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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