Strength of Materials

Cornel Sandi BIŢ

STRENGTH OF MATERIALS

2014

CONTENTS 1. Introduction……………………………………………………………………. 1.1 Generalities…………………………………………………………………. 1.2 The object and problems of strength of materials. Connections with other engineering sciences………………………………………………... 1.3 A short review of the history of strength of materials…………………….. 1.4 Bodies schematic representation in strength of materials…………………. 1.5 Loads schematic representation in strength of materials calculus………….. 1.6 Supports and reactions……………………………………………………… 1.7 Equilibrium equations………………………………………………………

2 2 2 4 9 12 16 18

2. Internal forces in statically determinate members……………………………... 2.1 Internal forces computation………………………………………………… 2.2 Differential relations among load, shear and bending moment…………... 2.3 Examples concerning the main types of diagrams…………………………. 2.3.1. Axial forces diagrams………………………………………………... 2.3.2 Shear and bending-moment diagrams………………………………... 2.3.3 Torque diagrams……………………………………………………… 2.4 Superposition method………………………………………………………. 2.5 Moving loads……………………………………………………………….. 2.6 Internal forces diagrams for plane structures (2d structures - frames) and spacial (3d) structures……………………………………………………. Problems to be assigned………………………………………………………...

23 23 25 28 28 29 40 41 42

3.First moments and second moments of an area………………………………… 3.1 First moments of an area. Centroid of an area…………………………….. 3.2 Second moments of an area………………………………………………… 3.3 Parallel - axis theorem (Steiner’s relations)………………………………... 3.4 Moments of inertia of simple surfaces…………………………………….. 3.5 Moments of inertia of complex surfaces (composite areas)………………... 3.6 Moments of inertia with respect to inclined axes…………………………... 3.7 Radius of gyration. Ellipse of inertia……………………………………….. Problems to be assigned………………………………………………………...

57 57 59 60 62 65 66 70 74

4. Displacements, stress and strain……………………………………………….. 4.1 Displacements………………………………………………………………. 4.2 The concept of strain……………………………………………………….. 4.3 The concept of stress……………………………………………………….. 4.4 Relationships among internal forces and stresses within a beam (member) cross section………………………………………………………………..

75 75 75 77

5. Strength of materials basic assumptions………………………………………..

83

6. Axial loading…………………………………………………………………...

87

I

43 53

82

6.1 Stresses and strains…………………………………………………………. 6.2 Poisson's ratio………………………………………………………………. 6.3 Stress concentrations……………………………………………………….. 6.4 Own weight effect. Members of constant strength…………………………. 6.5 Stress-strain diagram……………………………………………………….. 6.5.1 Generalities……………………………………………………………. 6.5.2 Safety coefficients. Allowable strengths……………………………… 6.6 Statically indeterminate problems………………………………………….. 6.6.1 Members with unhomogeneous cross sections………………………... 6.6.2 Straight members fixed at both ends………………………………….. 6.6.3 Systems of parallel members………………………………………….. 6.6.4. Systems of concurrent bars…………………………………………… 6.6.5 Problems involving stresses produced by assembling imperfections…. 6.6.6 Problems involving temperature changes……………………………... 6.6.7 Two more numerical examples concerning statically indeterminate problems……………………………………………………………… Problems to be assigned………………………………………………………...

87 98 100 102 107 107 110 111 113 116 118 120 121 123

7. Shearing stresses in members of small cross sections…………………………. 7.1 Shearing stress and shearing strain………………………………………... 7.2 Strength of a simple riveted joint…………………………………………... 7.3 Bearing stress in connections………………………………………………. 7.4 Welded connections………………………………………………………… Problems to be assigned………………………………………………………...

135 136 137 138 139 141

8. Basic elements of the theory of elasticity............................................................ 8.1 Plane state of stress…………………………………………………………. 8.2 Spatial state of stress……………………………………………………….. 8.3 Monoaxial state of stress…………………………………………………… 8.4 Pure shear state……………………………………………………………... 8.5 Generalized Hooke’s law…………………………………………………. 8.6 Strain energy………………………………………………………………... 8.6.1 Elastic strain energy under axial loading………………………………. 8.6.2 Shearing stresses elastic strain energy…………………………………. 8.6.3 Strain energy for a general state of stress……………………………… 8.6.4 Elastic strain-energy density associated with a change in volume. Elastic strain-energy density associated with a distortion (a change in shape)…………………………………………………………………... 8.6.5 Relation among E, G and ν…………………………………………….

143 143 148 150 151 151 155 156 158 158

9. Torsion…………………………………………………………………………. 9.1 Torsion of circular members……………………………………………….. 9.1.1 Stresses and strains…………………………………………………….. 9.1.2 Statically indeterminate shafts…………………………………………. 9.2 Torsion of noncircular (rectangular) members……………………………...

163 164 164 175 180

II

125 128

160 161

9.3 Thin-walled hollow shafts (tubes)………………………………………….. Problems to be assigned………………………………………………………..

185 191

10. Bending………………………………………………………………………. 10.1 Prismatic members in pure bending………………………………………. 10.2 Shearing stresses in beams subjected to simple bending…………………. 10.3 Prevention of longitudinal sliding in case of composite sections………. 10.4 Beams of constant strength………………………………………………... 10.5 Unsymmetric bending…………………………………………………….. 10.6 General bending…………………………………………………………… Problems to be assigned………………………………………………………...

195 196 203 210 213 218 221 222

Strength of materials typical key words and phrases…………………………….. Apendix I…………………………………………………………………………. Apendix II……………………………………………………………………...…. Apendix III…………………………………………………………………….….

225 247 248 249

III

PREFACE Today, more than ever, engineering applications are often interdisciplinary, involving the interrelationship of the basic engineering sciences (mechanics, metallurgy, chemistry, physics, mathematics, etc.). This is the reason why the modern engineer must have a fundamental knowledge in each of these particular areas. There is no doubt that one of the most important matter within the preparing of the engineering students is represented by the course of Strength of Materials, which is a very important link of this knowledge. Furthermore, for a successful machine or structural design, a thorough mastery of strength of materials is a must. This book is intended to be a basic course of Strength of Materials which to develop in the engineering student the ability to analyze a given problem in a simple and logical manner and to apply to its solutions a few fundamental and wellunderstood principles. But there is also another important reason this book has been written for: to accustom the Romanian engineering students with the main concepts, symbols and trends belonging to the English Engineering Life. As we know, the world of modern science has progressed swiftly across hitherto impossible barriers. There has been more change in the last twenty years than in the previous fifty and more in fifty years than in the previous 200. Never was it more important to have available information on the countless facts of modern scientific knowledge and achievements. Every science-oriented engineer should equip himself with the most up-to-date achievements of science and technique. But for this, he must know the other man’s language. Research people need knowledge of reports and publications in many languages. You can not have a proper access to the impressive quantity of Internet information if you don’t master a foreign language, especially English. The barrier that is set up by a difference in language may constitute a serious obstacle in pure science or in the more practical world of technology. There are three main ways in which scientific information may be acquired: - by adequate translation; - by persuading scientists to publish their work in the better-known languages; - by teaching scientists to read, to write and to speak foreign languages. From these three solutions, the third one is, of course, much more convenient to anyone. Only from the above mentioned ideas one can conclude that it is so important to master a foreign language, especially English which is spread all over the world. Engineering English is no longer a problem nowadays. Every student can easily and rapidly master it, provided he is set on learning it. This is the conclusion to be drawn as far as the difficulties in the search for technical information are concerned. A great philosopher used to say: “ The knowledge of a language is another weapon in life.”

Strength of Materials

1. INTRODUCTION

1.1 GENERALITIES During the whole history of the human society, the practical needs have been those which fundamentally contributed to the development of sciences like mathematics, physics, chemistry, astronomy, biology, etc. In the same manner, the Strength of Materials has arisen from the necessity of building safely, economically and aesthetically. Through the ages, the application of materials in engineering design has posed difficult problems to mankind. In the Stone Age the problems were mainly in the shaping of material. In the early days of the Bronze Age and the Iron Age the difficulties were both in production and shaping. For many centuries metal-working was laborious and extremely costly. Estimates go that the equipment of a knight and horse in the thirteenth century was of the equivalent price of a Centurion tank in Word War II. With the improving skill of metal working, applications of metals in structure increased progressively. Then it was experienced that structures built of these materials did not always behave satisfactorily and unexpected failures often occurred. Detailed descriptions of castings and forgins produced in the Middle Age exist. When judged with present day knowledge, this production methods must have been liable to build important technical deficiencies into the structures. The vastly increasing use of metals in the nineteenth century caused the number of accidents and casualties to reach unknown levels. For example, the number of people killed in railway accidents in Great Britain was in the order of two hundred per year during the decade 1860-1870. Most of the accidents were a result of derailing caused by fractures of wheels, axles or rails. Most of these accidents were certainly due to a poor design, which mainly means a poor understanding and use of Strength of Materials, when design a certain mechanical structure. Within the context of modern engineering design, Strength of Materials has continued to occupy a very important place, providing a useful and necessary tool in understanding the behaviour of certain mechanical structures subjected to external loads.

1.2 THE OBJECT AND PROBLEMS OF STRENGTH OF MATERIALS. CONNECTIONS WITH OTHER ENGINEERING SCIENCES Three fundamental areas of engineering mechanics are statics, dynamics and strength of materials. While statics and dynamics are devoted primarily to the study of the external effects of forces on rigid bodies (i.e. bodies which do not deform 2

Introduction

under the action of the external loads), strength of materials deals with the relations between externally applied loads and their internal effects on bodies. Moreover, the bodies are no longer assumed to be rigid; the deformations, however small, are of major interest. In mechanical design, the engineer must consider both dimensions and material properties to satisfy requirements of strength and rigidity. When loaded, a machine component or structure should neither break nor deform excessively. Let us now consider for example a body subjected to several external loads (forces and couples) – Fig.1.1. Before the action of the loads, one can find a certain state inside the body. But, due to the action of the external loads, the mechanical state inside the body will change. Each particular point of the involved body “feels” in a certain way the action of the external loads, depending upon the position of the point inside the body and the physical nature of the material at that level. Within the above mentioned context Strength of Materials tries to offer reasonable answers to some Fig. 1.1 important questions such as: - what does a certain point of the body “feel” if the external loads act? - how does the involved body deform under the action of the external loads? - which are the mathematical connections between the mechanical effects occured at the level of the points of the body and the values of the external loads? - which are the critical values of the external loads for which the body fails or deforms excessively ? These are only some matters Strength of Materials is concerning with. In its obvious intention to give valuable answers to the above mentioned questions, Strength of Materials has ordered its problems in three important classes: Dimensioning problems: - input data : values and physical nature of the external loads; geometry of the body; mechanical properties of the material the body is made of; - output data : the required dimensions so that the body would not fail or deform excessively. Determination of the allowable external loads: - input data: body dimensions and geometry; mechanical properties of the material; - output data: the maximum allowable values of the external loads so that the body would not fail or deform excessively. 3


Checking problems: - input data : body dimensions and geometry; mechanical properties of the material; external loads values and orientation; - output data : a checking calculus which to conclude if the involved body would fail or not within such conditions. Strength of Materials provides the engineers with the mathematical means of getting inside the bodies, of understanding the intimate behavior of matter, in order to give valuable answers to the questions mentioned above. As a branch of science, Strength of Materials belongs to the Mechanics of Solid Deformable Bodies together with: The Theory of Elasticity, Theory of Plasticity and Statics, Dynamics and Stability of Constructions. Although the Theories of Elasticity and Plasticity concern with the same problems as Strength of Materials, they use only a limited number hypothesis (assumptions), are much more exact but, many times, exceedingly complicated. The Theories of Elasticity and Plasticity use complex mathematical tools, in many cases less adequate to the engineering applications. On the contrary, Strength of Materials uses a large number of theoretical simplifying hypothesis, approximations and experimental analysis methods, providing a useful tool for immediate engineering applications. Through its practical and theoretical investigations Strength of Materials interferes with a lot of fundamental sciences such as: mathematics, physics, technology, chemistry, etc. even with biology, proving again – if it is still the case – the absence of once called hard lines between the sciences.

1.3 A SHORT REVIEW OF THE HISTORY OF STRENGTH OF MATERIALS From the earliest times when people started to build, it was found necessary to have information regarding the strength of materials so that rules for determining safe dimensions of different bodies or structural elements could be experienced. No doubt the Egyptians had some empirical rules of this kind, for without them it would have been impossible to erect their monuments, temples, pyramids and obelisks, some of which still exist. The Greeks further advanced the art of building. They developed statics, which underlies the mechanics of materials. Archimedes (287-212 B.C.) gave a rigorous proof of the conditions of equilibrium of a lever and outlined methods of determining centers of gravity of bodies. He used his theory in the construction of various hoisting devices. The Romans were great builders. Not only some of their monuments and temples remain, but also roads, bridges and fortifications. We know something of their building methods from the book of Vitruvius, a famous Roman architect and engineer of the time of Emperor Augustus. Most of the knowledge that the Greeks and Romans accumulated in the way of 4

Introduction

structural engineering was lost during the Middle Ages and only since the Renaissance has it been recovered. Thus when the famous Italian architect Fontana (1543-1607) erected the Vatican obelisk at the order of Pope Sixtus V, this work attracted wide attention from European engineers. But we know that the Egyptians had raised several such obelisks thousands of years previously, after cutting stone from the quarries of Syene and transporting it on the Nile. Indeed, the Romans had carried some of the Egyptians obelisks from their original sites and erected them in Rome; thus it seems that the engineers of the sixteenth century were not as well equipped for such difficult tasks as their predecessors. During the Renaissance there was a revival of interest in science and art leaders appeared in the field of architecture and engineering. Leonardo da Vinci (1452-1519) was a most outstanding man of that period. He was not only the leading artist of his time but also a great scientist and engineer. Leonardo da Vinci was greatly interested in mechanics and, in one of his notes, he states: “Mechanics is the paradise of mathematical science because here we come to the fruits of mathematics.” The first attempts to find the safe dimensions of structural elements analytically were made in the Leonardo da Vinci seventeenth century. Galileo Galilei’s (1564-1642) famous book Two New Sciences shows the writer’s efforts to put the methods applicable in stress analysis into a logical sequence. It represents the beginning of the science of Strength of Materials. In 1678, the paper Of Spring was published, a paper whose author was Robert Hooke (1635-1703). It contains the results of Hooke’s experiments with elastic bodies. This is the first published paper in which the elastic properties of materials are discussed. The linear relation between the force and the deformation is the so-called Hooke’s law, which later on was used as the foundation upon which further development of the Strength of Materials of elastic bodies was built. Bernoulli family produced outstanding mathematicians for more than a hundred years: Jacob, Nicholas, John, Daniel Bernoulli. Besides mathematics these outstanding mathematicians were also attracted by Mechanics and Strength of Materials. Galileo Galilei For example, Daniel Bernoulli was the first to derive the differential equations governing lateral vibrations of prismatic bars, using it to study particular modes of this motion and to John Bernoulli belongs the well-known hypothesis of plane sections for beams in bending. 5


Leonard Euler (1707-1783), Daniel Bernoulli’s pupil, was one of the greatest mathematicians of all times. In the field of Strength of Materials he was principally interested in the geometrical forms of elastics curves. His contributions in the buckling phenomenon, for example, is essential.

Daniel Bernoulli Leonard Euler

Navier (1785-1836) published in 1826 the first real book on Strength of Materials where his main achievements in this field were incorporated. If we compare this book with those of the eighteenth century, we clearly see the great progress made in mechanics of materials during the first quarter of the nineteenth. Engineers of the eighteenth century used experiments and theory to establish formulas for the calculation of ultimate loads. Navier, from the very beginning, states that it is very important to know the limit up to which structures behave perfectly elastically and suffer no permanent deformations. Within the elastic range, deformation can be Navier assumed proportional to force and comparatively simple formulas can be established for calculating these quantities. Beyond the elastic limit the relation between forces and deformations becomes very complicated and no simple formulas can be derived for estimating ultimate loads. Navier suggests that the formulas derived for the elastic conditions should be applied to existing structures, which had proved to be sufficiently strong, so that safe stresses for various materials could be determined which later could be used in selecting proper dimensions for new structures. S. D. Poisson (1781-1840) was born in a small town near Paris in a poor family and, until he was fifteen years old, he had no chance to learn more than to read and write. In 1796 he was sent to his uncle at Fountainebleau, and there he was able to visit mathematics classes. His excellent progress made him became in 1812 a member of the French Academy. In the field of Strength of Materials, the principal results obtained by Poisson are incorporated in two memoirs which he published in 6

Introduction

1829 and 1831, and in his course of mechanics (Traité de Mécanique) published in 1833. The two outstanding engineers, G. Lamé (1795-1870) and B.P.E. Clapeyron (1799-1864), graduated from the École Polytechnique of Paris in 1818. They had important contributions to the Strength of Materials within their help offered to the new Russian engineering school of that time – the Institute of Engineers of Ways of Communication in St. Petersburg. This new Russian school was later to have a powerful influence upon the development of engineering science in Russia. Lamé and Clapeyron had to teach applied mathematics and physics in this school and they had also to help with the design of S. D. Poisson various important structures in which the Russian government was interested: for example, several suspension bridges were designed and erected in St. Petersburg at that time. These bridges (constructed between 1824 and 1826) were the first suspension bridges build on the European Continent. After their return to Paris, Lamé and Clapeyron continued their work and, as a recognition of their fundamental contributions within the engineering world, they were elected members of the French Academy of Sciences. Barré de Saint-Venant (1797-1886) was born in the castle de Fortoiseau (Seine – et – Marne). His talent in mathematics was noticed very early and he was given a careful coaching by his father who was G. Lamé a well-known expert in economy. Later he studied at the Lycée of Bruges and in 1813, when sixteen years of age, he entered the École Polytechnique after taking the competitive examinations. Here he showed his outstanding ability and became the first in his class. Unfortunately, the political events of 1814 (when referring to Napoleon, Saint-Venant said : “My conscience forbids me to fight for an usurper.”) forced him to interrupt his academic studies – being proclaimed a deserter and never again allowed to resume his study at the École Polytechnique. After nine years, the government permitted him to enter the École des Ponts et Chaussées without examination. After graduating, he devoted all his life to engineering (especially to Barré de Saint-Venant 7


the Theory of Elasticity and Strength of Materials), in 1868 being elected a member of the Academy of Sciences. D.J. Juravski (1821-1891) graduated in 1842 from the Institute of Engineers of Ways of Communication in St. Pertersburg. His career was closely related with the development of railroad construction in Russia. A. Wöhler (1819-1914) was born in the family of a schoolmaster in the province of Hannover and he received his engineering education at the Hannover Polytechnical Institute. Being an outstanding student, he won a scholarship after his graduation, which enabled him to get a practical training, both at D.J. Juravski the Borsing locomotive works in Berlin and in the construction of the Berlin-Anhalter and Berlin-Hannover railways. In this way he had to solve many problems concerning the mechanical properties of materials and started his famous investigations upon the fatigue strength of metals. It can justly be said that scientific investigation in the field of fatigue of materials began with Wöhler’s work. For each kind of fatigue test, Wöhler designed and built all the necessary machines and measuring instruments. In designing these machines, he imposed stringent requirements upon the accuracy with which forces and deformations were to be measured so that his machines represent an important advance in the technique of structural materials. The above mentioned scientists were only a few of those who fundamentally contributed to the development of Strength of Materials. Of course, there are much more names involved in this problem and it would be unfair if we did not even mention them here: C. A. Coulomb (1736 –1806), Augustin Cauchy A. Wöhler (1789 – 18557), Thomas Young (1773 – 1829), J. C. Maxwell (1831 – 1879), Otto Mohr (1835 – 1918), Alberto Castigliano (1837 – 1884) etc. The Romanian school has also offered excellent researchers and scientists to the Strength of Materials engineering world: C. C. Teodorescu, Gh. Buzdugan, Ştefan Nădăşan, Radu Voinea, D. R. Mocanu, Petre Augustin, etc. Nowadays, the courses of Strength of Materials occupy a fundamental place within the Romanian academic engineering education, with a strong tradition and hope for the future.

8

Introduction

1.4 BODIES SCHEMATIC REPRESENTATION IN STRENGTH OF MATERIALS It is to be recalled from Section 1.2 that the main task of Strength of Materials concerns the mechanical behaviour of different bodies or structural elements subjected to external loads. Within its theoretical and practical investigations, Strength of Materials operates with three important types of bodies: a) Bars: represent those bodies whose length is much greater than the other two dimensions (Fig.1.2). Examples of this class: beams, members, rods, columns, pins, rivets, shafts, etc. Fig. 1.2

square cross section

rectangular cross section

circular cross section

annular cross section

I - Shapes

U - Shapes

T – Shapes

Angles

Fig. 1.3

The cross section of a bar is defined as the plane section of minimum area obtained through the intersect between the involved bar and a plane passed through the bar at the some arbitrary point. The cross section of a bar has a certain area generally denoted by A [mm2] and may have different geometric shapes (Fig.1.3). The cross section of a bar may be constant or variable, as shown in Fig.1.4. 9


a.

b. Fig. 1.4

The bar axis is defined as the geometric locus of the cross sections centers of gravity (centroids). It is also called the medium curve. Along its different portions the bar axis may be straight (Fig.1.5a) or curved (Fig.1.5b).

a.

b. Fig. 1.5

In Strength of Materials calculus schemes the bars are usually represented through their longitudinal axes (Fig 1.6). Simplified representation

Fig. 1.6 10

Introduction

It is important to be noted that, in most cases, a bar is geometrically represented within an Oxyz coordinate system, the bar cross section being located in the zOy plane while Ox is the longitudinal axis, i.e. the bar axis or the medium curve. b) Plates: represent those bodies whose two dimensions (length and width) are much greater than the third dimension (thickness), Fig. 1.7.

Fig. 1.7

Fig. 1.8

The midsurface of a plate is defined as the geometric locus of the plate thickness midpoints. Depending upon the shape of the midsurface, the plates may be classified in two important types: plane plates (Fig. 1.7) and curved plates (Fig 1.8).

Fig. 1.9

Fig. 1.10

Depending upon the thickness variation there are two main distinct types of plates: with constant thickness (Fig. 1.9) and with variable thickness (Fig. 1.9). c) Blocks: represent those bodies whose dimensions are of the same size order (Fig 1.10) Although the above classification has been done on geometrical criteria, it does also remain available for Strength of Materials calculus. On the other hand anyone can observe the high variety of the surrounding bodies, objects or structural elements. It is the engineering designers' practical experience which has to choose, for each particular body or structural element, the appropriate class from those presented above. The fundamental Strength of Materials problems, concepts and theories have been constructed on the bars class. Many of these theoretical and practical investigations results have been then extended to the other two classes: plates and blocks. 11


1.5 LOADS SCHEMATIC REPRESENTATION IN STRENGTH OF MATERIALS CALCULUS Since Strength of Materials deals with the relations between externally applied loads and their internal effects on bodies, the clear understanding of the external loads action represents a first and necessary step in the analysis of a given mechanical structure. In fact, a load represents the action of a certain body on another body. It models therefore the interaction between two or more bodies or structural elements. There are two main classes the loads may be classified in: forces and couples (moments of forces). When International System metric units are being used, forces (usually denoted by P or F) are expressed in newtons (N) and couples (moments of forces - usually denoted by M) are expressed in newton-meters (N m). However, when one finds that these units are exceedingly small or big quantities, multiples or submultiples of these units may be also used. Within the framework of different criteria, several specific types of loads have to be differentiated: a) Depending on the size of the interacting bodies contact surface area, we may have: concentrated forces: when the contact surface area may be theoretically reduced to a point, Fig 1.11.

Fig. 1.11

distributed forces: when the interaction of the bodies in contact is transmitted through a surface of a certain area, Fig 1.12.

Fig. 1.12

Furthermore, the forces may be distributed on surfaces (Fig 1.13) or, if a dimension of such surfaces is too small, one can talk about forces distributed on lines, Fig.1.14. 12

Introduction

Fig. 1.13

Fig. 1.14

In Fig. 1.15 some examples of forces distributed on lines in case of a simply supported beam have been represented.

Fig. 1.15

13


The above mentioned concentrated and distributed forces may be encountered in a wide range of practical situations or engineering applications, Fig.1.16.

a.

A liquid in a vessel. The lateral vessel walls are subjected to a linearly distributed force (pressure is the liquid density, g is the p ρ g x , where b. gravitational acceleration and x is the depth at the level where the pressure is measured. A pressurized gas in a thin – walled vessel. The force p acting on the vessel walls is uniformly distributed. * A uniform snow layer acts on a house root with a uniformly distributed force. * A railway vagon acts on the railways with mobile concentrated forces.

c.

d.

Fig. 1.16 14

Introduction

It is to be noted that the moments of forces may be classified in the same manner. b) Depending on the loads variation in time we may have:

constant loads: the loads (P or M) remain constant in time (Fig. 1.17). Fig. 1.17

loads with a periodical variation in time (Fig. 1.18). In such cases the loads do periodically vary in time, between a maximum and a minimum value. Fig. 1.18

loads with a random variation in time (Fig. 1.19).

Fig. 1.19

c) Depending on the time in which a load is applied to a certain body we may have: static loads: which do slowly vary from zero up to the nominal value, remaining then constant in time (Fig. 1.20a).

a.

b. Fig. 1.20

15


dynamic loads: which vary in a very short time from zero up to the maximum value (Fig. 1.20b). An example of this type has been represented in Fig.1.21. The rod BD of uniform cross section is hit at its end B by a body of mass m, moving with a velocity v0. The rod deforms under the impact with . After vibrating for a while, the rod will come to rest and all its internal stresses will disappear. Such a sequence of events is referred to as an impact loading (this load is an example of dynamic loads class). Fig. 1.21

1.6 SUPPORTS AND REACTIONS The mechanical connections between bodies – subjected to different loads and the surrounding environment are provided by the mechanical supports. From our knowledge of statics we have to recall three basic types of mechanical supports: simple supports: which prevent the supported element from moving linearly along a direction perpendicular to the supporting base ( Fig. 1.22).

Fig. 1.22

In Fig. 1.23 an example of a simply supported beam has been represented. Usually the supporting points are denoted by capital letters A , B , C …

Fig. 1.23

Fig. 1.24

16

Introduction

pin-connections: which prevent the supported element from moving linearly along two perpendicular directions at the supporting point (Fig. 1.24). In Fig. 1.25 an example of a beam, simply supported at A and pin– connected at B has been shown. Fig. 1.25

fixed connections: which prevent the supported element from moving linearly or rotating at the supporting point. In Fig.1.26 a column with one fixed end B and one pin-connected end A has been represented. If a structural element is fixed at a certain point, it is usually said that the element is embedded at that point. For any of the above presented types of connections, if the support prevents the supported element from moving or rotating along a certain direction then, along that direction a reaction occurs.

Fig. 1.26

Depending upon the number of prevented linear movements or rotations, in case of plane problems, the following reactions will develop (Fig. 1.27).

simple support

pin–connection

fixed connection:

Fig. 1.27

As shown in Fig.1.27, the horizontal reactions are denoted by X while the vertical reactions by Y, with subscripts representing the involved supporting point. The reactions represent in fact the action of the surrounding environment on bodies or structural elements subjected to external loads. These reactions can be then calculated using different equations representing the bodies or structural elements conditions of mechanical equilibrium. 17


Let us consider now, for example, a beam, simply supported at end A and pin–connected at end B, subjected to a uniformly distributed force p and a concentrated force P (Fig. 1.28). Due to the action of the external loads p and P, at the level of the supporting points A and B the reactions YA, XB and YB will develop. Fig. 1.28

These reactions may be computed from the beam conditions of mechanical equilibrium. After the reactions have been computed, there will be no difference between the external loads p and P and these reactions, all representing a global system of loading for a body in mechanical equilibrium (Fig.1.29).

Fig. 1.29

1.7 EQUILIBRIUM EQUATIONS The equilibrium equations represent the mathematical expression of equilibrium for a body or a structural element subjected to external loads (original loads and reactions). In case of plane problems (2D problems) the conditions of equilibrium consist in two equations representing the projections of all forces about two arbitrary perpendicular directions in the plane and one equation representing a summation of moments about an arbitrary point of the plane. Let us consider for example a beam, simply supported at A and pin–connected at B (Fig. 1.30), supporting a load P. 18

Introduction

Fig. 1.30

Due to the action of force P, reactions YA, XB, YB will develop. In such cases any strength of materials problem has to start with the computation of the unknown reactions YA, XB, YB . In writing the mathematical conditions of equilibrium a coordinate system must be firstly chosen (Oxy), Fig. 1.30. After that, three necessary equilibrium equations have to be written as follows: summation of forces about Ox axis equals zero : X

0

X

0,

B

summation of forces about Oy axis equals zero : , summation of moments about the pin support B equals zero: Y

M

0

B

YA

0

YB

P(a

P

0

b) - Y A b

YA

YB

0

YA

P

P(a

b)

.

b

Substituting for YA into ∑Y =0, we write: YB

P

YA

P a

P

b

Pb

Pa

b

b

Pb

Pa

.

b

We may write therefore: X

B

0;Y A

P a b

b

;YB

Pa

.

b

The negative sign of YB expresses that the real physical sense of reaction YB is opposite than that shown in Fig. 1.30. Once the reactions have been computed, strength of materials doesn’t make any differentiation between the original loading (represented by force P) and these reactions, all being treated as a global system of forces acting upon a body in mechanical equilibrium (Fig. 1.31). Following the same reasoning, in case of 3D problems six equilibrium equations are required: three equations of forces and three equations of moments.

Fig. 1.31

19


SAMPLE PROBLEMS For the 2D mechanical structural elements presented below, supported and loaded as shown, determine the values of reactions using the required mathematical equilibrium equations: X

0

Y

X

0

M

0;

B

YA

P

YA

0

B

YB

0;

P b

0.

We write therefore: P b

YA



Fig. 1.32 X

0

Y M

X

0 B

0

P a

YB

;



X

0.

B

0;

B

YA

;

YB

3P

Y A 800

0;

600

P 400

600

2 P 300

0;

We may finally write: YA

114 , 28 N ;

YB

185 , 72 N ;

X

X

0

X

Y

0

10

A

X

Fig. 1.33

10

X

M

X

0

A

0;

5 2

20 kN .

20

10

2

2

3

5 2 1

M

2

0

Y

2

2

2

M

Fig. 1.34

0;

10 kN .

A

YA

0

A

2

2

2

YA M

0.

B

YA

0

A

20 kN

A

20

2

m.

0

X

2 YB

10

YA

YB

Y B 1500

YB

0;

A

2

20

14 ,142 kN ;

A

0

;

2

24 ,142 kN . 10 1000

20

500

2

0;

2

11 , 38 kN .

Finally we may write: X

Fig. 1.35

20

A

14 ,142 kN ; Y A

12 , 762 kN ; Y B

11 , 38 kN ;

2. INTERNAL FORCES IN STATICALLY DETERMINATE MEMBERS 2.1 INTERNAL FORCES COMPUTATION Consider a body of arbitrary shape acted upon by several external loads (Fig. 2.1). In statics, we would start by determining the resultant of the applied loads to determine whether or not the body remains at rest. If the resultant is zero, we have static equilibrium – a condition generally prevailing in structures. If the resultant is not zero we may apply inertia forces to bring about dynamic equilibrium. Such cases will be discussed later under dynamic loading. For the present, we consider only cases involving static equilibrium. Generally speaking, when loads are applied to a certain mechanical structure or machine, each component of such a Fig. 2.1 structure or machine is subjected to external loads of different values (Fig. 2.1). Under the action of the external loads, internal forces occur inside the involved component (assimilated to the arbitrary body represented in Fig. 2.1). If these internal forces reach critical values the body (component) will fail. One of the methods commonly used for the determination of internal forces in strength of materials is known as the method of sections. In fact the problem remains the same like that presented in the previous chapter: what does every point of the body (generically represented in Fig. 2.1) ,,feel’’ – when the body is subjected to external loads in mechanical equilibrium?

Fig. 2.2

Fig. 2.3

The method of sections consists in passing an exploratory plane π through an arbitrary point of the body and with an arbitrary orientation (Fig. 2.2). In this way two distinct segments of the body will occur (Fig.2.3), the left surface (SL) and the


right surface (SR) representing the internal plane surfaces of the body, originally in contact. Since the body represented in Fig. 2.2 is in equilibrium, neither of the two segments of Fig. 2.3 can be in equilibrium. If we want to bring the segment II for example in the same state it is in Fig. 2.2, the action of segment I on segment II (which actually exists inside the body represented in Fig. 2.2) has to be considered. This action may be reduced at the centroid O of surface SR to a resultant force R and a resultant moment M (Fig. 2.4). In other words R and M represent the action of segment I on segment II, as a global mechanical effect occured at the level of the entire section SR. In fact this mechanical effect develops inside the body represented in Fig. 2.2. Furthermore it is to be noted that M and Fig. 2.4 R represent the effect of the external loads acting on the segment I (i.e. P1, Pn, M1 – Fig. 2.2) which develops inside the body at the level of SR. The resultant force R and the resultant moment M are called internal forces. Under the action of M, R, P2, Pk, Mk the segment II of the body is now in mechanical equilibrium (as it really is in the actual state of Fig. 2.2). Now using the adequate equilibrium equations for segment II, the values of internal forces R and M can be derived. A similar reasoning may be also applied to the segment I of the body, on which internal forces R’ and M’ develop (Fig. 2.5). From action and reaction mechanical law we may write: R’ = - R, M’ = - M. Let us now apply the above reasoning to a loaded statically determinate member (Fig. 2.6). It is to be mentioned that a statically determinate member is a member for which all reactions can be completely computed from statics alone. Fig. 2.5 After computing the specific reactions (YA, YB, ZB…etc) corresponding to the supporting points A and B, the member represented in Fig. 2.6 is in fact a body subjected to several external loads in mechanical equilibrium. Passing an exploratory plane at some arbitrary point of the member, perpendicular to the axis of the member, and considering only a segment of

22

Internal forces in statically determinate members

the member (just like in the preceding discussion) the internal forces R and M are revealed (Fig. 2.7). We do also attach to the segment considered in Fig. 2.7 a coordinate system whose origin is taken at the centroid O of the exploratory cross section (SR). Ox is the axis of the member while Oz and Oy represent the axes to which the exploratory cross section of the member is reported. For convenience, the internal Fig. 2.6 forces M and R are resolved into components that are normal and tangent to the cross section considered, within the chosen coordinate system (Fig. 2.8) - R resolved into components N, Ty and Tz, while M into components Mx, Miy , Miz.

Fig. 2.7

Fig. 2.8

Each component reflects a certain effect of the applied loads on the member and is given a special name as follows: N: axial force (R component along Ox axis, or, more briefly, x axis). This component measures the pulling (or pushing) action perpendicular to the section considered. A pull represents a tensile force that tends to elongate the member, whereas a push is a compressive force that tends to shorten it. Ty, Tz: shearing forces (R components along y axis and z axis respectively). These are components of the total resistance to sliding the portion to one side of the exploratory section past the other. The resultant shearing force (acting on zOy plane) is usually denoted by T, and its components by Ty and Tz to identify their directions. Mx: twisting couple (twisting moment or torque) (M component along x axis). This component measures the resistance to twisting the member and is commonly given the symbol Mt. Miy, Miz: bending moments. These components measure the resistance to bending the member about the y or z axes and are often denoted merely by Miy and Miz. 23


The quantities N, Ty, Tz, Mx, Miy, Miz are also called internal forces. Each of them produces a certain type of mechanical effect on the involved member: N : axial

loading

T y , T z : shearing M t : torsion M

iy

,M

iz

; loading

;

;

: bending

.

The simultaneous presence on the current member cross section of two or more types of internal forces determines a combined loading. Although the type of coordinate system used within such analysis is, in a way, controversial, we shall use the following sign convention: N, Ty and Tz should be considered positive if orientated to the opposite sense of the axes; Mt, Miy and Miz should be considered positive if orientated to the sense of the axes. From the preceding discussion, it is obvious that the internal effect of a given loading depends upon the selection and orientation of the exploratory section. In particular, if the loads act in a single plane, say the xy plane as is frequently the case, the six components of Fig. 2.8 reduce to only three namely, the axial force (N), the shearing force (T) and the bending moment Miz. This in why, in case of plane problems (when plane members are subjected to loads contained in the same plane) the internal forces refer to only three components whose positive sign convention should be taken as follows:

Fig. 2.9

The positive sign convention represented in Fig. 2.9 should be used for plotting the axial forces, shearing forces and bending moments diagrams. As it will be explained later, the positive sign convention corresponding to the face SR is used when the member is covered from the left to the right while the positive sign convention corresponding to the face SL is used when the member is covered from the right to the left. 24


2.2 DIFFERENTIAL RELATIONS AMONG LOAD, SHEAR AND BENDING MOMENT Let us now consider a simply supported beam AB of span , carrying a distributed load p per unit length (Fig. 2.10a), and let C and C’ be two points of the

a.

b. Fig. 2.10

beam at an infinitely small distance dx from each other. We shall detach the portion CC’ of the beam and draw its free body diagram (Fig. 2.10b). The forces exerted on the free body include a load of magnitude pdx and the internal forces at C and C’ as shown. The shear and bending moment at C will be denoted by T and M respectively, and will be assumed positive while the shear and bending moment at C’ will be denoted by T + dT and M + dM respectively. Since the shear and bending moment are assumed to be positive, the internal forces will be directed as shown in Fig. 2.10 b. It is also to be mentioned that since the distance dx between C and C’ is considered infinitely small, the load p may be assumed uniformly distributed per length dx and may be replaced by a resultant pdx, (Fig. 2.11).

Fig.2.11.

From the mechanical equilibrium of the detached segment CC’ we write: the summation of forces about the vertical direction is zero: Fy

0

T

dT

25

pdx

T

0

.


We write . Dividing by dx the two members of the equation, we have: pdx

dT

dT

p

.

(2.1)

dx

writing now that the summation of moments about C’ is zero, we have: M

C '

0

M

dM

dx

pdx

M

Tdx

0

.

2 p dx

The third member of this equation

2

being an infinitely small quantity

2

compared with the others, it may be neglected and we may write: dM

Tdx

.

Dividing now the two members of the equation by dx we obtain: dM

T

.

(2.2)

dx

Relations (2.1) and (2.2) may be written in a single one as follows: 2

d M dx

2

dT

.

p

(2.3)

dx

The above presented relations may be successfully used for plotting the shear and the bending moment diagrams. Generally speaking, internal forces diagrams (i.e. diagrams of axial forces, shearing forces, torsion and bending moments) are a graphical representation of the successive values of axial force N, shearing force T, torque Mt and bending moment Mi in the various sections against the distance measured from one end of the involved member. In particular, relations (2.1), (2.2) and (2.3) bring us several important rules concerning the shear and bending moment diagrams: The distributed force p measures the tangent slope of the shear curve (shear diagram). If p = 0, the shearing force will be constant; It should be observed that Eq. (2.1) is not valid at a point where a concentrated force is applied. At such a point the shear curve is discontinues and a sudden change occurs in the diagram. The value of the sudden change in the shear diagram, when a concentrated force is applied, equals the value of that concentrated force; Equation (2.2) indicates that the slope

dM

i

of the bending moment diagram is

dx

equal to the value of the shearing force. This is true at any point where the 26


shearing force has a well defined value, i.e. at any point where no concentrated load is applied; Equation (2.2) does also show that T = 0 at points where M is maximum. This property facilitates the determination of the points where the beam (a member in bending is often referred to a beam) is likely to fail under bending; If a concentrated couple is applied at an arbitrary point of the beam, a sudden change in the bending moment diagram occurs, the change value being equal to the applied concentrated moment (couple); Equation (2.3) shows that the shear and the bending moment curves will always be, respectively, one or two degrees higher than the load curve. For example if the load curve is a horizontal straight line (the case of an uniformly distributed load p), the shear curve is an oblique straight line and the bending moment curve is a parabola. If the load curve is an oblique straight line (first degree), the shear curve is a parabola (second degree) and the bending moment curve is a cubic (third degree). With the above rules in mind, we should be able to sketch the shear and the bending moment diagrams without actually determining the function T(x) and M(x) along the member, once a few values of the shear and the bending moment have been computed. The sketches obtained will be more accurate if we make use of the fact that, at any points where the diagrams are continuous, the slope of the shear curve is equal to (- p) and the slope of the bending moment curve is equal to T. For plotting the internal forces diagrams, the following steps have to be covered: a) Denoting of the important points. An important point of a member is a point where a certain change (geometrical, loading, etc) occurs. The supporting points are usually denoted by capital letters A, B, C, etc. and the other important points by figures 1, 2, 3 etc.; b) Two successive important points define a portion of the member; c) Determination (when necessary) the magnitude of the reactions at the supports; d) A covering sense of the member has to be chosen (from the left to the right, from the right to the left or both); e) For each distinct portion of the member a current cross section at distance x from one end of the involved portion has to be considered; f) For the current cross section considered, each distinct internal force (N, T, Mt, Mi) has to be mathematically expressed as a function of x: N(x), T(x), Mt(x), Mi(x); g) Plotting the functions N(x), T(x), Mt(x), Mi(x) along the entire member, the internal forces diagrams are finally obtained.

27


2.3

EXAMPLES CONCERNING THE MAIN TYPES OF DIAGRAMS

2.3.1 AXIAL FORCES DIAGRAMS Example 1 Draw the axial force diagram for the horizontal member with one fixed end and uniform cross section, shown in Fig. 2.12.

a.

b. Fig. 2.12

Step 1 Step 2 Step 3

important (main) points: 1, 2 and A; main portions of the member: 1 - 2; 2 - A; the magnitude of reactions may be determinated using the condition of mechanical equilibrium: Fx = 0

P + 2P - XA = 0

Step 4

XA = 3P ;

the covering sense of the member: let us say it is from the left to the right; Step 5 we first consider the first portion of the member (1 -2) and an exploratory current cross section located at distance x from end 1 of the portion. Looking to the left one can conclude that the single axial force component acting upon the current cross section considered is equal to P. For any value of x this component remains constant. This is why, for portion 1 - 2, the axial force will be constant (N P). The corresponding axial force diagram of portion 1 - 2 has to be hachured perpendicularly to a reference horizontal line. Since the covering sense of the member was chosen from the left to the right, the positive sign convention I has been used (Fig. 2.12a). In the same manner, the axial force for the second portion 2 - A of the member is: N2-A = P + 2P = 3P = ct. 28


It is to be mentioned that portion 2 - A for example, could have been covered from the right to the left as well. In such a case the current cross section is taken at distance x from A and, looking to the right, we have: NA-2 = XA = 3P (the same value as above). When covering the member from the right to the left, the positive sign convention II should be taken (Fig.2.12b). The above presented algorithm for plotting the axial forces diagrams remains unchanged even if the loading or the geometry are much more complicated or the internal forces are not axial but shearing forces or bending moments. The following examples will be accompanied by no supplementary explanations. Example 2 Draw the axial force diagram for the member supported and axially loaded as shown in Fig. 2.13. Portion 1 - 2 or, more simple, 1 - 2: N(x) = 0; Portion 2 - A: N(x) = 3P = ct.

Fig. 2.13

Example 3 Draw the axial force diagram for the member shown in Fig. 2.14. Fx = 0; XA - 20 - 10 - 5 2 = 0; XA = 40 kN. Portion 1 - 2: N(x) = 20 kN; Fig. 2.14

Portion 2-A: x

N(x) = 20 + 10 + 5 x = 30 +5 x ; x

0;

N

2

30 kN ;

2 m;

N

A

40 kN .

2.3.2 SHEAR AND BENDING-MOMENT DIAGRAMS Fig. 2.15 shows a simply supported beam that carries a concentrated load P, being held in equilibrium by the reactions YA and YB. For the time being we neglect the mass of the beam and consider only the effect of load P. Applying the method of sections, let us assume that a cutting plane d - d, located at a distance x from point A, divides the beam into two segments. 29


Fig.2.15

The free-body diagram of the left segment (Fig. 2.16) shows that the externally applied load is YA . To maintain equilibrium in this segment of the beam the internal forces occurring at the level of the exploratory section d - d must supply the resisting forces necessary to satisfy the conditions of static equilibrium. In this case, the external load is vertical, so the condition Fx= 0 (the x axis is horizontal) is automatically satisfied.

Fig. 2.16

Since the left segment of the beam is in equilibrium, the resisting shearing force T acting on the left segment has to be numerically equal to YA. In other words, the shearing force in the beam may be determined from the summation of all vertical components of the external loads acting on either side of the section. However, it is simpler to restrict this summation to the loads that act on the segment to the left of the section. This definitions of the shearing force (also called vertical shear or just shear) may be expressed mathematically as: T

Fy

L

,

(2.4)

the subscript L emphasizing that the vertical summation includes only the external loads acting on the beam segment to the left of the section being considered. In computing T, when the beam is covered from the left to the right, upward acting forces and loads are considered as positive (see also the sign convention presented in the preceding section). This rule of sign produces the effect shown in Fig. 2.17, in which a positive shearing force tends to move the left segment upward with respect to the right segment, and vice versa.

Fig. 2.17 30


For a complete equilibrium of the free-body diagram in Fig. 2.15 and Fig. 2.16 the summation of moments must also balance. In this discussion YA and T are equal, thereby producing a couple Mi that is equal to YA x and is called the bending moment because it tends to bend the beam. Analogous to the computation of T at the current cross section, the bending moment is defined as the summation of moments about the centroidal axis of any selected cross section of all loads acting either to the left or to the right side of the section, being expressed mathematically as: M

i

M

M

L

R

,

(2.5)

where the subscript L indicates that the bending moment is computed in terms of the loads acting to the left of the section, while the subscript R referring to loads acting to the right of the section. Why the centroidal axis of the exploratory section must be chosen as the axis of bending moment may not be clear at this moment; this will be explained later. To many engineers, bending moment is positive if it produces bending of the beam concave upward, as in Fig. 2.18.

Fig. 2.18

We prefer to use an equivalent convention, which states that the upward acting external forces cause positive bending moments with respect to any section while downward forces cause negative bending moments. Therefore, if the left segments of the beam is concerned (Fig. 2.16), this is equivalent to taking clockwise moments about the bending axis as positive, as indicated by the moment sense of YA. With respect to the right segment of the beam (Fig. 2.16) this convention means that the moment sense of the upward reaction YB is positive in counterclockwise direction. This convention has the advantage of permitting a bending moment to be computed, without any confusion in sign, in terms of the forces to either the left or the right of a section, depending on which requires the least mathematical work. We never need think about whether a moment is clockwise or counterclockwise; upward acting forces always cause positive bending moments regardless of whether they act to the left or to the right of the exploratory section. The definition of shearing force and bending moment may be summarized mathematically as follows: T Fy Fy ; L R M

i

M

M

L

31

R

,


in which positive effects are produced by upward forces and negative effects by downward forces. This rule of sign will be used exclusively hereafter. To avoid conflict with this rule, we must compute vertical shear in terms of the forces lying to the left of the exploratory section. If the forces acting to the right of the section were used, it would be necessary to take downward forces as positive so as to agree with the sign convention shown in Fig. 2.17. Example 1 Draw the shear and bendingmoment diagrams for the cantilever beam shown in Fig. 2.19. (A cantilever beam is a beam with a fixed end, subjected at its free end to a single concentrated force P). We observe that the internal forces exerted on a current cross section at distance x from the free end 1 are represented by: a shearing force T of magnitude T = P (see the positive sign convention); a bending moment

Fig. 2.19

Mi = - P x:

x

0

Mi 1

x



Mi A

0; P .

We note (Fig. 2.19) that the negative values corresponding to the bending-moment diagram are represented above the reference line. In this way the bending-moment diagram shows us how the involved beam deforms under the action of the external loads. Example 2 Draw the shear and bendingmoment diagram for a simply supported beam AB, of span  subjected to a single concentrated load P (Fig. 2.20) – the case Fig. 2.20 of Fig. 2.15. We first determine the reactions at the supports from the free-body diagram of the entire beam (Fig. 2.20); we find that: YA

P b 

;

YB

32

P

a 

.


For the portion A - 1, cutting the beam at distance x from end A, we have: T = YA = constant; x

0

Mi A

0;

x

a

Mi 1

YA

Mi = YA x:

P b

a

a

Pab



While the bending moment increases linearly from M = 0 at x = 0 to

.

 M

Pab 

at x =

a, we note that the shear has a constant value. Even if the problem is quite simple, it is more convenient to cover the second portion of the beam from the right to the left. Therefore, cutting the beam at distance x from end B and using the adequate sign convention we have: B - 1: T

YB

P a

;

 x

0

M i B

x

b

M i 1

Mi = YB x:

0; Pab

.



We can now complete the shear and bending-moment diagrams (Fig. 2.20). For portion B - 1 the shear has a negative constant value while the bending moment increases linearly from M = 0 at B to

M

Pab

at 1 (for x = b).



Remarks If a concentrated traverse force acts at a section of the beam, a sudden change in the shear diagram at that section occurs, the sudden change value being equal to that concentrated force. In our case of Fig. 2.20, at point 1, the sudden change is: Pb

Pa





P a

b



P



P.



If, for a certain portion of the beam, the shear is constant, the bending-moment diagram is linear; Covering the beam from the left to the right within the portion A - 1 and then from the right to the left within the portion B - 1, and since at point 1 there is no concentrated moment, there will be no sudden change in the bending-moment diagram at point 1. This is why we have obtained the same value of the maximum bending moment at 1; The covering sense of the beam, when plotting such diagrams, has no importance. It may be chosen from the left to the right, or from the right to the left or combined, as it is convenient to us; 33


When designing a beam like that presented in Fig. 2.20, we must note that the strength of the beam is usually controlled by the maximum absolute value Mi max of the bending moment in the beam (in our case

Pab

M i max

).



We note from the foregoing example that, when a beam is subjected only to concentrated forces, the shear is constant between the applied forces while the bending-moment varies linearly between the forces. In such situations, therefore, the shear and bending-moment diagrams may easily be drawn, once the values of T and Mi have been obtained at sections selected just to the left and just to the right of the points where the loads and reactions are applied. Numerical examples 1. Draw the shear and bending-moment diagrams for a simply supported beam subjected to two concentrated loads (forces) as shown in Fig. 2.21. Determination of the reactions at the supports Fy= 0 ;

YA -5 - 10 + YB = 0

YA + YB = 15 kN ;

MA = 0 ;

YB 4 - 10 3 - 5 1 = 0

YB = 8,75 kN ;

MB = 0 ;

YA 4 - 5 3 - 10 1 = 0

YA = 6,25 kN .

Portion A - 1 T = YA = 6,25 kN ; Mi = YA x;

x

0

M iA

x

1m

0;

Mi 1

6 , 25 kN

m.

Portion 1 - 2 T = YA - 5 = 6,25 - 5 = 1,25 kN ; Mi = YA (1 + x) - 5x . This means that x

0

Mi

6 . 25 kN

m;

1

x

2m

Mi

6 , 25

1

2

8 , 75 kN

m.

For the last portion it is much more convenient to cover the beam from the right to the left.

Fig. 2.21 Portion B - 2 T = - YB = - 8,75 kN; x

Mi = YB x

5 2

2

x

0

M i

1m

0; B

M i 2

8 , 75 1

8 , 75 k N

m.

We obtain therefore the shear and bending-moment diagrams shown in Fig. 2.21. 34


2. Draw the axial force, shear and bending - moment diagrams for the beam shown in Fig. 2.22. The beam represented in Fig. 2.22 can be drawn in a simplified manner as shown in Fig. 2.23. As in preceding example, the reactions are determined by considering the entire beam as a free body, they are: XB = 10 kN; YA = 22,5 kN; YB = 2,5 kN. Portion 1 - A Fig. 2.22

N

10

T

Mi

2

10

cos 45

2

2

1 A

x



2 sin 45

10



10 kN ;

10 kN ;

x

10 x

:

2

0

Mi

0;

1

x

1m

Mi

10 kN

m.

A

Portion A - 2 N

10

T

10

10 M i A

2

2

2 sin 45

22 , 5 10



10 kN ;

YA

12 , 5 kN ; 2

2

1

x

YA

2

10 1

Fig. 2.23



cos 45

x

x

0

M i A

x

1m

M i 2

22 , 5 x 10

:

kN

2 ,5 k N

x

m; m.

It is more convenient to us to cover the last portion of the beam from the right to the left. Portion B – 2 N

X

T

YB

M i B

2

B

10 kN ; 2 , 5 kN ;

YB

x

2 ,5 x

:

x

0

M i B

0;

x

1m

M i 2

2 ,5 k N

m.

We can now complete the axial force, shear and bending-moment diagrams of Fig. 2.23. We note that the axial force has a constant value along the beam; the shear has also constant values between the important points of the beam while the bending moment varies linearly. At points (sections) where concentrated forces act, sudden changes in the shear diagram occur (whose values equals the applied concentrated forces). Since there are no concentrated moments on the beam there will be no sudden changes in the bending-moment diagram. 35


Example 3 Draw the shear and bending-moment diagrams for a beam, simply supported at its ends and subjected to a uniformly distributed load p (Fig. 2.24). Due to the symmetry of loading and geometry, the reactions are: YA

p 

YB

.

2

As usually, we cut the beam at distance x from A and note that: T

p

YA

px

x

0;

:

px

2



x

p

TA

;

2

; T

0;

2

M

i

YA

x

p x

x

p 

2

2

px

x

2

:

2

p

 ; TB

x

Fig. 2.24

.

2

x

0;

M

A

0;

x

;

M

B

0.

Within the calculus, the distributed load over the current portion of the beam has been replaced by its resultant px applied at the midpoint of the involved portion. Since at the midpoint of the beam the shear equals zero, the bending moment reaches a maximum value at that point: M

max

M

i



p



p



2

2

2

2

2

2

p

2

.

8

We do also note that the shear diagram is represented by an oblique straight line (Fig. 2.24), while the bending-moment diagram by a parabola. In the section where T = 0, the bending-moment has a maximum value. Example 4 Draw the shear and bending-moment diagrams for a beam, simply supported at its ends and subjected to a linearly distributed load (Fig. 2.25). The entire beam is taken as a free body, and, from the conditions of equilibrium, we write:

36


Fy = 0 ;

YA

MB = 0 ;

YA



;

2 

p0



2 

p0

YA

MA = 0 ; Y B

p0

YB



0;

3

;

6 

p0



2

p0

YB



2



0;

3

.

3

Fig. 2.25

Using the first equation of equilibrium ( Fy = 0) we check that the values already obtained for YA and YB are correct. Now writing the mathematical expressions of the shear and bending-moment at an arbitrary section at distance x from end A, we have: T M

px

YA

;

2

YA

i

x

px

x

x

x

2

.

3

From similar triangles we may write: px

x

p0



px

x

p0

,



which substituted in the preceding expressions of T and Mi, leads to : T

YA

x

0;

x = ;

M

i

YA

x

px 2

x

px

p0

x

2

p0

TA =

p0

3

6

x

p0

p0 x



2

6

2

2

;

;

6

TB =

x

p0

6

x

x

p0

p0

6

2

p0

2

= -

;

3 2

x

x



6

37

p0

p0 x

p0 x

6

6

3

x :

0;

M

x = ;

M

iA iB

= 0; = 0;


The shear curve is thus a parabola while the bending-moment curve is a third degree function. The shear curve intersects the x axis at a distance given by equation: T

0

p0

p0 x

6

2

2

0



x

.

3

Therefore, the maximum value of the bending moment occurs at



x

, since T

3

(and thus

dM

i

) is zero for this value of x:

dx

M

 i

3

p0 6

 3

p0 6



3

3

p0 9

2

.

3

Example 5 Draw the shear and bending-moment diagrams for a simply supported beam, subjected to a concentrated moment M0 applied at point 1 (Fig. 2.26).

The entire beam is taken as a free body and we have: YA

Fig. 2.26

M 

0

;

YB

M

0

.



The negative sign of YB indicates that the real sense of this reaction is opposite to that represented in Fig. 2.26. The shear at any section is constant and equal to M0 / . Since a concentrated moment (couple) is applied at 1, the bending-moment diagram is discontinuous at 1; the bending-moment decreases suddenly by an amount equal to M0. Remark The concentrated moment in Fig. 2.26 symbolizes for example the action of two equal and opposite concentrated forces as shown in Fig. 2.27, where M0 = P d.

Fig. 2.27

38


A complex sample problem Sketch the shear and bending-moment diagrams for the simply supported beam shown in Fig. 2.28.

Considering the entire beam as a free body, we determine the reactions as follows: Fy = 0 ;YA + YB + 5 - 10 1 = 0; YA + YB = 5; MB = 0 ; 5 3 -10 1 2,5 + YA 2 + 15 = 0; YA = - 2,5 kN; MA = 0 ; 5 1 -10 1 0,5 + 15 - YB 2 = 0; YB = 7,5 kN. Using the first equation of equilibrium ( Fy = 0) we check that the Fig. 2.28 two values obtained for YA and YB are valid. Next we draw the shear and bending-moment diagrams. The sketches obtained will be more accurate if we make use of the fact that, at any point where the curves are continuous, the slope of the shear curve is equal to -p while the slope of the bending-moment curve is equal to T. Portion 1 - A T

5

10

x

0;

T1

x

1m;

TA

x:

5 kN ; 5 kN .

This means that at the midpoint between 1 and A (for x = 0,5 m) the shear is zero, and, therefore, the bending-moment reaches a maximum value. It is to be mentioned that this point of maximum for the bending moment is valid only for the involved portion (i.e. 1 - A). Within the other portions of the beam the bending-moment could reach grater values as well. This is why, the maximum value of the bending reached within a certain portion of the beam is called a local maximum. There are cases in which a local maximum does also represent a global maximum too. x M i

5x

10

x

x

5x

5x

2

:

0;

M i

x

1 m;

M i A

x

0 ,5 m ;

M i MAX

2

0;

1

0; 1 , 25 k N

Portion A - 2 T

M i

5

10 1

5 1

x

YA

10 1

5

10

0 ,5

2 ,5

x :

39

7 , 5 kN

;

x

0;

M i A

x

1 m;

M i 2

0; 7 ,5 k N

m .

m .

Strength of Materials M i 2

= - 7,5 kN m tells us that close to the end 2 of the portion A - 2, the bending moment reaches

such a value. It will be more convenient to us to cover the last portion of the beam from the right to the left (i.e. from B to 2): Portion B - 2

This time,

M i 2

T

YB

M i

YB

= - 7,5 kN

7 , 5 kN ;

x

7 ,5 x :

x

0;

M i B

x

1 m;

M i 2

0; 7 ,5 k N

m .

m tells us that close to the end 2 of the portion B - 2, the bending

moment reaches such a value. In this way we have obtained two values for the bending-moment at point 2: one for the portion A - 2, close to the point 2 to the left and one for the portion B - 2 close to the same point 2 but to the right. Since at point 2 there is a concentrated moment acting on the beam (equal to 15 kN m), the sudden change in the bending-moment diagram at point 2 is correct.

2.3.3 TORQUE DIAGRAMS In the preceding sections we have discussed about axial forces, shear and bending-moment diagrams. Here we shall consider members which are in torsion. More specifically we shall learn to draw internal forces diagrams for members subjected to twisting couples or torques. We say that a member is subjected to torsion if at any cross section of the member, the internal forces are represented by a torque vector directed along the axis of the member. To sketch the torque diagrams the method of sections may be used, as presented in the preceding sections. Draw the torque diagram for a member fixed at one end and subjected to concentrated and uniform distributed torques as shows in Fig. 2.29. Considering the entire member as a free body we obtain the reactions at A. M

M

A

M

0

0

x

2M

M 0



0



4M

0

;

Using the method of sections and covering the member from 1 to A we have: 1-2 Mt M 0; 2-3 M t M 0 2M 0 3M 0 ;

Fig. 2.29 40


3-A M

t

M

0

2M

0

m x

M

0

2M

M 0



0

x

;

x

0; M

3

3M

x

; M

A

4M

0

0

; .

We note that, in such a case, the sign used for torques is not so important. As soon as a certain sign has been adopted for the first met torque, the signs for the other torques have to be adopted consequently. Therefore, the torque diagrams may be sketched above or below the reference line. Like in the preceding examples, if a concentrated torque acts at a certain section of the member, at that point a sudden change in the torque diagram occurs (the change being equal to that concentrated torque). The torque diagrams are usually hachured as shown in Fig. 2.29.

2.4 SUPERPOSITION METHOD The superposition principle is a consequence of the material linear-elastic behaviour: the effect at any point of a linear-elastic mechanical structure subjected to several loads represents the summation of the effects produced by each of these loads acting separately. Using the superposition method, a complicated problem may be solved through a summation of simple problems. For example, the shear and bendingmoment diagram for the beam shown in Fig. 2.30a, may be sketched as an algebraical summation of the three diagrams of Fig. 2.30b.

a. Fig. 2.30

Important remark The drawing of the internal forces diagrams (axial forces, shear, bendingmoment or torque diagrams) may be performed in a unique, simple and logical manner: the involved member is cut at an arbitrary point, the internal force of a certain type representing the summation of all corresponding external loads (or b. moments) acting to the left or to the right of the cross section considered (and using the adequate sign convention). 41


In case of shear and bending moment diagrams, a particular case may also arise. The presence of one, two or more intermediate pin connections between different segments of a beam, offers one, two or more additional conditions for the computation of the external reactions. Draw the shear and bending-moment diagrams for the beam shown in Fig. 2.31. Due to the presence of the intermediate pin connection at point 2, the bending moment (as internal force) at that section is zero. On the other hand, the bending moment at point 2, represents the summation of all bending moments given by the loads acting to the left side of section 2. We obtain therefore: M

YA

4a

i2

0

p 2 a 3a

0

YA

1, 5 ap .

From now on the shear and bending moment diagrams may be sketched as if the support A and the intermediate pin connection would have not existed, the beam being subjected at A by an upward vertical concentrated external Fig. 2.31 force equal to 1,5ap. We finally obtain the shear and bending moment diagrams shown in Fig. 2.31.

2.5 MOVING LOADS A truck or other vehicle rolling across a beam or girder constitutes a system of concentrated loads at fixed distance from each other. For beams carrying only concentrated loads the maximum bending moment occurs under one of the loads. Therefore the problem here is to determine the bending moment under each load when each load is in a position to cause a maximum moment to occur under it. The largest of these various values is the maximum moment that governs the design of the beam. In Fig. 2.32, P1 ,P2, P3 and P4 represent a system of loads at fixed distances a, b and c from each other; the loads move as an unit across the simply supported beam with span . Let us locate the position of P2 when the bending moment under this load is maximum. If we denote the resultant of the loads on the span by R and its position from P2 by e, the value of the left reaction is: 42


R

YA

(

x) .

e



The bending moment under P2 is then:

Fig. 2.32

M

i

M

2

(

M )L

R 

(

e

x) x

P1 a .

To compute the value of x that will give the maximum M2, we set the derivative of M2 with respect to x equal to zero: dM

2

R

(



dx

e

2 x)

0

;

from which: x



e

2

2

.

(2.6)

This value of x is independent of the number of loads to the left of P2, since the derivative of all terms of the form P1a with respect to x will be zero. Equation (2.6) may be expressed in terms of the following rule: the bending moment under a particular load is a maximum when the center of the beam is midway between that load and the resultant of all loads then on the span. With this rule we locate the position of each load when the moment at that load is a maximum and compute the value of each such maximum moment. The maximum shearing force occurs at, and is equal to, the maximum reaction. The maximum reaction for a group of moving loads on a span occurs either at the left reaction, when the leftmost load is over that reaction, or at the right reaction when the rightmost load is over it. In other words, the maximum reaction is the reaction to which the resultant load is nearest.

2.6 INTERNAL FORCES DIAGRAMS FOR PLANE STRUCTURES (2D structures - FRAMES) AND SPATIAL (3D) STRUCTURES The principle presented above for sketching the straight beams internal forces diagrams may be easily extended to the plane or spacial structures. Let us consider for example the plane beam shown in Fig. 2.33, for which we have to draw the axial force, shear and bending-moment diagrams. An observer "O” covering the beam from 1 to A (or from A to 1, as it is easier from the mathematical point of view) sees each straight portion of the beam as a beam for which applies the rules presented in the preceding sections.

43


Therefore, for portion 1-2, at a current section at distance x from 1, we have (Fig. 2.33): N

0;

T

P;

M

i

P

x

0; M

x

a; M

x:

i1 i2

0; Pa .

Fig. 2.33

Fig.2. 34

The second straight portion of the beam may be covered from 2 to A or A to 2 as well. Let us suppose that the second case is being used.

Fig.2. 35

The effect of the concentrated force P is transmitted up to the current cross section located at distance x from point 2, where the observer is placed. Therefore, at that section we have: 44


N

P;

T M

0; P

i

a.

We note that all internal forces corresponding to the portion 2 - A are constant. We are now in the position to draw the internal forces diagrams (N, T, Mi). It is to be mentioned that, in such cases the diagrams are sketched with respect to a reference line representing the x axis of the beam (the axis directed along the beam). Analogous to the straight beams, for N and T diagrams + means above the reference line. For the bending-moment diagrams, "+" means below and "-" means above the reference line (from the observer's point of view). With these remarks, the internal forces diagrams have been represented in Fig. 2.36.

Fig. 2.36

The diagrams represented in Fig 2.36 tell us what does the plane beam feel (as a global effect) at each particular cross section, when subjected to the external load P. We also note that it was not necessary to compute the reactions XA, YA, MA for sketching the internal forces diagrams. SAMPLE PROBLEMS a) Draw the axial force, shear and bending moment diagrams for the frame and the loading shown in Fig. 2.37. Considering the frame as a free body we first determine the reactions: Fx

0

xA

0;

Fy

0

YA

YB

M

A

P;

0

YB

2

P

0

YB

P 2

Fig. 2.37 45


YA

P

YB

.

2

Applying the above presented principle and letting an observer to cover the beam from A to 1 and then from B to 1 we have: A - 1: N T M

P

YA X

A

X

i

;

2 0; x

A

0.

B - 2: N

0;

T

M

P

YB

i

YB

;

2

x

P

x

0; M

x

; M

iB

x;

2

0; P

i2

P



2

.

2

2 - 1: N

0;

T

M

YB

i

Y B (

P

P

P

P

2

x)

;

2

Px

P

(

2

x x)

0; M

Px ; x

; M

P i2 i1

;

2 0.

The axial force, shear and bending moment diagrams have been represented in Fig. 2.38.

Fig. 2.38

b) Draw the axial force, shear and bending-moment diagrams for the frame shown in Fig.2.39. 46


In Fig. 2.39b the simplified form of the frame together with the reactions have been represented. Considering the entire beam (frame) as a free body and using the external reference coordinate system Oxy, we determine the reactions as follows:

Fig. 2.39 F

x

0

X

F

y

0

YA

5 ap

A

YB

M

A

0

2 aY B

M

B

0

YA

0

X

2 ap

0

p 2a a

2a

5 ap

2a

A

5 ap ; YA

5 ap

YB

2a

p 2a a

2 ap ;

0

YB 0

YA

6 ap ; 4 ap .

We have found therefore that: X

A

YA YB

5 ap ; 4 ap ; 6 ap .

The second equation of equilibrium (∑Fy = 0) may be used as a checking equation when the reactions YA and YB were computed from ∑MA = 0 and ∑MB = 0. The N, T and Mi diagrams are shown in Fig. 2.40.

Fig. 2.40 47


c) Draw the axial force, shear and bending-moment diagrams for the curved beam of radius R shown in Fig. 2.41. Although the axis of the beam is not a straight line, the principle presented above for sketching the N, T, Mi diagrams remains valid. The problem consists in determining the internal forces corresponding to each particular cross section of the beam. In order to locate the current cross section an angular parameter must be used (instead of the linear parameter “x” which has been used up the now) for each particular portion of the curved beam.

Fig. 2.42

Fig. 2.43

Let us now consider the first portion 1–2 of the built-in arch shown in Fig. 2.41. The axial force, shear and bending moment equations for segment 1 – 2 are obtained similarly by passing a cross section a–a anywhere between 1 and 2. As discussed above, the cross section is located by the parameter . When varies between 0˚ and 90˚ the whole portion 1 - 2 of the curved beam is covered. The concentrated load P which acts at section 1, transmits its effect through the segment BCDE up to the current cross section D'E’ located by parameter (Fig. 2.42). This means that a vertical downward load P will act at the centroid O of the current cross section D’E’. It is in fact the internal force exerted on the current cross section and may be resolved into two components: one component perpendicular to the current cross section D’E’ and the other one contained within the plane of the cross section. The first component represents the current axial force (N) while the second component represents the corresponding shearing force (T). An observer “O” placed at the current cross section, using the proper positive sign convention will see that: 0;N 0 ;T

N

T

P cos

:

P sin

:

;N

P; 0; 0;

2 ;T 2

48

P.


The bending moment exerted by the concentrated load P, acting at point 1, with respect to the centroid O of the current cross section is (Fig. 2.43): M

P(R

i

R cos

)

PR (1

0;M

):

0;

i1

;M 2

cos

PR .

i2

The same reasoning may be applied when the second portion of the curved beam is to be covered. This time it will be more convenient to us to cover Fig. 2.43 the beam from A to 2. But in this case we have to compute the reactions YA and MA at first. This will be done by considering the entire curved beam as a free body and using the corresponding equations of equilibrium, Fig.2.42. Fy

M

0

YA

0

A

M

P

2P

P

A

2R

0

3P ;

YA

2P

R

0

M

4 PR .

A

We write: N

Y A cos

3 P cos

0;N ;N T

3P ;

A

2

2

Y A sin

0. 3 P sin

0 ;T A

2

;T2

;

;

0; 3P .

Fig. 2.44 M

M

i

0; M

A

iA

;M 2

i2

YA (R

M

A

R cos

)

4 PR

3 PR (1

cos

);

4 PR ;

PR .

The axial force, shear and bending moment diagrams have been represented in Fig. 2.45. It is to be noted that all the established properties of the N, T, Mi diagrams of straight beams remains valid. 49


Fig. 2.45

We shall now consider a simple spatial structure (a 3D structure, i.e. a three dimensional structure) represented by a beam, fixed at one end, and subjected to two concentrated loads P and Q at the other end, Fig. 2.46.

Fig. 2.46

d) Draw the axial force, shear, bending moment and torque diagrams for the 3D beam shown in Fig. 2.46. Although the problem seems to be a little bit more completed the basic principle for sketching the internal force diagrams remains unchanged. Before solving the problem there are still some important remarks to be done: As it will be discussed later, the sign of the shearing force has no physical consequences in designing a beam or a certain mechanical structure. This is why, in case of complicate structures, we shall give the T sign up and we shall represent the T diagrams as they are convenient to us; The sign of the bending moment does also depend upon the relative position of the observer, Fig. 2.47. Although the two beams represented in Fig. 2.47 are entirely identical from the geometrical and loading point of view, the two corresponding bending - moment diagrams have different signs. These signs are, after all, simple conventions. On the other hand, it is to be observed that in both cases of Fig 2.47, the bending-moment diagrams occupy the same position with respect to the reference line. In other words, this means that the position of the bending-moment diagrams do not depend upon the observer's position. The 50


location of the bending moment diagram with respect to the reference line corresponds to the position of the beam fibres in tension.

Fig. 2.47

For this reason, in many cases, we shall not note the sign of the bending moment diagrams and we shall represent these diagrams on the side corresponding to the beam fibres in tension, Fig. 2.48. Let us now return to the original problem regarding the simple 3D structure shown in Fig. 2.46. Covering the beam from 1 to A and attaching a proper coordinate system (whose Ox axis is usually directed along the beam) to each main portion of the beam (Fig. 2.46) we obtain the axial force, shearing force, bending - moment and torque diagrams shown in Fig. 2.49. It should Fig. 2.48 be noted that, for each main portion of the beam, two shearing forces and two bending-moments could exist simultaneously (about Oy and Oz axes). We shall now conclude our analysis concerning the main types of internal forces by observing that these diagrams are in fact a graphical representation of a global mechanical effect occured at any particular cross section of a given member subjected to external loads. 51


Fig. 2.49

While these diagrams represent a first and necessary step in the analysis of a given structural member, they do not tell us whether the external loads may be safely supported. Whether or not a given structural member will break under the external loading clearly depends upon the ability of the material to withstand the corresponding elementary forces occurred at the level of each particular point of the member cross sections. This is why, after a short study of the moments of inertia within the next chapter, some other chapters of the text will be devoted to the analysis of the stresses and of the corresponding deformations in various structural members, considering axial loading, shearing loading, torsion and bending successively. Each analysis will be based upon a few basic concepts, namely, the conditions of equilibrium of the forces exerted on the member, the relations existing between stress and strain in the material and the conditions imposed by the supports and loading of the member. The study of each type of loading will be complemented by examples, sample problems and problems to be assigned, all designed to strengthen the students understanding of the subject.

52


PROBLEMS TO BE ASSIGNED

P2 P.2.1 Draw the axial force, shear and bending – moment diagrams for the members, frames and loading shown (Fig. P.2.1).

a.

b.

c.

d.

e.

f.

g.

h.

i.

j. Fig. P.2.1

53


k.

l.

m.

n.

o.

p.

Fig. P.2.1 (continued)

54


r.

s.

t.

u. Fig. P.2.1 (continued)

P.2.2 Draw the torque diagrams for the members and loading shown (Fig. P.2.2).

Fig. P.2.2 P.2.3 Draw the axial force, shear, bending – moment and torque diagrams for the 3-D structures and loading shown (Fig. P.2.3).

55


a.

b.

c.

d.

e.

f. Fig. P.2.3

56

3.FIRST MOMENTS AND SECOND MOMENTS OF AN AREA

Many engineering formulas and applications such as those relating to strength of beams, columns, shafts, etc., involve the use of different mathematical expressions which describe, from the mathematical point of view, the shape and dimensions of the cross sections. These mathematical expressions are called: geometrical characteristics. For members under axial loading (tension or compression) the single involved geometrical characteristic is represented by the cross-sectional area A of the member. A higher value of the cross-sectional area A of the member means a higher strength of the member under axial loading (Fig.3.1). For structural elements in bending, torsion etc. there are also other geometrical characteristics involved within the strength calculus: the static moments (first moments of an area) and the moments of inertia (second moments Fig. 3.1 of an area).

3.1 FIRST MOMENTS AREA

OF AN AREA. CENTROID OF AN

Consider an area A located in the zOy plane (Fig.3.2). Denoting by z and y the coordinates of an element of area dA, we define the first moment of area A with respect to z axis as the integral: S

y dA

z

.

(3.1)

A

Similarly, the first moment of area A with respect to the y axis is defined as the integral: S

z dA

y

.

(3.2)

A

We note that each of these integrals may be positive, negative or zero, depending upon the position of the coordinate axes. The first moments of area, Sz and Sy, are expressed in mm3, cm3, m3, etc.


Since in almost all cases the area A of Fig.3.2 is assimilated to the crosssectional area of a beam, a shaft etc., it is presented as being located in the zOy plane, the Ox axis being directed along the beam, shaft etc. The centroid of area A is defined as the point G of coordinates zG and yG (Fig.3.2), which satisfy the relations: Fig. 3.2 z dA S

A

zG

A

y

S

A

y

A zG ;

(3.3)

y dA Sz

A

yG

A

Sz

A

A yG .

Comparing (3.1) and (3.2) with (3.3), we note that the first moments of area A may be expressed as the products of the area and the coordinates of its centroid: Sy

A zG ;

Sz

A

(3.4)

yG .

When an area possesses an axis of symmetry, the first moment of the area with respect to that axis is zero. Indeed, considering the area A of Fig.3.3, which is symmetric with respect to the Oy axis, we observe that to every element of area dA of abscissa z corresponds an element of area dA’ of abscissa -z. It follows that the integral in (3.2) is zero and, thus, Fig. 3.3 Sy=0. It does also follow from the first of the relations (3.3) that zG = 0. Thus, if an area A possesses an axis of symmetry, its centroid G is located on that axis. If an area possesses two axes of symmetry (Fig.3.4) the centroid G coincides with its geometric center.The coordinate axes passing through the centroid of a given area are called centroidal (or central) axes. It is to be observed that the integrals involved in relations (3.1) and (3.2) are actually double integrals, which have to be calculated with the well known mathematical methods (Fig. 3.5). S

y dz dy ; S

y dA

z A

z dA

y A

(D )

58

z dz dy (D )

.

First moments and second moments of an area

Fig. 3.4

Fig. 3.5

3.2 SECOND MOMENTS OF AN AREA Consider again an area A located in the zOy plane (Fig. 3.6) and an element of area dA of coordinates z and y. The second moment, or moment of inertia, of area A with respect to the Oz axis, and the second moment, or the moment of inertia, of area A with respect to the Oy axis are defined, respectively, as: 2 2 (3.5) Iy z dA . Iz y dA ; A

A

While each of the above integrals is actually a double integral, it is possible in Fig. 3.6 many applications to select elements of area dA in the shape of thin horizontal or vertical strips, and thus reduce the computation to simple integration. This will be illustrated later. We now define the centrifugal moment of inertia (or the product of inertia) of area A with respect to Oz and Oy axes (Fig. 3.6) as the integral: I zy

zy d A .

(3.6)

A

Relations (3.5) show that the moments of inertia of an area are positive quantities and are expressed in mm4, cm4, m4 etc. On the other hand, relation (3.6) shows that the centrifugal moment of inertia may be positive, negative or zero, depending upon the locations of the area relative to the involved axes. It is positive if the area lies principally in the first or third quadrants and negative if the area lies principally in the second or fourth quadrants. We define the polar moment of inertia of area A with respect to point O (Fig. 3.6) as the integral: 59


I

r

p

2

dA,

(3.7)

A

where r is the distance from O to the element dA. While this integral is again a double integral, it is possible in the case of a circular area to select elements of area dA in the shape of thin circular rings, and thus reduce the computation of Ip to a simple integration. This will be illustrated later. It is to be noted that the polar moment of inertia is also a positive quantity, being expressed in mm4, cm4, m4 etc. An important relation may be established between the polar moment of inertia Ip of a given area and the moments of inertia Iz and Iy of the same area. Noting that r2 = z2 + y2 (Fig. 3.6) we write: I

r

p

2

dA

z

A

2

y

2

A

dA

z A

2

dA

y

2

dA

A

or I

p

Iz

Iy

.

(3.8)

If an area has an axis of symmetry, this axis together with any axis perpendicular to it will form a set of axes for which the centrifugal moment of inertia is zero. Consideration of the symmetrical section shown in Fig. 3.7 will disclose that, for any differential area dA, there is a symmetrically placed equal differential area dA’. With respect to the Oy axis of symmetry, the z coordinates of dA and dA’ are equal but of opposite sign, whereas their y coordinates are equal and of the same sign regardless of the Fig. 3.7 position of the Oz axis. Hence the sum of the products zy dA for each such pair of symmetrically placed elements dA and dA’ will be zero. It follows, therefore, that the value of zy d A for the entire area will be zero if either or both reference axes are axes of A

symmetry.

3.3 PARALLEL - AXIS THEOREM (STEINER’S RELATIONS) Consider the moments of inertia Iz and Iy and the centrifugal moment of inertia Izy of an area A with respect to two arbitrary perpendicular axes Oz and Oy (Fig. 3.8). We assume to know the quantities Iz, Iy and Izy, where 60


Iz

y

2

dA ;

I

z

y

A

2

dA ;

A

zy d A .

I zy A

Let us now consider another coordinate system z1O1y1, translated with quantities a and b with respect to the axes Oy and Oz of the first coordinate system. The problem which arises consists in determining the quantities I z , I y and Iz

1

y

Fig. 3.8

1

1

of the same area A but with respect to the axes of the new coordinate system. 1

We write: Iz

2

y1 d A

1

A

y

2

dA

y dA

2b

2

A

a

y

dA

dA

z

z dA

2a

a

2

A

y

A

Iz

2 bS

2

2 az

dA

I

2

b A;

z

a

2

dA

b

A

a A;

y

a dA

z

zy

zb

ay

dA

I zy

bS

ab

dA

A

z dA

b

2

2 aS

y

A

zy d A

dA

A

z1 y1d A

1

2

b

A

A

1

2

A

2

2 by

A

z

z dA

Iz

2

b

A

z1 d A

1

2

y A

A

y

dA

A

y

I

2

b

y dA

a

A

ab

A

y

aS

z

A

Thus, the mathematical connection between the moments of inertia I zy

of an area A and the same quantities

abA .

Iz

, 1

I

y

and 1

Iz

1

y

Iz

,

I

y

and

calculated with respect to 1

the translated coordinate system z1O1y1, is described by the following relations: Iz I

1

y1

Iz

1

y1

2

Iz

2 bS

z

b A

I

2 aS

y

a A

y

I zy

bS

2

y

aS

where a is the distance between axes Oy and O1y1, b is the distance between axes Oz and O1z1, 61

z

(3.9) abA ,


Sz, Sy are the static moments (first moments) of area A with respect to axes Oz and Oy. If the point O is the centroid of area A, it follows from relations (3.4) that Sz = Sy = 0 and we may write: Iz I

y

Iz

Iz

1

1

1

1

y

1

2

Iz

b A;

I

a A;

2

y

I zy

(3.10)

abA .

Relations (3.10) are known as Steiner’s formulas. For example, the first relation of (3.10) expresses that the moment of inertia of an area with respect to an arbitrary Oz1 axis is equal to the moment of inertia Iz

of the same area with respect to the centroidal Oz axis parallel to the Oz1, plus the product b2A of area A and the square of the distance b between the two axes. This result is also known as the parallel-axis theorem. It makes it possible to determine the moment of inertia of an area with respect to a given axis, when its moment of inertia with respect to a centroidal axis of the same direction is known. Conversely, it makes it possible to determine the moment of inertia Iz of an area A with respect to a centroidal axis Oz, when the moment of inertia I z of A with respect to a parallel axis 1

2

is known, by subtracting from I z the product b A. We should note that the parallel1

axis theorem may be used only if one of the two axes involved is a centroidal axis.

3.4 MOMENTS OF INERTIA OF SIMPLE SURFACES a) Rectangular area For the rectangular area A shown in Fig. 3.9, determine the moments of inertia Iz, Iy and Izy with respect to the centroidal Oz and Oy axes. As mentioned before: 2

y dA

Iz

.

A

We select as an element of area (dA) a horizontal strip of length b and thickness dy (Fig.3.9). We write: dA = b dy. Fig. 3.9

It follows that: 62

First moments and second moments of an area h

h

2 2

Iz

2

y dA

y b dy

by

2

1

b

3

h

A

3

3

h

h

3

h

8

3

b 2h

8

3

bh

24

3

.

12

2

2

Thus, the moment of inertia Iz of a rectangular area with respect to the centroidal Oz axis is: bh

Iz

3

.

(3.11)

.

(3.12)

12

In the same manner, it follows that : 3

hb

Iy

12

Since Oz and Oy are axes of symmetry, we have: I zy

0.

b) Circular area For the circular area shown in Fig. 3.10 determine the polar moment of inertia I p and the moments of inertia Iz, Iy and Izy.

Fig. 3.10

Fig. 3.11

We select as an element of area (dA) a ring of radius r and thickness dr, (Fig.3.11). The polar moment of inertia of area A is:

I

d

d

2

2

2

r dA

p A

r 0

2

2 r dr

2

3

r dr

2

r

4

d

4

0

2 0

63

d

2

4

16

0

d 32

4

.


Thus d

I

p

4

.

(3.13)

32

Due to the symmetry of the circular area, we have Iz = Iy. Recalling (3.8), we write: I

Iz

p

I

2I

y

z

2I

d y

4

32

and, thus Iz

I

d y

4

.

(3.14)

64

c) Triangular area Determine the moments of inertia for a triangle of base b and altitude h with respect to an axis coinciding with its base and a centroidal axis parallel to its base. Select the differential element as shown in Fig. 3.12. From b h

m

y

similar

triangles,

we

have

. The moment of inertia with respect

h

to z1 axis is obtained from: h

Iz

h

2

y dA

1

y

A

2

m dy

0

bh 3

3

b h

2

y

b h

h

4

bh

3

bh

3

dy

h

0 4

y

3

4

bh

3

.

12

We have thus obtained: Fig. 3.12

Iz

bh 1

3

.

To determine the centroidal moment of inertia Iz

1

(3.15)

12 Iz

, we transfer the known value of

, from the base axis z1 to the parallel axis z. Since the transfer distance is

shown in Fig. 3.12, we write: 64

h 3

as


Iz

1

Iz

h

2

bh

3

.

2

It follows that Iz

Iz

h 1

3

2

bh

bh

2

12

3

bh 18

3

bh

3

.

(3.16)

36

3.5 MOMENTS OF INERTIA OF COMPLEX SURFACES (COMPOSITE AREAS) To determine of the moments for inertia of a complex surface the following steps have to be covered: - the complex surface (area) A has to be divided into several component parts of areas A1, A2 ...; - determination of the centroidal point G of the complex area; - since the integral representing the moment of inertia of area A may be subdivided into integrals extending over A1, A2 ..., the moment of inertia of A with respect to a given axis will be obtained by adding the moments of inertia of areas A1, A2 ... , with respect to the same axis. Before adding the moments of inertia of the component areas, however, the parallel-axis theorem should be used to transfer each moment of inertia to the desired axis. This is shown in the following example. Determine the moments of inertia Iz, Iy and Izy of area A shown in Fig. 3.13, with respect to the centroidal axes.

Fig. 3.13

Fig. 3.14

We first divide the complex area A into the two rectangular areas A1 and A2 (Fig 3.14) and denote their centroids and their own centroidal axes by G1 ,G2 , z1 , y1 , z2, y2 respectively. 65


We may now determine the coordinates zG and yG of the centroid G of the composite area A, using, for example, the coordinate system z1 G1 y1 as follows: y i Ai

3 a 1, 5 a

i

yG

Ai

4a

4a 1, 5 a ;

2

(i = 1,2)

4 a 1, 5 a

i

1, 5 a 2a

z i Ai

4 a 1, 5 a 2

i

zG

Ai

4a

2

0 , 75 a

.

4 a 1, 5 a

i

Recalling the formulas (3.11) and (3.12) and using the parallel-axis theorem we may write the moments of inertia of the composite area A as follows: 4a Iz

a

1, 5 a

2

4a

2

1, 5 a

Iy

4a

3

0 , 75 a

2

4a 4a

1, 5 a

2

4 a 1, 5 a

23 , 33 a

4

;

2

3

a

1, 5 a 2a

12

0

1, 5 a

12

4a

10 , 2 a

3

2 ,5 a

12

a

I zy

3

12 4

0 , 75 a

4 a 1, 5 a

2

;

1 , 565 a

0 , 652 a

4a

2

1, 5 a 0

2 ,5 a

1 , 565 a

2a

0 , 652 a

4 a 1, 5 a

2 4

7 , 43 a .

3.6 MOMENTS OF INERTIA WITH RESPECT TO INCLINED AXES In some cases, it is necessary to determine the moments of inertia with respect to axes that are inclined to the usual axes. The moments of inertia in such cases can be obtained by formal integration, but a general formula is usually easier to use. The problem may be stated as follows: assuming the values Iz, Iy and Izy with respect to the Oz and Oy axes to be known, determine the values of

Fig. 3.15

66

First moments and second moments of an area I

,

z1

I

and

y1

Iz

y

1

with respect to the Oz1 and Oy1 axes inclined at an angle

1

with

Oz and Oy axes, as shown in Fig. 3.15. The coordinates for a typical differential area dA are given by z and y with respect to the y and z axes, and by y1 and z1 relative to the z1 and y1 axes. The relation between these coordinates can be obtained by projecting the coordinates z and y on the z1 and y1 axes. This gives (Fig. 3.15): z1

z cos

y sin

;

y1

y cos

z sin

.

By definition, the values of Iz

Iz

and 2

;

y1 d A

1

I

y

1

(3.17)

are: 1

2

I

y

z1 d A

1

A

;

Iz

1

y

z1 y1 d A

1

A

.

A

Replacing the values of z1 and y1 from (3.17) we have: Iz

2

y1 d A

1

y cos

A

I zy sin 2

z1 d A

1

z cos

A

I

Iz

1

y

I

y

sin 2

y sin

2

cos

2 yz cos

2

sin

2

z sin

dA

dA

z

2

2

cos

2 zy cos

2

sin

y sin

2

dA

A

2

I z sin

z1 y1 d A

1

2

cos

;

A

y

2

y A

2

2

y

dA

A

I z cos

I

2

z sin

2

z cos

A

;

I zy sin 2

y sin

y cos

z sin

dA

A

zy cos

2

2

2

z sin

cos

y sin

cos

I zy cos

cos

yz sin

2

dA

A

Iz

I

y

sin

2

sin

2

.

We, thus, obtain: Iz I Iz

y

1

I z cos

1

I z sin

1

y

1

Iz

2

I

2

I I

y

If the relations 67

sin

y y

sin cos

2 2

cos

2 I zy sin

cos

;

2 I zy sin

cos

;

I zy cos

2

sin

2

(3.18) .


sin

1

2

cos 2

,

1

2

cos

cos 2

2

2

are substituted in (3.18), we may write: Iz I

y

Iz

1

Iz

1

Iz

1

cos 2 2 cos 2

1

2 I

Iz 1

y

I

y

1

cos 2

1

2 cos 2

y

I

y

2

sin 2

;

I zy sin 2

;

I zy cos 2

2

1

I zy sin 2

or Iz

Iz I

I 2

1

Iz y

1

Iz

y

2 Iz y

I

y

cos 2

I zy

sin 2

;

cos 2

I zy

sin 2

;

2 I

1

Iz

Iz

y

I

y

2 I

y

sin 2

I zy

2

1

cos 2

(3.19)

.

When the values of Iz, Iy and Izy are known, relations (3.19) permit the values of I z , I y and I z y with respect to the Oz1 and Oy1 axes, inclined at an angle to the 1

1

1

1

Oz and Oy axes, to be determined without further integration. In a sense, these relations do for inclined axes what the Steiner’s formula does for parallel axes. A simple analysis of relations (3.19) tells us that I z , I y and I z y are 1

1

1

1

functions of angle . One could ask: which are the values of angle that make these quantities ( I z , I y and I z y ) maximum or minimum? The angles defining the 1

1

1

1

maximum and the minimum moments of inertia, also called the principal moments of inertia, may be found by differentiating (3.19) with respect to and setting the derivative equal to zero: d Iz

1

2 sin 2

Iz

d dI d

I

y

2 cos 2

2 y

1

2 sin 2

Iz

I 2

We find that:

68

I zy

2Iz

1

y

0; 1

(3.20) y

2 cos 2

I zy

2Iz

1

y

0. 1


2 I zy

tg 2

Iz

Equation (3.21) gives us two values of I

y

.

I

(

(3.21)

y

1 and

2

1

) for which 2

Iz

and 1

have extreme values. This is why the equation (3.21) is always written as: 1

2 I zy

tg 2

The extreme conditions for inertia

Iz

1

y

Iz

,

1, 2

I

I

.

(3.22)

y

– (3.20 ) - mean in fact that the product of

y

1

Iz

1

equals zero. In the same time, a second differentiation of (3.20) shows 1

that: d

2

d

Iz

2

d

I

1

2

Iz

1

2

d

which means that a maximum value of

y

(3.23)

,

implies a minimum value of 1

I

y

and 1

vice versa. Substituting for from equation (3.21) into (3.19) we obtain the extreme values of quantities I z and I y , called the principal moments of inertia: 1

1

I 1, 2

Iz

I

y

2

1 2

Iz

with respect to the axes Oz1 and Oy1, rotated with angle

I

2 y

2

4 I zy

(3.24)

1.

In this way we found two perpendicular directions given by 1 and

2

1

for which the moments 2

of inertia of the original area have extreme values (a maximum value I1 with respect to one of these directions and a minimum value I2 with respect to the other direction). Usually, the axis of maximum is denoted by 1 while the axis Fig. 3.16 of minimum by 2. It is important to be mentioned again that the product of inertia of the original area with respect to the coordinate system 1O2 (Fig. 3.16) is zero. 69


Axes 1 and 2 are called principal axes. One could demonstrate that if Izy < 0 the axis of maximum is placed in the first quadrant while, if Izy > 0, the axis of maximum is placed in the second quadrant.

3.7 RADIUS OF GYRATION. ELLIPSE OF INERTIA The term radius of gyration is used to describe another mathematical expression and occurs most frequently in column formulas. Radius of gyration is usually denoted by the symbol i and is defined as: I

i

,

(3.24)

A

where I is the moment of inertia and A the area. Thus, we have: iz

Iz

;

A

I iy

y

;

A

(3.25) i1

I1 A

;

i2

I2

.

A

The following is a geometric interpretation of this relation. Assume the area of Fig. 3.2 to be squeezed into a long narrow strip as shown in Fig. 3.17. Each differential element of area dA will then have the same distance iz from the Oz axis. The moment of inertia is given by: 2

Iz

2

y dA

iz dA

A

A

2

iz

Fig. 3.17

dA

2

iz

A.

A

The strip may be placed on either side of the reference axis, since if iz is negative, squaring it will automatically make it plus. Also, part of the strip may be at a distance iz from one side of the reference axis and the remainder of the strip at equal distance iz from the other side of the axis. In view of this discussion, the radius of gyration is frequently considered to be the uniform distance from the reference axis at which the entire area may be assumed to be distributed. For an area whose dimensions perpendicular to a reference axis are negligibly small compared with its distance from that axis, the radius of gyration is practically equivalent to the centroidal location of the area. 70


The ellipse of equation z

2

2

y

2

1

2

i2

(3.26)

0

i1

represents the centroidal principal ellipse of inertia with respect to a certain area A.

Sample problems 1. For the area shown in Fig. 3.18 determine: (a) the centroidal point G of the area A; (b) the moments of inertia Iz, Iy and Izy with respect to the centroidal reference system zGy; (c) the principal axes of inertia 1 and 2; (d) the principal moments of inertia I1 and I2; (e) the principal radii of inertia i1 and i2 and (f) draw the ellipse of inertia. Solution We first divide the area A of the whole surface into three rectangular areas with centroidal points G1, G2 and G3 (Fig. 3.18). We observe that the centroid of the second rectangular area G2 coincides with the centroid G of the whole area A. Thus, axes z2 and y2 coincide with the centroidal axes Gz and Gy of the whole area A. Recalling (3.11) and (3.12) and using the parallel-axis theorem we may write the moments of inertia of the composite area A as follows: 165 Iz

30

3

30

2

135

30

165

240

2

12

2,1573

3

12

10

8

mm

4

Fig. 3.18

;

30 Iy

165

3

82 , 5

2

240 165

30

135

3

2

0 , 9038

12

I zy

30

10

8

mm

4

;

12

82 , 5

30

165

2

1 ,1026

10

8

mm

4

.

The principal directions of inertia are given by

tg 2 1 , 2

2 I zy Iz

Iy

2 1 ,1026 2 ,1573

1

0 , 9038

30 ,19



;

2

1

 59 , 81 . 2

Since Izy > 0, the principal axis of maximum is placed in the second quadrant (Fig. 3.18).

71


The principal moments of inertia are:

I 1, 2

Iz

Iy

1

2

2 ,1573

Iz

2

10

8

0 , 9038

Iy

2

10

2

4 I zy 8

1 2 ,1573

2

10

8

0 , 9038

8

10

2

4 1 ,1026

10

8

2

.

2

We finally have: I1 I2

2 , 799

10

0 , 262

10

8 8

mm

4 4

mm

; .

The principal centroidal radii of inertia are: i1

i2

I1 A

2 , 799 30

I2 A

30

165

8

10

2

30

0 , 262

10

165

2

127 , 94

mm ;

39 ,14

mm .

240 8

30

240

The principal centroidal ellipse of inertia has been represented in Fig. 3.18. 2. For the composite area of Fig. 3.19, composed of two U-shaped profiles, determine the same quantities like in the previous example. Solution We first divide the area A of the whole surface into two areas 1 and 2 (the two U shapes) having the centroidal points at G1 and G2 (Fig. 3.19). From the appropriate tables containing the geometrical characteristics of rolled-steel shapes (APPENDIX III) we may get all the necessary data: I

1:

I

z

y

A

Fig 3.19

248 10

mm

4

;

1

3600 10

4

mm

4

;

1

42 , 3 10

1

Iz

In these tables we may also find the location of the centroidal point with respect to the Ushaped section (22,3 mm - Fig. 3.19). Using the coordinate system z1 G1 y1 , we can compute now the position of the centroidal point G of the entire area as follows:

4

3600

2

mm

10

4

2

mm

; 4

;

2

2:

Iy

248

10

4

mm

4

;

2

A2

72

42 , 3 10

2

mm

2

.

First moments and second moments of an area y i Ai

yG

120

22 , 3

Ai

2

z i Ai

zG

22 , 3

Ai

2

48 , 85 mm ;

2

42 , 3 10

120

2

42 , 3 10

2

42 , 3 10

71 ,15 mm .

2

42 , 3 10

We have thus located the centroidal reference coordinate system of the composite area: zGy (Fig.3.19). Recalling again the formulas (3.11) and (3.12) and using the parallel –axis theorem we may write the moments of inertia of the composite area as follows: Iz

248

10

4

58 , 668

I

y

3600

81 , 30 10

I zy

0

6

10

10

2

48 , 85

mm

4

71 ,15

6

4

mm

48 , 85

42 , 3 10

4

2

2

42 , 3 10

3600

10

4

120

22 , 3

48 , 85

2

42 , 3 10

2

;

2

42 , 3 10

2

248

10

4

120

71 ,15

22 , 3

2

42 , 3 10

2

;

71 ,15

29 , 4 10

42 , 3 10

6

mm

4

2

0

120

22 , 3

48 , 85

120

71 ,15

22 , 3

.

The principal directions of inertia are given by

tg 2 1 , 2

2 I zy Iz

Iy

2

29 , 4 10

58 , 668

6 1

81 , 30 10

6 2

34 , 47 124 , 47

 

; .

Since Izy > 0, the principal axis of maximum is placed in the second quadrant (Fig 3.19). The principal moments of inertia are

I 1, 2

Iz

Iy

1

2

Iz

2

Iy

We finally have I1 I2

101 , 48 38 , 48

The principal centroidal radii of inertia are:

73

10 10

6 6

mm mm

4 4

; .

2

2

4 I zy

.


i1

I1

101 , 48

A

2

10

42 , 3 10

6 2

109 , 52 mm ;

i2

I2 A

38 , 48 2

10

6

42 , 3 10

2

67 , 44 mm .

The principal centroidal ellipse of inertia is shown in Fig. 3.19.

PROBLEMS TO BE ASSIGNED P.3 For the areas shown in the figures below, locate the centroids of the areas and then determine: the second moments of the involved areas ( Iz, Iy and Izy ) with respect to the centroidal axes; the principal axes of inertia 1 and 2 ; the principal moments of inertia I1 and I2; the principal radii of inertia i1 and i2, and, finally, draw the ellipse of inertia.

Fig. P.3.1

Fig. P.3.2

Fig. P.3.3

Fig. P.3.4

Fig. P.3.5

Fig. P.3.6

Fig. P.3.7

Fig. P.3.8

Fig. P.3.9

74

Fig. P3.10

4. DISPLACEMENTS, STRESSES AND STRAINS 4.1 DISPLACEMENTS Consider a body subjected to several external loads P1, P2,…, Pk, Pn in mechanical equilibrium, (Fig. 4.1). The body has been represented in a rectangular    coordinate system Oxyz, with i , j and k the versors of the reference axes. Let us assume that, before the action of the external loads, a certain point of the body occupies the position A (Fig. 4.1). Due to the action of the external loads the body deforms and, after deformation of the body, the point A is displaced to A . The length A A is called the total displacement of point A. Denoting the displacement Fig. 4.1 components of point A by u, v and w, we may write AA

 u i

v

 j

 w k

(4.1)

The components u, v, w are called linear displacements. 4.2 THE CONCEPT OF STRAIN Consider again a body subjected to several external loads in mechanical equilibrium, P1,…,Pn. Before the action of the loads the involved body occupies a certain position in space. Due to the action of the external loads the body deforms (Fig. 4.2) and all its points are displaced to the other positions. Two points occupying the positions A and B, within the initial undeformed state of the body, are displaced to A and B respectively. In this way the distance between points A and B changes due to the action of the external loads. Denoting by the original length of Fig. 4.2 the segment AB and by ’


the length of A’B’ we define the linear absolute deformation as ’- .



(4.2)

Letting  approach zero, we define the deformation per unit length at point A (or the normal strain at point A) along direction AB as  lim 



0

.

(4.3)

Since Δ and are expressed in the same units, the normal strain ε obtained by dividing  by is a dimensionless quantity. Returning now to Fig. 4.2, let us consider three points (O, C, D) of the body in its original, undeformed state, where segments OC and OD are perpendicular. After deformation of the body, the points O, C, D are displaced to O , C , D the segments O C and O D losing their perpendicularity. In the same time a change of the right angle between OC and OD occurs ( COD C O D ). Letting the segments OC and OD approach zero, we define the shearing strain as COD

lim OC OD

C 'O ' D '

0 0

.

(4.4)

In fact represents the change of the original right angle COD. In case of a rectangular element of infinitely small sides a, b and c the shearing strain corresponding to the x and y directions is, (Fig. 4.3):

Fig. 4.3

If we attach a rectangular coordinate reference system Oxyz to the body represented in Fig. 4.2, the knowledge of the body deformation state implies the knowledge, for each particular point of the body, of all components contained in the table bellow.

T

x

xy

xz

yx

y

yz

zx

zy

z

.

(4.5)

is called the deformations tensor. In other words, the deformation state of a body subjected to certain external loads is defined by the nine components of table (4.5). Since xy , xz , yz , only six components of table (4.5) are distinct. yx zy zx This is why the deformation state is a tensorial quantity. T

76

Displacements, stresses and strains

4.3 THE CONCEPT OF STRESS While the method of sections (i.e. internal forces diagrams) presented in chapter II represents a first and necessary step in the analysis of a certain mechanical structure (member, beam, etc) it does not tell us whether the loads may be safely supported. This is why we have to introduce the concept of stress. Let us now return for a while to the method of sections, considering a body subjected to several external loads in mechanical equilibrium (Fig. 4.4a).

b.

a. Fig. 4.4

Passing an exploratory plane through the body at some arbitrary point we obtain the two segments shown in Fig. 4.4b. (I and II). In this way we expose the internal forces acting on the exploratory section that are necessary to maintain the equilibrium of either segment. As presented in chapter II, in general, the internal forces reduce to a force and a couple that, for convenience are resolved into their components that are normal and tangent to the section, (Fig. 4.5). 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 get ready to show why this is so, we shall prove it as we progress. Anyhow, the internal forces defined above (R, M, R’, M’ or their components) represent a global effect occurred within the Fig. 4.5 exploratory section considered. But, if the considered body is to fail under the action of the external loads, this failure begins at the level of a certain point of the exploratory section. This is why we have to analyze the local effect occurred at the level of the exploratory sectional points, when the involved body is subjected to several external loads in mechanical equilibrium. Such an analysis requires the concept of stress.

77


Let us now consider the Ist segment of the body shown in Fig.4.4, for which we take an elementary area ΔA of the section surface considered, around an arbitrary point C, (Fig.4.6). We denote by P the surface force acting on ΔA. This force models the mechanical interaction between the points of ΔA and the points of the area ΔA’ of the second segment of the body with which ΔA comes in contact. The vector quantity

Fig. 4.6

P

pm

(4.6)

A

is called the total average stress developed on ΔA. Now letting ΔA approach zero (which means that ΔA reduces to the point C), the vector quantity p

P lim A

(4.7)

A

0

is called the total stress at point C or simply the stress at point C, (Fig.4.6). Generally speaking, the stresses on different sections passing though the same point in a body are different. Usually, the stress is resolved into a component (called the normal stress) acting along the normal direction n to the ΔA surface and a tangential component (called the shearing stress) acting along a tangential direction t , contained within the plane of surface ΔA (Fig.4.6). Since the force P is expressed in newtons (N) and ΔA in square millimeters, the stress p and the stress components and will be expressed in N/mm2. This unit is called a megapascal (MPa). We have 1 MPa

1

N mm

2

.

If we take a rectangular coordinate reference system xyz with its origin at point C (with Cx axis perpendicular to ΔA and Cz and Cy axes contained within the ΔA plane), the shearing stress may be resolved into other two components acting along Cy and Cz respectively, Fig. 4.7.To indicate the acting plane and the direction of a normal stress, we associate the stress with a coordinate subscript.

78


For instance, indicates the normal stress acting on a plane perpendicular to the x axis and also in the x direction. As to a shearing stress, we should associate it with two coordinate subscripts. For instance, indicates the shearing stress acting on a plane perpendicular to the x axis, but in the y direction. The stress p shown in Fig. 4.7 may be thus expressed as x

xy

Fig.4.7  i x

p

where

  i ,j

xy

 k xz

 j

,

(4.7)



and k are the versors of the reference axes. The stress components positive sign convention has been shown in Fig. 4.8.

Fig. 4.8

Fig. 4.9

Returning to the exploratory plane π considered, it is to be noted that there is an infinit number of planes (sections), which may be passed through the point C of the body. For each plane passed through point C, we find a certain stress p . Although there is an infinit number of stresses which may be associated to the point C, one could demonstrate (and this will be done later) that the stress state at point C is defined by the stresses corresponding to the three perpendicular planes at C (Fig.4.9), where px

x

i

py

yx

i

pz

zx

i

xy

y

zy

The quantity 79

j

xz

k

j

yz

k

j

z

k

(4.8)


T

x

xy

xz

yz

y

yz

zx

zy

z

(4.9)

is called the stress tensor at point C. To facilitate the visualization of the stress condition at point C, we shall consider a small cube of side a centered of C and the stresses exerted an each of the six faces of the cube (Fig.4.10).

Fig. 4.10

Fig. 4.11

The stress components shown in the figure are , and , which represent the normal stresses of faces respectively perpendicular to the x, y and z axes, and the six shearing stress components , , etc. We recall that, according to the definition y

x

xy

z

xz

of the shearing stress components, represents the y component of the shearing stress exerted on the face perpendicular to the x axis, while 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.4.10 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 C, the error involved is small and vanishes as side a of the cube approaches zero. Important relations between the shearing stress components will now be derived. Let us consider the free-body diagram of the small cube centered at point C (Fig.4.11). The normal and shearing forces acting on the various faces of the cube are obtained by multiplying the corresponding stress components by the area ΔA of each face. Selecting coordinate axes centered at C, we may write the six equilibrium equations: xy

yx

Fx M

0 x

;

Fy

0;

M

80

0 y

; 0

0;

Fz

;

M

z

(4.10) 0

.

(4.11)


Since forces equal and opposite to the forces actually shown in Fig. 4.11 are acting on the hidden faces of the cube, it is clear that equations 4.10 are satisfied. Turning now to the equations (4.11), we first consider the last of these equations, M 0 . Using a projection on the xy plane (Fig.4.12), 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 A a , counterclockwise moment the other of clockwise moment A a . We write therefore z

yx

xy

M

z

0

(

A) a

xy

(

A) a

yx

0

from which we conclude that xy

yx

.

(4.12)

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 Fig. 4.12 shearing stress exerted on a face perpendicular to the y axis. From the remaining two equations (4.11) , we derive in a similar manner the relations yz

zy

;

zx

xz

.

(4.13)

Fig. 4.13

We conclude from equations (4.12) and (4.13) that only six stress components are required to define the condition of stress at a given point C, instead of nine as originally assumed in (4.9). These stress components are , , , , , . 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 (Fig.4.13). The two shearing stresses have the same orientation with respect to the common edge. This property is called the shearing stresses duality law. x

81

y

z

xy

xz

yz


4.4 RELATIONSHIPS AMONG INTERNAL FORCES AND STRESSES WITHIN A BEAM (MEMBER) CROSS SECTION The method of sections tells us that, in the most general state, the global internal force at the level of a given cross section of a beam or member, may be expressed through six components (Fig.4.14): N , T , T , M M , M and M . y

z

x

Fig. 4.14

iy

t

iz

Fig. 4.15

These components do also represent a global effect corresponding to the entire cross section. In the same time these global internal forces determine the development of normal and shearing stresses at the level of the cross - sectional points, (Fig. 4.15). For example, at the level of a certain element of area dA, of coordinates z and y, the normal stress and the shearing stresses and may xy

x

xz

M , M and M represent in fact the occur. The internal forces N , T , T , M summation of the elementary effects given by normal and shearing stresses at the level of the cross - sectional elements of area dA. We may write therefore the following relationships: y

z

x

iy

t

N

x

dA;

xy

dA;

xz

dA;

iz

A

Ty A

Tz A

M

x

(4.14) M

t

xy A

M

iy

x

z dA;

A

M

iz

x A

82

y d A.

z

xz

y dA;

5. STRENGTH OF MATERIALS BASIC ASSUMPTIONS

To evaluate the stress, strains and displacements in a strength of materials problem, we must derive a series of basic equations. During the process of derivation, however, if we consider all the influential factors in an all-round way, the results obtained will be so complicated that practically no solutions can be found. Therefore, we have to make some basic assumptions about the properties of the body considered. Under such assumptions, we can neglect some of the influential factors of minor importance temporarily, thus simplifying the strength of materials calculus. In this text we shall comply with the following assumptions in classical strength of materials: 1. The body is continuous, i.e., the whole volume of the body is filled with continuous matter, without any void. Only under this assumption, can the physical quantities in the body, such as stresses, strains and displacements, be continuously distributed and thereby expressed by continuous functions of coordinates in space. In reality, all engineering materials are composed of elementary particles and do not accord with the assumption of continuity. However, it may be conceived that this assumption will lead to no significant errors so long as the dimensions of the body are very large in comparison with those of the particles and with the distances between neighboring particles. 2. The body is homogeneous: the properties are the same throughout the body. Under this assumption we may analyse an elementary volume isolated from the body and then apply the results of analysis to the entire body. 3. The body is isotropic so that the material properties are the same in all directions. Thus, the strength study will be independent of the orientation of coordinate axes. Most engineering materials do not satisfy the above last two assumptions completely. Structural steel, for instance, when studied with a microscope, is seen to consist of crystals of various kinds and various orientations. It seems that the material is far from being homogeneous and isotropic. However, since the dimensions of any single crystal are very small in comparison with those of the entire body, and since the crystals are orientated at random, the behavior of a piece of steel, on average, appears to justify the assumptions of homogeneity and isotropy. This is the reason why the strength of materials calculus based on these assumptions can be applied to steel structures with very great accuracy so long as none of the members has been


subjected to the process of rolling which may produce a definite orientation of the crystals. In contrast with steel, wood is definitely not isotropic, since the properties of wood in the direction of the grain differ greatly from those in the perpendicular directions. In assuming isotropic material, we shall of course exclude the treatment of wooden structures. 4. The displacements and strains are small, i.e., the displacement components of all points of the body during deformation are very small in comparison with its original dimensions and the strain components and the rotations of all line elements are much smaller than unity. Thus when we formulate the equilibrium equations relevant to the deformed state, we may use the lengths and angles of the body before deformation. In addition, when we formulate the geometrical equations involving strains and displacements, we may neglect the squares and products of small quantities. For example, when writing the Fig. 5.1 moment given by the force P at point A (Fig. 5.1) we choose the arm  instead of ’, the difference between the two lengths being a very small quantity. This is why we may write all the equilibrium equations on the undeformed state of the body.

a.

b. Fig. 5.2

Following the same idea, if the two members of Fig. 5.2a are subjected to a force P applied at B, within the context of the above mentioned assumption the angle made by each member with the vertical direction remains approximately equal to  (Fig. 5.2b). 84

Strength of Materials Basic Assumptions

A direct consequence of the small displacements and strains assumption is the method (principle) of superposition. 5. The body is perfectly elastic, i.e., it wholly obeys Hooke’s law of elasticity, which shows the linear relations between the stress components and the strain components (Fig. 5.3). Under this assumption, the elastic constants will be independent of the magnitudes of these components. The justifications for this assumption lies in the physical behavior of nearly all materials in engineering construction. In the other words, if the strains caused in a certain body by the application of a given load disappear when the load is removed, the material is said to behave elastically.

Fig. 5.3

The linear relations between stress and strain in their simplest way, may be expressed by Hooke’s law as follows :

 =E

and

 = G  ,

(5.1)

where E is called the modulus of elasticity of the material involved (or Young’s modulus, after the English scientist Thomas Young (1773-1829) and G is called the modulus of rigidity or shear modulus of the material. For structural steel we have: E = 2,1105 MPa ; G = 8 104 MPa. Fig. 5.3 does also tell us that the involved material behaves linearly. 6. Saint-Venant’s principle: except in the immediate vicinity of the points of application of the loads, the stress distribution in a body may be assumed independent of the actual mode of application of the loads. This statement practically applies to any type of load. While Saint-Venant’s principle makes it possible to replace a given loading by a simpler one for the purpose of computing the stresses in a structural member, we should keep in mind two important points when applying this principle: a. The actual loading and the loading used to compute the stresses must be statically equivalent; 85


b. Stresses cannot be computed in this manner in the immediate vicinity of the points of application of the loads. Advanced theoretical or experimental methods must be used to determine the distribution of stresses in these areas. Applying the above principle at a section K of a beam, for example, (where K is not placed in the immediate vicinity of the points of application of the loads), the effect occured at K is the same for both types of loading (Fig.5.4a,b).

a.

b. Fig. 5.4

7. Bernoulli’s hypothesis For structural members in tension or compression, the plane and parallel cross sections before the deformation of the members, remain plane and parallel after deformation (Fig.5.5). For beams (structural elements in bending), the plane sections, perpendicular to the beam axis before the deformation of the beam, remain plane and perpendicular to the beam axis after deformation (Fig.5.6). Fig. 5.5

Fig. 5.6

The above presented assumptions let us simplify the strength of materials calculus without loosing the accuracy in a fundamental manner. 86

6. AXIAL LOADING

One of the basic problems facing the engineer is to select the proper material and proportion it to enable a structure or machine to perform its function efficiently. For this purpose, it is essential to determine the strength, stiffness and other properties of materials. These properties are then compared with those computed using the strength of materials models. From this comparison the desired unknown elements may be found.

6.1 STRESS AND STRAIN A bar is said to be under axial loading (tension - Fig. 6.1a or compression Fig. 6.1b) if, at any cross section, the internal forces are represented by a single component: the axial force N.

a.

b. Fig. 6.1

To determine the stress and strain in a bar under axial loading, the following basic assumptions will be considered: The material is continuous, homogeneous, isotropic and linear-elastic; The displacements and strains are small; Bernoulli’s hypothesis is available; Hooke’s law is expressed through: σ x E ε x . Static considerations Suppose that a cutting plane , perpendicular to the bar axis (Ox), isolates the right segment of the bar (Fig. 6.2a). Then, as shown in Fig. 6.2a, the resisting internal force over the cut section must balance the applied load P. From the condition of equilibrium we have: Fx

0

N

P

0

N

P

.


Fig. 6.2

a.

The internal axial force N represents a global effect occurred at the level of the entire cut cross section considered. On the other hand, at the level of the cross sectional points, normal stresses develop (Fig. 6.2b). Thus, considering an element of area dA around an arbitrary cross sectional point Q, the elementary axial force developed b. at this level is (Fig. 6.3). (6.1) dN dA . This means that the relation between the global axial force N and the normal stress over the cross section considered may be written as follows: dA

N

,

(6.2)

A

where A is the bar cross-sectional area. Geometrical considerations Since Bernoulli’s hypothesis is available, we may note that all points of a certain cross section displace with the same quantity u, the section remaining plane after deformation (Fig. 6.4).

Fig. 6.3

We may write therefore  x

constant



.

(6.3)

From Hooke’s law we have x

E

x

constant

.

Since the stress distribution over the cross section considered is constant, written outside the integral (6.2), obtaining 88

(6.4) may be

Axial loading

a.

b. Fig. 6.4 dA

N

dA

A

A

(6.5)

A

and therefore N

P

A

A

.

(6.6)

We should also note that, in formula (6.6) is obtained by dividing the magnitude P of the resultant of the internal forces, distributed over the cross-section, by area A; it represents, therefore, the average value of the stress over the cross section, rather than the stress at a specific point of the cross section. To define the stress at a given point Q of a current cross section, we should consider a small area dA (Fig. 6.3) around Q. Dividing the magnitude of dN by dA, we obtain the average value of the stress over dA. In general the value obtained for the stress at a given point Q of the cross section is different from the value of the average stress, given by formula (6.6), and σ is found to vary across the section. In a slender rod subjected to equal and opposite concentrated forces P (Fig. 6.5a) this variation is small in a section away from the points of application of the a. b. c. d. concentrated forces (Fig. 6.5c), but Fig. 6.5. it is quite noticeable in the neighborhood of these points (Fig. 6.5b,d). In practice, we shall assume that the distribution of normal stresses in an axially loaded member is uniform, except in the immediate vicinity of the points of application of the loads. The value of the stress is then equal to σ average and may be obtained from formula (6.6). However, we should realize that, when we assume a uniform distribution of stresses over a given cross section, the concentrated forces P must be applied at the centroid of the section (Fig. 6.1). In other words, an uniform 89


distribution of stresses is possible only if the line of action of the applied concentrated loads P passes through the centroid G of the section considered. This type of loading is called centric loading (Fig. 6.6).

a. Fig. 6.6

b. Fig. 6.7

However, if a two-force member is axially loaded but eccentrically, as shown in Fig. 6.7a, we find from the conditions of equilibrium of the portion of member shown in Fig 6.7b, that the internal forces at a given section must be equivalent to a force P applied at the centroid of the section and a couple M = aP. In such a case the distribution of stresses cannot be uniform. This point will be discussed in detail later, within another chapter. Let us now return to formula (6.6). To determine whether the bar of Fig. 6.1 may support the given load P, we must compare the value obtained for σ under this loading with the maximum value of the stress which may be safely applied to steel (in case the bar is made of steel). From a table of properties of materials we find the maximum allowable stress in the type of steel to be used ( σ a  ). Let us suppose we have to design a simple member (like that represented in Fig. 6.1, for example) so that it would not fail under specified loading conditions. In such a case, on one hand, we must compute the actual stresses σ in the involved member using formula (6.6): σ

N

P

A

A

.

On the other hand we have to find out the allowable value of the normal stresses σ a  for the involved material (using different engineering tables containing materials properties). In the end, the strength condition of the member considered is P a 

A

.

(6.7)

The condition (6.7) may be used for all three specific types of strength of materials problems. 90

Axial loading

Another important aspect of the analysis and design of structures relates to the deformations caused by the loads applied to the structures. Clearly, it is important to avoid deformations so large that they may prevent the structure for fulfilling the purpose for which it was intended. Let us now consider a bar BC, of length  and uniform cross-sectional area A, which is suspended from B (Fig. 6.8a). If we apply a load P to end C, the bar elongates (Fig. 6.8b), with  . Since  x



,

from Hooke's law, we have: E

x

Substituting for write:

E

 

.

(6.8)

from (6.6) into (6.8) we may

N

E

 

A

a.

x

,

b. Fig. 6.8

from which it follows that



N

P

EA

EA

,

(6.9)

where E is the modulus of elasticity of the material involved and N is the axial internal force which is constant along the member considered. Equation (6.9) may be used only if the bar is homogeneous (constant E), has a uniform cross section of area A and is loaded at its ends. If the bar is loaded at other points, or if it consists of several portions of various cross sections and possibly of different materials, we must divide it into component parts which satisfy individually the required conditions for the application of formula (6.9). Denoting respectively by Ni, i, Ai and Ei the internal force, the length, the cross sectional area and the modulus of elasticity corresponding to the part i, we may express the deformation of the entire bar as: N i i

 i

.

(6.10)

E i Ai

For example, let us determine the total deformation under the given loading conditions.

91



of the rods shown in Fig. 6.9,


a.

b. Fig. 6.9

a)



1

B

1

2

2

3

3

P B

2a

3 Pa

2 Pa

7 Pa

EA

EA

EA

EA

Thus, in the case a of Fig 6.9, the total deformation

 of

the bar shown is



. 7 Pa

.

EA

b)



1

P

B

2a EA

1

2

2

3

3

2 Pa

2 Pa

5 Pa

EA

2 EA

2 EA

4

4

7 ,5

B

Pa

.

EA

In case of a rod of variable cross section and/or variable axial internal force (when N=N(x) and A=A(x), x being the parameter along the rod considered), formula (6.9) must be applied only to an element of the bar with an infinitely small length dx (Fig. 6.10). Since such an element has an infinitely small length dx, the variation of the cross sectional area A and/or that of the axial internal force N may be neglected along the element considered, formula (6.9) becoming available. Thus, in the case of the bar shown in Fig. 6.10, we may write the deformation of the element of length dx as: N x

dx

E

dx

(6.11)

A x

which is in fact the formula (6.9). The total deformation of the bar shown in Fig. 6.10 is therefore obtained by integrating (6.11) over the length  of the bar: 



Fig. 6.10

0

92

N (x) EA ( x )

dx .

(6.12)

Axial loading

Example 6.1 For the rod shown in Fig. 6.11, draw the axial force diagram, determine the required diameter d and the total displacement of point 1, knowing that the allowable normal stress for the involved material is a= 150 MPa and the modulus of elasticity is E = 2,1 105 MPa. Considering the free body of rod 1,2,3,B, we note that the reaction at B is : YB = 250 kN. It follows that: N1-2 = 125 kN = constant; N2-3 = 125 kN = constant; N3-B = 125 + 125 = = 250 kN = constant. The axial force diagram has been represented in Fig. 6.11. To determine the required diameter d so that the rod would not fail under the specified loading, we must use the condition (6.7): max a. The maximum value of normal stresses along the rod considered may be established only after the computation of normal stress for each particular portion of the rod. We write therefore: Fig. 6.11

1 2

N1

2

A1

2

125

10 d

3

4 125

2

10

3

;

2

d

4

N 2 3

125

2 3

A2

10

3

(1 , 5 d )

3

4 125 2

10

(1 , 5 d )

3 2

;

4 3 B

N3

B

A3

B

250

10

(1 , 5 d )

3

4 250

2

10

(1 , 5 d )

2

3

.

4

We note that the maximum value of the normal stresses corresponds to the first portion of the rod: 4 125 max

1

2

d

10

3 a 

2

It follows that: d

4 125

10

3

32 , 57

150

We write: d = 32,57 mm

93

32,6 mm .

mm .

150 .


The total vertical displacement of point 1 (i.e. the deformation  of the rod) is obtained from (6.10) after the rod has been divided into four portions, as shown in Fig. 6.11. We write: N i

 i

125

10

2 ,1 10

i

1

E i Ai 3



2

2000

125

32 , 6

5



2 3

2

2 ,1 10

10

5

3

3 B

1000

250

(1 , 5 32 , 6 )

2

2 ,1 10

5

1500

(1 , 5 32 , 6 )

4

4

3

10

2 , 69 mm . 2

4

Example 6.2 The rigid bar BCD is supported by two vertical rods 1 and 2 (Fig. 6.12). The two rods are made of steel with E = 2,1 105 MPa and a=150 MPa. For the 76,5 kN force shown, determine the required diameters of the two rods (d1 and d2) and the vertical displacement of point C.

Fig. 6.12

Fig. 6.13

The axial forces in the two vertical rods may be determined by cutting the two rods and using the equilibrium equations for the free body obtained (Fig. 6.13). We write: M

0

(D )

Fy

0

N 1 2 , 25 N1

N

2

76 , 5 1 , 25

76 , 5 k N

0

N

2

N1

42 , 5 k N ;

34 k N .

It is important to be mentioned that the determination of the axial forces represents a first and necessary step within axial loading problems. We do also note that, in our case, the two axial forces in rods 1 and 2 are positive, which means that the two rods are in tension. The strength conditions for the two rods are: N1 1

42 , 5 10

3 a 

2

A1

d1

150

42 , 5 10

d1

3

4

18 , 99 mm ;

150

4

N 2

2

A2

34

10

d

2 2

3 a 

150

4

94

d2

34

10

3

150

4

16 , 98 mm .

Axial loading

We adopt d1 = 19 mm; d2 = 17 mm. The deformations of the two rods are (Fig. 6.14.):

Fig. 6.14 N 1 1

1

42 , 5 10

EA 1

2 ,1 10

3

1000 19

5

0 , 713 mm ;

2

4 

N 2 2

EA

2

2

34 10

3

2 ,1 10

1 , 25 1000 5

17

2

0 , 891 mm .

4

From similar triangles, we may finally find that the vertical displacement of point C is CC"= 0,792 mm. Example 6.3 Two rods of uniform circular cross-sections BC and BD support a 30 kN vertical force (Fig. 6.15). Knowing that the allowable stress of the rods material is a = 150 MPa and E = 2,1 105 MPa, determine the required values of the diameters d1 and d2 and the vertical displacement of point B. From the geometry of the structure shown in Fig. 6.15 one can find that the lengths of the two rods are respectively:

Fig. 6.15

1

0 , 956 mm ; 

2

0 , 743 mm . 

Passing a cross-section through each of the two rods, the two corresponding internal axial forces N1 and N2 are revealed. From the mechanical equilibrium of point B we have (Fig. 6.16):

95

Strength of Materials X

0

B

YB

0

N

1

cos 30

N

1

cos 60





N

2

cos 40

N

2

cos 50





0;

30 .

We find: N1 = 24,46 kN; N2 = 27,64 kN . Applying the strength condition for the two rods, we Fig. 6.16

write: N1 1

24 , 46 10

3

A1

150

a 

2 d1

d1

24 , 46 10

3

4

14 , 41 mm ;

150

4

N 2

27 , 64 10

2

A2

d

3

150

a 

2 2

d2

27 , 64 10 150

4

The deformations of the two rods are (Fig. 6.17):

1

N 1 1

24 , 46 10

EA 1

2 ,1 10

3

0 , 956

10

14 , 41

5

3

0 , 682

2

mm ;

4 

N 2 2

EA

2

2

27 , 64 10 2 ,1 10

5

3

0 , 743 15 , 31

10 2

3

0 , 531 mm .

4

It is to be observed that, although the concentrated load P acts vertically, due to the lack of symmetry of the system considered, the point B displaces along direction BB’ which is not vertical. In the same time, the deformation of the two rods, 1 and 2, are revealed by sketching a perpendicular line from the initial position of point B to the two rods after deformations (although the method is approximate, it models with a sufficient accuracy the deformations of such bars when the hypothesis of small deformations is available). Using the same hypothesis of small deformations, we may also consider that, after deformation, the two rods 1 and 2 form with the vertical direction the same angles of 60 and 50 respectively. Assuming that the angle between BB’ and the vertical direction is , we may write (Fig. 6.17):

Fig. 6.17

96

3

4

15 , 31 mm .

Axial loading 1

BB

cos 60

cos 50

 cos 60





; BB '



2

1





cos 60



1

cos



sin 60

sin

cos 50



cos

cos 60

 1 cos 50

sin 60 tg

cos



 1 sin 50 tg

 1 sin 50

tg



sin 50



sin

2 



cos 50

2

cos

1 

2



cos 50

 2 cos 60

 2 sin 60







sin 50 tg





 2 sin 60 tg

 2 cos 60



 1 cos 50



.

We have: tg

 1 cos 50



 1 sin 50



 

2 2



cos 60



sin 60

0 , 682 cos 50 0 , 682 sin 50



0 , 531 cos 60



0 , 531 sin 60

 

10 .



The vertical displacement of point B is therefore (Fig. 6.18): v

B

BB

1

B B cos cos

0 , 682

Fig. 6.18

cos ( 60



cos 10



60 

cos



1, 045 mm .

10 )

Example 6.4 The rigid bar BOC is supported by the rod CD, having a circular uniform cross section of diameter d. Rod CD is made of steel with a = 150 MPa and E = 2,1 105 MPa. For the 15,7 kN force shown determine the required value of diameter d and the vertical displacement of point B. Free Body: the bar BOC M

(O )

0

N 1 0 ,4

P 0 ,3

0

N 1 0 ,4

15 , 7 0 , 3

0

N1

11 , 775

10

11 , 775 kN ( tension );

Strength condition: N1 1

A1

11 , 775 d

10 2

3 a 

150 MPa

d

3

4

10

mm .

150

4

Vertical displacement of point B, v B (Fig. 6.20). We denote by C’ and B´ the displaced positions of points C and B respectively. Since the bar BOC is rigid, the segments OC and OB rotate with the same angle. From similar triangles we may write: CC BB

v

1

0 ,4

B

0 ,3

97

.


Fig. 6.19

Fig. 6.20

Thus: v

B

1

0 ,3

N 1 1

0 ,3

0 ,4

EA 1

0 ,4

11 , 775

10

3

10

5

2 ,1 10

0 , 7 10 2

3

0 , 4997

mm .

4

6.2 POISSON'S RATIO When a homogeneous slender bar is axially loaded, the resulting stress and strain satisfy the Hooke’s law as long as the elastic limit of the material is not exceeded.

a.

b. Fig. 6.21

Assuming that the load P is directed along the x axis (Fig. 6.21a), we have where A is the cross-sectional area of the bar, and from Hooke’s law: X

x

,

x

= P A,

(6.13)

E

where E is the modulus of elasticity of the material. We do also note that the normal stresses on faces respectively perpendicular to the y and z axes are zero: y = z = 0 (Fig. 6.21b). It would be tempting to conclude that the corresponding strains y and z are also zero. This, however, is not the case. In all engineering materials, the elongation produced by an axial tensile force P in the direction of the force is accompanied by a contraction in any transverse direction (Fig. 6.22). 98

Axial loading

Since the material under consideration is assumed to be homogeneous and isotropic, the strain must have the same value for any transverse direction: . This value is referred to as the lateral strain. The absolute value of the ratio of the lateral strain over the axial strain is called Poisson’s ratio, denoted by the Greek letter . We write: y

Fig. 6.22 y

z

x

x

z

(6.14)

.

For the loading condition represented in Fig. 6.21, an increase of dimension along the x axis, means a decrease of transverse dimensions, so that we have: . y

x

and

z

x

(6.15)

.

It follows that the relations which fully describe the condition of strain under an axial load applied along a direction parallel to the x axis are: x x

x

;

y

E

z

(6.16)

.

E

The above presented phenomenon is also referred to as the transverse contraction. During such a contraction the original dimensions of the bar under consideration change as follows (Fig. 6.22): h

h

h

h

h

b

b

b

b

b

0

0

0

h1

y

b 1

z

0

0

h 1

y

b 1

z

0 1

x

x

;

x

x

;

.

The new volume of the bar after deformation is: V

h 1

x

b 1

0 1

x

x

,

the original volume (before deformation) being: b h 0.

V0

The variation of the volume due to the axial load applied is therefore: V

V

V0

2

b h  0 (1

b h 0 1

2

b h 0 1

x

99

2

2 x

x

2

x

2

x

) (1

x

1

x

1

2 x

2

2 x

b h 0

)

2

3 x

1

(6.17)


b h l0

x [1

2

2

2 x  0

2

2

x

x



0

].

0

Since the strains , , are much smaller than unity, the last three terms in the above relation may be omitted. We have, therefore: x

y

z

V

V

V0

b h 0

x

1

(6.18)

.

2

On the other hand, it has been experimentally observed that a bar in tension does always increase its volume. This means that: V

0

b h 0

x

1

2

0

1

2

0

.

Since the very definition of Poisson’s ratio requires it to be a positive quantity, we may conclude that, for any engineering material: 0

0 ,5 .

(6.19)

In numerous cases (for numerous materials)

= 0,3.

6.3 STRESS CONCENTRATIONS The exact computation methods of the theory of elasticity and the experimental photoelastic methods show that, for members in tension or compression having a variable cross section, the corresponding normal stresses are not uniformly distributed over the cross sections. Therefore, Bernoulli’s hypothesis does loose its validity. When a structural member contains a discontinuity, such as a hole or a sudden change of the cross section, high localized stresses may also occur near the discontinuity. Fig. 6.23 and 6.24 show the distribution of stresses at the level of the critical sections corresponding to two such cases.

Fig. 6.23

Fig. 6.24

100

Axial loading

The ratio between the maximum stress ( max) and the average stress ( ave) computed at the level of the critical (narrowest) section of the involved discontinuity is called the stress - concentration factor: max k

.

(6.20)

ave

It is to be noted that the average stress ( ave) is computed with the basic formula (6.6) while the maximum stress ( max) represents the real maximum stress occured due to the stress-concentration phenomenon. To determine the maximum stress occurring near a discontinuity in a given member in tension or compression, the designer needs only to compute the average stress ave = P A in the critical section and multiply the result obtained by the appropriate value of the stress concentration factor k. On the other hand, the stress-concentration factor is given in form of tables or graphs, as shown in Fig. 6.25 for the two cases represented in Fig. 6.23 and Fig. 6.24.

Fig. 6.25

It is also to be noted that the above presented procedure remains valid only if max does not exceed the proportional limit of the involved material (the concept of proportional limit of a material will be explained later).

101


6.4 OWN WEIGHT EFFECT. MEMBERS OF CONSTANT STRENGTH As we saw in the preceding sections, the strength calculus for different types of members in tension or compression does neglect the own weight of the members. This procedure remains valid only if the length of the involved members has low values. Otherwise, if the length of the members has important values, the own weight cannot be neglected. Consider for example a homogeneous rod BC of length  and uniform cross-sectional area A, which is suspended from B, hanging under its own weight, (Fig. 6.26). Let us also assume that the rod material specific weight is . is defined as the ratio between the weight G and the corresponding volume of the material, V. We may Fig. 6.26 write therefore: G

or

G

(6.21)

V .

V

At an arbitrary cross section of the involved bar, located at distance x from end C we write: N x

x

P

G x

N x

P

P

Ax

A

x

0; NC

P;

x

;

P

x

0;

A x;

P

A

N

B

A.

P C

1

x;

A

;

x

;

A P

B

2

,

A

where G(x) represents the weight of the bar portion below the cross section considered. The diagrams of the axial forces and the normal stresses have been shown in Fig. 6.26. Note that the maximum values of axial force and normal stress occur within the upper cross section of the rod (for x = ). The strength condition is: P max

B

A



a 

.

We may write therefore that, the required cross-sectional area of the rod is:

102

(6.22)

Axial loading

P

A a 

where of.

all

(6.23)

,



is the allowable value of the normal stress for the material the rod is made

In the absence of force P (P = 0) we have: (6.24)

.

max

This means that, in such a case, if the length of the rod reaches a critical value c, the rod breaks under its own weight. For  = c, the maximum stress in the rod equals the breaking strength of the material ( B) the rod is made of. We write: max

c

B

B

c

(6.25)

.

The quantity B will be defined later. The total deformation  of the rod shown in Fig. 6.26 is obtained by integration over the length  of the rod: 

N x

 0

dx

EA x

1



1

N x dx

EA

Ax

Px

EA

1

0



2

2

EA

0

1



P

A x dx

0

A

P

EA

2

.

2

Since: A

G

V

G ,

V

we may write: 

1

A

P

E A

2

2

1 E A

P

G 



2

E A

P

G

.

2

Thus, it follows that under its own weight the total deformation of the rod subjected to the concentrated force P is: P 

G  2 , E A

(6.26)

where G is the total weight of the involved rod. We note that the rod shown in Fig. 6.26 has a constant cross-section. It follows from (6.22) that the normal stress is maximum at section B. 103


Fig. 6.28

Fig. 6.27

This means that only at B, the rod can be dimensioned in a proper way, all other cross-sections having bigger dimensions than necessary. This is the reason why the engineers have tried to design a bar so that each cross-section of the bar to be equally loaded (i.e. at every cross-section of the bar the normal stress has the same value). Thus the concept of MEMBERS OF CONSTANT STRENGTH has arisen. Obviously, such a member of constant strength must have a continue variable cross section, increasing from the lower end to the upper end (Fig. 6.27). At distance x from end 1, the cross-sectional area is A(x). The problem is to determine the mathematical expression of the function A(x), finding in this way the cross-sectional area variation mode along the involved member. Let us now consider again the member shown in Fig. 6.27. We denote by G(x) the weight of the member portion of length x. We do also consider two cross-sections (mn and m'n') at an infinite small distance dx from each other (Fig. 6.28). The cross-sectional area increases from mn to m'n' with an infinite small quantity dA(x). At the level of the cross-section mn we may write: x

N x

P

A x

G x

.

(6.27)

A x

Recalling that a member of constant strength must have the same stress at each particular cross-section (equal to a) we write: x

P

G x a 

A x

,

and thus A x

a 

P

G x

.

(6.28)

Applying formula (6.28) for the cross-section m'n' (for which the area is increased with dA(x) and the weight of the lower member portion with dG(x)) we have: 104

Axial loading

dA x

A x

P

a 

dG x

G x

or

(6.29) A x

dA x

P

a 

A x dx

G x

,

Substracting (6.28) from (6.29) we write: A x

dA x

d A x

a 

A x

P

a 

A x dx

G x

P

G x

,

or A x dx

.

This means that d A x

,

dx

A x

(6.30)

a 

Integrating member by member we have: ln A x

x

,

C

a 

or x C

A x

e

(6.31)

,

a 

where C is an integration constant. We have obtained therefore an exponential variation of the cross-sectional area along the involved member. The constant C may be obtained from (6.27) and (6.31) making x = 0 and = a. We write: N 0

P

P

A 0

A 0

e

C

ln

a 

C

.

Thus e

C

P

P

a 

.

(6.32)

a 

It follows that A x

e

P

ln

x a 

x

a 

e

a 

ln

e

P a 

P

x

e

a 

a 

and thus A x

P

x

e

a 

a 

105

.

(6.33)


Using (6.28), the corresponding weight of the member portion below the cross section mn is x

G x

A x

P

a 

a 

P e

(6.34)

1 .

The total deformation of the member is obtained by integration over the length  of the member: 

N x



1

dx

EA x

0

E



N x

dx

A x

0

1 E

 a 

dx

a 



(6.35)

.

E

0

Although the relation (6.33) insures the condition of equal strength, such a member is enough complicated from the technological point of view. A much more practical solution is shown in Fig. 6.28. We may observe that the member shown in Fig. 6.29 has three distinct portions of cross-sectional areas A1, A2, A3 and lengths 1,  2,  3 respectively. For each portion the maximum stress is to be reached at the upper end. For the first portion we may write: A1  1

P m ax

P

A1

1

 1.

A1

Since this stress has to be equal to the allowable stress ( member is made of, we have P m ax

1

A1

1

a 

a)

of the material the

.

The necessary value of the cross-sectional area A1, so that the first portion of the member would not fail, is P

A1

(6.36)

. 1

a 

In the same manner, for the second portion we may write A1  1

P m ax

A2

2

1 m ax

A2

A2  2

a 

2

A1  1

P a 

A1  1

P

A2

a 

2

a 

106

P

A1  1

A2

A2

2

A1 2

a 

P a 

 2.

1

a  a 

. 2

Axial loading

For the third portion we have

Fig. 6.29

P

A3

1

a 

a 

2 a 

2

. 3

a 

In general, for a member with n distinct portions, the required cross-sectional area Ai of the ith portion is therefore: P

Ai

1

a 

a 

i -1 a 

2 

. a 

i

(6.37)

In particular, if all portions of the member have the same length  we may write: Ai

P a 

i -1 a 



i

.

(6.38)

6.5 STRESS-STRAIN DIAGRAM 6.5.1 GENERALITIES Although apparently simple, the tensile tests of the material provide the strength of materials with necessary and useful information. It is to be noted that, the strength of a material is not the only criterion that must be considered in engineering design. The stiffness of a material is frequently of equal importance. On the other hand, the mechanical properties such as hardness, toughness and ductility are generally those which determine the selection of a certain material within a certain engineering context. These properties are determined by making tests on the materials and comparing the results with established standards. Although a complete description of these tests is the province of materials testing and hence will not be 107


given here, one of the tests (the tension test of steel) and its results will be considered because it helps to develop several important basic concepts. The tensile tests stress-strain diagram represents the graphical visualization of the relation between stress and strain in a given material, being a very important characteristic of the material. To obtain the stress-strain diagram of a material, one usually conducts a tensile test on a specimen made of that material. One type of specimen commonly used is shown in Fig. 6.30. In a tensile test, the specimen is gripped between the jaws of a testing machine (Fig. 6.31). Before the applying of the tensile load P, two marks are inscribed on the specimen (k and k’ - Fig. 6.30), at a distance 0 from each other. After the specimen is placed in the testing machine, as the load P increases, the distance between the two marks does Fig. 6.30 also increase. At a given moment “i”, the axial force at any cross-section of the cylindrical portion of the specimen is Ni (equal to P), while the distance between k and k’ is i. This means that, at any moment, we have two values: - the axial force in the involved specimen Ni = P; - the instant distance between the two marks k and k’. The distance i is measured with a dial gage and the elongation i = i -0 is recorded for each value of P. We then convert each pair of readings (Ni and i) at a certain moment “ i” into the instant normal stress and strain of the member as

Fig. 6.31

follows: N i

A

i

and

 i

108



0

i





i



0

0

,

(6.39)

Axial loading 2

where A is the original cross-sectional area of the specimen

A

d0

.

4

Plotting these data on a graph with the ordinate representing the stress i and the abscissa representing the strain i of the specimen at different moments throughout the test, we obtain the diagram called the stress-strain diagram. Fig. 6.32 represents such a graph for structural steel. The diagram obtained by plotting the stress versus the strain, reflects the behavior of the involved material. It is to be mentioned that the following conditions have to be fulfilled within a tensile test: - the specimen must have a constant cross section; - the material must be homogeneous; - the load must be axial, Fig. 6.32 i.e. produces uniform stress. Finally, note that since strain represents a change in length divided by the original length, strain is a dimensionless quantity. However, it is common to use units of meters per meter (m / m) or millimeters per millimeter (mm/mm) when referring to strain. In engineering work, strains of the order of 1 10-3 are frequently encountered. The main points and concepts developed from the stress-strain curve (Fig. 6.32) are: The proportional limit (P) represents the maximum value of the stress p up to which the stress is directly proportional to the strain . From the origin O up to the proportional limit, the stress-strain diagram is a straight line. We may write: =E

(6.40)

,

a relation known as Hooke’s law, after the English mathematician Robert Hooke. The quantity E is called the modulus of elasticity of the material involved, or Young’s modulus. The elastic limit (E) represents the stress e up to which the material will return

Fig. 6.33 109


to its original shape when unloaded; The yield limit (Y) is the stress y at which there is an appreciable elongation or yielding of the material without any corresponding increase of the load. However, the phenomenon of yielding is peculiar to structural steel; other grades of steel and steel alloys or other materials do not possess it, as is indicated by the typical stressstrain curves of these materials shown in Fig. 6.33. The yield strength is closely associated with the yield point. The ultimate stress, or ultimate strength ( u) as it is more commonly called, is the highest ordinate on the stress-strain curve. The rupture strength ( r) is the stress at failure. For structural steel it is somewhat lower that the ultimate strength because the rupture strength is computed by dividing the rupture load by the original cross-sectional area, which, although convenient is incorrect. The error is caused by a phenomenon known as necking. As failure occurs, the material stretches very rapidly and simultaneously narrows down, as shown in Fig. 6.34, so that the rupture load is actually distributed over a smaller area.

Fig. 6.34

If the rupture area is measured after failure occurs, and divided into the rupture load, the result is a truer value of the actual failure stress. Although this is considerably higher than the ultimate strength, the ultimate strength is commonly taken as the maximum stress of the material.

6.5.2 SAFETY COEFFICIENTS. ALLOWABLE STRENGTHS The working stress, also called the allowable stress, is the maximum safe stress a material may carry. In design the allowable stress all should be limited to values not exceeding the proportional limit so as not to invalidate the stress-strain relation of Hooke’s law on which all subsequent theory is based. However, since the proportional limit is difficult to determine accurately, it is customary to base the allowable stress on either the yield point or the ultimate strength, divided by a suitable number c, called the factor of safety (or the safety coefficient): y a 

or

c

u a 

.

(6.41)

c

The determination of the factor of safety which should be used for various applications is one of the most important engineering tasks. On the one hand, if a factor of safety is chosen too small, the possibility of failure becomes unacceptably large; on the other hand, if a factor of safety is chosen unnecessarily large, the result 110

Axial loading

is an uneconomical or nonfunctional design. The choice of the factor of safety which is appropriate for a given design application requires engineering judgement based on many considerations, such as the following: - variations which occur in material properties; - the number of loadings which may be expected during the life of the structure or machine; - the type of loadings which are planned for in the design, or which may occur in the future; - the type of failure which may occur; - uncertainty due to the methods of analysis; - deterioration which may occur in the future due to poor maintenance or due to unpreventable natural causes; - the importance of a given member to the integrity of the whole structure. For the majority of structural and machine applications, the factors of safety are specified by design specifications or building codes written by committees or experienced engineers working with professional societies, with industries or agencies.

6.6 STATICALLY INDETERMINATE PROBLEMS In the problems considered in the preceding sections, we could always use the conditions of mechanical equilibrium to determine the internal forces produced in the various portions of a member under given loading conditions. But there are many problems, however, in which the internal forces cannot be determined from statics alone (i.e. the equations of static equilibrium are not sufficient for a solution). Such cases are called statically indeterminate and require the use of additional relations that depend on the elastic deformations Fig. 6.35 in the members. The following examples will show how to handle this type of problems. Consider, for example, two rods BC and DC, pin-connected at C and subjected to a vertical concentrated force P, applied at C, Fig. 6.35. The axial forces (N1 and N2) in the two rods may be completely computed from statics alone. We write Fx

0

N 1 sin

Fy

0

N 1 cos

111

N

sin

2

N

2

cos

0

, P

0

.


from which we obtain the values of N1 and N2. This is a typical example of statically determinate problem. But what happens if another rod EC is attached (Fig. 6.36)? In such a case a part of the internal forces N1 and N2 will be taken over by the supplementary rod EC. So that, the internal forces in the three rods will be N 1' , N 2' and N 3' . We have therefore three unknowns in the problem ( N 1' , N 2' , N 3' ). On the other hand, from statics we can use only two equations of mechanical equilibrium as in the preceding case: Fx

0

and

Fy

0

.

Clearly, the two above equations are not sufficient to determine the three unknown internal forces N 1' , N 2' and N 3' . We have therefore, a statically indeterminate problem. The difference between the number of unknown quantities (3 in this case) and the number of equations available from statics is called the indetermination degree. In our example, the Fig. 6.36 indetermination degree is 1. It is to be noted that 2, 3 or more supplementary rods may be attached to the system shown in Fig. 6.35. In consequence, the indetermination degree will be 2, 3, …etc. Let us now consider another example. A rigid bar is pin-connected at O, and suspended from a vertical rod BC, as shown in Fig. 6.37. Determine the internal axial force N1 in the rod caused by the load P applied at D. Since, from the summation of moments at O, we can determine the value of N1, the problem is statically determinate. We write M

Fig. 6.37

0

N1 a

P

0

P

N1

.

a

But a supplementary vertical rod B’C’ attached to the system shown in Fig. 6.37, transforms the problem into a statically indeterminate one, Fig. 6.38. This time, we have two unknowns and only a single equilibrium equation available from statics. We write M

Fig. 6.38

O

O

0

N

2

a' N1 a

P 

0

.

Thus the indetermination degree is 1. In the same time the presence of the supplementary attached rod determines the decrease of the original axial 112

Axial loading

force N1 shown in Fig. 6.37. If a new supporting rod is attached, the indetermination degree increases with 1 unit (becoming 2). Such statically indeterminate problems (in which the reactions and the internal forces cannot be determined from statics alone) may be resolved by using adequate additional relations involving deformations which are obtained from the geometry of the problem (as presented below). Several specific types of statically indeterminate problems may be differentiated.

6.6.1 MEMBERS WITH UNHOMOGENEOUS CROSS SECTIONS A rod (1) of length , cross-sectional area A1, and modulus of elasticity E1, has been placed inside a tube (2) of the same length, but of a cross sectional area A2 and modulus of elasticity E2, as shown in Fig 6.39. Determine the internal forces in the two members when the load P is applied through a rigid plate as shown. We denote by N1 and N2 the unknown axial forces in the two involved members. The single equation which may be written from statics is: N1

N

P

2

.

(6.42)

Clearly, one equation is not sufficient to determine the two unknown internal forces N1 and N2.. Thus, the problem is statically indeterminate. However, the geometry of the problem shows that the two members have the same axial deformation. We may write, therefore Fig. 6.39

1

2

N 1

N 2

N1

N

E 1 A1

E 2 A2

E 1 A1

E 2 A2

(6.43)

2

Equations (6.42) and (6.43) may be solved simultaneously for N1 and N2: N1

A1 E 1 P A1 E 1

A2 E 2

;

N

A2 E 2 P 2

A1 E 1

A2 E 2

.

(6.44)

The above presented example may be also extended to the general case shown in Fig 6.40, where a member is composed of other "n" members, all being subjected to a concentrated load, applied through a rigid plate. Denoting by N1, N2,...,Nn the unknown axial forces in the n members we may write N1 + N2 + N3 +...+ Nn = P

113

(6.45)


But one equation (written from statics) is not sufficient to determine the n unknown axial forces. The problem is (n-1) degree statically indeterminate. However, since the axial deformations of the n members have the same value, we may also write another (n-1) supplementary equations as follows: 1

Fig. 6.40 N 1 1

N 2 2

N 3 3

E 1 A1

E 2 A2

E 3 A3

...

N n n

2

.

E n An

3

n

(6.46)

Equations (6.45) and (6.46) may be solved for N1, N2, N3,...,Nn. Numerical example A rod (1) with a variable circular cross section has been placed inside a tube (2) of the same length . The rod is made of steel with E E 1 2 ,1 10 5 MPa while the tube is made of aluminum with E E 2 0 , 7 10 5 MPa . The two members are compressed by a concentrated load P acting through a rigid plate, as shown in Fig. 6.41. Draw the axial forces and normal stresses diagrams for the two members.

Fig. 6.41 Denoting by N1 and N2, respectively, the axial forces in the rod and in the tube, we may write from statics (6.47) N1 N 2 P . The geometry of the problem shows that the axial deformations of the two members are equal. We write therefore: 1

2.

(6.48) Since the rod has a variable circular cross section, its total axial deformation may be expressed as 114

Axial loading 

1 0

6d

N1

N1

dx

E 1 A1 ( x )

;

dx

(6.49)

E 1 A1 ( x )

0

where A1(x) represents the current cross-sectional area of the rod, at distance x, as shown in Fig. 6.42. We write A1 ( x )

BC

2

BC

;

4

2 BB '

B' C '

2 BB ' d

.

From similar triangles we may write BB ' 2d

x d

6d

2 2 BB '

x

d

6d x

BC

6 x

d

6d

6

Fig. 6.42

6 x

A1 ( x )

x

2 BB '

4

2

6d

(6 d

36

x)

.

2

144

In other words, the rod circular cross sectional area at distance x from the rod end is A1 ( x )

(6 d

x)

2

.

(6.50)

144

Now, the rod axial deformation may be expressed as follows: 6d

1 0

N1

N1

dx

E 1 A1 ( x )

144 N 1

(6 d

E1

E1

x)

( 1)

1

6d

0

6d

144 (6 d

0

x)

144

dx 2

144 N 1

E1

1

E1

6d

N1

6d

6d

0

1 (6 d

x)

144 N 1

dx 2

E1

6d

0

(6 d

x )'

(6 d

x)

1

144 N 1

1

1

144 N 1

6d

E1

6d

12 d

E 1 12 d

2

dx

12 N 1 E1d

.

In the same time, the tube axial deformation is 2

N 2 E 2 A2

N

2

6d

E 2 A2

N E2

2

9d

6d 2

N 4d

2

24 d

E 2 5d

2

24 2

5

N

2

.

E 2d

4

Since the two axial deformations have the same value, we may write 1



12 N 1 2

E1d

24 N 5

2

,

E2d

obtaining that N

2

0 ,833

N1.

(6.51)

Equations (6.47) and (6.51) may be solved simultaneously for N1 and N2. We shall finally have: N1

36 k N ;

N

30 k N .

2

The axial forces and the normal stresses diagrams have been sketched in Fig. 6.41. 115


6.6.2 STRAIGHT MEMBERS FIXED AT BOTH ENDS A horizontal member BC is fixed at its both ends before the external axial concentrated force P is applied. The member is of length , uniform cross section of area A, and modulus of elasticity E (Fig. 6.43). Compute the stresses corresponding to the portions B-1 and 1-C due to the application of load P at point 1? Due to the action of the concentrated load P, applied at Fig. 6.43 point 1, two horizontal reactions at points B and C develop (Fig. 6.44). From statics we may write X

X

B

P

C

.

(6.52)

Thus, the problem is statically indeterminate (the indetermination degree being equal to 1). However, the unknown reactions XB and XC may be determined if we observe from the geometry that the total elongation  of the member must be zero (points B and C are fixed from the beginning). Denoting by  B 1 and  1 C , respectively, the elongations of the portions B-1 and 1-C, we write 

X



B C

B

a



P

EA

1

B 1

X

B

b

C

0

EA

or X

a

B

X

P b

Pb B

a

X

Pb b



B

.

b

0

(6.53)

Carrying this value into (6.52) we write Fig. 6.44 P b

X



C

P

X

C

P 

P b

P





We may write therefore X

X

Pb B

116

;

 Pa

C



b

.

P a 

.

(6.54)

Axial loading

With this values in mind we are now in the position to draw the axial forces diagram. This diagram has been sketched in Fig. 6.44. The normal stress for each specific portion of the member is Pb N B 1

B 1

A



Pb

A

A

;

Pa N1 1 C

C

A



Pa

A

A

.

The normal stresses diagrams have been also sketched in Fig. 6.44. Numerical example A tube of length , with the inner diameter of 50 mm and the outer diameter of 80 mm, has been fixed at its both ends as shown in Fig. 6.45. Determine the reactions at B and C and draw the axial forces and normal stresses diagrams for the loading shown. Following the same procedure as presented above, we write Fx

0





X

B

X

B



C

180

180

B

10

3

B 1

80

1

80 10

3

X

2

0

C

2

1, 2 10

X

B

X

B

C

X

260

C

3

0 , 9 10

;

(6.55)

X

180

B

10

EA

3

1, 2 10

3

EA

3

0

.

(6.56)

EA

Equations (6.55) and (6.56) may be solved for XB and XC: X

B

160 k N ;

X

C

100 k N .

With these values, the axial forces diagrams have been sketched in Fig. 6.45. The corresponding normal stresses for each distinct portion of the member are: N B 1

160

B 1

A

10 2

80

3

72 , 75 MPa

;

9 , 09 MPa

;

2

60

4 N1 1 2

20 10

2

A

80

2

3

60

2

4 N 2 C

Fig. 6. 45

100

2 C

A

80 4

117

2

10

3

60

45 , 47 MPa 2

.


If the tube is made of steel, for example, with

a 

not fail under the action of the two concentrated forces (

= 160 MPa, we can conclude that it does =72,75 MPa

max

a 

160 MPa

).

6.6.3 SYSTEMS OF PARALLEL MEMBERS A horizontal rigid bar B1...Bn+1 is suspended from "n" vertical members (bars) as shown in Fig. 6.46. Determine the internal axial forces for the n bars and for the loading shown. Since from statics there are only two equations available ( F 0 M 0 ) the problem is and statically indeterminate (the indetermination degree being "n-2" ). Due to the action of the vertical concentrated forces P1, P2,..., Pn the rigid bar displaces vertically and rotates with a certain angle as shown in Fig. 6.46. We denote by ' ' ' B 1 , B 2 ,..., B n 1 the displaced Fig. 6.46. positions of points B1, B2,..,Bn+1. ' ' Since the bar B1 B2 B3... Bn+1 is rigid, points B 1 , B 2 , B 3' ,..., B n' 1 lie in a straight line. In the same time the vertical displacements of the supporting points B2, B4, Bi+1,...,Bn+1 coincide with the elongations of the vertical bars 1, 2 , 3, ...n respectively. This is why, the geometry of the problem lets us write other n-2 relations among  1 ,  2 ,...,  n , and, thus n-2 relations among N 1 , N 2 , N 3 ,..., N n . In this way, n distinct equations may be written among the axial forces in the n vertical bars: N 1 , N 2 , N 3 ,..., N n . Solving this system we reach to the values of N 1 , N 2 , N 3 ,..., N n as functions of the external applied loads P1, P2, ...,Pn. The normal stress corresponding to each of the n vertical bars may be then computed using the formula: Y

Ni i

,

i

1 ,..., n

.

Ai

Numerical example 1 The rigid bar BCD is suspended from three vertical bars 1, 2 and 3 as shown in Fig. 6.47. Knowing that the length of the three vertical bars are 1 = 1 m, 2 = 0,75 m, 3 = 1,25 m respectively, and the allowable stress of the material the bars are made of is a =120 MPa, determine the required value of the diameter d if a concentrated force P = 180 kN is applied at K.

118

Axial loading

Considering the free body of the rigid bar BCD, two equilibrium equations may be written from statics: Fy

M

0

N

N

2

3

P

; (6.57)

0

A

N1 1 N

N1

P 1, 5 2 N3

2

N3 2 1, 5 P

0

.

(6.58)

On the other hand, we may also write a geometrical relation among  1 ,  2 and  3 as follows: Fig. 6.47 2

2

3

1

3

2

1

or

2

.

We may write therefore

2

N 2 2

N 3 3

N 1 1

EA

EA

EA

2N

2

0 , 75

N 3 1, 25

N1 1

N1

1, 5 N

Equations (6.57), (6.58) and (6.59) may be solved simultaneously for N1

1, 9 k N ;

N

2

36 , 2 k N ;

N

3

41 , 9 k N .

N1, N

1, 25

2

2

and

N

N3

3

0

(6.59)

:

Since the three vertical bars 1, 2 and 3 have identical cross-sections (i.e. identical cross sectional areas A), the maximum normal stress develops in the bar for which the axial force is maximum (i.e. the bar 3). We write therefore N max

3

3

A

41 , 9 10 d

2

3 all

120

d

4 41 , 9 10

3

21 mm

.

120

4

Thus, the minimum acceptable value for the diameter d, so that the system would not fail, is 21 mm.

Numerical example 2 The rigid bar OBCD is supported as shown by two steel vertical bars 1 and 2 of 10 mm diameter ( E 2 ,1 10 5 MPa ). Determine the normal stress in each vertical bar caused by the load P = 19 kN applied at C . (Fig. 6.48). 119


From statics we may write M

0

O

N 1 2a

P 3 , 25 a

N 2 5a

2N1

5N

0

3 , 25 P

2

.

(6.60) Thus, the problem is statically indeterminate (indetermination degree is 1). However, from similar triangles we have: BB '

2a

1

2a

DD '

5a



5a

5 N 1 1

2 N 2 2

EA

EA

2

5

5 N 1 750

Fig. 6.48

N

2

1, 5 N 1 .

(6.61) From (6.60) and (6.61) we finally have N1

6 ,5 k N ;

N

9 , 75 k N .

2

With these values, the corresponding normal stresses in the two vertical bars are: N1 1

6 , 5 10

A

d

3

2

10

4 N 2

2

10 4

6.6.4. SYSTEMS OF CONCURRENT BARS Numerical example

120

2

82 , 76 MPa

4

9 , 75 10

A

3

6 , 5 10

2

3

124 MPa

.

,

1

2N

2

2

1250



2

Axial loading

Three members made of the same material and having the same cross-sectional areas are used to support a load P, as shown in Fig. 6.49. Knowing that E 2 ,1 10 5 MPa and A 600 mm 2 determine the axial force and stress in each member  when 30 and compute the vertical displacement of point B. From statics we may write Fx

0

N 1 sin 30

Fy

0



N 3 sin 30 

N 1 cos 30



0

N

2

N1

N3

N 3 cos 30

; 

(6.62) 0

 Fig. 6.49 (6.63) 2 N 1 cos 30 N 2 120 . Due to the symmetry of the system, the point B displaces to B' as shown in Fig. 6.49. We have

1 cos 30



BB '



2

.

(6.64)

The lengths of the three bars are 1200

1

cos 30

2

1200

3

1385 , 6 mm



;

.

mm

With these values equation (6.64) becomes: N 1 1 EA cos 30

N 2 

2

N

EA

N 1 1 2



N1

1385 , 6

1,15

N1

1200

2

.

(6.64')

Equations (6.62), (6.63) and (6.64') may be now solved for N1, N2, N3: N1

41 , 63 k N

N

1 ,15

2

N1

N 3; 47 , 87 k N .

The corresponding normal stress in each member are: N1 1

A

41 , 63 10 3

3

69 , 38 MPa

600

and

N 2

2

A

47 , 87 10

3

79 , 78 MPa

.

600

The vertical displacement of point B is v

B



N 2 2

EA

2

47 , 87 10 2 ,1 10

5

3

1200

0 , 456 mm

.

600

NOTE: The statically indeterminate non-symmetrical systems of concurrent bars will be discussed later, using energetic methods.

6.6.5 PROBLEMS INVOLVING STRESSES PRODUCED BY ASSEMBLING IMPERFECTIONS 121


Let us consider a straight bar of uniform cross-section of area A and modulus of elasticity E. This bar has to be fixed between two supporting points B and C as shown in Fig. 6.50a. During the assembling process it comes out that the bar is shorter than necessary with a small quantity .

a.

b. Fig. 6.50

To fix the bar B'C' between points B and C, one must pull with an external axial force N until point C' reaches the position C. Only at that moment the assembly is possibly. But, at that moment, we find an internal axial force in the bar, which is equal to N. The necessary axial force used for the compensation of the imperfection is: N



N

EA 

EA

.

(6.65)

We may write that, the normal stress in the bar due to the compensation of the assembling imperfection is N A

E 

.

(6.66)

This assembling normal stress will be then added to the stress resulting from the functional use the bar has been designed for. This is why such assembling stresses are undesirable stresses, in many cases being unknown when starting the operation of the involved member. Numerical example The rigid bar OBCD is supported as shown by three vertical steel roads 1, 2 and 3 (with the crosssectional areas A 1 100 mm 2 , A 2 150 mm 2 and A 3 300 mm 2 respectively and E 2 ,1 10 5 MPa ) and pin connected at point O. The vertical rods have the same length l 3 m . During the assembling it comes out that a clearance exists between the supporting point D and the end of the rod 3, Fig. 6.51. Knowing that 0 , 6 mm and a 1 m determine the normal stresses in the three rods and the rotation angle of the rigid bar OBCD if the imperfection is removed.

122

Axial loading

The clearance is compensated by pulling downward from point D and pulling upward from the end F of rod 3, such as point D displaces to D' with quantity y while point F displaces to D' with quantity  3 . When points D and F are pin connected at D', the three vertical rods 1, 2 and 3 will be in tension, the internal axial forces in the three rods being N1, N2 and N3 respectively. Since the compensated clearance is a very small quantity the geometry of the system will not change essentially, the points B, C and D displacing vertically to Fig. 6.51 B', C' and D' respectively. In the same time, since the bar OBCD is rigid, points O, B', C' and D' will be in a straight line after compensation of . Considering the free body of bar OBCD, from statics we may write: M

O

0

N1 a

N

2a

2

N

3

3a

0

N1

2N

2

3N

3

.

(6.67)

With three unknown quantities (N1, N2, N3) and only a single equilibrium equation from statics, the problem is statically indeterminate (the indetermination degree being 2). However, two more equations may be added from the geometry of the deformed system. From similar triangles we may write: 1

BB '



CC '

a



2a

2

N

N 2

1

2

2

EA

2 N 1

N

EA 1

2

2

2N1

A2 A1

150 100

(6.68)

3N1

2

2N1

and 1 y

a

y

3

3a

1

.

But, on the other hand, 3

y

3

N 3

1

3

EA 3

3

N 1

.

EA 1

(6.69)

Equations (6.67), (6.68) and (6.69) may be solved simultaneously for N1, N2 and N3: N1

1058 N ;

N

2

3174 N ;

N

3

2478 N .

With these values, the corresponding normal stresses in the three vertical rods due to the compensation of are

123


1

N1

1058

A1

100

N 2

3174

A2 N

3

2

10 , 58 MPa ;

21 ,16 MPa ;

150

3

2478

A3

300

8 , 26 MPa .

The angle of rotation of the rigid bar OBCD is

tg

BB ' a

1 a

N 1 EA 1

1

1058

a

3000

2 ,1 10

5

1

1, 58 10

4

rad .

1000

100

6.6.6 PROBLEMS INVOLVING TEMPERATURE CHANGES The mechanical structures elements which have been considered so far were assumed to remain at the same temperature while being loaded. We shall now consider the situation involving changes in temperature. Let us consider for example a homogeneous member BC, of length 0 and uniform cross section, fixed at its both ends as shown in the figure below. If the member temperature is raised by t the member tends to elongate with the quantity 

0

t

,

(6.70)

where is a constant characteristic of the material, called the coefficient

Fig. 6.52 

1

of thermal expansion (expressed in C ). Since the member BC is fixed at its both ends, it cannot elongate, the supports exerting equal and opposite axial forces N on the member after the temperature has been raised, to keep it from elongating (Fig. 6.53). It follows that a state of stress occurs in the member. Since N cannot be computed from statics, the problem is statically indeterminate (the indetermination degree being 1). Fig. 6.53 To compute the unknown value of reaction N, we detach the member from its support at C (Fig. 6.54a) and let it elongate freely as it undergoes the temperature change t (Fig. 6.54b). Equation (6.70) tells that, in such conditions, the member elongates with the quantity 

124

0

t

.

Axial loading

After the elongation has been produced we may return to the original state of Fig. 6.53 by applying the force N at point C and making the elongation become zero again, (Fig. 6.54c). From Fig. 6.54b,c we may write

a.

b. 0

N0

t

EA  0 EA

N

t

0

c.

EA

t.

(6.71) The normal stress in the member due to the temperature change is therefore

Fig. 6.54

N

EA

A

t

E

t

(compression).

(6.72)

A

Numerical example Determine the normal stresses 1 and 2 corresponding to the two portions BC and CD of the bar shown in Fig. 6.55, when the temperature is raised by t 30  C . Use the values: 5 5 2 2 ; 1=500 mm; 2=600 mm; E1 2 ,1 10 MPa ; E 2 0 , 7 10 MPa ; A 1 200 mm ; A 2 400 mm 6  1 6  1 . 12 10 C ; 2 24 10 C 1 Since the temperature  increases with t C , the bar tends to elongate but cannot because of the restrains imposed on its ends. The elongation of the rod is thus zero. Equalizing the elongation produced freely (in case the bar is detached from its support at B) due to the temperature change with that

Fig. 6.55 produced by the action of N, we have 1

1

t

2



2

t

N 1

N 

E 1 A1

E 2 A2

We may write therefore N

1 E 1 A1



2

t

E 2 A2

125

1

1

2



2

2

.

(6.73)

Strength of Materials t

N

1

1

2

1



E 1 A1



6

30 12 10

2

500

500

2

2 ,1 10

E 2 A2

5

6

24 10 600

200

0 , 7 10

5

Noting that the axial forces in the two distinct portions of the bar are equal to obtain the following values of the stress in the portions BC and CD of the bar:

1

BC

2

CD

N

18360

A1

200

N

18360

A2

400

600

18 , 360 k N

400 N

18 , 360 kN

, we

91 , 8 MPa ;

45 , 9 MPa .

6.6.7 TWO MORE NUMERICAL EXAMPLES CONCERNING STATICALLY INDETERMINATE PROBLEMS Example 1 The rigid bar OBC is supported as shown by two steel rods of cross-sectional area A=200 mm and modulus of elasticity E 2 ,1 10 5 MPa . Determine the stresses in the two rods caused by the load P applied at C. Considering the free body of the bar OBC and using statics we may write 2

M

0

O

N 1 sin

1

N

2

sin

2

2

P 2

0

,

(6.74)

Fig. 6.56 where 1

2

atg

1, 5



56 , 30 ;

1

atg

1, 5

36 ,86



.

2

Since we may write only a single equation from statics, the problem is statically indeterminate (with two unknowns: N1 and N2). However, the geometry of the problem shows that (Fig. 6.56) the points

126

.

Axial loading

B and C displace to B' and C' respectively, due to the action of force P. From similar triangles we may write: 1 BB '

1

CC '

2



cos 90

1

1



2

2



cos 90

2

1 sin

1

 sin

2

1

 1 sin

2

1

2



1

2

2

sin



2

sin

1

2

 1 sin

2

.

(6.75)

2

Equations (6.74) and (6.75) may be solved simultaneously for N1 and N2: N1

12 , 035 k N ;

N

12 , 49 k N .

2

With these values, the corresponding stresses in the two rods are: N1 1

A N

2

12 , 035

3

10

60 MPa ;

200 2

A

12 , 49 10

3

62 MPa .

200

Example 2 Draw the axial force and stress diagrams for the steel member shown in Fig. 6.57, where a 0 , 3 mm clearance exists between the end of the member and the supporting point C before the application of the loads. Assume that E 2 ,1 10 5 MPa .

Fig. 6.57 127


Due to the action of forces P1 and P2 the end C' of the member reaches the support C and thus a reaction XC occurs at this support. This is why we shall have two unknown reactions (XB and XC) and only a single equation of mechanical equilibrium which may be written from statics: FX

0

X

B

X

C

250

125

(6.76)

375 kN

The second necessary equation involving XB and XC will be determined from the condition that the total deformation of the member corresponding to the final state (after the loads are applied) is equal to . We write  BC B 1 1 2 2 3 3 C 0 , 3 mm . (6.77) Thus, X

B

600

EA 1

X

B

250

600

X

250

B

EA 1

EA

600

X

B

250 EA

2

125

600

0 , 3 mm

,

(6.77')

2

where 70

A1

2

3848 , 45 mm

2

4

and 50

A2

2

1963 , 49 mm

2

4

are the cross-sectional areas of the portions B-2 and 2-C respectively. Equations (6.76) and (6.77') may be solved simultaneously for XB and XC : X

B

317 , 413 k N ;

X

C

57 , 587 k N .

We are now in the position to draw the axial force diagram. This has been done in Fig. 6.57. Dividing the axial force (N) by the corresponding cross-sectional areas, we obtain the normal stress diagram also sketched in Fig. 6.57.

PROBLEMS TO BE ASSIGNED P6 P.6.1 A solid cylindrical rod and a solid rectangular rod are welded together at C as shown (Fig. P.6.1). Draw the axial force diagram and determine the corresponding normal stresses for each particular portion of the member and the total displacement of point B, knowing that the modulus of elasticity for the two rods is E

2 ,1 10

5

MPa

.

128

Axial loading

Fig. P.6.1 P.6.2 Determine the total deformations of the steel rods shown (Fig. P.6.2a,b), under the given loads, if E 2 ,1 10 5 MPa .

a.

b. Fig. P.6.2

P.6.3 For the rod shown in the figure below: a) draw the axial force diagram; b) determine the required value of area A; c) compute the displacement of point 2; d) compute the total elongation of the rod. The rod is made of steel with the allowable normal stress elasticity

E

2 ,1 10

5

MPa

. It is also known that

a

250 mm

150 MPa

a 

and

P

100 kN

and the modulus of .

Fig. P.6.3 P.6.4 The two rigid bars O1BC and O2 DE are pin - connected at points O1 and O2, respectively. The bars are supported by two rods 1 and 2 as shown in Fig. P.6.4. Knowing that a 0 , 5 m , the 150 MPa , the modulus of elasticity allowable normal stress for the two rods a  5

and the applied force P = 50 kN , determine: a) the axial forces (N1 and N2) in the two rods; b) the required value of area A; c) the elongations of the two rods  1 and  2 . E

2 ,1 10

MPa

129


Fig. P.6.4 P.6.5 Compute the total elongation of a tapered rod subjected to an external load P as shown in Fig. P.6.5.

Fig. P.6.5 P.6.6 A rigid bar OBC is supported by a vertical rod of uniform annular cross – section as shown (Fig. P.6.6). The vertical rod 1 is made of aluminum with E 0 , 7 10 5 MPa . For the 40 kN force shown, determine the required values of the diameters d and D and the vertical displacement of point C if the allowable stress of the material the rod is made of is a  120 MPa .

Fig. P.6.6

130

Axial loading

P.6.7 Two cylindrical rods, one made of steel (with

ES

2 ,1 10

5

MPa

) and the other one made of

5

aluminum (with E A  0 , 7 10 MPa ) support a 60 kN force as shown (Fig. P.6.7). Knowing that 180 MPa the normal stress allowable values of the two materials are and a  steel

, determine the required values of the diameters d1 and d2, and the vertical displacement of point B. a 

A

110 MPa

Fig. P.6.7 P.6.8 Two cylindrical rods, one made of aluminum ( E A  brass ( E B P

5

0 , 7 10

5

MPa

) and the other one made of

) are joined at point 1. The composite rod is fixed at one end (B) while a 0 , 3 mm gap exists between the other end and the right wall (C). Determine the value of force P 0 required to cancel the gap between the end 3 of the composite rod and the wall. If P 2 P0 : 1, 05 10

MPa

Determine the reactions at B and C; Draw the axial force diagram for the composite rod; Draw the normal stresses diagram; Determine the horizontal displacement of point 2.

Fig. P.6.8

P.6.9 A rigid bar OBCDF is supported as shown by three vertical cylindrical rods of diameters 16 mm . The rods are made of steel with 150 MPa and d 1 10 mm , d 2 12 mm and d 3 a  5

. Determine the maximum value of force P which may be safely applied at F and the vertical displacement of points C and F. E

2 10

MPa

131


Fig. P.6.9

P.6.10 A rod consisting of the two portions, BC and CF is fixed at its both ends and subjected to an axial force P as shown (Fig. P.6.10). Knowing that D 2 d 60 mm , a  180 MPa and E 2 ,1 10 MPa , determine the maximum value of force P which may be safely applied at C and the horizontal displacement of point C.

Fig. P.6.10

P.6.11 A rigid bar BCDF is suspended from four identical wires as shown (Fig. P.6.11). Determine the axial forces in the wires (N1, N2, N3 and N4) when a force P is applied at D.

Fig. P.6.11 P.6.12 The rigid bar OBCD is supported by two cylindrical rods of diameters d2 16 mm respectively (Fig. P.6.12). The two rods are made of steel with a  E

2 ,1 10

5

MPa

d1 160

10 mm MPa

. Determine the maximum value of force P which may be safely applied at D.

132

and and

Axial loading

Fig. P.6.12 P.6.13 Two cylindrical rods, one of steel and the other one of aluminum are joined at C. The composite rod is fixed at its both ends and then subjected to the loading shown. Knowing that 5 5 180 MPa , 120 MPa , E Steel 2 ,1 10 MPa , E A  0 , 7 10 MPa , P 10 kN and a  Steel a  A

, determine the reactions at B and F, draw the axial force diagram and determine the required values of the diameters d1 and d2 of the composite rod. 

0 ,5 m

Fig. P.6.13 P.6.14 A steel rod of circular cross section of diameter d is placed inside a steel tube of diameters d1 and D1. The rod is shorter with 0 ,1 mm than the tube. A rigid plate is attached to the tube and then an external axial force P is applied as shown (Fig. P.6.14). Knowing that 5 190 MPa , E 2 ,1 10 MPa , d 50 mm , D 1 70 mm and  40 mm , d 1 300 mm , a  determine the maximum value of force P which may be safely applied. Steel

Fig. P.6.14

133


P.6.15 Four identical cylindrical rods of diameter (with

E

2 ,1 10

5

MPa

and

a 

200 MPa

B and B’ (Fig. P.6.15). Knowing that rod when the gap is compensated.

d

10 mm

and length

have to be joined, while a 25



and

35





0 ,15 mm

1m

, made of steel

gap exists between

, determine the normal stress in each

Fig. P.6.15 P.6.16 Four identical members are used to support the load P  normal stress in each member when 20 , d 15 mm ,  the vertical displacement of point O.

(Fig.P.6.16). Determine the and E 2 10 5 MPa . Compute

40 kN 1m

Fig. P.6.16 P.6.17 A rigid bar OBC is supported by three steel rods as shown (Fig. P.6.17). Knowing that 5 2 ,1 10 MPa determine the normal stress in each 10 mm and E d1 40 mm , d 2 30 mm , d 3 rod when an external force P 50 kN is applied at B.

134

Axial loading

Fig. P.6.17 P.6.18 A rod consisting of two cylindrical portions of distinct materials is fixed at its both ends as shown (Fig. P.6.18). Determine the axial force in the rod and the corresponding normal stresses induced in the two portions by a temperature rise of t  .

Fig. P.6.18 P.6.19 Two horizontal rigid beams are suspended from three identical steel rods as shown (Fig. P.6.19). Knowing that E 2 ,1 10 5 MPa , A 10 mm 2 and  1 m determine the maximum value of force P, which may be safely applied at C’ and the vertical deflection of points B’, C and D’.

Fig. P.6.19

135

7. SHEARING STRESSES IN MEMBERS OF SMALL CROSS SECTIONS

Bolts, pins, rivets and welds used within different machine components and technical systems are considered pieces of small dimensions, providing certain simplifying assumptions in Strength of Materials calculus. In many cases, these elements are subjected to transverse forces, as shown in Fig. 7.1.

Fig. 7.1

If the distance e between the two shearing forces application points 1 and 2 is too small, the bending moment at any cross section between the two points may be neglected, and thus the single internal force in the member remaining the shearing force T = P. The shearing force T determines the development of the shearing stresses  at any cross section of the member considered (between points 1 and 2). Shearing stresses are commonly found in bolts, pins, rivets and welds used to connect various structural members and machine components. Consider for example, the two plates B and C, connected by a rivet as shown in Fig. 7.2. Due to the action of the opposite forces P, the cross section FG of the rivet is sheared by a shearing force T equal to P. The analysis of riveted, bolted and welded connections involve so many indeterminate factors that exact computation solutions are impossible. Nevertheless, by making certain Fig. 7.2 simplifying assumptions, we can easily obtain practical solutions.


7.1 SHEARING STRESS AND SHEARING STRAIN Let us consider for example, a current cross section of a member subjected to shear. The shearing force T may have a certain orientation in the section considered, as shown in Fig.7.3. The shearing force T may be resolved into two components Tz and Ty, acting along the axes Oz and Oy respectively. Due to the action of Tz and Ty at the level of an arbitrary element of area dA the shearing stresses xz and xy develop. For the entire cross section considered, the following relations may be written:   xy d A  T y ;  A     xz d A  T z . A

(7.1)

The actual distribution of the shearing stresses in the section considered is therefore statically indeterminate. However, since the rivets, bolts and welds are in general components of small dimensions, we may assume that the shearing stresses are uniformly distributed over any cross section.

Fig. 7.3

This means that we may write:  xy  constant

;

 xz  constant

.

(7.2)

Relations (7.1) become therefore:  xy     xz 

 dA

 Ty ;

A



dA  T z ;





 

xy

A  Ty ;

xz

A  Tz ;

A



Ty  ;  xy  A  Tz  , xz   A

(7.3)

or, in general,  

T

(7.4)

.

A

The strength condition is therefore:  

T A

136

  a  ,

(7.5)

Shearing Stresses in Members of Small Cross Sections

where A is the cross-sectional area (associated with the area of the plane surface shown in Fig. 7.3) and a – the allowable value of the involved material shearing stress. In general: a= (0,5  0,8) a. (7.6) Relation (7.5) may be used for all three types of strength of materials characteristic problems: - dimensioning problems; - checking problems; - calculus of the allowable external loads. It should be emphasized that the value obtained by using the relation (7.4) is an average value of the shearing stress over the entire section. As we shall see later, the actual value  of the shearing stress varies from zero at the surface of the member to a maximum value max which may be much larger than the average value. Nevertheless, the accuracy obtained using the relation (7.4) is sufficiently high when the cross-sectional areas of the members considered have low values. The shearing strain is not of a great importance in such cases. It consists in fact in a relative displacement v of the sheared cross sections located at distance e from each other, as shown in Fig.7.4. Fig. 7.4 If the material obeys the Hooke’s law, we write (Fig. 7.4): tg    

v

 v   e 

e



e 

G

Te

.

GA

Thus, the relative displacement of the sheared cross sections 1 and 2 is: v 

Te

(7.6)

GA

where T is the shearing force, e is the distance between the applied external forces, G is the shear modulus while A represents the cross-sectional area of the involved member.

7.2 STRENGTH OF A SIMPLE RIVETED JOINT The rivet shear is produced at those sections where it withstands the relative displacement tendency of the joined elements, as shown in Fig.7.5. We may observe that the shearing force at the rivet sheared cross section T equals P. The allowable maximum applied force P is therefore: 137


T m ax  P  A s   a  

 d 4

2

  a  ,

where a represents the allowable shearing stress of the rivet involved material. If the rivet does join a number of i + 1 elements, then i sheared sections exist, with a cumulated area: Fig. 7.5 As  i

 d

2

.

(7.8)

4

7.3 BEARING STRESS IN CONNECTIONS A rivet (pin, bolt etc.) is also subjected to bearing stresses. Consider for example the three plates of thickness t connected by a rivet as shown in Fig. 7.6. Since there are two sheared cross –sections in this case, the rivet is said to be in double shear. The contact between the rivet and the joined elements develops on half –cylinder surfaces as shown in Fig. 7.6 Fig. 7.7.

Fig. 7.7

Fig. 7.8

Since the actual distribution of the contact forces – and of the corresponding stresses- is quite complicated (Fig. 7.8), one uses in practice an average nominal value b of the stress, called the bearing stress, obtained by dividing the load by the area of the rectangle representing the projection of the rivet on the plate section (Fig. 7.9) 138


Since this area is equal to t d in the case of element 2 (for example) of Fig. 7.7 (Fig. 7.9), where t is the plate thickness and d the diameter of the rivet, we write: 

b



P



A

P

(7.9)

.

t d

Fig. 7.9

This value has to be equal to or less than  b a  (the allowable value of the involved material bearing stress). We write therefore: 

b



P td

 

b a 

.

(7.10)

In conclusion, a rivet has to be designed considering two important matters: the shearing stress and the bearing stress. Both computed stresses must be equal to or less than the allowable corresponding stress values, so that the involved structure would not fail under the action of external loads.

7.4 WELDED CONNECTIONS The reliability of welded connections has increased to the point where they are used extensively to supplement or replace riveted or bolted connections in structural and machine design. It is frequently more economical to fabricate a member by welding simple component parts together than to use a complicated casting. Welding is a method of joining metals by fusion. With heat from either an electric arc or an oxyacetilene torch, the metal at the joint is melted and fuses with additional metal from a welding rod. When cool, the weld material and the base material form a continuous and almost homogeneous joint. To protect the weld from excessive oxidation, a heavily coated welding rod is used that releases an inert gas that envelopes the arc stream; this technique is called the shielded arc process. The welded connections offer several important advantages: - the connected members strength is not diminished through additional drilling 139


like the riveted or bolted connections do; - the welded connections require a relatively simple technological process, leading to a low price of the manufacturing; - the maintenance technological process require a minimum effort. Here are the main types of welds: a) Butt welds (Fig. 7.10) In such a case, the weld is subjected to tension, the normal stress being computed with the formula:  

P (b  2t ) t

(7.11)

.

We note that a length b – 2t is used instead of b. This happens due to the technological flaws, which usually occur at the two ends of the weld. The strength condition is therefore:

Fig. 7.10  

P (b  2t )  t



a  w

(7.12)

,

where  a  w represents the allowable stress of the weld, usually taken to be: 

a  w

 0 ,8 

a 

(  a  = the allowable stress of the base material). b) Transverse fillet welds (Fig. 7.11) The strength of transverse fillet welds is assumed to be determined by the shearing resistance of the weld throat regardless of the direction of the applied load. In the 45 fillet weld (Fig. 7.11), with the leg equal to t, the shearing area through the throat is the length of weld b times the throat depth. Considering the real length of the weld (as specified above) equal to b 2a, the two welds shearing stress is (Fig. 7.11): w 

Fig. 7.11

140

P 2 a b  2 a 

.

(7.13)


The strength condition may be written therefore as follows: w 

P

  a 

2 a b  2 a 

w

,

(7.14)

where  a  represents the allowable shearing stress of the weld, usually taken to be: w

 a 

w

 0 , 65 

a 

.

(  a  – the allowable stress of the base material). c) Side fillet welds (Fig. 7.12)

Fig. 7.12

Using the same reasoning we may write: w 

P 2 a   2 a 

  a 

w

.

(7.15)

PROBLEMS TO BE ASSIGNED P.7 P.7.1 Two steel plates are joined together by four rivets of 35 mm diameter as shown (Fig. P.7.1). Knowing that the allowable values of the shearing and bearing stresses of the material the rivets are made of are a = 70 MPa,  b = 120 MPa respectively, and the allowable normal stress of a 

the plates is a = 160 MPa, determine the maximum value of force P which may be safely applied at C. P.7.2 A gusset plate is riveted to a larger plate by three rivets of 30 mm diameter as shown (Fig. P.7.2). Determine the shearing stresses developed in the rivets. P.7.3 Two plates are welded to resist a load P as shown (Fig. P.7.3). What maximum value of P can be applied if  a  weld = 70 MPa and  a  plate = 160 MPa ? P.7.4. Determine the maximum value of force P which may be applied to the welded plates shown (Fig. P.7.4), if  a  plates = 160 MPa and  a  weld  0 , 8  a  plates .

141


Fig. P.7.1

Fig. P.7.2

Fig. P.7.3

Fig. P.7.4

Fig. P.7.5

Fig. P.7.6

P.7.5 A load P is applied to a steel rod supported as shown (Fig. P.7.5). Knowing that 

a 

steel

=

190 MPa;  a  steel = 110 MPa and  b a  = 250 MPa determine the maximum value of force P which may be safely applied to the steel rod. P.7.6 Two wooden planks are joined by the glued joint shown (Fig. P.7.7). Determine the required value of length  so that a 6 kN load to be safely supported if  a  glue = 1,5 MPa.

142

8. BASIC ELEMENTS OF THE THEORY OF ELASTICITY

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. Even in the case of members under axial or transverse loading, the normal and shearing stresses on planes which are not perpendicular to the axis of the involved members, obey complex and difficult mathematical and physical matters. This is why, before going further, we have to understand several important basic elements concerning the theory of elasticity. They will help us to go deeper inside the investigated mechanical elements or machine components and to find reasonable answers to the problems of Strength of Materials.

8.1 PLANE STATE OF STRESS Every elastic body is spatial and, in general, every external loads system is spatial. Hence, strictly speaking, any strength of materials problem (or elasticity problem) is a spatial problem. For its solution, we have to consider all the components of stress, strain and displacement. However, if the body has a particular shape and the external loads are distributed in a particular manner, we may consider the spatial problem as a plane one and neglect some of the components. This will greatly simplify the mathematical aspect of solution while the results may still be applied in engineering design with sufficient accuracy. Let us now consider a thin plate of uniform thickness, subjected to loads acting within the plate middle plane as shown in Fig. 8.1. In such a case the thin plate is Fig. 8.1 said to be in a plane stress condition. From the thin plate we isolate an infinitely small element in shape of a prism (Fig. 8.2). For convenience the plate dimension (or that of the infinitely small element considered) in the Oz direction is taken to be unity. The problem which arises consists in the determination of stresses at the arbitrary point O of the plate, on planes which are perpendicular to the plate middle


plane. In other words, we have to determine the normal and shearing stresses on arbitrary planes BCB’C’ inclined with angle  with respect to the Oy axis, as shown in Fig. 8.2. We assume the elementary area of the rectangular surface BCB’C’ to be equal to dA. While the stresses acting on surface BCB’C’ differ slightly from the stresses at O, the error involved is small and vanishes as sides OB and OC approach zero.

Fig. 8.2

On the other hand, due to the action of the external loads P1, P2,...,Pn, normal and shearing stresses develop on the faces OBO’B’, OCO’C’ and BCB’C’ of the element shown in Fig. 8.2. These stresses are in fact a direct consequence of the interaction between the element and the surrounding material of the plate. For convenience, the element OBCO’B’C’ has been represented in Fig. 8.3 in a simplified manner. In Fig. 8.3, since the plane BC approaches point O, the stress acting on BC will become the stress acting on the inclined plane considered passing through the point O. Now assuming to know the values of stresses acting on the planes OB and OC (i.e. σy, τyx, σx, τxy ), we shall determine the stresses σ and τ on the inclined plane BC. From the summation of Fig. 8.3 moments about point D (Fig. 8.3) we write:



M

D

 0  τ yx  d A  sin α 

BC 2

cos α  τ xy d A  cos α 

This leads to a relation already known: τ xy  τ

144

yx

.

BC 2

sin α  0

.

Basic elements of the theory of elasticity

From the mechanical equilibrium of all forces projected on σ and τ directions respectively we have:



Fσ  0 

σ  d A  σ x d A  cos α cos α  σ y d A  sin α sin α  τ xy d A  cos α sin α  τ yx d A  sin α cos α  0

 τ  d A  σ x d A  cos α sin α  σ

y

;

Fτ  0 

d A  sin α cos α  τ xy d A  cos α cos α  τ yx d A  sin α sin α  0 .

We may write therefore:  σ  σ x cos 2 α  σ y sin 2 α  2 τ xy sin α cos α ;   2 2 τ  σ x  σ y   sin α cos α  τ xy cos α  sin α .  





(8.1)

Using the relations sin

2

1  cos 2 α

α 

;

cos

2

α 

1  cos 2 α

2

,

2

(8.1) becomes 1  cos 2 α 1  cos 2 α  σ  σx   σy   τ xy sin 2 α ;  2 2  σ x  σ y   sin 2 α τ   τ xy cos 2 α .  2

(8.1’)

or σx  σ     σx  τ   

 σ

y

σ



2

x

 σ 2

 σ

y

y

 cos 2 α  τ xy sin 2 α ;

(8.1’’)

 sin 2 α  τ xy cos 2 α .

2

A simple analysis of relations (8.1’’) tells us that the normal and the shearing stresses σ and τ on surface BCB’C’ inclined with angle α (and thus the normal and the shearing stresses at point O about a surface parallel to BCB’C’) are functions of angle α. One could ask: which are the values of angle α that make these quantities (σ and τ) maximum or minimum? These values may be found by differentiating (8.1’’) with respect to α and setting the derivatives equal to zero. dσ

 2

σ

 σ

x

dx σ  2  dα 

dσ

y

 sin 2 α  2 τ xy  cos 2 α  0 

2 x

 σ 2

145

y

  sin 2 α  τ xy  cos 2 α   0  

.

(8.2)


We find that tg 2 α 

2 τ xy σ

.

 σ

x

(8.3)

y



Equation (8.3) gives us two values of α (α1 and α2 = α 1+ ) for which the normal 2

stress σ has extreme values. This is why, the equation (8.3) is always written as 2 τ xy

tg 2 α 1 , 2 

σ

x

 σ

(8.3’)

. y

In other words, the two values of α (α1 and α2) do give us two perpendicular directions in the plane about which the normal stress σ has extreme values (a maximum value σ1 with respect to one direction and a minimum value σ2 with respect to the other one). The two extreme stresses σ1 and σ2 are thus perpendicular to each other. They may be obtained by substituting α from (8.3) into the expression of σ from (8.1’’). This gives

σ x  σ   1   σ x  σ2   

 σ



y

2  σ



y



σ x

1 2



2

1 2

 σ

y

2

2

 4 τ xy ;

(8.4)

σ x

 σ

y



σ x

 σ

y

2

2

 4τ

2 xy

.

or σ 1 ,2 

σ x

 σ



y

2



1 2

2

 4 τ xy

.

(8.5)

The two extreme normal stresses are called principal stresses, while the corresponding directions (given by α1 and α2) are called principal directions. Returning now to the second equation of (8.1’’), differentiating it with respect to α and setting the derivative equal to zero, we have dτ dα

 2

σ

x

 σ

y

2

cos 2 α  2 τ xy sin 2 α  0

.

(8.6)



We thus find two perpendicular directions (α3 and α4 = α3+ ) for which the shearing 2

stress τ has extreme values (a maximum value τ3 with respect to one direction and a minimum value τ4 with respect to the other one). The two directions are given by equation: dτ dα

 0  tg 2 α 3 , 4 

146

σ

y

 σ

2 τ xy

x

.

(8.7)


Substituting these values of α into the second equation of (8.1’’) we get to

σ x

1

τ 3 ,4  

2

 σ

y

2

2

 4 τ xy

,

(8.8)

or τ 3 ,4  

1 2

σ 1

 σ2

.

(8.9)

Since tg 2 α 1 , 2  

1 tg 2 α 3 , 4

,

(8.10)

the shearing stress τ reaches its extreme values on planes inclined at 45° with the planes for which the normal stress σ reaches its extreme values σ1 and σ2. It is to be noted that the extreme condition (8.2) coincides with τ 

σ

x

 σ 2

y

sin 2 α  τ xy cos 2 α  0

.

In other words, on planes coinciding with the principal directions, the shearing stress τ is zero. Returning now to the original thin plate of Fig. 8.2, we may facilitate the visualization of the stress condition at point O by considering an infinitely small cube of different orientations, centered at O and the stresses exerted on each of the lateral faces of the cube, (Fig. 8.4). For a certain orientation of the cube B1B2B3B4 (given by the two perpendicular directions 1 and 2) the stresses acting on the Fig. 8.4 lateral faces of the cube reduce to the principal normal stresses σ1 and σ2, while the corresponding shearing stresses become zero (Fig. 8.4). Rotating this cube by 45o within the plate plane, we get to the cube C1C2C3C4 on whose lateral faces act both the extreme values of the shearing stress ( τ max

 

σ1  σ 2 2

) and the normal stresses denoted by σ3 and σ4. In other words, on

147


planes corresponding to the action of the extreme shearing stresses, the normal stresses do not become zero.

8.2 SPATIAL STATE OF STRESS Consider now a body subjected to several external loads P1, P2,…,Pi, Pn in mechanical equilibrium (Fig. 8.5). To understand the stress condition created by these loads at some arbitrary point O within the body, we isolate an infinitely small element in form of a parallelepiped, with its edges parallel to the coordinate axes and having the lengths dx, dy and dz (Fig. 8.6a). We sketch this element at a larger scale (Fig. 8.6b). Generally speaking, the Fig. 8.5 stress components acting on the element faces are functions of x, y and z. Thus, the stress components on a pair of parallel faces are not equal but differ by a differential quantity.

a.

b. Fig. 8.6

For instance, if the average normal stress component on a face is σx, then, that on the parallel face, due to the variation of x, will be σ

x



σ

x

x

148

dx

.


Applying the same reasoning like in the preceding section, one could find a certain spatial orientation of the element considered, so that the shearing stresses on the element faces to be zero and, hence, the corresponding normal stresses to become the principal stresses σ1, σ2 and σ3 (Fig. 8.7). One could demonstrate that the principal stresses σ1, σ2 and σ3 (where σ 1  σ 2  σ 3 ) may be computed as the real roods of equation

Fig. 8.7 σ

              

3

2

 I 1σ

I1  σ

 I 2σ  I3  0

 σ

x

I2  σ xσ

σ

y

x

I 3  τ yx τ zx

 σ

z

 σ xσ

z

τ xy

τ xz

σ

τ yz

y

y

τ zy

σ

(8.11)

,

;  σ yσ

2

z

2

2

 τ xy  τ xz  τ yz

; (8.12)

.

z

We get to this result letting dx, dy and dz approach zero (Fig. 8.6b), so that the element in form of a parallelepiped is contracted to point O. Using the same analogy we may also reach to the expressions of the extreme shearing stresses: τ 1 ,2  

σ 1

 σ2 2



;

τ 1 ,3  

σ 1

 σ3 2

;

τ 2 ,3  

σ 2

 σ3

.

(8.13)

2

These shearing stresses correspond to the planes passing through each of the principal axes of stress and bisecting the right angle between the other two principal axes of stress. Thus we see that:  the magnitude of the largest shearing stress is equal to half the difference between the largest and the smallest principal stresses;  the largest shearing stress acts on the plane bisecting the angle between these two principal stresses and passing through the line of action of the third principal stress. 149


8.3 MONOAXIAL (UNIAXIAL) STATE OF STRESS With the above presented concepts in mind, we may now return to the monoaxial state of stress. This is in fact a plane state of stress for which σ

y

 0

τ xy  τ yx  0

and

.

Let us consider again the case of a member under axial loading (Fig. 8.8).

a.

b. Fig. 8.8

From this member we detach an infinitely small element in shape of a prism and, for convenience, we represent it in plane, separately (Fig. 8.8b). This element is in fact the one represented in Fig. 8.3, making σy = τyx = τxy= 0. We note therefore that the conditions of stress in the member may be described as shown in Fig. 8.8b. The only stresses exerted on the face OC of the prism (which is perpendicular to the x axis) are the normal stresses σx. However, on planes inclined with an angle α, normal and shearing stresses (σ, τ) develop. We thus conclude that the same loading condition may lead to different interpretations of the stress condition at a given point, depending upon the orientation of the plane BC (Fig. 8.8b). Using equation (8.3) tg 2 α 1,2 

with

τ xy  σ

y

 0

2 τ xy σ

x

 σ

, y

, we find that 1  0



;

 1 

2

 2





.

2

In other words, the two principal directions are parallel to the axes Ox and Oy. The corresponding principal stresses σ1 and σ2 are (Eq. 8.5): σ 1 ,2 

σ

x

 σ

y

2



1 2

σ x

σ

y

  4 τ xy2



σ

x

2



1 2

σ

x

,

which finally gives: σ1  σ x ;  σ 2  0 .

(8.14)

150


In the same time, using equation (8.9), the maximum shearing stress is given by τ max  

σ1  σ 2

 

2

σx

,

(8.15)

2

acting on a plane inclined at 45o with Oy axis (α = 45o, Fig. 8.8).

8.4 PURE SHEAR STATE Let us assume that a pure state of stress exists at an arbitrary point O of a body subjected to several external loads in mechanical equilibrium. In such a case σx= σy = 0, the pure state of stress being defined by the stress components τxy= τyx as shown in Fig. 8.9. Using equation (8.3’) tg 2 α 1 , 2 

2 τ xy σ

x

 σ

, y

with σx = σy= 0, we find that π  α  ;  1 4  α  π  π  3π .  2 4 2 4

Fig. 8.9

The principal stresses are given by: σ 1 ,2 

σ

x

 σ 2

y



1 2

σ x

 σ

y

2

2

 4 τ xy   τ xy

.

In other words, this means that a pure shear state is equivalent to a state of stress consisting in a tensile stress σ1 and a compressive stress σ2 of the same magnitude (σ1=τxy; σ2 = -τxy). The principal stresses σ1 and σ2 act on planes inclined at 45o and 135o with Ox axis as shown in Fig. 8.9.

8.5 GENERALIZED HOOKE’S LAW First of all we have to recall that, for a member in tension or compression which undergoes small deformations, involving only the straight-line portion of the corresponding stress-strain diagram, the Hooke’s law may be written in its simple form as follows:   E 

151

.

(8.16)


As mentioned in the preceding sections, E is called the modulus of elasticity of the material involved. In other words, the validity of such a law means that the stress σ is directly proportional to the strain ε. In the same manner, for values of the shearing stress (τ) which do not exceed the proportional limit in shear (in case of members subjected to shear) the Hooke’s law for shearing stress and strain may be written as   G 

,

(8.17)

where G is called the shear modulus of the material involved. If a rectangular coordinate system Oxyz is being used, the quantities in the above two relations may become σx, σy, σz, εx, εy, εz, τxy, τxz, τyz, γxy, γxz, γyz and  x    y   z    xy  xz   yz

 E x ;  E

;

y

 E z ;  G

xy

;

 G

xz

;

 G

yz

.

(8.18)

Let us now consider a body (made of a homogeneous isotropic material) subjected to several external loads in mechanical equilibrium P1, P2,…, Pk, Pn (Fig. 8.10a).

a.

b. Fig. 8.10

We assume that the most general state of stress occurs within the body, due to the action of the external loads. 152


This means that, if we isolate an elementary parallelepiped from the body, with its edges parallel to the coordinate axes, normal and shearing stresses develop on each face of such a parallelepiped (Fig. 8.10b). For convenience, the variation of stresses from one face to the other parallel face has been neglected. Let us now consider first only the effect of the stress component σx (Fig. 8.11).

Fig. 8.11

We recall that σx causes a normal strain to



γσx

εx 

σx

in the Ox direction and strains equal

E

in each of the Oy and Oz directions. We may write therefore:

E

action of σx

 x 



;  y   

x

E

x

 



x

E

;  z   

 

x



x

Similarly, the stress component σy, if applied separately will cause a strain Oy direction and strains





y

.

(8.19)

E



y

in the

E

in the other two directions. Finally, the stress

E

component σz causes a strain



z



in the z direction and strains



E

z

in the Ox and

E

Oy directions. We write action of σy

action of σz

 x   

y

 

 x    z  

 

y

;  y 

E

 E

z



;  y  

y

E

 E

;  z  

z

;  z 

 

y

;

(8.20)

E



z

.

(8.21)

E

Combining the results obtained, we may write that the components of strain corresponding to a multiaxial loading of the parallelepiped involved (the state in which only the normal stresses σx, σy and σz act, being all different from zero is referred to as a multiaxial loading – Fig. 8.12) are

153


  x   y   z 1     x   y   z ;   x   x   x   x  E E E E   y     x 1  z                    y    x   z  ;  y y y y E E E E     x   y  1                  z   z   x   y . z z z  z E E E E 















(8.22)



These relations are referred to as the generalized Hooke’s law. They are valid only as long as the stresses do not exceed the proportional limit, and as long as the deformations involved remain small. It is important to mention again that a positive value for a normal stress component signifies tension, while a negative value signifies compression. We are now in the position to complete the generalized Hooke’s law corresponding to the state of stress of the parallelepiped sketched in Fig. 8.10b, by adding the shearing strains γxy, γxz, and γyz

Fig. 8.12

as follows (Fig. 8.13): 

xy





xy

; 

G





xz

xz

G

; 

yz





yz

G

.

(8.23)

We conclude therefore that relations (8.22) and (8.23) represent the complete form of the generalized Hooke’s law for a homogeneous isotropic material. For a plane state of stress (σz = 0, τxz= τyz = 0) the generalized Hooke’s law becomes:   



x



y



z

xy

1



x

 

y

;



y

 

x

;

E 1 E 

(8.24)

1

 

E





xy



x

;

.

G

A simple examination of the relations (8.22) and (8.23) might lead us to believe that three distinct constants E, G and ν must first be determined experimentally, if we are to predict the deformation caused in a given material by an arbitrary combination of stresses. Actually, only two of these constants need to be 154


determined experimentally for any given material. But this will be discussed later, when a relation among E, G and ν will be found.

Fig. 8.13

8.6 STRAIN ENERGY Consider a solid deformable body subjected to several external loads P1, P2,…, Pk, Pn in mechanical equilibrium (Fig. 8.14). If the loads applied increase slowly from 155


zero up to their nominal values, the loads application points A, B, K, N displace to different positions A', B', K', N' respectively, an external work L being done in this way (Fig. 8.14).

Fig. 8.14

The work done by the applied loads P1, P2,…,Pk, Pn must result in the increase of some energy associated with the deformation of the body. This energy is referred to as the strain energy of the body involved. We denote this energy by U. If the deformations of the body do not exceed the elastic limit of the material, then, the storaged strain energy U will be completely released to the surrounding environment when the external loads P1, P2,…, Pk, Pn are removed. In such a case U is called the elastic strain energy. On the other hand, we may consider that the whole performed external work is completely transformed into strain energy. We write therefore L U

.

(8.25)

8.6.1 ELASTIC STRAIN ENERGY UNDER AXIAL LOADING Consider a straight rod BC of length , fixed at one end and subjected to an axial external force P as shown in Fig. 8.15. We assume that the rod material obeys the Hooke’s law and the proportional limit is not exceeded. We do also assume that the force P is applied statically (P does slowly Fig. 8.15 increase from zero up to its nominal value, as shown in Fig. 8.16). Since the material does not exceed the proportional limit, during the application of force P, the axial force N in the rod will increase directly proportional to the horizontal displacement u of point C as shown in Fig. 8.17a.

156


When N reaches the value P, the displacement u reaches the final value of the rod elongation Δ. In the same manner the normal stress in the rod, during the application of force P, does slowly increase and directly proportional to the normal strain ε, as shown in Fig. 8.17b.

Fig. 8.16

a.

b. Fig. 8.17

The work done by force P, as the rod elongates is equal to the hachured area located under the force deformation diagram (Fig. 8.17a). We may write therefore: 

L 



N du 

P 

.

(8.26)

2

0

The elastic strain energy under axial loading is U  L 

P  2



 A

 

 A

2







  A  



2

2



 

V

,

(8.27)

2

where V  A   represents the volume of the rod. Since the Hooke’s law is valid (   E   ), we may also write U 

 

V 

2





2



V 

E



2

V

.

(8.28)

2E

Work and energy are expressed in the same units, obtained by multiplying units of length by units of force. Thus, if the International System metric units are used, work and energy are expressed in N  m , this unit being called a joule (J). Returning now to the relations (8.27) and (8.28), we observe that the strain energy depends upon the dimensions of the rod involved. To eliminate the effect of size and to direct our attention to the properties of the material we shall define the concept of elastic strain energy per unit volume. This quantity is referred to as the elastic strain-energy density and is denoted by U . We write therefore D

U

D



U V



  2

157





2

2E

.

(8.29)


In this way the elastic strain energy absorbed by an infinitely small element of material of volume dV is dU  U

D

 

 dV 

dV 

2



2

dV

(8.30)

.

2E

Thus, the total strain energy Ut of a body under axial loading is

U

t

U D dV





V



 

dV 

2

V



2

 2E

dV

.

(8.31)

V

8.6.2 SHEARING STRESSES ELASTIC STRAIN ENERGY If a material is subjected to plane shearing stresses (Fig. 8.18) the elastic strain energy may be computed in a similar manner.

Fig. 8.18

For such a case, the elastic strain energy density may be written therefore as

U

D

 



2





2

(8.32)

.

2G

Thus, the total elastic strain energy Ut of a body under shearing stresses is U

t



 V

 

dV 

2



2

 2G

dV

.

(8.33)

V

where V is the volume of the body involved.

8.6.3 STRAIN ENERGY FOR A GENERAL STATE OF STRESS In the preceding sections we have determined the expression of the elastic strain energy of a body under normal and shearing stresses. Let us now consider a body subjected to several external loads in mechanical equilibrium (Fig. 8.19a). 158


We do also assume that the most general state of stress develops in the body due to the action of the external loads. Now we shall isolate an elementary parallelepiped from the body (Fig. 8.19b). In such a case the general state of stress is characterized

a.

b. Fig. 8.19

by the six stress components σx, σy, σz, τxy, τxz, and τyz (for convenience, in Fig. 8.19b, the stresses have been represented only on the visible faces of the elementary parallelepiped considered). If the body behaves linear elastically, the elastic-strain energy density for a general state of stress may be obtained by adding the expressions given within the preceding sections. In this way, the elastic strain-energy density may be expressed as follows: U

D



 x



x

 y

2

y



 z

2



z

2



xy

 2

xy





xz

 2

xz





yz



yz

.

(8.34)

2

Recalling the expressions representing the Hooke’s law for a homogeneous, elastic and isotropic body:           



xy

x



y



z



1



x

  

y



z

 ;



y

  

x



z

 ;



z

  

y

 .

E 1 E 1 E





xy

G

;

xz





x

xz



;

G

yz





yz

G

and substituting for the strain components εx, εy, εz, γxy, γxz, γyz into (8.34), we obtain the elastic strain-energy density as follows: 159


U

D



1 2E



2 x



2 y



2 z

    E

x



  x

y

z

  y

z



1 2G





2 xy

2 xz



2 yz

.

(8.35)

If the principal axes are used as coordinate axes, the shearing stresses become zero and (8.35) reduces to U

D

1





2E

2 1



2 2



2 3





  1

E

2

  1

3

  2

3

.

(8.36)

Thus, the total elastic strain energy (Ut) of a body under the most general stress condition may be written as U

t



U D dV ,

(8.37)

V

where V is the volume of the body involved and UD the elastic strain-energy density mentioned above.

8.6.4 ELASTIC STRAIN-ENERGY DENSITY ASSOCIATED WITH A CHANGE IN VOLUME. ELASTIC STRAIN-ENERGY DENSITY ASSOCIATED WITH A DISTORTION (A CHANGE IN SHAPE) Due to the action of the external loads a solid body changes both its volume and its shape. In this way we may separate the elastic-strain energy density at a given

a.

b.

c.

Fig. 8.20

point of the body into two components:  a component Uv associated with a change in volume of the material at that point;  a component US associated with a change in shape (a distortion) of the material at the same point. A given state of stress may be obtained by superposing two states of stress as shown in Fig. 8.20, where σ1, σ2, σ3 are the principal stresses and σave (or  ) is the average value of the principal stresses: 

ave



1 

2

3

160



3



.

(8.38)


The state of stress described in Fig. 8.20b tends to change the volume of the element, but not its shape, since all the faces of the element are subjected to the same stress σave (or  ). The state of stress described in Fig. 8.20c tends to change the shape of the element. Recalling (8.36), the elastic-strain energy density associated with the state of stress described in Fig. 8.20b may be written as

1



2E



2E





1

UV 

3



2 1

2





2 2



2 2

3 

2E





2 1

 1

6E

E

 

2 3

  

3 1  2  

2



 1

2

2E



2

2

  1

3

  2

3





     

E

E

1  2

 

2 3



 1  2    2E  3

1



2



3

3

 . 2

3

  

2



(8.39)

In the same manner, the elastic strain-energy density associated with the state of stress described in Fig. 8.20c, may be written as:

U

S

U

D

UV 

1 2E



2 1



U

S



S



2 2



2 3

 

E

 1

  1

3



3



  1

2

2

  2

3



1  2 6E



2 1



2 2



2 3

which finally gives 1



1

6E



2



2



2

 

1

 

2



  1

3

  2

2

3



2



(8.40)

.

(8.41)

or

U

1 3E



2 1

2



8.6.5 RELATION AMONG E, G AND

2 3

3

ν

Consider a rectangular plate of uniform thickness, made of a homogeneous and isotropic material (Fig. 8.21). The plate is assumed to be in a plane stress condition, being subjected to tension about Ox axis (  x    0 ) and to compression about Oy axis (  y    0 ) as shown in the figure. 161

,


Recalling now the pure shear state properties described in section 8.4, the stress condition mentioned above is equivalent to a pure shear state at 45o, as shown in Fig. 8.21. Since the elasticstrain energy densities associated with the two states of stress have the same value, we may write U

Fig. 8.21

D

U

I

D

II

,

(8.42)

where U



2E



1 2E







1



I

D

2



0

2 1





E



2 1

2 2

E

2 0









E



2 2



2 3

 1

 E

2

 1



1



2E

2

2 0

  1

  

  2

3

0



2

 

E

3



 0      0  

2

1    ,

0

E

and (8.35) U

D

II



1 2E



2 x



2 y



2 z



 E



x



y

  x



0

z

  y

z



1 2G



2 xy



2 xz



2 yz





2 xy



2G



2 0

2G

We may write therefore: 

2

1    

E

2 0

.

2G

It thus follows that G 

E 2 1  



.

(8.43)

This relation may be used to determine one of the constants E, G or ν from the other two.

162

.

9. TORSION

Members in torsion are encountered in many engineering applications, when there is necessary to transmit a twisting couple Mt from one plane to another parallel plane (Fig. 9.1). The twisting couples are also called torques.

Fig. 9.1

These couples (Fig. 9.1) have a common magnitude and opposite senses. A case in which torsion is dominant is provided by the transmission shafts, which are used to transmit power from one point to another, as from a motor to a machine tool, from an engine to the rear axle of an automobile or from a steam turbine to an electric generator, etc. (Fig. 9.2).

Fig. 9.2

These shafts may either be solid (Fig. 9.3a) or they may be hollow (Fig. 9.3b).


a.

b. Fig. 9.3

If we know the power P which is to be transmitted through the shaft (expressed in horsepowers - HP) and the frequency of the rotation n (number of revolutions per minute - rpm ), one could demonstrate that the twisting couple (or torque) the shaft is subjected to is given by the formula M t

7 , 02

P

kN

(9.1)

m .

n

If we express the power in kW (kilowatts), formula (9.1) becomes: M t

9 , 55

P

kN

(9.2)

m .

n

9.1 TORSION OF CIRCULAR MEMBERS 9.1.1 STRESSES AND STRAINS Consider a circular shaft AB of diameter d, subjected to equal and opposite torques Mt , as shown in Fig. 9.4. We pass a section perpendicular to the axis of the shaft through some arbitrary point C.

a.

b. Fig. 9.4

The cross section at C will be subjected to elementary shearing forces dF acting perpendicular to the radius r of the shaft. The elementary shearing force dF developed at the level of an elementary area dA of the cross-section considered, is given by the shearing stress at that level multiplied by the area dA (Fig. 9.4b) 164

Torsion

Expressing that the sum of the moments given by the shearing forces dF with respect to the axis of the shaft is equal in magnitude to the externally applied torque Mt, we have: r dF M t, (9.3) A

where A is the cross-sectional area. Since dF = dA, (9.3) becomes: r dA

(9.4)

M t.

A

While the relation (9.4) gives us an important condition which has to be satisfied by the shearing stresses at any given cross section of the shaft, it does not tell us how these stresses are distributed over the cross section. This means that the problem is statically indeterminate, i.e., the distribution of stresses cannot be determined from statics alone. The integral of (9.4) may be solved only if the shearing stress distribution law is found. In deriving the torsion formulas, we make the following assumptions: - the member in torsion has a constant cross-section; - the member material is continuous, homogeneous and isotropic; - Hooke’s law is available;

a.

b. Fig. 9.5

- the member is loaded by twisting couples in planes that are perpendicular to the axis of the member; - stresses do not exceed the proportional limit. Under the above mentioned assumptions, let us now consider a circular member subjected to torsion (Fig. 9.5). The deformation in the circular member subjected to torsion may be easily illustrated if we trace several parallel circles and straight generatrix lines on the surface of the circular member as shown in Fig. 9.5. When the circular member represented in Fig. 9.5a is subjected to torsion it will 165


deform (Fig. 9.5b). Analysing the mode in which the circular member considered does deform under the specified loading, we note that: - when a circular member is subjected to torsion, every cross section remains plane and undistorted; - a straight radial line in the section (for example O1A) remains straight after deformation and rotates with a certain angle ; - the distance between two cross-sections remains unchanged after deformation. This means that x = 0, so that, applying the Hooke`s law, we have: x

0.

(9.5)

- due to the action of Mt the rectangle abcd (Fig. 9.5a) changes into a parallelogram (a`b`c`d` - Fig. 9.5b), the original right angles of abcd modifying with the quantity (the shearing strain). It follows therefore that shearing stresses develop on the faces of the elementary volume a`b`c`d`a``b``c``d`` (Fig. 9.5b). Let us consider now a point C located on the circumference of a given cross section of the circular member in torsion (Fig. 9.6).

Fig. 9.6

As mentioned above, at the level of point C, a shearing stress develops, contained within the cross-sectional plane. We do also assume that has a certain orientation within the cross-sectional plane (Fig. 9.6). But may be resolved into two components ( 1 and 2), respectively tangential to the circumference at C and radial. On the other hand we know 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. This means that the shearing stress component 2 will be accompanied by an equal shearing stress 2` acting along Ox direction and located on the external surface of the member (Fig. 9.6). Since this surface is not loaded, 2` = 0 and, therefore, 2 = 2` = 0. 166

Torsion

In other words, the shearing stress at point C will be directed along the tangent to the circumference at C, perpendicular to the radius OC. We assume that this property remains available for any other point of the cross-section (Fig. 9.7). We shall now determine the distribution of the shearing stresses in a given cross section of a circular shaft (Fig. 9.8a).

Fig. 9.7

The shaft of diameter d and length , is attached to a fixed support at one end. When the torque Mt is applied at section O, the shaft will twist with an angle called the angle of twist.

a.

b. Fig. 9.8

Let us now detach from the shaft an element of length dx, located between two cross sections of the shaft (I and II – Fig. 9.8a). We have sketched this element separately (Fig. 9.8b).The element considered is also subjected to torsion by the same torque Mt, the two ends of the element rotating reciprocally with an infinitely small angle of twist d . For convenience, we shall consider that the end II is fixed while the end I rotates with angle d . In this way, an arbitrary point C of the end I will displace to C``, while a point B, located on the same radius, will displace to B``, (Fig. 9.8b). We may write therefore: tg

C C `` C C`

O 1C d dx

where is the corresponding shearing strain (Fig. 9.8b). 167

r

d dx

,

(9.6)


Denoting by

(the angle of twist per unit length) the quantity

d

, the shearing

dx

strain may be expressed as follows: (9.7) where r is the distance between the arbitrary investigated point C and the axis x of the shaft. Since Hooke`s law applies, we have r

,

G

G r

(9.8)

,

where G is the shear modulus. Recalling relation (9.4), we may write: M

r dA

t

G r

A

Since G and

r dA

A

G

r

2

(9.9)

dA .

A

are constant, we have: M

t

G

r

2

dA

G

I

(9.10)

p

A

or M

t

GI

p

(9.11)

,

where Ip is the polar moment of inertia of the shaft cross-section

I

d p

4

. With

32

this expression of , relation (9.8) becomes: G r

G r

M GI

Fig. 9.9

t

M

p

I

t

r.

(9.12)

p

Equation (9.12) is called the elastic torsion formula. It shows a linear distribution of the shearing stresses over an arbitrary cross-section of a shaft in torsion, (Fig. 9.9). Fig. 9.9 shows the shearing stress distribution in a solid circular shaft for the cross-sectional points located on Oy axis. Due to the symmetry of the cross section, we shall find the same stress distribution for any other cross-sectional points located on the same diameter. The shearing stress at the level of an arbitrary point A of the shaft cross-section is therefore (Fig. 9.9): 168

Torsion

M A

I

t

r.

p

Since the shearing stress in the shaft varies linearly with the distance r from the axis of the shaft, the maximum shearing stress develops on the circumferencial points of the shaft cross-section. We write therefore: M m ax

I

t

M

r m ax

I

p

t

M

R

I

p

t p

d

(9.13)

.

2

The maximum shearing stress may be also written as M m ax

I

t p

M

t

W

p

(9.14)

,

rm ax

where Wp is called the polar strength modulus of the involved cross-section. For a circular cross section of diameter d, the polar strength modulus is d I W

p

p

rm ax

4

d

32 d

3

(9.15)

.

16

2

To determine whether the shaft may support the applied external torque Mt, we must compare the maximum value of the shearing stress given by relation (9.14) with the maximum value of the stress which may be safely applied to the material the shaft is made of. The strength condition is therefore m ax

M

t

W

p

a 

(9.16)

,

where ais the maximum allowable shearing stress in the type of the material the involved shaft is made of. Relation (9.16) applies for circular members in torsion and it may be used for all specific strength of materials problems. Even if the formula (9.16) has been derived for a shaft of uniform circular cross section, it may be also used for a shaft of variable cross section (Fig. 9.10a) or for a shaft subjected to torques at locations other than its ends (Fig. 9.10b).

a.

b. Fig. 9.10 169


Let us now return for a while to the concept of angle of twist, to derive a relation between this angle and the torque exerted on the shaft (Fig. 9.8a). We recall that the angle of twist per unit length is given by (9.11): M

t

G I

.

p

On the other hand, this quantity has been defined as d

,

d x

where

is the angle of twist. We may write therefore d

M

d x

G I

t

d x , p

which gives 

0

M G I

t

d x,

(9.17)

p

where  is the length of the shaft involved. Since the circular shaft shown in Fig. 9.8 has a uniform cross-section (I p

d 32

4

constant) and is subjected to a constant torque equal to Mt (Mt =

constant along the shaft), relation (9.17) becomes: M GI

Fig. 9.11

 t p

d x

M t GI

0

(expressed in radians).

(9.17')

p

The relation obtained shows that, within the elastic range, the angle of twist is proportional to the torque Mt applied to the shaft and to the length  of the shaft. This relation (9.17') may be used only if the material of the shaft is homogeneous (G = constant), the shaft is loaded only at its ends and has a uniform cross section. If the shaft consist of several portions with various values of the cross sections diameters and /or is subjected to torque at locations other than its ends (Fig. 9.11.), we must divide the shaft into component 170

Torsion

parts which satisfy individually the required conditions of relation (9.17'). For the shaft represented in Fig. 9.11, we may write therefore: M 0 A

D

A

B

B

C

C

D

G I

M 0 GI

M

1

G I

p1

3M 0

1

GI

p1

2M

0

GI

p2

M

2

2M

0

G I

p2

3M 0

2



0

3

0



3

p3

,

p3

where: 4

I

d1 p1

4

4

;

I

32

d2 p2

;

I

32

d3 p3

.

32

In other words, the total angle of twist of the shaft represented in Fig. 9.11, i.e. the angle through which end A rotates with respect to end B, is obtained by adding algebraically the angles of twist of each component part. In case of a shaft with a variable circular cross section (Fig. 9.12a), relation (9.17') may be applied only to an element of infinitely small length dx of the shaft. Since such an element has an infinitely small length (dx) we may neglect the variation of its cross section from one end to the other. (Fig. 9.12). This means that we may apply the formula (9.17') for this element, as follows: a.

d

b.

M

0

dx

G I

p

(x)

,

(9.18)

where:

Fig. 9.12

- d is the infinitely small angle of twist representing the angle through which end A of the shaft element of Fig. 9.12b rotates with respect to the end B; - Ip(x) is the polar moment of inertia corresponding to the diameter d(x) of the cross section of the shaft, located at distance x from the end of the shaft (Fig. 9.12a). In other words, Ip is a function of x: I

p

I

p

x

d

4

32

171

x

.


Now integrating (9.18) in x, from 0 to , we obtain the total angle of twist of the shaft: 

0

M G I

0

p

(9.19)

dx

(x)

or, in general: 

0

M G I

p

t

dx

.

(9.20)

(x)

Formula (9.20) does also apply when Mt has a continuous variation along a circular member in torsion (Mt = Mt(x), Fig. 9.13a) or when both Mt and Ip vary along the member involved (Fig. 9.13b).

a.

b. Fig.9.13

Since the shearing stresses over the solid cross sections of circular shafts in torsion have a linear distribution, in many cases one chooses hollow cross sections, thus reducing the own weight of the shafts (Fig. 9.14a). The distribution of the shearing stresses in a hollow cylindrical shaft of inner diameter d and outer diameter D has been represented in Fig. 9.14b. In such cases, the maximum shearing stress does also develop on the circumferential points of the shaft cross section. We may apply therefore the formula (9.14): m ax

M

t

W

p

,

where the polar strength modulus Wp is given by: 172

Torsion

a.

b. Fig. 9.14

D I W

p

I

p

r m ax

D

4

d

32

p

D

4

4

d

1

32

32

D

D

D

D

2

2

2

3

1

16

d D

4

D

4

3

1

m

4

,

(9.21)

16

where m expresses the ratio d / D. Important remarks Since the shearing stress cannot take place in one plane only (section 4.3), in case of a circular shaft in torsion, a shearing stress must be also exerted on other plane, perpendicular to the first one. This is why the shearing stresses over the shaft cross sections in torsion are always accompanied by equal shearing stresses directed along the shaft (Fig. 9.15). These longitudinal shearing stresses explain the specific longitudinal failure of members in torsion, made of different materials, whose strength is much smaller about longitudinal direction than about normal direction (normal to the axis of the member). This is for example the case of rods made of wood, whose failure takes place longitudinally, along the fibres of the material (Fig. 9.16). Consider now an element a located on the surface of a circular shaft subjected to torsion (Fig. 9.17). As presented above, on the faces of such an element, the only stresses which develop are the shearing stresses max = . 173


Fig. 9.15

Fig. 9.16

The element is said to be in pure shear. Another similar element b, located on the same surface but rotated with 45 is subjected to a tensile stress ( 1 = ) over two of its faces and to a compressive stress ( 2 = - ) on the other two faces (see the previous chapter). In general, the ductile materials fail in shear.

Fig.9.17

Therefore, when subjected to torsion, a member made of a ductile material breaks along a plane perpendicular to its longitudinal axis (Fig. 9.18a). On the other hand, brittle materials are weaker in tension than in shear. Thus, when subjected to torsion

a.

Fig. 9.19

Fig. 9.18

a member made of a brittle material tends to break along surfaces which are b. perpendicular to the direction about which tension is maximum ( 1–Fig. 9.18b), i.e. along surfaces forming a 45 angle with the axis of the member. If a circular shaft subjected to c. torsion presents an abrupt change in the diameter of its cross section, stress concentrations occur within the immediate vicinity of the discontinuity (Fig. 9.19). 174

Torsion

In such cases, the shearing stresses corresponding to the discontinuity area may be NOT computed with the above presented formulas. The maximum shearing stress corresponding to the portion of the shaft with diameter d (and for the immediate vicinity of the discontinuity) - Fig. 9.19, is given by: max

k

n

,

(9.22)

where n is the nominal value of the maximum shearing stress for the portion involved (computed with formula (9.14)), and k is a stress –concentration factor. The stress-concentration factor k is usually given in the form of tables or graphs, as shown in Fig. 9.20.

Fig.9.20

9.1.2 STATICALLY INDETERMINATE SHAFTS The statically indeterminate problems of torsion represent that kind of problems in which the internal torques cannot be determined from statics alone (i.e. the equations of static equilibrium are not sufficient for a solution). In such cases, the equilibrium equations must be complemented by relations involving the deformations of the shaft and obtained by considering the geometry of the problem. The following examples will show how to analyze statically indeterminate shafts. 175


Example 1 A shaft AB is attached to fixed supports at both ends and subjected to the torques shown (Fig. 9.21). Knowing that the entire shaft is made of steel for which G = 8 104 MPa and that M0 = 500 N m, d = 30 mm and = 500 mm, determine the maximum value of the shearing stress in the shaft ( max) and the angle of twist of section 1.

Fig.9.21 Due to the action of the two concentrated twisting couples (M0 and 3M0), two reactions develop at the supports (i.e. the torques MA and MB). The single equilibrium equation which may be written from statics is: M

0

t

M A

M 0

3M 0

M B

0

M A

M B

2M 0

.

(9.23)

Since this equation is not sufficient to determine the two unknown torques MA and MB , the problem is statically indeterminate. However, MA and MB may be determined if we note that the total angle of twist of shaft AB must be zero, since both of its ends are restrained. We write therefore: 0

A B

M A GI p

A 1

2

(M A

1 2

M 0)  GI p

1

2 B

(M A

3M 0 ) 

M 0 GI p

2

(9.24)

0,

2

where: I p

d 1

32

4

and

Ip

(2d ) 2

4

32

16

d 32

176

4

16 I p

. 1

Torsion

Thus, (9.24) becomes: 2

M A G I p

M 0) 

(M A

G 16 I p

1

M 0) 

(M A

G 16 I p

1

0,

1

or: M A

2M A

M 0

M A

16

M 0

0

16

which finally gives: 1

M A

34

(9.25)

M 0 .

Substituting this value into the original equilibrium equation, we have: 67

M B

34

(9.26)

M 0.

With these values, we are now in the position to draw the torques diagram. This has been done in Fig. 9.21. For the computation of the maximum shearing stress in the shaft, we have to apply formula (9.14) for each particular portion of the shaft. We write: 1 M t A 1 A 1 m ax

M 0

34

W p A 1

d

1

500

67 2

B m ax

34

Wp 2 B

M t 2 B

M t 1 2

2 , 77

3

30

(9.27)

MPa ;

16

67

M0

(2d )

500

10

3

34

3

60

16

Since

3

34

3

16

Mt 2 B

10

(9.28)

23 , 23 MPa .

3

16

, at the same value of Wp, it follows that

2

B m ax

1 2 m ax ,

and thus it is

not necessary to compute the value of 1-2 max , when we are looking for the maximum value of the shearing stress along the shaft. Comparing (9.27) with (9.28) we conclude that the maximum shearing stress develops at portion 2 - B, and has the value: m ax

2

23 , 23

B m ax

(9.29)

MPa .

The angle of twist of section 1 is: 1 M A 1

A 1

G I

2 p

34 G

2

M 0

1

500

10

3

2 500

34 d

4

8 10

4

30 32

32

177

4

2 , 3 10

3

rad .

(9.30)


Example 2 A shaft AB is attached to fixed supports at both ends and subjected to a uniformly distributed torque m, as shown in Fig. 9.22. Knowing that the entire shaft is made of steel for which G = 8 104 MPa and that a = 40 MPa, d = 40 mm and  = 500 mm, determine: a) The maximum allowable value of torque m so that the shaft would not fail; b) The angle of twist of section 1. a) From statics we write: Mt

0

M A

(9.31)

m .

M B

Since the total angle of twist of shaft AB is zero, we have: A

B

0

A 1

1 B



M A 0

mx

m 

M A

dx

GI p 1

0

,

(9.32)

GI p 2

where: 2d

Ip 1

4

16

32

d

4

16 I

32

p

;

I

2

Fig 9.22 We write therefore: 

M A

mx

G 16 I P 2 0

178

M dx

A

m 

G IP 2

0

d p

2

32

4

.

Torsion 

16

mx d x

M A

m 

M A

0

0

1

M A

16

2

m

M A

2

m

2

0 ,

which finally gives: (9.33)

0 , 97 m  .

M A

Substituting this value into the original equilibrium equation, we obtain: (9.34)

0 , 03 m  .

M B

The torques diagram has been represented in Fig. 9.22. The strength condition is: M t m ax m ax

A 1

Wp

0 , 97 m 

A 1

2d

A 1

40 ,

a 

3

16

which gives: 2d

m

3

16

40

2 40

0 , 97 

16

3

40 0 , 97

8291

N

mm / mm ,

500

and Mt m ax m ax

1

B

1

0 , 03 m 

B

Wp 1 B

d

40 ,

a 

3

16

which gives:

m

d a 

3

1

40

0 , 03 

16

40

3

16

1

33510

N

mm / mm .

0 , 03 500

The allowable value of m is given by the smallest of the two quantities computed above. Therefore, we finally write: m = 8291 N mm / mm .

(9.35)

b) The angle of twist of section 1 is: M B 1

GI p 2

0 , 03 m  G

d



0 , 03 8291

4

32

179

8 10

4

500 40 32

500 4

3 , 09 10

3

rad . (9.36)


9.2 TORSION OF NONCIRCULAR (RECTANGULAR) MEMBERS The formulas presented in the preceding section, for the distribution of strain and stress under a torsional loading apply only to circular members. This happens because the derivation of these formulas has been based upon the assumption that the cross sections of the member in torsion remain plane and undistorted. But this is not the case if the cross sections of the member in torsion are noncircular. For a rectangular member in torsion (for example): the plane and undistorted cross sections of member before deformation do not remain plane and undistorted after deformation, Fig. 9.23. the shearing stresses over the cross sections of a rectangular member in torsion may be not assumed to vary linearly with the distance from the axis of the member. More than that, the shearing stress at the corners of the cross section in such a case is zero. Consider for example a rectangular member in torsion (Fig.9.24). Due to the action of the torque Mt , shearing stresses develop over a current cross section of the Fig. 9.23 member. We detache a small cubic element of the rectangular member in torsion as shown in Fig. 9.24.

Fig. 9.24

We assume that the shearing stress developed on the face of the element considered has an arbitrary orientation within the section. However it may be resolved into two components ( xy and xz) as shown in Fig. 9.24. On the other hand we know that shear 180

Torsion

cannot take place in one plane only; an equal shearing stress must be exerted on another plane perpendicular to the first one. This means that the shearing stress components xy and xz will be accompanied by equal shearing stresses yx and zx respectively acting along the Ox direction. But, since the external surfaces of the member are not loaded, we may write: yx

0 ;

xy

yx

0;

zx

0;

zx

(9.37)

0.

It follows therefore that :

xz

(9.38)

and thus: (9.39)

0.

We conclude that there is no shearing stress at the corners of the cross section of a rectangular member in torsion. The determination of stresses in rectangular members subjected to torsion may be done only through the Theory of Elasticity. Since this determination is not very simple, involving a mathematical reasoning which is beyond the scope of this text, we shall indicate here only the final results concerning the distribution of the shearing stresses over the cross sections of rectangular members in torsion. Consider for example a rectangular member in torsion (Fig. 9.25a).

a.

b. Fig. 9.25

We denote by the length of the member, by b and h, respectively, the narrower and the wider side of the cross section and by Mt the magnitude of the torque applied to the member. One could demonstrate that, the shearing stresses over a current cross section of the member, at points located on the sides and on the axes 181


Oz and Oy, have the distribution shown in Fig. 9.25b. The maximum shearing stress develops at the midpoints of the wider sides of the cross section (points B and B`), being equal to: M m ax

M

t

k hb

2

t

(9.40)

,

Wt

where Wt is called the torsional strength modulus of the cross section involved. We note that: 2

Wt

(9.41)

k hb ,

where k is a coefficient which depends only upon the ratio h / b. The shearing stress developed at the midpoints of the narrower sides of the cross section is given by (Fig. 9.25b): ' max

k2

max

(9.42)

,

where k2 is also a coefficient depending only upon the ratio h / b. The angle of twist per unit length is: M

M

t

G It

t

G k1 h b

3

(9.43)

,

where G is the shear modulus and : It

k1 h b

3

(9.44)

is the torsional moment of inertia, with k1 a coefficient depending upon the ratio h/b. The angle of twist may be therefore expressed as: M

t



(9.45)

,

G It

or, in general, analogous to the formula (9.16): 

0

M

t

(9.45’)

d x.

G It

The coefficients k, k1 and k2 are called Saint-Venant’s coefficients. They are given in table 9.1 for a number of values of the ratio h/b. Table 9.1 Saint-Venant’s coefficients h/b k k1 k2

1,00 0,208 0,141 1,000

1,50 0,231 0,196 0,859

1,75 0,239 0,214 0,820

2,00 0,246 0,229 0,795

2,50 0,258 0,249 0,766

3 0,267 0,263 0,753

182

4 0,282 0,281 0,745

6 0,299 0,299 0,743

8 0,307 0,307 0,742

10 0,313 0,313 0,742

0,333 0,333 0,742

Torsion

From table 9.1 we note that, for higher values of the ratio h/b, coefficients k and k1 approach 1 . With such values, formula (9.40) becomes: 3

M m ax

M

t

Wt

M

t

k hb

2

1

3M

t

hb

2

hb

t 2

(9.46)

.

3

Thus, for a thin-walled member of uniform thickness and arbitrary shape, the maximum shearing stress is the same as for a rectangular member with a very large value of h / b and may be determined from (9.46) - Fig. 9.26.

a.

b. Fig. 9.26

Generally speaking, the maximum shearing stress and the angle of twist per unit length in case of noncircular members in torsion (within the elastic range), respectively, may be expressed as: M m ax

t

;

Wt M

t

GI

t

,

where Wt and It are given in form of tables, for different kinds of sections. The strength condition is therefore: M m ax

where

a 

t

Wt

a  ,

(9.47)

is the allowable value of the shearing stress for the material involved. 183


Numerical example A torque M0 is applied, as shown in Fig. 9.27, to a solid steel shaft with a built –in end. The shaft has two distinct portions: a rectangular portion (with sides h and b) of length 1 and a circular portion (with diameter d) of length 2. Knowing that: b

10 mm , h

15 mm , d

25 mm , G

8 10

4

MPa ,  1

1m,  2

0 ,5 m ,

a 

50 MPa ,

determine: a) the maximum allowable value of torque M0 which may be safely applied to the shaft; b) the angle of twist of section 1, 1.

Fig. 9.27 (a) As shown in Fig. 9.27, the torque Mt is constant along the shaft. Since the shaft has two distinct portions, we must compute the maximum allowable torque which may be safely applied to each portion separately ( M

' t

and

''

M t

). Finally, the required value of the torque will be: M t

'

M 0

"

(9.48)

min( M t , M t )

The rectangular portion (1-2): m ax

M t

'

M t

k hb

2 a 

M t

M t

Wt

k hb

0 , 231

15 10

The circular portion (2-B):

184

2

2

50

a 

17325

N

mm

17 , 325

N

m.

Torsion Mt m ax

Mt

Wp

d

16 M t

3

d

a 

3

16 3

d

"

Mt

Mt

25 a 

16

3

50

153398

N

mm

153 , 398 N

m.

16

Thus, the allowable torque which may be safely applied to the entire shaft is: M t

M

'

"

min( M t , M t )

0

17 , 325 N

m.

(b) The angle of twist of section 1 is:

1

1

2

2

M t  1 2 1

M t  2 2 B

G It

G Ip

B

M t  1 2 1 G k1 h b

M t  2 2 B

3

G

d

4

32 17 , 325 8 10

4

10

3

0 ,196

1000 15 10

17 , 325

10

3

3

8 10

4

500 25

4

0 , 076 rad .

32

9.3 THIN-WALLED HOLLOW SHAFTS (TUBES) As discussed in the preceding section, the torsion of noncircular shafts requires advanced mathematical methods. However, a simple approximate solution is possible for the special case of thin-walled hollow shafts. Let us now consider for example a hollow member of noncircular section subjected to torsion (Fig. 9.28a). The computation of the shearing stresses equivalent to the applied torque Mt , requires the following assumptions:

a.

b. Fig. 9.28 185


the thickness t of the wall is small compared to the other dimensions of the member in torsion; the thickness t of the wall may vary within a cross section of the member, but remains constant along the axis of the member; the shape of the member cross section is arbitrary; since the thickness of the wall is small compared to the other dimensions of the member in torsion, we may admit that the shearing stresses are uniformly distributed over the wall thickness, but with a certain variation law along the contour of the cross section; since the inner and outer walls of the hollow shaft are free surfaces, the stresses on them are equal to zero. Thus the shearing stress at any point of a cross section of the tube is directed along the tangent to the center line – Fig. 9.28a. This reasoning is similarly to that presented in Fig. 9.6. For the determination of the shearing stress variation law along the center line of the member cross section, we shall detach from the member an element of the wall (121’2’aba’b’ - Fig. 9.28b), bounded by two transverse planes at a distance dx from each other and by two longitudinal planes perpendicular to the wall, at distance ds from each other (measured along the center line of the wall). The shearing stress 1 across the thickness t1 induces a numerically equal longitudinal stress, as discussed within section (4.3). In the same manner, across thickness t2, the corresponding shearing stress 2 is accompanied by a numerically equal longitudinal stress directed along the x axis, (Fig. 9.28b). Since the element considered is in mechanical equilibrium, we may write that the summation of all forces acting along x axis is zero. This means that: 1 t1

dx

2

t2 dx

0

1 t1

2

t2

t

constant.

(9.49)

Since the element represented in Fig. 9.28b has been chosen arbitrarily, the above relation tells us that the product t of the shearing stress at a given level (point) of the cross section and of the wall thickness t at that level is a constant throughout the member. The term t is called the shear flow. In other words, at points where the wall thickness has a minimum value, the corresponding shearing stress is maximum and vice versa. To relate the shear flow to the applied torque Mt , we consider a small element of the wall section, of length ds (Fig. 9.29). Since this small element has an infinitesimal length ds, we may consider that the wall thickness remains approximately constant along the element (and equal to t).

186

Torsion

Fig. 9.29

The elementary force which develops at the level of this element is expressed through the product of the shearing stress at this level and the area (tds) of the element (Fig. 9.29). We write: dF

(9.50)

t ds .

The moment of this elementary force about an arbitrary point O within the cavity of the member is: dM

dF

h

(9.51)

t h ds

Since the externally applied torque Mt represents the sum of all such elementary moments around the wall section, we may write: M

t

dM

dF

h

(9.52)

t h ds

The shear flow t being a constant, we have: M

t

t

h ds

t 2

(9.53)

,

where is the area bounded by the center line of the wall cross section (Fig. 9.30). It follows that: M 2

Fig. 9.30

t

,

(9.54)

t

which is called the Bredt’s first formula.

In (9.54) t is the wall thickness at a given point of the member cross section and the area bounded by the center line (Fig. 9.30). The maximum shearing stress max develops at points where t has a minimum value. We write therefore: 187


M m ax

2

M

t

t m in

t

(9.55)

,

Wt

where Wt is the torsional strength modulus for the involved member in torsion. The strength condition becomes: M m ax

where

2

M

t

t m in

t

Wt

a 

(9.56)

,

is the allowable value of the shearing stress for the material involved. The angle of twist of a thin-walled hollow shaft may be obtained by using the method of energy: the work done by the external torque M t will be equal to the elastic strain energy accumulated by the member in torsion. Consider for example an element of length dx detached from the hollow member of noncircular section represented in Fig. 9.28 (Fig. 9.31a). a

a.

b. Fig. 9.31

We denote by V the volume of this element of length dx. The elementary work done by the statically applied torque Mt is: 1

dL

2

M

t

d

,

(9.57)

where d is the elementary angle of twist (i.e. the angle of twist corresponding to the elementary shaft of Fig. 9.31a). On the other hand this work is equal to the elastic strain energy accumulated while the torque Mt is applied. As shown within the preceding chapter, the elastic strain energy density in case of pure shear is: 2

U

D

. 2G

188

Torsion

Thus, the elastic strain energy accumulated by the entire element of Fig. 9.31a is: 2

U

U

dV

D

(9.58)

dV ,

V

V

2G

where G is the shear modulus and dV is the elementary volume of the element shown in Fig. 9.31a. This elementary volume has been represented in Fig. 9.31b separately. Since this elementary volume has an infinitesimal length ds, we may consider that the wall thickness at its level is approximately constant and equal to t (Fig. 9.31b). It follows that: (9.59) dV t ds dx . We write therefore : dL

1 2

2

M td

U

dV V

2

1

2G

M td

2

dV 2G

V

2

M td

2

1 2G

t dx ds

M

1

t

2

t

M td

2

2

1

2

1

2 t

M

t dx ds

2G

2

4

1 t

2

t dx ds 2G

t dx ds

2G

M 8G

2 t dx 2

1

d s.

t

We have: 1 2

M

M td

2 t dx

1

2

8G

(9.60)

ds .

t

The angle of twist per unit length is therefore: d

1

M t

dx

2

4G

M

ds

t

M

t 2

4G 1

t

,

(9.61)

G It

ds

t

where: It

2

4

(9.62)

.

ds t

Within the above calculus, Bredt’s first formula has been applied. Relation (9.61) is called the Bredt’s second formula. If the hollow shaft has a constant thickness t, it follows that: It

2

4

2

4

ds

1

t

t

s being the length of the center line. 189

s

2

4 s

t

,

(9.63)


Numerical example How many times does the hollow shaft strength of Fig. 9.32,a decrease if a longitudinal split is done (Fig. 9.32b) and a = 6 t ? While both shafts are subjected to the same torque Mt , their strength (given by Wt) is completely different.

a.

b. Fig. 9.32

For the hollow shaft represented in Fig. 9.32a we write : M t m ax

1

Wt

M t 2

1

M t 2

t

2a t

M t 2 Gt

M t 2

t

72 t

3

.

In the second case (Fig. 9.32b) we have in fact a thin-walled member of uniform thickness whose maximum shearing stress is the same as for a rectangular bar with a very large value of the ratio h/b. Therefore, we have to use formula (9.46). We write: M t m ax

2

Wt

M t 1 hb

2

3

M t 2

3M t

1 4a

t

2

4 at

2

3M t 4G t t

M t 2

8t

3

Thus, Wt Wt

1 2

72 t 8t

3

3

9.

In other words the strength does decrease 9 times if a longitudinal split is done in the shaft.

190

3

.

Torsion

PROBLEMS TO BE ASSIGNED P.9 P.9.1 A shaft ABC is fixed at one end and subjected to a torque Mt = M0 at the other end as shown (Fig. P.9.1). Portion AB, having a square cross section, is made of steel while portion BC, with a circular cross section, is made of aluminum. Knowing that a = =40 mm, d = 70 mm , the allowable shearing stress of steel all s = 100 MPa, the allowable shearing stress of aluminum all Al = 70 MPa, determine the largest torque M0 which may be applied at A. Neglect the effect of stress concentration.

Fig. P.9.1 P.9.2 A steel shaft ABC is fixed at one end and subjected to the torques shown. Knowing that  = =0,5 m, all = 90 MPa, M0 = 1,5 kN m and G = 8 104 MPa, draw the torque diagram, determine the required value of the diameter d and the angle of twist at A.

Fig. P.9.2

Fig. P.9.3

P.9.3 A solid tapered shaft AB, made of steel, is subjected to the torques shown. Knowing that  = 0,6 m, all = 100 MPa, d = 30 mm, G = 8 104 MPa: (a) draw the torque diagram; (b) determine the largest value of M0 which may be safely applied; (c) determine the angle of twist at A. 191


P.9.4 A shaft ABC having two distinct portions of circular cross section, is fixed at its both ends and loaded as shown. Determine the maximum value of torque M0 which may be safely applied at B if d = 30 mm and all = 90 MPa. Do also determine the angle of twist at B if  = 0,8 m and G = 8 10 4 MPa. P.9.5 Two solid shafts made of the same material, are subjected to the same torque M0 as shown (Fig. P.9.5). Determine the ratio of d and a so that the maximum shearing stresses in both shafts would be of the same value. P.9.6 A brass shaft, having two distinct portions (of annular and circular cross sections respectively), is subjected to a uniformly distributed twisting couple m and a torque M0 = 5ml as shown (Fig. P9.6). Knowing that d = 40 mm, all = 80 MPa, = 0,6 m, Fig. P.9.4 G = 3,9 104 MPa: (a) Determine the torques exerted on the shaft by each of the fixed supports; (b) Draw the torque diagram; (c) Determine the largest value of m which may be applied without exceeding the allowable shearing stress all ; (d) Determine the angle of twist between 1 and 2.

Fig. P.9.5

Fig. P.9.6

P.9.7 Two tapered shafts AB and BC are bonded together at B and attached to fixed supports at A and C (Fig. P.9.7). Shaft AB is made of aluminum (with all Al = 75 MPa and GAl = 2,6 104 MPa) while shaft BC is made of brass (with

all

B

=90 MPa and GB = 3,9 104 MPa). Knowing that  =

0,8m and M0 = 2 kN m: (a) Determine the torques exerted on the composite shaft by each of the fixed supports; (b) Draw the torque diagram; 192

Torsion

(c) Determine the required value of d; (d) Determine the angle of twist at B.

Fig. P.9.7

Fig. P.9.8

P.9.8 A steel shaft of square cross section and an aluminum tube are fixed at one end and connected to a rigid plate at the other end, as shown (Fig. P.9.8). Determine the maximum value of torque M0 which may be safely applied to the rigid plate and the angle of twist at B if: GAl = 2,6 104 MPa, GSteel = 8 104 MPa, all Al = 90 MPa, all Steel =100 MPa and  = 1,5 m.

Fig. P.9.9

Fig. P.9.10

P.9.9 A steel thin-walled tube of square cross section AB and an aluminum tube BC of annular cross section are fixed at one end and connected to a rigid plate at B as shown (Fig. P.9.9). Determine the largest value of torque M0 which may be safely applied at 1 if: d = 80 mm, a = 40 mm, t = 5 mm, all Steel = 100 MPa, all Al = 80 MPa, Gsteel = 8 104 MPa, GAl = 2,6 104 MPa and  = 0,8 m. Do also determine the angle of twist at B. P.9.10 A solid tapered shaft ABC made of steel is fixed at one end and connected to two steel rods as shown (Fig. P.9.10). Knowing that d = 30 mm, all Steel =90 MPa, all Steel =180 MPa, d1 = 10 mm, d2 = 16 mm, ESteel = 2 105 MPa, GSteel = 8 104 MPa and  = 1 m, determine the largest value of torque M0 which may be safely applied at B.

193


194

10. BENDING

A member is said to be in bending if at its current cross section a bending moment develops. A member in bending is called beam. The bending may be classified using different criteria: a) After the forces position in space, we have: plane bending: the forces acting on the beam are located within the same plane, this plane containing the beam axis and one of the principal axes of inertia of the beam cross sections (Fig. 10.1). unsymmetric bending: the forces acting on the beam are located within the same plane, this plane contains the beam axis but contains neither of the principal axes of inertia of the beam cross sections (Fig. 10.2).

Fig. 10.1

Fig. 10.2

general bending: the forces acting on the beam are not located within the same plane, but each force does intersect the axis (Ox) of the beam (Fig. 10.3). b) After the types of internal forces developing at the current cross section we have: pure bending: the internal forces reduce to a bending moment (of a constant value, orientation and sense along the beam), the shearing force being zero (Fig. 10.4a and Fig. 10.4b – between points 1 Fig. 10.3 and 2).


simple bending: the bending moment Mi developed at the current cross section of the beam is accompanied by the shearing force T (Fig. 10.4b, between points A-1 and B-2).

a.

b. Fig. 10.4

10.1 PRISMATIC MEMBERS IN PURE BENDING As discussed above, an example of a member in pure bending is furnished by the portion 1-2 of the beam shown in Fig. 10.4b. In deriving the pure bending formulas, we make the following assumptions: the beam material is continuous, homogeneous, isotropic and elastic; material modulus of elasticity in tension is equal to that in compression; the beam cross section is constant along the member involved; stresses do not exceed the proportional limit (i.e. Hooke’s law is available); the beam is in pure bending; Bernoulli’s hypothesis is also valid. We shall now detach a portion of infinitely small length dx, located between points 1 and 2, from the beam shown in Fig. 10.4b (Fig. 10.5). This portion is bounded by the planes (I and II) perpendicular to the axis of the beam. Due to the action of the external loads the portion considered will bend, but will remain symmetric with respect to the plane containing the loads applied to the beam. Moreover, since the bending moment Miz is the same at any cross section, the portion considered will bend uniformly (Fig. 10.5b). Thus the line C’’D’’ (for example) along which the upper face of the member intersects the plane of the external forces P will have a constant curvature. In other words, the line C’’D’’, which was originally a straight line, will be transformed into a circle of center O1 and so will the lines CD, C’D’ etc. 196

Bending

On the other hand, we note that, due to the action of the external loads, the longitudinal fibers of the portion considered will deform: the fibres of the upper part of the portion considered decrease in length while the fibers of the lower part increase in length. It follows that there must exist a surface parallel to the upper and the lower faces of the beam whose fibers do neither increase nor decrease in length. This surface is called the neutral surface. The neutral surface intersects the plane of forces P along an arc of circle CD (Fig. 10.5b) which is called the neutral axis of the beam.

a. (Scheme of Fig. 10.4b)

b. Fig. 10.5

Since the longitudinal fibres of the member in pure bending do modify their length along the longitudinal direction, we conclude that normal stresses x develop at any cross section of such a member. Using the theory of elasticity or certain experimental methods, one may verify that the normal component x is the only nonzero stress component exerted on the current cross-sectional points of the beam. Let us now return for a while to the bended portion of the beam represented in Fig. 10.5b. We denote by d the angle made by the two sectional planes I and II which bound the beam portion considered and by the radius of curvature corresponding to the neutral axis CD. We shall now consider a fibre C’D’ located at a distance y bellow the neutral surface. As discussed above, the neutral fibre CD will not modify its length in bending. The length of fibre CD will therefore remain equal to dx. We write (10.1) dx d , On the other hand, the arbitrary longitudinal fibre C’D’ considered will change its original length dx with quantity dx . We have dx

dx

y d

197

,

(10.2)


It follows that d

(d x )

dx

yd

d

,

yd

or (10.3)

.

The strain of fibre C’D’ considered becomes therefore x

 x

dx



yd

dx

y

.

(10.4)

d

We conclude that the longitudinal normal strain varies linearly, throughout the member, with the distance y from the neutral surface. Moreover, since Hooke’s law for uniaxial stress applies, we have: x

E

E

x

y

x

,

(10.5)

where E is the modulus of elasticity of the material involved. Formula (10.5) shows that, in the elastic range, the normal stress varies linearly with the distance from the neutral surface (Fig. 10.6). Recalling the relations between the internal forces and stresses, applied to the beam shown in Fig. 10.6, we write: N

d A

E y

0

A

Fig. 10.6 d A

E

0

yd A

A

E

0

Sz

0

A

Sz

0

(10.6)

.

(This means that Oz is a centroidal axis.) M

zd A

iy

Ey

0

A

zd A

0

E

A

I zy

y

zd A

0

E

I zy

0

A

0

.

(10.7)

(This means that Oz and Oy are the principal axis of inertia of the beam cross section)

M

Ey

yd A

iz A

yd A

M

A

E

y

iz A

198

2

d A

M

E iz

Iz

M

iz

Bending

1

M

iz

EI

z

(10.8)

Formula (10.8) (usually called Euler-Bernoulli's formula) gives us the expression of the curvature of the beam neutral surface as a function of the applied bending moment Miz, modulus of elasticity E and the second moment of the beam cross sectional area A with respect to the axis about which the bending moment is applied (Iz). The curvature is defined as the reciprocal of the radius of curvature . It measures the deformation of the member caused by the bending moment Miz. Substituting

1

for

M

iz

EI

z

into (10.5), we write: E

x

y

1

E

y

M EI

iz z

M

iz

We have obtained therefore the expression of the normal stress from the neutral axis (Oz) as: M

iz

y

,

Iz

y

.

Iz x

at any distance y

(10.9)

called Navier’s formula or elastic flexure formula. We note that the stress is compressive above the neutral axis (y < 0) and tensile below the neutral axis (y > 0). This happens only if the applied bending moment is positive (Fig. 10.7a). If the bending moment at the cross section considered is negative then we shall have compression bellow and tension above the neutral axis (Fig. 10.7b). From formula (10.9) it follows that the maximum normal stress develops at cross sectional points for which the distance y is maximum.

a.

b. Fig. 10.7 199


We write M m ax

iz

M

y m ax

Iz

iz

Iz

M

iz

W

z

,

(10.10)

y m ax

where

W

Iz z

is called the elastic cross section modulus with respect to the Oz

y m ax

axis. The strength condition becomes therefore M m ax

where

i m ax

W

a 

,

(10.11)

z

is the allowable value of the normal stress for the material involved. The elastic cross section modulus depends upon the shape and dimensions of the cross - section: a 

rectangular cross section (Fig. 10.8):

bh W

Iz z

y m ax

3 2

bh

12 h

(10.12)

.

6

2

Fig. 10.8

circular cross – section (Fig.10.9):

d W

Iz z

y m ax

64 d 2

Fig. 10.9

200

4

d 32

3

.

(10.13)

Bending

annular cross-section (Fig. 10.10): D W

Iz z

4

d

64

4

64 D

y m ax

2

D

d

1

64

D D

d

D

3

1

c

4

,

32

2

Fig. 10.10

where the ratio

(10.14)

4

4

has been denoted by c.

D

IMPORTANT REMARK The relatively small number of engineering application where pure bending is encountered does not justify a complex study in this matter. Fortunately the results obtained for pure bending may be also applied to simple bending or to other types of loading as well. An example is presented below. Numerical example A steel beam of the cross-section shown, is subjected to several external loads (Fig. 10.11). Knowing that a  150 MPa : a) Draw the shear and bending moment diagrams;

Fig. 10.11 201


b) Determine the values of Iz and Wz ; c) Determine the minimum allowable value of dimension b so that the beam would not fail due to the action of the external loads. a) For the shear and bending moment diagrams we write M

0

10

21 kN

.

A

YB

M

0

B

YA

2

20

10 10

30

YA 8

2

8 4 4

20

30

6

20

YB

8 4 4

8

0

20

;

0

;

.

51 kN

With these values, the shear and bending moment diagrams have been sketched in Fig. 10.11. b) The second moment of the beam cross - sectional area is: 4

b b

Iz

2b

3

4

12

0 , 666

b

4

.

64

It follows that Iz

Wz

0 , 666

y max

b

4

0 , 666

b

3

.

b

c) The maximum normal stress is M max

89 , 56 10

i max

Wz

0 , 666

b

6 3

all

150 MPa

.

Thus 89 , 56 10

b

6

3

0 , 666

97 mm

150

.

It is important to note that formula (10.9) has been derived for a member with a plane of symmetry and a uniform cross section. As presented in chapter 6 or 9, for beams heaving a variable cross section, the corresponding normal stresses will not agree with Navier’s formula. In such cases a stress concentration occurs. For example higher stresses develop if the beam cross-section presents a sudden change (Fig. 10.12). The ratio between the maximum actual normal stress ( ' ) and the maximum stress ( max ) computed with formula (10.10) is called the stress concentration factor: max

max

(10.15)

k max

202

Bending

The stress concentration factor is given in form of tables or graphs for different particular cases. The actual value of the maximum stress at the critical cross section may be then expressed as m ax

where

m ax

k

m ax

,

Fig. 10.12

is computed with Navier’s formula.

10.2 SHEARING STRESSES IN BEAMS SUBJECTED TO SIMPLE BENDING Let us consider a beam with a vertical plane of symmetry, in simple bending. This means that, at any cross section, the bending moment (Miz or Miy) is accompanied by the corresponding shearing force (Ty or Tz respectively) - Fig. 10.13. Since a shearing force develops at any cross section of the beam, besides the normal stresses x , shearing stresses do also develop at any elementary area dA of the cross section. It may be shown that, excepting some particular points of the cross section, the direction of (with components xy and xz ) cannot be determined using the strength of materials methods only.

a.

b. Fig. 10.13 203


For example, let us detache the element abcda'b'c'd' from the beam represented in Fig. 10.13b, Fig. 10.14. We assume that, at an arbitrary point B located on the circumference of the beam cross section, the shearing stress has an arbitrary direction. It may be resolved however into two components 1 and as shown in the figure. But we know that shear cannot take Fig. 10.14 place in one plane only, an equal shearing stress occuring on another plane perpendicular to the first one. This means that 1 must be accompanied by 1 ' which is perpendicular to 1 and contained within the surface aba'b' . Since the face aba'b' of the element considered is a part of the free surface of the beam, all stresses on this face must be zero. Thus, 2

' 1

0

(10.16)

.

It follows that 1

1

0.

(10.17)

Since the stress component equals zero, we conclude that the shearing stress at point B acts along the tangent at the cross section circumference. This conclusion remains valid for any particular point of the cross section circumference. Let us now to return to the current cross section of the beam represented in Fig. 10.13a. As mentioned above, at the level of this cross section the internal forces are represented by the bending moment Miz and the shearing force Ty. Since the effect of Miz (which gives the normal stresses ) has been discussed previously (Navier's formula) we shall now concentrate our attention on the shearing stresses. We shall try to investigate the shearing stresses occuring at points located on an arbitrary segment mn, parallel to Oz axis, at distance y from the neutral surface (i.e. from Oz axis) (Fig. 10.15). As demonstrated above, at points m and n the shearing stress is tangent to the cross section circumference. On the other hand, due to the symmetry of the cross section with respect to the plane Oxy, the tangents at points m and n will intersect the Oy axis at the Fig. 10.15 same point O1 (Fig. 10.15). Within such a 1

x

204

Bending

context, two basic assumptions (usually called Juravski's assumptions) have to be considered: for any other point m1 of segment mn, the direction of passes through point O1; the shearing stress remains constant for all points located on the segment mn. Its value depends only upon the distance y from the Oz axis (i.e. from the neutral surface). From Fig. 10.15 we have xy

tg

(m1 )

xz 1

(m1 )

xy

At point C, for which

1

0

xz

(m1 )

1

xy

xy

( m 1 ) tg

1

.

(10.18)

, we write xz

C

xy

C

tg

C

0

0

.

(10.19)

In other words, the shearing stress component xz at any particular point of segment mn may be expressed as a function of The problem is, xy . therefore, to find out the mathematical connection between the shearing stress component xy and the shearing force Ty over the cross section considered. This will be done bellow. We shall now detache from the beam of Fig. 10.13 a portion of infinitely small length dx, bounded by two planes perpendicular to the axis of the beam (sections S1 and S2) (Fig. 10.16a). We do also cut this portion with a longitudinal plane passing through an arbitrary segment mn (analogous to that of Fig. 10.15) of the beam cross section. This plane

a.

b. Fig. 10.16 205


(mnm'n') will be, therefore, parallel to the beam neutral surface (Fig. 10.16b). We shall now retain only that part of the beam located bellow the sectional plane mnm'n' (Fig. 10.16b). We denote by A1 the area of surface mnq (this area remaining constant along Ox axis) and by b the width of the beam of distance y from the neutral axis Oz. As represented in Fig. 10.16b, at the level of the cross section S1, M i M iz and Ty develop, while, at the level of the cross section S2, M i d M i and T y develop as cross sectional internal forces. The retained element mnqm'n'q' is subjected to: the cross sectional shearing stresses xy developed at the level of the crosssectional points located on segment mn. As discussed above, these stresses are accompanied by equal longitudinal shearing stresses yx . the normal stresses developed at the level of the cross-sectional surfaces mnq and m'n'q' of area A1. The axial force N1 developed at the level of the entire mnq surface is therefore Navier’s formula N1

M

d A A1

A1

i

M

y1 d A

Iz

i

M

y1 d A

Iz

i

Iz

A1

Sz,

(10.20)

where S is the first moment of A with respect to the neutral axis Oz while Iz is the moment of inertia of the entire cross sectional area of the beam about the neutral axis. Since an increase of the bending moment ( d M i ) occurs at the level of section S , the axial force developed at surface m’n’q’ is therefore z

1

2

N

N1

2

d N1

M

i

dM

i

Sz

.

(10.21)

Iz

We shall now write the equilibrium equation of element mnqm’n’q’ about Ox axis as follows: M

i

dM Iz

i

Sz yx

bd x

M Iz

i

Sz

0

.

We note that yx

is the corresponding axial force given by the average shearing

bd x

stress

yx

over the differential area of width b and length dx.

represents the differential change in bending moment distance dx. dM

i

We write:

206

M

i

within the

Bending

M

i

Sz

dM

Iz

yx

yx

Iz dM

bd x

Sz

bd x

M Iz

Sz

i

dM yx

Iz Ty

yx

Sz

i

d x

i

i

Sz

0

Sz b Iz

.

b Iz

In other words, the shearing stress developed at the level of the beam cross – sectional points located on segment mn, at distance y from the neutral axis Oz, may be expressed through: xy

Ty xy

yx

Sz

(Jurawski’s formula),

(10.22)

b IZ

where Ty: the shearing force at the beam cross-section considered; b: the width of the beam cross-section at the level where the shearing stresses must be derived; Iz: the moment of inertia of the entire cross-section with respect to Oz neutral axis; Sz: the first moment of the area located either above or bellow the level where the shearing stresses have to be computed.

a.

b. Fig. 10.17

It is important to note that the cross–sectional shearing stresses xy (which depend upon the quantities described in formula (10.22)), are always accompanied by equal longitudinal shearing stresses which tend to shear the beam about a longitudinal plane (Fig. 10.17a). In other words, at each such level a reciprocal sliding tendency occurs (Fig. 10.17b). 207


RECTANGULAR SECTIONS The distribution of shearing stresses ( xy ) in a rectangular section may be sketched using formula (10.22) – Fig. 10.18. We shall compute the shearing stress value, at a current level mn located at distance y1 from the neutral axis and then we shall represent the shearing stress variation along the rectangular section depth as a Fig. 10.18 function of y1. Thus, at the level mn (Fig. 10.18) we write Ty xy

Sz

T

b Iz

T bh

b

z

b Iz

h

b

3

S

1

y1

2

2

h 2

T

y1 b

12

bh

h

2

4

y1

2

12

6T

=

bh

3

2

b

h 3

2

y1

4

2

.

We write therefore 6T xy

bh

h 3

2

y1

4

2

( y1 )

xy

.

(10.23)

This shows that the shearing stresses have a parabolical distribution along the rectangular section depth. The maximum shearing stress develops at the level of the neutral axis and may be found by substituting y1 with zero in (10.23). We have xy

xy max

xy

(0 )

6T bh

3

h

2

4

3

T

3

T

2

bh

2

A

.

This means that, the maximum shearing stress developing at the level of the neutral axis Oz is max

3

T

2

A

,

(10.24)

where A is the rectangular area ( A b h ). In other words the maximum shearing stress is 50 % greater than the average shearing stress. 208

Bending

We do also note that, for

h

y1

, the shearing stress

xy

2

becomes zero.

CIRCULAR SECTIONS The distribution of shearing stresses in a circular section may be derived in a similar manner. We shall compute the value of the shearing stress at a current level mn, at distance y1 from the neutral axis and then we shall represent the shearing stresses variation along the circular section depth as a function of y1 (Fig. 10.19). For convenience we shall first derive the current expression of the first moment of the area A1 located below the current Fig. 10.19 level mn (hachured area of Fig. 10.19) with respect to the neutral axis Oz as a function of distance y1. We write S z ( y1 )

ydA

mn

.

(10.25)

A1

We select the element of area dA in the shape of a thin horizontal strip and thus reduce the computation of the above double integral to integration in a single variable (Fig. 10.19). We write dA

b y dy

2

r

2

2

,

y dy

where r is the radius of the circular section. We have therefore r

S z y1

ydA

r

2y

A1

r

xy

bI

Z

y dy

r

3

r

2

S z y1

2

y

2

r

2

y1

y

2

2

3

Ty Sz

2

y1

2

With this expression of

2

2

r y1

3

r

2

2

y1

3

2

.

, the shearing stress becomes

T bI

Sz Z

T 2

r

2

2 d

2 y1

(d

2r

209

64

)

4

3

3

r

2

y1

2

2

,

2

y dy


which reduces to 4 T xy

3

1

A

y1 r

2 xy

2

y1

,

(10.26)

where A is the area of the entire circular section considered. This shows that the shearing stresses have a parabolic distribution along the circular section depth. The maximum shearing stress does also develop at the level of the neutral axis and may be found by substituting y with zero in (10.26). We have xy

1

xy max

xy

O

4

T

3

A

.

(10.27)

This indicates that the maximum shearing stress in a circular section is 33 % greater r , the shearing stress than the average shearing stress. We do also note that, for y 1 becomes zero (Fig. 10.19). xy

10.3 PREVENTION OF LONGITUDINAL SLIDING IN CASE OF COMPOSITE SECTIONS For beams with a long span or subjected to high loads, a high value of the elastic cross-section modulus is required. In many such cases beams with composite sections are chosen. If a beam, for example, were composed of two or more layers placed on each other, bending would produce the effect shown in Fig. 10.20.

Fig. 10.20

The separate layers 1 and 2 would slide past each other and the total strength of the beam would be the sum of the strengths of the layers. The beam is considerably weaker than a solid beam of equivalent dimensions. To increase the strength of the beam shown in Fig. 10.20, the layers are gripped together by means of several rivets or bolts. These rivets or bolts will prevent the layers from sliding when bended (Fig.10.21). The built-up beam will work like a solid beam of equivalent dimensions, and a considerably more effort is required to make it fail. On the other hand the rivets or bolts which prevent the layers from sliding reciprocally will be sheared by longitudinal forces developed at the layers separation plane. It is important to mention that this horizontal reciprocal sliding tendency does 210

Bending

Fig. 10.21

also occur even if the beam is solid. This may be explained through the longitudinal shearing stresses yx which develop in simple bending (Fig. 10.17).

Numerical example The composite beam shown in Fig. 10.22, is made by discontinuous welding. It is subjected to a uniformly distributed force q 42 kN / m and three concentrated forces P 420 kN . Knowing that the weld throat depth is a 7 mm , the length of weld c 100 mm and the weld allowable shearing stress

a 

80 MPa

:

Fig. 10.22 a) b) c) d)

Draw the shear and bending-moments diagrams; Compute the values of Iz and Wz for the beam section involved; Determine the maximum values of the normal and shearing stresses ( max and max ); Determine the required value of length e at which the discontinuous welds have to be placed so that the composite section would not fail in bending; e) Draw the normal and shearing stresses distribution over the cross section 2left ; f) Compute the values of the principal stresses 1 and 2 at point K of section 2right. 211


Solution a. The shear and bending moments diagrams have been sketched in Fig. 10.22. b. Within the computation of the cross-sectional Iz and Wz we shall neglect the weld area. We write therefore 240

Iz

W

3

20

410

2

20

240

10

2

12

z

800

3

20 , 40 10

8

mm

4

;

12

Iz

20 , 40 10

y m ax

420

8

48 , 57 10

5

mm

3

.

c. The maximum normal stress is M max

735

imax

W

10

6

48 , 57 10

z

151 , 32 MPa

5

.

The maximum shearing stress develops at the level of the neutral surface of the beam (Oz axis) at the cross section where the shear is maximum. From Juravski’s formula we have: T max max

Sz

462

10

3

20

b Iz

240 10

410

20 , 40 10

400 8

10

200

62 , 68 MPa

.

d. The welds are subjected to longitudinal shearing forces, which are a direct consequence of the shearing stresses yx which develop on longitudinal planes at levels mn or m’ n’ – Fig. 10.22. As we already know, the longitudinal shearing stresses Juravski’s formula: T yx

xy

Sz

yx

xy

may be computed with

.

b Iz

For a covering computation we shall chose: T

T max

b

b min

462

20 , 40 10

S

20

z

240

8

;

N

;

10 mm

Iz

3

10

mm

4

;

410 mm

3

.

Two welds (the upper right and left welds for example) have to cover the longitudinal shearing force developed on a rectangular area of dimensions b e . This force is obtained from the shearing stress yx multiplied by this area: F

b e

yx

T

S

z

T

b e

b Iz

Sz

e

.

Iz

On the other hand, this force has to be supported by the two welds of length c. We write T

S

z

e

2a c

2a

Iz

which finally gives: 212

a 

,

(10.28)

Bending e

.

216 mm

e. The normal and shearing stresses distribution at section 2left may be obtained using Navier’s and Juravsky's formulas. The results are given bellow.

Fig. 10.23 At the level of point K (Fig. 10.22) two kinds of stresses develop: A normal stress

x

, perpendicular to the cross section at K;

A shearing stress xy , acting within the cross-sectional plane, and of a constant value for all cross-sectional points located at level m’’n’’ – Fig. 10.22. The principal stresses at K for the cross section 2right are: x

y

1, 2

2

1

2 x

2

4

y

2 xy

,

where .

0

y

We write x

xy

(k )

M

i

Iz (k )

T

735

y

10

6

20 , 40 10 S

z

210

b Iz

10 240

3

410

8

10

240

20 , 40 10

147 , 72 MPa

;

0 , 42 MPa

.

415 8

It follows that 147 , 72 1, 2

2

1

147 , 72

2

4 0 , 42

2

,

2

which reduces to 1 2

0 , 00119

MPa ;

147 , 721 MPa .

10.4 BEAMS OF CONSTANT STRENGTH The above presented matters concerning bending have been limited so far to prismatic beams of constant cross sections. If the applied bending moment is constant 213


along the beam involved, the stresses developed at any cross section will have the same value and distribution (Fig. 10.24). Since the maximum normal stress max usually controls the design of the beam, the strength condition max

a 

(where a  is the allowable value of the normal stress for the material involved) for the beam shown in Fig. 10.24 is met in the same manner at each particular cross section. If the applied bending moment varies along the beam, the strength condition must insure that the stresses in the critical section(s) are at most

Fig. 10.24

equal to the allowable values of the normal and shearing stresses (Fig. 10.25). It follows that, in all other sections, the stresses will be smaller (or much smaller) than the allowable stresses.

Fig. 10.25

A prismatic beam therefore, is almost always overdesigned, and considerable savings of material may be realized by using 214

Bending

nonprismatic beams of variable cross sections. The problem is, therefore, to design such beams for which the elastic cross sectional modulus W z (or W y ) to vary along the beam ( W z = W z (x) or W z = W z (x)), so that at any cross section, to have M W

i

x

z

x

(10.29)

.

a 

It follows that M

W z (x)

x

i

.

(10.30)

a 

If we know the variation law of the bending moment as a function of x ( M i M i ( x ) ) and the stress allowable value of the material involved - a  , we may thus design the geometry of the beam which to lead to an optimum use of the material. A beam designed in this manner is referred to as a beam of constant strength. Let us now return to the beam represented in Fig. 10.26 in a simplified manner. The bending moment diagram shows a linear variation of M i with respect to x: M

i

P

x

.

This beam may be designed as a beam of constant strength if we vary the cross section in a continuous way along the length of the member. Usually, there are two ways to do this in practical

Fig. 10.26

applications: to vary the depth of the beam at constant width or to vary the width of the beam at constant depth. Let us now consider the first case (Fig. 10.26). The strength condition at any cross section of the beam is: M i (x) max

W z (x)

215

a 

.


It follows therefore that P

W z (x)

2

x

b h (x)

,

6

a 

which gives 6P

h(x)

x

b

.

(10.31)

a 

Formula (10.31) shows us a parabolic variation of the beam cross section depth as a function of x. The geometry of such a beam has been represented in Fig. 10.26. On the other hand, it is important to note that, for cross sections located in the vicinity of the external force P application point, the cross sectional area decreases too much and does not meet the shearing stresses strength condition requirement (

max

3T

3P

2A

2A

a 

, A: the cross-sectional area;

: the allowable value of the

a 

shearing stresses). This is why we have to adopt a constant cross section which to insure the shearing stresses strength condition for a certain length x of the beam (Fig. 10.26). We denote by y the depth of the beam throughout this portion. We write therefore 0

0

max

3 T

3

P

2 A

2 b

y0

a 

,

which gives 3P

y0

2b

h x0 a 

6P

x0

b

a 

.

This means that 9P 4b

2

2

6P 2 a 

b

x0 a 

x0

9P 4b

2

2

b 2 a 

3 P

a 

6P

8 b

a  2 a 

.

In conclusion, the beam of constant strength (with a variable depth and a constant width) must have a constant cross section for a length

x0

3 P 8 b

a  2 a 

,

following than a parabolic variation of the depth, as shown in Fig. 10.26. The beam of constant strength designed in this manner provides savings of material of aprox. 30 %, with respect to a prismatic beam (Fig. 10.27). The volume of the beam of constant strength is (Fig. 10.27) Fig. 10.27 216

Bending

2

V

.

abc

3

In the second case we may keep the depth constant and vary the width of the

beam (Fig. 10.28). We shall follow the same reasoning. The only difference compared with the preceding case consists in the variation of the width instead of that of the depth. The strength condition at any cross section of the beam is:

Fig. 10.28

max

M

i

x

W

z

x

a 

.

It follows therefore that P

W z (x)

x

b x

h

2

,

6

a 

which gives 6P

b x

h

x

2

.

(10.32)

a 

Formula (10.32) shows us a linear variation of the beam cross section width as a function of x. The geometry of such a beam has been represented in Fig. 10.28. On the other hand, as specified in the preceding case, the shearing stresses strength condition requirement must also be met. We write therefore (Fig. 10.28): max

3 T

3

P

2 A

2 h b1

P

6P

a 

which gives b1

3 2 h

a 

217

h

x1

2 a 

.

,


This means that x1

h

3P 2h

a 

2 a 

6P

h

a 

4

a 

.

In conclusion, the beam of constant strength (with a variable width and a constant depth) must have a constant cross section for a length

x1

h

a 

4

,

a 

following then a linear variation of the width, as shown in Fig. 10.28. The beam of constant strength designed in this manner provides savings of material of aprox. 50 % with respect to a prismatic beam (Fig. 10.29). The volume of the beam of constant strength is (Fig. 10.29). V

Fig. 10.29

1

abc

.

2

10.5 UNSYMMETRIC BENDING A beam is said to be under unsymmetric bending if at any cross section the bending moment vector is not directed along a principal centroidal axis of the cross section (Fig. 10.30).

Fig. 10.30

GENERAL CASE OF UNSYMMETRIC BENDING STATE OF STRESS Consider a straight beam of constant cross section subjected to unsymmetric bending. A centroidal rectangular coordinate system zOy is attached to the beam cross sections (Fig. 10.31). Due to the action of the applied bending moment M i (with components M iz and M iy ), normal stresses develop at the level of each particular elementary area d A of the beam cross section. To derive the unsymmetric bending formulas, we make the following assumptions: 218

Bending

stresses do not exceed the proportional limit (i.e. Hooke’s law is available; the beam is in unsymmetric bending; Bernoulli’s hypothesis is also valid. The stresses developed may be obtained by superposing the stresses corresponding to the bending moment components M iz and M iy , as long as the conditions of applicability of the principle of superposition are satisfied. Within the context of the above presented assumptions, we may consider that the normal strain of the beam longitudinal fibres does Fig. 10.31 linearly depends upon the coordinates z and y. We may write therefore: C1 z

x

C2 y

C3

,

(10.33)

where C1’, C2’ and C3’ are constants. Since the Hooke ‘s law conditions are satisfied we have x

E

x

E C1 z

C2 y

C3

(10.34) E C1 z

E C2 y

E C3

C1 z

C2 y

C3

.

where C1, C2 and C3 are also constants. The relationships among internal forces and stresses within the involved beam cross section may be written as follows N

d A

0;

zd A

A

We may now substitute

C1z

M

iy

;

A

C2y

C3

A

into (10.44) and write

C1z

C2y

C3 d A

C1z

C2y

C3 zd A

C1z

C2y

C3 y d A

0;

A

M

iy

A

A

219

yd A

M

iz

.

;

M

iz

.

(10.35)


C1

zd A

C2

A

C1

yd A

z

2

d A

d A

C2

yz d A

C3

A

zy d A

C2

A

0;

A

A

C1

C3

A

zd A

M

iy

;

(10.36)

A

y

2

d A

C3

A

yd A

M

iz

.

A

It follows that C 1 I zy

C2Iz

C 3S z

C1I

y

C 2 I zy

C3S

C1S

y

C2S z

C3A

M

y

iz

;

M

iy

;

(10.37)

0.

Since Oz and Oy are centroidal axes we have Sz

yd A

0;

zd A

0.

A

S

y A

In these conditions, the last relation of (10.37) gives C3

A

0

(10.38)

0.

C3

We may now retain the first two relations of (10.37), with C3 = 0, and (10.34) in a single system of three equations: C 1 I zy

C2Iz

C1I

C 2 I zy

C1z

y

M

iz

M

C2y

; ;

iy

(10.39)

.

This may be considered a system of three equations with two unknowns (C1 and C2). The condition of compatibility in accordance with Rouche’s theorem is therefore I zy

Iz

I

I zy

y

z

M

iz

M

iy

y

220

0

,

(10.40)

Bending

where I zy

Iz

Iy

I zy

Iy

I zy

z

y

0

.

It follows that M

iz

I zy

Iz

z

y

M

iy

I zy

Iz

Iy

I zy

0

,

which finally reduces to: yI

IzI

The condition

0

z I zy

y

2

y

M

I zy

y I zy iz

IzI

z Iz 2

y

I zy

M

iy

,

(10.41)

, leads to the equation of the cross section neutral axis (N.A): I

y

y

I zy

z M

iz

I zy

y

Iz

z M

iy

0

.

(10.42)

10.6 GENERAL BENDING A beam is said to be under general bending if the forces acting on the beam are not located within the same plane, but the support of each force does intersect the Ox axis of the beam (Fig. 10.3). In such cases the beam deforms after a certain curve in space. The neutral axes of different cross sections will not be located within the same plane and there will be NO neutral plane. To resolve such cases the following steps have to be covered: The external forces have to be resolved about two principal planes (Oz and Oy); The bending moment diagrams must be then drawn for each of the two planes (Oz and Oy); The critical sections (where the resultant bending moment has maximum values) must be established; For each critical section the maximum normal stress has to be computed and then compared with the normal stress allowable value of the material involved.

221


PROBLEMS TO BE ASSIGNED P.10 P.10.1 For the beams shown (Fig. P.10.1): (a) Draw the shearing force and bending moment diagrams; (b) Determine the cross-sections centroidal points, Iz and Wz ; (c) Determine the required dimensions of the cross-sections (t=?) if

a.

b.

c.

d. Fig. P.10.1 222

a

= 180 MPa.

Bending

P.10.2 For the beam shown in Fig. P.10.2, determine the largest value of the uniformly distributed force q which may be applied without exceeding either of the following allowable stresses: a = 90 MPa and a = + 30 MPa.

Fig. P.10.2 P.10.3 For the beam shown in Fig. P.10.3: (a) Draw the shear and bending moment diagram; (b) Determine the centroid, Iz and Wz of the cross-section; (c) Determine the allowable uniformly distributed load q if 120 MPa in compression.

a =

+ 40 MPa in tension and

a

=-

Fig. P.10.3 P.10.4 A steel I - shaped beam must support the loading shown (Fig. P.10.4). Knowing that 180 MPa, select the lightest I- shaped profile required.

Fig. P.10.4 P.10.5. For the beam and loading shown (Fig. P.10.5): (a) Draw the shear and bending moment diagrams; (b) Determine the required value of t if a = 200 MPa ; (c) Draw the normal and shearing stresses diagrams at sections Aleft and Bleft. 223

all

=


Fig. P.10.5 P.10.6 A cantilever beam AB has a constant depth h = 20 mm and a variable width b as shown (Fig. P.10.6). Locate the cross section where the normal stress has a maximum value and determine this value.

Fig. P.10.6 P.10.7 Three beams of the same cross section are pin connected at B and C and loaded as shown (Fig. P.10.7). Determine the required value of the uniformly distributed load q knowing that aЀ  = 160 MPa.

Fig. P.10.7

224

Appendix I - TYPICAL MOMENTS OF INERTIA Cross-sectional shape

Computation formulas

Iz

bh

Cross-sectional shape

Computation formulas

3

A

;

hb

3

Iz

;

12

2

a

Iy

Iz

4

;

Iz

2 e

h 2

Iy

A

bh

;Iz

hb

3

Iy

36

a

h

(a

Iz Iy

2

;

b

b )( a

2

h (B

2

2

b ).

2 ab )

;

36 B 2

B (B

h 36 B

2

ab )

d (B

h

d ) (B

(B

2 d )( B

2

2

2 ab )

2 ab ) .

12 B 3

bh

;

12 bh ( b

2

2

c )

;

12 2

h bc

I zy

.

12

Hexagon ; 3

Iz

;

5

Iy

5

4

a ;

16

36 hb

b

3

36

Polygon with n equal sides A

bh

;

I

2 Iz

bh

4

nR

sin

(2

cos

sin

(2

cos

);

24 3

;

I

12

na

4

96

n = 3, I

247

;

b

48

2 bh

4 ab

36

3

48

bh

2

a

;

12

I zy

bh

3

h

b

3 a

3

a ;

Iy

A

h 2a

;e

2

Iy

Iz

b)h

12

Iy

A

(a

(1 a

4

cos 3 / 96

)

) 2

Appendix II - It AND Wt FORMULAS FOR DIFFERENT TYPES OF CROSS-SECTIONAL SHAPES OF MEMBERS IN TORSION No

1


It

R

4



Wt

d

2

32



d

2

 max 3

occurs on circumference

16

 (D  d )

32

16 D

4

a b 3

3

R

3

 (D  d ) 4

2

4

Remarks

2

3

4

 max

occurs on circumference

3

a  b

n b

4

 ab

2

 nb

2



2

3

2

We denote by a

2

n 1

 n ( b o  b1 )

 n ( b o  b1 )

n 1

2 bo

3

4

4

2

4

5

4

4

We denote by ao bo

4



 n 1

b

2

2  t min

2

ds t(s)



a1

 n 1

b1

 - the area bounded by

the center line of the wall  cross section;  max occurs where t is minimum (t = tmin) We admit that

6

2 r t 3

2 r t 2

t  r

 max = const.

2

7

2b h b t1



 max occurs where t is

2

h t2

249

2 bht

min

minimum (t = tmin) tmin = min (t1, t2)

Appendix II (continued) No

It

Wt

Remarks

0,296R4

0,348R3

  0 - at corners


 max - at the diameter midpoint

8

 max - at sides midpoint

4

0,0216a

a

9

3

20


10

0,141a4

0,208a3

0,977a3=0,188b3

11

1,04a4= = 0,108b4

Khb3

K1hb2

  0 at corners



  0 at corners

12 h/ b k k1 k2

1,0

1,5

2,0

3,0

4,0

8,0

10,0

0,141 0,208 1,0

0,196 0,231 0,859

0,229 0,246 0,796

0,263 0,267 0,753

0,281 0,282 0,745

0,298 0,299 0,743

0,307 0,307 0,743

0,31 0,31 0,74

t

1

13

3

3

1

14

ht

1

3

ht

3

h

It

3

 [ t ( s )] ds

t max

0

249

2

 0 h t  h

t max  h

0,33 0,33 0,33

Appendix II (continued) No

It


Wt

Remarks

15 2

 rt

2

3

   hit i

3

It

i

t max

Profile

c1

17

4

 c1

2

r/D c1 c2

18

0 1 0,5



[ 1,12

L 0,99

η R

d

D

4

c2

32

0,05 0,98 0,52

4

4

 0 ,17 dD

3

19

249

T 1,12

R

0,1 0,93 0,52

32

 0 , 01 D

t  h

2

3

3

16

 rt

2

3

 c2

0,2 0,78 0,49

I 1,31 D

┼ 1,17

3

16

0,3 0,59 0,42

0,4 0,5 0,40 0,24 0,33 0,24

APPENDIX III - PROPERTIES OF ROLLED-STEEL SHAPES

ROLLED-STEEL SHAPES

I SHAPED PROFILE STAS 564 – 80 (dimensions and geometrical characteristics) h – Depth b – Width d – Web thickness t – Thickness

Designation

I8 I 10 I 12 I 14 I 16 I 18 I 20 I 22 I 24 I 26 I 28 I 30 I 32 I 36 I 40

Dimensions [mm] H

b

t

d=R

r

80 100 120 140 160 180 200 220 240 260 280 300 320 360 400

42 50 58 66 74 82 90 98 106 113 119 125 131 143 155

5,9 6,8 7,7 8,6 9,5 10,4 11,3 11,92 12,80 13,77 14,85 15,82 16,92 19,05 21,10

3,9 4,5 5,1 5,7 6,3 6,9 7,5 8,1 8,7 9,4 10,1 10,8 11,5 13,0 14,4

2,3 2,7 3,1 3,4 3,8 4,1 4,5 4,9 5,2 5,6 6,1 6,5 6,9 7,8 8,6

Crosssectional area A[cm2] 7,58 10,6 14,2 18,3 22,8 27,9 33,5 39,6 46,1 53,4 61,1 69,1 77,8 97,1 118

R – Flange inner radius r – Flange outer radius Iz, Iy – Axial moments of inertia Wz, Wy – Strength moduli iz, iy – Radii of gyration Sz - First moment of the half-cross-sectional area

Designation Iz[cm4]

Axis z - z Wz[cm3]

77,8 171 328 573 935 1450 2140 3060 4250 5740 7590 9800 12510 19610 29210

19,5 34,2 54,6 81,9 117 161 214 278 354 442 542 653 782 1090 1460 249

iz[cm]

Iy[cm4]

Axis y -y Wy[cm3]

3,2 4,01 4,81 5,61 6,4 7,2 8,0 8,80 9,59 10,4 11,1 11,9 12,7 14,2 15,7

6,29 12,2 21,5 36,2 54,7 81,3 117 162 221 288 364 451 555 818 1160

3,0 4,88 7,41 10,71 14,8 19,8 26,0 33,1 41,7 51,0 61,2 72,2 84,7 114 149

Sz[cm3] iy[cm] 0,91 1,07 1,23 1,40 1,55 1,71 1,87 2,02 2,20 2,32 2,45 2,56 2,67 2,90 3,13

11,4 19,9 31,8 47,7 68,0 93,4 125 162 206 257 316 381 457 638 857

I8 I 10 I 12 I 14 I 16 I 18 I 20 I 22 I 24 I 26 I 28 I 30 I 32 I 36 I 40

249


ROLLED-STEEL SHAPES ANGLES - Equal Legs STAS 424 – 80 (dimensions and static quantities for bending) a – Leg Width g – Leg Thickness C –Centroid e – Distance between the centroid and the outer side z, y – Centroidal Axes ξ, η or 1,2 - principal centroidal axes

Size and Thickness [mm]

Cross-sectional dimensions [mm] a g r r1

20 x 20 x 3 20 x 20 x 4 25 x 25 x 3 25 x 25 x 4 25 x 25 x 5 30 x 30 x 4 30 x 30 x 5 35 x 35 x 4 35 x 35 x 5 40 x 40 x 4 40 x 40 x 5 45 x 45 x 5 45 x 45 x 6

20 20 25 25 25 30 30 35 35 40 40 45 45

3 4 3 4 5 4 5 4 5 4 5 5 6

3,5 3,5 3,5 3,5 3,5 5 5 5 5 6 6 7 7

2 2 2 2 2 2,5 2,5 2,5 2,5 3 3 3,5 3,5

Crosssectional area

[cm2] 1,12 1,45 1,42 1,85 2,26 2,27 2,78 2,67 3,28 3,08 3,79 4,30 5,08

Distance between axes, [mm] e u v1 v2 0,60 0,64 0,72 0,76 0,80 0,88 0,92 1,00 1,04 1,12 1,16 1,28 1,32

1,41 1,41 1,77 1,77 1,77 2,12 2,12 2,47 2,47 2,83 2,83 3,18 3,18

0,84 0,90 1,02 1,08 1,13 1,24 1,30 1,42 1,48 1,64 1,64 1,81 1,87

0,70 0,71 0,87 0,89 0,91 1,05 1,07 1,24 1,25 1,42 1,42 1,58 1,59

z–z Iz=Iy [cm4] 0,39 0,41 0,80 1,01 1,20 1,80 2,16 2,95 3,56 5,43 5,43 7,84 9,16

251

y-y Wz=Wy [cm3] 0,28 0,36 0,45 0,58 0,71 0,85 1,04 1,18 1,45 1,91 1,91 2,43 2,88

r – Legs inner radius r1 – Legs outer radius Iz, Iy – Axial moments of inertia Wz, Wy – Strength moduli Iz, Iy – Radii of gyration Iξ or I1, Iη or I2 – Principal moments of inertia, maximum and minimum Sz - First moment of the half-cross-sectional area

Inertial Geometrical Quantities ξ–ξ 1–1 η–η 2-2 iz=iz Iξ=I1 iξ Iη=I2 Wη [cm] [cm4] [cm] [cm4] [cm3] 0,59 0,61 0,74 0,16 0,19 0,58 0,77 0,73 0,21 0,23 0,75 1,26 0,94 0,33 0,32 0,74 1,60 0,93 0,43 0,40 0,73 1,89 0,91 0,52 0,46 0,89 2,85 1,12 0,75 1,61 0,88 3,41 1,41 0,92 0.70 1,05 4,68 1,33 1,23 0,86 1,04 5,64 1,31 1,49 1,01 1,20 8,60 1,51 2,26 1,37 1,20 8,60 1,51 2,26 1,37 1,35 12,4 1,70 3,25 1,80 1,34 14,5 1,69 3,82 2,04

iη [cm] 0,38 0,38 0,48 0,48 0,48 0,58 0,57 0,68 0,67 0,78 0,77 0,87 0,87

Izy [cm4] 0,25 0,28 0,465 0,585 0,685 1,05 1,245 1,725 2,075 2,62 3,17 4,575 5,34

APPENDIX III (continued) Size and Thickness [mm]

Cross-sectional dimensions [mm] a g r r1

50 x 50 x 5 50 x 50 x 6 50 x 50 x 7 60 x 60 x 5 60 x 60 x 6 60 x 60 x 8 60 x 60 x 10 70 x 70 x 6 70 x 70 x 7 70 x 70 x 8 70 x 70 x 10 80 x 80 x 6 80 x 80 x 8 80 x 80 x 10 90 x 90 x 8 90 x 90 x 9 90 x 90 x 11 100 x 100 x 8 100 x 100 x 10 100 x 100 x 12 120 x 120 x 10 120 x 120 x 12 130 x 130 x 12 130 x 130 x 14 140 x 140 x 12 140 x 140 x 14 150 x 150 x 14 150 x 150 x 16 160 x 160 x 12 160 x 160 x 14

50 50 50 60 60 60 60 70 70 70 70 80 80 80 90 90 90 100 100 100 120 120 130 130 140 140 150 150 160 160

5 6 7 5 6 8 10 6 7 8 10 6 8 10 8 9 11 8 10 12 10 12 12 14 12 14 14 16 12 14

7 7 7 8 8 8 8 9 9 9 9 10 10 10 11 11 11 12 12 12 13 13 14 14 15 15 16 16 17 17

3,5 3,5 3,5 4 4 4 4 4,5 4,5 4,5 4,5 5 5 5 5,5 5,5 5,5 6 6 6 6,5 6,5 7 7 7,5 7,5 8 8 8,5 8,5

Crosssectional area

[cm2] 4,80 5,69 6,56 5,82 6,91 9,63 11,1 8,13 9,40 10,60 13,10 9,35 12,30 15,1 13,9 15,5 18,7 15,5 19,2 22,7 23,2 27,5 30,3 34,7 32,5 37,6 40,3 45,7 37,4 43,3

Distance between axes, [mm] e u v1 v2 1,40 1,45 1,49 1,64 1,69 1,77 1,85 1,93 1,97 2,01 2,09 2,17 2,26 2,34 2,50 2,54 2,62 2,74 2,82 2,90 3,31 3,40 3,64 3,72 3,90 3,98 4,21 4,29 4,39 4,47

3,54 3,54 3,54 4,24 4,24 4,24 4,24 4,95 4,95 4,95 4,95 5,66 5,66 5,66 6,36 6,36 6,36 7,07 7,07 7,07 8,49 8,49 9,19 9,19 9,90 9,90 10,6 10,6 11,3 11,3

1,98 2,01 3,10 2,32 2,39 2,50 2,61 2,73 2,79 2,85 2,96 3,07 3,19 3,30 3,53 3,59 3,70 3,87 3,99 4,11 4,69 4,80 5,15 5,26 5,50 5,61 5,95 6,07 6,19 6,30

1,76 1,77 1,78 2,11 2,11 2,14 2,17 2,46 2,47 2,49 2,52 2,82 2,82 2,85 3,17 3,18 3,21 3,52 3,54 3,57 4,23 4,26 4,60 4,63 5,04 5,07 5,31 5,34 5,74 5,77

z–z Iz=Iy [cm4] 12,80 12,80 14,60 19,4 22,8 29,2 34,9 36,9 42,4 47,5 57,2 55,8 72,2 87,5 104 116 138 145 177 207 313 368 472 540 602 689 845 949 913 1046

251

y-y Wz=Wy [cm3] 3,61 3,61 4,16 4,45 5,29 6,89 8,41 7,27 8,41 9,52 11,70 9,57 12,6 15,4 16,1 18,0 21,6 19,9 24,6 29,1 36,0 42,7 50,4 58,2 59,7 68,8 78,2 88,7 78,6 90,8

Inertial Geometrical Quantities ξ–ξ 1–1 η–η 2-2 iz=iz Iξ=I1 iξ Iη=I2 Wη [cm] [cm4] [cm] [cm4] [cm3] 1,50 17,4 1,90 4,54 2,59 1,50 20,4 1,89 5,33 2,61 1,49 23,1 1,88 6,10 2,91 1,82 30,7 2,30 8,02 3,45 1,82 36,2 2,29 9,43 3,95 1,80 46,2 2,26 12,1 4,86 1,78 55,1 2,23 14,8 5,67 2,13 53,5 2,68 15,2 5,69 2,12 67,1 2,67 17,5 6,27 2,11 75,3 2,66 19,7 6,91 2,09 90,5 2,63 23,9 8,09 2,44 88,5 3,08 23,1 7,55 2,43 115 3,06 29,8 9,36 2,41 139 3,03 36,3 11,0 2,74 166 3,45 43,1 12,2 27,4 184 3,45 47,8 13,3 2,72 218 3,41 57,1 15,4 3,06 230 3,85 59,8 15,4 3,04 280 3,83 72,9 18,3 3,02 328 3,80 85,7 20,9 3,67 497 4,63 129 27,5 3,65 584 4,60 151 31,5 3,97 750 5,00 194 37,7 3,94 857 4,97 223 42,4 4,31 957 5,43 248 44,9 4,30 1094 5,42 284 50,5 4,58 1340 5,77 347 58,3 4,56 1510 5,74 391 64,4 4,94 1450 6,23 376 60,5 4,92 1662 6,20 431 68,1

iη [cm] 0,97 0,97 0,96 1,17 1,17 1,16 1,16 1,37 1,36 1,36 1,35 1,56 1,55 1,55 1,76 1,76 1,75 1,96 1,95 1,94 2,36 2,35 2,54 2,53 2,76 2,74 2,94 2,93 3,17 3,16

Izy [cm4] 6,43 7,535 8,5 11,34 13,385 17,05 20,15 19,15 24,8 27,8 33,3 32,7 42,6 51,35 61,45 68,1 80,45 85,1 103,55 121,15 184 216,5 278 317 354,5 405 496,5 559,5 537 615,5


ROLLED-STEEL SHAPES ANGLES - Unequal Legs STAS 425 - 80 (dimensions and static quantities for bending) a – Big leg size b – Small leg size g – Legs thickness C – Centroid ez, ey – Distance between the centroid and the outer leg side z, y – Centroidal axes ξ, η or 1,2 - Principal centroidal axes

Size and Thickness [mm] 30x20x3 30x20x4 40x20x3 40x20x4 45x30x4 45x30x5 60x30x5 60x30x6 60x40x5 60x40x6 60x40x7 65x50x6 65x50x7 65x50x8

Cross-sectional dimensions [mm] a

b

30

20

40

20

45

30

60

30

60

40

65

50

g 3 4 3 4 4 5 5 6 5 6 7 6 7 8

Crosssectional area A [cm2]

r

r1

3,5

2

3,5

2

4,5

2

6

3

6

3

6,5

3,5

1,43 1,86 1,73 2,26 2,86 3,52 4,29 5,08 4,79 5,68 6,55 6,58 7,60 8,60

r – Legs inner radius r1 – Legs outer radius Iz, Iy – Axial mements of inertia Iξ or I1, Iη or I2 – principal moments of inertia, maximum şi minimum Wz, Wy – Strength moduli iz, iy – Radii of gyration Sz - First moment of the half-cross-sectional area

Inertial Geometrical Quantities

Distance between axes, [mm] tgα ez

ez

u1

u2

v1

v2

v3

0,99 1,03 1,42 1,47 1,48 1,52 2,15 2,20 1,96 2,00 2,04 2,04 2,08 2,11

0,50 0,54 0,44 0,48 0,74 0,78 0,68 0,72 0,97 1,01 1,05 1,29 1,33 1,37

2,05 2,02 2,61 2,58 3,06 3,04 3,89 3,86 4,10 4,08 4,06 4,52 4,50 4,49

1,51 1,52 1,77 1,80 2,23 2,26 2,67 2,69 3,01 3,02 3,03 3,60 3,62 3,64

0,86 0,91 0,79 0,83 1,21 1.27 1,20 1,25 1,68 1,72 1,77 2,15 2,19 2,23

1,04 1,04 1,19 1,17 1,58 1.58 1,77 1,75 2,1 2,10 2,09 2,39 2,39 2,39

0,56 0,58 0,46 0,50 0,80 0,83 0,72 0,74 1,10 1,12 1,14 1,50 1,52 1,54

253

0,427 0,427 0,257 0,252 0,434 0,429 0,256 0,252 0,434 0,431 0,427 0,575 0,572 0,569

Iy [cm4] 1,25 1,59 2,80 3,59 5,77 6,98 15,6 18,2 17,2 20,1 22,9 27,2 31,1 34,8

y-y Wy [cm3] 0,62 0,81 1,09 1,42 1,91 2,35 4,04 4,78 4,25 5,03 5,79 6,1 7,03 7,93

iy [cm] 0,93 1,92 0,27 1,26 1,42 1,41 1,90 1,89 1,89 1,88 1,87 2,03 2,02 2,01

Iz [cm4] 0,44 0,55 0,47 0,60 2,05 2,47 2,60 3,02 6,11 7,12 8,07 14,0 15,9 17,7

z-z Wz [cm3] 0,29 0,38 0,30 0,39 0,91 1,11 1,12 1,32 2,02 2,38 2,74 3,77 4,34 4,89

iz [cm] 0,55 0,55 0,52 0,51 0,85 0,84 0,78 0,.77 1,13 1,12 1,11 1,46 1,45 1,44

- I i [cm4] [cm] 1,43 1,00 1,81 0,99 2,96 1,31 3,80 1,30 6,63 1,52 8,00 1,51 16,5 1,96 19,2 1,95 19,8 2,03 23,1 2,02 26,3 2,00 33,8 2,27 38,5 2,25 43,0 2,23

- I i [cm4] [cm] 0,26 0,42 0,33 0,42 0,31 0,42 0,39 0,42 1,19 0,65 1,45 0,64 1,69 0,63 1,99 0,63 3,54 0,86 4,15 0,86 4,75 0,85 7,43 1,06 8,51 1,06 9,57 1,50

Izy [cm4] 0,43 0,54 0,64 0,81 1,98 2,37 3,56 4,06 5,99 6,91 7,86 11,37 13,9 14,3

APPENDIX III (continued) Size and Thickness [mm] 65x50x9 75x50x7 80x60x7 80x65x6 80x65x8 80x65x10 90x60x6 90x60x8 100x50x8 100x50x10 100x75x7 100x75x9 100x75x11 120x80x8 120x80x10 120x80x12 150x90x10 150x90x12 150x100x10 150x100x12 150x100x14

Cross-sectional dimensions [mm] a

b

75 80

50 60

90

60

100

50

100

75

120

80

150

90

150

100

Crosssectional area A [cm2]

g

r

r1

9 7 7 6 8 10 6 8 8 10 7 9 11 8 10 12 10 12 10 12 14

6,5 6,5 8

3,5 3,5 4

8

4

9

4,5

10

5

11

5,5

12,5

6,5

13

6,5

9,58 8,31 9,38 8,41 11,0 13,6 8,69 11,4 11,4 14,1 11,9 15,1 18,2 15,5 19,1 22,7 23,2 27,5 24,5 28,7 33,2

Inertial Geometrical Quantities

Distance between axes, [mm] tgα ez

ez

u1

u2

v1

v2

v3

2,15 2,48 2,51 2,39 2,47 2,55 2,89 2,97 3,59 3,67 3,06 3,15 3,23 3,83 3,92 4,00 5,00 5,08 4,80 4,89 4,97

1,49 1,25 1,52 1,65 1,73 1,81 1,41 1,49 1,12 1,2 1,83 1,91 1,99 1,87 1,95 2,03 2,04 2,12 2,34 2,42 2,50

4,48 5,10 5,55 5,61 5,59 5,56 6,14 6,11 6,49 6,43 6,96 6,91 6,87 8,23 8,18 8,14 10,1 10,1 10,3 10,2 10,2

3,63 3,77 3,79 4,63 4,65 1,68 4,50 4,54 4,44 4,40 5,42 5,45 5,49 5,99 6,03 6,06 7,05 7,10 7,50 7,53 7,56

2,28 2,13 2,17 2,69 2,79 2,90 2,46 2,56 2,00 2,08 3,10 3,22 3,32 3,27 3,37 3,46 3,60 3,70 4,10 4,19 4,28

2,36 2,63 2,92 2,94 9,94 2,95 3,16 3,15 2,96 2,93 3,61 3,63 3,65 4,23 4,21 4,20 5,03 5,00 5,25 5,24 5,23

1,57 1,38 1,40 2,01 2,05 2,11 1,60 1,69 1,18 1,22 2,18 2,22 2,27 2,16 2,19 2,25 2,24 2,30 2,68 2,73 2,77

253

0,567 0,433 0,546 0,649 0,645 0,640 0,442 0,437 0,257 0,253 0,553 0,549 0,545 0,437 0,435 0,432 0,360 0,358 0,442 0,439 0,435

Iy [cm4] 38,2 46,4 59,0 52,9 68,1 82,2 71,7 92,5 116 141 118 148 176 226 276 323 533 627 552 650 744

y-y Wy [cm3] 8,77 9,24 10,7 9,41 12,3 15,1 11,7 15,4 18,1 22,2 17,0 21,5 25,9 27,6 34,1 40,4 53,3 63,3 54,1 64,2 74,1

iy [cm] 2,00 2,36 2,51 2,51 2,49 2,46 2,87 2,85 3,18 3,16 3,15 3,13 3,11 3,82 3,80 3,77 4,80 4,77 4,78 4,76 4,73

Iz [cm4] 19,4 16,5 28,4 31,2 40,1 48,3 25,8 33,0 19,5 23,4 56,9 71,0 84,0 80,8 98,1 114 146 171 198 232 264

z-z Wz [cm3] 5,39 4,39 6,34 6,44 8,41 10,3 5,61 7,31 5,04 6,17 10,0 12,7 15,3 13,2 16,2 19,1 21,0 24,8 25,8 30,6 35,2

iz [cm] 1,42 1,41 1,74 1,93 1,91 1,89 1,72 1,70 1,31 1,29 2,19 2,17 2,15 2,28 2,26 2,24 2,51 2,49 2,86 2,84 2,82

- I i [cm4] [cm] 47,0 2,25 53,3 2,53 72,00 2,77 68,5 2,85 88,0 2,82 10,6 2,79 82,8 3,09 107 3,06 123 3,28 149 3,25 145 3,49 181 3,47 214 3,44 260 4,10 317 4,07 371 4,04 591 5,05 694 5,02 637 5,13 719 5,10 856 5,07

- I i [cm4] [cm] 10,5 1,05 9,57 1,07 15,4 1,28 15,6 1,35 20,3 1,36 24,8 1,35 14,6 1,30 19,0 1,29 12,7 1,05 15,4 1,05 30,1 1,59 37,8 1,59 45,1 1,58 46,6 1,73 56,8 1,72 66,6 1,71 88,3 1,95 104 1,94 112 2,15 132 2,15 152 2,14

Izy [cm4] 15,7 15,09 23,7 24,2 30,9 36,8 25,2 32,1 26,7 31,6 48,7 60,5 71,3 79,4 96,5 111 161 189 191 227 257


ROLLED-STEEL SHAPES

U SHAPED PROFILE STAS 564 – 80 (dimensions and geometrical characteristics)

h – Depth b – Width t – Thickness

Dimensions [mm] Designation U5 U6,5 U8 U 10 U 12 U 14 U 16 U 18 U 20 U 22 U 24 U 26 U 30

h

b

d

t

50 65 80 100 120 140 160 180 200 220 240 260 300

38 42 45 50 55 60 65 70 75 80 85 90 100

5 5,5 6 6 7 7 7,5 8 8,5 9 9,5 10 10

7 7,28 7,76 8,26 8,72 9,72 10,20 10,68 11,16 12,14 12,62 13,60 15,60

R 7,5

r 3,5 4 4 4,5 4,5 5 5,5 5,5 6 6,5 6,5 7 8

Crosssectional area 2

A[cm ] 7,12 9,03 11,0 13,5 17,0 20,4 24,0 28,0 32,2 37,4 42,3 48,3 58,8

Iz[cm4] 26,4 57,5 106 205 361 605 925 1350 1910 2690 3600 4820 8030 254

R – Flange inner radius r – Flange outer radius Iz, Iy – Axial moments of inertia Wz, Wy – Strength moduli iz, iy – Radii of gyration Sz - First moment of the half-cross-sectional area

Static quantities for bending z-z y -y 3 4 Wz[cm ] iz[cm] Iy[cm ] Wy[cm3] 10,6 17,7 26,5 41,2 60,7 86,4 116 150 191 245 300 371 535

1,92 2,52 3,10 3,91 4,62 5,45 6,21 6,95 7,70 8,48 9,22 9,99 11,7

9,12 14,1 19,4 29,3 43,2 62,7 85,3 114 148 197 248 317 495

3,75 5,07 6,36 8,49 11,1 14,8 18,3 22,4 27,0 33,6 39,6 47,7 67,8

iy[cm] 1,13 1,25 1,33 1,47 1,59 1,75 1,89 2,02 2,14 2,30 2,42 2,56 2,90

Sz[cm3]

ey[cm]

15,9 24,5 36,3 51,4 68,8 89,6 114 146 179 221 316

1,37 1,42 1,45 1,55 1,60 1,75 1,84 1,92 2,01 2,14 2,23 2,36 2,70

STRENGTH OF MATERIALS TYPICAL KEY WORDS AND PHRASES ENGLISH - ROMANIAN

ROMANIAN - ENGLISH

A

A

ABSCISSA – abscisă ALGEBRAIC SUM - însumare algebrică ALLOWABLE LOAD - sarcină capabilă ALLOWABLE STRESS – tensiune admisibilă ALLOWABLE - STRESS METHOD metoda tensiunii admisibile ALLOY - aliaj ALLOYING - aliere ALUMINUM - aluminiu ALUMINUM PIPE – ţeavă de aluminiu ALUMINUM SHELL - manta (cămaşă) de aluminiu AMPLITUDE – amplitudine ANALYSIS – analiză ANALYTICAL FUNCTION – funcţie analitică ANGLE - unghi ANGLE OF TWIST - unghi de torsiune ANGULAR VELOCITY - viteza unghiulară ANNEALED STEEL – oţel recopt ANNULAR CROSS SECTION –secţiune inelară APPENDIX – anexă ARC OF CIRCLE - arc de cerc AVERAGE STRESS - tensiune medie AVERAGE VALUE - valoare medie AXIAL LOAD - forţă axială AXIAL LOADING - solicitare axială AXLE - ax

ABSCISĂ - abscissa ALAMĂ - brass ALIAJ - alloy ALIERE - alloying ALUMINIU - aluminum ALUNECARE - slip ALUNGIRE - elongation AMPLITUDINE - amplitude ANALIZA - analysis ANALOGIE CU MEMBRANA (TORSIUNE) - membrane analogy ANEXA - appendix ANSAMBLU MOMENT – FORŢĂ (TORSOR) - force-couple system APLICAREA FORMULEI LUI EULER PENTRU ALTE CAZURI DE REZEMARE ALE BARELOR extension of Euler’s formula to columns with other end conditions APLICATĂ DIRECT LA VALOAREA NOMINALĂ (DESPRE O FORŢĂ, ETC) - fully applied ARBORE COTIT – crankshaft ARBORE DE TRANSMISIE transmission shaft ARBORE DE TRANSMISIE CILINDRIC MASIV (PLIN) – solid cylindrical transmission shaft ARBORI STATIC NEDETERMINAŢIstatically indeterminate shafts ARC CU SPIRĂ STRÂNSĂ - close coiled spring ARC DE CERC - arc of circle ARC DE TORSIUNE – torsional spring ARC LAMELAR - leaf spring

B BALL SOCKET - cuzinet, locaş sferic BAR –bară


ROMANIAN – ENGLISH

ENGLISH - ROMANIAN BAUSCHINGER EFFECT - efectul Bauschinger BEAM –grindă BEAM OF CONSTANT STRENGTH grindă de egală rezistenţă BEAM UNIT WIDTH - grindă de lăţime unitate BEARING – rulment BEARING FRAME - cadru portant BEARING STRESS - tensiune (presiune) de contact BEARING SURFACE – suprafaţă de contact BENDING – încovoiere BENDING MOMENT - moment de încovoiere BENDING-MOMENT DIAGRAM diagramă de moment încovoietor BIAXIAL STRESS CONDITION - stare de tensiune biaxială BINDING –fretare BOLT – bolţ BOUNDARY CONDITIONS – condiţii de contur (de rezemare) BRASS – alamă BRASS LAYER - strat de alamă BREAKING STRENGTH – rezistenţa la rupere BRITTLE MATERIALS – materiale fragile BRONZE – bronz BUCKLE (TO-) - flamba (a-) BUCKLING – flambaj BUCKLING COEFFICIENT – coeficient de flambaj

ARCE DE CERC CONCENTRICEconcentric arcs of circle ARIA SECŢIUNII TRANSVERSALEcross-sectional area ARTICULAŢIE - pin-connection ARTICULAŢIE PLASTICĂ - plastic hinge AX-axle AXA DE COORDONATE - coordinate axis AXA NEUTRĂ - neutral axis AXE CENTRALE - centroidal axes AXE CENTRALE PRINCIPALE DE INERŢIE ALE SECŢIUNII TRANSVERSALE - principal centroidal axes of the cross section AXE DE COORDONATE – coordinate axes AXE PRINCIPALE - principal axes

B

BANDĂ, CUREA - strap BARĂ - bar BARĂ ZVELTĂ - slender member (bar) BARĂ (ZVELTĂ CE POATE PREZENTA PERICOLUL PIERDERII STABILITĂŢII), COLOANĂ, STÂLP – column BARĂ (ZVELTĂ) ÎNCASTRATĂ LA CAPETE - column with two fixed ends BARĂ CONICĂ - tapered bar BARĂ CURBĂ - curved member BARĂ DE LUNGIME MICĂ - stubby column BARĂ DREAPTĂ - straight bar C BARĂ PRISMATICĂ - prismatic bar CABLE – cablu BARĂ ZVELTĂ DUBLU CANTILEVER BEAM - grindă în ARTICULATĂ LA CAPETE -pin-ended consolă column CAP - capac, calotă BARE (ZVELTE) CE FLAMBEAZĂ ÎN CAST IRON – fontă DOMENIUL ELASTIC - long columns 226

Strength of Materials typical key words and phrases

ENGLISH – ROMANIAN


CASTIGLIANO’S THEOREM - teorema lui Castigliano CENTRE OF CURVATURE - centru de curbură CENTRIC AXIAL LOAD - forţă axială aplicată în centrul de greutate CENTRIC LOAD - sarcină (forţă) aplicată în centrul de greutate CENTROID - centru de greutate CENTROIDAL AXES - axe centrale CHECK (TO-) - verifica (a-) CIRCULAR CROSS SECTION – secţiune circulară CIRCULAR HOLE - gaură circulară CIRCULAR WIRE – sârmă (fir) de secţiune circulară CLEARANCE – interstiţiu CLOCKWISE - în sensul acelor de ceasornic CLOSE-COILED SPRING - arc cu spiră strânsă COEFFICIENT OF FRICTION – coeficient de frecare COEFFICIENT OF THERMAL EXPANSION - coeficient de dilatare liniară COLD-WORKED - prelucrat la rece COLLAR – manşon, colier COLUMN - bară (zveltă ce poate prezenta pericolul pierderii stabilităţii), coloană, stâlp COLUMN WITH TWO FIXED ENDS bară (zveltă) încastrată la capete COMPRESSED-AIR TANK - recipient cu aer comprimat COMPRESSIVE STRESS - tensiune de compresiune CONCENTRATED LOAD – forţă concentrată CONCENTRIC ARCS OF CIRCLE arce de cerc concentrice CONCRETE - beton CONDITIONS OF EQUILIBRIUM -

BARE (ZVELTE) CE FLAMBEAZĂ ÎN DOMENIUL ELASTO-PLASTIC intermediate columns BARE (ZVELTE) DE LUNGIME MICĂ, LA CARE NU APARE PERICOLUL DE PIERDERE A STABILITĂŢII - short columns BARE CU PEREŢI SUBŢIRI - thinwalled members BETON – concrete BOLŢ - bolt BRAD - fir BRONZ - bronze

C CABLU - cable CABLU DE OŢEL - steel cable CADRAN – dial gage CADRU – frame CADRU PORTANT - bearing frame CAL PUTERE - horsepower CAPAC, CALOTĂ - cap CAPACITATEA DE A ACUMULA ENERGIE – energy-absorbing capacity CAPĂT LIBER - free end CAUZE NATURALE IMPREVIZIBILE – unpreventable natural causes CENTRU DE CURBURĂ - centre of curvature CENTRU DE GREUTATE - centroid CENTRU DE FORFECARE - shear center CERCUL LUI MOHR - Mohr’s circle CERCUL LUI MOHR PENTRU STAREA PLANĂ DE DEFORMAŢIEMohr’s circle for plane strain CERCUL LUI MOHR PENTRU STAREA PLANĂ DE TENSIUNEMohr’s circle for plane stress CHERESTEA – timber CILINDRU TUBULAR -hollow cylinder

227




condiţii de echilibru CONE - con CONNECTION – legatură CONSTANTS OF INTEGRATION constante de integrare COORDINATE AXES - axe de coordonate CORNERS OF THE SECTION colţurile secţiunii CORROSION - coroziune CORROSION RESISTANCE – rezistenţa la coroziune COUNTERCLOCKWISE –în sens invers acelor de ceasornic (sens trigonometric) COUPLE - cuplu, moment CRACK – fisură CRACK GROWTH – creşterea fisurii CRADLE -jgheab, cuzinet de reazem CRANE - macara CRANKSHAFT - arbore cotit CREEP - fluaj CRITICAL LOAD - sarcină critică CRITICAL STRESS - tensiune critică CROSS-SECTIONAL AREA - aria secţiunii transversale CURVATURE – curbură CURVE – curbă CURVED MEMBER - bară curbă CUT (TO-) –secţiona (a-) CYCLIC LOADING - sarcină ciclică CYLINDRICAL BRASS VESSEL - vas cilindric din alamă CYLINDRICAL PRESSURE VESSELS -vase de presiune cilindrice

COEFICIENT (FACTOR) DE CONCENTRARE A TENSIUNILOR stress-concentration factor COEFICIENT DE DILATARE LINIARĂ – coefficient of thermal expansion COEFICIENT DE FLAMBAJ - buckling coefficient COEFICIENT DE FRECARE-coefficient of friction COEFICIENT DE SIGURANŢĂ - factor of safety COEFICIENT DE ZVELTEŢE slenderness ratio COEFICIENT EFECTIV DE ZVELTEŢE - effective slenderness ratio COEFICIENŢI DE INFLUENŢĂ influence coefficients COEFICIENTUL LUI POISSONPoisson’s ratio COLŢURILE SECŢIUNII - corners of the section COMPONENTELE TENSIUNII - stress components CON - cone CONCENTRATORI DE TENSIUNE stress concentrators CONDIŢII DE CONTUR (DE REZEMARE) - boundary conditions CONDIŢII DE ECHILIBRU - conditions of equilibrium CONDUCTA DE EVACUARE-penstock CONSTANTE DE INTEGRARE constants of integration CONTRACŢIE TRANSVERSALĂ D transverse contraction DASHED LINE - linie punctată COORDONATE RECTANGULAREDECAY – putrezire rectangular coordinates DEFLECTION - săgeată CORNIER DIN OŢEL - steel angle DEFLECTION AND SLOPE AT POINT COROZIUNE – corrosion A - săgeata şi rotirea în punctul A CORP RIGID – rigid body DEFLECTION OF BEAMS – deformaţia CORP SOLID – solid body grinzilor CREŞTEREA FISURII - crack growth 228




DEFLECTION OF BEAMS – deformaţia grinzilor (supuse la încovoiere) DEFORMABLE STRUCTURE structură deformabilă DEFORMATION – deformaţie DEGREE OF ACCURACY - grad de precizie DEGREES CELSIUS (FAHRENHEIT) grade Celsius (Fahrenheit) DENSITY – densitate DESIGN – proiectare DESIGN OF TRANSMISSION SHAFTS - proiectarea arborilor de transmisie DIAGRAM –diagramă DIAL GAGE – cadran DIAMETER – diametru DIEDRAL ANGLE - unghi diedru DIMENSIONLESS QUANTITY mărime adimensională DIRECTLY PROPORTIONAL - direct proporţional DISCONTINUITY - discontinuitate DISPLACEMENT - deplasare DISTANCE - distanţă DISTRIBUTED FORCES – forţe distribuite DISTRIBUTED LOADS – sarcini distribuite DOUBLE INTEGRAL - integrală dublă DOUBLE SHEAR - forfecare dublă DRAWING, SKETCH – desen DUCTILE MATERIALS – materiale ductile DYNAMIC LOAD - sarcină dinamică

CRIC HIDRAULIC - hydraulic jack CRITERIUL LUI MOHR - Mohr’s criterion CUI – nail CUPLU, MOMENT - couple CURBĂ - curve CURBURĂ - curvature CUZINET, LOCAŞ SFERIC -ball socket CURGERE (A MATERIALULUI) yielding

D

DECHIDERE (A UNEI GRINZI) - span DEFECT DE STRUCTURĂ - flaw DEFORMAŢIA GRINZILOR (SUPUSE LA ÎNCOVOIERE) - deflection of beams DEFORMAŢIA PERMANENTĂ permanent deformation DEFORMAŢIE - deformation DEFORMAŢIE PLASTICĂ - plastic deformation DEFORMAŢIE SPECIFICĂ - strain DEFORMAŢIE SPECIFICĂ CAUZATĂ DE VARIAŢII DE TEMPERATURĂthermal strain DEFORMAŢIE SPECIFICĂ LONGITUDINALĂ - normal strain DEFORMAŢIE SPECIFICĂ TRANSVERSALĂ - lateral strain DEFORMAŢIE SPECIFICĂ TRANSVERSALĂ (LUNECARE SPECIFICĂ) - shearing strain DEFORMAŢIE TOTALĂ - total deformation DENSITATE – density E DEPLASARE – displacement ECCENTRIC AXIAL LOAD – forţă DEPLASARE RELATIVĂ - relative axială aplicată excentric displacement ECCENTRICAL - excentric DERIVATĂ PARŢIALĂ - partial ECCENTRICITY - excentricitate derivative EDGE OF THE BEAM - marginea DESCĂRCARE - unloading (muchia ) grinzii DESEN - drawing, sketch 229




EFFECTIVE LENGTH – lungime de flambaj EFFECTIVE SLENDERNESS RATIO coeficient efectiv de zvelteţe ELASTIC CORE - miez elastic ELASTIC CURVE – fibră medie deformată ELASTIC FLEXURE FORMULA relaţia lui Navier (încovoiere) ELASTIC LIMIT – limită de elasticitate ELASTIC RANGE – domeniu elastic ELASTIC SECTION MODULUS modulul de rezistenţă la încovoiere al secţiunii (în domeniul elastic) ELASTOPLASTIC MATERIAL material elastoplastic ELECTRIC GENERATOR - generator electric ELEMENTARY FORCES – forţe elementare ELEMENTARY INTERNAL FORCES eforturi elementare ELEMENTARY WORK - lucru mecanic elementar ELLIPSE – elipsă ELLIPTIC CROSS SECTION – secţiune transversală eliptică ELONGATION – alungire EMPIRICAL FORMULAS - formule empirice ENDURANCE LIMIT – rezistenţa la oboseală ENERGY - energie ENERGY-ABSORBING CAPACITY – capacitatea de a acumula energie ENGINEERING PRACTICE - practica inginerească EQUATION – ecuaţie EQUILIBRIUM – echilibru EQUILIBRIUM POSITION – poziţie de echilibru EQUIVALENCE – echivalenţă

DIAGRAMĂ - diagram DIAGRAMĂ DE FORŢE TĂIETOAREshear diagram DIAGRAMĂ DE MOMENT ÎNCOVOIETOR - bending-moment diagram DIAGRAMA TENSIUNEDEFORMAŢIE SPECIFICĂ-stress-strain diagram DIAMETRU – diameter DIAMETRU EXTERIOR-outer diameter DIAMETRU INTERIOR- inner diameter DIRECT PROPORŢIONAL - directly proportional DISCONTINUITATE - discontinuity DISPUNEREA PE LUNGIME A SUDURILOR SE FACE DIN...ÎN… mm - longitudinal spacing of the welds is of…mm DISTANŢĂ - distance DISTRIBUŢIE LINEARĂ - linear distribution DISTRIBUŢIE PARABOLICĂ parabolical distribution DISTRUGERE – failure DOMENIU ELASTIC - elastic range

E ECHILIBRU – equilibrium ECHILIBRU INIŢIAL - original equilibrium ECHIVALENŢĂ - equivalence ECONOMII DE MATERIAL -savings of material ECRUISARE – strain-hardening ECUAŢIE - equation ECUAŢIE DIFERENŢIALĂ DE ORDINUL DOI - second-order linear differential equation ECUAŢIE DIFERENŢIALĂ NEOMOGENĂ LINIARĂ DE ORDINUL DOI, CU COEFICIENŢI CONSTANŢI-

230




EQUIVALENT STATIC LOAD –sarcină statică echivalentă EULER’S FORMULA - formula lui Euler EXPERIMENTAL EQUIPMENT instalaţie experimentală EXTENSION OF EULER’S FORMULA TO COLUMNS WITH OTHER END CONDITIONS – aplicarea formulei lui Euler pentru alte cazuri de rezemare ale barelor EXTERNAL FORCES - forţe exterioare

F

- linear nonhomogeneous differential equation of the second order with constant coefficients ECUAŢIE DIFERENŢIALĂ OMOGENĂ LINIARĂ - linear homogeneous differential equation EFECTUL BAUSCHINGER Bauschinger effect EFORTURI ELEMENTARE elementary internal forces ELEMENT (COMPONENT AL UNEI STRUCTURI) - member ELEMENTE DIN LEMN - wooden members ELICE, SPIRALĂ (ÎN SPAŢIU) - helix ELIPSĂ-ellipse ENERGIA DE DEFORMAŢIE LA ÎNCOVOIERE - strain energy in bending ENERGIA DE DEFORMAŢIE LA TORSIUNE - strain energy in torsion ENERGIE SPECIFICĂ DE DEFORMAŢIE - strain energy density ENERGIE - energy ENERGIE CINETICĂ - kinetic energy ENERGIE DE DEFORMAŢIE - strain energy ENERGIE POTENŢIALĂ - potential energy EPRUVETĂ - specimen EPRUVETĂ CU SUPRAFAŢĂ LUSTRUITĂ - polished specimen EROARE PROCENTUALĂ - percentage error EVIDENŢIEREA GRAFICĂ A TUTUROR SARCINILOR CE SOLICITĂ UN ANUMIT CORP ÎN ECHILIBRU MECANIC, INCLUSIV REACŢIUNILE - free-body diagram EXCENTRIC - eccentrical EXCENTRICITATE - eccentricity

FACTOR OF SAFETY – coeficient de siguranţă FAILURE – ruptură, rupere, avarie, distrugere, cedare, defectare, întrerupere, etc. FATIGUE – oboseală FATIGUE CRACK - fisura de oboseală FICTITIOUS OR DUMMY LOAD sarcină fictivă FIGURE – figură FIR – brad FIRST MOMENT (OF AN AREA) moment static FIBER – fibră FIXED SUPPORT – încastrare FLANGED JOINT - îmbinare cu flanşe FLAT END - terminaţie aplatizată FLAW - defect de structură FLEXURAL RIGIDITY - produsul EI (încovoiere) FLUID - fluid FLYWHELL - volant FORCE – forţă FORCE-COUPLE SYSTEM - ansamblu moment-forţă (torsor) FORGED COMPONENT - piesă forjată F FORMULA - formulă FRACTURE – rupere FALCĂ (DE MENGHINĂ, ETC) - jaw 231



ENGLISH - ROMANIAN FRACTURE CRITERIA - teorii de rupere FRACTURE MECHANICS - mecanica ruperii FRAME - cadru FREE END - capăt liber FREE SURFACE - suprafaţă liberă FREE-BODY DIAGRAM – evidenţierea grafică a tuturor sarcinilor ce solicită un anumit corp în echilibru mecanic, inclusiv reacţiunile FREQUENCY – frecvenţă FREQUENCY OF ROTATION - viteza de rotaţie FREQUENTLY ENCOUNTERED – frecvent întâlnit FRUSTRUM OF A CIRCULAR CONE - trunchi de con FULLY APPLIED – aplicată direct la valoarea nominală (despre o forţă ,..,etc) FUNCTION OF X – funcţie de x

G GAP - interstiţiu, imperfecţiune de montaj GEAR - roată dinţată GENERAL STATE OF STRESS - stare complexă de tensiune GENERALIZED HOOKE’S LAW legea lui Hooke generalizată GLASS – sticlă GLUED JOINT – îmbinare prin lipire GRAIN - grăunte GRAPH – grafic GROUND – sol

H HALF-CYLINDER –jumătate de cilindru HEAT TREATMENT – tratament termic HEIGHT – înălţime HELIX - elice, spirală (în spaţiu) HEXAGON – hexagon HIGH-CARBON STEEL – oţel cu

FIABILITATE – reliability FIBRĂ - fiber FIBRA MEDIE DEFORMATĂ - elastic curve FIER PUR - pure-iron FIGURĂ - figure FIR – wire FIR DE NYLON - nylon thread FISURĂ - crack FISURĂ DE OBOSEALĂ - fatigue crack FISURI MICROSCOPICE - microscopic cracks FLAMBA (A-) -buckle (to-) FLAMBAJ – buckling FLUAJ - creep FLUID – fluid FLUXUL TENSIUNILOR TANGENŢIALE - shear flow FONTĂ - cast iron FORFECARE - shear FORFECARE DUBLĂ - double shear FORFECARE SIMPLĂ - single shear FORMULĂ - formula FORMULA LUI EULER - Euler’s formula FORMULE EMPIRICE - empirical formulas FORŢĂ - force FORŢĂ AXIALĂ - axial load FORŢĂ AXIALĂ APLICATĂ ÎN CENTRUL DE GREUTATE – centric axial load FORŢĂ AXIALĂ APLICATĂ EXCENTRIC - eccentric axial loading FORŢĂ CAPABILĂ - largest allowable force FORŢĂ CONCENTRATĂ-concentrated load FORŢĂ CRESCĂTOARE - increasing load FORŢĂ ORIZONTALĂ -horizontal load FORŢĂ TĂIETOARE - shearing force FORŢE DISTRIBUITE - distributed

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conţinut ridicat de carbon HIGH-STRENGTH STEEL – oţel de rezistenţă ridicată HINGED BEAM - grindă cu articulaţie HINT - indicaţie (în rezolvarea unei probleme) HOLE – gaură HOLLOW CYLINDER - cilindru tubular HOMOGENEOUS MATERIAL material omogen HOOKE’S LAW - legea lui Hooke HOOP STRESS - tensiune circumferenţială (vase cu pereţi subţiri) HORIZONTAL DEFLECTION – săgeata pe direcţie orizontală HORIZONTAL LOAD - forţă orizontală HORIZONTAL PLANE - plan orizontal HORSEPOWER - cal putere HYDRAULIC JACK - cric hidraulic HYDROSTATIC PRESSURE – presiune hidrostatică HYPOTENUSE –ipotenuză

I IMPACT FACTOR – multiplicator de impact IMPACT LOADING – sarcină aplicată cu şoc IMPULSE - impuls INCREASING LOAD - forţă crescătoare INERTIA - inerţie INFLUENCE COEFFICIENTS coeficienţi de influenţă INNER DIAMETER - diametru interior INTEGRATION – integrare INTERMEDIATE COLUMNS - bare (zvelte) ce flambează în domeniul elastoplastic INTERSECT (TO-) – intersecta (a-) INVERSELY PROPORTIONAL - invers proporţional ISOTROPIC MATERIAL - material izotrop

forces FORŢE ELEMENTARE - elementary forces FORŢE EXTERIOARE - external forces FORŢE TRANSVERSALE - transverse forces; transverse load FRECVENT ÎNTÂLNIT - frequently encountered FRECVENŢĂ - frequency FRETARE – binding FUNCŢIE ANALITICĂ - analytical function FUNCŢIE DE GRADUL 2 - second degree function FUNCŢIE DE GRADUL 3 - third degree function FUNCŢIE DE x - function of x FUNCŢIE TREAPTĂ - step function FUNCŢII SINGULARE - singularity functions

G GAURĂ - hole GAURĂ CIRCULARĂ - circular hole GAURĂ CONICĂ - taper hole GENERATOR ELECTRIC - electric generator GRINDĂ EXTRUDATĂ CU PEREŢI SUBŢIRI- thin-walled extruded beam GÂTUIRE - necking GRAD DE PRECIZIE - degree of accuracy GRADE CELSIUS (FAHRENHEIT) – degrees celsius (fahrenheit) GRAFIC - graph GRĂUNTE - grain GREUTATE PROPRIE - own weight GREUTATE SPECIFICĂ - specific weight GRINDĂ - beam GRINDĂ CU ARTICULAŢIE - hinged beam GRINDĂ CU INIMĂ PLINĂ -web beam

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J JAW - falcă (de menghină, etc) JUST TO THE RIGHT OF POINT B – în secţiunea B dreapta

K KERN – sâmbure central KINETIC ENERGY - energie cinetică

L LABORATORY – laborator LARGEST ALLOWABLE FORCE forţă capabilă LATERAL STRAIN – deformaţie specifică transversală LAW – lege LEAF SPRING - arc lamelar LENGTH – lungime LEVER – pârghie LINEAR DISTRIBUTION – distribuţie liniară LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION - ecuaţie diferenţială omogenă liniară LINEAR NONHOMOGENEOUS DIFFERENTIAL EQUATION OF THE SECOND ORDER WITH CONSTANT COEFFICIENTS - ecuaţie diferenţială neomogenă liniară de ordinul doi, cu coeficienţi constanţi LOAD POINT OF APPLICATION punct de aplicaţie a sarcinii LOADING – încărcare LOCATION - loc poziţie LOGARITHMIC SCALE – scară logaritmică LONG COLUMNS - bare (zvelte) ce flambează în domeniul elastic LONG CRACK – macrofisură LONGITUDINAL SHEARING STRESSES - tensiuni tangenţiale longitudinale

GRINDĂ CU ZĂBRELE - truss GRINDĂ DE EGALĂ REZISTENŢĂbeam of constant strength GRINDĂ DE LĂŢIME UNITATE-beam unit width GRINDĂ DE LEMN - wooden beam GRINDĂ DE SECŢIUNE DREPTUNGHIULARĂ CU LĂŢIME MICĂnarrow rectangular beam GRINDĂ DIN BETON ARMAT reinforced concrete beam GRINDĂ DIN PROFIL CU TALPĂ LATĂ- wide-flanged beam GRINDĂ EXTRUDATĂ CU PEREŢI SUBŢIRI - thin-walled extruded beam GRINDĂ ÎN CONSOLĂ - cantilever beam GRINDĂ SIMPLU REZEMATĂ -simply supported beam GRINDĂ SIMPLU STATIC NEDETERMINATĂ - statically indeterminate beam to the first degree GRINDĂ SUPRAÎNCĂRCATĂ overstressed beam GRINZI STATIC NEDETERMINATE statically indeterminate beams GRINZI ZVELTE - slender beams GROSIME – thickness

H HEXAGON – hexagon

I IMPULS - impulse INDICAŢIE (ÎN REZOLVAREA UNEI PROBLEME) - hint INEL - ring INERŢIE – inertia INSTABIL - unstable INSTALAŢIE EXPERIMENTALĂ experimental equipment INTEGRALĂ DUBLĂ - double integral INTEGRARE – integration INTEGRARE ÎN RAPORT CU O

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LONGITUDINAL SPACING OF THE WELDS IS OF…mm - dispunerea pe lungime a sudurilor se face din...în...mm LOW - CARBON STEEL - oţel cu conţinut scăzut de carbon LOW - STRENGTH STEEL - oţel de rezistenţă scăzută L-SHAPED MACHINE ELEMENT piesă în formă de L

M MACHINE COMPONENTS - organe de maşini MACHINE TOOL - maşină unealtă MACROSCOPIC CRACKS - fisuri macroscopice MASS - masa MATERIAL - material MAXIMUM ALLOWABLE SPEED viteză maximă admisibilă MAXIMUM DEFLECTION - săgeata maximă MAXIMUM-DISTORTION-ENERGY CRITERION - teoria energiei maxime modificatoare de formă MAXIMUM-NORMAL-STRESS CRITERION -teoria tensiunii normale maxime MAXIMUM-SHEARING-STRENGTH CRITERION - teoria tensiunii tangenţiale maxime MAXWELL’S RECIPROCAL THEOREM - teorema reciprocităţii deplasărilor (Maxwell) MECHANICAL WORKING - prelucrare mecanică MEMBER - element (component al unei structuri) MEMBRANE - membrană MEMBRANE ANALOGY - analogie cu membrana (torsiune) METHODS OF ANALYSIS - metode de analiză

SINGURĂ VARIABILĂ - simple integration INTERSECTA (A-) - intersect (to-) INTERSTIŢIU - clearance INTERSTIŢIU, IMPERFECŢIUNE DE MONTAJ – gap INVERS PROPORŢIONAL - inversely proportional IPOTENUZA – hypotenuse IPOTEZA SIMPLIFICATOARE simplifying assumption

Î ÎMBINARE CAP LA CAP - splice ÎMBINARE CU FLANŞE - flanged joint ÎMBINARE PRIN LIPIRE - glued joint ÎMPINGE (A-) - push (to-) ÎN SECŢIUNEA B DREAPTA - just to the right of point B ÎN SENS INVERS ACELOR DE CEASORNIC (SENS TRIGONOM.) - counterclockwise ÎN SENSUL ACELOR DE CEASORNIC - clock-wise ÎNĂLŢIME - height ÎNCĂRCARE - loading ÎNCASTRARE - fixed support ÎNCERCARE - test ÎNCERCARE LA TRACŢIUNE - tensile test ÎNCOVOIERE - bending ÎNCOVOIERE OBLICĂ - unsymmetric bending ÎNCOVOIERE PURĂ - pure bending ÎNFĂŞURARE - wrap ÎNSUMARE ALGEBRICĂ - algebraic sum ÎNTINDE (A-), TRAGE (A-) - stretch ÎNVELIŞ - shell

J JGHEAB, CUZINET DE REAZEM – cradle

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MICROSCOPIC CRACKS - fisuri microscopice MILD STEEL - oţel moale (cu conţinut scăzut de carbon) MODULUS OF ELASTICITY - modul de elasticitate MODULUS OF RIGIDITY - modul de rigiditate MOHR’S CIRCLE - cercul lui Mohr MOHR’S CIRCLE FOR PLANE STRESS - cercul lui Mohr pentru starea plană de tensiune MOHR’S CRITERION - criteriul lui Mohr MOHR’S CIRCLE FOR PLANE STRAIN - cercul lui Mohr pentru starea plană de deformaţie MOMENT OF INERTIA OF THE SECTION ABOUT THE PRINCIPAL CENTROIDAL AXIS - moment de inerţie al secţiunii în raport cu axa centrală principală a secţiunii

N NAIL - cui NARROW RECTANGULAR BEAM grinda de secţiune dreptunghiulară cu lăţime mică NECKING - gâtuire NEGATIVE (POSITIVE) SIGN - semn negativ (pozitiv) NEGLECTING THE TERM CONTAINING - neglijînd termenul ce-l conţine pe… NEUTRAL AXIS - axă neutră NEUTRAL SURFACE - suprafaţă neutră NORMAL STRAIN - deformaţie specifică longitudinală NORMAL STRESS - tensiune normală NUMERICAL VALUE - valoare numerică NUT - piuliţă NYLON THREAD - fir de nylon

JUMĂTATE DE CILINDRU - halfcylinder

L LABORATOR - laboratory LĂŢIME - width LEGĂTURĂ - connection LEGE - law LEGEA LUI HOOKE GENERALIZATĂ - generalized Hooke’s law LEGEA LUI HOOKE - Hooke’s law LIMITA DE ELASTICITATE - elastic limit LIMITA DE PROPORŢIONALITATE proportional limit LINIE DREAPTĂ - straight line LINIE PUNCTATĂ - dashed line LOC POZIŢIE - location LOVI (A-) - strike (to-) LUCRU MECANIC - work LUCRU MECANIC ELEMENTAR elementary work LUNGIME - length LUNGIME DE FLAMBAJ - effective length

M MACARA - crane MACROFISURĂ - long crack MANŞON, COLIER - collar MANTA (CĂMAŞĂ) DE AL-aluminum shell MARCĂ TENSOMETRICĂ -strain gage MARGINEA (MUCHIA ) GRINZII-edge of the beam MĂRIME ADIMENSIONALĂ dimensionless quantity MĂRIME SCALARĂ - scalar quantity MĂRIME VECTORIALĂ - vectorial quantity MASA - mass

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ENGLISH - ROMANIAN

O OBLIQUE PLANE - plan înclinat OBLIQUE SECTION - secţiune oblică ORDINATE - ordonată ORIGINAL EQUILIBRIUM - echilibru iniţial OUTER DIAMETER - diametru exterior OVERSTRESSED BEAM - grindă supraîncărcată OWN WEIGHT - greutate proprie

P PARABOLA - parabolă PARABOLICAL DISTRIBUTION distribuţie parabolică PARABOLOID OF REVOLUTION paraboloid de revoluţie PARALLEL-AXIS THEOREM -teorema axei paralele (relaţiile lui Steiner) PARALLELOGRAM - paralelogram PARENTHESE -paranteză PARTIAL DERIVATIVE - derivată parţială PEG - ştift, cep PENSTOCK - conducta de evacuare PERCENTAGE ERROR - eroarea procentuală PERMANENT DEFORMATION deformaţie permanentă PERPENDICULAR - perpendicular PIN-CONNECTION - articulaţie PIN-ENDED COLUMN - bară zveltă dublu articulată la capete PIPE - ţeavă PLANE OF ARBITRARY ORIENTATION - plan cu o orientare oarecare PLANE OF SYMMETRY - plan de simetrie PLANE STATE OF STRAIN - stare plană de deformaţie PLANE STATE OF STRESS -stare plană de tensiune

MAŞINĂ DE ÎNCERCĂRI LA TORSIUNE - torsion testing machine MAŞINA UNEALTĂ - machine tool MATERIAL - material MATERIAL ELASTOPLASTIC elastoplastic material MATERIAL IZOTROPIC - isotropic material MATERIAL OMOGEN - homogeneous material MATERIALE DUCTILE - ductile materials MATERIALE FRAGILE - brittle materials MECANICA RUPERII - fracture mechanics MEMBRANA - membrane MENGHINĂ - vice (vise) METODA TENSIUNII ADMISIBILE allowable - stress method METODE DE ANALIZĂ - methods of analysis MICROFISURĂ - short crack MIEZ DE OŢEL - steel core MIEZ ELASTIC - elastic core MODUL DE ELASTICITATE -modulus of elasticity MODUL DE ELASTICITATE TRANSVERSAL (G) -shear modulus MODUL DE REZISTENŢĂ LA ÎNCOVOIERE AL SECŢIUNII (ÎN DOMENIUL ELASTIC) - elastic section modulus MODUL DE RIGIDITATE - modulus of rigidity MOMENT DE ÎNCOVOIERE-bending moment MOMENT DE INERŢIE AL SECŢIUNII ÎN RAPORT CU AXA CENTRALĂ PRINCIPALĂ A SECŢIUNII - moment of inertia of the section about the principal centroidal axis

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PLANE STRAIN STATE - starea plană de deformaţie PLANE STRESS STATE - starea plană de tensiune PLANK - scândură PLASTIC DEFORMATION - deformaţie plastică PLASTIC HINGE - articulaţie plastică PLASTIC ZONE - zona plastică PLATE - placă PLOT (TO-) - trasa (a-) PLYWOOD - placaj POINT OF INFLECTION - punct de inflexiune POISSON’S RATIO - coeficientul lui Poisson POLAR MOMENT OF INERTIA moment de inerţie polar POLISHED SPECIMEN - epruvetă cu suprafaţă lustruită POLYSTYRENE - polistiren POOR MAINTENANCE - proasta întreţinere (a utilajului, structurii, etc. ) PORTION - porţiune (de bara, etc) POST - reazem, suport POTENTIAL ENERGY - energie potenţială POWER - putere PRINCIPAL AXES - axe principale PRINCIPAL CENTROIDAL AXES OF THE CROSS SECTION - axe centrale principale de inerţie ale secţiunii transverasale PRINCIPAL PLANES AND PRINCIPAL STRESSES - plane principale şi tensiuni principale PRINCIPAL STRESSES - tensiuni principale PRISMATIC BAR - bară prismatică PROBLEMS TO BE ASSIGNED probleme propuse spre rezolvare

MOMENT DE INERŢIE POLAR - polar moment of inertia MOMENT DE TORSIUNE - torque (twisting couple) MOMENT STATIC - first moment (of an area) MULTIPLICATOR DE IMPACT-impact factor

N NEGLIJÎND TERMENUL CE-L CONŢINE PE…- neglecting the term containing… NIT - rivet NIVELUL TENSIUNII - stress level

O OBOSEALĂ - fatigue ORDONATĂ - ordinate ORGANE DE MAŞINI - machine components OŢEL - steel OŢEL ALIAT, CĂLIT ŞI REVENIT quenched and tempered alloy steel OŢEL CĂLIT - quenched-steel OŢEL CU CONŢINUT RIDICAT DE CARBON - high-carbon steel OŢEL DE REZISTENŢĂ RIDICATĂhigh-strength steel OŢEL DE REZISTENŢĂ SCĂZUTĂ low-strength steel OŢEL LAMINAT - rolled steel OŢEL MOALE (CU CONŢINUT SCĂZUT DE CARBON) - mild steel OŢEL RECOPT - annealed steel OŢEL REVENIT - tempered steel

P PARABOLĂ - parabola PARABOLOID DE REVOLUŢIE paraboloid of revolution

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PROJECTION - proiecţie PARALELOGRAM - parallelogram PROPAGATE (TO-) - propaga (a se -) PARANTEZĂ - parenthese PROPORTIONAL LIMIT - limită de PĂTRAT CU LATURA a =…- square of proporţionalitate side a =… PULL (TO-) -(a) trage PERETE DE GROSIME CONSTANTĂPULLEY - roată de transmisie uniform wall thickness PURE BENDING - încovoiere pură PERPENDICULAR - perpendicular PURE-IRON - fier pur PIATRĂ - stone PUSH (TO-) - a împinge PIESĂ FORJATĂ - forged component PIESĂ ÎN FORMĂ DE L - L shaped Q machine element QUENCHED, TEMPERED ALLOY PIESĂ LAMINATĂ - rolled component STEEL - oţel aliat, călit şi revenit PÂRGHIE - lever QUENCHED-STEEL - oţel călit PIULIŢĂ - nut PLACĂ - plate R PLACAJ - plywood RADIAN - radian PLAN CU O ORIENTARE OARECARE RADIUS - rază RADIUS OF CURVATURE - rază de - plane of arbitrary orientation PLAN DE SIMETRIE - plane of curbură RADIUS OF GYRATION - rază de symmetry PLAN ÎNCLINAT - oblique plane inerţie PLAN ORIZONTAL - horizontal plane RATIO - raport PLAN VERTICAL - vertical plane REACTION - reacţiune PLANE PRINCIPALE ŞI TENSIUNI REASONING - raţionament RECTANGULAR COORDINATES - PRINCIPALE - principal planes and principal stresses coordonate rectangulare RECTANGULAR CROSS SECTION - POLISTIREN - polystyrene PORŢIUNE (DE BARĂ, ETC) - portion secţiune dreptunghiulară REDUNDANT REACTION - reacţiune POZIŢIE DE ECHILIBRU - equilibrium position suplimentară (static nedeterminată) INGINEREASCĂ REINFORCED CONCRETE BEAM - PRACTICA engineering practice grindă din beton armat MECANICĂ RELATIONS AMONG LOAD, SHEAR PRELUCRARE AND BENDING MOMENT - relaţii mechanical working între sarcină, forţă tăietoare şi moment PRELUCRAT LA RECE - cold-worked PRESIUNE HIDROSTATICĂ încovoietor RELATIVE DISPLACEMENT - hydrostatic pressure PRINCIPIUL LUI SAINT-VENANT deplasare relativă Saint-Venant’s principle RELIABILITY - fiabilitate REPEATED LOADING - sarcină ciclică PRINCIPIUL SUPRAPUNERII DE RESIDUAL STRESSES - tensiuni EFECTE - superposition method remanente 239




RESILIENCE - rezilienţă RESULTANT - rezultantă RHOMBUS - romb RIGHT-HAND SUPPORT - suportul din dreapta RIGID BODY - corp rigid RING - inel RIVET - nit ROD - tijă ROD UNDER ITS OWN WEIGHT - tijă sub greutate proprie ROLLED COMPONENT -piesă laminată ROLLED STEEL - oţel laminat ROLLED-STEEL SHAPES - profile din oţel laminat ROLLING - rostogolire RULE - regulă RUPTURE - rupere RUSTING - ruginire

PROASTA ÎNTREŢINERE (A UTILAJULUI, STRUCTURII, ETC. ) poor maintenance PROBLEMĂ REZOLVATĂ - sample problem PROBLEME PROPUSE SPRE REZOLVARE - problems to be assigned PRODUSUL EI (ÎNCOVOIERE) flexural rigidity PROFIL CU TALPĂ LATĂ - wideflanged PROFILE DIN OŢEL LAMINAT rolled-steel shapes PROIECTARE - design PROIECTAREA ARBORILOR DE TRANSMISIE - design of transmission shafts PROIECŢIE - projection PROPAGA (A SE -) - propagate (to-) PUNCT DE INFLEXIUNE - point of inflection S SAINT-VENANT’S PRINCIPLE - PUNCTUL DE APLICAŢIE AL SARCINII - load point of application principiul lui Saint-Venant SAMPLE PROBLEM – problemă PUTERE - power PUTREZIRE - decay rezolvată SAVINGS OF MATERIAL - economii de material R SCALAR QUANTITY - mărime scalară RADIAN - radian SECOND DEGREE FUNCTION - RAPORT - ratio funcţie de gradul 2 RAŢIONAMENT - reasoning SECOND ORDER LINEAR RAZĂ - radius DIFFERENTIAL EQUATION - ecuaţie RAZĂ DE CURBURĂ - radius of diferenţială de ordinul doi curvature SEMIMAJOR AXIS - semiaxa mare (a RAZĂ DE INERŢIE - radius of gyration unei elipse) REACŢIUNE - reaction SEMIMINOR AXIS - semiaxa mică (a REACŢIUNE SUPLIMENTARĂ unei elipse) (STATIC NEDETERMINATĂ) SHAFT-DISK-BELT ARRANGEMENT redundant reaction -sistem arbore - roată de curea - curea REAZEM, SUPORT - post SHEAR - forfecare RECIPIENT CU AER COMPRIMATSHEAR CENTER - centrul de forfecare compressed-air tank SHEAR DIAGRAM - diagramă de forţe REGULĂ - rule tăietoare 240




SHEAR FLOW - fluxul tensiunilor tangenţiale SHEAR MODULUS - modul de elasticitate transversal (G) SHEARING FORCE - forţă tăietoare SHEARING STRAIN - deformaţie specifică transversală (lunecare specifică) SHEARING STRESS tensiune tangenţială SHEARING STRESSES ON TRANSVERSE SECTION - tensiuni tangenţiale pe secţiuni transversale SHELL - înveliş SHORT COLUMNS - bare (zvelte) de lungime mică, la care nu apare pericolul de pierdere a stabilităţii SHORT CRACK - microfisură SIMILAR TRIANGLES - triunghiuri asemenea SIMPLE INTEGRATION - integrare în raport cu o singură variabilă SIMPLIFYING ASSUMPTION - ipoteză simplificatoare SIMPLY SUPPORTED BEAM - grindă simplu rezemată SINGLE SHEAR - forfecare simplă SINGULARITY FUNCTIONS - funcţii singulare SLENDER BEAMS - grinzi zvelte SLENDER MEMBER - bară (element) zveltă SLENDERNESS RATIO - coeficient de zvelteţe SLIP - alunecare SLOPE - pantă SLOWLY INCREASING LOAD sarcină uşor crescătoare SMOOTH HORIZONTAL SURFACE suprafaţă orizontală netedă SOLID BODY - corp solid SOLID CYLINDRICAL SHAFT - arbore de transmisie cilindric masiv (plin) SOLUTION - soluţie (la o problemă)

RELAŢIA LUI NAVIER (ÎNCOVOIERE) - elastic flexure formula RELAŢII ÎNTRE SARCINĂ, FORŢĂ TĂIETOARE ŞI MOMENT ÎNCOVOIETOR - relations among load, shear and bending moment REZILIENŢĂ - resilience REZISTENŢA LA COROZIUNE corrosion resistance REZISTENŢA LA CURGERE - yield strength REZISTENŢA LA OBOSEALĂ endurance limit REZISTENŢA LA RUPERE - breaking strength REZISTENŢA MATERIALELOR strength of materials REZULTANTĂ - resultant RIGIDITATE - stiffness ROATĂ - wheel ROATĂ DE TRANSMISIE - pulley ROATĂ DINŢATĂ - gear ROMB - rhombus ROSTOGOLIRE - rolling ROTIRE - slope ROZETĂ TENSOMETRICĂ - strain rosette RUGINIRE - rusting RULMENT - bearing RUPERE - fracture; rupture RUPTURA, RUPERE, AVARIE, DISTRUGERE, CEDARE, DEFECTARE, ÎNTRERUPERE, ETC. - failure

S SĂGEATĂ - deflection SĂGEATĂ MAXIMĂ - maximum deflection SĂGEATĂ PE DIRECŢIE ORIZONTALĂ - horizontal deflection SĂGEATĂ PE DIRECŢIE VERTICALĂ -vertical deflection

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SPAN - deschidere (a unei grinzi) SPECIFIC WEIGHT - greutate specifică SPECIMEN - epruvetă SPHERICAL PRESSURE VESSELS vase de presiune sferice SPLICE - îmbinare cap la cap SQUARE CROSS SECTION - secţiune transversală pătratică SQUARE OF SIDE a=…- pătrat cu latura a=… STABILITY - stabilitate STABILITY OF STRUCTURES stabilitatea structurilor STABLE - stabil (ă) STRAIN GAGE - marcă tensometrică STATICALLY INDETERMINATE static nedeterminat (ă) STATICALLY INDETERMINATE BEAMS - grinzi static nedeterminate STATICALLY INDETERMINATE BEAM TO THE FIRST DEGREE grinda simplu static nedeterminate STATICALLY INDETERMINATE SHAFTS - arbori static nedeterminaţi STATICS - statica STEAM TURBINE - turbină cu abur STEEL - oţel STEEL ANGLE - cornier din oţel STEEL CABLE - cablu de oţel STEEL CORE - miez de oţel STEP FUNCTION - funcţie treaptă STIFFNESS - rigiditate STONE - piatră STRAIGHT BAR - bară dreaptă STRAIGHT LINE - linie dreaptă STRAIN - deformaţie specifică STRAIN ENERGY - energie de deformaţie STRAIN ENERGY DENSITY - energie specifică de deformaţie STRAIN ENERGY IN BENDING energie de deformaţie la încovoiere

SĂGEATA ŞI ROTIREA ÎN PUNCTUL A - deflection and slope at point A ŞAIBĂ - washer SARCINĂ (FORŢĂ) APLICATĂ ÎN CENTRUL DE GREUTATE-centric load SARCINĂ APLICATĂ CU ŞOC -impact loading SARCINĂ CAPABILĂ - allowable load SARCINĂ CICLICĂ - cyclic load; repeated loading SARCINĂ CRITICĂ - critical load SARCINĂ DINAMICĂ - dynamic load SARCINĂ FICTIVĂ - fictitious or dummy load SARCINĂ NECUNOSCUTĂ - unknown load SARCINĂ STATICĂ ECHIVALENTĂ equivalent static load SARCINĂ UŞOR CRESCĂTOARE slowly increasing load SARCINI DISTRIBUITE - distributed loads SCARĂ LOGARITMICĂ - logarithmic scale SCÂNDURĂ - plank SECŢIONA (A-) - cut (to-) SECŢIUNE CIRCULARĂ -circular cross section SECŢIUNE DREPTUNGHIULARĂ rectangular cross section SECŢIUNE INELARĂ - annular cross section SECŢIUNE OBLICĂ - oblique section SECŢIUNE TRANSVERSALĂ CONSTANTĂ - uniform cross-section SECŢIUNE TRANSVERSALĂ DE FORMĂ TRAPEZOIDALĂ - trapezoidal cross section SECŢIUNE TRANSVERSALĂ DE FORMĂ TRIUNGHIULARĂ - triangular cross section

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SECŢIUNE TRANSVERSALĂ ELIPTICĂ - elliptic cross section SECŢIUNE TRANSVERSALĂ PĂTRATICĂ - square cross section SECŢIUNE TRANSVERSALĂ VARIABILĂ - variable cross section SEMIAXA MARE (A UNEI ELIPSE) semimajor axis SEMIAXA MICĂ (A UNEI ELIPSE) semiminor axis SEMN NEGATIV (POZITIV) - negative (positive) sign SÂMBURE CENTRAL - kern SIMETRIE - symmetry SÂRMĂ (FIR) DE SECŢIUNE CIRCULARĂ - circular wire SISTEM ARBORE - ROATĂ DE CUREA-CUREA shaft-disk-belt arrangement SOL - ground SOLICITARE AXIALĂ - axial loading SOLUŢIE (LA O PROBLEMĂ)-solution STABIL - stable STABILITATE - stability STABILITATEA STRUCTURILOR stability of structures STARE COMPLEXĂ DE TENSIUNE general state of stress STARE DE TENSIUNE BIAXIALĂ biaxial stress condition STARE PLANĂ DE DEFORMAŢIE plane state of strain STARE PLANĂ DE TENSIUNE - plane T state of stress TABLE - tabel STARE SPAŢIALĂ DE TENSIUNE TAPER HOLE - gaură conică three-dimensional state of stress TAPERED BAR - bară conică TEMPERATURE CHANGES - variaţii STARE PLANĂ DE DEFORMAŢIE plane state of strain de temperatură STATIC NEDETERMINAT (Ă) TEMPERED STEEL - oţel revenit TENSILE STRESS -tensiune de tracţiune statically indeterminate STATICA – statics TENSILE TEST - încercare la tracţiune STICLĂ - glass TEST - încercare STRAIN ENERGY IN TORSION energie de deformaţie la torsiune STRAIN ROSETTE -rozetă tensometrică STRAIN-HARDENING - ecruisare STRAP - bandă, curea STRENGTH OF MATERIALS rezistenţa materialelor STRESS - tensiune STRESS AT A POINT - tensiunea dintrun punct STRESS COMPONENTS -componentele tensiunii STRESS CONCENTRATORS concentratori de tensiune STRESS LEVEL - nivelul tensiunii STRESS-CONCENTRATION FACTOR - coeficient (factor) de concentrare a tensiunilor STRESSES UNDER COMBINED LOADINGS - tensiuni în cazul solicitărilor compuse STRESS-STRAIN DIAGRAM diagrama tensiune-deformaţie specifică STRETCH (TO-) - întinde, trage (a-) STRIKE (TO-) -lovi (a-) STRUCTURE - structură STUBBY COLUMN - bară de lungime mică SUPERPOSITION METHOD -principiul suprapunerii de efecte SYMMETRY - simetrie

243




THEORY OF ELASTICITY - teoria elasticităţii THERMAL STRAIN - deformaţie specifică cauzată de variaţii de temperatură THICKNESS - grosime THIN-WALLED EXTRUDED BEAM grindă extrudată cu pereţi subţiri THIN-WALLED HOLLOW SHAFTS tuburi cu pereţi subţiri THIN-WALLED MEMBERS - bare cu pereţi subţiri THIRD DEGREE FUNCTION - funcţie de gradul 3 THREE-DIMENSIONAL STATE OF STRESS - stare spaţială de tensiune TIMBER - cherestea TORQUE (TWISTING COUPLE) moment de torsiune TORSION - torsiune TORSION TESTING MACHINE maşină de încercări la torsiune TORSIONAL SPRING - arc de torsiune TOTAL DEFORMATION - deformaţie totală TRANSMISSION SHAFT - arbore de transmisie TRANSVERSE FORCES - forţe transversale TRANSVERSE CONTRACTION contracţie transversală TRANSVERSE LOAD - sarcină transversală TRAPEZOIDAL CROSS SECTION secţiune transversală de formă trapezoidală TRIANGULAR CROSS SECTION secţiune transversală de formă triunghiulară TRUSS – grindă cu zăbrele

STRAT DE ARAMĂ - brass layer STRUCTURĂ - structure STRUCTURĂ DEFORMABILĂ deformable structure STRUCTURI BIDIMENSIONALE-twodimensional structures SUDURĂ - weld SUPORTUL DIN DREAPTA -right-hand support SUPRAFAŢĂ DE CONTACT - bearing surface SUPRAFAŢĂ LIBERĂ - free surface SUPRAFAŢĂ NEUTRĂ -neutral surface SUPRAFAŢĂ ORIZONTALĂ NETEDĂ - smooth horizontal surface Ş ŞTIFT, CEP – peg T TABEL - table TENSIUNE - stress TENSIUNE (PRESIUNE) DE CONTACT - bearing stress TENSIUNE ADMISIBILĂ - allowable stress TENSIUNE CIRCUMFERENŢIALĂ (VASE CU PEREŢI SUBŢIRI) - hoop stress TENSIUNE CRITICĂ - critical stress TENSIUNE DE COMPRESIUNE compressive stress TENSIUNE DE TRACŢIUNE - tensile stress TENSIUNE MEDIE - average stress TENSIUNE NORMALĂ - normal stress TENSIUNE TANGENŢIALĂ - shearing stress TENSIUNEA DINTR-UN PUNCT-stress at a point TENSIUNI ÎN CAZUL SOLICITĂRI LOR COMPUSE - stresses under combined loading

244




TWO-DIMENSIONAL STRUCTURES - TENSIUNI PRINCIPALE - principal structuri bidimensionale stresses TENSIUNI REMANENTE - residual stresses U UNIFORM CROSS-SECTION - secţiune TENSIUNI TANGENŢIALE LONGI TUDINALE - longitudinal shearing transversală constantă UNIFORM WALL THICKNESS - perete stresses TENSIUNI TANGENŢIALE PE SEC de grosime constantă TIUNI TRANSVERSALE - shearing UNIT - unitate UNKNOWN LOAD - sarcină stresses on transverse sections TEOREMA AXEI PARALELE necunoscută (RELAŢIILE LUI STEINER) - parallelUNLOADING - descărcare axis theorem UNPREVENTABLE NATURAL TEOREMA LUI CASTIGLIANO CAUSES - cauze naturale imprevizibile Castigliano’s theorem UNSTABLE - instabil (ă) UNSYMMETRIC BENDING - TEOREMA RECIPROCITĂŢII DEPLASĂRILOR (MAXWELL)- Maxwell’s încovoiere oblică reciprocal theorem V TEORIA ELASTICITĂŢII - theory of VARIABLE CROSS SECTION -secţiune elasticity transversală variabilă TEORIA ENERGIEI MAXIME MODIVECTOR - vector FICATOARE DE FORMĂ - maximumVECTORIAL QUANTITY - mărime distortion-energy criterion vectorială TEORIA TENSIUNII NORMALE VELOCITY - viteză MAXIME - maximum-normal-stress VERTICAL DEFLECTION - săgeata pe criterion direcţie verticală TEORIA TENSIUNII TANGENŢIALE VERTICAL PLANE - plan vertical MAXIME - maximum-shearing-strength VIBRATION - vibraţie criterion VICE (VISE) - menghină TEORII DE CURGERE - yield criteria VOLUME - volum TEORII DE RUPERE - fracture criteria TERMINAŢIE APLATIZATĂ -flat end W TIJĂ-rod WASHER - şaibă TIJĂ SUB GREUTATE PROPRIE - rod WEB BEAM - grindă cu inimă plină under its own weight WELD - sudură TORSIUNE - torsion WHEEL - roată TRAGE (A-) -pull (to-) WIDE-FLANGED - profil cu talpă lată TRASA (A-) -plot (to-) WIDE-FLANGED BEAM - grindă din TRATAMENT TERMIC - heat treatment profil cu talpă lată TRIUNGHIURI ASEMENEA - similar WIDTH - lăţime triangles WIRE - fir TRUNCHI DE CON - frustum of a WOODEN BEAM – grindă de lemn circular cone 245




WOODEN MEMBERS - elemente din TUBURI CU PEREŢI SUBŢIRI - thinlemn walled hollow shafts WORK - lucru mecanic TURBINĂ CU ABUR - steam turbine WRAP - înfăşurare

Ţ

Y

ŢEAVĂ - pipe ŢEAVĂ DE ALUMINIU-aluminum pipe

YIELD CRITERIA - teorii de curgere YIELD STRENGTH - rezistenţa la U curgere UNGHI - angle YIELDING - curgere (a materialului) UNGHI DE TORSIUNE - angle of twist UNGHI DIEDRU - diedral angle UNITATE - unit

V VALOARE MEDIE - average value VALOARE NUMERICĂ - numerical value VARIAŢII DE TEMPERATURĂ temperature changes VAS CILINDRIC DIN ALAMĂ cylindrical brass vessel VASE DE PRESIUNE CILINDRICE cylindrical pressure vessels VASE DE PRESIUNE SFERICE spherical pressure vessels VECTOR - vector VERIFICA (A-) - check (to-) VIBRAŢIE - vibration VITEZĂ - velocity VITEZĂ DE ROTAŢIE - frequency of rotation VITEZĂ MAXIMĂ ADMISIBILĂ maximum allowable speed VITEZĂ UNGHIULARĂ - angular velocity VOLANT - flywhell VOLUM - volume

Z ZONA PLASTICĂ - plastic zone

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BIBLIOGRAFIE 1. BEER, P., F., JHONSON, R., E. – Mechanics of materials, Mc Graw – Hill Inc., SUA, 1992. 2. BIA, C., ILIE, V. – Rezistenţa Materialelor şi Teoria elasticităţii, Editura didactică şi Pedagogică, Bucureşti, 1983. 3. BIŢ, C., RADU, N., Gh., CIOFOAIA, V. – Elemente de mecanica ruperii, Editura Macarie, Târgovişte, 1997. 4. BIŢ, C. – Puncte de vedere asupra oboselii mecanice, Editura Universităţii Transilvania, Braşov, 2001. 5. BOLFA, T., ROŞCA, C., DUMITRIU, N., BIŢ, C. – Rezistenţa Materialelor, Braşov, 1996. 6. BROEK, D. – Elementary EngineeringFracture Mechanics, Martinus Nijhoff Publishers, London, 1982. 7. BUZDUGAN, Gh. – Rezistenţa materialelor, Editura Academiei, Bucureşti, 1986. 8. CIOFAIA, V. – Rezistenţa Materialelor şi elemente de construcţii industriale, Reprografia Universităţii din Braşov, 1987. 9. CIOFOAIA, V., BOTIŞ, M., DOGARU, F., CURTU, I. – Metoda elementelor finite, Editura Infomarket, Braşov, 2001. 10. CIOFOAIA, V., CURTU, I. – Teoria elasticităţii corpurilor izotrope şi anizotrope, Universitatea Transilvania, Braşov, 2000. 11. CIOFOAIA, V., TALPOŞI, A., BIŢ, C. – Teoria elasticităţii şi plasticităţii, Braşov, 1995. 12. CIOFOAIA, V., ULEA, M. - Teoria elasticităţii şi rezistenţa materialelor, Braşov, 1992. 13. CURTU, I., CRIŞAN, R. - Rezistenţa materialelor şi teoria elasticităţii, curs şi aplicaţii, partea I, Reprografia Universităţii Transilvania, Braşov, 1997. 14. CURTU, I., ROŞCA, C. – 2288 probleme de rezistenţa materialelor, Reprografia Universităţii Transilvania, Braşov, 1991. 15. CURTU, I., CRIŞAN, L. R., BIŢ, C. - Rezistenţa materialelor şi teoria elasticităţii, curs şi aplicaţii, partea a II - a, Universitatea Transilvania, Braşov, 1998. 16. CURTU, I., BIŢ, C. - Rezistenţa materialelor şi teoria elasticităţii, curs şi aplicaţii, partea a III - a, Universitatea Transilvania, Braşov, 2000. 17. CURTU, I., BIŢ, C. - Rezistenţa materialelor şi teoria elasticităţii, curs şi aplicaţii, partea a IV - a, Universitatea Transilvania, Braşov, 2001. 18. DEUTSCH, I., GOIA, I., CURTU, I., NEAMŢU, T., SPERCHEZ, Fl. – Probleme de rezistenţa Materialelor, Ediţia I, E.D.P., Bucureşti,1979. 19. DEUTSCH, I., GOIA, I., CURTU, I., NEAMŢU, T., SPERCHEZ, Fl. – Probleme de rezistenţa Materialelor, Ediţia a II - a, E.D.P., Bucureşti,1983. 20. DEUTSCH, I., GOIA, I., NEAMŢU, T., SPERCHEZ, Fl. – Probleme de rezistenţa Materialelor, E.D.P., Bucureşti,1980. 21. DIETER, G., E. – Metalurgie mecanică, Editura Tehnică, 1970. 22. GOIA, I., A. – Rezistenţa materialelor, Vol. I, Editura Transilvania, Braşov, 2000. 23. MUNTEANU, M., Gh., RADU, N., Gh., POPA, Al., V. - Rezistenţa materialelor I, Reprografia Universităţii din Braşov, 1989. 24. MUNTEANU, M., Gh., RADU, N., Gh., POPA, Al., V. - Rezistenţa materialelor II, Reprografia Universităţii din Braşov, 1989. 25. NĂSTĂSESCU, V., BÂRSAN, Gh. - Rezistenţa materialelor – Probleme – vol. 1 şi 2, Editura Academiei Tehnice Militare, Bucureşti, 1997. 26. RADU, Gh., N., MUNTEANU, M., Gh., BIŢ, C. - Rezistenţa materialelor şi elemente de teoria elasticităţii, Vol. I, Editura Macarie, Târgovişte, 1995. 27. RADU, Gh., N., MUNTEANU, M., Gh., BIŢ, C. - Rezistenţa materialelor şi elemente de teoria elasticităţii, Vol. II, Editura Macarie, Târgovişte, 1995. 28. TIMOSHENKO, S. – History of Strength of Materials, Mc. Graw – Hill, Book Company Inc., SUA, 1953. 29. ZHILUN, XU. – Applied Elasticity, John Wiley & Soons, SUA, 1992.

Strength of Materials

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