Course Content

Properties of Matter  Lessons

Properties of Matter  Quiz 1 and 2

Properties of Matter  Quiz 3 and 4
Mechanical Properties of Solids
Introduction
In earlier chapter, we studied the rotation of the bodies and then realised that the motion of a body depends on how mass is distributed within the body. We restricted ourselves to simpler situations of rigid bodies. A rigid body generally means a hard solid object having a definite shape and size. But in reality, bodies can be stretched, compressed and bent. Even the rigid steel bar can be deformed when a sufficiently large external force is applied on it. This means that even solid bodies are not perfectly rigid.
A solid has a definite shape and size. In order to change (or deform) the shape or size of a body, a force is required. If we stretch a helical spring, by gently pulling its ends, the length of the spring increases slightly. When you leave the ends of the spring, it regains its original size and shape. The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity and deformation caused is known as elastic deformation. However, if you apply force to a lump of putty or mud, they have no gross tendency to regain their shape, and they have no gross tendency to regain their previous shape, and they get permanently deformed. Such substances are called plastic and this property is called plasticity. Putty and mud are close to ideal plastics.
The elastic behaviour of materials plays a vital role in engineering design. For example, while designing a building, knowledge of elastic properties of materials like steel, concrete etc. is essential. The same is true in the design of bridges, automobiles, rope ways etc. One can also askCan we design an aeroplane which is very light but sufficiently strong? Can we design an artificial limb which is lighter but stronger? Why does a railway track have a particular shape like I? Why is glass brittle, while brass is not not? Answers to these questions begin with the study of how relatively simple kinds of loads or forces act to deform different solid bodies. In this chapter, let us get in to the details of the elastic behaviour and mechanical properties of solids which would answer most of such questions.
Elastic Behaviour of Solids
We know that in a solid, each atom or molecule is surrounded by neighbouring atoms or molecules. These are bonded together by interatomic or intermolecular forces and stay in a stable equilibrium position. When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic (or inter molecular) distances. When the deforming force is removed, the interatomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size. The restoring mechanism can be visualised by taking a model of springball system shown in Fig. 1. Here the balls represent atoms and springs represent the interatomic forces.
If you try to displace any ball from its equilibrium position, the spring system tries to restore the ball back to its original position. Thus elastic behaviour of solids can be explained in terms of microscopic nature of the solid. Robert Hooke, an English physicist (16351703 AD)performed experiments on springs and found that the elongation (change in length) produced in a body is proportional to the applied force or load. In 1676, he presented his law of elasticity, now called Hooke’s law. We shall study about in a separate section. This law, like Boyle’s law, is one of the earliest quantitative relationships of science. It is very important to know the behaviour of the materials under various kinds of load from the context of engineering design.
Stress and Strain
When a force is applied on body, it is deformed to a small or large extent depending upon the nature of the material of the body and the magnitude of the deforming force. The deformation may not be noticeable visually in many materials but it is there. When a body is subjected to a deforming force, a restoring force is developed in the body. This restoring force per unit area is known as stress. If F is the force applied and A is the area of cross section of the body.
Magnitude of the stress = F/A …………………………………………………(1)
The SI unit of stress is Nm^{2} or pascal (Pa) and its dimensional formula is [ML^{1}T^{2}].
There are three ways in which a solid may change its dimensions when an external force acts on it. These are shown in Fig. (2). In Fig. 2(a), a cylinder is stretched by two equal forces applied normal to its crosssectional area. The restoring force per unit area in this case is called tensile stress. If the cylinder is compressed under the action of applied forces, the restoring force per unit area is known as compressive stress. Tensile or compressive stress can also be termed as longitudinal stress.
In both cases, there is a change in the length of cylinder. The change in the length ΔL to the original length L of the body (cylinder in this case) is known as longitudinal strain.
Longitudinal strain =ΔL/L ……………………………………………(2)
However, if two equal and opposite deforming forces are applied parallel to the crosssectional area of the cylinder, as shown in Fig. 2(b), there is relative displacement between the opposite faces of the cylinder. The restoring force per unit area developed due to the applied tangential force is known as tangential or shearing stress.
As a result of applied tangential force, there is a relative displacement Δx between opposite faces of the cylinder as shown in Fig. 2(b). The strain so produced is known as shearing strain and is defined as the ratio of relative displacement of the forces Δx to the length of the cylinder L.
Shearing strain x/L = tan θ ………………………………………………………………….(3)
where θ is the angular displacement of the cylinder from the vertical (original position of the cylinder). Usually θ is very small, tan θ is nearly equal to angle θ, (if θ = 10°, for example, there is only 1% difference between θ and tan θ).
(a) Cylinder subjected to tensile stress stretches it by an amount ΔL. (b) A cylinder subject to shearing (tangential) stress deforms by an angle θ (c) A book subject to a shearing stress (d) A solid sphere subjected to a uniform hydraulic stress shrinks in volume by an amount ΔV.
It can also be visualised, when a book is pressed with the hand and pushed horizontally, as shown in Fig. 2(c). Thus, shearing strain = tan θ≈θ …………………………………….(4)
In Fig. 2(d), a solid sphere placed in the fluid under high pressure is compressed uniformly on all sides. The force applied by the fluid acts in perpendicular direction at each point of the surface and the body is said to be under hydraulic compression. This leads to decrease in its volume without any change in its geometric shape.
The body develops internal restoring forces that are equal and opposite to the forces applied by the fluid (the body restores its original shape and size when taken out from the fluid). The internal restoring force per unit area in this case is known as hydraulic stress and in magnitude is equal to the hydraulic pressure(applied force per unit area).
The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of change in volume (ΔV) to the original volume (V).
Volume strain = ΔV/V ……………………………………………………………………………(5)
Since the strain is a ratio of change in dimension to the original dimension, it has no units or dimensional formula.
Hooke’s Law
Stress and strain take different forms in the situations depicted in Fig. (2). For small deformations the stress and strain are proportional to each other. This is known as Hooke’s law.
Thus,
stress ∝ strain
stress = k × strain ………………………………………………………………………………..(6)
where k is the proportionality constant and is known as modulus of elasticity.
Hooke’s law is an empirical law and is found to be valid for most materials. However, there are some materials which do not exhibit this linear relationship.
StressStrain Curve
The relation between the stress and the strain for a given material under tensile stress can be found experimentally. In a standard test of tensile properties, a test cylinder or a wire is stretched by an applied force. The fractional change in length (the strain) and the applied force needed to cause the strain are recorded. The applied force is gradually increased in steps and the change in length is noted. A graph is plotted between stress (which is equal in magnitude to the applied force per unit area) and the strain produced. A typical graph for a metal is shown in Fig. (3). Analogous graphs for compression and shear stress may also be obtained. The stressstrain curves vary from material to material. These curves help us to understand how a given material deforms with increasing loads. From the graph, we can see that in the region between O to A, the curve is linear. In this region, Hooke’s law is obeyed. The body regains its original dimensions when the applied force is removed. In this region, the solid behaves as an elastic body.
A typical stressstrain curve for a metal
In the region from A to B, the stress and strain are not proportional. Nevertheless, the body still returns to its original dimension when the load is removed. The point B in the curve is known as yield point (also known as elastic limit) and the corresponding stress is known as yield strength (S_{y}) of the material.
If the load is increased further, the stress developed exceeds the yield strength and strain increases rapidly even for a small change in the stress. The portion of the curve between B and D shows this. When the load is removed, say at some point C between B and D, the body does not regain its original dimension. In this case, even when the stress is zero, the strain is not zero. The material is said to have permanent set. The deformation is said to be plastic deformation. The point D on the graph is the ultimate tensile strength (S_{u}) of the material. Beyond this point, additional strain is produced even by a reduced applied force and fracture occurs at point E. If the ultimate strength and fracture points D and E are close, the material is said to be ductile.
As stated earlier, the stressstrain behaviour varies from material to material. For example, rubber can be pulled to several times its original length and still returns to its original shape. Fig. (4) shows stressstrain curve for the elastic tissue of the aorta, present in the heart. Note that although elastic region is very large, the material does not obey Hooke’s law over most of the region.
Stressstrain curve for the elastic tissue of Aoria, the large tube (vessel) carrying blood from the heart.
Secondly, there is no well defined plastic region. Substances like tissue of aorta, rubber etc., which can be stretched to cause large strains are called elastomers.
Elastic Moduli
The proportional region within the elastic limit of the stressstrain curve (region OA in Fig. 3) is of much importance of structural and manufacturing engineering designs. The ration of stress and strain, called modulus of elasticity, is found to be characteristic of the material.
Young’s Modulus
Experimental observation show that for a given material, the magnitude of the strain produced is same whether the stress is tensile or compressive. The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as Young’s modulus and is denoted by the symbol Y.
Y = σ/ε ……………………………………………………………………….(7)
From Eqs. (1) and (2), we have
Y = (F/A)/(ΔL/L)
= (F× L)/(A ×ΔL) ………………………………………………………………………….(8)
Since strain is dimensionless quantity, the unit of Young’s modulus is the same as that of stress i.e., N m^{2} or Pascal (Pa). Table 1 gives the values of Young’s modulii and yield strengths of some materials.
From the data given in Table 1, it is noticed that the metals Young’s modulii are large. Therefore, these materials require large force to produce small change in length. To increase the length of a thin steel wire of 0.1 cm^{2} cross sectional area by 0.1%, a force of 2000N is required. The force required to produce the same strain in aluminium, brass and copper wires having the same crosssectional area are 690 N, 900 N and 1100 N respectively. It means that steel is more elastic than copper, brass and aluminium. It is for this reason that steel is preferred in heavy duty machines and in structural designs, Wood, bone, concrete and glass have rather small Young’s modulii.
Table 1 – Young’s modulii and yield strengths of some materials
Substance  Density ρ (kg m^{3})  Young’s modulus Y (10^{9} N m^{2})  Ultimate strength, S_{u} (10^{6} N m^{2})  Yield strength, S_{y} (10^{6} N m^{2}) 

Aluminium  2710  70  110  95 
Copper  8890  110  400  200 
Iron (wrought)  78007900  190  330  170 
Steel  7860  200  400  250 
Glass#  2190  65  50  – 
Concrete  2320  30  40  – 
Wood#  525  13  50  – 
Bone#  1900  9  170  – 
Polystyrene  1050  3  48  – 
# substance tested under compression
Example 1
A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel is 2.0 × 10^{11} N m^{2}.
Answer
We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod. Then the stress on the rod is given by
Stress = F/A = F/πr²
= 100 × 10³N/3.14 ×10² m²
= 3.18×10^{8} N m^{2}
The elongation, ΔL = (F/A)L/Y= 3.18×10^{8} N m^{2} × 1m/(2×10^{11} N m^{2})
= 1.59×10^{3} m
= 1.59 mm
The strain is given by
Strain = ΔL/L = (1.59×10^{3} m)/(1 m) = 1.59 ×10^{3}
= 0.16%
Example 2
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.
Answer
The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of crosssection A. From. Eq. (7) we have stress = strain × Young’s modulus. Therefore
W/A = Y_{c} × (ΔL_{c}/L_{c}) = Y_{s} × (ΔL_{s}/L_{s})
where the subscripts c and s refer to copper and stainless steel respectively. Or,
ΔL_{c}/ΔL_{s }= (Y_{s}/Y_{c}) × Y_{s} ×(L_{c}/L_{s})
Given L_{c }= 2.2 m, L_{s }= 1.6 m
From Table 1 Y_{c }= 1.1 × 10^{11} Nm^{2 } and
Y_{s }= 2.0 × 10^{11} Nm^{2}
ΔL_{c}/ΔL_{s }= (2.0 × 10^{11} / 1.1 × 10^{11}) × (2.2/1.6) = 2.5
The total elongation is given to be
ΔL_{c }+ ΔL_{s }= 7.0 × 10^{4 }m
Solving the above equations,
ΔL_{c }= 5.0 × 10^{4 }m and ΔL_{s }= 2.0 × 10^{4 }m
Therefore
W = (A × Y_{c }× ΔL_{c })/L_{c}
= π(1.5 × 10^{3 })² × [(5.0 × 10^{4 }× 1.1 × 10^{11 }/2.2
=1.8 × 10^{2 }N
Example 3
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig. 5). The combined mass of all the persons performing the act, and the tables, plaques etc., involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.
Answer
Total mass of all performers, tables, plaques etc. = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid = 280 – 60 = 220 kg
Weight of this supported mass = 220 kg wt. = 220 × 9.8 N = 2156 N
Weight supported by each thighbone of the performer = 1/2 (2156) N = 1078 N.
From Table 1, the Young’s modulus for bone is given by,
Y = 9.4 × 10^{9} Nm^{2}
Length of each thighbone L = 0.5 m
the radius of thighbone = 2.0 cm
Thus the crosssectional area of the thighbone A = π × (2 × 10^{2})² m² = 1.26 × 10^{3} m²
Using Eq. (8), the compression in each thighbone (ΔL) can be computed as
ΔL = [(F × L)/(Y × A)]
= [(1078 × 0.5)/(9.4× 10^{9} × 1.26 × 10^{3})]
= 4.55 × 10^{5}m or 4.55 × 10^{3 }cm.
This is very small change! The fractional decrease in the thighbone is ΔL/L = 0.000091 0r 0.0091%.
Determination of Young’s Modulus of the Material of a Wire
A typical experimental arrangement to determine the Young’s modulus of a material of wire under tension is shown in Fig 6. It consists of two long straight wires of same length and equal radius suspended side by side from a fixed rigid support. The wire A (called the reference wire) carries a millimeter main scale M and a pan to place a weight. The wire B (called the experimental wire) of uniform area of crosssection also carries a pan in which known weights can be placed. A vernier scale V is attached to a pointer at the bottom of the experimental wire B, and the main scale M is fixed to the reference wire A. The weights placed in the pan exert a downward force and stretch the experimental wire under a tensile stress. The elongation of the wire (increase in length) is measured by the vernier arrangement. The reference wire is used to compensate for any change in length that may occur due to change in room temperature, since any change in length of the reference wire due to temperature change will be accompanied by an equal change in experimental wire.
Both the reference and experimental wires are given an initial small load to keep the wires straight and the vernier reading is noted. Now the experimental wire is gradually loaded with more weights to bring it under a tensile stress and the vernier reading is noted again. The difference between two vernier readings gives elongation produced in the wire. Let r and L be the initial radius and length of the experimental wire, respectively. Then the area of crosssection of the wire would be πr². Let M be the mass that produced an elongation ΔL in the wire. Thus the applied force is equal to Mg, where g is the acceleration due to gravity. From Eq. (8), Young’s modulus of the material of the experiment is given by
Y = σ/ε = (Mg/πr²).(L/ΔL)
= Mg × L/(πr²×ΔL) ……………………………………………………………….(9)
Shear Modulus
The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material and is represented by G. It is also called modulus of rigidity.
G = shearing stress (σ_{s}) / shearing strain
G = (F/A)/(Δx/L)
= (F ×L)/(A × Δx) ………………………………………………………………………..(10)
Similarly, from Eq. (4)
G = (F/A)/θ
= (F/(A×θ) ………………………………………………………………………….(11)
The shearing stress σ_{s }can also be expressed as
σ_{s }= G ×θ ………………………………………………………………………….(12)
SI unit of shear modulus is Nm^{2 } or Pa. The shear modulii of a few common materials are given in Table 2. It can be seen that shear modulus (or modulus of rigidity) is generally less than Young’s modulus (Table 1). For most materials G ≈ Y/3.
Table 2 – Shear modulii (G) of some common materials
Material  G (10^{9} N m^{2} or GPa) 

Aluminium  25 
Brass  36 
Copper  42 
Glass  23 
Iron  70 
Lead  5.6 
Nickel  77 
Steel  84 
Tungsten  150 
Wood  10 
Example 9.4
A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 10^{4} N. The lower edge is riveted to the floor. How much will the upper edge be displaced?
Answer
The lead slab is fixed and the force is applied parallel to the narrow face as shown in Fig. 7. The area of the face parallel to which this force is applied is
A = 50 cm × 10 cm
= 0.5 m × 0.1 m
= 0.5 m³
Therefore, the stress applied is
= (9.4 × 10^{4} N/0.05 m²)
= 1.80 × 10^{6} N.m^{2}
We know that shearing strain =(Δx/L ) = Stress/G,
Therefore the displacement Δx = (Stress×L)/G = 1.8 × 10^{6} N.m^{2}× 0.5 m)/(5.6 × 10^{9} N.m^{2}× 10^{6} N.m^{2})
= 1.6 × 10^{4 }m = 0.16 mm
Bulk Modulus
We have seen that when a body is submerged in a fluid, it undergoes a hydraulic stress (equal in magnitude to the hydraulic pressure). This leads to decrease in the volume of the body thus producing a strain called volume strain (Eq. 5). The ratio of hydraulic stress to the corresponding hydraulic strain is called bulk modulus. It is denoted by symbol B.
B = p/(ΔV/V) …………………………………………………………………………………..(13)
The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs. That is, if p is positive, ΔV is negative. Thus for a system in equilibrium, the value bulk modulus B is always positive. SI unit of bulk modulus is the same as that of pressure i.e., N.m^{2 }or Pa. The bulk modulii of a few common materials are given in Table 3.
The reciprocal of the bulk modulus is called compressibility and is denoted by k. It is defined as the fractional change in volume per unit increase in pressure.
k = (1/B) = (1/Δp) ×(ΔV/V) …………………………………………………………………………………(14)
It can be seen from the data given in Table 3 that the bulk modulii for solids are much larger than for liquids, which are again much larger than the bulk modulus for gases (air).
Table 3 – Bulk modulii (B) of some common materials
Material  B (10^{9} N m^{2} or GPa) 

Solids  
Aluminium  72 
Brass  61 
Copper  140 
Glass  37 
Iron  100 
Nickel  260 
Steel  160 
Liquids  
Water  2.2 
Ethanol  0.9 
Carbon disulphide  1.56 
Glycerine  4.76 
Mercury  25 
Gases  
Air (at STP)  1.0 × 10^{4} 
Thus solids are least compressible whereas gases are most compressible. Gases are about a million times more compressible than solids! Gases have large compressibilities, which vary with pressure and temperature. The incompressibility of the solids is primarily due to the tight coupling between the neighbouring atoms. The molecules in liquids are also bound with their neighbours but not as strong as in solids. Molecules in gases are very poorly coupled to their neighbours. Table 4 shows the various types of stress, strain, elastic modulii and the applicable state of matter at a glance.
Table4 Stress, strain and various elastic modulii
Type of stress  Stress  Strain  Change in  Elastic modulus  Name of modulus  State of Matter  

shape  volume  
Tensile or compressive  Two equal and opposite forces perpendicular to opposite faces (σ = F/A)  Elongation or compression parallel to force direction (ΔL/L) (longitudinal strain)  Yes  No  Y = (F× L)/(A ×ΔL)  Young’s modulus  Solid 
Shearing  Two equal and opposite forces parallel to opposite surfaces [forces in each case such that total force and total torque on the body vanishes (σ_{s} = F/A)]  Pure shear θ  Yes  No  G = (F × θ )/A  Shear modulus  Solid 
Hydraulic  Forces perpendicular everywhere to the surface, force per unit area (pressure) same everywhere  Volume change (compression or elongation (ΔV/V)  No  Yes  B = p/(ΔV/V)  Bulk modulus  Solid, liquid and gas 
Example 5
The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, (ΔV/V), of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 10^{9} N m^{2}. (Take g = 10 m s^{2}).
Answer:
The pressure exerted by a 3000 m column of water on the bottom layer
p = hρg = 3000 m × 1000 kg m^{3}× 10 m s^{2}
= 3 × 10^{7} kg m^{1} s^{2}
= 3 × 10^{7} N m^{2}
Fractional compression ΔV/V, is
ΔV/V = stress/B = (3 × 10^{7} N m^{2})/(2.2 × 10^{9} N m^{2})
Poisson’s Ratio
Close observations with the Young’s modulus experiment (explained in section 9.6.2), show that there is also a slight reduction in the cross section(or in the diameter) of the wire. The strain perpendicular to the applied force is called lateral strain. Simon Poisson pointed out that within the elastic limit, lateral strain is directly proportional to the longitudinal strain. The ratio of the lateral strain to the longitudinal strain in a stretched wire is called Poisson’s ratio. If the original diameter of the wire is d and the contraction of the diameter under stress is Δd, the lateral strain is Δd/d. If the original length of the wire is L and the elongation under stress is ΔL, the longitudinal strain is ΔL/L. Poisson’s ratio is then (Δd/d)/(ΔL/L) or (Δd/ΔL) × (L/d). Poisson’s ratio is a ratio of two strains; it is a pure number and has no dimensions or units.Its value depends only on the nature of material. For steels the value is between 0.28 and 0.30, and for aluminium alloys it is about 0.33.
Elastic Potential Energy in a stretched wire
When a wire is put under a tensile stress, work is done against the interatomic forces. This work is stored in the wire in the form of elastic potential energy. When a wire of original length L and area of crosssection A is subjected to a deforming force F along the length of the wire, let the length of the wire be elongated by l. Then from Eq. (8), we have F = YA × (l/L). Here Y is the Young’s modulus of the material of the wire. Now for a further elongation of infinitesimal small length dl, work done dW is F × dl or YAldl/L. Therefore, the amount of work done (W) in increasing the length of the wire from L to L + l, that is from l = 0 to l = l is
= 1/2 × stress × strain x volume of the wire
This work is stored in the wire in the form of potential energy (U). Therefore the elastic potential energy per unit volume (u) is
u = 1/2 σε …………………………………………………………………..(15)
Applications of Elastic Behaviour of Materials
The elastic behaviour of materials play an important role in everyday life. All engineering designs require precise knowledge of the elastic behaviour of materials. For example, while designing 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 crosssection of the type I? Why does a heap of sand or 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 lifting capacity of 10 tonnes or metric tons (1 metric ton = 1000 kg), how thick the steel rope should be? We obviously want that the load does not deform the rope permanently. Therefore, the extension should not exceed the elastic limit. From Table 1, we find that mild steel has a yield strength (S_{y})of about 300× 10^{6} Nm^{2}. Thus, the area of crosssection (A) of the rope should at least be
A ≥ W/S_{y }= Mg/S_{y ……………………………………………………………………………………………………}_{(15)}
= (10^{4} kg × 10 ms^{2})/(300× 10 N m^{2})
= 3.3 × 10^{4} m^{2}
corresponding to a radius of about 1 cm for a rope of circular crosssection. 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 rigid rod. So, the ropes are always made 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 weight. Similarly, in the design of buildings use of beams and columns is very common. In both 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. 8. A bar of length l, and breadth b, and depth d when loaded at the centre by a load W sags by an amount given by
δ = W l³/(4bd³Y) …………………………………………………………………………..(16)
This relation can be derived using what you have already learnt and a little calculus. From Eq. (16), we see that to reduce the bending for a given load, one should use a material with a large Young’s modulus Y. 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. 9 (b). This is called buckling. To avoid this, a common compromise is the crosssectional shape shown in Fig. 9(c). This section provides a large loadbearing surface and enough depth to prevent bending. This shape reduces the weight of the beam without sacrificing the strength and hence reduces cost.
Fig. 9 Different crosssectional 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:
Use of pillars or columns is also a very common in buildings and bridges. A pillar with rounded ends as shown in Fig. 10(a) supports less load than with a distributed shape at the ends [Fig. 10(b)]. The precise design of a bridge or a building has to take in to account the conditions under which it will function, the cost and long period, 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. 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}
Or. h = 30 × 10^{7} N m^{2}/(3 × 10^{3} kg m^{3}x 10 ms^{2})
=10 km
which is more than the height of Mt. Everest!