Strength Of Materials - Simple Stresses and Strains PDF

1 CHAPTER SIMPLE STRESSES AND STRAINS 1.1. INTRODUCTION.. When an external force acts on a body, the body tends to undergo some deformation. Due to cohesion between the molecules, the body resists deformation. This resistance by which materia...

1 CHAPTER SIMPLE STRESSES AND STRAINS 1.1. INTRODUCTION.. When an external force acts on a body, the body tends to undergo some deformation. Due to cohesion between the molecules, the body resists deformation. This resistance by which material of the body opposes the deformation is known as strength of material. Within a certain limit (i.e., in the elastic stage) the resistance offered by the material is proportional to the deformation brought out on the material by the external force. Also within this limit the resistance is equal to the external force (or applied load). But beyond the elastic stage, the resistance offered by the material is less than the applied load. In such a case, the deformation continues, until failure takes place. Within elastic stage, the resisting force equals applied load. This resisting force per unit area is called stress or intensity of stress. 1.2. STRESS.. The force of resistance per unit area, offered by a body against deformation is known as stress. The external force acting on the body is called the load or force. The load is applied on the body while the stress is induced in the material of the body. A loaded member remains in equilibrium when the resistance offered by the member against the deformation and the applied load are equal. P Mathematically stress is written as, σ = A where σ = Stress (also called intensity of stress), P = External force or load, and A = Cross-sectional area. 1.2.1. Units of Stress. The unit of stress depends upon the unit of load (or force) and unit of area. In M.K.S. units, the force is expressed in kgf and area in metre square (i.e., m2). Hence unit of stress becomes as kgf/m2. If area is expressed in centimetre square (i.e., cm2), the stress is expressed as kgf/cm2. In the S.I. units, the force is expressed in newtons (written as N) and area is expressed as m. Hence unit of stress becomes as N/m2. The area is also expressed in millimetre square 2 then unit of force becomes as N/mm2 1 N/m2 = 1 N/(100 cm)2 = 1 N/104 cm2 F∵ I = 10–4 N/cm2 or 10–6 N/mm2 GH 1 cm 2 = 2 10 mm 1 2 JK 1 STRENGTH OF MATERIALS ∴ 1 N/mm2 = 106 N/m2. Also 1 N/m2 = 1 Pascal = 1 Pa. The large quantities are represented by kilo, mega, giga and terra. They stand for : Kilo = 103 and represented by...... k Mega = 106 and represented by...... M Giga = 109 and represented by...... G Terra = 1012 and represented by...... T. Thus mega newton means 106 newtons and is represented by MN. The symbol 1 MPa stands for 1 mega pascal which is equal to 106 pascal (or 106 N/m2). The small quantities are represented by milli, micro, nano and pico. They are equal to Milli = 10–3 and represented by...... m Micro = 10–6 and represented by...... µ Nano = 10–9 and represented by...... η Pico = 10–12 and represented by...... p. Notes. 1. Newton is a force acting on a mass of one kg and produces an acceleration of 1 m/s2 i.e., 1 N = 1 (kg) × 1 m /s2. 2. The stress in S.I. units is expressed in N/m2 or N/mm2. 3. The stress 1 N/mm2 = 106 N/m2 = MN/m2. Thus one N/mm2 is equal to one MN/m2. 4. One pascal is written as 1 Pa and is equal to 1 N/m2. 1.3. STRAIN.. When a body is subjected to some external force, there is some change of dimension of the body. The ratio of change of dimension of the body to the original dimension is known as strain. Strain is dimensionless. Strain may be : 1. Tensile strain, 2. Compressive strain, 3. Volumetric strain, and 4. Shear strain. If there is some increase in length of a body due to external force, then the ratio of increase of length to the original length of the body is known as tensile strain. But if there is some decrease in length of the body, then the ratio of decrease of the length of the body to the original length is known as compressive strain. The ratio of change of volume of the body to the original volume is known as volumetric strain. The strain produced by shear stress is known as shear strain. 1.4. TYPES OF STRESSES.. The stress may be normal stress or a shear stress. Normal stress is the stress which acts in a direction perpendicular to the area. It is represented by σ (sigma). The normal stress is further divided into tensile stress and compressive stress. 1.4.1. Tensile Stress. The stress induced in a body, when subjected to two equal and opposite pulls as shown in Fig. 1.1 (a) as a result of which there is an increase in length, is known as tensile stress. The ratio of increase in length to the original length is known as tensile strain. The tensile stress acts normal to the area and it pulls on the area. 2  SIMPLE STRESSES AND STRAINS Let P = Pull (or force) acting on the body, A = Cross-sectional area of the body, L = Original length of the body, dL = Increase in length due to pull P acting on the body, σ = Stress induced in the body, and e = Strain (i.e., tensile strain). Fig. 1.1 (a) shows a bar subjected to a tensile force P at its ends. Consider a section x-x, which divides the bar into two parts. The part left to the section x-x, will be in equilibrium if P = Resisting force (R). This is shown in Fig. 1.1 (b). Similarly the part right to the section x-x, will be in equilibrium if P = Resisting force as shown in Fig. 1.1 (c). This resisting force per unit area is known as stress or intensity of stress. X P P X (a) P Resisting force (R) (b) P Resisting force (R) (c) P P R R (d) Fig. 1.1 Resisting force ( R) Tensile load ( P) ∴ Tensile stress = σ = = (∵ P = R) Cross-sectional area A P or σ=...(1.1) A And tensile strain is given by, Increase in length dL e= =....(1.2) Original length L 1.4.2. Compressive Stress. The stress induced in a body, when subjected to two equal and opposite pushes as shown in Fig. 1.2 (a) as a result of which there is a decrease in length of the body, is known as compressive stress. And the ratio of decrease in length to the original length is known as compressive strain. The compressive stress acts normal to the area and it pushes on the area. Let an axial push P is acting on a body in cross-sectional area A. Due to external push P, let the original length L of the body decreases by dL. 3 STRENGTH OF MATERIALS Fig. 1.2 Then compressive stress is given by, Resisting Force ( R) Push ( P) P σ=  . Area ( A) Area ( A) A And compressive strain is given by, e = Decrease in length  dL. Original length L 1.4.3. Shear Stress. The stress induced in a body, when subjected to two equal and opposite forces which are acting tangentially across the resisting section as shown in Fig. 1.3 as a result of which the body tends to shear off across the section, is known as shear stress. The corresponding strain is known as shear strain. The shear stress is the stress which acts tangential to the area. It is represented by τ. P P (a) P P (b) Fig. 1.3 4  SIMPLE STRESSES AND STRAINS Consider a rectangular block of height h, length L and width unity. Let the bottom face AB of the block be fixed to the surface as shown in Fig. 1.4 (a). Let a force P be applied tangentially along the top face CD of the block. Such a force acting tangentially along a surface is known as shear force. For the equilibrium of the block, the surface AB will offer a tangential reaction P equal and opposite to the applied tangential force P. P P D C D C Resistance R X X X X X X R h Resistance A P B A P B L (a) (b) (c) Fig. 1.4 Consider a section x-x (parallel to the applied force), which divides the block into two parts. The upper part will be in equilibrium if P = Resistance (R). This is shown in Fig. 1.4 (b). Similarly the lower part will be in equilibrium if P = Resistance (R) as shown in Fig. 1.4 (c). This resistance is known as shear resistance. And the shear resistance per unit area is known as shear stress which is represented by τ. Shear resistance R ∴ Shear stress, τ = = Shear area A P = (∵ R = P and A = L × 1)...(1.3) L×1 Note that shear stress is tangential to the area over which it acts. As the bottom face of the block is fixed, the face dl dl ABCD will be distorted to ABC1D1 through an angle φ D D1 C C1 P as a result of force P as shown in Fig. 1.4 (d). And shear strain (φ) is given by, h Transversal displacement φ φ φ= Distance AD A B DD1 dl or φ= =...(1.4) L AD h Fig. 1.4 (d) 1.5. ELASTICITY AND ELASTIC LIMIT.. When an external force acts on a body, the body tends to undergo some deformation. If the external force is removed and the body comes back to its original shape and size (which means the deformation disappears completely), the body is known as elastic body. This property, 5 STRENGTH OF MATERIALS by virtue of which certain materials return back to their original position after the removal of the external force, is called elasticity. The body will regain its previous shape and size only when the deformation caused by the external force, is within a certain limit. Thus there is a limiting value of force up to and within which, the deformation completely disappears on the removal of the force. The value of stress corresponding to this limiting force is known as the elastic limit of the material. If the external force is so large that the stress exceeds the elastic limit, the material loses to some extent its property of elasticity. If now the force is removed, the material will not return to its original shape and size and there will be a residual deformation in the material. 1.6. HOOKE’S LAW AND ELASTIC MODULII.. Hooke’s Law states that when a material is loaded within elastic limit, the stress is proportional to the strain produced by the stress. This means the ratio of the stress to the corresponding strain is a constant within the elastic limit. This constant is known as Modulus of Elasticity or Modulus of Rigidity or Elastic Modulii. 1.7. MODULUS OF ELASTICITY (OR YOUNG’S MODULUS).. The ratio of tensile stress or compressive stress to the corresponding strain is a constant. This ratio is known as Young’s Modulus or Modulus of Elasticity and is denoted by E. Tensile stress Compressive stress ∴ E= or Tensile strain Compressive strain σ or E=...(1.5) e 1.7.1. Modulus of Rigidity or Shear Modulus. The ratio of shear stress to the corresponding shear strain within the elastic limit, is known as Modulus of Rigidity or Shear Modulus. This is denoted by C or G or N. Shear stress τ ∴ C (or G or N) = =...(1.6) Shear strain φ Let us define factor of safety also. 1.8. FACTOR OF SAFETY.. It is defined as the ratio of ultimate tensile stress to the working (or permissible) stress. Mathematically it is written as Ultimate stress Factor of safety =...(1.7) Permissible stress 1.9. CONSTITUTIVE RELATIONSHIP BETWEEN STRESS AND STRAIN.. 1.9.1. For One-Dimensional Stress System. The relationship between stress and strain for a unidirectional stress (i.e., for normal stress in one direction only) is given by Hooke’s law, which states that when a material is loaded within its elastic limit, the normal stress developed is proportional to the strain produced. This means that the ratio of the normal 6  SIMPLE STRESSES AND STRAINS stress to the corresponding strain is a constant within the elastic limit. This constant is represented by E and is known as modulus of elasticity or Young’s modulus of elasticity. Normal stress σ ∴ = Constant or =E Corresponding strain e where σ = Normal stress, e = Strain and E = Young’s modulus σ or e=...[1.7 (A)] E The above equation gives the stress and strain relation for the normal stress in one direction. 1.9.2. For Two-Dimensional Stress System. Before knowing the relationship between stress and strain for two-dimensional stress system, we shall have to define longitudinal strain, lateral strain, and Poisson’s ratio. 1. Longitudinal strain. 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). 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. 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. 1.5. 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 =...[1.7 (B)] L δb δd and lateral strain = or...[1.7 (C)] b d b P P d (d – δd) (b – δb) L L + δL Fig. 1.5 7 STRENGTH OF MATERIALS 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. 3. 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 =...[1.7 (D)] Longitudinal strain or Lateral strain = µ × Longitudinal strain As lateral strain is opposite in sign to longitudinal strain, hence algebraically, lateral strain is written as Lateral strain = – µ × Longitudinal strain...[1.7 (E)] 4. Relationship between stress and strain. Consider a σ2 two-dimensional figure ABCD, subjected to two mutually perpendicular stresses σ1 and σ2. A D Refer to Fig. 1.5 (a). Let σ1 = Normal stress in x-direction σ2 = Normal stress in y-direction σ1 σ1 Consider the strain produced by σ1. B C The stress σ1 will produce strain in the direction of x and also in the direction of y. The strain in the direction of x will be σ2 σ longitudinal strain and will be equal to 1 whereas the strain Fig. 1.5 (a) E σ in the direction of y will be lateral strain and will be equal to – µ × 1 E (∵ Lateral strain. = – µ × longitudinal strain) Now consider the strain produced by σ2. The stress σ2 will produce strain in the direction of y and also in the direction of x. The σ2 strain in the direction of y will be longitudinal strain and will be equal to whereas the E σ strain in the direction of x will be lateral strain and will be equal to – µ × 2. E Let e1 = Total strain in x-direction e2 = Total strain in y-direction σ1 σ Now total strain in the direction of x due to stresses σ1 and σ2 = −µ 2 E E σ2 σ Similarly total strain in the direction of y due to stresses σ1 and σ2 = −µ 1 E E σ1 σ ∴ e1 = −µ 2...[1.7 (F)] E E σ2 σ e2 = −µ 1...[1.7 (G)] E E 8  SIMPLE STRESSES AND STRAINS The above two equations gives the stress and strain relationship for the two-dimensional stress system. In the above equations, tensile stress is taken to be positive whereas the compressive stress negative. 1.9.3. For Three-Dimensional Stress System. Fig. 1.5 (b) shows a three-dimensional body subjected to three orthogonal normal stresses σ1, σ2, σ3 acting in the directions of x, y and z respectively. Consider the strains produced by each stress Y σ2 separately. The stress σ1 will produce strain in the direction of x and also in the directions of y and z. The strain in the σ σ1 direction of x will be 1 whereas the strains in the direction E σ of y and z will be – µ 1. σ3 X E σ Similarly the stress σ2 will produce strain 2 in Z E σ Fig. 1.5 (b) the direction of y and strain of – µ 2 in the direction of x E and y each. σ σ Also the stress σ3 will produce strain 3 in the direction of z and strain of – µ × 3 in E E the direction of x and y. σ σ σ Total strain in the direction of x due to stresses σ1, σ2 and σ3 = 1 − µ 2 − µ 3. E E E Similarly total strains in the direction of y due to stresses σ1, σ2 and σ3 σ2 σ σ = −µ 3 −µ 1 E E E and total strains in the direction of z due to stresses σ1, σ2 and σ3 σ3 σ σ = −µ 1 −µ 2 E E E Let e1, e2 and e3 are total strains in the direction of x, y and z respectively. Then σ1 σ σ e1 = −µ 2 −µ 3...[1.7 (H)] E E E σ2 σ σ e2 = −µ 3 −µ 1...[1.7 (I)] E E E σ3 σ σ and e3 = −µ 1 −µ 2...[1.7 (J)] E E E The above three equations give the stress and strain relationship for the three orthogonal normal stress system. Problem 1.1. A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kN. If the modulus of elasticity of the material of the rod is 2 × 105 N/mm2 ; determine : (i) the stress, (ii) the strain, and (iii) the elongation of the rod. 9 STRENGTH OF MATERIALS Sol. Given : Length of the rod, L = 150 cm Diameter of the rod, D = 2.0 cm = 20 mm π ∴ Area, A= (20)2 = 100π mm2 4 Axial pull, P = 20 kN = 20,000 N Modulus of elasticity, E = 2.0 × 105 N/mm2 (i) The stress (σ) is given by equation (1.1) as P 20000 σ= = = 63.662 N/mm2. Ans. A 100π (ii) Using equation (1.5), the strain is obtained as σ E=. e σ 63.662 ∴ Strain, e= = = 0.000318. Ans. E 2 × 10 5 (iii) Elongation is obtained by using equation (1.2) as dL e=. L ∴ Elongation, dL = e × L = 0.000318 × 150 = 0.0477 cm. Ans. Problem 1.2. Find the minimum diameter of a steel wire, which is used to raise a load of 4000 N if the stress in the rod is not to exceed 95 MN/m2. Sol. Given : Load, P = 4000 N Stress, σ = 95 MN/m2 = 95 × 106 N/m2 (∵ M = Mega = 106) = 95 N/mm2 (∵ 106 N/m2 = 1 N/mm2) Let D = Diameter of wire in mm π 2 ∴ Area, A= D 4 Load P Now stress = = Area A 4000 4000 × 4 4000 × 4 95 = = or D2 = = 53.61 π 2 π D2 π × 95 D 4 ∴ D = 7.32 mm. Ans. Problem 1.3. Find the Young’s Modulus of a brass rod of diameter 25 mm and of length 250 mm which is subjected to a tensile load of 50 kN when the extension of the rod is equal to 0.3 mm. Sol. Given : Dia. of rod, D = 25 mm π ∴ Area of rod, A = (25)2 = 490.87 mm2 4 Tensile load, P = 50 kN = 50 × 1000 = 50,000 N Extension of rod, dL = 0.3 mm Length of rod, L = 250 mm 10  SIMPLE STRESSES AND STRAINS Stress (σ) is given by equation (1.1), as P 50,000 σ= = = 101.86 N/mm2. A 490.87 Strain (e) is given by equation (1.2), as dL 0.3 e= = = 0.0012. L 250 Using equation (1.5), the Young’s Modulus (E) is obtained, as Stress 101.86 N/mm 2 E= = = 84883.33 N/mm2 Strain 0.0012 = 84883.33 × 106 N/m2. Ans. (∵ 1 N/mm2 = 106 N/m2) 9 2 2 = 84.883 × 10 N/m = 84.883 GN/m. Ans. (∵ 109 = G) Problem 1.4. A tensile test was conducted on a mild steel bar. The following data was obtained from the test : (i) Diameter of the steel bar = 3 cm (ii) Gauge length of the bar = 20 cm (iii) Load at elastic limit = 250 kN (iv) Extension at a load of 150 kN = 0.21 mm (v) Maximum load = 380 kN (vi) Total extension = 60 mm (vii) Diameter of the rod at the failure = 2.25 cm. Determine : (a) the Young’s modulus, (b) the stress at elastic limit, (c) the percentage elongation, and (d) the percentage decrease in area. π 2 π Sol. Area of the rod, A= D = (3)2 cm2 4 4 LM∵ FG 1 mIJ 2 PPO = 7.0685 cm2 = 7.0685 × 10–4 m2. MN cm 2 = H 100 K Q (a) To find Young’s modulus, first calculate the value of stress and strain within elastic limit. The load at elastic limit is given but the extension corresponding to the load at elastic limit is not given. But a load of 150 kN (which is within elastic limit) and corresponding extension of 0.21 mm are given. Hence these values are used for stress and strain within elastic limit Load 150 × 1000 ∴ Stress = = N/m2 (∵ 1 kN = 1000 N) Area 7.0685 × 10 −4 = 21220.9 × 104 N/m2 Increase in length (or Extension) and Strain = Original length (or Gauge length) 0.21 mm = = 0.00105 20 × 10 mm ∴ Young’s Modulus, Stress 21220.9 × 10 4 E= = = 20209523 × 104 N/m2 Strain 0.00105 11 STRENGTH OF MATERIALS = 202.095 × 109 N/m2 (∵ 109 = Giga = G) = 202.095 GN/m2. Ans. (b) The stress at the elastic limit is given by, Load at elastic limit 250 × 1000 Stress = = Area 7.0685 × 10 −4 4 = 35368 × 10 N/m 2 = 353.68 × 106 N/m2 (∵ 106 = Mega = M) = 353.68 MN/m2. Ans. (c) The percentage elongation is obtained as, Percentage elongation Total increase in length = × 100 Original length (or Gauge length) 60 mm = × 100 = 30%. Ans. 20 × 10 mm (d) The percentage decrease in area is obtained as, Percentage decrease in area (Original area − Area at the failure) = × 100 Original area FG π × 3 π IJ H4 K × 100 2 − × 2.25 2 4 = π × 32 4 F3 2 − 2.25 2 I = GH 3 2 JK × 100 = (9 − 5.0625) 9 × 100 = 43.75%. Ans. Problem 1.5. The safe stress, for a hollow steel column which carries an axial load of 2.1 × 103 kN is 125 MN/m2. If the external diameter of the column is 30 cm, determine the internal diameter. Sol. Given : Safe stress*, σ = 125 MN/m2 = 125 × 106 N/m2 Axial load, P = 2.1 × 103 kN = 2.1 × 106 N External diameter, D = 30 cm = 0.30 m Let d = Internal diameter ∴ Area of cross-section of the column, π π A= (D2 – d2) = (.302 – d2) m2 4 4 P Using equation (1.1), σ = A *Safe stress is a stress which is within elastic limit. 12  SIMPLE STRESSES AND STRAINS 2.1 × 10 6 4 × 2.1 × 10 6 or 125 × 106 = or (.302 – d2) = π π × 125 × 10 6 (.30 2 − d 2 ) 4 or 0.09 – d2 = 213.9 or 0.09 – 0.02139 = d2 ∴ d= 0.09 − 0.02139 = 0.2619 m = 26.19 cm. Ans. Problem 1.6. The ultimate stress, for a hollow steel column which carries an axial load of 1.9 MN is 480 N/mm2. If the external diameter of the column is 200 mm, determine the internal diameter. Take the factor of safety as 4. Sol. Given : Ultimate stress = 480 N/mm2 Axial load, P = 1.9 MN = 1.9 × 106 N (∵ M = 106) = 1900000 N External dia., D = 200 mm Factor of safety = 4 Let d = Internal diameter in mm ∴ Area of cross-section of the column, π π A= (D2 – d2) = (2002 – d2) mm2 4 4 Using equation (1.7), we get Ultimate stress Factor of safety = Working stress or Permissible stress 480 ∴ 4= Working stress 480 or Working stress = = 120 N/mm2 4 ∴ σ = 120 N/mm2 Now using equation (1.1), we get P 1900000 1900000 × 4 σ= or 120 = = A π 2 2 π (40000 − d 2 ) (200 − d ) 4 1900000 × 4 or 40000 – d2 = = 20159.6 π × 120 or d2 = 40000 – 20159.6 = 19840.4 35 kN ∴ d = 140.85 mm. Ans. Problem 1.7. A stepped bar shown in Fig. 1.6 is subjected to an 2 cm axially applied compressive load of 35 kN. Find the maximum and DIA minimum stresses produced. 3 cm Sol. Given : DIA Axial load, P = 35 kN = 35 × 103 N Fig. 1.6 Dia. of upper part, D1 = 2 cm = 20 mm 13 STRENGTH OF MATERIALS π ∴ Area of upper part, A1 = (202) = 100 π mm2 4 π π Area of lower part, A2 = D22 = (302) = 225 π mm2 4 4 The stress is equal to load divided by area. Hence stress will be maximum where area is minimum. Hence stress will be maximum in upper part and minimum in lower part. Load 35 × 10 3 ∴ Maximum stress = = = 111.408 N/mm2. Ans. A1 100 × π Load 35 × 10 3 Minimum stress = = = 49.5146 N/mm2. Ans. A2 225 × π 1.10. ANALYSIS OF BARS OF VARYING SECTIONS.. A bar of different lengths and of different diameters (and hence of different cross- sectional areas) is shown in Fig. 1.6 (a). Let this bar is subjected to an axial load P. Section 3 Section 2 Section 1 A1 A2 A3 P P L1 L2 L3 Fig. 1.6 (a) Though each section is subjected to the same axial load P, yet the stresses, strains and change in lengths will be different. The total change in length will be obtained by adding the changes in length of individual section. Let P = Axial load acting on the bar, L1 = Length of section 1, A1 = Cross-sectional area of section 1, L2, A2 = Length and cross-sectional area of section 2, L3, A3 = Length and cross-sectional area of section 3, and E = Young’s modulus for the bar. Then stress for the section 1, Load P σ1 = =. Area of section 1 A1 Similarly stresses for the section 2 and section 3 are given as, P P σ2 = and σ3 = A2 A3 Using equation (1.5), the strains in different sections are obtained. σ1 P FG∵ P IJ ∴ Strain of section 1, e1 = E = A1 E H σ1 = A1 K 14  SIMPLE STRESSES AND STRAINS Similarly the strains of section 2 and of section 3 are, σ2 P σ P e2 = = and e3 = 3 =. E A2 E E A3 E Change in length of section 1 But strain in section 1 = Length of section 1 dL1 or e1 = L1 where dL1 = change in length of section 1. ∴ Change in length of section 1, dL1 = e1L1 PL1 FG∵ P IJ = A1 E H e1 = A1 E K Similarly changes in length of section 2 and of section 3 are obtained as : Change in length of section 2, dL2 = e2 L2 PL2 FG∵ P IJ = A E 2 H e2 = A2 E K and change in length of section 3, dL3 = e3 L3 F∵ I = PL3 A3 E GH e3 = P A3 E JK ∴ Total change in the length of the bar, PL1 PL2 PL3 dL = dL1 + dL2 + dL3 = + + A1 E A2 E A3 E LM P L1 L2 + L + 3 OP = N E A1 A2 A3 Q...(1.8) Equation (1.8) is used when the Young’s modulus of different sections is same. If the Young’s modulus of different sections is different, then total change in length of the bar is given by, LM L 1 + L2 + L3 OP dL = P NE A 1 1 E2 A2 E3 A3 Q...(1.9) Problem 1.8. An axial pull of 35000 N is acting on a bar consisting of three lengths as shown in Fig. 1.6 (b). If the Young’s modulus = 2.1 × 105 N/mm2, determine : (i) stresses in each section and (ii) total extension of the bar. Section 3 Section 2 Section 1 35000 N 35000 N 2 cm DIA 3 cm DIA 5 cm DIA 20 cm 25 cm 22 cm Fig. 1.6 (b) 15 STRENGTH OF MATERIALS Sol. Given : Axial pull, P = 35000 N Length of section 1, L1 = 20 cm = 200 mm Dia. of section 1, D1 = 2 cm = 20 mm π ∴ Area of section 1, A1 = (202) = 100 π mm2 4 Length of section 2, L2 = 25 cm = 250 mm Dia. of section 2, D2 = 3 cm = 30 mm π ∴ Area of section 2, A2 = (302) = 225 π mm2 4 Length of section 3, L3 = 22 cm = 220 mm Dia. of section 3, D3 = 5 cm = 50 mm π ∴ Area of section 3, A3 = (502) = 625 π mm2 4 Young’s modulus, E = 2.1 × 105 N/mm2. (i) Stresses in each section Axial load Stress in section 1, σ1 = Area of section 1 P 35000 = = = 111.408 N/mm2. Ans. A1 100 π P 35000 Stress in section 2, σ2 = = = 49.5146 N/mm2. Ans. A2 225 × π P 35000 Stress in section 3, σ3 = = = 17.825 N/mm2. Ans. A3 625 × π (ii) Total extension of the bar Using equation (1.8), we get FL I Total extension = P E GH A 1 1 + L2 L + 3 A2 A3 JK 35000 FG 200 + 250 + 220 IJ = 2.1 × 10 5 H 100 π 225 × π 625 × π K 35000 = (6.366 + 3.536 + 1.120) = 0.183 mm. Ans. 2.1 × 10 5 Problem 1.9. A member formed by connecting a steel bar to an aluminium bar is shown in Fig. 1.7. Assuming that the bars are prevented from buckling sideways, calculate the magnitude of force P that will cause the total length of the member to decrease 0.25 mm. The values of elastic modulus for steel and aluminium are 2.1 × 105 N/mm2 and 7 × 104 N/mm2 respectively. Sol. Given : Length of steel bar, L1 = 30 cm = 300 mm 16  SIMPLE STRESSES AND STRAINS Area of steel bar, A1 = 5 × 5 = 25 cm2 = 250 mm2 P Elastic modulus for steel bar, E1 = 2.1 × 105 N/mm2 5 cm × 5 cm Steel bar Length of aluminium bar, 30 cm L2 = 38 cm = 380 mm Area of aluminium bar, 10 cm × 10 cm A2 = 10 × 10 = 100 cm2 = 10000 mm2 Aluminium bar 38 cm Elastic modulus for aluminium bar, E2 = 7 × 104 N/mm2 Total decrease in length, dL = 0.25 mm Fig. 1.7 Let P = Required force. As both the bars are made of different materials, hence total change in the lengths of the bar is given by equation (1.9). FG L + L IJ HE A E A K 1 2 ∴ dL = P 1 1 2 2 0.25 = P G F I H 2.1 × 10 × 2500 7 × 10 × 10000 JK 300 380 or 5 + 4 = P (5.714 × 10–7 + 5.428 × 10–7) = P × 11.142 × 10–7 0.25 0.25 × 10 7 ∴ P= = 11.142 × 10 −7 11.142 = 2.2437 × 105 = 224.37 kN. Ans. Problem 1.10. The bar shown in Fig. 1.8 is subjected to a tensile load of 160 kN. If the stress in the middle portion is limited to 150 N/mm2, determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.2 mm. Young’s modulus is given as equal to 2.1 × 105 N/mm2. Sol. Given : Tensile load, P = 160 kN = 160 × 103 N Stress in middle portion, σ2 = 150 N/mm2 Total elongation, dL = 0.2 mm Total length of the bar, L = 40 cm = 400 mm Young’s modulus, E = 2.1 × 105 N/mm2 Diameter of both end portions, D1 = 6 cm = 60 mm ∴ Area of cross-section of both end portions, π A1 = × 602 = 900 π mm2. 4 160 kN 160 kN 6 cm DIA 6 cm DIA 40 cm Fig. 1.8 17 STRENGTH OF MATERIALS Let D2 = Diameter of the middle portion L2 = Length of middle portion in mm. ∴ Length of both end portions of the bar, L1 = (400 – L2) mm Using equation (1.1), we have Load Stress =. Area For the middle portion, we have P π σ2 = where A2 = D2 A2 4 2 160000 or 150 = π D 22 4 4 × 160000 ∴ D22 = = 1358 mm2 π × 150 or D2 = 1358 = 36.85 mm = 3.685 cm. Ans. ∴ Area of cross-section of middle portion, π A3 = × 36.85 = 1066 mm2 4 Now using equation (1.8), we get LM P L1 L2 + OP Total extension, dL = N E A1 A2 Q LM 160000 (400 − L 2 ) L2 OP or 0.2 = 2.1 × 10 5 N900 π + 1066 Q [∵ L1 = (400 – L2) and A2 = 1066] 0.2 × 2.1 × 10 5 (400 − L2) L2 or = + 160000 900 π 1066 1066(400 − L2) + 900 π L 2 or 0.2625 = 900 π × 1066 or 0.2625 × 900π × 1066 = 1066 × 400 – 1066 L2 + 900π × L2 or 791186 = 426400 – 1066 L2 + 2827 L2 or 791186 – 426400 = L2 (2827 – 1066) or 364786 = 1761 L2 364786 ∴ L2 = = 207.14 mm = 20.714 cm. Ans. 1761 1.10.1. Principle of Superposition. When a number of loads are acting on a body, the resulting strain, according to principle of superposition, will be the algebraic sum of strains caused by individual loads. While using this principle for an elastic body which is subjected to a number of direct forces (tensile or compressive) at different sections along the length of the body, first the free body diagram of individual section is drawn. Then the deformation of the each section is obtained. The total deformation of the body will be then equal to the algebraic sum of deformations of the individual sections. 18  SIMPLE STRESSES AND STRAINS Problem 1.11. A brass bar, having cross-sectional area of 1000 mm2, is subjected to axial forces as shown in Fig. 1.9. A B C D 50 kN 80 kN 10 kN 20 kN 600 mm 1m 1.20 m Fig. 1.9 Find the total elongation of the bar. Take E = 1.05 × 105 N/mm2. Sol. Given : Area, A = 1000 mm2 Value of E = 1.05 × 105 N/mm2 Let dL = Total elongation of the bar. The force of 80 kN acting at B is split up into three forces of 50 kN, 20 kN and 10 kN. Then the part AB of the bar will be subjected to a tensile load of 50 kN, part BC is subjected to a compressive load of 20 kN and part BD is subjected to a compressive load of 10 kN as shown in Fig. 1.10. 50 kN 50 kN A B 20 kN 20 kN B C 10 kN 10 kN B D Fig. 1.10 Part AB. This part is subjected to a tensile load of 50 kN. Hence there will be increase in length of this part. ∴ Increase in the length of AB P1 = × L1 AE 50 × 1000 = × 600 (∵ P1 = 50,000 N, L1 = 600 mm) 1000 × 1.05 × 10 5 = 0.2857. Part BC. This part is subjected to a compressive load of 20 kN or 20,000 N. Hence there will be decrease in length of this part. ∴ Decrease in the length of BC P2 20,000 = × L2 = × 1000 (∵ L2 = 1 m = 1000 mm) AE 1000 × 1.05 × 10 5 = 0.1904. 19 STRENGTH OF MATERIALS Part BD. This part is subjected to a compressive load of 10 kN or 10,000 N. Hence there will be decrease in length of this part. ∴ Decrease in the length of BD P3 10000 = × L3 = = 2200 AE 1000 × 1.05 × 10 5 (∵ L3 = 1.2 + 1 = 2.2 m or 2200 mm) = 0.2095. ∴ Total elongation of bar = 0.2857 – 0.1904 – 0.2095 (Taking +ve sign for increase in length and –ve sign for decrease in length) = – 0.1142 mm. Ans. Negative sign shows, that there will be decrease in length of the bar. Problem 1.12. A member ABCD is subjected to point loads P1, P2, P3 and P4 as shown in Fig. 1.11. B C D 2 A 2500 mm P1 2 P2 P3 2 P4 625 mm 1250 mm 120 cm 60 cm 90 cm Fig. 1.11 Calculate the force P2 necessary for equilibrium, if P1 = 45 kN, P3 = 450 kN and P4 = 130 kN. Determine the total elongation of the member, assuming the modulus of elasticity to be 2.1 × 105 N/mm2. Sol. Given : Part AB : Area, A1 = 625 mm2 and Length, L1 = 120 cm = 1200 mm Part BC : Area, A2 = 2500 mm2 and Length, L2 = 60 cm = 600 mm Part CD : Area, A3 = 1250 mm2 and Length, L3 = 90 cm = 900 mm Value of E = 2.1 × 105 N/mm2. Value of P2 necessary for equilibrium Resolving the forces on the rod along its axis (i.e., equating the forces acting towards right to those acting towards left), we get P1 + P3 = P2 + P4 20  SIMPLE STRESSES AND STRAINS But P1 = 45 kN, P3 = 450 kN and P4 = 130 kN ∴ 45 + 450 = P2 + 130 or P2 = 495 – 130 = 365 kN The force of 365 kN acting at B is split into two forces of 45 kN and 320 kN (i.e., 365 – 45 = 320 kN). The force of 450 kN acting at C is split into two forces of 320 kN and 130 kN (i.e., 450 – 320 = 130 kN) as shown in Fig. 1.12. From Fig. 1.12, it is clear that part AB is subjected to a tensile load of 45 kN, part BC is subjected to a compressive load of 320 kN and part CD is subjected to a tensile load 130 kN. A B 45 kN 45 kN 320 kN 320 kN B C 130 kN 130 kN C D Fig. 1.12 Hence for part AB, there will be increase in length ; for part BC there will be decrease in length and for part CD there will be increase in length. ∴ Increase in length of AB P 45000 = × L1 = × 1200 (∵ P = 45 kN = 45000 N) A1 E 625 × 2.1 × 10 5 = 0.4114 mm Decrease in length of BC P 320,000 = × L2 = × 600 (∵ P = 320 kN = 320000) A2 E 2500 × 2.1 × 10 5 = 0.3657 mm Increase in length of CD P 130,000 = × L3 = × 900 (∵ P = 130 kN = 130000) A3 E 1250 × 2.1 × 10 5 = 0.4457 mm Total change in the length of member = 0.4114 – 0.3657 + 0.4457 (Taking +ve sign for increase in length and –ve sign for decrease in length) = 0.4914 mm (extension). Ans. Problem 1.13. A tensile load of 40 kN is acting on a rod of diameter 40 mm and of length 4 m. A bore of diameter 20 mm is made centrally on the rod. To what length the rod 21 STRENGTH OF MATERIALS should be bored so that the total extension will increase 30% under the same tensile load. Take E = 2 × 105 N/mm2. Sol. Given : 40 kN 40 kN 4m Fig. 1.12 (a) Tensile load, P = 40 kN = 40,000 N Dia. of rod, D = 40 mm π ∴ Area of rod, A = (402) = 400π mm2 4 (4 – x)m xm d D 4m Fig. 1.12 (b) Length of rod, L = 4 m = 4 × 1000 = 4000 mm Dia. of bore, d = 20 mm π ∴ Area of bore, a = × 202 = 100 π mm2 4 Total extension after bore = 1.3 × Extension before bore Value of E = 2 × 105 N/mm2 Let the rod be bored to a length of x metre or x × 1000 mm. Then length of unbored portion = (4 – x) m = (4 – x) × 1000 mm. First calculate the extension before the bore is made. The extension (δL) is given by, P 40000 × 4000 2 δL = ×L= = mm AE 400π × 2 × 10 5 π Now extension after the bore is made = 1.3 × Extension before bore 2 2.6 = 1.3 × = mm...(i) π π The extension after the bore is made, is also obtained by finding the extensions of the unbored length and bored length. For this, find the stresses in the bored and unbored portions. Stress in unbored portion Load P 40000 100 = = = = N/mm2 Area A 400π π ∴ Extension of unbored portion Stress = × Length of unbored portion E 22  SIMPLE STRESSES AND STRAINS 100 (4 − x) = × (4 – x) × 1000 = mm π × 2 × 10 5 2π Stress in bored portion Load P 40000 40000 = = = = Area ( A − a) (400π − 100π) 300π ∴ Extension of bored portion Stress = × Length of bored portion E 40000 4x = × 1000 x = mm 300π × 2 × 10 5 6π ∴ Total extension after the bore is made (4 − x) 4 x = +...(ii) 2π 6π Equating equations (i) and (ii), 2.6 4 − x 4 x = + π 2π 6π 4 − x 4x or 2.6 = + or 2.6 × 6 = 3 × (4 – x) + 4x 2 6 or 15.6 = 12 – 3x + 4x or 15.6 – 12 = x or 3.6 = x ∴ Rod should be bored upto a length of 3.6 m. Ans. Problem 1.14. A rigid bar ACDB is hinged at A and supported in a horizontal position by two identical steel wires as shown in Fig. 1.12 (c). A vertical load of 30 kN is applied at B. Find the tensile forces T1 and T2 induced in these wires by the vertical load. E F T1 T2 1m A C D B d1 d2 C¢ A C D B 1m 1m 1m D¢ B¢ 30 kN Fig. 1.12 (c) Fig. 1.12 (d) Sol. Given : Rigid bar means a bar which will remain straight. Two identical steel wires mean the area of cross-sections, lengths and value of E for both wires is same. ∴ A1 = A2, E1 = E2 and L1 = L2 Load at B = 30 kN = 30,000 N Fig. 1.12 (c) shows the position of the rigid bar before load is applied at B. Fig. 1.12 (d) shows the position of the rigid bar after load is applied. 23 STRENGTH OF MATERIALS Let T1 = Tension in the first wire T2 = Tension in the second wire δ1 = Extension of first wire δ2 = Extension of second wire Since the rigid bar remains straight, hence the extensions δ1 and δ2 are given by δ1 AC 1 = = δ2 AD 2 ∴ 2δ1 = δ2...(i) But δ1 is the extension in wire EC FG T IJ × L HA K 1 1 Stress in EC × L1 1 T1 × L1 ∴ δ1 = = = E1 E1 A1 × E1 T2 × L2 Similarly δ2 = A2 × E2 Substituting the values of δ1 and δ2 is equation (i), T1 × L1 T2 × L2 2× = A1 × E1 A2 × E2 But A1 = A2, E1 = E2 and L1 = L2. Hence above equation becomes ∴ 2T1 = T2...(ii) Now taking the moments of all the forces on the rigid bar about A, we get T1 × 1 + T2 × 2 = 30 × 3 or T1 + 2T2 = 90...(iii) Substituting the value of T2 from equation (ii), into equation (iii), we get T1 + 2(2T1) = 90 or 5T1 = 90 90 ∴ T1 = = 18 kN. Ans. 5 From equation (ii), T2 = 2 × 18 = 36 kN. Ans. Note. After calculating the values of T1 and T2, the stresses in the two wires can also be obtained as : Load T1 Stress in wire EC = = Area A1 T2 and Stress in wire FD =. A2 1.11. ANALYSIS OF UNIFORMLY TAPERING CIRCULAR ROD.. A bar uniformly tapering from a diameter D1 at one end to a diameter D2 at the other end is shown in Fig. 1.13. Let P = Axial tensile load on the bar L = Total length of the bar E = Young’s modulus. 24  SIMPLE STRESSES AND STRAINS D1 D2 P P x dx L Fig. 1.13 Consider a small element of length dx of the bar at a distance x from the left end. Let the diameter of the bar be Dx at a distance x from the left end. FG D − D2 IJ x H K 1 Then Dx = D1 – L D1 − D2 = D1 – kx where k = L Area of cross-section of the bar at a distance x from the left end, π π Ax = D2= (D1 – k.x)2. 4 x 4 Now the stress at a distance x from the left end is given by, Load σx = Ax P 4P. = = π 2 π ( D − k. x) 2 ( D1 − k. x) 1 4 The strain ex in the small element of length dx is obtained by using equation (1.5). Stress σ x ∴ ex = = E E 4P 1 4P = × = π ( D1 − k. x) 2 E π E( D1 − k. x) 2 ∴ Extension of the small elemental length dx = Strain. dx = ex. dx 4P =. dx...(i) π E( D1 − k. x) 2 Total extension of the bar is obtained by integrating the above equation between the limits 0 and L. 25 STRENGTH OF MATERIALS ∴ Total extension, dL = z 0 L 4 P. dx π E( D1 − k. x) 2 = 4P πE z L 0 (D1 – k.x)–2. dx = 4P πE z0 L ( D1 − k. x) −2 × (− k) (− k). dx [Multiplying and dividing by (– k)] 4P LM (D − k. x) OP 4P LM −1 L 1 OP L MN (− 1) × (− k) PQ = π Ek N ( D 1 = πE 0 1 − k. x) Q 0 4P L OP = M 1 − π Ek N D − k. L D − k × 0 Q 1 1 1 4P L 1 O = π Ek N M D − 1 k. L − D 1 PQ 1 D1 − D2 Substituting the value of k = in the above equation, we get L Total extension, LM OP dL = F 4P D −D I M M F D 1− D I − D1 PP πE.G H L JK MN D − GH L JK. L PQ 1 2 1 2 1 1 4 PL LM 1 − 1 O P π E. (D − D ) N D − D + D D Q = 1 2 1 1 2 1 4 PL LM 1 − 1 OP π E. (D − D ) N D D Q = 1 2 2 1 4 PL ( D − D2 ) 4 PL = × 1 =...(1.10) π E. ( D1 − D2 ) D1 D2 π ED1 D2 If the rod is of uniform diameter, then D1 – D2 = D 4 PL ∴ Total extension, dL =...(1.11) π E. D2 Problem 1.15. A rod, which tapers uniformly from 40 mm diameter to 20 mm diameter in a length of 400 mm is subjected to an axial load of 5000 N. If E = 2.1 × 105 N/mm2, find the extension of the rod. Sol. Given : Larger diameter, D1 = 40 mm Smaller diameter, D2 = 20 mm Length of rod, L = 400 mm Axial load, P = 5000 N Young’s modulus, E = 2.1 × 105 N/mm2 Let dL = Total extension of the rod 26  SIMPLE STRESSES AND STRAINS Using equation (1.10), 4 PL 4 × 5000 × 400 dL = = π E D1 D2 π × 2.1 × 10 5 × 40 × 20 = 0.01515 mm. Ans. Problem 1.16. Find the modulus of elasticity for a rod, which tapers uniformly from 30 mm to 15 mm diameter in a length of 350 mm. The rod is subjected to an axial load of 5.5 kN and extension of the rod is 0.025 mm. Sol. Given : Larger diameter, D1 = 30 mm Smaller diameter, D2 = 15 mm Length of rod, L = 350 mm Axial load, P = 5.5 kN = 5500 N Extension, dL = 0.025 mm Using equation (1.10), we get 4 PL dL = π E D1 D2 4 PL 4 × 5500 × 350 or E= = π D1 D2 dL π × 30 × 15 × 0.025 = 217865 N/mm2 or 2.17865 × 105 N/mm2. Ans. 1.12. ANALYSIS OF UNIFORMLY TAPERING RECTANGULAR BAR.. A bar of constant thickness and uniformly tapering in width from one end to the other end is shown in Fig. 1.14. X a b P P X x dx L t Fig. 1.14 27 STRENGTH OF MATERIALS Let P = Axial load on the bar L = Length of bar a = Width at bigger end b = Width at smaller end E = Young’s modulus t = Thickness of bar Consider any section X-X at a distance x from the bigger end. Width of the bar at the section X-X (a − b) x =a– L a−b = a – kx where k = L Thickness of bar at section X-X = t ∴ Area of the section X-X = Width × thickness = (a – kx)t ∴ Stress on the section X-X Load P = = Area (a − kx)t Extension of the small elemental length dx = Strain × Length dx Stress FG∵ IJ Stress = E × dx H Strain = E K FG P IJ H (a − kx)t K FG∵ Stress = P IJ = E × dx H (a − kx)t K P = ax E(a − kx)t Total extension of the bar is obtained by integrating the above equation between the limits 0 and L. ∴ Total extension, dL = z 0 L P E(a − kx)t dx = P Et z0 L dx (a − kx) P LM FG 1IJ = – P [log (a – kL) – log a] OP L = Et. loge (a − kx) N H kK Etk Q 0 × − e e P L F a IJ OP = P Etk [log a – log (a – kL)] = e e Etk N M log G H a − kL K Q e LM F I OP P M G a JJ P log G FG∵ k = a − bIJ = F IJ MM GG a − FG IJ L JJ PP Et G a − b ae − b H L K H L K N H H L K KQ 28  SIMPLE STRESSES AND STRAINS PL a = log e....(1.12) Et(a − b) b Problem 1.17. A rectangular bar made of steel is 2.8 m long and 15 mm thick. The rod is subjected to an axial tensile load of 40 kN. The width of the rod varies from 75 mm at one end to 30 mm at the other. Find the extension of the rod if E = 2 × 105 N/mm2. Sol. Given : Length, L = 2.8 m = 2800 mm Thickness, t = 15 mm Axial load, P = 40 kN = 40,000 N Width at bigger end, a = 75 mm Width at smaller end, b = 30 mm Value of E = 2 × 105 N/mm2 Let dL = Extension of the rod. Using equation (1.12), we get PL a dL = log e Et(a − b) b 40000 × 2800 FG 75 IJ = 2 × 10 5 × 15(75 − 30) log e H 30 K = 0.8296 × 0.9163 = 0.76 mm. Ans. Problem 1.18. The extension in a rectangular steel bar of length 400 mm and thickness 10 mm, is found to be 0.21 mm. The bar tapers uniformly in width from 100 mm to 50 mm. If E for the bar is 2 × 105 N/mm2, determine the axial load on the bar. Sol. Given : Extension, dL = 0.21 mm Length, L = 400 mm Thickness, t = 10 mm Width at bigger end, a = 100 mm Width at smaller end, b = 50 mm Value of E = 2 × 105 N/mm2 Let P = axial load. Using equation (1.12), we get PL FG IJ a dL = Et(a − b) log e H K b P × 400 FG 100 IJ or 0.21 = 2 × 10 5 × 10(100 − 50) log e H 50 K = 0.000004 P × 0.6931 0.21 ∴ P = = 75746 N 0.000004 × 0.6931 = 75.746 kN. Ans. 29 STRENGTH OF MATERIALS 1.13. ANALYSIS OF BARS OF COMPOSITE SECTIONS.. A bar, made up of two or more bars of equal lengths but of different materials rigidly fixed with each other and behaving as one unit for extension or compression when subjected to an axial tensile or compressive loads, is called a composite bar. 1 2 L For the composite bar the following two points are important : 1. The extension or compression in each bar is equal. Hence deformation per unit length i.e., strain in each bar is equal. 2. The total external load on the composite bar is equal P to the sum of the loads carried by each different material. Fig. 1.15 Fig. 1.15 shows a composite bar made up of two different materials. Let P = Total load on the composite bar, L = Length of composite bar and also length of bars of different materials, A1 = Area of cross-section of bar 1, A2 = Area of cross-section of bar 2, E1 = Young’s Modulus of bar 1, E2 = Young’s Modulus of bar 2, P1 = Load shared by bar 1, P2 = Load shared by bar 2, σ1 = Stress induced in bar 1, and σ2 = Stress induced in bar 2. Now the total load on the composite bar is equal to the sum of the load carried by the two bars. ∴ P = P1 + P2...(i) Load carried by bar 1 The stress in bar 1, =. Area of cross-section of bar 1 P1 ∴ σ1 = or P1 = σ1 A1...(ii) A1 P2 Similarly stress in bar 2, σ2 = or P2 = σ2 A2...(iii) A2 Substituting the values of P1 and P2 in equation (i), we get P = σ1A1 + σ2 A2...(iv) Since the ends of the two bars are rigidly connected, each bar will change in length by the same amount. Also the length of each bar is same and hence the ratio of change in length to the original length (i.e., strain) will be same for each bar. Stress in bar 1 σ But strain in bar 1, = = 1. Young’s modulus of bar 1 E1 σ2 Similarly strain in bar 2, =. E2 30  SIMPLE STRESSES AND STRAINS But strain in bar 1 = Strain in bar 2 σ σ ∴ = 1 = 2...(v) E1 E2 From equations (iv) and (v), the stresses σ1 and σ2 can be determined. By substituting the values of σ1 and σ2 in equations (ii) and (iii), the load carried by different materials may be computed. E1 Modular Ratio. The ratio of is called the modular ratio of the first material to the E2 second. Problem 1.19. A steel rod of 3 cm diameter is enclosed centrally in a hollow copper tube of external diameter 5 cm and internal diameter of 4 cm. The composite bar is then subjected to an axial pull of 45000 N. If the length of each bar is equal to 15 cm, determine : (i) The stresses in the rod and tube, and (ii) Load carried by each bar. Take E for steel = 2.1 × 105 N/mm2 and for copper = 1.1 × 105 N/mm2. Sol. Given : Dia. of steel rod = 3 cm = 30 mm ∴ Area of steel rod, π Copper As = (30)2 = 706.86 mm2 tube 4 External dia. of copper tube 15 cm Steel rod = 5 cm = 50 mm Internal dia. of copper tube = 4 cm = 40 mm 3 cm ∴ Area of copper tube, 4 cm π 5 cm Ac = [502 – 402] mm2 = 706.86 mm2 4 P = 45000 N Axial pull on composite bar, P = 45000 N Fig. 1.16 Length of each bar, L = 15 cm Young’s modulus for steel, Es = 2.1 × 105 N/mm2 Young’s modulus for copper, Ec = 1.1 × 105 N/mm2 (i) The stress in the rod and tube Let σs = Stress in steel, Ps = Load carried by steel rod, σc = Stress in copper, and Pc = Load carried by copper tube. Now strain in steel = Strain in copper σs σ FG∵ σ IJ or Es = c Ec H E = strain K Es 2.1 × 10 5 ∴ σs = × σc = × σc = 1.909 σc...(i) Ec 1.1 × 10 5 31 STRENGTH OF MATERIALS Load Now stress = , ∴ Load = Stress × Area Area Load on steel + Load on copper = Total load σs × As + σc × Ac = P (∵ Total load = P) or 1.909 σc × 706.86 + σc × 706.86 = 45000 or σc (1.909 × 706.86 + 706.86) = 45000 or 2056.25 σc = 45000 45000 ∴ σc = = 21.88 N/mm2. Ans. 2056.25 Substituting the value of σc in equation (i), we get σs = 1.909 × 21.88 N/mm2 = 41.77 N/mm2. Ans. (ii) Load carried by each bar As load = Stress × Area ∴ Load carried by steel rod, Ps = σs × As = 41.77 × 706.86 = 29525.5 N. Ans. Load carried by copper tube, Pc = 45000 – 29525.5 = 15474.5 N. Ans. Problem 1.20. A compound tube consists of a steel tube 140 mm internal diameter and 160 mm external diameter and an outer brass tube 160 mm internal diameter and 180 mm external diameter. The two tubes are of the same length. The compound tube carries an axial load of 900 kN. Find the str