Mechanical Properties of Solids PDF

CHAPTER NINE MECHANICAL PROPERTIES OF SOLIDS 9.1 INTRODUCTION In Chapter 7, we studied the rotation of the bodies and then realised that the moti...

CHAPTER NINE MECHANICAL PROPERTIES OF SOLIDS 9.1 INTRODUCTION In Chapter 7, we studied the rotation of the bodies and then realised that the motion of a body depends on how mass is 9.1 Introduction distributed within the body. We restricted ourselves to simpler 9.2 Elastic behaviour of solids situations of rigid bodies. A rigid body generally means a 9.3 Stress and strain hard solid object having a definite shape and size. But in 9.4 Hooke’s law reality, bodies can be stretched, compressed and bent. Even 9.5 Stress-strain curve the appreciably rigid steel bar can be deformed when a 9.6 Elastic moduli sufficiently large external force is applied on it. This means 9.7 Applications of elastic that solid bodies are not perfectly rigid. behaviour of materials A solid has definite shape and size. In order to change (or deform) the shape or size of a body, a force is required. If Summary you stretch a helical spring by gently pulling its ends, the Points to ponder length of the spring increases slightly. When you leave the Exercises ends of the spring, it regains its original size and shape. The Additional exercises 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 the 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 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 an important 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, ropeways etc. One could also ask — Can 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? Answers to such questions begin with the study of how relatively simple kinds of loads or forces act to deform different solids bodies. In this chapter, we shall study the 2020-21 236 PHYSICS elastic behaviour and mechanical properties of elasticity, now called Hooke’s law. We shall solids which would answer many such study about it in Section 9.4. This law, like questions. Boyle’s law, is one of the earliest quantitative relationships in science. It is very important to 9.2 ELASTIC BEHAVIOUR OF SOLIDS know the behaviour of the materials under various kinds of load from the context of We know that in a solid, each atom or molecule engineering design. is surrounded by neighbouring atoms or molecules. These are bonded together by 9.3 STRESS AND STRAIN interatomic or intermolecular forces and stay in a stable equilibrium position. When a solid is When forces are applied on a body in such a deformed, the atoms or molecules are displaced manner that the body is still in static equilibrium, from their equilibrium positions causing a it is deformed to a small or large extent depending change in the interatomic (or intermolecular) upon the nature of the material of the body and distances. When the deforming force is removed, the magnitude of the deforming force. The the interatomic forces tend to drive them back deformation may not be noticeable visually in to their original positions. Thus the body regains many materials but it is there. When a body is its original shape and size. The restoring subjected to a deforming force, a restoring force mechanism can be visualised by taking a model is developed in the body. This restoring force is of spring-ball system shown in the Fig. 9.1. Here equal in magnitude but opposite in direction to the balls represent atoms and springs represent the applied force. The restoring force per unit area interatomic forces. is known as stress. If F is the force applied normal to the cross–section and A is the area of cross section of the body, Magnitude of the stress = F/A (9.1) The SI unit of stress is N m–2 or pascal (Pa) and its dimensional formula is [ ML–1T–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. 9.2. In Fig.9.2(a), a cylinder is stretched by two equal forces applied normal to its cross-sectional 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 the cases, there is a change in the length of the cylinder. The change in the length Fig. 9.1 Spring-ball model for the illustration of elastic ∆L to the original length L of the body (cylinder behaviour of solids. in this case) is known as longitudinal strain. If you try to displace any ball from its ∆L Longitudinal strain = (9.2) equilibrium position, the spring system tries to L restore the ball back to its original position. Thus However, if two equal and opposite deforming elastic behaviour of solids can be explained in forces are applied parallel to the cross-sectional terms of microscopic nature of the solid. Robert area of the cylinder, as shown in Fig. 9.2(b), Hooke, an English physicist (1635 - 1703 A.D) there is relative displacement between the performed experiments on springs and found opposite faces of the cylinder. The restoring force that the elongation (change in the length) per unit area developed due to the applied produced in a body is proportional to the applied tangential force is known as tangential or force or load. In 1676, he presented his law of shearing stress. 2020-21 MECHANICAL PROPERTIES OF SOLIDS 237 Robert Hooke (1635 – 1703 A.D.) Robert Hooke was born on July 18, 1635 in Freshwater, Isle of Wight. He was one of the most brilliant and versatile seventeenth century English scientists. He attended Oxford University but never graduated. Yet he was an extremely talented inventor, instrument-maker and building designer. He assisted Robert Boyle in the construction of Boylean air pump. In 1662, he was appointed as Curator of Experiments to the newly founded Royal Society. In 1665, he became Professor of Geometry in Gresham College where he carried out his astronomi- cal observations. He built a Gregorian reflecting telescope; discovered the fifth star in the trapezium and an asterism in the constellation Orion; suggested that Jupiter rotates on its axis; plotted detailed sketches of Mars which were later used in the 19th century to determine the planet’s rate of rotation; stated the inverse square law to describe planetary motion, which Newton modified later etc. He was elected Fellow of Royal Society and also served as the Society’s Secretary from 1667 to 1682. In his series of observations presented in Micrographia, he suggested wave theory of light and first used the word ‘cell’ in a biological context as a result of his studies of cork. Robert Hooke is best known to physicists for his discovery of law of elasticity: Ut tensio, sic vis (This is a Latin expression and it means as the distortion, so the force). This law laid the basis for studies of stress and strain and for understanding the elastic materials. As a result of applied tangential force, there It can also be visualised, when a book is is a relative displacement ∆x between opposite pressed with the hand and pushed horizontally, faces of the cylinder as shown in the Fig. 9.2(b). as shown in Fig. 9.2 (c). The strain so produced is known as shearing Thus, shearing strain = tan θ ≈ θ (9.4) strain and it is defined as the ratio of relative In Fig. 9.2 (d), a solid sphere placed in the displacement of the faces ∆x to the length of fluid under high pressure is compressed the cylinder L. uniformly on all sides. The force applied by the fluid acts in perpendicular direction at each ∆x point of the surface and the body is said to be Shearing strain = = tan θ (9.3) under hydraulic compression. This leads to L decrease in its volume without any change of where θ is the angular displacement of the its geometrical shape. cylinder from the vertical (original position of The body develops internal restoring forces the cylinder). Usually θ is very small, tan θ that are equal and opposite to the forces applied is nearly equal to angle θ , (if θ = 10°, for by the fluid (the body restores its original shape example, there is only 1% difference between θ and size when taken out from the fluid). The and tan θ). internal restoring force per unit area in this case (a) (b) (c) (d) Fig. 9.2 (a) A cylindrical body under tensile stress elongates by ∆L (b) Shearing stress on a cylinder deforming it by an angle θ (c) A body subjected to shearing stress (d) A solid body under a stress normal to the surface at every point (hydraulic stress). The volumetric strain is ∆V/V, but there is no change in shape. 2020-21 238 PHYSICS is known as hydraulic stress and in magnitude The body regains its original dimensions when is equal to the hydraulic pressure (applied force the applied force is removed. In this region, the per unit area). solid behaves as an elastic body. The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of change in volume (∆V) to the original volume (V ). ∆V Volume strain = (9.5) V Since the strain is a ratio of change in dimension to the original dimension, it has no units or dimensional formula. 9.4 HOOKE’S LAW Stress and strain take different forms in the situations depicted in the Fig. (9.2). For small deformations the stress and strain are proportional to each other. This is known as Fig. 9.3 A typical stress-strain curve for a metal. Hooke’s law. In the region from A to B, stress and strain Thus, are not proportional. Nevertheless, the body still stress ∝ strain returns to its original dimension when the load stress = k × strain (9.6) is removed. The point B in the curve is known where k is the proportionality constant and is as yield point (also known as elastic limit) and known as modulus of elasticity. the corresponding stress is known as yield Hooke’s law is an empirical law and is found strength (σy ) of the material. to be valid for most materials. However, there If the load is increased further, the stress are some materials which do not exhibit this developed exceeds the yield strength and strain linear relationship. increases rapidly even for a small change in the stress. The portion of the curve between B and 9.5 STRESS-STRAIN CURVE D shows this. When the load is removed, say at The relation between the stress and the strain some point C between B and D, the body does for a given material under tensile stress can be not regain its original dimension. In this case, found experimentally. In a standard test of even when the stress is zero, the strain is not tensile properties, a test cylinder or a wire is zero. The material is said to have a permanent stretched by an applied force. The fractional set. The deformation is said to be plastic change in length (the strain) and the applied deformation. The point D on the graph is the force needed to cause the strain are recorded. ultimate tensile strength (σu ) of the material. The applied force is gradually increased in steps Beyond this point, additional strain is produced and the change in length is noted. A graph is even by a reduced applied force and fracture plotted between the stress (which is equal in occurs at point E. If the ultimate strength and magnitude to the applied force per unit area) fracture points D and E are close, the material and the strain produced. A typical graph for a is said to be brittle. If they are far apart, the metal is shown in Fig. 9.3. Analogous graphs material is said to be ductile. for compression and shear stress may also be As stated earlier, the stress-strain behaviour obtained. The stress-strain curves vary from varies from material to material. For example, material to material. These curves help us to rubber can be pulled to several times its original understand how a given material deforms with length and still returns to its original shape. increasing loads. From the graph, we can see Fig. 9.4 shows stress-strain curve for the elastic that in the region between O to A, the curve is tissue of aorta, present in the heart. Note that linear. In this region, Hooke’s law is obeyed. although elastic region is very large, the material 2020-21 MECHANICAL PROPERTIES OF SOLIDS 239 9.6.1 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. σ Y= (9.7) ε From Eqs. (9.1) and (9.2), we have Y = (F/A)/(∆L/L) = (F × L) /(A × ∆L) (9.8) Fig. 9.4 Stress-strain curve for the elastic tissue of Since strain is a dimensionless quantity, the Aorta, the large tube (vessel) carrying blood unit of Young’s modulus is the same as that of from the heart. stress i.e., N m–2 or Pascal (Pa). Table 9.1 gives does not obey Hooke’s law over most of the the values of Young’s moduli and yield strengths region. Secondly, there is no well defined plastic of some material. region. Substances like tissue of aorta, rubber From the data given in Table 9.1, it is noticed etc. which can be stretched to cause large strains that for metals Young’s moduli are large. are called elastomers. Therefore, these materials require a large force to produce small change in length. To increase 9.6 ELASTIC MODULI the length of a thin steel wire of 0.1 cm2 cross- sectional area by 0.1%, a force of 2000 N is The proportional region within the elastic limit required. The force required to produce the same of the stress-strain curve (region OA in Fig. 9.3) strain in aluminium, brass and copper wires is of great importance for structural and having the same cross-sectional area are 690 N, manufacturing engineering designs. The ratio 900 N and 1100 N respectively. It means that of stress and strain, called modulus of elasticity, steel is more elastic than copper, brass and is found to be a characteristic of the material. aluminium. It is for this reason that steel is Table 9.1 Young’s moduli and yield strenghs of some material # Substance tested under compression 2020-21 240 PHYSICS preferred in heavy-duty machines and in where the subscripts c and s refer to copper structural designs. Wood, bone, concrete and and stainless steel respectively. Or, glass have rather small Young’s moduli. ∆Lc/∆Ls = (Ys/Yc) × (Lc/Ls) Given Lc = 2.2 m, Ls = 1.6 m, u Example 9.1 A structural steel rod has a From Table 9.1 Yc = 1.1 × 1011 N.m–2, and radius of 10 mm and a length of 1.0 m. A Ys = 2.0 × 1011 N.m–2. 100 kN force stretches it along its length. ∆Lc/∆Ls = (2.0 × 10 /1.1 × 1011) × (2.2/1.6) = 2.5. 11 Calculate (a) stress, (b) elongation, and (c) The total elongation is given to be strain on the rod. Young’s modulus, of ∆Lc + ∆Ls = 7.0 × 10-4 m structural steel is 2.0 × 1011 N m-2. Solving the above equations, ∆Lc = 5.0 × 10-4 m, and ∆Ls = 2.0 × 10-4 m. Answer We assume that the rod is held by a Therefore clamp at one end, and the force F is applied at W = (A × Yc × ∆Lc)/Lc the other end, parallel to the length of the rod. = π (1.5 × 10-3)2 × [(5.0 × 10-4 × 1.1 × 1011)/2.2] Then the stress on the rod is given by = 1.8 × 102 N t F F Stress = = 2 u Example 9.3 In a human pyramid in a A πr circus, the entire weight of the balanced 100 × 10 N 3 group is supported by the legs of a = 2 performer who is lying on his back (as 3.14 × 10 ( −2 m ) shown in Fig. 9.5). The combined mass of = 3.18 × 108 N m–2 all the persons performing the act, and the The elongation, tables, plaques etc. involved is 280 kg. The mass of the performer lying on his back at ( F/A ) L the bottom of the pyramid is 60 kg. Each ∆L = Y thighbone (femur) of this performer has a length of 50 cm and an effective radius of (3.18 × 10 8 N m –2 ) (1m ) 2.0 cm. Determine the amount by which each thighbone gets compressed under the = 11 –2 2 × 10 N m extra load. = 1.59 × 10–3 m = 1.59 mm The strain is given by Strain = ∆L/L = (1.59 × 10–3 m)/(1m) = 1.59 × 10–3 = 0.16 % t u Example 9.2 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 cross-section A. From Eq. (9.7) we have stress = strain × Young’s modulus. Therefore W/A = Yc × (∆Lc/Lc) = Ys × (∆Ls/Ls) Fig. 9.5 Human pyramid in a circus. 2020-21 MECHANICAL PROPERTIES OF SOLIDS 241 Answer Total mass of all the performers, tables, will be accompanied by an equal change in plaques etc. = 280 kg experimental wire. (We shall study these Mass of the performer = 60 kg temperature effects in detail in Chapter 11.) 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 = ½ (2156) N = 1078 N. From Table 9.1, the Young’s modulus for bone is given by Y = 9.4 × 109 N m–2. Length of each thighbone L = 0.5 m the radius of thighbone = 2.0 cm Thus the cross-sectional area of the thighbone A = π × (2 × 10-2)2 m2 = 1.26 × 10-3 m2. Using Eq. (9.8), the compression in each thighbone (∆L) can be computed as ∆L = [(F × L)/(Y × A)] = [(1078 × 0.5)/(9.4 × 109 × 1.26 × 10-3)] = 4.55 × 10-5 m or 4.55 × 10-3 cm. This is a very small change! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%. t 9.6.2 Determination of Young’s Modulus of the Material of a Wire Fig. 9.6 An arrangement for the 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 Both the reference and experimental wires are tension is shown in Fig. 9.6. It consists of two given an initial small load to keep the wires long straight wires of same length and equal straight and the vernier reading is noted. Now radius suspended side by side from a fixed rigid the experimental wire is gradually loaded with support. The wire A (called the reference wire) more weights to bring it under a tensile stress carries a millimetre main scale M and a pan to and the vernier reading is noted again. The place a weight. The wire B (called the difference between two vernier readings gives experimental wire) of uniform area of cross- the elongation produced in the wire. Let r and L section also carries a pan in which known be the initial radius and length of the weights can be placed. A vernier scale V is experimental wire, respectively. Then the area attached to a pointer at the bottom of the of cross-section of the wire would be πr2. Let M experimental wire B, and the main scale M is be the mass that produced an elongation ∆L in fixed to the reference wire A. The weights placed the wire. Thus the applied force is equal to Mg, in the pan exert a downward force and stretch the experimental wire under a tensile stress. The where g is the acceleration due to gravity. From Eq. (9.8), the Young’s modulus of the material elongation of the wire (increase in length) is measured by the vernier arrangement. The of the experimental wire is given by reference wire is used to compensate for any Mg L σ change in length that may occur due to change Y= = π r 2. ∆L in room temperature, since any change in length ε of the reference wire due to temperature change = Mg × L/(πr2 × ∆L) (9.9) 2020-21 242 PHYSICS 9.6.3 Shear Modulus Therefore, the stress applied is The ratio of shearing stress to the corresponding = (9.4 × 104 N/0.05 m2) shearing strain is called the shear modulus of = 1.80 × 106 N.m–2 the material and is represented by G. It is also called the modulus of rigidity. G = shearing stress (σs)/shearing strain G = (F/A)/(∆x/L) = (F × L)/(A × ∆x) (9.10) Similarly, from Eq. (9.4) G = (F/A)/θ = F/(A × θ) (9.11) The shearing stress σs can also be expressed as σs = G × θ (9.12) aaaaaaaaaaaaaaaaaaaaa SI unit of shear modulus is N m–2 or Pa. The aaaaaaaaaaaaaaaaaaaaa shear moduli of a few common materials are Fig. 9.7 given in Table 9.2. It can be seen that shear modulus (or modulus of rigidity) is generally less We know that shearing strain = (∆x/L)= Stress /G. than Young’s modulus (from Table 9.1). For most Therefore the displacement ∆x = (Stress × L)/G materials G ≈ Y/3. = (1.8 × 106 N m–2 × 0.5m)/(5.6 × 109 N m–2) = 1.6 × 10–4 m = 0.16 mm t Table 9.2 Shear moduli (G) of some common materials 9.6.4 Bulk Modulus 9 –2 Material G (10 Nm In Section (9.3), we have seen that when a body or GPa) is submerged in a fluid, it undergoes a hydraulic Aluminium 25 stress (equal in magnitude to the hydraulic Brass 36 pressure). This leads to the decrease in the Copper 42 volume of the body thus producing a strain called Glass 23 volume strain [Eq. (9.5)]. The ratio of hydraulic Iron 70 stress to the corresponding hydraulic strain is Lead 5.6 called bulk modulus. It is denoted by symbol B. Nickel 77 B = – p/(∆V/V) (9.13) Steel 84 The negative sign indicates the fact that with Tungsten 150 an increase in pressure, a decrease in volume Wood 10 occurs. That is, if p is positive, ∆V is negative. Thus for a system in equilibrium, the value of u Example 9.4 A square lead slab of side 50 bulk modulus B is always positive. SI unit of cm and thickness 10 cm is subject to a bulk modulus is the same as that of pressure shearing force (on its narrow face) of 9.0 × i.e., N m–2 or Pa. The bulk moduli of a few 104 N. The lower edge is riveted to the floor. common materials are given in Table 9.3. How much will the upper edge be displaced? The reciprocal of the bulk modulus is called compressibility and is denoted by k. It is defined Answer The lead slab is fixed and the force is as the fractional change in volume per unit applied parallel to the narrow face as shown in increase in pressure. Fig. 9.7. The area of the face parallel to which k = (1/B) = – (1/∆p) × (∆V/V) (9.14) this force is applied is It can be seen from the data given in Table A = 50 cm × 10 cm 9.3 that the bulk moduli for solids are much = 0.5 m × 0.1 m larger than for liquids, which are again much = 0.05 m2 larger than the bulk modulus for gases (air). 2020-21 MECHANICAL PROPERTIES OF SOLIDS 243 Table 9.3 Bulk moduli (B) of some common Gases have large compressibilities, which vary Materials with pressure and temperature. The incompressibility of the solids is primarily due Material B (109 N m–2 or GPa) to the tight coupling between the neighbouring Solids atoms. The molecules in liquids are also bound Aluminium 72 with their neighbours but not as strong as in Brass 61 solids. Molecules in gases are very poorly coupled to their neighbours. Copper 140 Table 9.4 shows the various types of stress, Glass 37 strain, elastic moduli, and the applicable state Iron 100 of matter at a glance. Nickel 260 u Example 9.5 The average depth of Indian Steel 160 Ocean is about 3000 m. Calculate the Liquids fractional compression, ∆V/V, of water at Water 2.2 the bottom of the ocean, given that the bulk Ethanol 0.9 modulus of water is 2.2 × 109 N m–2. (Take Carbon disulphide 1.56 g = 10 m s–2) Glycerine 4.76 Answer The pressure exerted by a 3000 m Mercury 25 column of water on the bottom layer Gases p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2 Air (at STP) 1.0 × 10–4 = 3 × 107 kg m–1 s-2 = 3 × 107 N m–2 Thus, solids are the least compressible, whereas, Fractional compression ∆V/V, is gases are the most compressible. Gases are about ∆V/V = stress/B = (3 × 107 N m-2)/(2.2 × 109 N m–2) a million times more compressible than solids! = 1.36 × 10-2 or 1.36 % t Table 9.4 Stress, strain and various elastic moduli Type of Stress Strain Change in Elastic Name of State of stress shape volume Modulus Modulus Matter Tensile Two equal and Elongation or Yes No Y = (F×L)/ Young’s Solid or opposite forces compression (A×∆L) modulus compressive perpendicular to parallel to force (σ = F/A) opposite faces direction (∆L/L) (longitudinal strain) Shearing Two equal and Pure shear, θ Yes No G = F/(A×θ) Shear Solid (σs = F/A) opposite forces modulus parallel to oppoiste or modulus surfaces forces of rigidity in each case such that total force and total torque on the body vanishes Hydraulic Forces perpendicular Volume change No Yes B = –p/(∆V/V) Bulk Solid, liquid everywhere to the (compression or modulus and gas surface, force per elongation) unit area (pressure) (∆V/V) same everywhere. 2020-21 244 PHYSICS 9.6.5 POISSON’S RATIO 1 = × stress × strain × volume of the Careful observations with the Young’s modulus 2 experiment (explained in section 9.6.2), show wire that there is also a slight reduction in the cross- This work is stored in the wire in the form of section (or in the diameter) of the wire. The strain elastic potential energy (U). Therefore the elastic perpendicular to the applied force is called potential energy per unit volume of the wire (u) is lateral strain. Simon Poisson pointed out that within the elastic limit, lateral strain is directly 1 proportional to the longitudinal strain. The ratio u= ×σε (9.15) 2 of the lateral strain to the longitudinal strain in a stretched wire is called Poisson’s ratio. If the 9.7 APPLICATIONS OF ELASTIC original diameter of the wire is d and the BEHAVIOUR OF MATERIALS contraction of the diameter under stress is ∆d, The elastic behaviour of materials plays an the lateral strain is ∆d/d. If the original length important role in everyday life. All engineering of the wire is L and the elongation under stress designs require precise knowledge of the elastic is ∆L, the longitudinal strain is ∆L/L. Poisson’s behaviour of materials. For example while ratio is then (∆d/d)/(∆L/L) or (∆d/∆L) × (L/d). designing a building, the structural design of Poisson’s ratio is a ratio of two strains; it is a the columns, beams and supports require pure number and has no dimensions or units. knowledge of strength of materials used. Have Its value depends only on the nature of material. you ever thought why the beams used in For steels the value is between 0.28 and 0.30, construction of bridges, as supports etc. have and for aluminium alloys it is about 0.33. a cross-section of the type I? Why does a heap of sand or a hill have a pyramidal shape? 9.6.6 Elastic Potential Energy Answers to these questions can be obtained in a Stretched Wire from the study of structural engineering which When a wire is put under a tensile stress, work is based on concepts developed here. is done against the inter-atomic forces. This Cranes used for lifting and moving heavy work is stored in the wire in the form of elastic loads from one place to another have a thick potential energy. When a wire of original length metal rope to which the load is attached. The L and area of cross-section A is subjected to a rope is pulled up using pulleys and motors. deforming force F along the length of the wire, Suppose we want to make a crane, which has let the length of the wire be elongated by l. Then a lifting capacity of 10 tonnes or metric tons (1 from Eq. (9.8), we have F = YA × (l/L). Here Y is metric ton = 1000 kg). How thick should the the Young’s modulus of the material of the wire. steel rope be? We obviously want that the load Now for a further elongation of infinitesimal does not deform the rope permanently. small length dl, work done dW is F × dl or YAldl/ Therefore, the extension should not exceed the L. Therefore, the amount of work done (W) in elastic limit. From Table 9.1, we find that mild increasing the length of the wire from L to L + l, steel has a yield strength (σy) of about 300 × that is from l = 0 to l = l is 106 N m–2. Thus, the area of cross-section (A) of the rope should at least be l YAl YA l 2 A ≥ W/σy = Mg/σy (9.16) W= ∫0 L dl = 2 × L = (104 kg × 9.8 m s-2)/(300 × 106 N m-2) = 3.3 × 10-4 m2 2 1 l  corresponding to a radius of about 1 cm for W = × Y ×   × AL a rope of circular cross-section. Generally 2 L  a large margin of safety (of about a factor of ten in the load) is provided. Thus a thicker 1 rope of radius about 3 cm is recommended. = × Young’s modulus × strain2 × 2 A single wire of this radius would practically volume of the wire be a rigid rod. So the ropes are always made of a number of thin wires braided together, 2020-21 MECHANICAL PROPERTIES OF SOLIDS 245 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 (a) (b) (c) consider the case of a beam loaded at the centre and supported near its ends as shown in Fig. 9.9 Different cross-sectional shapes of a beam. Fig. 9.8. A bar of length l, breadth b, and depth d (a) Rectangular section of a bar; (b) A thin bar and how it can buckle; when loaded at the centre by a load W sags by (c) Commonly used section for a load an amount given by bearing bar. δ = W l 3/(4bd 3Y) (9.17) The use of pillars or columns is also very common in buildings and bridges. A pillar with rounded ends as shown in Fig. 9.10(a) supports less load than that with a distributed shape at the ends [Fig. 9.10(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. 9.8 A beam supported at the ends and loaded at the centre. 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. For a given material, (a) (b) increasing the depth d rather than the breadth b is more effective in reducing the bending, since Fig. 9.10 Pillars or columns: (a) a pillar with rounded δ is proportional to d -3 and only to b-1(of course ends, (b) Pillar with distributed ends. the length l of the span should be as small as The answer to the question why the maximum height of a mountain on earth is ~10 km can possible). But on increasing the depth, unless also be provided by considering the elastic the load is exactly at the right place (difficult to properties of rocks. A mountain base is not under arrange in a bridge with moving traffic), the uniform compression and this provides some deep bar may bend as shown in Fig. 9.9(b). This shearing stress to the rocks under which they is called buckling. To avoid this, a common can flow. The stress due to all the material on compromise is the cross-sectional shape shown the top should be less than the critical shearing in Fig. 9.9(c). This section provides a large load- stress at which the rocks flow. bearing surface and enough depth to prevent At the bottom of a mountain of height h, the bending. This shape reduces the weight of the force per unit area due to the weight of the beam without sacrificing the strength and mountain is hρg where ρ is the density of the hence reduces the cost. material of the mountain and g is the acceleration 2020-21 246 PHYSICS due to gravity. The material at the bottom 30 × 107 N m-2. Equating this to hρg, with experiences this force in the vertical direction, ρ = 3 × 103 kg m-3 gives and the sides of the mountain are free. Therefore, hρg = 30 × 107 N m-2. this is not a case of pressure or bulk compression. h = 30 × 107 N m-2/(3 × 103 kg m-3 × 10 m s-2) There is a shear component, approximately hρg = 10 km itself. Now the elastic limit for a typical rock is which is more than the height of Mt. Everest! SUMMARY 1. Stress is the restoring force per unit area and strain is the fractional change in dimension. In general there are three types of stresses (a) tensile stress — longitudinal stress (associated with stretching) or compressive stress (associated with compression), (b) shearing stress, and (c) hydraulic stress. 2. For small deformations, stress is directly proportional to the strain for many materials. This is known as Hooke’s law. The constant of proportionality is called modulus of elasticity. Three elastic moduli viz., Young’s modulus, shear modulus and bulk modulus are used to describe the elastic behaviour of objects as they respond to deforming forces that act on them. A class of solids called elastomers does not obey Hooke’s law. 3. When an object is under tension or compression, the Hooke’s law takes the form F/A = Y∆L/L where ∆L/L is the tensile or compressive strain of the object, F is the magnitude of the applied force causing the strain, A is the cross-sectional area over which F is applied (perpendicular to A) and Y is the Young’s modulus for the object. The stress is F/A. 4. A pair of forces when applied parallel to the upper and lower faces, the solid deforms so that the upper face moves sideways with respect to the lower. The horizontal displacement ∆L of the upper face is perpendicular to the vertical height L. This type of deformation is called shear and the corresponding stress is the shearing stress. This type of stress is possible only in solids. In this kind of deformation the Hooke’s law takes the form F/A = G × ∆L/L where ∆L is the displacement of one end of object in the direction of the applied force F, and G is the shear modulus. 5. When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, the Hooke’s law takes the form p = B (∆V/V), where p is the pressure (hydraulic stress) on the object due to the fluid, ∆V/V (the volume strain) is the absolute fractional change in the object’s volume due to that pressure and B is the bulk modulus of the object. POINTS TO PONDER 1. In the case of a wire, suspended from celing and stretched under the action of a weight (F) suspended from its other end, the force exerted by the ceiling on it is equal and opposite to the weight. However, the tension at any cross-section A of the wire is just F and not 2F. Hence, tensile stress which is equal to the tension per unit area is equal to F/A. 2. Hooke’s law is valid only in the linear part of stress-strain curve. 3. The Young’s modulus and shear modulus are relevant only for solids since only solids have lengths and shapes. 4. Bulk modulus is relevant for solids, liquid and gases. It refers to the change in volume when every part of the body is under the uniform stress so that the shape of the body remains unchanged. 2020-21 MECHANICAL PROPERTIES OF SOLIDS 247 5. Metals have larger values of Young’s modulus than alloys and elastomers. A material with large value of Young’s modulus requires a large force to produce small changes in its length. 6. In daily life, we feel that a material which stretches more is more elastic, but it a is misnomer. In fact material which stretches to a lesser extent for a given load is considered to be more elastic. 7. In general, a deforming force in one direction can produce strains in other directions also. The proportionality between stress and strain in such situations cannot be described by just one elastic constant. For example, for a wire under longitudinal strain, the lateral dimensions (radius of cross section) will undergo a small change, which is described by another elastic constant of the material (called Poisson ratio). 8. Stress is not a vector quantity since, unlike a force, the stress cannot be assigned a specific direction. Force acting on the portion of a body on a specified side of a section has a definite direction. EXERCISES 9.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper? 9.2 Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material? Fig. 9.11 9.3 The stress-strain graphs for materials A and B are shown in Fig. 9.12. Fig. 9.12 2020-21 248 PHYSICS The graphs are drawn to the same scale. (a) Which of the materials has the greater Young’s modulus? (b) Which of the two is the stronger material? 9.4 Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young’s modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus. 9.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. Fig. 9.13 9.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face? 9.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. 9.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? 9.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support ? 9.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension. 9.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path. 9.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. 9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3? 9.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. 9.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 106 Pa. 9.16 How much should the pressure on a litre of water be changed to compress it by 0.10%? 2020-21 MECHANICAL PROPERTIES OF SOLIDS 249 Additional Exercises 9.17 Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil? Fig. 9.14 9.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires. Fig. 9.15 9.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10-2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid- point. 9.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one quarter of the load. 9.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom? 2020-21