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HTM811S - Heat Treatment of Metals 414 - 1st Opp - June 2023 |
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nAmI BIA unIVERSITY
OF SCIEn CE Ano TECHn OLOGY
FACULTY OF ENGINEERING AND THE BUILT ENVIRONMENT
DEPARTMENT OF CIVIL, MINING AND PROCESSENGINEERING
QUALIFICATION: BACHELOROF ENGINEERING IN METALLURGY
QUALIFICATION CODE: 08BEMT
LEVEL: 8
COURSECODE: HTM 811S
COURSENAME: HEAT TREATMENT OF METALS 414
SESSION:June 2023
DURATION: 3 HOURS
PAPER: THEORY
MARKS: 100
EXAMINER{S)
FIRST OPPORTUNITY QUESTION PAPER
Mrs Jaquiline T. Kurasha
MODERATOR:
Prof Josias Van der Merwe
INSTRUCTIONS
1. Answer all questions.
2. Read all the questions carefully before answering.
3. Marks for each questions are indicated at the end of each question.
4. Please ensure that your writing is legible, neat and presentable.
PERMISSIBLE MATERIALS
1. Examination paper.
2. Non-programmable calculator.
THIS QUESTION PAPER CONSISTS OF 6 PAGES (Including this front page and appendix)
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Question 1 [25 marks]
(a)Two samples of steel contain 95% pearlite. With the aid of Fe-Fe3CEquilibrium Diagram
(see Appendix 1) and Lever rule, estimate the carbon content of each sample if one is
known to be:
(i)
hypoeutectoid and the other
[10]
(ii) hypereutectoid.
[10]
(b) Figure Ql features a Time Temperature Transformation (TTT) diagram for a hypoeutectoid
plain carbon steel. The Al critical temperature and A3 critical temperature can be clearly
distinguished in the TTT diagram. With the aid of the Fe-CEquilibrium diagram (see Appendix
1) determine the carbon content (wt.%) in the steel grade featured in Figure Ql.
[5]
600
~00
I,q- :
ffi<100
0.
w
I- 300
400
20
OQ
j
i
!I ~47 ,
I
--1--4-~-!---j.--:l"""'...:;-.,.:?"l,,.....-.,.....-' ;~·-·--+'""1-'--r--;~~
80
100
0
.., ,, ..... 1...
0.5 I 2 5 IO
iot
10)
104
TIME- SECONDS
:o~ l01
Figure Ql. A Time Temperature Transformation (TTT) diagram for a hypoeutectoid plain
carbon steel.
Question 2 [25 marks]
Excellent combinations of hardness, strength, and toughness are provided by bainite. With
the aid of heat treatment facility available, you austenitized a eutectoid steel at 750°C,
quenched it to 250°C and held the steel at 250°C for two hours, and finally permitted the
steel to cool to room temperature. With the aid of the time-temperature diagram (Figure
Q2), answer the following questions:
(i)
Was the required bainitic structure produced in the eutectoid steel? Briefly explain.
Sketch the relevant time-temperature path line.
[5]
(ii) Briefly indicate the phase transformations observed in the steel upon cooling.
[5]
2
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(iii) Indicate the structural phases that will be available in steel at the end of the
treatment.
[5]
(iv) Explain the meaning of the figures (from 14 to 66) on the right vertical axis of the
diagram. How can these figures be used?
[10]
900
800
700
t 600
E 500
::,
r,,
E 400
<.>
0..
E 300
200
100
£>.1
MI s
A-ff M+ r,,
M
10
102
)03
Timc(s)-
+Tp
14
38
40
I
B
42
1 52
57
I
66
M
I
J04
105
J06
Figure Q2. TTT diagram for a eutectoid steel. Note: Yu - is unprocessed (retained) austenite.
Question 3 [25 marks]
(a) A batch of quenched and tempered steel gears was rejected by the customer because of
surface cracks (see a gear after the heat treatment in Figure Q3). The gears were made of
1080 steel grade and quenched in water. With the aid of Time-Temperature Transformation
Diagram in Figure Q3, suggest at least five ways to minimize the risk of cracking upon
treatment.
[15]
3
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Type: 1080
Composition: Fe - 0,79% C - 0.76% Mn Grain sh:e: 6
Au,"nitized at 899°c (1650°F)
eoo
-
'T"'I I II 1111 II 1111
A
111111
1100
I
e---•
I---"-• lOO
1200 .,__A
I,,""
1" -J~ /
I
·- -·r-
,.,.,:;-
-·-
- II 11111 II 11111
-.,-
II
- 38
W
500
1000 I II
'A+
\\,
'-
F+C
40
- 40
- ffi400
D.
- ...'. ... 300
800
iOO --A
...,,_F•C .......,...:..'..r... ,,"...r.i-.,,,,._...
41
- 43
60
- . \\ ...........
- $5
-- -- -·-JJ,.~;..._.
r
=="'»~ - ·-- ---- 200 400
I
-t--f-
H)O 200
- 0
11.o1
1-T DIAGRAM
1 III I
.. ,.t,.t._..l,f.f.,.,,.I1MN7.i(;,
11111111111
I 1111
II
11ill111II
67
-
H
o.a 1 2 5 10
TIME - SECONDS
Figure Q3. A 1080 steel gear after heat treatment and the TTT diagram for a 1080 steel.
(b) Another batch of gears was rejected by the customer due to insufficient surface hardness.
The gears are made of steel grade 1040, quenched and tempered. You prepare a cross
section and examine the harness as a function of distance from the surface (see Table Q3).
What is the reason for the heat treatment defect? Is it possible to rectify the problem?
(Hint: plot the hardness as a function of distance.)
[10]
Table Q3. Hardness distribution along the cross section
Distance from the quenched end, mm
1,5
3
5
15
20
Hardness HRC
40
50
55
52
50
Question 4 [25 marks]
(a) A stainless steel contains 12% Cr, 1.0% Si, and 0.05% C. With the aid of Schaeffler
diagram (Figure Q4), estimate the content of nickel required produce paramagnetic
steel.
[10]
4
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30
i:::
'cf?_ 25
t.n
z·-u 0
+ 'cf?_ 20
0
'cf?_M+
II
.µ
15
i:::
·-:>
::,
10
-O"'
Q)
Q)
z·u- 5
00
Austenite
Martensite
+
Ferrite
10
20
30
40
Chron1ium equivalent = % Cr + % lvlo + 1.5% Si
+ 0.5% Nb
Figure Q4. Schaeffler diagram
(b) Some stainless steels are prone to sensitization due to formation of chromium
carbides. You have observed that your grandmother uses a frying pan made of
stainless steel 304 with 18%Cr, 8% Ni, and 0,10% C. This is what is indicated in the
label. How soon should you expect the failure of the frying pan due to sensitization if
your grandmother cooks pancakes for the whole family every other day?
[10]
(c) In order to efficiently recycle stainless steel scrap, we wish to separate the high-nickel
stainless steel from the low-nickel stainless steel. Design a method for doing this with
the aid of Schaeffler diagram (Figure Q4).
[5]
............................................... End..........................................................
Appendix 1
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2 Pages 11-20 |
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Tcmpcmlure (cC)
1700
1500
1536°c
I.
1300
1100
,-solid solution
(J\\ustcnitc)
J\\ustcnitc +
Lcdcburilc +
Ccn\\1:nlitc
'I
r-solid solu1ion + Fc1C
Ccmcnlilc +
l.cdcburitc
1147"C
a -solid solution (Ferrite)
500
Pcnrlilc +
Ferri le
00 Pc.11litc_/. '
I
1009fF. e '1
I0 llypo-
1 cutcctoid
Pcarli1c +
Ccmcntilc
Pcarli1c +
Ccnxntilc +
l.c&:buritc
(tnuu formed)
r a-solid solution + 1C
21
3
I lypcn:utcdoid
I
I
Steel
723 °C
Ccmcn1i1c +
l.cdcburi1c-
l.cdcburitc
5
Cns1iron
6
6.67
Carbon (\\\\1 %)
I
I
I
Fe-FeJC Equilibrium Diagram
6
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