Final Exam Wednesday, March 21, noon to 3:00 pm (160 points total) 1. TTT Diagrams A U.S. steel producer has four quench baths, used to quench plates of eutectoid steel to 700 C, 590 C, 350 C, and 22 C respectively. Using the TTT diagram below, advise the company on how they can produce steel with the following microstructures. Assume that each bath will instantaneously allow the steel to reach the bath temperature. (8 pts., 2 pts. each) a. 50% fine pearlite, 12.5% bainite, 37.5% martensite. 590 C for 5 seconds, 350 C for 50 seconds, cool to room temperature. b. 50% coarse pearlite, 50% martensite. 700 C for 8,000 seconds, cool to room temperature. c. 50% bainite, 50% coarse pearlite. 700 C for 8,000 seconds, 350 C for at least 200 seconds, cool to room temperature. d. 85-95% fine pearlite, 5-15% coarse pearlite. 700 C for 1,000 seconds, 590 C for at least 10 seconds, cool to room temperature. e. The company makes a cylinder of eutectoid steel of radius L. Due to heat transfer, the entire cylinder will not cool at the same rate. The cooling rates are as follows: at the surface (a), the cooling rate is 200 C /s, at a distance (b) into the cylinder, the cooling rate is 140 C /s, and at a distance (c) into the cylinder the cooling rate is 30 C /s. Please draw the microstructure as a function of distance into the cylinder. (12 pts)
a b c 100% Martensite 100% Pearlite 100% Martensite Mixture of Martensite and Pearlite 2. Phase diagram: Sketch the following diagram in your bluebook. ε + β a. Determine the melting point of Bi and Pb. Mark it on the diagram. (2 pts) T m (Bi)~ 270C T m (Pb)~325C
b. Write down any invariant points. State what type of invariant point it is, and write down the reaction and temperature (ex: L >α+β). (4 pts) Peritectic: L + (Pb) -> (epsilon Pb) 185.0 o C Eutectic: L -> (epsilon Pb) + (Bi) 125.9 o C c. Label all the phase fields. (5 pts) d. Calculate the number of degrees of freedom for all phase fields. (5 pts) Single phase field = α, β, ε, and Liquid C=2 (Bi + Pd = 2 components), P=1, N=1 F = 2-1+1 = 2 Two phase field = α + ε, α+ Liquid, β+liquid, ε+ Liquid, β+ ε C=2, P=2, N=1 F = 2-2+1 = 1 e. Determine the phase composition(s) at 100 C for an overall composition of 50%Pb. (4 pts) The tie line is illustrated in the above figure. α phase is 99%Bi-1%Pd ε phase is 39%Bi-61%Pd 3. Stress and Strain A metal rod with the length 0.4 m and the radius 0.01 m has the following stress-strain curve: a. Determine the Young s Modulus. (4 pts) The slope of the linear portion of the plot gives the elastic modulus. From the inset figure, E = Δσ/Δε ~ 60MPa/.001 = 60GPa b. Determine the yield strength for a strain offset of 0.002. (4 pts) The yield strength is given by the intersection of the curve with the parallel offset line featured in the figure above. From the figure, σ y = 90MPa c. Determine the maximum load that can be sustained by the cylinder. (6 pts)
The tensile strength is given by the maximum of the curve. The figure gives a value of σ ~ 1100MPa. The maximum load is then given by F= σa 0 = (1100MPa)*π(0.01m) 2 = 3.45 10 5 N d. The rod experiences a tension perpendicular to the circular end with a stress of 160MPa. Using the modulus of elasticity you found in part a, and assuming entirely elastic deformation, calculate the elongation. (6 pts) σ = εe = (Δl/l 0 )*E Δl = σl 0 /E = (160MPa)(0.4m)/60GPa = 0.0016m 4. Failure a. Describe qualitatively the difference between brittle failure and ductile failure at the fracture interface of a metal. (2 pts) Brittle failure usually results in a relatively flat interface, with a cross sectional area comparable to that of the original sample; while ductile failure usually involves necking and a very small fracture interfacial area. b. For the two cracks below under equal nominal applied tensile stress along the y axis, compute the ratio of the maximum stress at the tip of crack A to crack B, and state which one is larger. (8 pts) y-axis ( ) ( ) ( ) x-axis Crack B has a larger maximum stress.
c. Fracture toughness in a brittle material is given by the expression: where E is the modulus of elasticity, the specific surface energy, the critical stress required for crack propagation, and is one half the length of an internal crack. Suppose now that crack A in part b is in material A with E = 100 MPa, and crack B from part b is in material B with E = 50 MPa. If the two cracks retain the same geometry as in part b (length, radius of curvature), and all other parameters not stated here being equal, which material will have a higher critical stress, and which one will have higher fracture toughness? Explain your reasoning. (10 pts) The fracture toughness is independent of the length of the internal crack which is obvious once the expression for critical stress is substituted into the expression for fracture toughness, thus material A will have a higher fracture toughness due to its higher elastic modulus. ( ) Material A also has a higher critical stress. 5. Bonding Answer the following questions with arguments related to bonding. a. Why is H 2 0 more dense as a liquid than a solid? (8 pts) There is extensive intermolecular hydrogen bonding between water molecules in the solid phase. The hydrogen bonds give an ordered and rigid, but open final structure of ice. The open structure is why ice is less dense than liquid water. Below is the structure of Kevlar, a high strength, low density material commonly used for applications in body armor. In bold represent is a monomer unit, dashed lines indicate hydrogen bonds.
a. What is the dominant form of bonding between carbon atoms in the backbone? (2 pts) Covalent, sp 2 hybridized HC=CH b. How does this bonding along the backbone contribute to the high tensile strength of Kevlar? (2 pts) Hybridized bonding, strong bonds, highly directional leading to rigid backbone c. In the C-N bond, will the electron density be higher on the nitrogen or the carbon? (2 pts) High electron density on N because more electronegative d. Do you expect Kevlar to have a more crystalline or amorphous intermolecular structure? (2 pts) Crystalline, ordered e. What kind of intermolecular bonding is responsible for the structure you chose in part (e)? (2 pts) Large number of hydrogen bonds between the N- and O- groups provide ordering between chains f. Do you expect that Kevlar is a thermoplastic or a thermoset? (2 pts) Thermoset 6. Planes a. Draw the following planes ( ), ( ) and (210). For reach plane draw the corresponding perpendicular direction on the same coordinate axis. (3 pts for each plane and direction)
- - - - - - b. For a BCC crystal system, list the primary and secondary families of slip planes. (4 pts) Primary={110}, Secondary={211} ( {311} also acceptable) c. Draw the BCC crystal with the secondary family of slip planes. Draw the shortest atom translation in this slip plane. What is the family of slip directions in this slip plane? (7 pts) < 11> drawn above in red 7. Deformation Cold Working A non-cold-worked 1040 steel cylindrical bar having an initial diameter of 25.1 mm is to be cold worked by drawing in order to reduce the cross-sectional area such that the final diameter is 22.5mm. Figures 7.19a (left) and 7.19c (right) from the text are provided.
a. Determine the ductility (%EL) and yield strength of the steel rod after a single drawing process reducing the diameter from the initial 25.1mm to the final 22.5mm. (8 pts) First determine the percent cold work,%cw: 2 2 do dd 2 2 Ao A d 2 2 do d d % CW 100% 100% 100% 2 2 Ao d d o o 2 25.1mm 22.5mm 2 25.1mm 2 2 % CW 100% 19.64% Reading the figures provided with the %CW found gives us the ductility and yield strength. For %CW = 19.64% (~20%) we find: %EL = 12.5% and σ y = 740MPa b. If a recrystallization heat treatment were performed on the bar after the processing in part a, to what would the mechanical properties (ductility and yield strength) change? (2 pts) The recrystallization process returns the mechanical properties to the non-cold-worked values (those corresponding to 0%CW) %EL = 26% σ y = 450MPa c. The final 1040 steel rod with a 22.5mm diameter must have a yield strength of at least 600 MPa and a ductility of at least 20%. The yield strength found in part a from a single drawing process is above the required value. However, the ductility is below the required 20%. In order to produce the 1040 steel rod with a 22.5mm diameter with the required properties an alternative process must be used. In this case we will draw the steel to some intermediate diameter, perform a recrystallization heat treatment and then draw the
steel rod to the desired final diameter. Determine the intermediate diameter to which the steel must be drawn before recrystallization in order for it to have the required mechanical properties after the second cold-work draw. (10 pts) The effect of the first draw process is not of concern due to the recrystallization heat treatment which follows the draw returning the mechanical properties to the non-coldworked values. We must examine the figures to determine a %CW which results in both desired values for the mechanical properties for the second drawing process. The yield strength is not at or above 600MPa until at %CW > 4%. The ductility is not at or above 20% until at a %CW < 4% This shows us that the second draw process must be conducted from an intermediate diameter to the final 22.5mm diameter such that %CW = 4% 2 2 do dd 2 2 Ao A d 2 2 do d d % CW 100% 100% 100% 2 2 Ao d d o o 2 dd 22.5mm 22.5mm do 22.96mm % CW 4% 1 0.04 1 1 100% 100% The entire process would be a cold-work draw from 25.1mm to 22.96mm followed by a recrystallization heat treatment which is finally followed by a cold- work draw from 22.96mm to 22.5mm. 8. Dislocations and Plastic Deformation a. If the total length of the dislocations contained in 10 cm 3 sample of a material is equal to the equatorial circumference of the Earth, what is the dislocation density of that sample? The circumference of the Earth around the equator is about 4 x 10 4 km. (3 pts) Dislocation density is defined as the number of dislocation intersecting a randomly chosen area in a sample (number per area) which is the same as the sum of the length of all dislocations contained in a given volume (total length per volume). We find: b. You have an ideal material that has no dislocations. Can this material elastically deform? (1 pt) Can it plastically deform? (1 pts) Elastic deformation does not depend on the presence of dislocations, so the material can elastically deform. However, plastic deformation can only occur if dislocations are present and able to move, so this material cannot plastically deform unless new dislocations are formed. c. You have a material that has many dislocations. The material has been processed such that fracture occurs before there is enough stress to cause the dislocations to move. Can this material elastically deform? (1 pt) If the nucleation of new dislocations is suppressed can the material plastically deform? (1 pts)
Elastic deformation occurs without the motion of dislocations, so it may deform elastically. However, due to fracture before dislocation movement the material cannot plastically deform. d. You have a ceramic material with some non-zero concentration of dislocations. You also have a metal with the same non-zero concentration of dislocations. Which material do you expect to plastically deform easier (with a lower applied load)? (1 pt) Briefly explain why? (1 pts) I would expect the metal to deform easier. Since there is an equal amount of dislocations, the material which allows for easier dislocation motion will plastically deform more easily. Due to covalent or ionic bonding in ceramics which hinder dislocation motion greater than metallic bonding, dislocations are generally harder to move than in metals. Use the following figure showing a plane of a crystal with a dislocation for parts e h. The dislocation line runs perpendicular to the plane shown. e. What type of dislocation is depicted in the figure? (1 pt) This is an edge dislocation. f. Does the Burgers vector for this dislocation run parallel or perpendicular to the dislocation line? (2 pts) The Burgers vector for edge dislocations run perpendicular to the dislocation line. In this case the Burgers vector would point to the right in the plane (using a clockwise Burgers loop) while the dislocation line is perpendicular to the plane. g. Imagine there is an applied load such that the dislocation feels a net force to the right. Draw the figure depicting the same crystal and dislocation such that the dislocation has moved 2 steps to the right. (2 pts) h. Draw the figure depicting the same crystal after the dislocation has escaped to the right side assuming no more atoms exist past the edge of the figure. (2 pts)
i. A Force, F, is applied to the crystal from parts e-h at an angle inclined 30 from the normal to the slip plane (see diagram below). Slip can occur as described above along the slip plane with a slip direction to the right (90 off the normal to the slip plane ). F is the only force which has an effect on the dislocation. If the load is removed the dislocation feels no shear stress. The cross-section over which the force is applied has an area of 100 mm 2. The yield strength of the material is 200MPa. Find the critical resolved shear stress for the slip system? (2pts) If the applied force is 10,000N will the sample deform plastically? (1 pt) For plastic deformation to occur, the resolved shear stress on the slip plane must meet or exceed the critical resolved shear stress. We are given the inclination of the plane, ɸ=30. We can find the inclination of the slip direction for this system: λ = 90-30 = 60. crss y coscos max crss 200MPa cos 30 cos 60 86.6MPa F 10000N 100MPa 2 A 100mm The applied load is half the yield strength so the sample will not plastically deform due under the 10000N load. We can test this comparing resolved shear stress as well: R 100MPa cos 30cos 60 43.3MPa From R crss we can again see that plastic deformation will not occur for the applied force of 10000N. j. A simple cubic crystal with lattice parameter a = 0.3 nm is subject to a shear stress which causes 150 edge dislocations moving along the same slip plane to escape the crystal to
the right along a lattice direction. How far does the new step which has formed protrude beyond the original boundary of the crystal? (1 pts) Every dislocation that escapes increases the protrusion size by the lattice parameter in the direction of dislocation movement. For a simple cubic crystal all the lattice parameter are the same so the new step protrudes 150 x a = 1