3, MSE 791 Mechanical Properties of Nanostructured Materials

Module 3: Fundamental Physics and Materials Design Lecture 1 1. What is strain (work) hardening? What is the mechanism for strain hardening?

Strain hardening is the flow stress increase with increasing strain. Its magnitude is !"!"

, where σ is the true stress and ! is the true strain. The strain hardening is caused by the accumulation of crystalline defects: primarily dislocations, and also stacking faults and twins in some cases.

2. What are the factors that determine the strength of a metal? List some factors that can

enhance the strength. Strength is determined by dislocation obstacles that block dislocation slip. Factors that can increase the strength include: a) smaller grain size (follows the Hall-Petch relationship: ! = !!+kd-1/2, b) solute atoms (solution hardening), c) second phase-particles, d) twins, e) stacking faults, f) entangled dislocations, g) phase boundaries

3. What are the factors that determine the ductility of a metal? List two mechanical factors

that can enhance the ductility. 1) Strain hardening rate (primary); 2) Strain rate sensitivity (secondary). High steain hardening rate and high strain rate sensitivity will increase the uniform elongation (i.e. ductility).

4. Nanostructured metals usually have low ductility. Explain why in terms of mechanical

behavior and deformation mechanisms. Nanostructurd metals usually have low ductility because they have low strain hardening rate. The low strain hardening rate is caused by low and even zero accumulation of dislocations and other defects. When the grain size becomes very small, e.g. <100 nm, there is very few if any dislocation source in the grain interior. Grain boundaries act as both dislocation source and sink, so that dislocation is not stored/accumulated in the grain interior. Dislocations grain boundaries are attracted and absorbed by the grain boundary.

5. The figure below shows that the nanostructured Ti becomes unstable (necking) at very low

strain under tensile test. Can you conclude that the nanostructured Ti is brittle? Justify your answer. No. A material is brittle if it has no plasticity under any deformation mode. Nanostructured Ti is unstable under tensile test, but is stable under compression test. Therefore, it is not brittle. It only does not have high ductility under tension.

Fig. 3-1. Tensile stress-strain curves of nanostructured Ti. 6. How are atoms stacked in bcc, fcc and hcp structures? Explain with the help of sketches.

Check textbook or wikipedea.

7. Draw bcc, fcc and hcp crystal structures. Draw all close-packed direction, and close-

packed planes in fcc and hcp structures. Sketch the {110} and {112} slip planes in bcc structure. Check textbook for MSE200 or MSE201.

8. The figure below shows a screw dislocation. Assuming the sense (vector) of the dislocation line is pointing downward, draw a circuit to determine its Burgers vector. What if the sense of the dislocation line is reversed?

Fig. 3-2. Schematics of a screw dislocation.

The Burgers vector will be reversed if the sense of the dislocation line is reversed.

9. The figure below shows an edge dislocation. Assuming the sense (vector) of the dislocation line is pointing into the paper, draw a circuit to determine its Burgers vector. What if the sense of the dislocation line is reversed?

Fig. 3-3. Schematics of an edge dislocation.

10. The figure below shows a dislocation (dotted line) and its Burgers vectors at three locations. Are these three Burgers vector the same? Mark the senses of the dislocation line at the three locations.

Fig. 3-4. Schematics of a dislocation in a cubic crystal.

All 3 Burgers vectors are the same. The sense of the dislocations line goes from left to right.

11. The figure below shows three dislocations meeting at one point. What is the relationship among b1, b2, and b3.

Fig. 3-5. Schematics of a net work of three dislocations meeting at one dislocation node. b1 = b2 + b3

12. The figure below shows three dislocations meeting at one point. What is the relationship among b1, b2, and b3?

Fig. 3-6. Schematics of a net work of three dislocations meeting at one dislocation node.

b1 + b2 + b3 = 0

13. The figure below shows atom packing on the (111) plane. A, B, and C marks the possible stacking positions for all close-packed layers. a) Draw a lattice dislocation and show how it can dissociate into two partial dislocations, b) b1, b2, and b3 represent the three Burgers vectors of partial dislocations on the (111) plane. The slip of a partial dislocation generates a stacking fault in its wake. Do b1, b2, and b3 produce the same type of stacking faults in terms of stacking sequence? c) Do b1, and -b3 produce the same type of stacking faults in terms of stacking sequence? Show your steps in reaching your conclusion.

Fig. 3-7. Schematics of close packed (111) plane in an fcc crystal, the possible stacking positions and three partial dislocations, b1, b2, and b3.

(a) see the drawing in the figure. (b) Yes, they produce the same type of stacking faults. (c) No. –b3 won’t work. It produce a ball on ball situation that is not possible energetically.

14. The figures below illustrate the relationship between an fcc unite cell and a Thompson tetrahedron. List the Burgers vectors of all lattice dislocations, partial dislocations and stair-rod dislocations in the Thompson tetrahedron.

Fig. 3-8. Thompson tetrahedron

Burgers vector for full dislocations are: AB, BC, CA, AD, BD, CD and BA CB, AC, DA, DB, DC. Note that AB means the vector start at A and end at B. Therefore, BA =-AB Burgers vector for partial dislocations are: On the BCD plane: Bα , Cα , Dα; and αB, αC, αD. On the ADC plane: Aβ , Cβ , Dβ; and βA, βC, βD. On the ABD plane: Aγ , Bγ , Dγ; and γA, γB, γD. On the ABC plane: Aδ , Bδ , Cδ; and δA, δB, δC. Stair-rod dislocations: αβ , βγ , γδ , δα; αδ , δγ , γβ , βα

15. Using the above Thompson tetrahedron, a lattice dislocation Burgers vector can be dissociated into pairs of partials. Write the dissociation equations.

One example only here: BC = Bα + αC (on BCD plane), BC = Bδ + δC (on ABC plane).

Lecture 2: 1. See Figure 3-7, verify that the slip of b1, b2, and b3 on successive (111) planes will produce a

deformation twin. Can the mixture of b1, b2, and b3 and -b1, -b2, and -b3 produce a twin? Demonstrate it using sequence of b2, -b1, b3, b3, -b2, b2 and -b3. From Fig. 3-7, it can be seen that: For (, b2, and b3 , the change of stacking sequence is : A→B, B→C, and C→A For -b1, -b2, and -b3 , the change of stacking sequence is : A→C, B→A, and C→B For b1, b2, and b3 to slip on adjacent slip planes, the sequence change produce a twin C A B C B C A B A B C A C A B →b3 C B C →b2 A A A→b1 B B B C C C C B B B B A A A A For b2, -b1, b3, b3, -b2, b2 and -b3. It can be seen that –b1 does not work because it produces B on B situation. In addition, if the –b1 slips on the twin boundary, it can reverse the stacking sequence produced by an earlier twinning partial to result in detwinning. C A C B C B A B A C A C B C -b1 B A b2 B B C C C B B B A A A

2. The two white lines in the figure represent two close packed planes. Calculate the angle between these two white lines using fcc crystal structure.

Fig. 3-9. The high-resolution image of a twin in a fcc crystal. The white lines represent {111} close packed planes. The twin boundary is also a close packed plane. The angle between two close packed planes can be calculated as (e.g. (111) plane with (1-11) plane: Cosα = [1x1+1x(-1) +1x1)/[(12 + 12 +12)1/2 x )x(12 + (-1)2 +12)1/2] =1/3 α = 70.53° The angle between the white lines = 2α≈ 141°

3. Prove that the glides of three partials, b1, b2, and b3, on three successive slip planes produce a twin without producing macroscopic strain.

As shown in Fig. 3-7, b1 + b2 + b3 = 0. Therefore no strain will be produced.

Lecture 3 1. Describe the dislocation emission process of a Frank-Read source.

See the lecture ppt and video. 2. Describe the pole twinning mechanism. Does this mechanism produce a macroscopic strain?

In pole twinning mechanism, a screw dislocation has a Burgers vector that links one slip plane to the adjacent slip plane. A partial dislocation has one end pinned by the screw dislocation and the other end free to slip. Under an applied stress, the partial dislocation slips around the screw dislocation, and move to the next slip plane after every revolution to generate a deformation twin. Because the partial has a fixed Burgers vector, each slip plane swiped by the partial shears to a direction and magnitude that equal the Burgers vector of the partial, which produces a macroscopic strain.

3. Describe the three-layer twinning mechanism. Does this mechanism produce a macroscopic strain?

See the figure, two dislocations with Burgers vector of AB and AC meet to have a dislocation reaction AB + AC = 3Aδ . If the dislocation line is parallel to Aδ , then the three partials with a Burgers vector of Aδ can slip on three adjacent slips planes to form a twin nucleus. Multiple nuclei next to each other can form a twin more than 3 layers thick.

4. Describe the double-pole twinning mechanism. Does this mechanism produce a macroscopic strain?

See slide 6 of lecture 3.

5. Explain why the coarse-grained metals are harder to deform by twinning with decreasing grain size?

This is because for coarse-grained metals, the Hall-Petch slope for twinning is higher than the Hall-Petch slope for dislocation slip.

Lecture 4 1. Why does nano-meter sized grains often contains no dislocations?

Because the dislocations are too close to the grain boundaries to be stable in the grain interior. They are attracted to the grain boundaries.

2. Why grain boundary sliding/grain rotation become more important at very small grain sizes, e.g. < 20 nm?

This because of two reasons with decreasing grain size: 1) the area of grain boundaries increases dramatically, 2) the critical stress for dislocation slip increases dramatically, making it more difficult for dislocation to slip.

3. Why dislocation emission from grain boundaries may become dominate deformation mechanism in nanocrystalline metals?

Because there is almost no dislocation source in the grain interior, and grain boundaries have become the source and sink for dislocations.

4. List possible deformation twinning mechanisms in nanocrystalline fcc metals.

a) Partial emission from grain boundaries, b) accidental overlapping of dissociated dislocations, c) grain boundary splitting and migration, d) Random emission of partials from grain boundaries, e) Coordinated slip of three partials,

5. Define non-equilibrium grain boundary.

A non-equilibrium grain boundary contains more dislocations than needed to accommodate the misorientation across the boundary, i.e. it has higher energy than an equilibrium grain boundary.

6. Describe two dislocation reaction processes on dislocation multiplication on the grain boundaries.

See slides 11 and 12 in the ppt for lecture 4.

Lecture 5 1. Calculate the shear strain produced by a deformation twin that is produced by the glide of partial dislocations with the same Burgers vector in an fcc metal.

For a fcc metal, the interplane spacing of slip plane (111) is

d = a12 +12 +12

=a3

.

The Burgers vector of the twinning partial is a/6[112], its magnitude is

b = a612 +12 + 22 = a

6.

The shear strain is

2. Calculate the shear strain produced by a deformation twin that is produced by alternative gliding of partial dislocations with two different Burgers vectors in an fcc metal.

For a fcc metal, the interplane spacing of slip plane (111) is

d = a12 +12 +12

=a3

.

The Burgers vector of the twinning partial is a/6[112], its magnitude is

b = a612 +12 + 22 = a

6.

For alternative Burgers vectors on alternative planes, one Burgers vector shear is produced on two slip planes. Therefore

! =b2d

=a 62a 3

=24

3. Explain how random activation of partials can produce a twin without producing macroscopic strain.

Because on average, bii=1

n

! = 0

! =bd=a 6a 3

=22

4. Explain how cooperative slip of three partial mechanism produce a twin without macroscopic strain.

Three partials with Burgers vectors: b1 + b2 + b3 = 0. Under an appropriate applied stress, two partials (e.g. b1 and b3) move forward, while the third partial cannot move because no applied stress can drive these partials simultaneously. The first two partials create two stacking faults and they also attract the third partial. Therefore, when the applied stress fluctuates to become weaker, the third partial will be dragged to be closer to the first two partials. Therefore the 3 partials moves forward in a stop-go coordinated way.

5. What roles do twins without macroscopic strain play in the deformation of nanomaterials

a) reorient the crystal to make it easier to deform by dislocations, b) relieve local stress with the slip of partials, c) smoother grain boundary to facilitate grain boundary sliding/grain rotation.

6. Why is it easier for nanocrystalline grains to produce twins with no macrostrain?

Because Because the grain boundaries act as the source of partial dislocation emission. This allows different types of partials to be emitted.

Lecture 6 1. Describe the grain size effect on deformation twinning for fcc metals (draw a graph).

2. Explain how and why the detwinning during deformation is affected by grain size. What is the implication of this to the stability of microstructures.

Detwinning usually occur by the reaction of dislocations with grain boundaries. Such a reaction usually has high energy barriers, which need high applied stresses to overcome. With decreasing grain sizes, the materials flow at higher stresses, which makes it statistically more likely for the detwinning to occur.

3. Define the stable and unstable stacking fault energies on a generalized stacking fault energy curve.

Twin

ning

diff

icul

ty

Grain Size

Lecture 7

Fig. 7-1 1. Using Fig. 7-1, shows how a 30º partial dislocation (e.g. Cα) cross-slip onto the twin boundary. Write the dislocation reactions.

Cα = Cδ + δα 2. Using Fig. 7-1, shows how a 90º partial dislocation (e.g. Dα) cross-slip onto the twin boundary. Write the dislocation reactions.

Dα = Aδ because the fcc twin is of compound type. 3. Using Fig. 7-1, shows how a 30º dislocation (e.g. Cα) transmit through the twin boundary to glide in the twin. Write the dislocation reactions.

Cα = Cαʹ′ + αʹ′α 4. Using Fig. 7-1, shows how a 90º dislocation (e.g. Dα) transmit through the twin boundary to glide in the twin. Write the dislocation reactions.

Dα = Dδ + δα Dδ = δDʹ′ = δαʹ′ + αʹ′Dʹ′ Therefore: Dα = δα + δαʹ′ + αʹ′Dʹ′ =(4/9)Aδ + αʹ′Dʹ′

6. Using Fig. 7-1, shows how a screw dislocation (e.g. BC ) cross-slip onto the twin boundary. Write the dislocation reactions.

No dislocation reaction is needed for a screw dislocation with a Burgers vector of BC to cross-slip from the BCD plane to ABC plane, because BC is on both BCD plane and ABC plane.

7. Using Fig. 7-1, shows how a screw dislocation (e.g. CΒ) transmit through the twin boundary to glide in the twin. Write the dislocation reactions.

No dislocation reaction is needed. Reason is the same as in the problem 6.

Lecture 8 1. What is difference in the strain sensitivity behaviors of nanocrystalline bcc and fcc metals as a function of grain size?

For a nanocrystalline bcc metal, its strain rate sensitivity decreases with decreasing grain size. For an fcc metal, the trend is opposite.

2. Compare the cores of screw dislocations in fcc and bcc metals.

A screw dislocation in a bcc metal can dissociate into three partials on three planes, while a screw dislocation in an fcc metal will dissociate on a 2D slip plane.

3. Compare the cores of screw and edge dislocations in bcc metals.

A screw dislocation in a bcc metal can dissociate into three partials on three planes, while an edge dislocation in a bcc metal will dissociate on a 2D slip plane.

4. In a bcc metal, why is the deformation behavior controlled by screw dislocations?

Because the core of the screw dislocation is 3 dimensional, which makes it hard to slip on any slip plane. It needs thermal activation to form kinks first. Therefore, screw dislocations are slow moving and controls the deformation behavior (e.g. strain rate sensitivity).

5. How does a screw dislocation move in a bcc metal? How about an edge dislocation?

Because the core of the screw dislocation is 3 dimensional, which makes it hard to slip on any slip plane. It needs thermal activation to form kinks first. The kinks are edge dislocations and move fast, resulting in the screw dislocation to move forward.

6. Why does a screw dislocation in a bcc metal needs high thermal activation to move?

Because the core of the screw dislocation is 3 dimensional, which makes it hard to slip on any slip plane. It needs thermal activation to form kinks first. The kinks are edge dislocations and move fast, resulting in the screw dislocation to move forward.

7. Compare a coarse-grained bcc metal with a coarse-grained fcc metal, which one has higher strain rate sensitivity? Why?

Coarse-grained bcc metals have much higher strain rate sensitivity than coarse-grained fcc metals, because the slip of screw dislocations in bcc metals has high thermal activation energy.

8. How grain size affect the strain rate sensitivity of an nanocrystalline fcc metal? Why?

For a nanocrystalline fcc metal, the strain rate sensitivity increases with decreasing grain sizes because grain boundary sliding/grain rotation contribute more to deformation. These grain boundary process involves diffusion and has high thermal activation energy.

9. How grain size affect the strain rate sensitivity of a bcc metal? Why?

For a nanocrystalline fcc metal, the strain rate sensitivity decreases with decreasing grain sizes because edge and mixed dislocations contribute more to deformation screw dislocaiton contribute less. The slip of edge and mixed dislocations have low strain rate sensitivity.

Lecture 9 1. Why do nanocrystalline materials usually have low ductility?

Because they have low strain hardening rate. 2. Why do nanocrystalline materials usually have low strain hardening rate?

Because they are ineffective in defects (e.g. dislocation) accumulation during plastic deforamtion.

3. Derive the Considère's Criterion.

P = Aσ, where P is the load, A is the cross-section area and σ is the true normal stress dP = σ dA + A dσ During a tensile test, for the sample to be mechanically stable (no necking), we need dP ≥ 0, i.e., σ dA + A dσ ≥ 0, or d!!

+dAA! 0 ,

During the plastic deformation, the volume is conserved, i.e. AL = constant, where L is the sample length. We have A dL + LdA = 0, i.e. dAA= !

dLL= !d! , where ε is the true strain.

Therefore d!!

! d" " 0 . We have

d!d"

!! . This is the Considère's Criterion for determining the necking point on a true stress-

strain curve. For engineering stress-strain curves, the necking occurs at the highest engineering stress.

4. List 3 possible approaches to simultaneously increase the strength and ductility of nano fcc metals.

a) second phase particles, b) twin boundaries, c) high angle grain boundaries, d) deformation twinning

5. How grain size affect twinning in hcp metals?

Twinning becomes more difficult with decreasing grain size.