Answer Final Exam Sem i 1213-Edit

7/28/2019 Answer Final Exam Sem i 1213-Edit

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UNIVERSITI TEKNIKAL MALAYSIA MELAKA

PEPERIKSAAN AKHIR SEMESTER IFINAL EXAMINATION SEMESTER I

SESI 2012/2013SESSION 2012/2013

FAKULTI KEJURUTERAAN MEKANIKAL

ANSWER SCHEME

KOD MATAPELAJARAN : BMCB 1423SUBJECT CODE

MATAPELAJARAN : SAINS BAHANSUBJECT MATERIALS SCIENCE

PENYELARAS : HAMZAH BIN MOHD DOMCOORDINATOR

KURSUS : BMCS/BMCDCOURSE

MASA : 2 JAM 30 MINIT SAHAJATIME 2 HOURS AND 30 MINUTES ONLYTARIKH : 9 JANUARI 2013

DATE 9 JANUARY 2013TEMPAT : KOMPLEKS SUKANVENUE KOMPLEKS SUKANARAHAN KEPADA CALON

INSTRUCTION TO CANDIDATES

(1) Jawab hanya EMPAT (4) dari ENAM (6) soalan.Answer only FOUR (4) out of SIX (6) questions.

(2) Markah keseluruhan bagi peperiksaan ini adalah 100 markah.Total marks for this examination is 100 marks.

KERTAS SOALAN INI TERDIRI DARIPADA 21 MUKA SURAT SAHAJA

(TERMASUK MUKA SURAT HADAPAN)THIS QUESTION PAPER CONTAINS 21 PAGES INCLUSIVE OF FRONT PAGE

SULITCONFIDENTIAL

SULITCONFIDENTIAL


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Q1 The atom is a basic unit of matter that consists of a dense central nucleus surrounded

by a cloud of negatively charged electrons. The atomic nucleus contains a mix of

positively charged protons and electrically neutral neutrons (except in the case of

hydrogen-1, which is the only stable nuclide with no neutrons). The electrons of an

atom are bound to the nucleus by the electromagnetic force. Likewise, a group of

atoms can remain bound to each other by chemical bonds based on the same force,

forming a molecule. An atom containing an equal number of protons and electrons is

electrically neutral; otherwise it is positively or negatively charged and is known as an

ion. An atom is classified according to the number of protons and neutrons in its

nucleus: the number of protons determines the chemical element, and the number of

neutrons determines the isotope of the element. In chemistry and physics the idea of

the atom is a key concept. To understand many of the other concepts in chemistry

some knowledge of the atom is necessary.

(a) Primary Inter-atomic Bonds

i. Briefly cite the main differences between ionic, covalent, and metallic

bonding.

(2 marks)

ii. State the Pauli Exclusion Principle.

(2 marks)

(b) Calculate the force of attraction between a K+ and an O2 ion the centers of

which are separated by a distance of 1.6 nm.

(6 marks)


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(c) Compute the percents ionic character of the interatomic bonds for the

following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2.

(10 marks)


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(d) Explain type(s) of bonding would be expected for each of the following

materials: brass (a copper-zinc alloy), rubber, barium sulfide (BaS), solid

xenon, bronze, nylon, and aluminum phosphide (AlP).

(5 marks)


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Q2 Structural engineering depends on the knowledge of materials and their properties in

order to understand how different materials support and resist loads.

(a) Explain why hydrogen fluoride (HF) has a higher boiling temperature than

hydrogen chloride (HCl) (19.4 C vs. 85 C), even though HF has a lower

molecular weight.

(6 marks)

(b) Show that the atomic packing factor for BCC is 0.68.

(5 marks)


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(c) Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic

weight of 55.85 g/mol. Compute and compare its theoretical density with the

experimental value found in Appendix A.

(8 marks)


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(d) Calculate the radius of an iridium atom, given that Ir has an FCC crystal

structure, a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.

(6 marks)


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Q3 (a) Any deviation from the perfect atomic arrangement in a crystal is said to

contain imperfections or defects. Adding alloying elements to a metal is one

way of introducing a crystal defect. Crystal imperfections have strong

influence upon many properties of crystals, such as strength and electrical

conductivity. Thus some important properties of crystals are controlled by as

much as by imperfections and by the nature of the host crystals.

i. Describe what are interstitial and substitutional type defects? Providing

such illustrations is a must in supporting your answer.

(6 marks)

Answer

Interstitial defectimpurity atom (smaller size) lie at the position in between of host

atom (larger size).


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Example of interstitial solid solution defect of Carbon (orange) in Ferum (grey).

Substitutional defectSome impurity atom (prox same size w/host ) replace the

position of some of the host atom.

Example of substitutional solid soln. defect of Cu (orange) in Ni (grey).

ii. List and explain TWO (2) types of Planar Defects in solids.

(4 marks)

Answer

1. Twin BoundryEssentially a reflection of atom positions across the twin plane.

2. Stacking FaultsFor FCC metals an error in ABCABC packing sequence

become ABCABABC.

(b) Calculate the activation energy for vacancy formation in aluminum, given that

the equilibrium number of vacancies at 500 C (773 K) is 7.55 1023 m3. The

atomic weight and density (at 500 C) for aluminum are, respectively,

26.98 g/mol and 2.62 g/cm3.

(6 marks)


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Answer

(c) Crystal imperfections are present in ceramic materials. These imperfections are

classified according to their geometry and shape. Explain and illustrate the

following imperfections that can exist in ceramics crystal lattices:

i. Frenkel imperfection.

(2 marks)

ii. Schottky imperfection.

(2 marks)

Answer

Frenkel imperfection

a cation vacancy-cation interstitial pair.

Schottky imperfection.

a paired set of cation and anion vacancies.


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Schottkydefect

Frenkel

defect

(d) Using the data in the Table 3, predict the relative degree of solid solubility of

the following elements in aluminum:

(i) copper (ii) manganese (iii) zinc (iv) magnesium

(v) silicon

Use the scale as very high (70 - 100%); high (30 - 70%); moderate (10 - 30%);

low (1 - 10%); and very low (


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Answer

dx

dCDJ

ii. Explain on the factors that affect the diffusion rate in solid metal crystals?

(3 marks)

Answer

Diffusing SpeciesThe magnitude of the diffusion coefficient, D is indicative

of the rate at which atoms diffuse.

TemperatureIt has a most profound influence the coefficients and the

diffusion rates.

(b) With reference to Tabulation of Diffusion Data shown in Appendix B,

i. Calculate the diffusion coefficient for magnesium in aluminum at 450 C.

(2 marks)

Answer

D = D0 exp Qd

RT

= (1.2 10-4 m2/s)exp 131,000 J /mol

(8.31 J /mol- K)(450 273 K)

= 4.08 10-14 m2/s

ii. Propose the time required at 550 C to produce the same diffusion result

(interms of concentration at a specific point) as for 15 h at 450 C.

(4 marks)

Answer

This portion of the problem calls for the time required at 550 C to produce the

same diffusion result as for 15 h at 450C. Equation 6.7 is employed as

D450t450 = D550t550

Now, from Equation 6.8 the value of the diffusion coefficient at 550C is

calculated as


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)273550)(-/31.8(

/000,131exp/sm101.2= )( 24-550

KKmolJ

molJD

= 5.76 10-13 m2/s

Thus,

550

450450550 =

D

tDt

=(4.08 1014 m2 /s) (15h)

(5.76

10

13

m

2

/s)

= 1.06 h

(c) Atomic diffusion occurs in metallic solids mainly by a vacancy or substitution

mechanism and an interstitial mechanism. In fact, the atoms in solid materials

are in constant motion, rapidly changing positions.

i. Explain TWO (2) conditions that allow these atoms motion.

(3 marks)

1. Vacancies or other crystal defects are present which means there must be

an empty adjacent site.

2. There is enough activation energy. The atom must have sufficient energy

to break bonds with its neighbour atoms and then cause some lattice

distortion during the displacement.

ii. Describe why are the rates of diffusion in polymer materials are higher

than the metallic materials.(3 marks)

Rates of diffusion are greater through amorphous regions than through

crystalline regions; the structure of amorphous material is more open for

which diffusive movements occur through small voids between polymer

chains from one open amorphous region to an adjacent open one. Whilst in

metals, the diffusion mechanism maybe considered to be analogous to

interstitial diffusion.


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(d) A steel gasket of 2.5 mm thick is to be case-hardened in nitrogen atmosphere at

900 oC at a steady-state diffusion condition. It is known that the concentration

of nitrogen in the steel at the high pressure surface is 2 kg/m3

. The diffusion

coefficient for nitrogen in steel at this temperature is 1.2 x 10 -10 m2/s, and the

diffusion flux is found to be 1.0 x 10-7 kg/m2-s. How far into the gasket from

this high-pressure side will the concentration be 0.5 kg/m3? How long will it

take for the nitrogen to completely reach the low pressure surface of the

gasket? Assume a linear concentration profile. (8 marks)

Q5 Mechanical properties of materials refer to the relationship between its response to an

applied load or force. It plays important roles in the structural applications and

processing of materials. Several material testing techniques, such as tensile test,

impact test and hardness test are used to measure mechanical properties of materials.

(a) Distinguish between:

i. Engineering stress and true stress.

(2 marks)

ii. Engineering strain and true strain.


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(2 marks)

iii. Elastic and plastic deformation.

(2 marks)

Answer

(a)

i. Engineering stress: average uniaxial force divided by original cross-sectional area.

= F/Ao.

True stress: average uniaxial force divided instantaneous minimum cross-

sectional area.

= F/Ai.

ii. Engineering strain: change in length of sample divided by the original lengthof sample.=l/loTrue strain:

iii. Elastic deformation: if a metal deformed by a force returns to its originaldimensions after the force is removed, the metal is said to be elastically

deformed.

Plastic deformation: if the metal deformed by a force not returns to its

original dimensions after the force is removed, the metal is said to be

elastically deformed.

(b) A bar of steel alloy that exhibits the stress-strain behavior shown in Figure 5 is

subjected to a tensile load; the specimen is 375 mm long and of square cross

section 5.5 mm on a side.

Figure 5 Tensile stress-strain behavior for alloy steel.


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i. Compute the magnitude of the load necessary to produce an elongation of

2.25 mm.

(4 marks)

ii. Predict the deformation would be after the load has been released.

(3 marks)

Answer

(i)

This is within the elastic region; from the inset of Figure 5, this corresponds toa stress of about 1250 MPa. Now,

in which b is the cross-section side length. Thus,

(ii) After the load is released there will be no deformation since the material

was strained only elastically.

(c) A cylindrical specimen of some metal alloy 10 mm in diameter is stressed

elastically in tension. A force of 15,000 N produces a reduction in specimen

diameter of 7 103 mm. Compute Poissons ratio for this material if its

elastic modulus is 100 Gpa.

(6 marks)

Answer

From Equations 7.5 and 7.1

z

=

E=

F

A0E=

F

d0

2

2

E

=4F

d02E

Since the transverse strain x is just

x =d

d0

and Poissons ratio is defined by Equation 7.8, then


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= xz

= d/d0

4F

d02E

= d0dE

4F

= (10 103 m)(7 106 m) ()(100 109 N /m2)

(4)(15,000N)= 0.367

(d) Why design/safety factor is important in engineering? List THREE (3) criteria

for which design/safety factors to be based on.

(6 marks)

Answer

It is important in order to avoid, thus to protect the design against

unanticipated failure.

The criteria upon which factors of safety are based are (1) consequences of

failure, (2) previous experience, (3) accuracy of measurement of mechanical

forces and/or material properties, and (4) economics.


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Q6 (a) Iron-carbon phase diagram contains ferrite, austenite, cementite (Fe3C) and

ferrite as solid phases is shown in Figure 6.1. The phase diagram may be

divided into two parts; an iron-rich portion (0 wt% C to 6.70 wt% C) and the

other (not shown) for composition between 6.70 wt% C to 100 wt% C (pure

graphite).

Figure 6.1 Iron-carbon phase diagram.

i. Define austenite and cementite phases that exist in the iron-carbon phase

diagram.

(4 marks)

ii. Distinguish between hypoeutectoid and hypereutectoid plain carbon

steel.

(4 marks)

(b) Referring to the Pb-Sn phase diagram in Figure 6.2, consider the very slow

cooling (i.e. equilibrium cooling) of a 40 wt% Sn alloy.

+Fe3C

, Austenite

+L

L

+Fe3C, Ferrite Cementite (Fe3C)

+

727oC

1147oC

1493oC

4.302.14

0.76

0.022

1394oC

1538oC

912oC

15 20105 250

2500

2000

1500

1000

400

600

800

1000

1200

1400

1600

Composition (at% C)

Composition (wt% C)

1 2 3 4 5 6 6.700(Fe)

Temperature(oC)

Temperature(oF)


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i. Draw schematic sketches of the microstructures that would be observed

at 300 C, 200 C, 180 C and 100 C. Be sure to label all of the phases

that are present and indicate the composition of each phase.

(4 marks)

ii. Indicate the pro-eutectic (or primary) phase in this alloy.

(2 marks)

iii. At 180 C, determine the phase amounts (or relative amounts) of the:

a) pro-eutectic (or primary) phase

b) eutectic

c) eutectic

(3 marks)

Figure 6.2 PbSn Phase Diagram


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(c) Refer to the isothermal transformation diagram (or TTT curve) for eutectoid steel

is a showed in Figure 6.3. For each of the two cooling sequences outlined below,

indicate the approximate percentages of each phase present after every cooling

step. Assume that the starting microstructure is 100% austenite held at 800 C.

Figure 6.3 Isothermal Transformasi Diagram

i. Rapidly cool to 650 C and hold for 20 seconds, then rapidly cool to 350 C

and hold for 10 seconds, then quench to room temperature.

(4 marks)

ii. Rapidly cool to 630 C and hold for 10 seconds, then rapidly cool to 450 C

and hold for 10 seconds, and then quench to room temperature.

(4 marks)

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APPENDIX A/LAMPIRAN A

Characteristics of Selected Elements

Aluminum Al 13 26.98 2.71 FCC 0.143 0.053 3+ 660.4Argon Ar 18 39.95 - - - - Inert -189.2Barium Ba 56 137.33 3.5 BCC 0.217 0.136 2+ 725Beryllium Be 4 9.012 1.85 HCP 0.149 0.095 3+ 1278Cadmium Cd 48 112.41 8.65 HCP 0.149 0.095 2+ 321Calcium Ca 20 40.08 1.55 FCC 0.197 0.100 2+ 839Carbon C 6 12.011 2.25 Hex. 0.071 ~0.016 4+ (Sublimes

@3367)Fluorine F 9 19.00 - - - 0.133 1- -220Gold Au 79 196.97 19.32 FCC 0.144 0.137 1+ 1064Helium He 2 4.003 - - - - Inert -

272(@26atm)Hydrogen H 1 1.008 - - - - 1+ -259Iodine I 53 126.91 4.93 Ortho. 0.136 0.220 1- 114Iron Fe 26 55.85 7.87 BCC 0.124 0.077 2+ 1538Lead Pb 82 207.2 11.35 FCC 0.175 0.120 2+ 327Lithium Li 3 6.94 0.534 BCC 0.152 0.068 1+ 181Magnesium Mg 12 24.31 1.74 HCP 0.160 0.072 2+ 649Manganese Mn 25 54.94 7.44 Cubic 0.112 0.067 2+ 1244Mercury Hg 80 200.59 - - - 0.110 2+ -38.8


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Molybdenum Mo 42 95.94 10.22 BCC 0.136 0.070 4+ 2617APPENDIX B/LAMPIRAN B

Tabulation of Diffusion Data

Diffusing

Species Host Metal Do(m2/s)

Activation Energy Qd Calculated Values

kJ/mol eV/atom T(oC) D (m2/s)

Fe -Fe (BCC) 2.8 x 10-4 251 2.60 500 3.0 x 10-21

900 1.8 x 10-15

Fe -Fe (FCC) 5.0 x 10-5 284 2.94 900 1.1 x 10-17

1100 7.8 x 10-16

C -Fe 6.2 x 10-7 80 0.83 500 2.4 x 10-12

900 1.7 x 10-10

C -Fe 2.3 x 10-5 148 1.53 900 5.9 x 10-12

1100 5.3 x 10-11

Cu Cu 7.8 x 10-5 211 2.19 500 4.2 x 10-19

Zn Cu 2.4 x 10-5 189 1.96 500 4.0 x 10-18

Al Al 2.3 x 10-4 144 1.49 500 4.2 x 10-14

Cu Al 6.5 x 10-5 136 1.41 500 4.1 x 10-14

Mg Al 1.2 x 10-4 131 1.35 500 1.9 x 10-13

Cu Ni 2.7 x 10-5 256 2.65 500 1.3 x 10-22


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EQUATIONS:

% Ionic Character = { 1exp[-(0.25)(XAXB)2] } x 100

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Answer Final Exam Sem i 1213-Edit