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 crosssection 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 10^{6} Nm^{2}. Therefore, the area of crosssection of the rope should at least be
A W / S_{y} = Mg / S_{y}
=
=
= 3.3 x 10^{4} m^{2}
corresponding to a radius of about 1 cm for a rope of circular crosssection. 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 =
Use of pillars or columns is also very common in buildings and bridges.


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 10^{7} N m^{2}, equating this to h ρ g gives
h ρ g = 30 x 10^{7} N m^{2}
Or h =
=
= 10 km