188 6 Mechanical b ehaviour of metals ¿ ¿ l 2 l 1 Fig. 6.22. Straight dislocation line subjected to a shear stress on an area l 1 l 2 . When the dislocation has moved by l 2 , the upper half of the crystal has slipped by one Burgers vector b. This requires a work of E = F ext · b = τ l 1 l 2 b . (6.13) The dislocation has moved by a distance l 2 . The work needed can also be calculated by E = F d l 2 , when F d denotes the force on the dislocation. As both energies are equal, the force on the dislocation is F d = τl 1 b . (6.14) Here we used the fact that the force is perpendicular on the dislocation line. If the orientation between the stress tensor σ, the dislocation line l 1 and the Burgers vector b is arbitrary, the Peach-Koehler equation F d = (σ · b) × l 1 (6.15) holds. Equation (6.15) can be derived in a similar way to equation (6.14) by calculating the energy. If the dislocation line is displaced by l 2 , the crystal above the covered area has slipped. The normal vector in this area is given by the cross product l 1 × l 2 /|l 1 × l 2 |. The stress in this area is σ ·(l 1 ×l 2 )/|l 1 ×l 2 |, resulting in a force of σ ·(l 1 ×l 2 ). Because the crystal has slipp ed by a Burgers vector, the work is E = ` σ · (l 1 × l 2 ) ´ · b . Using rules for scalar and vector products, this equation can be re- written due to symmetry of the stress tensor: E = (σ · b) ·(l 1 × l 2 ) = ` ( σ · b) ×l 1 ´ · l 2 . This energy equals the force on the dislocation, multiplied by l 2 : E = F d · l 2 . 6.3 Overcoming obstacles 189 As the energies must agree for arbitrary l 2 , the Peach-Koehler equation results: F d = (σ · b) ×l 1 . Peierls force We already saw in section 6.2.3 that atomic bonds have to flip for a dislocation to move. This requires stretching of the bonds and therefore n eed s energy. The resulting force fixes the dislocation at its momentary position and has to be overcome to move it. Thus, if the applied stress is too small, no dislocation movement is possible and the crystal cannot deform plastically. Figure 6.13 above illustrates this using the sphere model of atoms. This retaining f orce is called Peierls force (or Peierls-Nabarro force). It determines the yield strength (or critical resolved shear stress, see section 6.2.5) of single crystals if their impurity content is small. In face-centred cubic or hexagonal close-packed metals, the Peierls stress is about 10 −5 G (where G is the shear modulus) and can therefore not explain the strength of engineering alloys. In these, other obstacles for the movement of dislocations play a role, to be discussed in section 6.3. In body-centred cubic metals, the Peierls force is larger than in the close-packed structures, espec ially at low temperatures, and influe nce s the yield strength significantly. This will be explained in section 6.3.2. After the dislocation has moved by half a Burgers vector, the Peierls force pushes it forwards and moves it to the position of the next energy minimum. The stored energy is usually dissipated as he at (i. e., as random crystal vibra- tion) in the crystal. The Peierls force thus acts as a kind of frictional force and reduces the effective stress that can be used to drive the dislocation to overcome other obstacles. Other inner stresses, caused for example by other obstacles (see the next section), can counteract the external stress τ in a similar way to the Peierls stress. They can thus also be considered as inner frictional forces or stresses. The stress τ ∗ that is effectively available to move the dislocation is thus τ ∗ = τ −τ i . If a certain kind of obstacle is investigated, it is often useful to combine the contributions of all other obstacles to a single frictional stress τ i and to assume that the dislocation is driven by the effective stress τ ∗ . 6.3 Overcoming obstacles Dislocations can be retarded by different kinds of obstacles . We already know one of these, the Peierls force. Other types, such as precipitates of a sec- ond phase, grain boundaries, or impurity atoms, will be discussed below in section 6.4 when we look at strengthening mechanisms. Here we want to un- derstand in what ways a dislocation can overcome an obstacle. As we will see, 190 6 Mechanical b ehaviour of metals µ 2¸ R F R F R F (a) Complete figure µ F T T µ µ (b) Equilibrium of forces for the dislo- cation line Fig. 6.23. Deflection of a dislocation line pinned by obstacles (only part of the dislocation be tween two obstacles is shown) it is important whether the dislo cation movement is aided by the temperature or not. The first case is called a thermally activated process, the second an athermal one. 6.3.1 Athermal processes Let there be several obstacles in our material with a distance of 2λ between them (figure 6.23). Consider a dislocation pinned on these obstacles. When the external stress τ acts on the dislocation, it tries to move on and bows out. Its shape is a segment of a circle because this covers the greatest area with the least-most energy to create new length of dislocation line. The component of the force in the direction of movement is, according to equation (6.14), F = 2λbτ . Therefore, each obstacle exerts a retaining force F R with opposite orientation and identical magnitude, f or each obstacle takes half of the force F from two dislocation segments. If T is the line tension of the dislocation (see equation (6.3)), this force is F R = 2T sin θ = 2T λ R = Gb 2 λ R , where G is the shear modulus, b the Burgers vector, and R the radius of the dislocation segment. Equalling F and F R yields τ = Gb 2λ sin θ . (6.16) Here it is crucial that the obstacle cannot bear arbitrarily large forces. If F max is the maximum force the obstacle can bear, the dislocation can detach 6.3 Overcoming obstacles 191 t 1 t 2 b b (a) Before the annihilation. The dislocation segments with oppo- site orientation attract t 1 t 2 b b ' ' (b) After the annihilation. Since both disloca- tion segments lie on the same lattice plane they can annihil ate, resulting in a dislocation loop around the obstacle and a free dislocation Fig. 6.24. Annihilation of dislocation segments in the Orowan mechanism a. b. c. d. Fig. 6.25. Overcoming an obstacle by cutting of dislocations (after [74]) from the obstacle if F R exceeds F max . Therefore, there is a critical value sin θ = F max /Gb 2 , and sin θ can be considered as dimensionles s measure of the obstacle strength. It may seem contradictory that sin θ takes only a limited range of values but F max does not. This is resolved by realising that at sin θ = 1, the dislocation will have bowed out so far that it becomes a semi-circle, resulting in an annihilation of neighbouring dislocation segments (figure 6.24). The dislocation can move on, regardless of the strength of the obstacle. During this process, small dislocation loops remain around the obstacles. The region they enclose did thus not slip by a Burgers vector. This process of overcoming an ob stacle is called Orowan mechanism, and the required Orowan stress is τ = Gb 2λ . (6.17) If F max /Gb 2 < 1, the strength of the obstacle is not sufficient to retain the dislocation until the Orowan mechanism starts. In this case, the dislocation passes through the obstacle, thus cutting it and shearing one part of the obstacle against the other as shown in figure 6.25. This can only happen if the obstacle can slip in the same slip system as the surrounding material. This is always the case when the obstacle is another dislocation. If the obstacle is a 192 6 Mechanical b ehaviour of metals location dislocation energy location of the obstacle (a) Valley dislocation energy location of the obstacle location (b) Hill Fig. 6.26. Obstacles with different energy particle (for instance, a precipitate), it can slip in the same slip system if it is coherent. A semi-coherent particle can also slip in the same slip s ystem as the surrounding material, but the dislocation may have to move to another slip system by climb or cross slip because the matrix material and the particle have s ome, but not all, of the slip systems in common. Incoherent particles cannot be cut by a dislocation. In summary, the following parameters are the main factors in determining the stress needed to overcome obstacles by cutting or the Orowan me chanism: The strength of the obstacle, the distance between the obstacles, and the elas- tic stiffness of the material. If we use aluminium as an example (G = 25.4 GPa, b = 2.86×10 −10 m), we immediately see that the obstacles can only be effective when their distance is significantly smaller than a micrometre (cf. exercise 21). Obstacles must thus be distributed finely to increase the stress needed to move a dislocation appreciably. We can also see from this consideration that materials with a small shear modulus, like magnesium or aluminium, can never be as strong as materials with high modulus. For instance, precipitation- hardened 5 aluminium alloys (G = 26 500 MPa) have a yield strength R p of 600 MPa at most. If the same strengthening method is used in nickel-base alloys (G = 74 500 MPa) the yield strength can be as high as 1400 MPa, in good agreement with the value expected from the shear moduli. It does not matter for the efficiency of an obstacle whether the energy of the dislocation is increased or decreased within it (see figure 6.26). In the first case, energy is needed for the dislocation to penetrate the obstacle i. e., the dislocation is stopped in front of the obstacle. In the second case, the dislocation easily enters the obstacle – releasing some energy as heat –, but additional en ergy is required to detach it again. Screw dislocations can use another mechanism, cross slip, to overcome ob- stacles (see section 6.2.4). As their slip plane is not fixed, they can evade to another plane not blocked by the obstacle as illustrated in figure 6.27. 5 In precipitation hardening, finely distributed particles of a second phase are cre- ated by a special heat treatment. This method will be explained in section 6.4.4. 6.3 Overcoming obstacles 193 b slip plane cross slip plane?? dislocation line obstacle Fig. 6.27. Overcoming obstacles by cross slip of a screw dislocation. The resulting edge dislocations in the cross slipped segment cannot continue to slip in the original direction b ecause they are oriented inappropriately Although the external shear stress on this so-called secondary slip plane or cross slip plane is smaller than on the primary one, moving along this path can be easier than trying to overcome the obstacle by cutting or the Orowan mechanism. This is the cas e if the effective shear stress τ ∗ (see section 6.2.9) on the secondary slip plane is larger than on the primary one due to the ab- sence of the obstacle force. Because screw dislocations can use this additional mechanism, they are frequently able to overcome obstacles more easily than edge dislocations. In general, it is important to notice that it is not simply the more mo- bile type of dislocation that determines plastic deformation: If we consider a dislocation loop with different mobility of the segments as an example, we see that the more mobile type will at first cover a greater distance, but only a small amount of slip is caused by this movement. During the process , the dislocation line reorients itself, increasing the amount of the less mobile type. Thus, the importance of the more mobile one is reduced (see section 6.2.3 and figure 6.11). Because the majority of the slip has now to be performed by the less mobile type, it considerably affects the resistance against plastic deformation. 6.3.2 Thermally activated process es Aided by thermal energy, dislocations may overcome obstacles even when the external stress is not sufficient to exert a force that exceeds the strength of the obstacle. This is called a thermally activated process (appendix C.1 provides a gen eral introduction to this concept). Consider a dislocation trying to move through an arrangement of obstacles as sketched in figure 6.28. We assume that the energy of the dislocation is larger within the obstacle than far away from it. 6 The stress need ed to move 6 As explained above, the obstacles are still obstacles if they attract the dislocation b e cause energy is needed to leave the obstacle. All arguments made here can easily be converted to this case. 194 6 Mechanical b ehaviour of metals ¿ ¿ i ¿ m ¿ ¿ ¤ d ¤ x 2¸ b b b Fig. 6.28. Critical shear stress for overcoming obstacles by a dislocation. Further explanations in the text the dislocation through the obstacle is plotted in the figure. The position of the dislocation is characterised by a single coordinate x because it moves from left to right in the figure. Far away from the obstacle, a frictional stress τ i is required to move the dislocation (see section 6.2.9). In the region of the obstacle, the required stress increases and then decreases again behind it. If we assume th at the effect of the obstacle is restricted to its vicinity, the required stress increases steeply. To simplify the calculations, we approximate the resulting stress curve by a rectangular one with the appropriate height and width. The width d ∗ of the rectangle is then a measure of the width of the obstacle. A stress of τ m has to b e exerted to move the dislocation through the obsta- cle. The work Q done by this stress can be calculated, using equation (6.14) for the force on a dislocation: Q = (τ m − τ i ) · 2bλd ∗ . (6.18) We subtracted the frictional stress τ i because it does not describe the effect of the obstacle, but of the material without it. Q is the obstacle energy, the energy b arrier the dislocation has to overcome. If the effective stress τ ∗ is larger than τ m −τ i , the dislocation can overcome the obstacle. If it is smaller, a certain amount of energy is missing, given by ∆E = Q − 2λbd ∗ τ ∗ . This can be provided by thermal activation. The probability P for this is, according to appendix C.1, given by P ∝ exp  − ∆E kT  = exp  − Q − 2λbd ∗ τ ∗ kT  . (6.19) 6.3 Overcoming obstacles 195 ¾ F T fcc bcc, " 2 >" 1 bcc, " 1 Fig. 6.29. Schematic illustration of the tem- p e rature dependence of the yield strength of face-centred and body-centred cubic metals Here k is Boltzmann’s constant and T the absolute temperature. The quantity b · 2λd ∗ has the unit of a volume und is thus frequently called activation volume V ∗ . Equation (6.19) states that overcoming obstacles becomes easier, the higher the temperature is, and the smaller the energy barrier. It is valid for any kind of obstacle. If kT is larger than the obstacle energy Q, the effect of the obstacle is negligible. If we consider the Peierls force from section 6.2.9 as obstacle, it can also be overcome by thermal activation. This is especially relevant if the Peierls force is large i. e., when slip is along planes that are not close-packed, for example in body-centred cubic lattices. For this reason, the yield strength of body-centred cubic lattices is strongly depende nt on the temperature, different from face- centred cubic metals (figure 6.29). The Peierls stress can reach values of up to se veral hundred megapascal. It may seem contradictory that the Peierls stress is on the one hand able to determine the yield strength of a metal and can nevertheless b e overcome by thermal activation already at room temperature. The reason for this is that its activation volume is rather small. The stress τ m needed to athermally overcome the barrier is large, but due to the small size of the activation volume, the obstacle energy Q is still small enough to be provided by thermal activation at room temperature. The stronger dependence of the flow stress on the in body-centred cubic metals can also be explained by equation (6.19). To see this, we have to take a closer look at the meaning of the equation. So far, we talked only about the probabil- ity of the dislocation overcoming the obstacle, but not about the time needed to do so. Intuitively, it is rather obvious that the probability has to increase with time, but it is not so obvious how this can be seen from equation (6.19). The equation has to be interpreted as stating the probability to overcome the obstacle in a single ‘trial’. Thermal fluctuations cause the dislocation to vibrate with a characteristic frequency. Each vibration can be considered as one trial to overcome the obstacle. This explains that with increasing strain rate ˙ε, the available number of trials becomes smaller. The yield strength must therefore increase with increasing ˙ε; this is more pronounced in body-centred cubic metals. This agrees with experimental observation (see figure 6.29). 196 6 Mechanical b ehaviour of metals ¾ III ¾ II ¾ I ¾ T yield strength cleavage fracture ¿ F (T 1 ) ¡¿ F (T 1 ) ¾ ¿ (a) Ductile behaviour at temp erature T 1 ¾ III ¾ II ¾ I ¾ T yield strength cleavage fracture ¿ F (T 2 ) ¡¿ F (T 2 ) ¾ ¿ (b) Transiti on temperature T 2 < T 1 ¾ III ¾ II ¾ I ¾ T yield strengh cleavage fracture ¿ F (T 3 ) ¡¿ F (T 3 ) ¾ ¿ (c) Brittle behaviour at T 3 < T 2 Fig. 6.30. Illustration of the brittle-ductile transition using Mohr’s circle 6.3.3 Ductile-brittle transition As we saw above, thermal activation is needed to overcome the Peierls barrier in body-centred cubic metals. This does not only cause strong hardening with decreasing temperature, it can also lead to a transition between ductile and brittle behaviour in a rather narrow temperature range. Figure 6.30 shows the transition between ductile and brittle fracture, using Mohr’s circle. At elevated temperatures, the material flows plastically before the maximum ten- sile stress has reached the c leavage strength. At low temperatures, the yield strength has increased, but the cleavage strength is almost unchanged, so the material fractures before plastic flow starts. There is a transition regime between these two regions, the so-called ductile-brittle transition. It is not a material parameter because it depends on the stress state and the strain rate. As the equivalent stress, governing the onset of plastic flow (see section 3.3.1), is independent of the hydrostatic stress state, while b rittle fracture depends on the maximal principal stress, brittle fracture is especially easy if the state is one of triaxial tension. 6.3.4 Climb So far, we assumed that the dislocation segment considered stays in its slip plane. This is not always true as we already saw for the case of a cross-slipping screw dislocation. We also saw that an edge dislocation cannot by-pass an ob- stacle in this way. However, they can leave their slip plane by another mech- anism, the thermally activated climb process. During climb, the dislocation either incorporates vacancies or emits them, see figure 6.31. The dislocation thus moves perpendicularly to its slip plane. For this process to b e relevant, the vacancy density and mobility within the crystal must be large. As explained in appendix C.1, the vacancy density and mobility increase exponentially with the temperature. Therefore, significant climb can occur only at high temper- 6.3 Overcoming obstacles 197 (a) Initial stage (b) Step 1 (c) Step 2 Fig. 6.31. Climb process of an edge dislo cation. By incorporating or emitting va- cancies, the dislo cation line can leave its original slip plane atures (approximately above 40% of the melting temperature). This process will be discussed in more detail in section 11.2.2. 6.3.5 Intersection of dislocations Dislocations are a particularly important type of obstacles for the movement of other dislocations. Dislocations oriented in parallel interact and exert forces on each other as we already learned in section 6.2.7. Repulsive forces hinder the approach of the dislocations, attractive forces hinder their separation. Both f orces impede dislocation movement. If the dislocations are not parallel, their movement can nevertheless be influenced. If one dislocation by-passes the other, they create, depending on their Burgers vectors, kinks or jogs in the other dislocation [40, 61]. The dif- ference between kinks and jogs is that kinks are within the slip plane while jogs leave it. Figure 6.32 shows the effect of a vertically drawn dislocation on a passing, horizontally drawn one for different configurations, illustrating the generation of a kink or a jog. It has to be noted that the passing dislocation will also create a kink or jog in the vertical dislocation, but for clarity this has not been included in the figure. Kinks and jogs create edge-like segments in screw dislocations and vice versa. The length of the dislocation grows in many configurations by one Burgers vector of the other dislocation. Due to the energy stored in a dislocation, energy has to be provided by the passing dislocation so that the dislocation is an energy barrier. Additional energy is needed because of the interaction of the stress fields. Dislocations that are not parallel to the moving dislocation and act as obstacles are descriptively called forest dislocation. The additional edge segments created in a screw dislocation have another consequence: A jog in a screw dislocation (figure 6.33) can move only in the original slip plane by incorporating or emitting vacancies, thus reducing the mobility. This is the reason why screw dislocations are slower than edge dis- locations at low temperatures [40]. [...]... reduction of the life time of a component (see chapter 10) Table 6. 6 summarises the strength of some particle-strengthened aluminium alloys Precipitation hardening Strengthening of materials by fine particles is frequently obtained by precipitation hardening This process will now be explained, using the example of the alloy system aluminium-copper 214 6 Mechanical behaviour of metals Table 6. 6 Effect of particle... For example, long-time application of precipitation-hardened aluminium alloys at temperatures above 200℃ is impossible due to excessive coarsening of the precipitates To use high-strength materials at high temperatures, another method of particle strengthening can be achieved by 218 6 Mechanical behaviour of metals Fig 6. 49 Micrograph of an aluminium reinforced with aluminium-oxide particles [124] introducing... hardening, these materials are easy to deform and nevertheless exhibit high strength after deformation has finished Although coarse second-phase particles do not strengthen a material as efficiently as fine-grained particles, they offer a multitude of ways to influence material properties 6. 4 Strengthening mechanisms 211 Fig 6. 43 Calculation of the average spacing of randomly distributed particles within... area (2a)2 of the plane divided by the number of spheres intersecting it, N · 2r/2a, yielding 212 6 Mechanical behaviour of metals ¿ ¿ ¿ ¿ Fig 6. 44 Formation of an anti-phase boundary during slip of a dislocation through an ordered lattice structure containing two elements (2λ)2 = (2a)2 1 (2a)3 4πr3 /3 2πr2 /3 2r2 = = ≈ 3 N · 2r/2a fV 2r(2a) fV fV (6. 27) Thus, the mean spacing 2λ of the particles is... r fV (6. 28) The particles act as obstacles in a similar way to the solid solution atoms discussed in section 6. 4.3 They elastically distort the crystal and thus interact with the dislocations, and they also change the line tension of a dislocation, for the elastic stiffness of the particle is usually different from that of the matrix If the dislocation cuts through the particle, one part of the particle...198 6 Mechanical behaviour of metals (a) Cutting of an edge dislocation Depending on the orientation of the edge dislocation, a kink forms in the cutting dislocation b (b) Cutting of a screw dislocation A jog forms in the cutting dislocation Fig 6. 32 Cutting of dislocations of various types and orientations (after [40]) The type and orientation of the moving dislocation is... T6, naturally aged ≤ 140 250 400 10 10 6 alloy state Al 99.5 O, annealed AlSi 1 MgMn O, annealed T4, naturally aged T61, artificially aged T6, artificially aged AlCu 4 MgSi 700 T °C 60 0 single phase melt 1 500 melt + solid T 1 400 300 two-phase 200 3 3 100 2 2 0 0 4 8 12 16 20 24 28 32 36 40 wt-% Cu t (a) Phase diagram of the alloy system Al-Cu (b) Heat treatment for precipitation hardening Fig 6. 46. .. example, Young’s modulus of the second phase is larger than that of the matrix, load will be transferred from the matrix to the particle if the material is stressed elastically, and the stiffness increases In the context of fibre- and plate-shaped particles this will be discussed further in chapter 9 The amount of load transfer depends strongly on the shape of the particles and their arrangement Other... introduce a new proportionality constant kHP , the amount of grain boundary strengthening is kHP ∆σgbs = √ d (6. 25) This is the Hall-Petch equation, containing the Hall-Petch constant kHP Its value is 3.5 N/mm3/2 for copper, 12 .6 N/mm3/2 for titanium, and 22 N/mm3/2 for a low-alloy steel [55] Figure 6. 36 shows the dependence of the yield strength of a low-alloy steel on the grain size Strengthening by grain... 6. 46 Precipitation hardening of an aluminium-copper alloy Figure 6. 46( a) shows part of the phase diagram of the Al-Cu system One prerequisite for precipitation hardening is the existence of a two-phase region where the matrix phase (in the example, aluminium with copper in solid solution) is in equilibrium with the precipitation phase (a copper-rich phase in the example), a so-called miscibility gap (see . 2λbd ∗ τ ∗ kT  . (6. 19) 6. 3 Overcoming obstacles 195 ¾ F T fcc bcc, " 2 >" 1 bcc, " 1 Fig. 6. 29. Schematic illustration of the tem- p e rature dependence of the yield strength of face-centred. ductility for different states of deformation of pure aluminium and an aluminium alloy. One advantage of work hardening is that it is simple to achieve and is often a by-product of the manufacturing process,. τ  . (6. 23) The number m of dislocations piled up in a grain is on the one hand pro- portional to the diameter of the grain, on the other it is also proportional to 202 6 Mechanical b ehaviour of