IntroductionExternal loads Surface forcesBody ForcesEquilibriumInternal loading / Internal ForcesSign ConventionsAxial forces, SF, BMGeneral State of stressGeneral condition of loading Components of stress

Analysis and design of any structure or machine involve two major questions: (a) Is the structure strong enough to withstand the loads applied to it and (b) is it stiff enough to avoid excessive deformations and deflections? In Statics, the members of a structure were treated as rigid bodies; but actually all materials are deformable and this property will henceforth be taken into account. Thus Strength of Materials may be regarded as the statics of deformable or elastic bodies.

BRANCHES OF ENGINEERING MECHANICS

Both the strength and stiffness of a structural member are functions of its size and shape and also of certain physical properties of the material from which it is made. These physical properties of materials are largely determined from experimental studies of their behavior in a testing machine. The study of Strength of Materials is aimed at predicting just how these geometric and physical properties of a structure will influence its behavior under service conditions. The applications of the subject are broad in scope and will be found in all branches of engineering.

The primary objective of the Strength of material / Mechanic of Material is the development of relationships between the loads applied to a non-rigid body and the internal forces and deformations induced in the body.Mechanics of materials is a branch of mechanics that develops relationships between the external loads applied to a deformable body and the intensity of internal forces acting within the body. This subject is also concerned with computing the deformations of the body, and it provides a study of the body's stability when the body is subjected to external forces.

General State of Stress at a PointWhen forces are applied to a structural member, the forces are transmitted through the member as internal forces. Qualitatively, the intensity of the internal force at any point is called the stress at that point (see Chapter 3). Stresses in loaded members result from two basic types of forces, namely, surface forces and body forces (see chapter 1). Surface forces are those that act on the surface of a body or member, for example when one body or member comes in contact with another body. Body forces act throughout the volume of the membe3r; examples are gravitational, centrifugal, and magnetic forces.

Gravitational body forces are the most common body forces in static or quasistatic structural members and are generally much smaller than surface forces. For this reason, body forces are often neglected in comparison to surface forces without introducing a significant error (However, there are many applications where this is not true. Consider, for example, the mirror in a large space telescope. With diameters on the order of a few meters or a few tens of meters, the self-weight of the telescope can cause enough gravity sag to distort the mirror considerably from its intended shape.)

To determine the nature of internal forces, we divide the member into two parts by passing a cutting plane through the point of interest. Each of the two resulting parts may be considered a free body. The internal forces acting on the exposed cross-sectional area may have a distribution such that the internal force F (which is a vector and thus represented in boldface) varies in both magnitude and direction from point to point.)

The forces that act on a structure include the applied loads and the resulting reaction forces.The applied loads are the known loads that act on a structure. They can be the result of the structures own weight, occupancy loads, environmental loads, and so on. The reactions are the forces that the supports exert on a structure. They are considered to be part of the external forces applied.

Forces on a structure can arise from many sources, such as the structure's own weight, any objects placed on it, wind pressure and so forth. Force is a vector quantity, that is, it has both magnitude and direction. The SI unit of force is the Newton (N), which is defined as the force required to impart an acceleration of one metre per second per second to a mass of one kilogram (that is, 1 N = 1 kg m/s2). An object placed on a structure will thus impart a vertical force equal to its mass multiplied by the acceleration due to gravity (g = 9.81 m/s2).

Surface Forces. As the name implies, surface forces are caused by the dir contact of one body with the surface of another. In all cases these forces distributed over the area of contact between the bodies, Fig. 1-1a. In particular, if this area is small in comparison with the total surface area of the body then the surface force may be idealized as a single concentrated force, which is applied to a point on the body, Fig. 1-1a. For example, this might be done to represent the effect of the ground on the wheels of a bicycle when studying the loading on the bicycle. If the surface loading is applied along a narrow area, the loading may be idealized as a linear distributed load, w(s). Here the loading is measured as having an intensity of force/length along the area and is represented graphically by a series of arrows along the line s

(a)Fig. 1-1

(b)Fig. 1-1

s, Fig. 1-1a. The loading along the length of a beam is a typical example of where this idealization is often applied, Fig. 1-1b. The resultant force FR of w(s) is equivalent to the area under the distributed loading curve, and this resultant acts through the centroid C or geometric center of this area. Body Force. A body force occurs when one body exerts a force on another body without direct physical contact between the bodies. Examples include the effects caused by the earth's gravitation or its electromagnetic field. Although body forces affect each of the particles composing the body, these forces are normally represented by a single concentrated force acting on the body. In the case of gravitation, this force is called the weight of the body and acts through the body's center of gravity.

The surface forces that develop at the supports or points of support between bodies are called reactions. For two-dimensional problems, i.e., bodies subjected to coplanar force systems, the supports most commonly encountered are shown in Table 1-1. Note the symbol used to represent each support and the type of reactions it exerts on its contacting member. One possible way to determine a type of support reaction is to imagine the attached member as being translated or rotated in a particular direction. If the support prevents translation in a given direction, then a force must be developed on the member in that direction. Likewise, if rotation is prevented, a couple moment must be

exerted on the member. For example, a roller support only prevents translation in the contact direction, perpendicular or normal to the surface. Hence, the roller exerts a normal force F on the member at the point of contact. Since the member is free to rotate about the roller, a couple moment cannot be developed by the roller on the member at the point of contact. Remember that the concentrated forces and couple moment shown in Table 1-1 actually represent the resultants of distributed surface forces that exist between each support and its contacting member. Although it is these resultants that are determined in practice, it is generally not important to determine the actual surface load distribution, since the area over which it acts is considerably smaller

2.8 SUPPORTS AND EXTERNAL REACTIONS In structural analysis terminology the consequence of subjecting a structure to an action is termed the "response" of the structure. A structure subjected to an action (whether static or dynamic) will move (translate and/or rotate) indefinitely unless the action is resisted in some way. "Supports" are provided as the means of preventing free movement of the structure. A statically loaded structure will deform into a new shape but remain attached to its supports and at rest in its new position. A dynamically loaded structure will remain attached to itsthan the total surface area of the contacting member.

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 is subjected to simultaneously applied loads and come to rest instantaneously. Supports prevent free motion of the structure as a whole by developing forces to counteract the load actions. The counteracting forces are commonly termed "external reactions" or simply "reactions."

The forces on a body can also give rise to moments, which tend to cause the body to rotate about an axis. The moment of a force about an axis is simply equal to the magnitude of the force multiplied by the perpendicular 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 effect of the force P acting at B is to impart both a direct force P and a moment 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 in the resultant of three forces Fx, Fy and Fz and similarly each moment is the resultant of three couples Mx, My and Mz.

Type of force systemCollinear concurrent , coplanar .parallel , coplanar. Nonconcurrent , nonparallel ,coplanar concurrent , noncoplanarparallel , noncoplanar .Nonconcurrent , nonparallel , noncoplanar..

possible resultantsForceForceForce or a coupleForce or a coupleForceForce or a coupleForce or a couple , or a force and a coupleSUMMARY

Equilibrium of a body requires both a balance of forces, to prevent the body from translating or moving along a straight or curved path, and a balance of moments, to prevent the body from rotating. These conditions can be expressed mathematically by the two vector equations

Here, F represents the sum of all the forces acting on the body, and MO is the sum of the moments of all the forces about any point O either on or off the body. If an x, y, z coordinate system is established with the origin at point O, the force and moment vectors can be resolved into components along the coordinate axes and the above two equations can be written in scalar form as six equations, namely,

Often in engineering practice the loading on a body can be represented as a system of coplanar forces. If this is the case, and the forces lie in the xy plane, then the conditions for equilibrium of the body can be specified by only three scalar equilibrium equations; that is,

In this case, if point O is the origin of coordinates, the moments will be directed along the z axis, which is perpendicular to the plane that contains the forces. Successful application of the equations of equilibrium requires complete specification of all the known and unknown forces that act on the body. The best way to account for these forces is to draw the body's free-body diagram. Obviously, if the free-body diagram is drawn correctly, the effects of all the applied forces and couple moments can be accounted for when the equations of equilibrium are written.

Consider a body of arbitrary shape acted upon by the forces shown in Fig. 1-2. In statics, we would start by determining the resultant of the applied forces to determine whether or not the body remains at rest. If the resultant is zero, we have static equilibriuma condition generally prevailing in structures. If the resultant is not zero, may apply inertia forces to bring about dynamic equilibrium. Such cases are discussed later under dynamic loading. For the present, we consider only cases involving static equilibrium. In strength of materials, we make an additional investigation of the internal distribution of the forces. This is done by passing an

Figure 1-2 Exploratory section a-a through loaded member.

exploratory section a-a through the body and exposing the internal forces acting on the exploratory section that are necessary to maintain the equilibrium of either segment. In general, the internal forces reduce to a force and a couple that, for convenience, are resolved into components that are normal and tangent to the section, as shown in Fig. 1-3. The origin of the reference axes is always taken at the centroid which is the key reference point of the section. Although we are not yet ready to show why this is so, we shall prove it as we progress; in particular, we shall prove it for normal forces in the next article. If the x axis is normal to the section, the section is known as the x surface or, more briefly, the x face.

Figure 1-3Components of internal effects on exploratory section a-a.

The notation used in Fig. 1-3 identifies both the exploratory section and the direction of the force or moment component. The first subscript denotes the face on which the component acts; the second subscript indicates the direction of the particular component. Thus Pxy is the force on the x face acting in the y direction. Each component reflects a different effect of the applied loads on the member and is given a special name, as follows: 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. Pxx

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.Torque. This component measures the resistance to twisting the member and is commonly given the symbol T. 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. Pxy, PxzMxy, MxzMxx

2.3 Internal forces in structures So far, we have concentrated on the external forces applied to structures the applied loads and the support reactions. In order for the structure to transmit the external loads to the ground, internal forces must be developed within the individual members. The aim of the design process is to produce a structure that is capable of carrying all these internal forces, which may take the form of axial forces, shear forces, bending moments or torques. Consider first a two-dimensional beam where the applied forces and reactions all lie in a single plane (Figure 2.14(a)). The internal forces at a point in the structure can be found by splitting it at that point

and drawing free body diagrams for the two sides (Figure 2.14(b)). The requirements of equilibrium state that not only must the resultant force on the entire structure be zero, but the resultant on any segment of it must also be zero. It is therefore clear that there must be forces acting at the cut point, as shown. These are drawn on the free body diagrams of the segmented structure as though they were external loads, but they are in fact the internal forces in the beam. The forces can be thought of as the external forces that would have to be applied to the cut beam in order 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 a shear force, and a bending moment M.

For equilibrium at the cut point, the forces acting on the faces either side of the cut must be equal and opposite; this means that, when the two segments are put together to form the complete structure, there is no resultant external load at that point. For a member in three-dimensional space, a total of six internal forces must be considered, as shown in Figure 2.15. Here, there is again an axial force T in the x direction, and the resultant shear force has been resolved into components Sy and Sz parallel to the y and z axes respectively. There are also moments about each of the three axes: My tends to cause the structure to bend in the horizontal (x - z) plane; Mz causes bending in the vertical (x - y) plane; Mx causes the member to twist about its longitudinal axis, and is called a torque. For the time

Figure 2.15

being, however, we will restrict ourselves to two-dimensional systems.2.3.1 Sign convention for internal forces Before defining the internal forces more fully, we need to extend the sign convention introduced earlier. For a two-dimensional system, positive forces act in the positive x and y directions, and a positive moment about the z-axis acts from the positive x towards the positive y-axis, that is anti-clockwise. We can also define a positive face of a member as one whose outward normal is in a positive axis direction. Thus, for the beam segment in Figure 2.16(a), the right-hand face is a positive x-face, the top surface is a positive y-face and the other two faces are negative.

We can now define a positive internal force as one which acts either in a positive direction on a positive face or in a negative direction on a negative face. Conversely, a negative force either acts in a negative direction on a positive face or vice versa.

EQUILIBRIUM AND STRESS We will not attempt a thorough review of equilibrium methods of statics, since presumably the reader is familiar with analytical concepts that lead to computation of equivalent-force systems and reactions for load-carrying members. For now the discussion will be conceptual. Later, certain topics of statics will be reviewed as considered pertinent. The beam-type structure of Fig. 1.1 is considered to be in equilibrium with six actions occurring at each end of the beam. There are six and only six possible actions that can occur: three forces and three couples, which are evaluated using the equilibrium equations of statics. A section passed through the beam anywhere along its axis can be viewed as shown in Fig. 1.2. Internal forces and couples act on the cut section as illustrated and have magnitudes necessary to produce equilibrium for the free body. Specifically, we classify the forces and couples as in Fig. 1.3. The force

zxFIGURE 1.1 Beam-type structure showing the six actions of statics.

y

vector acting along the beam axis is shown in Fig. 1.3a and causes either tension or compression on the section, depending upon its direction. The remaining two force vectors (Fig. 1.3b) produce shear loading on the cut section that is characterized by the forces acting tangent to the cut section as opposed to acting normal to the cut section as in the case of the axial force. The three couples of Fig. 1.2 are illustrated in Figs. 1.3c-e as vectors. The axial-couple vector of Fig. 1.3c represents a twisting couple whose direction is determined using the right-hand-screw rule. The twisting couple, 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 vector representation is interpreted as illustrated. It turns out that these couples cause a combination of tension and compression on the cut section. The early chapters of this text are devoted to analytical methods for computing the magnitude and direction of these six actions and then the computation of the corresponding stresses. In particular, Chapter 2 deals with the action of Fig.

1.3a. Chapter 4 is devoted to the twisting couple of Fig. 1.3c applied to members of circular cross section. Analytical methods for computing shear forces and bending moments of Figs. 1.3b, d, and e are discussed in Chapter 5. Chapter 6 deals with the methods for computing stress caused by the bending moments of Figs. 1.3d and e, and Chapter 7 is devoted to the derivation of computational methods for the shear stress produced by the shear forces of Fig. 1.3b. Load-carrying members subject to various combinations of the forces and couples are analyzed in Chapter 8. This analysis is the culmination of the study of equivalent force systems and their associated stresses.

INTERNAL STRESSES AND STRESS RESULTANTS (INTERNAL FORCES) Framed members subjected to load must develop internal stresses to resist the loads and prevent a material failure. The integrated effects of stresses are force quantities of a particular magnitude and direction. A frequent terminology used to refer to these force quantities is stress resultants. Internal forces is an alternative terminology. Internal force, that can be created in framed members are categorized into four types.

Axial force shear force flexural moment torsional moment The determination of these force quantities is one of the basic aims of structural analysis. The particular force quantities created in a given structure depend upon its behavior. For a planar frame member, the significant internal forces are the hear and axial direct forces and the flexural moment. Figure 2.11 depicts the manner in which these generalized forces are developed by the internal stresses. The resultants of the shear stress distribution , and the axial stress distribution 1,

also a major cause of deflections, and this subject is also discussed as is deflection due to shear. Internal forces developed in each type of framed structure are presented in Table 2.1.

Table 2.1.Internal Forces (Stress Resultants) for the Basic Structural Types

Structural TypeSchematic of Assumed Member ForcesDescriptionPlane truss and space trussOnly axial Force, F1, is significant.Beam and plane frameOnly in-plane shear force F1 , flexural moment F2 , and axial force F3, are significant. For beams, axial force generally does not exist.

Structural TypeSchematic of Assumed Member ForcesDescriptionGridOnly shear force F1 and flexural moment F2 (both transverse to the plane of the grid) and the torsional moment F3 are significant. Space FrameAll stress resultants are significant: axial force F1, shear forces F2 and F3 torsional moments F4, and flexural moments F5 and F6

PROBLEM: ANALYZE THE TRUSS BY STIFFNESS METHOD

ANALYSIS OF GRIDS

Definition:

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

when integrated over the member cross section, are the transverse shear force V and the longitudinal (or axial) force P, respectively. Flexural stresses are the source of two internal forces. Tensile and compressive region of this stress distribution,2 produce the compressive force C and tensile force T, respectively. Unlike the resultant P. which acts at the centroid. the resultants T and C are separated by a distance, and neither of their lines of action coincides with the centroid. Consequently, the net effect of T and C is the creation of an internal resisting moment, M. Shear and moment resultants that develop in an individual framed member exhibit a relationship to each other and to the transverse loads applied to the member. These fundamental interrelationships will be formulated in Chapter 3. Flexural moment is

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 of application. Each of the six types of framed structures were illustrated in Fig. 1.4. Inspection of that figure indicates the applied load types that are included in each of the six idealized models. Planar truss. Concentrated loads applied in two orthogonal directions at the joints. Space truss. Concentrated loads applied in three ol1hogonal directions at the joints.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

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 outof-plane 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 in-plane moments applied at the joints and in the plane of the frame.Space frame. a. Transverse loads and flexural moments that are

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. The weight of objects (either dead or live load) and the hydrostatic pressure of water are two examples of applied loads. In each case there is contact between the load source and the loaded structure. In a strict technical sense. loads arc not always applied to the structure. Frequently. structures are subjected to phenomena not commonly referred to as loads. Temperature change and shrinkage of

material are two examples. Each of these phenomena cause a structure to experience strain and stress and consequently, to deform. These are the same kinds of effects as caused by applied loads. When such effects are included, it is conventional to refer to the general category of loads as "actions." In this text the terms "load" and "action" are treated as synonomous, both meaning any effect that causes stress and/or strain in a structure. any action must satisfy Newtons first law, i.e., it must cause a reaction. The fundamental concepts of how a structure an action are developed in the balance of this chapter.

Internal Loadings. One of the most important applications of statics in the analysis of problems involving mechanics of materials is to be able to determine the resultant force and moment acting within a body, which are necessary to hold the body together when the body is subjected to external loads. For example, consider the body shown in Fig. 1-2a, which is held in equilibrium by the four external forces.* In order to obtain the internal Loadings acting on a specific region within the body, it is necessary to use the method of sections. This requires that an imaginary section or "cut" be made through the region where the internal loadings are to be determined. The two parts of the body are * The body's weight is not shown, since it is assumed to be quite small, and therefore negligible compared with the other loads.

(a)Fig. 1-1

(b)Fig. 1-2

then separated, and a free-body diagram of one of the parts is drawn. If we consider the section shown in Fig. 1-2a, then the resulting free-bodydiagram of the bottom part of the body is shown in Fig. 1-2b. Here it can be seen that there is actually a distribution of internal force acting on the "exposed" area of the section. These forces represent the effects of the material of the top part of the body acting on the adjacent material of the bottom part. Although this exact distribution may be unknown, we can use statics to determine the resultant internal force and moment, FR and MRo this distribution exerts at a specific point O on the sectioned area, Fig. 1-2c. Since the entire body is in equilibrium, then each part of the body is also in equilibrium. Consequently, FR and MRo can be determined by

applying Eqs. 1-1 to anyone of the two parts of the sectioned body. When doing so, note that FR acts through point 0, although its computed value will not depend on the location of this point. On the other hand, MRo does depend on this location, since the moment arms must extend from O to the line of action of each force on the free-body diagram. It will be shown in later portions of the text that point O is most often chosen at the centroid of the sectioned area, and so we will always choose this location for O, unless otherwise stated. Also, if a member is long and slender, as in the case of a rod or beam, the section to be considered is generally taken perpendicular to the longitudinal axis of the member. This section is referred to as the cross section.

(c)Fig. 1-2

(d)Fig. 1-2

Later in this text we will show how to relate the resultant internal force and moment to the distribution of force on the sectioned area, Fig. 1-2b, and thereby develop equations that can be used for analysis and design of the body. To do this, however, the components of FR and MRo acting both normal or perpendicular to the sectioned area and 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; 1-2d, then FR and MRo can each be resolved into three components. Four different types of loadings 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 the two segments of the body.

V is called the shear force, and it can be determined from its two components using vector addition, V = Vx + Vy. The shear force lies in the plane of the area and is developed when the external loads tend to cause the two segments of the body to slide over one another.Tz is called the torsional moment or torque. It is developed when the external loads tend to twist one segment of the body with respect to the other.M is called the bending moment. It is determined from the vector addition of its two components, M = Mx + My. The bending moment is caused by the external loads that tend to bend the body about an axis lying within the plane of the area.

In this text, note that representation of a moment or torque is shown in three dimensions as a vector with an associated curl, Fig. I-2d. By the right-hand rule, the thumb gives the arrowhead sense of the vector and the fingers or curl indicate the tendency for rotation (twist or bending).

Fig. 1-3

Fig. 1-3

Each of the six unknown x, y, z components of force and moment shows in Fig. 1-2d can be determined directly from the six equations of equilibrium, that is, Eqs. 1-2, applied to either segment of the body. If, however, the body is subjected to a coplanar system of forces, then only normal-force, shear, and bending-moment components will exist at the section. To show this, consider the body in Fig. 1-3a. If it is in equilibrium, then the internal resultant components, acting at the indicated section, can be determined by first "cutting" the body into two parts, as shown in Fig. 1-3b, and then applying the equations of equilibrium to one of the parts. Here the internal resultants consist of the normal force N, shear force V, and bending moment Mo. These loadings must be equal in magnitude and opposite in direction on each

of the sectioned parts (Newton's third law). Furthermore, the magnitude of each unknown is determined by applying the three equations of equilibrium to either one of these parts, Eqs. 1-3. If we use the x, y, z coordinate axes, with origin established at point O, as shown on the left segment, then a direct solution for N can be obtained by applying Fx = 0, and V can be obtained directly from Fy = 0. Finally, the bending moment Mo can be determined directly by summing moments about point O (the z axis), Mo = 0, in order to eliminate the moments caused by the unknowns N and V.

The method of sections is used to determine the internal loadings at a point located on the section of a body. These resultants are statically equivalent to the forces that are distributed over the material on the sectioned area. If the body is static, that is, at rest or moving with constant velocity, the resultants must be in equilibrium with the external loadings acting on either one of the sectioned segments of the body. We will now present a procedure that can be used for applying the method of sections to determine the internal resultant normal force, shear force, bending moment, and torsional moment at a specific location in a body.

Support Reactions. First decide which segment of the body is to be considered. Then before the body is sectioned, it will be necessary to determine the support reactions or the reactions at the connections only on the chosen segment of the body. This is done by drawing the free-body diagram for the entire body, establishing a coordinate system, and then applying the equations of equilibrium to the body. Free-Body Diagram. Keep all external distributed loadings, couple moments, torques, and forces acting on the body in their exact locations, then pass an imaginary section through the body at the point where the internal resultant loadings are to be determined. If the body represents a member of a structure or mechanical device, this section is

often taken perpendicular to the longitudinal axis of the member. Draw a free-body diagram of one of the' 'cut" segments and indicate the unknown resultants N, V, M, and T at the section. In most cases, these resultants are placed at the point representing the geometric center or centroid of the sectioned area. In particular, if the member is subjected to 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 the resultant components acting along the axes. Equations of Equilibrium. Apply the equations of equilibrium to obtain the unknown resultants. Moments should be summed at the section, about the axes where the resultants act. Doing this eliminates the

unknown forces N and V and allows a direct solution for M (and T). If the solution of the equilibrium equations yields a negative value for a resultant, the assumed directional sense of the resultant is opposite to that shown on the free-body diagram. The following examples illustrate this procedure numerically and also provide a review of some of the important principles of statics. Since statics plays such an important role in the analysis of problems in mechanics of materials, it is highly recommended that one solves as many problems as possible of those that follow these examples.

General State of stressAs we have stated, the six actions can combine to produce a combination of stresses. The question of exactly what is stress should be answered in a very simplistic manner that is also quite exact in a theoretical sense. We idealize a load-carrying body as shown in Fig. 1.4 and say that the body is a continuum. Essentially, a continuum is a

FIGURE 1.4 Idealized body.

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 definite molecular structure. On a microscopic scale a material is composed of a space with atoms at specific locations. The continuum model is a collection of many molecules and is large enough that the individual molecular interactions for the material can be ignored and the total of all molecular interactions can be averaged and the continuum can be assigned some overall gross property to describe its behavior. The continuum is considered to be quite large compared to an atomistic model; however, it can still be imagined to be small enough to be of differential size. In other words, we can effect the mathematical concept of the limit of some quantity with respect to a length dimension. We assume that as we find the limiting value of a quantity as a length parameter approaches zero that we do not violate the material assumption of a continuum; the continuum still exists even though it may be of differential size.

Consider the continuum of Fig. 1.4 to be in statical equilibrium and imagine a slice taken through the continuum, as shown in Fig. 1.5. The continuum is still in equilibrium since internal forces and couples at the slice can be viewed as external balancing forces and couples. We are concerned with forces acting on the smooth, cut surface that are illustrated as acting on individual, small surface areas. On a small scale these forces are considered to be acting in a direction either normal or tangent to their respective areas. The forces originate from the molecular interactions at the microscopic level; however, as we have stated, we assume that microscopic forces are averaged and their resultants act on the individual areas. We do not consider small couples acting on each element even though they may exist. It has been demonstrated, both experimentally and analytically, that small-body couples can be ignored for the general theory of mechanics of materials. Every force depicted in Fig. 1.5 can be resolved into one normal component and two shear components. The definition of normal and shear arises naturally; normal

yFIGURE 1.5

forces act normally to their area and shear forces act tangentially to their area. Visualize one small area A as shown in Fig. 1.5b. The force F is shown in components Fx, Fy, and Fz. Normal stress is denoted by (sigma) and is defined as the normal force per unit area; hence, the terminology normal stress. Normal stress is given as (1.1)The Greek letter (tau) is usually used for shear stress and is the shear force divided by the area, or (1.2)

and (1.3)A more complete description is given in Chapter 9 concerning shear and normal stresses 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 produced either a normal stress or shear stress. In every case we begin with a definition of stress, either normal or shear, as given by Eqs. (1.1), (1.2), and (1.3), and establish a theory that describes how a particular action causes a stress. The distribution of stress that occurs on the cross section of the member because of the action of forces and couples is dependent upon the way the load-carrying member deforms. In the case of the axial force, Fig. 1.3a, the member either elongates or shortens in the axial direction. If we assume a uniform deformation, meaning that every point on the cross

section deforms an equal amount parallel to the beam axis, then we must investigate the limitation of the theory subject to that assumption. It turns out that each of the six actions produces some corresponding deformation of the member, and prior to developing a theory for stress distribution we must establish the deformation characteristics of the member when subjected to a particular load. Chapters 2-9 are devoted to the study of concepts that have been briefly introduced in this section.

In Sec. 1.2 we showed how to determine the internal resultant force and moment acting at a specified point on the sectioned area of a body, Fig. 1-9a. It was stated that these two loadings represent the resultant effects of the actual distribution of force acting over the sectioned area, Fig. 1-9b. Obtaining this distribution of internal loading, however, is one of the major problems in mechanics of materials.

Later it will be shown that to solve this problem it will be necessary to study how the body deforms under load, since each internal force distribution will deform the material in a unique way. Before this can be done, however, it is first necessary to develop a means for describing the internal force distribution at each point on the sectioned 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 A shown shaded in Fig. 1-9b. As we reduce A to a smaller and smaller size, we must make two assumptions regarding the properties of the material. We will consider the material to be continuous, that is, to consist of a continuum or uniform distribution of matter having no voids, rather than being composed of a finite number

of distinct atoms or molecules. Furthermore, the material must be cohesive, meaning that all portions of it are connected together, rather than having breaks, cracks, or separations. Now, as the subdivided area A of this continuous-cohesive material is reduced to one of infinitesimal size, the distribution of force acting over the entire sectioned area will consist of an infinite number of forces, each acting on an element A located at a specific point on the sectioned area. A typical finite yet very small force F, acting on its associated area A, is shown in Fig. 1-9c. This force, like all the others, will have a unique direction, but for further discussion we will replace it by two of its components, namely, Fn and Ft, which are taken normal and tangent to the area, respectively. As the area A approaches zero, so do the

force F and its components; however, the quotient of the force and area will, in general, approach a finite limit. This quotient is called stress, and as noted, it describes the intensity of the internal force on a specific plane (area) passing through a point.

Fig. 1.9(a)

Fig. 1.9(b)(c)

Normal Stress. The intensity of force, or force per unit area, acting normal to A is defined as the normal stress, (sigma). Mathematically it can be expressed as If the normal force or stress "pulls" on the area element A as shown in Fig. I-9c, it is referred to as tensile stress, whereas if it "pushes" on A it is called compressive stress.Shear Stress. Likewise, the intensity of force, or force per unit area, acting tangent to A is called the shear stress, (tau). This component is expressed mathematically as (14)

Fig. 1.9

(15)In Fig. I-9c, note that the orientation of the area A completely specifies the direction of Fn, which is always perpendicular to the area. On the other hand, each shear force Ft can act in an infinite number of directions within the plane of the area. Provided, however, the direction of F is known, then the direction of Ft can be established by resolution of F as shown in the figure. Cartesian Stress Components. To specify further the direction of the shear stress, we will resolve it into rectangular components, and to do this we will make reference to x, y, z coordinate axes, oriented as shown

in Fig. 110a. Here the element of area A = x y and the three Cartesian components of F are shown in Fig. 1-10b. We can now express the normal-stress component as

Fig. 1.10(a)

Fig. 1.10(c)(b)

Fig. 1.11(b)

Fig. 1.11(c)(d)

The subscript notation z in z is used to reference the direction of the outward normal line, which specifies the orientation of the area ~A. Two subscripts are used for the shear-stress components, zx and zy. The z specifies the orientation of the area, and x and y refer to the direction lines for the shear stresses.and the two shear-stress components as

To summarize these concepts, the intensity of the internal force at a point in a body must be described on an area having a specified orientation. This intensity can then be measured using three components of stress acting on the area. The normal component acts normal or perpendicular to the area, and the shear components act within the plane of the area. These three stress components are shown graphically in Fig. 1-l0c. Now consider passing another imaginary section through the body parallel to the xz plane and intersecting the front side of the element shown in Fig.1-10a. The resulting free-body diagram is shown in Fig. 1-11a. Resolving the force acting on the area A = x z into its rectangular components, and then determining the intensity of these

force components, leads to the normal stress and shear-stress components shown in Fig. 1-11b. Using the same notation as before, the subscript y in y, yx, and yz refers to the direction of the normal line associated with the orientation of the area, and x and z in yx and yz refer to the corresponding direction lines for the shear stress. Lastly, one more section of the body parallel to the yz plane, as shown in Fig. 1-11c, gives rise to normal stress x and shear stresses xy and xz, Fig. 1-11d. If we continue in this manner, using corresponding parallel x planes, we can "cut out" a cubic volume element of material that represents the state of stress acting around the chosen point in the body, Fig. 1-12.

Equilibrium Requirements. Although each of the six faces of the element in Fig. 1-12 will have three components of stress acting on it, if the stress around the point is constant, some of these stress components can be related by satisfying both force and moment equilibrium for the element. To show the relationships between the components we will consider a free-body diagram of the element, Fig. 1-13a. This element has a volume of V = x y z, and in accordance with Eqs. 1-4 and 1-5, the forces acting on each face are determined from the product of the average stress times the area of the face. For simplicity, we have not labeled the "dashed" forces acting on the "hidden" sides of the element. Instead, to view, and thereby label, some of these forces, the element is shown from a front view in Fig. 1-13b. Here it should be noted that the

Fig. 1.13 (a) Element free-body diagram

Fig. 1.13 (b) Element free-body diagram

force components on the "hidden" sides of the element are designated with stresses having primes, and these forces are shown in the opposite direction to their counterparts acting on the opposite faces of the element.If we now consider force equilibrium in the y direction, we have

CONCEPT OF STRESS AT A GENERAL POINT IN AN ARBITRARILY LOADED MEMBER In Arts. 1-3 and 1-4 the concept of stress was introduced by considering the internal force distribution required to satisfy equilibrium in a portion of a bar under centric load. The nature of the force distribution led to uniformly distributed normal and shearing stresses on transverse planes through the bar. In more complicated structural members or machine components the stress distributions will not be uniform on arbitrary internal planes; therefore. a more general concept of the state of stress at a point is needed.

Consider a body of arbitrary shape that is in equilibrium under the action of a system of applied forces. The nature of the internal force distribution at an arbitrary interior point O can be studied by exposing an interior plane through O as shown in Fig. 1-13a. The force distribution required on such an interior plane to maintain equilibrium of the isolated part of the body, in general, will not be uniform; however, any distributed force acting on the small area A surrounding the point of interest O can be replaced by a statically equivalent resultant force Fn through O and a couple Mn. The subscript n indicates that the resultant force and couple are associated with a particular plane through O-namely, the one having an outward normal in the n direction at O. For any other plane through O the values of F

and M could be different. Note that the line of action of Fn or Mn may not coincide with the direction of n. If the resultant force Fn is divided by the area A, an average force per unit area (average resultant stress) is obtained. As the area A is made smaller and smaller, the couple Mn vanishes as the force distribution becomes more and more uniform. In the limit a quantity known as the stress vector4 or resultant stress is obtained. Thus, 4 The component of a tensor on a plane is a vector; therefore, on a particular plane, the stresses can be treated as vectors.

FIG. 113

FIG. 113

FIG. 113

In Art. 1-3 it was pointed out that materials respond to components of the stress vector rather than the stress vector itself. In particular, the components normal and tangent to the internal plane were important. As shown in Fig. 1-13b the resultant force Fn can be resolved into the components Fnn and Fnt. A normal stress n and a shearing stress n are then defined as

For purposes of analysis it is convenient to reference stresses to some coordinate system. For example, in a Cartesian coordinate system the stresses on planes having outward normals in the x, y. and z directions are usually chosen. Consider the plane having an outward normal in the x direction. In this case the normal and shear stresses on the plane will be x and x, respectively. Since x, in general, will not coincide with the y or z axes, it must be resolved into the components xy and xz, as shown in Fig. 1-13c. Unfortunately the state of stress at a point in a material is not completely defined by these three components of the

stress vector since the stress vector itself depends on the orientation of the plane with which it is associated. An infinite number of planes can be passed through the point, resulting in an infinite number of stress vectors being associated with the point. Fortunately it can be shown (see Art. 1-9) that the specification of stresses on three mutually perpendicular planes is sufficient to describe completely the state of stress at the point. The rectangular components of stress vectors on planes having outward normals in the coordinate directions are shown in Fig. 1-14. The six faces of the small element are denoted by the directions of their outward normals so that the positive x face is the one whose outward normal is in the direction of the positive x axis. The coordinate axes x, y, and z are arranged as a right-hand system. The

FIG. 114

sign convention for stresses is as follows. Normal stresses (indicated by the symbol and a single subscript to indicate the plane on which the stress acts) are positive if they point in the direction of the outward normal. Thus normal stresses are positive if tensile. Shearing stresses are denoted by the symbol followed by two subscripts; the first subscript designates the plane on which the shearing stress acts and the second the coordinate axis to which it is parallel. Thus, xy is the shearing stress on an x plane parallel to the z axis. A positive shearing stress points in the positive direction of the coordinate axis of t he second subscript if it acts on a surface with an outward normal in the positive direction. Conversely, if the outward normal of the surface is in the negative direction, then the positive shearing stress points in the

negative direction of the coordinate axis of the second subscript. The stresses shown on the element in Fig. 1-14 are all positive.

STRESS UNDER GENERAL LOADING CONDITIONS; COMPONENTS OF STRESS The examples of the previous sections were limited to members under axial loading and connections under transverse loading. Most structural members and machine components are under more involved loading conditions. Consider a body subjected to several loads P1, P2, etc. (Fig. 1.32). To understand the stress condition created by these loads at some point Q within the body, 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 original loads, and to normal and shearing forces distributed over the section. We shall denote by Fx and Vx, respectively, the normal and the shearing forces acting on a small

Fig. 1.32

Fig. 1.33(a)(b)

area A surrounding point Q (Fig. 1.33a). Note that the superscript x is used to indicate that the forces Fx and Vx act on a surface perpendicular to the x axis. While the normal force Fx has a well-defined direction, the shearing force Vx may have any direction in the plane of the section. We therefore resolve Vx into two component forces, Vxy and Vxz in directions parallel to the y and z axes, respectively (Fig. 1.33 b). Dividing now the magnitude of each force by the area A, and letting A approach zero, we define the three stress components shown in Fig. 1.34: (1.18)

zFig. 1.34

We note that the first subscript in x, xy and xz is used to indicate that the stresses under consideration are exerted on a surface perpendicular to the x axis. The second subscript in xy and xz identifies the direction of the component. The normal stress x is positive if the corresponding arrow points in the positive x direction, i.e., if the body is in tension, and negative otherwise. Similarly, the shearing stress components xy and xz are positive if the corresponding arrows point, respectively, in the positive y and z directions. The above analysis may also be carried out by considering the portion of body located to the right of the vertical plane through Q (Fig. 1.35). The same magnitudes, but opposite directions, are obtained for the normal and shearing forces Fx, Vxy and Vxz Therefore, the same values are also obtained for the corresponding stress components, but since the section in Fig. 1.35 now faces the negative x axis, a positive sign for x will indicate that the corresponding arrow points in the negative x direction. Similarly, positive signs for xy and xz will indicate that the

corresponding arrows point, respectively, in the negative y and z directions, as shown in Fig. 1.35. zFig. 1.35

Passing a section through Q parallel to the zx plane, we define in the same manner the stress components, y, yz, and yx Finally, a section through Q parallel to the xy plane yields the components z, zx and zy.To facilitate the visualization of the stress condition at point Q, we shall consider a small cube of side a centered at Q and the stresses exerted on each of the six faces of the cube (Fig. 1.36). The stress components shown in the figure are x, y and z which represent the normal stress on faces respectively perpendicular to the x, y, and z axes, and the six shearing stress components xy xz etc. We recall that, according to the definition of the shearing stress components, xy represents the y component of the shearing stress exerted on the face perpendicular to the x axis, while yx represents the x component of the shearing stress exerted on the face perpendicular to 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 stresses acting on the faces of the cube differ slightly from the stresses at Q the error

Fig. 1.36

Fig. 1.37

involved is small and vanishes as side a of the cube approaches zero. Important relations among the shearing stress components will now be derived. Let us consider the free-body 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 are obtained by multiplying the corresponding stress components by the area A of each face. We first write the following three equilibrium equations: (1.19)Since forces equal and opposite to the forces actually shown in Fig. 1.37 are acting on the hidden faces of the cube, it is clear that Eqs. (1.19) are satisfied. Considering now the moments of the forces about axes Qx', Qy', and Qz' drawn from Q in directions respectively parallel to the x, y, and z axes, we write the three additional equations (1.20)

Fig. 1.38

Using a projection on the x'y' plane (Fig. 1.38), we note that the only forces with moments about the z axis different from zero are the shearing forces. These forces form two couples, one of counterclockwise (positive) moment (xy A)a, the other of clockwise (negative) moment (xy A)a. The last of the three Eqs. (1.20) yields, therefore, from which we conclude that The relation obtained shows that the y component of the shearing stress exerted on a face perpendicular to the x axis is equal to the x component of the shearing stress exerted on a face perpendicular to the y axis. From the remaining two equations (1.20), we derive in a similar manner the relations (1.22)(1.21)

We conclude from Eqs. (1.21) and (1.22) that only six stress components are required to define the condition of stress at a given point Q, instead of nine as originally assumed. These six components are ux, uy, uz, T XY' Tyz, and T ZX. We also note that, at a given point, shear cannot take place in one plane only; an equal shearing stress must be exerted on another plane perpendicular to the first one. For example, considering again the bolt of Fig. 1.29 and a small cube at the center Q of the bolt (Fig. 1.39a), we find that shearing stresses of equal magnitude must be exerted on the two horizontal faces of the cube and on the two faces that are perpendicular to the forces P and P' (Fig. 1.39b). Before concluding our discussion of stress components, let us consider again the case of a member under axial loading. If we consider a small cube with faces respectively parallel to the faces of the member and recall the results obtained in Sec. 1.11, we find that the conditions of stress in the member may be described as shown in Fig. lAOa; the only stresses are normal stresses U x exerted on the faces of the cube which are perpendicular to the x axis. However, if the small cube

Fig. 1.39(a)(b)

Fig. 1.40

is rotated by 45 about the z axis so that its new orientation matches the orientation of the sections considered in Fig. 1.31c and d, we conclude that normal and shearing stresses of equal magnitude are exerted on four faces of the cube (Fig. 1.40b). We thus observe that the same loading condition may lead to different interpretations of the stress situation at a given point, depending upon the orientation of the element considered. More will be said about this in Chap 7.

Stress, Strain, and Their Relationships 2.1 Introduction The concepts of stress and strain are two of the most important concepts within the subject of mechanics of materials or mechanics of deformable bodies. They are discussed in detail in this chapter, particularly as they relate to two-dimensional situations. In the case of stress as well as in the case of strain, emphasis is placed on the use of the semigraphical procedure known as the Mohrs circle solution. The underlying mathematical concepts leading to Mohr's circle are developed and discussed. Examples are solved to illustrate the use of this powerful semigraphical method of solution, which will be used throughout this book whenever problems are encountered dealing with stress and strain analysis.

The treatment given to the concepts of stress and strain in this book differs from that in other books in several important respects, the most significant of which is the fact that the sign convention adopted for strain is compatible with the sign convention for stress as they relate to the construction of the corresponding Mohr's circles. This approach is advantageous in that it makes the construction of the stress and strain Mohr's circles identical. The discussion relating stress to strain in this chapter is limited to the range of material behavior within which the strain varies linearly with stress. This procedure frees the students from information which, although very important, is extraneous for the time being. A more complete discussion of material behavior is provided in Chapters 3 and 4.

2.2 Concept of Stress at a Point If a body is subjected to external forces, a system of internal forces is developed. These internal forces tend to separate or bring closer together the material particles that make up the body. Consider, for example, the body shown in Figure 2.1(a), which is subjected to the external forces Fl, F2, ... , Fi. Consider an imaginary plane that cuts the body into two parts, as shown. Internal forces are transmitted from one part of the body to the other through this imaginary plane. Let the free-body diagram of the lower part of the body be constructed as shown in Figure 2,1 (b). The forces F1, F2, and F3 are held in equilibrium by the action of an internal system of forces 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/or by a couple. For the sake of simplicity in introducing the concept of stress, only the force R is assumed to exist. In general, the force R may be decomposed into a

Figure 2.1

Figure 2.1

component Fn perpendicular to the plane and known as the normal force, and a component 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 area under consideration, and it is therefore desirable to be able to determine the magnitude of both the normal and shear stresses at any point within the area. If the normal and shear forces acting over a differential element of area A in the neighborhood of point O are Fn and Ft respectively, as shown in Figure 2.1 (b), then the normal stress and the shearing stress, are given by the following expressions: (2.1)

In the special case where the components Fn and Fr are uniformly distributed over the entire area A, then = Fn/A and = Ft/A. Note that a normal stress acts in a direction perpendicular to the plane on which it acts and it can be either tensile or compressive. A tensile normal stress is one that tends to pull the material particles away from each other, while a compressive normal stress is one that tends to push them closer together. A shear stress, on the other 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 ( or ) consist of units of force divided by units of area. Thus, in the British gravitational system of measure, such units as pounds per square inch (psi) and kilopounds per square inch (ksi) are common. In the metric (SI) system of measure, the unit that has been proposed for stress is the Newton per square meter (N/m2), which is called the pascal 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 by the symbol MPa. This unit may also be written as MN/m2.

Components of Stress In the most general case, normal and shear stresses at a point in a body may be considered to act on three mutually perpendicular planes. This most general state of stress is usually referred to as triaxial. It is convenient to select planes that are normal to the three coordinates axes x, y, and z and designate them as the X, Y, and Z planes, respectively. Consider these planes as enclosing a differential volume of material in the neighborhood of a given point in a stressed body. Such a volume of material is depicted in Figure 2.2 and is referred to as a three-dimensional stress element. 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 two perpendicular components. The notation for stresses used in this text consists of affixing one subscript to a normal stress, indicating the plane on which it is acting, and two subscripts to a

shear stress, the first of which designates the plane on which it is acting and the second its direction. For example, x is a normal stress acting on the X plane, xy is a shear stress acting on the X plane and pointed in the positive y direction, and xz is a shear 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 of the three mutually perpendicular planes that define the stress element. Thus there exists a total of nine stress components that must be specified in order to define completely the state of stress at any point in the body. By considerations of the equilibrium of the stress element, it can easily be shown that, xy = yx, xz = zx, and yz = zy so that the number of stress components required to completely define the state of stress at a point is reduced to six.

By convention, a normal stress is positive if it points in the direction of the outward normal to the plane. Thus a positive normal stress produces tension and a negative normal stress produces compression. A component of shear stress is positive if it is pointed along the positive direction of the coordinate axis and if the outward normal to its plane is also in the positive direction of the corresponding axis. If, however, the outward normal is in the negative direction of the coordinate axis, a positive shear stress will also be in the negative direction of the corresponding axis. The stress components shown in Figure 2.2 are all positive. It should be noted, however, that such a sign convention for shear stress is rather cumbersome. It is only used in the analysis of triaxial stress problems that are usually dealt with in advanced courses such as the theory of elasticity. A complete study of the triaxial or three-dimensional state of stress is beyond the scope of this chapter, and the analysis that follows is limited to the special case in which the stress components in one direction are all zero. For example, if all the stress

components in the z direction are zero (i.e., xz = yz = z = 0), the stress condition reduces to a biaxial or two-dimensional state of stress in the xy plane. This state of stress is referred to as plane stress. Fortunately, many of the problems encountered in practice 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 in which all the stress components in one direction vanish. Thus, if it is assumed that all the components in the z direction shown in Figure 2.2 are zero (i.e., xz = yz = z = 0), the stress element shown in Figure 2.3(a) is obtained and it is the most general plane stress condition that can exist. It should be observed that a stress element is in reality a schematic representation of two sets of perpendicular planes passing through a point and that the element degenerates into a point in the limit when both dx and dy approach zero.

From considerations of the equilibrium of forces on the stress element in Figure 2.3(a), it can be shown that xy = yx. Thus assume that the depth of the stress element into the paper is a constant equal to h. Since, by definition, force is the product of stress and the area over which it acts, then a summation of moments of all forces about a z axis through point O leads to the following equation: from which

STRESS AT A POINT Stress 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 C are to be computed. The point must be drawn large enough for it to be visualized. Therefore the concept of a stress block or material element is necessary. The stress block is the point enlarged for the practical purpose of drawing it and is referred to as a stress element since its actual size is elemental. Point C of Fig. 8.2 is shown enlarged in Fig. 8.la. Physically, the element can be visualized as a small square near the outside surface of the beam located at point C. The stresses are identified as acting on the edge of the elemental square and is the standard concept of stress at a point in two dimensions.

FIGURE 8.1 Element viewed along the z axis;(b) positive stresses acting on an element; (a)(b)

FIGURE 8.1 (c) element viewed along the x axis;(d) element viewed along the y axis. (c)(d)

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 direction of the stress at the face, or area, of the cube on which it acts. For instance, the stress xx 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 second subscript identifies the direction of the stress. A shear stress on the face of the cube normal to the y axis and acting in the x direction would be yx as shown in Fig. 8.1b. The complete three-dimensional description of stress at a point is illustrated in the figure. The discussion in the previous chapters concerning shear stresses on mutually perpendicular planes would imply that xy = yx yz = zy and xz = zx. The subscript notation lends itself toward identifying this equivalence. The nine components shown in Fig. 8.lb are conveniently presented using the following format.

(8.5)The use of the boldface will represent the stress at a point. Equation (8.5) represents the stress, , as a nine-component quantity and is referred to as a stress tensor, which has certain mathematical properties that are useful in advanced 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 the element, they are actually applied to a surface that is perpendicular to the plane of the page. The stress components of Fig. 8.la will be written as

*