) p (5.2.8) The limit values for t → ∞ derive from qɺ0 r* ( p) qɺ0 = = Q∞ p →0 p A lim Q (t ) = lim Q* ( p) = lim p →0 t →∞ since we know that (5.2.9) r* ( p) 1 ≈ when p → 0 . With qp = AQ0τ , we get p A qp = Q0 qɺ0τ Q∞ (5.2.10) The general relation t q (t ) = Q (t )f (0) + ∫ Q (τ )f ′(t − τ )dτ 0 (5.2.11) reads, for t > θ , t >θ θ ⇒ q(t ) = ∫ Q (τ )f ′(t − τ ) dτ 0 (5.2.12) So, when t → ∞, f ′(t − τ ) → A and then θ lim q(t ) = A∫ Q(τ ) dτ = qp′ t →∞ 0 (5.2.13) Exercise 5.3 Tensile and relaxation tests A tensile test at constant axial strain rate qɺ0 is performed on a linear non-ageing viscoelastic material described by the parameters Q (load) and q (deformation). The load response has the form Q(t ) = qɺ0 α (1 − exp(− at ) ) + β where α, β and a are constants obtained experimentally. (5.3.1) 187 1) Find the differential equation obeyed by this material. Calculate the creep and relaxation functions and give the stiffness and viscosity characteristics of two equivalent rheological models which can be used to represent this viscoelastic behaviour. 2) The tensile test is performed up to time t0 and then followed by a relaxation test. Calculate Q(t) for t ≥ t0 . Solution 1) Using the Laplace-Carson transforms of q(t) and Q(t), we get q(t ) = qɺ0tY(t ) ⇒ Q (t ) = qɺ0 α (1 − exp(−at ) ) + β * qɺ0 p ⇒ Q* ( p) = qɺ0 (α + β ) − qɺ0α q ( p) = p p+a (5.3.2) Thus, the relation between q* ( p ) and Q* ( p) may read Q* = (α + β ) pq* − α p2 * q p+a (5.3.3) or pQ* + aQ* = β p 2 q* + a(α + β ) pq* (5.3.4) whose original is the differential equation Qɺ + aQ = β qɺɺ + a (α + β )qɺ (5.3.5) From (5.3.4), we can write ( p + a)Q* = p ( β p + a(α + β ) ) q* and then (5.3.6) 188 α 1 * q p + a β (α + β ) α +β f * ( p) = * = = + α Q p ( β p + a (α + β ) ) p p + a(1 + ) β * p ( β p + a(α + β ) ) * Q p = β p + aα r ( p) = * = q p+a p+a (5.3.7) The original creep and relaxation functions then read 1 α α tY(t ) + 1 − exp −a (1 + )t Y(t ) f (t ) = 2 α +β β a (α + β ) r* ( p) = βδ(t ) + aα exp −at Y(t ) ( ) (5.3.8) Referring to the creep functions of a Kelvin model (EK, ηK) and of a dashpot (ηD), we conclude that the studied viscoelastic behaviour can be represented by a series assemblage of the dashpot ( η D = α + β ) and the Kelvin model defined by a β the parameters ( EK = (α + β ) 2 , η K = (α + β ) ). Equivalently, referring to the α α relaxation functions of a Maxwell model (EM, ηM) and of a dashpot (ηD’), we conclude that the studied viscoelastic behaviour can be represented as well by a parallel assemblage of the dashpot ( η D′ = β ) and the Maxwell model defined by the parameters ( EM = aα , η M = α ). 2) The whole deformation path is described by the relation q (t ) = qɺ0 [ tY(t ) − (t − t0 )Y(t − t0 ) ] (5.3.9) which transforms into q* ( p) = qɺ0 1 − exp ( − pt0 ) p (5.3.10) Thus we have Q* ( p) = r* ( p)q* ( p) = qɺ0 ( β + aα ) (1 − exp(− pt0 ) p+a (5.3.11) The original function Q(t) has then the form Q(t ) = qɺ0 [ Φ (t ) − Φ (t − t0 ) ] (5.3.12) 189 with Φ (t ) defined by its Laplace-Carson transform Φ * ( p ) given by aα p = α + β −α p+a p+a (5.3.13) Φ (t ) = (α + β )Y(t ) − α exp(−at )Y(t ) (5.3.14) Φ* ( p ) = β + So, we find and then we get for t ≥ t0 Q(t ) = qɺ0α exp(−at ) [ exp(at0 ) − 1] (5.3.15) Thus we note that there is a load discontinuity at t = t0 with + Q (t0 ) = qɺ0α [1 − exp(−at0 ) ] − Q (t0 ) = qɺ0α [1 − exp(−at0 ) ] + β qɺ0 ⇒ ∆Q (t0 ) = − β qɺ0 (5.3.16) Exercise 5.4 Complex modulus 1) A preliminary approach to the modelling of the 1-D response of a linear nonageing viscoelastic material uses a parallel assemblage of one Kelvin model and one Maxwell model, as studied in Exercise 5.1. Using the solution of this exercise (part 2) for the relaxation function, show that the parameters EK, ηK, EM and ηM of this model can be identified from the sole following information obtained from the normalised response Q(t )/qɺ0 to the stepwise deformation rate qɺ0 Y(t ) : value and time derivative at t = 0+ , ordinate at t = 0 and slope of the asymptote for t →∞. 2) In order to get an improved modelling of the observed response, we consider now the parallel assemblage of one Kelvin model ( E∞ ,η0 ) and n Maxwell models ( Ei ,ηi , i = 1 to n) and we compare the corresponding complex modulus to the experimental one. Calculate this modulus as a function of the parameters E∞ ,η0 , Ei , the relaxation times τ i = ηi /Ei and the pulsation ω. 3) A further improvement consists in the consideration of a continuous relaxation spectrum g(τ ) . Calculate the corresponding complex modulus and find its limits for ω → 0 and ω → ∞ . Draw qualitatively the associated Cole and Cole diagram. Find the parameters EK, ηK, EM and ηM of a scheme with only one Maxwell 190 model (see question 1) which would lead to the best approximation to this diagram when, at once, ω → 0 and ω → ∞ . Solution 1) From (5.1.11) in Exercise 5.1, we get r *( p) = E η p Q*( p ) = EK + η K p + M M * q ( p) EM + η M p (5.4.1) The normalised response Q(t )/qɺ0 to the stepwise deformation rate qɺ0 Y(t ) can be derived from its Laplace-Carson transform through the relations q* ( p ) = qɺ0 Q* ( p) = p r* ( p) ⇒ EMη M Q * r * EK = = + ηK + qɺ0 p p EM + η M p (5.4.2) The limit values for t → 0, ∞ are obtained from the limits of the LaplaceCarson transforms for p → ∞,0 , namely Q* EK + EM Q (t ) , ≈ η K + ( EK + EM )t t →0+ ⇔ p →∞ ⇒ ɺ ≈ η K + q p qɺ0 0 (5.4.3) * t →∞ ⇔ p →0 ⇒ Q ≈ η + η + EK , Q(t ) ≈ η + η + E t K M K M K qɺ0 p qɺ0 Consequently, the four model parameters EK, ηK, EM and ηM can be obtained Q (t ) from the experimental curve using the values ηK and ( η K + ηM ) and the qɺ0 slopes ( EK + EM ) and EK. 2) The Laplace-Carson transform of the relaxation function now reads r *( p) = E∞ + η0 p + ∑ i Eiηi p Ei + ηi p (5.4.4) and the complex modulus r *(iω ) , with i 2 = −1 , is given by r *(iω ) = E∞ + iη0ω + ∑ i iEiηiω Ei + iηiω This can also be written in the classical form, with τ i = ηi /Ei : (5.4.5) 191 Eiω 2τ i2 ′ G ( ) = E + ω ∑i 1 + ω 2τ 2 ∞ i r *(iω ) = G′(ω ) + iG′′(ω ) with τ E G ′′(ω ) = ω η + i i 0 ∑ 2 2 i 1+ ω τi (5.4.6) 3) For a continuous relaxation spectrum g(τ ) , with τ varying from 0 to ∞, the Maxwell models with a relaxation time lying between τ and ( τ + dτ ) have the equivalent stiffness dE = g(τ )dτ . Consequently, by analogy with (5.4.6), we get r *(iω ) = E∞ + ω 2 ∫ ∞ 0 ∞ τ g(τ ) τ 2 g(τ ) dτ + iω η 0 + ∫ dτ 2 2 2 2 0 1+ ω τ 1+ ω τ (5.4.7) When ω → 0 , we have ∞ ∞ r *(iω )ω ≈0 ≈ E∞ + iω η0 + ∫ τ g(τ )dτ + ω 2 ∫ τ 2 g(τ )dτ + ... 0 0 (5.4.8) whereas, when ω → ∞ , ∞ r *(iω )ω →∞ ≈ E∞ + ∫ g(τ )dτ + iη0ω + ... 0 (5.4.9) So, the associated Cole and Cole diagram qualitatively looks like suggested in Fig. 5.4.1. For the simple model of Question 1, the complex modulus would read, according to (5.4.6) with only one Maxwell model r1*(iω ) = EK + EMω 2τ M2 E τ + iω η K + M 2M 2 1 + ω 2τ M2 1 + ω τM (5.4.10) Fig. 5.4.1 Qualitative sketch of the Cole and Cole diagram for a continuous relaxation spectrum 192 At very low and very high frequencies, we have * η M2 2 ω + ... r1 (iω )ω ≈0 ≈ EK + iω (η K + η M ) + EM r*(iω ) ω →∞ ≈ EK + EM + iη Kω + ... 1 (5.4.11) So, the best approximation to (5.4.7) by a parallel assemblage of one Kelvin model and one Maxwell model when, at once, ω → 0 and ω → ∞ , is obtained when the following conditions are fulfilled: E = E η + η = η + ∞τ g(τ )dτ K M 0 ∞ ∫0 K ∞ EK + EM = E∞ + ∫ g(τ )dτ η K = η0 0 (5.4.12) EK = E∞ η K = η0 ∞ ∞ EM = ∫0 g(τ )dτ η M = ∫0 τ g(τ )dτ (5.4.13) that is when 5.2 Local viscoelastic constitutive equations and response of viscoelastic structures The Boltzmann superposition principle can still be used to derive the local expression of the constitutive equations for linear viscoelastity. Their general form is unchanged, except that the scalar functions must be changed into fourth-order tensors. Advantage may be taken of the assumption of nonageing and of the use of the Laplace-Carson transform to reduce viscoelastic to elastic structural design through the correspondence theorem. This approach may also be used to derive estimates for the overall properties of heterogeneous viscoelastic materials (see Exercise 5.8). Exercise 5.5 Identification of constitutive equations 193 An experimental investigation is performed in view of identifying the local constitutive equation of an isotropic linear non ageing viscoelastic material in isothermal conditions. This equation can be written (see Eq. 5.110 in Volume I) σ (t ) = λ (t ) ⊙ Tr ε (t ) δ + 2µ (t ) ⊙ ε (t ) E (t ) ⊙ ε (t ) = [1 + ν (t )] ⊙ σ (t ) −ν (t ) ⊙ Tr σ (t ) δ (5.5.1) Two tests (simple shearing and hydrostatic pressure) are conducted in order to identify the functions E(t) and ν(t). 1) Simple shear: in its principal axes, the stress tensor σ (t ) reads 0 σ (t ) 0 (σ (t ) ) = 0 −σ (t ) 0 0 0 0 (5.5.2) Show that this test can be characterised by only two parameters, namely Q(t ) = σ (t ) for the load and q(t) for the deformation. What will be q(t)? Express * the Laplace-Carson transform rSS ( p ) of the relaxation function for this test and for these parameters as a function of E *( p) and ν *( p) . A “dynamic” investigation (sinusoidal vibration q = q0 cos ω t with the stationary response with the amplitude Q0 , the same pulsation ω and the phase lag ϕ (ω ) ) yields the relations Q0 Bω with B > 0 R(ω ) = q = ( A2 + ω 2 )1/2 0 π tan ϕ (ω ) = A with A > 0, 0 ≤ ϕ ≤ ω 4 (5.5.3) Calculate the complex modulus in the form G′(ω ) + iG′′(ω ) . Check that it has the same form as the one associated to a parallel assemblage of one Kelvin model (EK, ηK) and one Maxwell model (EM, ηM) and find the associated relaxation function rSS (t ) . 2) Hydrostatic pressure: the stress field σ (t ) = − P(t )δ (with P > 0 ) is uniform. Find the deformation parameter q(t) which can be associated with the load pa* ( p ) of the rerameter Q(t ) = − P(t ) . Express the Laplace-Carson transform rHP laxation function for this test and for these parameters as a function of E *( p) and ν *( p) . This test is performed at constant deformation rate, i.e., qɺ = qɺ0Y (t ) and the response is 194 Q(t ) = qɺ0 (Ct + D ) C >0 D>0 (5.5.4) * Calculate rHP ( p ) as a function of C and D and find a rheological model consistent with this relaxation function. 3) Application: Calculate E *( p) and ν *( p) and comment qualitatively on their variations. A horizontal cantilever beam, with a square cross-section, made of this material, is fixed at one end and subjected at the other end to a constant vertical force F parallel to two edges of the cross-section. Calculate the vertical displacement v(t) at the free end of the beam and find a rheological model yielding the same response for the parameters (F, v). Solution 1) Simple shear: the deformation work variation per unit volume must read dW = Qdq = σ ij dε ij → q = ε11 − ε 22 = σ d(ε11 − ε 22 ) (5.5.5) Application of the Laplace-Carson transform of (5.5.1) yields 1 +ν * * ν * * 1 +ν * * * σ 11 − * (σ 11 + σ 22 σ )= * 2(1 + ν * ) * E E E* * ⇒ q = σ E* 1 +ν * * ν * * 1 +ν * * * σ σ σ σ = − ( + ) = − 22 11 22 E* E* E* ε11* = * ε 22 (5.5.6) With Q* = σ * , the corresponding (transformed) relaxation function is then * rSS ( p) = Q* ( p ) E* ( p) = = µ * ( p) * * q ( p) 2 (1 +ν ( p) ) (5.5.7) * From R(ω) and tan ϕ , the complex modulus rSS (iω ) can be derived by the relations * rSS (iω ) = G′(ω ) + iG ′′(ω ) = R(ω ) ( cos ϕ (ω ) + isin ϕ (ω ) ) = R(ω ) cos ϕ (ω ) (1 + i tan ϕ (ω ) ) Since 0 ≤ ϕ ≤ π , one has simply 4 (5.5.8) 195 cos ϕ = 1 2 1 + tan ϕ 1 = 1+ A 2 ω2 = ω (A 2 1/2 + ω2 ) (5.5.9) and then * rSS (iω ) = Bω ω A 1 + i ( A2 + ω 2 )1/2 ( A2 + ω 2 )1/2 ω (5.5.10) Bω 2 ABω = 2 +i 2 A + ω2 A + ω2 As already shown in Exercise 4 (see Eq. 5.4.6), the complex modulus for a parallel assemblage of one Kelvin model and one Maxwell model reads * rK/ / M (iω ) = EK + EMω 2τ M2 E τ + iω ηK + M 2M 2 with τ M = η M /EM (5.5.11) 2 2 1+ ω τM 1+ ω τM So, (5.5.10) and (5.5.11) coincide if EK = 0 η K = 0 EM = B η M = B / A (5.5.12) which means that one single Maxwell model is needed to model the studied response in simple shear. The associated relaxation function reads rSS (t ) = EM exp(− EM ηM t ) = B exp(− At ) (5.5.13) 2) Hydrostatic pressure: the deformation work variation per unit volume must read dW = Qdq = σ ij dε ij → q = ε kk = − Pdε kk (5.5.14) From (5.5.1), we get E *( p) ε kk* ( p) = 1 − 2ν * ( p) σ kk* ( p) = −3 1 − 2ν * ( p ) P* ( p) (5.5.15) E *( p) q* ( p) = 3 1 − 2ν * ( p) Q* ( p) (5.5.16) and then 196 or * rHP ( p) = Q* ( p ) E * ( p) = q* ( p) 3 1 − 2ν * ( p) (5.5.17) With qɺ = qɺ0Y (t ) and Q(t ) = qɺ0 (Ct + D) , we find E * ( p) * ⇒ r ( p ) = C + Dp = C HP Q* = qɺ0 ( + D ) 3 1 − 2ν * ( p) p pq* = qɺ0 (5.5.18) i.e., the transformed relaxation function of the Kelvin model ( EK = C ,η K = D ). 3) Application: from (5.5.7) and (5.5.17), we get * * * 3rHP − 2rSS ν ( p) = * * 2 ( rSS + 3rHP ) * * 9rSS rHP * E ( p) = r* + 3r* SS HP (5.5.19) From (5.5.13) and (5.5.18), namely Bp * rSS ( p) = p +A r* ( p) = C + Dp HP (5.5.20) 3(C + Dp )( p + A) − 2 Bp * ν ( p) = 2 ( 3(C + Dp)( p + A) + Bp ) 9 Bp(C + Dp ) E * ( p) = 3(C + Dp)( p + A) + Bp (5.5.21) E *( p) and ν *( p) are given by Without trying to perform the complete inversion of these Laplace-Carson transformed functions, we can at least easily derive the limit values of the original functions and their time derivatives when t → 0+ and t → ∞ from the limit values of ν * ( p ), p ν * ( p) −ν (0+ ) , E * ( p) and p E * ( p) − E (0+ ) when p → ∞ and p → 0 , respectively. We find that ν (t ) starts from ½ at t = 0 and decreases 197 (ν ′(0+ ) = − B / 2 D < 0 ) whereas it tends to the asymptotic value ½ when t → ∞ . As for E (t ) , it starts from 3B and decreases ( E ′(0+ ) = − B( B + 3 AD ) / D < 0 ) and it tends to 0 when t → ∞ . The problem of the cantilever beam can be solved by using the elastic solution and the correspondence theorem (see Sect. 5.3.2.1 in Volume I). For linear isotropic elasticity, the displacement at the free end of the beam vel is known to be vel = Fl 3 3EI with I = a4 12 (5.5.22) where E is the Young modulus, l is the beam length, I is the principal moment of inertia and a is the edge length of the square cross-section. Consequently, the Laplace-Carson transformed viscoelastic solution v(t ) will be v* = F *l 3 3E * I with I = a4 12 (5.5.23) With F constant, we find v* = 4 Fl 3 1 a4 E* ⇒ v(t ) = 4 Fl 3 1 J (t ) with J *( p) = * 4 a E ( p) (5.5.24) According to (5.5.21), we have 3(C + Dp)( p + A) + Bp 9 Bp(C + Dp ) 1 A 1 = + + 3B 3Bp 9 D( p + C ) D (5.5.25) 1 A 1 C + t+ 1 − exp − t 3B 3B 9C D (5.5.26) J * ( p) = and then J (t ) = and hence v(t ) by (5.5.24). 198 Obviously, J (t ) is the creep function of a Burgers model, i.e. of the series as3B semblage of the Maxwell model ( EM = 3B, η M = ) and of the Kelvin model A ( EK = 9C , ηK = 9 D ). Exercise 5.6 Responses of a circular thin tube A circular thin tube (height h, mean radius R0, thickness e ≪ R0 ), open at its ends, is made of a linear isotropic non ageing viscoelastic material. 1) This tube is subjected to a uniform internal pressure P, considered as the load parameter Q for this test. What should be the associated deformation parameter q? The creep function for this test fIP relating Q and q is found to read f IP (t ) = a + b t (5.6.1) Why must the constants a and b be positive? Does the studied behaviour exhibit for this test instantaneous elasticity? limited or unlimited creep flow? complete or partial relaxation? complete or partial creep recovery? (NB: remind that α * (t ) π 1 ! = 2 2 = α ! p −α (0 ≤ α ≤ 1) ) (5.6.2) Show that such a response is not consistent with a finite and discrete relaxation spectrum. The stress state which results from the uniform internal pressure applied to this circular cylindrical open tube reads, in cylindrical polar coordinates (r, θ, z) (σ ) ( r ,θ , z ) 0 0 = 0 Σ 0 0 0 0 0 (5.6.3) Prove that Σ = PR0 /e and derive from that the relation between the LaplaceCarson transformed creep (or relaxation) function f IP* ( p) (or rIP* ( p) ) and the transformed viscoelastic Young modulus E * ( p). 2) The same tube is subjected to axial sinusoidal vibrations with the pulsation ω. Calculate the associated complex compliance J * (iω ) = 1 / E * (iω ) = E (ω )exp ( iϕ (ω ) ) −1 (5.6.4) 199 Show that the Cole and Cole diagram for the complex modulus E * (iω ) is a circular arc starting at the origin for ω = 0 . 3) A static tensile test is performed on the same tube. It is observed that, during this test, the lateral shortening ε 22 is always three times smaller that the axial strain ε 11 . Deduce from this property and from what precedes the constituve equations of the studied material. Solution 1) The increment of strain energy by unit volume reads dW = P d(πR 2 h) dR =P = Pdq 2πRhe e ⇒q= R − R0 e (5.6.5) The creep function must be positive and non decreasing. Consequently a and b must be positive. Since f (0+ ) = a ≠ 0, the response exhibits instantaneous elasticity and since f (t ) → ∞ when t → ∞, the creep flow is unlimited. The LaplaceCarson transformed creep function reads f IP* ( p) = a + b π 1 = * 2 p rIP ( p) (5.6.6) A relaxation test exhibits no singularity at t = 0+ (since lim rIP* ( p ) = 1/a ≪ ∞ ). p →∞ * IP In addition, the relaxation is complete (since lim r ( p ) = 0 ). A creep recovery test p →0 is ruled by the following equations P(t ) = P0 [ Y(t ) − Y(t − τ )] ⇒ P* ( p) = P0 [1 − exp(− pτ )] b π q* ( p) = f IP* ( p) P* ( p ) = P0 a + [1 − exp(− pτ ) ] 2 p (5.6.7) b π lim q(t ) = lim q* ( p) = lim P0 pτ a + =0 t →∞ p →0 p →0 2 p So, the creep recovery is complete. For a viscoelastic body exhibiting instantaneous elasticity and unlimited creep flow defined by a finite and discrete relaxation spectrum, the transformed creep function reads (see Sect. 5.1.3.4 in Volume I) 200 f * ( p) = 1 1 Js + + ∑s E0 η ∞ p 1 + pτ s (5.6.8) with E0 ≠ 0 and η∞ ≪ ∞ . The creep recovery response is then lim q(t ) = lim q* ( p ) = lim P0 pτ f * ( p) = P0 p →0 t →∞ p →0 τ η∞ (5.6.9) which means that the recovery is only partial. Consequently, we can state that the studied behaviour cannot be described by a finite and discrete relaxation spectrum. Using (5.6.3), we can write the strain energy density (5.6.5) in the form dW = P dR dR = σ : dε = Σ e R0 ⇒ Σ=P R0 e (5.6.10) The transformed constitutive equations yield 1 + ν * * ν * * Σ * R0 P* Σ − *Σ = * = E* E E e E* * R − R0 R r e = = q* = 0 IP* q* R0 R0 e E * = ε θθ (5.6.11) so that we find rIP* ( p) = 1 e2 e2 1 = 2 E * ( p) = 2 * f ( p ) R0 R0 J ( p) * IP (5.6.12) 2) According to (5.6.12), the complex compliance reads J * (iω ) = e2 * e2 b π f (iω ) = 2 a + 2 IP R0 R0 2 iω (5.6.13) For physical reasons (namely the positivity of G’(ω) and G”(ω), which will be 1 1− i checked later on), we have to choose = instead of its opposite and i 2 (5.6.13) becomes J * (iω ) = e2 b π b π a+ −i 2 R0 2 2ω 2 2ω (5.6.14) 201 If ψ is the phase lag for J * (iω ) , we have ϕ = −ψ b π 2 2ω ⇒ tan ϕ = − tanψ = b π a+ 2 2ω (5.6.15) From this expression, several simplifications can be derived, namely b π a tan ϕ = 2 2ω 1 − tan ϕ E = E * (iω ) = ⇒ J = J * (iω ) = ae 2 1 + tan 2 ϕ ae 2 = R02 (1 − tan ϕ ) R02 (cos ϕ − sin ϕ ) R02 (cos ϕ − sin ϕ ) ae 2 R02 (cos ϕ − sin ϕ )(cos ϕ + isin ϕ ) ae 2 R02 ′ ω G ( ) cos ϕ (cos ϕ − sin ϕ ) = R2 ae 2 ⇒ G ′2 + G ′′2 − 02 (G ′ − G ′′) = 0 2 ae G ′′(ω ) = R0 sin ϕ (cos ϕ − sin ϕ ) 2 ae E * (iω ) = (5.6.16) The last relation between G’(ω) and G”( ω) is the equation of the Cole and Cole curve (with the restriction G′ ≥ 0, G′′ ≥ 0 ): it is the equation of an arc or circle, centred at ( R02 R02 , − ) , starting from the origin when ω → 0 and tending to 2ae 2 2ae 2 R02 ,0) when ω → ∞ . ae 2 3) The described property implies the relation ( 1 3 1 3 * ε 22 = −ν *ε11* = − ε11* ⇒ ν * ( p) = = ν (t ) (5.6.17) The constitutive equations for the studied material then read 4 * 1 * * * E ( p)ε ( p) = 3 σ ( p) − 3 Tr σ ( p) δ or 4 1 ε (t ) = J (t ) ⊙ σ (t ) − J (t ) ⊙ Tr σ (t ) δ 3 3 (5.6.18) 202 with J * ( p) = 1 e2 e2 π = 2 f IP* ( p ) = 2 a + b E ( p) R0 R0 2 p * (5.6.19) 5.3 Miscellaneous Exercise 5.7 Ageing viscoelasticity A concrete slab has been poured on a horizontal rigid substratum with conditions of perfect bonding at the interface. The slab, with the thickness 0.25m, is considered as infinite in the horizontal directions x2 and x3 and there are no exterior forces. When it is not subjected to stresses, the concrete material suffers at age u a SH shrinkage isotropic contraction ε (u ) = −a(u )δ . It obeys an isotropic ageing linear viscoelastic behaviour. In Table 5.7.1 are given, for several values of the age u - the delayed Young relaxation function E(t,u) for t = 200 days - the instantaneous Young relaxation function E(u,u) Table 5.7.1 Observation data u days a(u) mm/m E(200,u) MPa E(u,u) MPa 0 0 5,000 20,000 7 0.1 10,000 32,000 28 0.2 15,000 38,000 77 0.3 18,000 39,000 163 0.4 22,000 40,000 200 0.425 40,000 40,000 We aim at predicting the risk of cracking, given that the tensile strength of the used concrete is about 4 MPa. It is assumed that - the rigid substratum forces the resulting displacement field to reduce to its vertical component, depending on x1 only, say v1 ( x1 ). - the sole non zero components of the stress field are σ 22 and σ 33 . - the Poisson ratio is constant and equal to 0.2. 1) Draw a qualitative plot of the time variations of the relaxation functions E(t,u) for several values of u. 2) Use first an elastic approximation of the concrete behaviour to estimate the tensile stress σ 22 for t = 200 days with the elastic Young modulus given by 203 E (u, u ) for u = 0, 7, 28, 77, 163 and 200 days. What can be concluded from that concerning the risk of crack formation? 3) Going back to the actual ageing viscoelastic behaviour, give the theoretical expression of σ 22 (t ) and find numerically an estimate of it for t ≤ 200 days by use of a stepwise approximation of a (u ) . Comment on the influence of viscosity on the risk of cracking. Solution 1) See Fig. 5.7.1. E (t , u ) E (u, u ) E (t , u4 ) E (t , u3 ) E (t , u2 ) 0 u1 u2 u3 u4 E (t , u1 ) u, t Fig. 5.7.1 Qualitative plot of the time variations of E (t, u) for several values of u 2) According to an elastic treatment, we would have, with ε ije the elastic strain ε ij = ε ijSH + ε ije = −a(u )δij + 1 +ν ν σ ij − σ kk δij E E (5.7.1) By symmetry, σ kk = σ 22 + σ 33 = 2σ 22 . From (5.7.1) and ε 22 = 0 , we get ε 22 = 0 = −a(u ) + and then 1 −ν σ 22 E (5.7.2) 204 σ 22 = E a(u ) 1 −ν (5.7.3) At t = 200 days , we find σ 22 (200) = .425x10−3 E = .53,125x10−3 E .8 (5.7.4) With E = E (u, u ) for u = 0, 7, 28, 77, 163 and 200 days, we get σ 22 (200) = 10.6, 17.0, 20.2, 20.7, 21.3 and 21.3 MPa , respectively (5.7.5) So, with σ 22 (200) ≫ 4 MPa , an elastic treatment would unambiguously predict the formation of shrinkage induced cracks. 3) According to the general expression (5.101) in Volume I and taking the shrinkage deformation and the isotropy into account, we can write, with σ (0+ ) = 0 t t t 0 0 0 ∫ E (t , u)d (ε (u) + a(u)δ ) = ∫ (1 +ν (t , u ) ) d (σ (u ) ) − ∫ ν (t , u )d ( Trσ (u )δ ) (5.7.6) Since ν (t , u ) is constant, this relation reads t ∫ E (t , u)d (ε (u ) + a(u )δ ) = (1 +ν )σ (t ) −ν Trσ (t )δ (5.7.7) 0 As before in 2), σ kk = σ 22 + σ 33 = 2σ 22 and ε 22 = 0 so that (5.7.7) yields σ 22 (t ) = 1 1 −ν t ∫ E (t , u )da(u ) (5.7.8) 0 Table 5.7.2 Stepwise approximation for a numerical integration of (5.7.8) ∆u1 = 7days ∆a1 = 10 −4 E1 = 104 ∆u2 = 21days ∆a1 = 10 ∆u3 = 49days −4 E2 = 1.5x104 ∆a1 = 10 ∆u4 = 86days −4 ∆a1 = 10 E3 = 1.8x104 −4 E4 = 2.2x104 ∆u5 = 37days ∆a1 = .25x10−4 E2 = 4x104 This integral can be approximated numerically by σ 22 (t ) ≃ 1 1 −ν ∑ 5 i =1 Ei ∆ai (5.7.9) 205 with the stepwise approximation of Table 5.7.2. This treatment yields an estimate of σ 22 (t ) for different values of t, namely σ 22 (7days) ≃ 1.25MPa σ (28days) ≃ 3.125MPa 22 σ 22 (77days) ≃ 5.375MPa σ (163days) ≃ 8.125MPa 22 σ 22 (200days) ≃ 9.375MPa (5.7.10) These results show that the risks of crack formation are lowered by the action of viscosity but that these risks remain present (here after about 40 days) due to the ageing behaviour of concrete. Exercise 5.8 Viscoelastic two-phase material An isotropic two-phase viscoelastic material is constituted of two isotropic incompressible non-ageing linear viscoelastic maxwellian phases (i), with the volume fractions c1 and c2, which obey the constitutive equations eɺ = sɺ Ei + s ηi (5.8.1) where e and s are the strain and stress deviators, respectively. Due to a particulate morphology of the composite, the Mori-Tanaka model is used to get an estimate of the overall behaviour, with phase (2) considered as a spherical inclusion in an infinite matrix made of phase (1). Show that the effective behaviour can be described by one scalar creep function f MT (t ) . Prove that f MT* ( p ) is a rational fraction in p and, without trying to derive f MT (t ) explicitly, give the conditions for it to describe the creep response of a parallel assemblage of one Kelvin model (EK, ηK) and one Maxwell model (EM, ηM). Derive the corresponding transformed effective relaxation function r MT* ( p) and show that the associated relaxation spectrum reduces to two single lines. Comment qualitatively on the differences between the relaxation spectra as predicted by the self-consistent scheme (see Sect. 5.3.3.3 in Volume I) and the Mori-Tanaka model. Solution 206 Due to the constitutive behaviour of both phases, the overall behaviour of the twophase material is necessarily isotropic incompressible non-ageing linear viscoelastic and obeys the constitutive equations S (t ) = 2µ MT (t ) ⊙ E (t ) or * * S ( p ) = 2µ MT* ( p )E ( p) (5.8.2) where S (t ) and E (t ) are the macroscopic stress and strain deviators. These equations show that the effective behaviour can be described by one scalar creep function f MT (t ) whose Laplace-Carson transform is given by f MT* ( p) = 1 1 = MT* r ( p) 2µ ( p) MT* (5.8.3) The elastic Mori-Tanaka estimate of the shear modulus is known to be (see Eq. 2.194), with f = c2 : µ MT = µ1 5 [ (1 − f ) µ1 + f µ 2 ] + 2(1 − f )( µ 2 − µ1 ) 5µ1 + 2(1 − f )( µ 2 − µ1 ) (2 + 3 f )( µ 2 − µ1 ) = µ1 1 + 5µ1 + 2(1 − f )( µ2 − µ1 ) (5.8.4) According to the correspondence theorem, we can then write for the viscoelastic problem µ MT* 5 (1 − f ) µ1* ( p ) + f µ 2* ( p ) + 2(1 − f ) ( µ2* ( p) − µ1* ( p ) ) ( p) = µ ( p) * 5µ1 ( p) + 2(1 − f ) ( µ2* ( p) − µ1* ( p) ) * 1 5µ1* ( p) + 2(1 − f ) ( µ2* ( p) − µ1* ( p) ) 1 f MT* ( p) = * * * * * 2µ1 ( p ) 5 (1 − f ) µ1 ( p) + f µ 2 ( p) + 2(1 − f ) ( µ2 ( p) − µ1 ( p ) ) (5.8.5) or µ2* ( p) µ1* ( p) 1 f MT* ( p) = * µ * ( p) 2µ1 ( p ) 3(1 − f ) + (2 + 3 f ) 2* µ1 ( p) (3 + 2 f ) + 2(1 − f ) with, from (5.8.1) (5.8.6) 207 * s ( p) 1 1 * * e ( p) = + s ( p) = * 2 µi ( p ) Ei ηi p (5.8.7) and then 1 1 1 = + 2µ ( p ) Ei ηi p * i or 2µi* ( p ) = Eiηi p Ei + ηi p ⇓ (5.8.8) * 2 * 1 µ ( p) E2η 2 E1 + η1 p = µ ( p) E1η1 E2 + η2 p We then have f MT* ( p) = = ( E1 + η1 p ) [ (3 + 2 f ) E1η1 ( E2 + η 2 p) + 2(1 − f ) E2η 2 ( E1 + η1 p)] E1η1 p [3(1 − f ) E1η1 ( E2 + η2 p) + (2 + 3 f ) E2η 2 ( E1 + η1 p )] (η1 p + E1 )( Ap + B ) E1η1 p(Cp + D) (5.8.9) E1 BC ( E1C − η1 D)( AD − BC ) A η1 AD η1 ACD = + 1 + D E1C p p+ C This result proves first that f MT* ( p ) is a rational fraction in p. After decomposition into simple elements, it happens then to read as the sum of two contributions: the two first terms in the brackets express the creep behaviour of a Maxwell model (EM, ηM) with EM = E1C ηD ηM = 1 A B iff C ≥ 0 and A D ≥0 B (5.8.10) It is easy to check in (5.8.9) that both these conditions are satisfied, since f ≤1. The third term in the brackets expresses the creep behaviour of a Kelvin model (EK, ηK) with 208 E1η1CD 2 E = K ( E1C − η1 D )( AD − BC ) E1η1C 2 D η = K ( E1C − η1 D )( AD − BC ) iff ( E1C − η1 D )( AD − BC ) ≥ 0 (5.8.11) This condition is fulfilled too since A =η1η 2 [ (3 + 2 f ) E1 + 2(1 − f ) E2 ] B =E1 E2 [ (3 + 2 f )η1 + 2(1 − f )η 2 ] E1C−η1 D = 3(1 − f ) E1η1 ( E1η 2 − E2η1 ) ⇒ (5.8.12) C =η1η 2 [3(1 − f ) E1 + (2 + 3 f ) E2 ] AD−BC = 25 fE1 E2η1η 2 ( E1η 2 − E2η1 ) D =E1 E2 [3(1 − f )η1 + (2 + 3 f )η 2 ] From (5.8.9), the Laplace-Carson transformed relaxation function reads D p+ E η p ( Cp + D ) E C C 1 1 r MT* ( p ) = = 1 p (η1 p + E1 )( Ap + B) A ( p + E1 )( p + B ) A η1 (5.8.13) After a decomposition of this rational fraction into simple elements, we find A( E C − η D ) p η ( AD − BC ) p E 1 1 1 r MT* ( p ) = + 1 E1 B A( E1 A − η1 B) p+ p+ η1 A (5.8.14) We know that E1/η1 and B/ A are positive and ( E1C − η1 D ) and ( AD − BC ) are given by (5.8.12); In addition we have E1 A −η1 B = E1η1η 2 [ (3 + 2 f ) E1+ 2(1 − f ) E2 ] − E1 E2η1 [ (3 + 2 f )η1+ 2(1 − f )η 2 ] = (3 + 2 f ) E1η1 ( E1η 2 − E2η1 ) (5.8.15) Consequently we get E1C − η1 D 3(1 − f ) E1η1 ( E1η 2 − E2η1 ) 3(1 − f ) E A − η B = (3 + 2 f ) E η ( E η − E η ) = (3 + 2 f ) 1 1 1 1 1 2 2 1 AD − BC 25 f E E η η ( E η − E 25 f E2η 2 1 2 1 2 1 2 2η1 ) = = E1 A − η1 B (3 + 2 f ) E1η1 ( E1η2 − E2η1 ) 3+ 2 f (5.8.16) 209 and (5.8.14) reads simply r MT* ( p ) = 3(1 − f ) E1 p 25 f E1 E2η1η 2 p + 3 + 2 f p + E1 (3 + 2 f ) A p + B A η1 (5.8.17) The original relaxation function r MT (t ) is then r MT (t ) = E 25 f E1 E2η1η 2 3(1 − f ) E1 B exp − 1 t + exp − t η 3+ 2 f (3 + 2 f ) A A 1 (5.8.18) This is the relaxation function of a parallel assemblage of two Maxwell models with the relaxation times τ 1 = η1 E1 and τ 2 = A B and with the amplitudes 3(1 − f ) E1 25 f E1 E2η1η2 G1 = and G2 = , respectively, which corresponds to a re3+ 2 f (3 + 2 f ) A laxation spectrum reducing to two single lines. The prominent morphological role of the matrix according to the Mori-Tanaka model is expressed by the fact that one of these relaxation times, τ1, is the one of the matrix, with an amplitude which only depends of the elastic modulus of the matrix, E1 (and, of course, of the volume fractions). These properties strongly differ from those which would have derived from the use of the self-consistent scheme, which makes both phases play analogous morphological roles and then forces the associated relaxation spectrum to be continuous (see Sect. 5.3.3.3 in Volume I). This is also reflected by the fact that the “long range memory effect”, which reduces to the additional presence of a Kelvin-type response (5.8.11) to the basic Maxwellian one according to the MoriTanaka model, is much more effective (see the square root in Eq. 5.133 in Volume I) when the self-consistent scheme is used. 210 Chapter 6 Exercises of Appendix I-A (Annex 1): Atomic and Molecular Structures Exercise 6.1 Binding energy Calculate the repulsion potential of NaCl, given the lattice parameter 3.96x10-10m and the binding energy 777.9 kJ/mole. Solution The repulsion potential is the difference between the binding energy and the Coulomb attractive potential, which is: Uc = N e2 1 M 4πε0 r0 (6.1.1) where N is the Avogadro number 6x1023, e the charge of the electron 1.6x10-19 C, M the Madelung constant 1.75 for NaCl structure and 4πε0 = 1.112 C2/Jm. The distance between ions r0 is 3.96x10-10/√2 = 2.8x10-10 m. Thus the Coulomb attractive potential is equal to −863 kJ/mole and the repulsive potential is -778 + 863 = 85 kJ/mole. Exercise 6.2 Miller indices Which {110} planes contain the direction [111]? Solution (1 10) (10 1 ) (01 1 ) for which hu + kv + lw = 0 Exercise 6.3 Miller indices Is the direction [123] in the plane (111)? 211 Solution No: 1x1 + 2x1 + 3x1 ≠ 0 Is the direction [123] in the plane (11 1 ) ? Solution Yes: 1x1 + 2x1 + 3x(−1) = 0 Exercise 6.4 Densest planes Find the densest planes and directions in FCC, BCC and CPH structures. Solution In FCC, the closest atom to the one at the origin of coordinates is at ½a, ½a, 0. Therefore the densest directions are <110>. 110 , 10 1 and 0 11 directions are perpendicular to the 111 one. Thus the (111) plane, which contains three densest directions, is a densest plane. The densest planes are {111} planes. In BCC, the closest atom to the one at the origin of coordinates is at ½a, ½a, ½a. Therefore the densest directions are <111>, 111 and 1 11 directions are ( ) perpendicular to the 110 one. Thus the 110 plane, which contains two densest directions, is a densest plane. The densest planes are {110} planes. ( ) In CPH, the densest directions < 10 10 > are in the basal plane 0001 , which is the densest plane. Exercise 6.5 FCC, BCC and CPH structures a) For the three structures FCC, BCC and CPH, what is the number of atoms in the unit cell? Solution FCC: 8 atoms at the cube corners, each belonging to 8 unit cells + 6 atoms at the face centres, each belonging to 2 cells = 4 atoms per unit cell. BCC: 8 atoms at the cube corners, each belonging to 8 unit cells + 1 atom at the cell centre = 2 atoms per unit cell. 212 CPH: Twice 6 atoms at the hexagons summits, each belonging to 6 unit cells + twice 1 atom at the hexagon centre, each belonging to 2 cells + 3 atoms within the cell = 6 atoms per unit cell. (We consider really super unit cells). b) Find the ratio of the atomic volume to the unit cell volume. Solution FCC: In the unit cube of side a, along a diagonal of length a 2 there are 2 close packed atoms; their radius being a 2 4 , the atomic volume is π 2a 3 24 . With 4 atoms per unit cell, the sought ratio is π 2 6 = 0.74. BCC: In the unit cube of side a, along a diagonal of length a 3 there are 2 close packed atoms; their radius being a 3 4 , the atomic volume is π 3a3 16 . With 2 atoms per unit cell, the sought ratio is π 3 8 = 0.68. CPH: close packed like FCC. c) Find the dimension of the insertion site. Solution FCC: The insertion site is at ¼a, ¼a, ¼a.The distance to the centres of the surrounding atoms is a 3 16 . The radius of the insertion site is therefore ( ) 3 4 − 2 4 a = 0.079a . Its ratio to the atomic radius is 0.225. BCC: The insertion site at the centre of a cube face (or similarly at the centre of a side) is not symmetrical (Fig.6.5.1). Its small axis is, r being the atomic radius: ( ) ( a 2−r = 1 2− 3 4 a = 2 Fig.6.5.1. Insertion site in BCC ) 3 − 1 r = 0.154r 213 CPH: Close packed: same as FCC. d) What are the reciprocal lattices? Solution FCC: The reciprocal lattice is BCC with a*. a = 1 BCC: The reciprocal lattice is FCC with a*. a = 1 CPH: The reciprocal lattice is hexagonal rotated through 30° about the c axis with c* = 2π/c and a∗ = 4π a 3 e) How many (a) nearest (b) second nearest neighbours are there? Solution FCC: (a) 12 nearest neighbours (b) 12 second nearest neighbours. BCC: (a) 8 nearest neighbours (b) 6 second nearest neighbours CPH: closed packed as FCC Exercise 6.6 The c/a ratio in hexagonal structures What is the value of the c/a ratio for close packing in the hexagonal system? Solution The height of the Thomson tetrahedron is a 3 . Therefore c a = 2 3. Exercise 6.7 Angle between directions Find the angle between the directions [123] and [110] as well as between [111] and [122] in cubic systems. Solution If θ is the angle between [123] and [110], we have cos θ = and so θ = 55.46°. 1⋅1 + 2 ⋅1 + 3⋅ 0 12 12 (1 + 4 + 9) (1 + 1) = 3 28 (6.7.1) 214 If θ is the angle between [111] and [122], we find cos θ = 1x1 + 1x2 + 1x2 12 12 (1 + 1 + 1) (1 + 4 + 4) = 5 27 (6.7.2) and so θ = 15.79° Exercise 6.8 Planes in zone Find the condition that must be satisfied by the indices h, k, l, for planes in zone whose direction is [uvw]. Solution hu + kv + lw = 0 (6.8.1) Exercise 6.9 Orientation of slip lines Using the stereographic projection, find the possible orientation of the grains on which the slip lines of Fig.A1.30 (Volume I, p. 540) are seen. Solution Figure A1.30a shows single slip lines. Aluminium being FCC, they correspond to the activation of a (111) slip plane. The number of possible orientations of the grain is infinite. Figure A1.30b shows two families of slip line making an angle of 73° (Fig.6.9.1 below). Now two (111) planes make an angle arcos(1/3) = 70.5°. It can be assumed that the slip lines result from the activation of two (111) slip planes perpendicular to the surface on which the slip lines are seen. The tensile stress acting on the main slip system increased this angle, explaining the difference between 70.5° and 73°. The surface is perpendicular to a [110] direction. 215 Fig. 6.9.1. The two families of slip lines make an angle of 73°, slightly larger than arcos (1/3), owing to tensile deformation. Exercise 6.10 Laue spots Explain the shape of the Laue spots in Fig.A1.31 (Volume I). Solution The spots are elongated. This corresponds to a continuous distribution of diffraction angles denoting a strain-hardened crystal. Exercise 6.11 Laue transmission Explain the Laue transmission photograph of a steel sheet in Fig.A1.32 (Volume I). Solution 216 Each ring corresponds to diffraction by a family of (hkl) planes. They are not continuous owing to a limited number of grains in the annealed sheet. Exercise 6.12 Diffraction spectrum Interpret the diffraction spectrum of Fig.A1.33 (Volume I). Solution The factor Q = h 2 + k 2 + l 2 = ( 4a 2 λ 2 ) sin 2 θ (refer to p.522, Volume I) can be found from the approximate diffraction angles of the various rays. Table 6.12.1 shows that they can be interpreted as the diffraction spectrum of a BCC structure. The lattice parameter is deduced given the wave length of Cu Kα radiation: λ = 1.6x10-10 m. Table 6.12.1. Interpretation of the diffraction spectrum given in figure A1.33 (Volume I) Angle 2θ (°) 40 58 75 88 101 115 132 155 sin2θ 0.117 0.235 0.371 0.483 0.595 0.711 0.834 0.953 sin2θ/sin22θ 1 2 3.17 4.13 5.08 6.07 7.12 8.14 Q 2 4 6 8 10 12 14 16 {hkl} {110} {200} {112} {220} {310} {222} {123} {400} a (10-10m) 3.30 3.23 3.22 3.26 3.28 3.29 3.28 3.28 The average value of the lattice parameter is 3.27x10-10m. Elements Ta and Nb with a lattice parameter of 3.30x10-10m could be the diffracting elements. It can be checked that this lattice parameter gives diffraction angles 2θ which fit well the ones on the figure A1.33 (Volume I). Exercise 6.13 Internal stress measurements a) Find the change in the diffraction angles for a polycrystal subjected to a tension E/105, where E is the Young modulus. Solution From the Bragg relation: 217 2d sin θ = nλ (6.13.1) we have σ E = ∆d ∆θ =− d tan θ (6.13.2) Hence ∆θ = −10−5 tan θ . b) Show how X-ray diffraction can be used to measure an applied tension. Solution Fig.6.13.1. X-ray diffraction for measuring the tensile stress. Measure the diffraction angle θ for {hkl} at an angle φ (Fig.6.13.1). Then: ∆d σ = (1 +ν ) sin 2 φ − 1 d E (6.13.3) The operation can be repeated for various angles φ. Plotting ∆d/d, from (6.13.3) as a function of sin2φ yields σ/E. Exercise 6.14 Parity rules Prove the parity rules for the indices of reflecting planes for BCC and FCC structures. Solution 218 The structure factor for atom hxj, kyj, lzj is: Fj ( hkl ) = e ( 2 πi hx j + ky j + lz j ) . The reflection is extinct if this factor is equal to 0. For BCC: Fj ( hkl ) = e =e ( ( 2 πi hx j + ky j + lz j 2 πi hx j + ky j + lz j ) 1 + e ) +e πi ( h + k + l ) ( ) ( ) ( ) 2 πi h x j +1 2 + k y j +1 2 + l z j +1 2 (6.14.1) Hence the reflection is extinct if: e πi ( h + k + l ) = −1 (6.14.2) implying (h + k + l) is odd. The sum of the indices for reflecting planes is even for the BCC structure {110}, {200}, {112}, etc. For FCC the reflection is extinct if h + k +2l, h + 2k +l, 2h + k +l are odd. Thus h, k and l are of different parities. The indices of the reflecting planes are of the same parities for the FCC structure {111}, {200}, {220} etc. 219 Chapter 7 Exercises of Appendix I-B (Annex 2): Phase Transformations Phase transformations in materials play a key role on their mechanical properties. The basis of the study of phase transformations is briefly introduced in Appendix B of Volume I. A number of exercises (A2.11.1 to 9) are proposed at the end of this Appendix. The solutions of these exercises are given below. Initial numbers of these exercises are indicated in brackets Exercise 7.1 (A2.11.1) Equilibrium diagram. Purification by zone melting A beam of an alloy AB, whose equilibrium diagram is given below, is placed in a furnace which is moved slowly so that a moving liquid region, of length l, is formed from the left to the right (Fig.7.1.1). Use the phase diagram (Fig. 7.1.2) to explain the redistribution of the solute B between the liquid and solid phases. Can the operation be repeated? If so, in what circumstances? This method is used to purify Ga and Si semiconductors and to produce certain high-purity metals. Fig.7.1.1 Principle of zone refining. Solution In a binary alloy in which the addition of the element B lowers the melting temperature of solute A, the phase diagram shows the existence of a different solubility of the element B between the solid and the liquid phases (Figs 7.1.2 a and b). This difference in concentration is defined with the partitioning coefficient, k, of element B in A such as k = CS/CL. In Fig. 7.1.2a, k <1, while in Fig. 7.1.2b, k > 1. 220 Fig. 7.1.2 Phase diagrams (a) k<1; (b) k>1. It can easily be shown that with a diagram such as that of Fig. 7.1.2a, it is possible to purify the solid phase when moving the melting device from the left to the right. This can be shown by writing the mass balance during a small translation of the melted zone by dx. The melted zone gives rise to the formation of a small slice of solid dx with a concentration CS such as CS dx of impurities is lost (for a bar with a unit section) as indicated in Fig. 7.1.3. But the melting of another slice of thickness dx in the initial solid located on the right leads to an increase of impurity by C0 dx Fig. 7.1.3 Mass balance. The solute equilibrium is such that: C0dx − CS dx = ldCL (7.1.1) Using the relation between CS and CL (k = CS/CL), and the boundary condition: x = 0, CS = kC0, Eq. 7.1.1 can easily be integrated to give: kx CS ( x ) = C0 1 + ( k − 1) exp − l (7.1.2) 221 The variation of the ratio CS/C0 is schematically shown in Fig. 7.1.4 for different values of the partitioning factor, k. These curves show a horizontal asymptote CS/C0 = 1. Fig. 7.1.4. Theoretical profiles of concentration along a bar as a function of the value of the partitioning coefficient, k. In (7.1.2) the length of the bar does not appear because we have assumed that the bar is infinitively long. The asymptotical regime is reached more rapidly when k increases and when the length of the melted zone, l , is small. In practice zone melting with multisource is used in order to accelerate the process of purification (Fig. 7.1.5). Fig. 7.1.5 Zone refining with muti-sources. Exercise 7.2 (A2.11.2) Steel microstructures Steels of various carbon contents are cooled slowly from the austenite state. Fig. A2.52 (Volume I) gives micrographs of the structures. What is the carbon content of each? Use the Fe-Fe3C part of the iron-carbon equilibrium diagram to justify your answers. Solution 222 Remind the lever rule. Using this rule, the composition of the four steels can be given approximately. Steel a contains some pearlite (~ 10%). This means that the carbon content is larger than 0.02 (see Fig. A.26). Using the lever rule, this leads to C~ 0.08. Steel b contains a larger amount of pearlite (~ 0.40). This means that C ~ 0.33. Steel c is fully pearlitic. This indicates that C ~ 0.80. Steel d contains some intergranular cementite. This steel is slightly hypereutectoid, i.e. C > 0.80 Exercise 7.3 (A2.11.3) Martensitic transformation Two steels with 12% Cr and 5% and 7% Ni respectively are thermally treated. They are quenched from 1,000°C: what can be predicted from the Fe-Cr-Ni equilibrium phase diagram given in Fig. A2.53 (Volume I)? (martensite laths are formed in both cases). They are then subjected to a repeated annealing: 1h at 700°C – cooling in air – cooling to –196°C – 5h at 500°C, after which the residual austenite content is measured at ambient temperature. The data are given in the table. Explain what has happened; would you expect an increase in the nickel content in the austenite? Electron microscopy shows that the γ phase is distributed in bands: how do you explain this? Figure A2.54 (Volume I) gives the results of tensile tests to determine the yield strength of the two specimens: how do you explain the rapid variation of this with temperature? Solution Figure A2.53 (Volume I) shows that at 1,000°C these materials are fully austenitic. Quenching produces lath martensite with eventually some retained austenite. The equation giving MS page 583 (Volume I) predicts that MS(12Cr−5Ni) = 272°C, MS(12Cr−7Ni) = 238°C. Annealing at 500°C will produce a two-phase microstructure: austenite reverted from tempered martensite (or ferrite) + ferrite (or tempered martensite). The austenite will be enriched in nickel (which is a γ former). This reverted austenite can partly be transformed into martensite during cooling, but residual austenite will remain present. This austenite is metastable and will be transformed into martensite (strain induced phase transformation) during a tensile test. This explains the anomalous variation of the yield strength of these materials when tested at 0°C and below. Exercise 7.4 (A2.11.4) Steel microstructures Various heat treatments have been applied to a Cr-Mo steel (0.39C, 1.5Cr, 0.5Mo), the TTC curve for which is given in Fig. A2.55 (Volume I). This steel is 223 much used in mechanical engineering for items such as gears. Why is such care taken over the conditions for austenitising (maintaining in the γ phase) and the resulting grain size of the austenite? Four structures resulting from various cooling rates are shown in the figure. Identify these microstructures. Solution Low austenitising temperatures are used to avoid a detrimental growth of the austenite grain size. The material with a hardness of 634 DPH (DPH: diamond pyramid hardness is the same as Vickers hardness (HV)) was obtained with a high cooling rate (~ 25°C/s). The microstructure is fully martensitic. The material with a hardness of 327 DPH was obtained with a cooling rate of about 1°C/s. The CCT diagram (see Sect. A.2.6.2, Volume I) shows that the microstructure is fully bainitic. The material with a hardness of 281 DPH was obtained with a cooling rate of about 10°C/min: it contains some ferrite (white phase) in a bainite matrix. The material corresponding to a hardness of 227 DPH was slowly cooled at about 40°C/h. It contains a significant amount of pro-eutectoid ferrite (white phase). Exercise 7.5 (A2.11.5) Surface treatment: Cementation Carbon cementation: Cementation is a thermo-chemical treatment, the aim of which is to increase the carbon content of the sample at the surface. This enrichment is achieved by keeping the surface in contact with a carbon-donating material in powder, paste or liquid form. The diffusion process is always followed by the thermal treatment of quenching: why is this? The quenching creates residual compressive stresses in the surface: why? Suppose the operation takes place in the gaseous phase (Fig. A2.56, Volume I); show that the solution to Fick’s second equation is: x C − C0 = erf CS − C0 2 Dt (7.5.1) 2 x exp −u 2 du , with C0 the initial carbon content of the steel π 0 and CS the saturation value for the γ phase at the temperature of the treatment. Figure A2.56 (Volume I) gives carbon concentration profiles for a steel with C0 = 0.15%, CS = 1.3% at 925°C. Use these to find the diffusion coefficient of carbon in the γ phase and compare your value with that given in the tables in Sect. A2.3.2 (Table A2.4, Volume I). What do you think of the result? Why is it necessary to conduct this process at a temperature above that of the eutectoid? where erf ( x ) = Solution ∫ ( ) 224 Quenching has to be applied for the formation of martensite. The formation of martensite occurs preferentially at the free surface (which is cooler than the bulk). As this phase transformation is accompanied by an expansion (a few %, see p. 580, Volume I) this volume change produces compressive residual stresses at the free surface, which is beneficial for the fatigue properties of the components. Modeling of these residual stresses is complex. In particular the “transformation plasticity” (see Sect. 3.4.4.4 in Volume I) must be taken into account. See that the expression: x C − C0 = erf CS − C0 2 Dt (7.5.2) is demonstrated in classical books on diffusion (see for example Crank 1956). Numerical calculations must be made to identify the diffusion coefficient from the results reported in Fig.A2.56 (Volume I). Simplified calculations can be made using an approximation for the erf function which is valid for small distances, x from the free surface and for large times, t, i.e.: 1/2 1 x2 C − C0 ≃ 1 − exp − π Dt CS − C0 (7.5.3) Taking, for instance x = 0.5 mm, one obtains with Eq. 7.5.2, at t = 2 hours, D ~ 1.20x10-10 m2/s; t = 4 hours, D ~ 0.30x10-10 m2/s. Using Table A2.4 (Volume I, p. 558), one obtains at 925°C (1,198 K), D = 0.28 10-10 m2/s, which is a value very close to those derived from the carbon profiles shown in Fig. A2.56 (Volume I). This cementation heat-treatment must be performed in the γ field (see Fe−C diagram in Fig. A2.6 in Volume I) where the solubility of carbon is sufficiently high. This diagram shows that, at 925°C, the material is in the γ field. Exercise 7.6 (A2.11.6) Solidification Solidification of a Fe-C-Ni alloy has resulted in segregation of the additive elements C, Ni. The distances over which these segregations occur depend on the conditions during the solidification and on the growth of dendrites. Taking the data in Table A2.4 (Volume I) concerning the diffusion coefficients for these elements in Feγ , show that there is a possibility of homogenising this alloy by holding it in the austenitic phase for a long time. What quicker methods can you suggest 225 for the process, bearing in mind that band structures such as those shown in Fig. A2.37 (Volume I), are to be avoided? Solution The dendrites spacing is between 0.1 mm and 1 mm in many solidification processes. A heat treatment of homogenisation is efficient if the alloying elements (C, Ni) have enough time to diffuse over these distances. This heat treatment has to be performed at elevated temperature. However the temperature must be below the local melting point which is, most often, decreased by the presence of alloying elements. See Fe-C phase diagram (Fig. A2.6, Volume I) and Fe-Ni diagram (Fig. A2.7, Volume I). A simple relationship can be used to estimate the diffusion distance, x: x= Dt (7.6.1) where D is the diffusion coefficient and t, the time for homogenisation. Assuming that the heat treatment is made at 1,200°C (1,473K) and using the values of the diffusion coefficients for C and Ni given in Table A2.4 (Volume I), it is found that: - For x = 0.1 mm, tNi = 30 hours, and tC = 26s. Carbon which is an interstitial element is known to diffuse must faster than Ni which a substitutional element in Fe. The inhomogeneities can therefore be largely reduced with a heat treatment lasting about one day. - This is not the case when x = 1mm. In these conditions it is found that tNi = 2,800 hours (which is clearly not realistic) and tC = 0.8 hour. One way to reduce the times of homogenisation is to predeform the material to reduce the mean distance between the dendrites. This predeformation must be made at elevated temperature (1,000 – 1,100°C). The existence of bands can be related to some segregation. Exercise 7.7 (A2.11.7) Hardening by precipitation and coalescence of the precipitates in a ferritic stainless steel Experiments have been made with steels of the following percentage composition: (a) Fe – 19.6Cr – 2.03Ni – 0.97Al (b) Fe – 19.6Cr – 4.15Ni – 1.87Al 1) Explain the choice of these compositions, knowing that the hardening phase produced by the precipitation has the composition NiAl (structure B2, CSCl, see Appendix A, Volume I). Is it normal for these steels to have a BCC structure? 2) Samples of the steels have been held at 1,150°C for 1 h; what structure will they have? After this they are cooled quickly and then aged at various tempera- 226 tures. The curves of Fig. A2.57 (Volume I) give the variations of hardness with time of ageing and the micrographs of Fig. A2.58 (Volume I) show the precipitation of the NiAl phase. a) Is the form of the hardness curves what you would expect? b) Comment on the legend on the micrographs. Why is there a diffraction spot of structure of the type (001) in the [001] section? Solution 1) These materials containing about 20%Cr are stainless. The presence of Ni and Al leads to the precipitation of the ordered B2 (NiAl) phase. Electron diffraction shows the existence of this phase after ageing: an extra spot (100) is observed (Fig. A2.58d in Volume I). These precipitates are ordered particles whose lattice parameter is close to that of the matrix and they are in cube-cube orientation relationship with the matrix: (100)matrix // (100)precipitate, and [001]matrix // [001]precipitate The materials have a BCC structure as indicated by the Schaeffler diagram. This is due to the fact that Cr and Al are strong ferrite stabilizers. 2) At 1,150°C, the materials keep a BCC structure and Ni and Al atoms are in solid solution in a high chromium ferrite. Figure A2.57 (Volume I) shows a typical behaviour for precipitation hardening with the existence of a peak. This suggests that at the early beginning of ageing small precipitates are sheared by dislocations with a Burgers vector ½ <110>. The dislocations are then grouped in pairs separated by a stacking fault (see Fig. 3.93 in Volume 1). After the maximum of the hardness, the particles are bigger and they are by-passed by the Orowan process with the formation of dislocations loops left around the precipitates (see Fig. 3.90 in Volume 1). Exercise 7.8 (A2.11.8) Precipitation Figure A2.59 (Volume I) shows the variation of precipitate size (d) with time at various temperatures, suggesting a growth law of the form: A= 64γ S DCE∞ Ω 2 9kT (7.8.1) Deduce the value of the apparent activation energy Qd for diffusion of Al and Ni in α−iron (Fe−20Cr), assuming that γS and Ω are independent of temperature. The variation of the solubility has been determined – how could this be done? – and the results tabulated; discuss the values found for Qd. Solution 227 The growth law: d 3 − d 03 = A ( t − t0 ) with A = 64γ S DCE∞ Ω 2 9kT (7.8.2) corresponds to the LSW theory (see Sect. A2.5.3.2 in Volume I). In this expression t0 is the time at which the system enters the coalescence regime, when the mean particle size is d0, CE∞ is the concentration of the solute at infinity, γS is the surface energy between the precipitate and the matrix, D is the diffusion coefficient, Ω is the molar volume of the species entering into the particles. The results shown in Fig. A2.59 (Volume I) indicate that this law is well obeyed in the alloys presented in Exercise 7.7. The concentration CE∞ is slightly dependent on temperature. The activation energy of diffusion can be determined, assuming that γS and Ω are independent of temperature. This equation can be written as: log AT Q = cste − C∞ kT (7.8.3) The values of A can be determined from the results of Fig. A2.59 (Volume I), knowing the value of t0 (for more details, see Taillard, Pineau, 1982). Fig. 7.8.1. Coarsening of NiAl precipitates in a Fe-Cr ferritic steel (Taillard, Pineau, 1982). 228 The representation of Eq. 7.8.3 is shown in Fig. 7.8.1 where a straight line can be drawn through the data points. The slope of this line leads to a value of Q = 302 ± 81KJmol−1 . This value is compatible with the diffusion activation energies of the elements present in the alloys which are given in Table 7.8.1. This table also includes the estimated values for the diffusion coefficients of aluminium, nickel, chromium and iron in α−Fe. From this table it can be concluded that chromium probably does not intervene as the element controlling the coarsening rate but it is impossible to distinguish between the other elements, although the apparent energy Q is closer to that of Al. Table 7.8.1. Values of the effective diffusion coefficients in Fe − 19.6Cr − 4.15Ni − 1.8Al and DAl, DNi, DCr and DFe in α−Fe (Taillard, Pineau, 1982). Temperature (K) DCr (m2s-1) DFe (m2s-1) DNi (m2s-1) DAl (m2s-1) D (m2s-1) 823 5.0 x 10-22 2.3 x 10-20 3.6 x 10-20 9.3 x 10-18 1.7 x 10-19 873 8.9 x 10 -21 -19 -19 -17 3.9 x 10-18 923 1.1 x 10-20 1.9 x 10-16 7.2 x 10-17 1.9 x 10 1.2 x 10-18 2.8 x 10 1.8 x 10-18 4.6 x 10 Exercise 7.9 (A2.11.9) Effect of applied stress on the morphology of precipitates in nickel-based alloys As Fig. A2.60 (Volume I) shows, the morphology of single-crystal materials containing a large volume fraction of precipitates (~ 50−60%) is considerably changed by the application of stress. In this case (alloy Udimet 700), tension leads to a rafting process and a redistribution of precipitates, initially cubic and distributed isotropically, into layers perpendicular to the direction of the stress. In contrast to this, compression rearranges them into rods or needles aligned parallel to the stress. These changes have an important effect on the creep behaviour of the material. Give a qualitative explanation of these changes, bearing in mind the following: - the precipitates are coherent with the matrix - any departure from coherence, measured by the relative difference in the values of the parameters for the γ’ phase and the matrix, δ = (aγ’ − aγ)/aγ, can be positive, negative or zero. - the elastic constants, Young’s modulus in particular, are different fro the precipitates and the matrix. Further information can be found in: Pineau A (1976) Influence of uniaxial stress on the morphology of coherent precipitates during coarsening-elastic energy considerations. Acta Metall 24:559564 Nabarro FRN (1996) Rafting in superalloys. Metall Mater Trans 27A:513-530 229 Solution The most advanced nickel-based alloys used for the fabrication of turbine blades are single crystalline with the <001> axis parallel to the longitudinal direction of the blades. These materials contain a very high volume fraction (~60%) of γ’ precipitates (Ni3Al, L12 structure, see Figure 1.41 in Volume I). These particles have initially a cubic shape. The blades are submitted to complex thermo-mechanical loads, in particular an axial load due to the centrifugal force. As these materials are working at very elevated temperatures (~1,000°C), the precipitates are not stable. They grow and they can rearrange themselves to adopt a plate-like morphology (Fig. A2.60a in Volume I) or a needle shape (Fig. A.2.60b in Volume I), depending on the orientation of the applied stress and the misfit parameter δ = (aγ’ – am)/am, where aγ’ and am are the lattice parameters of the precipitates and that of the matrix, respectively. The stress free misfit is a pure dilatation. This partly explains why the particles adopt initially a cube shape in the stress free conditions. Many studies have been devoted to the ageing of these materials under an applied stress. The rearrangement of the particles in the presence of an applied stress is named “rafting” in the literature (see e.g. Nabarro, 1996). In this exercise an attempt is made to explain qualitatively these shape changes in terms of the elastic theory of inclusions (see Sect.2.7 in Volume 1). The matrix and the precipitates are assumed to be elastically isotropic. Equation 2.174 in Volume I can be extended to include the inhomogeneity effect ( Em ≠ Ep ) , where E m and Ep are the Young modulus of the matrix and that of the precipitates. The total energy per unit volume, ξ t (ξ =Wst /V in Eq.2.174 ) can be written as the sum of three terms (see Pineau, 1967): ξt = ξincl + ξint + ξinh (7.9.1) where: 1 2 ξincl = − σ ij I eij T ξint = −σ A e33T 1 2 ξinh = − σ A e33T ∗ (a ) (b) (7.9.2) (c) In this expression, σA is the applied stress (σA > 0 for a tensile stress; σA< 0 for a compressive stress; eijT* is the transformation strain for the inclusion effect ∗ ∗ ∗ ∗ ( ) e11T = e22T = e33T = aɺɺ and eijT i ≠ j = 0 ; σ ijI is the stress inside the inclusion in 230 T is the equivalent strain for an inhomogeneous the absence of applied stress; e33 ( Ep ≠ Em ) inclusion. This equivalence was defined by Eshelby (1961). The inhomogeneity effect (ξinh ) is only due to the fact that, in the absence of coherency misfit, the particles have elastic constants different from those of the matrix. The inclusion effect (ξincl ) is related to the misfit. The interaction effect (ξint ) is related to the fact that the precipitation occurs under an applied stress. After some lengthy calculations it can be shown that: 2 σ σ = A A + B A + C 2 Em aɺɺ Em aɺɺ Em aɺɺ ξt (7.9.3) In this equation, the coefficients A, B and C are dependent on the particle shape and on the ratio Ep/Em. The principle of the calculations is to determine the shape (sphere, plates perpendicular to the applied stress, or needles aligned with the stress axis) which minimizes the value of the total strain energy given by Eq. 7.9.3. It should be realised that, due to the difference in elastic moduli between the matrix and the precipitates, anisotropy in the misfit is produced, as schematically shown in Fig. 7.9.1 where the deformations of the matrix (solid lines) and that of the precipitates (dotted lines) are indicated. Let us consider for instance the case where σA > 0, δ > 0 and Ep < Em. Figure 7.9.1b shows that under a tensile stress the deformation which has to be applied to the inclusion to be relocated within the matrix is larger along the stress axis than in the other directions. The situation is thus similar to that corresponding to a particle located in a matrix without any apT T plied stress but such as: e11 =e22 T T and e33 >e11 . Inclusion theory shows that in these conditions the strain energy for a plate perpendicular to the stress axis is lower than that of a cube. Other qualitative variations for the shape of the particles based on this energy analysis are shown in Figs 7.9.1c, d and e. In particular the needle-like morphology observed in Fig. A2.60 is in qualitative agreement with the predicted shape indicated in Fig. 7.9.1d. During coarsening the variation in the surface energy has also to be considered. When the shape change occurs by the agglomeration of particles, the surface energy is always decreased when initially independent cube-shaped particles are transformed into needles or plates. The observations reported in Fig. A2.60 were made by Tien and Copley (1971) on single crystals of Udimet 700 in which δ = +2 x10-4. 231 Fig. 7.9.1 Stress-induced coarsening of precipitates in nickel-based superalloys. Sketch showing the influence of the orientation of the applied stress and of the difference in Young modulus of the matrix and the precipitates on the relative shape change of both phases. References Crank J (1956) The mathematics of diffusion. Oxford Science Publications, Oxford University Press. Eshelby JD (1961) Elastic inclusions and inhomogeneities in Progress in solid mechanics II. In: Sneddon I.N., Hill R. (ed)),vol 3, North-Holland, Amsterdam, 87-140 Nabarro FRN (1996) Rafting in superalloys. Metall Mater Trans 27A:513-530 Pineau A (1976) Influence of uniaxial stress on the morphology of coherent precipitates during coarsening-elastic energy considerations. Acta Metall 24:559-564 Taillard R, Pineau A (1982) The precipitation of the intermetallic compound NiAl in Fe19Wt%Cr alloys. Mater Sci Eng 54:209-219 Tien JK, Copley SM (1971) The effect of uniaxial stress on the periodic morphology of coherent gamma prime precipitates in nickel-base superalloy crystals. Metall Trans 2:215-219 232 Chapter 8 Exercises of Appendix I-C (Annex 3): Continuum Mechanics: Basic Concepts and Equations 8.1 Deformations Exercise 8.1 Calculation of strain fields in Cartesian coordinates Calculate the strain tensors for the following transformations T, first for finite transformations and second assuming small perturbations. Interpret these transformations and find the principal directions and strains for the case of small perturbations. In orthonormal Cartesians T is defined by 1) x = a + λa y = b + µb z = c +ν c 2) x = a y = b + 2γ a z = c (8.1.1) Solution 1) We have dx = (1 + λ )da d x = dy = (1 + µ )db dz = (1 +ν )dc da d a = db dc (8.1.2) and then 2 2 2∆ij dai da j = d x − d a = λ (2 + λ )da 2 + µ (2 + µ )db 2 + ν (2 + ν )dc 2 (8.1.3) Equating coefficients, we find for the Green-Lagrange strain tensor 0 0 λ (1 + λ 2) 0 µ (1 + µ 2) (∆) = 0 0 0 ν (1 + ν 2) (8.1.4) 233 For small perturbations, we get 0 0 µ 0 0 ν λ (ε ) = 0 0 (8.1.5) So, the principal directions coincide with the coordinate system and λ is the dilatation in the direction e1, etc. 2) Similarly, we find dx = da d x = dy = 2γ da + db dz = dc da d a = db dc (8.1.6) and then 2 2 2∆ij dai da j = d x − d a = 4γ 2 da 2 + 4γ dadb (8.1.7) So, the Green-Lagrange strain tensor is 2γ 2 (∆) = γ 0 0 0 0 0 0 γ (8.1.8) which for small perturbations reduces to 0 γ 0 0 0 0 0 0 (ε ) = γ (8.1.9) This is a shear strain fied; the principal strains are ±γ and 0 and the principal directions e1 + e 2 , e1 − e2 , e3 . Exercise 8.2 Calculation of strain fields in cylindrical polars In cylindrical polars (e r , eθ , e z ) , with M = M ( r,θ , z ) , T takes M to M' with 234 1) 2) M ′ = M + u( r )e r + w( z )e z M ′ = M + v ( r )e θ 3) M ′ = M + aze θ 4) M′= M + (8.2.1) bθ ez 2π Calculate the corresponding strain tensors, first for finite transformations and second assuming small perturbations. Interpret these transformations and find the principal directions and strains for the case of small perturbations. Solution 1) Using the same method as in Exercise 8.1, we can write ∂M ∂M ∂M dr + dθ + dz ∂r ∂θ ∂z = e r dr + re θdθ + e z dz dM = With (8.2.2) der de = e θ and θ = − e r , we find from (8.2.1) dθ dθ d M ′ = d M + u′( r )e r d r + u( r )e θdθ + w′( z )e z d z (8.2.3) = [1 + u′( r )] e r d r + [ r + u( r )] e θdθ + [1 + w′( z ) ] ez d z Thus we get 2 2 2 2 2 d M ′ − d M = [1 + u′] dr 2 + [ r + u ] dθ 2 + [1 + w′] dz 2 − ... ... − dr 2 − r 2dθ 2 − dz 2 (8.2.4) = 2u′ + u′ dr + 2ur + u dθ + 2 w′ + w′ dz 2 2 2 2 2 2 According to (A3.8) in Volume I, this is to be identified with 2 2 d M ′ − d M = 2 E i E j ∆ij dθi dθ j (8.2.5) where (θi ) = ( r ,θ , z ) and ( E i ) = (1, r ,1) . The non zero components of ∆ are then easily found to be 235 ∆rr = u′( r ) + u′( r ) 2 2 ∆θθ = u( r ) u( r ) 2 + r 2r 2 ∆zz = w′( z ) + w′( z )2 2 (8.2.6) We could have as well used the general intrinsic definition of the GreenLagrange tensor, namely T 2∆ = F .F − δ (8.2.7) which leads to (A3.7) in Volume I, that is 2∆ij = 1 ∂M ′ ∂M ′ (θ k , t ). (θ k , t ) − δij ∂θi E i E j ∂θ j (8.2.8) with (θi ) = ( r ,θ , z ) and ( E i ) = (1, r ,1) in cylindrical polars. We then find, with use of (8.2.3) 2∆rr = (1 + u′)2 − 1 = 2u′ + u′2 1 u2 2 2∆θθ = 2 ( r + u ) − 1 = 2ru + 2 r r 2∆zz = (1 + w′)2 − 1 = 2 w′ + w′2 (8.2.9) and all other components vanish. Hence u′2 ( r ) u′( r ) + 2 ∆ = 0 0 ( ) 0 u( r ) u 2 ( r ) + r 2r 2 0 0 w′2 ( z ) w′( z ) + 2 0 (8.2.10) and for small perturbations u′( r ) ε = 0 0 () 0 u( r ) r 0 0 0 w′( z ) (8.2.11) 236 So, the principal directions coincide with the coordinate system (e r , eθ , ez ) and the non zero components are the principal strains. 2) Instead of (8.2.3), dM ′ is now given by d M ′ = d M − v ( r )e r d θ + v′( r )e θdr = [e r + v′( r )e θ ] d r + [ re θ − v( r )e r ] dθ + e z d z (8.2.12) and then we have ∂M ′ ∂r = e r + v′( r )e θ ∂M ′ = re θ − v ( r )e r ∂θ ∂M ′ ∂z = e z (8.2.13) so that, according to (8.2.8), we find 2 2∆rr = v′ ( r ) v2 (r) 2∆θθ = 2 r v( r ) 2∆rθ = 2∆θr = − r + v′( r ) (8.2.14) and all other components vanish. Hence v′2 ( r ) 2 v ( r ) v′( r ) ∆ = − + 2 2r 0 ( ) and for small perturbations − v( r ) v′( r ) + 2r 2 v2 (r ) + 2r 2 0 0 0 0 (8.2.15) 237 0 v ( r ) v′( r ) ε = − + 2r 2 0 − v ( r ) v′( r ) + 2r 2 () 0 0 0 0 0 (8.2.16) v( r ) v′( r ) This represents a pure shear. The principal strains are ± − and 0; 2 2r the principal axes e r + e θ , e r − eθ , e z . Such a transformation can be achieved, for example, by first preventing the internal surface of a tube from moving by soldering it to a rigid cylinder and then applying a torque over the external surface. 3) The increment dM ′ is given by d M ′ = d M − aze r d θ + ae θdz = e r d r + [ re θ − aze r ] dθ + [ae θ + e z ] d z (8.2.17) and then ∂M ′ = er ∂r ∂M ′ = re θ − aze r ∂θ ∂M ′ = ae θ + e z ∂z (8.2.18) From (8.2.8), we find az 2∆rθ = 2∆θr = − r 2∆ = 2∆ = a θz zθ a2 z2 r2 2 2∆zz = a 2∆θθ = (8.2.19) and all other components vanish. Hence; the strain tensor is 0 az ∆ = − 2r 0 ( ) az 2r a2 z2 2r 2 a 2 − 0 a 2 a2 2 For small perturbations ( a ≪ 1 ), this becomes (8.2.20) 238 0 az ε = − 2r 0 − () 0 a 2 0 az 2r 0 a 2 (8.2.21) a r 2+ z2 . 2r 4) Using the same method as in the foregoing questions, we find whose eigenvalues are 0 and ± bθ ez 2π b = e r d r + re θ + e z dθ + e z d z 2π dM ′ = dM + ∂M ′ = er ∂r ∂M ′ b = re θ + ez ∂θ 2π 0 0 b2 ∆ = 0 8π 2 r 2 b 0 4 πr ( ) 0 ε = 0 0 () 0 0 b 4 πr 0 b 4 πr 0 0 b 4 πr 0 ∂M ′ = ez ∂z (8.2.22) (8.2.23) (8.2.24) (8.2.25) As we already know from Sect. 3.3.7.1 in Volume I, this strain field is associated with a screw dislocation. 239 Exercise 8.3 Strain compatibility (small perturbations) 1) Show that the tensor field (α ( x, y ) ) = ex + fy ax + by ex + fy cx + dy (8.3.1) where a, b, c, d, e and f are constants can represent a plane strain tensor field and deduce the components of the associated displacement vector field. 2) In an orthonormal fixed system (e1 , e 2 , e 3 ) , the non-zero components of a strain field are ε12 = ε 21 = g( x1 , x2 , x3 ) . Find the general form of g( x1 , x2 , x3 ) . 3) The given unit vectors n and m are such that the base (n, m, e 3 ) is orthonormal and fixed. The strain tensor has the form ε = h( x1 , x2 , x3 )(n ⊗ m + m ⊗ n ) . Find the most general form of the function h( x1 , x2 , x3 ) . Solution 1) For a plane problem the compatibility equations reduce to 2 ∂ 2ε xy ∂x∂y = 2 ∂ 2ε xx ∂ ε yy + ∂y 2 ∂x 2 (8.3.2) This involves only second order partial derivatives; since α ( x, y ) is affine with respect to x and y, it satisfies the equation (8.3.2) identically. For the corresponding displacement ( u ( x, y ), v( x, y ) ) we have ∂u a = ax + by ⇒ u = x 2 + bxy + h( y ) ε xx = ∂x 2 ∂v d 2 = cx + dy ⇒ v = y + cxy + g( x) ε yy = ∂y 2 ∂u ∂v + = bx + cy + h′( y ) + g′( x) 2ε xy = ∂y ∂x (8.3.3) Hence, according to (8.3.1), we must have bx + cy + h ′( y ) + g′( x) = 2ex + 2 fy or bx − 2ex + g′( x) = 2 fy − cy − h′( y ) = −G (8.3.4) 240 where G is a constant, since it is the common value of two functions depending only on x and y, respectively. So, we find b 2 g( x) = (e − 2 ) x − Gx + B h( y ) = ( f − c ) y 2 + Gy + A 2 (8.3.5) where A and B are constants and the displacement vector field is a 2 c 2 u ( x, y ) = 2 x + bxy + ( f − 2 ) y + Gy + A v( x, y ) = d y 2 + cxy + (e − b ) x 2 − Gx + B 2 2 (8.3.6) Note that this displacement field is defined to within an arbitrary rigid body displacement OM ∧ G + K where G = (0,0, G ) and K = ( A, B,0). 2) The compatibility equations give ε12,13 = ε12,23 = ε12,33 = 0 ε12,12 = 0 (8.3.7) The three first equations imply g( x1 , x2 , x3 ) = ax3 + b( x1 , x2 ) (8.3.8) where a is a constant. Then from the fourth equation we get g( x1 , x2 , x3 ) = ax3 + b( x1 ) + c( x2 ) (8.3.9) 3) In the base (n, m, e 3 ) , with coordinates n,m,x3 the only non-vanishing components of the strain tensor are ε nm = ε mn = h( x1 , x2 , x3 ). So, according to (8.3.9) in question 2), we must have ε nm = ε mn = ax3 + b(n) + c(m) (8.3.10) Therefore, changing the axes and expressing n and m in terms of x1 and x2, we obtain h( x1 , x2 , x3 ) = ax3 + b [ n( x1 , x2 )] + c [ m( x1 , x2 )] (8.3.11) 241 which is the most general expression for glide on the slip system (n, m). Exercise 8.4 Deformation of curves and surfaces 1) A wire in the form of a curve in 3-dimensional space, for which we take the arc length s as parameter, is subjected to the transformation M ( s ) → M ′( s ) = M ( s ) + u ( s ) (8.4.1) where u(s) is a vector field in R3. Determine the associated strain tensor. 2) A canvas sheet acting as a roof is held firmly by its four sides. As a result of a fall of snow it is subjected to a displacement field u such that M ( x, y,0) → M ′( x, y + v( x), w( x, y )) (8.4.2) Find the associated strain tensor. 3) We consider a straight beam which in its undeformed state is a (long) cylinder with generators parallel to Ox and has a plane of symmetry containing Ox. We take for Ox the inertial axis of the beam, also called the mid fibre. We seek to express the strain at any point of the beam in terms of the quantities that describe the change of shape of the mid fibre only, with the following assumptions: H1: a point M on the mid fibre is displaced to M' such that M ( x,0,0) → M ′ [ x + u ( x), v( x),0] (8.4.3) H2: plane cross-sections of the beam are not deformed and remain normal to the axis (the plane of symmetry is conserved). (a) Find the strain tensor ε ( M ) for the axis Ox. (b) Find an expression for the displacement of an arbitrary point P( x, y,0) and the associated strain tensor ε ( P ) : for simplicity, u' and v' can be taken as negligible relative to 1. Solution With t the unit tangent vector at s, we write dM 2 2 d M = ds ds = tds ⇒ d M = ds d M ′ = tds + du ds ⇒ d M ′2 = 1 + 2t. du + ( du ) 2 ds 2 ds ds ds (8.4.4) 242 from which we find the single non zero component of the strain tensor to be ∆tt ( s ) = t. du 1 du 2 + ( ) ds 2 ds (8.4.5) du . ds 2) With i , j and k the unit vectors of the orthonormal base, we have For small perturbations this becomes ε tt ( s ) = t. d M = dxi + dy j ⇒ d M 2 = dx 2 + dy 2 ∂w ∂w dy k d M ′ = dxi + [ dy + v′( x)dx ] j + dx + ∂ ∂y x 2 ⇒ d M ′2 = dx 2 + dy + v′( x)dx 2 + ∂w dx + ∂w dy [ ] ∂y ∂x (8.4.6) and then ′2 ∂w 2 v + ( ∂x ) 1 ( ∆ ) = 2 ∂w ∂w v′ + ∂x ∂y ∂w ∂w ∂x ∂y ∂w 2 ( ) ∂y v′ + (8.4.7) a second-order tensor in R2 3) (a) On the axis Ox, we have d M = dxi ⇒ d M 2 = dx 2 2 2 2 2 2 d M ′ = (1 + u′)dxi + v′dx j ⇒ d M ′ = (1 + u′) dx + v′ dx (8.4.8) and the single non-zero component of the strain tensor is ∆xx given by 2 2 2∆ xx ( M )dx 2 = d M ′ − d M = (2u′ + u ′2 + v′2 )dx 2 1 ⇒ ∆ xx ( M ) = u ′ + (u′2 + v′2 ) 2 (8.4.9) (b) The vector OP( x, y,0) is displaced to OP′ = OM + MM ′ + yn , with n the unit vector normal to the tangent vector t : (1 + u′, v′,0) , that is 243 n= k ∧t 1 : (−v′,1 + u′,0) k ∧t (1 + u′)2 + v′2 (8.4.10) Therefore we have OP′ : ( x + u, v,0) + y (1 + u′) 2 + v′2 (−v′,1 + u ′,0) (8.4.11) 2 With the assumptions made, y does not vary and then dP = dx 2 . The exact 2 calculation of dP′ is made difficult by the need to evaluate (d/dx)n(x); but taking u' and v' to be small relative to 1 gives the simplified expression P′ : ( x + u − yv′, y + v,0) ⇒ dP ′ : (1 + u′ − yv′′, v′,0) dx (8.4.12) So dP′ 2 = (1 + u′ − yv′′) 2 + v′2 = (1 + u′) 2 + v′2 − 2 yv′′(1 + u ′) + y 2 v′′2 dx (8.4.13) 1 1 2 2 2 2 ′ ′′ ′ ′′ ⇒ ∆xx ( P) = 2dx 2 (dP − dP ) = ∆xx ( M ) − yv (1 + u ) + 2 y v Neglecting the terms of higher order of smallness we have ε xx ( P) = ε xx ( M ) − yv′′ (8.4.14) where, to the approximation we are making, v'' is the change in curvature, here the curvature of the beam in its final state. The linearised strain component tensor ε xx ( P) at an arbitrry point with ordinate y is ε xx ( P) = u′ − yv′′ 8.2 Stresses Exercise 8.5 Statically admissible stress fields (8.4.15) 244 1) For homogeneous plates acted on by forces in their own planes, find in each case a uniform statically admissible stress state and give the associatedMohr representation: (a) a rectangular plate ABCD with shear τ applied at each edge and compression p at a pair of opposite edges. (b) a parallelogram plate ABCD, with acute internal angle (π/2 − α ) , with compression p1 and p2 parallel to the opposite edges. 2) The wall of a dam has the form of a prism whose section is a right-angled triangle OAB (see Fig. 8.5.1) with OA, facing up-stream, vertical, OB, facing downstream inclined at 45° to the vertical and AB fixed. The density of the wall material is ρ, that of the water in the dam, which is full, is ϖ (atmospheric pressure can be neglected). Show that the equilibrium equations for the dam are satisfied by a stress field as follows: ax + by ex + fy cx + dy (σ ) = ex + fy (8.5.1) where a, b, c, d, e and f are constants to be calculated. y 0 x π/4 water dam A B Fig. 8.5.1 Sketch of the studied dam 3) A semi-infinite, heavy homogeneous body is subjected to atmospheric pressure at the surface; investigate the stress field and show that there are an unlimited number of statically admissible fields. Solution 1) (a) The plane stress tensor has the form σ xx σ xy xy σ yy (σ ) = σ On AB n : (1,0) and σ .n = σ xx i + σ xy j = −τ j . (8.5.2) 245 On BC n : (0,1) and σ .n = σ xy i + σ yy j = −τ i − p j . Hence a statically admissible tensor is 0 τ p (σ ) = − τ (8.5.3) a uniform tensor field which clearly satisfies the equilibrium equations without a body force term. It is easily shown that the boundary conditions are indeed satisfied on the edges CD ( n : (−1,0)) and DA ( n : (0, −1)). The associated Mohr representation is indicated in Fig. 8.5.2. τ τ -p -p/2 0 σ Fig. 8.5.2 Mohr representation for question 1(a) of Exercise 8.5. (b) We reason as before with, on AB, n AB : (cos α , − sin α ) and on CB, n CB : (0,1) . Hence σ xx σ xy cos α − p1 σ = − p1 cos α = with σ σ yy − sin α 0 τ = p1 sin α xy σ xx σ xy 0 − p2 sin α σ = − p2 cos α with = τ = − p2 sin α σ xy σ yy 1 − p2 cos α (8.5.4) Looking for a uniform stress field, we have from (8.5.4) σ xx cos α − σ xy sin α = − p1 σ xy cos α − σ yy sin α = 0 σ = − p sin α σ yy = − p2 cos α 2 xy arriving finally at the following (uniform) statically admissible stress field (8.5.5) 246 p1 + p2 sin 2 α − (σ ) = cos α − p sin α 2 − p2 sin α − p2 cos α (8.5.6) The Mohr circle, which is centred on the σ axis, must go, according to (8.5.4), through the points σ = − p1 cos α AB τ = p1 sin α σ = − p2 cos α CB τ = − p2 sin α (8.5.7) Hence the diagram in Fig. 8.5.3. Fig. 8.5.3 Mohr representation for question 1(b) of Exercise 8.5. 2) From the equilibrium equations we get σ xx , x + σ xy, y = 0 ⇒ a + f = 0 σ xy, x + σ yy, y − ρ g = 0 ⇒ e + d − ρ g = 0 (8.5.8) with g the gravitational acceleration. The boundary conditions on OB ( x = − y ) are and on OA ( x = 0 ) 1 (σ xx + σ xy ) = 0 ⇒ a − b + e − f = 0 2 1 (σ xy + σ yy ) = 0 ⇒ e − f + c − d = 0 2 (8.5.9) 247 −σ xx = −by = −ϖy ⇒ b = ϖ −σ xy = − fy = 0 ⇒ f = 0 (8.5.10) From the 6 equations (8.5.8-10), the considered stress field is statically admissible if a = f = 0 b = e = ϖ d = ρ g − ϖ c = ρ g − 2ϖ (8.5.11) 3) With Oy in the direction of the upward vertical the problem is invariant with respect to x and z, so the stress field is a function of y only. The equilibrium equations are ∂σ xy ∂y = 0 ∂σ yy ∂y − ρ g = 0 ∂σ yz ∂y = 0 (8.5.12) whence σ xy = a σ yz = b σ yy = ρ gy + c (8.5.13) The conditions at the boundary y = 0 are σ xy = 0 σ yy = − p0 σ yz = 0 ⇒ a = b = 0 c = − p0 (8.5.14) Thus the equilibrium equations enable us to determine σ xy , σ yy and σ yz uniquely, but only these components. From the nature of the problem it seems reasonable to infer that σ xz = 0 , but this still leaves σ xx and σ zz arbitrary until the constitutive equations for the medium have been specified. 8.3 Problems in linear elasticity Exercise 8.6 Notches and cracks It is known that, near the origin of a crack in an elastic medium that is linear, homogeneous and isotropic, the stress field σij, (and therefore the deformation field εij) has a singularity of the form r-½: for example, in the symmetric crack opening mode (Mode I) 248 σ ij (r ,θ ) = KI f ij (θ ) r (8.6.1) where KI is the stress-intensity factor for Mode I. This exercise is to prove the existence of such a singularity, considering the crack as a speci al case of a notch (see also Chapter 2, Volume II). We assume that perturbations are small and that at a point M (r ,θ ) near to the origin of the notch of angle 2ϕ symmetrical with respect to the x1 axis (see Fig. 8.6.1) the displacement field u ( M ) has the form ui = r α fi (θ ) in the cylindrical coordinate system (er, eθ), with uz = 0 in plane strain. Fig. 8.6.1 Coordinates and geometry of the notch 1) Express the strain ε (r ,θ ) and stress σ (r ,θ ) tensors at point M as functions of f, g and α. Solution In cylindrical polars we have in plane strain conditions (see Appendix F in Volume I) ur,r (ε ) = 1 (ru + u − u ) θ, r r,θ θ 2r 1 (ruθ,r + ur,θ − uθ ) 2r 1 (uθ,θ + ur ) r (8.6.2) 249 With ur = r α f (θ ) and uθ = r α g(θ ) , we find 1 1 αf (α − 1)g + f ′ 2 2 (ε (r ,θ ) ) = rα −1 1 (α − 1)g + 1 f ′ ′ g + f 2 2 (8.6.3) Using the constitutive equations we get 1 0 0 σ = λε kk δ + 2µε = λ r α −1 ( (α + 1)f + g′ ) 0 1 0 + ... 0 0 1 1 1 αf (α − 1)g + f ′ 0 2 2 1 α −1 1 ... + 2µ r (α − 1)g + f ′ g′ + f 0 2 2 0 0 0 (8.6.4) that is (σ ( r , θ ) ) = r α −1 (λ + 2µ )α f + λ (f + g′) µ ( f ′ + (α − 1)g ) λα f + (λ + 2 µ )(f + g′) µ ( f ′ + (α − 1)g ) (8.6.5) with σ rz = σ θz = 0 and σ zz = λε kk = λ r α −1 ( (α + 1)f + g′ ) . 2) From the equilibrium equations for the stresses find a system of two linear differential equations satisfied by the functions f and g. Deduce from that the differential equation, independent of the elastic constants, for f(θ) and find the general solution. Find f and g for the particular case of a load that is symmetrical with respect to the notch, i.e. to the x1 axis. Solution Expanding the equilibrium equation in cylindrical polars (see Appendix F in Volume I for the expression of divσ ), we find 250 ∂σ r r 1 ∂σ r θ ∂σ rz σ r r − σ θθ + + + + ρ fr = 0 r ∂θ ∂z r ∂r σ ∂σ r θ 1 ∂σ θθ ∂σ z θ + + + 2 rθ + ρ fθ = 0 r ∂θ ∂z r ∂r ∂σ z r 1 ∂σ z θ ∂σ z z σ z r + + + + ρ fz = 0 r ∂θ ∂z r ∂r (8.6.6) from which in this case, with (8.6.5) 2 µ f ′′ + [ (α − 1)(λ + µ ) − 2µ ] g′ + (α − 1)(λ + 2µ )f + = 0 2 (λ + 2µ )g′′ + [α (λ + µ ) + λ + 3µ ] f ′ + (α − 1) µ g = 0 (8.6.7) which do not depend on r, or, with use of E and ν instead of λ and µ α − 3 + 4ν (α 2 − 1)(1 −ν ) ′′ ′ f + g + 2 f =0 1 − 2ν 1 − 2ν 2(1 −ν ) g′′ + α + 3 − 4ν f ′ + (α 2 − 1)g = 0 1 − 2ν 1 − 2ν (8.6.8) which do not depend on E. Elimination of g between the equations (8.6.8) for f and g gives the following fourth-order homogneous differential equation for f(θ): f (4) + 2(α 2 + 1)f ′′ + (α 2 − 1)2 f = 0 (8.6.9) which does not involve any of the elastic constants. Putting f (θ ) = exp ( kθ ) we have the characteristic equation k 4 + 2(α 2 + 1)k 2 + (α 2 − 1) 2 = 0 (8.6.10) with roots ± (α + 1)i and ± (α − 1)i where i 2 = −1 ; the general solution is therefore f (θ ) = a1 cos(α + 1)θ + a2 cos(α − 1)θ + a3 sin(α + 1)θ + a4 sin(α − 1)θ (8.6.11) For symmetric loads the displacements must be symmetric, with u (r ,θ ) = u (r , −θ ) ; f is therefore an even function of θ, so the sine terms vanish, giving f (θ ) = a1 cos(α + 1)θ + a2 cos(α − 1)θ (8.6.12) 251 Substituting this in the first of the equations (8.6.8) gives an equation for g' which after simplifying is g′(θ ) = − a1 (α + 1)cos(α + 1)θ − a2 (α − 1) α + 3 − 4ν cos(α − 1)θ α − 3 + 4ν (8.6.13) Integrating, with g'(0) = 0, we have g(θ ) = −a1 sin(α + 1)θ − a2 α + 3 − 4ν sin(α − 1)θ α − 3 + 4ν (8.6.14) 3) The equations found for the functions f and g have the trivial solution f = g = 0 (a1 = a2 = 0) . Using the fact that the stress vector is zero at the edges of the notch, find the condition for a non-trivial solution to exist. Deduce that this solution can be either regular or singular and that for a crack we have α = 1/2 . Solution The condition that the edges of the notch are free of stress is expressed by stating that the stress vector components are zero for θ = ±ϕ . Taking account of the parity of f and g this gives, for θ = ϕ σ rθ θ =ϕ = 0 ⇒ f ′(ϕ ) + (α − 1)g(ϕ ) = 0 σ θθ θ =ϕ = 0 ⇒ [ (α + 1)λ + 2µ ] f (ϕ ) + (λ + 2µ )g′(ϕ ) = 0 (8.6.15) f ′(ϕ ) + (α − 1)g(ϕ ) = 0 (1 −ν + αν )f (ϕ ) + (1 −ν )g′(ϕ ) = 0 (8.6.16) that is Substituting (8.6.12) and (8.6.14) for f and g we get two linear homogeneous equations for a1 and a2, namely (α − 1)sin(α − 1)ϕ =0 a1 sin(α + 1)ϕ + a2 α − 3 + 4ν −a1α (1 − 2ν )cos(α + 1)ϕ + a2 (1 −ν + αν + A)cos(α − 1)ϕ = 0 with A = − (1 −ν )(α − 1)(α + 3 − 4ν ) α − 3 + 4ν (8.6.17) 252 The condition for the existence of a non-zero solution for a1 and a2 is that the determinant of the coefficients vanishes, that is sin(α + 1)ϕ −α (1 − 2ν )cos(α + 1)ϕ (α − 1)sin(α − 1)ϕ =0 α − 3 + 4ν (1 −ν + αν + A)cos(α − 1)ϕ (8.6.18) with A given in (8.6.18). This equation reduces to (α + 1) cos(α − 1)ϕ sin(α + 1)ϕ − (α − 1) cos(α + 1)ϕ sin(α − 1)ϕ = 0 (8.6.19) sin 2αϕ + α sin 2ϕ = 0 (8.6.20) or This can be solved graphically by studying the intersection, for any fixed value of ϕ lying between 0 and π, of the straight line y = −(sin 2ϕ ) x and the sine wave y = sin(2ϕ x) with x > 0 ; there are two cases to be considered. (a) 0 < ϕ < π 2 , i.e. π 2ϕ > 1 This corresponds to a salient, not a notch. The construction for the graphical solution shows that in this case the positive abscissa α0 of the intersection is larger than π 2ϕ > 1 . Since the radial variation of the stress and strain solution is rα-1, it follows that this remains finite at the origin and there is no singularity. (b) π 2 < ϕ < π , i.e. π 2ϕ < 1 and sin 2ϕ < 0 The solution is now such that α 0 < π 2ϕ < 1 , so that ( α − 1 ) is negative and there is a singularity at the apex of the notch. For a crack ( ϕ = π ), sin 2ϕ = 0 , giving the root α 0 = π 2π = 1 2 . The strains and stresses then have a singularity of the order r −1 2 and the displacements are like r 1 2 . Exercise 8.7 The Green tensor for the elastic isotropic infinite medium The problem (see Sect. 2.7.1 in Volume I) is to find the displacement field u(M) in an infinite linearly elastic isotropic medium due to a unit point force F applied at a point P, that is to prove Eq. 2.154 in Volume I. As we know, since the system is linear, the Green tensor G ( M , P ) for the infinite homogeneous medium, with zero displacement at infinity, obeys the relation u ( M ) = G ( M , P ).F ( P ) (8.7.1) 253 With this, the principle of superposition can be used again to construct the solution for any distribution of applied forces. 1) Show that the Navier equations (A3.17) in Appendix C of Volume I are satisfied identically by a displacement field of the form u = ∇2 B − 1 grad(div B) 2(1 −ν ) (8.7.2) where ν is the Poisson ratio and B satisfies an equation of the form ∇4 = ρ f µ (8.7.3) with ρ, µ the density and elastic shear modulus of the medium. Solution In rectangular Cartesians, the Navier equations read ui ,kk + 1 ρ uk ,ki + f i = 0 1 − 2ν µ (8.7.4) According to (8.7.2), we have ui = Bi ,kk − α Bk ,ki α= 1 2(1 −ν ) (8.7.5) and then ui ,kk = Bi ,kkll − α Bk ,ikll uk ,k = Bk ,kll − α Bk ,kll = (1 − 2ν )α Bk ,kll (8.7.6) so that (8.7.4) becomes Bi ,kkll − α Bk ,ikll + α Bk ,ikll = Bi ,kkll = − ρ f µ i or ∇ 4 B = − ρ f µ (8.7.7) 2) Find the field B due to a point force F applied at the origin; deduce the corresponding dispacement field u and the Green tensor, assuming an infinite medium and zero displacement at infinity. 254 Solution The body forces f must be taken in the form ρ f = F δ(r ) (8.7.8) where δ(r) is the Dirac function and r is the distance of a point from the origin; so the equation to be solved is ∇4 B = − F µ δ( r ) (8.7.9) We can use for that a classical result of the theory of Coulomb potentials, namely ∇ 2 (1 r ) = −4πδ(r ) (8.7.10) Now r,k = xk r r,kl = δ kl xk xl − 3 r r ⇒ ∇ 2 r = r,kk = 2 r (8.7.11) whence 2 ∇ 4 r = ∇ 2 ( ) = −8πδ(r ) r (8.7.12) So ∇4 B = − F µ δ(r ) = F 4 r ∇ r ⇒ ∇4 B − F=0 8πµ 8 µ π (8.7.13) from which B= r F+K 8πµ (8.7.14) where K is any biharmonic vector field (i.e., satisfying ∇ 4 K = 0 ). Taking account that u depends on the second derivatives of B and vanishes at infinity, one can show that it suffices to take K = 0, giving 255 B= r F 8πµ (8.7.15) From (8.7.5) and (8.7.15) we find Bi ,kk = Fi 4πµ r Bk ,k = xk Fk 8πµ r ⇒ Bk ,ki = Fi x xF − k i 3k 8πµ r 8πµ r (8.7.16) and then ui = xi x j xi x j 1 1 (2 − α )δij + α 2 Fj = (3 − 4ν )δij + 2 Fj (8.7.17) 8πµ r r 16πµ (1 −ν )r r So, referring to (8.7.1), we have proved that Gij ( M , O) = xi x j 1 (3 − 4ν )δij + 2 16πµ (1 −ν )r r (8.7.18) If now we put e = ( x − x′ ) x − x′ , we have, for a unit force applied at P( x′) Gij ( M , P ) = Gij ( x, x′) = Gij ( x − x′) = (3 − 4ν )δij + ei e j 16πµ (1 −ν ) x − x′ (8.7.19) which is Eq. 2.154 in Volume I. Exercise 8.8 Spherical inhomogeneity The problem is to find directly, i.e. without use of the general Eshelby’s solution of the ellipsoidal inhomogeneity problem, the strain state in a spherical elastic isotropic inhomogeneity (I) with properties (µI, νI) and perfet bonding at the interface to an infinite elastic isotropic medium with properties (µm, νm) which is subjected to a pure shear γ at infinity. We solve this by assuming a general form for the displacement field and determining the details by setting up the Navier equations with the boundary conditions at the interface and at infinity. We use spherical co-ordinates with origin at the centre of the sphere. 1) Write the Navier equations for the problem, assuming no body forces. Solution The Navier equations in the absence of body forces read 256 (λ + µ )grad ( div(u ) ) + µ∆(u ) = 0 (8.8.1) Using spherical polars, we have, according to Appendix F of Volume I (Sect. F.3) 1 u 1 uθ divu = ur,r + uθ,θ + 2 r + uϕ,ϕ + r r r sin θ r tan θ (8.8.2) (divu ),ϕ 1 eϕ grad(divu ) = (divu ),r e r + (divu ),θ eθ + r r sin θ (8.8.3) 1 u 1 uθ X = grad(divu ) = (ur,r + uθ,θ + 2 r + uϕ,ϕ + ), r e r r r r sin θ r tan θ uθ 1 1 u 1 + (ur,r + uθ,θ + 2 r + uϕ,ϕ + ),θ eθ r r r r sin θ r tan θ uθ u 1 1 1 + (ur,r + uθ,θ + 2 r + uϕ,ϕ + ),ϕ eϕ r sin θ r r r sin θ r tan θ (8.8.4) and then We find u u ru − uϕ,ϕ ruθ,r − uθ 1 1 X = (ur,rr + uθ,θ r − 2 uθ,θ + 2 r,r − 2 2r + ϕ,2ϕ r + 2 ) er r r r r r sin θ r tan θ u u cos θ u u uθ 1 1 + (ur,rθ + uθ,θθ + 2 r,θ + ϕ,ϕθ − ϕ,ϕ 2 + θ,θ − ) eθ (8.8.5) r r r r sin θ r sin θ r tan θ r sin 2 θ u u 1 1 1 + (ur,rϕ + uθ,θϕ + 2 r,ϕ + uϕ,ϕϕ + θ,ϕ ) eϕ r sin θ r r r sin θ r tan θ On the other hand, we have from Sect F.3 in Volume I (note that in the second term of the component on eϕ , sin 2θ must be changed into sin θ , as follows) 2u 2u 2(uθ sin θ ),θ Y = ∆u = ∆ur − 2r − − 2 ϕ,ϕ e r 2 r r sin θ r sin θ 2u cos θ 2u u + ∆uθ + 2r,θ − 2 θ 2 − 2ϕ,ϕ 2 eθ r r sin θ r sin θ 2u u 2u cos θ + ∆uϕ + 2 r,ϕ − 2 ϕ 2 + 2θ,ϕ 2 eϕ r sin θ r sin θ r sin θ (8.8.6) 257 with ∆ur , ∆uθ , ∆uϕ to be calculated according to the relation ∆f = f ,rr + 2 f ,r r + f,θθ r 2 + f,θ 2 r tan θ + f ,ϕϕ (8.8.7) 2 r sin 2 θ Thus, the Navier equations will read in this case (λ + µ ) X + µ Y = 0 or X + (1 − 2ν )Y = 0 (8.8.8) with X and Y given by (8.8.5) and (8.8.6), respectively. 2) The boundary conditions at infinity are as follows: 1 ux∞ = γ x 2 1 uy∞ = − γ y 2 uz∞ = 0 (8.8.9) or in spherical polars ur∞ = γr 2 sin 2θ cos 2ϕ uθ∞ = γr 2 sin θ cos θ cos 2ϕ uϕ∞ = − γr 2 sin θ sin 2ϕ (8.8.10) Look for the solution in the form ur = U r (r )sin 2θ cos2ϕ uθ = U θ (r )sinθ cosθ cos2ϕ uϕ = U ϕ (r )sinθ sin2ϕ (8.8.11) Solution To determine these functions we first substitute (8.8.11) into the Navier equations. This gives the following equations: 2 2 cos 2ϕ sin θ (r U r′′ + 2rU r′ − 2U r − 3rU θ′ + 3U θ ) er r 2 +2(rU ′ − U ) + 2(rU ′ − U ) ϕ ϕ θ θ 2sin θ cos θ cos 2ϕ ′ + rU + 2U r − 3U θ eθ r r2 2sin 2ϕ 2 − 2 sin θ (rU r′ + 2U r − 3U θ ) + 2(U θ + U ϕ ) eϕ r sin θ X= and (8.8.12) 258 u 2ur,r ur,θθ u + 2 + 2 r,θ + 2 r,ϕϕ2 ur,rr + r r r tan θ r sin θ Y = er 2uϕ,ϕ 2ur 2(uθ sin θ ),θ − 2 − 2 − r 2 sin θ r sin θ r u 2 u u u θ, r + θ,2θθ + 2 θ,θ + 2 θ,ϕϕ2 uθ,rr + θ θ r r r tan r sin eθ + 2uϕ,ϕ cos θ 2ur,θ uθ + 2 − 2 2 − 2 2 r sin θ r sin θ r 2uϕ,r uϕ,θθ u u + 2 + 2 ϕ,θ + 2 ϕ,ϕϕ2 uϕ,rr + r r r tan θ r sin θ + eϕ uϕ 2uθ,ϕ cos θ 2ur,ϕ − + 2 2 + 2 r sin θ r sin θ r 2 sin 2 θ (8.8.13) that is cos 2ϕ sin 2θ (r 2U r′′+ 2rU r′− 8U r + 6U θ ) − 4(U θ + U ϕ ) e r r2 4(U θ + U ϕ ) cosθ cos 2ϕ 2 + sin θ (r U θ′′+ 2rU θ′+ 4U r − 6U θ ) − eθ (8.8.14) 2 r sin θ Y= + ( ) 4(U θ + U ϕ ) sin θ sin 2ϕ 2 ′′ r U ϕ + 2rU ϕ′− 2U ϕ − 4U r + 4U θ − eϕ 2 r sin 2 θ Now, we set separately to zero those expressions that depend on θ and those that do not. We are then left with 3 equations which read U θ + U ϕ = 0 2 2(1 −ν )(r U r′′+ 2rU r′ ) − 2(5 − 8ν )U r − 3rU θ′+ 3(3 − 4ν )U θ = 0 2 (1 − 2ν )(r U θ′′ + 2rU θ′ ) + 2rU r′ + 4(1 −ν )(2U r − 3U θ ) = 0 (8.8.15) We now look for the solutions as linear functions of r, r3, r-2 and r-4, in the following form: 3 A3 5 − 4ν A4 6ν 3 U r = A1r − 1 − 2ν A2 r + r 4 + 1 − 2ν r 2 U = −U = A r − 7 − 4ν A r 3 − 2 A3 + 2 A4 θ ϕ 1 2 1 − 2ν r4 r2 (8.8.16) 259 where ν is either νI or νm according as r <a or r >a, the radius of the inclusion, and the constants Ai have different values in the inclusion and the matrix. Let us call them Ai in (I) and Bi in the matrix. Now (i) in the inclusion (r ≤ a) there is no singularity at r = 0 ; so A3 = A4 = 0 (ii) in the matrix the conditions at infinity give U r (∞) = U θ (∞) ≈ γ r/2 ; so B1 = γ /2 and B2 = 0 . The remaining coefficients A1, A2, B3 and B4 are determined.by stating continuity of the diplacement and the stress vector across the matrix-inclusion interface: (i) displacement: U r (a− ) = U r (a+ ) and U θ (a− ) = U θ (a+ ) , giving 6ν I 3B3 5 − 4ν m B4 γ 3 A1a − 1 − 2ν A2 a = 2 a + a 4 + 1 − 2ν a 2 I m A a − 7 − 4ν I A a 3 = γ a − 2 B3 + 2 B4 1 1 − 2ν I 2 2 a4 a2 (8.8.17) (ii) stress vector: σ rr (a− ) = σ rr (a+ ), σ rθ (a− ) = σ rθ (a+ ), σ rϕ (a− ) = σ rϕ (a+ ) with 2µ σ rr = 1 − 2ν ( (1 −ν )ε rr + ν (ε θθ + ε ϕϕ ) ) = 2µ sin 2 θ cos 2ϕ (1 −ν )U ′ + ν (2U − 3U ) r r θ 1 − 2ν r σ µε = 2 rθ rθ ′ 1 = µ sin θ cos θ cos 2ϕ U θ + r (2U r − U θ ) σ rϕ = 2µε rϕ 1 = − µ sin θ sin 2ϕ U θ′ + (2U r − U θ ) r (8.8.18) from which, after simplification, we have the two further equations: γ B 3ν I 5 −ν m B4 A2 a 2 = µ m − 12 53 − 2 µ I A1 + 1 − 2 2 a 1 − 2ν m a3 ν I µ A − 7 + 2ν I A a 2 = µ γ + 8 B3 + 2 1 + ν m B4 m I 1 1 − 2ν 2 a5 1 − 2ν m a 3 I 2 Solving (8.8.17) and (8.8.19), we find (8.8.19) 260 15(1 −ν m ) µm γ A1 = A2 = 0 2(4 − 5 ν ) µ + (7 − 5 ν ) µ m I m m 2 B 3( µ m − µ I ) γ 3 5 =− 2 ( 2(4 − 5ν m ) µ I + (7 − 5ν m ) µm ) 2 a B 5(1 − 2ν m )( µm − µ I ) γ 34 = a 2 2(4 5 ) (7 5 ) − + − ν µ ν µ ( ) m I m m 2 (8.8.20) The vanishing of A2 implies that in the inclusion U r = U θ = −U ϕ = A1r ; hence ur (I) = 2 A1 γ uθ (I) = ur∞ 2 A1 γ uθ∞ uϕ (I) = 2 A1 γ uϕ∞ (8.8.21) Thus the strain (and therefore the stress) is uniform in the inhomogeneity; the tensor is proportional to that of the prescribed shear strain. Putting 2 4 − 5ν m 15 1 −ν m (8.8.22) µm γ µm + β ( µ I − µm ) 2 (8.8.23) β= we get A1 = and hence, for the strain deviators eI = Since a deviatoric strain E µm ∞ E µm + β ( µ I − µm ) ∞ (8.8.24) can be decomposed into two orthogonal shear ∞ strains, (8.8.24) still holds for arbitrary E : it is the Eshelby relation for a spherical inhomogeneity (see Sect. 2.7.3.4 in Volume I). 8.4 Miscellaneous Exercise 8.9 Tensile test at finite strain 261 A circular cylindrical rod (initial length l0, initial radius r0) deforms uniformly, without volume change, under the axial load Q. In a Cartesian rectangular coordinate system with 1 as the rod axis, any point of the rod with the initial coordinates (X1, X2, X3) moves to (x1, x2, x3) with x1 = l X1 l0 x2 = r X2 r0 x3 = r X3 r0 (8.9.1) Let A = (l − l0 ) l0 be the corresponding axial elongation. 1) Using the polar decomposition of the transformation gradient F of a finite transformation T (see Sect. 1.4.3 in Volume I), give the general relation between the Lagrangian (Green-Lagrange) and Eulerian (Almansi-Euler) strain tensors and the symmetric part S of F . Show that with ε defined by ε = S − δ , one has 1 2 −1 L = Fɺ .F . Express all the foregoing tensors for the studied tensile test. Compare ∆ = ε + ε 2 . Express the Almansi-Euler strain rate tensor d as a function of l dλ l0 λ the numerical values of A, ∆11, ε11, A11 and the axial logarithmic strain e = ∫ for A = 50% . Solution The polar decomposition of F reads F = U .S with S=S T T and U = U −1 (8.9.2) From d x = F .d X , d x′ = F .d X ′ , one has T T 2 d x.d x′ = d X .F .F .d X ′ = d X .S .U .U .S .d X ′ = d X .S .d X ′ (8.9.3) So, for d x = d x′ and d X = d X ′ , we get 2 2 ds = d X .S .d X 2 ds0 = d X .d X 2 ⇒ ds 2 − ds02 = d X .( S − δ).d X = 2d X .∆.d X (8.9.4) 2 and then the relation between the Green-Lagrange ( ∆ ) and the Cauchy ( S ) strain tensors, namely 262 ∆= 1 2 S −δ 2 ( ) (8.9.5) With ε = S − δ , we find ∆= 1 1 2 (δ + ε ) 2 − δ ) = ε + ε ( 2 2 (8.9.6) −1 −1 Similarly, for a Eulerian description, we have d X = F .d x, d X ′ = F .d x′ with F −1 −1 = S .U T and then d X .d X ′ = d x.F −1T −1 −1 .F .d x′ = d x.F .F −1T −1 T −1 −2 .d x′ = d x.S .U .U .S .d X ′ = d x.S .d x′ (8.9.7) Hence, for d x = d x′ and d X = d X ′ ds 2 = d x.d x −2 ⇒ ds 2 − ds02 = d x.(δ − S ).d x = 2d x.Α.d x 2 −2 ds0 = d x.S d x (8.9.8) with the Almansi-Euler strain tensor Α given by 1 2 Α = (δ − S −2 ) (8.9.9) From (8.9.4), the Lagrangian strain rate tensor is simply ∆ɺ , whereas, from (8.9.8), we find for a Eulerian description d d −1 (d x) = ( F .d X ) = Fɺ .d X = Fɺ .F .d x = L.d x dt dt (8.9.10) d d d T (d x.d x) = (d x).d x + d x. (d x) = d x.( L + L ).d x = 2d x.d .d x dt dt dt (8.9.11) and then Hence, the Eulerian strain rate tensor d is given by 263 1 T d = (L + L ) 2 with L = Fɺ .F −1 (8.9.12) For the problem under investigation, we find l l0 F = ( ) 0 0 0 r r0 0 0 0 = (S ) ⇒ U = δ r r0 (8.9.13) and l 0 0 −1 l 0 r (ε ) = ( S − δ ) = 0 r − 1 0 0 r 0 − 1 0 r0 (8.9.14) Note that, since there is no volume change, we must have J = det( F ) = lr 2 =1 l0 r02 (8.9.15) From (8.9.13) we get (S ) 2 l 2 l02 = 0 0 0 2 2 0 r r 0 0 0 r 2 r02 (8.9.16) and then l2 0 0 2 −1 l0 1 2 1 r2 (∆ ) = 2 S − δ = 2 0 r 2 −1 0 0 2 r 0 0 − 1 r02 ( ) (8.9.17) 264 For a Eulerian description, we find (S ) −2 l02 l 2 = 0 0 0 2 0 r2 r 0 l02 1 − 2 l 0 1 0 ⇒ ( A) = 0 2 r02 r 2 0 0 1− r02 r2 0 0 0 (8.9.18) r02 1− 2 r As for the strain rate tensors, we find for the Lagrangian description llɺ 2 l0 ( ∆ɺ ) == 0 0 rrɺ r02 0 0 rɺ r0 (F ) 0 0 0 0 rrɺ r02 (8.9.19) and for the Eulerian one lɺ l0 ɺ (F ) = 0 0 0 rɺ r0 0 −1 l0 l = 0 0 0 r0 r 0 0 0 r0 r (8.9.20) and then, due to the symmetry ( L ) = ( Fɺ .F ) −1 lɺ l = 0 0 0 rɺ r 0 0 0 = (d ) rɺ r (8.9.21) For A = 50% , we have l l0 = 1.5 and then ∆11 = .625 , ε11 = .5 , Α11 = .278 and e = log1.5 = .405 . 2) Let C = Q S0 the nominal axial stress, with S0 the initial cross-section area. Calculate in the same axes system the components of the Cauchy stress tensor σ , referred to the current configuration, and those of the Piola-Khirchhoff II stress tensor Π , referred to the initial configuration (see Sect. 1.4.3 in Volume I), as 265 functions of C and A. Compare, for A = 50% , the numerical values of σ 11 C and Π11 C. Solution The Cauchy stress tensor for this tensile test reads Q S (σ ) = 0 0 0 0 0 0 0 0 (8.9.22) with, according to (8.9.15) σ 11 = r2 Q Q S0 l = = C 02 = C = C (1 + A) S S0 S r l0 (8.9.23) The Lagrangian Piola-Khirchhoff II stress tensor Π is linked with the Cauchy stress tensor by Π = J F −1.σ .( F T )−1 (8.9.24) Ql02 2 Sl (Π ) = 0 0 (8.9.25) which gives, with J = 1 0 0 0 0 0 0 Hence we have Π11 = Q S0 l02 Q l l02 C = = 2 2 S0 S l S0 l0 l 1+ A (8.9.26) Thus, comparing (8.9.23) and (8.9.26), we find for A = 50% σ 11 C = 1.5 and Π11 C = 1 1.5 = .667.