Mp4f01 sem 1 1011

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee MP4F01 – MATERIALS ENGINEERING SEMESTER 1 EXAMINATION 2010-2011 (Dec 2010) Question 1 1 (a) The constant Ďƒo is called the friction stress and it is sensitive to the test temperature and chemical composition. Ďƒo basically represent the stress required to move free dislocations along the slip planes in the bcc crystal.

At higher temperatures, the vibrational energy within the bcc lattice will be higher. Consequently, propagation of free dislocation within the lattice will be easier as slip occur easily at elevated temperature. At lower temperature, the lattice will be relatively stagnant and any changes along the slip plane may be limited.

At different compositions, the lattice strain created in the crystal will be different. For example, movement of free dislocation within the lattice may be impeded by the large lattice strain created around a larger atom.

1 (b) Carbon content in maraging steels are kept low for two main reasons.

Low carbon content enables the formation of a relatively soft but heavily dislocated martensite. The soft martensite drastically enhance the toughness of maraging steels, which is crucial at such high strength.

Besides that, limited carbon content would also prevent the formation of carbide. Formations of carbide in maraging steels are unfavourable because it would deplete the amount of free Mo and Ti particles in the solution. Mo and Ti are very crucial in the precipitation of intermetallic compound, such as Ni 3Mo and Ni3Ti. The predominant strengthening mechanism of maraging steels is the precipitation of intermetallic compounds which gives rise to precipitation hardening.

1 (c) W and Mo are added to high speed steels mainly because they are strong carbide formers. Since high speed steels are used under elevated temperature, red hardness of the steel needs to be enhanced. Besides that, the wear resistant carbides will also enhance the wear resistant and useful life of the steel. The carbide forming elements can promote finer grain size and hence improve the toughness, by providing some undissolved carbide particles at the quenching temperature.

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee 1(d) Predominant strengthening mechanism in cold work steels is martensitic hardening.

Grain refinement will not lead to significantly higher strength because majority of the material’s strength comes from the hard martensite. The effect of grain refinement will be overshadowed by the high strength created by the hard martensite.

However, it is still desirable to refine grains of the steel. Smaller grains will give rise to larger grain boundary areas and hence higher toughness. High toughness is crucial in cold work steels since they are often used in high load bearing applications.

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee Question 2 2 (a) The method used to synthesize powders in Figure 1 is suggested to be plasma synthesis. Firstly, the powders are very fine (around 40nm) and this is only achievable by plasma synthesis. Besides, the powders are obviously unaggregated.

The method used to synthesize powders in Figure 2 is suggested to be spray drying. The powder size of around 10Âľm is achievable by this method. Besides the particles are largely granulated and flowable.

The method used to synthesize powders in Figure 3 is suggested to be precipitation-filtration methods. The particle size of 1Âľm is achievable. Besides, the particles have well defined size distribution and morphologies.

The sintering technique to consolidate these powders can be spark plasma sintering. The process is able to achieve high theoretical density by usage of temperature, pressure and electric field effect. The quick process also reduces grain coarsening.

2 (b) The process to fabricate high performance ceramic matrix composites is referred to as chemical vapour infiltration. The pre-formed reinforced fibre is used as the starting point. This structure acts as the skeleton of the composite. Then, CVD process is used to allow vapour to flow into the pores and fill the matrix. An application for such a composite is rocket nozzle.

2 (c) (i)

Plasma spraying involves high temperature where the temperature can reach 10000 K. At such elevated temperature, decarburization of WC-Co occur. As a result, large amounts of brittle W 2C and other carbon-deficient phases are present in the coating. The resulting coating may be too brittle.

(ii)

Zirconia ceramics typically have low thermal conductivity and high melting temperatures. Since HVOF processes involves lower operating temperature of around 3000°C, heat transfer may not be sufficient to thoroughly melt the zirconia particles before coating occurs. As a result, the coating formed may not be too strong.

(iii) Plasma spraying is only suitable to form thin walled coatings. Spray formed ceramics are too thick to be formed by plasma spray. Besides, The high operating temperature (around 10000K) of plasma spray would have caused monoclinic to tetragonal transformation and results in a more brittle ceramic.

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee Question 3 3 (a) The polymer labeled A should be a crosslinked polymer undergoing degradation. Over a wide range of temperature, the crosslinked structure of the polymer stayed intact and maintained a high modulus. However, at elevated temperature, possible bond scission might have cause the modulus to drop significantly. The degradation process was accelerated when the thermal energy present exceeds the binding energy of the covalent bonds.

The polymer labeled B should be a polymerizing acrylic resin. Over time, the acrylic resin starts to experience crosslinking. Elevated temperature helps to promote the process of crosslinking. As a result, the modulus of the material starts to rise significantly.

The polymer labeled C should be a thermoplastic. At lower temperature, the plastic acts like a rigid glass and maintain high modulus. However, when the temperature rise past possible glass-transition temperature, the modulus of the plastic started to drop.

3 (b) Glass Transition Temperature (Tg) is the temperature below which the long-range segmental motion in Amorphous polymers ceases and the material essentially becomes a glass. 

Polarity – Polar substituent groups enhance intermolecular interactions, increasing the structure’s rigidity and hence increasing Tg.

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Molar Mass – Tg increases with molecular mass at low molar mass but reaches a point at moderate molar mass where further increase has little effect on Tg.

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Main chain flexibility – presence of double bonds or cyclic or aromatic functional groups will increase the overall stiffness of the polymer chain and hence increase the Tg.

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Plasticizers – Plasticizers are low molecular weight compounds capable of getting in between polymer chains to reduce intermolecular interactions. As a result, the Tg of the polymer will be reduced.

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee 3 (c) The difference of shrinkage and densities mainly occur depends on the ability of the chains in the samples to close pack. Stereo-regularity promotes close packing. As a result of which, syndiotactic polymers are able to close pack more than atactic polymers due to its stereo-regularity. Hence, the chains which are more closely packed results in a polymer which has greater shrinkage and higher density.

ATACTIC POLYMER

SYNDIOTACTIC POLYMER 3 (d) (i)

Molecular molar mass = 3(12) + 6(1) = 42 Number average relative molar mass = degree of polymerization x molecular molar mass = 20000 x 42 = 840000 g/mol

(ii)

Number average relative molar mass = 0.9x20000 = 18000 g/mol

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee Question 4 4 (a) Based on the Maxwell Model, G(t) = k e-t/т. G(t) is known as relaxation modulus and is representative of the instantaneous modulus of a polymer at any point of time. When time t increases, the value of G(t) will decrease. As a result, the stress of the polymer and the pressure that it can withstand starts to decrease. Since the valve in a tap serve to withstand the pressure of the water in the pipe, taps will start to leak when the G(t) of the valves drop below the water pressure.

The first step is to prevent over-turning the pipe or locking it too tight. This would prevent forming a large initial strain on the valve.

The second step is to keep the water pressure at an adequate level. Some

household tend to use main water pumps, to increase the water pressure around the house, unnecessarily. The better alternative is to only use built-in pumps in devices that require high pressures, such as the shower tap.

4 (b) a/W = 10/20 = 0.5 Y = 1.09 – 1.73(0.5) + 8.2(0.5)2 – 14.17(0.5)3 +14.55(0.5)4 = 1.413 σF = (3*0.08/2*0.008*0.022) 390 = 14.625 MPa KIC = Y σF√(πa) = 3.66 MPa m1/2 For a tension specimen, the stress is simply force per area(F/A). Fracture toughness, KIC remains the same for the same material. σF

= KIC/Y√(πa) = 3.66/(7.631x√(0.025π) = 1.711 MPa

Force needed to fracture = σFA = 1.711 x106 x (0.05x0.01) = 855.5 N

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee 4 (c) Polymer nano-composites are not used extensively due to the problem of agglomeration. Agglomeration happens in nano-composites because nano-size fillers tend to stick together and are difficult to be distributed evenly. As a result of clumping, more regions are significantly weaker than the other, hence failure regions increase.

The key issues in producing polymer nano-composites are the effectiveness of exfoliation and dispersion of the layers of the nano-particles. The level of exfoliation can be manipulated to fit the application, trading off between mechanical property and cost. Dispersion can be induced by increasing the organophilic capability of the nano-particles.

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.