Mats University Distance Education Assignments Meaning

Mat E 202 Fall 2013 Assignment #3 Assigned: Oct 4, 2013 Due: Oct 18, 2013 Department of Chemical and Materials Engineering 1. (a) Calculate the atomic Packing Factor (APF) of the unit cell filled by each atom in: (i) Simple Cubic (SC) (ii) Body Centered Cubic (BCC) (iii) Face Centered Cubic (FCC) (iv) Diamond Cubic (to determine structure, see appendix for help) (b) Can you increase the APF of the above crystal structure by changing the atomic radius? Why or why not? (c) What are the planes of highest planar packing atomic factor (PPF) in each of the structures (except (iv) diamond structure)? And what are the magnitudes of PPF for each? (d) What are the directions of highest linear atomic packing factor (LPF) in each of the structures (except (iv) diamond structure)? (e) Calculate the radius of an iridium atom (Hint: FCC; density=22.4g/cm3; atomic weight=192.2 g/mol) (a) (i) SC: There is 1 atom per unit cell. The maximum radius that a hard sphere can have is r = 1 Mat E 202 Assigned: Oct 4, 2013 Fall 2013 Assignment #3 Due: Oct 18, 2013 a/2, where a is the lattice constant. When the radius is this value, then the spheres on the corners of the unit cell just touch each other. Hence the packing fraction is: (ii) BCC: There are 2 lattice points per unit cell. The maximum radius that a hard sphere can have is , where a is the lattice constant (see drawing below and solutions for 1(c) & (d)). When the radius is this value, then the spheres on the corners of the unit cell just touch the body-centered sphere. Hence the packing fraction is: 2R (because atoms are touching each other) a/ a/2 a/2 a/2 a/2 a/2 a/2  (iii) FCC: There are 4 lattice points per unit cell. The maximum radius that a hard sphere can have is , where a is the lattice constant (see drawing below in solutions for 1(c) & (d)). When the radius is this value, then the spheres on the corners of the unit cell just touch the face-centered sphere. Hence the packing fraction is: (iv) Diamond: There are 8 lattice points per unit cell. The maximum radius that a hard sphere can have is (see drawing below), where a is the lattice constant. When the radius is this value, then the spheres on the corners of the unit cell just touch the nearest neighbor sphere. Hence the packing fraction is: 2 Mat E 202 Fall 2013 Assigned: Oct 4, 2013 Due: Oct 18, 2013 Assignment #3 (b) No. APF is determined by the ‘fraction’ of occupied volume by atoms divided by total unit cell volume. Therefore, atomic radius a term is annihilated by definition. In other words, APF is not dependent upon the radius of atoms. (c) & (d) (i) R SC {100} family of planes = <100> a = 2R = 0.785 <100> family of directions: LPF = 1 (ii) 3 Mat E 202 Fall 2013 Assigned: Oct 4, 2013 Due: Oct 18, 2013 Assignment #3 R BCC {110} family of planes = = 0.833 <111> family of directions: LPF = 1 (iii) FCC {111} family of planes (2 atoms in plane)  3 x half + 3 x (1/6) atoms = = 0.907 <110> family of directions: LPF = 1 (e)  Density 3 where ρ = density [g/cm ], M=molar mass [g/mol], n= Number of atoms per unit cell [atoms/unit cell], NA = Avogardo’s number [atoms/mol], and v = volume per unit cell [cm3/unit cell] By plugging in the numbers, 4 192.2 22.4 For FCC, √  6.02 √ 10 √ . 5.70 2.72 10 10 2.72 Å 2. Consider a three-dimensional lattice system. Using miller indices, (a) Draw a cubic unit cell with lattice constant ao. (b) Show on this drawing the [ 1 0 1 ] direction. (c) List all of the members in the < 1 0 1 > family of directions and show each of them on the figure. (d) Draw a three-dimensional rectangular lattice, where x, y axes have lattice constant of a1 4 Mat E 202 Fall 2013 Assignment #3 and z has that of a2 (a1 < a2) Assigned: Oct 4, 2013 Due: Oct 18, 2013 (e) Repeat questions (b) and (c) for the rectangular lattice in (d). (a) z a0 a0 y a0 x (b) z The direction is defined as the line from origin to (a0, 0, -a0) (See solid green line). Alternatively, the line can be transposed onto the planes on the unit cell (See dashed green lines) y a0 x -a0 (c) z z z [101] y x [101] [101] y x [110] [101] y [110] x 5 Mat E 202 Fall 2013 Assigned: Oct 4, 2013 Due: Oct 18, 2013 Assignment #3 z z z [110] y [011] [011] [110] y x [011] x [011] y x Total = 12 (d) z a2 a1 y a1 x (e) z a2 z a2 a1 [101] [101] [101] [101] a1 y z a2 [011] y a1 a1 a1 x x z a2 [011] y a1 x [011] [011] a1 y a1 x Total = 8 3. A silicon wafer (shown below) is a thin disk-shaped single crystal of silicon. This is the most essential ingredient of semiconductor industry. Conventionally, the wafers are denoted as the miller index of the surface plane; typically, (100), (110), and (111) wafers are produced. 6 Mat E 202 Fall 2013 Assignment #3 Assigned: Oct 4, 2013 Due: Oct 18, 2013 (a) Si wafers have diamond cubic structure with atomic weight and mass density of 28.086 g/mol and 2.33 g/cm3, respectively. Calculate the lattice parameter. (Again, see appendix for help) (b) Sketch (100), (110) and (111) planes (showing atom positions & whether they touch each other). Calculate planar atomic packing factor (PPF). (c) Based on the nature of bonding, is silicon wafer brittle or ductile? Why? (d) When silicon wafer is subjected to a sharp and intense impulse at the center of wafer, sketch the direction of cleaving for (100) and (111) wafers, respectively. (Hint: In diamond structure, catastrophic fracture is known to propagate through <110> family of directions) (a) Density  where ρ = density [g/cm3], M=molar mass [g/mol], n= Number of atoms per unit cell [atoms/unit cell], NA = Avogardo’s number [atoms/mol], and v = volume per unit cell [cm3/unit cell] By plugging in the numbers, 8 2.33  5.43 10 28.086 6.02 10 1.60 10 5.43 Å 7 Mat E 202 Fall 2013 (b) Assignment #3 Assigned: Oct 4, 2013 Due: Oct 18, 2013 (1,1,1) (1,1,0) a (1,1/2,1/2) PPF = (1,0,1) a (1,0,0) (100) plane 0.295 /√ (1,0,1) (1/2,1/2,1) (0,1,1) a (3/4,1/4,1/4) (1/4,3/4,1/4) (0,1,0) (1/2,1/2,0) (1,0,0) a (110) plane PPF = √ √ 0.417 /√ (0,0,1) a (1/2,0,1/2) (0,1/2,1/2) (1/2,1/2,0) (1,0,0) (1,1,0) a PPF = (111) plane √ / √ √ / √ /√ 0.340  NOTE: All circles and lines are drawn TO SCALE (c) Single crystalline silicon is a good example of 100% covalent bonding, where strongly anisotropic nature of the atomic bonding precludes plastic deformation of the material. Therefore, silicon wafer is brittle. The brittleness of silicon wafer can be visualized in the following movie (NOTE: The wafer in the movie has a surface of (100) plane) http://www.youtube.com/watch?v=kivkEir4RcM 8 Mat E 202 Fall 2013 Assigned: Oct 4, 2013 Due: Oct 18, 2013 Assignment #3 (d) By looking at (100) and (111) planes, <110> directions may look as following, respectively. Family of <110> direction on (100) plane z Family of <110> direction on (111) plane z x y x y Therfore, by a strong impulse pointed at the centre of wafer, wafers will be cleaved as following: Note that the pictures were taken as tilted and the angles between the fracture lines are 90 (left; (100) wafer) and 60 (right; (111) wafer) degrees, respectively. < NOTES FOR CONCEPT CLARIFICATION – NOT FOR SCORE MARKING > It is the basic concepts of fracture mechanics that cleavage takes place on certain “planes” along certain “directions” on those planes. Specifically in diamond cubic structure, cleavage takes place on {111} and {110} planes. In both cases, crack preferentially propagates along <110> family of directions. (For fundamental mechanisms, see: R. Perez & P. Gumbsch, Phys. Rev. Lett. (2000) 84 (23), 5347-5350) In this problem, the cleavage planes were not mentioned to reduce confusion in solving the problem. 4. The diagram below shows four different point defects, different elements being different shades. 9 Mat E 202 Fall 2013 Assignment #3 (a) Name the four defects shown in the image. Assigned: Oct 4, 2013 Due: Oct 18, 2013 (b) Briefly explain which of the above listed defects requires the most energy to create. (c) Atomic radius, crystal structure, electronegativity, and the most common valence are tabulated in the following table for several elements; for those that are nonmetals, only atomic radii are indicated. Consider carbon ( C ) and chrome ( Cr ). Do you expect them to mix well with iron (Fe)? Why or why not? If so, will they be located at substitutional or interstitial sites? Why? (d) Cu and Ni are known to mix well with each other. State logical reasons for them to mix well. (e) Imagine an alloy with a 50:50 atomic ratio between Cu and Ni. Construct a (100) planar view of the lattice of the 50:50 Cu:Ni alloy (HINT: show 6 unit cells). (f) Convert the atomic ratio in (e) to weight ratio between Cu and Ni. (a) 1. Vacancy; 2. Self-interstitial; 3. Interstitial impurity atom; 4. Substitutional impurity atom (b) Of the above defects, the self-interstitial (2) would require the most energy to create, because it causes the most lattice strain (long ranging strain field due to large distortions in the lattice). In order to insert a relatively large atom (i.e. a solvent/lattice atom) into a relatively small void (i.e. an interstitial space), the lattice around the void must be distorted extensively. Moreover, there is a vacancy created when the solvent atom moves into the interstitial space, which also requires energy to form. (c) (i) Chrome (Cr) in iron (Fe):  Cr and Fe have comparable atomic radii. Therefore, its Cr may be substitutional impurity 10 Mat E 202 Fall 2013 Assignment #3 atom in Fe if mixing is allowable. Following four steps of empirical mixing rule, Step 1: Atomic Size Factor: % . 100% . . Assigned: Oct 4, 2013 Due: Oct 18, 2013 100% 0.6% ≪ % …FAVORABLE Step 2: Crystal structure  both are BCC … FAVORABLE Step 3: Electronegativity  EFe = 1.8, ECr = 1.6; thus, DE%=11% … FAVORABLE Step 4: Valence of Cr (solute) is +3, whereas that of Fe (solvent) is +2. … NOT FAVORABLE : Therefore, Chromium is expected to have limited solubility in iron. In fact, the solubility value is measured to be ~8at% at 300°C (Bergner et al., Scripta Materialia (2009), 61, 1060). (NOTE: only count the first two steps in homework evaluation. The ones expected a good solubility from steps 1 & 2 must receive full credit.) (ii) Carbon (C) in iron (Fe):  Simple answer: Carbon atoms are so small that they can fit into the interstitial site between BCC iron. Therefore, carbon is expected to mix well with iron as interstitial impurity atoms. (NOTE: This simple answer must receive full credit.)  In-depth answer: (Not required; Just for educational purpose) The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situated in the plane of this (100) face, midway between the two vertical unit cell edges, and one quarter of the distance between the bottom and top cell edges. From the right triangle that is defined by the three arrows we may write a 2 a 2   +   = 2  4  However, from Equation 3.3, a = 4R , 3 (R  r) 2 and, therefore, making this substitution, the above equation  takes the form  4R 2  4R 2 2 2   +   = R + 2R r + r 2 3  4 3   After rearrangement the following quadratic equation results: r 2 + 2R r  0.667 R 2 = 0  And upon solving for r: r  (2R)  (2R) 2  (4)(1)(0.667R 2 ) 2 11  Mat E 202 Fall 2013 Assigned: Oct 4, 2013 Due: Oct 18, 2013 Assignment #3 2R  2.582R 2  And, finally r()  2R  2.582R  0.291R 2 Of course, only the r(+) root is possible, and, therefore, r = 0.291R. For Fe atoms, R=0.1241 [nm], thus interstitial atomic radius of 0.361 [nm] can be easily accommodated. Carbon has atomic radius of 0.071 [nm], which is far smaller than what is allowed as an interstitial impurity atoms for BCC iron. Therefore, carbon is expected to mix well with iron as interstitial impurity atoms. (d) Copper (Cu) in Nickel (Ni):  Cu and Ni have comparable atomic radii. Therefore, its Cu may be substitutional impurity atom in Ni if mixing is allowable. Following four steps of empirical mixing rule, Step 1: Atomic Size Factor: . . % 100% 100% 2.6% ≪ % …FAVORABLE . Step 2: Crystal structure  both are FCC … FAVORABLE Step 3: Electronegativity  EFe = 1.9, ECr = 1.8; thus, DE%=5.6% … FAVORABLE Step 4: Valence of Cu (solute) and Ni (solvent) are both known to be +2 … FAVORABLE : Therefore, all the empirical rules expect a good mixing between the two metals. (e) Cu Ni Overall structure is the same as pure atom case. However, slight local distortion is expected due to the atomic radii difference between Cu and Ni (somewhat exaggerated in the picture above). Indeed, the local distortion of atomic arrangement is the mechanism of hardening by mixing Cu and Ni (compared to pure metals). (f) Atomic weight of Cu and Ni are 63.55 [g/mol] and 58.69 [g/mol], respectively, at room temperature. Therefore, . . . . . . 100 52.0% 100 48.0% 12 Mat E 202 Fall 2013 Assigned: Oct 4, 2013 Due: Oct 18, 2013 Assignment #3 5. Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of vacancies at 500°C is 7.571023 m-3. The atomic weight and density (@ 500°C) for aluminum are, respectively, 26.98 g/mol and 2.62 g/cm3. From equilibrium vacancy equation,    From above equation, k = (Boltzmann constant) = 1.38 x 10-23 [J/K] T = 500°C = 773°K Nv = (# of vacancy per unit volume) = 7.571023 [atoms/m3] And N (total # of atoms per unit volume) needs to be calculated. We know the equation for density where ρ = density [g/cm3], M=molar mass [g/mol], n= Number of atoms per unit cell [atoms/unit cell], NA = Avogardo’s number [atoms/mol], and v = volume per unit cell [cm3/unit cell] From definition, (n / v) = N. Therefore, . . . 5.85 10 10 5.85 10 Now, we have all the numbers. 1.38 1.20 10 10 773 1/1.6 10 . . 0.75 13 Mat E 202 Fall 2013 Assignment #3 Assigned: Oct 4, 2013 Due: Oct 18, 2013 INTENDED LEARNING GOALS FOR THIS ASSIGNMENT Declarative knowledge Identify the structural differences between amorphous and crystalline solids Identify structures of metals (e.g., BCC, FCC) Identify different cases of imperfections in crystals. Procedural knowledge Determine atomic packing factor, mass density, linear density, planar density Convert Miller indices for directions and planes into graphical representations Convert composition in atomic % to weight % for binary alloys Use proper units, nomenclature, and vocabulary Conditional knowledge Compare materials with different properties and rationalize their differences based on bond energy fundamentals and curves Reflective knowledge Explain the relationships between structure and density, elastic modulus, and ductility. Explain the different origins for terms in the bond energy function, and how 14

Help

Mat E 202 Fall 2013 Assignment #4 Solution Assigned: Oct. 18, 2013 Due: Nov. 1, 2013 Department of Chemical and Materials Engineering (/72) 1. a) (resolved) shear stress b) a slip plane 2. a. Temperature Diffusivity increases exponentially with temperature – this relationship can be expressed in terms of an Arrhenius relation, i.e.: Diffusion is a kinetic process, and depends on the movement of atoms. At higher temperatures, atoms have more energy and thus move more quickly with larger vibrational amplitude, resulting in higher values of diffusivity. b. Concentration gradient The concentration gradient does not affect diffusivity. The concentration gradient does, of course, significantly influence the speed and direction of diffusion itself, as defined by Fick’s first law: The concentration gradient has no effect on the diffusion coefficient. c. Crystal structure / atom type In general, the smaller the diffusing atom and the lower the atomic packing density of the matrix crystal structure, the greater the diffusivity. Every time a diffusing atom jumps from one lattice/ interstitial position to the next, it must distort the lattice. This distortion has an activation energy barrier associated with it: unless the diffusing atom has more than a minimum threshold of energy, it cannot jump. The more lattice distortion that is required for the atom to jump, the greater this energy barrier will be. As the energy barrier is decreased, the rate at which a diffusing atom can move – i.e. the diffusivity – increases (because a lower energy barrier means that is more likely that the atom has enough energy to overcome the barrier at any given time). The amount of lattice distortion caused by a diffusing atom decreases with decreasing size of the atom and with decreasing atomic packing of the matrix crystal structure – hence, smaller atoms and lower atomic packing lead to higher diffusivities 1 Mat E 202 Fall 2013 Assignment #4 Solution Assigned: Oct. 18, 2013 Due: Nov. 1, 2013 3 The diffusivity of C in α-ferrite (BCC iron) and diffusivity of C in γ-austenite (FCC iron) is given at two temperatures: Temperature [°C] 500 900 1100 Diffusivity [m2/s] C in γ-Fe --5.9 x 10-12 5.3 x 10-11 Diffusivity [m2/s ] C in α-Fe 2.4 x 10 -12 1.7 x 10-10 --- a. What are the activation energies of diffusion for C in α-Fe and C in γ-Fe (in units of eV/atom)? (/) b. Explain why the diffusivity of C in α-Fe is higher than the diffusivity of C in γ-Fe. (/2) Solution: a. In order to calculate activation energies, we need to use: where: = diffusivity = pre-exponential diffusion constant = activation energy = universal gas constant = 8.314 J/mol·K = absolute temperature First, however, we need to rewrite the equation by taking the natural logarithm of both sides: Now, we can calculate the activation energy using and for C in γ-Fe. – in both cases, the slope will be equal to (in Kelvin!) for C in α-Fe and and the intercept will be equal to . As we have two data points for each condition we can calculate the activation energy: For C in α-Fe: 2 Mat E 202 Fall 2013 For C in γ-Fe, Assignment #4 Solution Assigned: Oct. 18, 2013 Due: Nov. 1, 2013 b. Carbon has a higher diffusivity in α-Fe than γ-Fe because the crystal structure of α-Fe is BCC whereas the crystal structure of γ-Fe is FCC. FCC has greater atomic packing than BCC (i.e. the atoms are more densely packed in FCC). Carbon occupies interstitial sites in the structure; thus, carbon diffuses through interstitial diffusion. Therefore, the lower atomic packing factor in a BCC structure allows carbon to diffuse more easily – i.e. there is more empty space for diffusion. As a result, all else held constant, diffusion is slower through an FCC material (i.e. γ-Fe) than through a BCC material (i.e. α-Fe). C x − Co  x  vs.   for a gas carburizing process (i.e. diffusion of C in γ-Fe for C s − Co  2 Dt  surface hardening). The initial concentration of C in the γ-Fe is 0.1 wt%. Use this plot to answer the following questions. Assume the T = 1000 °C , Do = 2.3x10-5 m2/s and Qd = 148 kJ/mol. 4. Below is a plot of a) What is carbon concentration at 1mm below the surface after carburizing for 1 hour. Assume that Cs = 0.8wt%. b) Determine the time necessary for the carbon content to be 0.4wt% at a distance of 0.5 mm below the surface.  x t= 3600 s, D = 1.94x10-11 x =0.001 m, therefore   2 Dt C − Co From the graph this corresponds to x = 0.2 therefore C1mm = 0.24 C s − Co a) C x − Co  x  =  C s − Co  2 Dt  C x − Co = 1.1 from the graph C s − Co therefore t = 2657 s or 0.74 hrs. b)  x   2 Dt   = 0.3    = 1.89  where D = 1.94x10-11 and x =0.0005m 3 Mat E 202 Assigned: Oct. 18, 2013 Fall 2013 Assignment #4 Solution Due: Nov. 1, 2013 5. a. List four strengthening mechanisms and describe how strengthening is achieved. (/8) Grain boundary strengthening: achieved by decreasing grain size, which increases grain boundary area (i.e. more grain boundaries per unit area). Adjacent crystals on either side of a grain boundary have different orientations and hence the slip planes are not continuous across the grain boundary. Because the slip planes do not line up, a dislocation cannot easily move from one grain to another. Therefore, as the grain boundary area increases, slip becomes increasingly difficult. Solution strengthening (hardening): achieved by adding solute atoms to the metal. The addition of a solute atom in a lattice, be it interstitial or substitutional, results in the creation of strain fields. Dislocations also have strain fields, and the strain fields of dislocations interact with the strain fields associated with substitutional solute atoms, resulting in pinning of dislocations (dislocations can minimize their energy by reducing their strain fields by being close to solute atoms). Interstitial solute atoms can easily diffuse to dislocation cores and fill in the empty space directly below the dislocation, and hence also pin the dislocation. Strain hardening: achieved by plastically deforming the metal. Plastic deformation creates dislocations and increases the dislocation density. As the dislocation density increases, so do dislocation-dislocation interactions. The strain fields around dislocations can result in repulsion, and thus dislocations will not want to come too close together. As such, when the dislocation density becomes too high, dislocations will eventually stop moving. Precipitation hardening: achieved by adding small, hard particles of a secondary phase (i.e. precipitates) using heat treatments. Precipitates resist the motion of dislocations (they have a very high strength). These particles are essentially impenetrable obstacles to dislocation motion, and cause the dislocation to require more energy to either cut through the precipitate (particle shear) or to bow (dislocation bowing). 5. The following data were obtained for the yield stength as a function of grain size for a carbon steel and an aluminum alloy. Carbon steel Aluminum alloy d (µm) σy (MPa) d (µm) σy (MPa) 406 93 42 223 106 129 16 225 75 145 11 225 43 158 8.5 226 30 189 5.0 231 16 233 3.1 238 a) determine the material constants, σo and ky, in this equation for the carbon steel b) determine the material constants, σo and ky, in this equation for the aluminum c) At what grain size will both the steel and aluminum have a σy = 300 MPa. 4 Mat E 202 Assigned: Oct. 18, 2013 Fall 2013 Assignment #4 Solution Due: Nov. 1, 2013 Solution: a. To demonstrate that the behavior of the alloys is consistent with the Hall-Petch equation the data must be plotted as σy versus 1/√d. If the alloys are consistent the subsequent plot will be linear. In this case, the behaviour is consistent with the Hall-Petch Equation The intrinsic strength and Hall-Petch coefficient can be determined from the intercept and slope respectively. From the plot above: Alloy type Intrinsic strength, σo [MPa] Hall-Petch Coefficient, ky [MN/m3/2] Carbon 60.5 0.69 Steel Aluminum 215.5 0.036 Alloy b) for σy = 300 MPa, d= 8.30 µm for steel and d = 0.18 µm aluminum 7. Consider annealing of a heavily cold worked specimen. a. Draw a schematic curve of the tensile strength and ductility (%EL) as a function of annealing temperature for cold worked brass (indicate the three stages of annealing – assume 1hr of annealing time). b. Explain how and why the tensile strength and ductility change in each of the three stages of annealing (HINT: discuss grain size and shape changes and dislocation density). 5 Mat E 202 Fall 2013 Solution: Assignment #4 Solution Assigned: Oct. 18, 2013 Due: Nov. 1, 2013 Tensile Strength Ductility b. These mechanisms operate at different zones: Recovery: In this zone no significant change in σuts or %El. occurs. Dislocations move and rearrange via diffusion of atoms. A very small decrease in dislocation density occurs. Recrystallization: Large change in σuts and %El. occurs In this zone, movement and annihilation of dislocations takes place. New dislocation-free grains are formed which leads to restoration of ductility and decrease in strength. Grain growth: Diffusion and reduction of high energy grain boundaries (interfacial defects) leads to grain growth. Larger grains will be less strong and more ductile but the effects are not as drastic as those due to the recrystallization. 8 .A cylindrical rod of fully annealed 1040 steel (originally 15.2 mm in diameter) is to be cold worked by drawing; the circular cross section will be maintained during deformation but the final diameter must be 10 mm. A cold-worked tensile strength in excess of 850 MPa and a ductility of at least 12% EL are desired. Explain how this may be accomplished with cold working and annealing treatments only. Be as quantitative as possible. (/6) • Assume that 1040 steel experiences cracking at ~50% cold work. • Identify the amount of cold work in each stage and the final diameter of each stage as appropriate. • You can only cold work and anneal the specimen (no hot work allowed!). (See figures below) 6 Mat E 202 Fall 2013 Assignment #4 Solution Assigned: Oct. 18, 2013 Due: Nov. 1, 2013 Solution: Required cold-worked steel’s mechanical properties: df = 10 mm, Tensile strength ≥ 850 MPa, %Elongation ≥ 12 Given: d0 = 15.2 mm, %CW0 = 0 (fully annealed), %CWmax = 50 For the required reduction in area: This is around 2.62 times maximum possible cold work so the operation should be done in multiple stages. From the attached graphs: For UTS ≥ 850 MPa: %CW > 25 and For Elongation ≥ 12%: %CW < 25 Therefore solve for intermediate diameter for a part with: df = 10 mm and %CW = 25: di = 11.55 mm Checking the %cold work for drawing the original diameter to the intermediate diameter: Which is less than 50 the maximum allowable %CW. Therefore the following forming process can be used to shape the rod: 1. CWi = 42.3% : d0 = 15.2 mm di = 11.6 mm 2. Full annealing: di = 11.6, CW = 42.26% 0% 3. CWf = 25% : di = 11.6 mm df = 10.0 mm; TS ≥ 850 MPa, %El ≥ 12 7 Mat E 202 Fall 2013 Assignment #4 Solution Assigned: Oct. 18, 2013 Due: Nov. 1, 2013 INTENDED LEARNING GOALS FOR THIS ASSIGNMENT YOU SHOULD BE ABLE TO: Declarative knowledge Identify four types of point defects. Describe dislocation motion. Identify variables which affect diffusion. State the four strengthening mechanisms. Define hot and cold working. Procedural knowledge Show how an applied stress is resolved into a shear stress. Calculate the diffusion constant and activation energy for tabulated diffusivity data. Use the effective depth approximation for non-steady state diffusion. Calculate strength based on grain size (Hall-Petch relation). Draw strength and ductility curves as a function of annealing temperature. Determine amount of cold work required to produce desired properties. Conditional knowledge Analyze relative importance of strengthening mechanisms in different metal alloys. Determine the effect of certain variables on diffusivity. Potential reversibility of strengthening mechanisms. Reflective knowledge Explain the correlation between atomic structure and energy of defects and diffusivity. Explain relationships between atomistic processes and property changes during strengthening. 8

Help

0 thoughts on “Mats University Distance Education Assignments Meaning”

    -->

Leave a Comment

Your email address will not be published. Required fields are marked *