Stiff (1)
The stiffer a material is, the more difficult it is to be deformed (have its shape or size changed) by a force.
Plastic (1)
A material is said to be plastic if, when you deform it, it stays in the new shape after the force is released.
Elastic (1)
If a material is elastic, then it will return to its original shape after being deformed. This is the opposite of plastic.
Brittle (1)
A brittle material does not change shape at all when a deforming force is applied. It eventually breaks without warning, and the pieces can be fitted back together.
Ductile (1)
A material is ductile when it can be deformed by a large, steadily applied force. Most economic metals are ductile- they can be drawn out into wire, for example.
Tensile Strength (2)
The Tensile Strength, or Ultimate Tensile Strength (UTS) of a material, is the amount of tensile stress a material can take just before snapping. It is measured in N m-2 or Pa.
Yield Stress (1)
The Yield Stress is the amount of tensile stress required for a material to yield. Certain materials, for example mild steel, when under sufficient stress, begin to extend at a very fast rate without additional pressure being applied. This is caused by the material’s internal structure losing its integrity. The crystal planes within the metal can slide past each other, and the material becomes completely plastic for a time, but then breaks if any more force is applied.
Information Sources
1: A student’s review notes, found on the college intranet.
Address:
2: Microsoft Encarta 99 Encyclopedia – “Tensile Strength”
All of my evidence is in Appendix 1 of the assignment.
Information about a metal
Aluminium is a very light metal, with a melting point of 660°C and a boiling point of 2467°C.
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 = π d2 / 4 A = π (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.
Part 3
To broaden my knowledge of materials, I asked my classmate from Physics to send me some information on a metal that he had researched, and the Young Modulus, calculated in Part 2. In return, I sent him the information and Young Modulus of my metal (Aluminium).
I sent the information in the main body of the e-mail rather than as an attached document, as it was just a paragraph summarising my research.
The e-mail I received is in Appendix 2, and the one I sent is in Appendix 3.
Part 4
Aluminium
Minimum Cross Sectional Area and Diameter
The safety factor for the suspension bridge is 30%.
30% of the Tensile strength of Aluminium is 80 x 0.30 = 24 MPa.
I am given that the wire length is 20m, and that the (maximum) load is 40.2 tonnes.
40.2 tonnes exerts a force of 40.2 x 9.81 x 1000 = 1962 kN
(the 9.81 is the gravitational field strength)
Tensile stress is calculated thus- σ = F / A Rearranging this gives A = F / σ.
A = 1962x103 / 24x106 = 81.75x10-3 m2.
Cross sectional area is calculated thus: A = ( π d2 ) / 4
Rearranging this gives d = √ (4A / π )
d = √ ( 4 x 81.75x10-3 / 3.142 ) = 0.3226 m. or 32.26 cm.
Extension produced by Maximum Load
By rearranging the equation for Young Modulus, E = F l / A e, we can arrive at e = F l / A E.
From here on it is just a matter of supplying numbers to the equation:
F = 1962x103 N
l = 20 m
A = 81.75x10-3 m2
E = 71.1x109 Pa (from Part 2)
e = 1962x103 x 20 / 81.75x10-3 x 71.1x109 = 6.75x10-3 m or 6.75 mm.
Copper
Minimum Cross Sectional Area and Diameter
30% of the Tensile strength of Copper is 150 x 0.30 = 45 MPa.
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 / σ.
A = 1962x103 / 45x106 = 43.6x10-3 m2.
d = √ (4A / π )
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.
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