Chapter 11: 상태도 (Phase Diagrams) 학습목표 2 개의원소가결합이되면평형상태는어떤상태일까? 특히, 우리가다음과같은요소를정한다면.. -- 조성 (e.g., wt% Cu - wt% Ni), -- 온도 (T, temperature ) 몇개의상이형성될까? 각각의상은어떤조성이될까? 각각의상의양은얼마일까? Phase A Phase B Nickel atom Copper atom Chapter 11-1
Phase Equilibria: Solubility imit 용액 (Solution) 고체, 액체, 가스, 단일상 혼합물 (Mixture) 1개이상의상으로구성 용해한도 (Solubility imit): 단일상에서다른원소가존재할수있는최대조성 Q: 20 C 수용액에서설탕이녹을수있는용해한도는? Temperature ( C) 10 0 8 0 6 0 4 0 20 Answer: 65 wt% sugar. At 20 C, if C < 65 wt% sugar: syrup At 20 C, if C > 65 wt% sugar: syrup + sugar Sugar/Water Phase Diagram Solubility imit (liquid solution i.e., syrup) (liquid) + S (solid sugar) 0 20 40 6065 80 100 C = Composition (wt% sugar) Water Adapted from Fig. 11.1, Callister & Rethwisch 9e. Sugar Chapter 11-2
성분과상 (Components and Phases) 성분 (Components) : 합금에존재하는원소또는화합물 (e.g., Al and Cu) 상 (Phases) : 물리적및화학적으로서로달리구성하고있는재료의영역 (e.g., and β). Aluminum- Copper Alloy Adapted from chapteropening photograph, Chapter 9, Callister, Materials Science & Engineering: An Introduction, 3e. β (lighter phase) (darker phase) Chapter 11-3
Temperature ( C) 100 조성과온도의영향 온도를변화하면상의개수는 : path A to B. 조성을변화하면상의개수는 : path B to D. watersugar system Fig. 11.1, Callister & Rethwisch 9e. 80 60 40 20 0 0 ( liquid solution i.e., syrup) B (100 C,C = 70) 1 phase (liquid) + S (solid sugar) 20 40 60 70 80 100 C = Composition (wt% sugar) D (100 C,C = 90) 2 phases A (20 C,C = 70) 2 phases Chapter 11-4
고용도 ( Solid Solubility) 의기준 Simple system (e.g., Ni-Cu solution) 결정구조전기음성도 r (nm) Ni FCC 1.9 0.1246 Cu FCC 1.8 0.1278 결정구조, 전기음성도, 원자반경이비슷할경우, 높은상호용해도를갖는다. (W. Hume Rothery rules) 니켈과구리는모든요소에서서로에대해완전히용해된다. Chapter 11-5
상태도 (Phase Diagrams) T, C, P 의함수로서상을나타냄.. 기본으로 : - 2 원계시스템 (binary systems): 2 개의성분으로구성. - 독립적변수 : 온도와조성 (T and C) (P = 1 atm 대기압. ) Phase Diagram for Cu-Ni system 1600 1500 1400 1300 1200 1100 1000 0 (liquid) (FCC solid solution) 20 40 60 80 100 2 phases: (liquid) (FCC solid solution) 3 different phase fields: + Fig. 11.3(a), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) wt% Ni Chapter 11-6
전율고용 2원계상태도 Isomorphous Binary Phase Diagram 상태도 : Cu-Ni system. System is: -- 2 원계 (binary) i.e., 2 성분 : Cu and Ni. -- isomorphous i.e., 다른원소에대하여완전한용해도를보여주는상태 ; 0 ~ 100 wt% Ni 에서 상이연장된다. 1600 1500 1400 1300 1200 1100 1000 0 (liquid) (FCC solid solution) 20 40 60 80 100 Fig. 11.3(a), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) Cu-Ni phase diagram wt% Ni Chapter 11-7
상태도 (Phase Diagrams): Rule 1: 온도 (T ) 와조성 (C o ) 를알면 : -- 상태도에서어떤상이존재할수있는지를정의할수있다. Examples: A(1100 C, 60 wt% Ni): 1 phase: B (1250 C, 35 wt% Ni): 2 phases: + Fig. 11.3(a), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) 존재하는상의정의 1600 1500 1400 1300 1200 1100 1000 0 (liquid) B (1250ºC,35) (FCC solid solution) A(1100ºC,60) 20 40 60 80 100 Cu-Ni phase diagram wt% Ni Chapter 11-8
Phase Diagrams: Determination of phase compositions Rule 2: If we know T and C 0, then we can determine: -- the composition of each phase. Examples: Consider C 0 = 35 wt% Ni At T A = 1320 C: Only iquid () present C = C 0 ( = 35 wt% Ni) At T D = 1190 C: Only Solid () present C = C 0 ( = 35 wt% Ni) At T B = 1250 C: Both and present C = C liquidus ( = 32 wt% Ni) C = C solidus ( = 43 wt% Ni) T A 1300 T B 1200 T D 20 (liquid) Cu-Ni system A B D 3235 tie line (solid) 43 30 40 50 C C 0 C Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) wt% Ni Chapter 11-9
Phase Diagrams: Determination of phase weight fractions Rule 3: T 와 C 0 를정의하면 : -- 각상의질량비를정의할수있다. Examples: Consider C 0 = 35 wt% Ni T A 에서 : Only iquid () present W = 1.00, W = 0 T D 에서 : Only Solid ( ) present W = 0, W = 1.00 T B 에서 : W = W = Both and present S R + S R R + S = 0.27 T A 1300 T B 1200 T D 20 (liquid) Cu-Ni system A B R S D 3235 tie line (solid) 43 C 30 40 50 C C 0 Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) wt% Ni Chapter 11-10
The ever Rule 타이라인 (Tie line) 서로열적평형상태의상을연결하는선 때로는등온선이라고도한다. 1300 T B 1200 (liquid) R B tie line S (solid) 각상의비율은? 지랫대법칙을생각하자. (teeter-totter) M M 20 30 C 40 50 C 0 C wt% Ni Adapted from Fig. 11.3(b), Callister & Rethwisch 9e. R S Chapter 11-11
Ex: Cu-Ni Alloy 의냉각 Phase diagram: Cu-Ni system. A 조성의합금에대하여냉각을동반미세구조변화를고려한다. C 0 = 35 wt% Ni alloy (liquid) 1300 : 35 wt% Ni : 46 wt% Ni 120 0 35 32 (solid) A B C 24 D 36 E : 35 wt%ni 43 46 Cu-Ni system : 32 wt% Ni : 43 wt% Ni : 24 wt% Ni : 36 wt% Ni : 35 wt% Ni 110 0 20 Adapted from Fig. 11.4, Callister & Rethwisch 9e. 35 30 40 50 C 0 wt% Ni Chapter 11-12
기계적성질 : Cu-Ni System 고용강화 (solid solution strengthening) 에따른효과 : -- Tensile strength (TS) -- Ductility (%E) Tensile Strength (MPa) 400 300 TS for pure Cu 200 0 20 40 60 80 100 Cu Ni Composition, wt% Ni Adapted from Fig. 11.5(a), Callister & Rethwisch 9e. TS for pure Ni Elongation (%E) 60 50 40 30 %E for pure Cu 20 0 20 40 60 80 100 Cu Ni %E for pure Ni Composition, wt% Ni Adapted from Fig. 11.5(b), Callister & Rethwisch 9e. Chapter 11-13
2 원계공정시스템 (Binary-Eutectic Systems) 2 components Ex.: Cu-Ag system 3 개의단일상지역존재 (,, β) 제한된용해한도 : : mostly Cu β: mostly Ag T E : T E 이하에서는액상X C E : T E 에서의조성 has a special composition with a min. melting T. 1200 1000 600 400 200 0 공정반응 (Eutectic reaction) (C E ) (C E ) + β(c βe ) cooling + (liquid) + β Cu-Ag system + β β T 800 779 C E 8.0 71.9 91.2 C E 20 40 60 80 100 C, wt% Ag Fig. 11.6, Callister & Rethwisch 9e [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]. heating Chapter 11-14
EX 1: Pb-Sn 공정상태도 150 C 에서 40 wt% Sn-60 wt% Pb alloy: 상을정의하시오. Answer: + β -- 각상의조성은? Answer: C = 11 wt% Sn C β = 99 wt% Sn -- 각상의상대적으로존재하는양은? Answer: S C W β - C 0 = = R+S C β - C = W β = = 99-40 99-11 R = R+S 40-11 99-11 = 59 = 0.67 88 C 0 - C C β - C = 29 = 0.33 88 300 200 150 100 0 + 183 C 18.3 R (liquid) + β Pb-Sn system + β 61.9 97.8 11 20 40 60 80 99100 C C 0 Fig. 11.7, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] S C, wt% Sn C β β Chapter 11-15
220 C 에서 40 wt% Sn-60 wt% Pb alloy: 상을정의하시오. Answer: + -- 각상의조성은? Answer: C = 17 wt% Sn C = 46 wt% Sn -- 각상의상대적으로존재하는양은? Answer: W = C - C 0 C - C = = 6 29 EX 2: Pb-Sn 공정상태도 = 0.21 W = C 0 - C C - C = 23 29 46-40 46-17 = 0.79 300 220 200 100 0 + (liquid) 183 C + β Pb-Sn system + β 17 20 40 46 60 80 100 C 0 C R S C Fig. 11.7, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] C, wt% Sn β Chapter 11-16
공정상태도를이용한 미세구조개발 I 예를들어, C 0 < 2 wt% Sn 결과적으로 : 실온에서는 -- 조성 C 0 를갖는 상이다결정으로존재한다. 400 300 200 T E : C 0 wt% Sn : C 0 wt% Sn + (Pb-Sn System) 100 + β Fig. 11.10, Callister & Rethwisch 9e. 0 C 0 10 20 2 (room T solubility limit) 30 C, wt% Sn Chapter 11-17
공정상태도를이용한 미세구조개발 II 예를들어, 2 wt% Sn < C 0 < 18.3 wt% Sn Result: + β 의온도에서는 -- 결정립의다결정이생성되고작은 β- 상의입자들이석출된다. 400 300 200 T E 100 + + β : C 0 wt% Sn : C 0 wt% Sn β Pb-Sn system Fig. 11.11, Callister & Rethwisch 9e. 0 10 20 2 C 0 (sol. limit at T room ) 18.3 (sol. limit at T E ) 30 C, wt% Sn Chapter 11-18
공정상태도를이용한 미세구조개발 III 합금조성이공정조성일때 (C 0 = C E ) 결과 : 공정미세구조를형성 (lamellar structure) -- 와 β 상이판상구조로교대로반복됨. 300 Pb-Sn system 200 T E + 183 C : C 0 wt% Sn + β β Micrograph of Pb-Sn eutectic microstructure 100 0 Fig. 11.12, Callister & Rethwisch 9e. + β β: 97.8 wt% Sn : 18.3 wt%sn 20 40 60 80 100 18.3 C E 97.8 61.9 C, wt% Sn 160 μm Fig. 11.13, Callister & Rethwisch 9e. (From Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.) Chapter 11-19
amellar Eutectic Structure Figs. 11.13 & 11.14, Callister & Rethwisch 9e. (Fig. 11.13 from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.) Chapter 11-20
공정상태도를이용한 미세구조개발 IV 예를들어, 18.3 wt% Sn < C 0 < 61.9 wt% Sn 결과 : 상의초정상와공정미세구조가형성된다. 300 Pb-Sn system 200 T E 100 0 + Fig. 11.15, Callister & Rethwisch 9e. R R + β : C 0 wt% Sn S 20 40 60 80 100 18.3 61.9 97.8 S + β C, wt% Sn β primary eutectic eutectic β T E 의바로위 : C = 18.3 wt% Sn C = 61.9 wt% Sn W S = = 0.50 R + S W = (1- W ) = 0.50 T E 의바로아래 : C = 18.3 wt% Sn C β = 97.8 wt% Sn S W = R + S = 0.73 W β = 0.27 Chapter 11-21
아공정 (Hypoeutectic) & 과공정 (Hypereutectic) Fig. 11.7, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 300 200 T E 100 + + β + β β (Pb-Sn System) (Figs. 11.13 and 11.16 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.) hypoeutectic: C 0 = 50 wt% Sn 0 175 μm Fig. 11.16, Callister & Rethwisch 9e. 20 40 60 80 100 eutectic 61.9 eutectic: C 0 = 61.9 wt% Sn 160 μm eutectic micro-constituent Fig. 11.13, Callister & Rethwisch 9e. C, wt% Sn hypereutectic: (illustration only) β β β β β Adapted from Fig. 11.16, Callister & Rethwisch 9e. (Illustration only) β Chapter 11-22
중간화합물 (Intermetallic Compounds) Fig. 11.19, Callister & Rethwisch 9e. [Adapted from Phase Diagrams of Binary Magnesium Alloys, A. A. Nayeb-Hashemi and J. B. Clark (Editors), 1988. Reprinted by permission of ASM International, Materials Park, OH.] Mg 2 Pb Note: 금속간화합물은상태도에라인으로존재한다 화학양론적조성이므로면이아님 - (i.e. 화합물의조성은고정값이다 ) Chapter 11-23
공석 (Eutectoid) 하나의고상은두개의다른고상으로변환 intermetallic compound S 2 S 1 +S 3 - cementite γ cool + Fe 3 C (For Fe-C, 727 C, 0.76 wt% C) heat 공정, 공석 & 포정 Eutectic, Eutectoid, & Peritectic 공정 (Eutectic): 액체는두개의고체상으로변환 cool + β (For Pb-Sn, 183 C, 61.9 wt% Sn) heat 포정 (Peritectic) - 액체및하나고체상이제 2 의고상으로변환 S 1 + S 2 cool δ + γ (For Fe-C, 1493 C, 0.16 wt% C) heat Chapter 11-24
Eutectoid & Peritectic Cu-Zn Phase diagram Peritectic transformation γ + δ Eutectoid transformation δ γ + ε Fig. 11.20, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 2, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] Chapter 11-25
Iron-Carbon (Fe-C) 상태도 2 important points - Eutectic (A): γ + Fe 3 C - Eutectoid (B): γ + Fe 3 C 1600 δ 1400 1200 1000 800 600 γ γ + (austenite) B γ γ γ γ 1148 C A γ +Fe 3 C 727 C = T eutectoid +Fe 3 C +Fe 3 C Fe 3 C (cementite) 120 μm Result: Pearlite = alternating layers of and Fe 3 C phases Fig. 11.26, Callister & Rethwisch 9e. (From Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.) 400 0 1 2 3 4 5 6 6.7 (Fe) 0.76 4.30 Fig. 11.23, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] C, wt% C Fe 3 C (cementite-hard) (ferrite-soft) Chapter 11-26
γ γ γ γ 아공석강 (Hypoeutectoid Steel) γ γ γ γ γ γ γ γ δ 1600 1400 1200 1000 800 600 γ γ + (austenite) 727 C 400 0 1 2 3 4 5 6 6.7 (Fe) C 0 C, wt% C pearlite 0.76 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C Fe 3 C (cementite) (Fe-C System) Adapted from Figs. 11.23 and 11.28, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] pearlite 100 μm Hypoeutectoid steel 초석페라이트 Adapted from Fig. 11.29, Callister & Rethwisch 9e. (Photomicrograph courtesy of Republic Steel Corporation.) Chapter 11-27
γ γ γ γ W = s/(r + s) W γ =(1 - W ) 아공석강 (Hypoeutectoid Steel) δ 1600 1400 1200 1000 800 600 W pearlite = W γ W = S/(R + S) γ γ + (austenite) r s R S 727 C pearlite 400 0 1 2 3 4 5 6 6.7 (Fe) C 0 0.76 W Fe 3 C =(1 W ) pearlite 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C C, wt% C Fe 3 C (cementite) (Fe-C System) 100 μm Hypoeutectoid steel Adapted from Figs. 11.23 and 11.28, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 초석페라이트 Adapted from Fig. 11.29, Callister & Rethwisch 9e. (Photomicrograph courtesy of Republic Steel Corporation.) Chapter 11-28
Fe 3 C 과공석강 (Hypereutectoid Steel) γ γ γ γ γ γ γ γ γ γ γ γ δ 1600 1400 1200 1000 800 600 γ γ + (austenite) 727 C 400 0 1 C 0 2 3 4 5 6 6.7 (Fe) C 0 C, wt% C pearlite 0.76 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C Fe 3 C (cementite) (Fe-C System) Adapted from Figs. 11.23 and 11.31, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] pearlite 60 μm Hypereutectoid steel proeutectoid Fe 3 C Adapted from Fig. 11.32, Callister & Rethwisch 9e. (Copyright 1971 by United States Steel Corporation.) Chapter 11-29
Fe 3 C γ γ γ γ W γ =x/(v + x) W =(1-W γ ) Fe 3 C W pearlite = W γ W = X/(V + X) δ 1600 1400 1200 1000 W =(1 - W ) Fe 3 C 과공석강 (Hypereutectoid Steel) 800 γ γ + (austenite) V 727 C 600 pearlite + Fe 3 C 400 0 1 C 0 (Fe) C 0 2 3 4 5 6 6.7 0.76 v x pearlite X 1148 C γ + Fe 3 C +Fe 3 C C, wt% C Fe 3 C (cementite) (Fe-C System) 60 μm Hypereutectoid steel proeutectoid Fe 3 C Adapted from Figs. 11.23 and 11.31, Callister & Rethwisch 9e. [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] Adapted from Fig. 11.32, Callister & Rethwisch 9e. (Copyright 1971 by United States Steel Corporation.) Chapter 11-30
Example 11.4 99.6 wt% Fe-0.40 wt% C 강은공석반응선바로밑에존재한다. 다음을정의하시오 : a) Fe 3 C 및 ferrite () 의조성은?. b) 100g 의강에서 cementite (Fe 3 C) 는몇 gram? c) 100g 의강에서 pearlite 및공석페라이트 () 는몇 gram? Chapter 11-31
Solution to Example Problem a) Using the RS tie line just below the eutectoid C = 0.022 wt% C C Fe3 C = 6.70 wt% C b) 지랫대법칙사용 100 g 에서 Fe 3 C 의양 1600 δ 1400 1200 1000 800 600 γ γ + (austenite) R Fig. 11.23, Callister & Rethwisch 9e. [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 727 C 1148 C γ + Fe 3 C S + Fe 3 C +Fe 3 C Fe C (cementite) = (100 g)w Fe3 C = (100 g)(0.057) = 5.7 g 400 0 1 2 3 4 5 6 6.7 C, wt% C C C 0 C Fe C 3 Chapter 11-32
Solution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and realizing that C 0 = 0.40 wt% C C = 0.022 wt% C C pearlite = C γ = 0.76 wt% C 100 g에서의펄라이트양 = (100 g)w pearlite = (100 g)(0.512) = 51.2 g 1600 δ 1400 1200 1000 800 600 γ γ + (austenite) V X 727 C 400 0 1 2 3 4 5 6 6.7 C C γ C, wt% C C 0 Fig. 11.23, Callister & Rethwisch 9e. [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 1148 C 공석페라이트 = 펄라이트양 Fe3C 의양 = 51.2-5.7 = 45.6g 초석페라이트 = 전체페라이트 - 공석페라이트 = 94.3-45.6= 48.8 γ + Fe 3 C + Fe 3 C +Fe 3 C Fe C (cementite) Chapter 11-33
Alloying with Other Elements T eutectoid changes: C eutectoid changes: T Eutectoid (ºC) Ti Mo Ni Si W Cr Mn C eutectoid (wt% C) Ti Si Mo Ni Cr W Mn wt. % of alloying elements Fig. 11.33, Callister & Rethwisch 9e. (From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.) wt. % of alloying elements Fig. 11.34,Callister & Rethwisch 9e. (From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.) Chapter 11-34
Summary 상태도 (Phase diagrams) 다음을결정하는데유용한데이터 : -- 존재하는상의종류및숫자, -- 각상의조성 ( composition), -- 각상의존재질량비 (weight fraction) ( 어떤시스템에서온도및조성주어진상태에서 ) 합금의미세구조에대한영향 -- 합금의조성 -- 열적평형상태를유지하는냉각속도의여부 상태도에서중요한상변태 (phase transformation) 는공정 (eutectic), 공석 (eutectoid) 그리고포정 (peritectic) 이다. Chapter 11-35