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Matter and Materials Physics Assignment

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Introduction

Physics Key Skills assignment PHX1 Robert Moss Matter and Materials Part 1 Here is a list of definitions of important terms, with the sources of information in brackets: Tensile Stress (1) Tensile stress is the tensional force acting on a solid per unit cross-sectional area. It is represented by the symbol ? and is measured in N m-2 or Pa. (They are the same thing). Formula: ? = F / A Strain (1) The tensile strain is calculated by dividing the extension produced in a solid (i.e. a wire) under tension, by its original length. It is represented by the symbol ? and has no units. (It is a length divided by a length). Formula: ? = e / l The Young Modulus (1) The Young Modulus is a means by which we can measure a material's relative resistance to tensional force. If a material obeys Hooke's Law (as most of them do, at least to an extent) then the tensile stress is proportional to the tensile strain, and the ratio of stress to strain is a constant, which is characteristic of a material. This is called the Young Modulus, and it is calculated by dividing tensile stress by tensile strain. It is represented by the symbol E, and is measured in N m-2 or Pa, as we divide a measurement in Pa by one with no units. Formula: E = ? / ? , or E = F l / A e. ...read more.

Middle

It is highly electropositive and reactive, and it is this property that makes it extremely resistant to corrosion- on contact with air it rapidly forms a 'skin' of aluminium oxide, which resists any further chemical corrosion. Aluminium is the most abundant metal in the Earth's crust, but most of it is contained within complex silicate minerals, from which it is far too expensive to extract. The main source of Aluminium is Bauxite, which is an impure aluminium oxide ore. Part 2 I have been given data for the force and extension from stretching a wire, and the original length and diameter as constants. The formula for the Young Modulus is E = F l / A e , so I will have to manipulate the formula into a graph. If I made a graph of F against e, the gradient would be equal to F / e. Multiplying this by the l / A would give me the Young Modulus. Calculation for the Area... A = ? d2 / 4 A = ? (1.00x10-3)2 / 4 = 7.85x10-7 m2 Therefore l / A = 3.00 / 7.85x10-7 = 3.82x106 m-1 Below is the spreadsheet data I downloaded. To the right is the graph I constructed from the data. Load (N) Extension (mm) 0 0 1 0.05 2 0.11 3 0.16 4 0.22 5 0.27 6 0.32 7 0.37 8 0.43 9 0.48 10 0.54 11 0.59 12 0.65 13 0.7 14 0.75 15 0.8 ...read more.

Conclusion

The wire length is 20m, and that the (maximum) load is 40.2 tonnes. Force = 40.2 x 9.81 x 1000 = 1962 kN A = F / ?. A = 1962x103 / 45x106 = 43.6x10-3 m2. d = ?? (4A / ? ) d = ? ( 4 x 43.6x10-3 / 3.142 ) = 0.2356 m. or 23.56 cm. Extension produced by Maximum Load e = F l / A E. F = 1962x103 N l = 20 m A = 43.6x10-3 m2 E = 117x109 Pa. (from Part 3) e = 1962x103 x 20 / 43.6x10-3 x 117x109 = 0.00769 m, or 7.69 mm. Conclusion The maximum extensions produced in the wires are fairly similar, so we should look at the necessary diameters of the wires. A diameter of 32.26 cm for Aluminium is very large. Large enough of be infeasible, in fact. I doubt that any civil engineer has ever considered Aluminium for this purpose- true, it is lightweight, but it is also very weak and also very expensive to produce. Copper is not much better, having a minimum diameter of 23.56 cm. This is still very big. Copper is rather less expensive to produce, but you would just need too much of it to use it for suspension wires. Of the two metals, I would favour copper, but it really is the best of a bad bunch. ?? ?? ?? ?? Matter and Materials Assignment Robert Moss SC1 C:\Documents and Settings\ckd\My Documents\essays\doc\after\10637.doc - 1 - 1 May 2007 ...read more.

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