Mechanical Behaviour of Engineering Materials - Metals, Ceramics, Polymers and Composites 2010 Part 12 potx

432 13 Solutions 155 − λ det @ 55 55 155 − λ ˆ ˜ A = −λ (155 − λ)2 − 552 = , −λ ⇒ λ1 = 0, λ2 = 210, λ3 = 100 The result is: σI = 210 MPa, σII = 100 MPa, σIII = MPa b) σeq,T = σq σIII = 210 MPa > Rp0.2 The material yields I − ˆ ˜ c) σeq,M = (σI − σII )2 + (σII − σIII )2 + (σIII − σI )2 = 181.93 MPa < Rp0.2 The material does not yield d) It is not possible to decide because both yield criteria are only approximately true e) τ = σeq,M /M = 58.7 MPa < τF No significant activation of dislocation movement f) The deviator can be calculated from σ = σ − σhyd , using σhyd = tr σ/3 = (155 + 155 + 0)/3 MPa = 103.¯ MPa This results in ¯ 51.6 σ = @ 55 0 55 51.¯ A MPa −103.¯ Solution 11: The parameter m from equation (3.36) is m = 50 MPa/40 MPa = 1.25 a) The hydrostatic stress state is characterised by σ11 = σ22 = σ33 = σhyd and σ23 = σ13 = σ12 = Parabolic: According to equation (3.37), we find at yielding s» –2 m−1 m−1 Rp = · σeq,pM,F + σeq,pM,F + 2m 2m m−1 = · σeq,pM,F = σeq,pM,F m This results in a ‘hydrostatic yield strength’ of σeq,pM,F = 5/3 · Rp = 66.7 MPa Conical: According to equation (3.39), we find at yielding Rp = · [(m − 1) · σeq,cM,F + 0] = σeq,cM,F 2m 10 This results in a ‘hydrostatic yield strength’ of σeq,cM,F = 10/3·Rp = 133.3 MPa b) Parabolic: s» –2 0.25 0.25 σeq,pM = σ11 + σ11 + σ 2.5 2.5 2.5 11 r 321 = σ11 + σ = 1.89 σ11 = 1.058 Rp 10 100 11 The material yields 13 Solutions σ/MPa 433 200 150 100 50 0.00 0.01 0.02 0.03 0.04 ε/− Fig 13.4 Neuber’s hyperbola Conical: σeq,cM = » – q 0.25 σ11 + 2.25 4σ11 = 1.9 σ11 = 1.064 Rp 2.5 The material yields c) Parabolic: σeq,pM 0.25 = (−σ11 ) + 2.5 s» 0.25 σ11 2.5 –2 + σ = 1.69 σ11 = 0.947 Rp 2.5 11 The material does not yield Conical: » – q σeq,cM = 0.25 (−σ11 ) + 2.25 · 4σ11 = 1.7 σ11 = 0.952 Rp 2.5 The material does not yield Solution 12: a) Reading off from diagram 4.3, we find a stress concentration factor of Kt = 1.67 b) The nominal stress at the notch root is σnss = F = 198.94 MPa π(d/2)2 According to equation (4.1), we find σmax = Kt σnss = 332 MPa σmax is above Rp and Rm Thus, the component could not be used c) Using equation (4.5) yields σmax εmax = σnss Kt = 1.623 MPa E The corresponding Neuber’s hyperbola is shown in figure 13.4 d) The values can be read off the diagram: σmax = 210 MPa, εmax = 0.008 = 0.8% 434 13 Solutions e) The component can be used because the maximum strain is significantly smaller than the strain at necking Solution 13: a) Since the stress is defined as force per area, we have to look at the stress over the width of a lattice constant and have to integrate the stress field in x1 direction over a distance of one lattice constant The force is thus Z aNaCl KIc KIc ˆ √ ˜aNaCl KIc 3/2 √ √ = aNaCl √ F = aNaCl r = √ · 2aNaCl 2π r 2π 2π b) The force is F = kx if the bond is strained by a distance x The force at a strain of aNaCl /10 must, according to the assumption, equal the force from subtask a) The fracture toughness can thus be calculated as follows: k aNaCl KIc 3/2 = √ · 2aNaCl , 10 2π k KIc 1/2 = √ · 2aNaCl , 10 2π √ √ k 2π KIc = = 0.634 MPa m 1/2 10 a NaCl This value is of the correct order of magnitude for a ceramic crystal c) Because we simply used the stress field calculated from continuum mechanics to find the force at one atom, the calculation is incorrect since there can be no stresses in between the atomic positions The calculation could be improved by using the elastic stress field at some distance from the crack tip and by calculating the displacements of all atoms inside this region using the force law Furthermore, it would be necessary to quantify the fracture strain of a bond more precisely Assuming a simple spring force is also a severe approximation because the potential curve is not parabolic if the displacements are large (see, for example, figure 2.6) Calculations accounting for all this can yield realistic values for the fracture toughness of a material The calculation as presented here can be accepted as a very coarse approximation that mainly serves to show why a stress singularity does not imply a force singularity at the position of the crack tip Solution 14: We start by adding the elastic line and the 95% line to the diagram (figure 13.5) Reading off the forces yields F5 = 13.5 kN, Fmax = 14.5 kN F5 is to the left of Fmax , corresponding to the case from figure 5.16(b), yielding FQ = F5 The condition (5.32) has to be met: Fmax /FQ = 1.07 ≤ 1.1 (true) The geometry factor for the initial crack length is f = 9.66 Using equation (5.30) √ yields KQ = 23.3 MPa m We finally have to check the inequality (5.33) The righthand side is 2.5(KQ /Rp )2 = 5.2 mm All required dimensions (B, a, W − a) fulfil this condition 13 Solutions 435 Fmax F / kN 14 12 F5 10 linear elastic 95% line 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ∆s / mm Fig 13.5 Determination of the forces F5 and Fmax in the load-displacement diagram √ Thus, the fracture toughness is KIc = 23.3 MPa m Solution 15: The solution is based on section 5.2.3 a) Design against yielding: The stress state is uniaxial with σ = pD/(2t) Thus, the condition σ = pD/(2t) < Rp0.2 must be met: Since σ = 1200 MPa < 1420 MPa, there is no yielding Design against cleavage fracture: σI < σC : 1200 MPa < 2200 MPa Cleavage fracture is not to be expected √ Design against crack propagation: σI < KIc / πa, where a = 1.5 mm is the maximum half crack length to be expected: 1200 MPa < 1311 MPa There will be no crack propagation The tube can be used b) Since Rp was stated for uniaxial loading, the result is independent of the yield criterion because the service load is also uniaxial.1 c) Yielding occurs at a pressure p = 2tRp0.2 /D = 14.2 MPa Cleavage fracture will be observed at a pressure p = 2tσC /D = 22 MPa From the fracture toughness, the stress can be calculated using σ = √ √ KIc / πa The resulting failure pressure is p = 2tKIc /( πa D) = 13.1 MPa Thus, the tube will fail by crack propagation at a pressure p = 13.1 MPa if a crack of length a = 1.5 mm is present d) The yield strength and the fracture toughness are reached simultaneously at a √ value ac from equation (5.28): ac = (90 MPa m/1420 MPa)2 /π = 1.28 mm e) See figure 13.6 f) According to equation (5.3), the crack opening is v0 = 2σa = 0.036 mm = 36 µm , E If the yield criteria had been assumed equal for pure shear, there would be a difference in uniaxial tension 436 13 Solutions KI/MPa m 100 crack propagation 80 60 idealised realistic 20 0 yield 40 200 400 600 800 1000 1200 1400 σ/MPa Fig 13.6 Failure-Assessment diagram for exercise 15 where the additional factor is necessary because equation (5.3) uses the displacement of one crack surface which is half of the crack opening Solution 16: a) At small displacements a with shear strain γ = x/a: τ (x) = G x, Hooke’s law from equation (2.5) can be used x a (13.17) The shear stress is given by (see figure 12.4) “ x” τ (x) = τmax sin 2π a for all x between ≤ x ≤ a For small arguments α approximated as sin(α) ≈ α The result is τ (x) ≈ τmax · 2π 1, the sine can be x a Equalling this with equation (13.17) yields τmax = G , 2π (13.18) where τmax is equal to the theoretical shear stress τF In aluminium, with a ˜ shear modulus of G = 26 500 MPa, this results in the estimate τF = 4218 MPa ˜ In reality, pure aluminium has a yield strength of only Rp0.2 ≈ 50 MPa, corresponding to a maximum shear stress of τF = 25 MPa The simple estimate is thus too large by two orders of magnitude From this, we can conclude that slip does not occur by shifting layers of atoms simultaneously b) The equilibrium position at x = a/2 is unstable because the atoms of one layer are situated between those of the other layer, resulting in a maximum strain of the bonds An infinitesimal displacement from this position would result in the atoms moving to either x = or x = a This is due to the fact that the stiffness C = dτ /dx is negative at this point: ˛ dτ ˛ ˛ C= dx ˛ x=a/2 13 Solutions „ « a/2 = 2πτmax cos 2π = 2πτmax cos π = −2πτmax a 437 Solution 17: a) A dislocation moving from one side of the crystal to the other causes a slip of one Burgers vector b To shear the crystal by a length s, N = s/b = 3.5 × 105 dislocations have to move through the crystal Because the dislocations are not all at one side of the crystal initially, they can, on average, cover only half the length of the crystal, resulting in twice this value, N = × 105 b) The dislocation density is the dislocation length per volume To completely shear the crystal over its width, the dislocation has to extend throughout the crystal and thus have a length of 10 mm The resulting dislocation density is thus = aN N = = × 109 m−2 a3 a Not all of the dislocations can contribute because of their orientation Assume that the shear is in the x direction If we consider screw dislocations, all dislocations with a line vector in the y direction can contribute (one third of all screw dislocations) Of the edge dislocations, only those with line vector in the y direction can contribute that have the additional half-plane in the z direction (one sixth of all edge dislocations) As we are interested in an estimate only, we can assume that about one fifth of all dislocations contribute to the deformation This results in a final estimate for the dislocation density of about 3.5 × 1010 m−2 c) If the grain size is d, the dislocations not move throughout the crystal, but are limited to one grain because, due to the small amount of deformation, the stresses can be expected to be too small to allow dislocations to pass grain boundaries The number of dislocations thus increases by a factor s/d = 100, resulting in a dislocation density of = 3.5 × 1012 m−2 d) The total length of all dislocations is L = a3 = 3.5 × 106 m = 3500 km Solution 18: The energy per length of a dislocation line is T ≈ Gb2 /2 according to equation (6.3) Strictly speaking, this is only valid for a straight segment, but we will see that the required energy is so large that this is irrelevant The energy E of a dislocation loop of length 6b is E = 6Gb3 /2 = 1.8 × 10−18 J The probability to form such a ` ´ dislocation loop is P = exp −E/(kT ) , resulting in P300 ≈ 1.5 × 10−189 at 300 K and P900 ≈ 1.1 × 10−63 at 900 K The thermally activated generation of dislocation loops is thus practically impossible Solution 19: To calculate the increase in strength, we can use equation (6.20), with the increase being the difference of the strengthening contribution in both states: 438 13 Solutions Table 13.1 Determination of the Weibull modulus i ˜ Pf,i ` ´ ˜ ln ln 1/(1 − Pf,i ) σi /MPa ln(σi /MPa) 0.125 0.375 0.625 0.875 −2.013 −0.755 −0.019 0.732 54.60 90.02 244.69 665.15 4.0 4.5 5.5 6.5 √ ∆σ = kd M Gb ( − √ 0) = 230 MPa Solution 20: Using the Hall-Petch equation (6.25), the contribution to strengthening is ∆σcoarse = √ √ k/ dcoarse in the coarse-grained and ∆σfine = k/ dfine in the fine-grained material Reducing the grain size strengthens by the difference of these two contributions: k k −√ ∆σ = √ dfine dcoarse Solving for the new grain size, we find for ∆σ = 80 MPa „ dfine = ∆σ +√ k dcoarse «−2 = 1.48 µm Solution 21: a) The Orowan stress is τ = Gb/2λ according to equation (6.17) The normal stress σ and the shear stress τ are related by the Taylor factor which takes a value of M = 3.1 in face-centred cubic metals This results in 2λ = GbM = 39 nm ∆Rp0.2 b) According to equation (6.28), we find r fV r = 2λ = 5.5 nm Solution 22: We start by sorting the data with increasing size and assigning approximate failure probabilities according to equation (7.14) (see table 13.1) We also enter the quanti´ ` ˜ ties ln ln 1/(1 − Pf,i ) and ln(σi /MPa) into the table to enable drawing the diagram in figure 13.7 From this diagram, we can read off the Weibull modulus m = which equals the slope The intersection with the axis is −m ln(σ0 /MPa) = −5.6, yielding σ0 = 270 MPa 13 Solutions ln ln1−P f −1 −2 −3 −4 −5 −6 −7 439 ln σ m –m ln σ0 Fig 13.7 Graphical determination of m and σ0 in a diagram analogous to figure 7.17 Solution 23: a) The pressure is p = F/A = mg/(LB) = 6.131 25 × 10−3 N/mm2 b) Equation (7.6) yields » „ «m – V σlimit − Pf = exp − , V0 σ0 „ «m V σlimit ln(1 − Pf ) = − , V0 σ0 r σlimit V0 m = − ln(1 − Pf ) σ0 V (13.19) Using V /V0 = 1, we thus find σlimit = 0.541 σ0 Relating this to the pressure, σlimit = 2pL2 /d2 , we find for the thickness r 2pL2 dmin = = 15.1 mm (13.20) 0.541 σ0 c) From equation (13.19), we find, using V1 /V0 = LBdmin /Vspec = 12.08, the new value of the maximum stress σlimit,1 = 0.458 σ0 With the help of equation (13.20), the thickness is calculated to be dmin,1 = 16.35 mm This, however, changes the specimen volume V , changing the allowed stress σlimit from equation (13.19) Using the current thickness value dmin,1 = 16.35 mm yields a permitted stress of σlimit,2 = 0.456 σ0 Using again equation (13.20) results in the new thickness dmin,2 = 16.40 mm The change from dmin,1 to dmin,2 is rather small, making further iterations of the procedure unnecessary d) Production needs not to be stopped because all of the material volume is maximally stressed in tension, but only a small part of it in bending The strength in a tensile test is thus smaller than in a bending test Thus, the safety of the product is increased by this error p e) d = 2pL2 /Rp = 11.07 mm 440 13 Solutions Solution 24: a) We want to determine the parameters B ∗ and n from equation (7.2) To get a system of linear equations for the parameters, we write this equation as ` ´ tf = B ∗ σ −n ⇒ ln tf = ln B ∗ − n ln σ Using σ1 = 140 MPa, tf1 = 375.2 h, σ2 = 150 MPa, and tf2 = 94.4 h allows to write the system of equations: ` ´ ` ´ ln tf1 = ln B ∗ − n ln σ1 and ln tf2 = ln B ∗ − n ln σ2 , tf1 σ2 ln = n ln tf2 σ1 The result is n = 20.0 and B ∗ = 3.1529×1045 MPa20 h Using the provided value n−2 of the inert strength, we find B = B ∗ /σc = 0.3912 MPa2 h Note: Due to the large exponents in this calculation, your results may differ from those stated here by several percent The values here result if the exact values are used b) The failure probability can be calculated from equation (7.10), with V /V0 = 1, m∗ = m/(n − 2) = 1.2222, and t0 (σ) = B ∗ σ −n = 3.1529 × 1045 MPa20 h · (100 MPa)−20 = 314 016 h: " „ «m∗ # tf Pf (25 000 h) = − exp − t0 (σ) " „ «1.2222 # 25 000 h = − exp − = 4.4% 314 016 h The failure probability is larger than the design value 0.5% The component cannot be used c) The failure probability can be reduced using a proof test d) The calculation is analogous to the derivation of equation (7.16): ( " „ «m∗ #)  » „ «m –ff tf σp − exp − − − exp − t0 (σ) σ0  » „ «m –ff Gf (tf , σ) = σp − − exp − σ0 " „ «m∗ „ «m # tf σp = − exp − + (13.21) t0 (σ) σ0 The proof stress can be calculated from equation (13.21): " σp = σ0 ln(1 − Gf ) + „ tf t0 (σ) «m∗ #1/m 13 Solutions " = 375 MPa ln(1 − 0.005) + „ 25 000 h 314 016 h «1.2222 441 #1/22 The result is σp = 324.1 MPa e) The fraction of scrapped parts is calculated using equation (7.3): " „ «22 # 324.1 MPa Pf (324.1 MPa) = − exp − = 4.0% 375 MPa Solution 25: a) Because of the parallel connection of the elements, the strain ε is the same in both of them The stress σS (t) in the spring and σD (t) in the dashpot element are σ = σS (t) + σD (t) The strain rate in the dashpot element is thus dε σD σ − Eε = = dt η η This first-order differential equation can be solved by separation of variables: dε dt = , σ − Eε η Z Z dε dt = , σ − Eε η t − ln (σ − Eε) = + C , E η where C is a constant of integration Solving for ε yields » „ « – E ε= σ − exp − t C E η The constant of integration C can be determined by the fact that the strain ε is zero at time t = because the dashpot element cannot react immediately to the stress Thus, we find C = σ and » „ «– σ E ε= − exp − t E η The strain increases with time and approaches a value σ/E because the dashpot element will have relaxed completely and all of the stress is transferred by the spring b) In a relaxation experiment, the strain is to be increased discontinuously by a finite value This causes an infinite stress in the dashpot element in this model A relaxation experiment can therefore not be modelled with this approach c) The three elements in the three-parameter model are denoted as follows: Element is the spring element in series, element is the parallel spring element, and element is the dashpot element This yields the following relations for stresses and strains: A.6 Important constants and tensor operations 457 Using equation (A.6), we can calculate the transformation matrix √ √ 2/2 √2/2 √ (gi i ) = − 2/2 2/2 If we perform the coordinate transformation of a, (ai ) = √ √ √ 2/2 √2/2 2/2 √ √ (ai ) = (gi i ) (ai ) = = − 2/2 2/2 2/2 T , we find A.6 Important constants and tensor operations In this section, some important rules, conventions, and constants are summarised The Kronecker delta δij represents a tensor with invariant components that not change in any coordinate rotation It is defined as δij = for i = j, for i = j (A.7) Written in the component notation, the Kronecker delta is nothing but the unit tensor:   0 (δij ) = =   0 A tensor of any order can be transposed by reverting the sequence of the indices: (Cijkl )T = (Clkji ) In a tensor of second order, this means exchanging the rows and columns The trace of a tensor of second order is the sum of its diagonal elements: tr A = Aii = A11 + A22 + A33 (A.8) The determinant det A of a tensor of second order is the determinant of its matrix representation In three dimensions, it is det A = − A11 A22 A33 + A12 A23 A31 + A13 A21 A32 − A11 A23 A32 − A12 A21 A33 − A13 A22 A31 The positive terms are formed by the components on the diagonals that point downwards and to the right, the negative by those pointing upwards and to the right.2 This simple rule is only valid in three dimensions 458 A Using tensors A.7 Invariants Each tensor representation is characterised by some properties that are not changed by a coordinate transformation The length of a vector, for example, is not changed by a coordinate transformation It can be computed by the contraction of the vector with itself: |a|2 = a2 + a2 + a2 = a2 + a2 + a2 , 3 or, in index notation, |a|2 = ai = ai (A.9) Since |a| is defined by a contraction, it does not change during a coordinate transformation The length of a vector is thus called an invariant If we choose a special xi coordinate system where the vector representation is (ai ) = |a| T , the vector is completely specified by stating its length and the coordinate system Like vectors, tensors of any order possess invariants, quantities that not change during a coordinate transformation A second-order tensor has three invariants, called its eigenvalues (from the German word ‘eigen’ meaning ‘own, peculiar, particular’) λ(k) For a tensor (Aij ), they can be calculated from the characteristic equation (k) Aij − δij λ(k) vj =0 (A.10) or A − 1λ(k) v (k) = , (k) where the trivial solution vj = is ignored This equation always has three (not necessarily distinct) solutions λ(k) with the associated eigenvectors v (k) In general, the eigenvalues and -vectors calculated this way are complex and thus cannot be interpreted physically In the common case of a symmetric tensor, fulfilling Aij = Aji , the eigenvalues are real numbers and the eigenvectors are perpendicular to each other This, for example, is the case in stress or strain states in a classical continuum The eigenvalues in this case are called principal stresses and principal strains, respectively; the eigenvectors are the principal directions or axes Similar to a vector, a symmetric second-order tensor can be completely characterised by stating its eigenvalues and -vectors To so, we form a new xi coordinate system of the eigenvectors, normalised to length In this system, the tensor is diagonal: A.8 Derivations of tensor fields gi = v (i ) , λ(1) = 0  (Ai j ) = δi j 459 λ(i ) λ(2)  0  λ(3) In the representation using the principal axes, the component matrix contains only diagonal elements There is a special set of invariants, the principal invariants Jk , that can be formed from the eigenvalues: J1 = λ(1) + λ(2) + λ(3) , J2 = −λ(1) λ(2) − λ(1) λ(3) − λ(2) λ(3) , J3 = λ(1) λ(2) λ(3) Written for the tensor itself, they can be calculated as follows: J1 = Aii J2 = [Aij Aji − Aii Ajj ] J3 = det(Aij ) = tr A , = tr A AT − tr A = det A , These principal invariants are important in the context of yielding of materials and are used in section 3.3.1 A.8 Derivations of tensor fields Physical quantities are frequently not defined by a single tensor, but by assigning a tensor to each point in space, thus defining a field Examples are a temperature field (a scalar field), where a temperature is specified at each point in space, a velocity field in a flowing fluid (a vector field), where the flow direction and speed are stated at each point, or the stress field in a material (a tensor field of second order), assigning a value of the stress tensor to each point Because the value of the field may differ at different points in space, fields can be derived in the different spatial directions The derivative of a scalar field f (x) with respect to x is a vector, calculated by the rule   ∂f (x)/∂x1 ∂f (x)  ∂f (x)/∂x2  = ∂x ∂f (x)/∂x3 Each component states how the scalar field changes in the respective spatial direction This vector field is called the gradient of f 460 A Using tensors The gradient of a vector field v(x) can be calculated in a similar way It states how the vector field changes in each spatial direction and is thus a tensor field of second order The rule to calculate this gradient can be most easily written in component notation: ∂v(x) ∂x = ij ∂vj ∂xi It is important to note that the first index of the gradient, i, is in the denominator, not in the numerator of the right-hand side In the same way, higher-order tensors can be derived with respect to a vector, always using the first index from the denominator Occasionally, tensor fields have to be derived with respect to scalar quantities To so, each component of the field is simply derived separately, for example dv(α) = dα dv1 (α) dα dv2 (α) dα dv3 (α) dα T The order of the tensor remains unchanged in this operation B Miller and Miller-Bravais indices In many cases, it is necessary to specify directions and planes in a crystal lattice It is most sensible to so using a crystallographic coordinate system, with axes parallel to the edges of the chosen unit cell All parallel directions and planes in a crystal are equivalent, rendering it unnecessary to state the origin of the direction vector or plane B.1 Miller indices For specifying directions and planes in a crystal, the origin of the crystallographic coordinate system is positioned in a lattice point, and the axes are scaled so that the length of every edge of the unit cell is one Thus, for noncubic lattice types, this coordinate system is non-Cartesian Using the so-called Miller indices, a direction is specified by a straight line through the origin of the coordinate system The coordinates describing the line are called the indices and written with square brackets: [hkl] They result from the intersection of the line with the nearest lattice point e g., [112] in figure B.1(a) If negative values occur, they are denoted by a bar on top of the coordinate, for example [110] If we not want to specify a certain direction, but all cristallographically equivalent directions, we use indices in angle brackets: hkl Cristallographically equivalent directions are those that can be transformed into each other by using a crystal symmetry In a cubic crystal, for example, all space diagonals ([111], [111], [111], [111]) are equivalent A plane is also specified by three numbers They are determined in the following way: Choose the origin of the coordinate at any point not in the plane considered Determine the intersections m, n, and p of the plane with the coordinate axes as shown in figure B.1(b) (here: m = 1, n = 1, p = 2) If the plane is parallel to one of the axes, the intersection is assumed to occur at ˜ infinity We now calculate the reciprocal values of the intersections: h = m−1 , ˜ = n−1 , ˜ = p−1 (here: h = 1, k = 1, ˜ = 0.5) Next, we form the smallest ˜ ˜ k l l 462 B Miller and Miller-Bravais indices [112] l p (221) n k a c a c b b h m (a) For a direction (b) For a plane Fig B.1 Determination of Miller indices l i k h Fig B.2 Coordinate system used for MillerBravais indices ˜ ˜ l triple consisting of integers, h : k : l, with the same ratios as h : k : ˜ This triple characterises the plane and is written in parentheses: (hkl), for example (221) If the set of all equivalent planes is to be specified, curly braces are used: {hkl} In the case of a cubic lattice, the indices of a plane specify its normal vector B.2 Miller-Bravais indices In hexagonal crystals, where the base plane exhibits a 120° symmetry, the Miller-Bravais system is used to specify planes and directions, using a coordinate system with four axes: Three of these, with angles of 120° between them, lie in the base plane and are equivalent, the fourth is perpendicular to the base plane: [hkil] (figure B.2) In this way, the symmetry of the crystal is reflected in the indices The first three indices obey the additional constraint h + k + i = (B.1) B.2 Miller-Bravais indices 463 Apart from this peculiar way of defining the coordinate system, the calculation of the indices of directions and planes is the same as for the Miller indices However, the normal vector of a plane does not correspond to the plane’s indices C A crash course in thermodynamics In this appendix, we will explain two important thermodynamic concepts needed in this book in different places: The first concept is thermal activation of processes, the second the concept of free energy A detailed discussion of thermodynamics can be found, for example, in Reif [116] C.1 Thermal activation We are looking for the probability that a process needing an energy ∆E occurs in a certain system The system may take this energy from its stored thermal energy If the temperature of the system is above absolute zero, its components (for example, the atoms it consists of) are in permanent, irregular motion, the so-called Brownian motion Slightly simplified, we can consider the thermal activation of a process as being caused by these random movements of the atoms acting together and enabling the process The probability for such an event will become the larger, the higher the temperature To understand thermal activation in greater detail, we need one basic principle of thermodynamics If a system can exist in a number of different states Z1 , Z2 , with energies E1 , E2 , and if it is in thermal equilibrium, Boltzmann’s law states that the probability P (Zi ) to find the system in state Zi is given by P (Zi ) ∝ exp − Ei kT (C.1) Here T is the system temperature and k is Boltzmann’s constant (k = 1.38 × 10−23 J/K) The constant of proportionality in the equation can be calculated from the fact that the sum over all probabilities must be From this law, the probability that the system changes its state by thermal activation to a state with an energy that is larger by ∆E is P (∆E) ∝ exp − ∆E kT (C.2) 466 C A crash course in thermodynamics As an example, we can estimate the density of vacancies in a metallic crystal If a vacancy is formed, all atoms adjacent to the vacancy thus possess unsaturated bonds, thus increasing the energy A typical value for the activation energy required for this process is ∆E ≈ 10−19 J If we put this number into equation (C.2),1 we can calculate the probability of a vacancy being at a certain lattice site as × 10−12 at 0℃ and as 10−4 at 500℃ Due to the large number of atoms in a crystal, these probabilities also correspond to the vacancy density As can be seen, the number of vacancies strongly increases with temperature Chemists frequently use a different form of Boltzmann’s law, replacing Boltzmann’s constant with the so-called gas constant R = 8.314 J/mol K Using this, the energy in equations (C.1) and (C.2) has to be given per mole The formula is then to be interpreted as giving the probability that the process of interest does not happen only once, but 6.022 × 1023 times The activation energy in the example above then takes a value of 60 kJ/mol C.2 Free energy and free enthalpy All of thermodynamics is based on the following two laws: • First law of thermodynamics: The energy U of a closed system is constant • Second law of thermodynamics: The entropy S of a closed system takes its maximum value in thermal equilibrium A closed system in this context is a system with constant volume that can exchange neither heat nor particles with its environment The statement of the first law is nothing but the law of the conservation of energy To understand the second law, we need to think about the meaning of entropy The entropy of a system is a measure of the probability that the system is in a certain macroscopic state In a closed system, in which the energy is constant, all microscopic states are equally probable according to Boltzmann’s law If we observe the system macroscopically, we will find that state most often that can be obtained by the largest number of distinct microscopic states Therefore, it is highly improbable that all gas molecules inside a container will gather in one corner, leaving a vacuum everywhere else The process is not impossible, but there are only a very small number of possibilities to arrange the gas molecules in the corner, compared to the number of possibilities to distribute them evenly over the whole volume Simplifying, it can be said that In this calculation we assume that each lattice site can exist in one of two states, either occupied with an atom or vacant As the probability for a vacancy is small, the probability of the site being occupied is close to one The proportionality constant in equation (C.2) can therefore be set to in very good approximation C.2 Free energy and free enthalpy 467 the entropy of a system is a measure of its disorder because an ordered state can only be created in a comparably small number of ways These considerations were made for a closed system In practice, we usually have to deal with systems in contact with their environment According to the previous section, the probability for a system to be in a certain state Zi in thermal equilibrium at a temperature T is given by equation (C.1) The most probable state is therefore the state with the lowest energy, and the higher the energy of a state, the lower is the probability to find the system in this state This statement seems to contradict every-day experience: If we consider again the example of the gas molecules in a container, now at a fixed temperature, it seems to imply that all gas molecules should lie at rest on the floor of the container because this would minimise their potential and kinetic energy This, however, is not observed This seeming contradiction can be resolved by considering the number of different states the gas molecules can be in to obtain a certain macroscopic state There are only a small number of possibilities to produce the state of lowest energy described above, but a very large number of configurations in which the gas molecules are irregularly distributed everywhere in the container For this reason we will almost certainly observe one of the irregular configurations A simple example can serve to illustrate the distinction: A die is loaded so that it shows the number with a probability of 25%, each other number with a probability of only 15% If we throw the die ten times, the probability to throw ten sixes is larger than the probability of any other exactly specified sequence of numbers Nevertheless, the probability for this event is only (1/4)10 ≈ 10−6 , for there is only one possibility to get ten sixes, but, for example, already 50 ways to throw nine sixes together with another number If we consider a system S held at a certain temperature by bringing it in contact with a heat bath W, the two can exchange energy The entropy is maximised for the complete system i e., for the system S and the heat bath W together In this case, the entropy of S itself is not necessarily maximised because the complete system may increase its entropy by a process that diminishes the entropy of S but increases the entropy of W by a greater amount To describe the system S, we introduce a new quantity, the free energy F It is defined as F = U − TS , (C.3) where U is the internal energy of the system S and S is its entropy F is minimised when the system is in contact with a heat bath and has a fixed volume.2 F is not maximised because the entropy enters with a minus sign 468 C A crash course in thermodynamics If the system can also change its volume, being held under constant pressure p,3 it can also change its inner energy by increasing or decreasing its volume V against the pressure In this case, the free enthalpy or Gibb’s energy G takes the place of F : G = U − T S + pV (C.4) In a system at constant temperature and pressure, the free enthalpy is minimised If we are interested in whether a certain process will take place under these conditions, we have to look at changes in free enthalpy: If it decreases, the process can take place One example for this is the investigation of nucleation in section 6.4.4 In solids, the distinction between free energy and free enthalpy is usually unnecessary because the volume change on changes in pressure is small C.3 Phase transformations and phase diagrams As explained in the previous section, a system with constant volume in contact with a heat bath (i e., at constant temperature) minimises its free energy At low temperatures, the influence of the entropy is small according to equation (C.3) so that the system tends to minimises its inner energy With increasing temperature, the entropy becomes more and more important, and the system will not be in the state of lowest energy This temperature dependence of the state is the reason for the occurrence of phase transformations as will be explained in the following As an example, consider a metal: At low temperatures, it is crystalline because this arrangement minimises the energy by ensuring strong bonds between the atoms (see section 1.2) The entropy of the crystal, however, is small due to its long-range order, for the positions of all atoms in the crystal are (almost) fixed In the liquid state, on the other hand, the metal atoms are more weakly bound, resulting in a higher inner energy The entropy is larger, however, because there is no long-range order and the atoms can move about freely Figure C.1 shows curves of the free energy for the liquid and solid state of a material According to equation (C.3), both free energies depend linearly on temperature5 so that the curves intersect at a certain temperature value Below this value, the crystalline state is the state of lowest free energy, above this temperature, the liquid state is favoured If we heat the system beyond The pressure p is the negative hydrostatic stress σhyd and is thus negative under tensile stresses One counter-example is the transformation toughening of ceramics discussed in section 7.5.4 Here we make the simplifying assumption that neither the internal energy nor the entropy themselves are temperature-dependent C.3 Phase transformations and phase diagrams F 469 solid liquid Fliquid Fsolid Tm T Fig C.1 Free energy in the liquid and solid state as function of the temperature this temperature, a phase transformation between the solid and the liquid state will take place This temperature is thus the melting temperature In general, a phase transformation does not occur exactly at the transformation temperature Directly above the melting temperature, the reduction in free energy that can be obtained by the transformation is small Furthermore, some additional energy is needed to form a liquid phase because the interface between the two phases is in a high-energy state This makes it possible to heat a system beyond its melting temperature without a phase transformation The phase transformation can only occur if the increase in free energy is large enough to compensate for the additional interface energy If the process occurs infinitely slow, however, the phase transformation can occur directly at the transformation temperature because there is always a finite probability of the additional interface energy being provided by thermal activation If we look at an alloy instead of a pure material, the situation becomes more complicated Depending on the solubility of the alloying elements, the system can reduce its free energy by either mixing the components or by separating different phases Consider a completely solid system made of two elements A and B The entropy of the system is largest if both elements form a solid solution because separating the atoms in two phases would reduce the number of possibilities to arrange the atoms At sufficiently high temperatures, we therefore expect complete solubility of the two elements At low temperatures, the behaviour of the system depends on the strength of the bond between the elements A and B: If it is stronger than the bond between A–A and B– B, a solid solution is favoured at low temperatures as well If the bond is weaker, it is better to separate the phases at low temperatures The result is a miscibility gap, a region in the phase diagram (see below) where the elements are not completely soluble Figure C.2 shows a phase diagram of a system with a miscibility gap In this kind of diagram, the concentration of the elements is put on the horizontal axis, the temperature on the vertical Within the diagram, the regions consisting of different phases are marked to allow determining the phases of the system as function of temperature and concentration of the alloying elements The regions are separated by boundary lines which denote a phase transformation 470 C A crash course in thermodynamics T single phase A 20 miscibility gap, two-phase 40 60 80 100 %B Fig C.2 Sketch of a phase diagram of a system with a miscibility gap in the solid state At low temperatures, two separate phases coexist (point 1), at high temperatures the elements mix in solid solution (point 2) In some regions, the alloy is in a single phase; in others, it is in a two-phase state.6 In phase diagrams of binary alloys, regions with one phase always adjoin to regions with two phases, except for single points This rule is helpful in reading more complicated phase diagrams In the two-phase region, an A-rich and a B-rich phase coexist The concentration within the two phases can be read off by drawing a horizontal line from the point characterising the actual system state At the intersection of this line with the boundaries of the two-phase region, the concentration within the two phases can be found The amount of the two phases can also be determined because the overall concentration must equal the known concentration in the alloy If c denotes the total concentration of B, cA the B concentration in the A-rich phase, and cB the B concentration in the B-rich phase, the masses mA and mB of the A- and B-rich phase are given by the lever rule mA cB − c = , mA + mB cB − cA mB c − cA = mA + mB cB − cA (C.5) If we cool down a system with a miscibility gap from the high-temperature region, phases will separate similarly to the transition between solid and liquid phases Again, the transformation will only occur exactly at the transformation temperature shown in the diagram if the cooling is extremely slow To separate the two phases, diffusion processes in the solid have to occur which are slow at low temperatures Thus, it is possible to supercool the system by In real systems, even at higher temperatures, a solid solution may not form because the melting temperature may be reached before the increase in entropy is large enough C.3 Phase transformations and phase diagrams 471 melt T melt + solid solution solid solution A 20 40 60 80 100 %B Fig C.3 Schematic phase diagram of a binary system with complete solubility of the components cooling down so fast that no diffusion can occur If we keep the system at low temperatures, the diffusion coefficient may be so small that the system state is practically stable, although it is not the thermodynamically favoured state of lowest free energy This is called a metastable state As already explained, a solid will transform to a liquid at high temperatures The liquid state can also be drawn in the phase diagram As an example, figure C.3 shows the phase diagram of a binary alloy with complete solubility in the liquid and the solid state At low temperatures, a solid solution forms; at high temperatures, the system is liquid If the system is cooled down from the liquid phase, a two-phase region is encountered in which melt and solid phase coexist As we can see by drawing a horizontal line, the solid regions are richer in the component with the higher melting point, whereas the concentration of the lower-melting component is larger in the melt With decreasing temperature, the concentrations in the melt and the solid change until the single-phase solid solution forms when the second boundary line is crossed Frequently, the components A and B are only partially soluble in the solid state In this case, the phase diagram is more complicated One possible diagram is shown in figure C.4 At high temperatures, in the liquid state, both components are completely soluble, but at low temperatures, the solubility is small so that two phases can coexist, one A-rich phase with dissolved B atoms, usually called the α phase, and one B-rich phase with dissolved A atoms, usually called the β phase The lower part of the phase diagram is similar to figure C.2, for in this case there is also a miscibility gap with a two-phase region in the solid state However, the gap ends at a certain temperature because the material starts to melt At higher temperatures, a mixture of melt and α or β phase forms, depending on the concentration There is a certain concentration, called the eutectic concentration, where the solid transforms directly into a melt and becomes liquid at a certain temperature, not in a temperature 472 C A crash course in thermodynamics T melt S+α S+β α β eutectic α+β A 20 40 60 80 100 %B Fig C.4 Schematic phase diagram of a binary system with limited solubility of the components regime In this case, the temperature is below the melting temperature of both components Depending on the binding energy between the atoms and on the possibility of additional phases that may form, real phase diagrams can be much more complex than the simple examples shown here One example is the phase diagram of the iron-carbon system in figure 6.50 ... denotes the total concentration of B, cA the B concentration in the A-rich phase, and cB the B concentration in the B-rich phase, the masses mA and mB of the A- and B-rich phase are given by the... averaged over fibre and matrix and will not occur at any point in the component Af /A and Am /A are the area fractions of fibre and matrix and thus equal the volume fractions ff and fm because the... three-dimensional space It is thus one-dimensional, and the tensor is a tensor of first order The components of the matrix can be characterised by a single index In a scalar, which is coordinate-independent,