Mechanical Properties of Solids PDF

CHAPTER EIGHT MECHANICAL PROPERTIES OF SOLIDS 8.1 INTRODUCTION In Chapter 6, we studied the rotation of the bodies and then rea...

CHAPTER EIGHT MECHANICAL PROPERTIES OF SOLIDS 8.1 INTRODUCTION In Chapter 6, we studied the rotation of the bodies and then realised that the motion of a body depends on how mass is 8.1 Introduction distributed within the body. We restricted ourselves to simpler 8.2 Stress and strain situations of rigid bodies. A rigid body generally means a 8.3 Hooke’s law hard solid object having a definite shape and size. But in 8.4 Stress-strain curve reality, bodies can be stretched, compressed and bent. Even 8.5 Elastic moduli the appreciably rigid steel bar can be deformed when a 8.6 Applications of elastic sufficiently large external force is applied on it. This means behaviour of materials that solid bodies are not perfectly rigid. A solid has definite shape and size. In order to change (or Summary deform) the shape or size of a body, a force is required. If Points to ponder you stretch a helical spring by gently pulling its ends, the Exercises 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 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 2024-25 168 PHYSICS elastic behaviour and mechanical properties of ∆L solids which would answer many such Longitudinal strain = (8.2) questions. L However, if two equal and opposite deforming 8.2 STRESS AND STRAIN forces are applied parallel to the cross-sectional When forces are applied on a body in such a area of the cylinder, as shown in Fig. 8.1(b), manner that the body is still in static equilibrium, there is relative displacement between the it is deformed to a small or large extent depending opposite faces of the cylinder. The restoring force upon the nature of the material of the body and per unit area developed due to the applied the magnitude of the deforming force. The tangential force is known as tangential or deformation may not be noticeable visually in shearing stress. many materials but it is there. When a body is As a result of applied tangential force, there subjected to a deforming force, a restoring force is a relative displacement ∆x between opposite is developed in the body. This restoring force is faces of the cylinder as shown in the Fig. 8.1(b). equal in magnitude but opposite in direction to The strain so produced is known as shearing the applied force. The restoring force per unit area strain and it is defined as the ratio of relative is known as stress. If F is the force applied normal displacement of the faces ∆x to the length of the to the cross–section and A is the area of cross cylinder L. section of the body, ∆x Magnitude of the stress = F/A (8.1) Shearing strain = = tan θ (8.3) The SI unit of stress is N m–2 or pascal (Pa) L and its dimensional formula is [ ML–1T–2 ]. where θ is the angular displacement of the There are three ways in which a solid may cylinder from the vertical (original position of the change its dimensions when an external force cylinder). Usually θ is very small, tan θ acts on it. These are shown in Fig. 8.1. In is nearly equal to angle θ , (if θ = 10°, for Fig.8.1(a), a cylinder is stretched by two equal example, there is only 1% difference between θ forces applied normal to its cross-sectional area. and tan θ). The restoring force per unit area in this case is It can also be visualised, when a book is called tensile stress. If the cylinder is pressed with the hand and pushed horizontally, compressed under the action of applied forces, as shown in Fig. 8.2 (c). the restoring force per unit area is known as Thus, shearing strain = tan θ ≈ θ (8.4) compressive stress. Tensile or compressive In Fig. 8.1 (d), a solid sphere placed in the fluid stress can also be termed as longitudinal stress. under high pressure is compressed uniformly on In both the cases, there is a change in the all sides. The force applied by the fluid acts in length of the cylinder. The change in the length perpendicular direction at each point of the ∆L to the original length L of the body (cylinder surface and the body is said to be under in this case) is known as longitudinal strain. hydraulic compression. This leads to decrease (a) (b) (c) (d) Fig. 8.1 (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. 2024-25 MECHANICAL PROPERTIES OF SOLIDS 169 in its volume without any change of its compression and shear stress may also be geometrical shape. obtained. The stress-strain curves vary from The body develops internal restoring forces material to material. These curves help us to that are equal and opposite to the forces applied understand how a given material deforms with by the fluid (the body restores its original shape increasing loads. From the graph, we can see and size when taken out from the fluid). The that in the region between O to A, the curve is internal restoring force per unit area in this case linear. In this region, Hooke’s law is obeyed. 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 = (8.5) V Since the strain is a ratio of change in dimension to the original dimension, it has no units or dimensional formula. 8.3 HOOKE’S LAW Stress and strain take different forms in the situations depicted in the Fig. (8.1). For small deformations the stress and strain are proportional to each other. This is known as Fig. 8.2 A typical stress-strain curve for a metal. Hooke’s law. Thus, In the region from A to B, stress and strain stress ∝ strain are not proportional. Nevertheless, the body still stress = k × strain (8.6) returns to its original dimension when the load where k is the proportionality constant and is is removed. The point B in the curve is known known as modulus of elasticity. as yield point (also known as elastic limit) and Hooke’s law is an empirical law and is found the corresponding stress is known as yield to be valid for most materials. However, there strength (σy ) of the material. are some materials which do not exhibit this If the load is increased further, the stress linear relationship. developed exceeds the yield strength and strain increases rapidly even for a small change in the 8.4 STRESS-STRAIN CURVE stress. The portion of the curve between B and 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) and fracture points D and E are close, the material the strain produced. A typical graph for a metal is said to be brittle. If they are far apart, the is shown in Fig. 8.2. Analogous graphs for material is said to be ductile. 2024-25 170 PHYSICS 8.5 ELASTIC MODULI The proportional region within the elastic limit of the stress-strain curve (region OA in Fig. 8.2) is of great importance for structural and manufacturing engineering designs. The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the material. 8.5.1 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 Fig. 8.3 Stress-strain curve for the elastic tissue of Young’s modulus and is denoted by the symbol Y. Aorta, the large tube (vessel) carrying blood from the heart. σ Y= (8.7) As stated earlier, the stress-strain behaviour ε varies from material to material. For example, From Eqs. (8.1) and (8.2), we have rubber can be pulled to several times its original length and still returns to its original shape. Fig. Y = (F/A)/(∆L/L) 8.3 shows stress-strain curve for the elastic = (F × L) /(A × ∆L) (8.8) tissue of aorta, present in the heart. Note that Since strain is a dimensionless quantity, the although elastic region is very large, the material unit of Young’s modulus is the same as that of does not obey Hooke’s law over most of the region. stress i.e., N m–2 or Pascal (Pa). Table 8.1 gives Secondly, there is no well defined plastic region. the values of Young’s moduli and yield strengths Substances like tissue of aorta, rubber etc. of some material. which can be stretched to cause large strains From the data given in Table 8.1, it is noticed are called elastomers. that for metals Young’s moduli are large. Table 8.1 Young’s moduli and yield strenghs of some material # Substance tested under compression 2024-25 MECHANICAL PROPERTIES OF SOLIDS 171 Therefore, these materials require a large force Answer The copper and steel wires are under to produce small change in length. To increase a tensile stress because they have the same the length of a thin steel wire of 0.1 cm2 cross- tension (equal to the load W) and the same area sectional area by 0.1%, a force of 2000 N is of cross-section A. From Eq. (8.7) we have stress required. The force required to produce the same = strain × Young’s modulus. Therefore strain in aluminium, brass and copper wires W/A = Yc × (∆Lc/Lc) = Ys × (∆Ls/Ls) having the same cross-sectional area are 690 N, where the subscripts c and s refer to copper 900 N and 1100 N respectively. It means that and stainless steel respectively. Or, steel is more elastic than copper, brass and ∆Lc/∆Ls = (Ys/Yc) × (Lc/Ls) aluminium. It is for this reason that steel is Given Lc = 2.2 m, Ls = 1.6 m, preferred in heavy-duty machines and in From Table 9.1 Yc = 1.1 × 1011 N.m–2, and structural designs. Wood, bone, concrete and Ys = 2.0 × 1011 N.m–2. glass have rather small Young’s moduli. ∆Lc/∆Ls = (2.0 × 10 /1.1 × 1011) × (2.2/1.6) = 2.5. 11 The total elongation is given to be u Example 8.1 A structural steel rod has a ∆Lc + ∆Ls = 7.0 × 10-4 m radius of 10 mm and a length of 1.0 m. A Solving the above equations, 100 kN force stretches it along its length. ∆Lc = 5.0 × 10-4 m, and ∆Ls = 2.0 × 10-4 m. Calculate (a) stress, (b) elongation, and (c) Therefore strain on the rod. Young’s modulus, of W = (A × Yc × ∆Lc)/Lc structural steel is 2.0 × 1011 N m-2. = π (1.5 × 10-3)2 × [(5.0 × 10-4 × 1.1 × 1011)/2.2] = 1.8 × 102 N ⊳ Answer We assume that the rod is held by a clamp at one end, and the force F is applied at uExample 8.3 In a human pyramid in a the other end, parallel to the length of the rod. circus, the entire weight of the balanced Then the stress on the rod is given by group is supported by the legs of a performer F F who is lying on his back (as shown in Fig. Stress = = A πr 2 8.4). The combined mass of all the persons performing the act, and the tables, plaques 100 × 10 N 3 = etc. involved is 280 kg. The mass of the ( ) −2 2 3.14 × 10 m performer lying on his back at the bottom of = 3.18 × 108 N m–2 the pyramid is 60 kg. Each thighbone (femur) The elongation, of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the ( F/A ) L amount by which each thighbone gets ∆L = Y compressed under the extra load. (3.18 × 10 8 N m –2 ) (1m ) = 11 –2 2 × 10 N m = 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 % ⊳ u Example 8.2 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. Fig. 8.4 Human pyramid in a circus. 2024-25 172 PHYSICS Answer Total mass of all the performers, tables, Table 8.2 Shear moduli (G) of some common materials plaques etc. = 280 kg Mass of the performer = 60 kg Material G (109 Nm–2 Mass supported by the legs of the performer or GPa) at the bottom of the pyramid Aluminium 25 = 280 – 60 = 220 kg Brass 36 Weight of this supported mass Copper 42 = 220 kg wt. = 220 × 9.8 N = 2156 N. Glass 23 Weight supported by each thighbone of the Iron 70 performer = ½ (2156) N = 1078 N. Lead 5.6 From Table 9.1, the Young’s modulus for bone Nickel 77 is given by Steel 84 Y = 9.4 × 109 N m–2. Tungsten 150 Length of each thighbone L = 0.5 m Wood 10 the radius of thighbone = 2.0 cm Thus the cross-sectional area of the thighbone u Example 8.4 A square lead slab of side 50 A = π × (2 × 10-2)2 m2 = 1.26 × 10-3 m2. cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × Using Eq. (9.8), the compression in each 104 N. The lower edge is riveted to the floor. thighbone (∆L) can be computed as How much will the upper edge be displaced? ∆L = [(F × L)/(Y × A)] = [(1078 × 0.5)/(9.4 × 109 × 1.26 × 10-3)] Answer The lead slab is fixed and the force is = 4.55 × 10-5 m or 4.55 × 10-3 cm. applied parallel to the narrow face as shown in This is a very small change! The fractional Fig. 8.6. The area of the face parallel to which decrease in the thighbone is ∆L/L = 0.000091 or this force is applied is 0.0091%. ⊳ A = 50 cm × 10 cm = 0.5 m × 0.1 m 8.5.2 Shear Modulus = 0.05 m2 The ratio of shearing stress to the corresponding Therefore, the stress applied is shearing strain is called the shear modulus of = (9.4 × 104 N/0.05 m2) the material and is represented by G. It is also = 1.80 × 106 N.m–2 called the modulus of rigidity. G = shearing stress (σs)/shearing strain G = (F/A)/(∆x/L) = (F × L)/(A × ∆x) (8.10) Similarly, from Eq. (9.4) G = (F/A)/θ = F/(A × θ) (8.11) The shearing stress σs can also be expressed as aaaaaaaaaaaaaaaaaaaaa σs = G × θ (8.12) aaaaaaaaaaaaaaaaaaaaa SI unit of shear modulus is N m–2 or Pa. The Fig. 8.5 shear moduli of a few common materials are given in Table 9.2. It can be seen that shear We know that shearing strain = (∆x/L)= Stress /G. modulus (or modulus of rigidity) is generally less Therefore the displacement ∆x = (Stress × L)/G than Young’s modulus (from Table 9.1). For most = (1.8 × 106 N m–2 × 0.5m)/(5.6 × 109 N m–2) materials G ≈ Y/3. = 1.6 × 10–4 m = 0.16 mm ⊳ 2024-25 MECHANICAL PROPERTIES OF SOLIDS 173 8.5.3 Bulk Modulus Table 8.3 Bulk moduli (B) of some common Materials In Section (8.3), we have seen that when a body is submerged in a fluid, it undergoes a hydraulic Material B (109 N m–2 or GPa) stress (equal in magnitude to the hydraulic Solids pressure). This leads to the decrease in the volume of the body thus producing a strain called Aluminium 72 volume strain [Eq. (8.5)]. The ratio of hydraulic Brass 61 stress to the corresponding hydraulic strain is called bulk modulus. It is denoted by symbol B. Copper 140 B = – p/(∆V/V) (8.12) Glass 37 The negative sign indicates the fact that with Iron 100 an increase in pressure, a decrease in volume occurs. That is, if p is positive, ∆V is negative. Nickel 260 Thus for a system in equilibrium, the value of Steel 160 bulk modulus B is always positive. SI unit of bulk modulus is the same as that of pressure Liquids i.e., N m–2 or Pa. The bulk moduli of a few common Water 2.2 materials are given in Table 8.3. The reciprocal of the bulk modulus is called Ethanol 0.9 compressibility and is denoted by k. It is defined Carbon disulphide 1.56 as the fractional change in volume per unit increase in pressure. Glycerine 4.76 k = (1/B) = – (1/∆p) × (∆V/V) (8.13) Mercury 25 It can be seen from the data given in Table 8.3 that the bulk moduli for solids are much Gases larger than for liquids, which are again much Air (at STP) 1.0 × 10–4 larger than the bulk modulus for gases (air). Table 8.4 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. 2024-25 174 PHYSICS Thus, solids are the least compressible, whereas, 8.5.5 Elastic Potential Energy gases are the most compressible. Gases are about in a Stretched Wire a million times more compressible than solids! When a wire is put under a tensile stress, work Gases have large compressibilities, which vary is done against the inter-atomic forces. This with pressure and temperature. The work is stored in the wire in the form of elastic incompressibility of the solids is primarily due potential energy. When a wire of original length to the tight coupling between the neighbouring L and area of cross-section A is subjected to a atoms. The molecules in liquids are also bound deforming force F along the length of the wire, with their neighbours but not as strong as in let the length of the wire be elongated by l. Then solids. Molecules in gases are very poorly from Eq. (8.8), we have F = YA × (l/L). Here Y is coupled to their neighbours. the Young’s modulus of the material of the wire. Table 8.4 shows the various types of stress, Now for a further elongation of infinitesimal strain, elastic moduli, and the applicable state small length dl, work done dW is F × dl or YAldl/ of matter at a glance. L. Therefore, the amount of work done (W) in increasing the length of the wire from L to L + l, u Example 8.5 The average depth of Indian that is from l = 0 to l = l is Ocean is about 3000 m. Calculate the YAl YA l 2 fractional compression, ∆V/V, of water at W= ∫0 l dl = × L 2 L the bottom of the ocean, given that the bulk 2 modulus of water is 2.2 × 109 N m–2. (Take 1 l  g = 10 m s–2) W = × Y ×   × AL 2 L  1 = × Young’s modulus × strain2 × Answer The pressure exerted by a 3000 m column of water on the bottom layer 2 p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2 volume of the wire = 3 × 107 kg m–1 s-2 1 = 3 × 107 N m–2 = × stress × strain × volume of the 2 Fractional compression ∆V/V, is wire ∆V/V = stress/B = (3 × 107 N m-2)/(2.2 × 109 N m–2) This work is stored in the wire in the form of = 1.36 × 10-2 or 1.36 % ⊳ elastic potential energy (U). Therefore the elastic potential energy per unit volume of the wire (u) is 1 8.5.4 POISSON’S RATIO u= ×σ ε (8.14) 2 The strain perpendicular to the applied force is called lateral strain. Simon Poisson pointed out 8.6 APPLICATIONS OF ELASTIC that within the elastic limit, lateral strain is BEHAVIOUR OF MATERIALS directly proportional to the longitudinal strain. The elastic behaviour of materials plays an The ratio of the lateral strain to the longitudinal important role in everyday life. All engineering strain in a stretched wire is called Poisson’s designs require precise knowledge of the elastic ratio. If the original diameter of the wire is d behaviour of materials. For example while and the contraction of the diameter under stress designing a building, the structural design of is ∆d, the lateral strain is ∆d/d. If the original the columns, beams and supports require length of the wire is L and the elongation under knowledge of strength of materials used. Have stress is ∆L, the longitudinal strain is ∆L/L. you ever thought why the beams used in Poisson’s ratio is then (∆d/d)/(∆L/L) or (∆d/∆L) construction of bridges, as supports etc. have × (L/d). Poisson’s ratio is a ratio of two strains; a cross-section of the type I? Why does a heap it is a pure number and has no dimensions or of sand or a hill have a pyramidal shape? units. Its value depends only on the nature of Answers to these questions can be obtained material. For steels the value is between 0.28 and from the study of structural engineering which 0.30, and for aluminium alloys it is about 0.33. is based on concepts developed here. 2024-25 MECHANICAL PROPERTIES OF SOLIDS 175 Cranes used for lifting and moving heavy loads This relation can be derived using what you from one place to another have a thick metal have already learnt and a little calculus. From rope to which the load is attached. The rope is Eq. (8.16), we see that to reduce the bending pulled up using pulleys and motors. Suppose we for a given load, one should use a material with want to make a crane, which has a lifting a large Young’s modulus Y. For a given material, capacity of 10 tonnes or metric tons (1 metric increasing the depth d rather than the breadth ton = 1000 kg). How thick should the steel rope b is more effective in reducing the bending, since be? We obviously want that the load does not δ is proportional to d -3 and only to b-1(of course deform the rope permanently. Therefore, the the length l of the span should be as small as extension should not exceed the elastic limit. From Table 8.1, we find that mild steel has a possible). But on increasing the depth, unless yield strength (σy) of about 300 × 106 N m–2. Thus, the load is exactly at the right place (difficult to the area of cross-section (A) of the rope should arrange in a bridge with moving traffic), the at least be deep bar may bend as shown in Fig. 8.7(b). This A ≥ W/σy = Mg/σy (8.15) is called buckling. To avoid this, a common = (104 kg × 9.8 m s-2)/(300 × 106 N m-2) compromise is the cross-sectional shape shown = 3.3 × 10-4 m2 in Fig. 8.7(c). This section provides a large load- corresponding to a radius of about 1 cm for a bearing surface and enough depth to prevent rope of circular cross-section. Generally a bending. This shape reduces the weight of the large margin of safety (of about a factor of ten beam without sacrificing the strength and in the load) is provided. Thus a thicker rope of hence reduces the cost. 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. 8.6. A bar of length l, breadth b, and depth d when loaded at the centre by a load W sags by (a) (b) (c) an amount given by Fig. 8.7 Different cross-sectional shapes of a beam. (a) Rectangular section of a bar; δ = W l 3/(4bd 3Y) (8.16) (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. 8.9(a) supports less load than that with a distributed shape at the ends [Fig. 8.9(b)]. The precise design of a bridge or a building has to take into account the conditions under which it will function, the Fig. 8.6 A beam supported at the ends and loaded cost and long period, reliability of usable at the centre. material, etc. 2024-25 176 PHYSICS 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, (a) (b) this is not a case of pressure or bulk compression. There is a shear component, approximately hρg Fig. 8.8 Pillars or columns: (a) a pillar with rounded itself. Now the elastic limit for a typical rock is ends, (b) Pillar with distributed ends. 30 × 107 N m-2. Equating this to hρg, with The answer to the question why the maximum ρ = 3 × 103 kg m-3 gives height of a mountain on earth is ~10 km can hρg = 30 × 107 N m-2. also be provided by considering the elastic h = 30 × 107 N m-2/(3 × 103 kg m-3 × 10 m s-2) properties of rocks. A mountain base is not under = 10 km uniform compression and this provides some 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. 2024-25 MECHANICAL PROPERTIES OF SOLIDS 177 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. 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 8.1 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? 8.2 Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material? Fig. 8.9 2024-25 178 PHYSICS 8.3 The stress-strain graphs for materials A and B are shown in Fig. 8.10. Fig. 8.10 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? 8.4 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. 8.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. 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. 8.11 8.6 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? 8.7 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. 8.8 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? 8.9 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 ? 8.10 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. 8.11 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. 2024-25 MECHANICAL PROPERTIES OF SOLIDS 179 8.12 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. 8.13 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? 8.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. 8.15 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. 8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%? carry one quarter of the load. 2024-25