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M202 F2023
Assignment #6
Due: Dec 6th, 2023 at 4pm
1.
Using the heat treatments and the TTT diagram (
eutectoid
steel) shown below:
Draw
and
Label
i) the final microstructure including microconstituent(s) and phase(s) present
ii) the correct phase morphology (e.g. lamellar or spherical etc.)
Determine
:
iii) the approximate percentages of each microconstituent and phase
3
marks
for each heat treatment (21marks total)
*Assume that each specimen was held at 760°C sufficiently long to reach equilibrium before
applying the cooling conditions shown below*
a)
Cool rapidly to 650
o
C, hold for 1000 s, and then quench to room temperature.
b)
Reheat the specimen in part a) to 700
o
C for 2 days (Hint Spheroidization)
c)
Cool rapidly to 570
o
C, hold for 3000 s and then quench to room temperature.
d)
Cool rapidly to 375
o
C, hold for 100 s, then quench to room temperature.
e)
Rapidly cool to 630
o
C, hold for 10 s, rapidly cool to 265
o
C, hold for 1000 s, and then
quench to room temperature.
f)
Rapidly cool to 250
o
C, hold for 100 s, then rapidly quench to room temperature.
Reheat to 371
o
C for 1 hour and slowly cool to room temperature.
g)
Which of the above heat treatments will give a steel with:
i.
the highest UTS, and why.
ii.
the highest ductility, and why.
iii.
the highest Young’s modulus of elasticity, and why.
NOTE
* Bainite, Tempered Martensite and Pearlite are microconstituents composed of ferrite and
cementite.
Martensite is considered both a microconstituent and a metastable phase*
a)
b)
c)
d)
e)
f)
g)
i)
ii)
iii)
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2
.
A 1045 steel is subjected to the following heat treatments.
Determine
for each heat treatment
(Note
: you do
NOT
have to draw and label)
i) the microconstituent(s) and phase(s) present
- treat primary/proeutectoid ferrite as a
distinct phase.
ii) the approximate percentage (%) of each
PHASE
(consider only the total amount of
ferrite where applicable ) in each heat treatment.
2 marks each (8 marks total)
a) Heat to 900°C and hold long enough for the sample to come to equilibrium. Cool
rapidly to 700°C and hold for 30 s. Quench to room temperature.
b) Heat to 900°C and hold long enough for the sample to come to equilibrium. Cool
rapidly to 400°C and hold for 10 s. Quench to room temperature.
c) Heat to 700°C and hold long enough for the sample to come to equilibrium. Quench to
room temperature and then reheat to 250°C for 5 minutes and then air cool. (Hint: 1b in
first question)
d) Heat to 900°C and hold long enough for the sample to come to equilibrium. Cool
rapidly to 600°C and hold for 30 s. Quench to room temperature.
Weld metal
HAZ
Base metal
3
.
A schematic of a weld is made with a hybrid laser weld is shown below. Depending on
the alloy and welding procedure, the HAZ may undergo
significant
microstructure and
property changes. If martensite is present in the HAZ, the weld is usually considered
unacceptable (martensite is brittle), and many different solutions must be implemented (e.g.,
pre or post weld heat treatment
) to avoid its formation.
Answer
the following questions assuming the base metal is 4340 (left) and 1080 (right)
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4340
(2 marks each) (6 marks total)
a.
Assume the average cooling rate of the heat affected zone (HAZ) in the vicinity of
a weld is 10°C/s.
Estimate
the microstructures (microconstituents and/or phase(s)
and their approximate amounts) that will be present in the heat affected zone (HAZ).
b.
What
would the answer to part a) be if the cooling rate was 0.1°C/s?
c.
Why
would you classify 4340 as a difficult alloy to weld?
1080
(
2 marks each) (6 marks total)
a.
Assume the average cooling rate of the heat affected zone (HAZ) in the vicinity of
a weld is 10°C/s.
Estimate
the microstructures (microconstituents and/or phase(s)
and their approximate amounts) that will be present in the heat affected zone (HAZ).
b.
What
would the answer to part a) be if the cooling rate was 0.1°C/s?
c.
Why
would you classify 1080 as a difficult alloy to weld?
Q3 (2 marks) Suppose each of these alloys was welded with too high a cooling rate, and
martensite formed in the HAZ. Which of the two alloys would be harder in the HAZ?
M202 F2022
Assignment #6
Due: Dec 2nd
2022 at 4pm
8
M202 F2022
Assignment #6
Due: Dec 2nd
2022 at 4pm
4.
Imagine you inherit a 1970 Camaro such as the one pictured above. You manage to break the
crankshaft and you are too poor to buy a brand new one. You find a deal on eBay for a used
racing crankshaft ($200), but there is one catch – the crankshaft was in a fire! Luckily it wasn’t
heavily oxidized or distorted but the heat treatment is no longer to specifications. You also have
no idea from which alloy the crankshaft was made. After some research you find that aftermarket
GMC 350 crankshafts for racing are typically made from 5140 or 4130 steel, and in some cases
4340 (if you have $3000 to spend on a crankshaft!).
A 1970 Camaro Z-28 (at left) with a standard Chevy 350 Block Engine (center) with an after-market replacement
crankshaft (right).
Answer the following
(2 marks each) (8 marks total)
a.
Explain
if the
hardness
or
hardenability
changed for the crankshaft that was in the fire.
b.
Explain
how you would determine if the crankshaft has high hardenability or low
hardenability.
9
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M202 F2022
Assignment #6
Due: Dec 2nd
2022 at 4pm
c.
The original surface hardness of the crankshaft, before the fire was 55-57 HRC - to be strong,
resist wear, and prevent crack initiation to prolong fatigue life. Yet, the center was tougher (e.g., a
hardness between 32 and 38 HRC). For such a specification,
which
alloy must the crankshaft be:
5140 or 4340 steel?
How
do you know?
d.
The thickest circular section of the crankshaft is between 2.5 and 3 inches.
Describe
the heat
treatment process to restore the required specifications for your used crankshaft as noted in part (d).
10
M202 F2022
Assignment #6
Due: Dec 2nd
2022 at 4pm
11
M202 F2022
Assignment #6
Due: Dec 2nd
2022 at 4pm
5
.
Consider the heat treatment of a precipitation hardenable aluminum alloy (AA2014). The
property curves as a function of aging time and temperature for this alloy are shown below. This
alloy primarily forms hard CuAl
2
precipitates (the phase diagram is also shown).
4 marks each
(12 marks total)
a.
Is it
possible
to heat treat a such an alloy and achieve a yield strength of 400 MPa with a
ductility of 14% El? If so,
explain
how you would heat treat the alloy.
b.
What
range of Copper composition would you specify for this alloy while still maintaining the
ability to precipitation harden aluminum? Why?
c.
Imagine you are working at a company called the “Fly by Night Heat Treating”. The
BOSS tells
you that you have an emergency heat treating job - a small temporary structure just arrived that
will only be in use for a
maximum
of 1 month.
What
heat treatment schedule is necessary to
meet all the required specifications?
12
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M202 F2022
Assignment #6
Due: Dec 2nd
2022 at 4pm
Assume the following:
The structure is made from aluminum alloy 2014.
The structure must be able of withstanding a static tension of 350 MPa, and cannot
have <10%El at any time during service.
The operating temperature will be a consistent 150°C.
You have up to
2 days
to get the component heat treated and on the truck for
shipping.
The furnaces you will need are at temperature already, and that your job takes
precedence in the shop
13
Related Questions
1.50% Pearlite, 50% Tempered Martensite 2.50% Bainite, 45% Martensite, 5% Austenite
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form 50% Fine Pearlite + 50% Coarse Bainite.
800
A
1400
Eutectoid temperature
700
A
1200
600
1000
500
B
800
400
A
300
600
M(start)
200
M + A
50%
400
M(50%)
М190%)
100
200
10-1
1
10
102
103
104
105
Time (s)
Temperature (°C)
Temperature (°F)
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heat treatment do you apply? Using this graph, how is it possible to increase the
strength and hardness of this alloy? Consider this question within the scope of
heat treatments.
Composition (at % Cu)
10
20
30
700
1200
600
L
a+ L
0 +L
1000
a
500
(CuAl,)
a + 0
800
400
600
300
10
20
30
40
50
(AI)
Composition (wt% Cu)
Temperature (°C)
Temperature (°F)
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1- Temperature, phases and carbon weight for Eutectic.
2- Temperature, phases and carbon weight for Eutectoid
Thecementite,
3-
austenite, ferrite, delta phases with their
4- Types of ferrous alloys on different carbon rates.
5- Draw the phase at T= 700 CD and 0.7 wt% C what is its name ?
temperatures and carbon weight.
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Material science
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Can anyone help with this question
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Please explain ths graph from Materials Science and EngineeringAn Introduction
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13- Copper-rich copper-beryllium alloys are precipitation hardenable. After consulting the portion of the phase
diagram shown in the figure below, do the following:
(a) Specify the range of compositions over which these alloys may be precipitation hardened.
Temperature (°C)
(b) Briefly describe the heat-treatment procedures steps (in terms of temperatures) that would be used to
precipitation harden an alloy having a composition of 1 wt% Be.
1000
800
600
400
0
(Cu)
0
1
Composition (at% Be)
10
15
a + L
2
α + 1/2
Liquid
Composition (wt% Be)
866°C
-620°C
3
20
x + 1₁
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800
A
Eutectoid temperature
H 1400
700
A
H 1200
600
1000
500
B
800
400
A
300
600
M(start)
200
M + A
50%
400
M(50%)
M(90%)
100
200
10-1
1
10
102
103
104
105
Time (s)
Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition, specify the final microstructure and
approximate amount of each. Assume a small specimen has been held long enough to have achieved a complete and homogeneous
austenitic structure prior to treatment.
Sample (1): Quickly cool specimen from 800°C to 575°C, hold for 10 s, then quench to room temperature.
Sample (2): Quickly cool specimen from 800°C to 500°C, hold for 100 s, then quench to room temperature.
O after treatment, sample 1 is 50% pearlite, 50% austenite
O after treatment, sample 2 is bauxite
O after treatment, sample 1 is 25% pearlite
O after treatment, sample 2 is bainite
O after treatment, sample 2 is austenite
O None of the answers is correct.
O after treatment, sample 2 is coarse pearlite
O after treatment, sample 1 is…
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a: Quench to from 800oC to 350oC and hold for 102 seconds:b: Quench the alloy to room temperaturec: Reheat the alloy to 700oC and hold for 105 seconds
arrow_forward
Copper-rich copper-beryllium alloys are precipitation hardenable. After consulting
the portion of the phase diagram shown in Figure below, do the following:
Composition (at% Be)
10
15
20
Liquid
1000
866°C
800
-620°C
600
a+
400
2
3.
(Cu)
Composition (wt% Bo)
(1) Indicate the range of compositions over which these alloys be precipitation
hardened.
(ii) Describe the heat-treatment procedures (in terms of temperatures) that would be
used to precipitation harden an alloy having a composition of your choosing but lying
within the range given for part (i).
Temperature ("O
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Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid
composition, sketch (mark the microconstituents, eg. a-Fe, Fe,C) and specify the
name of the final microstructure (e.g. austenite, pearlite, bainite etc.) of a small
specimen that has been subjected to the following time-temperature treatment. In
each case assume that the specimen begins at 760 Cand that is has been held at this
temperature long enough to have achieved a complete and homogeneous austenite
structure.
800
-Eutectoid temperature
1400
700
1200
600
1000
500
800
400
300
600
Mistart)
200
50%
400
ME50%)
M90%)
100
200
10
10
10
10
10
10
Time (
eoha ileySons, lnc. Al righta reserved
(a) Cool rapidly to 350 S, hold for 10' s, then quench to room temperature
(b) Cool rapidly to 665 S, hold for 10's, then quench to room temperature
(c) Cool rapidly to 150gin 5 seconds, then cool to room temperature
(d) Cool rapidly to 300gin 1 second, then cool to room temperature in 10s
(3J anjedueI
Temperature F)
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Part C. Which type of phase transformation is important in heat treatment of steels? (circle one)Peritectic Eutectic Eutectoid Incongruent meltingAt what temperature and overall composition wt% C, does it occur?Write the reaction equation for this transformation, and briefly describe the nature of each of the three important phases that are involved.
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For each question, provide an explanation, identifying the correct choice ( it is marked in red) and explaining why it's the right answer, as well as why the other options are incorrect.
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Using the TTT diagram for eutectoid steel, draw the specified cooling path on the diagram.
Indicate what phases you expect in the final product with percentage of each phase. Assume the material has been fully austenitized before cooling.
a. Water cool to room temperature
b. Hot quench in molten salt to 690°C and then cooled isothermally for 2 hours:
c. Hot quench to 610°C hold 3 minutes and water quench
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1. Draw a diagram of the gray and nodular gray cast iron's substructure.
2. Comment on the gray wrought iron's mechanical characteristics.
3. Describe a scenario in which cast iron would be used in manufacturing such in engineering
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Below is a continuous cooling transformation curve for steel (upper) and for a silicate melt (lower).
Describe the similarities and differences as follows.
a. What is the critical cooling rate in each case?
b. What phase(s) form in each case when the critical cooling rate is exceeded?
c. What phase(s) form in each case if the cooling rate is very slow?
800
1400
Ae lemperature
Austenite
*peorlite begins
10FI
700
1200
Austenitepearlite complete 1600
1000
5009
800
Tronsformation stops
400
600
300
400
Austenite -mortenstic structure
200
-Isathermal dogram
Continuous transformaton
diogrom
Constant rate coding curves - 100
200
Findl structure
Mortensite Morlensite
i peorlile
Peorite (softer, coorser-
0.1
10?
10
Tronsformation time, seconds
Continuous Cooling-Transformation (C-T)Diagram
(Derived from the isothermal-transformation diagram for a plain-carbon eutectoid steel)
1200
Glass
Crystallization
begins
1100
1000
Critical
cooling
rate
Glass-ceramic
900
Crystallization
ends
800
700
102
10…
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2. Why does gray cast iron lack tensile strength?
3. What can you do to give gray wrought iron a little toughness?
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