introduction to material science may 2011

5/28/2018 Introduction to Material Science May 2011

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UN IVERSITITEKNOLOC IPETRONAS

COURSE :

,.

MCBI I 21MBBI Q52 I MCBI D12-- INTRODUCTION TOMATERIAL SCIENCE I INTRODUCTION TOMATERIAL SCIENCE AND ENGINEERING

DATE 8THSEPTEMBER 2011 THURSDAY)TIME 2.30 PM - 4,30 PM 2 hours)

INSTRUCTIONS TO CANDIDATES

I Answer ALL questions in the Answer Booklet.2. Begin EACH answer on a new page.3. Indicate clearly answers that are cancelled, if any.4. Where applicable show clearly steps taken in arriving at the solutions and

indicate ALL assumptions.5. Do not open this Question Booklet until instructed.

Note : There are TEN 10) pages in this Question Booklet including the coverpage and appendices.

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UNIVERSITI

Universiti Teknologi PETRONAS


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I. Aluminum is an FCC metal with an atomic weight of 26.98 glmol, an atomicradius of 0.143 nm, and a density of 2.71 glcm3 at 20C.

a. Compare the planar densities of the {11O} and {111} planes ofaluminum. Classify which of the planes you would expect to have thehigher surface enemy.

[9 marks]

b. By constructing appropriate diagrams, identity how slip occurs in FCCaluminum. Sketch and label at least one slip plane from the FCC slipsystem and indicate the corresponding slip directions on that plane.

[8 marks]

C. By constructing appropriate diagrams, formulate the size in nm) of animpurity atom that can just fit into the interstitial site of the aluminumunit cell. State the name of this interstitial site and the coordinates ofthe interstitial sites.

[8 marks]

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MCBI012 MBB 10521MAB1003 w

2. A carburizing heat treatment of a steel component was carried out byintroducing 1.10 wt% carbon at the surface of the component. The initialconcentration of carbon within the component is 0.20 wt%. This processproduces a 2.5-mm case depth (depth of carburized layer) after a 2-hourcarburizing treatment at 820C.

a. Calculate the carbon content at 2.5-mm beneath the surface. (Gasconstant, R=8.31 Jlmole0 K).

[9 marks]b. it was decided to increase the case depth to 3.5-mm by increasing

the carburizing time. Determine the time required to achieve the newcase depth.

[6 marks]

C. If instead it was decided to increase the carburizing temperature butmaintain the 2-hour treatment time, evaluate the change intemperature required to achieve a case depth of 3.5-mm.

[10 marks]

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MCBIOI2 MBB 10521MAB1063

b. An isothermal transformation diagram for an iron-carbon alloy ofeutectoid composition is shown in FIGURE Q4b. The steel washeated to 754C and cooled under various conditions. For eachheat treatment, specify and sketch the final microstructure formed.

FIGURE Q4bi. Quenched to 600C and held for 120 seconds then rapidly

cooled to room temperature.[4 marks]

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MCBI0I21 MBB 10521MABI063

900 III-I

1040 Steel

140

a

7000 600.2Il 500

400

300

00 10

...

20 30 40Percent cold work

FIGURE Q3a

FIGURE Q3b

50 60

120

100

80

60

40

70

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MCB10121 MBB 10521MAB1063

4. a. 8-kg of a steel alloy containing 0.55 wt C is slowly cooled from800C to slightly below 77C. Using the iron-iron carbide phasediagram as shown in FIGURE Q4a, answer the fallowingquestions.

Composition (at C)19 15 25

AC)H

FIGURE Q4a

i. Determine the number of phases present at thistemperature and the weight of each phase.

[6 marks]

ii. Sketchthe microstructureof the phasesformed.[3 marks]

iii. Write the eutectic and eutectoid reactions for an iron-ironcarbide system.[4 marks]

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ii. Quenched to 670C and held at this temperature for 104seconds then rapidly cooled to room temperature.

[4 marks]rrlIir Quenched to 050C and held for 20 seconds then rapidly

cooled to 400C and held for 1O seconds followed byquenching to room temperature.

[4 marks]

-END OF PAPER-

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APPENDIX - Tables and Charts

Table Al Tabulation of Error Function Values

00.0250.05

0.100.150.200.250.300.350.400.454.50

erf(z) z erf(z)00.02820.05640.1 250.16840.22270.27630.32860.37940.42840.47550.5205

erf(z)0.93400.95230.96610.9'7630.98380.98910.99280.99530.99810.99930.99980.9999

Table A2: Tabulation of Diffusion Data

a-Fe 2.8 x 1-4 251 2.60(BCC)130(rn2/s) k. /mol eV/atom

DiffusingSpecies Dost Metal

FeFe 7-Fe 5.0 X 1U-S 284 2.94(FCC)C a-Fe 6.2 x10' 80 0.83C yFe 2.3 x IO 148 1.53

Cu Cu 7.8 x :Os 211 2.19Zn Cu 2.4 x 1O- 189 1.96Al Al 2.3 x iQ-d 144 1.49Cu Al 6.5 xiO W 136 1.41Mg Al 1.2 x 1Q131 1.35Cu Ni 2.7 x 1Qs 256 2.65

0.550.600.650.700.750.800.850.900.951.01.11.2

0.56330.60390.64200.67780.71120.7421.0.77070.79700.820.94.54270.88020.9103

z1.31..1.51.61.71.81,.2.02.22.42.62.8

Activation Energy Qd

.,w...,.,.... .....,.. ..,,. ..., . ,.,..,,w..,. .., . ..w..

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Temperature C)

Reciprocalemperature1000/K)Figure Al : Plot of the logarithm of diffusion coefficient versus the reciprocal ofabsolutetemperaturefor diffusion pairs.

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introduction to material science may 2011

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