학습목표
• 2개의 원소가 결합이 되면 평형상태는 어떤 상태일까?
• 특히, 우리가 다음과 같은 요소를 정한다면..
-- 조성 (e.g., wt% Cu - wt% Ni), -- 온도 (T, temperature)
몇 개의 상이 형성될까 ?
각각의 상은 어떤 조성이 될까 ? 각각의 상의 양은 얼마일까?
Chapter 11: 상태도 (Phase Diagrams)
Phase B Phase A
Nickel atom
Phase Equilibria: Solubility Limit
Q:
20°C 수용액에서 설탕이 녹을 수 있는 용해한도는?Answer: 65 wt% sugar .
At 20°C, if C < 65 wt% sugar:
syrup
At 20°C, if C > 65 wt% sugar:
syrup + sugar
65
• 용해한도(Solubility Limit):
단일 상에서 다른 원소가 존재할 수 있는 최대 조성
Sugar/Water Phase Diagram
S uga r
T em p er at u re ( ° C)
0 20 40 60 80 100 C = Composition (wt% sugar)
L
(liquid solution i.e., syrup)
Solubility
Limit L
(liquid)
+ S
(solid sugar) 20
4 0 6 0 8 0 10 0
W at er
Adapted from Fig. 11.1, Callister & Rethwisch 9e.
• 용액 (Solution) – 고체, 액체, 가스, 단일상
• 혼합물 (Mixture) – 1개 이상의 상으로 구성
• 성분 (Components) :
합금에 존재하는 원소 또는 화합물
(e.g., Al and Cu)• 상 (Phases) :
물리적 및 화학적으로 서로 달리 구성하고 있는 재료의 영역 (e.g.,
α and β).
Aluminum- Copper Alloy
성분과 상 (Components and Phases)
α (darker phase)
β (lighter phase)
Adapted from chapter- opening photograph, Chapter 9, Callister, Materials Science &
Engineering: An
Introduction, 3e.
70 80 100 60
40 20
0
T em per atur e ( ° C)
C = Composition (wt% sugar)
L
(
liquid solution i.e., syrup)
20 100
40 60 80
0
L
(liquid)
+ S
(solid sugar)
조성과 온도의 영향
• 온도를 변화하면 상의 개수는 : path A to B.
• 조성을 변화하면 상의 개수는 : path B to D.
water- sugar system
Fig. 11.1, Callister &
Rethwisch 9e.
D (100°C,C = 90)
2 phases
B (100°C,C = 70)
1 phase
A (20°C,C = 70)
2 phases
고용도( Solid Solubility)의 기준
결정구조 전기음성도 r (nm)
Ni FCC 1.9 0.1246
Cu FCC 1.8 0.1278
• 결정구조, 전기음성도, 원자 반경이 비슷할 경우, 높은 상호 용해도를 갖는다. (W. Hume – Rothery rules)
Simple system (e.g., Ni-Cu solution)
• 니켈과 구리는 모든 요소에서 서로에 대해 완전히 용해된다.
상태도 (Phase Diagrams)
• T, C, P 의 함수로서 상을 나타냄. .
• 기본으로:
- 2원계 시스템 (binary systems): 2개의 성분으로 구성.
- 독립적 변수: 온도와 조성 (T and C) (P = 1 atm 대기압. )
Phase Diagram
for Cu-Ni systemFig. 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.)
• 2 phases:
L (liquid)
α (FCC solid solution)
• 3 different phase fields:
L
L + α α
wt% Ni
20 40 60 80 100 1000 0
1100 1200 1300 1400 1500 1600
T(°C)
L (liquid)
α
(FCC solid
solution)
Cu-Ni phase diagram
전율 고용 2원계 상태도
Isomorphous Binary Phase Diagram
• 상태도:
Cu-Ni system.
• System is:
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.)
-- 2원계 (binary)
i.e., 2성분 :Cu and Ni.
-- isomorphous
i.e., 다른 원소에대하여 완전한 용해도를 보여주는 상태; 0 ~ 100 wt% Ni에서 α상이
연장된다.
1000 0 20 40 60 80 100 wt% Ni
1100 1200 1300 1400 1500 1600
T(°C)
L (liquid)
α
(FCC solid
solution)
wt% Ni
20 40 60 80 100 1000 0
1100 1200 1300 1400 1500 1600
T(°C)
L (liquid)
α
(FCC solid
solution)
Cu-Ni phase diagram
상태도 (Phase Diagrams) : 존재하는 상의 정의
• Rule 1: 온도(T )와 조성 (C o )를 알면:
-- 상태도에서 어떤 상이 존재할 수 있는지를 정의할 수 있다.
• Examples:
A(1100°C, 60 wt% Ni):
1 phase: α
B (1250°C, 35 wt% Ni):
2 phases: L + α
B ( 1250º C ,35)
A(1100ºC,60)
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
20 1200 1300
T(°C)
L (liquid)
α (solid)
30 40 50
Cu-Ni system
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:
T A A
35 C 0 32 C L
At TA
= 1320°C:Only Liquid (L) present C
L = C 0
( = 35 wt% Ni)
At T
B
= 1250°C:Both
α and L present
C
L
= C liquidus( = 32 wt% Ni) C α
= C solidus( = 43 wt% Ni)
At TD
= 1190°C:Only Solid (
α) present
Cα = C 0 ( = 35 wt% Ni) Consider C 0 = 35 wt% Ni
T D D
tie line
4 C 3 α
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.)
T B B
• Rule 3: T 와 C 0 를 정의하면:
-- 각 상의 질량비를 정의할 수 있다.
• Examples:
T A 에서
: Only Liquid (L) presentW L = 1.00, W
α
= 0T D 에서
: Only Solid (α ) present
W L
= 0, Wα
= 1.00Phase Diagrams :
Determination of phase weight fractions
wt% Ni
20 1200 1300
T(°C)
L (liquid)
α (solid)
3 0 4 0 5 0
Cu-Ni system
T A A
35 C 0 32 C L T B B
T D D
tie line
4 C 3 α R S
T B 에서
: Bothα and L present
= 0.27
W L = S R + S
W α = R R + S
Consider C 0 = 35 wt% Ni
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.)
• 타이 라인 (Tie line) –서로 열적 평형상태의 상을 연결하는 선– 때로는 등온선이라고도 한다.
The Lever Rule
각 상의 비율은 ?
지랫대 법칙을 생각하자.
(teeter-totter)
M L M α
R S
wt% Ni
20 1200 1300
T(°C)
L (liquid)
α (solid)
3 0 4 0 5 0
T B B
tie line
C 0
C L
C α S R
Adapted from Fig. 11.3(b),
Callister & Rethwisch 9e.
wt% Ni
20 120 0 130 0
3 0 4 0 5 0
110 0
L (liquid)
α (solid) T(
°C)
A
35 C 0
L: 35 wt%Ni
Cu-Ni system
• Phase diagram:
Cu-Ni system.
Adapted from Fig. 11.4, Callister & Rethwisch 9e.
• A 조성의 합금에 대하여 냉각을 동반 미세 구조 변화를 고려한다.
C 0 = 35 wt% Ni alloy
Ex: Cu-Ni Alloy의 냉각
35 46 32 43
α: 43 wt% Ni L: 32 wt% Ni α: 46 wt% Ni L: 35 wt% Ni B
C
E L: 24 wt% Ni
α: 36 wt% Ni
24 D 36
α: 35 wt% Ni
기계적 성질 : Cu-Ni System
• 고용 강화(solid solution strengthening)에 따른 효과:
-- Tensile strength (TS) -- Ductility (%EL)
Adapted from Fig. 11.5(a), Callister & Rethwisch 9e.
T ens ile S tr engt h (M P a)
Composition, wt% Ni
Cu Ni
0 20 40 60 80 100 200
300 400
TS for pure Ni
TS for pure Cu
E longat ion (% EL )
Composition, wt% Ni
Cu 0 20 40 60 80 100 Ni 20
30 40 50 60
%EL for pure Ni
%EL for pure Cu
Adapted from Fig. 11.5(b),
Callister & Rethwisch 9e.
2 components
has a special composition with a min. melting T.
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.].
2원계 공정시스템
(Binary-Eutectic Systems)
• 3 개의 단일상 지역 존재 (L, α, β)
• 제한된 용해 한도:
α: mostly Cu β: mostly Ag
• T E : T E 이하에서는 액상X : T E 에서의 조성
• C E
Ex.: Cu-Ag system
Cu-Ag system
L (liquid)
α L + α L + β
β
α + β
C, wt% Ag
20 40 60 80 100
200 0 1200
T(°C)
400 600 800 1000
C E
T E 8.0 71.9 91.2
779°C
cooling
• 공정 반응 ( Eutectic reaction)
L(C E ) α(C αE ) + β(C βE )
L + α
L + β
α + β
200
T(°C)
18.3
C, wt% Sn
20 60 80 100
0 300
100
L (liquid)
α 183°C
61.9 97.8 β
• 150°C에서 40 wt% Sn-60 wt% Pb alloy: 상을 정의하시오.
Pb-Sn system
EX 1: Pb-Sn 공정 상태도
Answer:
α+ β -- 각 상의 조성은?
-- 각상의 상대적으로 존재하는 양은?
150
40 C 0 11
C α
99 C β
R S
Answer: C α = 11 wt% Sn C β = 99 wt% Sn
W α = C β - C 0 C β - C α
= 99 - 40
99 - 11 = 59
88 = 0.67 S
R+S =
W β = C 0 - C α C β - C α R =
R+S
= 29
= 0.33
= 40 - 11 Answer:
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
Answer: C α = 17 wt% Sn -- 각 상의 조성은?
L + β
α + β
200
T(°C)
C, wt% Sn
20 60 80 100
0 300
100
L (liquid)
α β
L + α
183°C
• 220°C에서 40 wt% Sn-60 wt% Pb alloy: 상을 정의하시오.
Pb-Sn system
EX 2: Pb-Sn 공정 상태도
-- 각상의 상대적으로 존재하는 양은?
W α = C L - C 0
C L - C α = 46 - 40 46 - 17
= 6
29 = 0.21
W L = C 0 - C α
C L - C α = 23
29 = 0.79
40 C 0
46 C L 17
C α
220 R S
Answer: α + L
C L = 46 wt% Sn
Answer:
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.]
• 예를 들어,
C 0 < 2 wt% Sn
• 결과적으로: 실온에서는 -- 조성 C 0 를 갖는 α상이 다결정으로 존재한다.
공정 상태도를 이용한 미세 구조 개발 I
0
L + α
200
T( ° C)
C, wt% Sn
10 2
20 C
0300
100
L α
30
α + β
400
(room T solubility limit)
T
E(Pb-Sn
System)
α L L: C
0wt% Sn
α : C
0wt% Sn
Fig. 11.10, Callister &
Rethwisch 9e.
• 예를 들어,
2 wt% Sn < C 0 < 18.3 wt% Sn
• Result:
α + β 의 온도에서는 -- α 결정립 의 다결정이 생성되고
작은 β-상의 입자 들이 석출된다.
Fig. 11.11, Callister &
Rethwisch 9e.
공정 상태도를 이용한 미세 구조 개발 II
Pb-Sn system
L + α
200
T(°C)
C , wt% Sn
10
18.3 20 0
C
0300
100
L
α
30
α + β
400
(sol. limit at T
E) T
E2
(sol. limit at T room )
L α L: C
0wt% Sn
α β
α : C
0wt% Sn
• 합금 조성이 공정 조성일 때 (C 0 = C E )
• 결과: 공정 미세구조를 형성 (lamellar structure) -- α 와 β 상이 판상 구조로 교대로 반복됨.
공정 상태도를 이용한 미세 구조 개발 III
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.)
160 μm
Micrograph of Pb-Sn eutectic
microstructure
Pb-Sn system
L + β
α + β
200
T(°C)
20 60 80 100
0
300
100
L
α β
L + α
183°C
40 T
E18.3
α : 18.3 wt%Sn
97.8 β: 97.8 wt% Sn
C
EL: C
0wt% Sn
Lamellar 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.)
• 예를 들어, 18.3 wt% Sn < C 0 < 61.9 wt% Sn
• 결과: α 상의 초정상와 공정 미세구조가 형성 된다.
공정 상태도를 이용한 미세 구조 개발 IV
18.3 61.9
S R
97.8 R S
primary α eutectic α
eutectic β
W L = (1- W α ) = 0.50 C α = 18.3 wt% Sn C L = 61.9 wt% Sn
S R + S
W α =
= 0.50• T E 의 바로 위:
• T E 의 바로 아래 : C α
= 18.3 wt% SnC β = 97.8 wt% Sn
S R + S
W α =
= 0.73W β = 0.27
Fig. 11.15, Callister &
Pb-Sn system
L + β
200
T(°C)
20 60 80 100
0
300
100
L
α β
L + α
40
α
+β
T
EL: C 0 wt% Sn α L
L α
L + α
L + β α + β
200
C, wt% Sn
20 60 80 100
0 300
100
L
α β
T
E40
(Pb-Sn System)
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.]
160 μm eutectic micro-constituent
Fig. 11.13, Callister &
hypereutectic: (illustration only)
β β β
β β
β
Adapted from Fig. 11.16, Callister & Rethwisch 9e.
(Illustration only)
(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.)
175 μm
α α
α α α
α
hypoeutectic: C
0= 50 wt% Sn
Fig. 11.16, Callister &
Rethwisch 9e.
T(°C)
61.9 eutectic
eutectic: C
0= 61.9 wt% Sn
중간 화합물
(Intermetallic Compounds)
Mg 2 Pb
Note: 금속 간 화합물은 상태도에 라인으로 존재한다 – 화학양론적조성 이므로 면이 아님
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.]
• 공석 (Eutectoid) – 하나의 고상은 두 개의 다른 고상으로 변환
S 2 S 1 +S 3
γ α + Fe 3 C (For Fe-C, 727°C, 0.76 wt% C)
intermetallic compound - cementite
cool heat
공정, 공석& 포정
Eutectic, Eutectoid, & Peritectic
• 공정 (Eutectic): 액체는 두 개의 고체 상으로 변환
L cool α + β (For Pb-Sn, 183°C, 61.9 wt% Sn)
heat
cool heat
• 포정 (Peritectic) - 액체 및 하나 고체상이 제2의 고상으로 변환
S 1 + L S 2
δ + L γ (For Fe-C, 1493°C, 0.16 wt% C)
Eutectoid & Peritectic
Cu-Zn Phase diagram
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.]
Eutectoid transformation δ γ + ε
Peritectic transformation γ + L δ
Iron-Carbon (Fe-C) 상태도
• 2 important points
- Eutectoid (B):
γ ⇒ α + Fe 3
C- Eutectic (A):
L ⇒ γ + Fe 3
CFig. 11.23, Callister
& Rethwisch 9e.
[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted
Fe 3 C ( c em ent it e)
1600 1400 1200
1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite)
γ +L
γ +Fe 3 C
α +Fe 3
Cδ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C = T eutectoid
4.30
Result: Pearlite = alternating layers of α and Fe
3C phases
120 μm
Fig. 11.26, Callister & Rethwisch 9e.
(From Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, 1985.
0.76
B
γ γ γ γ
A L+Fe
3 C
Fe 3 C (cementite-hard)
α (ferrite-soft)
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite)
γ +L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
0 .7 6
아공석 강 (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.]
초석 페라이트 pearlite
100 μm Hypoeutectoid steel α
pearlite
γ γ γ α γ
α α
γ γ γ γ
γ γ γ γ
Chapter 11 - 28
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite)
γ +L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
0 .7 6
아공석 강 (Hypoeutectoid Steel)
γ γ γ α γ
α α
s r
W α = s/(r + s) W γ =(1 - W α
)R S
α
pearlite
W pearlite = W γ W α ’
= S/(R +S) W Fe =(1 – W α’
)3 C
Adapted from Fig. 11.29, Callister & Rethwisch 9e.
(Photomicrograph courtesy of Republic Steel Corporation.)
초석 페라이트 pearlite
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.]
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite)
γ +L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
과공석 강 (Hypereutectoid Steel)
0. 76
C
0Fe
3C
γ γ γ γ γ γ γ γ γ γ γ γ
proeutectoid Fe 3 C
60 μm Hypereutectoid steel
pearlite
pearlite
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.]
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800
600
400 0 1 2 3 4 5 6 6.7
L γ
(austenite)
γ +L
γ + Fe 3 C
α + Fe 3 C
L+Fe 3 C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
Adapted from Fig. 11.32, Callister & Rethwisch 9e.
proeutectoid Fe 3 C
60 μm Hypereutectoid steel
pearlite
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.]
과공석 강 (Hypereutectoid Steel)
0. 76 C
0pearlite Fe
3C
γ γ γ γ
v x V X
W pearlite = W γ W α = X/(V + X)
W Fe =(1 - W α )
3
C’
W =(1-W γ )
W γ =x/(v + x)
Fe
3C
Example 11.4
99.6 wt% Fe-0.40 wt% C 강은 공석반응선 바로 밑에 존재한다. 다음을 정의하시오:
a) Fe 3 C 및 ferrite ( α)의 조성은?.
b) 100g의 강에서 cementite (Fe 3 C) 는 몇 gram?
c) 100g의 강에서 pearlite 및 공석 페라이트 ( α)
는 몇 gram?
Solution to Example Problem
b) 지랫대 법칙 사용
a) Using the RS tie line just below the eutectoid
C α = 0.022 wt% C C Fe
3 C = 6.70 wt% C
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800 600
400 0 1 2 3 4 5 6 6.7
L
γ
(austenite) γ +L
γ + Fe
3C α + Fe
3C
L+Fe
3C
δ
C , wt% C 1148°C
T(°C)
727°C
C 0
R S
C Fe C C α 3
100 g 에서 Fe 3 C의 양
= (100 g)W Fe
3 C
= (100 g)(0.057) = 5.7 g
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.]
Fe 3 C ( c em ent it e)
1600 1400 1200 1000 800 600
400 0 1 2 3 4 5 6 6.7
L
γ
(austenite) γ +L
γ + Fe
3C α + Fe
3C
L+Fe
3C
δ
C , wt% C 1148°C
T(°C)
727°°C
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 α = 0.022 wt% C
C pearlite = C γ = 0.76 wt% C
C 0 V X
C γ C α
100 g에서의 펄라이트양
= (100 g)W pearlite
= (100 g)(0.512) = 51.2 g
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.]
공석 페라이트= 펄라이트 양 – Fe3C 의 양 = 51.2-5.7 = 45.6g
Alloying with Other Elements
• T eutectoid changes:
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.)
T Eu te c to id (º C)
wt. % of alloying elements
Ti
Ni
Mo Si
W
Cr Mn
• C eutectoid changes:
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.)
wt. % of alloying elements