Saturday, 29 April 2017





  1. Stress
  2. Strain
  3. Torsion
  4. Shear and Moment in Beams
  5. Stresses in Beams
  6. Deflection of Beams
  7. Statically Indeterminate Beams
  8. Stresses due to combined loads
  9. Composite Beams
  10. Columns
  11. Additional Beam Topics
  12. Special Topics
  13. Inelastic Action


This textbook is intended for use in a first course in mechanics of materials. Programs of instruction relating to the mechanical sciences, such as mechanical,
civil, and aerospace engineering, often require that students take this course in the second or third year of studies. Because of the fundamental nature of the subject matter, mechanics of materials is often a required course, or an acceptable technical elective in many other curricula. Students must have completed courses in statics of rigid bodies and mathematics through integral calculus as prerequisites to the study of mechanics of materials. 



The three fundamental areas of engineering mechanics are statics, dynamics, and mechanics of materials. Statics and dynamics are devoted primarily to the study of the external effects upon rigid bodies—that is, bodies for which the change in shape (deformation) can be neglected. In contrast, mechanics of materials deals with the internal effects and deformations that are caused by the applied loads. Both considerations are of paramount importance in design. A machine part or structure must be strong enough to carry the applied load without breaking and, at the same time, the deformations must not be excessive.

The differences between rigid-body mechanics and mechanics of materials can be appreciated if we consider the bar. The force P required to support the load W in the position shown can be found easily from equilibrium analysis. After we draw the free-body diagram of the bar, summing moments about the pin at O determines the value of P. In this solution, we assume that the bar is both rigid (the deformation of the bar is neglected) and strong enough to support the load W. In mechanics of materials, the statics solution is extended to include an analysis of the forces acting inside the bar to be certain that the bar will neither break nor deform excessively.

Bearing Stress

If two bodies are pressed against each other, compressive forces are developed on the area of contact. The pressure caused by these surface loads is called bearing stress. Examples of bearing stress are the soil pressure beneath a pier and the contact pressure between a rivet and the side of its hole. If the bearing stress is large enough, it can locally crush the material, which in turn can lead to more serious problems. To reduce bearing stresses, engineers sometimes employ bearing plates, the purpose of which is to distribute the contact forces over a larger area.



So far, we have dealt mainly with the strength, or load-carrying capacity, of structural members. Here we begin our study of an equally important topic of mechanics of materials—deformations, or strains. In general terms, strain is a geometric quantity that measures the deformation of a body. There are two types of strain: normal strain, which characterizes dimensional changes, and shear strain, which describes distortion (changes in angles). Stress and strain are two fundamental concepts of mechanics of materials. Their relationship to each other defines the mechanical properties of a material, the knowledge of which is of the utmost importance in design.

Axial Deformation- Stress-Strain Diagram

The strength of a material is not the only criterion that must be considered when designing machine parts or structures. The stiffness of a material is often equally important, as are mechanical properties such as hardness, toughness, and ductility. These properties are determined by laboratory tests. Many materials, particularly metals, have established standards that describe the test procedures in detail. We will confine our attention to only one of the tests—the tensile test of steel—and use its results to illustrate several important concepts of material behavior.

Elastic Limit : A material is said to be elastic if, after being loaded, the material returns to its original shape when the load is removed. The elastic limit is, as its name implies, the stress beyond which the material is no longer elastic. The permanent deformation that remains after the removal of the load is called the permanent set. The elastic limit is slightly larger than the proportional limit. However, because of the diffculty in determining the elastic limit accurately, it is usually assumed to coincide with the proportional limit. 

Yield Point : The point where the stress-strain diagram becomes almost horizontal is called the yield point, and the corresponding stress is known as the yield stress or yield strength. Beyond the yield point there is an appreciable elongation, or yielding, of the material without a corresponding increase in load. Indeed, the load may actually decrease while the yielding occurs. However, the phenomenon of yielding is unique to structural steel. Other grades of steel, steel alloys, and other materials do not yield, as indicated by the stress-strain curves of the materials. Incidentally, these curves are typical for a first loading of materials that
contain appreciable residual stresses produced by manufacturing or aging processes. After repeated loading, these residual stresses are removed and the stress-strain curves become practically straight lines.

Ultimate Stress : The ultimate stress or ultimate strength, as it is often called, is the highest stress on the stress-strain curve. 

Rupture Stress : The rupture stress or rupture strength is the stress at which failure occurs. For structural steel, the nominal rupture strength is considerably lower than the ultimate strength because the nominal rupture strength is computed by dividing the load at rupture by the original crosssectional area. The true rupture strength is calculated using the reduced area of the cross section where the fracture occurred. The difference in the two values results from a phenomenon known as necking. As failure approaches, the material stretches very rapidly, causing the cross section to narrow. Because the area where rupture occurs is smaller than the
original area, the true rupture strength is larger than the ultimate strength. However, the ultimate strength is commonly used as the maximum stress that the material can carry. 

Statically Indeterminate Problems : 
If the equilibrium equations are suffcient to calculate all the forces (including support reactions) that act on a body, these forces are said to be statically determinate. In statically determinate problems, the number of unknown forces is always equal to the number of independent equilibrium equations. If the number of unknown forces exceeds the number of independent equilibrium equations, the problem is said to be statically indeterminate.

Static indeterminacy does not imply that the problem cannot be solved; it simply means that the solution cannot be obtained from the equilibrium equations alone. A statically indeterminate problem always has geometric restrictions imposed on its deformation. The mathematical expressions of these restrictions, known as the compatibility equations, provide us with the additional equations needed to solve the problem (the term compatibility refers to the geometric compatibility between deformation and the imposed constraints). Because the source of the compatibility equations is deformation, these equations contain as unknowns either strains or elongations. We can, however, use Hooke’s law to express the deformation measures in terms of stresses or forces. The equations of equilibrium and compatibility can
then be solved for the unknown forces. 

Procedure for Solving Statically Indeterminate Problems 
In summary, the solution of a statically indeterminate problem involves the following steps:

Draw the required free-body diagrams and derive the equations of equilibrium
Derive the compatibility equations
To visualize the restrictions on deformation, it is often helpful to draw a sketch that exaggerates the magnitudes of the deformations
Use Hooke’s law to express the deformations (strains) in the compatibility equations in terms of forces (or stresses)
Solve the equilibrium and compatibility equations for the unknown forces.


Introduction : In many engineering applications, members are required to carry torsional loads. In this chapter, we consider the torsion of circular shafts. Because a
circular cross section is an effcient shape for resisting torsional loads, circular shafts are commonly used to transmit power in rotating machinery. We also discuss another important application—torsion of thin-walled tubes. Torsion is our introduction to problems in which the stress is not uniform, or assumed to be uniform, over the cross section of the member. Another problem in this category, which we will treat later, is the bending of beams. Derivation of the equations used in the analysis of both torsion and bending follows these steps: 
Make simplifying assumptions about the deformation based on experimental evidence
Determine the strains that are geometrically compatible with the assumed deformations
Use Hooke’s law to express the equations of compatibility in terms of stresses
Derive the equations of equilibrium(These equations provide the relationships between the stresses and the applied loads.)

Shear and Moment in Beams

Introduction : The term beam refers to a slender bar that carries transverse loading; that is, the applied forces are perpendicular to the bar. In a beam, the internal force system consists of a shear force and a bending moment acting on the cross section of the bar. As we have seen in previous chapters, axial and torsional loads often result in internal forces that are constant in the bar, or over portions of the bar. The study of beams, however, is complicated by the fact that the shear force and the bending moment usually vary continuously along the length of the beam. The internal forces give rise to two kinds of stresses on a transverse section of a beam: (1) normal stress that is caused by the bending moment and (2) shear stress due to the shear force. This chapter is concerned only with the variation of the shear force and the bending moment under various combinations of loads and types of supports. Knowing the distribution of the shear force and the bending moment in a beam is essential for the computation of stresses and deformations, which will be investigated in subsequent chapters

Procedure for determining shear force and bending moment diagrams :

The following is a general procedure for obtaining shear force and bending moment diagrams of a statically determinate beam:

Compute the support reactions from the FBD of the entire beam
Divide the beam into segments so that the loading within each segment is continuous. Thus, the end-points of the segments are discontinuities of loading, including concentrated loads and couples.

Perform the following steps for each segment of the beam: 

Introduce an imaginary cutting plane within the segment, located at a distance x from the left end of the beam, that cuts the beam into two parts
Draw a FBD for the part of the beam lying either to the left or to the right of the cutting plane, whichever is more convenient. At the cut section, show V and M acting in their positive directions
Determine the expressions for V and M from the equilibrium equations obtainable from the FBD. These expressions, which are usually functions of x, are the shear force and bending moment equations for the segment
Plot the expressions for V and M for the segment. It is visually desirable to draw the V-diagram below the FBD of the entire beam, and then draw the M-diagram below the V-diagram.
The bending moment and shear force diagrams of the beam are composites of the V- and M-diagrams of the segments. These diagrams are usually discontinuous and/or have discontinuous slopes at the end-points of the segments due to discontinuities in loading.

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