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Question One
The understanding of the four classes of materials is very crucial since it leads to a clear understanding of the properties of such materials which plays a vital role when it comes to the selection of materials for specific applications (Ashby, Hugh and Cebon, 2007). This is mainly because materials used in some particular applications are required to posses’ specific properties hence the understanding of such properties ensures that the right material is chosen for the correct application. For instance, among the four classes of materials such as the metal alloys, polymers, ceramics and glasses and composite materials each of them is utilized for particular functions that are closely related to the specific properties of that material (Askeland and Phulé, 2005).
For example, considering steel which a metal alloy it usually has diverse applications that are highly depended on the properties of the steel. Thus, the properties of steel which is an alloy of iron and carbon are determined by the amount of carbon present which is thereby used to indicate the hardness and tensile strength of the steel (Callister, 2000). However, the two properties of steel are very crucial in determining specific applications of the steel. Hence, the applications of steel vary from light to mild and then heavy duty on the basis of the properties of the steel which are highly determined by amount of carbon present.
References
Ashby, M., Hugh, S. and Cebon, D. 2007. Materials: engineering, science, processing and design (1st ed.).New York: Butterworth-Heinemann.
Askeland, D. R. and Phulé, P.P. 2005. The Science & Engineering of Materials (5th ed.). Boston, MA: Thomson-Engineering.
Callister, W.D. 2000. Materials Science and Engineering – An Introduction (5th ed.). Hoboken, NJ: John Wiley and Sons.
Question Two
Yield strength is a very crucial indictor used in almost every engineering design and influenced by a variety of factors such as chemical composition, raw material quality, heat treatment process, forming process, and so on. Hence, it is one of the most important value which must be determined in every structural design. Therefore, if a component being designed must support force during its use then yield strength is used to ensure that the component is not capable of plastically deforming (Davies and Oelmann, 1983). Thus, the yield strength becomes essential in making sure that the selected component has high yield strength so that the produced stress does not go beyond yield strength.
On the other hand, the tensile strength is relatively not crucial for selection and application of the ductile materials mostly because many of the plastic deformation usually occurs before it is reached. Hence, despite giving some information on the hardness and material defects when it is compared to yield strength it is less effective in determining the right material for certain designs (Askeland, 1988). Yield strength is usually determined through a stress-strain curve as indicated in the diagram below:
Fig. 1 Stress – strain curve
References
Askeland, D.R. 1988. The Science and Engineering of Materials. Dordrecht: Chapman and Hall.
Davies, D.J. and Oelmann, L.A. 1983. The Structure, Properties and Heat Treatment of Metals. London: Pitman Books Ltd.
Question Three
Iron is usually the main component of steel. However, any component that is added to the composition of basic steel ends up affecting the properties of the steel as well as how it reacts to various processes of fabrication. In addition, the variations and additions in steel compositions are usually accountable for a great variety of steel grades thereby resulting to varied steel properties (Smith, 1990). For instance, an effective steel composition leads to an enhanced toughness and/ or strength and a chance of using cost-effective as well as straightforward methods of heat-treatment.
However, the extent of steel alloying results to varied elasticity modulus of the steel depending on the composition. This is also indicative that steel with high levels of carbon tend to be resistant to heat treatment and records higher elastic modulus (Smith, 1990).
Young Modulus of Elasticity of various steel compositions at different temperatures (oC)
| -73oC | 21oC | 93oC | 149oC | 204oC | 260oC | 316oC | 371oC | 427oC | |
| Carbon steel C <= 0.3% | 30.0 | 29.3 | 28.6 | 28.1 | 27.5 | 27.1 | 26.5 | 25.3 | 24.0 |
| Carbon-moly steels | 29.9 | 29.2 | 28.5 | 28.0 | 27.4 | 27.0 | 26.4 | 25.3 | 23.9 |
| Nickel steels Ni 2% – 9% | 28.5 | 27.8 | 27.1 | 26.7 | 26.1 | 25.7 | 25.2 | 24.6 | 23.0 |
| Cr-Mo steels Cr 1/2% – 2% | 30.4 | 29.7 | 29.0 | 28.5 | 27.9 | 26.9 | 26.3 | 25.5 | 24.8 |
| Cr-Mo steels Cr 2 1/4% – 3% | 31.4 | 30.6 | 29.8 | 29.4 | 28.8 | 27.7 | 27.1 | 26.3 | 25.6 |
| Cr-Mo steels Cr 5% – 9% | 31.7 | 30.9 | 30.1 | 29.7 | 29.0 | 28.0 | 27.3 | 26.1 | 24.7 |
| Chromium steels Cr 12%, 17%, 27% | 30.1 | 29.2 | 28.5 | 27.9 | 27.3 | 26.1 | 25.6 | 24.7 | 23.2 |
| Austenitic steels (TP304, 310, 316, 321, 347) | 29.1 | 28.3 | 27.6 | 27.0 | 26.5 | 25.3 | 24.8 | 24.1 | 23.5 |
Reference
Smith, W.F. 1990. Principles of Materials Science and Engineering. New York: McGraw-Hill.
Question Four
Hardness testing is often done in preference to tensile testing mainly because it is an effective way which is usually used to rapidly and non-destructively test the hardness of a material (Francois, 2008). Hence, hardness testing can actually be used to determine the strength of a material through a non-destructive way thereby providing an easier way of ensuring that the strength of that particular material is determined without necessarily testing the elongation of the material which is mostly used in tensile testing (Francois, 2008). However, the tensile testing deforms the material in comparison to hardness testing which determines the strength of the material non-destructively.
In addition, the hardness testing apart from determining the hardness of a material it also determines its strength while tensile testing can only be used to test the strength of a material. Moreover, after the determination of the hardness of a material through hardness testing it is possible to use it in approximating the tensile strength of various materials (Francois, 2008). Finally, due to the ease at which the hardness testing is conducted in comparison to complex tensile tests makes the former to be more preferred and often used in preference for the latter.
Reference
Francois, D. 2008. Structural components: mechanical tests and behavioural laws. Hoboken, NJ: John Wiley & Sons, Inc.
Question Five
In selecting the right material for a particular application determining of the fracture toughness (Kc) is often advantageous compared to Charpy impact test results. This is mainly because the former indicates the ability of a material having a crack to resist fracture thereby being one of the most significant property for any material used in almost every design applications (Anderson, 1995). However, despite the Charpy impact test being a quick, cheap and easy method of determining the hardness of a material its results remains to be comparative in comparison to the specific ones obtained in fracture toughness (Anderson, 1995).
In addition, the fracture toughness can be used quantitativelyto express a resistance of a material to brittle fracture in presence of a crack. Therefore, if a material consists of more fracture toughness then it will possibly undergo ductile fracture. Moreover, brittle fracture is indicative of a material with less fracture toughness. On the other hand, charpy impact test can also used to determine these factors but only on percentage basis (Anderson, 1995).
Reference
Anderson, T.L. 1995.Fracture Mechanics: Fundamentals and Applications. Boston, MA: CRC Press.
Question Six
Yield Strength =Force (Newton) Area (m2)
= (100 1000) ÷ (πr2 )
= 100000 ÷ π ×0.004×0.004
= 1988071570 Pa
= 1.9 × 109 Pa
Elasticity Modulus
E = (F)(L 1)/(A)(L 2)
Where E = Elastic Modulus
F = force applied
A = the cross-sectional area
L 2 = change in length
L 1 = original length
Where: F= ( 100×1000 Newton); A= (πr2 0.004×0.004= 0.0000503) m2 ; L 2 ( 1.0-0.2=0.8) ; L 1 (50 mm= 0.05 m)
E = 5000 ÷ 0.00004024
E = 124254473.16
E = 1.2 × 108 Psi
Elongation
= 1.0 mm- 0.2 mm
= 0.8 mm
Therefore; 50 + 0.8 = 50.8 mm
Reference
Smith, W.F. 1990. Principles of Materials Science and Engineering. New York: McGraw-Hill.
