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Exam3Sol - ME 083 Spring 2001 Structure and Properties of...

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Unformatted text preview: ME 083 Spring 2001 Structure and Properties of Solids Hourly Exam #3 W SOZWZONS 0 Time: 50 minutes 0 Closed book, closed notes, closed neighbor o Calculator is allowed 0 Partial credit will be given. Show your work! 0 Useful information and phase diagrams on last three pages. 0 Use the back of sheets for additional space 0 Good Luck! 0 I have neither given nor received unauthorized information while completing this examination Signed: ______________________ Problem 1 ____ [30%] Problem 1 Extra Credit -__... [5%] Problem 2 ____ [20%] Problem 3 ____ [50%] Total .____ [100%] [30%] Problem 1: Heat Treatment and Microstructure Shown in Figure 1 are isothermal transformation diagrams for a hypoeutectoid, a hypereutectoid and a eutectoid steel. 3 (a) [5%] Which of the three diagrams is associated with a composition that is: C (i) Hypoeutectoid: e (ii) Hyperentectoid: A» Cb (iii) Eutectoid: [Extra Credit 5%] How does the start temperature of the martensitic transformation de- pend on alloy carbon content? It is sufficient to indicate the trend. (b) [5%] If a thin specimen of Alloy II (Fig. 1(b)) is quenched instantaneously from 900 "C to 550 "C, held for 1 second, and quenched to room temperature, what phases will be present in the microstructure‘? ”um; cow—2m MAL 99000;: ~52; % 7%; WM? (WHA/«K 50/75 BMW/Z“), 7%; gem/Wm Ami? ML 794w my 75 mew/372':- Mm 77/E’A/Ao/V /s @U/QJCMQD 75 £20” 737me 4,? 50 % aux—2' W/QMZ’ 50 % #4931823” (c) [5%] Sketch the resulting microstructure resulting from the heat treatment in item (0), and label important features. (d) [5%] Calculate the maximum amount of proeutectoid phase that can form when heat treating a plain carbon steel containing 0.4 wt.% C. What are the constituents of this proeutectoid phase? Z745" /% W30 Pfig‘gg‘ fmflr/NS 0? *@&E’ AND Aosrauzrz‘ (y) _ .Q’iéLQLLQ’O, 1 £0.57 foé'I-E— 0,76~0.022— [Lg/f1? (e) [5%] Calculate the maximum amount of proeutectoid phase that can form when heat treating a plain carbon steel containing 1.2 wt.% C. What are the constituents of this proeutectoid phase? WE /% #‘W/Efgfado WM (ZWWAA Fe 3 c ([57491/77/2’)/4N0 swam/2— (y) [(2- 0'?6 = Ha: t 5%“ O. 2 {5:36 5,?- 016 O'OW/ {07 q 6 (f) [5%] Design two heat treatments that would result in a maximum difference in the amount of proeutectoid phase that would be present in the microstructure of the hypereutectoid steel after heat treating. (—5— 1:76. / [6.) Hum WAX/”UH cum/777 a: Pfiamfl Was; WWO H3197 /F ONE flows 7%? EM? Mat/E 72.7 "a #56 A6007 Moo/x Aw max! $404M (”ems F9 Pam EV/QBAWEE (a) M‘n-zl ' , (b) mm," 63'! JUII' _ 1 mi» F tiny ” . . . = . n {1.1 I m m3 10‘ :n3 In‘ 1! I I m m: “‘3‘ In1 rm Tum: (Wt? Tm»: Mrs! (C) Aim m 'l' ...... We 1m . ' ; Lg: " lm‘m ihwr 1.1., u . . ' 11E 1 m In2 151‘ H? In‘ I'mu- MY: Figure 1: Isothermal transformation diagrams for three plain carbon steels: a hypoeutectoid, a hypereutectoid, and a eutectoid. [20%] Problem 2: Line Defects (a) [15%] Calculate the critical resolved shear stress in a mystal if a stress of 170 MPa appiied in the [100] direction is required to move a dislocation in the [111] direction on the (101) plane. fcge=<9yéc¢®v2 [ jfj / __ Dooj.E//T_T _ 11‘ g :30 'L=___ @¢”W ”WW2 W57 V27 @885: I70 7%? #Paiflé’ M65” (b) [5%] Is this a ECG or an FCC crystal? Why? @PIQ-ANE.‘ (/0!) SHP 0/23/700: [H71 {IIO§(///> W”! swarms gag 60: damage W1? 5 HOE Me: ME 77/; Ewes MW wee—2:7 flick/No Dave/W/ ext/0 7Z1; 4 ”0,91%— 77,5— gag/2M Mr?! MMESZ‘ mus/r7 [50%] Problem 3: Phase Diagrams (a) [10%] With reference to the Fe—C phase diagram (Fig. 2), estimate the carbon content of a slowly-cooled (normalized) steel which contains 80 wt.% pearlite and 20 wt.% proeutectoid ferrite. Show you calculations, and state any assumptions. £330”; 77/47 MA M4353 my; £009; GEMS/fl WEN 80 M Z, Peggy? 6<WE W (cow/0a 8920i% Away/IE. May/No. we? 1.70/5: 7an MOVE” 6.90.022. . 3» {at = 0,9;— #Looza/ Q 0,6/ I?» 77/; gag/80w (mm? at 72/537553 /s 4/3907— 0,6/ %&t C, (b) Three invariant reactions occur in the Fe—Fe3C binary system. Write equations which describe these invariant reactions upOn cooling- Give the approximate compositions of the phases involved in each reaction. Name the type of each invariant reaction and write down the temperature at which it occurs. (i) [5%] Fifi/23377; / M 673 ”c L + 95:; .1... (ewes/77w 0,53% 0,00] % 0.x? % (AW (i) [5%] 5/ch , [M ”c L W...... 3; 7L 6; 3C (mas/[7w 4,3 % 214% 6.70 % aesa/ [i] [5%] Lam/[q V272. y ”E? 04+ Fag c (magma/u 0.76% 0.022% 6.20% am (c) [5%] Apply the Gibbs phase rule to any one of the three reactions in (1)). Define your variables, and state the significance of the result for the phase equilibrium occurring at the chosen reaction. 6/883 FZKASE @AE' [OMS/T We: /——>§+f—g3c /' F= C~P+/ $533.) F; @3655: figs—mrv szZ—37L/eo 73: it ”#853 We maze—3 r2442” WSEW/aC/ZFM C: a [gr/arufiU/‘S ME @229 ,4? W75 EM? AND M7 Amy maSWW/N ME We He ((1) [5%] Briefly explain the difference between alloy cemposition and phase composition. Un— der what cenditions are the two equal? MW; Meg/4AA (was/27w 0/: 725’ AWV/oswmy w M: %) _W%}$E (.5V7_,_/@$/7ZQ';U_;; (WPOS/f/‘W 6%: W may/m; fl/M Passe/7‘ films 925%; WWW. 26/407 (ayes/”w AND flax/355‘ @775 37/7044 we; MEL/P764}; //u we: M55“ FOB/ms 0M. (e) Now consider the simp ified (high temperature/ low carbon content) Fe—C phase diagram in Figure 3. (i) [3%] Label all phases in Figure 3. (ii) {7%} Determine the composition and relative amounts of each phase just below the peritectic isotherm for an alloy with composition X3=0.35 wt.% C. (WPOSJ/‘OA/ yi/QU/O .' 0.53 % c, 0' 353;: = ”g.% {I OJ; %C ”C‘- Org—OJ?- I i}: /“0t5—= 0:5" 05’ (iii) [3%] What is the range of alloy compositions that will peritectically transform? 00?; @5053 (ML-'70 C) (iv) [2%] What is the maximum solid solubility of C in J—Fe? 009‘ Mt% G" Ms "C, Useful Information Constants (a) Gas constant: R=8.314 [J [Incl—K] (b) Boltzmann’s constant: k3 = 1.38 x 10‘23 [J /atom—K] Useful Formulas Vector Product ‘ Dot Product Schmid’s Law Twas 1“ ineld (303035) 0050“: Where qt is the angle between the normal to the slip plane and the applied stress direction, and A is the angle between the slip and stress directions. Gibbs Phase Rule F=C—P+2 At constant Pressure: F = C — P + 1 10 The Iron-Carbon Phase Diagram Composition (at% C) 10 15 20 25 2500 1200 1, Austenite 2000 Temperature (’C) l—fl O 8 Temperature PF) 1500 600 Cementite (F930) 1000 400 o ' 1 2 3 4 5 6 6.70 (Fe) l Composition (wt% C) Figure 2: The iron—carbon phase diagram. 11 Tcmpcmmrc (9' ‘i I; : 0. I25 ml. 54- (.‘ X2 = (i. I '1 M. ‘3? C X; = 0.35 “1. rei- (7 I530 . [56“ _____ 1540 'x532;i';::_ ‘ _ S" .: L. 5': 1495’ 2* 7 7 -- 14m g; I440 I420 13' - [yr—Fe) Z ' Auflgflile 14m .330 S 5+ 1 13m Fe OHS 0.“) [HS “.20 [1.25 ”.30 I135 {140 (145 ".50 (1.55 0.“) ".65 Weight perccnlagc autism Figure 3: Simplified peritectic iron—carbon phase diagram. 12 UJU (175 ...
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