STRENGTH OF MATERIAL.pdf

Contents CHAPTER ONE.............................................................................................................................. 1 SIMPLE STRESS AND STRAIN..............................................................

Contents CHAPTER ONE.............................................................................................................................. 1 SIMPLE STRESS AND STRAIN.................................................................................................. 3 1.1 INTRODUCTION............................................................................................................. 3 1.2 Tension and Compression (Direct Stresses).................................................................... 3 1.3 Strain.................................................................................................................................. 6 1.4 Tensile Test........................................................................................................................ 7 1.4.2 Factor of Safety............................................................................................................ 11 1.4.3 Relationship between Stress and Strain........................................................................ 12 1.5 Poisson’s Ratio (ν)........................................................................................................... 14 1.6 Volume Change................................................................................................................ 15 Problems 1................................................................................................................................ 16 CHAPTER TWO........................................................................................................................... 17 AXIALLY LOADED SPRINGS/ BARS...................................................................................... 17 2.1 Displacement of Axially Loaded Members................................................................... 18 2.2 Statically Indeterminate Structures/ Bars.................................................................... 22 2.3 Strain Energy................................................................................................................... 24 2.4 Pre-Stress or Strain (Initial Stress)................................................................................ 26 Problems 2................................................................................................................................ 27 CHAPTER 3.................................................................................................................................. 29 TEMPERATURE AND SHEAR STRESSES.............................................................................. 29 3.1 Temperature Stresses...................................................................................................... 29 3.1.1 Change in Length of Unrestrained Bar......................................................................... 30 3.1.2 Statically Indeterminate Bars....................................................................................... 31 3.1.3 Creep of Materials under Sustained Stresses............................................................... 33 3.1.4 Fatigue under Repeated Stresses.................................................................................. 34 3.2 Shearing Stresses............................................................................................................. 35 3.2.1 Complementary Shearing Stress.................................................................................. 37 3.3 Shear Strain..................................................................................................................... 38 3.4 Relationship between Shear Stress and Shear Strain.................................................. 39 3.5 Shear Strain Energy........................................................................................................ 40 Problems 3................................................................................................................................ 40 TREBLA CHAPTER 4.................................................................................................................................. 42 TORSION...................................................................................................................................... 42 4.1 Fundamental Concepts................................................................................................... 42 4.1.1 Relationship between Torque and Angle of Twist....................................................... 43 4.1.2 Angle of Twist.............................................................................................................. 46 4.2 Non- Uniform Torsion..................................................................................................... 46 4.3 Transmission of Power by Circular shafts/ Solid Circular Bars................................ 49 4.4 Strain Energy................................................................................................................... 50 Problems 5................................................................................................................................ 52 CHAPTER FIVE........................................................................................................................... 55 BEAMS, SHEAR FORCE AND BENDING MOMENTS.......................................................... 55 5.1 Beams................................................................................................................................ 55 5.1.1 Statically Determinate and Indeterminate Beams........................................................ 56 5.1.2 Loads on Beams........................................................................................................... 57 5.2 Shear Force and Bending Moment................................................................................ 58 5.2.1 Shear Force................................................................................................................... 58 5.2.2 Bending Moment.......................................................................................................... 59 5.2.3 Relationship between Bending Moment, Shear Force and Load Intensity.................. 60 5.3 Shear Force and Bending Moment Diagrams.............................................................. 61 5.3.1 Drawing of Shear Force and Bending Moment Diagrams........................................... 61 6.3.2 Point of Contra flexure................................................................................................. 62 6.3.3 Principle of Superposition............................................................................................ 62 Problems 5................................................................................................................................ 64 CHAPTER SIX............................................................................................................................. 66 BENDING..................................................................................................................................... 66 6.1 INTRODUCTION........................................................................................................... 66 6.2 The Bending Equation...................................................................................................... 66 6.3 Position of the Neutral Axis............................................................................................ 67 6.4 The General Bending Formular..................................................................................... 68 6.5 Strain Energy of Bending................................................................................................. 69 Problems 6................................................................................................................................ 72 TREBLA  CHAPTER ONE SIMPLE STRESS AND STRAIN 1.1 INTRODUCTION Structural and solid body mechanics is concerned with analysing the effects of applied loads. These are external to the material of the structure or body and result in internal reacting forces, together with deformations or displacements conforming to the principles of Newtonian Mechanics. The strength of a material is defined largely by the internal stress or intensities or force in the material. Thus the concern of this subject will be internal effects of forces acting on the body. The bodies themselves will no longer be considered to be perfectly rigid as was assumed in statics, instead the calculation of deformation of various bodies under a variety loads will be one of primary concern. Usually, the objectives of our analysis will be the determination of stress, strains and deflections produced by the loads. If these quantities can be found for all values of load then we will have a complete picture of the mechanical behaviour of the body. Knowledge of these stresses is essential to the safe design of a machine, an aircraft or any type of structure. Most of these practical structures consist of a complex arrangement of many component members. The detail stress analysis is a difficult task even when the loading conditions are simple. The problem is further complicated because the loads experienced by a structure are variable and sometimes unpredictable. Forces acting on the body result in four basic deformations or displacements of structure or solid bodies and these are tension, compression, bending and twisting. 1.2 Tension and Compression (Direct Stresses). The fundamental concepts of stress and strain can be illustrated by considering a straight metal bar loaded at its ends by collinear (or axial) forces P coinciding and acting through the centroid of each cross – section as shown in Figure 1-1. TREBLA  a P P a Fig. 1 -1 Bar in tension The axial forces produce a uniform stretching of the bar, hence the bar is said to be in tension. To investigate the internal stress produced in the bar by the axial force we make an imaginary cut at the section a-a. This section is taken perpendicular to the longitudinal axis of the bar hence it is known as cross section. a PP a σ We now isolate the right part of the bar as a free- body. The tensile load P acts at the right hand end of the free body. At the other end are forces representing the hatching of the removed part of the bar or the part that remains. These forces are continuously distributed over the cross-section. The intensity of these forces (Force /Area) is called stress and is represented by σ (sigma). Assuming that the load has a uniform distribution over the cross-section, then it is readily seen that its resultant is equal to the intensity σ x area of cross section of the bar. From the equilibrium of the body shown, the resultant must be equal in magnitude and opposite in direction to the applied load P, hence σ = P/A Units: Nm-2 (Pascal) where σ = stress P = applied load, TREBLA  A = cross-sectional area. The above equation is the equation for uniform stress in an axially loaded bar of arbitrary crosssectional shape. When the bar is stretched by a force P, the resulting stresses are tensile. If the forces are reversed in direction at each end of the bar i.e directed towards the bar, the bar is said to be in a state of compression and the resulting stresses are compressive stresses (Figure 1-2). PP Fig. 1-2 Bar in compression Tensile and compressive stresses are together referred to as Direct or Normal (Perpendicular) stresses. They are referred to as normal stresses because the stresses act in a direction perpendicular to the cut surface. Customarily, tensile stresses are defined to be positive and the compressive stresses to be negative. It is important to realise that the stress equation gives the average normal stress acting on the cross-sectional area. With concentrated loads or where the material being loaded changes in shape i.e. a hole through the bar, then the stress will generally not be the same over each segment of the cross section. The maximum stress will depend on the bar’s geometry and the type of discontinuity. Tables and graphs are available which enable stress concentration factor K to be determined. The maximum stress is then being K times the average stress. Example 1.1 A steel bar of rectangular cross-section 25 mm x 20 mm carries an axial tensile load of 3 kN. Estimate the average tensile stress in the cross-section. Example 1.2 A steel bolt 25 mm in diameter carries a load of 4 kN in tension. Estimate the tensile stress of the section A and at the screw section B when the diameter of the thread is 20 mm. TREBLA 1.3 Strain L Fig 1-3 An axially loaded bar δ undergoes a P P change in length becoming L+δ longer when in tension and shorter when in compression. The change in length can be denoted by δ (delta), Figure 1-3, for the bar in tension. This elongation is the cumulative result of the stretching of the material throughout the length of the bar. Now assuming that the material is the same everywhere in the bar then, if we consider half of the bar, it will have an elongation equal to δ/2. Similarly, if we consider a unit length of the bar, it will have an elongation of 1/L times the total elongation δ. In this manner we arrive at the concept of elongation per unit length or strain and it is denoted by ε (Epsilon) and is given by the equation: ε = δ/ L. where δ = elongation (change in length) L = original/ initial length ε= strain. Strain is a measure of the deformation produced in the member by the load. Direct stresses produce a change in length in the direction of the stress. The above definition of strain is useful only for small distortions in which the extension δ is small compared to the original length L. It is however, adequate for most engineering problems where values of strain are of order of 0.001. If the bar is in tension the strain is called a tensile strain representing an elongation or stretching of the material. If the bar is in compression the strain is compressive and the bar shortens. Tensile strain is taken as positive and the compressive, negative. The strain ε is call Axial or Normal strain because it is associated with normal stresses. Normal strain is a dimensionless quantity since it is a ratio of two lengths and hence has no units. Example 1.3 A cylindrical block of concrete is 300 mm long and has a circular cross section of 100 mm in diameter. It carries a total compressive load of 67 kN and under this load contracts 0.2 mm. Estimate the compressive stress and the compressive strain. TREBLA Example 1.4 A bar with a rectangular cross section 20 mm x 40 mm and length L = 2.8 m is subjected to an axial tensile force of 70 kN. The measured elongation of the bar δ is 1.2 mm. Calculate the tensile stress and the strain in the bar. 1.4 Tensile Test (Stress –Strain Diagrams) TREBLA TREBLA The mechanical properties of materials used in engineering are determined by tests performed on small specimens of the material. These tests are conducted in material- testing laboratories equipped with testing machines capable of loading the specimens in a variety of ways including static and dynamic loading in tension and compression. Most common material test is the tensile test, in which tensile loads are applied to the ends of a cylindrical specimen. As the specimen is pulled, the load P is measured and recorded either automatically or by reading from the dial gauge. The elongation over the gauge length is measured simultaneously with the load, usually by mechanical gauges or extensometers. The load is applied very slowly. The axial stress in the test specimen is calculated by dividing the load P by the cross-sectional area A. When the initial area of the bar is used in this calculation, the resulting stress is called the Nominal Stress (Conventional or Engineering Stress). More exact value of the axial stress known as the true stress can be calculated by using the actual area of the bar, which can become significantly less than the initial area for some materials. The average axial strain in the bar is found from the measured elongation δ between the gauge marks by dividing δ by the gauge length L. If the initial gauge length is used then the nominal strain is obtained. If the actual is used in calculating the strain, the true strain or natural strain is obtained. After performing tension and the compression test in determining the stress and strain of various magnitudes of a load, the plot of stress versus strain diagram can be made. Such a diagram is a characteristic of the material and conveys important information about the mechanical properties and types of behaviour of the material. The stress – strain diagram for typical mild steel (low carbon steel) in tension is shown in Figure 1-4. Fig 1-4 Stress- strain diagram for typical mild-steel TREBLA Strains are plotted on the horizontal axis and stresses on the vertical axis. The diagram begins with a straight line from O to A. In this region, the stress and strain are directly proportional and the behaviour of the material is said to be linear. Beyond point A, the linear relationship between stress and strain no longer exists, although the material may still be in the elastic state, in the sense that, if the load were removed, the strain also would return to zero. Hence A is the proportional limit and B the elastic limit. With an increase in load beyond the proportional limit, the strain begins to increase more rapidly for each increment in stress. The stress – strain curve has a smaller and smaller slope, until at point B, the curve becomes horizontal. Beginning at this point, considerable elongation occurs with no noticeable increase in tensile force from B to C in the diagram. This phenomenon is known as yielding of the material and the stress at point B is called the Yield Stress. From B to C the material becomes perfectly plastic which means the material can deform without an increase in the applied load. The elongation of the mild steel specimen in a perfectly plastic region is typically 10 – 15 times the elongation that occurs between the onset of loading and the proportional limit. After undergoing the large straining that occurred during yielding in the region BC the steel begins to strain harden. During strain hardening the material undergoes changes in its atomic and crystalline structure, resulting in increase resistant of the material to further deformation. Thus, additional elongation requires an increase in tensile load and the stress – strain diagram has a positive slope from C to D. The load eventually reaches its maximum value and the corresponding stress at point D is called the Ultimate Stress. Further stretching of the bar is actually accompanied by a reduction in the load and fracture finally occurs at a point such as E. Lateral contraction of the specimen occurs when it is stretched resulting in a decrease of the cross sectional area. If the actual cross sectional area of the narrow part of the neck is used to calculate the stress, the true stress –strain curve will follow the dashed line CE′. In the vicinity of the ultimate stress, the reduction in area of one bar becomes visible. A pronounced necking of the bar occurs. Brittle Material is one showing relatively little elongation at fracture in the tensile test i.e. they fail in tension at relatively low values of strain. Examples are: Concrete, Stone, Cast Iron, Ceramic Materials Ductile Material is one that undergoes large strains before failure, thus stretch appreciably before breaking. Examples are: Mild steel, Copper, Aluminium, Magnesium, Manganese, Brass (Zn Cu), Bronze (Zn S) TREBLA Ductility and Brittleness The capacity for being drawn out plastically before breaking is called the ductility of the material and is measured by the following two quantities; - Percentage elongation and percentage reduction in area (contraction). PERCENTAGE ELONGATION is defined as: 𝐿𝑓 −𝐿𝑜 × 100% 𝐿𝑜 Where Lf – Distance between the gauge marks at fracture. Lo – Original gauge length. PERCENTAGE REDUCTION is defined as: 𝐴𝑜 −𝐴𝑓 ×100% 𝐴𝑜 where Af – final area of the fracture section Ao – Original cross-sectional area. A material is generally classified as brittle if the percentage elongation is less than 5 in a gauge length of 50 mm. 1.4.2 Factor of Safety When designing a structure a safety factor has to be taken into account in order to ensure the working stresses keep within safe limits. It has been pointed out that stress is calculated from knowledge of the magnitude and position of application of the load, the dimensions of the member, and the properties of the material. In practice none of these factors is known exactly, and possible errors arise from various sources. In spite of all these approximations and assumptions, a body of theory has been developed which in many cases can be shown to agree with experimental results within a reasonable margin of error, and forms the basis for sound design. TREBLA The maximum permissible stress, or working stress or design stress, is determined from a consideration of the above factors, taking into account the social and economic consequences of failure, and the factor of safety is normally defined as the ratio between the failure stress and the working stress. Factor of safety = Failure stress Working stress To avoid failure, the following conditions must be satisfied. Maximum Induced Stress ≤ Allowable Stress The allowable stress can be obtained easily if the material properties and the desired safety factor are known. For ductile materials subjected to static loading, the allowable stress is often defined as: Yield Stress Allowable stress = Factor of safety For those ductile materials with no well-defined yield stress, the allowable stress is defined as: Proof Stress Allowable stress = Factor of safety For brittle materials it is often defined as: Allowable stress = Ultimate Stress Factor of safety 1.4.3 Relationship between Stress and Strain (Hooke’s Law) TREBLA Within the elastic region, there is a linear relationship between stress and strain. The linear relationship between stress and strain for a bar in simple tension or compression can be expressed as: 𝝈 = 𝑬𝜺 where σ – stress ε – strain E – constant of proportionality known as Modulus of Elasticity or Young’s Modulus for the material The above equation is commonly known as Hooke’s law. Deductions: σ = Eε σ = P/A ε = δ/L E = σ/ε 𝑃𝐿 𝐸= 𝐴𝛿 or 𝛿 = 𝑃𝐿 𝐴𝐸 Also, 𝐹 = 𝑘𝑥 𝐹 𝐾= 𝑋 Implying, P/ δ = k = AE/L The modulus of elasticity is the slope of the stress – strain diagram in the linearly elastic region (within the proportional limit) and its value depends upon the particular material being used. The unit of E is N/m2. Typical Values of E 200 GPa (steel) 70 GPa (aluminium) 11 GPa (wood) TREBLA Example 1.5 A tensile test is carried out on a bar of mild steel of diameter 20mm. The bar yields under a load of 8 kN. It attains a maximum load of 15 kN and breaks finally at a load of 7 kN. Estimate: (i) the tensile stress at the yield point. (ii) the ultimate stress. (iii) the average stress at the breaking point if the diameter of the neck is10 mm. Example 1.6 A concrete cube 150 mm x 150 mm x 150 mm is loaded in a compression-testing machine. If the compressive force acting normal to one face of the cube is 250 kN, calculate the compressive stress in the concrete. 1.5 Poisson’s Ratio (ν) When a bar is loaded in tension, the axial elongation is accompanied by a lateral contraction (normal to the direction of the applied load), Figure 1-6, in which the dashed lines represent the shape before loading and solid lines give the shape after loading. The lateral strain is proportional to the axial strain in the linear elastic range. The ratio of the strain in the lateral direction to the strain in the axial direction is known as Poisson’s ratio, and is denoted by ν (xu). PP Fig. 1-6 Axial elongation and lateral contraction of a bar in tension 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛, 𝑦 𝑉=− 𝐴𝑥𝑖𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛, 𝑥 The minus sign is because when one strain is tensile the other is compressive. 𝑉=− 𝑦 𝑥 or 𝑦= −𝑉 𝑥 TREBLA  or 𝑦= 𝑉𝑥 The lateral strain εy is also given as εy = ∆d/d or ∆t/t where d and t are diameter and thickness respectively of a material. The lateral strain represents a decrease in width (negative strain) and the axial strain represent elongation (positive strain). For compression, the opposite situation occurs with the bar becoming shorter (negative axial strain) and wider (positive lateral strain). Therefore, the Poisson’s ratio has a positive value for most materials. Poisson’s ratio ranges from 0.25 to 0.35 for many metals. 0.3 is commonly used for metals. Concrete has values from 0.1 to 0.2, and cork approximately 0.0. The theoretical maximum value is 0.5. 1.6 Volume Change Because the dimensions of the bar in tension or compression are changed when the load is applied, the volume of the bar also changes. The change in volume can be calculated from the axial and lateral strains. The unit volume change ℮ is defined as: the change in volume, ∆V ∆V e = = Original volume, VO Vo e = σ (1 − 2V) E But ∆v is given as; ∆𝑉 = 𝑉𝑓 − 𝑉𝑜 = 𝜎𝐸 𝑉𝑜(1 − 2𝑉) Hence 𝑒= (1 − 2𝑉) Example 1.7 A bar of circular cross-section is loaded by tensile forces P = 85 kN. The bar has length L = 3.0 m and diameter, D = 30 mm. It is made of aluminum with Modulus of Elasticity E = 70 GPa and TREBLA Poisson’s ratio, ν = 1/3. Calculate the elongation δ, the decrease in diameter ∆d, and the increase in volume ∆V of the bar. Assume the proportional limit of the material is not exceeded. Example 1.8 A steel pipe of length L = 1.2 m, outside diameter d2 = 150 mm, and inside diameter d1 = 110 mm is compressed by and axial force P = 620 kN. The material has modulus of elasticity E = 200 GPa and Poisson’s ratio ν = 0.3. Determine the following quantities for the pipe: (a) the shortening δ (b) the lateral strain (c) the increase ∆d2 in the outer diameter and the increase ∆d1 in the inner diameter (d) the increase ∆t in the wall thickness (e) the increase ∆V in the volume of the material and, (f) the dilation ℮. Problems 1 1.1 Figure 1.1 shows one end of a beam resting upon a bearing pad made of a rubber compound and which is 250 mm square in plan area. If the vertical reaction at the end of the beam is 3000 kN, calculate the compressive stress in the bearing pad. [Ans: – 48.0 N/mm2] 250 mm Fig. 1.1 1.2 A metal bar ABC having two different cross-sectional areas is loaded by an axial force P (Figure 1.2). Parts AB and BC are circular in cross-section with diameters 50 mm and 38 mm, respectively. If normal stress in part AB is 40 MPa, what is the normal stress σBC in part BC? [Ans: 69.3 MPa] TREBLA  Fig. 1.2 1.7 A prismatic bar is loaded in tension by axial forces. Find Poisson’s ratio for the material if the ratio of the unit volume change to the unit change in cross-sectional area is equal to – 2/3. [Ans: 3/10] 1.9 A bar 3 m long is made of two bars, one of copper having E = 105 GN/m2 and the other of steel having E = 210 GN/m2. Each bar is 25 mm broad and 12.5 mm thick. This compound bar is stretched by a load of 50 kN. Find the increase in length of the compound bar and the stress produced in the steel and copper. The length of copper as well as steel is 3 m each. [Ans: 1.52 mm, 106 MPa, 53 MPa] CHAPTER TWO AXIALLY LOADED SPRINGS/ BARS This chapter is devoted to the behaviour of axially loaded members, which are structural elements having straight longitudinal axes and carrying only axial forces (tensile or compressive). Their cross-sections may be solid, hollow or tubular or thin-walled and open. TREBLA 2.1 Displacement of Axially Loaded Members An axially loaded spring is similar to a bar in tension. Consider a spring axially loaded by a force P. P Under the action of the force, P, the spring elongates by an amount δ so that its total length becomes (L+ δ), where L is the original length. The spring constant, K is given as; K = P⁄δ The compliance of the spring is the reciprocal of the spring constant or the deflection produced by a load of unit value and is given as 𝑓 = 𝛿⁄𝑃 The terms stiffness and flexibility are commonly used rather than spring constant and compliance in structural analysis. The stiffness, K of an axially loaded bar below is defined as force required to produce a unit deflection, hence stiffness of the bar, K is; K = P⁄δ = EA L P The flexibility, f is defined as the deflection given to a unit load. Thus, the flexibility of an axially loaded bar is L f= EA A prismatic bar of length L is loaded in tension by axial forces P as shown in Figure 2.1. The uniform stress is given by the equation σ = P/A and the axial strain is ε = δ/ L. Assuming Hooke’s law is being obeyed then σ = Eε. The above expressions are combined to give the following equation for the elongation of the bar: 𝑃𝐿 𝛿= TREBLA  𝐸𝐴 L δ PP Fig. 2.1 Axially loaded bar in tension Thus, when a prismatic bar of linearly elastic material is loaded only at the ends, we can obtain its change in length from the above equation. Suppose, that a prismatic bar is loaded by one or more axial loads acting at intermediate points along the axis of the bar Figure 2.2 and 2.3. A P1 a P1 A B a b C P2 B P2 c D P3 b C Fig. 2.2, 2.3 Bar with intermediate axial loads/ Bar consisting of prismatic bars We can determine the axial force in each part of the bar (that is in part AB, BC, and CD, say, by statics). We then calculate the elongation or shortening of each part separately. Finally, the changes in lengths can be added algebraically to obtain the total elongation. This method can be used if the bar consists of several prismatic segments, each having different axial forces, different dimensions, and different materials. The change in length may be obtained from the equation TREBLA In which the subscript i is a numbering index for the various segments of the bar. Note especially that Pi is not an external load but is the internal axial force in segment i. If either the axial force or cross-sectional area varies continuously along the axis of the bar, the above equation is not suitable. Instead the elongation can be found by considering a differential element of the bar, obtaining an expression for its elongation and then integrating over the length of the bar. Consider a tapered bar shown in Figure 2.4(a). The bar is subjected to a continuously distributed load, as shown by the arrows in the Figure. A B A C B P L x dx P Fig. 2.4a Fig. 2.4b A continuous axial load of this kind may be produced by centrifugal forces, magnetic forces or the weight of a bar hanging vertically. Now consider any intermediate cross-section C, at a distance x from the left- hand end (Figure 2.4b). The internal axial force P(x) acting at this section may be determined from static equilibrium using part of the bar AC or CB as a free body. Thus the force P(x) is known in terms of x. Also the cross-sectional area A(x) at section C may be expressed as a function of x. Next we consider an element of the bar of length dx at section C, the elongation dδ of this element may be obtained as follows; The elongation of the entire bar is obtained by integrating over the length L as; This same procedure is applied to a prismatic bar hanging vertically under its own weight, Figure 2.5. TREBLA  A x dx L B Fig. 2.5 A bar hanging vertically under its own weight The elongation of the entire bar AB is given as; Example 2.1 A steel bar 2.5 m long has circular cross-section of diameter d1 = 20 mm over one-half of its length and diameter d2 = 13 mm over the other half. (a) How much will the bar elongate under a tensile load P = 22 kN? (b) If the same volume of the material is made into a bar of constant diameter d and length 2.5 m, what will be the elongation under the same load P? (Assume E = 210 GPa). d1 = 20 mm d2 = 13 mm P P = 22 kN 1.25 m 1.25 m Example 2.2 A steel bar AD (see figure) has a cross-sectional area of 260 mm2 and is loaded by forces P1 = 12 kN, P2 = 8 kN, and P3 = 6 kN. The lengths of the segments of the bar are a = 1.5 m, b = 0.6 m, and c = 0.9 m. (a) Assuming that the modulus of elasticity E = 210 GPa, calculate the change in length δ of the bar. Does the bar elongate or shorten? (b) By what amount P should the load P3 be increased so that end D of the bar does not move when the loads are applied? TREBLA  P1 P2 P3 A B C D a b c 2.2 Statically Indeterminate Structures/ Bars In the preceding section we deal with axially loaded bars and other simple structures that could be analysed by static equilibrium. Such structures are classified as statically determinate. For many structures, however, the equations of static equilibrium alone are not sufficient for the calculation of axial forces and reactions in the members; these structures are called statically indeterminate. Structures of this type can be analysed by supplementing the equilibrium equations with additional equations pertaining to the displacement of the structure. Consider a prismatic bar AB of cross-sectional area A attached to rigid supports at both ends and axially loaded by a force P at an intermediate point C (Figure 2.6a). From the free-body diagram Figure 2.6b, the reactions RA and RB cannot be found by statics alone, because only one equation of equilibrium is available. RA A A P a C C P L C b B RB B Fig. 2.6a Fig. 2.6b ∑Fvert = 0 RA + RB − P = 0 --------------------- (1) We need an additional equation to be able to solve for the two unknowns. This equation is obtained based on the observation that a bar with both ends fixed does not change in length. That is δAB = 0. --------------------------- (2) This also means that the total change in length of the bar caused by the three forces is zero. This equation is called Compatibility Equation expressing the fact that the change in length of the bar must be compatible with the conditions at the supports. TREBLA Considering Figure 2.6b, the changes in lengths of the upper and lower segments of the bar are respectively, δAC = REAAa and δCB = −REABb where the minus sign indicates a shortening of the bar. Equation (2) now becomes: RAa RBb δAB = δAC + δCB = − = 0 EA EA Or RAa RBb − = 0… … … (3) EA EA We now solve simultaneously equations (1) and (3) to get; 𝑅𝐴 = 𝑃𝑏 𝐿 and 𝑅𝐵 = 𝑃𝑎 𝐿 With the reactions known, all other forces and displacement equations can be found. Stresses in the two segments of the bar can also be found directly from the internal axial forces. Note: The analysis of structurally indeterminate structures involves setting up and solving equations of equilibrium and equations of compatibility. Example 2.3 The axially loaded bar ABCD shown in the figure is held between rigid supports. The bar has crosssectional area Ao from A to C and 2Ao from C to D. (a) Obtain formulars for the reactions RA and RD at the ends of the bar. (b) Determine the displacement δB and δC at points B and C respectively. TREBLA  Ao 2Ao P A B C D L/4 L/4 L/2 2.3 Strain Energy The strain energy of an axially loaded bar of a linearly elastic material is given as P2L U= ---------------------- (1) 2EA For linearly elastic spring the strain energy is obtained by replacing the stiffness EA/L of the prismatic bar by the stiffness k of the spring, and is given as; 𝑃2 𝑈= 2𝐾 The total strain energy U of a bar consisting of several segments is equal to the sum of strain energies of the individual segments. Considering Figure 2.7, the total strain energy is the strain energy of segment AB plus the strain energy of segment BC. A P1 B C P2 Fig. 2.7 Bar consisting of prismatic segments having different cross-sectional areas and different axial forces TREBLA Therefore for a bar of any prismatic segment and constant internal axial force within each segment, and made up of different materials, the total strain energy is given as; n Pi2Li ∑ 2EiAi i=1 in which Pi is the axial force acting in segment i and Li, Ei and Ai are properties of segment i. For a non-prismatic bar – Figure 2.8 with continuously varying axial force, equation (1) is applied to a differential element shown in the figure and integrating along the length of the bar gives the strain energy of the bar as; 𝐿 𝑃(𝑥)2𝑑𝑥 𝑈=∫ 0 2𝐸𝐴(𝑥) P(x) and A(x) are the axial force and cross-sectional area at distance x from the end of the bar. A dx L B P Fig. 2.8 Non-prismatic bar with varying axial force The same procedure is applied to a bar hanging vertically under its own weight. For such a bar – Figure 2.9, the strain energy is also given as; 𝐿 𝑃(𝑥)2 𝑈 ) 2 L P(x) dx TREBLA  x dx L Fig. 2.9 Bar hanging vertically under its own weight Example 2.4 The bar ABC shown in the figure is loaded by a force P acting at end C and by a force Q acting at the midpoint B. The bar has constant axial rigidity EA. (a) Determine the strain energy U1 of the bar when the force P acts alone (Q = 0). (b) Determine the strain energy U2 when the force Q acts alone (P = 0). (c) Determine the strain energy U when the forces P and Q act simultaneously upon the bar. Example 2.5 Determine the strain energy stored in a bar suspended from one end due to its own weight. Assume linear elastic behaviour. 2.4 Pre-Stress or Strain (Initial Stress) Sometimes, it happens that before any load is applied to any part of a machine or structure, it is already in state of stress. For example, the bolts holding down the heads of the cylinder of a steam TREBLA engine are put into tension by tightening up the units. The same applied to the bolt in a flanged coupling of a steam pipe, or the big end bolts of the connecting rod. In some cases, relative rigidity of the bodies has to taken into account while in others it is not taken into account. For example in the case of a cylinder head, there will be little deformation, under the pressure, of nuts compared with the deformation of the bolts, so that the head can be treated as rigid and the stress in the bolt when there is steam in the cylinder will be the same as the initial stress before steam come to the cylinder. On the other hand, in dealing with flanged joints, the elasticity of the packing must be considered. Any tension applied to the joints will be taken up partly by extra tension in the bolt and partly by reduced compression in the packing. Problems 2 2.1 A high strength steel rod with diameter d = 25 mm and modulus of elasticity E = 200 GPa must transmit a tensile load P = 134 kN. a) If the length L of the rod is 1 m, what is its final length? b) What is the ratio of the length of the rod to the increase in length? c) If the increase in length is limited to 3.8 mm, what is the maximum load P max that can be permitted? d P P L [1.00136m, 735, 373.4] 2.2 A prismatic bar AD is subjected to loads P1, P2, and P3 acting at points B, C, and D, respectively, as shown in the figure. Each segment of the bar is 500 mm long. The bar has cross-sectional area A = 900 mm2 and is made of copper with E = 120 GPa. a) Determine the displacement δD at the free end of the bar. b) What should be the load P3 if it is desired to reduce the displacement at end D to half of its original value? TREBLA  [0.755 mm, 22.8 kN] 2.3 A bar ACB having two different cross-sectional areas A1 and A2 is held between rigid supports at A and B. A load P acts at point C, which is distance b1 from end A and b2 from end B. a) Obtain formulars for the reactions RA and RB at supports A and B, respectively due to load P. b) Obtain a formular for the downward displacement δC of point C. AP b1 A1 C b2 A2 B 2.4 A bar AD of length L, cross-sectional area A, and modulus of elasticity E is subjected to loads 5P, 3P, and P acting at points B, C, and D, respectively. Segments AB, BC, and CD have lengths L/6, L/2, and L/3, respectively. a) Obtain a formular for the strain energy U of the bar. b) Calculate the strain energy if P = 21 kN, L = 1.2 m, A = 1,780 mm2, and the material is aluminium with E = 72 GPa. 23P2L [U = , 8 kJ] 12EA TREBLA  CHAPTER 3 TEMPERATURE AND SHEAR STRESSES 3.1 Temperature Stresses External loads are not the only sources of stresses and strains in a structure. When the temperature of a body is raised or lowered the material expands or contracts. If the expansion and contraction is wholly or partially restricted stresses and strains are set up in the structure. Consider a long bar AB of a material – Figure 3.1 at a temperature θo. The bar is now subjected to an increase ∆θ in temperature. A B TREBLA  Fig. 3.1 A bar of material subjected to an increase in temperature As the bar is unrestrained, when it is heated, every element of the material will undergo thermal strains in all directions resulting in increase in the dimensions of the bar. If A is taken as the reference point and side AB maintained its original position, the block will have the shape shown by the dashed lines. For most homogeneous structural materials if the temperature changes the uniform thermal strains εt undergone by the materials is seen to be proportional to the temperature change ∆θ. The strain is thus give as εt = α (∆θ) ---------------------- (1) α is a property of the material known as the coefficient of thermal expansion or coefficient of linear expansivity. α has unit of 1/K or 1/ oC. Conventionally the thermal strain resulting from expansion is taken to be positive and that resulting from contraction is taken to be negative. If the expansion of the bar is prevented, it is as if the bar is compressed to its original shape and dimensions above. The compressive strain is given by equation 1 above. The corresponding stress is; σ = Eε = α E (∆θ) By a similar argument the tensile stress set up in a constrained bar subjected to a temperature difference ∆θ is given as σ = α E (∆θ) It is assumed that the material remains elastic. 3.1.1 Change in Length of Unrestrained Bar Consider a prismatic bar of length L (Figure 3.2) subjected to a temperature difference ∆θ. δt is the elongation of the bar due to the temperature change ∆θ. ∆θ L δt Fig. 3.2 Increase in length of a prismatic bar subjected to temperature change ∆θ. TREBLA If the bar is homogeneous and isotropic and the temperature increase is uniform throughout the bar, the increase in length δt is given by δt = εtL = α (∆θ) L Thus the change in any dimensions of the bar can be obtained by multiplying the original dimensions by the thermal strain. For a bar of length L subjected to a temperature increase ∆θ, the increase in length is δt and therefore the final length Lf becomes, Lf = L + δt = L + α (∆θ) L = L [1 + α (∆θ)] Example 3.1 A steel bar of length 200 mm is at a temperature of 10 oC. If the material properties are; E = 210 GPa and α = 12 x 10-6/K. Find: i The thermal strain induced in the bar ii The new length of the bar when it is heated to 23 oC. Example 3.2 A 150 mm diameter steam pipe is laid in a trench at a temperature of 13 oC. When steam passes through the pipe, its temperature rises to 120oC; a) What is the increase ∆d in the diameter of the pipe if the pipe is free to expand in all directions? b) (b) What is the axial stress σ in the pipe if the trench restrains the pipe so that it lengthens only one-third as much as it would if it could expand freely? (Note: The pipe is made of steel with modulus of elasticity E = 200 GPa and coefficient of thermal expansion α = 12 x 10-6 /oC). 3.1.2 Statically Indeterminate Bars In the previous discussion the bars and structures considered are free to expand or contract. There are, however, certain structures that can be restrained and depending on their character and nature of the temperature changes may or may not develop temperature stresses. The analysis of statically indeterminate structures that develop temperature stresses is analysed by establishing equilibrium equations, compatibility equations- temperature displacement relations together with force- displacement relations. TREBLA Consider a prismatic bar AB (Figure 3.3a) of linear elastic material, rigidly constrained at the two ends and subjected to a temperature difference of ∆θ. Figure 3.3b is a free-body diagram of Figure 3.3a and 3.3c, 3.3d are the increase and the decrease in the length of the bar due to temperature change ∆θ and external load RA respectively. As the temperature is raised the bar tends to elongate but is restrained by the supports A and B. Consequently reactions RA and RB are developed at the supports and the bar is subjected to compressive stresses. Considering Figure 3.3b, the equilibrium equation is determined as; ΣFvert = 0, RA – RB = 0 --------------- (1) The compatibility equation is given as; δAB = 0 In Figure 3.3c, when only the temperature change is acting, the bar elongates by an amount δt. In Figure 3.3d when only the reaction RA acts the bar shortens by an amount δRA. The compatibility equation then becomes 𝛿𝐴𝐵 = 𝛿𝑡 − 𝛿𝑅𝐴 = 0 𝑅𝐴𝐿 = 𝛼(∆𝜃)𝐿 − = 0… … …. (2) 𝐸𝐴 Solving equations (1) and (2) gives TREBLA  𝑅𝐴 = 𝑅𝐵 = 𝐸𝐴𝛼(∆𝜃) From which the thermal stresses σt can be obtained. I.e. σt = RAA = RBB = Eα(∆θ) Example 3.3 A composite bar made up of aluminum bar and steel bar, is firmly held between two unyielding supports as shown in figure below. An axial load of 200 kN is applied at B at 47 0C. Find the stress in each material, when the temperature is 97 0C. Take Ea = 70 GPa; Es = 210 GPa; αa = 24 x 10-6/ 0C and αs =12 x 10-6/ 0C. 3.1.3 Creep of Materials under Sustained Stresses At ordinary temperatures, most metals will sustain stresses below the limit of proportionality for long periods without showing additional measurable strains. At these temperatures metals deform continuously when stressed above the elastic range. This process of continuous inelastic strain is called creep. At high temperatures metals lose some of their elastic properties and creep under constant stress takes place more rapidly. Strain Primary Secondary Tertiary c εo a b Fracture Instantaneously Plastic TREBLA  Time Fig. 3.4 Creep Curve for a Material in the Inelastic Range When a tensile specimen of a metal is tested at a high temperature under a constant load, the strain assumes instantaneously some values εo; if the initial strain is in the inelastic range of the material then creep takes place under constant stress. At first, the creep rate is fairly rapid but diminishes until a point (b) is reached on the strain –time curve. The point ‘b’ is a point of inflexion in this curve, and continued application of the load increases the creep rate until fracture of the specimen occurred at ‘c’. 3.1.4 Fatigue under Repeated Stresses When a material is subjected to repeated tensile stresses within the elastic range, it is found that the material tires and fractures rather suddenly after a large but finite number of repetitions of stress; the material is said to fatigue. σ Endurance Limit 50 Medium strength 40 Titanium alloy 30 Mild Steel 20 Aluminium light alloy 10 104 105 106 107 108 109 No. of cycles to failure Fig. 3.5 Forms of the stress-endurance curve for steel, aluminium light alloy and titanium alloy. The endurance to fatigue at a given stress level is the number of complete cycles of loading to that stress level required to bring about fracture of the materials. Failure of the material after a large number of cycles of tensile stress occurs with little or no permanent set. Fractures show the characteristics of brittle materials. Fatigue is primarily a problem of repeated tensile stresses; this is due, probably, to the fact that microscopic cracks in a material can propagate more easily when TREBLA the material is stressed in tension. In the case of steels, it is found that there is a critical stress called the endurance limit below which fluctuating stresses cannot cause a fatigue failure. Titanium alloys show a similar phenomenon. No such endurance limit has been found for other non-ferrous metals and other materials. 3.2 Shearing Stresses A part from the direct stresses (tensile and compressive stresses), and thermal stresses, there is another type of stress which plays a vital role in the behaviour of the materials and structures, especially metals. This stress acts parallel or tangential to a surface. Consider a thin P block of a material, which A B is glued to a table. Suppose a Block of material thin plate is now glued to the upper surface of the block (Figure 3.6). Thin plate Fig. 3.6 A block glued to a table on top of which is glued a thin plate If a horizontal force P is applied to the plate, the plate will tend to slide along the top of the block of the material, and the block itself will tend to slide along the table. So that, provided the glued surfaces remain intact, the table resists the sliding of the block, and the block resists the sliding of the plate on its upper surface. This is equivalent to applying two equal and opposite parallel forces not in the same line to two parts of a structure. There is a tendency for one part of the body to slide over or shear from the other part across any section. Assuming the block is to be divided by any imaginary horizontal plane such as AB, the part of the block above this plane will be trying to slide over the part below the plane. The material on each side of this plane is said to be subjected to a shearing action, the stresses arising from these actions are called Shearing Stresses. Shear stress is denoted by the symbol τ (thaw), and shearing stress on any surface is defined as the intensity of shearing force tangential to the surface. The average shear stress is obtained by dividing the total shear force V by the area A over which it acts. 𝑉 TREBLA  𝜏= 𝐴 In many cases, the shear force is not distributed uniformly over the section; if δV is the shear force on any elemental area δA of a section, the shear stress on that elemental area is; Shear stress arises in many practical problems. For example, Figure 3.7 (a) shows two flat plates held together by a single rivet and carries a tensile force P. P a b P Fig. 3.7 (a) A rivet in a single shear We imagine the rivet divided into two portions by the plane ab, then the upper half of the rivet is tending to slide over the lower half, and a shearing is set up in the plane ab. The shearing takes place at one cross-section of the rivet. The rivet is therefore said to be in a single shear. Figure 3.7 (b) shows three flat plates held together by a single rivet and carry a tensile force P. a b c d P/2 P P/2 Fig. 3.7 (b) A rivet in double shear In the above diagram, shearing takes place over two cross-sections of the rivet. That is in plane ab and in plane cd. The rivet is therefore said to be in double shear. TREBLA Example 3.3 A punch of diameter 19 mm is used to punch a hole in a 6 mm steel plate. A force P= 116 kN is required. What are the average shear stress in the plate and the average compressive stress in the punch? Example 3.4 Three steel plates are held together by a (sixteen) 16 mm diameter rivet. If the load transmitted is 5 kN, estimate the shear stress in the rivet. a b 2.5 kN c d 5 kN 2.5 kN 3.2.1 Complementary Shearing Stress Consider now the equilibrium of one of the elementary blocks of a rectangular block in a state of shear – Figure 3.8. τ yx τxy b τ xy τ yx a Fig. 3.8 An elementary block in a state of shear Let τyx be the shear stress on the horizontal face of the element and τxy, be the shear stress on the vertical faces of the element. Suppose ‘a’ is the length of the element and ‘b’ its height and that it has unit breath. The total shear force on the upper and lower faces is then aτyx while the total shear forces on the end faces bτxy, so that for rotational equilibrium; 𝑎𝜏𝑦𝑥𝑏 = 𝑏𝜏𝑦𝑥𝑎 TREBLA  or 𝜏𝑦𝑥 = 𝜏𝑥𝑦 This implies that whenever there is a shearing stress over a plane passing through a given line, there must be an equal complementary shearing stress on a plane perpendicular to the given plane and passing through the given line. The direction of the two shearing stresses must be either both towards or both away from the line of intersection of the two planes in which they act. NOTE: In direct stress there is a change in volume but in shear there is no change in volume. An element subjected to shear stresses only is said to be in pure shear. 3.3 Shear Strain τ γ τ Consider a rectangular block of material, subjected to shear stresses τ in one plane. The shearing stresses distort the rectangular face of the block into a parallelogram as shown above. If the right angles at the corners of the faces change by amounts γ, then γ is the shear strain. The shear strain γ can be defined as the change in the right angle. S I G N C O N VE N T I O N The shear stress acting on a positive face of an element is positive if it acts in the direction of one of the coordinate axes and negative if it acts in the negative direction of the axis. The shear stress acting on a negative face of an element is positive if it acts in the negative direction of an axis, and negative if it acts in the positive direction of an axis. TREBLA  Positive shear stress Negative shear stress The sign convention for shear strain is related to that of shear stresses. Shear strain in an element is positive when the angle between two positive or (two negative) faces is reduced. The strain is negative when the angle between two positive or (two negative) faces is increased. Positive shear strain Negative shear strain NOTE: y A face is said to be positive if it has its – ve face +ve face outward normal directed in the x positive – ve face direction of a coordinate axis. The opposite faces are negative faces. +ve face 3.4 Relationship between Shear Stress and Shear Strain The τ versus γ diagram is similar in shape to the σ versus ε diagram. For many materials the initial part of the shear stress–strain diagram is a straight line. Within the linear elastic region, the shear stress and shear strain are directly proportional. That is; τ=Gγ TREBLA where G is the shearing modulus of elasticity (Modulus of Rigidity). G is the ratio of the shear stress to the shear strain, and it is also the slope of the linear portion of the shear stress- strain diagram. 3.5 Shear Strain Energy For a member in pure shear the shear strain energy is given as; U = 1 2 × Volume of the material Gγ 2 The shear strain per unit volume also known as the Shear Resilience is given as; 𝑈= 𝐺𝛾2 Problems 3 3.1 Find the minimum size of a hole that can be punched in a 20 mm thick mild steel plate having ultimate shear strength of 300 N/mm2. The maximum permissible compressive stress in the punch material is 1200 N/mm2. [20 mm] 3.2 A block of wood is tested in direct shear using the loading frame and test specimen shown in the figure. The load P produces shear in the specimen along plane AB. The height h of plane AB is 50 mm and its width (perpendicular to the plane of the drawing) is 100 mm. If the load P = 16 kN, what is the average shear stress τaver in the wood? TREBLA  [3.2 MPa] 3.3 A plastic bar ACB having two different solid circular cross sections is held between rigid supports as shown in Figure Q3.4. The diameters in the left- and right- hand parts are 50 mm and 76 mm respectively. The corresponding lengths are 230 mm and 300 mm. Also, the modulus of elasticity E is 6 GPa, and the coefficient of thermal expansion α is 100 x 10-6 /oC. The bar is subjected to a uniform temperature increase of 50 oC. Calculate the following quantities: (a) The compressive force P in the bar; (b) The maximum compressive stress σc; and (c) The displacement δc of the point C. 50 mm 76 mm A C B 230 mm 300 mm Fig. Q3.4 [86.885 kN; 44.26 MPa; – 0.544 mm (to the left)] TREBLA  CHAPTER 4 TORSION 4.1 Fundamental Concepts This refers to the twisting of structural members when it is loaded by couples that produce a rotation about its longitudinal axis. We are concerned with circular bars subjected to torsion. For example, axles drive shafts in machines, propeller shafts, steering rods etc. P1 P2 T1 T2 P1 P2 TREBLA Consider a bar of circular cross- section twisted by a torque T (Figure below). From consideration of symmetry it can be shown that cross-sections of the bar rotate as rigid bodies about the longitudinal axis with radii remaining straight with cross-section remaining plane and circular. γ T B A θ radius r L A longitudinal line on the surface of the shaft will rotate through a small angle γ while a radius at the end of the face of the shaft turns through an angle θ (Angle of Twist). Hence Lγ=rθ γ is the shear strain experienced by the shaft. Thus; The same relationship is obtained for surfaces at a radial distance r, which is internal in the shaft. The maximum values of the shear stress and shear strain for a circular shaft will thus occur at surface of the shaft. For linearly elastic material; Shear stress varies linearly with the distance from the centre of the bar. This is a consequence of Hooke’s law. If stress- strain relation is non-linear other methods of analysis are needed. 4.1.1 Relationship between Torque and Angle of Twist Consider an element in the cross-section of radius r and area dA shown below. If at this radius the shear stress is τ, then the shearing force = τdA. TREBLA  τ r τ dr D The moment of this force about the axis of the bar or shaft axis is dM = τdAr substituting for τ from (i), Total moment for the shaft is the torque and is the sum of the moments of shearing forces on all areas of the ring segments that made up the shaft section. But From equations (i) and (ii); Variation of Shear Stress τmax TREBLA  r τmax This equation is known as torsional formular. It shows that the shear stress is proportional to the applied torque and inversely proportional to the polar moment of inertia J. Example For a bar of solid circular cross section For a circular tube or a hollow shaft The quantity J/r is known as polar sectional modulus. TREBLA 4.1.2 Angle of Twist The angle of twist of a bar of linearly elastic material is written as; Example 4.1 A solid shaft has diameter 20 mm, length 1 m and shear modulus of elasticity G = 70 GPa. The bar is subjected to torques T acting at the ends. (a) If the torque T = 300 Nm, what is the maximum shear stress in the bar? What is the angle of twist between the ends? (b) If the allowable shear stress is 50 MPa and the allowable angle of twist is 2.2o, what is the maximum permissible torque? 4.2 Non- Uniform Torsion In the previous section we dealt with bars in pure torsion. Pure torsion refers to torsion of a prismatic bar subjected to torques acting only at the ends. For non-uniform torsion the bars need not be prismatic and the applied torques may act anywhere along the axis of the bar. 1. Bar Consisting of Prismatic Segments with Constant Torque throughout the Segment TREBLA The bar shown has two different diameters and loaded by torques acting at A, B, C and D. There are three segments AB, BC and CD. We determine the magnitude and direction of the internal torque in each segment by inspection or by cutting sections through the bar, drawing free-body diagrams and solving equations of equilibrium. When finding the shear stress in each segment, we need only the magnitudes of the torque but not the sign. That is the direction of the stresses is not important. However, for the angle of twist for the entire bar we need to know the direction of twist in each segment so that we can combine correctly. SIGN CONVENTION An internal torque is positive if its vector points away from the cut surface/section and negative when it points towards the section. The maximum shear stress in each segment of the bar is obtained from the torsional formula, 𝜏𝑚𝑎𝑥 = 𝑇𝑟 𝐽. The maximum stress in the entire bar is the largest stress from among the J stresses calculated for each of the segments. The angle of twist in each segment is found from 𝜃 = 𝑇𝐿 , The total at one end of the bar with GJ 𝐺𝐽 respect to the other end is obtained by summation. θ = θ1 + θ2 + θ3 + ---------- + θn Ti is the internal torque found from equilibrium. TREBLA 2. Bar with Varying Cross- section and Constant Torque T T When the A B x dx torque is constant, the L maximum shear stress in a solid bar always occurs at the cross- section having the smallest diameter. That is 𝜏𝑚𝑎𝑥 = 16𝜋 𝑑𝑇3. So, applying the equation to point B, we can find the maximum shear stress in the bar. For angle of twist, we consider a differential element of length dx at distance x from one end of the bar. The differential angle of rotation for this element is; J(x) is the polar moment of inertia of the cross-section distance x from the end. 3. Bar with continuously varying Cross- section and continuously Varying Load t TA B T A B x dx L TREBLA There is distributed torque of intensity t per unit distance along the axis of the bar. Internal torque therefore varies along the axis. Knowing the internal torque T(x) and polar moment of inertia J(x) as functions of x, we can use the torsional formular to determine how the shear stress varies along the axis of the bar. The cross-section of the maximum shear stress can then be identified, and the maximum shear stress can be determined. The angle of twist of the bar is given as; 4.3 Transmission of Power by Circular shafts/ Solid Circular Bars Shafts are used to transmit power from one device to another. The amount of power transmitted depends on magnitude of the torque and speed of rotation. The power P is given as, P = Tω P is measured in Watts (W) and ω is measured in rad/s. The commonly used speed is n measured in number of revolutions per minute (rpm). ω is therefore written as; Example 4.2 A steel shaft 35 mm in diameter and 1.2 m long held rigidly at one end has a hand wheel 500 mm diameter keyed to the other end. The G of the material is 80GPa. (i) What load applied tangential to the rim of the wheel will produce a torsional shear of 60 MPa? (ii) How many degrees will the wheel turn when this load is applied? TREBLA Example 4.3 A shaft is transmitting 97.5 kW at 180 rpm. The allowable shear stress in the material is 60 MPa, find the suitable diameter for the shaft. The shaft is not to twist more than 1o in a length of 3 metres. Take G = 80 GPa. 4.4 Strain Energy A B L For a bar in pure torsion under the action of torque T, the bar will twist and the free end rotates through angle θ, given as; The work done by the torque is equal to the area under the torque deflection curve. Torque T T dθ Angle of Twist θ TREBLA  Non- uniform Torsion 1. For a prismatic bar/ prismatic segments with constant torque in each segment; Ui = strain energy of segment i, implying U = summation of the strain energies in each segment Ti is the internal torque in segment i, Li, Gi and Ji are the properties of the ith segment. TREBLA  2. For a bar of varying internal torque or varying cross-section, we determine the strain energy of an element and then integrate along the axis to find that of the entire bar. Strain energy density u in pure shear is given as; Example 4.4 A solid circular bar AB of length L is fixed at one end and free at the other. Three loading different conditions are to be considered. (a) Torque Ta acting at the free end, (b) Torque Tb acting at the midpoint of the bar, and (c) Torque Ta and Tb acting simultaneously. For each case of loading obtain a formular for the strain energy stored in the bar. Then evaluate the strain energy for the following data: Ta = 100 Nm, Tb = 150 Nm, L = 1.6 m, G = 80 GPa and J = 79.52 x 103 mm4. A C B Tb Ta L/2 L/2 Problems 5 4.1 (a) Determine the diameter of a solid steel shaft which can transmit 45 kW at 400 rev/ min if the maximum shear stress is not to exceed 45 N/mm2. TREBLA  (b) If the weight of the shaft is reduced by 50 % by drilling an axial hole through the centre, what will be the maximum stress if the power and speed remains unchanged? [49.5 mm, 60.8 MN/ m2] 4.2 (a) Determine the maximum allowable power which can be transmitted by 150 mm diameter shaft running at 240 rev/min when the permissible shear stress is 50 N/ mm2. (b) The shaft shown has a coupling on it which has 6 bolts on a 260 mm diameter pitch circle. Determine the diameter of the bolts if the maximum shear stress in the bolts must not exceed 100 N/ mm2, and the power transmitted is 314 kW at 500 rev/ min. D = 150 mm [833 kW; 10 mm] 4.3 Shaft AB has a 30-mm diameter and is made of a steel with an allowable shearing stress of 90 MPa, while shaft BC has a 50-mm diameter and is made of an aluminum alloy with an allowable shearing stress of 60 MPa. Neglecting the effects of stress concentration, determine the largest torque T that can be applied at A. TREBLA  [T = 477 N.m] 4.4 The design specifications of a 1.2-m-long solid shaft require that the angle of twist of the shaft not exceed 4o when a torque of 750 N.m is applied. Determine the required diameter of the shaft, knowing that the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77.2 GPa. [d = 36.1 mm] TREBLA  CHAPTER FIVE BEAMS, SHEAR FORCE AND BENDING MOMENTS 5.1 Beams A beam is a structural member that is designed to resist forces acting transverse to its axis. Types of Beams 1. Simple Beam A beam which has a pin support at one end and a roller support at the other is called a simple supported beam. 2. Cantilever (Built-in, fixed end) A beam which is completely supported at one end by being framed in a solid wall (built – in) and free at the other end. TREBLA  P P M Reactions at supports RH A B A B RV 3. Overhang Beam (Overhanging beam) If one or both ends of a beam projects beyond its supports and is free at its ends it is called an overhang beam. 5.1.1 Statically Determinate and Indeterminate Beams A beam is said to be statically determinate if all the support reactions can be determined by the laws of static equilibrium. That is; If all the support reactions cannot be determined by the laws of static equilibrium, then the beam is said to be indeterminate. A cantilever beam supported at its unrestrained end is called a supported cantilever TREBLA  A simple beam extended continuously over more than two supports is called a continuous beam A supported cantilever and a continuous beam are statically indeterminate. 5.1.2 Loads on Beams Loads acting on beams may be concentrated, distributed or couple. (a) Concentrated/ Point Load It is one which is supported on an area of a material relatively small that it may for convenient calculation be assumed to be a point. P1 P2 P3 (b) Distributed Load Distributed loads act over a distance. Such loads are measured by their intensity which is expressed in units of force per unit distance, N/m along the axis of the beam. Distributed loads may be uniform or non-uniform. (i) Uniformly distributed loads (udl) have constant intensity q. q N/m q N/m TREBLA  (ii) Non- uniformly distributed loads are varying loads that have intensity q that changes with distance along the axis. (c) Couple or Moment P M P b P D a a M = Pb ≡ 5.2 Shear Force and Bending Moment When a beam is loaded, internal stresses or strain are created. To determine these stresses and strains, the internal forces or couples that act across a cross- section on the beam must be found. 5.2.1 Shear Force TREBLA Consider a simply supported beam shown. The algebraic sum of all the forces including reactions is zero and the algebraic sum of moments of all the forces about any point x- x is zero. If, however only the forces to any side of the section at x- x are considered, the algebraic sum will not in general be zero. x W1 W2 W3 R1 x R2 A shearing force at a section of the beam such as x- x is the algebraic sum of the forces on either side of that section. The sum of the forces to the left of the section must be numerically equal but opposite in direction to the sum of those to the right. SIGN CONVENTION CORRESPONDING DEFORMATION A positive shear stress tends to deform element by causing the right hand phase to move downward with respect to the left hand phase. 5.2.2 Bending Moment The bending moment at a section x- x is the algebraic sum of the moments of the forces on either side of the section. Since the moment of the forces on the whole beam about x- x must be zero, it implies that moment of those forces to the left of x- x must be numerically equal but opposite to the moment of those on the right of x- x. TREBLA SIGN CONVENTION CORRESPONDING DEFORMATION A positive moment elongates the lower part of the beam and compresses the upper part, and vise- versa. 5.2.3 Relationship between Bending Moment, Shear Force and Load Intensity At any section, distance x from some datum on a beam, TREBLA q= intensity of loading V= shear force M = bending moment Bending moment is maximum or minimum if 𝑑𝑀 = 0 it follows that as 𝑑𝑀 = 𝑉, then the 𝑑𝑥 𝑑𝑥 bending moment is maximum or minimum if V = 0. 5.3 Shear Force and Bending Moment Diagrams A shear force diagram is one which shows the variation of shear force along the length of the beam. A bending moment diagram is one which shows the variation of bending moments along the length of the beam. These variations provide necessary information for the design analysis of the beam. In particular the maximum value of the bending moment is usually the primary consideration in the design and selection of beams. 5.3.1 Drawing of Shear Force and Bending Moment Diagrams STEPS 1. Establish the values of all external reactions on the beam 2. Isolate a portion of the beam either to the left or to the right of any transverse section and apply the equations of equilibrium. 3. Get expressions for V and M. 4. Draw the shear force and bending moment diagrams. NOTE: In drawing bending moment diagrams, it is advisable to always sketch the deflected shape before any calculations and using this as a guide to the corrected shape of the bending moment diagrams. Always draw the bending moment diagram on the side of the beam which is in tension. Plotting this way enables the values of shearing force (SF) and bending moments (BM) at any point in the beam to be rapidly determined by vertical projections. In drawing bending moment diagram, one can work from left to right and or from right to left but for shearing force diagram it is better to work symmetrically from left to right. TREBLA 6.3.2 Point of Contra flexure If the direction of curvature of a beam changes at a point, the point is called point of contra flexure or point of inflexion. This coincides with the point on the BM diagram where the bending moment changes sign from positive bending (sagging) to negative bending (hogging) and vise-versa. Bending moment at the point of contra flexure is zero. Therefore, to locate the point of contra flexure, we write down the equation for the bending moment at a point to the left/ right of the point of contra flexure within the appropriate interval. The bending moment equation is then equated to zero and solving for x, specifying this distance with respect to the left or the right end of the beam. 6.3.3 Principle of Superposition This principle helps in more rapid construction of the SF and BM diagrams when the beam is subjected to complex loading. It states that: If a structure is made of linear elastic material and is loaded by a combination of loads which do not strain the structure beyond the linear elastic range, then the resulting shearing force and bending moment are equal to the algebraic sum of the shearing forces and bending moments which would have been produced by each of the loads acting separately. The principle is illustrated below: Consider a beam having a single concentrated load at mid span and self weight w kN/m as shown. W w kN/m A B By superposition, L we consider the loads to be applied separately. TREBLA The maximum bending moment under the action of the total load is equal to the maximum bending moment WL/4 under the action of concentrated load plus maximum bending load under the action of uniformly distributed self-weight (WL2/8). Example 5.1 Draw the shear force (SF) and bending moment (BM) diagrams of the beam below: P a b C A B L Example 5.2 Draw the shear force (SF) and bending moment (BM) diagrams of the beam below: q N/m TREBLA  A B L Example 5.3 Draw the shear force (SF) and bending moment (BM) diagrams of the beam below: P L Example 5.4 Draw the shear force (SF) and bending moment (BM) diagrams of the beam below: q N/m L Problems 5 5.1 A beam 10 m long is simply supported at its ends and carries constant loads of 30 kN and 80 kN at distances of 3 m from each ends. Draw the shear force and bending moment diagrams. 5.2 In the following problem, sketch the shearing force and bending moment diagrams marking the values of all important points. TREBLA 5.3 Use the diagram below to answer the following questions: 4 kN/m 2 kN/m 4m 4m (a) Sketch the deflection curve, (b) Draw the shear force diagram, (c) Draw the bending moment diagram. 5.4 A beam L m long is supported symmetrically with equal overhangs at each end of 2 m. Each overhang carries a concentrated load of W at its extreme end, and there is a concentrated load W in the centre of the beam. If the bending moment at mid-span is equal to that at each support, calculate the distance between the supports. Sketch the SF and BM diagrams and find the position of the points of contraflexure. [16 m; 6 m and 14 m from the end A] TREBLA  CHAPTER SIX BENDING 6.1 INTRODUCTION When a beam is loaded and bends then its axis is bent into a curve with the upper surface becoming extended (in tension) and lower surface shortened (in compression). There is therefore a plane intermediate between the surfaces experiencing no tension or compression i.e. no change in length. This plane is called the Neutral Plane. The intersection of the neutral plane and any cross section of the beam is the Neutral Axis. 6.2 The Bending Equation Consider a beam that is bent into an arc of a circle. The cross section of the beam is symmetrical about the axis of bending. The beam is bent by an application of a couple of moment M (Pure bending). The radius of the neutral axis is R. y S T Neutral axis For segment PQ the neutral axis does not P Q M M change in length and is the length of R the section and hence layer ST before it was θ bent. ST is at a distance y from the neutral axis. TREBLA Initial length of ST ST = PQ = RQ The change in length on bending is yθ The strain of layer ST If the material is within the limit of proportionality, the stress acting normally to the cross section of the beam on layer ST, will be −σb NA The stress is proportional to the distance from the neutral axis which implies the maximum stress will occur at the beam surface. 6.3 Position of the Neutral Axis The stress acting on the cross section of layer ST of the beam is given as TREBLA If the layer has a cross sectional area dA then the longitudinal force acting on the layer The total longitudinal force acting on the beam is the sum of the forces acting on all such layers forming the beam. i.e. But if the beam is acted on only by bending moments, there is no net longitudinal force; the beam will not suffer any overall extension or contraction longitudinally. 𝑦𝑑𝐴 = 0; This is the first moment of area about the neutral axis and is zero if 𝑅 y = 0, since dA ≠ 0. This can be zero if the moment has been taken about the centroid. This means the neutral axis must pass through the centroid. 6.4 The General Bending Formular For the beam bent into the arc of radius R considered previously, Moment of F about the neutral axis TREBLA  But Since the stress (σ) on a layer a distance y from the neutral axis is given by We combine * and ** to have Where, M = Bending moment acting at a given section; σ = Bending stress; I = Second moment of area of the cross section about the neutral axis; y = Distance from the neutral axis to the extreme fibre; E = Young’s modulus of the material; R = Radius of curvature of the beam. Assumptions for the derivation: 1. Material is linearly elastic 2. E is same in tension and compression 3. The material is homogeneous (same physical properties throughout) 4. Plane sections at right angles to the longitudinal axis remain plane after bending. 6.5 Strain Energy of Bending Consider a beam in pure bending subjected to couples M. TREBLA  θ M M L86 Let the radius of curvature be R and angle subtended by the arc θ As the bending couples are gradually increased in magnitude from zero to their maximum values M, they perform work