NEET Physics Notes - Chapter 9 Elasticity - PDF

60 446 Elasticity E3 Chapter 9 Elasticity (i) When two atoms are at very large distance, the potential energy is negative and becomes more negative as r is decreased. (ii) When the distance between the two atoms becomes r , the potential energy of the system of two atoms becomes minimum (i.e. attain...

60 446 Elasticity E3 Chapter 9 Elasticity (i) When two atoms are at very large distance, the potential energy is negative and becomes more negative as r is decreased. (ii) When the distance between the two atoms becomes r , the potential energy of the system of two atoms becomes minimum (i.e. attains maximum negative value). As the state of minimum potential energy is the state of equilibrium, hence the two atoms at separation r will be in a state of equilibrium. ID Interatomic Forces The forces between the atoms due to electrostatic interaction between the charges of the atoms are called interatomic forces. These forces are electrical in nature and these are active if the distance between the two atoms is of the order of atomic size i.e. 10 metre. (1) Every atom is electrically neutral, the number of electrons (negative charge) orbiting around the nucleus is equal to the number of protons (positive charge) in the nucleus. So if two atoms are placed at a very large distance from each other then there will be a very small (negligible) interatomic force working between them. (2) When two atoms are brought closer to each other to a distance of the order of 10 m, the distances between their positive nuclei and negative electron clouds get disturbed, and due to this, attractive interatomic force is produced between two atoms. (3) This attractive force increases continuously with decrease in r and becomes maximum for one value of r called critical distance, represented by x (as shown in the figure). Beyond this the attractive force starts decreasing rapidly with further F decrease in the value of r. x (4) When the distance between O r r0 the two atoms becomes r , the Attraction interatomic force will be zero. This distance r is called normal or equilibrium distance. U (r = 0.74 Å for hydrogen). (5) When the distance between r the two atoms further decreased, the O r0 interatomic force becomes repulsive in nature and increases very rapidly with decrease in distance between two atoms. Fig. 9.1 (6) The potential energy U is related with the interatomic force F by the following relation. 0 U Repulsion –10 D YG U –10 0 0 ST 0 0 F dU dr ( U0  7.2  10 19 Joule for hydrogen). (iii) When the distance between the two atoms is further decreased (i.e. r < r ) the negative value of potential energy of the system starts decreasing. It becomes zero and then attains positive value with further decrease in r (as shown in the figure). 0 Intermolecular Forces The forces between the molecules due to electrostatic interaction between the charges of the molecules are called intermolecular forces. These forces are also called Vander Waal forces and are quite weak as compared to inter-atomic forces. These forces are also electrical in nature and these are active if the separation between two molecules is of the order of molecular size i.e.  10 m. (1) It is found that the force of attraction between molecules varies inversely as seventh power of the distance between them i.e. –9 a 1 or Frep  7 7 r r The negative sign indicates that the force is attractive in nature. (2) When the distance between molecules becomes less than r , the forces becomes repulsive in nature and is found to vary inversely as ninth power of the distance between them i.e. Fatt  0 Frep  1 r9 Therefore F  Fatt  Frep or Frep  b. r9 force between a b  7  9 r r two molecules is given by The value of constants a and b depend upon the structure and nature of molecules. Elasticity 447 (4) Potential Energy : Potential energy can be approximately A B expressed by the formula U  n  m r r A represents repulsive contribution and term rn B represents the attractive contribution. Constants A, B and numbers m rm and n are different for different molecules. For majority of solids n = 12 and m = 6. So potential energy can be expressed as U  Comparison Between Intermolecular Forces A B  r 12 r 6 Interatomic States of Matter and The three states of matter differ from each other due to the following two factors. (1) The different magnitudes of the interatomic and intermolecular forces. (2) The extent of random thermal motion of atoms and molecules of a substance (which depends upon temperature). ID (1) Similarities (i) Both the forces are electrical in origin. (ii) Both the forces are active over short distances. (iii) General shape of force-distance graph is similar for both the forces. 0 E3 where the term (iv) Both the forces are attractive up to certain distance between atoms/molecules and become repulsive when the distance between them become less than that value. (2) Dissimilarities (i) Interatomic force depends upon the distance between the two atoms, whereas the intermolecular force depends upon the distance between the two molecules as well as their relative orientation. (ii) Interatomic forces are about 50 to100 times stronger than intermolecular forces. (iii) The value of r for two atoms is smaller than the corresponding value for the molecules. Therefore one molecule is not restricted to attract only one molecule, but can attract many molecule. It is not so incase of atoms, since the atoms of one molecule cannot bind the atoms of other molecules. 60 (3) Intermolecular forces between two molecules has the same general nature as shown in the figure for interatomic forces. Comparison Chart of Solid, Liquid and Gaseous States Property Solid Liquid Gas Not definite Not definite Definite Volume Definite Definite Not definite Density Maximum Less than solids but more than gases. Minimum Incompressible Less than gases but more than solids. Compressible Crystalline Non-crystalline Constant Not constant Not constant K U K >> U Strongest Less than solids but more than gases. Weakest Molecules vibrate about their mean position but cannot move freely. Molecules have limited free motion. Molecules are free to move. Matter remains in solid form below a certain temperature. Liquids are found at temperatures more than that of solid. These are found at temperatures greater than that of solids and liquids. Compressibility Crystallinity D YG U Shape Interatomic or intermolecular distance Relation between kinetic energy K and potential energy (U) U Intermolecular force ST Freedom of motion Effect of temperature Note :  The fourth state of matter in which the medium is in the form of positive and negative ions, is known as plasma. Plasma occurs in the atmosphere of stars (including the sun) and in discharge tubes. Types of Solids A solid is that state of matter in which its constituent atoms or molecules are held strongly at the position of minimum potential energy and it has a definite shape and volume. The solids can be classified into two categories, crystalline and glassy or amorphous solids. Comparison chart of Crystalline and Amorphous Solids Crystalline solids Amorphous or glassy solids The constituent atoms, ions or molecules are arranged in a regular repeated three dimensional pattern, within the The constituent atoms, ions or molecules are not arranged in a regular repeated three dimensional pattern, within the 448 Elasticity solid. (7) Elastic fatigue : The temporary loss of elastic properties because of the action of repeated alternating deforming force is called elastic fatigue. solid. Definite shape. external geometric No regularity in external shape. Due to elastic fatigue : They are anisotropic. They are isotropic. They have sharp melting point. They don't have sharp melting point. They have a long-range order of atoms or ions or molecules in them. They don’t have a long-range order. They are considered true and stable solids. They are not regarded as true and stable solids. Elastic Property of Matter (1) Elasticity : The property of matter by virtue of which a body tends to regain its original shape and size after the removal of deforming force is called elasticity. (ii) Spring balances show wrong readings after they have been used for a long time. (iii) We are able to break the wire by repeated bending. (8) Elastic after effect : The time delay in which the substance regains its original condition after the removal of deforming force is called elastic after effect. It is the time for which restoring forces are present after the removal of the deforming force, it is negligible for perfectly elastic substance, like quartz, phosphor bronze and large for glass fibre. Stress When a force is applied on a body, there will be relative displacement of the particles and due to property of elasticity, an internal restoring force is developed which tends to restore the body to its original state. The internal restoring force acting per unit area of cross section of the deformed body is called stress. At equilibrium, restoring force is equal in magnitude to external force, stress can therefore also be defined as external force per unit area on a body that tends to cause it to deform. ID (2) Plasticity : The property of matter by virtue of which it does not regain its original shape and size after the removal of deforming force is called plasticity. (i) Bridges are declared unsafe after a long time of their use. 60 All the bonds are not equally strong. E3 All the bonds in ions, or atoms or molecules are equally strong. (3) Perfectly elastic body : If on the removal of deforming forces the body regain its original configuration completely it is said to be perfectly elastic. Stress  U A quartz fibre and phosphor bronze (an alloy of copper containing 4% to 10% tin, 0.05% to 1% phosphorus) is the nearest approach to the perfectly elastic body. If external force F is applied on the area A of a body then, D YG (4) Perfectly plastic body : If the body does not have any tendency to recover its original configuration, on the removal of deforming force, it is said to be perfectly plastic. Force F  Area A Unit : N / m 2 (S.I.) , dyne / cm 2 (C.G.S.) Dimension : [ML1 T 2 ] Paraffin wax, wet clay are the nearest approach to the perfectly plastic body. Stress developed in a body depends upon how the external forces are applied over it. Practically there is no material which is either perfectly elastic or perfectly plastic and the behaviour of actual bodies lies between the two extremes. On this basis there are two types of stresses : Normal and Shear or tangential stress (5) Reason of elasticity : In a solids, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to neighbouring molecules. These forces are known as intermolecular forces. U For simplicity, the two molecules in their equilibrium positions (at intermolecular distance r = r ) are shown by connecting them with a spring. 0 (1) Normal stress : Here the force is applied normal to the surface. It is again of two types : Longitudinal and Bulk or volume stress (i) Longitudinal stress (a) It occurs only in solids and comes in to picture when one of the three dimensions viz. length, breadth, height is much greater than other two. (b) Deforming force is applied parallel to the length and causes increase in length. (6) Elastic limit : Elastic bodies show their property of elasticity upto a certain value of deforming force. If we go on increasing the deforming force then a stage is reached when on removing the force, the body will not return to its original state. The maximum deforming force upto which a body retains its property of elasticity is called elastic limit of the material of body. (e) Longitudinal stress produced due to decrease in length of a body under a deforming force is called compressive stress. ST In fact, the spring connecting the Fig. 9.2 two molecules represents the intermolecular force between them. On applying the deforming forces, the molecules either come closer or go far apart from each other and restoring forces are developed. When the deforming force is removed, these restoring forces bring the molecules of the solid to their respective equilibrium position (r = r ) and hence the body regains its original form. (c) Area taken for calculation of stress is the area of cross section. (d) Longitudinal stress produced due to increase in length of a body under a deforming force is called tensile stress. 0 Elastic limit is the property of a body whereas elasticity is the property of material of the body. (ii) Bulk or Volume stress (a) It occurs in solids, liquids or gases. (b) In case of fluids only bulk stress can be found. Elasticity 449 (d) Deforming force is applied normal to surface at all points. (e) Area for calculation of stress is the complete surface area perpendicular to the applied forces. (f) It is equal to change in pressure because change in pressure is responsible for change in volume. (2) Shear or tangential stress : It comes into picture when successive layers of solid move on each other i.e. when there is a relative displacement between various layers of solid. (i) Here deforming force is applied tangential to one of the faces. (ii) Area for calculation is the area of the face on which force is applied.  F Note :  When a beam is bent both compression strain as well as an extension strain is produced. Breaking strength Fig. 9.3 Difference between Pressure and Stress P normal Stress ID be E or May be compressive or tensile in nature. O Plastic region A B C  Strain (1) When the strain is small (< 2%) (i.e., in region OP) stress is Fig. 9.8 proportional to strain. This is the region where the so called Hooke’s law is obeyed. The point P is called limit of proportionality and slope of line OP gives the Young’s modulus Y of the material of the wire. If  is the angle of OP from strain axis then Y = tan. U Always compressive in nature. Stress can tangential. Elastic region Elastic limit Stress Pressure is always normal to the area. Fig. 9.7 If by gradually increasing the load on a vertically suspended metal wire, a graph is plotted between stress (or load) and longitudinal strain (or elongation) we get the curve as shown in figure. From this curve it is clear that : E3 Fixed face Pressure x L Stress-strain Curve A (iii) It produces change in shape, volume remaining the same. produces a change in the shape of the body without changing its volume, strain produced is called shearing strain. It is defined as angle in radians through which a plane perpendicular to the fixed surface of the cubical body gets turned under the effect of tangential force. 60 (c) It produces change in volume and density, shape remaining same. Strain D YG The ratio of change in configuration to the original configuration is called strain. Being the ratio of two like quantities, it has no dimensions and units. Strain are of three types : (1) Linear strain : If the deforming force produces a change in length alone, the strain produced in the body is called linear strain or tensile strain. Linear strain  l Change in length(l) Originallength(l) U Linear strain in the direction of deforming force is called longitudinal strain and in a direction perpendicular to force is called lateral strain. (2) Volumetric strain : If the deforming force produces a change in volume alone the strain produced in the body is called volumetric strain. Change in volume( V ) Original volume( V ) ST Volumetric strain  (3) Shearing strain : If the deforming force l F Fig. 9.4 (V – V) (2) If the strain is increased a little bit, i.e., in the region PE, the stress is not proportional to strain. However, the wire still regains its original length after the removal of stretching force. This behaviour is shown up to point E known as elastic limit or yield-point. The region OPE represents the elastic behaviour of the material of wire. (3) If the wire is stretched beyond the elastic limit E, i.e., between EA, the strain increases much more rapidly and if the stretching force is removed the wire does not come back to its natural length. Some permanent increase in length takes place. (4) If the stress is increased further, by a very small increase in it a very large increase in strain is produced (region AB) and after reaching point B, the strain increases even if the wire is unloaded and ruptures at C. In the region BC the wire literally flows. The maximum stress corresponding to B after which the wire begins to flow and breaks is called breaking or ultimate tensile strength. The region EABC represents the plastic behaviour of the material of wire. (5) Stress-strain curve for different materials are as follows : Fig. 9.5 x F   L  Fixed face Fig. 9.6 Brittle material Ductile material Elastomers 450 Elasticity O P O Strain The plastic region between E and C is small for brittle material and it will break soon after the elastic limit is crossed. C Stress P C E C Stress Stress E O Strain The material of the wire have a good plastic range and such materials can be easily changed into different shapes and can be drawn into thin wires Example : Glass, cast iron. 60 Example. Mild steel Hooke’s law and Modulus of Elasticity (ii) Increment in the length of wire l  stress  constant  E strain ID Strain Fig. 9.9 FL Mg(L / 2) MgL L2 dg =   2Y AY AY 2 AY [As mass (M) = volume (AL) × density (d)] (iv) Thermal stress : If a rod is fixed between two rigid supports, due to change in temperature its length will change and so it will exert a normal stress (compressive if temperature increases and tensile if temperature decreases) on the supports. This stress is called thermal stress.  Elongation l  U O L , greater will be the elongation in the wire. r2 (iii) Elongation in a wire by its own weight : The weight of the wire Mg act at the centre of gravity of the wire so that length of wire which is stretched will be L/2. i.e., greater the ratio D YG There are three modulii of elasticity namely Young’s modulus (Y), Bulk modulus (K) and modulus of rigidity () corresponding to three types of the strain. Young's Modulus (Y) It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality. Normal stress F / A FL Y    longitudinal strain l / L Al U If force is applied on a wire of radius r by hanging a weight of mass M, then  change in length(l) initiallength(L) stress strain  As by definition, coefficient of linear expansion   Fig. 9.10 l L  l  thermal strain   L Note :  final length  initiallength 2 L  L  1 Initiallength L  Young’s modulus = F [As Y = stress/strain] And tensile or compressive force produced in the body = YA  (i) If the length of a wire is doubled, Then longitudinal strain = L F So thermal stress = Y  MgL r 2 l ST Y  FL    As Y  Al    So if same stretching force is applied to different wires of same L material, l  2 [As F and Y are constant] r Stress The constant E is called modulus of elasticity. (1) It’s value depends upon the nature of material of the body and the manner in which the body is deformed. (2) It's value depends upon the temperature of the body. (3) It’s value is independent of the dimensions (length, volume etc.) of the body. FL r 2 Y E3 According to this law, within the elastic limit, stress is proportional to the strain. i.e. stress  strain or Strain Stress-strain curve is not a straight line within the elastic limit for elastomers and strain produced is much larger than the stress applied. Such materials have no plastic range and the breaking point lies very close to elastic limit. Example rubber In case of volume expansion Thermal stress = K Where K = Bulk modulus,  = coefficient of cubical expansion (v) Force between the two rods : Two rods of different metals, having the same area of cross section A, are placed end to end between two massive walls as shown in figure. L1 L2 The first rod has a length L , 1 2 coefficient of linear expansion  Y2 Y1 and young’s modulus Y. The corresponding quantities for second Fig. 9.11 rod are L ,  and Y. If the temperature of both the rods is now raised by T degrees. 1 Y = stress [As strain = 1] So young’s modulus is numerically equal to the stress which will double the length of a wire. 1 1 2 2 2 Elasticity 451 Increase in length of the composite rod (due to heating) will be equal to [As l = L  ] and due to compressive force F from the walls due to elasticity, decrease in length of the composite rod will be equal to  L1 L2  F     Y1 Y 2  A UV  F A  L1 L2    [L11  L2 2 ] T    Y1 Y 2  F l Fl 2 volume 1  stress  strain  volume 2 1  stress  strain 2 1  Y  (strain)2  volume 2 1  Y  (strain) 2 2 ID but from the definition of young’s modulus 1  (stress) 2 2Y Note :  If the force on the wire is increased from F to F and the elongation in wire is l then energy stored in the wire 1 (F1  F2 ) U  l 2 2 …(ii) 1 2 U YA from (i) and (ii) k  L D YG It is clear that the value of force constant depends upon the dimension (length and area of cross section) and material of a substance. (vii) Actual length of the wire : If the actual length of the wire is L, then under the tension T , its length becomes L and under the tension T , its length becomes L. 1 2 1 and L2  L  l2  L 2  L  …(i) T2 k …(ii) L1 T2  L 2 T1 T2  T1 U From (i) and (ii) we get L  Work Done in Stretching a Wire ST In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy. If a force F acts along the length L of the wire of cross-section A and stretches it by x then stress F / A FL YA  F   x strain x / L Ax L So the work done for an additional small increase dx in length, YA dW  Fdx  x. dx L Hence the total work done in increasing the length by l, 1 YA 2 l l YA W  0 dW  0 Fdx  0. x dx  l 2 L L l This work done is stored in the wire.  Thermal energy density = Thermal energy per unit volume = 1  2 Thermal stress  strain 2 T1 k L1  L  l1  L1  L  Y  Energy stored in per unit volume of wire (U ) 1 1  (stress)2  volume 2Y …(i) F YA  l L [As AL = volume of wire] Total energy stored in wire (U) 1 Fl 2 (vi) Force constant of wire : Force required to produce unit elongation in a wire is called force constant of material of wire. It is denoted by k.  k 1 (stress)2 2Y V A[L11  L2 2 ]T F  L1 L2      Y1 Y 2  or  1 F l 1 1     stress  strain   Y  (strain)2 2 2 2 A L E3 compression i.e. Dividing both sides by volume of the wire we get energy stored in unit volume of wire. FL    As l  AY    as the length of the composite rod remains unchanged the increase in length due to heating must be equal to decrease in length due to YAl  1 YAl 2 1   Fl  As F  L  2 L 2  60 l1  l2  [L11  L2 2 ] T  Energy stored in wire U  = 1 1 F l 1 = (Y  )( ) = Y  2 ( )2 2 A L 2 2 Breaking of Wire When the wire is loaded beyond the elastic limit, then strain increases much more rapidly. The maximum stress corresponding to B (see stress-strain curve) after which the wire begin to flow and breaks, is called breaking stress or tensile strength and the force by application of which the wire breaks is called the breaking force. (i) Breaking force depends upon the area of cross-section of the wire i.e., Breaking force  A  Breaking force = P × A Here P is a constant of proportionality and known as breaking stress. A F (ii) Breaking stress is a constant for a given Fig. 9.12 material and it does not depend upon the dimension (length or thickness) of wire. (iii) If a wire of length L is cut into two or more parts, then again it's each part can hold the same weight. Since breaking force is independent of the length of wire. (iv) If a wire can bear maximum force F, then wire of same material but double thickness can bear maximum force 4F (v) The working stress is always kept lower than that of a breaking stress. 452 Elasticity i.e., adiabatic elasticity is equal to  times pressure. Cp ] [Where   Cv breaking stress , may have large value. working stress (vi) Breaking of wire under its own weight. Breaking force = Breaking stress  Area of cross section Note :  Weight of wire = Mg = ALdg = PA [P =Breaking stress] [As mass = volume  density = ALd] E P  Ldg  P  L  dg Then the ratio of normal stress to the volumetric strain within the elastic limits is called as Bulk modulus. This is denoted by K.  pV F/A   V / V V If a liquid of density  , volume V and bulk modulus K is compressed, then its density increases. As density   (V – V)  V m so  V  V VP  V From (i) and (ii)    V P  V K  ID volume or      P  K  P      1   [1  CP] K   D YG 2 –1 A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. Volume of the spherical body V  2 Gases have two bulk moduli, namely isothermal elasticity E and adiabatic elasticity E. (1) Isothermal elasticity (E) : Elasticity possess by a gas in isothermal condition is defined as isothermal elasticity. (Boyle’s law) U PdV + VdP = 0  PdV = – VdP  R 1 V  R 3 V Bulk modulus K   V  ST dP stress  E  E = P P  (dV / V ) strain (Poisson’s law) Differentiating both sides, P  V  1 dV  V  dP  0   PdV  VdP  0 E =  P Fig. 9.14 P V …..(ii) mg    As P  A    (2) Adiabatic elasticity (E) : Elasticity possess by a gas in adiabatic condition is defined as adiabatic elasticity. dP stress P   E   dV  strain    V  …(i) V P mg   V K AK i.e., Isothermal elasticity is equal to pressure. PV  = constant m 4 R 3 3 V R 3 V R Differentiating both sides For adiabatic process, 1    As K  C    Fractional Change in the Radius of Sphere S.I. unit of compressibility is N m and C.G.S. unit is dyne cm. For isothermal process, PV = constant [As  =   –  ] A massless piston of area A floats on the surface of the liquid. 1 V  K pV –1 …(ii) U The negative sign shows that with increase in pressure p, the volume decreases by V i.e. if p is positive, V is negative. The reciprocal of bulk modulus is called compressibility. …(i) But by definition of bulk modulus K Fig. 9.13 where p = increase in pressure; V = original volume; V = change in C = compressibility =    1  E > E 60 When a solid or fluid (liquid or gas) is subjected to a uniform pressure all over the surface, such that the shape remains the same, then there is a change in volume. K P Density of Compressed Liquid Bulk Modulus Normal stress volumetric strain P i.e., adiabatic elasticity is always more than isothermal elasticity. This is the length of wire if it breaks by its own weight. K  E Ratio of adiabatic to isothermal elasticity E3 So that safety factor = Substituting the value of get V from equation (ii) in equation (i) we V R 1 mg  R 3 AK Modulus of Rigidity Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of rigidity of the material of the body and Shearing stress is denoted by , i.e.   Shearing strain Elasticity 453 In this case the shape of a body changes but its volume remains unchanged. Consider a cube of material fixed at its lower face and acted upon by a tangential force F at its upper surface having area A. The shearing stress, then, will be A Q' Q R F   L   Fixed face P dr / r dr dL    As   dL / L  r   L    S Fig. 9.15 F A (iii) Theoretical value of Poisson’s ratio 1    0.5. (iv) Practical value of Poisson’s ratio 0    0.5 Relation between Y, k,  and  Poisson’s Ratio L r r – dr L + dL D YG Lateral strain : The ratio of change in radius or diameter to the original radius or diameter is called lateral strain. Longitudinal strain : The ratio of change in length to the original length is called longitudinal strain. F Fig. 9.16 The ratio of lateral strain to longitudinal strain is called Poisson’s ratio (). Lateral strain i.e.   Longitudinal strain U Negative sign indicates that the radius of the bar decreases when it is stretched. ST Relation Between Volumetric Strain, Lateral Strain and Poisson’s Ratio If a long bar have a length L and radius r then volume V  r 2 L Differentiating both the sides dV  r dL   2rL dr 2 both the sides dV r dL  2rL dr dL dr    2 V L r r 2 L r 2 L by volume Y  3 K(1  2 ) …(i) Y  2(1   ) …(ii) Eliminating  or Y between these, we get Y  9 K 3K   …(iii)   3 K  2 6 K  2 …(iv) Torsion of Cylinder If the upper end of a cylinder is clamped and a torque is applied at the lower end the cylinder gets twisted by angle . Simultaneously shearing strain  is produced in the cylinder. P r Poisson’s ratio is a dimensionless and a unitless quantity. 2 Moduli of elasticity are three, viz. Y, K and  while elastic constants are four, viz, Y, K,  and . Poisson’s ratio  is not modulus of elasticity as it is the ratio of two strains and not of stress to strain. Elastic constants are found to depend on each other through the relations : U When a long bar is stretched by a force along its length then its length increases and the radius decreases as shown in the figure. Dividing  Volume = constant or K =  i.e. the material is incompressible. E3 shear stress F / A F   shear strain  A dr / r dL / L dV dL =0  [1  2 ] V L (ii) If a material having  = 0, then lateral strain is zero i.e. when a substance is stretched its length increases without any decrease in diameter e.g. cork. In this case change in volume is maximum. QQ' x  PQ L Only solids can exhibit a shearing as these have definite shape.   (i) If a material having  = – 0.5 then ID So   1  dV   1  2  AdL  [where A = cross-section of bar] or   This shearing force causes the consecutive horizontal layers of the cube to be slightly displaced or sheared relative to one another, each line such as PQ or RS in the cube is rotated through an angle  by this shear. The shearing strain is defined as the angle  in radians through which a line normal to a fixed surface has turned. For small values of angle, Shearing strain    dV dL dL dL   2  (1  2 ) V L L L 60 Shearing stress   Volumetric strain = longitudinal strain + 2(lateral strain) x of bar Q (i) The angle of twist  is directly proportional to the distance from the fixed end of the cylinder. l  At fixed end   0 o and at free end  = maximum. O (ii) The value of angle of shear  is directly proportional to the radius of the cylindrical shell. Fig. 9.17  A B At the axis of cylinder  = 0 and at the outermost shell  = maximum. (iii) Relation between angle of twist () and angle of shear () 454 Elasticity r l (iv) Twisting couple per unit twist or torsional rigidity or torque required to produce unit twist. r 4 2l  C  r4  A2 (v) Work done in twisting the cylinder through an angle  is Factors Affecting Elasticity (1) Hammering and rolling : Crystal grains break up into smaller units by hammering and rolling. This results in increase in the elasticity of material. r 4  2 1 C 2  2 4l W Interatomic Force Constant Behaviour of solids with respect to external forces is such that if their atoms are connected to springs. When an external force is applied on a solid, this distance between its atoms changes and interatomic force works to restore the original dimension. The ratio of interatomic force to that of change in interatomic F distance is defined as the interatomic force constant. K  r –2 Note :  0 Important Facts About Elasticity (1) The body which requires greater deforming force to produce a certain change in dimension is more elastic. Example : Ivory and steel balls are more elastic than rubber. U separation r is N  A / r02. Example : (i) For same load, more elongation is produced in rubber wire than in steel wire hence steel is more elastic than rubber. 0 Elastic Hysteresis D YG When a deforming force is applied on a body then the strain does not change simultaneously with stress rather it lags behind the stress. The lagging of strain behind the stress is defined as elastic hysteresis. This is the reason why the values of strain for same stress are different while increasing the load and while decreasing the load. Hysteresis loop : The area of the stress-strain curve is called the hysteresis loop and it is numerically equal to the work done in loading the material and then unloading it. For B ST Stress U For A Stress (4) Impurities : Due to impurities in a material, elasticity can increase or decrease. The type of effect depends upon the nature of impurities present in the material. (2) When equal deforming force is applied on different bodies then the body which shows less deformation is more elastic. 0 The number of atoms in area A of wire having interatomic Strain Name of substance (3) Temperature : Intermolecular forces decreases with rise in temperature. Hence the elasticity decreases with rise in temperature but the elasticity of invar steel (alloy) does not change with change of temperature. The number of atoms having interatomic distance r in length l of a wire, N = l/r. O (2) Annealing : The metals are annealed by heating and then cooling them slowly. Annealing results in decrease in the elasticity of material. ID It is also given by K  Y  r0 [Where Y = Young's modulus, r0 = Normal distance between the atoms of wire] Unit of interatomic force constant is N/m and Dimension MT 60 C If we have two tyres of rubber having different hysteresis loop then rubber B should be used for making the car tyres. It is because of the reason that area under the curve i.e. work done in case of rubber B is lesser and hence the car tyre will not get excessively heated and rubber A should be used to absorb vibration of the machinery because of the large area of the curve, a large amount of vibrational energy can be dissipated. E3 AB = r = l    O (ii) Water is more elastic than air as volume change in water is less for same applied pressure. (iii) Four identical balls of different materials are dropped from the same height then after collision, balls rises upto different heights. The order of their height can be given by h > h > h > h because ivory steel rubber clay Y >Y >Y >Y. ivory steel rubber clay (3) The value of moduli of elasticity is independent of the magnitude of the stress and strain. It depends only on the nature of material of the body. (4) For a given material there can be different moduli of elasticity depending on the type of stress applied and resulting strain. Strain Fig. 9.18Young’s modulus (Y) 10 N/m 10 2 Bulk modulus (K) Modulus of rigidity () 10 N/m 10 10 N/m 10 2 Aluminium 6.9 7.0 2.6 Brass 9.0 6.7 3.4 Copper 11.0 13.0 4.5 Iron 19.0 14.0 4.6 Steel 20.0 16.0 8.4 Tungsten 36.0 20.0 15.0 Diamond 83.0 55.0 34.0 2 Elasticity 455 Water – 0.22 – Glycerin – 0.45 – Air – 1.01 – (5) The moduli of elasticity has same dimensional formula and units as that of stress since strain is dimensionless.  Dimensional formula is ] while units dyne/cm or Newton/m. 2 To minimize the depression in the beam, it is designed as I-shaped girder. 2 (vi) For a beam with circular cross-section depression is given by (6) Greater the value of moduli of elasticity more elastic is the material. But as Y  (1/l), K  (1/V) and   (1/) for a constant stress, so smaller change in shape or size for a given stress corresponds to greater elasticity. (7) The moduli of elasticity Y and  exist only for solids as liquids and gases cannot be deformed along one dimension only and also cannot sustain shear strain. However K exist for all states of matter viz. solid, liquid or gas.   W L3 12 r 4 Y (vii) A hollow shaft is stronger than a solid shaft made of same mass, length and material. Torque required to produce a unit twist in a solid shaft liquid 2l gas  hollow  (r24  r14 ) 2l ID  hollow r24  r14 (r22  r12 )(r22  r12 )    solid r4 r4 r l Since two   (r22  r12 )l U D YG (12) In transmitting power, an automobile shaft is sheared as it rotates, so shearing strain is set up, hence modulus of rigidity is involved. (13) The shape of rubber heels changes under stress, so modulus of rigidity is involved. Practical Applications of Elasticity (i) The metallic parts of machinery are never subjected to a stress beyond elastic limit, otherwise they will get permanently deformed. …(ii) From (i) and (ii), (10) In a suspension bridge there is a stretch in the ropes by the load of the bridge. Due to which length of rope changes. Hence Young’s modulus of elasticity is involved. (11) In an automobile tyre as the air is compressed, volume of the air in tyre changes, hence the bulk modulus of elasticity is involved. …(i) and torque required to produce a unit twist in a hollow shaft (9) For a rigid body l, V or  = 0 so Y, K or  will be , i.e. elasticity of a rigid body is infinite. Diamond and carborundum are nearest approach to rigid bodies. r 4  solid  (8) Gases being most compressible are least elastic while solids are most i.e. the bulk modulus of gas is very low while that for liquids and solids is very high. K > K > K solid 60 [ ML T 2 E3 1 2 Substituting  hollow   solid r22  r12 r2 shafts are  r  2 this made …(iii) from equal volume  we get,  r12 r22 value in equation (iii) 1   >  hollow solid i.e., the torque required to twist a hollow shaft is greater than the torque necessary to twist a solid shaft of the same mass, length and material through the same angle. Hence, a hollow shaft is stronger than a solid shaft. (ii) The thickness of the metallic rope used in the crane in order to lift a given load is decided from the knowledge of elastic limit of the material of the rope and the factor of safety.  Metals are polycrystalline materials. (iv) Maximum height of a mountain on earth can be estimated from the elastic behaviour of earth. become plastic. ST U (iii) The bridges are declared unsafe after long use because during its long use, a bridge under goes quick alternating strains continuously. It results in the loss of elastic strength. At the base of the mountain, the pressure is given by P = hg and it must be less than elastic limit (K) of earth’s supporting material. K > P > hg  h  K or g K h max  g (v) In designing a beam for its use to support a load (in construction of roofs and bridges), it is advantageous to increase its depth rather than the breadth of the beam because the depression in rectangular beam.   W l3 4Ybd 3 l b d   Metals are elastic for small strains and for large strains, metals  The substances having large molecular structure (formed by the union of two to several thousand simple molecules) are called polymers.  Rubber is a polymer.  Rubber is elastic for very large strains.  It stretches easily at first but then becomes stiffer.  Young’s modulus is defined only for the solids.  Bulk modulus was first defined by Maxwell.  Bulk modulus is defined for all types of materials, solids, liquids and gases. W Fig. 9.19 456 Elasticity energy stored per unit volume.  Reciprocal of bulk modules is called compressibility.  Thermal stress in a rod  Y  . It is independent of the area of  Hooke’s law is obeyed only for small values of strain.  Higher value of the elasticity (modulus) means greater force is required for producing a given change. cross section or length of the wire.  Breaking stress for a wire of unit cross-section is called tensile strength.  The material which break as soon as the stress goes beyond the elastic limit are called brittle.  Breaking stress does not depend on the length or area of cross section of the wire. However it depends on the material of the wire.  The material which do not break well beyond the elastic limit are called ductile.  Breaking force depends on the area of cross section. Breaking stress of a wire is called tensile strength.  If we double the radius of rope its breaking force becomes four  Rubber sustains elasticity even when stretched several times its times. But the breaking stress remains unchanged. However it is not ductile. If breaks down as soon as the elastic limit is crossed.  Within elastic limit, the force constant for a spring is given by K YA L  Quartz is the best available example of perfectly elastic materials.  Isothermal elasticity = pressure (P)  Adiabatic elasticity = Ratio of specific heats × pressure =P the non rigid bodies. D YG  Diamond and carborundum are the nearest approach to the rigid  Elastic fatigue occurs, when a metal is subjected to repeated loading and unloading.  Theoretical value of Poisson’s ratio lies between –1 and +1/2 but practical value lies between zero and +1/2.  Negative value of poisson’s ratio means that if length increases then radius decreases.  If a beam of circular cross-section is loaded, its depression is inversely proportional to the fourth power of radius. i.e.   U  Elasticity is meaningless for the rigid bodies. It is the property of body. the beam is inversely proportional to the cube of thickness. ID  Elastic after effect is a temporary absence of the elastic properties.  If a beam of rectangular cross-section is loaded its depression at E3 length. 60  The deformation beyond elastic limit is called plasticity.  Stress and pressure have the same units and dimensions, but the U pressure is always normal to the surface but the stress may be parallel or perpendicular to the surface.  Normal stress is also called tensile stress when the length of the body tends to increase. ST  Normal stress is also called compressive stress when length of the body tends to decrease.  Tangential stress is also called shearing stress.  When the deforming force is inclined to the surface, both the tangential as well as normal stress are produced.  When a body is sheared, two mutually perpendicular strains are produced. They are called longitudinal strain and compressional strain. Both are equal in magnitude.  When a beam is bent, both extensional as well as compressional strain is produced.  The energy stored by an elastic material is the area under the force-extension graph. The area under the stress-strain graph gives the 1 r4