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MSE 566 Homework Module 3 Solution

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MSE 566 Homework Module 3 Solution 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: -1/2 ! = !!+kd , 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.
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