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9CHAPTER 1
10TEST YOUR UNDERSTANDING
11	The properties of materials are highly anisotropic when they are in the form of
12Sifat-sifat bahan adalah sangat tak isotrop apabila mereka adalah dalam bentuk
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14(b) Single crystalline
15Single crystalline- solid in which the regular periodic arrangement of atoms extends over the entire volume of the solid.
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17	In which of the following materials there is long range order?
18Di mana bahan berikut terdapat perintah jarak jauh?
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20	Single crystal
21	Single crystalline-solid in which the regular periodic arrangement of atoms extends over the entire volume of the solid. It posses long range order. 
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23	Which of the following materials normally exist(s) in amorphous forms?
24Antara bahan berikut biasanya wujud () dalam bentuk amorfus?
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26	Window glass
27Amorphous materials have no regular arrangement of atoms or molecules
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29	Among the crystal systems, the one with the least symmetry is
30Antara sistem kristal, dengan simetri-kurangnya ialah
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32	Triclinic
33Simple Triclinic
34a≠b≠c
35α≠β≠ϒ≠90⁰
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37	The direction along the face diagonal of a unit cell of a cubic crystal is denoted by
38Arah di sepanjang diagonal muka sel unit kristal padu ditandakan oleh
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40( c) [110]
41Lattice points at the corners and at the centers of all the six faces
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43	In a body-centered cubic lattice which one of the following directions has the maximum linear density?
44Dalam kekisi yang berpusatkan badan padu yang salah satu arahan berikut mempunyai ketumpatan linear yang maksimum?
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46	[111]
47Lattice points at the corners and at the body centre of the unit cell
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52	A crystal plane intercepts the crystal axes at 0.5 a,b, and is parallel to the c axis. The Miller indices of the plane is
53A memintas pesawat kristal kristal paksi pada 0.5 1, b, dan adalah selari dengan paksi c. Indeks Miller pesawat
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55(b) (210)
56The intercept of the plane : 1, 1/2b, ∞
57The reciprocal of the intercepts : 2, 1,0
58The Miller indices are: (210)
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60	Which one of the following planes in a FCC lattice has the highest planar density?
61Salah satu daripada pesawat yang berikut dalam kekisi FCC yang mempunyai ketumpatan satah tertinggi?
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63	(110)
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66	Which of the following structures have the highest Packing Factor?
67Antara struktur berikut, yang manakah mempunyai Faktor Pembungkusan tertinggi?
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69	FCC
70(d ) HCP
71There are 8 corner atoms and 6 atoms at the centre of the faces.
72The number of atoms per unit cell is: (1/8 X 8)+(1/2 X 6)=4
73Volume occupied by 4 atoms = 4X4 πR^3/3
74Volume of the unit cell=a^3= 〖(4R/√2)〗^3
75Packing factor of FCC lattice= (16πR^3)/((4R/√(2)〖^3〗)) = 0.74
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77	How many atoms per unit cell are there in diamond cubic structure?
78Berapa banyak atom per unit sel yang terdapat dalam struktur berlian padu?
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80( c) 8
81Number of atoms in the unit cell = (8 X 1/8) + (6 X ½) +4 = 8
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83	The coordination number in FCC structure is
84Bilangan koordinasi dalam struktur FCC
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86( d) 12
87The number of nearest neighbor= 12
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89	The ionic radius of Potassium is 1.33 and that of chlorine is 1.81. The stable configuration of KCl structure is
90Jejari ion Kalium adalah 1.33 dan klorin adalah 1.81. konfigurasi yang stabil struktur KCl
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92( d) cubic
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97	The number of nearest neighbor in Octahedral configuration is
98Bilangan jiran terdekat dalam konfigurasi oktahedral
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100	6
101The number of nearest neighbor=6
102Radius Ratio range= 0.414< x <0.732
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104	In a BCC lattice, in which of the following planes there is no Bragg reflection?
105Dalam kekisi BCC, di mana pesawat yang berikut tiada pantulan Bragg?
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107( d) (221)
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109	The crystal structure of germanium is 
110Struktur hablur germanium
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112( b) diamond cubic structure
113Besides diamond, the elemental semiconductors silicon and germanium crystallize in this structure. The structure belongs to the FCC Bravais lattice, with two atom forming the basis. Each atom has four nearest neighbor in a tetrahedral coordination with a bond of 109.5⁰. 
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115	Among the following bonds, which one is the weakest?
116Antara bon berikut, yang mana satu adalah yang paling lemah?
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118( d) van der Waals bonds
119The van der Waals bonds exist in crystals of inert gases and between molecules in organic molecular solids. 
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121	What types of bonding exists between water molecules and ice?
122Apakah jenis ikatan yang wujud di antara molekul air dan ais?
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124	Hydrogen bond
125Hydrogen molecule is a four- particles system. The two hydrogen nuclei A and B and the electron of the 1s orbital of nucleus A, referred to as electron 1 and the electron of the 1s orbital of nucleus B, referred to as electron 2. The four particles interact with each other. 
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127	Van der Waals bond is formed due to
128Van der Waals bon terbentuk disebabkan oleh
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130	Dipole- dipole interaction
131The electric field will cause an induced dipole moment in the neighboring atom. 
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133	The type of bonding in silicate crystal is 
134Jenis ikatan dalam kristal silikat
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136	Partially ionic and partially covalent
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139	The empirical expression for the repulsive force between two bonding atoms is expressed as A/R^12 or  λexp(-R)/ρ.
140Ungkapan empirik bagi daya tolakan antara dua atom ikatan dinyatakan sebagai A R / ^ 12 atau λ exp (-R) / ρ.
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142	Among the atomic bonds the one which has highly directional character is the covalent bond.
143Antara bon atom yang mempunyai watak yang sangat arah ikatan kovalen.
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145	In a metal crystal, the bonding is mainly due to the interaction between ion cores and conductions electrons. 
146Dalam kristal logam, ikatan adalah disebabkan oleh interaksi di antara teras ion dan elektron conductions.
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148	The unit of cohesive energy of crystal is kJ/mole.
149Unit tenaga padu kristal adalah kJ / mol.
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151	The type of bonding that occurs between molecules in a molecular solid is van der Waals. 
152Jenis ikatan yang berlaku antara molekul dalam pepejal molekul van der Waals.
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154	Example of crystal with perfect covalent bonding is diamond.
155Contoh kristal dengan ikatan kovalen yang sempurna adalah berlian.
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157	The number of orbital and the angles between their direction in sp,〖sp〗^2, and 〖sp〗^3 hybridizations are respectively 2 and 180⁰, 3 and 120⁰ and 4 and 109⁰.
158Bilangan orbital dan sudut antara arah mereka di sp, 〖sp〗 ^ 2, dan 〖sp〗 ^ 3 hybridizations adalah masing-masing 2 dan 180 ⁰, 3 dan 120 ⁰ dan 4 dan 109 ⁰.
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160	The Coulomb integral, exchange integral and the overlap integral are given by: 
161C=∫〖Ψ*〗_A(1)〖Ψ*〗_B(2)H’〖Ψ*〗_A(1)〖Ψ*〗_B(2) dτ_1dτ_2
162J=∫〖Ψ*〗_A(1)〖Ψ*〗_B(2)H’〖Ψ*〗_A(1)〖Ψ*〗_B(2) dτ_1dτ_2
163S=∫〖Ψ*〗_A(1)〖Ψ*〗_B(2)H’〖Ψ*〗_A(1)〖Ψ*〗_B(2) dτ_1dτ_2
164Kamiran Coulomb, pertukaran penting dan pertindihan penting diberi oleh:
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166	In the homonuclear diatomic molecule with molecular axis- along the x-axis, the p_z  orbital of the two atoms combine from π molecular bond (σ,π).
167Dalam molekul dwiatom homonuclear dengan molekul paksi di sepanjang paksi-x, orbit p_z dua atom bergabung dari bon molekul π (σ, π).
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169	The outermost filled molecular orbital in nitrogen molecule is bonding orbital. 
170Diisi orbit terluar molekul molekul nitrogen adalah ikatan orbit.
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176CHAPTER 2: IMPERFECTION OF CRYSTALL
177Part 1
1781. Schottky defects normally occur / Kecacatan Schottky biasanya berlaku
179Answer: (D) in metallic and ionic crystals / dalam kristal logam dan ion
180Reason: Schottky defects are normally found in ionic crystals. It involves vacancies of pairs of ions of opposite charges.
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1822. In Frenkel defects the ions that get transferred to interstitial positions mostly are
183Dalam kecacatan Frenkel ion yang dapat dipindahkan ke jawatan celahan kebanyakannya
184Answer: (A) Cations /Kation
185Reason:In ionic crystals since cation are generally smaller than anions, it is the cations that get transferred into interstitial sites.
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1873. If n is the number of vacancies in a crystal of N atoms at temperature T, and Ev is the energy of formation of a vacancy, the slope of the graph of In (n/N) versus 1/T is
188Jika n ialah bilangan kekosongan di dalam satu hablur atom N pada suhu T, dan Ev tenaga  
189  pembentukan kekosongan, kecerunan graf of In (n / N) berbanding 1 / T
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191Answer: (A) - E_v⁄k_B 
192Reason: Ev is the energy of formation of a vacancy to transfer an atom from the lattice site within a crystal lattice to surface of crystal. By using eq (2.1) and (2.3) in (2.2) 
193– Material Science book page 88
194∆F=nE_v-k_B T In  (N !)/(n !(N-n )!)
195Then, minimizing the free energy ( (d (∆F))/dn=0 ) we get
196n/(N-n)=exp⁡〖(-E_v/k_B T〗) ; using Stirling’s formula
197 In x! = x In x- x   
198∴ New equation:
199n/N=(-E_v/k_B T)
2004. The Burgers vector in an FCC crystal or lattice parameter α is
201    Vektor Burger di kristal FCC atau kekisi parameter α
202Answer: (B) (a/2) <110>
203Reason:
204Crystal Structure              	Possible slip plane	Slip direction	Burgers vector
205CS	{100}	<100>	a<100>
206FCC	{111}	<110>	a/2<110>
207BCC	{110}	<111>	a/2<110>
208HCP	{101}	<110>	a<110>
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2115. The imperfection between two crystallites in a polycrystalline material is classified 
212as
213   Ketidaksempurnaan antara dua crystallites dalam bahan polihablur dikelaskan sebagai
214Answer: (B) Plane defect / Plane kecacatan
215Reason: Surface imperfection is imperfections on a crystal surface having a thickness of a few atomic diameters.
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2176.  A crystal will be mechanically weak if / Kristal akan mekanikal lemah jika
218Answer: (B) there are a few dislocations in the crystal / terdapat beberapa kehelan 
219                  dalam kristal
220Reason: In crystal, with those close packed structures if one of the planes slips or one of planes is missing, then the arrangement has a fault.
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230Part 2
231	Presence of pair vacancies is energetically more favourable. / Kehadiran kekosongan pasangan lebih memihak kepada lebihan tenaga.
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233	The number of vacancies in a crystal decreases with decrease of temperature.Bilangan kekosongan ruang di dalam satu hablur berkurangan dengan penurunan suhu.
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235	In edge dislocation, Burgers vector is perpendicular to the dislocation line. / Dalam kehelan pinggir, Burger vektor adalah berserenjang dengan garis kepincangan.
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237	The slip planes in crystals are normally planes of highest atomic packing. / Satah gelincir dalam kristal biasanya satah pembungkusan atom tertinggi.
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239	The movement of dislocation is easier in FCC crystal. / Pergerakan kehelan adalah lebih mudah di FCC kristal.
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242Part 3
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244	For Schottky defect obtain Eq. 2.7 by minimizing the free energy (use Stirling’s formula). / Untuk kecacatan Schottky mendapatkan Persamaan 2.7 dengan meminimumkan tenaga bebas (gunakan formula Stirling).
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246Solution:
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248Formation of Schottky defect or a pair vacancy is energetically more favourable than the formation of single anion or cation vacancy.
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250The number of vacant pairs in a crystal of N ions at temperature T is given by
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252npair ≅N exp⁡〖(-E_p/2k_B T)〗
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254where: Ep is the energy required to create a vacant pair
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256This relation is obtained by taking the number ways n pairs may be formed as 〖((N !)/(n !(N-n)! ))〗^2and by using Stirling’s formula;
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258 In x! = x In x- x
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260∴Newequation:
261n/N=(-E_p/〖2k〗_B T)
262	The energy formation of vacancy on copper crystal is 0.90 eV. Find the factor by which the number of vacancies in a copper crystal would increase if it is heated from 300 to 500K. / Pembentukan tenaga kekosongan kristal tembaga adalah 0.90 eV. Cari faktor oleh bilangan kekosongan ruang dalam kristal kuprum akan meningkat jika ia dipanaskan dari 300 hingga 500K
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264Solution:
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266At 300K;
267k_B T=(〖(1.38 x 10〗^(-23)  x 300))/〖1.6 x 10〗^(-19) =0.0259 eV
268n/N=exp⁡〖(-E_v/(k_B T))=⁡█(e(-0.90eV/0.0259eV)@) 〗
269= 8.1037 x 〖10〗^(-16)/atom
270At 500K;
271k_B T=(〖(1.38 x 10〗^(-23)  x 500))/〖1.6 x 10〗^(-19) =0.0431 eV
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273n/N=exp⁡〖(-E_v/(k_B T))=⁡█(e(-0.90eV/0.0431eV)@) 〗
274=8.535 x 〖10〗^(-10)/atom
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276Factor by which the number of vacancies in copper crystal would increase from 300 to 500K;
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278〖10〗^(-10)/〖10〗^(-16) =〖10〗^6 times
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280	Average energy required to create a vacancy in a metal is 1 eV. Calculate the ratio of vacancies at 1000K to that at 500K. / Tenaga purata yang diperlukan untuk mewujudkan kekosongan dalam logam adalah 1 eV. Kira nisbah kekosongan pada 1000K itu pada 500K.
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282Solution:
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284By using formula;
285n=Nexp (-E_v/(k_B T))
286where; 
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288n – no. of vacancies
289N – total no. of metal vacancies
290Ev – energy to create the vacancies
291kB– Boltzman constant
292T – absolute temperature 
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295n_500=Nexp (-E_v/(500k_B ))andn_1000=Nexp (-E_v/(1000k_B ))
296Therefore, n_500/n_1000 =exp⁡[-E_v/(1000k_B )+E_v/(500k_B )]
297        =  exp⁡[E_v/(1000k_B )]
298Since Ev in eV therefore Boltzman constant also expressed in eV.
299k_B=(1.38x〖10〗^(-23))/(1.6 x 〖10〗^(-19) )
300= 8.625 x 10-5 eV/K
301n_500/n_1000 =exp⁡█([1/(1000)(8.625x〖10〗^(-5) ) ]@)
302In (n_500/n_1000 )=  1/((8.625x〖10〗^(-2) ) )
303〖       log〗_10 (n_500/n_1000 )=1/(2.303)(8.625x〖10〗^(-2) ) 
304=5.30344
305or                                n_500/n_1000 =1.082x〖10〗^5
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307	Iron is BCC with a lattice constant of 2.8Ã…. If slip occurs causing edge dislocation what would be the length of the Burgers vector? / Besi adalah BCC dengan kekisi tetap 2.8 Ã…. Jika slip berlaku menyebabkan kehelan pinggir apa yang akan menjadi panjang vektor Burger?
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309Solution:
310Iron – BCC
311a=2(R+r)
312a and R related by;
313√3 a=4R
314a=4R/√3
315                        =4 (2.8)/√3
316=6.47nm
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318R+r=a/2
319=6.47/3
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321=3.23
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323∴r=3.23-R
324=3.23-0.41
325=2.82nm
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328	Using the data in Table 2.1, indicate with a neat diagrams, the slip planes, the slip direction and the Burgers vectors in SC, FCC, BCC and HCP crystal. / Menggunakan data dalam Jadual 2.1, menunjukkan dengan gambarajah yang kemas, satah slip, slip arah dan vektor Burger di SC, FCC, BCC dan HCP kristal.
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330Solution:
331Crystal Structure	Possible slip plane	Slip direction	Burgers vector
332CS	{100}	<100>	a<100>
333FCC	{111}	<110>	a/2<110>
334BCC	{110}	<111>	a/2<110>
335HCP	{101}	<110>	a<110>
336Table 2.1
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339	Density of Schottky defect in a sample of NaCl is 5 x 1011 /m3 at 25oC. Na+ - Cl- distance is 2.82Ã…. Calculate the average energy required to create a Schottky defect. / Ketumpatan kecacatan Schottky dalam sampel NaCl 5 x 1011 / m3 pada suhu 25oC. Na + - Cl- jarak 2.82 Ã….Hitungkan tenaga purata yang diperlukan untuk mewujudkan kecacatan Schottky.
340Solution:
341The unit cells of Sodium Chloride without defects contains four-ion pairs and its volume will be (2 x 2.82)3 x 10-30= 1.794 x 10-28 m3.
342Therefore, 1 m3 of an ideal crystal will contain,
3434/(1.794 x 〖10〗^(-28) )=2.230 x 〖10〗^28  ion pairs
344Now, 
345n/N=exp⁡(-E_v/(〖2k〗_B T))
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347Therefore; 
3482.303〖log〗_10  n/N=exp⁡(-E_v/(k_B T))
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350or Ev = (2.303) (16.65) (2) (8.625x105) (298)
351    = 1.971 eV
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374CHAPTER 3: DIFFUSION IN SOLID
375Part 1
376	Fick’s law describe flow of atoms caused by / Undang-undang Fick menggambarkan aliran atom yang disebabkan oleh
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378Answer : (C). Concentration gradient / Kecerunan kepekatan
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380Reason : Consider system in which a concentration gradient of atom is maintained in the X-direction.
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383	Fick’s first law describes diffusion process when / Hukum pertama fick menerangkan proses resapan apabila
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385Answer : (D) concentration depends on space but independent of time / kepekatan bergantung pada ruang tetapi bebas masa
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387Reason : Fick’s first law describes flow under steady condition. For example, the concentration profile is maintained the same throughout the flow process. In this case, the flux will be independent of time.
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389	The error function has the property / Fungsi ralat mempunyai ciri
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391Answer : (C) erf (0) = 0
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393Reason : Initial concentration of the solute in the baris uniform throughout the length of the bar.
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395	Carburization of steel / Penyahkarbonan keluli
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397Answer : (A) Improves its fatigue resistance / Meningkatkan daya ketahanan 
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399Reason : The process of carburization works via the implantation of carbon atoms in to the surface layers of a metal. As metals are made up of atoms bound tightly into a metallic crystalline lattice, the implanted carbon atoms force their way into the crystal structure.
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404	Decarburization of steel occurs when steel occurs when steel is exposed to / Penyahkarbonan keluli berlaku apabila keluli berlaku apabila keluli terdedah kepada
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406Answer : (C) oxygen atmosphere / oksigen atmosfera
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408Reason : When steel is exposed to oxygen at an elevated temperature, it is subject to decarburization.
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411Part 2
412
413	The unit of diffusion coefficient is m2 s-1.
414Unit pekali resapan adalah m2 s-1.
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416	The temperature dependence of diffusion coefficient / Pergantungan suhu pekali resapan
417	In terms of activation energy given in unitsof J/mol is / Dari segi tenaga pengaktifan yang diberikan dalam unit  J / mol  D=D_o e^(-Q/RT)
418	In terms of activation energy given in units of J/atom is / Dari segi tenaga pengaktifan yang diberikan dalam unit J / atom adalahD= D_o e^(〖-E〗_(D/k_B T ) )
419
420	Diffusion coeffiecient increases with increasing temperature. / Pekali resapan yang meningkat dengan peningkatan suhu.
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422	Error function is defined by the definite integral / Fungsi kesilapan ditakrifkan oleh penting yang pasti
423
424	The diffusion of carbon in Fe (α) is lower than Fe (γ). / Penyebaran karbon dalam Fe (α) adalah lebih rendah daripada Fe (γ).
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438CHAPTER 4
439
440TEST YOUR UNDERSTANDING
441
442	When nickel is added to copper, at the temperature of about 1000⁰C 
443Apabila nikel ditambah tembaga,pada suhu kira-kira 1000⁰C
444
445	The alloy is in the FCC solid solution state for all compositions up to 100 wt% Ni
446Since Cu and Ni have nearly the same atomic radii and same crystal structure (FCC), the resultant alloy has the FCC structure for any composition of Ni in Cu from 0 to 100%
447
448	In copper- nickel alloy
449Dalam aloi tembaga-nikel
450
451       (b) there is no eutectic point
452Eutectic point, E is the point at which the liquidus line and the solidus line meet. Eutectic point also coexist three phases, there are liquid phase and two solid phases. If we see the phase diagram of copper-nickel alloy, it only has one solid phase. So, eutectic point does not exist in copper-nickel alloy.
453
454	A Tie- line is used for finding
455Satu talian Tie digunakan untuk mencari
456
457( c) composition of alloy in the two phases in equillibrium
458The Tie- line used to find the compositions of the alloy in the constituent phases in the binary alloy. 
459
460	In the phase diagram, the curve above which the alloy is in the liqiud state for all compositions is called the liquidus. 
461Dalam gambarajah fasa, lengkung di atas mana aloi di negeri liqiud untuk semua komposisi dipanggil liquidus.
462
463The line above which the system exists in the liquid state is called liquidus line. 
464Based on the phase diagram, the curve above which the system exists in the liquid state is called the liquidus line
465
466	At the eutectic point in a binary alloy, the number of degrees of freedom is zero.
467Pada titik eutektik dalam aloi perduaan, bilangan darjah kebebasan adalah sifar.
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478	Iron carbon alloy is ferromagnetic at room temperature
479Aloi besi karbon adalah feromagnetik pada suhu bilik
480
481( c) for compositions below 0.022 wt% of carbon
482In the lower temperature BCC phase, only a very small concentration of carbon is soluble. This is because carbon occupies the interstitial sites in the iron lattice and in the BCC structure the size and shape if interstitial sites are not very favourable for the carbon atoms. The maximum solubility is only 0.022 wt%. This phase is called the α phase. This is an important phase of Fe-C alloy because only in this phase the alloy is ferromagnetic below 768℃
483
484	In iron- carbon alloy there are two eutectic points. (true, false )
485Dalam aloi besi-karbon terdapat dua mata eutektik. (benar, palsu)
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488	Iron- carbon alloy of compositions between 0.008 wt% and 2.14 wt% of carbon above 1000⁰C exists in FCC structure and are classified as austenite.
489Besi-karbon aloi komposisi antara 0,008% berat dan 2.14% berat karbon di atas 1000 ⁰ C wujud dalam struktur FCC dan dikelaskan sebagai austenit.
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491Above 727⁰C the solubility of carbon becomes higher. Above 912⁰C iron is in FCC structure and the maximum solubility is 2.14 wt%. This phase is called the ϒ phase (or austenite) and in this phase the alloy is not ferromagnetic.
492
493	Iron- carbon alloy of compositions of carbon between 0.022 wt% and 6.76 wt% at temperatures below 727⁰C exists in two phase form α+Fe₃C.
494Aloi besi-karbon komposisi karbon antara 0,022% berat dan 6,76% berat pada suhu di bawah 727 ⁰ C wujud dalam dua fasa borang α + Fe ₃ C.
495
496In the α phase, if the concentration of the carbon is increased to more than 0.002 wt%, at the temperature below 727⁰C, a two phase system α+Fe₃C is stable. 
497
498	The eutectic temperature of Pb- Sn binary alloy is 183⁰C.
499Suhu eutektik Pb-Sn aloi binari 183 ⁰ C.
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501The soldering lead is made of 60- 40 composition lead- tin alloy so that it melts completely at a relatively low temperature of 183⁰C. 
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509CHAPTER 5 
510
511TEST YOUR UNDERSTANDING
512
513	Among the following types of materials which one has the highest modulus of elasticity?
514Antara jenis-jenis berikut bahan-bahan yang mempunyai modulus keanjalan yang tertinggi?
515
516(b) ceramics
517        The stiffness constant of ceramics materials are highest than polymers, metal, 
518diamond.
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521	Yield strength of the materials is the stress at which the material
522Kekuatan alah bahan adalah tekanan di mana bahan
523
524( c) becomes plastics
525Some materials exhibits a non-linear stress-strain curve. Such materials are said to be plastics. For low stress, the material are elastics(linear stress-strain relationship) and beyond a certain stress value the material becomes plastic(non-linear stress-strain relationship).
526
527	Materials that normally undergo creep are
528Bahan-bahan yang biasanya menjalani rayapan
529
530(b) metals at high temperature
531(c) low melting point materials
532Normally creep occurs in materials at high temperature. Low melting point materials undergo creep even at room temperature.
533
534	Creep curve is a plot of
535Lengkung rayapan adalah satu plot
536
537	Strain versus time
538	Creep curve for a material is drawn between strain and time for a fixed load and at a given temperature.
539
540	Among the following materials, which one has the highest hardness?
541Antara bahan berikut, yang mempunyai kekerasan yang tertinggi?
542
543( d) silicon carbide
544Hardness of material is measured by making a small dent in the material by forcing a small indenter on its surface under standard controlled condition of loading. The depth or the size of the resulting indentation is taken as a measure of hardness.
545
546
547
548
549	Brittle fracture occurs in materials due to
550Patah rapuh berlaku dalam bahan-bahan yang disebabkan oleh
551
552	Presence of cracks
553The fracture in brittle material occurs mainly due to pre-existence of tiny cracks on the surface of the materials, which get introduced during fabrication of the material. 
554
555	Time dependent deformation in which the materials does not recover its original dimension is called
556Ubah bentuk masa bergantung di mana bahan yang tidak pulih dimensi asalnya dipanggil
557
558	Viscoelasticity
559A material which deforms anelastically but does not completely recover its original form after any length of time is said to be viscoelastic. 
560
561 II FILL IN THE BLANKS
562
563	Ductility of a metal decreases with decreasing temperature.
564Kemuluran logam berkurang dengan suhu berkurangan.
565
566In ductile fracture the material undergoes substantial plastic deformation with high energy absorption before fracture and so toughness value of ductile materials is high. 
567
568
569	For a good fatigue resistance material,the slope of the S-N curve is low.
570Untuk bahan rintangan lesu baik, cerun keluk SN adalah rendah.
571
572For the highly fatigue resistant material the S-N curve will almost be horizontal (parallel) to the x-axis.
573
574
575	A material is said to be elastic if sterss-strain curve is linear.
576Bahan A berkata untuk menjadi elastik jika lengkung sterss-terikan adalah linear.
577For the highly fatigue resistant material the S-N curve will almost be horizontal (parallel) to the x-axis
578
579
580	Materials which exhibit non-linear sterss-strain curve have poor fatigue resistance.
581Bahan-bahan yang mempamerkan bukan linear sterss-terikan lengkung mempunyai rintangan lesu miskin.
582
583	Creep is more likely to occur in low melting point materials.
584Rayapan adalah lebih kerap berlaku dalam bahan titik lebur rendah.
585Low materials like lead, plastic, undergo creep even at room temperature.
586
587
588
589	Brittle type of fracture is accompanied by high energy absorbption.
590Jenis patah rapuh disertai oleh serapan tenaga yang tinggi.
591
592In brittle fracture the material remains perfectly elastic upto the fracture point. The energy of absorption accompanying brittle fracture is low,i.e toughness values of brittle materials are low. The fracture in materials occurs mainly due to pre-existence of tiny cracks on the surface of the material.
593
594	The failure of a material due to creep is normally called rupture.
595Kegagalan bahan akibat daripada rayapan biasanya dipanggil pecah.
596
597Stage 3 is called the tertiary creep. In this stage the creep rate increases with time and ultimately the material fails at the time. The materials is said to have ruptured and the time is called the rupture life time.
598
599	The area under the sterss-strain curve gives toughness of the materials. 
600Kawasan di bawah lengkung sterss-terikan memberikan keliatan bagi bahan tersebut.
601
602Toughness is defined as the ability of a material to absorb energy upto fracture. It is the area under the stress-strain curve. The unit of toughness is energy per unit volume.