Solutions to Homework Number FourDue Monday October 31, 2016

1) Prepare a report which explains how to use a phase diagram to the microstructure of various steelalloys. Include the following:

• A sample analysis of 1030 steel at room temperature based on both the micrograph and thephase diagram, explaining the components, constituents, phases, and states present.

• Calculations of the relative amounts of the phases present for 1040 (Fe-0.40w/oC) steel at thefollowing temperatures: Room Temperature, 600C, 750C, and 900C.

• Calculations of the relative amounts of the phases present (at room temperature) for steelsfor the following alloys: 1040 (Fe-0.40w/oC), 1077 (Fe-0.77w/oC) and 10100(Fe-1.0w/oC)

• Calculations of the relative amounts of the constituents present (at room temperature) forsteels for the following alloys: 1040 (Fe-0.40w/oC), 1077 (Fe-0.77w/oC) and 10100(Fe-1.0w/oC)

Use the Fe-C Phase Diagram ..

A sample analysis of 1030 steel at room temperature based on both the micrograph and the phasediagram, explaining the states, components, phases, and constituents present.

The micrograph for 1030 (Fe-0.30w/oC) and the Fe-C Phase Diagram are shown below.

The micrograph shows that the alloy is completely solid. However, there is no way to see thecomponents: iron (Fe) and carbon (C). That would require seeing individual atoms.

Micrograph of 1030 Steel

Fe (Iron) - C (Carbon) Phase Diagram

The micrograph for 1030 (Fe-0.30w/oC) and the Fe-C Phase Diagram are shown below.

Both the micrograph and the phase diagram show that there are two phases present. • The Phase Diagram shows that the two phases present are α-Fe and Fe3C.• The micrograph shows a white phase and a black phase. The white phase is α-Fe and the

black phase is Fe3C.

Micrograph of 1030 Steel

Fe (Iron)-C (Carbon) Phase Diagram- showingthe data point for Fe-0.45 w/oC at roomtemperature. Because the solubility of C in α-Feis essentially 0 at 400C, it will be 0 at roomtemperature.

The micrograph for 1030 (Fe-0.30w/oC) and the Fe-C Phase Diagram are shown below.

Both the micrograph and the phase diagram show that there are two constituents present. • The micrograph shows that the structure is proeutectoid(ic)-eutectoid(ic). One of the distinct

structures is the two phase zebra pattern which is the eutectoid(ic). The eutectoid(ic) is asingle constituent consisting of two phases. The second distinct structure are the large whiteblobs. This is the proeutectoid constituent which consists of one phase.

• The Phase Diagram shows that there are two constituents present are Proeutectid α-Fe andeutectoid which is made up of two solid state phases: α-Fe and Fe3C.

Micrograph of 1030 (Fe-0.30w/o C) Steel atRoom Temperature showing the differentconstituents.

Eutectoid Reaction Occursγ(Fe-0.77w/oC) 88w/o α(Fe-0.022w/oC)+ Fe 3C(Fe- 6.7w/oC)

ProEutectoid ReactionOccurs Before Eutectoid α-Fe makes up theProeutectoid Constituent

Fe (Iron) - C (Carbon) Phase Diagram showinghow the proeutectiod α-Fe and the eutectoidform during cooling.

X C CC C

X w oC w oCw oC w oC

X

Fe C

Fe C

Fe C

3 0

3

3

6 7 0 406 7 0

0 94

0 06

. / .. /. / /

.

.

Calculations of the relative amounts of the phases present for 1040 (Fe-0.40w/oC) steel at thefollowing temperatures: Room Temperature,600C, 750C, and 900C.

The Fe-C phase diagram showing the phasespresent and their compositions for 1040 steel(Fe-0.40w/oC) at both Room Temperatureand 600oC is shown on the right.

Note there are two phases present: α-Fe andFe3C.

The composition of α-Fe, Cα is 0 w/oC, and thecomposition of Fe3C, CFe3C is 6.7 w/oC.

Using the Lever Rule the relative amount of α-Fe can be calculated.

Thus, relative amounts of the phases present are 94w/o α-Fe and 6w/o Fe3C.

Iron (Fe)-Carbon(C) Phase Diagram showing theCompositions of α-Fe and Fe3C at Both RoomTemperature and 600C

The Fe-C phase diagram showing thephases present and their compositions for1040 steel (Fe-0.40w/oC) at 750oC is shownon the right.

Note there are two phases present, α with acomposition (Cα) of 0.01 w/oC and γ with acomposition (Cγ) of 0.65 w/oC.

Using the Lever Rule one can calculate therelative amounts of the phases present.

XC CC C

X w oC w oCw oC w oC

XX

0

0 65 0 400 65 0 010 390 61

. / . /

. / . /..

Thus the relative amounts of the phasespresent are 39w/o α and 61 w/o γ.

Iron (Fe)-Carbon(C) Phase Diagram showing theCompositions of α-Fe and γ-Fe at 750C

The Fe-C phase diagram showing thephases present and their compositions for1040 steel (Fe-0.40w/oC) at 900oC is shownon the right.

The only phase present is γ-Fe. So therelative amounts of the phases present100w/o γ-Fe.

Iron (Fe)-Carbon(C) Phase Diagram showing that only γ-Fe is present at 900C

X C CC C

X w oC w oCw oC w oC

X

Fe C

Fe C

Fe C

3 0

3

3

6 7 0 406 7 0

0 94

0 06

. / .. /. / /

.

.

XC CC C

X w oC w oCw oC w oC

X

E

P

0

0 40 0 020 7 0 02

0 49

0 51

. / .. /. / . /

.

..

Calculation of the relativeamounts of the phasespresent (at roomtemperature) for 1040 steel(Fe-0.40w/oC)

Note there are two phasespresent: α-Fe and Fe3C.

The composition of α-Fe, Cαis 0 w/oC, and thecomposition of Fe3C, CFe3Cis 6.7 w/oC.

Using the Lever Rule the relative amount of α-Fe can becalculated.

Thus, relative amounts of the phases present are 94w/o α-Feand 6w/o Fe3C.

Calculation of therelative amounts of theconstituents present (atroom temperature) for1040 steel (Fe-0.40w/oC)

Note that α-Feprecipitates from γ-Feprior to the eutectoidreaction. Thus there isproeutectoid α-Fe, aswell as eutectoid.

The amount of eutectoidcan be calculated bydetermining the amountof γ-Fe at 727C.

Thus there would be 51w/o proeutectoid α-Fe and 49w/oeutectoid constituent contains both α-Fe and Fe3C.

Solidification History of 1045 Steel (Fe-0.45w/oC)

Iron (Fe)-Carbon(C) Phase Diagramshowing the Compositions of α-Fe andFe3C at Room Temperature

Solidification of 1077 Steel

Calculation of therelative amounts of thephases present (atroom temperature) for1077 steel(Fe-0.77w/oC)

Note there are twophases present: α-Feand Fe3C.

The composition of α-Fe, Cα is 0 w/oC, andthe composition ofFe3C, CFe3C is 6.7 w/oC.

Using the Lever Rule the relative amount of α-Fe can becalculated.

Thus, relative amounts of the phases present are 88w/o α-Feand 12w/o Fe3C.

Calculation of therelative amounts of theconstituents present (atroom temperature) for1077 steel (Fe-0.77w/oC)

The cooling history for1077 (Fe-0.77w/oC) steelis shown on the right.The last reaction whichoccurs is γ–>α+Fe3C. Allthe γ-Fe is of theeutectoid composition.So there is only 1constituent: eutectoid.The eutectoid constituent is made up of two phases.

Thus there would be 100 w/o eutectoid constituent containsboth α-Fe and Fe3C.

Iron (Fe)-Carbon(C) Phase Diagramshowing the Compositions of α -Fe andFe3C at Room Temperature

XC CC Cw oC w oCw oC w oC

XX

Fe C

Fe C

Fe C

3

3

3

0

6 7 0 776 7 0

088012

. / . /. / /

.

.

XC CC Cw oC w oCw oC w oC

XX

Fe C

E

PFe C

0

3

3

10 0 776 7 0 770 960 04

. / . /

. / . /..

Calculation of therelative amounts of thephases present (at roomtemperature) for 10100steel (Fe-1.0w/oC)

Note there are twophases present: α-Fe andFe3C.

The composition of α-Fe,Cα is 0 w/oC, and thecomposition of Fe3C, CFe3Cis 6.7 w/oC.

Using the Lever Rule the relative amount of α-Fe can becalculated.

Thus, relative amounts of the phases present are 85w/o α-Feand 15w/o Fe3C.

Calculation of the relativeamounts of the constituentspresent (at roomtemperature) for 10100steel (Fe-1.0w/oC)

Note that Fe3C precipitatesfrom γ-Fe prior to theeutectoid reaction. Thusthere is proeutectoid Fe3C .The amount of this can becalculated by using theLever Rule at 727C

The amount of eutectoidcan be calculated bydetermining the amount ofγ-Fe at 727C.

Thus there would be 4w/o proeutectoid Fe3C and 96w/oeutectoid constituent contains both α-Fe and Fe3C.

Iron (Fe)-Carbon(C) Phase Diagramshowing the Compositions of α-Fe andFe3C at Room Temperature

XC CC Cw oC w oCw oC w oC

XX

Fe C

Fe C

Fe C

3

3

3

0

6 7 106 7 0

085015

. / . /. / /

.

.

2) Technical Memorandum for Lab Experiment

Introduction: It is very important that the introduction explains what information the phase diagramconveys. This should be followed by three one or two sentence justified predictions. The compositionranges for which a single phase or a combination of two phases must be included. In the range wheretwo phases should be present a description of how the relative amounts of the phases present areexpected to change with composition must be included. The ranges for expected microstructures:eutectic, granular, precipitate and proeutectic/eutectic and how when two constituents the relativeamounts will change with composition must be included. The following content is unacceptable: analloy-by-alloy prediction, statements of purpose and summaries of what was done.

Methodology: The data to be collected, alloy list must be included. The procedure portion of themethodology should be written to the intelligent person standard (3 maybe 4 sentences).

Results: The missing content from the Word Files, predicted amounts of phases present, predictedand observed microstructure must be included. An unaltered Word File from Google Drive iscompletely unacceptable. A partially completed Word File is only partially acceptable.

Discussion: There need to be enough specific references to the results to ensure that there is a goodcomparison between the results and the hypothesis.

3) Prepare a report explaining the difference between phases and constituents. Use examples fromboth Problems 1 and 2. Make sure your comparison includes micrograph analysis and calculationmethods.

From Problem 1:

A phase is a region homogeneous structure within a material (it is not a separate material). It has adistinct crystal structure. A constituent is a phase or combination of phases that have a distinctpattern. The eutectoid is a single constituent consisting of two phases.

To calculate the amount of phases, one first determines if there are two phases present. If so, on canuse the Lever Rule formulation at the temperature of interest.

To calculate the amount of constituents, one also uses the Lever Rule (if there are two present).However, if the two constituents, proeutectic (oid) and eutectic(oid); one performs the calculation atthe temperature that the eutectic(oid) reaction occurs. Thus for steel, we used the concentrations at727C, even though we were interested in room temperature.

From Problem 2:

One should have predicted and observed various combinations of proeutectic and eutectic. At theeutectic composition one should only have observed eutectic which is a combination of two phases.Both the Sn-30w/oBi and the Sn-45w/oBi one should have observed proeutectic β-Sn and eutectic.However, the amount of proeutectic β-Sn should be less in the Sn-45w/oBi than for the Sn-30w/oBi.Both the Sn-70w/oBi and the Sn-85w/oBi one should have observed proeutectic α-Bi and eutectic.However the amount of proeutectic α-Bi would be greater in the Sn-85w/oBi than in the Sn-70w/oBi.