SULIT Answer script Peperiksaan Semester Pertama Sidang Akademik 2013/2014 (Januari 2014)

ENT 145 – Materials Engineering Question 1 (a)

Identify the components and its specific materials used in a computer. (4 Marks / Markah)

Answer:

The keyboard, monitor, and tower housing –polymers (ABS, high impact polystyrene, blends) Tower casing – metal (aluminum alloy) Cable, cord covers – polymers (polyethylene, Teflon, PVC, etc.) Chip materials – metals, ceramics, electronic materials (silicon, silicon dioxide, copper, gold, silver, etc…) 5. Monitor (cathode-ray tube type) - Polymers and metals (Glass, steel, copper, PVC, rubber) 1. 2. 3. 4.

(b)

With the aid of sketches, explain the differences between ionic, covalent and metallic bonding. Give an example of chemical compound for each primary bonding. (6 Marks / Markah)

Answer:

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(c)

A small silver spoon contains 92.5 wt. % silver and 7.5 wt. % copper. The spoon has a mass of 100 grams. Given Avogadro’s number, NA = 6.022x1023 atoms/mol. (i) Determine the number of copper and silver atoms in the spoon. (8 Marks / Markah)

Answer:

𝑛𝐴 𝑉𝐶 𝑁𝐴 𝜌𝑉𝐶 𝑁𝐴 𝑛= 𝐴 𝜌=

Since 𝑚 = 𝜌𝑉 , so, number of atom, n is; 𝑚𝑁𝐴 𝑛= 𝐴

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SULIT (ENT 145) -2(ii) Explain the importance of adding copper to the silver spoon. (2 Marks / Markah) Answer:

Copper is added to silver to make the metal stronger and more durable.

Question 2 [Soalan 2]

(a)

Briefly explain conditions of atomic motion in diffusion. (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. (b)

Discuss the concept of non-steady state as it applies to diffusion. Give an example in your answer. (4 Marks / Markah)

Answer:

(c)

A FCC iron-carbon alloy initially contains 0.2 wt. % C is carburized at an elevated temperature and in an atmosphere where the surface carbon concentration is at 1.0 wt. %. (i)

If after 49.5 hour the concentration of carbon is 0.35 wt. % at a position 4.0 mm below the surface of the alloy, determine the temperature at which the treatment was carried out. Refer Table A1 and Table A2 in Appendix. (10 Marks / Markah)

Answer:

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(ii)

If the diffusion treatment is to be applied to an iron gear, propose a diffusion mechanism. Justify your answer. (4 Marks / Markah)

Answer:

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SULIT (ENT 145) -4Question 3 (a)

Discuss the differences between brittle and ductile metals with respect to its tensile stress-strain behaviour. (4 Marks / Markah)

Answer:

Ductile Metal experience plastic deformation upon fracture Brittle Metal very little or no plastic deformation

(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 -4mm. Compute the modulus of elasticity for this alloy, given that the Poisson’s ratio is 0.30. (6 Marks / Markah)

Answer:

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(c)

(ENT 145) -5Table 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. (i)

From Table 1, choose the material(s) that will not experience plastic deformation. Justify your choice(s). (6 Marks / Markah)

(ii)

By referring answer in (i), select the material(s) that will not experience a diameter reduction of more than 7.5 x 10-3 mm. (4 Marks / Markah)

Answer:

Answer:

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SULIT (ENT 145) -6-

Table 1 [Jadual 1]

Material Aluminum alloy Brass alloy Steel alloy Titanium alloy

Modulus of Elasticity (GPa) 70 101 207 107

Yield Strength (MPa) 200 300 400 650

Poisson’s Ratio 0.33 0.34 0.30 0.34

Question 4 (a)

Distinguish between ductile and brittle failures. Give an example for your answer. (4 Marks / Markah)

Answer:

Ductile Fracture: – Accompanied by significant plastic deformation

Brittle Fracture: – Little or no plastic deformation – Catastrophic

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(b)

(ENT 145) -7Figure 1 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 A3 in Appendix. (i) Compute the critical crack length (2a) if the vessel is made of 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. (6 Marks / Markah) Answer:

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SULIT (ENT 145) -8(iii) The vessel will fracture catastrophically if the crack lengths above the calculated value as shown in your answer in (i) and (ii), propose preventive measures need to be taken to increase the resistance to fatigue failure. (2 Marks / Markah) Answer:

Measures that may be taken to increase the fatigue resistance of the vessel: Polish the surface to remove stress amplification sites.  Reduce the number of internal defects (pores, etc.) by means of altering processing and fabrication techniques.  Modify the design to eliminate notches and sudden contour changes.  Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot peening.

t

σ 2a

r P

crack

σ Figure 1

Question 5 (a)

Discuss the difference between pearlite, spheroidite and martensite with respect to microstructures and mechanical properties. (6 Marks / Markah)

Answer:

Microstructure Alpha ferrite + cementite/ layer which alternate with one another Spherodite Alphaferrite + cementite / sphereshaped particles Martensite Alphaferrite + cementite / needle shaped grains Pearlite

(b)

Mechanical Properties Less ductile than spherodite. Harder than spherodite. extremely ductile. softest and weakest. Extremely hard but brittle

The complete isothermal transformation diagram of 0.76 wt% C steel alloy is shown in Figure 2. (i)

Develop the heat treatment process to produce a microstructure of 100 % martensite structure. (5 Marks / Markah)

Answer:

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(ii)

Develop the heat treatment process for producing a microstructure of 50% bainite and 50% martensite structure. (5 Marks / Markah)

(iii)

Alloy in question (i) is tested for bending fatigue test. It has experienced a fracture which occurs in brittle manner. Propose a suitable solution to improve mechanical properties of the alloy. Justify your answer. (4 Marks / Markah)

Answer:

Answer:

Ductility of martensite may be enhanced by heat treatment process known as tempering. Process of heating a martensitic steel to temperature between 250-650 oC for 1 hour and slowly cool to room temperature.

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SULIT (ENT 145) - 10 o

T ( C)

Pre-heat

Heating at 250-650 oC for 1 hour

Slow cooling at Tamb

t (h)

Figure 2 [Gambarajah 2]

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