Tangential or Shearing

UNIT-II PROPERTIES OF MATTER INTRODUCTION Elasticity is a branch of Physics which deals with the elastic property of materials. When an external force is applied to a body, there will be some change in its length,shape and volume. When this external force is removed, and if the body regains its original shape and size, then the body is said to be a Perfectly Elastic body. If the body does not regain its originai shape or size after removal of the applied force, then it is said to be Perfectly Plastic body. STRESS AND STRAIN: A body is said to be rigid body, if the distance between any two points in a body is unaltered due to application of the force. In practice no body is perfectly rigid. When a body is subjected to some external forces the body will offer some resistance to the deforming forces, as a result some work is done on the body and this work is stored as the elastic potential energy. Now if the deforming forces reremoved the energy stored brings back the body to its original condition. TYPESOFSTRESS Stress are classified into three types namely, Stress Tensile (or) Longitudinal Tangential or Shearing Hydrostatic i. Tensile or Longitudinal Stress Whentheforceisappliedparallel tothesurfaceofthebody,thenthestressis calledaslongitudinal stress or tensile stress. ii.tangential or Shearing stress Whentheforceisappliedalong thesurfaceofthebody,thenthe c o r r e s p o n d i n g stress exerted is calledastangential stress or shearing stress. iii. Hydrostatic Stress When a body is subjected to a uniform force from all sides, then the corresponding stress is called

hydrostatic stress. STRAIN: Strainisdefined asthe change indimension (fractional deformation)produced bythe externalforceofthe body.inother wayitcanalsobedefined asthe ration ofthechange in dimensiontotheoriginaldimension.it is a dimensionless quantity as it is a ratio between two quantities of same dimension. TYPESOFSTRESS Stress are classified into three types as follows, Strain Longitudinal Strain Volumetric Strain Shear Strain Longitudinal strain Longitudinal strain of a deformed body is defined as the ratio of the change in length of the body due to the deformation to its original length in the direction of the force. If l is the original length and dl the change in length occurred due to the deformation, the Longitudinal strain Changein Length l Longitudinal Strain original length l Linear strain may be a tensile strain, or a compressive strain according as l refers to an increase in length or a decrease in length of the body. If we consider one of these as +ve then the other should be considered as ve, as these are opposite in nature. Volumetric Strain Volumetric strain of a deformed body is defined as the ratio of the change in volume of the body to the deformation to its original volume. If V is the original volum and V the change in volume occurred due to the deformation, the volumetric strain is given by

Changein volume V VolumeStrain original volume V When a force is applied uniformly and normally to the entire surface of the body, then there will be a change in volume of the body, without any change in its shape. This strain is called bulk or volumetric strain. Shear strain Shear strain is defined as the strain accompanying a shearing action. It is the angle in radian measure through which the body gets distorted when subjected to an external shearing action. It is denoted by. Consider a cube ABCD subjected to equal and opposite forces Q across the top and bottom forces AB and CD. If the bottom face is taken fixed, the cube gets distorted through angle to the shape ABC D. Now strain or deformation per unit length is Shear strain () of cube = CC / CD = CC / BC = Deformation / Original length. Hooke s Law Hooke s law states that stress is proportional to strain upto elastic limit Stress α Strain Stress = E x Strain Modulus of elasticity (E) = Stress/ Strain

Modulus of Elasticity As there are three types of stress and strain, therefore, the modulus of elasticity is of three types 1.Young's modulus (Y).Bulk modulus or volume elasticity (k).modulus of rigidity or shear modulus (η) 1. Young's Modulus : The ratio of tensile or longitudinal stress to tensile or longitudinal strain of a material body is called Young's modulus. longitudinal stress Young ' s modulus longitudinal strain Y F / A FL L / L AL N/m Consider a wire of length L having uniform cross sectional area 'A'. One end of wire is tied to a rigid support while lower end or free end is loaded (or) stretched by a force F as shown in fig. The wi re elongates through l due to the applied force F. The longitudinal stress = F/A The longitudinal strain = i/l

F / A FL Y L / L AL It is different for different materials. N/m.Bulk modulus or volume elasticity (K): The ratio of volume stress to volume strain of a material body is calledbulk modulus. Not all deformations are linear. Sometimes an applied stress F/A results in a decrease of volume and the strain produced is a bulk modulus (K) of elasticity. The negative sign in the equation indicates that when pressure increases, volume decreases. Thus, the Bulk modulus can alternatively be defined as the product of volume and negative gradient of pressure with respect to volume. This property possess by solid, liquid and gases. It depends upon temperature and material. Compressibility : body. Compressibility is reciprocal of Bulk modulus. i.e. ratio of volume strain tovolume stress of a material The SI and CGS units are m /N and cm /dyne respectively. The dimensions are [M-1 L1 T].. Modulus rigidity (n) The ratio of tangential stress to shearing strain, within the elastic limits. Rigidity modulus (n) = Tangential stress / Shearing strain = F/A N/m

This property possesses by solids only. All three types of modulus of elasticity or elastic constants have same units and dimensions i.e. N/m and [M1 L-1 T-] respectively. Poisson's ratio: When a wire is stretched, its length increases, however at the same time its diameter decreases. The longitudinal elongation strain produced in the wire. The wire also gets contraction hence contraction strain is produced in the wire. The ratio of the lateral contraction strain to the longitudinal elongation strain is a constant and is called as the Poisson's ratio (σ) for the material. Consider a wire of length L and diameter D. If l is the increase in its length and d is decrease in its diameter, under the application of a force then, Longitudin al elongation strain l L

and Lateral contraction strain d D By definition of Poisson s ratio Lateral contraction strain Longitudinal elongation strain d / D l L Ld ld It is the ratio of two similar quantities, therefore it has no unit and dimensions. It is a pure number. It's maximum value is 0.5 and minimum value is -1. In most of the metals, its value is about 0.. Bending of Beams Beam: A beam is defined as a rod or bar. Circular or rectangular of uniform cross section whose length is very much greater than its other dimensions,such as breadth and thickness. It is commonly used in the construction of bridges to support roofs of the buildings etc. since the length of the beam is much greater than its other dimensions the shearing stresses are very small. Assumptions: While studying about the bending of beams, the following assumptions have to be made. 1. The length of the beam should be large compared to other dimensions.. The load(forces) applied should be large compared to the weight of the beam. The cross-section of the beam remains constant and hence the geometrical moment of inertia 1 9 also remains constant 4. The shearing stresses are negligible 5. The curvature ofthe beam is very small Bending of a Beam and neutral axis Let us consider a beam of uniform rectangular cross-section in the figure. A beam may be assumed to consist of a number of parallel longitudinal metallic fibers placed one over the other and are called as filaments as shown in the figure.

Let the beam be subjected to deforming forces at its ends as shown in the figure. Due to the deforming force the beam bends. We know the beam consists of many filaments. Let us consider a filament AB at the center of the beam. It is found that the filaments (layers) lying above AB gets elongated, while the filaments lying below AB gets compressed. Therefore the filament i.e. layer AB which remains u n a l t e r e d taken as the reference axis called as Neutral axis and the plane is called as neutral plane. Further, the deformation of any filament can be measured with reference to the neutral axis.

Bending moment of beam equation Q q The moment of couple due to elastic reactions (restoring couple) which balances the bending couple due to applied load is called the bending moment. Let us consider a beam under the action of deforming forces. The beam bends into a circular arc as shown in the fig. Let ABCD represent a small section of bent beam. Let PQ be the neutral axis of the beam and P Q be another filament at distance y from PQ. If R is the radius of curvature of the neutral axis of the and is the angle subtended by it at its centre of curvature C. Then we can write original length PQ = Radius x Angle = R...(1) If R+y is the radius of curvature of the filament P Q. Then we have P Q = (R+y)...() Extension Produced in the filament P Q Due to bending = P Q - PQ = (R+y)-R

= y...() The strain on the filament = Extensionproduced Originallength y = R y R...(4) The Young s modulus of the filament P Q Stress Y = Strain Stress Y xstrain y Yy z Y...(5) R R If A is the area of cross-section of the filament, then the force on the Filament. F Stressx Area Yy = xa R Y F xay...(6) R We know, moment of longitudinal force about the neutral axis = Force x distance = F x y = Y ay R The moment of all the forces about the Y Neutral axis ay...(7) R Here, I g = ay = AK is called as the geometrical moment of inertia. Where, A Total area of the beam K Radius of Gyration

Total moment of all the forces (or) YI g Internalbendingmoment...(8) R Case (i) Rectangular section Geometrical moment of inertia of rectangular section of a beam having breadth b and depth d, bd I g 1 Hence, bending moment for a rectangular section = Y bd R 1...(9) Case (ii) Circular section Geometrical moment of circular section of a beam of radius r, r I g 4 Hence bending moment for a circular section = Y r R 4 4...(10) Uniform Bending Elevation at the centre of the beam loaded at both ends. Fig.a

Let us consider a beam AB of negligible mass, supported symmetrically on two knife edges C and D. It is loaded with equal weight W at each end. Let l be the length between the two knife edges and a be the length between the knife edge and the load. CD = l and CA =DB = a Due to the applied load the beam bends into an arc of circle and procedures an elevation y.let P be any section of the beam. At the equilibrium position of the section PA of the beam two equal forces, the applied load W at A (download) and the normal reaction W at C (upward) are acting in the opposite direction constitute a couple. The External bending moment = Wa...(1) Internalbendingmoment Where, R is the radius of curvature. At Equilibrium, YI g...() External bending moment = Internal bending moment R YI g W a...() R Since, Wa is a constant, R is also constant. Therefore the beam bends into an arc of a circle of radius R. Hence the bending in this case is said to be uniform. From the figure (b), we have R y l l y Fig. b

If, R>>>y then R-y = R then l R 8 y Ry l 4...(4) Substituting (4) in () Y I g Wa l 8y The Elevation Wal y 8 YI g...(5) When, the Elevation h is measured, Young s modulus of the material of the beam can be calculated as, Y Wal 8yI g...(6) Case (i) Rectangular Section In the case of rectangular section beam. bd I g 1, where b is the breadth and d is the thickness of the Hence the elevation of the beam of rectangular section. Wal 1 y 8Y bd W al y...(7) Ybd Case (ii): Circular Section In the case of circular section where r I g 4 4, r is the radius of the beam.

Hence, the elevationh of the beam of circular section Wal 4 y 8Y r y Wal Y r...(8) Experiment 4 4 Construction: A rectangular beam AB of uniform section is supported horizontally on two knife edges A and B as shown in Figure. Two weight hangers of equal masses are suspended from the ends of the beam. A pin is arranged vertically at the mid-point of the beam. A microscope is focused on the tip of the pin. Procedure A dead loads are attached to the hangers. The microscope is adjusted such that the horizontal cross-wire coincides with the tip of the image of the pin and the readings on the vertical scale are taken. Equal weights i n s t e p s o f 5 0 g are added to both hangers simultaneously and the reading of the microscope in the vertical scale in noted. The experiment is repeated for decreasing order of magnitude of the equal masses. The observations are then tabulated and the mean elevation (y) at the mid point of the bar is determined. Observation

s.no Load Microscope reading mean (M/y) 1 W Loading Unloading Elevation(y) MSR VSC div TR div MSR VSC div TR div X10 - m metre Kg/m Cm Cm W+50 W+100 4 W+150 5 W+00 The mean elevation y produced by an addition of M say 50 gm is found by the formula. Wal y 8 YI g... (1) If the given beam is rectangular in shape bd I g 1... () Where b is the breadth and d is the thickness of the beam. Also, the weight W = Mg... () Substitute () and () in (1) we have, Mgal 1 y 8Y bd Mgal y Ybd The length of the bar between the knife edges l is measured. The distance of one of the weight hangers from the nearest knife edge a is measured. The breadth (b) and thickness (d) of the bar are measured by using vernier calipers and screw gauge. The young s modulus of the material of the beam is determined by the relation

Mgal Y ybd N m gal Y bd M y N m Graphical method (or) Dynamical method A graph is drawn between load (M) along x axis and elevation (y) along y axis.it is found to be a straight line as shown in fig. The slope of the straight line gives the value of y/m. Hence Young s modulus can be calculated as 1/slope = AC/BC = M/y gal bd 1 Slope N m Y DEPRESSIONOFACANTILEVER orpa Cantilever: It is a beam fixed horizontally at one end and loaded at the other end.

Theory Let us consider a beam fixed at one end and loaded at its other free end as shown in fig. Let AB is the neutral axis of a cantilever (a beam or rod)of length l is fixed at the end A and loaded at the free end B by a weight W.The end B is depressed to B. BB represents the vertical depression at the free end. Due to the load applied at the free end, a couple is created between the two forces. (i.e) (i) (ii) Force (load W ) applied at the free end towards downward direction and Reaction (R) acting in the upward direction at the supporting end. This external bending couple tends to bend the beam in the clockwise direction. But, since one end of the beam is fixed, the beam cannot rotate. Therefore the external bending couple must be balanced by another equal and opposite couple, created due to the elastic nature of the body called as internal bending moment. Consider the section of the cantilever P at a distance x from the fixed end A. Q is an other point at a distance dx form P i.e., PQ = dx. It is a distance (l-x) from the loaded end B.Considering the equilibrium of the portion PB,there is a force of reaction Wof P. The external bending moment=w xpb =W(l-x) (1) Internal bending moment of the cantilever = YI R Where Y Young s modulus of the cantilever. I- Geometricalmomentofinertia of its cross-section.r- Radius of the curvature of the neutralaxis at P. In the equilibrium position, External bending moment = Internal bending moment

YI g w( l x) R YI R w l g x..() Since P and Q are very near, we can assume that the radius of curvature R is practically the same. The tangents are drawn at P and Q meeting the vertical line BB at C and D. Let d be the angle between the tangents at P and Q. Length dx = radius x angle = Rd Then the angle POQ = dx d R.() Substituting the value of R from () in (), we have dxw ( l x) d.(4) YI g If dy is the depression due to the curvature at PQ dy l xd (5) Substituting value of d, dxw l x dy l x YI g W l x = Yi g dx (7) Total depression at the free end of the cantilever y l dy l 0 0 W ( l x) YI g dx W YI g l l x 0 dx

W YI l g 0 l x xl dx W YI g l x x xl l 0 W YI g l l l Depression of the cantilever at free end, y w YI g l If the depression y is meadured, Young s modulus of the material of the beam can be calculated as Wl Y yi Mgl g yi g Case (i) Rectangular Section In the case of rectangular section, bd I g, where b is the breadth and d is the thickness of the beam. 1 Hence, the depression of a cantilever of rectangular section

Case (ii) Circular section Wl 1 y Y bd 4wl y Ybd In the case of circular section, r I g 4 4, where r is the radius Hence, the depression of a cantilever of circular section, Wl 4 y Y r 4Wl y Yr 4 4 Experimental determination of Young s modulus by Cantilever Depression Construction: One end of the beam is rigidly clamped at one end to the edge of the tableusing G- clamp. A tall pin P is fixed vertically to the free end of the bar. A loop of cotton string or a hook is attached to this end of the bar and a weight hanger is suspended from it. A travelling microscope is focused on the tip of the pin as shown in fig. Procedure: A dead load without any slotted weights is attached to the hook. The microscope is adjusted such that the horizontal cross wire coincides with the tip of the image of the pin and the reading on the vertical scale is taken. Loads are added to the hanger in steps of 50g and every time, the readings are noted on the vertical scale.

These observations are also repeated while unloading in the same steps and the readings are tabulated.the mean depression y for a load M kg is found from the tabulated readings. The observations are tabulated as follows s.no Load (M) Kg 1 W Microscope reading mean (M/y) Loading Unloading Depression(y) MSR VSCdiv TRdiv MSR VSC div TR div X10 - m metre Kg/m Cm Cm W+50 W+100 4 W+150 5 W+00 Theoretically, we know the depression (y) produced by an addition of load Mkg (say 50g) is found by the formula. y Wl YI g (1) Where l is the length of the beam, b is the breadth of the beam and d is the thickness of the beam.,i g the geometric moment of inertia. If the given cantilever is rectangular in dimension,

I g bd 1 () Where b is the breadth and d is the thickness of the beam. Also, the weight W = Mg..() Substituting W and I g in (1) Mgl y 1 4Mgl Y bd Ybd 4Mgl Y bd y Young s modulus (or) N m Y 4gl bd M y N m.(4) Graphical method A graph is Plot between the load (M)and depression (y) along x and y axis respectively. The graph is a straight line as shown in fig. Equation (4) can also be written as 4gl bd 1. slope Y N m or Pa

By substituting the slope value from the graph in the above formula Young s modulus of the beam can be calculated. GIRDER A girder is a support beam used in construction. Girder is the term used to denote the main horizontal support of a structure, which supports smaller beams. A girder is commonly used more in the building of bridges and planes. I-Shape Girders The girders with upper and lower section broadened and the middle section tapered, so that it can withstand heavy loads over it is called as I-shape girders. The cross section of the girder takes the shape of the capital letter I as shown in fig. The vertical plate in the middle is known as the web and the top and bottom plates are referred to as flanges, steel is one of the most common material used to make I- beams, since it can withstand very heavy loads, although other materials, such as aluminium are sometimes used.

When a beam is used as a girder for a given load, depression must be minimum. We know that the depression at the mid point of a beam loaded at that point is given by 4Mgl y for a given load. Ybd Here, the depression can be minimised by, 1. Decreasing the load (Mg). Decreasing the length (l). Increasing the Young s modulus (Y) 4. Increasing the breadth (b) 5. Increasing the thickness of the girder (d) Since the length (l) and Young s modulus (Y) of the beam are the fixed quantity, it can not be altered. Therefore the breadth and thickness are adjusted to minimise the depression. Thus the girders are made of I shape and are called I shape girders. Applications of I- shape girders 1. Support beam for commercial and residential construction.. Support frames and coloumns for trolley ways, lifts and hoists.. Construction of platforms. 4. Trailer and truck bed framing. 5. Construction of bridges. 6. Machine bases. 7. Iron rails exclude in railway tracks. Advantages: 1. As the layers of the beam at top and bottom are subjected to maximum stress more material must be needed at these layers to withstand the strain.. As the stress around the neutral layer is small, material in these regions canbe removed without loss of efficiency. This would save cost of material of the girder.