mat101 w12 hw6 solutions

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MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012

Problem Set 6Due: Tuesday, March 6, 11:00 AM

1. TTT DiagramsBased on the transformation diagrams for eutectoid steel shown below, what microstructure

would result from the following cooling histories? Assume the steel starts above the eutectoidtemperature. Distiguish between coarse and fine pearlite when applicable.

a) Rapidly cooled to 600C, held for 7 seconds. Then rapidly cooled to 450C and held foran additional 10 seconds. Quenched to room temperature. (2 pts)80% Fine Pearlite, 10% Martensite, 10% Bainite.

b) Rapidly cooled to 600C, held for 1 minute, then rapidly cooled to 450C and held for 10seconds. Quenched to room temperature. (2 pts)

100% Fine Pearlite. Once this transformation has occurred, no further transformation

will.

c) Rapidly cooled to 600C, held for 7 seconds, then rapidly cooled to 170C and held for 1hour. Quenched to room temperature. (2 pts)

80% Fine Pearlite, 20% Martensite

. d) Rapidly cooled to 700C, held for an hour and a half, then rapidly cooled to 570C andheld for 10 seconds. Quenched to room temperature. (2 pts)

50% Coarse Pearlite, 50% Fine Pearlite

e) Cooled at a rate of 10C/s.(2 pts)100% Pearlite


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f) Cooled at a rate 125C/s.(2 pts)About 85% Martensite, 15% Pearlite.

g) For part (f) draw what the microstructure might look like. (8 pts)

2. Lever RulePlease refer to the Cu-Ag phase diagram on the following page to answer the following

questions.

a) Label all phase fields present below 32 atomic percent Cu. (5 pts)b) Determine the phase composition at 500 oC in weight percent for an overall composition

of C0=20at%Cu. (5 pts)

The tie line is illustrated in the figure bellow. phase is 48%Al-52%Cu

Al phase is 96%Al-4%Cu


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Al+L +L

+Al

A

c) Determine the weight percent of each phase at 500 oC for an overall composition ofC0=20at%Cu. (5 pts)Using weight percent Cu

W=(C0-CAl)/(C-CAl)=(38%-4%)/(52%-4%)=71%

WAl=100- W=29%

d) Calculate the overall weight percent of Aluminum for C0=20 at%Cu at 500 oC usingparts b and c, and check to make sure this agrees with the phase diagram. (5 pts)

Overall %Al = W(%Al in phase composition of )+ WAl(%Al in phase composition of Al)

Overall %Al = (.71)(.48)+(.96)(.29)=.34+.28=.62 or 62% Al.

The phase diagram shows 38% Cu.

e) Identify as many invariant points as you can! (extra credit!)


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3. Nucleation and Critical RadiusThe eutectoid temperature of iron/carbon alloy is 1000 K. This is the eutectoid between

austenite, and pearlite (cementite and ferrite.) Assume that the latent heat of fusion and surface

free energy for pearlite in the austenite phase are and , respectively.

a)

Compute the critical radius of nucleation for pearlite in austenite at the followingtemperatures of austenite: 950K, 850K, and 750K. State your answer in nanometers. (7

pts)

b) Compute the free energy of activation for pearlite in the austenite phase at the sametemperature undercoolings as above. (7 pts)

c) From your answers in parts a and b, sketch (qualitatively) schematics of the pearlitenucleation process (one at each temperature stated above). Your drawing does not need

to be up to scale as long as it captures the average number of stable pearlite nuclei and

their average critical radius at all three temperatures relative to one another. Recall that

the number of stable nuclei is exponentially proportional to the term (equation

10.8). Since this is the beginning of the nucleation process, the pearlite regions may be

represented as spheres instead of the equilibrium lamella structure. Be sure to label thedifferent phases in your drawings. (6 pts)


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4. Phase Diagram FunPlease refer to the X-Y phase diagram below to answer the following questions.

a) Label all regions on the phase diagram. Notice the two line compounds already labeled.Distinguish the single-phase regions with , , and L and the two-phase regions

between them as appropriate (7 pts)

b) Use the Gibbs phase rule to solve for the degree(s) of freedom in the two phase, singleline compound phase and single phase regions. (4 pts)Gibbs Phase Rule F = C- P + N

P is # of phases, F is degrees of freedom, C is # of components (which can be elements,

or stable compounds), N is the number of non-compositional variables. Since this phase

diagram is at a fixed pressure, the only non-compositional variable is temperature.

Single phase field(eg. )

C=2 (Ag + Mg = 2 components), P=1, N=1 F = 2-1+1 = 2

Two phase field(eg. + )

C=2 (Ag + Mg = 2 components), P=2, N=1 F = 2-2+1 = 1

Single phase line compound (eg. Ag3Mg)

C=1, P=1, N=1 F=1-1+1=1


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c) Label all invariant points on the phase diagram (eg. eutectics, peritectics, eutectiods,peritectiods, and congruent transformations). (7 pts)

Eutectic- start with liquid, end with 2 solid phases. (eg. L + )

Eutectoid-start with 1 solid phase, end with 2 solid phases (eg. + Ag3Mg )

Congruent transformation- start 1 liquid or solid phase, end 1 solid phase without a

change in composition. (eg. L )

d) Find the number of degrees of freedom for an eutectic point. (2 pts)Eutectic point

C=2, P=3, N=1 F = 2- 3 + 1 = 0

5. Making a Medieval SwordYou want to reproduce an authentic Medieval European sword. In order for the sword tomaintain a sharp edge the steel used must have sufficient hardness. However, increasing

hardness also will increase brittleness. A brittle sword will fracture under the heavy stresses ofbattle.

a) You decide to use AISI 1060 steel. What is the composition of this steel? (4 pts)

The first two digitis (10) indicate that this is a plain carbon steel, so it contains

predominantly iron with carbon, a little manganese and only residual concentrations ofother elements. The carbon concentration is given by the second two digits (60) as one

hundred times the weight percent of carbon, so this steel contains 0.60 wt% C.

b) How would the mechanical properties of the material be different if you used 1040 steeland 1080 steel as compared to the 1060 steel? (2 pts)

The tensile strength, yield strength and hardness increase with carbon concentrationwhile the ductility decreases. 1040 steel would be of lower strength and hardness and ofhigher ductility compared to the 1060. The 1080 steel would have increased strength and

hardness while being more brittle (lower ductility).

c) After forging the blade with a hammer and anvil into the desired shape, you perform a

normalizing process. Briefly describe the grain structure of the sword before and after

the process of normalizing. (4 pts)

After the steel is shaped via cold working, it contains large irregularly shaped grainswhich vary significantly in size.

The normalizing process decreases the grain size while making them more regular in

shape with a smaller distribution of grain sizes.

d) In order to add finishing details to the sword you would like to make it softer and more

ductile. This allows for easier grinding of the blade edge as well as any decoration you

may wish to engrave in the steel. What heat treatment process will result in a coarsepearlite grain structure which is soft and ductile? How is the cooling of this process

different than that of the normalizing process? (4 pts)


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The process described is a Full Annealing of the steel. The cooling step of a full anneal

is to turn off the furnace and let the furnace and steel cool together slowly. The cooling

step of the normalizing process is performed faster by removing the steel from the

furnace and letting it cool in air.

e) Finally, you have your sword shaped and decorated as desired. However, it is now toosoft to hold a sharp edge. The steel must be tempered to the desired hardness. Thisprocess converts some of the microstructure to what microconstituent? (2 pts)

The microconstituent formed when temper hardening the steel is martensite. The temper

process makes the steel more martensitic.

f) When performing the cooling step of the tempering process, sword makers have used

material such as clay to cover portions of the blade to slow the cooling rate of the covered

steel. Why might this technique be beneficial for making a sword which holds a sharpedge while remaining flexible enough to withstand the forces subjected to it in battle? (4

pts)

When tempering the steel, a faster cooling rate will have the result of making the steelmore martensitic which means that it will be harder and more brittle. By cooling

selective areas more slowly such as the middle of the blade running the length of the

sword, those areas will be more ductile and less brittle. This will result with the sword

having harder edges and more durable core. The blade will be more flexible whilemaintaining hard sharp edges. This happens somewhat naturally without using a

material to alter the cooling rate due to the cooling rate profile throughout the volume of

the steel (i.e. the middle of the steel will cool more slowly than the outer areas).

mat101 w12 hw6 solutions

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