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Applications of Elastic Behaviour of Materials

Elasticity of materials serve a lot of purposes to humans.

Engineering designs employ this behaviour of building materials while designing a building, the structural details of the columns, beams and supports require knowledge of the strength of materials used. Why do the beams used in construction of bridges, as supports etc have a cross-section of the type I? Why does a heap of sand or a hill have a pyramidal shape? Study these examples.

Cranes are used for lifting and moving heavy loads from one place to another. They have a thick metal rope to which the load is attached. It is pulled up using pulleys and motors. Suppose we want to construct a crane, which has a lifting capacity of 10 metric tons. How thick should the steel rope be? We obviously want that the load does not deform the rope permanently; therefore, the extension should not exceed the elastic limit. From Table 9.1, we find that steel has yield strength of 200 x 106 Nm-2. Therefore, the area of cross-section of the rope should at least be
A W / Sy = Mg / Sy
               = 3.3 x 10-4 m2

corresponding to a radius of about 1 cm for a rope of circular cross-section. The recommended radius would be about 10 cm taking into account the safety factor (~10). A single wire of this radius would practically be a rigid rod, so the ropes are always made of a number of thin wires braided together, like in pigtails, for ease in manufacture, flexibility and strength.

A bridge has to be designed such that it can withstand the load of the flowing traffic, the force of winds and its own weight. Similarly, in the design of buildings use of beams and columns is very common. In both these problems the bending of beams under a load is of prime importance. The beam should not bend too much or break. Let us, therefore, consider the case of a beam loaded at the centre and supported near its ends as shown in Figure (i). A bar of length l, breadth b, and thickness d when loaded at the centre by a load W sags by an amount given by
δ =
This relation can be derived using what you have already learnt and a little calculus. One should use a material with a large Young's modulus Y to reduce the bending for a given load. For a given material, increasing the thickness d rather than the breadth b is more effective in reducing the bending since δ is proportional to d-3 and only to b-1. However a deep bar may have a tendency to buckle as shown in Figure (ii) which shows different sectional shapes of a bar. To avoid this, a common compromise is the cross-sectional shape shown in Figure (ii)-(c). This section provides a large load bearing surface and enough depth to prevent bending. This shape reduces the weight of the beam without sacrificing the strength and hence reduces the cost.

Use of pillars or columns is also very common in buildings and bridges.

Figure (ii) Different cross-sectional shapes of a beam. (a) Rectangular section of a bar; (b) A thin bar and how it can tuckle; (c) Commonly used section for a load bearing bars

Figure (iii) - Pillars columns: (a) a pillar with rounded ends, 
(b) pillar with distributed ends

A pillar with rounded ends as shown in Figure (iii) - (a) supports less load than a distributed shape at the ends.

The elastic properties of rocks will help us to answer the question why the maximum height of a mountain on earth is ~10 km. . A mountain base is not under uniform compression; this provides some shearing stress to the rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height h the force per unit area due to the weight of the mountain is h ρ g where ρ is the density of the material of the mountain and g is the acceleration due to gravity. The material at the bottom experiences this force in the vertical direction, and the sides of the mountain are free. Therefore this is not a case of pressure or bulk compression. There is a shear component, approximately h ρ g  itself. Now the elastic limit for a typical rock is 30 x 107 N m-2, equating this to h ρ g gives
h ρ g = 30 x 107 N m-2
     Or h =
           = 10 km
Which is more than the height of Mt. Everest.

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