# Applications Of Elastic Behaviour Of Materials

The elastic behaviour of materials plays an important role in everyday life. All engineering designs require precise knowledge of the elastic behaviour of materials. For example while designing a building, the structural design of the columns, beams and supports require knowledge of strength of materials used. Have you ever thought why 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? Answers to these questions can be obtained from the study of structural engineering which is based on concepts developed here.

Cranes used for lifting and moving heavy loads from one place to another have a thick metal rope to which the load is attached. The rope is pulled up using pulleys and motors. Suppose we want to make a crane, which has a lifting capacity of 10 tonnes or metric tons (1 metric ton = 1000 kg). 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 mild steel has a yield strength ($\sigma_y$) of about 300 × 10$^6$ N.m$^{–2}$. Thus, the area of cross-section (A) of the rope should at least be

$A\geqslant \frac{w}{\sigma_y}$

= (104 kg × 9.8 m.s$^{-2}$)/(300 × 10$^6$ N.m$^{-2}$)

= 3.3 × 10$^{-4}$ m $^2$

corresponding to a radius of about 1 cm for a rope of circular cross-section. Generally a large margin of safety (of about a factor of ten in the load) is provided. Thus a thicker rope of radius about 3 cm is recommended. 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 the use of beams and columns is very common. In both the cases, the overcoming of the problem of bending of beam under a load is of prime importance. The beam should not bend too much or break. Let us consider the case of a beam loaded at the centre and supported near its ends as shown in Fig. 1. A bar of length l, breadth b, and depth d when loaded at the centre by a load W sags by an amount given by

$\delta =\frac{wl^3}{4bd^3Y} $

This relation can be derived using what you have already learnt and a little calculus. From Eq. (9.16), we see that to reduce the bending for a given load, one should use a material with a large Young’s modulus Y.

Fig.1 A beam supported at the ends and loaded at the centre. |

For a given material, increasing the depth d rather than the breadth b is more effective in reducing the bending, since δ is proportional to d$^{-3}$ and only to b$^{-1}$(of course the length l of the span should be as small as possible). But on increasing the depth, unless the load is exactly at the right place (difficult to arrange in a bridge with moving traffic), the deep bar may bend as shown in Fig. 2(b). This is called buckling. To avoid this, a common compromise is the cross-sectional shape shown in Fig. 2(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.

Fig.2 Different cross-sectional shapes of a beam. (a) Rectangular section of a bar; (b) A thin bar and how it can buckle; (c) Commonly used section for a load bearing bar. |

The use of pillars or columns is also very common in buildings and bridges. A pillar with rounded ends as shown in Fig. 3(a) supports less load than that with a distributed shape at the ends [Fig. 3(b)]. The precise design of a bridge or a building has to take into account the conditions under which it will function, the cost and long period, reliability of usable material, etc.

Fig. 3 Pillars or columns: (a) a pillar with rounded ends, (b) Pillar with distributed ends. |

The answer to the question why the maximum height of a mountain on earth is ~10 km can also be provided by considering the elastic properties of rocks. A mountain base is not under uniform compression and 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 hρg 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 hρg itself. Now the elastic limit for a typical rock is 30 × 10$^7$ N.m$^{-2}$.

Equating this to hρg, with ρ = 3 × 10$^3$ kg.m$^{-3}$ gives

hρg = 30 × 10$^7$ N.m$^{-2}$.

h = (30 × 10$^7$ N.m$^{-2}$)/(3 × 10$^3$ kg.m$^{-3}$ × 10 m.s$^{-2}$)

h = 10 km

which is more than the height of Mt. Everest!

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