Ch-1 Properties of Matter_KB

L. J. Institute of Engineering and Technology Subject: Physics (Group-1) Subject code: 3110011 F.Y. - B.E. Kamaldeep Bhatia

Properties of Matter Basic Definitions: Elasticity: The property of material body to regain its original condition (of length, volume, shape) on removal of the deforming forces is called elasticity. Perfect elastic Bodies: On removal of deforming forces, some bodies completely recover their original condition. These bodies are called perfect elastic bodies. Plastic Bodies: On removal of deforming forces, some bodies do not show any tendency to recover their original condition such bodies are called plastic bodies. Load: The combination of external forces acting on a body, in effect changing the form or dimensions of the body is known as load. This load is a deforming forces. Stress: Due to deforming forces or loads applied to a body, a reaction force is produced inside the body due to relative displacement of its molecules, tending to balance the load and restore the body to its original condition. The restoring force per unit area set up inside the deformed body is called stress. Strain: The change produced in the dimensions of body under a system of deforming forces in equilibrium is called strain. Longitudinal Strain and Stress: A body under the application of deforming force along a certain direction produces a change in length per unit length is called longitudinal strain and the deforming force applied per unit area of cross-section is called longitudinal stress. Young’s Modulus: The ratio of longitudinal stress to longitudinal strain, within the elastic limit. (Y) or (E). Young’s Modulus Y = longitudinal stress longituinal strain F/A Y = l/L F. L Y = a. l Where, F=force, L=original length a=area, l=change in length 2

Volume Strain and Stress: If the deforming force applied normally on the surface of a body is uniform, the volume of the body will change keeping the shape unchanged. The deforming force per unit area is volume stress (Normal Stress) and the change in volume per unit volume gives the volume strain. Bulk Modulus: The ratio of volume stress to volume strain is called bulk modulus or coefficient of cubical elasticity. Bulk Modulus (K) =VVoolluummee Stress strain FV K = (A) (v) FV K = Av PV K= v F A Where, = P is bulk pressure P, v is change in volume and V is original volume. Compressibility: The inverse of bulk modulus is called compressibility (C). 1 C=K Shearing Stress and Strain: If the deforming force is applied parallel to the surface of a body in such a way that it produces change in shape keeping the volume unchanged. The deforming force per unit area is called Shearing (tangential) Stress. Shear Stress =Ft A From the fig, it can be seen that after the force has been applied two parallel faces ABEF and DCGH slide over each other producing a shift of top face i.e. A’B’F’E’. The lines joining these two faces turn through an angle . 3

Thus, Shear strain is the relative displacement between two planes which are separated by a unit distance. Modulus of Rigidity: The modulus of rigidity of the material of body is defined as the ratio of shear (Tangential) Stress to shear Strain. Shear Stress η = Shear Strain ft/a η = l/L η = ftL al Poisson’s Ratio: The ratio of lateral strain to longitudinal strain is known as Poisson’s Ratio. lateral Strain −∆r/r σ = Longitudinal Strain = ∆l/l −∆r l σ = r . ∆l Working Stress & Factor of Safety: As Hooke’s law holds well in a very small region of stress, a safeguard is necessary to work with metals. The maximum stress applied on a given material within the elastic limit is called its working stress or working strength (or tenacity) and the load corresponding to it is called working load. Factor of safety is defined as the ratio of breaking stress to working stress. Factor of safety=bwroerakkiinngg stress stress Following considerations should be taken into account to decide factor of safety: (i) The static or dynamic loading conditions (ii) Maximum Expected load to be kept. (iii) Degree of expected wear and tear of material under use. 4

(iv) Possible outcomes of a breakdown. Hooke’s law: ‘For a small values of strain, the stress is proportional to strain’. Modulus of elasticity = ������������������������������������ ������������������������������������ The body is elastic if this linear relationship is observed. Stress-Strain Diagram:  A relationship between stress & strain is known as the stress-strain diagram tor a wire is shown in fig.  A linear past of the stress-strain curve indicates that the strain produced is directly proportional to the applied stress.  A Hooke’s law is obeyed up to the point A.  On removing the stress, the original condition of zero strain is recovered, denoted by point O.  A wire is perfectly elastic up to the stress A. Permanent Set:  Beyond the elastic limit, strain increases more rapidly with stress as shown by curve AB is big.  In this region, the elongation of wire consists of both elastic & plastic deformation. 5

 On removal of load, the wire does not recovers the original length but follows path BC & some remaining strain OC is crept in, which is known as permanent set. Yield Point:  In fig. the curve BD shows strain of large value without practically any increase in stress. The stress corresponding to D is less as compared to the point B. This point B at which large increase in strain starts is called yield point and the corresponding stress is known as yielding stress. Breaking Stress:  In this region the volume of the wire remains constant as the cross-section of the wire decreases uniformly with strain up to the point F.  The maximum deforming force on load applied to the wire per unit cross- section area is called the tensile stress on breaking stress. Breaking Point:  Beyond the Point F, without any increase in the load, the wire is elongated even if the load is reduced.  That is because, at some sections of the wire, the cross-sectional area increases at a faster rate & a local constriction, called ‘neck’ or ‘waist’ begins to develop.  At this point, load is decreased, so stress is reduced and the wire finally breaks at point E which represents breaking point. Ductility, Brittleness & Plasticity: Ductility: If a deforming force is applied on a material and it acquires a large permanent deformation without producing any fracture or rupture then property is called ductility. The breaking point for a ductile material will be vary much above its yield point B as D (in above fig). The portion DE will be vary large for the stress-strain diagram of ductile materials. Brittleness: Brittleness is the opposite to ductility. 6

Brittle substance ruptures without any appreciable amount of permanent deformation. Stress-Strain curve of a brittle material hardly shows any yield point & normally coincides with the breaking point. Plasticity: Some materials are capable of getting deformed continuously & permanently without producing any rupture. This property is called plasticity & the materials are called plastic. Elastic behaviours of Solids: Elastic After-effect:  Some solids regain their original condition on removal of load within the elastic limit but they take some time to do so, which is known as creep.  This (The) delay in coming in coming back to the original condition on removing the deforming force is called elastic after-effect. Elastic Hysteresis:  Due to elastic after-effect, the strain in a material like glass tends to remain present or lag behind the stress applied.  During a rapidly changing stress, the strain is smaller for a same value of stress when it is increasing.  From the fig, for same load P, while increasing or decreasing the stress, two different values of extensions ei and ed are observed respectively. This lag between load & extension is known as elastic Hystersis. 7

Elastic Fatigue:  It is observed that the torsional vibrations die off much faster in the case of wire kept vibrating continuously for a long time than that if a fresh wire is selected for vibration.  The continuously vibrating wire gets ‘fatigued’ and stops.  A body under respected strains above its elastic limit loses its elastic properties & break under a stress less than its normal breaking stress even within its elastic limit. Factors affecting Elasticity: The structure of a material greatly affects its elastic properties. Microscopic examination has established that metals are just an aggregation of a large number of crystallites which are arranged in a random fashion. Therefore, their cleavage planes are distributed at random, in all possible directions. However, single crystals of metals under deformation showed marked increase in their hardness. It is observed that if a single crystal of silver is stretched to a bit more than twice its length, its strength or stiffness increase to ninety-two times its original strength or stiffness. The following factors are influencing elasticity: Effect of Hammering, Rolling and Annealing: Due to hammering and rolling, the crystal grains break up into smaller units that increases their elastic properties; while annealing (a process of heating and then cooling gradually) tends to produce a uniform pattern of orientation of the crystallites by 8

orienting in one particular direction. This produces larger crystal grains, which decreases elastic properties, or increases the softness or plasticity of the material. This happens because slipping (or sliding between cleavage planes) starts at a weak spot and continues all through the crystal; while in the case of hammering and rolling, the slipping is confined to one crystal grain and stops at its boundary with the adjoining crystal. According to Sir Lawrence Bragg, “In order to be strong, a metal must be weak\", meaning thereby that metals with smaller grains are stronger than those with larger grains. Effect of Impurities: Sometimes, suitable impurities are intentionally added to metals to increase the binding of their crystal grains without affecting their orientation. To strengthen elastic properties, carbon in molten steel and potassium in gold are added in minute quantities. Such impurities greatly affect the elastic properties of the metal to which they are added. It may either enhance or impair them as they are themselves more elastic or plastic than the metal concerned. Effect of Change of Temperature: The elastic property of a material is influenced by a change in temperature. Elasticity decreases with rise in temperature for most of the cases and vice versa. One of the exceptions is invar steel, whose elasticity is not affected by any change in temperature. Lead becomes elastic and rings like steel when struck by a wooden mallet, if it is cooled in liquid air. At ordinary temperature, a carbon filament is highly elastic. When heated by a current through it, it become plastic so that it can be easily deformed by a magnet brought near it. Work Done in Stretching a wire: Let a force F be applied to a wire of length L and the stretch produced be l. if area of cross-section is A then, lF strain = L , stress = A Young’s Modulus Y = F.L A.l 9

Work done in stretching a wire by amount dl is given by, Amount dl is given by, dw = F dl Total Work, W = ∫ dW l = ∫ F dl 0 l YAl = ∫ L dl 0 l YA = L ∫ l dl 0 YAl2 W= L 1 W = 2 F. l 1 W = 2 (Stretching force × strain) Work done per unit volume=Potential energy stored in wire per unit volume. Twisting Couple on a solid cylinder or a Wire-Shaft: Consider a solid cylinder of length L and radius R to be clamped at its upper end, and let it be twisted by a couple applied to its lower end in a plane perpendicular to its length through an angle . 10

Due to elasticity of material of cylinder, a restoring couple tending to oppose the applied couple is set up in such a way that two balance each other in equilibrium position. To estimate this couple, consider the cylinder consisting of a large number of hollow coaxial cylinders, one inside the other and consider one such cylinder of radius x and thickness dx. It is seen that, radius of the base of cylinder turns through the same angle  but the displacement BB’ is maximum at the rim and gradually decreases to zero at the centre O. Hence, a straight line AB which was initially parallel to the axis OO’, will now take up the position AB’. So the angle of shear is equal to ∠BAB′ = . It is clear from the fig b that BB’=x. also from fig a  = x L If η be the co-efficient of rigidity of material of cylinder, shearing stress η = shear η = Ft  dA. η = FtL dAxθ ηdAxθ Ft = L Where dA =surface area of hollow cylinder = (2πx)dx 11

 Ft = (2πη) x2dx L Now, Moment of force about the axis OO’ of cylinder, dτ = Ft. x dτ = 2πηθ x3dx (L) Therefore, twisting couple on where cylinder is the integral taken over the radius x=0 to x=R τ = ∫ dτ 2πηθ x4 R τ= L [4] 0 2πηθ R4 τ= L . 4 πηθ R4 τ= L 2 Torsional rigidity of the material of cylinder is the twisting couple per unit twist. τ πηR4 C = θ = 2L In case of hollow cylinder: If instead of a solid cylinder, a hollow cylinder is taken whose inner and outer radii are Ri and Ro, respectively, the twisting couple, R0 2πηθ (L) τ = ∫ x3dx Ri 12

τ= 2πηθ x4 R0 [ ] L 4 Ri τ = 2πηθ [R440 − R44i ] L τ = πηθ (R40 − R4i ) …(1) 2L Therefore, torsional rigidity Ch or twisting couple per unit twist, Ch = τ = πη (R40 − R4i ) …(2) θ 2L Now considering the material of two cylinder are same, having density ρ, same length and mass m, Ms = Mh M πR2Lρ = π(R20 − R2i ) Lρ (as ρ = V ) R2 = R20 − R2i ..(3) Eq (2) can be written as, Ch = πη (R20 − R2i )(R20 + R2i ) … (4) 2L Substituting the value of R20 − R2i from eq (3) to eq (4) we have, Ch = πηR2 (R20 + R2i ) 2L Also from eq (3), R20 = R2i + R2 Ch = πηR2(2R2i + R2) … (5) 2L We know that πηR4 … (6) Cs = 2L Eq (5) and eq (6) clearly indicates that 2R2i + R2 > R2 13

Ch = 2R2i + R2 Cs R2 Ch > Cs Or Ch > 1 Cs i.e. torsional rigidity of a hollow cylinder is greater than that of a solid cylinder. Torsional Pendulum: (Time-Period) Consider a disk suspended from one end of a fine wire attached to its centre as shown in fig. this set up is called torsional pendulum. The torsion wire is inextensible and free to be twisted about is rotated in a horizontal plane and torsional vibrations are produced on releasing a twist. Let  be the angle of rotation and =0 represents the initial condition of wire. The wire is mechanically deformed due to twisting which sets up a restoring couple τ in the wire. For small angles of twist, the magnitude of this couple is directly proportional to the angle of twist θ. τ∝θ τ = −C … (1) Where, C is the torsional rigidity. 14

The equation of rotational motion of the system is, ∂2θ … (2) τ = I ∂t2 Where I is the moment of inertia of a disk about the perpendicular axis passing through its centre. From eq (1) and (2), I ∂2θ = −Cθ ∂t2 ∂2θ C … (3) ∂t2 = − I θ Angular acceleration, ∂W α = ∂t ∂2W α = ∂t2 α = −Cθ …(4) ( From eq (3) ) I Above equation represents that angular acceleration is proportional to the angular displacement or angle of twist θ and its motion is simple harmonic motion. ∝ = −w2θ …(5) ⇒ w = √C comparing eq (4) and eq (5) I Thus, periodic time, T = 2π√CI Depression of cantilever fixed at one end and loaded at other: In order to find the depression of a cantilever, let us first derive an expression of the bending moment of a beam. Let us consider a filament (ED) situated at a 15

distance r from the neutral filament (NN’) as shown in fig b. The length of filament AB and ED is same under unstrained condition. i.e. NN’ = ED =R Under the strained condition, the length of filament, ED = (R + r)∅ Change in length (R + r)∅ − R∅ r Strain = Original length = R∅ = R We know that, Stress = Y × strain Fr a=Y ×R Yar F= R Moment of force, Yar τ=r× R Yar2 τ= R 16

Now, total moment of force acting on the filaments due to straining (T): Yar2 T=∑ R = Y ∑ ar2 R Y T = R IG Where, IG = geometrical moment of inertia In equilibrium, the bending moment of beam is equal and opposite to the moment of bending couple due to load on one end. Y Bending amount of beam = R IG The quantity YIG is called the flexural rigidity of beam. From Fig (a), the force W=mg acting at B vertically downward produces an equal reaction at A acting vertically upward. These two forces generate a couple due to which a beam bends. The bending moment Mg (l-x) is balanced by a restoring couple formed by the forces acting on a part PB. Y Mg(l − x) = R IG Y is the depression of beam and radius of curvature R is given by, 1 ∂2y R = dx2 (ddyx)2]3⁄2 [1 + As depression is small slope is also very small hence (dy)2 is negligible. dx 1 ∂2y R = dx2 ∂2y = Mg(l − x) dx2 YIG 17

Integrating above equation twice, y = Mg lx2 − x3 YIG (2 6) At the free end B, x=l y = Mgl3 3YIG For rectangular beam having a cross-section of breadth b and thickness d: bd3 IG = 12 y = 4Mgl3 Ybd3 For circular beam, πr4 Ig = 4 y = 4Mgl3 3Yπr4 Therefore, (i) For a given beam, depression is directly proportional to load Mg. (ii) For a given load, the depression of a rectangular beam is directly proportional to the cube of length and inversely proportional to the breadth, cube of depth (Thickness) and Young’s Modulus. 18

Cantilever loaded uniformly:  Let us consider the uniform load distribution on the cantilever. Then the weight of the portion of the length (l-x) is w(l-x) which produces a bending moment about the section PQ. Here w is load per unit length.  The weight w(l-x) acts at a distance (l-x)/2 from the section PQ.  Bending moment = ������(������ − ������) · (������−������) = ������ ( ������ − ������)2 2  At equilibrium of the beam, 2 ������ ( ������ − ������)2 = ������������������ = ������������������ ������������ 2 ������ ������������ ������ ������������ = 2������������������ ( ������ − ������)2 ������������ Now, ������������ = (������ − ������)������������ ������ 2������������������ ������������ = (������ − ������) ( ������ − ������)2 ������������ ������ = ������ ( ������ − ������)3 ������������ 2������������������ ������ = ������ ∫0������( ������ − ������)3 ������������ 2������������������ ������ = ������������4 = ������������3 ( ∵ wl=W) 8������������������ 8������������������ Thus, Depression, ������ = ������������3 8������������������ 19

Cantilever Supported at Ends & loaded in Middle: Let us consider a beam of rectangular cross-section having length l, breadth b and depth d. It is horizontally supported at knife edges K1 and K2 and loaded at middle by weight W. The reaction of each of knife edges is W. 2 It may be considered as double cantilever each of length l having a load W2 . 2 Thus, the depression of middle point is given by, y = 4 (W2 ) (2l )3 Ybd3 Wl3 y = 4Ybd3 Mgl3 y = 4Ybd3 Where M is the mass suspended at middle. For a beam of circular cross-section, Mgl3 y = 12πYr4 I-shaped Girder: The girders with upper section broader and middle section tapered so that it can withstand heavy load over it is called I-shaped Girders. We know that depression ‘y’ in cause of rectangular cross-section is = ������������3 . Since ‘y’ is inversely 3������������������3 proportional to three powers of depth ‘d’ it is effective to increase ‘d’ instead of 20

breadth ‘b’. That is why, in girder of rectangular cross-section, longer side is taken as depth. When a girder is supported at its ends and is loaded, it is depressed in middle due to bending. As the compression and elongation are maxima for the uppermost and lowermost filaments w.r.t neutral axis, outer layers of girders are stronger than inner layers. Thus, girders are manufactured with their cross-section in the form of letter I. Viscosity: Viscosity is a property of fluid which describes the frictional resistance to flow of fluid and it measures the resistance of fluid and it measures the resistance of fluid to deform under shear stress. The backward dragging force or viscous force, 1 ������ ∝ ������; ������ ∝ −������ ; ������ ∝ ������ 21

−������������������ Or ∴ ������ = ������ ……….(1) ������������ ∴ ������ = −������������ ������������ |������| = ������������ ������������ ������������ ������ = ������������ − ������������������������������������������������������ ������������ ������������������������������������������������������ Unit – poise or Ns/m2 Also, eqn (1) can be written as, ������ = ������ ������������ ������������ Ostwald Viscometer: Procedure Fill the viscometer with the liquid being examined through tube to slightly above the mark G. Place the tube vertically in a water bath and attain the constant temperature. Through developing suction in the tube arm having bulb A such that the liquid rises 5 mm just above the mark C. Now, after releasing pressure or suction, measure the time (t) taken for the bottom of the meniscus to fall from the top edge of mark C to the top edge of mark D. Repeat this procedure for 3- 4 times to get higher accuracy. 22

Ostwald's viscometer works on the principle of Poiseuille's law for capillary tube. As per this law, the rate of flow of liquid (V) through a capillary rise having viscosity coefficient u, is given by, ������������ ∙ ������4 ������ = 8 ∙ ������ ∙ ������ Where, P = pressure difference across the tube, r= Radius of the tube: 1= length of capillary portion CD. Now, repeat the same procedure with the second liquid whose comparison is to be made having viscosity coefficient u', its rate of flow V', l he given by, ������������ ∙ ������4 ������′ = 8 ∙ ������′ ∙ ������ Since the pressure difference is directly proportional to the density of the fluids, we can replace pressure difference with density of the fluids. So, taking the ratio, of flow of rates of both fluids, we get: ������ ������ ∙ ������′ ������′ = ������′ ∙ ������ Due to variation in the pressure difference P and P’, the flow rates V and V also varies during the flow of the liquid but their times for the flow of unit volume of the two liquids varies inversely to V and V' respectively. Therefore, the total time t and t' for travelling the distance CD, will have relation like. ������ ������ ������′ = ������′ ������ ������′ ∙ ������ ������ ∙ ������ ������′ = ������ ∙ ������′ = ������′ ∙ ������′ From above equation, it is very clear the knowing the densities of fluids and by measuring the time to flow, ratio of viscosities can be easily measured. 23