Strength Of Materials - Elastic Constants PDF

2 HAPTER ELASTIC CONSTANTS 2.1. INTRODUCTION.. When a body is subjected to an axial tensile load, there is an increase in the length of the body. But at the same time there is a decrease in other dimensions of the body at right angles to the line of action of the app...

2 HAPTER ELASTIC CONSTANTS 2.1. INTRODUCTION.. When a body is subjected to an axial tensile load, there is an increase in the length of the body. But at the same time there is a decrease in other dimensions of the body at right angles to the line of action of the applied load. Thus the body is having axial deformation and also deformation at right angles to the line of action of the applied load (i.e., lateral deformation). This chapter deals with these deformations, Poisson’s ratio, volumetric strains, bulk modulus, relation between Young’s modulus and modulus of rigidity and relation between Young’s modulus and bulk modulus. 2.2. LONGITUDINAL STRAIN.. When a body is subjected to an axial tensile or compressive load, there is an axial deformation in the length of the body. The ratio of axial deformation to the original length of the body is known as longitudinal (or linear) strain. The longitudinal strain is also defined as the deformation of the body per unit length in the direction of the applied load. Let L = Length of the body, P = Tensile force acting on the body, δL = Increase in the length of the body in the direction of P. δL Then, longitudinal strain =. L 2.3. LATERAL STRAIN.. The strain at right angles to the direction of applied load is known as lateral strain. Let a rectangular bar of length L, breadth b and depth d is subjected to an axial tensile load P as shown in Fig. 2.1. The length of the bar will increase while the breadth and depth will decrease. Let δL = Increase in length, δb = Decrease in breadth, and δd = Decrease in depth. δL Then longitudinal strain =...(2.1) L δb δd and lateral strain = or...(2.2) b d 59 STRENGTH OF MATERIALS b P P d (d – δd) (b – δb) l l + δl Fig. 2.1 Note. (i) If longitudinal strain is tensile, the lateral strains will be compressive. (ii) If longitudinal strain is compressive then lateral strains will be tensile. (iii) Hence every longitudinal strain in the direction of load is accompanied by lateral strains of the opposite kind in all directions perpendicular to the load. 2.4. POISSON’S RATIO.. The ratio of lateral strain to the longitudinal strain is a constant for a given material, when the material is stressed within the elastic limit. This ratio is called Poisson’s ratio and it is generally denoted by μ. Hence mathematically, Lateral strain Poisson’s ratio, μ =...(2.3) Longitudinal strain or Lateral strain = μ × longitudinal strain As lateral strain is opposite in sign to longitudinal strain, hence algebraically, the lateral strain is written as Lateral strain = – μ × longitudinal strain...[2.3 (A)] The value of Poisson’s ratio varies from 0.25 to 0.33. For rubber, its value ranges from 0.45 to 0.50. Problem 2.1. Determine the changes in length, breadth and thickness of a steel bar which is 4 m long, 30 mm wide and 20 mm thick and is subjected to an axial pull of 30 kN in the direction of its length. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.3. Sol. Given : Length of the bar, L = 4 m = 4000 mm Breadth of the bar, b = 30 mm Thickness of the bar, t = 20 mm ∴ Area of cross-section, A = b × t = 30 × 20 = 600 mm2 Axial pull, P = 30 kN = 30000 N Young’s modulus, E = 2 × 105 N/mm2 Poisson’s ratio, μ = 0.3. Now strain in the direction of load (or longitudinal strain), Stress Load FG∵ Load IJ = E = Area × E H Stress = Area K P 30000 = = = 0.00025. A. E. 600 × 2 × 10 5 60  ELASTIC CONSTANTS δL But longitudinal strain =. L δL ∴ = 0.00025. L ∴ δL (or change in length) = 0.00025 × L = 0.00025 × 4000 = 1.0 mm. Ans. Using equation (2.3), Lateral strain Poisson’s ratio = Longitudinal strain Lateral strain or 0.3 = 0.00025 ∴ Lateral strain = 0.3 × 0.00025 = 0.000075. We know that δb FG δd IJδt Lateral strain = b or dH or K t ∴ δb = b × Lateral strain = 30 × 0.000075 = 0.00225 mm. Ans. Similarly, δt = t × Lateral strain = 20 × 0.000075 = 0.0015 mm. Ans. Problem 2.2. Determine the value of Young’s modulus and Poisson’s ratio of a metallic bar of length 30 cm, breadth 4 cm and depth 4 cm when the bar is subjected to an axial compressive load of 400 kN. The decrease in length is given as 0.075 cm and increase in breadth is 0.003 cm. Sol. Given : Length, L = 30 cm ; Breadth, b = 4 cm ; and Depth, d = 4 cm. ∴ Area of cross-section, A = b×d=4×4 = 16 cm2 = 16 × 100 = 1600 mm2 Axial compressive load, P = 400 kN = 400 × 1000 N Decrease in length, δL = 0.075 cm Increase in breadth, δb = 0.003 cm δL 0.075 Longitudinal strain = = = 0.0025 L 30 δb 0.003 Lateral strain = = = 0.00075. b 4 Using equation (2.3), Lateral strain 0.00075 Poisson’s ratio = = = 0.3. Ans. Longitudinal strain 0.0025 Stress P FG Load P IJ Longitudinal strain = E = A× E H ∵ Stress = = Area A K 400000 or 0.0025 = 1600 × E 400000 ∴ E= = 1 × 105 N/mm2. Ans. 1600 × 0.0025 61 STRENGTH OF MATERIALS 2.5. VOLUMETRIC STRAIN.. The ratio of change in volume to the original volume of a body (when the body is subjected to a single force or a system of forces) is called volumetric strain. It is denoted by ev. Mathematically, volumetric strain is given by δV ev = V where δV = Change in volume, and V = Original volume. 2.5.1. Volumetric Strain of a Rectangular Bar which is Subjected to an d Axial Load P in the Direction of its Length. Consider a rectangular bar of length L, width b P P and depth d which is subjected to an axial load P b in the direction of its length as shown in Fig. 2.2. L Let δL = Change in length, Fig. 2.2 δ b = Change in width, and δ d = Change in depth. ∴ Final length of the bar = L + δL Final width of the bar = b + δb Final depth of the bar = d + δd Now original volume of the bar, V = L.b.d Final volume = (L + δL)(b + δb)(d + δd) = L.b.d. + bdδL + Lbδd + Ld.δb (Ignoring products of small quantities) ∴ Change in volume, δV = Final volume – Original volume = (Lbd + bdδL + Lbδd + Ldδb) – Lbd = bdδL + Lbδd + Ldδb ∴ Volumetric strain, δV ev = V bdδL + Lbδ d + Ldδb = Lbd δL δ d δb = + +...(2.4) L d b δL δd δb But = Longitudinal strain and or are lateral strains. L d b Substituting these values in the above equation, we get ev = Longitudinal strain + 2 × Lateral strain...(i) From equation (2.3A), we have ∴ Lateral strain = – μ × Longitudinal strain. Substituting the value of lateral strain in equation (i), we get ev = Longitudinal strain – 2 × μ longitudinal strain 62  ELASTIC CONSTANTS = Longitudinal strain (1 – 2μ) δL = (1 – 2μ)...(2.5) L Problem 2.3. For the problem 2.1, determine the volumetric strain and final volume of the given steel bar. Sol. Given : The following data is given in problem 2.1 : L = 4000 mm, b = 30 mm, t or d = 20 mm, μ = 0.3. Original volume, V = L.b.d = 4000 × 30 × 20 = 2400000 mm3 FG IJ δL H The value of longitudinal strain i. e., K L in problem 2.1 is calculated δL as, = 0.00025 L Now using equation (2.5), we have δL Volumetric strain, ev = (1 – 2μ) L = 0.00025(1 – 2 × 0.3) = 0.0001. Ans. δV FG∵ δV IJ or V = 0.0001 H ev = V K ∴ δV = 0.0001 × V = 0.0001 × 2400000 = 240 mm3 ∴ Final volume = Original volume + δV = 2400000 + 240 mm3 = 2400240 mm3. Ans. Problem 2.4. A steel bar 300 mm long, 50 mm wide and 40 mm thick is subjected to a pull of 300 kN in the direction of its length. Determine the change in volume. Take E = 2 × 105 N/mm2 and μ = 0.25. Sol. Given : Length, L = 300 mm Width, b = 50 mm Thickness, t = 40 mm Pull, P = 300 kN = 300 × 103 N Value of E = 2 × 105 N/mm2 Value of μ = 0.25 Original volume, V = L×b×t = 300 × 50 × 40 mm3 = 600000 mm3 The longitudinal strain (i.e., the strain in the direction of load) is given by dL Stress in the direction of load = L E But stress in the direction of load P P = = Area b × t 63 STRENGTH OF MATERIALS 300 × 10 3 = = 150 N/mm2 50 × 40 dL 150 ∴ = = 0.00075 L 2 × 10 5 Now volumetric strain is given by equation (2.5) as dL ev = (1 – 2μ) L = 0.00075 (1 – 2 × 0.25) = 0.000375 dV Let δV = Change in volume. Then represents volumetric strain. V dV ∴ = 0.000375 V or dV = 0.000375 × V = 0.000375 × 600000 = 225 mm3. Ans. 2.5.2. Volumetric Strain of a Z Rectangular Bar Subjected to Three Forces Y which are Mutually Perpendicular. Consider a rectangular block of dimensions X X x, y and z subjected to three direct tensile stresses along three mutually perpendicular axis as shown in Fig. 2.3. Y Z Then volume of block, V = xyz. Fig. 2.3 Taking logarithm to both sides, we have log V = log x + log y + log z. Differentiating the above equation, we get 1 1 1 1 dV = dx + dy + dz V x y z dV dx dy dz or = + +...(2.6) V x y z dV Change of volume But = = Volumetric strain V Original volume dx Change of dimension x = x Original dimension x = Strain in the x-direction = ex dy Similarly, = Strain in y-direction = ey y dz and = Strain in z-direction = ez z 64  ELASTIC CONSTANTS Substituting these values in equation (2.6), we get dV = ex + ey + ez V Now, Let σx = Tensile stress in x-x direction, σy = Tensile stress in y-y direction, and σz = Tensile stress in z-z direction. E = Young’s modulus µ = Poisson’s ratio. σx Now σx will produce a tensile strain equal to in the direction of x, and a compressive E µ × σx strain equal to in the direction of y and z. Similarly, σy will produce a tensile strain E σy µ × σy equal to in the direction of y and a compressive strain equal to in the direction of x E E σ and z. Similarly σz will produce a tensile strain equal to z in the direction of z and a comp- E µ × σz ressive strain equal to in the direction of x and y. Hence σy and σz will produce E µ × σy µ × σz compressive strains equal to and in the direction of x. E E ∴ Net tensile strain along x-direction is given by σx µ × σy µ × σz σx σy + σz FG IJ ex = E − E − E = E −µ E. H K σyFG σ + σz IJ H −µ K x Similarly, ey = E E σ Fσ −µG + σy IJ H K z x and ez = E E Adding all the strains, we get 1 2µ e x + ey + ez = (σ + σy + σz) – (σx + σy + σz) E x E 1 = (σ + σy + σz)(1 – 2µ). E x dV But ex + ey + ez = Volumetric strain =. V dV 1 ∴ = (σx + σy + σz)(1 – 2µ)...(2.7) V E Equation (2.7) gives the volumetric strain. In this equation the stresses σx , σy and σz are all tensile. If any of the stresses is compressive, it may be regarded as negative, and the dV above equation will hold good. If the value of is positive, it represents increase in volume V dV whereas the negative value of represents a decrease in volume. V 65 STRENGTH OF MATERIALS Problem 2.5. A metallic bar 300 mm × 100 mm × 40 mm is subjected to a force of 5 kN (tensile), 6 kN (tensile) and 4 kN (tensile) along x, y and z directions respectively. Determine the change in the volume of the block. Take E = 2 ×105 N/mm2 and Poisson’s ratio = 0.25. Sol. Given : Dimensions of bar = 300 mm × 100 mm × 40 mm ∴ x = 300 mm, y = 100 mm and z = 40 mm ∴ Volume, V = x × y × z = 300 × 100 × 40 = 1200000 mm3 Load in the direction of x = 5 kN = 5000 N Load in the direction of y = 6 kN = 6000 N Load in the direction of z = 4 kN = 4000 N Value of E = 2 × 105 N/mm2 Poisson’s ratio, μ = 0.25 4 kN ∴ Stress in the x-direction, Load in x-direction 5 kN σx = y× z 40 mm m 5000 0m 300 mm 10 = = 1.25 N/mm2 100 × 40 6 kN Fig. 2.4 Similarly the stress in y-direction is given by, Load in y-direction σy = x×z 6000 = = 0.5 N/mm2 300 × 40 Load in z-direction And stress in z-direction = x× y 4000 or σz = 300 × 100 = 0.133 N/mm2 Using equation (2.9), we get dV 1 = (σx + σy + σz)(1 – 2μ) V E 1 = (1.25 + 0.5 + 0.113)(1 – 2 × 0.25) 2 × 10 5 1.883 = 2 × 10 5 × 2 1.883 ∴ dV = ×V 4 × 10 5 1.883 = × 1200000 4 × 10 5 = 5.649 mm3. Ans. 66  ELASTIC CONSTANTS Problem 2.6. A metallic bar 250 mm 4 MN × 100 mm × 50 mm is loaded as shown in Fig. 2.5. 400 kN Find the change in volume. Take E = 2 × 10 5 N/mm 2 and Poisson’s ratio 50 m = 0.25. m mm 0 Also find the change that should be 10 made in the 4 MN load, in order that there 250 mm should be no change in the volume of the 2 MN bar. Fig. 2.5 Sol. Given : Length, x = 250 mm, y = 100 mm and z = 50 mm ∴ Volume, V = xyz = 250 × 100 × 50 = 1250000 mm3 Load in x-direction = 400 kN = 400000 N (tensile) Load in y-direction = 2 MN = 2 × 106 N (tensile) Load in z-direction = 4 MN = 4 × 106 N (compressive) Modulus of elasticity, E = 2 × 105 N/mm2 Poisson’s ratio, µ = 0.25. Now σx = Stress in x-direction Load in x-direction = Area of cross-section 400000 400000 = = = 80 N/mm2 (tension). y× z 100 × 50 Load in y-direction Similarly, σy = x×z 2 × 10 6 = = 160 N/mm2 250 × 50 4000000 and σz = 250 × 100 = 160 N/mm2 (compression). Using equation (2.7) and taking tensile stresses positive and compressive stresses negative, we get dV 1 = (σx + σy + σz)(1 – 2µ) V E dV 1 or = (80 + 160 – 160)(1 – 2 × 0.25) V 2 × 10 5 80 = × 0.5 = 0.0002. 2 × 10 5 ∴ Change in volume, dV = 0.0002 × V = 0.0002 × 1250000 = 250 mm3. Ans. 67 STRENGTH OF MATERIALS Change in the 4 MN load when there is no change in volume of bar dV 1 Using equation (2.7), = (σx + σy + σz)(1 – 2μ) V E dV If there is no change in volume, then =0 V 1 ∴ (σ + σy + σz)(1 – 2μ) = 0. E x But for most of materials, the value of μ lies between 0.25 and 0.33 and hence the term (1 – 2μ) is never zero. ∴ σx + σy + σz = 0. The stresses σx and σy are not to be changed. Only the stress corresponding to the load 4 MN (i.e., stress in z-direction) is to be changed. ∴ σz = – σx – σy = – 80 – 160 = – 240 N/mm2 (compressive) Load Load Load But σz = = or 240 = Area x × y 250 × 100 ∴ Load = 240 × 250 × 100 = 6 × 106 N = 6 MN But already a compressive load of 4 MN is acting. ∴ Additional load that must be added = 6 MN – 4 MN = 2 MN (compressive). Ans. 2.6. VOLUMETRIC STRAIN OF A CYLINDRICAL ROD.. Consider a cylindrical rod which is subjected to an axial tensile load P. Let d = diameter of the rod L = length of the rod Due to tensile load P, there will be an increase in the length of the rod, but the diameter of the rod will decrease as shown in Fig. 2.6. L + δL P P d – δd d L Fig. 2.6 ∴ Final length = L + δL ∴ Final diameter = d – δd Now original volume of the rod, π 2 L= d ×L 4 π Final volume = (d – δd)2(L + δL) 4 π 2 = (d + δd2 – 2d × δd)(L + δL) 4 68  ELASTIC CONSTANTS π 2 = (d × L + δd2 × L – 2d × L × δd + d2 × δL 4 + δd2 × δL – 2d × δd × δL) π 2 = (d × L – 2d × L × δd + d2 × δL) 4 Neglecting the products and higher powers of two small quantities. ∴ Change in volume, δV = Final volume – Original volume π 2 π 2 = (d × L – 2d × L × δd + d2 × δL) – d ×L 4 4 π 2 = (d × δL – 2d × L × δd) 4 Change in volume δV ∴ Volumetric strain, ev = = Original volume V π 2 ( d × δ L − 2 d × L × δd) δL 2δd = 4 = −...(2.8) π 2 L d d ×L 4 δL δd where is the strain of length and is the strain of diameter. L d ∴ Volumetric strain = Strain in length – Twice the strain of diameter. Problem 2.7. A steel rod 5 m long and 30 mm in diameter is subjected to an axial tensile load of 50 kN. Determine the change in length, diameter and volume of the rod. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.25. Sol. Given : Length, L = 5 m = 5 × 103 mm Diameter, d = 30 mm π 2 π ∴ Volume, V = d ×L= (30)2 × 5 × 103 = 35.343 × 105 4 4 Tensile load, P = 50 kN = 50 × 103 Value of E = 2 × 105 N/mm2 Poisson’s ratio, μ = 0.25 Let δd = Change in diameter δL = Change in length δV = Change in volume Stress Now strain of length = E Load 1 FG∵ Load IJ = × Area E H Stress = Area K P 1 50 × 10 3 1 = × = × π E π 2 × 10 5 × d2 × 30 2 4 4 69 STRENGTH OF MATERIALS 0.4 × 50 × 10 3 = = 0.0003536 π × 30 2 × 2 × 10 5 δL But strain of length = L δL ∴ = 0.0003536 L ∴ δL = 0.0003536 × 5 × 103 = 1.768 mm. Ans. Lateral strain Now Poisson’s ratio = Longitudinal strain ∴ Lateral strain = Poisson’s ratio × Longitudinal strain FG∵ δL IJ = 0.25 × 0.0003536 H Longitudinal strain = L K = 0.0000884 δd But Lateral strain = d δd ∴ = 0.0000884 d ∴ δd = 0.0000884 × d = 0.0000884 × 30 = 0.002652 mm Now using equation (2.8), we get δV δL 2δd Volumetric strain, = − V L d = 0.0003536 – 2 × 0.0000884 = 0.0001768 ∴ δV = V × 0.0001768 = 35.343 × 105 × 0.0001768 = 624.86 mm3. Ans. 2.7. BULK MODULUS.. When a body is subjected to the mutually perpendicular like and equal direct stresses, the ratio of direct stress to the corresponding volumetric strain is found to be constant for a given material when the deformation is within a certain limit. This ratio is known as bulk modulus and is usually denoted by K. Mathematically bulk modulus is given by Direct stress σ K= Volumetric strain = FG IJ dV...(2.9) VH K 2.8. EXPRESSION FOR YOUNG’S MODULUS IN TERMS OF BULK MODULUS.. Fig. 2.7 shows a cube A B C D E F G H which is subjected to three mutually perpendicular tensile stresses of equal intensity. Let L = Length of cube dL = Change in length of the cube 70  ELASTIC CONSTANTS E = Young’s modulus of the material of the cube σ σ = Tensile stress acting on the faces E F μ = Poisson’s ratio. σ Then volume of cube, V = L 3 A B Now let us consider the strain of one of the sides of σ σ the cube (say AB) under the action of the three mutually H G perpendicular stresses. This side will suffer the following three strains : σ D C 1. Strain of AB due to stresses on the faces AEHD σ σ Fig. 2.7 and BFGC. This strain is tensile and is equal to. E 2. Strain of AB due to stresses on the faces AEFB and DHGC. This is compressive σ lateral strain and is equal to – μ. E 3. Strain of AB due to stresses on the faces ABCD and EFGH. This is also compressive σ lateral strain and is equal to – μ. E Hence the total strain of AB is given by dL σ σ σ σ = −μ× −μ× = (1 – 2μ)...(i) L E E E E Now original volume of cube, V = L3...(ii) If dL is the change in length, then dV is the change in volume. Differentiating equation (ii), with respect to L, dV = 3L2 × dL...(iii) Dividing equation (iii) by equation (ii), we get dV 3 L2 × dL 3dL = = V L3 L dL Substituting the value of from equation (i), in the above equation, we get L dV 3σ = (1 – 2μ) V E From equation (2.9), bulk modulus is given by σ σ LM∵ dV 3σ OP K= FG IJ dV = 3σ N = (1 − 2μ) Q H K (1 − 2μ) V E V E E =...(2.10) 3(1 − 2μ) or E = 3K (1 – 2μ)...(2.11) 3K − E From equation (2.11), the expression for Poisson’s ratio (μ) is obtained as μ =. 6K 71 STRENGTH OF MATERIALS Problem 2.8. For a material, Young’s modulus is given as 1.2 × 105 N/mm2 and 1 Poisson’s ratio 4. Calculate the Bulk modulus. Sol. Given : Young’s modulus, E = 1.2 × 105 N/mm2 1 Poisson’s ratio, μ = 4 Let K = Bulk modulus Using equation (2.10), E 1.2 × 10 5 1.2 × 10 5 K = 3(1 − 2μ) = FG2 = IJ 1 3 1− H4 K3× 2 2 × 1.2 × 10 5 = = 0.8 × 105 N/mm2. Ans. 3 Problem 2.9. A bar of 30 mm diameter is subjected to a pull of 60 kN. The measured extension on gauge length of 200 mm is 0.1 mm and change in diameter is 0.004 mm. Calculate : (i) Young’s modulus, (ii) Poisson’s ratio and (iii) Bulk modulus. Sol. Given : Dia. of bar, d = 30 mm π ∴ Area of bar, A = (30)2 = 225π mm2 4 Pull, P = 60 kN = 60 × 1000 N Gauge length, L = 200 mm Extension, δL = 0.1 mm Change in dia., δd = 0.004 mm (i) Young’s modulus (E) P 60000 Tensile stress, σ = = = 84.87 N/mm2 A 225π δL 0.1 Longitudinal strain = = = 0.0005 L 200 Tensile stress ∴ Young’s modulus, E = Longitudinal strain 84.87 = = 16.975 × 104 N/mm2 0.0005 = 1.6975 × 105 N/mm2. Ans. (ii) Poisson’s ratio (μ) Poisson’s ratio is given by equation (2.3) as Lateral strain Poisson’s ratio (μ) = Longitudinal strain FG δd IJ H dK FG∵ δL IJ = 0.0005 H Lateral strain = d K 72  ELASTIC CONSTANTS FG 0.004 IJ = H 30 K = 0.000133 = 0.266. Ans. 0.0005 0.0005 (iii) Bulk modulus (K) Using equation (2.10), we get E 1.6975 × 10 5 K = = 3(1 − 2μ) 3(1 − 0.266 × 2) = 1.209 × 105 N/mm2. Ans. 2.9. PRINCIPLE OF COMPLEMENTARY SHEAR STRESSES.. It states that a set of shear stresses across a plane is t always accompanied by a set of balancing shear stresses (i.e., D C of the same intensity) across the plane and normal to it. t t Proof. Fig. 2.8 shows a rectangular block ABCD, subjected to a set of shear stresses of intensity τ on the faces AB and CD. Let the thickness of the block normal to the plane A B t of the paper is unity. The force acting on face AB Fig. 2.8 = Stress × Area = τ × AB × 1 = τ. AB Similarly force acting on face CD = τ × CD × 1 = τ.CD = τ.AB (∵ CD = AB) The forces acting on the faces AB and CD are equal and opposite and hence these forces will form a couple. The moment of this couple = Force × Perpendicular distance = τ. AB × AD...(i) If the block is in equilibrium, there must be a restoring couple whose moment must be equal to the moment given by equation (i). Let the shear stress of intensity τ′ is set up on the faces AD and CB. The force acting on face AD = τ′ × AD × 1 = τ′. AD The force acting on face BC = τ′ × BC × 1 = τ′BC = τ′. AD (∵ BC = AD) As the force acting on faces AD and BC are equal and opposite, these forces also forms a couple. Moment of this couple = Force × Distance = τ′. AD × AB...(ii) For the equilibrium of the block, the moments of couples given by equations (i) and (ii) should be equal ∴ τ.AB × AD = τ′. AD × AB or τ = τ′. The above equation proves that a set of shear stresses is always accompanied by a transverse set of shear stresses of the same intensity. The stress τ′ is known as complementary shear and the two stresses (τ and τ′) at right angles together constitute a state of simple shear. The direction of the shear stresses on the block are either both towards or both away from a corner. 73 STRENGTH OF MATERIALS In Fig. 2.8, as a result of two couples, formed by the shear forces, the diagonal BD will be subjected to tension and the diagonal AB will be subjected to compression. 2.10. STRESSES ON INCLINED SECTIONS WHEN THE ELEMENT IS SUBJECTED TO SIMPLE SHEAR STRESSES Fig. 2.9 shows a rectangular block ABCD which is in a t state of simple shear and hence subjected to a set of shear D C stresses of intensity τ on the faces AB, CD and the faces AD q and CB. Let the thickness of the block normal to the plane of the paper is unity. t t It is required to find the normal and tangential stresses across an inclined plane CE, which is having inclination θ with E the face CB. A B t Consider the equilibrium of the triangular piece CEB of thickness unity. The forces acting on triangular piece CEB are shown in Fig. 2.10 and they are : C q (i) Shear force on face CB, Pn t P Q1 = Shear stress × area of face CB = τ × BC × 1 t × BC = Q1 = τ × BC acting along CB (ii) Shear force on face EB, E Q 2 = Shear stress × area of face EB B t × EB = Q2 = τ × EB × 1 = τ × EB acting along EB (iii) A force Pn normal to the plane EC Fig. 2.10 (iv) A force Pt tangential to the plane EC The force Q1 is acting along the face CB as shown in Fig. 2.11. This force is resolved into two components, i.e., Q1 cos θ and Q1 sin θ along the plane CE and normal to the plane CE respectively. The force Q2 is acting along the face EB. This force is also resolved into two components, i.e., Q2 sin θ and Q2 cos θ along the plane EC and normal to the plane EC respectively. For equilibrium, the net force normal to the plane CE qC should be zero. s co q n P Q ∴ Pn – Q1 sin θ – Q2 cos θ = 0 Q1 P t 1 s in or Pn = Q1 sin θ + Q2 cos θ q = τ × BC × sin θ + τ × EB × cos θ q n (∵ Q1 = τ × BC and Q2 = τ × EB) si Q1 2 (90–q) Q2 Q For equilibrium, the net force along the plane CE should E q B be zero. ∴ Pt – Q1 cos θ + Q2 sin θ = 0 Q2 or Pt = Q1 cos θ – Q2 sin θ co s q (– ve sign is taken due to opposite direction) Fig. 2.11 = τ × BC × cos θ – τ × EB × sin θ 74  ELASTIC CONSTANTS Let σn = Normal stress on plane CE σt = Tangential stress on plane CE Normal force on plane CE Then σn = Area of section CE Pn τ × BC × sin θ + τ × EB × cos θ = = CE × 1 CE × 1 BC EB = τ× × sin θ + τ × × cos θ CE CE = τ × cos θ × sin θ + τ × sin θ × cos θ FG∵ BC EB IJ H In triangle EBC, CE = cos θ and CE = sin θ K = 2τ cos θ × sin θ = τ sin 2θ...(2.12) Tangential force on plane CE and σt = Area of plane CE Pt τ × BC × cos θ − τ × EB × sin θ = = CE × 1 CE BC EB = τ× × cos θ – τ × × sin θ CE CE = τ × cos θ × cos θ – τ × sin θ × sin θ = τ cos2 θ – τ sin2 θ = τ [cos2 θ – sin2 θ] = τ cos 2θ...(2.13) For the planes carrying the maximum normal stress, σn should be maximum. But from equation (2.12) it is clear that σn will be maximum when sin 2θ = ± 1 π i.e., 2θ = ± 2 π or θ=± which means θ = 45° or – 45° 4 When θ = 45°, then from equation (2.12), we have σn = τ sin 90° = τ D C When θ = – 45°, then σn = – τ 45° (Positive sign shows the normal stress is tensile whereas negative sign shows the normal stress is compressive.) When θ = ± 45°, then from equation (2.13), we find 45° that σt = τ cos 2 × 45° A E B Fig. 2.12 = τ cos 90° = 0 This shows that the planes, which carry the maximum normal stresses, are having zero shear stresses. Now from equation 2.13, it is clear that shear stress will be maximum when cos 2θ = ± 1, i.e., 2θ = 0° or 180° or θ = 0° or 90° 75 STRENGTH OF MATERIALS When θ = 0° or 90°, the value of σn from equation (2.12), is zero. This shows that the planes, which carry the maximum shear stresses, are having zero normal stresses. These planes are known as planes of simple shear. Important points. When an element is subjected to a set of shear stresses, then : (i) The planes of maximum normal stresses are perpendicular to each other. (ii) The planes of maximum normal stresses are inclined at an angle of 45° to the planes of pure shear. (iii) One of the maximum normal stress is tensile while the other maximum normal stress is compressive. (iv) The maximum normal stresses are of the same magnitude and are equal to the intensity of shear stress on the plane of pure shear. 2.11. DIAGONAL STRESSES PRODUCED BY SIMPLE SHEAR ON A SQUARE BLOCK. Fig. 2.13 shows a square block ABCD of each side equal to ‘a’ and subjected to a set of shear stresses of intensity τ on the faces AB, CD and faces AD and CB. Let the thickness of the block normal to the plane of the paper is unity. τ τ τ D C D C D C θ 45° τ τ τ τ τ τ A B A B A B τ τ τ (a) (b) (c) Fig. 2.13 The normal stress (σn) on plane AC is given by equation (2.12) as σn = τ sin 2θ...(i) But as shown in Fig. 2.13 (b) the angle made by plane AC with face BC is given by, AB a tan θ = = [∵ ABCD is a square of side ‘a’] BC a =1 ∴ θ = 45° Substituting this value of θ in equation (i), we get σn = τ × sin 2 × 45° = τ × sin 90° = τ and σt = τ × cos 2θ = τ × cos 2 × 45° = τ × cos 90° = 0 Hence on the plane AC, a direct tensile stress of magnitude τ is acting. This tensile stress is parallel to the diagonal BD. Hence the diagonal BD is subjected to tensile stress of magnitude τ. 76  ELASTIC CONSTANTS Similarly it can be proved that on the plane BD, a direct compressive stress of magnitude τ is acting. This compressive stress is perpendicular to the plane BD or this compressive stress is along the diagonal AC. Hence the diagonal AC is subjected to compressive stress of magnitude τ. The pure direct tensile and compressive stresses active on the diagonal planes AC and BD are called diagonal tensile and diagonal compressive stresses. The stress on the diagonal plane AC (i.e., along diagonal BD) is tensile whereas on the diagonal plane BD i.e., along the diagonal AC is compressive. Hence the set of shear stresses τ on the faces AB, CD and the faces AD and CB are equivalent to a compressive stress τ along the diagonal AC and a tensile stress τ along the diagonal BD. 2.12. DIRECT (TENSILE AND COMPRESSIVE) STRAINS OF THE DIAGONALS.. In Art. 2.11, we have proved that when a square block ABCD of unit thickness is subjected to a set of shear stresses of intensity q on the faces AB, CD and the faces AD and CB, the diagonal BD will experience a tensile stress of magnitude q  whereas the diagonal AC will experience a compressive stress D1 D C1 C of magnitude q. Due to these stresses the diagonal BD will be elongated whereas the diagonal AC will be shorted. Let us E consider the joint effect of these two stresses on the diagonal BD. Due to the tensile stress q along diagonal BD, there will  be a tensile strain in diagonal BD. Due to the compressive stress  q along the diagonal AC, there will be a tensile strain in the diagonal BD due to lateral strain.* A B  Let µ = Poisson’s ratio Fig. 2.14 E = Young’s modulus for the material of the block Now tensile strain in diagonal BD due to tensile stress τ along BD Tensile stress along BD τ = = E E Tensile strain in diagonal BD due to compressive stress τ along AC µ×τ = E ∴ Total tensile strain along diagonal BD τ µ×τ τ = + = (1 + µ) E E E...(2.14) Similarly it can be proved that the total strain in the diagonal AC will be compressive and will be given by Total compressive strain in diagonal AC τ = (1 + µ). E *Please refer to Art. 2.4, in which it is proved that every strain in the direction of load is accom- panied by lateral strain of the opposite kind perpendicular to the direction of load. 77 STRENGTH OF MATERIALS The total tensile strain in the diagonal BD is equal to half the shear strain. This is proved as given below : Due to the shear stresses acting on the faces, the square block ABCD will be deformed to position ABC1D1 as shown in Fig. 2.14. Now increase in the length of diagonal BD = BD1 – BD ∴ Tensile strain in the diagonal BD Increase in length BD1 − BD = =...(i) Original length BD From D, draw a perpendicular DE on BD1. We know that the distortion DD1 is very small and hence angle DBD1 will be very small. Hence we can take BD = BE and ∠CDB = ∠C1D1B = 45° Now in triangle DD1E, ∠ DD1E = 45° ∴ Length D1E = DD1 cos (DD1E) DD1 = DD1 cos 45° = 2 In triangle ABD, BD = AB 2 + AD 2 = AD 2 + AD 2 = 2 × AD (∵ AB = AD) Now from equation (i), we have Tensile strain in diagonal BD1 − BD BD = BD BD1 − BE = [∵ BD = BE] BD D1 E = [∵ BD1 – BE = D1E] BD FG DD IJ H 2K 1 FG∵ DD1 IJ = 2 × AD H D1 E = 2 and BD = 2 × AD K 1 DD1 1 DD1 = × = 2× 2 AD 2 AD 1 FG∵ DD1 IJ = 2 Shear strain* H Shear strain = AD K...(2.15) 2.13.RELATIONSHIP BETWEEN MODULUS OF ELASTICITY AND MODULUS OF. RIGIDITY We have seen in the last article that when a square block of unit thickness is subjected to a set of shear stresses of magnitude τ on the faces AB, CD and the faces AD and CB, then *Please refer to Art. 1.4.3, for shear strain. 78  ELASTIC CONSTANTS the diagonal strain due to shear stress τ is given by equation (2.14) as τ Total tensile strain along diagonal BD = (1 + µ) E From equation (2.15) also we have total tensile strain in diagonal BD 1 1 Shear stress FG Shear stress = 2 shear strain = × 2 C H Shear strain = modulus of rigidity = C 1 τ = × (∵ Shear stress = τ) 2 C ∴ Equating the two tensile strain along diagonal BD, we get τ 1 τ (1 + µ) = × E 2 C τ 1 or (1 + µ) = (Cancelling τ from both sides) E 2C ∴ E = 2C (1 + µ)...(2.16) E or C=...(2.17) 2 (1 + µ) Problem 2.10. Determine the Poisson’s ratio and bulk modulus of a material, for which Young’s modulus is 1.2 × 105 N/mm2 and modulus of rigidity is 4.8 × 10 4 N/mm2. Sol. Given : Young’s modulus, E = 1.2 × 105 N/mm2 Modulus of rigidity, C = 4.8 × 104 N/mm2 Let the Poisson’s ratio =µ Using equation (2.16), we get E = 2C (1 + µ) or 1.2 × 105 = 2 × 4.8 × 104 (1 + µ) 1.2 × 10 5 or (1 + µ) = = 1.25 or µ = 1.25 – 1.0 = 0.25. Ans. 2 × 4.8 × 10 4 Bulk modulus is given by equation (2.10) as E 1.2 × 10 5 K = = (∵ µ = 0.25) 3 (1 − 2µ) 3(1 − 0.25 × 2) = 8 × 104 N/mm2. Ans. Problem 2.11. A bar of cross-section 8 mm × 8 mm is subjected to an axial pull of 7000 N. The lateral dimension of the bar is found to be changed to 7.9985 mm × 7.9985 mm. If the modulus of rigidity of the material is 0.8 × 105 N/mm2, determine the Poisson’s ratio and modulus of elasticity. Sol. Given : Area of section = 8 × 8 = 64 mm2 Axial pull, P = 7000 N Lateral dimensions = 7.9985 mm × 7.9985 mm Volume of C = 0.8 × 105 N/mm2 79 STRENGTH OF MATERIALS Let μ = Poisson’s ratio and E = Modulus of elasticity. Change in lateral dimension Now lateral strain = Original lateral dimension 8 − 7.9985 0.0015 = = = 0.0001875. 8 8 To find the value of Poisson’s ratio, we must know the value of longitudinal strain. But in this problem, the length of bar and the axial extension is not given. Hence longitudinal strain cannot be calculated. But axial stress can be calculated. Then longitudinal, strain will be equal to axial stress divided by E. P 7000 σ ∴ Axial stress, σ = = = 109.375 N/mm2 and longitudinal strain = Area 64 E σ But lateral strain = μ × longitudinal strain = μ × E μ × 109.375 or 0.0001875 = (∵ Lateral strain = 0.0001875) E E 109.375 ∴ = = 583333.33 μ 0.0001875 or E = 583333.33μ...(i) Using equation (2.17), we get E C = or E = 2C(1 + μ) 2(1 + μ) = 2 × 0.8 × 105 (1 + μ) (∵ C = 0.8 × 105) or 583333.33μ = 2 × 0.8 × 10 (1 + μ) 5 (∵ E = 583333.33μ) 583333.33μ or 1+μ = = 3.6458μ 2 × 0.8 × 10 5 ∴ 1 = 3.6458μ – μ = 2.6458μ 1 ∴ Poisson’s ratio = μ = = 0.378. Ans. 2.6458 Modulus of elasticity (E) is obtained by substituting the value of μ in equation (i). ∴ E = 583333.33μ 583333.33 ∴ E = = 2.2047 × 105 N/mm2. Ans. 2.6458 Problem 2.12. Calculate the modulus of rigidity and bulk modulus of a cylindrical bar of diameter 30 mm and of length 1.5 m if the longitudinal strain in a bar during a tensile stress is four times the lateral strain. Find the change in volume, when the bar is subjected to a hydrostatic pressure of 100 N/mm2. Take E = 1 × 105 N/mm2. Sol. Given : Dia. of bar, d = 30 mm Length of bar, L = 1.5 m = 1.5 × 1000 = 1500 mm π 2 π ∴ Volume of bar,V = d ×L= × 30 × 1500 4 4 = 1060287.52 mm3 80  ELASTIC CONSTANTS Longitudinal strain = 4 × Lateral strain Hydrostatic pressure, p = 100 N/mm2 Lateral strain 1 ∴ = = 0.25 Longitudinal strain 4 or Poisson’s ratio, μ = 0.25 Let C = Modulus of rigidity K = Bulk modulus E = Young’s modulus = 1 × 105 N/mm2 Using equation (2.16), we get E = 2C (1 + μ) or 1 × 105 = 2C(1 + 0.25) 1 × 105 ∴ C = = 4 × 104 N/mm2. Ans. 2 × 1.25 For bulk modulus, using equation (2.11), we get E = 3K (1 – 2μ) or 1 × 105 = 3K(1 – 2 × 0.25) (∵ μ = 0.25) 1 × 10 5 ∴ K = = 0.667 × 105 N/mm2. Ans. 3 × 0.5 Now using equation (2.9), we get p p K = Volumetric strain = dV FG IJ V H K where p = 100 N/mm 2 100 ∴ 0.667 × 105 = dV FG IJ V H K dV 100 or = = 1.5 × 10–3 V 0.667 × 10 5 ∴ dV = V × 1.5 × 10–3 = 1060287.52 × 1.5 × 10–3 = 1590.43 mm3. Ans. HIGHLIGHTS 1. Poisson’s ratio is the ratio of lateral strain to longitudinal strain. It is generally denoted by μ. 2. The tensile longitudinal stress produces compressive lateral strains. δl 3. If a load acts in the direction of length of a rectangular bar, then longitudinal strain = and l δb δd Lateral strain = or b d where δl = Change in length, δb = Change in width, δd = Change in depth. 81 STRENGTH OF MATERIALS 4. The ratio of change in volume to original volume is known as volumetric strain. 5. Volumetric strain (ev) for a rectangular bar subjected to an axial load P, is given by δl ev = (1 − 2μ). l 6. Volumetric strain for a rectangular bar subjected to three mutually perpendicular stresses is 1 given by, ev = (σ + σy + σz)(1 – 2μ) E x where σx, σy and σz are stresses in x, y and z direction respectively. 7. Principle of complementary shear stresses states that a set of shear stresses across a plane is always accompanied by a set of balancing shear stresses (i.e., of the same intensity) across the plane and normal to it. 8. Volumetric strain of a cylindrical rod, subjected to an axial tensile load is given by, ev = Longitudinal strain – 2 × strain of diameter δl δd =−2. l d 9. Bulk modulus K is given by, σ K= FG δV IJ. HVK 10. The relation between Young’s modulus and bulk modulus is given by, E = 3K (1 – 2μ). 11. When an element is subjected to simple shear stresses then : (i) The planes of maximum normal stresses are perpendicular to each other. (ii) The planes of maximum normal stresses are inclined at an angle of 45° to the plane of pure shear. (iii) One of the maximum normal stress is tensile while the other maximum normal stress is compressive. (iv) The maximum normal stresses are of the same magnitude and are equal to the shear stress on the plane of pure shear. 12. The relation between modulus of elasticity and modulus of rigidity is given by E E = 2C (1 + μ) or C =. 2(1 + μ) EXERCISE (A) Theoretical Questions 1. Define and explain the terms : Longitudinal strain, lateral strain and Poisson’s ratio. 2. Prove that the volumetric strain of a cylindrical rod which is subjected to an axial tensile load is equal to strain in the length minus twice the strain of diameter. 3. What is a bulk modulus ? Derive an expression for Young’s modulus in terms of bulk modulus and Poisson’s ratio. 4. Define volumetric strain. Prove that the volumetric strain for a rectangular bar subjected to an axial load P in the direction of its length is given by δl ev = (1 – 2μ) l δl where μ = Poisson’s ratio and = Longitudinal strain. l 82  ELASTIC CONSTANTS 5. (a) Derive an expression for volumetric strain for a rectangular bar which is subjected to three mutually perpendicular tensile stresses. (b) A test element is subjected to three mutually perpendicular unequal stresses. Find the change in volume of the element, if the algebraic sum of these stresses is equal to zero. 6. Explain briefly the term ‘shear stress’ and ‘complimentary stress’ with proper illustrations. 7. State the principle of shear stress. 8. What do you understand by ‘An element in a state of simple shear’ ? 9. When an element is in a state of simple shear then prove that the planes of maximum normal stresses are perpendicular to each other and these planes are inclined at an angle of 45° to the planes of pure shear. 10. Derive an expression between modulus of elasticity and modulus of rigidity. (B) Numerical Problems 1. Determine the changes in length, breadth and thickness of a steel bar which is 5 m long, 40 mm wide and 30 mm thick and is subjected to an axial pull of 35 kN in the direction of its length. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.32. [Ans. 0.0729 cm, 0.000186 cm, 0.000139 cm] 2. For the above problem, determine the volumetric strain and the final volume of the given steel bar. [Ans. 0.0000525, 6000317 mm3] 3. Determine the value of Young’s modulus and Poisson’s ratio of a metallic bar of length 25 cm, breadth 3 cm and depth 2 cm when the bar is subjected to an axial compressive load of 240 kN. The decrease in length is given as 0.05 cm and increase in breadth is 0.002. [Ans. 2 × 105 N/mm2 and 0.33] 4. A steel bar 320 mm long, 40 mm wide and 30 mm thick is subjected to a pull of 250 kN in the direction of its length. Determine the change in volume. Take E = 2 × 105 N/mm2 and m = 4. [Ans. 200 mm3] 5. A metallic bar 250 mm × 80 mm × 30 mm is subjected to a force of 20 kN (tensile), 30 kN (tensile) and 15 kN (tensile) along x, y and z directions respectively. Determine the change in the volume of the block. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.25. [Ans. 19.62 mm3] 6. A metallic bar 300 mm × 120 mm × 50 mm is loaded as shown in Fig. 2.15. Find the change in volume. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.30. 4.5 MN 500 kN 50 mm 120 mm 300 mm 2.5 kN Fig. 2.15 Also find the change that should be made in 4.5 MN load, in order that there should be no change in the volume of the bar. [Ans. 450 mm2, 4.5 MN] 7. A steel rod 4 m long and 20 mm diameter is subjected to an axial tensile load of 40 kN. Deter- mine the change in length, diameter and volume of the rod. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.25. [Ans. 2.5464, 0.05092, 5598 mm3] 83 STRENGTH OF MATERIALS 8. For a material, Young’s modulus is given as 1.4 × 105 N/mm2 and Poisson’s ratio 0.28. Calculate the bulk modulus. [Ans. 1.06 × 105 N/mm2] 9. A bar of 20 mm diameter subjected to a pull of 50 kN. The measured extension on gauge length of 250 mm is 0.12 mm and change in diameter is 0.00375 mm. Calculate : (i) Young’s modulus (ii) Poisson’s ratio and (iii) Bulk modulus. [Ans. (i) 1.989 × 105 N/mm2, (ii) 0.234, (iii) 1.2465 × 105 N/mm2] 10. Determine the Poisson’s ratio and bulk modulus of a material, for which Young’s modulus is 1.2 × 105 N/mm2 and modulus of rigidity is 4.5 × 104 N/mm2. [Ans. 0.33, 1.2 × 105 N/mm2] 11. A bar of cross-section 10 mm × 10 mm is subjected to an axial pull of 8000 N. The lateral dimen- sion of the bar is found to be changed to 9.9985 mm × 9.9985 mm. If the modulus of rigidity of the material is 0.8 × 105 N/mm2, determine the Poisson’s ratio and modulus of elasticity. [Ans. 0.45, 2.4 × 105 N/mm2] 12. Calculate the modulus of rigidity and bulk modulus of a cylindrical bar of diameter of 25 mm and of length 1.6 m, if the longitudinal strain in a bar during a tensile test is four times the lateral strain. Find the change in volume, when the bar is subjected to a hydrostatic pressure of 100 N/mm2. Take E = 1 × 105 N/mm2. [Ans. 4 × 104 N/mm2, 0.667 × 105 N/mm2, 1178 mm3] 13. A bar 30 mm in diameter was subjected to tensile load of 54 kN and the measured extension on 300 mm gauge length was 0.112 mm and change in diameter was 0.00366 mm. Calculate Poisson’s ratio and values of three modulii. [Ans. µ = 0.326, E = 204.6 kN/mm2, C = 77.2 kN/mm2, K = 196 kN/mm2] 14. Derive the relation between E and C. Using the derived relationship, estimate the Young’s modulus (E) when the modulus of rigidity (C) is 0.80 × 105 N/mm2 and the Poisson’s ratio is 0.3. [Hint. E = 2C (1 + µ) = 2 × 0.80 × 105 (1 + 0.3) = 2.08 × 105 N/mm2.] 84