answer schemes universiti malaysia perlis ent 145

17
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.]

Upload: truongque

Post on 12-Jan-2017

245 views

Category:

Documents


2 download

TRANSCRIPT

Page 1: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

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.]

Page 2: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

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)

Page 3: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

(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)

Page 4: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

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:

Page 5: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

(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 725oC.]

(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:

Page 6: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

Figure 1 [Gambarajah 1]

Page 7: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

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 Poisson’s 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 Poisson’s ialah 0.30.]

(6 Marks / Markah)

Ductile Metal experience plastic

deformation upon fracture

Brittle Metal very little or no

plastic deformation

Page 8: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

(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) Poisson’s 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

Page 9: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145
Page 10: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

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)

Page 11: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

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? Briefly explain the differences in these two

cases.

(Chap 9, Prob 93) [Bagi keluli karbon biasa dengan 1 % berat kandungan karbon pada 900 ° C. Pada purata, berapa

banyak atom karbon boleh anda dapati dalam 100 unit sel? Jika pada suhu bilik, kandungan karbon

ferit jatuh kepada 0.005 % berat, secara purata, berapa banyak sel unit anda perlukan untuk mencari

satu atom karbon? Terangkan secara ringkas perbezaan di dalam kedua-dua kes.] (5 Marks / Markah)

Page 12: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

Part B – Answer ONE (1) question [Bahagian B – Jawab SATU (1) soalan.]

Question 5 (CO3, C6) [Soalan 5]

(a) Describe ductile, brittle and fatigue failures. Explain in terms of failure surface. [Terangkan kegagalan mulur, rapuh dan kelelahan. Terangkan dari segi permukaan kegagalan]

(6 Marks / Markah)

Answer:

Ductile Fracture:

– Accompanied by

significant plastic

deformation

Brittle Fracture:

– Little or no plastic

deformation

– Catastrophic

Page 13: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

(b) A fatigue test is made with a maximum stress of 120 MPa and a stress amplitude of

165 MPa. Calculate the maximum and minimum stresses, the stress ratio and the

stress range.

(Chap 7, Prob 28) [Ujian kelelahan dibuat dengan tekanan maksima 120 MPa dan amplitud tekanan 165 MPa. Kirakan

tegasan maksima dan minima, nisbah tekanan dan julat tekanan.] (4 Marks / Markah)

Answer:

(c) An aircraft component is fabricated from an aluminium alloy that has plane strain

fracture toughness of 35 MPa√m. Fracture occurred at a stress of 250 MPa when the

maximum internal crack length is 2.0 mm.

(JW, Chap 8, Prob 6) [Satu komponen kapal terbang telah difabrikasi daripada aloi aluminum yang mempunyai keliatan

patah terikan satah 35MPa√m. Patah terjadi pada tegasan 250 MPa bila panjang retak dalaman

maksimum ialah 2.0mm.]

(i) Determine the value of dimensionless correction factor, Y. [Tentukan nilai bagi faktor pembetulan tak berdimensi, Y.]

(4 Marks / Markah)

Answer:

Page 14: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

(ii) If a stress level is increased to 325 MPa and the maximum internal crack length is

reduced to 1.1 mm, predict whether any fracture will occur. Compare your

answer with existing data. Justify your answer. [Jika aras tegasan ditingkatkan kepada 325MPa dan panjang retak dalaman maksimum

dikurangkan kepada 1.1mm, anggarkan sama ada patah akan terjadi. Bandingkan jawapan anda

dengan data yang tersedia. Justifikasi jawapan anda.] (6 Marks / Markah)

Answer:

Question 6 (CO3, C6) [Soalan 6]

(a) Pit, intergranular and stress corrosions are example of metallic form corrosion. [Kakisan bopeng, kakisan antara butir, kakisan tegasan adalah contoh bagi kakisan bentuk logam.]

(i) Describe conditions under which these corrosions occur. [Jelaskan dibawah keadaan apa berlakunya kakisan.]

(3 Marks / Markah)

Answer:

Pitting – Downward propagation of small pits and holes

Intergranular – Corrosion along grain boundaries often where precipitate particles

form

Stress corrosion - Corrosion at crack tips when a tensile stress is present and corrosive

environment

Page 15: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

(ii) Discuss the measures that may be taken to prevent and control these corrosions. [Bincangkan pengukuran yang akan diambil untuk mencegah dan mengawasinya kakisan

tersebut.] (3 Marks / Markah)

Answer:

Pitting – Alloying with 2 % molybdenum enhance their resistance

Intergranular – Lower the carbon content. Alloying with other metal, ex. Titanium,

which tendency to form carbide greater

Stress corrosion-Lower magnitude of stress and increase cross sectional area

(b) Figure 2 shows the circumferential stress, σ, (also called hoop stress) in a pressurized

cylindrical vessel is calculated by the equation,𝜎 = 𝑃𝑟/𝑡, where P is the internal

pressure, r is the radius of the vessel, and t is thickness. The vessel has 91.44 cm

diameter, 6.35 mm thickness and an internal pressure of 34.5 MPa. Assume the crack

occurred at the center of the vessel and Y = 1.0. For the fracture toughness KIC and

yield strength σY values of each vessel materials, refer Table A1 in Appendix.

(Chap 7, Prob 41) [Gambarajah 2 menunjukkan ketegasan lilitan σ (juga dipanggil tegasan berharap) di dalam sebuah

tangki silinder bertekanan dikirakan dengan menggunakan persamaan 𝜎 = 𝑃𝑟/𝑡, di mana P adalah

tekanan dalaman, r adalah jejari tangki, dan t adalah ketebalan. Tangki tersebut mempunyai garis

tengah 91.44 cm, ketebalan 6.35 mm dan tekanan dalaman 34.5 MPa. Anggap geometri retakan

berlaku di bahagian tengah tangki tersebut dan Y=1.0. Bagi nilai-nilai kekerasan patah, KIC dan

kekuatan alah σY bagi setiap bahan-bahan tangki tersebut, rujuk Jadual A1 di Lampiran.]

(i) Compute the critical crack length (2a) if the vessel is made of Al 7178-T651. [Kira panjang retakan kritikal (2a) sekiranya tangki tersebut diperbuat oleh Al 7178-T651.]

(8 Marks / Markah)

Answer:

(ii) If the material of alloy steel (17-7pH) to be used, compare the difference of the

critical crack length (2a) with your answer in (i). Justify your answer.

Page 16: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

[Sekiranya bahan keluli aloi (17-7pH) digunakan, bandingkan perbezaan retakan kritikal (2a)

dengan jawapan anda di (i). Justifikasi jawapan anda.]

(6 Marks / Markah)

Figure 2 [Gambarajah 2]

Answer:

r

P

2a

crack

σ

σ

t

Page 17: ANSWER SCHEMES UNIVERSITI MALAYSIA PERLIS ENT 145

Appendices [Lampiran]

Table A1 Typical Fracture Toughness Values for Selected Engineering Alloys

𝑚 = 𝜌𝑉

𝜌 = 𝑛𝐴

𝑉𝐶𝑁𝐴

𝐽 = −𝐷𝑑𝐶

𝑑𝑥

𝜕𝐶

𝜕𝑡= 𝐷

𝜕2𝐶

𝜕𝑥2

𝐶𝑥 − 𝐶0

𝐶𝑠 − 𝐶0= 1 − erf (

𝑥

2√𝐷𝑡)

ln 𝐷 = ln 𝐷0 −𝑄𝑑

𝑅 (

1

𝑇)

𝜎 = 𝐹 𝐴𝑜⁄

𝜖 =𝑙𝑖 − 𝑙𝑜

𝑙𝑜=

∆𝑙

𝑙𝑜

𝜎 = 𝐸𝜖

𝑣 = − ∈𝑥

∈𝑧

𝜎 = (𝑃𝑟 𝑡⁄ )

𝐾𝐼𝐶 = 𝑌𝜎√𝜋𝑎

𝐴𝑃𝐹 = 𝑉𝑆

𝑉𝑐