Questions and Answers: Steels, Microstructure and Properties

H. K. D. H. Bhadeshia

May 4, 2006

1 Question

The interaction between dislocations has consequences on deformation and transformation theory.

• In a cubic close–packed metal, a dislocation with Burgers vector b = a2[1 0 1], lying on slip

plane (1 1 1), is cut by a dislocation with b = a2[1 0 1] gliding on (1 1 1). Describe the jog

produced on the first dislocation.

• Explain why interfaces between martensite and austenite can only contain one set of glis-sile dislocations. Why must these interfacial dislocations lie along an invariant–line in theinterface?

1.1 Answer

• The jog will be along [1 0 1], which is not on (1 1 1) rendering the dislocation sessile.

• If the interface has more than one set of dislocations these may interfere in the manner ofjogs. This would cause the interface to become sessile. Note that if glissile jogs arise then itis not necessary to have more than one set of dislocations to accommodate the misfit.

• The dislocations must lie along the invariant–line because there is no distortion along that line.If there is any distortion along the dislocation line then it would have to be accommodatedby another set of misfit dislocations.

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2 Question

Comment on the following features of titanium alloys and steels:

• The eutectoid reaction in Fe–C can occur at temperatures as low as 400 C whereas that inTi–Cu alloys is very sluggish.

• The martensite in alloys which are β–Ti at high temperatures is not particularly hard whencompared with the parent phase. Steel martensite is much harder than the parent austenitefrom which it forms.

• Both titanium and steel melt at temperatures in excess of 1500 C. Steel can be used attemperatures as high as 1000 C but titanium cannot. Why is this?

• Both titanium and α–iron are embrittled by hydrogen but by different mechanisms.

2.1 Answers

• The eutectoid reaction in titanium alloys involves the long–range diffusion of substitutional

solutes.

• The martensite in titanium alloys has substitutional solute hardening. The substitution ofan atom in the lattice causes isotropic strains which mainly interact with the hydrostaticcomponents of dislocation strain fields. In body–centred cubic iron, the interstitial carboncauses tetragonal strains which also interact with the shear components of the dislocationstrain fields. This gives potent hardening. Note that this also explains why carbon has amuch smaller strengthening effect on the strength of austenite, where the octahedral holes inwhich the carbon resides are isotropic.

• Titanium tends to burn at high temperatures.

• Titanium embrittles by hydride formation which is associated with a large (18%) volumeexpansion. In steel the dissolved hydrogen reduces cohesion across grain boundaries, and ingeneral reduces the cohesive strength of iron.

2

3 Question

Why is Widmansttten ferrite able to grow by a displacive transformation mechanism at smallundercoolings below the equilibrium transformation temperature?

3.1 Answer

Widmanstatten ferrite plates observed optically consist of a pair of self–accommodating plateswhich grow cooperatively so that their shape deformations approximately cancel. Hence can format low undercoolings.

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4 Question

Martensite plates generally grow right across the austenite grain in which they nucleate, at leastduring the early stages of transformation. The strain energy per unit volume of martensite (thick-ness c and length r) is given by:

E =c

rµ(s2 + δ2) (1)

where the shear component of the shape change s = 0.18 and the dilatational component is δ = 0.05.The shear modulus is 5 × 1010 Pa. Calculate the thickness of martensite plates which form justbelow the MS temperature in two steels with austenite grain sizes 30 and 100µm respectively. Thefree energy change for martensitic transformation at MS is about −1000 Jmol−1 and the molarvolume 7 × 10−6 m3 mol−1.

4.1 Answer

Assume that the plates grow right across austenite grains and that all the driving force is used upin strain energy.

E =c

rµ(s2 + δ2) (2)

where E is the strain energy per unit volume, µ is the shear modulus of the austenite, c is thethickness of the martensite plate, r is the length of the martensite plate and, s and δ are the shearand dilatational strains of the shape deformation of martensite. r is set to the austenite grainsize since the plates are limited only by the austenite grain boundaries, giving 2.46 × 10−6 and8.18 × 10−6 m as the answers.

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5 Question

A c.c.p. crystal contains two intersecting twin variants, the twinning elements being (1 1 1) [1 1 2]and (1 1 1) [1 1 2] respectively. Find the direction of the line which remains invariant to both ofthese deformations.

5.1 Answer

Combination of two invariant–plane strains is equivalent to an invariant–line strain, with theinvariant–line at the intersection of the two invariant–planes, i.e. [1 1 0].

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6 Question

Compare mechanical twinning, annealing twins and martensitic transformations. Comment in eachcase on the nature of any permanent change in shape, the components of the strain associated withany shape deformation, on whether there is any change in crystallographic orientation, crystalstructure, or density, and finally on the morphology.

6.1 Answer

The table below assumes a metal with a cubic lattice.

Annealing Mechanical Martensite

Shape change none shear invariant–plane strainShear strain – 1√

2' 0.25

Dilatational strain – – ' 0.03 in steelDensity change – – yesCrystal structure change – – yesMorphology blocky lenticular lenticular

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7 Question

300M steel is used in the manufacture of aircraft undercarriage, in the quenched and temperedmartensitic condition. It has the chemical composition

Fe − 0.4C − 0.7Mn − 1.6Si − 0.8Cr − 1.8Ni − 0.25Mo wt.% (3)

When transformed into bainite, its large silicon concentration prevents the precipitation of cemen-tite, so that the bainite consists only of a mixture of bainitic ferrite and carbon–enriched residualaustenite. Use the T0 curve given below together with the empirical equation for the martensite–start temperature (MS) to calculate:

• the maximum fraction of bainitic ferrite that can form during isothermal transformation at400C (assume that the carbon concentration in ferrite is zero);

• the fraction of martensite in the final microstructure obtained by cooling to ambient temper-ature, given that

1 − f = exp−0.011(MS − T ) (4)

MS(C) = 561 − 474 × wC − 33 × wMn − 17 × wCr − 17 × wNi − 21 × wMo (5)

where wi represents the concentration of element i in wt.% and T is the temperature. f isthe fraction of martensite obtained when a fully austenitic sample is cooled to a temperatureT below MS .

7.1 Answer

A high silicon concentration prevents the precipitation of cementite during the bainite transforma-tion. The microstructure at the transformation temperature then consists of just bainitic ferriteand residual austenite. The transformation is diffusionless, but any excess carbon in the ferritesoon becomes partitioned into the austenite, which is thereby enriched.

Such a transformation must stop when the carbon concentration of the residual austenite, i.e. xγ

equals that given by the T0 curve. From mass balance,

Vαxα + Vγxγ = x (6)

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where Vα is the volume fraction of ferrite, Vγ is the volume fraction of residual austenite at thetransformation temperature with Vγ = 1 − Vα and x is the average carbon concentration in thealloy. Take xα to be zero since the solubility of carbon in ferrite is negligible, and set xγ = xT0

inwhich case,

Vα = 1 − x

xT0

≡ 1 − 0.40

0.70= 0.43 and Vγ = 0.57 (7)

Some of the residual austenite will transform to martensite on cooling to ambient temperature(25C). The martensite–start temperature MS of the carbon–enriched austenite is given by:

Ms = 561 − 474 × 0.70 − 33 × 0.70 − 17 × 0.8 − 17 × 1.8 − 21 × 0.25 = 157 C (8)

From the Koistinen and Marburger equation, the volume fraction of martensite formed by coolingto 25C is therefore

0.57 × [1 − exp−0.011(157 − 25)] = 0.44 (9)

To summarise,Vα = 0.43 Vγ = 0.13 Vα′ = 0.44 (10)

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8 Question

Sketch, steregrams centered on 0 1 1α, for the Kurdjumov–Sachs and Nishiyama–Wasserman ori-entation relationships. Hence deduce how one may be generated from the other (γ and α representaustenite and martensite respectively.)

Nishiyama–Wasserman1 1 1γ ||0 1 1α

< 1 0 1 >γ about 5.3 from < 1 1 1 >α towards < 1 1 1 >α

Kurdjumov–Sachs1 1 1γ ||0 1 1α

< 1 0 1 >γ || < 1 1 1 >α

8.1 Answer

The sketch below shows that an appropriate rotation of 5.26 about [011]α ≡ [111]γ generates oneorientation from the other. These orientation relationships are closely related because they bothoriginate from the Bain Strain.

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9 Question

Show diagramatically that it is not possible to obtain a fully coherent interface between austeniteand martensite in steel.

9.1 Answer

Suppose that the austenite is represented as a sphere with its unit cell edges denoted by the vectorsai with i = 1, 2, 3, as illustrated in Figure 1a,b. The Bain strain changes the sphere into an ellipsoidof revolution about a1. There are no lines in the (0 0 1)γ plane which are undistorted. However, itis possible to find lines such as wx and yz are undistorted by the deformation, but are rotated to thenew positions w′x′ and y′z′. One of these undistorted lines can be converted into an invariant–lineby adding a rigid body rotation as illustrated in Figure 1c.

Figure 1: (a) and (b) show the effect of the Bain strain on austenite, which when undeformed isrepresented as a sphere of diameter wx = yz in three–dimensions. The strain transforms it to anellipsoid of revolution. (c) shows the invariant–line strain obtained by combining the Bain strainwith a rigid body rotation.

However, there is no possible rotation which can create two non–parallel invariant–lines, i.e. aninvariant–plane. Therefore, austenite and martensite in steel cannot ever be joined at an interfacewhich is fully coherent and stress–free.

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10 Question

Give three essential characteristics of a martensitic transformation? Define an experiment by whicheach of these characteristics could be measured.

In the context of steels, explain clearly the difference between martensite, lower bainite, upperbainite and Widmanstatten ferrite.

10.1 Answer

A martensitic transformation is achieved by a deformation of the parent crystal structure; it there-fore leads to a shape deformation which can be detected by polishing the parent phase prior totransformation; the shape deformation leads to surface tilts which can be measured using atomicforce microscopy, interference optical microscopy or by the deflection of fiducial marks. Martensitictransformations are diffusionless and hence the measured composition of martensite must be thesame as that of the parent phase. The interface between martensite and the parent phase must beglissile, i.e. it must be able to move without diffusion. The glissile character can be established byelectron microscopy in which the Burgers vectors of the interfacial dislocations are measured andshown to lie out of the plane of the interface (although for pure screws they may lie in the interfaceplane).

Note that it is not correct to state that the transformation must occur at high speeds or at lowtemperatures, that it requires rapid quenching or that martensite is hard.

Martensitic transformation is diffusionless. Bainite grows without diffusion but because it formsat relatively high temperatures, the carbon can rapidly (a second or so) escape into the residualaustenite. It may then precipitate as cementite, giving an upper bainite microstructure. If thebainite transformation temperature is reduced then the carbon escapes less rapidly so that it ispossible for some of it to precipitate inside the ferrite plates, giving a lower bainite microstructure.Widmanstatten ferrite grows by a displacive paraequilibrium mechanism involving the diffusion ofcarbon. It thus grows at a rate which is controlled by the diffusion of carbon in the austenite aheadof the interface. Furthermore, there is little driving force available at the temperatures at whichWidmanstatten ferrite grows so two plates have to grow together in a self–accommodating mannerin order to reduce the overall strain energy. This is why it optically appears to be in the form of athin wedge, since the component plates have slightly different habit planes.

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11 Question

What are the mechanisms by which an ordered precipitate, which is coherent with the matrix, maystrengthen an alloy?

The addition of either aluminium or titanium to austenitic stainless steel causes the formation ofNi3Al or Ni3T i ordered precipitates. Why is it usual to use both aluminium and titanium toprecipitation harden the steel?

Explain in terms of the laws of thermodynamics why it is necessary to raise the maximum operatingtemperature in a power station in order to improve the efficiency of electricity production.

11.1 Answer

Order hardening, because pairs of dislocations have to enter the ordered phase. Also, modulushardening since the energy per unit length of a dislocation depends on the shear modulus of thecrystal. The modulus is likely to be different for the matrix and precipitate phases.

The aluminium to titanium ratio is controlled in order to get an optimum fit between the γ ′

(ordered) precipitate and γ f.c.c. matrix. A good fit is necessary to ensure a coherent interfacewhich gives long term microstructural stability.

The first law of thermodynamics states that energy is conserved i.e. the most that we can do whenwe burn fuel is to recover the energy completely. The second law says that complete recovery isonly possible when the lowest operating temperature is absolute zero since

Thermodynamic efficiency =Th − Tl

Th − 0(11)

where Th and Tl are the highest and lowest temperatures in the cycle. The third law says thatabsolute zero cannot be reached since in each heat extraction cycle only a fraction of the heat isremoved. In any case, reducing the temperature below ambient is impractical so the best way toimprove efficiency is to raise Th. Hence the need for high temperature materials.

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12 Question

Why does ferritic iron change from a ferromagnetic to paramagnetic state as the temperature israised?

12.1 Answer

Ferromagnetism involves the alignment of unspinpaired electrons. This is favoured by the accom-panying reduction in enthalpy, but there is also an increase in free energy due to a reduction inentropy. As the temperature increases, the free energy increase caused by the reduced entropyof ordering dominates, because it scales with T . Hence, magnetic ordering is destroyed, givingparamagnetic iron.

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13 Question

How would you prove that martensitic transformation is diffusionless?

In steels, the strain needed to change the structure of austenite into that of martensite is aninvariant–line strain. However, the shape deformation observed when martensite grows is aninvariant–plane strain. Show schematically how these different deformations may be reconciled,and hence explain why martensite frequently contains transformation twins.

Explain the role of the alloying elements in the following high–strength steel which has a quenchedand tempered martensitic microstructure:

Fe–2.0Mn–0.4C wt.%

State a typical heat–treatment for this alloy, and explain how the steel may be improved to achieveeven higher strength without sacrificing toughness.

13.1 Answer

Martensite is diffusionless because: (a) Its composition is identical to that of austenite, a necessarybut not sufficient condition; (b) it can grow at 1100 m s−1, a growth velocity which is far in excess ofany diffusion velocity; (c) it can grow at temperatures as low as 4 K where diffusion is inconceivablewithin the time frame of the experiment.

Referring to Fig. 2, the strain illustrated in going from (a) to (c) leaves the line normal to thediagram unchanged (i.e. an invariant–line strain). However, the observed change of shape is aninvariant–plane strain illustrated in the change from (a) to (b), the invariant plane being the verticalside of (a). The structure of (b) is therefore wrong, but its shape correct. The shape of (c) is wrongbut its structure correct.

This can be resolved by adding a further deformation which does not change the structure (i.e. slipor twinning), but which gives the correct macroscopic shape corresponding to (b).

The manganese is for hardenability, the carbon for strength. Typical heat–treatment - austenitiseat 850C for an hour, quench in oil, temper at about 400C for 1 hour. The low austenitisationtemperature avoids austenite grain coarsening. The high tempering temperature avoids a tough-ness minimum at about 350C, caused by coarse cementite. Add about 1 wt.% of silicon to thesteel to retard precipitation of cementite, so that the toughness dip is moved to higher temperingtemperatures. Thus, the modified steel can be tempered at only 300C for 1 hour to retain a higherlevel of strength.

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Figure 2: Crystallographic theory of martensite.

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14 Question

The Avrami equation gives the fraction ξ of transformation as

ξ = 1 − exp−πG3It4/3 (12)

when the product grows in the for of spheres at a constant rate G, with a constant nucleation rateI per unit volume during a time interval t. How would the exponent of time change if (a) growthis diffusion–controlled; (b) growth begins instantaneously from all of a fixed number of nucleationsites.

14.1 Answer

The Avrami equation comes from the integration of

− ln(1 − ξ) = (4π/3)

∫ t

0

G3I(t − τ)3dτ (13)

where τ is the incubation period for each particle.

When the dimension of the particle does not change as G(t − τ) but instead varies with t1/2 (i.e.diffusion–controlled growth), the time exponent in the final equation will be reduced to 5/2. This isbecause the volume of each particle scales with t3/2 and an additional t comes from the nucleationrate.

When there is no nucleation but growth starts instantaneously from a fixed number of sites, thetime exponent is obviously reduced to 3.

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15 Question

Explain what is meant by the term paraequilibrium. Illustrate schematically an isothermal sectionof the paraequilibrium phase diagram for an Fe–Mn–C alloy where austenite and ferrite can coexist.The sketch should include tie–lines.

15.1 Answer

Paraequilbrium is a constrained equilibrium. It occurs at temperatures where the diffusion of sub-stitutional solutes is not possible within the time scale of the experiment. Nevertheless, intersitialsmay remain highly mobile. Thus, in a steel, manganese does not partition between the ferrite andaustenite, but subject to that constraint, the carbon redistributes until it has the same chemicalpotential in both phases.

Therefore, the tie–lines in the phase diagram are all virtually parallel to the carbon axis, since Mndoes not partition between ferrite and austenite.

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16 Question

The surface of a steel M can be oxidised to form a layer of metal oxide MO. The thickness of thislayer is controlled by the rate at which oxygen atoms diffuse through the oxide layer. The oxygenatoms then combine with the metal at the M/MO interface.

Show that the thickness of the oxide layer varies with the square root of time at the oxidationtemperature.

16.1 Answer

The rate at which the oxide consumes oxygen is

(cOM − cMO)dz

dt(14)

where z is the oxide thickness, cMO is the oxygen concentration in the metal at the metal–oxideinterface and t is the time. This rate must equal the flux of oxygen arriving at the metal–oxideinterface, so that:

(cOM − cMO)dz

dt=

D(cOA − cOM )

z. (15)

It follows that

z2 = 2DtcOA − cOM

cOM − cMO. (16)

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17 Question

The diffusion controlled growth rate of a plate is given by

v =ΩD

r(1 − rc

r) (17)

where Ω is a representation of the undercooling below the equilibrium transformation temperature,r is the plate tip radius, rc is the critical plate tip radius at which the growth rate becomes zeroand D is the diffusivity in the parent phase.

Sketch the variation in v with r and show that the maximum rate of growth occurs at r = 2rc.

17.1 Answer

The velocity is zero at r = rc, and then peaks at r = 2rc, as can be demonstrated by differentiatingthe equation.

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18 Question

In an Fe–C alloy, the isothermal growth of ferrite occurs with equilibrium maintained locally at theinterface such that

(cαγ − cγα)v = D∂c

∂z(18)

where cαγ is the concentration of carbon in ferrite which is in equilibrium with austenite, cγα is thecorresponding concentration in austenite which is in equilibrium with ferrite, D is the diffusivity ofcarbon in austenite, v is the interface velocity and z is a coordinate defined normal to the interface.The concentration gradient is evaluated at the position of the interface.

Explain separately the meanings of the left and right hand sides of this equation, and hence theorigin of the equation.

How would similar conditions be derived for a ternary Fe–Mn–C alloy?

Hence describe two ways in which ferrite can grow in a ternary Fe–Mn–C alloy whilst maintaininglocal equilibrium at the interface, even though the diffusivities of Mn and C are different by manyorders of magnitude.

18.1 Answer

The boundary condition defined by the equation relates the interface velocity to the concentrationgradient at the interface. The rate at which solute is partitioned as the interface moves (i.e. lefthand side) must equal that at which it is carried away by diffusion (i.e. right hand side) if thecompositions at the moving interface are to remain constant.

For a ternary alloy it would be necessary to satisfy two such equations simultaneously, for each ofthe solutes:

(cγα1

− cαγ1

)v = −D1∇c1 (19)

(cγα2

− cαγ2

)v = −D2∇c2 (20)

where the subscripts refer to the solutes (1 for carbon and 2 for Mn).

Because D1 D2, these equations cannot in general be simultaneously satisfied for the tie–linepassing through the alloy composition c1, c2. It is, however, possible to choose other tie–lines whichsatisfy the equation. If the tie–line is such that cγα

1= c1 (e.g. line cd for alloy A of Fig. 3a), then

∇c1 will become very small, the driving force for carbon diffusion in effect being reduced, so thatthe flux of carbon atoms is forced to slow down to a rate consistent with the diffusion of manganese.Ferrite forming by this mechanism is said to grow by a ‘Partitioning, Local Equilibrium’ (or PLE)mechanism, in recognition of the fact that cαγ

2can differ significantly from c2, giving considerable

partitioning and long–range diffusion of manganese into the austenite (Coates, 1973c).

An alternative choice of tie–line could allow cαγ2

→ c2 (e.g. line cd for alloy B of Fig. 3b), so that∇c2 is drastically increased since only very small amounts of Mn are partitioned into the austenite.

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Figure 3: Schematic isothermal sections of the Fe–Mn–C system, illustrating ferrite growth occur-ring with local equilibrium at the α/γ interface. (a) Growth at low supersaturations (P–LE) withbulk redistribution of manganese, (b) growth at high supersaturations (NP–LE) with negligiblepartitioning of manganese during transformation. The bulk alloy compositions are designated bythe symbol • in each case.

The flux of manganese atoms at the interface correspondingly increases and manganese diffusioncan then keep pace with that of carbon, satisfying the mass conservation conditions. The growth offerrite in this manner is said to occur by a ‘Negligible Partitioning, Local Equilibrium’ (or NPLE)mechanism, in recognition of the fact that the manganese content of the ferrite approximatelyequals c2, so that little if any manganese partitions into austenite.

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19 Question

Explain why:

• martensite, Widmanstatten ferrite and bainite all occur in the form of thin plates;

• Widmanstatten ferrite grows at a rate which is much smaller than the speed of sound in themetal;

• an Fe–30Ni wt% martensite is weak compared with martensite in Fe–0.2C wt% alloy;

• alloying elements have a much greater effect on the kinetics of a reconstructive transformationwhen compared with the corresponding effect on displacive transformations.

19.1 Answer

All of these transformations are accompanied by a shape deformation which is an invariant–planestrain (IPS). When elastically accommodated, this causes a strain energy per unit volume givenby:

E =c

aµ(s2 + δ2) (21)

where µ is the shear modulus of the austenite and s and δ are respectively the shear and dilatationalcomponents of the IPS. c/a is the thickness to length ratio of the product so that a thin plate shapeminimises the strain energy.

Widmanstatten ferrite occurs at temperatures where diffusionless transformation is thermodynam-ically impossible. Thus, although the change in crystal structure is accomplished by a deformation,the displacements occur at a rate controlled by the diffusion of carbon in the austenite ahead ofthe interface.

Carbon causes a much greater degree of hardening than any substitutional solute. This is becauseit leads to a tetragonal distortion of the b.c.c. structure; a tetragonal strain can interact with allcomponents of the stress field of a dislocation. Substitutional solutes cause isotropic strains whichcan only interact with the small hydrostatic component of stress associated with edge dislocations.

The effect on displacive transformations is thermodynamic alone. On reconstructive transforma-tions the solute may partition between the parent and product phases by a process involvingdiffusion. This will have a further kinetic effect on reconstructive transformations.

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20 Question

Why is the lattice–invariant deformation unnecessary when austenite transforms by a martensiticmechanism to a hexagonal close–packed lattice?

20.1 Answer

In the case of the fcc–hcp transformation, the lattice deformation (Bain Strain) itself is an invariant–palne strain, i.e. the motion of Shockley partials on the alternate close–packed planes. B= P1 soa lattice invariant deformation is not needed. There are no transformation twins or slip steps to befound.

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21 Question

What are the mechanisms by which an ordered precipitate, which is coherent with the matrix, maystrengthen an alloy?

21.1 Answer

Order hardening, because pairs of dislocations have to enter the ordered phase. Also, modulushardening since the energy per unit length of a dislocation depends on the shear modulus of thecrystal. The modulus is likely to be different for the matrix and precipitate phases.

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22 Question

Upper bainite in conventional steels consists of a mixture of bainitic ferrite and cementite. Howcan this cementite be eliminated from the microstructure?

When the precipitation of cementite is prevented, the microstructure obtained by isothermal trans-formation in the bainite temperature range consists of just bainitic ferrite and carbon–enrichedretained austenite. State four potential advantages of this microstructure from the point of viewof toughness and strength.

One difficulty with the mixed microstructure of bainitic ferrite and retained austenite is that theaustenite, when present in large quantities, is unstable and hence forms high–carbon, brittle marten-site. Explain three ways in which this problem can be eliminated.

22.1 Answer

Silicon has an incredibly small solubility in cementite. Therefore, increasing the silicon concentra-tion of a steel to a value greater than about 1.5 wt% ensures the absence of cementite in upperbainite. Aluminium and to a lesser extent chromium, have the same effect. Note that silicon has asimilar effect in cast irons.

The six potential advantages of the microstructure can be listed as follows:

• Cementite is responsible for initiating fracture in high–strength steels. Its absence is expectedto make the microstructure more resistant to cleavage failure and void formation.

• The bainitic ferrite is almost free of carbon, which intensely strengthens ferrite and henceembrittles it.

• The microstructure derives its strength from the fine grain size of the ferrite plates, whichare less than 1 µm in thickness. It is the thickness of these plates which determines the meanfree slip distance, so that the effective grain size is less than a micrometer. This cannot beachieved by any other commercially viable process. Grain refinement is the only methodavailable for simultaneously improving the strength and toughness of steels.

• The ductile films of austenite which are intimately dispersed between the plates of ferrite havea crack blunting effect. They further add to toughness by increasing the work of fracture asthe austenite is induced to transform to martensite under the influence of the stress field ofa propagating crack. This is the TRIP, or transformation–induced plasticity effect.

• The diffusion of hydrogen in austenite is slower than in ferrite. The presence of austenite cantherefore improve the stress corrosion resistance of the microstructure.

• Steels with the bainitic ferrite and austenite microstructure can be obtained without the useof expensive alloying. All that is required is that the silicon concentration should be largeenough to suppress cementite.

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The problem of unstable retained austenite arises because the amount of bainite that can form islimited by the T0 curve on the phase diagram. Therefore, the three ways all relate to the T0 curve,as follows:

• By reducing the isothermal transformation temperature to increase xT0. The lower limit is

set by either the lower bainite or martensite–start temperature.

• By reducing the overall carbon concentration of the steel, so that the austenite reaches itslimiting composition at a later stage of reaction.

• By moving the T0 curves of the phase diagram to larger carbon concentrations. This can bedone by adjusting the concentration and type of substitutional solute.

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23 Question

Explain why martensite grows in the form of thin plates.

23.1 Answer

The shape deformation accompanying the growth of martensite is an invariant–plane strain with alarge shear component (Fig. ??). The shear strain is typically s = 0.25, with a dilatational strainof about δ = 0.03 normal to the habit plane. The strain energy per unit volume of martensite isgiven by (thickness c and length r) is given by:

E =c

rµ(s2 + δ2) (22)

where µ is the shear modulus of the parent phase. It follows that this strain energy is minimisedby adopting a plate shape. Another way of looking at this is illustrated in the figure below. Theabsolute magnitude of the displacements gets larger as the product gets thicker in the directionnormal to the habit plane (compare displacements at a and b. Therefore, the thinner the plate,the smaller the absolute displacements that have to be accommodated.

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24 Question

Explain what is meant by the term paraequilibrium. Illustrate schematically an isothermal sectionof the paraequilibrium phase diagram for an Fe–Mn–C alloy where austenite and ferrite can coexist.The sketch should include tie–lines.

Why does the austenite+ferrite phase field converge to a point when the carbon concentrationbecomes zero?

Define in thermodynamic terms, what is meant when an element is said to be trapped duringtransformation.

During the paraequilibrium growth of ferrite in an Fe–Mn–C alloy, which of the elements is trappedin the ferrite, and which in austenite?

24.1 Answer

Paraequilibrium is a constrained equilibrium. It occurs at temperatures where the diffusion of sub-stitutional solutes is not possible within the time scale of the experiment. Nevertheless, interstitialsmay remain highly mobile. Thus, in a steel, manganese does not partition between the ferrite andaustenite, but subject to that constraint, the carbon redistributes until it has the same chemicalpotential in both phases.

Therefore, the tie–lines in the phase diagram are all virtually parallel to the carbon axis, since Mndoes not partition between ferrite and austenite.

At zero carbon concentration, the austenite and ferrite have exactly the same chemical compositionsince Mn does not redistribute during transformation. Therefore, there is a unique concentrationat which the two phases have equal free energy, i.e. they are in equilibrium.

An element is said to be trapped when its chemical potential increases on transfer across the movinginterface.

The manganese is trapped in the ferrite since its chemical potential increases on transfer acrossthe interface, the iron is trapped in the austenite for the same reason. Manganese is an austenitestabiliser so its equilibrium concentration in austenite should be greater than that in ferrite.

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25 Question

Give three features of martensite that distinguish it from a reconstructive transformation such asallotriomorphic ferrite. Explain how you might characterise each of these features using experi-ments.

Giving reasons, explain which of the following deformations is an invariant–plane strain:

• simultaneous slip on two independent slip systems;

• mechanical twinning;

• elastic elongation of a material with a zero Poisson’s ratio;

• hydrostatic expansion.

25.1 Answer

A martensitic transformation is achieved by a deformation of the parent crystal structure; it there-fore leads to a shape deformation which has a large shear component, and which can be detectedby polishing the parent phase prior to transformation. The shape deformation causes surface tiltswhich can be measured using atomic force microscopy, interference optical microscopy or by thedeflection of fiducial marks. The formation of allotriomorphic ferrite is not accompanied by anyshear strain, simply a small volume change which does not deflect any fiducial marks.

Martensitic transformations are diffusionless and hence the measured composition of martensitemust always be the same as that of the parent phase. The chemical composition of allotriomor-phic ferrite corresponds to its equilibrium composition and hence in general will differ from thatof the parent phase. The chemical composition can be measured using energy dispersive X–raymicroanalysis, the atom–probe technique and X–ray diffraction.

The interface between martensite and the parent phase must be glissile, i.e. it must be able tomove without diffusion. The glissile character can be established by electron microscopy in whichthe Burgers vectors of the interfacial dislocations are measured and shown to lie out of the planeof the interface (although for pure screws they may lie in the interface plane). There are no suchrequirements for the structure of the interface between allotriomorphic ferrite and austenite (it maybe sessile or glissle, and will always require diffusion in order to translate).

Note that it is not correct to state that the transformation must occur at high speeds or at lowtemperatures, that it requires rapid quenching or that martensite is hard. Martensite can be soft,slow, and can take place at high temperatures.

• Slip is an invariant–plane strain (IPS), but the simultaneous operation of slip on two non–parallel planes leaves only the line of intersection undeformed and unrotated. This deforma-tion is therefore an invariant–line strain;

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• mechanical twinning is a shear on a plane which remains invariant (i.e. an IPS);

• elastic elongation of a material with a zero Poisson’s ratio is also an IPS since the deformationis only normal to the invariant–plane. Beryllium has virtually a zero Poisson’s ratio;

• hydrostatic expansion distorts all vectors and hence is not an IPS.

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26 Question

Explain the mechanism of the Widmanstatten ferrite transformation in steels, including a descrip-tion of the factors that determine its growth rate and shape.

Which solute is said to be trapped in the product phase during transformation of austenite in aFe–Mn–C alloy, in the following cases:

• the growth of Widmanstatten ferrite;

• the growth of martensite.

26.1 Answer

Widmanstatten ferrite grows by a displacive paraequilibrium mechanism involving the diffusion ofcarbon. It thus grows at a rate which is controlled by the diffusion of carbon in the austenite aheadof the interface. Furthermore, there is little driving force available at the temperatures at whichWidmanstatten ferrite grows so two plates have to grow together in a self–accommodating mannerin order to reduce the overall strain energy. This is why it optically appears to be in the form of athin wedge, since the component plates have slightly different habit planes.

• Manganese; carbon partitions during the paraequilibrium growth of Widmanstatten ferrite;

• both carbon and manganese since martensitic transformations are diffusionless.

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27 Question

Explain the origin of the following phenomena:

• Shape memory effect;

• rubber elasticity in metals;

• transformation induced plasticity (TRIP).

27.1 Answer

The shape memory effect arises because there exists an atomic correspondence between the parentand product phases during martensitic transformation (diffusionless). Consequently, any shapechange produced by transformation can be reversed on reverse transformation.

During martensitic transformation, several accommodating crystallographic variants of martensiteare produced from the parent grains. Each variant represents a large physical deformation (anIPS) on a different system. Under the influence of stress, a martensite variant which is more suitedto the stress grows by a diffusionless mechanism while consuming the others, thus causing thespecimen to become longer along the direction of the applied stress. Removal of the stress reversesthe martensite→martensite transformation and causes the specimen to return to its original form.Thus the metal behaves like a rubber.

The formation of martensite can be induced by stress (since the transformation involves a physicaldeformation). Consequently, if stress concentrations arise in a tensile specimen of a certain kindof an austenitic alloy, they can be relieved by the localised formation of martensite, thus allowingdeformation to continue without necking. This is the TRIP phenomenon, whereby necking isdelayed, thus increasing the effective plasticity.

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28 Question

Explain the difference between mechanical twinning and annealing twins. How would you distin-guish the two types of twinning using optical microscopy?

In cubic–close packed (c.c.p.) metals, the twin plane is 1 1 1 and the twin direction is < 1 1 2 >.Calculate the twinning shear.

What is the maximum number of different twin traces that you might expect to find in any givengrain of a severely deformed sample of a c.c.p. metal?

Explain why a shape memory metal (a) tends to lose its reversibility after many cycles; (b) whythere is a limit to the amount of reversible strain in any given cycle. State two applications ofshape memory metals.

28.1 Answer

Mechanical twinning is a deformation which involves the coordinated movement of atoms that leadsto a reorientation of a part of the crystal. Annealing twins grow by a reconstructive mechanismfrom a deformed microstructure in much the same way as conventional grain growth. They do notlead to any deformation. Mechanical twins are lenticular in shape with sharp tips in order minimisestrain energy. The shape of annealing twins is governed by the minimisation of interfacial energy.They are facetted on coherent interfaces with no sharp tips. Optical microscopy can also be usedto study displacements at the free surface. Surface relief will only be observed with mechanicaltwinning.

Twinning involves a shift of each of the close packed planes by a distance a6

< 1 1 2 >≡ (a/√

6)

where a is the lattice parameter. The spacing of the 1 1 1 planes is a/√

3. The twinning shearis the ratio of these two numbers, i.e. 1/

√2.

Four different traces (multiplicity of 1 1 1) planes.

(a) The shape memory effect is not perfectly reversible in that a few defects are created on eachcycle. This means that the defect density increases with the number of cycles, so much so that theinterfaces are eventually prevented from reversible motion.

(b) When a shape memory element is deformed, the strain is accomplished by the growth of certainmartensite orientations relative to others. Recoverable strain is exhausted when the element isconverted entirely into one set of orientations. Any further deformation is ordinary plastic yieldingand will damage the effect.

(c) Rings for joining, stents to keep arteries open, bendable spectacle frames . . .

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29 Question

Describe methods that you might use for the casting of the following items, justifying the equipmentnecessary and stating the reasons for the choice of the particular process:

• Some 3000 tonnes of molten steel which is subsequently to be rolled into plate.

• A propeller for a large ship.

• A ribbon of amorphous metal.

• A sample for the scientific study of homogeneous nucleation.

29.1 Answer

• Continuous casting. The metal is poured into a tundish from which is flows into an oscillatingwater–cooled copper mould, leaving it with a solidified surface and a partly solidified core. Thestrand then bends around with the help of supporting rollers and is sliced into manageablechunks for hot–rolling operations. This is a much more efficient process than casting intomolds in which pipe formation leads to considerable wastage.

• This would be sand cast. It is a specialised item so that the sand mould can be custom made.The surface quality is not too important since the propeller can be machined subsequently.

• Melt spinning since a high cooling rate is required. Involves directing a small jet of moltenmetal at a rapidly spinning water–cooled copper wheel.

• Two methods: drop tube in which the sample is allowed to solidify whilst its is droppingand hence has no contact with an external surface. The second method involves solidificationwhilst the sample is levitating in a conical RF coil.

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30 Question

What is the importance of the silicon concentration in the design of cast irons? How does thesilicon concentration affect the hardness of cast iron?

30.1 Answer

There are two major variants of cast iron: white and grey. Carbon is mostly precipitated asgraphite in grey iron whereas it occurs as cementite in white iron. Silicon retards the formation ofcementite in which it has a negligible solubility. Its presence in a concentration of about 2 wt.%therefore encourages the formation of grey cast iron. Therefore, silicon reduces the hardness greatlyby eliminating much of the cementite.

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31 Question

Explain how the shape memory effect associated with martensitic transformations works. Whyis the memory diminished by repeated cycling? What sets the limit to the maximum recoverablestrain?

Is it possible to have a shape memory polymeric material?

31.1 Answer

The shape deformation accompanying martensitic transformation can be reversed by transformingback to the parent phase. This, on its own, is NOT sufficient to explain how the shape memoryeffect works. Suppose a crystal of austenite is cooled to form many variants of martensite, in sucha way that they accommodate; hence, the overall shape is unaffected by transformation (the firststage in Figure 1). When a stress is applied, the favoured variant of martensite grows, leading toa shape change which complies with the stress (the second stage. On heating the shape change isreversed, thus regaining the original shape. This is the basis of the shape memory effect.

The memory effect relies on the reversibility of the shape deformation. It can be lost by theaccumulation of defects during transformation, i.e. by repeated cycling.

Excessive deformation, beyond that required to produce a single martensite variant, will lead toirreversible plastic strain so that the original shape will become impossible to recover.

A once–only shape memory effect is possible with polymers. If the polymer chains are straightenedduring the fabrication of a component (e.g. shrink wrap) then heat will cause them to curl up againvia the entropy spring effect.

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32 Question

Distinguish between recovery, recrystallisation and grain growth.

Determine the limiting grain size during grain growth in a metal containing 1 volume percent ofspherical inclusions whose average diameter is 600 nm. Explain any assumptions associated withyour calculation.

What are the advantages and disadvantages of a fine–grained microstructure?

32.1 Answer

In recovery, there is a change in the stored energy without any obvious change in the optical mi-crostructure. Excess vacancies and interstitials anneal out giving a drop in the electrical resistivitybut little change in hardness. Dislocations become mobile at a higher temperature, eliminate andrearrange to give polygonisation. Still higher temperatures can lead to recrystallisation, duringwhich new, relatively defect–free grains grow and consume the old deformed microstructure.

The recrystallised grain structure still contains a small amount of energy stored in the form ofgrain boundaries. The grains will tend to coarsen in order to reduce the amount of surface per unitvolume. This is grain growth.

Equate the Zener drag pressure PZ = 3γf/2r to the grain growth pressure PG = 3γ/D to find thelimiting grain size D:

PZ = PG orf

r=

2

D(23)

where f is the volume fraction of inclusion and r is the particle radius. It follows that D = 60µm.

The method assumes that the particles are all uniform in size, that all grain boundaries haveidentical energies per unit area, that the particles are randomly distributed.

Fine grains are desired for structural applications at low homologous temperatures, since they leadto an increase in strength via the Hall–Petch effect; grain size strengthening is the only way ofimproving toughness at the same time as strength. Of course, the surface per unit volume increasesas the grain size decreases so a fine grain structure is not appropriate for elevated temperatureapplications where creep is an issue. Grain boundaries are easy diffusion paths.

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33 Question

Explain why the diffusion in a solid solution of interstitial solutes is so much more rapid than thatof substitutional solutes. Vacancies are defects, so why does the vacancy concentration remainfinite at equilibrium?

33.1 Answer

The activation enthalpy of diffusion can be separated into two components, one the enthalpy ofmigration (there are distortions of the lattice as the atom jumps into an adjacent potential well)and the enthalpy of formation of a vacancy in an adjacent site. After all, for the atom to jumpit is necessary to have a vacant site. This latter component is unnecessary for interstitial atomswhich consequently have a small activation enthalpy, and hence diffuse far more rapidly than dosubstitutional solutes.

The introduction of a vacancy increases the enthalpy, but the −T∆S term due to configurationsentropy reduces the free energy. These two opposing effects lead to an equilibrium concentrationof vacancies at all but 0 K.

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34 Question

Explain how decarburisation causes a variation in the microstructure in a slowly cooled eutectoidsteel as a function of the distance from the exposed surface.

A Fe–0.7C wt.% steel is decarburised at 1200 K such that a constant carbon concentration of 0.1wt.% is maintained at the exposed surface. If the diffusion coefficient for carbon in austenite is2 × 10−5 mm2 s−1, how long will it take for the depth at which the concentration is 0.4 wt.% tobecome 2.5 mm?

How does your calculated time compare with an estimate made assuming that the diffusion distanceis 2

√Dt? Comment on why the two results are different.

How can decarburisation be prevented in practice?

34.1 Answer

By referring to the phase diagram, it can be deduced that the surface will be rich in ferrite, theunaffected regions away from the surface will be fully pearlitic. The intermediate regions will havea mixture of ferrite and pearlite according to the lever rule applied at the eutectoid temperature.

Using the error function solution given in the data book,

0.1 − 0.4

0.1 − 0.7= erf

(

2.5

2√

Dt

)

so that 0.475 =2.5

2√

Dtand t = 96.2 h (24)

On the other hand, withx = 2

√Dt we get t = 135.6 h (25)

The difference arises because this estimate assumes random walk, whereas diffusion in a concentra-tion gradient is driven along a particular direction.

Decarburisation can be prevented by heat treatment in an inert atmosphere, by wrapping thecomponent in stainless steel foil, or by painting with an isolating paint.

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35 Question

What is the three–dimensional shape of martensite, a pearlite colony and of flake graphite in greycast–iron? How can the shape of the flake graphite be modified into spheroidal graphite by alloying?

35.1 Answer

Martensite forms as thin plates (this minimises the elastic strain energy).

A pearlite colony is an interpenetrating bicrystal of ferrite and cementite. This can be visualisedby thinking of a cabbage (single crystal cementite) in a bucket of water (single crystal ferrite).

Each flake of graphite is a part of a rosette of leaves. The addition of traces of rare earth elementssuch as Mg and Ce poisons fast growth directions, thus giving spheroidal graphite.

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36 Question

Explain, in the context of binary solutions, what is meant by the term ‘chemical potential’. Hencejustify the fact that the common tangent construction, on a free energy versus concentration plot,gives the equilibrium compositions of the phases.

Distinguish between an ideal, a regular and a quasichemical solution thermodynamic model. Explainqualitatively why none of these models are used in their exact forms in the computer calculationof phase diagrams.

36.1 Answer

Consider an alloy consisting of two components A and B. For the phase α, the free energy will ingeneral be a function of the mole fractions (1 − X) and X of A and B respectively:

Gα = (1 − X)µA + XµB (26)

where µA represents the mean free energy of a mole of A atoms in α. The term µ is called thechemical potential of A, and is illustrated in Fig. 36.1a. Thus the free energy of a phase is simplythe weighted mean of the free energies of its component atoms. Of course, the latter varies withconcentration according to the slope of the tangent to the free energy curve, as shown in Fig. 36.1.

Consider now the coexistence of two phases α and γ in our binary alloy. They will only be inequilibrium with each other if the A atoms in γ have the same free energy as the A atoms in α,and if the same is true for the B atoms:

µαA = µγ

A (27)

µαB = µγ

B (28)

If the atoms of a particular species have the same free energy in both the phases, then there is notendency for them to migrate, and the system will be in stable equilibrium if this condition appliesto all species of atoms. Since the way in which the free energy of a phase varies with concentrationis unique to that phase, the concentration of a particular species of atom need not be identical inphases which are at equilibrium. Thus, in general we may write:

XαγA 6= Xγα

A (29)

XαγB 6= Xγα

B (30)

where Xαγi describes the mole fraction of element i in phase α which is in equilibrium with phase

γ etc.

The condition the chemical potential of each species of atom must be the same in all phases atequilibrium is quite general and obviously justifies the common tangent construction illustrated inFig. 36.1b.

An ideal solution is one where the atoms mix at random because there is no enthalpy change onmixing (∆HM ) the components (Table). The configurational entropy of mixing is (∆SM ) is easily

41

Figure 4: (a) Diagram illustrating the meaning of a chemical potential µ. (b) The common tangentconstruction giving the equilibrium compositions of the two phases at a fixed temperature.

derived because the probabilities can be estimated assuming a random distribution of atoms. Theenthalpy of mixing is finite for a regular solution, so that the atoms at low temperatures may not berandomly mixed. Nevertheless, as a convenient approximation, the entropy of mixing is assumed tobe ideal. A quasichemical model avoids this latter approximation. Note that the regular solutionmay be considered as a zeroth approximation quasichemical model.

The models describe above help in the understanding of phase equilibria and in the behaviour ofsolutions. The theories are, nevertheless, too complicated for general application in phase diagramcalculations. Computer methods are designed to enable the calculations to be implemented overthe entire periodic table, for any concentration, and in a seamless manner. The thermodynamicfunctions have to be constructed in such a way that the modification of one set of data does notentail the recreation of the entire dataset.

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Table 1: Elementary thermodynamic properties of solutions

Type ∆SM ∆HM

Ideal Random 0Regular Random 6= 0

Quasichemical — Not random 6= 0

37 Question

Give three examples of cases where the rate of energy dissipation is given by the product of the“‘flux” (J) and a corresponding “force”, (X), i.e.

Tσ = JX (31)

where T is the absolute temperature and σ is the rate of entropy production. How could thisrelationship be generalised for multiple dissipation processes?

Using the concept of forces and fluxes in the theory of irreversible thermodynamics, deduce the rela-tionship between the velocity V of a grain boundary and the free energy change ∆G accompanyingit motion. State any assumptions involved in this derivation.

Prove that the general relation between V and ∆G should in fact be as follows:

V ∝ exp−Q/kT[1 − exp−∆G/kT] (32)

where Q is the activation energy for the transfer of atoms across the grain boundary, k is theBoltzmann constant and T is the absolute temperature.

Is it possible to reconcile this equation with the relationship deduced from irreversible thermody-namics?

37.1 Answer

Table 2: Examples of forces and their conjugate fluxes. z is distance, φ is the electrical potentialin Volts, and µ is a chemical potential.

Force Flux

Electromotive force (e.m.f.) = ∂φ∂z Electrical Current

− 1

T∂T∂z Heat flux

−∂µi

∂z Diffusion flux

We have seen that in an irreversible process, the product of the force Z and the flux J gives therate of energy dissipation:

Tσ = JZ (33)

43

where T is the temperature, σ is the rate of entropy production. Tσ is therefore the rate of energydissipation. In many cases, it is found experimentally that J ∝ Z.

When there is more then one dissipative process, the total energy dissipation rate can still bewritten

Tσ =∑

i

JiXi. (34)

In the case of grain boundary motion, the rate of energy dissipation is simply V ∆G so that weimmediately get

V ∝ ∆G (35)

An alternative is to consider the transfer of atoms across a grain boundary (a barrier of height Q).The probability of forward jumps (i.e. jumps which lead to a reduction in free energy) is given by

exp−Q/kT (36)

whereas that of reverse jumps is given by

exp−(Q + ∆G)/kT = exp−Q/kT exp−∆G/kT (37)

.

The rate at which an interface moves is therefore given by

V ∝ exp−Q/kT[1 − exp−∆G/kT] (38)

Note that this relation is hardly that predicted from irreversible thermodynamics. However, theybecome identical when ∆G is small, i.e. there is not a great deviation form equilibrium. Note thatfor small x, expx ' 1 + x. Thus, at small driving forces,

Vi ∝ exp−Q/kT[∆G/kT ] (39)

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38 Essay Questions

• A weld is nothing but a defect.

• Thermomechanical processing has reached its limits.

• New materials only have niche applications.

• Austenite is non–magnetic.

• The future of safe automobiles lies in TWIP steels.

• Heat treatment is an avoidable expense.

• Nickel is the panacea for improved toughness.

• Longer Jominy specimens would be more revealing.

• Hydrogen can be used to dismantle large steel structures.

• Steels can be made transparent.

• Predictions in materials trends have been consistently wrong.

• Non–metallic inclusions can be beneficial to steels.

• Cementite does not have the formula Fe3C.

• Precipitates which are not spherical are desirable.

• At what density does iron become non–metallic?

• The modulus of iron is immutable.

• Ruthenium, osmium and hassium are iron analogues.

• Acoustic emissions tell us a lot about steel.

• Why is the thermal expansivity of γ greater than that of α?

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39 Question

Why does the carbon content of austenite which is in equilibrium with graphite, increase whentotal carbon concentration is increased, in ductile cast iron at 900C?

39.1 Answer

This would not happen if the cast iron is simply a binary iron–carbon alloy. However, when othersolutes (such as Mn, Si) are present, there are additional degrees of freedom. The phase rule saysP + F = C + 2 where P is the number of phases, F the degrees of freedom and C the number ofcomponents. For three components (Fe, Mn, C) and two phases (austenite and graphite), there arethen two degrees of freedom assuming constant pressure. If the temperature is also constant, thereis one degree of freedom left. So changing the concentration of any component, whilst keeping thetemperature constant, will alter equilibrium.

46

40 Question

Why are creep–resistant steels are severely tempered before service?

40.1 Answer

The tempering improves toughness and ductility. Another requirement is that material should fulfillthe manufacturing specifications since a very hard material makes life difficult during fabrication.In thin section materials (e.g. boiler tubes and pipes) this is achieved through tempering for fairlyshort times at high temperatures e.g. 770C for 2 h is typical for a 9Cr1Mo martensitic steel. Inthicker section materials (e.g. turbine rotor forgings and castings) a similar degree of temperingis achieved through longer periods at lower temperatures (longer times being necessary to achieveuniform temperature distribution).

This tempering also allows post–weld heat treatment to be carried out under similar conditionswithout significantly modifying the pre-welded properties of the welded components and in weld-ments the severity of the tempering is probably more critical, not just in terms of relaxing residualstresses, but also in tempering the brittle microstructures present in weld metal and heat–affectedzones. Without such tempering acceptable levels of toughness, crack growth resistance and stresscorrosion resistance in off-load conditions would not be achieved. Severe tempering certainly re-duces creep strength in the short term (and even in the longer term at lower temperatures) butevidence from long term creep programmes shows that after long term at the highest (and therforemost critical) operating temperatures there is little or no advantage of higher proof strength – thecreep rupture curves converge after long time. Evidence from the metallography shows that thehigher proof strength conditions are less stable than the lower proof strength conditions - they startwith higher dislocation density and smaller sub-grains but on ageing/creep testing these parametersrapidly converge with the structures observed in the lower proof strength conditions.

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41 Question

Comment on the use of aluminium in affecting the oxidation resistance of a steel destined forelevated temperature applications.

41.1 Answer

Aluminium helps form a potective alumina scale provided the concentration in solid solution exceedsabout around 5 wt%. This is the level normally added to the Fe20Cr ferritic steels (Fecralloy andthe oxide–dispersion strengthened alloy MA956).

Even with these concentrations, it is difficult to form alumina scales at low temperatures, sayup to 700C. This is because the diffusion of Al is slow and the only way to provide oxidationprotection is to give the material a high temperature pre-oxidation treatment in air at 1200C.Fe20Cr5Al steel retorts are sometimes used in creep and corrosion furnaces and while the materialoperates rather well at 1000-1300C in aggressive environments, the retorts undergo catastophicfailure (oxidation of Fe) at intermediate temperatures 600-800C. The problem can be alleviatedwith a high temperature pre–oxidation treatment.

48

42 Question

There is usually a incubation period in a time–temperature–transformation diagram of a steel.What is the reason for this? Can this period be regarded as the time for nucleation?

42.1 Answer

The incubation period cannot be regarded as a reflection of nucleation alone. All TTT diagramsmeasure the fraction of transformation, which is a combination of nucleation and growth. Therefore,the incubation time corresponds to the time taken to detect a given fraction of transformation. Thatfraction depends on the resolution of the experimental technique. Typically, 5% transformation.

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43 Question

Why are two separate tempering heat treatments given to martensitic creep–resistant steels afternormalising?

43.1 Answer

• After austenitisation and cooling, the steel may contain retained austenite whose decompo-sition can influence dimensional stability. Therefore, before machining, an initial temper isgiven at about 500C to induce the decomposition of the austenite. A second temperingtreatment may then be given after some other fabrication process. However, this does notexplain the very high temperatures used in many double tempering treatments.

• An initial temper to soften the material for fabrication, followed by a stress-relief heat treat-ment.

• A second temper may be required in order to stress relieve after a welding operation.

• The construction of safety–critical engineering structures usually follow strict, internationallyagreed specifications. These specifications are based on vast bodies of experience, but theway in which the rules have been decided are frequently not described. Therefore, the reasonsfor some of the historical procedures followed today are not understood. Any deviation fromspecifications can lead to unspecified risks.

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44 Question

Creep strength is frequently modelled using a large number of variables. Given the old principle thatone can fit an elephant with a fourteen parameter equation, does this procedure require justification?

44.1 Answer

An elephant should not be modelled as a sphere. The animal truly requires a large number ofvariables to describe it properly. Creep rupture strength is not a simple property - there are atleast 114 variables which influence the creep rupture strength of a ferritic steel. This is known fromphysical metallurgy experiments.

It is often the case in technological problems that simplification to reduce the number of variablesleads to a problem which bears little resemblance to the original issue.

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45 Question

Can martensite and bainite be distinguished using optical microscopy?

45.1 Answer

The best way in general is to use transmission electron microscopy because the scale of the mi-crostructure is much smaller than can be resolved using optical microscopy.

If only optical microscopy is available, then the sample should be etched lightly using nital. Themartensite, assuming that it has not been tempered, will appear light. The bainite will appearrelatively dark. This is because the bainite contains fine precipitates of cementite, which providelots of interfaces where an etchant can attack.

In the following actual images, the optical micrograph is marked “a”. The dark–etching regionsare bainite whereas the light background is martensite. Higher resolution transmission electronmicrographs are also provided in the same image.

The hardness of bainite is usually less than that of untempered martensite. Hardness measurementsare therefore useful in interpreting the microstructure.

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46 Question

Why does virgin martensite have a high dislocation density?

46.1 Answer

It is a misconception that all martensites have a high dislocation density. If the shape deformationdue to the transformation is elastically accommodated then there are in principle no defects intro-duced into the martensite as it grows. This is what happens with shape memory alloys, which relyon the absence of dislocation debris for reversibility.

The dislocations arise because the matrix is frequently unable to elastically accommodate theshape deformation. The resulting plastic relaxation of the shape deformation causes an increasein dislocation density in the parent phase, dislocations which are inherited by the martensite as itgrows.

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47 Question

During the deformation of a given microstructure, is the highest hardness reached at the ultimatetensile strength ?

47.1 Answer

No, because the ultimate tensile strength is the engineering stress, defined by the onset of necking inthe tensile specimen. The true stress, calculated by dividing the applied load by the instantaneouscross-sectional area, can be much larger. It is possible in a compression test to reach much higherstresses.

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48 Question

To what extent does the phenomenological theory of martensite crystallography apply to a diffusivephase transformation?

Would you consider a spread in orientation relationships to contradict the phenomenlogical theory?

48.1 Answer

The phenomenological theory of martensite crystallography, as applied to the transformation fromaustenite to ferrite, provides a unique relationship between the irrational crystallographic orienta-tion relation between the parent and product phases, the shape deformation, irrational habit planeand the lattice invariant shear.

To verify or contradict the theory, you therefore need to measure on a single plate, the four quantitieslisted above.

It does not therefore carry meaning to ask whether the theory applies to reconstructive transfor-mations since there is no shape deformation.

You could argue that an invariant-line is preserved during the nucleation of a reconstructive trans-formation product, but that is not the theory of martensite, simply an arbitrary condition whichmay help nucleation by ensuring a certain degree of coherency.

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