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Lecture no.1
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Lecture no.1

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  • 1. IntroductionExternal loadsSurface forcesBody ForcesEquilibriumInternal loading / Internal ForcesSign ConventionsAxial forces, SF, BMGeneral State of stressGeneral condition of loadingComponents of stress
  • 2. Analysis and design of any structure or machine involve two majorquestions: (a) Is the structure strong enough to with­stand the loadsapplied to it and (b) is it stiff enough to avoid excessive deformationsand deflections? In Statics, the members of a structure were treated asrigid bodies; but actually all materials are deformable and this propertywill henceforth be taken into account. Thus Strength of Materials maybe regarded as the statics of deformable or elastic bodies.
  • 3. BRANCHES OF ENGINEERING MECHANICS
  • 4. Both the strength and stiffness of a structural member are functions ofits size and shape and also of certain physical properties of the materialfrom which it is made. These physical properties of materials arelargely determined from experimental studies of their behavior in atesting machine. The study of Strength of Materials is aimed atpredicting just how these geometric and physical properties of astructure will influence its behavior under service conditions. Theapplications of the subject are broad in scope and will be found in allbranches of engineering.
  • 5. The primary objective of the Strength of material / Mechanic ofMaterial is the development of relationships between the loads appliedto a non­rigid body and the internal forces and deformations induced inthe body. Mechanics of materials is a branch of mechanics that developsrelationships between the external loads applied to a deformable bodyand the intensity of internal forces acting within the body. This subjectis also concerned with computing the deformations of the body, and itprovides a study of the bodys stability when the body is subjected toexternal forces.
  • 6. The forces that act on a structure include the appliedloads and the resulting reaction forces. The applied loads are the known loads that act on astructure. They can be the result of the structure’s ownweight, occupancy loads, environmental loads, and so on.The reactions are the forces that the supports exert on astructure. They are considered to be part of the externalforces applied.
  • 7. Forces on a structure can arise from many sources, such as thestructures own weight, any objects placed on it, wind pressure and soforth. Force is a vector quantity, that is, it has both magnitude anddirection. The SI unit of force is the Newton (N), which is defined asthe force required to impart an acceleration of one metre per second persecond to a mass of one kilogram (that is, 1 N = 1 kg m/s2). An objectplaced on a structure will thus impart a vertical force equal to its massmultiplied by the acceleration due to gravity (g = 9.81 m/s2).
  • 8. Surface Forces. As the name implies, surface forces are caused by thedir contact of one body with the surface of another. In all cases theseforces distributed over the area of contact between the bodies, Fig. 1­1a. In particular, if this area is small in comparison with the totalsurface area of the body then the surface force may be idealized as asingle concentrated force, which is applied to a point on the body, Fig.1­1a. For example, this might be done to represent the effect of theground on the wheels of a bicycle when studying the loading on thebicycle. If the surface loading is applied along a narrow area, theloading may be idealized as a linear distributed load, w(s). Here theloading is measured as having an intensity of force/length along thearea and is represented graphically by a series of arrows along the line s
  • 9. Concentrated force idealizationLinear distributedload idealization Fig. 1-1 (a)
  • 10. Fig. 1-1 (b)
  • 11. s, Fig. 1­1a. The loading along the length of a beam is a typicalexample of where this idealization is often applied, Fig. 1­1b. Theresultant force FR of w(s) is equivalent to the area under the distributedloading curve, and this resultant acts through the centroid C orgeometric center of this area.Body Force. A body force occurs when one body exerts a force onanother body without direct physical contact between the bodies.Examples include the effects caused by the earths gravitation or itselectromagnetic field. Although body forces affect each of the particlescomposing the body, these forces are normally represented by a singleconcentrated force acting on the body. In the case of gravitation, thisforce is called the weight of the body and acts through the bodys centerof gravity.
  • 12. The forces on a body can also give rise to moments, which tend to causethe body to rotate about an axis. The moment of a force about an axis issimply equal to the magnitude of the force multiplied by theperpendicular distance from the axis to the line of action of the force.Consider, for example, the lever AB shown in Figure 2.1 (a). The effectof the force P acting at B is to impart both a direct force P and amoment M = Pa on the hinge at A, as shown in Figure 2.1 (b). For a general case of forces and moments in space each force inthe resultant of three forces Fx, Fy and Fz and similarly each moment isthe resultant of three couples Mx, My and Mz.
  • 13. SUMMARYType of force system possible resultantsCollinear …………………………………………… Forceconcurrent , coplanar ……………………………. Forceparallel , coplanar…………………………………. Force or a couple Nonconcurrent , nonparallel ,coplanar ………… Force or a coupleconcurrent , noncoplanar………………………… Forceparallel , noncoplanar ……………………………. Force or a coupleNonconcurrent , nonparallel , noncoplanar…….. Force or a couple , or a force and a couple
  • 14. The surface forces that develop at the supports or points ofsupport between bodies are called reactions. For two­dimensionalproblems, i.e., bodies subjected to coplanar force systems, the supportsmost commonly encountered are shown in Table 1­1. Note the symbolused to represent each support and the type of reactions it exerts on itscontacting member. One possible way to determine a type of supportreaction is to imagine the attached member as being translated orrotated in a particular direction. If the support prevents translation in agiven direction, then a force must be developed on the member in thatdirection. Likewise, if rotation is prevented, a couple moment must be
  • 15. exerted on the member. For example, a roller support only preventstranslation in the contact direction, perpendicular or normal to thesurface. Hence, the roller exerts a normal force F on the member at thepoint of contact. Since the member is free to rotate about the roller, acouple moment cannot be developed by the roller on the member at thepoint of contact. Remember that the concentrated forces and couple momentshown in Table 1­1 actually represent the resultants of distributedsurface forces that exist between each support and its contactingmember. Although it is these resultants that are determined in practice,it is generally not important to determine the actual surface loaddistribution, since the area over which it acts is considerably smaller
  • 16. than the total surface area of the contacting member.2.8 SUPPORTS AND EXTERNAL REACTIONSIn structural analysis terminology the consequence of subjecting astructure to an action is termed the "response" of the structure. Astructure subjected to an action (whether static or dynamic) will move(translate and/or rotate) indefinitely unless the action is resisted in someway. "Supports" are provided as the means of preventing freemovement of the structure. A statically loaded structure will deforminto a new shape but remain attached to its supports and at rest in itsnew position. A dynamically loaded structure will remain attached to
  • 17. supports but its deformed shape will change as a function of time."Damping" will eventually bring it to rest but a finite length of time issubjected to simultaneously applied loads and come to restinstantaneously. Supports prevent free motion of the structure as awhole by developing forces to counteract the load actions. Thecounteracting forces are commonly termed "external reactions" orsimply "reactions."
  • 18. Equilibrium of a body requires both a balance of forces, to prevent thebody from translating or moving along a straight or curved path, and abalance of moments, to prevent the body from rotating. Theseconditions can be expressed mathematically by the two vectorequations ΣF =0 (1–1) Σ MO = 0
  • 19. Here, Σ F represents the sum of all the forces acting on the body, and ΣMO is the sum of the moments of all the forces about any point O eitheron or off the body. If an x, y, z coordinate system is established with theorigin at point O, the force and moment vectors can be resolved intocomponents along the coordinate axes and the above two equations canbe written in scalar form as six equations, namely, Σ Fx = 0 Σ Fy = 0 Σ Fz = 0 (1–2) ΣMx = 0 ΣMy = 0 ΣMz = 0
  • 20. Often in engineering practice the loading on a body can berepresented as a system of coplanar forces. If this is the case, and theforces lie in the x–y plane, then the conditions for equilibrium of thebody can be specified by only three scalar equilibrium equations; thatis, Σ Fx = 0 Σ Fy = 0 (1–3) Σ MO = 0
  • 21. In this case, if point O is the origin of coordinates, the moments will bedirected along the z axis, which is perpendicular to the plane thatcontains the forces. Successful application of the equations of equilibrium requirescomplete specification of all the known and unknown forces that act onthe body. The best way to account for these forces is to draw the bodysfree-body diagram. Obviously, if the free­body diagram is drawncorrectly, the effects of all the applied forces and couple moments canbe accounted for when the equations of equilibrium are written.
  • 22. Consider a body of arbitrary shape acted upon by the forces shown inFig. 1­2. In statics, we would start by determining the resultant of theapplied forces to determine whether or not the body remains at rest. Ifthe resultant is zero, we have static equilibrium—a condition generallyprevailing in structures. If the resultant is not zero, may apply inertiaforces to bring about dynamic equilibrium. Such cases are discussedlater under dynamic loading. For the present, we consider only casesinvolving static equilibrium. In strength of materials, we make an additional investigation ofthe internal distribution of the forces. This is done by passing an
  • 23. Figure 1­2 Exploratory section a-a through loaded member.
  • 24. exploratory section a­a through the body and exposing the internalforces acting on the exploratory section that are necessary to maintainthe equilibrium of either segment. In general, the internal forces reduceto a force and a couple that, for convenience, are resolved intocomponents that are normal and tangent to the section, as shown in Fig.1­3. The origin of the reference axes is always taken at the centroidwhich is the key reference point of the section. Although we are not yetready to show why this is so, we shall prove it as we progress; inparticular, we shall prove it for normal forces in the next article. If the xaxis is normal to the section, the section is known as the x surface or,more briefly, the x face.
  • 25. Figure 1-3Components of internal effects on exploratory section a-a.
  • 26. The notation used in Fig. 1­3 identifies both the exploratory section andthe direction of the force or moment component. The first subscriptdenotes the face on which the component acts; the second subscriptindicates the direction of the particular component. Thus Pxy is the forceon the x face acting in the y direction. Each component reflects a different effect of the applied loadson the member and is given a special name, as follows:Pxx Axial force. This component measures the pulling (or pushing) action perpendicular to the section. A pull rep­resents a tensile force that tends to elongate the member, whereas a push is a compressive force that tends to shorten it. It is often denoted by P.
  • 27. Pxy, Pxz Shear forces. These are components of the total resistance to sliding the portion to one side of the exploratory section past the other. The resultant shear force is usually designated by V, and its components by Vy and Vz to identify their directions.Mxx Torque. This component measures the resistance· to twisting the member and is commonly given the symbol T.Mxy, Mxz Bending moments. These components measure the resistance to bending the member about the y or z axes and are often denoted merely by My or Mz.
  • 28. 2.3 Internal forces in structuresSo far, we have concentrated on the external forces applied to structuresthe applied loads and the support reactions. In order for the structure totransmit the external loads to the ground, internal forces must bedeveloped within the individual members. The aim of the designprocess is to produce a structure that is capable of carrying all theseinternal forces, which may take the form of axial forces, shear forces,bending moments or torques. Consider first a two­dimensional beam where the applied forcesand reactions all lie in a single plane (Figure 2.14(a)). The internalforces at a point in the structure can be found by splitting it at that point
  • 29. and drawing free body diagrams for the two sides (Figure 2.14(b)). Therequirements of equilibrium state that not only must the resultant forceon the entire structure be zero, but the resultant on any segment of itmust also be zero. It is therefore clear that there must be forces acting atthe cut point, as shown. These are drawn on the free body diagrams ofthe segmented structure as though they were external loads, but they arein fact the internal forces in the beam. The forces can be thought of asthe external forces that would have to be applied to the cut beam inorder to produce the same deformations as in the original uncut beam.The forces shown are an axial force T, a transverse force S, known as ashear force, and a bending moment M.
  • 30. For equilibrium at the cut point, the forces acting on the faceseither side of the cut must be equal and opposite; this means that, whenthe two segments are put together to form the complete structure, thereis no resultant external load at that point. For a member in three­dimensional space, a total of six internalforces must be considered, as shown in Figure 2.15. Here, there is againan axial force T in the x direction, and the resultant shear force has beenresolved into components Sy and Sz parallel to the y and z axesrespectively. There are also moments about each of the three axes: Mytends to cause the structure to bend in the horizontal (x ­ z) plane; Mzcauses bending in the vertical (x ­ y) plane; Mx causes the member to
  • 31. y xz Figure 2.15
  • 32. being, however, we will restrict ourselves to two­dimensional systems.2.3.1 Sign convention for internal forcesBefore defining the internal forces more fully, we need to extend thesign convention introduced earlier. For a two­dimensional system,positive forces act in the positive x and y directions, and a positivemoment about the z-axis acts from the positive x towards the positive y-axis, that is anti­clockwise. We can also define a positive face of amember as one whose outward normal is in a positive axis direction.Thus, for the beam segment in Figure 2.16(a), the right­hand face is apositive x-face, the top surface is a positive y-face and the other twofaces are negative.
  • 33. We can now define a positive internal force as one which actseither in a positive direction on a positive face or in a negativedirection on a negative face. Conversely, a negative force either acts ina negative direction on a positive face or vice versa.
  • 34. EQUILIBRIUM AND STRESSWe will not attempt a thorough review of equilibrium methods of statics, sincepresumably the reader is familiar with analytical concepts that lead to computation ofequivalent­force systems and reactions for load­carrying members. For now thediscussion will be conceptual. Later, certain topics of statics will be reviewed asconsidered pertinent. The beam­type structure of Fig. 1.1 is considered to be in equilib­rium withsix actions occurring at each end of the beam. There are six and only six possibleactions that can occur: three forces and three couples, which are evaluated using theequilibrium equations of statics. A section passed through the beam anywhere alongits axis can be viewed as shown in Fig. 1.2. Internal forces and couples act on the cutsection as illustrated and have magnitudes necessary to produce equi­librium for thefree body. Specifically, we classify the forces and couples as in Fig. 1.3. The force
  • 35. z x FIGURE 1.1Beam­type structure showing the six actions of statics.
  • 36. y
  • 37. vector acting along the beam axis is shown in Fig. 1.3a and causes either tension orcompression on the section, depending upon its direction. The remaining two forcevectors (Fig. 1.3b) produce shear loading on the cut section that is characterized bythe forces acting tangent to the cut section as opposed to acting normal to the cutsection as in the case of the axial force. The three couples of Fig. 1.2 are illustrated inFigs. 1.3c­e as vectors. The axial­couple vector of Fig. 1.3c represents a twistingcouple whose direction is determined using the right­hand­screw rule. The twistingcouple, referred to as torque, causes a shear action to occur on the cut cross section.The couples of Figs. 1.3d and e are referred to as bending moments, and the vectorrepresentation is interpreted as illustrated. It turns out that these couples cause acombination of tension and com­pression on the cut section. The early chapters of this text are devoted to analytical methods forcomputing the magnitude and direction of these six actions and then the computationof the corresponding stresses. In particular, Chapter 2 deals with the action of Fig.
  • 38. 1.3a. Chapter 4 is devoted to the twisting couple of Fig. 1.3c applied to members ofcircular cross section. Analytical methods for computing shear forces and bendingmoments of Figs. 1.3b, d, and e are discussed in Chapter 5. Chapter 6 deals with themethods for computing stress caused by the bending moments of Figs. 1.3d and e, andChapter 7 is devoted to the derivation of computational methods for the shear stressproduced by the shear forces of Fig. 1.3b. Load­carrying members subject to variouscombina­tions of the forces and couples are analyzed in Chapter 8. This analysis is theculmination of the study of equivalent force systems and their associated stresses.
  • 39. INTERNAL STRESSES AND STRESS RESULTANTS (INTERNAL FORCES) Framed members subjected to load must develop internalstresses to resist the loads and prevent a material failure. The integratedeffects of stresses are force quantities of a particular magnitude anddirection. A frequent terminology used to refer to these force quantitiesis stress resultants. “Internal forces” is an alternative terminology.Internal force, that can be created in framed members are categorizedinto four types.
  • 40. 1.Axial force 2.shear force 3.flexural moment 4.torsional momentThe determination of these force quantities is one of the basic aims ofstructural analysis. The particular force quantities created in a givenstructure depend upon its behavior. For a planar frame member, thesignificant internal forces are the hear and axial direct forces and theflexural moment. Figure 2.11 depicts the manner in which thesegeneralized forces are developed by the internal stresses. The resultantsof the shear stress distribution τ, and the axial stress distribution σ1,
  • 41. also a major cause of deflections, and this subject is also discussed as isdeflection due to shear. Internal forces developed in each type offramed structure are presented in Table 2.1.
  • 42. Table 2.1. Internal Forces (Stress Resultants) for the Basic Structural Types Schematic of AssumedStructural Type Description Member ForcesPlane truss and Only axial Force, F 1, is space truss significant. F1 Only in­plane shear force F1 , flexural moment F2 , and axialBeam and plane frame force F3, are significant. For beams, axial force generally F3 does not exist. F2
  • 43. Schematic of AssumedStructural Type Description Member Forces F1 Only shear force F1 and flexural moment F2 (both Grid transverse to the plane of the grid) and the torsional moment F2 F3 F3 are significant. F2 F5 All stress resultants are signif­ icant: axial force F1, shear Space Frame forces F2 and F3 torsional F3 moments F4, and flexural F1 moments F5 and F6· F6 F4
  • 44. PROBLEM: ANALYZE THE TRUSS BY STIFFNESSMETHOD
  • 45. ANALYSIS OF GRIDSDefinition:A grid is a structure that has loads applied perpendicular to its plane.The members are assumed to be rigidly connected at the joints.A very basic example of a grid structure is floor system as shown
  • 46. y-axis Wzzj,δ zzj Wzzi,δ zzi Wzj,δ zj Wzi,δ zi x-axisWxxi,δ xxi Wxi,δ xi Wxxj,δ xxj Wxj,δ xj Wyi,δ yi is Wyj,δ yj x z-a Wyyi,δ yyi Wyyj,δ yyj FORCES AND DISPLACEMENT IN LOCAL COORDINATES
  • 47. when integrated over the member cross section, are the transverse shearforce V and the longitudinal (or axial) force P, respectively. Flexuralstresses are the source of two internal forces. Tensile and compressiveregion of this stress distribution,σ2 produce the compressive force C andtensile force T, respectively. Unlike the resultant P. which acts at thecentroid. the resultants T and C are separated by a distance, and neitherof their lines of action coincides with the centroid. Consequently, thenet effect of T and C is the creation of an internal resisting moment, M. Shear and moment resultants that develop in an individualframed member exhibit a rela­tionship to each other and to thetransverse loads applied to the member. These fundamentalinterrelationships will be formulated in Chapter 3. Flexural moment is
  • 48. It is common to refer to loads as “forces that are applied to a structure.”"Applied" is taken to mean “having an identifiable location or point ofapplication.” Each of the six types of framed structures were illustratedin Fig. 1.4. Inspection of that figure indicates the applied load types thatare included in each of the six idealized models.1. Planar truss. Concentrated loads applied in two orthogonal directions at the joints.2. Space truss. Concentrated loads applied in three ol1hogonal directions at the joints.3. Beams, a. Transverse loads and flexural moments that are applied along the member and in the plane of the beam. These can be concentrated or distributed in nature, b. concentrated moments
  • 49. applied at the joints and acting in the plane of the beam.• Grid. a. Concentrated or distributed loads (transverse to the plane of the grid) and flexural moments applied along the member length. b, Concentrated or distributed torsional moments acting along the member length. c, Concentrated loads (transverse to the plane of the frame) and out­of­plane moments applied at the joints.• Planar frame. a, Transverse loads and flexural moments that are applied along the member and in the plane of the frame. These can be concentrated or distributed in nature. b, Concentrated loads in two orthogonal directions and in­plane mo­ments applied at the joints and in the plane of the frame.• Space frame. a. Transverse loads and flexural moments that are
  • 50. applied along the member length. These may be concentrated or distributed in nature and act in any plane passing through the entire member. b, Concentrated or distributed torsional moments acting along the member length. c, Concentrated loads and moments applied at the joints and in any of the three orthogonal planes. Various load categories were described in Section 1.6. Theweight of objects (either dead or live load) and the hydrostatic pressureof water are two examples of applied loads. In each case there iscontact between the load source and the loaded structure. In a strict technical sense. loads arc not al­ways applied to thestructure. Frequently. structures are subjected to phenomena not com­monly referred to as loads. Temperature change and shrinkage of
  • 51. material are two examples. Each of these phenomena cause a structureto experience strain and stress and consequently, to deform. These arethe same kinds of effects as caused by applied loads. When such effectsare included, it is conventional to refer to the general category of loadsas "actions." In this text the terms "load" and "action" are treated assynonomous, both meaning any effect that causes stress and/or strain ina structure. any action must satisfy Newton’s first law, i.e., it must cause areaction. The fundamental concepts of how a structure an action aredeveloped in the balance of this chapter.
  • 52. General State of stress As we have stated, the six actions can combine toproduce a combination of stresses. The question of exactlywhat is stress should be answered in a very simplistic mannerthat is also quite exact in a theoretical sense. We idealize aload­carrying body as shown in Fig. 1.4 and say that the bodyis a continuum. Essentially, a continuum is a
  • 53. yz x FIGURE 1.4 Idealized body.
  • 54. collection of material particles, and its exact size is not important for this discussion.Materials are made up of clusters of molecules, and every material has a definitemolecular structure. On a microscopic scale a material is composed of a space withatoms at specific locations. The continuum model is a collection of many moleculesand is large enough that the individual molecular interactions for the material can beignored and the total of all molecular interactions can be averaged and the continuumcan be assigned some overall gross property to describe its behavior. The continuumis considered to be quite large compared to an atomistic model; however, it can stillbe imagined to be small enough to be of differential size. In other words, we caneffect the mathematical concept of the limit of some quantity with respect to a lengthdimension. We assume that as we find the limiting value of a quantity as a lengthparameter approaches zero that we do not violate the material assumption of acontinuum; the continuum still exists even though it may be of differential size.
  • 55. Consider the continuum of Fig. 1.4 to be in statical equilibrium and imaginea slice taken through the continuum, as shown in Fig. 1.5. The continuum is still inequilibrium since internal forces and couples at the slice can be viewed as externalbalancing forces and couples. We are concerned with forces acting on the smooth, cutsurface that are illustrated as acting on individual, small surface areas. On a smallscale these forces are considered to be acting in a direction either normal or tangent totheir respective areas. The forces originate from the molecu­lar interactions at themicroscopic level; however, as we have stated, we assume that microscopic forces areaveraged and their resultants act on the individual areas. We do not consider smallcouples acting on each element even though they may exist. It has been demonstrated,both experimentally and analytically, that small­body couples can be ignored for thegeneral theory of mechanics of materials. Every force depicted in Fig. 1.5 can be resolved into one normal componentand two shear components. The definition of normal and shear arises naturally;normal
  • 56. yFIGURE 1.5
  • 57. forces act normally to their area and shear forces act tangentially to their area.Visualize one small area ΔA as shown in Fig. 1.5b. The force ΔF is shown incomponents ΔFx, ΔFy, and ΔFz. Normal stress is denoted by σ (sigma) and is definedas the normal force per unit area; hence, the terminology normal stress. Normal stressis given as ∆Fx σx = (1.1) ∆AThe Greek letter τ (tau) is usually used for shear stress and is the shear force dividedby the area, or ∆Fy τy = (1.2) ∆A
  • 58. and ∆Fz τz = (1.3) ∆A A more complete description is given in Chapter 9 concerning shear and normalstresses that act at a point such as the small area of Fig. 1.5b. Previously, it was noted that each action illustrated in Fig. 1.3 producedeither a normal stress or shear stress. In every case we begin with a definition ofstress, either normal or shear, as given by Eqs. (1.1), (1.2), and (1.3), and establish atheory that describes how a particular action causes a stress. The distribution of stressthat occurs on the cross section of the member because of the action of forces andcouples is dependent upon the way the load­carrying member deforms. In the case ofthe axial force, Fig. 1.3a, the member either elongates or shortens in the axialdirection. If we assume a uniform deformation, meaning that every point on the cross
  • 59. section deforms an equal amount parallel to the beam axis, then we must investigatethe limitation of the theory subject to that assumption. It turns out that each of the six actions produces some correspondingdeformation of the member, and prior to developing a theory for stress distribution wemust establish the deformation characteristics of the member when subjected to aparticular load. Chapters 2­9 are devoted to the study of concepts that have beenbriefly introduced in this section.
  • 60. Fig. 1-1 (a)
  • 61. Fig. 1-2 (b)
  • 62. Internal Loadings. One of the most important applications of statics inthe analysis of problems involving mechanics of materials is to be ableto determine the resultant force and moment acting within a body,which are necessary to hold the body together when the body issubjected to external loads. For example, consider the body shown inFig. 1-2a, which is held in equilibrium by the four external forces.* Inorder to obtain the internal Loadings acting on a specific region withinthe body, it is necessary to use the method of sections. This requiresthat an imaginary section or "cut" be made through the region wherethe internal loadings are to be determined. The two parts of the bodyare * The bodys weight is not shown, since it is assumed to be quite small, and thereforenegligible compared with the other loads.
  • 63. then separated, and a free­body diagram of one of the parts is drawn. Ifwe consider the section shown in Fig. 1-2a, then the resulting free­bodydiagram of the bottom part of the body is shown in Fig. 1­2b. Here itcan be seen that there is actually a distribution of internal force actingon the "exposed" area of the section. These forces represent the effectsof the material of the top part of the body acting on the adjacentmaterial of the bottom part. Although this exact distribution may beunknown, we can use statics to determine the resultant internal forceand moment, FR and MRo this distribution exerts at a specific point O onthe sectioned area, Fig. 1-2c. Since the entire body is in equilibrium, then each part of the bodyis also in equilibrium. Consequently, FR and MRo can be determined by
  • 64. applying Eqs. 1­1 to anyone of the two parts of the sectioned body.When doing so, note that FR acts through point 0, although its computedvalue will not depend on the location of this point. On the other hand,MRo does depend on this location, since the moment arms must extendfrom O to the line of action of each force on the free­body diagram. Itwill be shown in later portions of the text that point O is most oftenchosen at the centroid of the sectioned area, and so we will alwayschoose this location for O, unless otherwise stated. Also, if a member islong and slender, as in the case of a rod or beam, the section to beconsidered is generally taken perpendicular to the longitudinal axis ofthe member. This section is referred to as the cross section.
  • 65. Fig. 1-2 (c)
  • 66. Fig. 1-2 (d)
  • 67. Later in this text we will show how to relate the resultantinternal force and moment to the distribution of force on the sectionedarea, Fig. 1­2b, and thereby develop equations that can be used foranalysis and design of the body. To do this, however, the componentsof FR and MRo acting both normal or perpendicular to the sectioned areaand within the plane of the area, must be considered. If we establish x,y, z axes with origin at point O, as shown in Fig; 1­2d, then FR and MRocan each be resolved into three components. Four different types ofloadings can then be defined as follows:Nz is called the normal force, since it acts perpendicular to the area.This force is developed when the external loads tend to push or pull on
  • 68. V is called the shear force, and it can be determined from its twocomponents using vector addition, V = Vx + Vy. The shear force lies inthe plane of the area and is developed when the external loads tend tocause the two segments of the body to slide over one another.Tz is called the torsional moment or torque. It is developed when theexternal loads tend to twist one segment of the body with respect to theother.M is called the bending moment. It is determined from the vectoraddition of its two components, M = Mx + My. The bending moment iscaused by the external loads that tend to bend the body about an axislying within the plane of the area.
  • 69. In this text, note that representation of a moment or torque isshown in three dimensions as a vector with an associated curl, Fig. I­2d.By the right-hand rule, the thumb gives the arrowhead sense of thevector and the fingers or curl indicate the tendency for rotation (twist orbending).
  • 70. Fig. 1-3
  • 71. F3Fig. 1-3
  • 72. Each of the six unknown x, y, z components of force andmoment shows in Fig. 1­2d can be determined directly from the sixequations of equilibrium, that is, Eqs. 1­2, applied to either segment ofthe body. If, however, the body is subjected to a coplanar system offorces, then only normal­force, shear, and bending­moment componentswill exist at the section. To show this, consider the body in Fig. 1­3a. Ifit is in equilibrium, then the internal resultant components, acting at theindicated section, can be determined by first "cutting" the body into twoparts, as shown in Fig. 1­3b, and then applying the equations ofequilibrium to one of the parts. Here the internal resultants consist ofthe normal force N, shear force V, and bending moment Mo. Theseloadings must be equal in magnitude and opposite in direction on each
  • 73. of the sectioned parts (Newtons third law). Furthermore, the magnitudeof each unknown is determined by applying the three equations ofequilibrium to either one of these parts, Eqs. 1­3. If we use the x, y, zcoordinate axes, with origin established at point O, as shown on the leftsegment, then a direct solution for N can be obtained by applying ΣFx =0, and V can be obtained directly from ΣFy = 0. Finally, the bendingmoment Mo can be determined directly by summing moments aboutpoint O (the z axis), Σ Mo = 0, in order to eliminate the moments causedby the unknowns N and V.
  • 74. The method of sections is used to determine the internal loadings at apoint located on the section of a body. These resultants are staticallyequivalent to the forces that are distributed over the material on thesectioned area. If the body is static, that is, at rest or moving withconstant velocity, the resultants must be in equilibrium with theexternal loadings acting on either one of the sectioned segments of thebody. We will now present a procedure that can be used for applyingthe method of sections to determine the internal resultant normal force,shear force, bending moment, and torsional moment at a specific
  • 75. Support Reactions. First decide which segment of the body is to beconsidered. Then before the body is sectioned, it will be necessary todetermine the support reactions or the reactions at the connections onlyon the chosen segment of the body. This is done by drawing the free­body diagram for the entire body, establishing a coordinate system, andthen applying the equations of equilibrium to the body.Free-Body Diagram. Keep all external distributed loadings, couplemoments, torques, and forces acting on the body in their exactlocations, then pass an imaginary section through the body at the pointwhere the internal resultant loadings are to be determined. If the bodyrepresents a member of a structure or mechanical device, this section is
  • 76. often taken perpendicular to the longitudinal axis of the member. Drawa free­body diagram of one of the cut" segments and indicate theunknown resultants N, V, M, and T at the section. In most cases, theseresultants are placed at the point representing the geometric center orcentroid of the sectioned area. In particular, if the member is subjectedto a coplanar system of forces, only N, V, and M act at the centroid. Establish the x, y, z coordinate axes at the centroid and show theresultant components acting along the axes.Equations of Equilibrium. Apply the equations of equilibrium toobtain the unknown resultants. Moments should be summed at thesection, about the axes where the resultants act. Doing this eliminates
  • 77. unknown forces N and V and allows a direct solution for M (and T). Ifthe solution of the equilibrium equations yields a negative value for aresultant, the assumed directional sense of the resultant is opposite tothat shown on the free­body diagram.The following examples illustrate this procedure numerically and alsoprovide a review of some of the important principles of statics. Sincestatics plays such an important role in the analysis of problems inmechanics of materials, it is highly recommended that one solves asmany problems as possible of those that follow these examples.
  • 78. In Sec. 1.2 we showed how to determine the internal resultant force andmoment acting at a specified point on the sectioned area of a body, Fig.1­9a. It was stated that these two loadings represent the resultant effectsof the actual distribution of force acting over the sectioned area, Fig. 1­9b. Obtaining this distribution of internal loading, however, is one ofthe major problems in mechanics of materials.
  • 79. Later it will be shown that to solve this problem it will benecessary to study how the body deforms under load, since eachinternal force distribution will deform the material in a unique way.Before this can be done, however, it is first necessary to develop ameans for describing the internal force distribution at each point on thesectioned area. To do this we will establish the concept of stress. Consider the sectioned area to be subdivided into small areas,such as the one ΔA shown shaded in Fig. 1­9b. As we reduce ΔA to asmaller and smaller size, we must make two assumptions regarding theproperties of the material. We will consider the material to becontinuous, that is, to consist of a continuum or uniform distribution ofmatter having no voids, rather than being composed of a finite number
  • 80. of distinct atoms or molecules. Further­more, the material must becohesive, meaning that all portions of it are connected together, ratherthan having breaks, cracks, or separations. Now, as the subdivided areaΔA of this continuous­cohesive material is reduced to one ofinfinitesimal size, the distribution of force acting over the entiresectioned area will consist of an infinite number of forces, each actingon an element ΔA located at a specific point on the sectioned area. Atypical finite yet very small force ΔF, acting on its associated area ΔA,is shown in Fig. 1­9c. This force, like all the others, will have a uniquedirection, but for further discussion we will replace it by two of itscomponents, namely, ΔFn and ΔFt, which are taken normal and tangentto the area, respectively. As the area Δ A approaches zero, so do the
  • 81. force ΔF and its components; however, the quotient of the force andarea will, in general, approach a finite limit. This quotient is calledstress, and as noted, it describes the intensity of the internal force on aspecific plane (area) passing through a point.
  • 82. Fig. 1.9 (a)
  • 83. (b) (c) Fig. 1.9
  • 84. Normal Stress. The intensity of force, or force per unit area, actingnormal to ΔA is defined as the normal stress, σ(sigma). Mathematicallyit can be expressed as ∆Fn σ = lim (1–4) ∆A→0 ∆AIf the normal force or stress "pulls" on the area element ΔA as shown inFig. I­9c, it is referred to as tensile stress, whereas if it "pushes" on ΔAit is called compressive stress.Shear Stress. Likewise, the intensity of force, or force per unit area,acting tangent to ΔA is called the shear stress, τ(tau). This componentis expressed mathematically as
  • 85. Fig. 1.9
  • 86. ∆Ft τ = lim (1–5) ∆A→0 ∆AIn Fig. I­9c, note that the orientation of the area ΔA completelyspecifies the direction of ΔFn, which is always perpendicular to the area.On the other hand, each shear force ΔFt can act in an infinite number ofdirections within the plane of the area. Provided, however, the directionof ΔF is known, then the direction of ΔFt can be established byresolution of ΔF as shown in the figure.Cartesian Stress Components. To specify further the direction of theshear stress, we will resolve it into rectangular components, and to dothis we will make reference to x, y, z coordinate axes, oriented as shown
  • 87. in Fig. 1–10a. Here the element of area ΔA = Δx Δy and the threeCartesian components of ΔF are shown in Fig. 1­10b. We can nowexpress the normal­stress component as ∆Fz σ z = lim ∆A→0 ∆A
  • 88. Fig. 1.10 (a)
  • 89. (b) (c) Fig. 1.10
  • 90. zF1 (b) Fig. 1.11
  • 91. z (d)(c) Fig. 1.11
  • 92. and the two shear­stress components as ∆Fx τ zx = lim ∆A→0 ∆A ∆Fy τ zy = lim ∆A→0 ∆AThe subscript notation z in σz is used to reference the direction of theoutward normal line, which specifies the orientation of the area ~A.Two subscripts are used for the shear­stress components, τzx and τzy.The z specifies the orientation of the area, and x and y refer to thedirection lines for the shear stresses.
  • 93. To summarize these concepts, the intensity of the internal forceat a point in a body must be described on an area having a specifiedorientation. This intensity can then be measured using threecomponents of stress acting on the area. The normal component actsnormal or perpendicular to the area, and the shear components actwithin the plane of the area. These three stress components are showngraphically in Fig. 1­l0c. Now consider passing another imaginary section through thebody parallel to the x–z plane and intersecting the front side of theelement shown in Fig.1­10a. The resulting free­body diagram is shownin Fig. 1­11a. Resolving the force acting on the area ΔA = Δx Δz into itsrectangular components, and then determining the intensity of these
  • 94. force components, leads to the normal stress and shear­stresscomponents shown in Fig. 1­11b. Using the same notation as before, thesubscript y in σy, τyx, and τyz refers to the direction of the normal lineassociated with the orientation of the area, and x and z in τyx and τyzrefer to the corresponding direction lines for the shear stress. Lastly,one more section of the body parallel to the y–z plane, as shown in Fig.1­11c, gives rise to normal stress σx and shear stresses τxy and τxz, Fig. 1­11d. If we continue in this manner, using corresponding parallel xplanes, we can "cut out" a cubic volume element of material thatrepresents the state of stress acting around the chosen point in the body,Fig. 1­12.
  • 95. Equilibrium Requirements. Although each of the six faces of theelement in Fig. 1­12 will have three components of stress acting on it, ifthe stress around the point is constant, some of these stress componentscan be related by satisfying both force and moment equilibrium for theelement. To show the relationships between the components we willconsider a free­body diagram of the element, Fig. 1­13a. This elementhas a volume of ΔV = Δx Δy Δz, and in accordance with Eqs. 1­4 and 1­5, the forces acting on each face are determined from the product of theaverage stress times the area of the face. For simplicity, we have notlabeled the "dashed" forces acting on the "hidden" sides of the element.Instead, to view, and thereby label, some of these forces, the element isshown from a front view in Fig. 1­13b. Here it should be noted that the
  • 96. zx Fig. 1.13 (a) Element free-body diagram
  • 97. zFig. 1.13 (b) Element free-body diagram
  • 98. force components on the "hidden" sides of the element are designatedwith stresses having primes, and these forces are shown in the oppositedirection to their counterparts acting on the opposite faces of theelement. If we now consider force equilibrium in the y direction, we have
  • 99. CONCEPT OF STRESS AT A GENERAL POINT IN AN ARBITRARILY LOADED MEMBERIn Arts. 1­3 and 1­4 the concept of stress was introduced by consideringthe internal force distribution required to satisfy equilibrium in aportion of a bar under centric load. The nature of the force distributionled to uniformly distributed normal and shearing stresses on transverseplanes through the bar. In more complicated structural members ormachine components the stress distributions will not be uniform onarbitrary internal planes; there­fore. a more general concept of the stateof stress at a point is needed.
  • 100. Consider a body of arbitrary shape that is in equilibrium underthe action of a system of applied forces. The nature of the internal forcedistribu­tion at an arbitrary interior point O can be studied by exposingan interior plane through O as shown in Fig. 1­13a. The forcedistribution required on such an interior plane to maintain equilibriumof the isolated part of the body, in general, will not be uniform;however, any distributed force acting on the small area ΔA surroundingthe point of interest O can be replaced by a statically equivalentresultant force ΔFn through O and a couple ΔMn. The subscript nindicates that the resultant force and couple are associated with aparticular plane through O­namely, the one having an outward normalin the n direction at O. For any other plane through O the values of ΔF
  • 101. and ΔM could be different. Note that the line of action of ΔFn or ΔMnmay not coincide with the direction of n. If the resultant force ΔFn isdivided by the area ΔA, an average force per unit area (averageresultant stress) is obtained. As the area ΔA is made smaller andsmaller, the couple ΔMn vanishes as the force distribution becomesmore and more uniform. In the limit a quantity known as the stressvector4 or resultant stress is obtained. Thus, 4 The component of a tensor on a plane is a vector; therefore, ona particular plane, the stresses can be treated as vectors.
  • 102. FIG. 1—13
  • 103. FIG. 1—13
  • 104. FIG. 1—13
  • 105. ∆Fn S n = lim ∆A→0 ∆AIn Art. 1­3 it was pointed out that materials respond to components ofthe stress vector rather than the stress vector itself. In particular, thecom­ponents normal and tangent to the internal plane were important.As shown in Fig. 1-13b the resultant force ΔFn can be resolved into thecomponents ΔFnn and ΔFnt. A normal stress σn and a shearing stress τnare then defined as ∆Fnn σ n = lim ∆A→0 ∆A
  • 106. ∆Fnt τ n = lim ∆A→0 ∆AFor purposes of analysis it is convenient to reference stresses to somecoordinate system. For example, in a Cartesian coordinate system thestresses on planes having outward normals in the x, y. and z directionsare usually chosen. Consider the plane having an outward normal in thex direc­tion. In this case the normal and shear stresses on the plane willbe σx and τx, respectively. Since τx, in general, will not coincide withthe y or z axes, it must be resolved into the components τxy and τxz, asshown in Fig. 1­13c. Unfortunately the state of stress at a point in a
  • 107. stress vector since the stress vector itself depends on the orientation ofthe plane with which it is associated. An infinite number of planes canbe passed through the point, resulting in an infinite number of stressvectors being associated with the point. Fortunately it can be shown(see Art. 1­9) that the specification of stresses on three mutuallyperpendicular planes is sufficient to describe com­pletely the state ofstress at the point. The rectangular components of stress vectors onplanes having outward normals in the coordinate directions are shownin Fig. 1­14. The six faces of the small element are denoted by thedirections of their outward normals so that the positive x face is the onewhose outward normal is in the direction of the positive x axis. Thecoordi­nate axes x, y, and z are arranged as a right­hand system. The
  • 108. FIG. 1—14
  • 109. sign convention for stresses is as follows. Normal stresses (indicated bythe symbol σ and a single subscript to indicate the plane on which thestress acts) are positive if they point in the direction of the outwardnormal. Thus normal stresses are positive if tensile. Shearing stressesare denoted by the symbol τ followed by two subscripts; the firstsubscript designates the plane on which the shearing stress acts and thesecond the coordinate axis to which it is parallel. Thus, τxy is theshearing stress on an x plane parallel to the z axis. A positive shearingstress points in the positive direction of the coordinate axis of t hesecond subscript if it acts on a surface with an outward normal in thepositive direction. Conversely, if the outward normal of the surface is inthe negative direction, then the positive shearing stress points in the
  • 110. negative direction of the coordinate axis of the second subscript. Thestresses shown on the element in Fig. 1­14 are all positive.
  • 111. STRESS UNDER GENERAL LOADING CONDITIONS; COMPONENTS OF STRESS The examples of the previous sections were limited to members under axialloading and connections under transverse loading. Most structural members andmachine components are under more involved loading conditions. Consider a body subjected to several loads P 1, P2, etc. (Fig. 1.32). Tounderstand the stress condition created by these loads at some point Q within thebody, we shall first pass a section through Q, using a plane parallel to the yz plane.The portion of the body to the left of the section is subjected to some of the originalloads, and to normal and shearing forces distributed over the section. We shall denoteby ΔFx and ΔVx, respectively, the normal and the shearing forces acting on a small
  • 112. Fig. 1.32
  • 113. (a) (b) Fig. 1.33
  • 114. area ΔA surrounding point Q (Fig. 1.33a). Note that the superscript x is used toindicate that the forces ΔFx and ΔVx act on a surface per­pendicular to the x axis.While the normal force ΔFx has a well­defined direction, the shearing force ΔVx mayhave any direction in the plane of the section. We therefore resolve ΔVx into twocomponent forces, ΔVxy and ΔVxz in directions parallel to the y and z axes,respectively (Fig. 1.33 b). Dividing now the magnitude of each force by the area ΔA,and letting ΔA approach zero, we define the three stress components shown in Fig.1.34: x ∆F σ x = lim ∆A→0 ∆A ∆V yx ∆Vzx τ xy = lim τ xz = lim (1.18) ∆A→0 ∆A ∆A→0 ∆A
  • 115. y τxy xz Fig. 1.34
  • 116. We note that the first subscript in σx, τxy and τxz is used to indicate that the stressesunder consideration are exerted on a surface perpendicular to the x axis. The secondsubscript in τxy and τxz identifies the direction of the component. The normal stress σxis positive if the corresponding arrow points in the positive x direction, i.e., if thebody is in tension, and negative otherwise. Similarly, the shearing stress componentsτxy and τxz are positive if the corresponding arrows point, respectively, in the positive yand z directions. The above analysis may also be carried out by considering the portion ofbody located to the right of the vertical plane through Q (Fig. 1.35). The samemagnitudes, but opposite directions, are obtained for the normal and shearing forcesΔFx, ΔVxy and ΔVxz Therefore, the same values are also obtained for thecorresponding stress components, but since the section in Fig. 1.35 now faces thenegative x axis, a positive sign for σx will indicate that the corresponding arrow points
  • 117. corresponding arrows point, respectively, in the negative y and z directions, as shownin Fig. 1.35. y σx Fig. 1.35 z
  • 118. Passing a section through Q parallel to the zx plane, we define in the samemanner the stress components, σy, τyz, and τyx Finally, a section through Q parallel tothe xy plane yields the components σz, τzx and τzy. To facilitate the visualization of the stress condition at point Q, we shallconsider a small cube of side a centered at Q and the stresses ex­erted on each of thesix faces of the cube (Fig. 1.36). The stress com­ponents shown in the figure are σx, σyand σz which represent the nor­mal stress on faces respectively perpendicular to the x,y, and z axes, and the six shearing stress components τxy τxz etc. We recall that, ac­cording to the definition of the shearing stress components, τxy represents the ycomponent of the shearing stress exerted on the face perpendicular to the x axis, whileτyx represents the x component of the shearing stress exerted on the face perpendicularto the y axis. Note that only three faces of the cube are actually visible in Fig. 1.36,and that equal and opposite stress components act on the hidden faces. While the
  • 119. yz x Fig. 1.36
  • 120. Fig. 1.37
  • 121. involved is small and vanishes as side a of the cube approaches zero. Important relations among the shearing stress components will now bederived. Let us consider the free­body diagram of the small cube centered at point Q(Fig. 1.37). The normal and shearing forces acting on the various faces of the cube areobtained by multiplying the corresponding stress components by the area ΔA of eachface. We first write the following three equilibrium equations: ∑F x =0 ∑F y =0 ∑F z =0 (1.19)Since forces equal and opposite to the forces actually shown in Fig. 1.37 are acting onthe hidden faces of the cube, it is clear that Eqs. (1.19) are satisfied. Considering nowthe moments of the forces about axes Qx, Qy, and Qz drawn from Q in directionsrespectively parallel to the x, y, and z axes, we write the three additional equations ∑M x =0 ∑M y =0 ∑M z =0 (1.20)
  • 122. Fig. 1.38
  • 123. Using a projection on the xy plane (Fig. 1.38), we note that the only forces withmoments about the z axis different from zero are the shear­ing forces. These forcesform two couples, one of counterclockwise (positive) moment (τxy ΔA)a, the other ofclockwise (negative) moment –(τxy ΔA)a. The last of the three Eqs. (1.20) yields,therefore, + ↑ ∑Mz = 0: (τ xy ∆A) a − (τ yx ∆A) a = 0 (1.21)from which we conclude that τ xy = τ yxThe relation obtained shows that the y component of the shearing stress exerted on aface perpendicular to the x axis is equal to the x component of the shearing stressexerted on a face perpendicular to the y axis. From the remaining two equations(1.20), we derive in a similar man­ner the relations similarly, τ yz = τ zy and τ yz = τ zy (1.22)
  • 124. We conclude from Eqs. (1.21) and (1.22) that only six stress com­ponents arerequired to define the condition of stress at a given point Q, instead of nine asoriginally assumed. These six components are ux, uy, uz, T XY Tyz, and T ZX. Wealso note that, at a given point, shear cannot take place in one plane only; an equalshearing stress must be exerted on another plane perpendicular to the first one. Forexample, considering again the bolt of Fig. 1.29 and a small cube at the center Q ofthe bolt (Fig. 1.39a), we find that shearing stresses of equal mag­nitude must beexerted on the two horizontal faces of the cube and on the two faces that areperpendicular to the forces P and P (Fig. 1.39b). Before concluding our discussion ofstress components, let us con­sider again the case of a member under axial loading. Ifwe consider a small cube with faces respectively parallel to the faces of the memberand recall the results obtained in Sec. 1.11, we find that the conditions of stress in themember may be described as shown in Fig. lAOa; the only stresses are normalstresses U x exerted on the faces of the cube which are perpendicular to the x axis.
  • 125. (a) (b) Fig. 1.39
  • 126. Fig. 1.40
  • 127. is ro­tated by 45° about the z axis so that its new orientation matches the orientationof the sections considered in Fig. 1.31c and d, we conclude that normal and shearingstresses of equal magnitude are exerted on four faces of the cube (Fig. 1.40b). Wethus observe that the same loading condition may lead to different interpretations ofthe stress situation at a given point, depending upon the orientation of the elementconsidered. More will be said about this in Chap 7.
  • 128. Stress, Strain, and Their Relationships2.1 Introduction The concepts of stress and strain are two of the most important concepts within thesubject of mechanics of materials or mechanics of deformable bodies. They arediscussed in detail in this chapter, particularly as they relate to two­dimensionalsituations. In the case of stress as well as in the case of strain, emphasis is placed on theuse of the semigraphical procedure known as the Mohr’s circle solution. Theunderlying mathematical concepts leading to Mohrs circle are developed anddiscussed. Examples are solved to illustrate the use of this powerful semigraphicalmethod of solution, which will be used throughout this book whenever problems areencountered dealing with stress and strain analysis.
  • 129. The treatment given to the concepts of stress and strain in this book differsfrom that in other books in several important respects, the most significant of which isthe fact that the sign convention adopted for strain is compatible with the signconvention for stress as they relate to the construction of the corresponding Mohrscircles. This approach is advantageous in that it makes the construction of the stressand strain Mohrs circles identical. The discussion relating stress to strain in this chapter is limited to the rangeof material behavior within which the strain varies linearly with stress. This procedurefrees the students from information which, although very important, is extraneous forthe time being. A more complete discussion of material behavior is provided inChapters 3 and 4.
  • 130. 2.2 Concept of Stress at a Point If a body is subjected to external forces, a system of internal forces isdeveloped. These internal forces tend to separate or bring closer together the materialparticles that make up the body. Consider, for example, the body shown in Figure2.1(a), which is subjected to the external forces Fl, F2, ... , Fi. Consider an imaginaryplane that cuts the body into two parts, as shown. Internal forces are transmitted fromone part of the body to the other through this imaginary plane. Let the free­bodydiagram of the lower part of the body be constructed as shown in Figure 2,1 (b). Theforces F1, F2, and F3 are held in equilibrium by the action of an internal system offorces distributed in some manner through the surface area of the imaginary plane.This system of internal forces may be represented by a single resultant force R and/orby a couple. For the sake of simplicity in introducing the concept of stress, only theforce R is assumed to exist. In general, the force R may be decomposed into a
  • 131. Figure 2.1
  • 132. Figure 2.1
  • 133. component Fn perpendicular to the plane and known as the normal force, and acomponent Ft, parallel to the plane and known as the shear force. If the area of the imaginary plane is to be A, then Fn / A and Ft / A represent,respectively, average values of normal and shear forces per unit area called stresses.These stresses, however, are not, in general, uniformly distributed throughout the areaunder consideration, and it is therefore desirable to be able to determine themagnitude of both the normal and shear stresses at any point within the area. If thenormal and shear forces acting over a differential element of area ΔA in theneighborhood of point O are ΔFn and ΔFt respectively, as shown in Figure 2.1 (b),then the normal stress σ and the shearing stress, are given by the followingexpressions: ∆Fn σ = lim ∆A→0 ∆A (2.1) ∆Ft τ = lim ∆A → 0 ∆A
  • 134. In the special case where the components Fn and Fr are uniformly distributed overthe entire area A, then σ = Fn/A and τ = Ft/A. Note that a normal stress acts in a direction perpendicular to the plane onwhich it acts and it can be either tensile or compressive. A tensile normal stress is onethat tends to pull the material particles away from each other, while a compressivenormal stress is one that tends to push them closer together. A shear stress, on theother hand, acts parallel to the plane on which it acts and tends to slide (shear)adjacent planes with respect to each other. Also note that the units of stress (σ or τ)consist of units of force divided by units of area. Thus, in the British gravitationalsystem of measure, such units as pounds per square inch (psi) and kilopounds persquare inch (ksi) are common. In the metric (SI) system of measure, the unit that hasbeen proposed for stress is the Newton per square meter (N/m 2), which is called thepascal and denoted by the symbol Pa. Because the Pascal is a very small quantity,another SI unit that is widely used is the mega Pascal (106 pascals) and is denoted bythe symbol MPa. This unit may also be written as MN/m2.
  • 135. Components of Stress In the most general case, normal and shear stresses at a point in a body maybe considered to act on three mutually perpendicular planes. This most general stateof stress is usually referred to as triaxial. It is convenient to select planes that arenormal to the three coordinates axes x, y, and z and designate them as the X, Y, and Zplanes, respectively. Consider these planes as enclosing a differential volume ofmaterial in the neighborhood of a given point in a stressed body. Such a volume ofmaterial is depicted in Figure 2.2 and is referred to as a three-dimensional stresselement. On each of the three mutually perpendicular planes of the stress element,there acts a normal stress, and a shear stress which is represented by its twoperpendicular components. The notation for stresses used in this text consists of affixing one subscript toa normal stress, indicating the plane on which it is acting, and two subscripts to a
  • 136. shear stress, the first of which designates the plane on which it is acting and thesecond its direction. For example, σx is a normal stress acting on the X plane, τxy is ashear stress acting on the X plane and pointed in the positive y direction, and τxz is ashear stress acting in the X plane and pointed in the positive z direction. It is observed from Figure 2.2 that three stress components exist on each ofthe three mutually perpendicular planes that define the stress element. Thus thereexists a total of nine stress components that must be specified in order to definecompletely the state of stress at any point in the body. By considerations of theequilibrium of the stress element, it can easily be shown that, τxy = τyx, τxz = τzx, and τyz= τzy so that the number of stress components required to completely define the stateof stress at a point is reduced to six.
  • 137. By convention, a normal stress is positive if it points in the direction of the outwardnormal to the plane. Thus a positive normal stress produces tension and a negativenormal stress produces compression. A component of shear stress is positive if it ispointed along the positive direction of the coordinate axis and if the outward normalto its plane is also in the positive direction of the corresponding axis. If, however, theoutward normal is in the negative direction of the coordinate axis, a positive shearstress will also be in the negative direction of the corresponding axis. The stresscomponents shown in Figure 2.2 are all positive. It should be noted, however, thatsuch a sign convention for shear stress is rather cumbersome. It is only used in theanalysis of triaxial stress problems that are usually dealt with in advanced coursessuch as the theory of elasticity.A complete study of the triaxial or three­dimensional state of stress is beyond thescope of this chapter, and the analysis that follows is limited to the special case inwhich the stress components in one direction are all zero. For example, if all the stress
  • 138. components in the z direction are zero (i.e., τxz = τyz = τz = 0), the stress conditionreduces to a biaxial or two­dimensional state of stress in the xy plane. This state ofstress is referred to as plane stress. Fortunately, many of the problems encountered inpractice are such that they can be considered plane stress problems.2.4 Analysis of Plane Stress As mentioned previously the state of stress known as plane stress is one inwhich all the stress components in one direction vanish. Thus, if it is assumed that allthe components in the z direction shown in Figure 2.2 are zero (i.e., τxz = τyz = τz = 0),the stress element shown in Figure 2.3(a) is obtained and it is the most general planestress condition that can exist. It should be observed that a stress element is in realitya schematic representation of two sets of perpendicular planes passing through a pointand that the element degenerates into a point in the limit when both dx and dyapproach zero.
  • 139. From considerations of the equilibrium of forces on the stress element in Figure2.3(a), it can be shown that τxy = τyx. Thus assume that the depth of the stress elementinto the paper is a constant equal to h. Since, by definition, force is the product ofstress and the area over which it acts, then a summation of moments of all forcesabout a z axis through point O leads to the following equation: τ xy = (h dy )dx − τ yx = (h dx)dy = 0from which τ xy = τ yx
  • 140. STRESS AT A POINTStress at a point is terminology that means exactly what it says. Refer to Fig.8.2 of Example 8.1. Stress at a point C would imply that the stresses at point Care to be computed. The point must be drawn large enough for it to bevisualized. Therefore the concept of a stress block or material element isnecessary. The stress block is the point enlarged for the practical purpose ofdrawing it and is referred to as a stress element since its actual size iselemental. Point C of Fig. 8.2 is shown enlarged in Fig. 8.la. Physically, theelement can be visualized as a small square near the outside surface of thebeam located at point C. The stresses are identified as acting on the edge of the elementalsquare and is the standard concept of stress at a point in two dimensions.
  • 141. (a)FIGURE 8.1(a)Element viewed along the z axis;(b) positive stresses acting on an x element; (b)
  • 142. (c)FIGURE 8.1(c) element viewed along the x axis;(d) element viewed along the y axis. (d)
  • 143. Actually, the elemental square is an elemental cube located at point C.The cube is shown in Fig. 8.1b. The subscript notation identifies the directionof the stress at the face, or area, of the cube on which it acts. For instance, thestress σxx is the normal stress acting on the face perpendicular to the x axis.The first subscript identifies the face, or area, of the cube. The secondsubscript identifies the direction of the stress. A shear stress on the face of thecube normal to the y axis and acting in the x direction would be τyx as shownin Fig. 8.1b. The complete three­dimensional description of stress at a point isillustrated in the figure. The discussion in the previous chapters concerning shear stresses onmutually perpendicular planes would imply that τxy = τyx τyz = τzy and τxz = τzx.The subscript notation lends itself toward identifying this equivalence. Thenine components shown in Fig. 8.lb are conveniently presented using thefollowing format.
  • 144. (8.5) The use of the boldface σ will represent the stress at a point. Equation(8.5) represents the stress, σ, as a nine­component quantity and is referred toas a stress tensor, which has certain mathematical properties that are useful inadvanced studies.Stress at a point will be viewed two dimensionally in this chapter. Figure 8.la is actually the cube as it is viewed along the z axis;therefore, even though the stresses appear to be applied along the edge of theelement, they are actually applied to a surface that is perpendicular to theplane of the page. The stress components of Fig. 8.la will be written as

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