This relation can be derived using what you have already learned and a little calculus. From Eq, we see that to reduce the bending for given load, one should use a material with large Young's modulus Y. For a given material, increasing the depth d rather than the breadth bis more effective in reducing the bending, since is proportional to d-3 and only to b (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. This is called buckling. To avoid this, a common compromise is the cross-sectional shape shown in This section provides a large load bearing surface and enough depth to prevent bending. This shape reduces the weight of the reduces the cost. beam without sacrificing the strength and hence
This relation can be derived using what you have already learned and a little calculus. From Eq. we see that to reduce the bending for given load, one should use a material with large Young's modulus Y. For a given material, increasing the depth d rather than the breadth bis more effective in reducing the bending, since S is proportional to d-3 and only to b (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. This is called buckling. To avoid this, a common compromise is a cross-sectional shape shown in This section provides a large load bearing surface and enough depth to prevent bending. This shape reduces the weight of then reduces the cost. beam without sacrificing the strength and hence
Use of pillars or columns is also very common in buildings and bridges. A pillar with rounded ends as shown in supports less load than that with a distributed shape at the ends 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, the reliability of usable materials etc.
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. 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 hpg where p 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 hpg itself.
Now the elastic limit fora typical rock is 30 x 107 N m2. Equating this to hpg, with
p =3 x 103 kg m-3 giveshpg = 30 x 107 Nm". Or
= 10 km
h = 30 x 10' N m/(3 x 10 kg mºx 10 ms)
which is more than the height of Mt. Everest
