Quiz November 2 with solution

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Dec 6, 2023

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1. T/F questions (28 points) True False 1) Cup-and-cone fracture can be observed for brittle failure. (for ductile) F 2) Ductile fracture is accompanied by significant plastic deformation. T 3) Stress is maximum at crack tip. T 4) Sharp corners are desired for reducing fracture in engineering design. ( sharp corner is not desirable. ) F 5) Plane strain fracture toughness is not dependent on thickness. T 6) Stress amplitude is defined as 𝜎 ?𝑎? . ( in the scope of fatigue, the stress amplitude is 𝜎 ?𝑎𝑥 −𝜎 ?𝑖? 2 ) F 7) Precipitation hardening uses fine precipitates to strengthen the material, and it prefers slow cooling. ( slow cooling → coarse precipitates ) F 8) Some metals have no fatigue limit, such as Aluminum. T 9) For metals without fatigue limit, it won’t fatigue under a low cyclic stress. (lecture 17 page 7) F 10) During phase change, the growth stage involves the appearance of very small particles of the new phase. (it is the nucleation stage) F 11) Generally, increasing the carbon content of a steel alloy increases its hardness. T 12) Vacancy diffusion is typically a slower process than interstitial diffusion. T 13) In the second stage heat treatment of precipitation hardening, the alloy is subjected to a constant elevated temperature for an extended period of time. T 14) By rapid cooling like water quenching, martensite will form in aluminum. ( Martensite is the iron carbon ) F 2. Non steady-state diffusion example problem (7 points) Consider the following concentration distribution within a solid solution, which is shown here at time t 0 = 0 seconds. The specimen is annealed at high temperature, which causes the solute to diffuse through the solvent. Draw the distribution at two later times, t 1 and t 2 . Label these curves. Note that t 2 > t 1 > t 0

Two key points: a). The beginning is different. The t3 curve start drops from C0 earlier than t2 b). The end is different. t3 should have reach a further location than t2. 3. Problem (15 points) Consider an iron-carbon alloy with a carbon composition of 1.2 wt% C. If the alloy has a temperature just below the eutectoid temperature of 727 degrees Celsius, calculate the following, making sure to submit your answers as percentages (i.e. values between 0 and 100) and using at least 3 significant figures a) The mass fractions of total ferrite and cementite 𝑊 𝛼 = ____ _82.36_ ____ wt % 𝑊 𝐹? 3𝐶 = ___ 17.64_ ______ wt %

𝑊 𝑎??ℎ𝑎 = 6.7 − 1.2 6.7 − 0.022 = 0.8236 𝑊 𝐹? 3 𝐶 = 1 − 𝑊 𝑎??ℎ𝑎 = 0.1764 b) The mass fraction of pearlite 𝑊 ??𝑎𝑟?𝑖?? = ___ 92.59 ___wt % 𝑊 ??𝑎𝑟?𝑖?? = 6.7 − 1.2 6.7 − 0.76 = 0.9259 c) The mass fraction of proeutectoid (primary) cementite 𝑊 𝐹? 3 𝐶,?𝑟𝑖?𝑎𝑟? = _____ 7.41 ____ wt% 𝑊 𝐹? 3 𝐶,?𝑟𝑖?𝑎𝑟? = 1 − 𝑊 ??𝑎𝑟?𝑖?? = 1 − 0.9259 = 0.0741 d) The mass fraction of eutectoid cementite 𝑊 𝐹? 3 𝐶,???????𝑖? = ___ 10.23 ___ wt % 𝑊 𝐹? 3 𝐶,???????𝑖? = 𝑊 𝐹? 3 𝐶 − 𝑊 𝐹? 3 𝐶,?𝑟𝑖?𝑎𝑟? = 0.1764 − 0.0741 = 0.1023

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4. (20 points) Shown below is the isothermal transformation diagram for a 0.45 wt% C steel alloy. 1) For each of the heat treatment procedure below, provide explanation of the phase transformation during each step. List the microconstituent(s) present in the final microstructure after the heat treatment procedure. (15 points) 2) For the final microstructures obtained from (a) – (e) , rank their tensile strength from low to high. (5 points) (a) Rapidly cool to 725 o C, hold for 10 s, and quench to room temperature. (b) Rapidly cool to 650 o C, hold for 100 s, rapidly cool to 400 o C, hold for 10 s, and quench to room temperature. (c) Rapidly cool to 700 o C, hold for 10 s, rapidly cool to 100 o C, hold for 100 s, rapidly reheat to 600 o C, hold for 10 4 s, and quench to room temperature. (d) Rapidly cool to 600 o C, hold for 1 s, rapidly cool to 400 o C, hold for 10 3 s, and quench to room temperature. (e) Rapidly cool to 400 o C, hold for 10 s, rapidly reheat to 600 o C, hold for 100 s, rapidly reheat to 720 o C, hold for 10 4 s, and quench to room temperature.

Solution: 1). (a) Heat treatment (a): Upon cooling to and being held at 725°C for about 10 seconds, proeutectoid ferrite begins to form. As it is rapidly cooled to room temperature, the remaining austenite begins to transform to martensite at about 300°C; all of this austenite will have transformed martensite at the end of this quench. Furthermore, there are no changes with the proeutectoid ferrite that has already formed; thus the final composition will consist of proeutectoid ferrite and martensite . Heat treatment (b): Upon rapid cooling and holding at 650°C for 100 s, all of the austenite is converted to proeutectoid ferrite and pearlite. By the subsequent rapid cooling and holding at about 400°C for 10 s, no changes occur. During the final cooling to room temperature no additional changes will occur. Thus the final microconstituent is proeutectoid ferrite and pearlite. Heat treatment (c): Upon cooling to and being held at 700°C for about 10 seconds, proeutectoid ferrite begins to form. Upon rapid cooling to and while at 100°C for 100 s, all of the remaining austenite is converted to martensite. During the subsequent rapid heating and holding at about 600°C for about 10 4 s, all the martensite will transform to tempered martensite; and during the final cooling to room temperature, no additional changes will occur. Thus the final microconstituent is proeutectoid ferrite and tempered martensite. Heat treatment (d): As the alloy is cooled to and held at 600°C for about 1 seconds, some of the alloy transforms to proeutectoid ferrite and pearlite. During the rapid cooling and holding at 400°C for about 10 3 seconds, all the remaining austenite will transform to bainite. Since, by this time all of the austenite has transformed, no changes will occur during the final rapid cooling to room temperature. Thus, the final microstructure will consist of bainite, proeutectoid ferrite, and pearlite. Heat treatment (e): As the alloy is cooled to and held at 400°C for about 10 seconds, approximately 50% of the alloy transforms to bainite. By the rapid heating to 600°C and maintaining the alloy for about 100 seconds at this temperature, all the remaining austenite will transform to proeutectoid ferrite and pearlite. By rapid heating and holding at 720 o C for about 10 4 s, all the bainite and pearlite will transform to spheroidite. No changes will occur during the final rapid cooling to room temperature. Thus, the final microstructure will consist of proeutectoid ferrite and spheroidite. (Full points can be given if the answer is spheroidite) 2). Ductility ranking from low to high: (e) < (b) < (d) < (c) < (a)

5. Problem (16 points) Two complete Jominy end-quench hardness curves for steels A and B are shown below. Thermal conductivity of steels A and B are the same. 1) Which steel has higher hardenability and why? (4 points) 2) Which steel is more likely to contain more alloying elements and why? (4 points) 3) Which steel will have more pearlite present at the center of a bar 2 inches in diameter and why? (4 points) 4) Which schematic below, (a) or (b), best describes the relative placement of steel A and B in isothermal transformation (TTT) diagram and why. (4 points) Solution: 1) A. At close to the quenched end, ‘A’ exhibits higher hardness, and A also maintains overall higher hardness vs. ‘B’. 2) A. More and higher alloying will increase hardenability. 3) B. B has lower hardenability and thus more pearlite might have formed. A B B A T t T t (a) (b)

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4) (b)

6. (14 points) An alloy plate contains five elliptical flaws as depicted in the figure below. Each flaw has the same radius of curvature at their tips. It is known that this metal alloy has a plane-strain fracture toughness of 90 MPa-m1/2 and that the value of Y (dimensionless geometry factor) is 1.2. Given the flaw sizes shown in the figure, determine the following questions. (1) Which crack is most likely to propagate under loading of F? (6 points) (2) Whether the plate will fracture when a tensile stress of 2000 MPa is applied along the direction of F? (8 points) Solution: (1) D is most likely to propagate, because flaws whose major axis is normal to tensile loads tend to propagate. Also, the magnified stress is higher at the tip of a longer flaw. (2) 𝐾 𝐼𝐶 = 𝑌𝜎√ 𝜋𝑎 a is equal to the crack length of flaw D, a = 0.5/2 mm 𝑌𝜎√ 𝜋𝑎 = 1 .2 × 2000 𝑀𝑃𝑎 × √ 3 .14 × 0 .25 × 10 −3 𝑚 = 67 .2 𝑀𝑃𝑎√ 𝑚 Since this value is less than K IC = 90 MPa-m 1/2 , fracture will not occur.

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