answer schemes universiti malaysia perlis ent 145

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  • ANSWER SCHEMES

    UNIVERSITI MALAYSIA PERLIS

    Peperiksaan Akhir Semester Pertama

    Sidang Akademik 2014/2015

    Januari 2015

    ENT 145 Material Engineering

    [Kejuruteraan Bahan]

    Masa: 3 jam

    Please make sure that this question paper has EIGHT (8) printed pages including this front

    page before you start the examination. [Sila pastikan kertas soalan ini mengandungi LAPAN (8) muka surat yang bercetak termasuk muka hadapan

    sebelum anda memulakan peperiksaan ini.]

    This question paper has SIX (6) questions. Answer ALL questions in PART A, and ONE (1)

    question in PART B. Each question contributes 20 marks. [Kertas soalan ini mengandungi ENAM (6) soalan. Jawab SEMUA soalan dalam BAHAGIAN A, dan SATU

    (1) soalan dalam BAHAGIAN B. Markah bagi tiap-tiap soalan adalah 20 markah.]

  • Part A Answer ALL questions [Bahagian A Jawab semua soalan]

    Question 1 (CO1, C4) [Soalan 1]

    (a) Identify four components and its specific materials used in a car. [Tentukan empat komponen-komponen dan bahan-bahan terperincinya digunakan dalam sebuah

    kereta.] (4 Marks / Markah)

    Answer:

    1- The engine metal (cast iron or aluminum alloys)

    2- Body metal (thin steel or aluminum alloys) also advanced composites (carbon fiber

    composites)

    3- Front panel mostly polymeric materials (polycarbonates)

    4- Tires polymeric composite (synthetic rubber, polyester fabric, steel belts)

    5- Light fixture polymeric glass (Plexiglass)

    6- Wires - metals (high conductivity copper)

    7- Windshield laminated glass (ceramic glass, acrylic and cellulose)

    8- Springs mostly steel alloys

    (b) With the aid of sketches, explain the main differences between ionic, covalent and metallic bonding. [Dengan bantuan lakaran, terangkan perbezaan utama antara ikatan ionik, ikatan kovalen dan ikatan

    logam.]

    (6 Marks / Markah)

  • (c) Atom X has an atomic radius of 0.1345 nm and density of 12.41 g/cm3. Given its atomic weight is 102.91 g/mol. [Atom X mempunyai jejari atom 0.1345 nm dan ketumpatan 12.41 g/cm

    3. Diberikan berat atomnya

    ialah 102.91 g/mol.]

    (i) Determine the crystal structure for atom X. [Nilai struktur kristal bagi atom X.]

    (6 Marks / Markah)

    (ii) Determine the atomic packing factor (APF) for crystal structure in (i).

    [Tentukan faktor pemadatan atom (APF) bagi struktur kristal pada (i).]

    (4 Marks / Markah)

  • Question 2 (CO1, C4) [Soalan 2]

    (a) Low-alloy steel can be categorized into low-carbon steel, medium-carbon steel and high carbon steel. For each of the low-alloy steel, describe the properties and its

    typical applications. [Keluli beraloi rendah dibahagikan kepada keluli berkarbon rendah, keluli berkarbon sederhana dan

    keluli berkarbon tinggi. Untuk setiap keluli beraloi rendah, terangkan sifat dan penggunaannya.] (6 Marks / Markah)

    Answer:

    Low carbon steel < 0.25wt%C Bridge,towers,auto

    structures sheet

    Medium carbon steel 0.25-0.6wt% C Piston,gears,etc

    High carbon steel 0.6 -1.4 wt% C Turbine,furnace etc

    (b) Briefly explain conditions of atomic motion in diffusion. [Terangkan secara ringkas syarat-syarat pergerakan atom dalam penyebaran.]

    (2 Marks / Markah) Answer:

    1) There must be empty adjacent site. 2) Must have sufficient energy to break bonds with its neighbour atoms and then cause some

    lattice distortion during the displacement.

    (c) Discuss the concept of non-steady state as it applies to diffusion. Give an example in your answer. [Bincangkan konsep keadaan tidak stabil apabila ianya digunapakai dalam penyebaran. Berikan satu

    contoh pada jawapan anda.]

    (4 Marks / Markah) Answer:

  • (d) A portion of the iron iron carbide (Fe-FeC) phase diagram is shown in Figure 1. Consider 3.0 kg of austenite containing 1.15 wt % C, cooled to below 725 C.

    [Satu bahagian bagi gambarajah fasa besi-besi karbida ditunjukkan dalam Gambarajah 1. Pertimbangkan 3.0 kg austenite yang mengandungi 1.15 wt % C, disejukkan ke bawah 725

    oC.]

    (JW Chap 9, Prob 50)

    (i) Determine the proeutectoid phase. [Tentukan fasa proeutektoid.]

    (2 Marks / Markah)

    Answer:

    The proeutectoid phase will be Fe3C since 1.15 wt % C is greater than the eutectoid

    composition (0.76 wt %).

    (ii) Calculate the mass of total ferrite and cementite form. [Kira jisim bagi jumlah ferit dan cementit yang terbentuk.]

    (3 Marks / Markah)

    Answer:

    (iii) Calculate the mass of the pearlite and proeutectoid phase form. [Kira jisim bagi pearlite dan fasa proeutektoid yang terbentuk .]

    (3 Marks / Markah)

    Answer:

  • Figure 1 [Gambarajah 1]

  • Question 3 (CO2, C6) [Soalan 3]

    (a) Discuss the differences between brittle and ductile metals with respect to its tensile stress-strain behaviour. [Bincangkan perbezaan antara logam rapuh dan logam mulur berdasarkan kepada kelakuan tegangan

    tegasan-terikan.]

    (4 Marks / Markah)

    Answer:

    (b) A cylindrical specimen of hypothetical metal alloy has a diameter of 8.0 mm. A tensile force of 1000 N produces an elastic reduction in diameter of 2.8 x 10

    -4 mm.

    Compute the modulus of elasticity for this alloy, given that the Poissons ratio is 0.30. [Satu spesimen bagi logam aloi hipotesis berbentuk silinder mempunyai garis pusat 8.0 mm. Daya

    tegangan sebanyak 1000 N menghasilkan pengurangan anjal bagi garis pusat sebanyak 2.8 x 10-4

    mm.

    Kira modulus anjal bagi aloi ini, diberi nisbah Poissons ialah 0.30.]

    (6 Marks / Markah)

    Ductile Metal experience plastic

    deformation upon fracture

    Brittle Metal very little or no

    plastic deformation

  • (c) Table 1 shows a list of materials and their mechanical properties. Each of the material will be tested as a cylindrical rod specimen with 100 mm long and having a diameter

    of 10 mm. If the tensile load is 27.5 kN, answer the following questions. [Jadual 1 menunjukkan senarai bagi bahan dan sifat mekanikalnya. Setiap bahan tersebut akan diuji

    sebagai satu specimen rod silinder dengan panjang 100 mm dan garis pusat 10 mm. Jika beban

    tegangan ialah 27.5 kN, jawab soalan-soalan berikut.]

    (i) From Table 1, choose the material(s) that will not experience plastic deformation. Justify your choice(s). [Daripada Jadual 1, pilih bahan yang tidak akan mengalami ubah bentuk plastik. Justifikasi

    pilihan anda.]

    (6 Marks / Markah)

    (ii) By referring to answer in (i), select the material(s) that will not experience a

    diameter reduction of more than 7.5 x 10-3

    mm. [Dengan merujuk kepada jawapan di (i), pilih bahan yang tidak akan mengalami pengurangan

    garis pusat tidak lebih daripada 7.5 x 10-3

    mm.]

    (4 Marks / Markah)

    Table 1 [Jadual 1]

    Material Modulus of

    Elasticity (GPa)

    Yield Strength

    (MPa) Poissons Ratio

    Aluminum alloy 70 200 0.33

    Brass alloy 101 300 0.34

    Steel alloy 207 400 0.30

    Titanium alloy 107 650 0.34

  • Question 4 (CO2, C6) [Soalan 4]

    (a) Briefly explain ductile iron and malleable iron. (Chap 9, Prob 37) [Terangkan secara ringkas besi mulur dan besi boleh tempa.]

    (4 Marks / Markah)

    Answer:

    (b) Cast irons are very important engineering materials. Briefly explain cast irons and

    their basic range of composition. Give their applications and four basic types of cast

    irons.

    (Chap 9, Prob 36) [Besi tuang adalah bahan kejuruteraan yang sangat penting. Terangkan secara ringkas besi tuang dan

    lingkup asas komposisi. Beri kegunaan-kegunaan mereka dan empat jenis asas besi tuang.] (6 Marks / Markah)

    Answer:

    1. Cast irons are a family of ferrous alloys intended to be cast into a desired shape rather than

    worked in the solid state.

    2. These alloys typically contain 2 - 4 % C and 1 3 % Si. Additional alloying elements may

    also be present to control or vary specific properties.

    3. Cast irons are easily melted and highly fluid and do not form undesirable surface films or

    shrink excessively; consequently, they make excellent casting irons. They also possess a wide

    range of strength and hardness values and can be alloyed to produce superior wear, abrasion,

    and wear resistance. In general, they are easy to machine.

    4. Their applications: engine cylinder blocks, gear boxes, connecting rods, valve and pump

    casings, gears, rollers, and pinions.

    5. The 4 basic types: white, gray, ductile and malleable.

    (c) Plain carbon and alloy steels are extensively used in manufacturing of bolts and screws. Give five reasons for this. (Chap 9, Prob 112) [Keluli karbon biasa dan keluli aloi digunakan dengan meluas dalam pembuatan bolt dan skru.

    Berikan lima sebab untuk ini.] (5 Marks / Markah)

    Answer:

    1. Bolts and screws must be made of materials that are high strength (to avoid failure)

    2. High modulus of elasticity (to allow high pre-loads without yielding)

  • 3. Resist fatigue failure (especially in applications that involve loading and unloading) even

    in the presence of threads.

    4. Be economical.

    5. Have ease of manufacturing steels satisfy all of these conditions.

    (d) For a plain carbon steel has carbon content of 1 wt % at 900 C. On average, how many carbon atoms can you find in 100 unit cells? If at room temperature, the carbon

    content of ferrite drops to 0.005 wt %, on average, how many unit cells would you

    have to search to find one carbon atom?