10.1 INTRODUCTION

In generalized strained body, stress at a point can be expressed as a stress tensor given by

The stress components σx, τyz, τzx, etc. due to external force may be fl exure, axial loads, torsion loads, or their combinations. In this chapter, we consider the stresses in a plane only. Consider stresses in X–Y plane.

σx and σy are the normal stresses τyz and τxy is the shearing or tangential stress τxy = τyx as τyx is complimentary shear stress to τxy. Denoting τxy = τyx, the generalized stress in 2D stress system is given by

The effects σx, σy, and τxy on inclined plane shown in Figure 10.2 can be represented as σn and τn, that is, normal stress and shear stress, respectively.

Often in the design problems, the dimensions of the stained element are proportioned depending on the maximum normal stress criterion or maximum shear stress criterion. Thus, it is required to quantify the maximum or minimum normal stress and maximum shear stress in terms of σx, σy, and τxy.

10.2 EXPRESSION FOR σn AND τn

Consider a generalized 2D stress system, shown in Figure 10.3(a).

Consider the triangular portion ABC and convert the stresses into forces, as shown in Figure 10.3(b). Take unit width for force calculations, that is, normal forces on the plane AC is σn × AC × 1.

Resolving the forces along the normal to the plane AC and applying equilibrium condition,

Resolving the forces along the plane and applying equilibrium condition,

The sign convention to be followed:

In clockwise direction, and −ve otherwise.

A tip for student: Always draw Figure 10.3(a) and compare that with the problem you have to solve. Then, ascertain proper sign for the given values in the problems you are solving.

10.3 EXPRESSION FOR MAXIMUM OR MINIMUM NORMAL STRESS

From the earlier discussion, we had the expression for σn (normal stress) and τn (shearing stress). We have to investigate maximum normal stress and minimum normal stress.

6.1 INTRODUCTION

Frameworks of axially loaded members arranged to resist external forces are called trusses. The members are generally welded or riveted at the joints, but for the sake of simplifi cation of calculations, the joints are assumed to be hinges. The general confi gurations of trusses are shown in Figure 6.1.

6.2 TYPES OF TRUSSES

The different types of trusses are classifi ed into three categories. They are

• 1. Statically determinate trusses

• 2. Statically indeterminate trusses

• 3. Statically unstable trusses or imperfect trusses.

This classifi cation is based on the members present in the truss and the number of equilibrium equations that can be used to analyze the truss. Analysis of statically determinate trusses only will be dealt in this chapter. Let us consider a simple triangulation shown in Figure 6.2.

In Figure 6.2(a), the truss has three members. Each member carries axial force (compression to tension), thus there are three internal forces. The three external reactions are namely two reactions (vertical and horizontal) at A and one reaction at B.

The total number of internal and external reactions is 6 = (3 + 3). There are three joints in the triangulation. The number of equilibrium equations at each and every joint is 2 (i.e., sum of vertical forces = 0). Hence the total equilibrium equations are 6 (i.e., 3 × 2). As the total number of reactive forces is equal to the total number of equilibrium equation, the structure can be analyzed using static equilibrium equations alone. Thus, these types of structures are called statically determinate structures. Consider the truss shown in Figure 6.3 (b).

Number of external reactions (R) = 4 (i.e., two at A and two at B)

Number of equilibrium equations = Number of unknown forces.

If the number of unknown forces is more than the number of equilibrium equations, the corresponding truss is called as statically indeterminate truss. If m + R > 2j, then the truss falls under the category of statically indeterminate truss. Consider the truss shown in Figure 6.3.

14.1 INTRODUCTION

The fl exure formula derived in the theory of simple bending assumes that the loading plane should pass through the axis of symmetry. This condition is not satisfi ed in some cases where the loading is gravity loading and the member rests on an inclined support such as purlins of a truss. In some cases, the cross-section may not have axis of symmetry such as angular sections. This chapter provides the solution for such problems where bending is referred as unsymmetrical bending.

14.2 UNSYMMETRICAL BENDING

If the plane of loading of a beam does not coincide with one of the principal axes of the crosssection, then the resulting bending is termed as unsymmetrical bending. Unsymmetrical bending may occur in two ways.

• 1. The section is symmetrical about two axes like I section, rectangular section, and circular section but the loading plane is inclined to both the principal axes.

• 2. The section itself is unsymmetrical like angle section or a channel.

The following examples illustrate unsymmetrical bending.

14.3 PRINCIPAL MOMENT OF INERTIA

Let us consider an arbitrary cross-section. The centroid of the cross-section is taken as the origin of coordinate axes system. Consider an elemental area δA at point P(x,y).

P(x,y) are coordinates of the elemental area about axes X and Y passing through the centroid of the cross-section.

Let the coordinate axes X and Y be rotated and be defi ned by the angle θ.

Then, the coordinates of the elemental area δA is given by transformation matrix.

Product of inertia may be positive or negative and it is zero if any one of the x or y axis is the axis of symmetry of the cross-section. A detailed calculation of the same is presented in the next coming topics of this chapter.

Principal axes are the axes about which the product of inertia of the cross-section is zero.

Hence for principal axes product of inertia IX'Y' be equated to zero. Let α be the angle that defi nes the principal axes.

9.1 INTRODUCTION

Torsion forms one of the basic structural actions besides axial force, bending, and shear. Rotating shafts, pulleys, and gears are the structural components, which are subjected to torsion/twisting moments. Torsion of circular shafts greatly differs from torsion of noncircular shafts. In this chapter, torsion of circular sections will be discussed. Torsion produces shearing stresses in the cross-section of the member. In case of circular cross-sections, the assumption that plane sections remain plane is valid, but the same is not valid in case of noncircular sections. However, solid noncircular sections are subjected to shearing stress alone. In case of open sections such as T, I, and C, the cross-section is subjected to shearing stresses and warping stresses. Warping stresses are normal stresses produced due to warping of the cross-section.

Assumption: The following assumptions were made to evaluate the shearing stress and tensional deformation (twist) in case of circular sections.

• 1. Material of the shaft is homogenous, elastic, and isotropic.

• 2. Plane sections remains plane, even after the application of the twisting moment.

Consider a circular shaft of length ‘L’, subjected to twisting moment ‘T’, as shown in Figure 9.1.

Let ‘D’ be the diameter of the shaft. Consider an angular portion of the shaft. Let its radius be ‘r’. Fixidity at one of the supports does not allow the cross-section to rotate but free rotation is possible at the free end. AB, a fi ber before the application of the twisting, takes the position AB' after the application of the twisting moment. The inclination of the line segment AB' with AB is same along the length of the shaft, as plane sections remain plane even after twisting.

The radial OB (shown in Figure 9.1) rotates by ‘θ’ and assumes the position OB'. ‘θ’ is called the angle of twist.

From geometrical considerations, assuming torsional deformations are very small.

This book Strength of Materials: Fundamentals and Applications is brought to the readers with an aim to provide suffi cient information and point out ways this information can be applied to practical problems in the domain of mechanics, as is done by design engineers in real life situations. It is structured in the form of a text book for undergraduates pursuing civil engineering, mechanical engineering and metallurgical engineering. Our experience in teaching ‘strength of materials’ over the past 30 years fi nds its place in this book. Many questions frequently raised by our students, and are also common problems faced by a lot of other students, while attempting to understand the subjects ‘strength of materials’ or ‘mechanics of solids’ are addressed in this book in the form of worked out examples and in the detailed treatment of the theoretical aspects.

Each chapter is provided with objectives and these objectives are mapped to the worked-out example problems and the exercise problems. This mapping strategy will also help the teaching faculty in deciding the course objectives and in evaluating the course objectives.

At the end of every chapter, previous GATE examination and UPSC competitive examination objective-type questions are provided with solutions. This book is useful for students preparing for competitive examinations.

The fi rst ten chapters are devoted to understanding the effects of basic structural actions, which would be suffi cient for an elementary treatment of the ‘strength of materials’ course. The remaining seven chapters focus on advanced topics wherein combined structural actions are considered.

Care is exercised so that mistakes or typographical errors are minimized. All the same we request the readers to comment and provide suggestions for improving the next edition of this book.

4.1 INTRODUCTION

In the previous chapter, we have seen how to draw shear force diagrams (SFDs) and bending moment diagrams (BMDs). The question that arises is that, how these bending moment (BM) and shear force (SF) should be handled in design and what type of stresses develops due to these internal forces? In this chapter, we will discuss about the quantifi cation of stress developed due to BM. The stress that develops due to bending is normal stress generally referred as fl exural stress or bending stress.

4.2 ASSUMPTIONS IN THEORY OF BENDING

• 1. Material of the beam is homogeneous, elastic, and isotropic.

• A material is said to be elastic, if the force–displacement relationship follows the same path during the loading as well as during the unloading. In homogeneous materials, the elastic properties such as modulus of elasticity and Poisson's ratio is same everywhere. A material is said to be isotropic, if the elastic properties are same in all directions.

• 2. Material of the beam obeys Hooke's law, that is, force–displacement relationship is linear.

• 3. Plane sections normal to the longitudinal remains plane even after bending.

• This is the most important assumption; it means that the strain variation is linear across the depth of the beam cross-section.

• In Figure 4.3, cantilever beam is subjected to moment M. AB is a plane section, before the application of the couple. After the application of the couple, the plane section AB has rotated about C, taken position A'B. As the A'B' is also a straight line, the displacement of any point on AB is proportional to its distance from C. Hence, the strain at any point on the cross-section AB is proportional to its distance from C.

• 4. The beam is subjected to pure bending.

• Pure bending means the beam should be subjected to constant BM and no SF and axial force act on it, for example,

• In the beam shown in Figure 4.4 (a), the portion BC of the beam ABCD is subjected to pure bending. In Figure 4.4 (b), the beam AB is subjected to pure bending.

1.1 INTRODUCTION

The study of strength of materials includes the understanding of internal stresses and deformations of members subjected to external loading. It also includes the study of failure criterion applicable for the solids subjected to loads. The major actions on the bodies subjected to external loading can be considered as axial force, which include axial compression and axial tension, shear force, bending moment, and torsion. Often members are subjected to the mentioned actions either individually or in combined state such as combined bending, torsion, and axial thrust. The internal reactions due to the external forces cannot be visualized, whereas the deformations can be observed, thus can be measured. Hence generally, the failure criterion of a body subjected to external loads depends not only on the internal actions but also on the deformations. To quantify the internal actions due to the above-said forces, the action of the forces and the corresponding deformations are to be studied in detail in the subsequent chapters. The different individual actions on members were presented in Figure 1.1(a)–(e). However, in this chapter concept of internal reactions and their effects will be discussed. In the subsequent chapters, the effects of mentioned individual actions and combined actions will be discussed in detail.

1.2 NORMAL STRESS AND SHEAR STRESS

Consider a body subjected to several forces and surface tractions as shown in Figure 1.2. Take a section 1-1, to observe the effect of all forces on the section considered. Let ‘P’ be the net resultant of the forces. The resistance developed by the body to this resultant at any point within the domain of the body is referred as stress.