Mechanical Behaviour of Engineering Materials - Metals, Ceramics, Polymers and Composites 2010 Part 6 ppt

The Peierls force thus acts as a kind of frictional forceand reduces the effective stress that can be used to drive the dislocation toovercome other obstacles.. Whenthe external stress τ

188 6 Mechanical behaviour of metals

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 l1 and theBurgers vector b is arbitrary, the Peach-Koehler equation

holds

Equation (6.15) can be derived in a similar way to equation (6.14) bycalculating the energy If the dislocation line is displaced by l2, thecrystal above the covered area has slipped The normal vector in thisarea is given by the cross product l1× l2/|l1× l2| The stress in thisarea is σ · (l1× l2)/|l1× l2|, resulting in a force of σ · (l1× l2) Becausethe crystal has slipped by a Burgers vector, the work is

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 needs energy Theresulting force fixes the dislocation at its momentary position and has to beovercome to move it Thus, if the applied stress is too small, no dislocationmovement is possible and the crystal cannot deform plastically Figure 6.13above illustrates this using the sphere model of atoms This retaining force iscalled 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 theirimpurity content is small In face-centred cubic or hexagonal close-packedmetals, the Peierls stress is about 10−5G (where G is the shear modulus) andcan therefore not explain the strength of engineering alloys In these, otherobstacles for the movement of dislocations play a role, to be discussed insection 6.3 In body-centred cubic metals, the Peierls force is larger than inthe close-packed structures, especially at low temperatures, and influences theyield strength significantly This will be explained in section 6.3.2

After the dislocation has moved by half a Burgers vector, the Peierls forcepushes it forwards and moves it to the position of the next energy minimum.The stored energy is usually dissipated as heat (i e., as random crystal vibra-tion) in the crystal The Peierls force thus acts as a kind of frictional forceand reduces the effective stress that can be used to drive the dislocation toovercome other obstacles

Other inner stresses, caused for example by other obstacles (see the nextsection), can counteract the external stress τ in a similar way to the Peierlsstress 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 − τi If a certain kind of obstacle is investigated, it is often useful to combinethe contributions of all other obstacles to a single frictional stress τi and toassume 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 knowone of these, the Peierls force Other types, such as precipitates of a sec-ond phase, grain boundaries, or impurity atoms, will be discussed below insection 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 behaviour of metals

µµ

(b) Equilibrium of forces for the cation line

dislo-Fig 6.23 Deflection of a dislocation line pinned by obstacles (only part of thedislocation between two obstacles is shown)

it is important whether the dislocation movement is aided by the temperature

or not The first case is called a thermally activated process, the second anathermal one

6.3.1 Athermal processes

Let there be several obstacles in our material with a distance of 2λ betweenthem (figure 6.23) Consider a dislocation pinned on these obstacles Whenthe 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 withthe least-most energy to create new length of dislocation line

The component of the force in the direction of movement is, according toequation (6.14), F = 2λbτ Therefore, each obstacle exerts a retaining force

FRwith opposite orientation and identical magnitude, for each obstacle takeshalf of the force F from two dislocation segments If T is the line tension ofthe dislocation (see equation (6.3)), this force is

FR= 2T sin θ = 2Tλ

R = Gb

2λ

R,where G is the shear modulus, b the Burgers vector, and R the radius of thedislocation segment Equalling F and FRyields

τ = Gb

Here it is crucial that the obstacle cannot bear arbitrarily large forces If

F is the maximum force the obstacle can bear, the dislocation can detach

(a) Before the annihilation The

dislocation segments with

oppo-site orientation attract

(b) After the annihilation Since both tion segments lie on the same lattice plane theycan annihilate, resulting in a dislocation looparound the obstacle and a free dislocationFig 6.24 Annihilation of dislocation segments in the Orowan mechanism

disloca-a

b

c

d

Fig 6.25 Overcoming an obstacle by cutting of dislocations (after [74])

from the obstacle if FR exceeds Fmax Therefore, there is a critical valuesin θ = Fmax/Gb2, and sin θ can be considered as dimensionless measure ofthe obstacle strength It may seem contradictory that sin θ takes only a limitedrange of values but Fmaxdoes 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 Duringthis process, small dislocation loops remain around the obstacles The regionthey enclose did thus not slip by a Burgers vector This process of overcoming

an obstacle is called Orowan mechanism, and the required Orowan stress is

τ = Gb

If Fmax/Gb2< 1, the strength of the obstacle is not sufficient to retain thedislocation until the Orowan mechanism starts In this case, the dislocationpasses through the obstacle, thus cutting it and shearing one part of theobstacle against the other as shown in figure 6.25 This can only happen if theobstacle can slip in the same slip system as the surrounding material This isalways the case when the obstacle is another dislocation If the obstacle is a

192 6 Mechanical behaviour of metals

location

dislocation

energy

location ofthe obstacle(a) Valley

dislocationenergy

location ofthe 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 system asthe surrounding material, but the dislocation may have to move to anotherslip system by climb or cross slip because the matrix material and the particlehave some, but not all, of the slip systems in common Incoherent particlescannot be cut by a dislocation

In summary, the following parameters are the main factors in determiningthe stress needed to overcome obstacles by cutting or the Orowan mechanism: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 = 2.86×10−10m), we immediately see that the obstacles can only be effectivewhen their distance is significantly smaller than a micrometre (cf exercise 21).Obstacles must thus be distributed finely to increase the stress needed tomove a dislocation appreciably We can also see from this consideration thatmaterials with a small shear modulus, like magnesium or aluminium, cannever be as strong as materials with high modulus For instance, precipitation-hardened5 aluminium alloys (G = 26 500 MPa) have a yield strength Rp of

600 MPa at most If the same strengthening method is used in nickel-basealloys (G = 74 500 MPa) the yield strength can be as high as 1400 MPa, ingood 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 thefirst 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, thedislocation easily enters the obstacle – releasing some energy as heat –, butadditional energy is required to detach it again

Screw dislocations can use another mechanism, cross slip, to overcome stacles (see section 6.2.4) As their slip plane is not fixed, they can evade

ob-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 ated by a special heat treatment This method will be explained in section 6.4.4

Although the external shear stress on this so-called secondary slip plane orcross slip plane is smaller than on the primary one, moving along this pathcan be easier than trying to overcome the obstacle by cutting or the Orowanmechanism This is the case 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 sence of the obstacle force Because screw dislocations can use this additionalmechanism, they are frequently able to overcome obstacles more easily thanedge dislocations

ab-In general, it is important to notice that it is not simply the more bile type of dislocation that determines plastic deformation: If we consider

mo-a dislocmo-ation loop with different mobility of the segments mo-as mo-an exmo-ample, wesee 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, thedislocation line reorients itself, increasing the amount of the less mobile type.Thus, the importance of the more mobile one is reduced (see section 6.2.3and 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 plasticdeformation

6.3.2 Thermally activated processes

Aided by thermal energy, dislocations may overcome obstacles even when theexternal stress is not sufficient to exert a force that exceeds the strength of theobstacle This is called a thermally activated process (appendix C.1 provides

a general 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 islarger within the obstacle than far away from it.6 The stress needed to move6

As explained above, the obstacles are still obstacles if they attract the dislocationbecause energy is needed to leave the obstacle All arguments made here caneasily be converted to this case

194 6 Mechanical behaviour of metals

Far away from the obstacle, a frictional stress τi is required to move thedislocation (see section 6.2.9) In the region of the obstacle, the required stressincreases and then decreases again behind it If we assume that the effect ofthe obstacle is restricted to its vicinity, the required stress increases steeply

To simplify the calculations, we approximate the resulting stress curve by arectangular one with the appropriate height and width The width d∗ of therectangle is then a measure of the width of the obstacle

A stress of τmhas to be exerted to move the dislocation through the cle The work Q done by this stress can be calculated, using equation (6.14)for the force on a dislocation:

We subtracted the frictional stress τi because it does not describe the effect

of the obstacle, but of the material without it Q is the obstacle energy, theenergy barrier the dislocation has to overcome

If the effective stress τ∗is larger than τm−τi, the dislocation can overcomethe 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 Theprobability P for this is, according to appendix C.1, given by



6.3 Overcoming obstacles 195

¾F

Tfcc

bcc, "2>"1

bcc, "1

Fig 6.29 Schematic illustration of the perature dependence of the yield strength offace-centred and body-centred cubic metals

tem-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 activationvolume V∗

Equation (6.19) states that overcoming obstacles becomes easier, thehigher the temperature is, and the smaller the energy barrier It is valid forany kind of obstacle If kT is larger than the obstacle energy Q, the effect ofthe obstacle is negligible

If we consider the Peierls force from section 6.2.9 as obstacle, it can also beovercome 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 inbody-centred cubic lattices For this reason, the yield strength of body-centredcubic lattices is strongly dependent on the temperature, different from face-centred cubic metals (figure 6.29) The Peierls stress can reach values of up

to several hundred megapascal

It may seem contradictory that the Peierls stress is on the one handable to determine the yield strength of a metal and can nevertheless

be overcome by thermal activation already at room temperature Thereason for this is that its activation volume is rather small The stress

τmneeded to athermally overcome the barrier is large, but due to thesmall size of the activation volume, the obstacle energy Q is still smallenough to be provided by thermal activation at room temperature

The stronger dependence of the flow stress on the in body-centred cubic metalscan also be explained by equation (6.19) To see this, we have to take a closerlook 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 increasewith 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 overcomethe obstacle in a single ‘trial’ Thermal fluctuations cause the dislocation tovibrate with a characteristic frequency Each vibration can be considered asone trial to overcome the obstacle This explains that with increasing strainrate ˙ε, the available number of trials becomes smaller The yield strength musttherefore increase with increasing ˙ε; this is more pronounced in body-centredcubic metals This agrees with experimental observation (see figure 6.29)

196 6 Mechanical behaviour of metals

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 withdecreasing temperature, it can also lead to a transition between ductile andbrittle behaviour in a rather narrow temperature range Figure 6.30 showsthe transition between ductile and brittle fracture, using Mohr’s circle Atelevated temperatures, the material flows plastically before the maximum ten-sile stress has reached the cleavage strength At low temperatures, the yieldstrength has increased, but the cleavage strength is almost unchanged, sothe material fractures before plastic flow starts There is a transition regimebetween these two regions, the so-called ductile-brittle transition It is not amaterial 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 brittle 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 slipplane This is not always true as we already saw for the case of a cross-slippingscrew 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 dislocationeither incorporates vacancies or emits them, see figure 6.31 The dislocationthus moves perpendicularly to its slip plane For this process to be relevant, thevacancy density and mobility within the crystal must be large As explained

in appendix C.1, the vacancy density and mobility increase exponentially withthe 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 dislocation By incorporating or emitting cancies, the dislocation line can leave its original slip plane

va-atures (approximately above 40% of the melting temperature) This processwill 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 ofthe dislocations, attractive forces hinder their separation Both forces impededislocation movement

If the dislocations are not parallel, their movement can nevertheless beinfluenced If one dislocation by-passes the other, they create, depending ontheir 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 whilejogs leave it Figure 6.32 shows the effect of a vertically drawn dislocation on

a passing, horizontally drawn one for different configurations, illustrating thegeneration of a kink or a jog It has to be noted that the passing dislocationwill also create a kink or jog in the vertical dislocation, but for clarity thishas 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 inmany configurations by one Burgers vector of the other dislocation Due tothe energy stored in a dislocation, energy has to be provided by the passingdislocation so that the dislocation is an energy barrier Additional energy isneeded because of the interaction of the stress fields Dislocations that are notparallel to the moving dislocation and act as obstacles are descriptively calledforest dislocation

The additional edge segments created in a screw dislocation have anotherconsequence: A jog in a screw dislocation (figure 6.33) can move only in theoriginal slip plane by incorporating or emitting vacancies, thus reducing themobility This is the reason why screw dislocations are slower than edge dis-locations at low temperatures [40]

198 6 Mechanical behaviour of metals

(a) Cutting of an edge dislocation Depending on the orientation of the edge cation, a kink forms in the cutting dislocation

dislo-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]) Thetype and orientation of the moving dislocation is not determined, here Depending

on its type and orientation, a kink or jog is created in the immobile dislocation

b1

b2 Fig 6.33 Screw dislocation with a edge-type seg-ment The only way to move the edge dislocation

segment in the original slip direction is by rating or emitting vacancies

incorpo-6.4 Strengthening mechanisms

Plastic deformation of metals is mainly determined by the mobility of cations To design engineering materials with high strength, dislocation move-ment has to be impeded In this section, we want to discuss possible mecha-nisms to do this by different obstacles and to see what amount of strengthening(or hardening, as it is also called) can be achieved

dislo-6.4.1 Work hardening

As explained above, dislocations are obstacles for other dislocations The moredislocations there are in a metal, the higher is its yield strength Dislocationsources, like the Frank-Read source or others described in section 6.2.8, createnew dislocations during plastic deformation and serve to increase the disloca-tion density This hardens the material, a process called work hardening, strain

The influence of the dislocation density on the strength of a metal can

be estimated: Consider a dislocation line moving through an array of tions perpendicular to it as sketched in figure 6.34 Let the distance betweenthe dislocation obstacles be 2λ If the dislocations were insurmountable, theywould have to be by-passed with the Orowan mechanism As they can be cutinstead, the necessary stress is smaller than the Orowan stress This results

disloca-in τcut= kdGb/2λ, with kd≈ 0.1 0.2

The spacing between the dislocation lines is determined by the dislocationdensity % If we simply assume all dislocations to be parallel and arrayed in aregular way, each penetration point in a plane perpendicular to the dislocationoccupies an area of 2λ·2λ The dislocation density is the number of penetrationpoints per unit area i e., √

% = 1/2λ Inserting this into the equation givenabove, we get

∆σd= kdM Gb√

as the contribution of the dislocations to the strength of the material Here

we used the Taylor factor M introduced in section 6.2.6 to convert from shear

to tensile stresses

The contribution of work hardening can, according to equation (6.20),amount to several hundred megapascal If we compare two materials (for in-stance, a low- and a high-strength steel) that differ strongly in their yieldstrength, the absolute contribution of work hardening is similar for both Rel-ative to the initial yield strength, the high-strength material thus has a smalleramount of hardening than the low-strength material In section 3.2.3, it wasexplained that this causes a lower elongation without necking (ductility).The strength of a material can thus be increased by simply deforming

it plastically This is used during rolling or wire drawing Table 6.4 shows

200 6 Mechanical behaviour of metals

Table 6.4 Effect of work hardening on the yield strength Rp0.2and fracture strain

One advantage of work hardening is that it is simple to achieve and is often

a by-product of the manufacturing process, for instance in deep drawing ofsteel sheets for car body parts However, increasing the dislocation density alsodecreases the ductility, so work hardening is only suitable for materials withhigh ductility Another disadvantage is that the strengthening is lost at hightemperatures (for instance during welding) due to recovery (see section 6.2.8)

6.4.2 Grain boundary strengthening

Grain boundaries are barriers for the movement of dislocations As the crystalorientation in the neighbouring grain is different, a dislocation cannot simplyenter it The stress field of the dislocation may initiate dislocation movement

in the neighbouring grain, but if the slip systems are less favourably orientedthere, a larger stress is needed to move dislocations than in the first grain

If a slip system is activated in a crystal, several dislocations are moving

on one slip plane in the same direction and can pile up at a grain boundary.Thus it is plausible, as will be explained below, that the strength of metalsincreases with decreasing grain size This strengthening mechanism is calledgrain boundary strengthening or strengthening by reduction of the grain size.The amount of grain boundary strengthening can be estimated using somesimplifying assumptions Consider a system of m dislocations, piled up at

a grain boundary and being numbered starting at the grain boundary (seefigure 6.35) This configuration may have been created by a dislocation sourcewithin the crystal that created several dislocations on the same slip plane

On each of these dislocations, the external stress τ acts to push it forwards,reduced by a frictional stress τi in the lattice (see section 6.2.9), resulting in

an effective stress τ∗ In addition, there is a forward-pushing stress on eachdislocation, caused by the interaction with the dislocations behind it Theforward acting stress τf on the jth dislocation is thus

6.4 Strengthening mechanisms 201

1 3

Two dislocations exert the same, oppositely oriented stress on each other:

τ(jk)= −τ(kj) Another stress acts on the first dislocation at the grain ary, namely the obstacle stress −τ0 created by the grain boundary that causesthe pile-up In equilibrium, forward- and backward-pushing stresses must bethe same on each dislocation: τf(j)+ τb(j) = 0 Summing over all dislocationsresults in

pro-202 6 Mechanical behaviour of metals

the stress τ∗, for the larger the stress, the smaller is the equilibrium distancebetween the dislocations.7 Introducing a proportionality constant k, we findfrom equation (6.23)

τ0= k(τ∗)2d

The retarding stress τ0 cannot exceed a critical value τc If the applied shearstress is larger than this, a slip system in the neighbouring grain will beactivated, which then starts to flow If τ0 = τc, the external stress takes thevalue

bound-kHP, the amount of grain boundary strengthening is

Strengthening by grain boundaries has another cause, already discussed

in section 6.2.6: During plastic deformation of a polycrystal, neighbouringgrains have to deform so that neither material overlaps nor gaps are created.Therefore, more slip systems have to be activated near the grain boundary

to enable compatible deformation of the grains Generally, some of these aremore difficult to activate and thus require a higher stress This effect is alreadyincluded in the measured values of the Hall-Petch constant

Grain boundary strengthening has the advantage that the ductility of thematerial does not decrease with decreasing grain size and increasing strength.One disadvantage is that, at elevated temperatures, grain boundaries softenand constitute a weak point of the material This will be discussed further

in chapter 11 Fine-grained materials are thus advantageous only in the temperature regime

low-In a material cooled from the melt, the grain size is determined mainly

by the cooling rate To produce a fine-grained material, the cooling rate must

be large, but this is technically difficult to achieve Fine-grained materials aretherefore usually produced in another way, by recrystallisation

7

This argument shows that m increases with τ∗ It is not so easy to show that thedependence is actually a proportionality

Here the material is heavily deformed at first, increasing the dislocationdensity to values of the magnitude 1015m−2 (see section 6.4.1) Due to theelastic distortion around the dislocations, the amount of stored elastic strainenergy in the crystal is large If the temperature is raised, the material recovers(see section 6.2.8) by re-ordering the dislocations and annihilating some ofthem, thus slightly reducing the dislocation density and the stored energy.Because of the large amount of stored elastic energy, the deformed state isthermodynamically unstable Favourably oriented regions, for instance neargrain boundaries or inclusions, serve as starting points or nuclei for the forma-tion of new, undeformed grains During recrystallisation, these nuclei grow bymoving their boundaries into the deformed material The newly created grainnow has a low dislocation density and thus a smaller amount of stored elasticenergy As the boundary between the nucleus and the already existing grains

is, like every grain boundary, a region of high energy, growing can only occur

if the increase in grain boundary energy is compensated by the decrease of thestored elastic strain energy from the decrease of the dislocation density Thehigher the dislocation density, the easier the nuclei can grow A large initialdislocation density finally produces a fine-grained structure, for the rate ofactivation of grains is large

To produce fine-grained metals, they are at first heavily deformed (byrolling, for example) and are then heat treated in a way leading to recrys-tallisation By controlling the amount of deformation and the heat treatmenttemperature, the resulting grain size can be adjusted rather precisely.6.4.3 Solid solution hardening

Another important way of strengthening metals is to alloy them with elementsthat are dissolved in the crystal lattice and form a solid solution Such atoms

204 6 Mechanical behaviour of metals

(a) Substitutional solid solution (b) Interstitial solid solution

Fig 6.37 Different types of solid solutions

(a) Smaller atom (b) Larger atom

Fig 6.38 Different sizes of dissolved atoms in solid solutions

elastically distort the crystal and can thus interact with the stress field of adislocation and impede its movement

Atoms in solid solutions can be situated on two different kinds of latticesites: They can either sit at the same position as the original atoms, thussubstituting one atom by another (substitutional solid solution), or they can

be placed in interlattice positions between the original atoms, forming aninterstitial solid solution Figure 6.37 sketches both cases An interstitial solidsolution can only form when the dissolved atoms are much smaller than those

of the host atoms Carbon in iron provides an example

Substitutional atoms act as obstacles for dislocation movement by differentmechanisms Most important is the elastic distortion of the lattice (figure 6.38)that interacts with the distortion around the dislocation If, for example, thedissolved atoms are larger than the host atoms, they produce compressivestresses in their vicinity An edge dislocation trying to enter this region withits own compressive region will thus be repelled and needs additional energy

to move on If the dislocation approaches the dissolved atom with its tensile gion, it will be attracted, and thus it becomes difficult to detach the dislocationfrom the solid solution atom, pinning the dislocation Smaller substitutionalatoms behave in the opposite way

re-A further interaction between the dislocation and the solid solution atom

is due to the different strength of the atomic bond between the dissolvedatom and its neighbours, resulting in a locally changed elastic modulus in thevicinity of the solid solution atom The line tension of the dislocation thuseither increases or decreases when it approaches the atom, causing anotherobstacle effect known as modulus interaction

Short-range order interaction [40] (sometimes called configurational action or Fisher effect) can also occur If, for example, the binding energy

between host atom A and dissolved atom B is larger than that between Batoms, it is energetically favourable to surround a B atom with A atoms.

If this short-range order is disturbed by slip, resulting in two neighbouring

B atoms, additional energy is needed (see figure 6.39) Dislocation movement

is thus impeded

Experimentally, the following relation between the contribution to ening ∆σsss and the concentration c of the impurity atoms is found:

The exponent n takes values of about 0.5 This is plausible because the spacing

of the obstacles decreases approximately with√

c as we will see in section 6.4.4(equation (6.28)).8

From what has been said so far, it might be presumed that it is best tochoose atoms with strongly differing radii as substitutional atoms to achieve

a large strengthening contribution This, however, is only partly true becausethe solubility of atoms decreases with increasing difference in the radii Forinstance, 100% nickel can be dissolved in copper because the radius difference

is only 2.7%, but copper can only dissolve 10% aluminium with a radiusdifference of 12% In general, a difference in the radii of less than about 15%

is required for good solubility The solubility is also larger if the elements arechemically similar and have the same crystal structure

A different case is the interstitial solid solution, for instance of carbon ornitrogen in steel Here a large difference in the radii is required because itenables the dissolved atoms to sit in an interstitial position

Substitutional solid solution strengthening has the advantage of beingrather temperature insensitive With increasing temperate, for instance dur-ing welding, the solubility of the atoms does not decrease, but increases, sostrengthening at room temperature is not impaired As long as the dissolvedatoms diffuse only slowly through the crystal and thus cannot move along8

There, the volume fraction of precipitates fV is used, which is equivalent to theconcentration c of the solid solution

206 6 Mechanical behaviour of metals

Table 6.5 Influence of solid solution strengthening on the yield strength Rp0.2andthe fracture strain A11.3in the annealed condition [18] The numbers in the materialnames state the approximate content of the particular alloying element in percentalloy Rp0.2/MPa A11.3/%

Another advantage of solid solution strengthened and thus single-phasealloys is their good corrosion resistance This is due to the absence of localisedgalvanic cells Localised galvanic cells are formed by two contacting phases ofdifferent chemical composition with a different position in the electrochemicalseries i e., one of them being less noble than the other If the material isattacked by a corrosive medium, the less noble metal can be dissolved

However, it has to be noted here that some solid solution strengthenedalloys are supersaturated at room temperature This is the case, forexample, for the alloy AlMg 4.5 Mn If the material is cooled too slowlyfrom elevated temperatures, precipitation reactions can occur (see sec-tion 6.4.4) If the precipitated particles are semi-coherent or incoherent,

as in the case above, and thus have a large nucleation barrier, neous nucleation at grain boundaries occurs preferredly This frequentlyleads to embrittlement and larger sensitivity to intercrystalline corro-sion

heteroge-Table 6.5 shows the effect of solid solution strengthening for aluminium If wecompare it to table 6.4, we see the favourable combination of strength andductility

One major disadvantage of solid solution strengthening is that those atomsthat would yield a large effect due to their large radius difference have only

a limited solubility (see above) Thus, this method can usually achieve only

a moderate strengthening The same is usually the case for interstitial solidsolutions, for relatively large interstitial atoms possess a limited solubility, also.One famous exception is carbon in ferritic steels Because the face-centredcubic γ phase can dissolve several percents of carbon at high temperature, it

is possible to ‘freeze in’ these large contents when cooling to the body-centredcubic α phase, although the elastic lattice distortion is large (hardening, seealso section 6.4.5)

Yield point phenomenon and strain ageing

As already discussed, the interaction between dissolved atoms and a tion can cause pinning of the dislocation Because the dissolved atoms canmove through the lattice by diffusion, they can pin dislocations even if these

disloca-do not move This is especially so for interstitial atoms, for they have a largediffusivity This is the cause of the apparent yield point of some metals andfor the so-called Portevin-Le-Châtelier effect (plc) as we will see now

If the diffusivity of the dissolved atoms is negligible, as assumed in theprevious section, dislocations can be pinned only when they move due toexternal stresses In this case, the presence of the dissolved atoms causesstrengthening (figure 6.40)

If we increase the diffusivity (for example by raising the temperature), weencounter the yield point phenomenon i e., we find an upper and a lower yieldstrength (see section 3.2) Solid solution atoms diffuse into the distorted re-gions near the dislocations line while the material is stress-free If externalstress is applied, the dislocation has to be ‘teared off’ its pinning points Thestress required for this defines the upper yield strength ReH of the material.After the dislocation has left its pinning points, it is more mobile than be-fore, and the yield strength reduces (lower yield strength ReL) Deformationlocalises in this region, with only a few grains participating Dislocations pile

up at the grain boundaries, thus increasing the stress in the neighbouringgrain, allowing dislocations there to become mobile as well Narrow bands oflocalised deformation, so-called Lüders bands, form within the material Thisalternation between local hardening by dislocation pile-up and removal ofthis deformation obstacle by tearing off dislocations in the neighbouring graincauses a strongly serrated flow curve Apart from these fluctuations, there is

no hardening Only after the Lüders bands have spread throughout the rial and all dislocations are removed from their pinning points does the yieldstrength increase beyond the lower yield strength by work hardening

mate-If the load is removed and the material stored for some time, the dissolvedatoms diffuse again to the dislocations and thus re-anchor them If the material

is deformed again, an upper and lower yield strength are again encountered.This is called strain ageing The temperature and time required for this ageing