Engineering Physics-I PH 6151 UNIT-II Properties of Matter Prepared by Dr.N.R.SHEELA (Asst.Prof) Dept. Of Applied Physics SVCE
Content of the UNIT-II (Properties of Matter) Elasticity-Hooke s Law- Relationship between three modulii of elasticity (ualitative)-stress- Strain diagram- Poisson s ratio-factors affecting elasticity-bending moment-depression of a cantilever-young s modulus by uniform bending- I-shaped Girders
Elastic Properties of Matter An elastic body is one that returns to its original shape after a deformation. Golf Ball Rubber Band Soccer Ball
Elastic Properties of Matter An inelastic body is one that does not return to its original shape after a deformation. Dough or Bread Clay Inelastic Ball
Elastic or Inelastic? An elastic collision loses no energy. The deform-ation on collision is fully restored. In an inelastic collision, energy is lost and the deformation may be permanent.
An Elastic Spring A spring is an example of an elastic body that can be deformed by stretching. x F A restoring force, F, acts in the direction opposite the displacement of the oscillating body. F = -kx
Stress and Strain Stress refers to the cause of a deformation, and strain refers to the effect of the deformation. The downward force F causes the displacement x. x F Thus, the stress is the force; the strain is the elongation.
Types of Stress A tensile stress occurs when equal and opposite forces are directed away from each other. F W Tension A compressive stress occurs when equal and opposite forces are directed toward each other. W F Compression
Types of Strain Longitudinal Strain 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. L A A F Changein Length l Longitudinal Strain original length l L
Types of Strain 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. Changein volume V VolumeStrain original volume V
Types of Strain Shear strain Shear strain is defined as the strain accompanying a shearing action. Shear strain () of cube = CC / CD = CC / BC = Deformation / Original length.
Longitudinal Stress and Strain L A A F For wires, rods, and bars, there is a longitudinal stress F/A that produces a change in length per unit length. In such cases: L Stress F A Strain L L
Example 1. A steel wire 10 m long and 2 mm in diameter is attached to the ceiling and a 200-N weight is attached to the end. What is the applied stress? L A A F First find area of wire: D (0.002 m) A 4 4 2 2 L A = 3.14 x 10-6 m 2 Stress F A 200 N 3.14 x 10 m -6 2 Stress 6.37 x 10 7 Pa
Example 1 (Cont.) A 10 m steel wire stretches 3.08 mm due to the 200 N load. What is the longitudinal strain? Given: L = 10 m; L = 3.08 mm L Srain L L 0.00308 m 10 m L Longitudinal Strain 3.08 x 10-4
The Elastic Limit The elastic limit is the maximum stress a body can experience without becoming permanently deformed. 2 m F 2 m Stress F A W W W Beyond limit If the stress exceeds the elastic limit, the final length will be longer than the original 2 m.
The Ultimate Strength The ultimate strength is the greatest stress a body can experience without breaking or rupturing. 2 m F Stress F A W W W W W If the stress exceeds the ultimate strength, the string breaks!
Hooke s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. x m F F = -kx K is the spring constant and is a property of the spring. 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
Poisson's ratio Longitudin al elongation strain and Lateral 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 0.3. contractionstrain Lateral contraction strain Longitudinal elongation strain d / D l L Ld ld l L d D
The Modulus of Elasticity Provided that the elastic limit is not exceeded, an elastic deformation (strain) is directly proportional to the magnitude of the applied force per unit area (stress). Modulus of Elasticity stress strain
Young s Modulus For materials whose length is much greater than the width or thickness, we are concerned with the longitudinal modulus of elasticity, or Young s Modulus (Y). Young ' s modulus longitudinal stress longitudinal strain Y F/ A FL L/ L AL
Shear Modulus A shearing stress alters only the shape of the body, leaving the volume unchanged. For example, consider equal and opposite shearing forces F acting on the cube below: A d A F l F The shearing force F produces a shearing angle. The angle is the strain and the stress is given by F/A as before. Rigidity Modulus (n) = Tangential Stress/ Shearing Strain N/m 2
Calculating Shear Modulus F l d A F Stress is force per unit area: Stress F A The strain is the angle expressed in radians: Strain d l The shear modulus (n) is defined as the ratio of the shearing stress F/A to the shearing strain The shear modulus: Units are in Pascals or N/m 2 n F A
Example 5. A steel stud (S = 8.27 x 10 10 Pa) 1 cm in diameter projects 4 cm from the wall. A 36,000 N shearing force is applied to the end. What is the defection d of the stud? 2 2 D (0.01 m) l A 4 4 F d Area: A = 7.85 x 10-5 m 2 S F A F A Fl ; d Fl dl Ad AS d (36, 000 N)(0.04 m) -5 2 10 (7.85 x 10 m )(8.27 x 10 Pa) d = 0.222 mm
Volume Elasticity Not all deformations are linear. Sometimes an applied stress F/A results in a decrease of volume. In such cases, there is a bulk modulus K of elasticity. K Volumestress Volumestrain F / A v/ V The bulk modulus is negative because of decrease in V.
The Bulk Modulus K Volumestress Volumestrain F / A v/ V Since F/A is generally pressure P, we may write: K Volumestress B Volumestrain F P/ A Vv / V/ V PV V Units remain in Pascals (Pa) since the strain is unitless.
Relationship between three modulii of Elasticity Relations connecting the lateral strain(),longitudinal strain(),poisson s ratio() and the three elastic modulii. (i) Relation between and Y is = (ii) Relation between and with the Bulk modulus is -2 = 1 Y (iii) Relation between and with the Rigidity modulus is (iv) Relation between Y, n and K is (v) Relation between n, K and is 9Kn y 3 K n 3K 2n 6K 2n (vi) Relation between Y, n and is 1 2n Y 1 3K 1 2n
ultimate tensile strength UTS Stress-Strain Diagram 3 necking yield strengt h y σ Eε E σ ε 1 E 2 ε Plastic Region Elastic Region 2 σy Strain ( ) (L/Lo) ε 1 4 Strain Hardening Fracture Elastic region slope =Young s (elastic) modulus yield strength Plastic region ultimate tensile strength strain hardening fracture 5
Stress-Strain Diagram Elastic Region (Point 1 2) - The material will return to its original shape after the material is unloaded( like a rubber band). - The stress is linearly proportional to the strain in this region. σ E ε σ Eε or : Stress : Elastic modulus (Young s Modulus) E : Strain σ ε Point 2 : Yield Strength : a point where permanent deformation occurs. ( If it is passed, the material will no longer return to its original length.)
Stress-Strain Diagram Tensile Strength (Point 3) The largest value of stress on the diagram is called Tensile Strength(TS) or Ultimate Tensile Strength (UTS). It is the maximum stress which the material can support without breaking. Fracture (Point 5) If the material is stretched beyond Point 3, the stress decreases as necking and non-uniform deformation occur. Fracture will finally occur at Point 5.
Stress-Strain Diagram Strain Hardening If the material is loaded again from Point 4, the curve will follow back to Point 3 with the same Elastic Modulus (slope). The material now has a higher yield strength of Point 4. Raising the yield strength by permanently straining the material is called Strain Hardening.
Factors Affecting Elasticity Body loses its elastic property even within elastic limit, due to elastic fatigue. Apart from elastic fatigue, some materials will have change in their property due to these factors Factors Effect of cyclic stress Effect of temperature Effect of impurities Effect of annealing Effect of nature of crystals
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.
Bending of a Beam and Neutral axis 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. Let the beam be subjected to deforming forces at its ends, due to which it bends. Let us consider a filament AB at the centre 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 unaltered 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 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.
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. P Q = (R+y)...(2) Extension Produced in the filament P Q Due to bending = P Q - PQ = (R+y)-R = y...(3) The strain on the filament Extension produced Original length
=...(4) The Young s modulus of the filament P Q Y = y R y R Stress Strain Stress YStrain Yy z...(5) R If A is the area of cross-section of the filament, then the force on the Filament. FStress Area = F Yy a R Y ay R We know, moment of longitudinal force about the neutral axis
= Force x distance = F x y = Y ay R 2 The moment of all the forces about the. Y Neutral axis ay R...(7) Here, I g = ay 2 = AK 2 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) Internal bending 2 moment YI R g
Case (i) Rectangular section bd I g 12 Hence, bending moment for a rectangular section 3 Y bd R 12 3 =...(9) Case (ii) Circular section r I g 4 Hence bending moment for a circular section 2 Y r R 4 4 =...(10)
Uniform Bending Elevation at the centre of the beam loaded at both ends. 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 produces 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) Internal bending moment Where, R is the radius of curvature. YI g...(2) R At Equilibrium, External bending moment = Internal bending moment YI W a R g...(3) 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, 2R y l l y 2 2 If, R>>>y then 2R-y = 2R 2Ry 2 l 4 2 l R 8 y Y I Wa...(4) Substituting (4) in (3) The Elevation g 2 8y l Wal y 8YI g 2
When, the Elevation y is measured, Young s modulus of the material of the beam can be calculated as, Y Wal 8yI g 2 Case (i): Rectangular Section I g y bd 12 3 2 Wal 12 8Y bd 3W al y 2Ybd 2 3 3 Case (ii): Circular Section r I g 4 4 2 Wal 4 y 8Y r y Wal 2Y r 2 4 4
Experiment 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 in steps of 50g are added to both hangers simultaneously and the reading of the microscope in the vertical scale is 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.
s.no Load Kg Procedure Microscope reading mean (M/y) Loading Unloading Elevation(y) MSR Cm VSC div TR div MSR cm VSC div TR div X10-2 m metre Kg/m 1 W 2 W+50 3 W+100 4 W+150 5 W+200
Procedure The mean elevation y produced by an addition of M say 50 gm is found by the formula. 2 Wal y 8 YI g... (1) If the given beam is rectangular in shape 3 bd I g... (2) 12 Where b is the breadth and d is the thickness of the beam. Also, the weight W = Mg Substitute (2) and (3) in (1) we have, 2 Mgal 12 y... (3) 3 8Y bd
3Mgal y 2Ybd 3 2 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. 3Mgal Y 3 2ybd 2 N m 2 3gal Y 2bd 3 2 M y N m 2
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. 1/slope = AC/BC = M/y Hence Young s modulus can be calculated as 3gal 2bd 2 1 Slope N Y 2 3 m orpa
DEPRESSION OF A CANTILEVER 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.
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 from P i.e., PQ = dx. It is at a distance (l-x) from the loaded end B. Considering the equilibrium of the portion PB, there is a force of reaction W of P. The external bending moment=w x PB = W (l-x) (1) R YI Internal bending moment of the cantilever =
Where Y Young s modulus of the cantilever. I - Geometrical moment of inertia of its cross-section. R - Radius of the curvature of the neutral axis at P. In the equilibrium position, External bending moment = Internal bending moment YI w( l x) R YI g R wl x g 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.
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 (2) in (3), we have dxw ( lx) d YI g.(4)
If dy is the depression due to the curvature at PQ dy l Substituting value of d, xd (5) dy lx W dxwl x YI g lx Yi g 2 dx (6) Total depression at the free end of the cantilever y l dy l 0 0 W ( lx) YI g 2 dx
W YI g l lx 0 2 dx W l YI W YI g g 0 l l 2 2 x x 2 2xldx 2 2 x 2xl 3 2 l 0 W YI g l 3 3 l 3 l 3 (7) Depression of the cantilever at free end y W YI g 3 l 3
Y Wl 3yI 3 Mgl 3 g yi g 3 If the depression y is measured, Young s modulus of the material of the beam can be calculated as Case (i) Rectangular Section In the case of rectangular section, bd I g 12 where b is the breadth and d is the thickness of the beam.hence, the depression of a cantilever of rectangular section y W YI g 4Wl y Ybd 3 3 l 3 3 3 Case (ii) Circular section In the case of circular section, r I g 4 where r is the radius Hence, the depression of a cantilever of circular section, 4 3 Wl 4 y 3Y r 4 4Wl y 3Yr 3 4
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. 1/slope = AC/BC = M/y Hence Young s modulus can be calculated as 3gal 2bd 2 1 Slope N Y 2 3 m orpa
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 table using 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. A travelling microscope is focused on the tip of the pin as shown in fig. 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
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 y Here, the depression can be minimised by, Decreasing the load (Mg) Decreasing the length (l) Increasing the Young s modulus (Y) Increasing the breadth (b) 4Mgl 3 Ybd Increasing the thickness of the girder (d) 3 for a given load.
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 Support beam for commercial and residential construction. Support frames and columns for trolley ways, lifts and hoists. Construction of platforms. Trailer and truck bed framing. Construction of bridges. Machine bases. Iron rails exclude in railway tracks.