Elasticity

MECHANICAL PROPERTIES OF MATTER A body is said to be rigid if the relative positions of its constituent particles remains unchanged when external deforming forces are applied to it. The nearest approach to a rigid body is diamond or carborundum. Actually no body is perfectly rigid and every body can be deformed more or less by the application of suitable forces. All these deformed bodies however regain their original shape or size, when the deforming forces are removed. The property of matter by virtue of which a body tends to regain its original shape and size after the removal of deforming forces is called elasticity. Some terms related to elasticity : 1. Deforming Force External force which try to change in the length, volume or shape of the body is called deforming force. 2. Perfectly Elastic Body The body which perfectly regains its original form on removing the external deforming force, is defined as a perfectly elastic body. Ex. : quartz – Very nearly a perfect elastic body. 3. Plastic Body (a) The body which does not have the property of opposing the deforming force, is known as a plastic body. (b) The bodies which remain in deformed state even after removed of the deforming force are defined as plastic bodies. 4. Internal restoring force When a external force acts at any substance then due to the intermolecular force there is a internal resistance produced into the substance called internal restoring force. At equilibrium the numerical value of internal restoring force is equal to the external force. STRESS The internal restoring force acting per unit area of cross–section of the deformed body is called stress. Stress  Internal restoring force Area of cross section = Fint ernal Fexternal = A A Stress depends on direction of force as well as direction of area of application so it is tensor. SI Unit : N–m–2 Dimensions : M1 L–1 T–2 There are three types of stress :(a) Longitudinal Stress : When the stress is normal to the surface of body, then it is known as longitudinal stress. There are two types of longitudinal stress (i) Tensile Stress : The longitudinal stress, produced due to increase in length of a body, is defined as tensile stress. (ii) Compressive stress : The longitudinal stress, produced due to decrease in length of a body, is defined as compressive stress. (b) Volume Stress If equal normal forces are applied every one surface of a body, then it undergoes change in volume. The force opposing this change in volume per unit area is defined as volume stress. (c) Tangential Stress or Shear Stress When the stress is tangential or parallel to the surface of a body then it is known as shear stress. Due to this stress, the shape of the body changes or it gets twisted. F A L D A B F C B L D F F C STRAIN The ratio of change of any dimension to its original dimension is called strain. Strain = change in size of the body original size of the body It is a unitless and dimensionless quantity. There are three types of strain : Type of strain depends upon the directions of applied force. (a) Longitudinal strain = (b) Volume strain = (c) Shear strain change in length of the body L = initial length of the body L V change in volume of the body = V original volume of the body When a deforming force is applied to a body parallel to its surface then its shape (not size) changes. The strain produced in this way is known as shear strain. The strain produced due to change of shape of the body is known as shear strain. tan  l l displacement of upper face or = = L L distance between two faces MECHANICAL PROPERTIES OF MATTER Relation Between angle of twist and Angle of shear When a cylinder of length '' and radius 'r' is fixed at one end and tangential force is applied at the other end, then the cylinder gets twisted. Figure shows the angle of shear ABA' and angle of twist AOA'. Arc r AA' = r and Arc AA' =  so r=     where = angle of l twist,  = angle of shear (i) When a material is under tensile stress restoring force are caused by intermolecular attraction while under compressive stress, the restoring force are due to intermolecular repulsion. (ii) If the deforming force is inclined to the surface at an angle such that and  90° then both tangential and normal stress are developed. (iii) Linear strain in the direction of force is called longitudinal strain while in a direction perpendicular to force lateral strain.  Stress – Strain Graph Breaking strength B Elastic Region Elastic Limit E Y C Pr op o L i m r t io n it P Plastic Region 0 Strain Stress Proportion Limit The limit in which Hook's law is valid and stress is directly proportional to strain is called proportion limit. Stress  Strain Elastic limit That maximum stress which on removing the deforming force makes the body to recover completely its original state. Yield Point The point beyond elastic limit, at which the length of wire starts increasing without increasing stress, is defined as the yield point. Breaking Point The position when the strain becomes so large that the wire breaks down at last, is called breaking point. At this position the stress acting in that wire is called breaking stress and strain is called breaking strain. Elastic after effect Elastic Fatigue : The loss of strength of the material due to repeated strains on the material is called elastic fatigue. That is why bridges are declared unsafe after a long time of their use. Creep : Elastic Hysteresis : The strain persists even when the stress is removed. This lagging behind of strain is called 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. load or stress loa di nc re as lo a i ng dd ec re as in g If a small force is applied for a long time then it causes breaking of metal. For example A fan is hung for 200 years then the shaft will break. extension or strain Breaking Stress : The stress required to cause actual facture of a material is called the breaking stress Breaking stress = F/A (i) Breaking stress also measures the tensile strength. (ii) Metals with small plastic deformation are called brittle. (iii) Metals with large plastic deformation are called ductile. (iv) Elasticity restoring forces are strictly conservative only when the elastic hysteresis is zero. i.e. the loading and unloading stress – strain curves are identical. The material which have low elastic hysteresis have also low elastic relaxation time. (v) Ex. Sol. Find out longitudinal stress and tangential stress on a fixed block. Longitudinal or normal stress Sol. 1 0 0 sin 3 0 0 = 5 N/m2 5 2 30 0 5m 1m 2m 1 00 cos 3 0 0  5 3N / m2 5 2 The breaking stress of aluminium is 7.5 × 108 dyne cm–2. Find the greatest length of aluminium wire that can hang vertically without breaking. Density of aluminium is 2.7 g cm–3. [Given : g = 980 cm s–2] Tangential stress Ex. 1  100N 2  Let  be the greatest length of the wire that can hang vertically without breaking. Mass of wire m = cross–sectional area (A) × length () × density (), Weight of wire = mg = Ag This is equal to the maximum force that the wire can withstand. Breaking stress =  7 .5 10 8 2 .7 98 0 l Ag  l g  7.5 × 108 =  × 2.7 × 980 A MECHANICAL PROPERTIES OF MATTER Hooke's Law Ex. Sol. If the deformation is small, the stress in a body is proportional to the corresponding strain, this fact is known stress as Hooke's Law. Within elastic limit : stress  strain  =constant strain This constant is known as modulus of elasticity or coefficient of elasticity. The modulus of elasticity depends only on the type of material used. It does not depend upon the value of stress and strain. Find out the shift in point B, C and D 1 0  1 0  0 .1 FL MgL  LB = LAB = = =4 × 10–3 m = 4mm 1 0 7  2.5  1 0 10 AY AY 1 00  0 .2 LC = LB + LBC =4 × 10–3 + =4 × 10–3 + 5 × 10–3 =9mm 1 0 7  4  10 1 0 A B 0.1m yAB = 2.5 × 10 N/m C 0.2m yBC = 4 × 10 N/m D 1 0 0  0.15 LD=LC + LCD = 9 × 10 + =9 × 10–3 + 15 × 10–3 =24 mm 1 0 7  1  10 10 –3 10 10 2 2 0.15m y = 1 × 1010 N/m2 CD 10kg A = 10–7 m2 Young's Modulus of Elasticity 'Y' Within elastic limit the ratio of longitudinal stress and longitudinal strain is called Young's modulus of elasticity. Y= longitudinal stress F / A F L = = longitudinal strain lA l/ L Within elastic limit the force acting upon a unit area of a wire by which the length of a wire becomes double, is equivalent to the Young's modulus of elasticity of material of a wire. If L is the length of wire, r is radius and  is the increase in length of the wire by suspending a weight Mg at its one end then Young's modulus  Mg / r  of elasticity of the material of wire Y= l / L   = 2 Unit of Y : N/m2 MgL r 2 l Dimensions of Y : M1L–1T–2 Increment of length due to own weight Let a rope of mass M and length L is hanged vertically. As the tension of different point on the rope is different. Stress as well as strain will be different at different point. (i) maximum stress at hanging point (ii) minimum stress at lower point  M Consider a dx element of rope at x distance from lower end then tension T    x g  L So stress =  M  xg T =   L A A Let increase in length of dx is dy then strain = dy dx M xg stress L A  M  xg So Young modulus of elasticity Y = =   dx = Y dy dy / dx strain  L A For full length of rope Mg LA  L 0 l xdx =Y  dy  0 Mg L2 MgL = Y = LA 2 2 AY [Since the stress is varying linearly we may apply average method to evaluate strain.] dx Ex. A thin uniform metallic rod of length 0.5 m and radius 0.1 m rotates with an angular velocity 400 rad/s in a horizontal plane about a vertical axis passing through one of its ends. Calculate (a) tension in the rod and (b) the elongation of the rod. The density of material of the rod is 104 kg/m3 and the Young's modulus is 2 × 1011 N/m2. Sol. (a) Consider an element of length dr at a distance r from the axis of rotation as shown in figure. The centripetal force acting on this element will be  dT  dmr2  Adr  r  2 . As this force is provided by tension in the rod (due to r elasticity), so the tension in the rod at a distance r from the axis of rotation will be due to the centripetal force due to all elements between x = r to x = L i.e., T=  L r A2 rdr  1 A2 L2  r 2  2 dr L ...(i)  1  2  6 1 2 2    r  = 8   1 0   r  N   4 2     Now if dy is the elongation in the element of length dr at position r then strain 1 So here T = × 104 ×  × 10–2 × (400)2 2 (b) dy stress T = = Y dr AY = 1 2 2 2 [L – r ] 2 Y So the elongation of the whole rod  2 L  2Y Ex.  L L 0 2 r 2  1  L3 1 10 4   400   0.5  dr = = × 3 3 Y 2  10 11 2 3 = 1  10 3 m 3 Find out the elongation in block. If mass, area of cross-section and young modulus of block are m, A and y respectively. x x dx Sol. T T m' F T+dT dx Acceleration, a= F m Elongation in element ‘dx’ = then T = ma Tdx , Ay where  m = m m F Fx x; T = x = l l m l l total elongation,  = l Tdx Fxdx Fl o Ay = o Al y = 2 Ay Bulk modulus of elasticity 'K' or 'B' Within elastic limit the ratio of the volume stress and the volume strain is called bulk modulus of elasticity. K or B = F/ A volume stress = V volume strain V  P V V The minus sign indicates a decrease in volume with an increase in stress. Unit of K : N m–2 or pascal MECHANICAL PROPERTIES OF MATTER Bulk modulus of an ideal gas is process dependence. For isothermal process PV = constant  PdV + VdP = 0  P =  dP So bulk modulus = P dV / V For adiabatic process PV = constant  PV  1 dV  V  dP  0  PdV  VdP  0  P  So bulk modulus = P For any polytropic process PVn = constant  nPVn–1dV|VndP=0  PdV+VdP=0  nP = So bulk modulus = nP  dP dV / V Compressibility : The reciprocal of bulk modulus of elasticity is defined as compressibility. C  Modulus of Rigidity ''  dP dV / V 1 K Within elastic limit the ratio of shearing stress and shearing strain is called modulus of rigidity of a material.  Ftan gential  shearing stress  A  = Ftan gential  =   shearing strain    A Note : Angle of shear '' always take in radian Poisson's Ratio ( In elastic limit, the ratio of lateral strain and longitudinal strain is called Poisson's ratio.  = lateralstrain L  D d  D = , where = = & = longitudinalstrain  D D L Work done in stretching a wire (Potential energy of a stretched wire) When a wire is stretched, work is done against the interatomic forces, which is stored in the form of elastic potential energy. For a wire of length Lo stretched by a distance x, the restoring elastic force is :  x  A  Lo  F = stress × area = Y  The work has to be done against the elastic restoring forces in stretching dx YA dW = F dx = L xdx 0 The total work done in stretching the wire from x = 0 to x =  is, then W l  0 1 1 × Y × (strain)2 × original volume= (stress) (strain) (volume) 2 2 (i) The value of K is maximum for solids and minimum for gases. (ii) For any ideal rigid body all three elastic modulus are infinite . (iii) is the characteristic of solid material only as the fluids do not have fixed shape. (iv) Potential energy density = area under the stress–strain curve. (v) Young's modulus = Slope of the stress–strain curve stress W= l YA YA  x 2  YA(l )2 xdx    = 2L o Lo L o  2 0 strain Ex. A steel wire of 4.0 m in length is stretched through 2.00 mm. The cross–sectional area of the wire is 2.0 mm2. If Young's modulus of steel is 2.0 × 1011 N/m2 find (i) the energy density of wire (ii) the elastic potential energy stored in the wire. Sol. (i) The energy density of stretched wire 2  2  1 0 3  1 1 1 = × stress × strain= × Y× (strain)2= × 2.0 × 1011×   = 2.5 × 104 J/m3 4 2 2 2   (ii) Elastic potential energy = energy density × volume = 2.5 × 104 × (2.0 × 10–6) × 4.0 J = 20 × 10–2 = 0.20J Ex. Find the depth of lake at which density of water is 1% greater than at the surface. Given compressibility K = 50 × 10–6 /atm. Sol. B= V P P  =– V V / V B We know P = Patm + h g and m = V = constant dV  dV  0  dV  P 1 h g  d  =– i.e. = = = V  B 100 B   [assuming  = constant]; h g  h= Ex. 1  1  10 5 B 1  h g  = 1 00 K 10 0  50  10 6 1 00 10 5 1 00  10 3 = m = 2km 5 00 0  10 6  1 00 0  10 50 MECHANICAL PROPERTIES OF MATTER Sol. Stress = F x =  A L 3 180 1 80 0 F = = = 0.3 rad 4 6 = 2 5  2 4 10 A 2 5  10  2.4  1 0 Strain =   as x = 0.3  x = 0.3 × 5 × 10–2 = 1.5 × 10–2 m = 1.5 mm L Analogy of Rod as a spring Y= stress Fl AY Y= or F =  strain Al l AY = constant, depends on type of material and geometry of rod. l F = k where k = AY = equivalent spring constant. l For the system of rods shown in figure (a), the replaced spring system is shown in figure (b) [two spring in series]. Figure (c) represents equivalent spring system. (a) (b) (c) Figure (d) represents another combination of rods and their replaced spring system. (d) Ex. (e) A mass ‘m’ is attached with rods as shown in figure. This mass is slightly stretched and released whether the motion of mass is S.H.M., if yes then find out the time period. k1 k2 m Sol. Here keq = k1 k 2 and T = 2 k1  k 2 m k eq = 2 A1 Y1 A 2 Y2 m(k1  k 2 ) where k1 = l and k2= l k1 k 2 1 2 Ex. Hanger is mass-less. A ball of mass ‘m’ drops from a height ‘h’, which sticks to hanger after striking. Neglect overturning, find out the maximum extension in rod. Assuming rod is massless. Let maximum extension be xmax. Sol. Applying energy conservation mg (h + xmax ) = where k1 = 1 k1 k 2 2 x 2 k1  k 2 A1 A 2 y1 y2 A1 y1 A y ; k2 = 2 2 & K eq  A y l  A y l l1 l2 1 1 2 2 1 1 therefore keqx2max – 2mgxmax – 2mgh = 0 xmax = 2 mg  4 m 2 g 2  8 mghk eq 2 k eq = mg m 2 g2 2mgh   k eq k eq k 2eq Otherway by S.H.M.  k eq m and v =   2 gh = keq m a2   y2  2mgh m 2 g2  2 k eq k eq = a mg Maximum extension = a + y = k + eq m 2 g2 2 mgh  k eq k eq Thermal Stress : If temperature of rod is increased by T, then change in length  =  T; strain =    T  But due to rigid support, there is no strain. Supports provide force on stresses to keep the length of rod same Y= stress strain Thermal stress = y strain = YT If T = (+) tive FF F  Y  T A FF If T = (–) tive F = AY T MECHANICAL PROPERTIES OF MATTER Ex. When composite rod is free, then composite length increases to 2.002 m for temperature 20°C to 120°C. When composite rod is fixed between the support, there is no change in component length find y and  of steel, if yc = 1.5 × 1013 N/m2 c = 1.6 × 10–5/°C. Sol.  = s s T + c cT 0.002 = [1.5 s + 0.5 × 1.6 ×10–5] × 100 s= 1 .2  1 0 5 = 8 × 10–6/°C 1 .5 there is no change in component length For steel x = s s T – F Fl s   s T  0  AYs AYs ....(i) For copper Fl c F x = AY – c c T = 0  Ay =  cT c c ...(ii) (ii)/(i) Ys = Yc c 1 .5  1 0 13  16  10 5 = s 8  10 6 Ys = 3 × 1013 N/m2 Applications of Elasticity : Some of the important applications of the elasticity of the materials are discussed as follows : The material used in bridges lose its elastic strength with time bridges are declared unsafe after long use. To estimate the maximum height of a mountain : The pressure at the base of the mountain = hg = stress. The elastic limit of a typical rock is 3 × 108 N m–2 The stress must be less than the elastic limits, otherwise the rock begins to flow. h< 3  108 3  108 < < 104 m ( = 3 × 103 kg m–3 ; g = 10 ms–2) g 3  103  10 or h = 10 km It may be noted that the height of Mount Everest is nearly 9 km. Torsion constant of a wire C= r 4 Where  is modulus of rigidity r and  is radius and length of wire respectively.. 2l (a) Toque required for twisting by angle  = C (b) Work done in twisting by angle , W = 1 C2. 2 Effect of Temperature on elasticity When temperature is increased then due to weakness of inter molecular force the elastic properties in general decreases i.e. elastic constant decreases. Plasticity increases with temperature. For example, at ordinary room temperature, carbon is elastic but at high temperature, carbon becomes plastic. Lead is not much elastic at room temperature but when cooled in liquid nitrogen exhibit highly elastic behaviour. For a special kind of steel, elastic constants do not vary appreciably temperature. This steel is called 'INVAR steel'. Effect of Impurity on elasticity Y is slightly increase by impurity. The inter molecular attraction force inside wire effectively increase by impurity due to this external force can be easily opposed.