------- -^^^ ^ REESE LIBRARY OF THK UNIVERSITY OF CALIFORNIA. -; Glass APPLIED MECHANICS. BY GAETANO LANZA, S.B., C. &M.E., ii PROFESSOR OF THEORETICAL AND APPLIED MECHANICS, MASSACHUSETTS INSTITUTE OF TECHNOLOGY. NINTH EDITION, REVISED. FIRST THOUSAND. UNIVERSITY OF NEW YORK: JOHN WILEY & SONS. LONDON : CHAPMAN & HALL, LIMITED. 1905. REESE c COPYRIGHT, 1885, 1900, 1905, BY GAETANO LANZA. RObBKT DRUMMOND, 1RIN1BK, NBW YORK. PREFACE. THIS book is the result of the experience of the writer in teaching the subject of Applied Mechanics for the last twelve years at the Massachusetts Institute of Technology. The immediate object of publishing it is, to enable him to dispense with giving to the students a large amount of notes. As, however, it is believed that it may be found useful by others, the following remarks in regard to its general plan are submitted. The work is essentially a treatise on strength and stabil- ity ; but, inasmuch as it contains some other matter, it was thought best to call it " Applied Mechanics," notwithstanding the fact that a number of subjects usually included in trea- tises on applied mechanics are omitted. It is primarily a text-book ; and hence the writer has endeav- ored to present the different subjects in such a way as seemed to him best for the progress of the class, even though it be at some sacrifice of a logical order of topics. While no attempt has been made at originality, it is believed that some features of the work are quite different from all pre- 147GG3 iv PREFACE. vious efforts ; and a few of these cases will be referred to, with the reasons for so treating them. In the discussion upon the definition of "force," the object is, to make plain to the student the modern objections to the usual ways of treating the subject, so that he may have a clear conception of the modern aspect of the question, rather than to support the author's definition, as he is fully aware that this, as well as all others that have been given, is open to objection. In connection with the treatment of statical couples, it was thought best to present to the student the actual effect of the action of forces on a rigid body, and not to delay this subject until dynamics of rigid bodies is treated, as is usually done. In the common theory of beams, the author has tried to make plain the assumptions on which it is based. A little more prominence than usual has also been given to the longi- tudinal shearing of beams. In that part of the book that relates to the experimental results on strength and elasticity, the writer has endeavored to give the most reliable results, and to emphasize the fact, that, to obtain constants suitable for use in practice, we must deduce them from tests on full-size pieces. This prin- ciple of being careful not to apply experimental results to cases very different from those experimented upon, has long been recognized in physics, and therefore needs no justifica- tion. The government reports of tests made at the Watertown Arsenal have been extensively quoted from, as it is believed PREFACE. that 'they furnish some of our most reliable information on these subjects. The treatment of the strength of timber will be found to be quite different from what is usually given ; but it speaks for itself, and will not be commented upon here. In the chapter on the " Theory of Elasticity," a combina- tion is made of the methods of Rankine and of Grashof. In preparing the work, the author has naturally consulted the greater part of the usual literature on these subjects ; and, whenever he has drawn from other books, he has endeavored to acknowledge it. He wishes here to acknowledge the assist- ance furnished him by Professor C. H. Peabody of the Massa- chusetts Institute of Technology, who has read all the proofs, and has aided him materially in other ways in getting out the work. GAETANO LANZA. MASSACHUSETTS INSTITUTE OF TECHNOLOGY, April, 1885. PREFACE TO THE FOURTH EDITION. THE principal differences between this and the earlier editions consist in the introduction of the results of a large amount of the experimental work that has been done during the last five years upon the strength of materials. The other changes that have been made in the book are not a great many, and have been suggested as desirable by the author's experience in teaching. September, 1890. PREFACE TO THE SEVENTH EDITION. THE principal improvements in this edition consist in the introduction, in Chapter VII, of the results of a considerable amount of the experimental work on the strength of materials that has been done during the last six years. A few changes have also been made in other parts of the book. October, 1896. PREFACE 'TO THE EIGHTH EDITION. IN this edition a considerable number of additional results of recent tests, especially upon full-size pieces, have been introduced, some of the older ones having been omitted to make room for them. September, 1900. PREFACE TO THE NINTH EDITION. THE principal improvements in the Ninth Edition consist in very extensive changes in Chapter VII, in order to bring the account of the experimental work that has been performed in various places up to date. Some changes have also been made in the mathematical portion of the book, especially in the Theory of Columns. TABLE OF CONTENTS. CHAPTER I. COMPOSITION AND RESOLUTION OF FORCES . , CHAPTER II. DYNAMICS . . 75 CHAPTER III. ROOF-TRUSSES **,* 138 CHAPTER IV. BBIDGE-TRUSSES 184 CHAPTER V. CENTRE OF GRAVITY 221 CHAPTER VI. STRENGTH OF MATERIALS 240 CHAPTER VII. STRENGTH OF MATERIALS AS DETERMINED BY EXPERIMENT ..... 350 viii TABLE OF CONTENTS. CHAPTER VIII. CONTINUOUS GIRDERS 743 r CHAPTER IX. EQUILIBRIUM CURVES. ARCHES AND DOMES . . 779 CHAPTER X. THEORY OF ELASTICITY, AND APPLICATIONS .... < 852 APPLIED MECHANICS, CHAPTER I. COMPOSITION AND RESOLUTION OF FORCES. i. Fundamental Conceptions. The fundamental con- ceptions of Mechanics are Force, Matter, Space, Time, and Motion. 2. Relativity of Motion. The limitations of our natures are such that all our quantitative conceptions are relative. The truth of this statement may be illustrated, in the case of motion, by the fact, that, if we assume the shore as fixed in position, a ship sailing on the ocean is in motion, and a ship moored in the dock is at rest ; whereas, if we assume the sun as our fixed point, both ships are really in motion, as both par- take of the motion of the earth. We have, moreover, no means of determining whether any given point is absolutely fixed in position, nor whether any given direction is an absolutely fixed direction. Our only way of determining direction is by means of two points assumed as fixed ; and the straight line joining them, we are accustomed to assume as fixed in direction. Thus, it is very customary to assume the straight line joining the sun with any fixed star as a line fixed in direction ; but if the whole visible universe were in motion, so as to change the absolute direction of this line, we should have no means of recognizing it. APPLIED MECHANICS. 3. Rest and Motion. - In order to define rest and motion, we have the following ; viz., When a single point is spoken of as having motion or rest, some other point is always expressed or understood, which is for the time being considered as a fixed point, and some direc- tion is assumed as a fixed direction : and we then say that the first-named point is at rest relatively to the fixed point, when the straight line joining it with the fixed point changes neither in length, nor in direction; whereas it is said to be in motion relatively to the fixed point, when this straight line changes in length, in direction, or in both. If, on the other hand, we had considered the first-named point as our fixed point, the same conditions would determine whether the second was at rest, or in motion, relatively to the first. A body is said to be at rest relatively to a given point and to a given direction, when all its points are at rest relatively to this point and this direction. 4. Velocity and Acceleration. When the motion of one point relatively to another, or of one body relatively to another, is such that it describes equal distances in equal times, however small be the parts into which the time is divided, the motion is said to be uniform and the velocity constant. The velocity, in this case, is the space passed over in a unit of time, and is to be found by dividing the space passed over in any given time by the time ; thus, if s represent the space passed over in time /, and v represent the velocity, we shall have When the motion is not uniform, if we divide the time into small parts, and then divide the space passed over in one of these intervals by the time, and then pass to the limit as these intervals of time become shorter, we shall obtain the velocity FORCE. Thus, if A.y represent the space passed over in the interval of time A^, then we shall have v = limit of as A/ diminishes, A/ or ds In this case the rate of change of velocity per unit of time is called the Acceleration, and if we denote it by/, we have 5. Force. We shall next attempt to obtain a correct defi- nition of force, or at least of what is called force in mechanics. It may seem strange that it should be necessary to do this ; as it would appear that clear and correct definitions must have been necessary in order to make correct deductions, and there- fore that there ought to be no dispute whatever over the mean- ing of the word force. Nevertheless, it is a fact in mechanics, as well as in all those sciences which attempt to deal with the facts and laws of nature, that correct definitions are only gradu- ally developed, and that, starting with very imperfect and often erroneous views of natural laws and phenomena, it is only after these errors have been ascertained and corrected by a long range of observation and experiment, and an increased range of knowledge has been acquired, that exactness and perspicuity can be obtained in the definitions. Now, this is precisely what has happened in the case of force. In ancient times rest was supposed to be the natural state of bodies ; and it was assumed that, in order to make them move, force was necessary, and that even after they had been set in motion their own innate inertia or sluggishness would cause them to come to rest unless they were constantly urgea APPLIED MECHANICS. on by the application of some force, the bodies coming to rest whenever the force ceased acting. It was under the influence of these vague notions that such terms arose as Force of Inertia, Moment of Inertia, Vis Viva or Living Force, etc. A number of these terms are still used in mechanics; but in all such cases they have been re -defined, such new mean- ings, having been attached to them as will bring them into accord with the more advanced ideas of the present time. Such definitions will be given in the course of this work, as the necessity may arise for the use of the terms. NEWTON'S FIRST LAW OF MOTION. Ideas becoming more precise, in course of time there was framed Newton's first law of motion ; and this law is as fol- lows : A body at rest will remain at rest, and a body in motion will continue to move uniformly and in a straight line, unless and until some external force acts upon it. The assumed truth of this law was based upon the observed facts of nature ; viz., When bodies were seen to be at rest, and from rest passed into a state of motion, it was always possible to assign some cause ; i.e., they had been brought into some new relationship, either with the earth, or with some other body: and to this cause could be assigned the change of state from rest to motion. On the other hand, in the case of bodies in motion, it wa<> seen, that, if a body altered its motion from a uniform rectilinear motion, there was always some such cause that could be .assigned. Thus, in the case of a ball thrown from the hand, the attraction of the earth and the resistance of the ait soon caused it to come to rest. In the case of a ball rolled along the ground, friction (i.e., the continual contact and collision with the ground) gradually destroyed its motion, and brought it to FORCE. rest ; whereas, when such resistances were diminished by rolling it on glass or on the ice, the motion always continued longer : hence it was inferred, that, were these resistances entirely removed, the motion would continue forever. In accordance with these views, the definition of force usually given was substantially as follows : Force is that which causes, or tends to cause, a body to change its state from rest to motion, from motion to rest, or to change its motion as to direction or speed. Under these views, uniform rectilinear motion was recog- nized as being just as much a condition of equilibrium, or of the action of no force or of balanced forces, as rest ; and the recognition of this one fact upset many false notions, destroyed many incorrect conclusions, and first rendered possible a science of mechanics. Along with the above-stated definition of force is ordinarily given the following proposition ; viz., Forces are proportional to the velocities that they impart, in a unit of time (i.e. to the accelerations that they impart), to the same body. The reasoning given is as follows : Suppose a body to be moving uniformly and in a straight line, and suppose a force to act upon it for a certain length of time t in the direction of the body's motion : the effect of the force is to alter the velocity of the body ; and it is only by this alteration of velocity that we recognize the action of the force. Hence, as long as the alteration continues at the same rate, we recognize the same force as acting. If, therefore,/ represent the amount of velocity which the force would impart in one unit of time, the total increase in the velocity of the body will be //; and, if the force now stop acting, the body will again move uniformly and in the same direction, but with a velocity greater by//. Hence, if we are to measure forces by their effects, it will follow that The velocity which a force will impart to a given (or standard) APPLIED MECHANICS. body in a unit of time is a proper measure of the force. And we shall have, that two forces, each of which will impart the same velocity to the same body in a unit of time, are equal to each other ; and a force which will impart to a given body twice the velocity per unit of time that another force will impart to the same body, is itself twice as great, or, in other words, Forces are proportional to the velocities that they impart, in a unit of time (i.e. to the accelerations that they impart), to the same body. MODERN CRITICISM OF THE ABOVE. The scientists and the metaphysicians of the present time are recognizing two other facts not hitherto recognized, and the result is a criticism adverse to the above-stated definition of force. Other definitions have, in consequence, been proposed ; but none are free from objection on logical grounds, and at the same time capable of use in mechanics in a quantitative way. The two facts referred to are the following ; viz., i. That all our ideas of space, time, rest, motion, and even of direction, are relative. 2. That, because two effects are identical, it does not follow that the causes producing those effects are identical. Hence, in the light of these two facts, it is plain, that, inas- much as we can only recognize motion as relative, we can only recognize force as acting when at least two bodies are con- cerned in the transaction ; and also that if the forces are simply the causes of the motion in the ordinary popular sense of the word cause, we cannot assume, that, when the effects are equal, the causes are in every way identical, although we have, of course, a perfect right to say that they are identical so far as the production of motion is concerned. I shall now proceed, in the light of the above, to deduce a definition of force, which, although not free from objection, seems as free as any that has been framed. It is one of the facts of nature, that, when bodies are by any FORCE. means brought under certain relations to each other, certain tendencies are developed, which, if not interfered with, will exhibit themselves in the occurrence of certain definite phe- nomena. What these phenomena are, depends upon the nature of the bodies concerned, and on the relationships into which they are brought. As an illustration, we know that if an apple is placed at a certain height above the surface of the earth, there is developed between the two bodies a tendency to approach each other ; and if there is no interference with this tendency, it exhibits itself in the fall of the apple. If, on the other hand, the apple were hung on the hook of a spring balance in the same posi- tion as before, the spring would stretch, and there would be developed a tendency of the spring to make the apple move upwards. This tendency to make the apple move upwards would be just equal to the tendency of the earth and apple to approach each other. This would be expressed by saying that the pull of the spring is just equal and opposite to the weight of the apple. As other illustrations of these tendencies developed in bodies when placed in certain relations to each other, we have the following cases : (a} When two bodies collide. (b} When two substances, coming together, form a chemical union, as sodium and water. (c) When the chemical union is entered into only by raising the temperature to some special point. Any of these tendencies that are developed by bringing about any of these special relationships between bodies might properly be called a force ; and the term might properly be, and is, used in the same sense in the mental and moral world, as well as in the physical. In mechanics, however, we have to deal only with the relative motion of bodies ; and hence we give the name force only to tendencies to change the relative 8 APPLIED MECHANICS. motion of the bodies concerned ; and this, whether these ten- dencies are unresisted, and exhibit themselves in the actual occurrence of a change of motion, or whether they are resisted by equal and opposite tendencies, and exhibit themselves in the production of a tensile, compressive, or other stress in the bodies concerned, instead of motion. DEFINITION OF FORCE. Hence our definition of force, as far as mechanics has to deal with it or is capable of dealing with it, is as follows; viz., - Force is a tendency to change the relative motion of the two bodies between which that tendency exists. Indeed, when, as in the illustration given a short time ago, the apple is hung on the hook of a spring balance, there still exists a tendency of the apple and the earth to approach each other ; i.e., they are in the act of trying to approach each other ; and it is this tendency, or act of trying, that we call the force of gravitation. In the case cited, this tendency is balanced by an opposite tendency on the part of the spring ; but, were the spring not there, the force of gravitation would cause the apple to fall. Professor Rarikine calls force "an action between two bodies, either causing or tending to cause change in their relative rest or motion ;" and if the act of trying can be called an action, my definition is equivalent to his. For the benefit of any one who wishes to follow out the discussions that have lately taken place, I will enumerate the following articles that have been written on the subject : (a) " Recent Advances in Physical Science," by P. G. Tait, Lecture XIV. (b) Herbert Spencer, "First Principles of Philosophy* (certain portions of the book). MEASURE OF FORCE. () Discussion by Messrs. Spencer and Tait, " Nature," Jan. 2, 9, 1 6, 1879. (d] Force and Energy, "Nature," Nov. 25, Dec. 2, 9, 16, 1880. 6. External Force. We thus see, that, in order that a force may be developed, there must be two bodies concerned in the transaction ; and we should speak of the force as that developed or existing between the two bodies. But we may confine our attention wholly to the motion or condition of one of these two bodies ; and we may refer its motion either to the other body as a fixed point, or to some body different from either; and then,' in speaking of the force, we should speak of it as the force acting on the body under consideration, and call it an external force. It is the tendency of the other body to change the motion of the body under con- sideration relatively to the point considered as fixed. 7. Relativity of Force. In adopting the above-stated definition of force, we acknowledge our incapacity to deal with it as an absolute quantity ; for we have defined it as a tendency to change the relative motion of a pair of bodies. Hence it is only through relative motion that we recognize force; and hence force is relative, as well as motion. 8. Newton's First Law of Motion. In the light of the above discussion, we might express Newton's first law of motion as follows : A body at rest, or in uniform rectilinear motion relatively to a given point assumed as fixed, will continue at rest, or in uni- form motion in the same direction, unless and until some external force acts either on the body in question, or on the fixed point, or on the body which furnishes us our fixed direction. This law is really superfluous, as it has all been embodied in the defini- tion. 9. Measure of Force. We next need some means of comparing forces with each other in magnitude ; and, subse- 10 APPLIED MECHANICS. quently, we need to select one force as our unit force, by means of which to estimate the magnitude of other forces. Let us suppose a body moving uniformly and in a straight line, relatively to some fixed point ; as long as this motion continues, we recognize no unbalanced force acting on it ; but, if the motion changes, there must be a tendency to change that motion, or, in other words, an unbalanced force is acting on the body from the instant when it begins to change its motion. Suppose a body to be moving uniformly, and a force to be applied to it, and to act for a length of time /, and to be so applied as not to change the direction of motion of the body, but to increase its velocity; the result will be, that the velocity will be increased by equal amounts in equal times, and if f represent the amount of velocity the force would impart in one unit of time, the total increase in velocity will 'be//. This results merely from the definition of a force ; for if the velocity pro- duced in one (a standard) body by a given force is twice as great as that produced by another given force, then is the ten- dency to produce velocity twice as great in the first case as in the second, or, in other words, the first force is twice as great as the second. Hence Forces are proportional to the velocities which they will impart to a given (or standard} body in a unit of time. We may thus, by using one standard body, determine a set of equal forces, and also the proportion between different forces. 10. Measure of Mass. After having determined, as shown, a set of equal (unit) forces, if we apply two of them to different bodies, and let them act for the same length of time on each, and find that the resulting velocities are unequal, these bodies are said to have unequal masses ; whereas, if the result- ing velocities are equal, they are said to have equal masses. Hence we have the following definitions : RELATION BETWEEN FORCE AND MOMENTUM. II I . Equal forces are those which, by acting 1 for equal times on tJie same or standard body, impart to it equal velocities. 2. Equal masses are those masses to which equal forces will impart equal velocities in equal times. 11. Suppose two bodies of equal mass moving side by side with the same velocity, and uniformly, let us apply to one of them a force F in the direction of the body's motion : the effect of this force is to increase the velocity with which the body moves ; and if we wish, at the same time, to increase the velocity of the other, so that they will continue to move side by side, it will be necessary to apply an equal force to that also. We are thus employing a force 2F to impart to the two bodies the required increment of velocity. If we unite them into one, it still requires a force 2F to impart to the one body resulting from their union the re- quired increment of velocity : hence, if we double the mass to which we wish to impart a certain velocity, we must double the force, or, in other words, employ a force which would impart to the first mass alone a velocity double that required. Hence Forces are proportional to the masses to which they will impart the same velocity in the same time. 12. Momentum. The product obtained by multiplying the number of units of mass in a body by its velocity is called the momentum of the body. 13. Relation between Force and Momentum. The number of units of momentum imparted to a body in a unit of time by a given force, is evidently identical with the number of units of velocity that would be imparted by the same force, in the same time, to a unit mass. Hence Forces are proportional to the momenta (or velocities per unit of mass) which they will generate in a unit of time. 12 APPLIED MECHANICS. Hence, if F represent a force which generates, in a unit of time, a velocity/" in a body whose mass is m, we shall have and, inasmuch as the choice of our units is still under our con- trol, we so choose them that F = mf; i.e., the force F contains as many units of force as mf contains units of momentum ; in other words, The momentum generated in a body in a unit of time by a. force acting in the direction of the body's motion, is taken as a measure of the force. 14. Statical Measure of Force. When the forces are prevented from producing motion by being resisted by equal and opposite fofces, as is the case in that part of mechanics known as Statics, they must be measured by a direct comparison with other forces. An illustration of this has already been given in the case of an apple hung on the hook of a spring balance. In that case the pull of the spring is equal in magni- tude to the weight of the apple : indeed, it is very customary to adopt for forces what is known as the gravity measure, in which case we take as our unit the gravitation, or tendemy to fall, of a given piece of metal, at a given place on the surface of the earth ; in other, words, its weight at a given place. The gravity unit may thus be the kilogram, the pound, or the ounce, etc. It is evident, moreover, from our definition of force, and the subsequent discussion, that whatever we take as our unit of mass, the statical measure of a force is proportional to its dynamical measure ; i.e., the numbers representing the magni- tudes of any two forces, in pounds, are proportional to the momenta they will impart to any body in a unit of time. 15. Gravity Measure of Mass. If we assume one pound as our unit of force, one foot as our unit of length, and NEWTON'S SECOND LAW OF MOTION 13 one second as our unit of time, the ratio between the number of pounds in any given force and the momentum it will impart to a body on which it acts unresisted for a unit of time, will depend on our unit of mass ; and, as we are still at liberty to fix this as we please, it will be most convenient so to choose it that the above-stated ratio shall be unity, so that there shall be no difference in the measure of a force, whether it is measured statically or dynamically. Now, it is known that a body falling freely under the action of its own weight acquires, every second, a velocity of about thirty-two feet per second : this number is denoted by g t and varies for different distances from the centre of the earth, as does also the weight of the body. Now, if W represent the weight of the body in pounds, and m the number of units of mass in its mass, we must have, in order that the statical and dynamical measures may be equal, W = mg. Hence m.y, g i.e., the number of units of mass in a body is obtained by divid- ing the weight in pounds, by the value of g at the place where the weight is determined. The values of W and of g vary for different positions, but the value of m remains always the same for the same body. UNIT OF MASS. If m = I, then W g; or, in words, The weight in pounds of the unit of mass (when the gravity measure is used} is equal to the value of g in feet per second for the same place. 16. Newton's Second Law of Motion. Newton's second law of motion is as follows : 14 APPLIED MECHANICS. " Change of momentum is proportional to the impressed mov- ing f rce > an d occurs along the straight line in which the force is impressed" Newton states further in his " Principia :" " If any force generate any momentum, a double force will generate a double, a triple force will generate a triple, momentum, whether simultaneously and suddenly, or gradually and successively impressed. And if the body was moving before, this momentum, if in the same direction as the motion, is added; if opposite, is subtracted; or if in an oblique direc- tion, is annexed obliquely, and compounded with it, according to the direction and magnitude of the two." Part of this law has reference to the proportionality between the force and the momentum imparted to the body ; and this has been already embodied in our definition of force, and illus- trated in the discussion on the measure of forces. The other part is properly a law of motion, and may be expressed as follows : If a body have two or more velocities imparted to it simulta- neously, it will move so as to preserve them all. The proof of this law depends merely upon a proper con- ception of motion. To illustrate this law when two velocities are imparted simultaneously to a body, let us suppose a man walking on the deck of a moving ship : he then has two motions in relation to the shore, his own and that of the ship. Suppose him to walk in the direction of motion of the ship at the rate of 10 feet per second, while the ship moves at 25 feet per second relatively to the shore : then his motion in relation to the shore will be 25 -|- 10 = 35 feet per second. If, on the other hand, he is walking in the opposite direction at the same rate, his motion relatively to the shore will be 25 10 15 feet per second. Suppose a body situated at A (Fig. i) to have two motions imparted to it simultaneously, one of which would carry it to B POLYGON OF MOTIONS 15 in one second, and the other to C in one second ; and that it is required to find where it will be at the end of one second, and what path it will have pursued. c Imagine the body to move in obedience to the first alone, during one second : it would thus arrive at B ; then suppose the second motion to be imparted to the body, instead of the first, it will arrive at the end of the next sec- ond at D, where BD is equal and parallel to AC. When the two motions are imparted simultaneously, instead of suc- cessively, the same point D will be reached in one second, instead of two; and by dividing AB and AC into the same (any) number of equal parts, we can prove that the body will always be situated at some point of the diagonal AD of the parallelogram, hence that it moves along AD. Hence follows the proposition known as the parallelogram of motions. PARALLELOGRAM OF MOTIONS. If there be simultaneously impressed on a body two velocities, which would separately be represented by the lines AB and AC, the actual velocity will be represented by the line AD. which is the diagonal of the parallelogram of which AB and AC are the adjacent sides. 17. Polygon of Motions. In all the above cases, the point reached by the body at the end of a second when the two motions take place simultaneously is the same as that which would be reached at the end of two seconds if the motions took place successively ; and the path described is the straight line joining the initial position of the body, with its position at the end of one second when the motions are simultaneous. The same principle applies whatever be the number of velocities that may be imparted to a body simultaneously. Thus, if we suppose the several velocities imparted to be (Fig. 2) AB, AC, AD, AE, and AF, and it be required to 1 6 APPLIED MECHANICS. determine the resultant velocity, we first let the body move with the velocity AB for one second ; at the end of that second it is found at B ; then let it move with the velocity AC only, and "at the end of another second it will be found at c ; then with AD only, and at the end of the third second it will be found at d; at the end of the fourth at e; at the end of the fifth at /. Hence the resultant velocity, when all are imparted simultaneously, is Af, or "the closing side of the polygon. This proposition is known as the polygon of motions. POLYGON OF MOTIONS. If there be simultaneously impressed on a body any number of velocities, the resulting velocity will be represented by the closing side of a polygon of which the lines representing tJie separate velocities form the other sides. 1 8. Characteristics of a Force A force has three characteristics, which, when known, determine it ; viz., Point of Application, Direction, and Magnitude. These can be repre- sented by a straight line, whose length is made proportional to the magnitude of the force, whose direction is that of the motion which the force imparts, or tends to impart, and one end of which is the point of application of the force ; an arrow-head being usually employed to indicate the direction in which the force acts. 19. Parallelogram of Forces. PROPOSITION. If two forces acting simultaneously at the same point be represented, in point of application, direction, and magnitude, by two adjacent sides of a parallelogram, their resultant will be represented by the diagonal of the parallelo- gram, drawn from the point of application of the two forces. PROOF. In the last part of 16 was proved the propo* PARALLELOGRAM OF FORCES. I/ sition known as the Parallelogram of Motions, for the state- ment of which the reader is referred to the close of that section. We have also seen that forces are proportional to the velo- cities which they impart, or tend to impart, in a unit of time, to the same body. Hence the lines representing the two impressed forces are coincident in direction with, and proportional to, the lines repre- senting the velocities they would impart in a unit of time to the same body ; and moreover, since the resultant velocity is represented by the diagonal of the parallelogram drawn with the component velocities as sides, the resultant force must coin- cide in direction with the resultant velocity, and the length of the line representing the resultant force will bear to the result- ant velocity the same ratio that one of the component forces bears to the corresponding velocity. Hence it follows, that the resultant force will be represented by the diagonal of the paral lelogram having for sides the two component forces. 20. Parallelogram of Forces : Algebraic Solution. PROBLEM. Given two forces F and F, acting at the same point A (Fig. 3), and inclined to each other at an angle ; required the magnitude and direction of the resultant force. Let AC represent F, AB represent F t , and let angle BAG ; then will R = AD A represent in magnitude and direction the resultant force. Also let angle DAC a; then from the tri~ angle DAC we have AD 2 = AC 2 + CD 2 - 2AC. CDcosACD. But ACD = 180 - .'. cosACD = -cos* .'. R 2 = F 2 + F 2 + 2FF, cos (9 + F 2 -f 2FF, cos (9. i8 APPLIED MECHANICS. This determines the magnitude of R. To determine its direc- tion, let angle CAD a. .'. angle BAD = a, and we shall have from the triangle DAC or and similarly CD : AD = sin CAD : smACD, F t : R = sin a : sin T? .*. sin a = -sin0, R sin(0-a) = sin0. R EXAMPLES. . Given F = 47-34, 75.46, = 73 14' 21"; find R and a. 2. Given ^ = 5.36, F l = 4.27, = 32 10' ; find R and a. 3. Given F = 42.00, F t = 31.00, = 150 ; find R and a. 4. Given F = 47.00, F t 75.00, 6 = 253 ; find R and a. 21. Parallelogram of Forces when 6 = 90. When the two given forces are at right angles to each other, the for- mulae become very much simplified, since the parallelogram becomes a rectangle. From Fig. 4 we at once deduce R = V^F* + ^;, sin a = ^, R COS a = . 1. Given ^ = 2. Given ^ = 3. Given ^ = 4. Given /? = 3.0, ^ = 3.0, F t = 5.0, F l = 23.2, F t = 5.0 ; find ^ and a. 5.0 ; find i? and a. 12.0 ; find ^ and a. 21.3 ; find R and a. DECOMPOSITION OF FORCES IN ONE PLANE. 19 22. Triangle of Forces. If three forces be represented* in magnitude and direction, by the three sides of a triangle taken in order, then, if these forces be simultaneously applied at one point, they will balance each other. Conversely, three forces which, when simultaneously applied at one point, balance each other, can be correctly represented in magnitude and direction by the three sides of a triangle taken in order. These propositions, which find a very extensive application, especially in the determination of the stresses in roof and bridge trusses, are proved as follows : If we have two forces, AC and AB (see Fig. 3), acting at the point A, their resultant is, as we have already seen, AD ; and hence a force equal in magnitude and opposite in direction to AD will balance the two forces AC and AB. Now, the sides of the triangle AC DA, if taken in order, represent in magnitude and direction the force AC, the force CD or AB, and a force equal and opposite to AD ; and these three forces, if applied at the same point, would balance each other. Hence follows the proposition. Moreover, we have AC : CD \ DA = sinAUC : sin CAD : smACZ>, or F-.F, \R = sin(0 -a) : sin a : sintf; or each force is, in this case, proportional to the sine of the angle between the other two. 23. Decomposition of Forces in one Plane. It is often convenient to resolve a force into two components, in two given directions in a plane containing the force. Thus, suppose we have the force R = AD (Fig. 3), and we wish to resolve it into two components acting respectively in the directions AC and AB ; i.e., we wish to find two forces acting respectively in these directions, of which AD shall be the resultant : we 20 APPLIED MECHANICS. determine these components graphically by drawing a parallelo- gram, of which AD shall be the diagonal, and whose sides shall have the directions AC and AB respectively. The algebraic values of the magnitudes of the compo- nents can be determined by solving the triangle ADC. In the case when the directions of the components are at right angles to each other, let the force R (Fig. 5), applied at O, make an angle a with OX. We may, by drawing the rect- angle shown in the figure, decompose R into two components, F and F u along OX and O Y respectively ; and we shall readily obtain from the figure, F = R cos a, Fi = R sin a. FIG. 6. EXAMPLES. i. The force exerted by the steam upon the piston of a steam-engine at the moment when it is in the position shown in the figure is AB = 1000 Ibs. The resistance of the guides upon the cross-head DE is vertical. Determine the force acting along the connecting-rod AC and the pressure on the guides ; also resolve the force acting along the connecting-rod into two components, one along, and the other at right angles to, the crank OC. 2. A load of 500 Ibs. is placed at the apex C of the frame ACB ': find the stresses in AC and CB respectively. 3. A load of 4000 Ibs. is hung at C, on the crane ABC: find the pressure in the boom BC, and the pull on the tie AC, where BC makes an angle of 60 with the horizontal, and AC an angle of 15. COMPOSITION OF FORCES IN ONE PLANE. 21 4. A force whose magnitude is 7 is resolved into two forces whose magnitudes are 5 and 3 : find the angles they make with the given; force. 24. Composition of any Number of Forces in One Plane, all applied at the Same Point. (a) GRAPHICAL SOLUTION. Let the forces be represented (Fig. 2) by AB, AC, AD, AE, and AF respectively. Draw Be || and = AC, cd || and = AD, de || and = AE, and ef j| and = AF; then will Af represent the resultant of the five forces. This solution is to be deduced from 17 in the same way as 19 is deduced from 1 6. c, (b) ALGEBRAIC SOLUTION. Let the given forces (Fig. 9), of which B, three are represented in the figure, be F, F t) F 2 , Fy F 4 , etc. ; and let the angles l made by these forces with the axis OX o 1 ^ < jj be a, a,, 02, a 3 , a 4 , etc., respectively. FlG -9- Resolve each of these forces into two components, in the directions OX and OY respectively. We shall obtain for the components along OX OA = Fcosa, OB = ^cosa,, OC F 2 cosa 2 , etc.; and for those along OY OA, = Fs'ma, OB, == ^sino,, OC, = J? 2 sma 2 , etc. These forces are equivalent to the following two ; viz., a force Fcos a -f F, cos a, + F 2 cos a 2 -f- F 3 cos a 3 + etc. along OX, and a force .Fsin a + F\ $ m a i + F* siri 2 + F z sin a 3 -f- etc. along OY. The first may be represented by ^Fcosa, and the second by ^Fsina, where 2, stands for algebraic sum. There remains only to find the resultant of these two, the magnitude of which is given by the equation R = V(2^cosa)2 -j- 22 APPLIED MECHANICS. and, if we denote by a^ the angle made by the resultant with OX, we shall have COS a r = sin OT R EXAMPLES. a 3 = 112 Find the result- ant force and its direction. Solution. F. a. COS a. sin a. F COS a. F sin a. 47 21 0-93358 o.35 8 37 43.87826 16.84339 73 4 8 0.66913 o-743i5 48.84649 54-24995 43 82 O.I39I7 0.99027 5-9843 1 42.58161 23 112 -0.37461 0.92718 -8.6l603 21.32414 90.09303 134.99909 *. 2^ cos a = 90.09303, -. R a = 134.99909, log ^F COS a = 1.954691 = 2.210331 + (2F sin a) 2 = 162.2976. log COS Or = 9.744360 Or = 56 I/- OBSERVATION. It would be perfectly correct to use the minus sign in extracting the square root, or to call R = 162.2976 ; but then we should have 050,= 90.09303 or 162.2976 i8o -f 56 - 134.99909 ? 162.2976 = 23 6 - 7'; COMPOSITION OF FORCES APPLIED AT SAME POINT. 2$ a result which, if plotted, would give the same force as when we call R = 162.2976 and a* == 56 if. Hence, since it is immaterial whether we use the plus or the minus sign in extracting the square root provided the rest of the computation be consistent with it, we shall, for convenience, use always plus. * = 77> 3. a,= 82, a 2 = 163, S= 275- a, = o, 2 = 90. 25. Polygon of Forces. If any number of forces be represented in magnitude and direction by the sides of a polygon taken in order, then, if these forces be simultaneously applied at one point, they will balance each other. Conversely, any number of forces which, when simultaneously applied at one point, balance each other, can be correctly repre- sented in magnitude and direction by the sides of a polygon taken in order. These propositions are to be deduced from 24 (a) in the same way as the triangle of forces is deduced from the parallelo- gram of forces. 26. Composition of Forces all applied at the Same Point, and not confined to One Plane. This problem can be solved by the polygon of forces, since there is nothing in the demonstration of that proposition that limits us to a plane rather than to a gauche polygon. The following method, however, enables us to determine algebraic values for the magnitude of the resultant and for its direction. 2 4 APPLIED MECHANICS. FIG. 10. We first assume a system of three rectangular axes, OX, OY, and OZ (Fig. 10), whose origin is at the common point of the given forces. Now, let OE = F be one of the given forces. First resolve it into two forces, OC and OD, the first of which lies in the z axis, and the second perpendicular to OZ, x or, as it is usually called, in the z plane ; the plane perpendicular to OX being the x plane, and that perpendicular to OY the y plane. Then resolve OD into two com- ponents, OA along OX, and OB along OY. We thus obtain three forces, OA, OB, and OC respectively, which are equivalent to the single force OE. These three components are the edges of a rectangular parallelepiped, of which OE = Fis the diagonal. Let, now, angle EOX = a, EOY = (3, and EOZ = y ; and we have, from the right-angled triangles EOA, EOB, and EOC respectively, OA = Fcosa, OB = Fcosp, OC = Fcosy. Moreover, OA 2 + OB 2 = OD 2 and OD 2 + OC 2 = OE 2 .'. OA 2 + OB 2 + OC 2 = OE 2 , and by substituting the values of OA, OB, and OC, given above, we obtain COS 2 a -j- COS 2 (3 + COS 2 y = I ; a purely geometrical relation existing between the three angles that any line makes with three rectangular co-ordinate axes. When two of the angles a, /3, and y are given, the third can be determined from the above equation. COMPOSITIOA r OF- FORCES APPLIED AT SAME POINT. 2$ Resolve, in the same way, each of the given forces into three components, along OX, OY, and OZ respectively, and we shall thus reduce our entire system of forces to the following three forces : i. A single force 2/7 cos a along OX. 2. A single force 2/7 cos ft along OY. 3. A single force 2/7 cosy along OZ. We next proceed to find a sin- gle resultant for these three forces. Let (Fig. ii) OA = 2/7 cos a OB = 2/7 cos ft OC = FIG. xx. Compounding OA and OB, we find OD to be their resultant ; and this, compounded with OC, gives OE as the resultant of the entire system. Moreover, OE 2 = OD 2 -4- OC 2 = OA 2 + OB 2 + OC 2 , or fc = (2/7 cos a)* 4- (S^cos^) 2 -h (2/? cosy)* ( 2.F cos 0) and if we let BOX a r , EOY = ft, and have (2/7 cosy)*; = y r , we shall cos a " = OA OE R 2/7 cos 8 2/7 co r = Y~^, and cosy r = ^ - This gives us the magnitude and direction of the resultant. The same observation applies to the sign of the radical for R as in the case of forces confined to one plane. 26 APPLIED MECHANICS. DETERMINATION OF THE THIRD ANGLE FOR ANY ONE FORCE. When two of the angles a, /3, and y are given, the cosine of the third may be determined from the equation, cos 2 a + cos 2 /? + cos 2 y = i ; but, as we may use either the plus or the minus sign in extract* ing the square root, we have no means of knowing which of the two supplementary angles whose cosine has been deduced is to be used. Thus, suppose a = 45, (3 = 60, then cosy = i -- i - J = f /. y = 60, or 1 20 ; but which of the two to use we have no means of deciding. This indetermination will be more clearly seen from the fol- lowing geometrical considerations : The angle a (Fig. 12), being given as 45, locates the line representing the force on a right circular cone, whose axis is OX, and whose semi-vertical angle is AOX-BOX = 4$. On the other hand, the statement that (3 = 60 locates the force on another right circular cone, having O Y for axis, and a semi-vertical angle of 60; both cones, of course, having their vertices at O. Hence, when a and (3 are given, we know that the line representing the force is an element of both cones ; and this is all that is given. (a) Now, if the sum of the two given angles is less than 90, the cones will not intersect, and the data are consequently inconsistent. DETERMINATION OF THE THIRD ANGLE. 2/ (b) If, on the other hand, one of the given angles being greater than 90, their difference is greater than 90, the cones will not intersect, and the data are again inconsistent. (c) If a + /? = 90, the cones are tangent to each other, and 7 = 90. (d) If a -f- J3 > 90, and a /? or /3 < 90, the cones intersect, and have two elements in common ; and we have no means of determining, without more data, which intersection is intended, this being the indetermination that arises in the algebraic solution. I. Given F = EXAMPLES. 63 a = 53 49 a = 8 7 2 = 70 ft = 42' 7 = 72' 7 = 45' Find the magnitude and direction of the resultant. Solution. p a. p. Y- COS a. cosp. COS Y . F COS a. /^COS/3. F COS y. 63 49 2 53 87 42 700 7 2 45 0.60182 0.05234 0.6l888 0.74314 0.94961 0.34202 0.29250 0.30902 0.70711 37.91466 2.56466 1.23776 46.81782 46.53089 0.68404 18.42750 15.14198 I.4I422 41.71708 2/^cos a 94-03275 2/^cos 3 34-98370 2.F cos y R = V(S^cosa) 2 -j- (XF cos/3) 2 + (S/? cosy) 2 = 108.6569. log 2/^cosa = 1.620314 log S^cos/^ = 1.973279 log 2/^cosy = 1.543866 log j? = 2.036057 log R = 2.036057 log R = 2.036057 log cos a r =9-584257 Iogcosj8 r =9.937222 log cos y r =9.507809 a r = 67 25' 20 X/ (3 r = 30 4' i4 /x =71 13' 5" 28 APPLIED MECHANICS. F. a. 0- F. a. V- 2. 4-3 47 2' 65 7' 3- 5 9 90 37.5 88 3' 10 5 ' 7 6.4 68 4' 8 3 2' 4 75 73 45 27. Conditions of Equilibrium for Forces applied at a Single Point. i. When the forces are not confined to one plane, we have already found, for the square of the resultant, But this expression can reduce to zero only when we have a = o, S/^cos (3 = o, and 2/^cos y = o ; for the three terms, being squares, are all positive quantities, and hence their sum can reduce to zero only when they are separately equal to zero. Hence : If a set of balanced forces applied at a single point be resolved into components along three directions at right angles to each other, the algebraic sum of the components of the forces along each of the three directions must be equal to zero, and con- versely. 2. When the forces are all confined to one plane, let that plane be the z plane ; then y = 90 in each case, and /. (3 = 90 - a /. cos (3 = sin a /. fc = (^F cos a) 2 4- Hence, for equilibrium we must have cos a) 2 4- CSJ? sin a) 2 = o; STATICS OF RIGID BODIES. 29 and, since this is the sum of two squares, o, and S/^sina = o. Hence : If a set of balanced forces, all situated in one plane \ and acting at one point, be resolved into components along two directions at right angles to each other, and in their own plane, the algebraic sum of the components along each of the tzvo given directions must be equal to zero respectively; and conversely. 28. Statics of Rigid Bodies. A rigid body is one that does not undergo any alteration of shape when subjected to the action of external forces. Strictly speaking, no body is absolutely rigid ; but different bodies possess a greater or less degree of rigidity according to the material of which they are composed, and to other circumstances. When a force is ap- plied to a rigid body, we may have as the result, not merely a rectilinear motion in the direction' of the force, but, as will be shown later, this may be combined with a rotary motion ; in short, the criterion by which we determine the ensuing motion is, that the effect of the force will distribute itself through the body in such a way as not to interfere with its rigidity. What this mode of distribution is, we shall discuss here- after ; but we shall first proceed to some propositions which can be proved independently of this consideration. 29. Principle of Rectilinear Transferrence of Force in Rigid Bodies. If a force be applied to a rigid body at the point A (Fig. 13) in the direction AB, whatever be the motion that this force would produce, it will be prevented from taking place if an equal and opposite force be applied at A, B, C, or D, or at FlG - I3 - any point along the line of action of the force : hence we have the principle that The point of application of a force acting on a rigid body, may be transferred to any other point which lies in the line of APPLIED MECHANICS. action of the force, and also in the body, without altering the resulting motion of the body, although it does alter its state of stress. 30. Composition of two Forces in a Plane acting at Different Points of a Rigid Body, and not Parallel to Each Other. Suppose the force F (Fig. 14) to be applied at A, and F, at B t both in the plane of the paper, and acting on the rigid body abcdef. Produce the lines of direction of the forces till they meet at <9, and suppose both F and F, to act at O. Con- struct the parallelogram ODHE, where OD = F and OE = F t ; then will OH R rep^ resent the resultant force in magnitude and in direction. Its point of application may be conceived at any point along the line OH, as at C, or any other point ; and a force equal and opposite to OH, applied at any point of the line OH, will balance F at A, and F, at B. The above reasoning has assumed the points A, B, C and O, all within the body : but, since we have shown, that when this is the case, a force equal and opposite to R at C will bal- ance Fat A, and F t at B, it follows, that were these three forces applied, equilibrium would still subsist if we were to remove the part bafeghc of the rigid body ; or, in other words, The same construction holds even when the point O falls out- side the rigid body. 31. Moment of a Force with Respect to an Axis Per- pendicular to the Force. DEFINITION. The moment of a force with respect to an axis perpendicular to the force, and not intersecting it, is the FIG. 14. EQUILIBRIUM OF THREE PARALLEL FORCES. FIG. 15. product 'of the force by the common perpendicular to (shortest distance between) the force and the axis. Thus, in Fig. 15 the moment of F about an axis through O and perpendicular to the plane of the paper is F(OA). The sign of the moment will depend on the sign attached to the force and that attached to the perpen- dicular. These will be assumed in this book in such a manner as to render the following true ; viz., The moment of a force with respect to an axis is called posi- tive when, if the axis were supposed fixed, the force would cause the body on which it acts to rotate around the axis in the direc- tion of the hands of a watch as seen by the observer looking at the face. It will be called nega- tive when the rotation would take place in the opposite direction. 32. Equilibrium of Three Parallel Forces applied at Different Points of a Rigid Body. Let it be required to find a force (Fig. 16) that will balance the two forces F at A, and -F t at B. Apply at A and B respectively, and in the line AB, the equal and opposite forces Aa and Bb. Their introduction will produce no alteration in the body's motion. The resultant of F and Aa is Af, that of F, and Bb is Bg. Compound these by the method of 30, and we obtain as result- ant ce. A force equal in magnitude and opposite in direction FIG. 16. 32 APPLIED MECHANICS. to cej applied at any point of the line cC, will be the force required to balance Fat A and F, at B ; and, as is evident from the construction, this line is in the plane of the two forces. Moreover, by drawing triangle fKl equal to Bbg, we can readily prove that triangles oce and Afl are equal : hence the angle oce equals the angle fAl, and R is parallel to /^and F t . Also R = ce = ch + he = ,4 AT + A7 = F + ^ __ _ AC fK Ad and CL~. M. =- -5.- BC~ Bb~ Bb' y .'. since ^4# = ^ BC F " BC AC AB where qr is any line passing through C. Hence we have the following propositions ; viz., If three parallel forces balance each other, 1. They must lie in one plane. 2. The middle one must be equal in magnitude and opposite in direction to the sum of the other two. 3. Each force is proportional to the fcj IB o distance between the lines of direction of the other two as measured on any line intersecting all of them. The third of the above-stated con- ditions may be otherwise expressed, thus : FIG. 17. The algebraic sum of the moments of the three forces about any axis perpendicular to the forces must be zero. RESULTANT OF A PAIR OF PARALLEL FORCES. 33 PROOF. Let F, F a and R (Fig. 17) be the forces ; and let the axis referred to pass through O. Draw OA perpendicular to the forces. Then we have F(OA) + Ft (OB) = F(OC + CA) + F t (OC - BC) = (F+F l )OC + F(AC) - But, from what we have already seen, F + F, = -R and JL^JH BC AC .-. F(AC) = ^(^C) .-. F(OA) + Ft(OB) = -R(OC) -f o F,(OB) + tf(0C) = o, or the algebraic sum of the moments of t\\Q forces about the axis through O is equal to zero. 33. Resultant of a Pair of Parallel Forces. In the preceding case, the resultant of any two of the three forces F y F iy and R, in Fig. 16 or Fig. 17, is equal and opposite to the third force. Hence follow the two propositions : I. If two parallel forces act in the same direction, their resultant lies in the plane of the forces, is equal to their sum, acts in the same direction, and cuts the line joining their points of application, or any common perpendicular to the two forces, at a point which divides it internally into two segments in- versely as the forces. II. If two unequal parallel forces act in opposite directions, their resultant lies in the plane of the forces, is equal to their difference, acts in the direction of the larger force, and cuts the line joining their points of application, or any common perpen- dicular to them, at a point which (lying nearer the larger force) 34 APPLIED MECHANICS. divides it externally into two segments which are inversely as the forces. Another mode of stating the above is as follows : i. The resultant of a pair of parallel forces lies in the plane of the forces. 2. It is equal in magnitude to their algebraic sum, and coin- cides in direction with the larger force. 3. The moment of the resultant about an axis perpendicu- lar to the plane of the forces is equal to the algebraic sum of the moments about the same axis. EXAMPLES. 1. Find the length of each arm of a balance such that i ounce at the end of the long arm shall balance i pound at the end of the short arm, the length of beam being 2 feet, and the balance being so propor- tioned as to hang horizontally when unloaded. 2. Given beam =28 inches, 3 ounces to balance 15. 3. Given beam = 36 inches, 5 ounces to balance 25 ounces. MODE OF DETERMINING THE RESULTANT OF A PAIR OF PARALLEL FORCES REFERRED TO A SYSTEM OF THREE RECTANGULAR AXES. Let both forces (Fig. 18) be parallel to OZ ' ; then we have, from what has preceded, F = = F_F> = be ab ac a But from the figure or .'. Fx 2 Fxt = FjX F^ 2 RESULTANT OF NUMBER OF PARALLEL FORCES. 35 and similarly we may prove that or i. The resultant of two parallel forces is parallel to the forces and equal to their algebraic sum. R=F+F, FIG. 18. 2. The moment of the resultant with respect to OX is equal to the algebraic sum of their moments with respect to OX ; and likewise when the moments are taken with respect to OY. 34. Resultant of any Number of Parallel Forces. Let it be required to find the resultant of any number of paral- lel forces. In any such case, we might begin by compounding two of them, and then compounding the resultant of these two with a third, this new resultant with a fourth, and so on. Hence, for the magnitude of any one of these resultants, we simply add to the preceding resultant another one of the forces ; and for the moment about any axis perpendicular to the forces, we add APPLIED MECHANICS. to the moment of the preceding resultant the moment of the new force. Hence we have the following facts in regard to the resultant of the entire system : I . The resultant will be parallel to the forces and equal to their algebraic sum. 2. The moment of the resultant about any axis perpendicular to the forces will be equal to the algebraic sum of the moments of the forces about the same axis. The above principles enable us to determine the resultant in all cases, except when the algebraic sum of the forces is equal to zero. This case will be considered later. 35. Composition of any System of Parallel Forces Y when all are in One Plane. Refer the forces to a pair of rect- angular axes, OX, OY (Fig. 19), and assume OY parallel to the forces. The forces and the co-ordinates of their lines of direction being as indicated in the figure, if we denote by R the resultant, and by X Q the co-ordinate of its line of direction, we shall have, from the preceding, R = ^F; ( i ) and if moments be taken about an axis through O, and perpendicular F, F, FIG. 19. to the plane of the forces, we shall also have Rx = -S.FX. Hence R = ^F and x (t = (2) determine the resultant in magnitude and in line of action, .except when %F = o, which case will be considered later. EQUILIBRIUM OF ANY SET OF PARALLEL FORCES. $? 36, Composition of any System of Parallel Forces not confined to One Plane. Refer the forces to a set of rect- angular axes so chosen that OZ is parallel to their direction. If we denote the forces by F iy F 2 , F y F 4 , etc., and the co-ordinates of their lines of direction by (* 7,), (x 2J jj> 2 ), etc., and if we denote their resultant by R, and the co-ordinates of its line of direction by (x m j^ ), we shall have, in accordance with what has been proved in 34, 1. The magnitude of the resultant is equal to the algebraic mm of the forces , or R = 2F. 2. The moment of the resultant about OY is equal to the mm of the moments of the forces about OY, or 3. The moment of the resultant about OX is equal to the of the moments about OX, or Hence determine the resultant in all cases, except when 2(x, y y z] be the intensity of the force per unit of volume at the point (x, y, z) ; then, if we represent by R the magnitude of the resultant, and by x , y , z m the co-ordinates of the centre of the distributed force, we shall have, from the principles explained in 38 and 39, the approx- imate equations R = and these give, on passing to the limit, the exact equations R - MV - - SpydV - ~ Jpd x ~' y ~' ~ 43. Centre of' Gravity. The weight of a body, or system of bodies, is the resultant of the weight of the separate parts or particles into which it may be conceived to be divided ; and the centre of gravity of the body, or system of bodies, is the centre of the above-stated system of parallel forces, i.e., the point through which the resultant always passes, no matter how the forces are turned. The weight of any one particle is the force which gravity exerts on that particle : hence, if we repre- FORCE APPLIED TO CENTRE OF STRAIGHT ROD. 43 sent the weight per unit of volume of a body, whether it be the same for all parts or not, by w, we shall have, as an approximation, and as exact equations, fwxdV fwydV fwzdV > (0 where W denotes the entire weight of the body, and x ot y m z , the co-ordinates of its centre of gravity. If, on the other hand, we let M = entire mass of the body, dM mass of volume dV t and m = mass of unit of volume, we shall have W = Mg, w = mg, wdV mgdV = gdM. Hence the above equations reduce to fxdM fydM fzdM Equations (i) and (2) are both suitable for determining the centre of gravity; one of the sets being sometimes most con- venient, and sometimes the other. 44. Centre of Gravity of Homogeneous Bodies __ If the body whose centre of gravity we are seeking is homogeneous, or of the same weight per unit of volume throughout, we shall have, that w ==. a constant in equations (i) ; and hence these reduce to 45. Effect of a Single Force applied at the Centre of a Straight Rod of Uniform Section and Material. If a straight rod of uniform section and material have imparted to it 44 APPLIED MECHANICS. a motion, such that the velocity imparted ima unit of time to each particle of the rod is the same, and if we represent this velocity by/, then if at each point of the rod, we lay off a line xy (Fig. 24) in the direction of the motion, and representing the velocity imparted to that point, the line bounding the other ends of the lines xy will be straight, and parallel to the rod. If we conceive the rod to be divided into any number of small equal parts, and denote the mass of one of these parts by <\M, then will contain as many units of momentum as there are units of force in the force required to impart to this particle the velocity f in a unit of time ; and hence f&M is the measure of this force. Hence the resultant of the forces which impart the velocity f to every particle of the rod will have for its measure fM, where M is the entire mass of the rod ; and its point of applica- tion will evidently be at the middle of the rod. It therefore follows that The effect of a single force applied at the middle of a straight rod of uniform section and material is to impart to the rod a motion of translation in the direction of the force, all points of , the rod acquiring equal velocities in equal times. 46. Translation and Rotation combined. Suppose that we.have a straight rod AB (Fig. 25), and suppose that such a force or such forces are applied to it as will impart to the point A in a unit of time the velocity Aa, and to the point B the (different) velocity Bb in a unit of time, both being perpendicu- lar to the length of the rod. It is required to determine the motion of any other point of the rod and that of the entire rod. TRANSLATION AND ROTATION COMBINED. 45 FIG. 25. Lay off Aa and Bb (Fig. 25), and draw the line ab t and pro- duce it till it meets AB produced in O : then, when these velocities Aa and Bb are imparted to the points A and B, the rod is in the act of rotating around an axis through O perpendicular to the plane of the paper ; for when a body is rotating around an axis, the linear velocity of any point of the body is perpendicular to the line joining the point in question with the axis (i.e., the perpendicular dropped from the point in question upon the axis), and proportional to the dis- tance of the point from the axis. Hence : If the velocities of two of the points in the rod are given, and if these are perpendicular to the rod, the motion of the rod is fixed, and consists of a rotation about some axis at right angles to the rod. Another way of considering this motion is as follows : Sup- pose, as before, the velocities of the points A and B to be represented by Aa and Bb respec- tively, and hence the velocity of any other point, as x (Fig. 26), to be represented by xy, or the length of the line drawn perpendicular to FIG. 26. AB, and limited by AB and ab. Then, if we lay off Aa, Bb, = \(Aa + Bb) = Cc, and draw a,b,, and if we also lay off Aa 2 a,a, and Bb 2 = bjb, we shall have the following relations ; viz., Aa = Aa, Aa 2 , Bb = Bb, + Bb 2 , xy = X y, xy 2 , etc., or we may say that the actual motion imparted to the rod in a unit of time may be considered to consist of the following two parts : 46 APPLIED MECHANICS. i. A velocity of translation represented by Aa J} the mean velocity of the rod ; all points moving with this velocity. 2. A varying velocity, different for every different point, and such that its amount is proportional to its distance from Cy the centre of the rod, as graphically shown in the triangles Aa 2 CBb 2 . In other words, the rod has imparted to it two motions : i. A translation with the mean velocity of the rod. 2. A rotation of the rod about its centre. 47. Effect of a Force applied to a Straight Rod of Uniform Section and Material, not at its Centre. If the force be not at right angles to the rod, resolve it into two com- ponents, one acting along the rod, and the other at right angles to it. The first component evidently produces merely a trans- lation of the rod in the direction of its length : hence the second component is the only one whose effect we need to study. To do this we shall proceed to show, that, when such a rod has imparted to it the motion described in 46, the single re- A cd B sultant force which is required to impart this motion in a unit of time is a force acting at right angles to the rod, at a point different from its centre ; and we shall de- (/ FIG. 27. termine the relation between the force and the motion imparted, so that one may be deduced from the other. Let A be the origin (Fig. 27), and let Ac = x, cd = dx. AB I = length of the rod. ce =f= velocity imparted per unit of time at distance x from A. Aa = / Bb = f 2 . w weight per unit of length. m mass per unit of length = ^!. g EFFECT OF FORCE APPLIED TO A STRAIGHT ROD. 47 W = entire weight of rod. M = entire mass of rod . g R = single resultant force acting for a unit of time to produce the motion. x distance from A to point of application of R. Then we shall have, Hence, from 42, AabB) = ^(/ +/,)/ = ^(/ + / 2 ). (i) 2 2 (2) . I /. + '/. , (-) " 3 /, +/, We thus have a force R, perpendicular to AB, whose mag- nitude is given by equation (i), and whose point of application is given by equation (3) ; the respective velocities imparted by the force being shown graphically in Fig. 27. EXAMPLES. i. Given Weight of rod = W = 100 Ibs., Length of rod = 3 feet, Assume g = 32 feet per second, Force applied = R = 5 Ibs., Point of application to be 2.5 feet from one end; determine the motion imparted to the rod by the action of the force for one second. 48 APPLIED MECHANICS. Solution. Equation (i) gives us, 5 = ( Equation (2) gives, < 2 '5)(5) - (^p) (3)C/ + /), or/ + 2/ 2 = 8 .-. / 2 = 4.8, / = -1.6. Hence the rod at the end nearest the force acquires a velocity of 4.8 feet per second, and at the other end a velocity of 1.6 feet per second. The mean velocity is, therefore, 1.6 feet per second; and we may consider the rod as having a motion of translation in the direc- tion of the force with a velocity of 1.6 feet per second, and a rotation about its centre with such a speed that the extreme end (i.e., a point | feet from the centre) moves at a velocity 4.8 1.6 = 3.2 feet per second. Hence angular velocity = ^ = 2.14 per second = 122. 6 per second. , 2. Given JF== 50 Ibs., /= 5 feet. It is desired to impart to it, in one second, a velocity of translation at right angles to its length, of 5 feet per second, together with a rotation of 4 turns per second : find the force required, and its point of application. 3. Assume in example 2 that the velocity of translation is in a direction inclined 45 to the length of the rod, instead of 90. Solve the problem. 4. Given a force of 3 Ibs. acting for one-half a second at a distance of 4 feet from one end of the rod, and inclined at 30 to the rod : determine its motion. 5. Given the same conditions as in example 4, and also a force of 4 Ibs., parallel and opposite in direction to the 3-lb. force, and acting also for one-half a second, and applied at 3 feet from the other end : determine the resulting motion. MOMENT OF THE FORCES CAUSING ROTATION. 49 6. Given two equal and opposite parallel forces, each acting at right angles to the length of the rod, and each equal to 4 Ibs., one being applied at i foot from one end, and the other at the middle of the rod ; find the motion imparted to the rod through the joint action of these forces for one-third of a second. 48. Moment of the Forces causing Rotation. Re- ferring again to Fig. 26, and considering the motion of the rod as a combination of translation and rotation, if we take moments about the centre C, and compare the total moment of the forces causing the rotation alone, whose accelerations are represented by the triangles aajbj), with the total moment of the actual forces acting, whose accelerations are represented by the trapezoid AabB, we shall find these moments equal to each other ; for, as far as the forces represented by the rectangle are concerned, every elementary force nt(xy^dx on one side of the centre C has its moment (Cx)\m(xy^dx\ equal and opposite to that of the elementary force at the same distance on the other side of C : hence the total moment of the forces represented graphically by the rectangle AaJb^B is zero, and hence The moment about C of those represented by the trapezoid equals the moment of those represented by the triangles. Hence, from the preceding, and from what has been pre- viously proved, we may draw the following conclusions : i. If a force be applied at the centre of the rod, it will impart the same velocity to each particle. 2. If a force be applied at a point different from the centre, and act at right angles to its length, it will cause a translation of the rod, together with a rotation about the centre of the rod. 3. In this latter case, the moment of the forces imparting the rotation alone is equal to the moment of the single resultant force about the centre of the rod, and the velocity of translation imparted in a unit of time is equal to the number of units of force in the force, divided by the entire mass of the rod. APPLIED MECHANICS. 49. Effect of a Pair of Equal and Opposite Parallel Forces applied to a Straight Rod of Uniform Section and Material. Suppose the rod to be AB (Fig. 28), and let the two equal and opposite parallel forces be Dd and Ee, each equal to F, applied at D and E respectively. The mean velocity imparted in a unit F of time by either force will be ; and, from what we have already seen, the trap- ezoid AabB will furnish us the means of representing the actual velocity imparted to any point of the rod by the force Dd. The relative magnitudes of Aa and Bb, the accelerations at the ends, will depend, of course, on the position of D ; but we shall 77 always have Cc l(Aa -f- Bb) = , a M quantity depending only on the magnitude of the force. So, likewise, the trapezoid AaJb^B will represent the velocities imparted by the force Ee ; and while the relative magnitude of Aa l and Bb l will depend upon the position of E, we shall always have Cc l = \(Aa l + Bb,) = . Hence, since Cc = Cc,, the centre C of the rod has no motion imparted to it by the given pair of forces, hence the motion of the rod is one of rotation about its centre C. The resulting velocity of any point of the rod will be the difference between the velocities imparted by the two forces ; and if these be laid off to scale, we shall have the second figure. Hence A pair of equal and opposite parallel forces, applied to a straight rod of uniform section and material, produce a rota- tion of the rod about its centre. Also, Such a rotation about the centre of the rod cannot be pro- FIG. 28. EFFECT OF STATICAL COUPLE ON STRAIGHT ROD. 51 duced by a single force, but requires a pair of equal and op- posite parallel forces. 50. Statical Couple. A pair of equal and opposite parallel forces is called a statical couple. 51. Effect of a Single Force applied at the Centre of Gravity of a Straight Rod of Non-Uniform Section and Material. In the case of a straight rod of non-uniform sec- tion and material, we may consider the rod as composed of a set of particles of unequal mass : and if we imagine each par- ticle to have imparted to it the same velocity in a unit of time,, then, using the same method of graphical representation as before (Fig. 24), the line ab, bounding the other ends of the lines representing velocities, will be parallel to AB ; but if we were to represent by the lines xy, not the velocities imparted, but the forces per unit of length, the line bounding the other ends of these forces would not, in this case, be parallel to AB. Moreover, since these forces are proportional to the masses, and hence to the weights of the several particles, their resultant would act at the centre of gravity of the rod. Hence A force applied at the centre of gravity of a straight rod will impart the same velocity to each point of the rod ; i.e., will im- part to it a motion of translation only. 52. Effect of a Statical Couple on a Straight Rod of Non-Uniform Section and Material. Let such a rod have imparted to it only a motion of rotation about its centre of gravity, and let us adopt the same modes of graphical repre- sentation as before. Let the origin be taken at O (Fig. 29), the centre of gravity of the rod. Let Aa = /, = velocity imparted to A. Bb = / 2 = velocity imparted to B. OA = a, OB = b, OC = x. CD = f = velocity imparted to C. dM = elementary mass at C. 52 APPLIED MECHANICS, Then, from similar triangles, we have /_4*-4r, a b and hence for the force acting on dM we have dF=(CE)dx = ^xdM. Hence the whole force acting on AO, and represented graph- ically by Aa^Oy is f (*x---a J - \ xdM, aj x = o and that acting on OB, and represented by B0b t , is / f*x = f (*x = o -V 2 / xdM = J - I xdM. bjx = -b ajx = -b Hence for the resultant, or the algebraic sum, of the two, we have But from 43 we have for the co-ordinate x of the centre of gravity of the rod f, x = a xdM M but, since the origin is at the centre of gravity, we have X = O, and hence \xdM = o .-. R = o. Jx=-6 Hence the two forces represented by Aa,O and Bb,O are equal in magnitude and opposite in direction : hence the rotation about the centre of gravity is produced by a Statical Couple. MEASURE OF THE ROTATORY EFFECT. 53 Now, a train of reasoning similar to that adopted in the case of a rod of uniform section and material will show that a single force applied at some point which is not the centre of gravity of the rod will produce a motion which consists of two parts ;. viz., a motion of translation, where all points of the rod have equal velocities, and a motion of rotation around the centre of gravity of the rod. 53. Moment of a Couple. The moment of a statical couple is the product of either force by the perpendicular dis- tance between the two forces, this perpendicular distance being called the arm of the couple. 54. Measure of the Rotatory Effect. Before proceed- ing to examine the effect of a statical couple upon any rigid body whatever, we will seek a means of measuring its effect in the cases already considered. The measure adopted is the moment of the couple ; and, in order to show that it is proper to adopt this measure, it will be necessary to show That the moment of the couple is proportional to the angu- lar velocity imparted to the same rod in a unit of time ; and from this it will follow That two couples in the same plane with equal moments will balance each other if one is right-handed and the other left-handed If we assume the origin of co-ordinates at C (Fig. 30), the centre of gravity of the rod, and if we denote by a the angular velocity imparted in a unit of time by the forces F and F, and let CD * CE = X M then we have for the linear velocity of a particle situated at a distance x from C the value CLT. FIG. 30. The force which will impart this velocity in a unit of time to the mass dM is axdM. 54 APPLIED MECHANICS. The total resultant force is afxtfM, which, as we have seen, is equal to zero. The moment of the elementary force about C is and the sum of the moments for the whole rod is and this, as is evident if we take moments about C, is equal t* Fxi - Fx 2 = F(x, - x,) = F(DE). Now, fx z dM is a constant for the same rod : hence any quan- tity proportional to F(DE) is also proportional to a. The above proves the proposition. Moreover, we have F(DE) = a whence it follows, that when the moment of the couple is given, and also the rod, we can find the angular velocity imparted in a unit of time by dividing the former by fx z dM. 55. Effect of a Couple on a Straight Rod when the Forces are inclined to the Rod. We shall next show that the effect of such a couple is the same as that of a couple of equal moment whose forces are perpen- r% =^^.^ dicular to the rod. /} ^"-^^^ In this case let AD and BC be the forces (Fig. 31). The moment of this couple is the product of AD by the per- c * pendicular distance between AD and BC, the graphical representation of this being the area of the parallelogram ADBC. EFFECT OF A STATICAL COUPLE ON A RIGID BODY. 55 Resolve the two forces into components along and at right angles to the rod. The former have no effect upon the motion of the rod : the latter are the only ones that have any effect upon its motion. The moment of the couple which they form is the product of Ad by AB, graphically represented by paral- lelogram AdBb ; and we can readily show that ADBC = AdBb. Hence follows the proposition. 56. Effect of a Statical Couple on any Rigid Body. Refer the body (Fig. 32) to three rectan- gular axes, OX, OY, and OZ, assuming the origin at the centre of gravity of the body, and OZ as the axis about which the body is rotating. Let the mass of the particle P be AJ/, and its co-ordinates be Then will the force that would impart FIG. 32. to the mass AJf the angular velocity a in a unit of time be where r =. perpendicular from P on OZ, or r ^x 2 + y 2 . This force may be resolved into two, one parallel to O Y an 33> at d will be equal in amount, and directly opposed to the resultant of F l at b and F t at e and both will act along the diagonal fh of the parallelogram fchg. For we have (fg)(ab) = (fc)(de) 9 each being equal to the area of the parallelogram. . . -.&. fg " f< ^ ~f* ' hence follows the proposition. Hence follows that for a couple we may substitute another in the same plane, having the same moment, and tending to rotate the body in the same direction. COUPLES IN THE SAME OR PARALLEL PLANES. 57 FIG. 33 (a). PROP. II. Two couples in parallel planes balance each other when their moments are equal, and the directions in which they tend to rotate the body are opposite. Let (Fig. 33 (a)) the planes of both couples, be perpendicular to OZ. Reduce them both so as to have their arms equal and transfer them, each in its own plane, till their arms are in the X plane. Let ab be the arm of one couple, and dc that of the other. Then will the two couples form an equilibrate system. For the resultant of the force at a and that at c acts at e, and is twice either one of its com- ponents, and hence is equal and di- rectly opposed to the resultant of the force at b and that at d. Hence we may generalize all our propositions in regard to the effect of statical couples and we may conclude that In order that two couples may have the same effect, it is necessary i. That they be in the same or parallel planes. 2. That they have the same moment. 3. That they tend to cause rotation in the same direction (i.e., both right-handed or both left-handed when looked at from the same side}. It also follows, that, for a given statical couple, we may sub- stitute another having the magnitudes of its forces different, provided only the moment of the couple remains the same. 57. Composition of Couples in the Same or Parallel Planes. If the forces of the couples are not the same, reduce them to equivalent couples having the same force, transfer them to the same plane, and turn them so that their arms shall lie in the same straight line, as in Fig. 34; the first couple consisting of the force F at A and F at B, and the second of F at B and F at C. L t FIG. APPLIED MECHANICS. The two equal and opposite forces counterbalance each other, and we have left a couple with force F and arm AC = AB + PC .*. Resultant moment = F. AC = F(AB) + F(BC). Hence : The moment of the couple which is the resultant of two or more couples in the same or parallel planes is equal t~ the algebraic sum of the moments of the component couples. EXAMPLES. i. Convert a couple whose force is 5 and arm 6 to an equivalent couple whose arm is 3. Find the resultant of this and another coupk in the same plane and sense whose force is 7 and arm 8 ; also find the force of the resultant couple when the arm is taken as 5. Solution. Moment of first couple = 5 x 6 = 30 When arm is 3, force = ^- = 10 Moment of second couple = 7 x 8 = 56 Moment of resultant couple = 30 -f 56 = 86 When arm is 5, force = - 8 ^ = 17 1. Given the following couples in one plane : Force. Arm. 12 17 1 3 8 5 7 6 9 12 12 10 9 14 6 J Force. 5 Arm. Convert to equivalent couples having the < following : 8 20 The first and the last three are right-handed ; the second, third, and fourth are left-handed. Find the moment of the resultant couple, and also its force when it has an arm n. COUPLES IN PLANES INCLINED TO EACH OTHER. 59 58. Representation of a Couple by a Line. From the preceding we see that the effect of a couple remains the same as long as i . Its moment does not change. 2. The direction of its axis (i.e., of the line drawn perpen- dicular to tJie plane of the couple} does not change. 3. The direction in which it tends to make the body turn (right-handed or left-handed) remains the same. Hence a couple may be represented by drawing a line in the direction of its axis (perpendicular to its plane), and laying off on this line a distance containing as many units of length as there are units of moment in the couple, and indicating by a dot, an arrow-head, or some other means, in what direction one must look along the line in order that the rotation may appear right-handed. This line is called the Moment Axis of the couple. 59. Composition of Couples situated in Planes inclined to Each Other. Suppose we have two couples situated neither in the same plane nor in parallel planes, and that we wish to find their resultant couple. We may proceed as fol- lows : Substitute for them equivalent couples with equal arms, then transfer them in their own plane respectively to such posi- tions that their arms shall coincide, and lie in the line of intersection of the two planes. This having been done, let OO, (Fig. 35) be the common arm, F and F the forces of one couple, F l and F t those of the other. The forces F and F, have for their resultant R, and F and F, have R,. Moreover, we may readily show that R and R, are equal and 60 APPLIED MECHANICS. parallel, both being perpendicular to <9<9 2 . The resultant of the two couples is, therefore, a couple whose arm is OO^ and force R, the diagonal of the parallelogram on F and F lt so that R = \F 2 -h F* + 2FF l cos 0, where is the angle between the planes of the couples. Now, if we draw from O the line Oa perpendicular to OO I and to F, and hence perpendicular to the plane of the first couple, and if we draw in the same manner Ob perpendicular to the plane of the second couple, so that there shall be in Oa as many units of length as there are units of moment in the first couple, and in Ob as many units of length as there are units of moment in the second couple, we shall have i. The lines Oa and Ob are the moment axes of the two given couples respectively. 2. The lines Oa and Ob lie in the same plane with F and F T , this plane being perpendicular to OO lt 3. We have the proportion Oa-.0b = F. 00, -.F^OO^F: F,. 4. If on Oa and Ob as sides we construct a parallelogram, it will be similar to the parallelogram on F and /v We shall have the proportion Oc : R = Oa : F = Ob\F*\ and since the sides of the two parallelograms are respectively perpendicular to each other, the diagonals are perpendicular to each other ; and since we have also Oc = R ' Oa and Oa = F. OO, .'. Oc = R . OO t , F it follows that Oc is perpendicular to the plane of the resultant couple, and contains as many units of length as there are units of moment in the moment of the resultant couple; in other COUPLE AND SINGLE FORCE IN THE SAME PLANE. 6l words, Oc will represent the moment axis of the resultant couple, and we shall have Oc = \Oa* -f Ob* + 2Oa . or, if we let Oa = Z, Ob = M, Oc = G, aOb = 0, G = VZ 2 + J/ 2 + 2ZJ/cos6>. This determines the moment of the resultant couple ; and, for the direction of its moment axis, we have and sin a Oc = sin 6 G sin0. Hence we can compound and resolve couples just as we do forces, provided we represent the couples by their moment axes EXAMPLES. 1. Given L = 43, M 15, 6 = 65; find resultant couple. 2. Given Z = 40, M = 30, # = 30 ; find resultant couple. 3. Given L i, M ' = 5, # = 45; find resultant couple. 60. Resultant of a Couple and a Single Force in the Same Plane. Let M (Fig. 36 or 37) be the moment of the wF FIG. 36. FIG. 37. given couple, and let OF = F be the single force. For the given couple substitute an equivalent couple, one of whose forces is F at O, equal and directly opposed to the single 62 APPLIED MECHANICS, force F, these two counterbalancing each other, and leaving only the other force of the couple, which is equal and parallel to the original single force F, and acts along a line whose M distance from O is OA = . F Hence The resultant of a single force and a couple in the same plane is a force equal and parallel to the original force, having its line of direction at a perpendicular distance from the original force equal to the moment of the couple divided by the force. Fig. 36 shows the case when the couple is right-handed, and Fig. 37 when it is left-handed. 61. Composition of Parallel Forces in General. In each case of composition of parallel forces ( 34, 35, and 36) it was stated that the method pursued was applicable to all cases except those where ^F= o. We were obliged, at that time, to reserve this case, because we had not studied the action of a statical couple ; but now we will ad pt a method for the composition of parallel forces which will apply in all cases. (a) When all the forces are in one plane. Assume, as we did in 35, the axis OY to be parallel to the forces ; assume the forces and the co-ordinates of their lines of direction, as shown in the figure (Fig. 38). Now place at the origin O, along OY, two equal and opposite forces, each equal to x F, ; then these three forces, viz., F, at D, OA, and OB, produce the same effect as F, at D alone ; but F, at D and OR form a couple (left-handed in the figure) whose moment is F,*,. Hence the force Ft is equivalent to COMPOSITION OF PARALLEL FORCES. i. An equal and parallel force at the origin, and 2. A statical couple whose moment is P\x^. Likewise the force F 2 is equivalent to (i) an equal and par- allel force at the origin, and (2) a couple whose moment is -F 2 x 2 , etc. Hence we shall have, if we proceed in the same way with all the forces, for resultant of the entire system a single force R = ^F along OY, and a single resultant couple (Observe that downward forces and left-handed couples are to be accounted negative.) Now, there may arise two cases. i. When ^F o, and 2. When 2F><0. CASE I. When ^F = o, the resultant force along Y van- ishes, and the resultant of the entire system is a statical couple whose moment is CASE II. When %F > < o, we can reduce the resultant to a single force. Let (Fig. 39) OB represent the resultant force along OY, R = %F. With this, compound the couple whose moment is M 2< O. 2. When $F = o. along OZ. in the y plane. in the x plane. CASE I. When < o, we can reduce to a single resultant force having a fixed line of direction. Lay off (Fig. 4 1 ) along OZ, OH $F. FIG. 4X . Combining this with the first of the above-stated couples, we R--SF COMPOSITION OF PARALLEL FOKCE.1. obtain a force R = 2' I. 5 4 3 2. 5 -4 3 3 2 i 2 2 i i 3 5 -3 3 5 Find the resultant in each example. 62. Resultant of any System of Forces acting at Dif- ferent Points of a Rigid Body, all situated in One Plane. Let CF = F (Fig. 43) be one of the given forces. Let all the forces be referred to a system of rectangular axes, as in the figure, and let a = angle made by F with FIG. 43. OX, etc. Let the co-ordi- M A nates of the point of application of F be AO = x, BO = SYSTEM OF FORCES ACTING ON RIGID BODY. 6? We first decompose CF = F into two components, parallel respectively to OX and O Y. These components are CD = Fcosa, CE = Fsina. Apply at O in the line O Y two equal and opposite forces, each equal to Fsin a, and at O in the line OX two equal and opposite forces, each equal to Fcosa. Since these four are mutually balanced, they do not alter the effect of the single force ; and hence we have, in place of Fat C, the six forces CD, OM, OK, CE, ON, OG. Of these six, CE and OG form a couple whose moment is (Fsma)x = Fxsina, CD and OK form a couple whose moment is (Fcosa)y = Fycosa. These two couples, being in the same plane, give as result- ant moment their algebraic sum, or F(y cos a x sin a) . We have, therefore, instead of the single force at C, the follow- ing: i. OM Fcos a along OX. 2. ON = Fsin a along O Y. 3. The couple M F(y cos a x sin a) in the given plane. Decompose in the same way each of the given forces ; and we have, on uniting the components along OX, those along OY, and the statical couples respectively, the following: i. A resultant force along OX, R x ^Fcos a. 2. A resultant force along OY, R y ^Fsm a . 3. A resultant couple in the plane, whose moment is M = %FLi> cos a x sin a}. 68 APPLIED MECHANICS. This entire system, on compounding the two forces at O t reduces to making with OX an angle a r , where ^F cos a cos OT = R 2. A resultant couple in the same plane, whose moment is M = *%F(y cos a x sin a) . Now compound this resultant force and couple, and we have, Y for final resultant, a single force equal and parallel to B E R, and acting along a line whose perpendicular dis- tance from O is equal to M R' G -\ ,-- A* R C I ^^ L E H FIG. 44 . Suppose (Fig. 44) the force OB = ^F cos a, 614 = OR = The equation of this line may be found as follows : + (S^ 1 sin a) 2 ; and let us suppose the resultant couple to be right-handed, and let then will the line ME parallel to OR be the line of direction of the single resultant force. CONDITIONS OF EQUILIBRIUM. 69 Assuming the force R to act at any point C (x r , y r ) of this line, if we decompose it in the same way as we did the single forces previously, we obtain i. The force R cos a r = 2^ cos a along OX. 2. The force R sina r = XFsina along OY. 3. The couple R(y r cos a r x r sin a^). Hence we must have R(y r cos a r x r sin a r ) = ^F(y cos a x sin a) = J/~. Hence for the equation of the line of direction we have M y r cos a r x r sin a r = . ( i j R Another form for the same equation is cos a) _ Xr (2J?sma) = M. (2) 63. Conditions of Equilibrium. If such a set of forces be in equilibrium, there must evidently be no tendency to h-ancosa ^sina) = o. Another and a very convenient way to state the conditions of equilibrium for this case is as follows : If the forces be resolved into components along two direction? at right angles to each other, then the algebraic sum of the com- ponents along each of these directions must be zero, and th* algebraic sum of the moments of the forces about any axis pendicular to the plane of the forces must equal zero. APPLIED MECHANICS. EXAMPLES. i. Given 2. Given p. X. y> 5 3 2 10 i 3 -7 4 2 P. X. * 12 27 3 -5 - 54 30 45 Find the resultant, and the equation of its line of direction. Find the resultant, and the equation of its line of direction. 64. Resultant of any System of Forces not confined to One Plane Suppose we have a number of forces applied at different points of a rigid body, and acting in different directions, of which we wish to find the resultant. Refer them all to a system of three rect- x angular axes, OX, OY, OZ (Fig. 45). Let PR = F be one of the given forces. Re- solve it into three components, PK, PH, and PG, parallel Let FIG. 45- respectively to the three axes. RPK = a, RPH = RPG Let OA x, OB = y, OC z, be the co-ordinates of the point of application of the force F. Now introduce at B and also at O two forces, opposite in direction, and each equal to PK. We now have, instead of the force PK, the five forces PK, BM, BN, OS, and OT. The two forces PK and BN form a couple in the y plane, whose axis is a line parallel to the axis OY, and whose moment is (PK)(EB) (Fcos a )z = Fzcosa. The FORCES NOT CONFINED TO ONE PLANE. fl forces Mand OT form a couple in the z plane, whose moment is (BM)(OB} = -Jycosa. Now do the same for the other forces PH and PG, and we shall finally have, instead of the force PR, three forces, F cos a, F cos ft F cos y, acting at O in the directions OX, O V, and OZ respectively, together with six couples, two of which are in the x plane, two in the y plane, and two in the z plane. They thus form three couples, whose moments are as fol- lows : Around OX, F(y cos y z cos /?) ; Around OY, F(zcosa #cosy); Around OZ, F(x cos fty cos a) . Treat each of the given forces in the same way, and we shall have, in place of all the forces of the system, three forces, ^F cos a along OX, ^F cos J3 along OY, along OZ; and three couples, whose moments are as follows : Around OX, M x ^F(y cos y z cos ft) ; Around O Y, M y = ^F(z cos a x cos y) ; Around OZ, M z 2F(xcos(3 jycosa). The three forces give a resultant at O equal to R = V(cosa) 2 -f (XFcos/3) 2 4- -&F cosy) 2 , (i) a ( . cosa r = - - , cos ft- = - ~ S cosy r = - - *-. ( 2 ) . K APPLIED MECHANICS. For the three couples we have as resultant --* COS /a = M' COS v = M z (3) (4) A, p, and v being the angles made by the moment axis of the resultant couple with OX, O Y, and OZ respectively. Thus far we have reduced the whole system to a single result- ant force at the origin, and a couple. Sometimes we can reduce the system still farther, and sometimes not. The following investigation will show when we can do so. Let (Fig. 46) OP R be the resultant force, and OC =M the moment axis of the resultant couple. Denote the angle between them by 6 (a quantity thus far undetermined). Pro- ject OP = R on OC. Its projection will be OD = RcosO; then project, in its stead, the broken line OABP on OC. By the principles of projections, the projection of this broken line will equal OD. Now OA, AB, and BP are the co-ordinates of P, and make with OC the same angle as the axes OX, OY, and OZ ; i.e., A., //,, and v respectively : hence the length of the projection is FIG. 46. But Hence OA = OAcosX + AB = R COS = R COS Or COS A. COS0 = COS Or COS A. + COS fi r COS fJL BP = cosy r . R cos p r cos p, -f ^cosy r cosv + cos y r cosv. (5) CONDITIONS OF EQUILIBRIUM. 73 This enables us to find the angle between the resultant force and the moment axis of the resultant couple. The following cases may arise : i. When cos o, or 6 90, the force lies in the plane of the couple, and we can reduce to a single force acting at a distance from O equal to , and parallel to R at O. R 2. When cos = I, or o, the moment axis of the couple coincides in direction with the force : hence the plane of the couple is perpendicular to the force, and no farther reduction is possible. 3. When is neither o nor 90, we can resolve the couple M into two component couples, one of which, McosO, acts in a plane perpendicular to the direction of R, and the other, J/sin 0, acts in a plane containing R. The latter, on being combined with the force R at the origin, gives an equal and parallel force whose line of action is at a distance from that of R at O, equal to MsmO R 4. When M = o, the resultant is a single force at O. 5. When R o, the resultant is a couple. 65. Conditions of Equilibrium. To produce equilibrium, we must have no tendency to translation and none to rotation. Hence we must have R = o and M = o. Hence we have, in general, six conditions of equilibrium ; viz., a = o, 2,J?cos/3 == o, ^F cos 7 == o. = o, My = o, M z = a 74 APPLIED MECHANICS. EXAMPLES. 1. Prove that, whenever three forces balance each other, they must lie in one plane. 2. Show how to resolve a given force into two whose sum is given, the direction of one being also given. 3. A straight rod of uniform section and material is suspended by two strings attached to its ends, the strings being of given length, and attached to the same fixed point : find the position of equilibrium of the rod. 4. Two spheres are supported by strings attached to a given point, and rest against each other : find the tensions of the strings. 5 . A straight rod of uniform section and material has its ends resting against two inclined planes at right angles to each other, the vertical plane which passes through the rod being at right angles to the line of intersection of the two planes : find the position of equilibrium of the rod, and the pressure on each plane, disregarding friction. 6. A certain body weighs 8 Ibs. when placed in one pan of a false balance of equal arms, and 10 Ibs. in the other : find the true weight of the body. 7. The points of attachment of the three legs of a three-legged table are the vertices of an isosceles right-angled triangle ; a weight of 100 Ibs. is supported at the middle of a line joining the vertex of one of the acute angles with the middle of the opposite side : find the pressure upon each leg. 8. A heavy body rests upon an inclined plane without friction : find the horizontal force necessary to apply, to prevent it from falling. 9. A rectangular picture is supported by a string passing over a smooth peg, the string being attached in the usual way at the sides, but one-fourth the distance from the top : find how many and what are the positions of equilibrium, assuming the absence of friction. 16. Two equal and weightless rods are jointed together, and form a right angle ; they move freely about their common point : find the ratio of the weights that must be suspended from their extremities, that one of them may be inclined to the horizon at sixty degrees. ii. A weight of 100 Ibs. is suspended by two flexible strings, one of which is horizontal, and the other is inclined at an angle of thirty degrees to the vertical : find the tension in each string. D YNAMICS. DEFINITIONS. ?$ CHAPTER II. DYNAMICS. 66. Definitions -- Dynamics is that part of mechanics which discusses the forces acting, when motion is the result. Velocity, in the case of uniform motion, is the space passed over by the moving body in a unit of time ; so that, if s repre- sent the space passed over in time t t and v represent the velocity, then Velocity, in variable motion, is the limit of the ratio of the space (AJ-) passed over in a short time (A/), to the time, as the latter approaches zero : hence r-* dt Acceleration is the limit of the ratio of the velocity ^A ; Im- parted to the moving body in a short time (A/), to the time, as the time approaches zero. Hence, if a represent the accelera- tion, * 76 APPLIED MECHANICS. 67. Uniform Motion In this case the acceleration is zero, and the velocity is constant ; and we have the equation s = vt. 68. Uniformly Varying Motion. In this case the ac- celeration is constant : hence a is a constant in the equation and we obtain by one integration ds v = - = +,,. where c is an arbitrary constant : to determine it we observe, that, if v represent the value of v when / = o, we shall have v = o -f c .'. c = v and by another integration s = -.vJiera s is the space passed over in time // the arbitrary con- s f ant vanishing, because, when / = o, s is also zero. 69. Measure of Force. It has already been seen, that, when a body is either at rest or moving uniformly in a straight line, there are either no forces acting upon it, or else the forces actr > upon it are balanced. If, on the other hand, the motion of -<>e body is rectilinear, but not uniform, the only unbalanced force acting is in the direction of the motion, and equal in mag- nitude to the momentum imparted in a unit of time in the direc- tion of the motion, or, in other words, to the limit of the ratio of the momentum imparted in a short time (A*), to the time, as the latter approaches zero. MECHANICAL WORK. UNIT OF WORK. // Thus, if F denote the force acting in the direction of the motion, m the mass, and a the acceleration, we shall have ,., dv d 2 s (i) F = ma = m = m . v ' dt dt 2 From (i) we derive mdv = Fdt; (2) and, if V Q be the velocity of the moving body at the time when / = 4, and 2/ x its velocity when / = t lt we shall have Xvi r> mdv = I Jto Fdt JV Q J t Q or m(v, - v ) = J *Fdt; (3) or, in words, the momentum imparted to the body during the time / = (/, / ) by the force F, will be found by integrating the quantity Fdt between the limits / x and t Q . 70. Mechanical Work. Whenever a force is applied to a moving body, the force is either used in overcoming resist- ances (i.e., opposing forces, such as gravity or friction), and leaving the body free to continue its original motion undis- turbed, or else it has its effect in altering the velocity of the body. In either case, the work done by the force is the prod- uct of the force, by the space passed through by the body *n the direction of the force. Unit qf Work. The unit of work is that work which is done when a unit of force acts through a unit of distance in the same direction as the force ; thus, if one pound and one foot are our units of force and length respectively, the unit of work will be one foot-pound. If a constant force act upon a moving body in the direction of its motion while the body moves through the space s, the work done by the force is Fs; APPLIED MECHANICS. and this, if the force is unresisted, is the energy, or capacity for performing work, which is imparted to the body upon which the force acts while it moves through the space s. Thus, if a lo-pound weight fall freely through a height of 5 feet, the energy imparted to it by the force of gravity during this fall is 10 X 5 = 50 foot-pounds, and it would be necessary to do upon it 50 foot-pounds of work in order to destroy the velocity acquired by it during its fall. If, on the other hand, the force is a variable, the amount of work done in passing over any finite space in its own direction will be found by in- tegrating, between the proper limits, the expression The power which a machine exerts is the work which it performs in a unit of time. The unit of power commonly employed is the horse-power, which in English units is equal to 33000 foot-pounds per minute, or 550 foot-pounds per second. 71. Energy. The energy of a body is its capacity for performing work. Kinetic or Actual Energy is the energy which a body pos- sesses in virtue of its velocity ; in other words, it is the work necessary to be done upon the body in order to destroy its velocity. This is equal to the work which would have to be done to bring the body from a state of rest to the velocity with which it is moving. Assume a body whose mass is m, and sup- pose that its velocity has been changed from V Q to v v Then if F be the force acting in the direction of the motion, we shall have, from equation (2), 69, that Fvdt = mvdv; (i) but vdt = ds /. Fds = mvdv. (2) ATWOOD'S MACHINE. 79 Hence, by integration, I mvdv = / Fds *Jvo *J /. \m(v* - V 2 ) = fFds; (3) but fFds is the work that has been done on the body by the force, and the result of doing this work has been to increase its velocity from v to v t . It follows, that, in order to change the velocity from v to v u the amount of work necessary to per- form upon the body is *(*,* - *>o 2 ) = i (z> x * - *>o 2 ). (4) 6 If v = o, this expression becomes \mv*, or ^ (5) 2g which is the expression for the kinetic energy of a body of mass m moving with a velocity v t . 72. Atwood's Machine. A particular case of uniformly accelerated motion is to be found in Atwood's machine, in which a cord is passed over a pulley, and is loaded with unequal weights on the two sides. Were the weights equal, there would be no unbalanced force acting, and no motion would ensue ; but when they are unequal, we obtain as a result a uniformly accelerated motion (if we disregard the action of the pulley), because we have a constant force equal to the difference of the two weights acting on a mass whose weight is the sum of the two weights. Thus, if we have a lo-pound weight on one side and a 5-pound weight on the other, the unbalanced force acting is F = io- 5 = 5 Ibs. SO APPLIED MECHANICS. T O " i_ f The mass moved is M == - 3UL : hence the resulting ac- celeration is 73. Normal and Tangential Components of the Forces acting on a Heavy Particle. If a body be in motion, either in a straight or in a curved line, and if at a certain instant all forces cease acting on it, the body will continue to move at a uniform rate in a straight line tangent to its path at that point where the body was situated when the forces ceased acting. If an unresisted force be applied in the direction of the body's motion, the motion will still take place in the same straight line; but the velocity will vary as long as the force acts, and, from what we have seen, the equation F=m* (i) dt 2 will hold. If an unresisted force act in a direction inclined to the body's motion, it will cause the body to change its speed, and also its course, and hence to move in a curved line. Indeed, if a force acting on a body which is in motion be resolved into two components, one of which is tangent to its path and the other normal, the tangential component will cause the body to change its speed, and the normal component will cause it to change the direction of its motion. The measure of the tangential component is, as we have seen, and we will proceed to find an expression for the normal com- ponent otherwise known as the Deviating Force. For this CENTRIFUGAL FORCE. 8 1 purpose we may substitute, for a small portion of the curve, a portion of the circle of curvature ; hence we will proceed to find an expression for the centrifugal force of a body which moves uniformly with a velocity v in a circle whose radius is r. CENTRIFUGAL FORCE. Let AC (Fig. 47) be the space described in the time A/. Then we have A B AC = The motion AC may be approximately consid- ered as the result of a uniform motion AB = z/A/ nearly, .and a uniformly accelerated motion PIG. 47 . BC = itf(A/) 2 = s, where a = acceleration due to centrifugal force. But (AB) 2 = BC . BD, or (vkty = %a(MY(2r + s) t where AO = OC = r /. v 2 = %a(2r + s) approximately 2V 2 .*. a = -- approximately. 2r + s For its true value, pass to the limit where s = o. Hence we have, for the acceleration due to the centrifugal force, the expression r' Hence the centrifugal force is -equal to gr 82 APPLIED MECHANICS. DEVIATING FORCE. If a body is moving in a curved path, whether circular or not, and the unbalanced force acting on it be resolved into tan- gential and normal components, the tangential component will be, as has already been seen, and the normal component will be mv 2 _ m/dsV r '- \dt)' where r is the radius of curvature of the path at the point in question. RESULTANT FORCE. Hence it follows that the entire unbalanced force acting on the body will be or F = m 74. Components along Three Rectangular Axes of the Velocities of, and of the Forces acting on, a Moving Rociy. If we resolve the velocity into three components along OX, OY, and OZ, we shall have, for these components respectively, dx dy , dz - aDd ' this being evident from the fact that dx, dy, and dz are respec- COMPONENTS OF VELOCITIES AND FORCES. 83 tively the projections of ds on the axes OX, OY, and OZ ; and, from the differential calculus, we have ds_ dt On the other hand, dx dy A dz *' it' and 7/ are not only the components of the velocity in the directions OX, OY, and OZ, but they are also the velocities of the body in these directions respectively. Now, the case of the accelerations is different ; for, while d 2 x d 2 y , d 2 z - are the accelerations in the directions OX, OY, and OZ respec- tively, they are not the components of the acceleration dt 2 along the three axes. That they are the former is evident from the fact that , dt -f-, and are the velocities in the directions of the axes, and at at d 2 x d 2 v d 2 z , ~~, are their differential co-efficients, and hence repre- sent the accelerations along the three axes. But if we consider the components of the force acting on the body, we shall have 84 APPLIED MECHANICS. for its components along OX t OY, and OZ, if a, ft, and y are the angles made by F with the axes respectively, Fcosa = m, F cos ft = m ^ Fcosy = m -. dt 2 dt 2 ^ 2 .-. F and we found ( 73) for F, the value Hence, equating these values of F, and simplifying, we shall have the equation Hence it is plain that - , ^-, and - can only be the com- dt 2 df dt 2 ponents of the actual acceleration when the last term f J vanishes, or when r = oo , i.e., when the motion is rectilinear. Moreover, we have the two expressions (i) and (2) for the force acting upon a moving body. The truth of the proposition just proved may also be seen from the following considerations : If a parallelopiped be constructed with the edges dx dy dz CENTRIFUGAL FORCE OF A SOLID BODY. 85 the diagonal will be the actual velocity ds df and will, of course, coincide in direction with its path. On the other hand, if a parallelepiped be constructed with the edges d 2 x d 2 y d 2 z dt 2 ' dt 2 ' dt 2 ' its diagonal must coincide in direction with the force and can coincide in direction with the path, and hence with the actual acceleration d 2 s dt 2 ' only when the force is tangential to the path, and hence when the motion is rectilinear. 75. Centrifugal Force of a Solid Body. When a solid body revolves in a circle, the resultant centrifugal force of the entire body acts in the direction of the perpendicular let fell from the centre of gravity of the body on the axis of rotation, and its magnitude is the same as if its entire weight were con- centrated at its centre of gravity. PROOF. Let (Fig, 48) the angular velocity = a, and the *eta' weight = W. Assume the axis of rotation perpendicular t the plane of the paper and passing through O ; assume, as axis of ;r, the perpendicular dropped from the centre of gravity upon the axis of rotation. The co-ordinates of the centre of gravity will then be (r , _^ ), and y will be equal to zero. FIG 8 If, now, P be any particle of weight w, where r = perpendicular distance from P on axis of rotatsoo, 86 APPLIED MECHANICS. and x OA, y = AP, we shall have for the centrifugal force of the particle at P w , -a. 2 r; g but if we resolve this into two components, parallel respectively to OX and OY, we shall have for these components and o.', = -wy, g sr g \g /r g ' and, for the resultant for the entire body we shall have, parallel to OX, (i) g and F y 2wy = Wy Q = o. (2) g g Hence the centrifugal force of the entire body is F-*-W*.; (3) ani if we let v = o,x = linear velocity of the centre of gravity, we have F- Wv * * ~~~ ) wnuh 13 the same as though the entire weight of the body ;cic concentrated at its centre of gravity. EXAMPLES. H. A lo-pound weight is fastened by a rope 5 feet long to the centre, aroun 1 which it revolves at the rate of 200 turns per minute ; hrd the pull on the cord. 2. A locomotive weighing 50000 Ibs., whose driving-wheels weigh toe Ibs., is running at 60 miles per hour, the diameter of the drivers UNIFORMLY VARYING RECTILINEAR MOTION. 8/ being 6 feet, and the distance from the centre of the wheel to the centre of gravity of the same being 2 inches (the drivers not being properly balanced) \ find the pressure of the locomotive on the track (a) when the centre of gravity is directly below the centre of the wheel, and (b) when it is directly above. 3. Assume the same conditions, except- that the distance between centre of the wheel and its centre of gravity is 5 inches instead of 2. 76. Uniformly Varying Rectilinear Motion. We have already found for this case ( 68) the equations - = a = a constant. (it* and we may write for the force acting, which is, of course, coin- cident in direction with the motion, F = m = ma = a constant. dr 77. Motion of a Body acted on by the Force of Gravity only. A useful special case of uniformly varying motion is that of a body moving under the action of gravity only. The downward acceleration due to gravity is represented by g feet per second, the value of g varying at different points on the surface of the earth according to the following law : g = gi(i 0.00284 cos 2X)(i ^ feet per second, where g, = 32.1695 feet, A = latitude of the place, h = its elevation above mean sea-level in feet, R 20900000 feet. 88 APPLIED MECHANICS. If, now, we represent by h the height fallen through by a descending body in time /, we shall have the equations, v. v + gt, h = v Q t + \gt*, where v is the initial downward velocity. If, on the other hand, we represent by v the initial upward velocity, and by h the height to which the body will rise in time / under the action of gravity only, we must write the equa- tions When v = o, the first set of equations gives v = gt y h = &/, which express the law of motion of a body starting from rest and subject to the action of gravity only. Eliminate / between these equations, and we shall have or h is called the height due to the velocity v, and represents the height through which a falling body must drop to acquire the velocity v ; and v = \2gh UNRESISTED PROJECTILE. 89 is the velocity which a falling body will acquire in falling through the height h. Thus, if a body fall through a height of 50 feet, it will, by that fall, acquire a velocity of about V 2 (3 2 i) (5) = V32i6.66 = 56.7 feet per second. Again : if a body has a velocity of 40 feet per second, we shall have v 2 1600 r , h = - - = 24.8 feet ; *g 64.3 and we say that the body has a velocity due to the height 24.8 feet, i. e., a velocity which it would acquire by falling through a height of 24.8 feet. EXAMPLES. 1. A stone is dropped down a precipice, and is heard to strike the bottom in 4 seconds after it started : how high is the precipice ? 2. How long will a stone, dropped down a precipice 500 feet high, take to reach the bottom ? 3. What will be its velocity just before striking the ground? 4. A body is thrown vertically upwards with a velocity of 100 feet per second ; to what height will it rise ? 5. A body is thrown vertically upwards, and rises to a height of 50 feet. With what velocity was it thrown, and how long was it in its ascent ? 6. What will be its velocity in its ascent at a point 15 feet above the point from which it started, and what at the same point in its descent ? 78. Unresisted Projectile. In the case of an unresisted projectile, we have a body on which is impressed a uniform APPLIED MECHANICS. motion in a certain direction (the direction of its initial motion), and which is acted on by the force of gravity only. Let OPC be the path (Fig. 49), OA the initial di- rection, and v the initial velocity, and the angle -4 CUT = K e. Then we shall FlG 49 have, for the hori- zontal and vertical components of the unbalanced force acting, when the projectile is at P (co-ordinates x and j), m = o along OX, and m = mg W along O Y. dP dt* Hence ^ = ' ^ Tip = ~ g ' ^ Integrating, and observing, that, when t o, the horizontal and the vertical velocities were respectively z; cos and z> sin 0, we have dx n , , = V Q cos 0, (3) ^ n t \ i- ***'-* W These equations could be derived directly by observing that the horizontal component of the initial velocity is V Q cos 0, and that this remains constant, as there is no unbalanced force act- ing in this direction, also that v sin 9 is the initial vertical velocity ; and, since the body is acted on by gravity only, this velocity will in time / be decreased by gt. UNRESISTED PROJECTILE. 91 Integrating equations (3) and (4), and observing that for / o, x and y are both zero, we obtain X = V Q COS O.t, (5) y = V Q sin O.t - \gt\ (6) Eliminate /, and we have ^ = *tan0 -- $* - (7) 2V * COS 2 as the equation of the path, which is consequently a parabola. Equations (i), (2), (3), (4), (5), (6), and (7) enable us to solve any problem with reference to an unresisted projectile. Equation (7) may be written / v 2 sin 2 0\ g / V ~~' ~~ ~ 2 Vo *ca*0 \ P sin0cos0 which gives for the co-ordinates of the vertex _ v 2 sin 2 _ z/o 2 sin cos y\ ~~~ ) x\ - - 2g g EXAMPLES. i. An unresisted projectile starts with a velocity of 100 feet pei* second at an upward angle of 30 to the horizon ; what will be its velocity when it has reached a point situated at a horizontal distance of icou teet from its starting-point, and how long will be required for it to that point? Solution. v = 100, = 30, v cos = 86.6, v sis as 50, g = 3 2 -i6. Equation (5) gives us 1000 = 86.6 / .'. / = = 11.55 seconds. 86,6 9 2 APPLIED MECHANICS. e> sin<9 - gt = 50 - 371.5 = -3 2I -5> v = V^(86.6) 2 + (3 2I -5) 2 = V75 + 103362 = 333. Hence the point in question will be reached in nj seconds after start- ing, and the velocity will then be 333 feet per second. 2. An unresisted projectile is thrown upwards from the surface of the earth at angle of 39 to the horizontal : find the time when it will reach the earth, and the velocity it will have acquired when it reaches the earth, the velocity of throwing being 30 feet per second. 3. A lo-pound weight is dropped from the window of a car when travelling over a bridge at a speed of 25 miles an hour. How long will it take to reach the ground 100 feet below the window, and what will be the kinetic energy when it reaches the ground ? 4. With what horizontal velocity, and in what direction, must it be thrown, in order that it may strike the ground 50 feet forward of the point of starting? 5. Suppose the same lo-pound weight to be thrown vertically up- wards from the car window with a velocity of 100 feet a minute, how long will it take to reach the ground, and at what point will it strike the ground ? 79. Motion of a Body on an Inclined Plane without Friction. If a body move on an inclined plane along the line of steepest descent, subject to the action of gravity only, and if we resolve the force acting on it (i.e., its weight) into two components, along and perpen- dicular to the plane respec- tively, the latter component will be entirely balanced by the resistance of the plane, and the former will be the only unbalanced force acting on the body. MOTION OF A BODY ON AN INCLINED PLANE. 93 Suppose a body whose weight is represented (Fig. 50) by HF = W to move along the inclined path AB under the action of gravity only. Let 9 be the inclination of AB to the horizon. Resolve W into two components, and HE = ^cos 9, respectively parallel and perpendicular to the plane. The former is the only unbalanced force acting on the body, and will cause it to move down the plane with a uniformly accel- erated motion ; the acceleration being (i) If the body is either at rest or moving downwards at the beginning, it will move downwards ; whereas, if it is first mov- ing upwards, it will gradually lose velocity, and move upwards more slowly, until ultimately its upward velocity will be de- stroyed, and it will begin moving downwards. The equations for uniformly varying motion are entirely applicable to these cases. Thus, suppose that the body has an initial downward velocity v ot this velocity will, at the end of the time /, become z> = ^ = z> + Crsintf)/ (2) at .-. s = v t -f k sin B . t*, (3) and, for the unbalanced force acting, we have F=ml = !(gsmO) = WsmO. (4) at 2 g 94 APPLIED MECHANICS. If, on the other hand, the body's initial velocity is upward, and we denote this upward velocity by v of we shall have the equations v =|= v - (g*m$)t (5) s = vt - fesinfl ./ 2 (6) F= -WsinO. (7) Again, if the initial velocity is zero, equations (2) and (3) become (8) From these we obtain, for this case, 2S do) and, substituting this value of / in (8), we have v = \2g(s sin 6), (n) or, if we let s sin 6 = h the vertical distance through which the body has fallen, we have v 2gh. (12) Hence, When a body, starting from rest, falls, under the action of gravity only, through a height h, the velocity acquired is \/2gh, whether the path be vertical or inclined. EXAMPLES. i. A body moves from the top to the bottom of a plane inclined to the horizon at 30, under the action of gravity only : find the time required for the descent, and the velocity at the foot of the plane. MOTION ALONG A CURVED LINE. 95 FIG. 51. 2. In the right-angled triangle shown in the figure (Fig. 51), given AB = 10 feet, angle BAC = 30: find the time a A body would require, if acted on by gravity only, to fall from rest through each of the sides respectively, AB being vertical. 3. Given inclination of plane to the horizon = 0, length of plane = /. compare the time of falling down the plane with the time of falling down the vertical. 4. A loo-pound weight rests, without friction, on the plane of example 3. What horizontal force is required to keep it from sliding down the plane. 5. Suppose 5 pounds horizontal force to be applied (a) so as to oppose the descent, () so as to aid the descent : find in each case how long it will take the weight to descend from the top to the bottom plane. 80. Motion along a Curved Line under the Action of Gravity only. We shall consider two questions in this regard : (a) the velocity at any point of the curve (b) the time of descent through any part of the curve. (a) Velocity at any point. Let us suppose the body to have started from rest at A, and to have reached the point P in time /, where AB = x (Fig. 52). Then, since the curved line AP may be considered as the limit of a broken line running from A to P, and as it has already been seen that the velocity acquired by falling through c a certain height depends only upon the height, and not upon the incli- nation of the path, we shall have for a curved line also FIG. 52. where v is the velocity at P. APPLIED MECHANICS. (b) Time down a curve. Referring to the same figure, let / denote the time required to go from A to P, and &t the time to go from P to f, where PP' = AJ, and BB ! =. kx ; then, as we have seen that the velocity at P is \2gx, we shall have approx- imately for the space passed over in time A/, the equation or, passing to the limit, This equation gives tis ds or / = c = r J^2gX J (2) v/here, of course, the proper limits of integration must be used. If / denote the time from A to P, we have = (""-*= J * ,,VV FK, 53 EXAMPLE. A body acted on by gravity only is constrained to move in the arc of a circle from A to C (Fig. 53), radius 10 feet. Find the time of describing the arc (quadrant) and the velocity acquired by the body when it reaches SIMPLE CIRCULAR PENDULUM. 97 8i. Simple Circular Pendulum. To find the time occu- pied in a vibration of a simple circu- ^c lar pendulum, we take D (Fig. 54) as origin, and DC as axis of x, and the axis of jj/at right angles to DC. Let AC /and BD = //, we shall have for the time of a single oscillation trom A to E /-, f J * = Now, from the equation of the circle AFDE, y 2 = 2/X X 2 , we have dy_ = I - x dx y ds I y s/2 ix - & Idx - x 2 )\_2g{h - *)] dx - x 2 V2/- or This can only be integrated approximately. Expanding f i J we obtain (-3T- 7 + ~Ta 4/ 32 / 2 98 APPLIED MECHANICS. The greatest value of x is //; and if h is so small that we may omit , we shall have as our approximate result t = J-f / dx = \m vQ ~*^T\ k = nA o> * g J yhx x 2 V g( h ) o V ^ o If, however, the value of h as compared with / is too large to render it sufficiently accurate to omit , but so small that 4/ we can safely omit the higher powers of ^, we shall have xdx h ' 4 / t h 4/[_2 or ' = V^ 1 + ^ (2) a nearer approximatioa The formula is the most used, and is more nearly correct, the smaller the value of h. EXAMPLES. i . Find the length of the simple circular pendulum which is to beat seconds at a place where g = 32^. Solution. SIMPLE CYCLOIDAL PENDULUM. 99 2. What is the time of vibration of a simple circular pendulum 5 feet long? 82. Simple Cycloidal Pendulum. The equation of the cycloid is x x y = # versin \- (2ax Jf 3 )^ a . dy_ \/ 2a ~ x dx V x ds_ = /2^\5 dx \ x I Hence we shall have, for the time of a single oscillation, dx or This expression is independent of //, so that the time of vibra- tion is the same whether the arc be large or small. A body can be made to vibrate in a cycloidal arc by suspend- ing it by a flexible string between two cycloidal cheeks. This is shown from the fact that the evolute of the cycloid is another cycloid (Fig. 55). To prove this, we have, from the equation of the cycloid, y = a versin - -j- (2ax dy _ t / dx ~ V 2a x ds a &.^_ <& *^ia - x I00 APPLIED MECHANICS. Hence the radius of curvature is and since we have for the evolute the relation ds' = dp, where ds f is the elementary arc of the evolute, f*x = za .-. /= I *; i/.*r^* and, observing that when x 2a p = o, we have If x l is the abscissa of the point of the evolute, - - x + d y = a - x ds and, transforming co-ordinates to B by putting x 2 . + 2a for we obtain which is the equation of another cycloid just like the first. The motion along a vertical cycloid may also be obtained by letting a body move along a groove in the form of a cycloid acted on by gravity alone ; and in this case the time of descent of the body to the lowest point is precisely the same at what- ever point of the curve the body is placed. 83. Effect of Grade on the Tractive Force of a Rail- way Train. -Asa useful particular case of motion on an inclined plane, we have the case of a railroad train moving up or down a grade. It is necessary that a certain tractive force EFFECT OF GRADE ON TRACTIVE FORCE. IOI be exerted in order to overcome the resistances, and keep the train moving at a uniform rate along a level track. If, on the other hand, the track is not on a level, and if we resolve the weight of the train into components at right angles to and along the plane of the track, we shall have in the latter component a force which must be added to the tractive force above referred to when we wish to know the tractive force re- quired to carry it up grade, and must be subtracted when we wish to know the tractive force required to carry it down grade. The result of this subtraction may give, if the grade is suffi- ciently steep and the speed sufficiently slow, a negative quan- tity ; and in that case we must apply the brakes, instead of using steam, unless we wish the speed of the train to increase. EXAMPLES. i. A railroad train weighing 60000 Ibs., and running at 50 miles per hour, requires a tractive force of 618 Ibs. on a level ; what is the tractive force necessary when it is to ascend a grade of 50 feet per mile? What when it is to descend? Also what is the amount of work per minute in each case ? Solution. The resolution of the weight will give (Fig. 50, 7?^ tor the com- ponent along the plane, (60000)^ = 568.2 nearly. Hence Tractive force for a level = 618.0, Tractive force for ascent = 1186.2, Tractive force for descent 49.8. To ascertain the work done per minute in each case, we have (a) For a level track, 6l8 x 5 6 o x 528 = 2719200 foot-lbs. (l>) Up grade, 2719200 + 6ooo ^ x 5 = 5219200 foot-lbs. (c) Down grade, 2719200 - 6ooo X J x 5 = 219200 foot-lbs. 102 APPLIED MECHANICS, 2. Suppose the tractive force required for each 2000 Ibs. of weight of train to be, on a level track, for velocities of 5.0 miles per hour, 10.0 20.0 30.0 40.0 50.0 60 6.1 Ibs., 6.6 8.3 ii. 2 15.3 20.6 27; find the tractive force required to carry the train of example i (a) Up an incline of 50 feet per mile at 30 miles per hour. (^) Down an incline of 50 feet per mile at 30 miles per hour. (<:) Down an incline of 10 feet per mile at 20 miles per hour. (//) What must be the incline down which the train must run to require no tractive force at 40 miles per hour? 3. If in the first example the tractive force remains 618 Ibs. while the train is going down grade, what will be its velocity at the end of one minute, the grade being 10 feet per mile? 84. Harmonic Motion If we imagine a body to be moving in a circle at a uniform rate (Fig. 56), and a second body to oscillate back and forth in the diameter AB, both starting from B, and if when the first body is * at C the other is directly un- der it at G, etc., then is the second body said to FIG. 56. move in harmonic motion. A practical case of this kind of mo- tion is the motion of a slotted cross-head of an engine, as shown in the figure ig- 57) i the crank moving at a form rate. In the case of the ordinary crank, and connecting-rod connecting the drive-wheel shaft of a stationary engine with the piston-rod, FIG. 57. HARMONIC MOTION. 1 03 we have in the motion of the piston only an approximation to harmonic motion. We will proceed to determine the law of the force acting upon, and the velocity of, a body which is con- strained to move in harmonic motion. Let the body itself and the corresponding revolving body be supposed to start from B (Fig. 56), the latter revolving in left-handed rotation with an angular velocity a, and let the time taken by the former in reaching G be t: then will the angle BOC at; and we shall have, if s denote the space passed over by the body that moves with harmonic motion, s = BG OB - OCcosat, or, if r=O= OC t s = r rcosa/, (l) the velocity at the end of the time t will be V = = arsina/, (2) and the acceleration at the end of time / will be (3) Hence the force acting upon the body at that instant, in the direction of its motion, is F = m = ma 2 r cos at = ma 2 (OG). (4) dt* The force, therefore, varies directly as the distance of the body from the centre of its path. It is zero when the body is at the IO4 APPLIED MECHANICS. centre of its path, and greatest when it is at the ends of its travel, as its value is then W ma 2 r = o?r; S this being the same in amount as the centrifugal force of the revolving body, provided this latter have the same weight as the oscillating body. On the other hand, the velocity is greatest when at = - (i.e., at mid-stroke) ; and its value is then v = ar, this being also the velocity of the crank-pin at mid-stroke. EXAMPLE. Given that the reciprocating parts of an engine weigh 10000 Ibs., the length of crank being i foot, the crank making 60 revolutions per minute ; find the force required to make the cross-head follow the crank, (i) when the crank stands at 30 to the line of dead points, (2) when at 60, (3) when at the dead point. 85. Work under Oblique Force. If the force act in any other direction than that of the motion, we must resolve it into two components, the component in the direction of the motion being the only one that does work. Thus if the force F is variable, and 6 equals the angle it makes with the direction of the motion, we shall have as our expression for the work done fFcosOds. Thus if a constant force of 100 Ibs. act upon a body in a direc- tion making an angle of 30 with the line of motion, then wil! the work done by the force during the time in which it moves through a distance of 10 feet be (100) (0.86603) (10) = 866 foot-lbs. ROTATION OF RIGID BODIES. 1 05 86. Rotation of Rigid Bodies -- Suppose a rigid body (Fig. 58) to revolve about an axis perpendicular to the plane of the paper, and passing through O ; imagine a particle whose weight is w to be situated at a perpendicular distance OA = r from the axis of rotation, and let the angular accel- eration be a : let it now be required to find the moment of the force or forces required to impart this ac- celeration ; for we know that, if the axis of rotation pass through the centre of gravity of the body, the motion can be imparted only by a statical couple ; whereas if it do not pass through the centre of gravity, the motion can be imparted by a single force. We shall have, for the particle situated at A, Weight = w. Angular acceleration = a. Linear acceleration = o.r. Force required to impart this acceleration to this particle w a-r. g 7ff Moment of this force about the axis = ar 2 . g Hence the moment of the force or forces required to impart to the entire body in a unit of time a rotation about the axis through O, with an angular velocity a, is 8 8 S where / is used as a symbol to denote the limit of ^wr 2 , and is called the Moment of Inertia of the body about the axis through O. 106 APPLIED MECHANICS. 87. Angular Momentum. This quantity,, which ex- g presses the moment of the force or forces required to impart to the body the angular acceleration a about the axis in question is also called the Angular Momentum of the body when rotat- ing with the angular velocity about the given axis. 88. Actual Energy of a Rotating Body. If it be re- quired to find the actual energy of the body when rotating with the angular velocity w, we have, for the actual energy of the particle at A, g 2 2g and for that of the entire body we shall have I z = limit of ^wr 2 \ but r* = x 2 + y 2 .'. I z = limit of So/C* 2 + y 2 } = limit of Saw 2 + limit of So/? 2 , (i) In the same way we have 7* = limit of ^wy 2 + limit of ^wz 2 , (2) I y = limit of Sow 2 + limit of ^wz 2 . (3) 99. Moments of Inertia of Solids around Parallel Axes. The moment of inertia of a solid body about an axis not passing through its centre of gravity is equal to its moment of inertia about a parallel axis passing through the centre of gravity, increased by the product of the entire weight of the body by the square of the distance between the two axes. PROOF. Refer the body to a system of three rectangular axes, OX, OY, and OZ, of which OZ is the one about which the moment of inertia is taken. Let the co-ordinates of the centre of gravity of the body with reference to these axes be (^oi Jo, #<>) Through the centre of gravity of the body draw a system of rectangular axes, parallel respectively to OX, OY, and OZ. Then we shall have for the co-ordinates of any point X = X o -\~ Xi, y = y* +y Z = Z + *,. APPLIED MECHANICS. Hence 7 2 = limit of 2w(x 2 + y 2 ) = limit of *Zwx 2 4 limit of = limit of *2w(x 4- x,) 2 4- limit of %w(y + y t ) 2 = x 2 limit of 2o> 4- 2 limit of 2w 4- 2x limit of 4- limit of 2 limit of 2o> 4- y 2 limit of 2w -f 2y limit of Soy^ -f limit of -f limit of = (* 2 -f 7o 2 ) -f limit of = r 2 W 4- // 4 2^ limit of But, since 6^ x is the centre of gravity, /. ^wx I = o and Hence 4 limit of limit of o. which proves the proposition. 100. Examples of Moments of Inertia. i . To find the moment of inertia of a sphere whose radius is r and weight per unit of volume w, about the axis OZ drawn through its centre. Solution. Divide the sphere into thin slices (Fig. 65) by planes drawn perpen- dicular to OZ. Let the distance of the slice shown in the figure, above O be z, and its thickness dz : then will its radius be Vr 2 z 2 ; and we can readily see, from ex- ample 2, 93, that its moment of inertia about OZ will be dz. FIG. 65. Hence the moment of inertia of the entire sphere about OZ will be w - f V - 2 J_ r V EXAMPLES OF MOMENTS OF INERTIA. which easily reduces to I z = 15 2. To find the moment of inertia of an ellipsoid (semi-axes a, b, c) about OZ (Fig. 66). SOLUTION. The equa- tion of the ellipsoid is Divide it into thin slices perpendicular to OZ, and let the slice shown in the figure be at a distance z from O. Then will this slice be elliptical, and its semi-axes will be FIG. 66. - V-**+ -*)-* Hence the new centre of percussion is at <9. Q. E. D. 103. Impact or Collision. Impact or collision is a pressure of inappreciably short duration between two bodies. The direction of the force of impact is along the straight line drawn normal to the surfaces of the colliding bodies at their point of contact, and we may call this line the line of impact. 124 APPLIED MECHANICS. The action that occurs in the case of collision may be de- scribed as follows : at first the bodies undergo compression ; the mutual pressure between them constantly increasing, until, when it has reached its maximum, the elasticity of the mate- rials begins to overpower the compressive force, and restore the bodies wholly or partially to their original shape and dimen- sions. Central impact occurs when the line joining the centres of gravity of the bodies coincides with the line of impact. Eccentric impact occurs when these lines do not coincide. Direct impact occurs when the line along which the relative motion of the bodies takes place, coincides with the line of impact. Oblique impact occurs when these lines do not coincide. CENTRAL IMPACT. 104. Equality of Action and Re-action. One funda- mental principle that holds in all cases of central impact is the equality of action and re-action ; in other words, we must have, that, at every instant of the time during which the impact is taking place, the pressure that one body exerts upon the other is equal and opposite to that exerted by the second upon the first. The direct consequence of this principle is, that the algebraic sum of the momenta of the two bodies before impact remains unaltered by the impact, and hence that this sum is just the same at every instant of, and after, the impact. If we let m lt m 2 , be the respective masses, c lt c 2 , their respective velocities before impact, v u v 2 , their respective velocities after impact, i/, v" , their respective velocities at any given instant during the time while impact is taking place, CO-EFFICIEA r T OF RESTITUTION. 125 then we must have the following two equations true ; viz., m 1 v l + m 2 v 2 = m l c l + m 2 c 2 , (i) mjf 4- m 2 v" = m l c l 4- m 2 c 2 . (2) 105. Velocity at Time of Greatest Compression. At the instant when the compression is greatest i.e., at the instant when the elasticity of the bodies begins to overcome the deformation due to the impact, and to tend to restore them to their original forms the values of v' and v" must be equal to each other; in other words, the colliding bodies must be moving with a common velocity v = v' = v". (i) To determine this velocity, we have, from equation (2), 104, combined with (i), v = m ^ + m * c \ (2) m l 4- m 2 106. Co-efficient of Restitution. In order to determine the values v lt v 2 , of the velocities after impact, we need two equations, and hence two conditions. One of them is fur- nished by equation (i), 104. The second depends upon the nature of the material of the colliding bodies, and we may dis- tinguish three cases : i. Inelastic Impact. In this case the velocity lost up to the time of greatest compression is not regained at all, and the velocity after impact is the common velocity ^ at the instant of greatest compression. In this case the whole of the work used up in compressing the bodies is lost, as none of it is restored by the elasticity of the material. 2. Elastic Impact. In this case the velocity regained after the greatest compression, is equal and opposite to that lost up to the time of greatest compression ; therefore v z/j = c v v. (i) v 2 v v c 2 . (2) 126 APPLIED MECHANICS. We may also define this case as that in which the work lost in compressing the bodies is entirely restored by the elasticity of the material, so that j . z2 , -- -- -- r 2222 Either condition will lead to the same result. 3. Imperfectly Elastic Impact. In this case a part only of the velocity lost up to the time of greatest compression is regained after that time. If, when the two bodies are of the same material, we call e the co-efficient of restitution, then we shall so define it that v v l c* v v c 2 or, in words, the co-efficient of restitution is the ratio of the velocity regained after compression to that lost previous to that time. In this case only a part of the work done in producing the compression is regained, hence there is loss of energy. Its amount will be determined later. Strictly speaking, all bodies belong to the third class ; the value of e being always a proper fraction, and never reaching unity, the value corresponding to perfect elasticity ; nor zero, the value corresponding to entire lack of elasticity. 107. Inelastic Impact. In this case the velocity after impact is the common velocity at the time of greatest com- pression ; hence v = v, = v 2 (i) (2) And for the loss of energy due to impact we have m 2 c 2 , ^v* 1 --- (M! -f- m 2 ) , 2 2 ELASTIC IMPACT. 12 J which, on substituting the value of v, reduces to ,,"'?) (' - <> (3) 2(m 1 +- m 2 ) 1 08. Elastic Impact. In this case we have, of course, the condition, equation (i), 104, m\v-L + ^2^2 == m^c-i -f* and, for second equation, we may use equation (3), 106 ; viz., w^! 2 , m 2 v 2 2 _ mj? m^c^ 2222 Combining these two equations, we shall obtain m l We can obtain the same result without having to solve an equation of the second degree, by using instead the equations (i) and (2) of 106, together with (i) of 104; i.e., m l v I -f- m z v z = m 1 f l + m^\ or and ( 105) l ~\~ As the result of combining these equations, and eliminating v, we should obtain equations (i) and (2), as above, for the values of z\ and v 2 . In this case the energy lost by the collision is zero. 128 APPLIED MECHANICS. 109. Special Cases of Inelastic Impact. (a) Let the mass m 2 be at rest. Then c 2 o, v = m ' c ' < + * .' . Loss of energy = m * m * . ( 2 ) () Let w 2 be at rest, and let m 2 = oo ; i.e., let the mass ;;/ r strike against another which is at rest, and whose mass is in- finite. We have m 2 = oo , c 2 = o, - = o, (3) m W. ^ r Wi^ r Loss of energy = - - -- - = -L, (4) or the moving body is reduced to rest by the collision, and all its energy is expended in compression. (c) Let m l c l = m 2 c 2 ; i.e., let the two bodies move towards each other with equal momenta : o, (5) and the loss of energy = ^^ -f ^2!, (6) 2 2 the entire energy being lost. 110. Special Cases of Elastic Impact. (a) Let the mass m 2 be at rest. Then c 2 =. o, EXAMPLES OF ELASTIC AND INELASTIC IMPACT. I2g (b) Let m 2 be at rest, and let m 2 oo . Then we have ' 2 =0, ^+ I m 2 V 2 = o. (4) Hence the moving body retraces its path in the opposite direc- tion with the same velocity. (c) Let m^, = m 2 c 2 . Then our equations of condition become WiVi + m 2 v 2 = o, 2222 and from these we readily obtain i.e., both bodies return on their path with the same velocity with which they approached each other. in. Examples of Elastic and of Inelastic Impact. 1. With what velocity must a body weighing 8 pounds strike one weighing 25 pounds in order to communicate to it a velocity of 2 feet per second, (a) when the bodies are perfectly elastic, (b) when wholly inelastic. 2. Suppose sixteen impacts per minute take place between two bodies whose weights are respectively 1000 and 1200 pounds, their initial velo- cities being 5 and 2 feet per second respectively : find the loss of energy, the bodies being inelastic. 112. Imperfect Elasticity. In this case we have the relations (see 106) V Vi _ v - c 2 130 APPLIED MECHANICS. where v = m + * f * ; and we have also m^, -h m 2 v 2 = m^ -f m 2 c 2 . Determining from them the values of z/ x and v z , we obtain Vs = (i + e) - ect, (i) v 2 = (i + - ec v (2) or, by substituting for v its value, These may otherwise be put in the form -- -'-(+) - O, (5) Moreover, we have for the loss of energy due to impact E = (^ 2 - ^ 2 ) + ^(r 2 2 - v 2 2 2 or but, from (5) and (6) respectively, f s - Vl = ( T + x = angular velocity of A after impact, , = e, - *,(,, - <* 2 e 2 ) 8 f 2 r ,(l + ') (0 /,#, -f/x W 2 = C 2 4- tfaOi*! ^2^2)7 - ' - (l + when these are resolved along and normal to the rafter. Hence normal load = x cos a -\ sin 45 4 This, resolved into components acting at each end of the rafter r gives a normal downward force at each end equal to -f- I 82 APPLIED MECHANICS. Hence, resolving all the forces acting at the left-hand support into components along and at right angles to the rafter, and imposing the condition of equilibrium that the algebraic sum of their normal components shall equal zero, we have, if we call upward forces positive, -f JFsin45 (%xcosa + ^-fFsin45) #sina = o; (i) but, since we have from (i) W 2#sina = sin 45 4 W o /. jfsma = sin 45 8 ( Then, proceeding to the apex of the roof, we have that the load , W cd = 4 gives, when resolved along the two rafters, a stress in each equal to 4 Hence the load to be supported in a direction normal to the rafter at the apex is sin 45 -f- (^ cos a -\ -- sin 45). 4 8 Hence, substituting for x its value, we have y = cl=dl= 5Tsin 4 5. (3) Then, proceeding to the left-hand support, and equating to zero the algebraic sum of the components along the rafter, we have bk = (ga 0)cos45 ~f~ -^cosa -f JWsii^ = f ^5^45. (4) SCISSOR-BEAM TRUSS WITHOUT HORIZONTAL TIE. 183 We have thus determined in (2), (3), and (4) the values of x y y, and bk eh. By way of verification, proceed to the middle of the left- hand rafter, and we find the algebraic sum of the components of be and x along the rafter to be and this is the difference between bk and cl, as it should be. We have thus obtained the direct stresses ; and we have, in addition, that the rafter itself is also subjected to a bending- moment from a normal load at the centre, this load being equal to xcosa H -- sin 45 = sin 45. 4 2 How to take this into account will be explained under the " Theory of Beams." 142. Examples. The following figures of roof -trusses may be considered as a set of examples, for which the stress diagrams are to be worked out. Observe, that, wherever there is a joint, the truss is to be supposed perfectly flexible, i.e., free to turn around a pin. FIG. 98. FIG. 99- FIG. too. FIG. 101. FIG. 102. FIG. 103. FIG. 104. FIG. FIG. 106. FIG. 107. FIG. 108 1 84 APPLIED MECHANICS. CHAPTER IV. BRIDGE-TRUSSES. 143. Method of Sections. It is perfectly possible to determine the stresses in the members of a bridge-truss graphically, or by any methods that are used for roof-trusses. In this work an analytical method will be used ; i.e., a method of sections. This method involves the use of the analytical con- ditions of equilibrium for forces in a plane explained in 63. These are as follows ; viz., If a set of forces in a plane, which are in equilibrium, be resolved into components in two directions at right angles to each other, then i. The algebraic sum of the components in one of these directions must be zero. 2. The algebraic sum of the components in the other of these directions must be zero. 3. The algebraic sum of the moments of the forces about any axis perpendicular to the plane of the forces must be zero. Assume, now, a bridge-truss (Figs. 109, no, in, 112, pages 186 and 187) loaded at a part or all of the joints. Conceive a vertical section ab cutting the horizontal members 6-8 and 7~9 and the diagonal 7-8, and dividing the truss into two parts. Then the forces acting on either part must be in equilibrium, in other words, the external forces, loads, and supporting forces, acting on one part, must be balanced by the stresses in the members cut by the section ; i.e., by the forces exerted by the other part of the truss on the part under consideration. Hence we must have the three following conditions ; viz., - SHEARING-FORCE AND BENDING-MOMENT. .185 i. The algebraic sum of the vertical components of the above-mentioned forces must be zero, 2. The algebraic sum of the horizontal components of these forces must be zero. 3. The algebraic sum of the moments of these forces about any axis perpendicular to the plane of the truss must be zero. 144. Shearing-Force and Bending-Moment. Assum- ing all the loads and supporting forces to be vertical, we shall have the following as definitions. The Shearing-Force at any section is the force with which the part of the girder on one side of the section tends to slide by the part on the other side. In a girder free at one end, it is equal to the sum of the loads between the section and the free end. In a girder supported at both ends, it is equal in magnitude to the difference between the supporting force at either end, and the sum of the loads between the section and that support- ing force. The Bending-Moment at any section is the resultant moment of the external forces acting on the part of the girder to one side of the section, tending to rotate that part of the girder around a horizontal axis lying in the plane of the section. In a girder free at one end, it is equal to the sum of the moments of the loads between the section and the free end, about a horizontal axis in the section. In a girder supported at both ends, it is the difference be- tween the moment of either supporting force, and the sum of the moments of the loads between the section and that sup- port ; all the moments being taken about a horizontal axis in the section. 145. Use of Shearing-Force and Bending-Moment. The three conditions stated in 143 may be expressed as fol- lows : i. The algebraic sum of the horizontal components of the stresses in the members cut by the section must be zero. 1 86 APPLIED MECHANICS. 2. The algebraic sum of the vertical components of the stresses in the members cut by the section must balance the shearing-force. 3. The algebraic sum of the moments of the stresses in the members cut by the section, about any axis perpendicular to the plane of the truss, and lying in the plane of the section, must balance the bending-moment at the section. As the conditions of equilibrium are three in number, they will enable us to determine the stresses in the members, pro- vided the section does not cut more than three ; and this determination will require the solution of three simultaneous equations of the first degree with three unknown quantities (the stresses in the three members). By a little care, however, in choosing the section, we can very much simplify the operations, and reduce our work to the solution of one equation with only one unknown quantity ; the proper choice of the section taking the place of the elimination. 146. Examples of Bridge-Trusses. Figs. 109-1 12 rep- resent two common kinds of bridge-trusses : in the first two the braces are all i 3 5 _7]ajk.n..i3 45 17 19 21 23 25 27 29 diagonal, in the last two they are partly vertical and partly diagonal. The first two are called Warren girders, or half-lattice girders ; since there is only one system of bracing, as in the figures. When, on the other hand, there are more than one system, so that the diagonals cross each other, they are called lattice girders. 147. General Outline of the Steps to be taken in determining the Stresses in a Bridge-Truss under a Fixed Load. i. If the truss is supported at both ends, find the sup- porting forces. VVV\/K/V\/\/\A/\/\/\A/ 2 4 6 b\ 8 *| 10 12 14 16 18 20 22 24 2628 FIG. 109. 1357 a 91 a- 11 13 vw 2157 2466 8 j 10 13 FIG. no. DETERMINING THE STRESSES IN A BRIDGE-TRUSS. 1 87 2. Assume, in all cases, a section, in such a manner as not to cut more than three members if possible, or, rather, three of those that 1 13 15 17 19 21 23 25 27 28 XlXIXbd/l/l/1// brought 7 a \ 10 12 14 16 18 20 22 24 26 FlG - 2 4 6 R 10 12 14 /\/]/\, / /MX 1357 9 11 13 FIG. are into action by the loads on the truss ; and it will save labor if we assume the section so as to cut two of the three very near their point of inter- section. 3. Find the shearing-force at the section. 4. Find the bending-moment at the section. 5. Impose the analytical conditions of equilibrium on all the forces acting on the part of the girder to one side of the section, the part between the section and the free end when the girder is free at one end, or either part when it is supported at both ends. In the cases shown in Figs. 109 and no, we may describe the process as follows ; viz., (a) Find the stress in the diagonal from the fact, that (since the stress in the diagonal is the only one that has a vertical component at the section) the vertical component of the stress in the diagonal must balance the shearing-force. (b) Take moments about the point of intersection of the diagonal and horizontal chord near which the section is taken ; then the stresses in those members will have no moment, so that the moment of the stress in the other horizontal must balance the bending-moment at the section. Hence the stress in the horizontal will be found by dividing the bending-moment at the section by the height of the girder. The above will be best illustrated by some examples. I 88 APPLIED MECHANICS. EXAMPLE I. Given the semi-girder shown in Fig. no, loaded at joint 13 with 4000 pounds, and at each of the joints l > 3> 5> 7> 9 an d ii with 8000 pounds. Suppose the length of each chord and each diagonal to be 5 feet. Required the stress in each member. Solution. For the purpose of explaining the method of procedure, we will suppose that we desire to find first the stresses in 8-10 and 9-10. Assume a vertical section very near the joint 9, but to the right of it, so that it shall cut both 8-10 and 9-10. If, now, the truss were actually separated into two parts at this section, the right-hand part would, in consequence of the loads acting on it, separate from the other part. This tendency to separate is counteracted by the following three forces : i. The pull exerted by the part <$-x of the bar 9-11 on the part x-\\ of the same bar. 2. The thrust exerted by the part 8-2 of the bar 8-10 on the part ^-10 of the same bar. 3. The pull exerted by the part 9-7 of the bar 9-10 on the part y-io of the same bar. The shearing-force at this section is 8000 -f- 4000 = 12000 Ibs., and this is equal to the vertical component of the stress in the diagonal. Hence T 2OOO Stress in 9-10 = = 12000(1.1547) = 13856 Ibs. This stress is a pull, as may be seen from the fact, that, in order to prevent the part of the girder to the right of the section from sliding downwards under the action of the load, the part 9-7 of the diagonal 9-10 must pull the part y-io of the same diagonal. Next take moments about 9 : and, since the moment of the stresses in 9-1 1 and 9-10 about 9 is zero, we must have that the moment of the stress in 8-10; i.e., the product of this stress by the height of the girder, must equal the bending-moment. DETERMINING THE STRESSES 2N A BRIDGE-TRUSS. 189 The bending-moment about 9 is 8000 x 5 4- 4000 x 10 = 80000 foot-lbs. 80000 Hence Stress in 8-10 4-33 80000(0.23094) = 18475 Proceed in a similar way for all the other members. The work may be arranged as in the following table ; the diagonal stresses being deduced from the shearing-forces by multiplying by 1.1547, and the chord stresses from the bending-moments by multiplying by 0.23094. 2_ Stresses in Diagonals cut Stresses in Chords opposite the JJ Shearing- by Section, in Ibs. Bending- respective Joints. c .S> Force Moment, in O * in Ibs. foot-lbs. * J Tension. Compression. Tension. Compression. I 44OOO 50806 72OOOO 166277 2 44OOO 50806 6lOOOO 140873 3 36000 4^69 500000 11547 4 36OOO 41569 4IOOOO 94685 5 28000 32331 \ 320000 73901 6 28OOO 32331 \ 250000 57735 7 2OOOO 23094 ' ISOOOO 41569 8 20000 23094 I3OOOO 30022 9 i 2000 13856 80000 18475 10 I2OOO 13856 5OOOO ,"547 II 4OOO 4619 2OOOO 4618 12 4OOO 4619 IOOOO 2309 EXAMPLE II. Given the truss (Fig. 109) loaded at each oi the lower joints with 10000 Ibs. : find the stresses in the members. The length of chord is equal to the length of diagonal = 10 ft. Throughout this chapter, tensions will be written with the minus, and compressions with the plus sign. Solution. Total load = 14(10000) = 140000 Ibs. Each supporting force = 70000 " The entire work is shown in the following tables: i go APPLIED MECHANICS. CO ^O O\ II II o o o to >-o to *t VO 00 CO * CO rt CO ^ CO x x to o ^o O to O H M ** M <-> o 4- + + + + + to o to o to O I I I I I I I I I I I I I O to o to O X X X X X X X 10 O to >-o O to O N co co Tj- X X X X X X 888 N CO CO 1 1 1 1 1 CO. CO N M >-> *- II II II II II II II II I I I I I I I I tt N ONOO t>sVO torfcoN "-I O ONOO txO MNNNNMNNMNH-,H4 i- M CO 00 ON O NH N CO DETERMINING THE STRESSES IN A BRIDGE-TRUSS, Numbers of Diagonals. Stresses in Diagonals, in Ibs. I- 2 28-29 70000 X I-I547 = 80829 2- 3 27-28 + 60000 X I.T547 = + 69282 3- 4 26-27 60000 X I.I547 = 69282 4- 5 25-26 + 50000 X I.I547 = + 57735 5-6 24-25 50000 X LI547 = -57735 6- 7 23-24 +40000 X I.I547 = +46188 7- 8 22-23 40000 X LI547 = -46188 8- 9 21-22 + 30000 X LI547 = + 34641 9-10 2O-2I 30000 X LI547 = 34641 IO-II I9-2O + 20000 X LI547 = + 23094 11-12 18-19 20000 X I.I547 = -23094 12-13 I7-I8 + IOOOO X LI547 = + II547 I3-H 16-17 i oooo x I.I547 = -H547 14-15 I5-I6 + LOWER CHORDS. IM umbers of Chords. Stresses in Chords, in Ibs. 2- 4 26-28 65OOOO X 0.11547 = - 755 6 4- 6 24-26 I2OOOOO X 0.11547 = 138564 6- 8 22-24 I65OOOO X 0.11547 = 190526 8-10 20-22 2OOOOOO X 0.11547 = -230940 10-12 I 8-20 225OOOO X 0.11547 = 259808 12-14 16-18 245OOOO X 0.11547 = -277128 I4-l6 245OOOO X 0.11547 = 282902 1 9 2 APPLIED MECHANICS. UPPER CHORDS. Numbers of Chords. Stresses in Chords, in Ibs. '- 3 27-29 350000 X 0.11547 = + 40415 3- 5 25-27 950000 x 0.11547 = + 109697 5- 7 23-25 I45OOOO X 0.11547 = + 167432 7- 9 21-23 1850000 X 0.11547 = + 213620 9-1 1 19-21 2I5OOOO X 0.11547 = +248261 ii 13 17-19 2350000 X 0.11547 = + 267355 'j -'5 i5-i7 245OOOO X 0.11547 = + 282902 EXAMPLE III. Given the same truss as in Example II., loaded at 2, 4, 6, 8, 10, and 12 with 10000 Ibs. at each point, the remaining lower joints being loaded with 50000 Ibs. at each joint : find the stresses in the members. EXAMPLE IV. Given a semi-girder, free at one end (Fig. 112), loaded at 2, 4, and 6 with 10000 Ibs., and at 8, 10, and 12 with 5000 Ibs. : find the stresses in the members. TRAVELLING-LOAD. 148. Half-Lattice Girder: Travelling-Load. When a girder is used for a bridge, it is not subjected all the time to the same set of loads. The load in this case consists of two parts, one, the dead load, including the bridge weight, together with any permanent load that may rest upon the bridge ; and the other, the moving or variable load, also called the travelling-load, such as the weight of the whole or part of a railroad train if it is a railroad bridge, or the weight of the passing teams, etc., if it is a common- road bridge. Hence it is necessary that we should be able to determine the amount and distribution of the loads upon the bridge which will produce the greatest tension or the greatest GREATEST DIAGONAL STRESSES IN GIRDER. 193 compression in every member, and the consequent stress pro- duced. 149. Greatest Stresses in Semi-Girder. Wherever the section be assumed in a semi-girder, it is evident that any- load placed on the truss at any point between the section and the free end increases both the shearing-force and the bending- momerit at that section, and that any load placed between the section and the fixed end has no effect whatever on either the shearing-force or the bending-moment at that section. Hence every member of a semi-girder will have a greater stress upon it when the entire load is on, than with any partial load. 150. Greatest Chord Stresses in Girder supported at Both Ends. Every load which is placed upon the truss, no matter where it is placed, will produce at any section whatever a bending-moment tending to turn the two parts of the truss on the two sides of the section upwards from the supports ; i.e., so as to render the truss concave upwards. Hence every load that is placed upon the truss causes com- pression in every horizontal upper chord, and tension in every horizontal lower chord. Hence, in order to obtain the greatest chord stresses, we assume the whole of the moving load to be upon the bridge. 151. Greatest Diagonal Stresses in Girder supported at Both Ends. To determine the distribution of the load that will produce the greatest stress of a certain kind (tension or compression) in any given diagonal, let us suppose the diag- onal in question to be 7-8 (Fig. 109), through which we take our section ab. Now it is evident that any load placed on the truss between ab and the left-hand (nearer) support will cause a shearing-force at that section which will tend to slide the part of the girder to the left of the section downwards with refer- ence to -the other part, and hence will cause a compressive stress in 7-8 ; while any load between the section and the right- 194 APPLIED MECHANICS. hand (farther) support will cause a shearing-force of the oppo- site kind, and hence a tension in the bar 7-8. Now, the bridge weight itself brings an equal load upon each joint ; hence, when the bridge weight is the only load upon the truss, the bar 7-8 is in tension. Hence, any load placed upon the truss between the section and the farther support tends to increase the shearing-force at that section due to the dead load (provided this is equally dis- tributed among the joints) ; whereas any load placed between the section and the nearer support tends to decrease the shear- ing-force at the section due to the dead load, or to produce a shearing-force of the opposite kind to that produced by the dead load at that section. Hence, if we assume the dead load to be equally distributed among the joints, we shall have the two following propositions true : (a) In order to determine the greatest stress in any diagonal which is of the same kind as that produced by the dead load, we must assume the moving load to cover all the panel points between the section and the farther abutment, and no other panel points. (b) In order to determine the greatest stress in any diagonal of the opposite kind to that produced by the dead load, we must assume the moving load to cover all the panel points between the section and the nearer abutment, and no others. This will be made clear by an example. EXAMPLE I. Given the truss shown in Fig. 113. Length of chord = length of diagonal = A A A g !Lu 10 feet. Dead load = 8000 Ibs. Y 4 Y Y Y Y Y Ypl applied at each upper panel point. FIG Moving load = 30000 Ibs. applied at each upper panel point. Find the greatest stresses in the members. EXAMPLE OF BRIDGE-TRUSS, 195 Solution, (a) Chord Stresses. Assume the whole load to be upon the bridge : this will give 38000 each (1) + 76788 (3) 4-. 20R423 (5) -f- 296181 (7; + 340059 (9) (2) -153575 (4) -263272(6) -329090 (8)- 3510251 Ibs. at eacn upper panel point ; i.e., omit- ting I and 17, where the load acts directly on the support, and not on the truss. FlG ' II4- Hence, considering the bridge so loaded, we shall have the fol- lowing results for the chord stresses : Each supporting force = sSooof-J 133000. Section at Bending-Moment, in foot-lbs. 2 16 133000 x 5 = 665000 3 15 133000 X IO = 1330000 4 14 133000 X 15 38000 x 5 = 1805000 5 13 133000 X 2O 38000 X 10 = 2280000 6 12 133000 X 25 - 3 8ooo( 5 -f- 15) = 2565000 7 ii 133000 X 30 - 38000(10 + 20) = 2850000 8 10 133000 x 35 38ooo( 5 + 15 -{- 25) = 2945000 9 133000 X 40 38000(10 4- 20 -j- 30) = 3040000 Numbers of Chords. Stresses in Upper Chords. i-3 I5-I7 665000 X 0.11547 = + 76788 3-5 I3-I5 1805000 X 0.11547 = +208423 5-7 11-13 2565000 X 0.11547 = +296181 7-9 9-1 1 2945000 X 0.11547 = +340059 APPLIED MECHANICS. Numbers of Chords. Stresses in Lower Chords. 2- 4 14-16 1330000 X O.II547 = -153575 4- 6 12-14 2280000 X O.II547 = -263272 6- 8 IO-I2 2850000 X O.II547 = 329090 8-10 3040000 X O.II547 = -351029 Next, as to the diagonals, take, for instance, the diagonal 7-8. When the dead load alone is on the bridge, the diagonal 7-8 is in tension. From the preceding, we see that the greatest tension is produced in this bar when the moving load is on the points 9, n, 13, and 15, and the dead load only on the points 3, 5, 7. Now, a load of 38000 Ibs. at 13, for instance, causes a shearing-force of (38000) = 9500 Ibs. at any section to the 10 left of 13; and this shearing-force tends to cause the part to the left of the section to slide upwards, and that to the right downwards. On the other hand, with the same load at the same place, there is produced a shearing-force of (38000) = 28500 Ibs. 16 at any section to the right of 13 ; and this shearing-force tends to cause the part to the left to slide downwards, and that to the right upwards. Paying attention to this fact, we shall have, when the loads are distributed as above described, a shearing- force at the bar 7-8 causing tension in this bar ; the magnitude of this shearing-force being 6 + 8) _ i6 16 Hence, we may arrange the work as follows : 6 ) = 41500. GREATEST DIAGONAL STRESSES IN GIRDER. 197 Greatest Stresses in Numbers of Greatest Shearing-Forces producing Stresses of Same Kind as Diagonals of Diagonals. Dead Load. as those due to Dead Load. 1-2 17-16 3^ (2 + 4+6+8+IO+I2+I4) = 133000 -'53575 2-3 16-15 ^(2+4+6+8+10+12+14) = I33 ooo + '53575 3-4 '5-H 3 ^(2+ 4 +6+8+io+ I2 )-~(2) = 98750 114027 4-5 14-13 ^(2+4+6+8+10+12) -^(2) = 98750 + i 14027 5-6 13-12 ^(2+4+6+8+10) -^(2+4) = 68250 - 78808 6-7 I 2-1 1 ^(2+4+6+8 + 10) -^(2+4) - 68250 + 78808 7-8 II-IO ^(2+4+6+8) -^(2+4+6) = 41500 - 47920 8-9 io- 9 ^(2+4+6+8) - ^(2+4+6) - 41500 + 479 20 Greatest Stresses in Numbers of Greatest Shearing-Forces producing Stresses of Kind Opposite Diagonals ^pf Diagonals. from Dead Load. Kind Oppo- site from Dead Load. 8-9 io- 9 ^(2+ 4 +6) - ^(2+4+6+8) - 18500 -21362 7-8 II-IO ^(2+4+6) - ?^( 2 + 4 +6+8) - 18500 + 21362 The diagonals 7-8, 8-9, 9-10, and 10-11 are the only ones that, under any circumstances, can have a stress of the kind opposite to that to which they are subjected under the dead load alone. I9 8 APPLIED MECHANICS. Fig. 114 exhibits the manner of writing the stresses on the diagram. 152. General Application of this Method. It is plain that the method used above will apply to any single system of bridge-truss with horizontal chords and diagonal bracing, what- ever be the inclination of the braces. When seeking the stress in a diagonal, the section must be so taken as to cut that diagonal ; and, as far as this stress alone is concerned, it may be equally well taken at any point, as well as near a joint, provided only it cuts that diagonal which is in action under the load that produces the greatest stress in this one, and no other. On the other hand, when we seek the stress in a horizontal chord, the section might very properly be taken through the joint opposite that chord. Taking it very near the joint, only serves to make one sec- tion answer both purposes simultaneously. 153. Bridge-Trusses with Vertical and Diagonal Bra- cing. When, as in Figs, in and 112, there are both vertical and diagonal braces, and also horizontal chords, we may deter- mine the stresses in the diagonals and in the chords just as before ; only we must take the section just to one side of a joint, and never through the joint. As to the verticals, in order to determine the stress in any vertical, we must impose the conditions of equilibrium between the vertical components of the forces acting at one end of that vertical: thus, if the loads are at the upper joints in Fig. in, then the stress in vertical 3-2 must be equal and opposite to the vertical component of the stress in diagonal 1-2, as these stresses are the only vertical forces acting at joint 2. Vertical 5-4 has for its stress the vertical component of the stress in 3-4, etc. Thus Stress in 3-2 = shearing-force in panel 1-3, Stress in 5-4 = shearing-force in panel 3-5, etc.- TRUSSES WITH VERTICAL AND DIAGONAL BRACING. 199 On the other hand, if the loads be applied at the lower joints, then Stress in 3-2 = shearing-force in panel 3-5, Stress in 5-4 = shearing-force in panel 5-7, etc. EXAMPLE. Given the truss shown in Fig. in. Given panel length = height of truss 10 feet, dead load per panel point = 12000 Ibs., moving load per panel point = 23000 Ibs. ; load applied at upper joints. Solution, (a) Chord Stresses. Assume the entire load on the bridge, i.e., 35000 Ibs. per panel point. Hence Total load on truss =13 (35000) = 455000 Ibs., Each supporting force = 227500 Ibs. Joint near which Section is taken. Bending-Moment at the Section very near the Joint, on Either Side of the Joint. I 28 3 27 227500 x 10 = 2275000 5 25 227500 X 20 35000 X 10 = 4200000 7 23 227500 X 30 35000(10 + 20) = 5775000 9 21 227500 X 40 35000 ( 10 + 20 + 30) 7000000 ir 19 227500 X 50 35000 (10 + 20 + 30 + 40) = 7875000 13 17 227500 X 60 35000(10 + 20 -h 30 + 40 + 50) = 8400000 IS 227500 X 70 35000 (10 + 20 + 30 + 40 + 50 -f- 60) = 8575000 To find any chord stress, divide the bending-moment at a section cutting the chord, and passing close to the opposite joint, by the height of the girder, which in this case is 10. Hence we have for the chord stresses (denoting, as before, com- pression by +, and tension by ) : 2OO APPLIED MECHANICS. Stresses in Upper Chords. Stresses in Lower Chords. i- 3 27-28 + 227500 2- 4 24-26 227500 3- 5 25-27 4-420000 4- 6 22-24 420000 5- 7 2 3- 2 5 + 5775 6- 8 20-22 -5775 7- 9 21-23 + 700000 8-10 18-20 700000 91 1 19-21 + 787500 IO-I2 1 6-1 8 -787500 11-13 17-19 + 840000 12-14 14-16 840000 i3-!5 iS- 1 ? + 8575 Diagonals. It is evident, that, for the diagonals, the same rule holds as in the case of the Warren girder : i.e., the greatest stress of the same kind as that produced by the dead load occurs when the moving load is on all the joints between the diagonal in question and the farther abutment ; whereas the greatest stress of the opposite kind occurs when the moving load covers all the joints between the diagonal in question and the nearer abutment. The work of determining the greatest shearing-forces may be arranged as in tables on p. 191. Counterbraces. If the truss were constructed with those diagonals only that slope downwards towards the centre, and which may be called the main braces, the diagonals 1 1-12, 13-14, 14-17, and 16-19 would sometimes be called upon to bear a thrust, and the verticals 12-13 and 17-16 a pull : this would necessitate making these diagonals sufficiently strong to resist the greatest thrust to which they are liable, and fixing the verticals in such a way as to enable them to bear a pull. In order to avoid this, the diagonals 10-13, 12-15, I 5~ I 6, and 17-18 are inserted, which are called counterbraces, and which come into action only when the corresponding main TRUSSES WITH VERTICAL AND DIAGONAL BRACING. 2OI braces would otherwise be subjected to thrust. They also prevent any tension in the verticals. Diagonals. Greatest Shearing- Forces of the Same Kind as those produced by Dead Load. I- 2 28-26 ^ I+2 + 3 + ... +I3 ) = 227500 3- 4 27-24 3J^ (l + 2 + 3+ ... + I2) _I5^( l) = I94H3 5-6 25-22 3 -^P(l + 2 + 3+ ... + II)- '-^(1 + 2) = 162429 7-8 23-20 ^(i + 2+3+ . . . + 10) - x -^?(i + 2+3) J-4 J-4 = 132357 9-10 2I-I8 ^(1 + 2+3+ ...+ 9)-^p(i + 2+.. + 4) = 103929 11-12 19-16 33222(1 + 2+3+...+ 8)-^(i + 2+.. + 5)= 77H3 I3T4 17-14 3^ (l + 2 + 3 +...+ 7) _^ I + 2+ .. + 6)= 52000 Diagonals. Greatest Shearing-Forces of the Opposite Kind to those produced by Dead Load. 3-14 17-14 ^ (I+2+3+ ... + 6) _^ I+2+ ... +7) = 28500 11-12 19-16 22f(i + 2+ ... + 5 )_H5p ( r + 2 + ... + 8) = 6643 9-10 2I-I8 ^(, + 2+ ... + 4) -^ (l + 2 + ... + 9) = ~i357i The main braces and counterbraces of a panel are never in action simultaneously. Hence we have, for the greatest stresses in the diagonals, the following results, obtained by multiplying the corresponding shearing-forces by - 1.414. cos 45 2O2 APPLIED MECHANICS. In the following I have used this number to three decimal places, as being sufficiently accurate for practical purposes. Stresses in Main Braces. Stresses in Counterbraces. I- 2 28-26 -321685 15-12 15-16 40299 3- 4 27-24 -274518 I3-IO I7-I8 - 9393 5-6 25-22 -229675 7- 8 23-20 -187153 9-10 21-18 146956 11-12 19-16 109080 i3~ I 4 17-14 - 735 28 Vertical Posts. Since the loads are applied at the upper joints, the conditions of equilibrium at the lower joints require that the thrust in any vertical post shall be equal to the vertical component of the tension in that diagonal which, being in action at the time, meets it at its lower end. Hence it is equal to the shearing-force in that panel where the acting diagonal meets it at its lower end. We therefore have, for the posts, the following as the greatest thrusts : STRESSES IN VERTICALS. 3- 2 27-26 + 2275OO 5- 4 25-24 + I94M3 7- 6 23-22 + 162429 9- 8 2I-2O + 132357 II-IO I9-I8 + 103929 13-12 I7-l6 + 77143 i5- J 4 + 52000 CONCENTRATING THE LOAD AT THE JOINTS. 203 X X X FIG. Fig. 115 shows the stresses marked on the diagram. 154. Manner of Concentrating the Load at the Joints. In using the methods given above, we are assuming that all the loads are concentrated at the joints, and that none are distributed over any of the pieces. As far as the mov- ing load is concerned, and also all of the dead load except the weight of the truss itself, this always is, or ought to be, effected ; and it is accomplished in a manner similar to that adopted in the case of roof-trusses. This method is shown in the figure (Fig. 1 1 6); floor-beams being laid across from girder to girder at the joints, on top of which are laid longi- tudinal beams, and on these the sleepers if it is a railroad bridge, or the floor if it is a road bridge. The weight of the truss itself is so small a part of what the bridge is called upon to bear, that it can, without appreciable error, be considered as concentrated at the joints either of the up- per chord, of the lower chord, or of both, according to the manner in which the rest of the load is distributed. 155. Closer Approxima- tion to Actual Shearing- Force. In our computations of greatest shearing-force, we FIG. 115. make an approximation which is generally considered to be APPLIED MECHANICS. sufficiently close, and which is always on the safe side. To illustrate it, take the case of panel 3-5 of the last example. In determining its greatest shearing-force, we considered a load of 35000 Ibs. per panel point to rest on all the joints from the right-hand support to joint 5, inclusive, and the dead load to rest on all the other joints of the~truss. Now, it is impossible, if the load is distributed uniformly on the floor of the bridge, to have a load of 35000 Ibs. on 5 and 12000 on 3 simultaneously ; for, if the moving load extended on the bridge floor only up to 5, the load on 5 would be only 12000 + ^-(23000) = 23500 Ibs., and that on 3 would then be 12000 Ibs. If, on the other hand, the moving load extends beyond 5 at all, as it must if the load on 5 is to be greater than 23500 Ibs., then part of it will rest on 3, and the load on 3 will then be greater than 12000 Ibs. ; for whatever load there is between 3 and 5 is supported at 3 and 5. Moreover, we know that the effect of increasing the load on 5 is to increase the shearing-force, provided we do not at the same time increase that on 3 so much as to destroy the effect of increasing that on 5. Hence, there must be some point between 3 and 5 to which the moving load must extend in order to render the shearing- force in panel 3-5 a maximum. Let the distance of this point from 5 be^r; then, if we let w = moving load per foot of length, Moving load on panel = wx, Part supported at 3 = -- , 20 Part supported at 5 = wx -. 20 Hence, portion of shearing-force due to the moving load on panel 3-5 equals CONCENTRATING THE LOAD AT THE JOINTS. I2/ WX 2 \ I WX 2 W I I*X 2 \ ( WX -- ) --- = ( I2X -- - ). i4\ 20 / 14 20 i4\ 20 / This becomes a maximum when its first differential co-efficient becomes zero, i.e., when therefore 12 - x = o X = 9.2 3 . Hence, when the moving load extends to a distance of 9.23 feet from 5, then the shearing-force in panel 3-5, and hence the stress in diagonal 3-4, is a maximum. Panels. Portion of Shearing-Force due to Moving Load on Panel. Value of x t in feet. Portion of Load at Joints named below. Portion of Load at Joints named below. i- 3 27-28 iv ( I 3 *A IO.OO 11500 3 11500 i4\ J 20 y 3- 5 5- 7 25-27 23-25 I4\ 20 / 9-23 8.46 3 5 9797 8230 5 7 11432 11227 I4\ 20 / 7- 9 21-23 I4\ 20 / 7.69 7 6801 9 10886 9-1 1 19-21 H( 9X ~ ^f?) 6.92 9 5507 n 10409 11-13 13-15 17-19 15-17 14 \ 20 / u Let angle 4-1-5 = i n Let angle 2-1-3 ** \ we shall have, if w -{- w^ entire load per panel point, Designation of Ties. EFFECT OF LOADS AT Resultant Tensions. 3 5 7 9 11 13 15 1-2 2-5 5-6 6-9 1-4 4-9 1-8 W + Wi O o o w + w l O W + Wi o o o w + Wi o o o 3 w-fw/t o o o W + Wi O o W + Wi 8 sin i W + Wi 2 sin i 2 w + w l 2 sin z' 2 W + Wj 2 sin z' 2 o w -f -Wi 2 sin z' 2 w + w t 2 sin z' 2 W + Wi 2 sin /' 2 w + w x 2 sin / 2 W + Wi 2 sin /' 2 W + Wi 4 sin * x W -\~ Wi 2 sin *! 7f -}- W x 4 sin z'i W + Wi sin /! ze/ + Wi 4 sin /i W + Wi 2 sin z'i W + Wi 4 sin /'x SW+Wi sin /j , 2(/+W,) 8 sin * 4 sin / 8 sin z 2 sin 2 8 sin* 4 sin / sin 2 i i The stresses in all the other members may be found in a similar manner. GENERAL REMARKS. 2I 9 159. Bollman's Truss. The description of this truss is made sufficiently clear by the figure. The upper chord is made in separate pieces ; and 1 3 5 7 9 11 12 the short diagonals 2-5, 3-4, 4-7, 5-6, 7-8, 6-9, 8-1 1, and 9-10 are only needed to prevent a bending of the upper chord at the joints. FlG - I24> When this is their only object, the stress upon them cannot be calculated : indeed, it is zero until bending takes place ; and then it is the less, the less the bending. Hence, in this case, the stress is wholly taken up by the principal ties ; and these have their greatest stress when the whole load is on the bridge. The computation of the stresses is made in a similar man- ner to that used in the Fink. 1 60. General Remarks. The methods already explained are intended to enable the student to solve any case of a bridge- truss where there is no ambiguity as to the course pursued by the stresses. In cases where a large number of trusses of one given type are to be computed, it would, as a rule, be a saving of labor to determine formulae for the stresses in the members, and then substitute in these formulae. Such formulae may be deduced by using letters to denote the load and dimensions, instead of inserting directly their numerical values ; and then, having deduced the formulae for the type of truss, we can apply it to any case by merely sub- stituting for the letters their numerical values corresponding to that case. Such sets of formulae would apply merely to specific styles of trusses, and any variation in these styles would require the formulae to be changed. 220 APPLIED MECHANICS. In order to show how such formulas are deduced, a few will be deduced for such a bridge as is shown in Fig. 1 1 1. Let the load be applied at the upper panel points only ; let dead load per panel point = w, moving load per panel point = w,. Let the whole number of panels be N, N being an even number. Let the length of one panel = height of truss = /. Then length of entire span = Nl. Consider the (n + i) th panel from the middle. The stress in the main tie is greatest when the moving load is on all the panel points from the farther abutment up to the panel in question, (n + th - Hence, for the n tb panel from the middle, the greatest shear' ing-force that causes tension in the main tie is equal to w-\-w 1 Hence stress in main tie N For the counterbrace, we should obtain, in a similar way, the formula _N H T \ ^~ 1 I \ fl I 2N LA 2 / 2 ] - wN(n + ,) } , which represents tension when it is positive. Proceed in a similar way for the other members. When there is more than one system, we must divide the truss into its component systems; and when there is ambiguity, we must use, in determining the dimensions of each member, the greatest stress that can possibly come upon it. CENTRE OF GRAVITY. 221 CHAPTER V. CENTRE OF GRAVITY. 161. The centre of gravity of a body or system of bodies, is that point through which the resultant of the system of parallel forces that constitutes the weight of the body or system of bodies always passes, whatever be the position in which the body is placed with reference to the direction of the forces. 162. Centre of Gravity of a System of Bodies. If we have a system of bodies whose weights are W iy W 2y W y etc., the co-ordinates of their individual centres of gravity being fo, y *i), (* y **}> (* y Iv * 3 ) etc., respectively, and if we denote by x m y , z , the co-ordinates of the centre of gravity of the system, we should obtain, just as in the determination of the centre of any system of parallel forces, i. By turning all the forces parallel to OZ> and taking moments about OY, (W, + W 2 + W 3 + etc.)* = W,x, + W 2 x 2 + W t x 3 + etc., or and, taking moments about OX, etc., or 222 APPLIED MECHANICS. 2. By turning all the forces parallel to OX, and taking moments about OY, (W* + W, + W z + etc.K = W& + W 2 z 2 + W z z z + etc., or Hence we have, for the co-ordinates of the centre of gravity of the system, EXAMPLES. I. Suppose a rectangular, homogeneous plate of brass (Fig. 125), where AD = 1 2 inches, AB = 5 inches, and whose weight is 2 Ibs., to have weights attached at the points A, B, C, and D respectively, equal to 8, 6, 5, and x $ Ibs. ; find the centre of gravity of the system. 4- Solution. Assume the origin of co-ordinates at the centre of the rectangle, and we have W, = 2, W 2 = 8, W, = 6, W 4 = 5, W s = 3, *, =o, x 2 = 6, x z = 6, * 4 = -6, ^ s = -6, Ji =o, ^ 2 = f, J 3 = -f, y 4 = -|, j s = f ; = o -f 48 -f 36 30.0 18.0 = 36, = o -f- 20 15 12.5 4- 7.5 = o, = 2 -f- 8 -f 6 + 5.0 4- 3.0 = 24; _ 3^> _ ^ = 2 4 = = 2 4 = Hence the centre of gravity is situated at a point E on the line OX, where OE = 1.5. CENTRE OF GRAVITY OF . HOMOGENEOUS BODIES. 22$ 2. Given a uniform circular plate of radius 8, and weight 3 Ibs. (Fig. 126). At the points A, B, C, and D, weights are attached equal to 10, 15, 25, and 23 Ibs. respectively, also given AB = 45, BC = 105'', CD = 120 ; find the centre of gravity of the system. 163. Centre of Gravity of Homogeneous Bodies. For the case of a single homogeneous body, the formulae have been already deduced in 44. They are fxdV ~ JdV and for the weight of the body, W = wfdV, where x& y , z m are the co-ordinates of the centre of gravity of the body, W its weight, and w its weight per unit of volume. From these formulae we can readily deduce those for any special cases ; thus, (a) For a volume referred to rectangular co-ordinate axes, d V dxdydz. x _ = fffxdxdydz _ = fffydxdydz = fffzdxdydz SSfdxdydz' y " Sffdxdydz* * == fffdxdydz (b) For a flat plate of uniform thickness, t, the centre of grav- ity is in the middle layer; hence only two co-ordinates are required to determine it. If it be referred to a system of rect- angular axes in the middle plane, dV '= tdxdy, _ ffxdxdy _ ffydxdy 224 APPLIED MECHANICS. The centre of gravity of such a thin plate is also called the centre of gravity of the plane area that constitutes the middle plane section ; hence (c) For a plane area referred to rectangular co-ordinate axes in its own plane, Sfxdxdy ffydxdy (d) For a slender rod of uniform sectional area, a, if x, y, z, be the co-ordinates of points on the axis (straight or curved) of the rod, we shall have dV ads a^(dx) 2 + (dy)* + (dzf> (I)* fyds = l- = fds A* (e) For a slender rod whose axis lies wholly in one plane, the centre of gravity lies, of course, in the same plane ; and if our co-ordinate axes be taken in this plane, we shall have z = c -=- =r o, and also Z Q = o. Hence we need only two co ax CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 22$ ordinates to determine the centre of gravity, hence dV =. ads m fxds _ J AMI) + } * 2 = BC = -O, FIG. 137. 5. Circular Arcs. (a) Circular Arc AB (Fig. 138). Angle A OB = 0,, radius = r, Use formula fyds FIG. i 3 i. " fds ' - fds ' but use polar co-ordinates, where ds = rdQ, sf = rcosO, y = rsinO, CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 229 r 2 cos OdO 2 f i Oi si sin OdB (i cos0,) Circular Arc AC (same figure). r sin #! , y<> = o. (c) Quarter-Arc of Circle AB, Radius r (Fig. 139). r 2 I 2 cos OdB Jo 2r Semi-circumference ABC (same figure). FIG- 13* 6. Combination of Circles and Straight Lines. Barlow Rail (Fig. 140). Two quadrants, radius r, and web, c , whose area = ^- the united area of the quadrants. Let united area of quadrants = A, area of web ; let AI IB FIG. 140. 230 APPLIED MECHANICS. 7. Areas. (a) T-Section (Fig. 141). Let length AB = B, EF = b, entire B height = H, GE = h. Let distance of centre x.*mmmmmm^i o f gravity below AB = x t ; therefore, taking moments about AB as an axis, -h(B-t)\ -$ - k(S - whence we can readily derive x t . (b) I-Section (Fig. 142). Let AB = B, GH = b, MN = b n entire height = H, BC = H h, EH = h t ; and let x t = distance of centre of gravity below AB. A Hence, taking moments about AB, we have Xl \B(H - h) B - h,) whence we can deduce x t (c} Triangle (Fig. 143). If we consider the triangle OBC as composed of an indefinite number of narrow strips parallel to the side CB, of which FLHK is one, the centre of gravity of each one of these strips will be on the line OD drawn from O to the middle point of the side CB ; hence the centre ^f gravity of the entire tri- c angle must be on the line OD. For a similar rea- son, it must be on the median line CE ; hence the centre of gravity mhst be at the intersection of the median lines, and hence BC . ODsv^ODC o FIG. 143. X Q = = ^OD. Moreover, area = CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. (d) Trapezoid (Fig. 144). First Solution. Bisect AB in 6>, and CE in >; let g^ be the centre of gravity of CEJB, and g t that of ABC. Then will 6 1 , the centre of gravity of the trape- zoid, be on the line g^^ and Gg, -. Gg, CEB* But it must be on the line OD; hence it is at their intersectioa From the similarity of GG l g l GG^g^ we have GG, GG ~ ABC BEC ~ CE B m b ; GG, andsince OD Second Solution. Fig. 144 (a). Let O be the point of intersection of the non-parallel sides AC and BE. Let OF = x lt OD * OG = x n . Take moments about an axis through O, and perpen- dicular to OF) and we readily obtain Fie. 232 APPLIED MECHANICS. (e) Parabolic Half-Segment OAB (Fig. 145). Let OA ~ x,, AB = jh ; let x ,y , be the co-ordinates of the centre of gravity ; let the equation of the parabola be y 2 = Y '.r, /-' 2 *M f Xl 3 Jo xdocd y 2a J ** d X = /*' rya t/o t/o ' r<^ /.r, ^ 3 ^ t/o y ~~ s i ~ 3v i MI Area (/) Parabolic Spandril OBC (Fig. 145). Let x ,y , be co-ordi- nates of centre of gravity of the spandril. xdxdy x^ (*y\ I _ l_?y /*Jr, /*y t / / L 1/0 ^ _ > /*, /^ / / 1/0 ^ 2 * Area = ^j, - CENTRE OF GRA VI TY OF HOMOGENEOUS BODIES. 233 (g) Circular Sector OAC (Fig. 146). Let OA = r, AOX = 0,, be the co-ordinates of the centre of gravity : . . ^o^O, Xo = /r r*V r"* x* /rcos0j /*jr tan 0, / *<**#+ / / JCflkfljK .ooggy-V^rra y ^-rtanfl, ^ Area c FIG. 146. Second Solution. Consider the sector to be made up of an indefinite number of narrow rings ; let p be the variable radius, and dp the thickness : Elementary area = 2pft l dp, and centre of gravity of this elementary area is on OX, at a distance from O equal to p ^ 1 [see Example 5 ()] ; X = (ti) Circular Half-Segment ABX (Fig. 146), f" f Q " xdxdy f" xVr* - x*dx Sector minus triangle %r*S l |r 2 sin 0, cos 0, , sn <, cos r r^ ^rcosgyo '"""' = . 4sin B ig 1 -sin a 0.cosg 1 , sin 0, cos 0,) ~~ 0, sin 0, cos 0, 234 APPLIED MECHANICS. 164. Pappus's Theorems. The following two theorems serve often to simplify the determination of the centres of gravity of lines and areas. They are as follows : THEOREM I. If a plane curve lies wholly on one side of a straight line in its own plane, and, revolving about that line, generates thereby a surface of revolution, the area of the sur- face is equal to the product of the length of the revolving line, and of the path described by its centre of gravity. Proof. Let the curve lie in the xy plane, and let the axis of y be the line about which it revolves. We have, from what fxds precedes, 163 (e\ X Q =- -' .*. x fds = fxds, where x equals the perpendicular distance of the centre of gravity of the curve from O Y, ds = elementary arc, 2irx fds = f(2irx)ds; or, reversing the equation, f(2irx)ds = But f(2irx)ds = surface described in one revolution, while s =. length of arc, and 2irx Q = path described by the centre of gravity in one revolution. Hence follows the proposition. THEOREM II. If a plane area lying wholly on the same side of a straight line in its own plane revolves about that line, and thereby generates a solid of revolution, the volume of the solid thus generated is equal to the product of the revolving area, and of the path described by the centre of gravity of the plane area during the revolution, PAPPUS'S THEOREMS. 235 Proof. Let the area lie in the xy plane, and let the axis OY be the axis of revolution. We then have, from what has preceded, if x = perpendicular distance of the centre of gravity of the plane area from OY t the equation, 163 (b), Sfxdxdy *- ffdxdy' Hence Xo ffdxdy = ffxdxdy; /. (2irx ) ffdxdy or ff(2trx)dxdy But ff(2irx)dxdy = volume described in one revolution, and 2iex Q = path described by the centre of gravity in one revolu- tion. Hence follows the proposition. The same propositions hold true for any part of a revolution, as well as for an entire revolution, since we might have multi- plied through by the circular measure 6, instead of by 2ir. It is evident that the first of these two theorems may be used to determine the centre of gravity of a line, when the length of the line, and the surface described by revolving it about the axis, are known ; and so also that the second theorem may be used to determine the centre of gravity of a plane area whenever the area is known, and also the volume described by revolving it around the axis. EXAMPLES. i. Circular Arc AC (Fig. 138). Length of arc = s = 2rO, sur- face of zone described by revolving it about O Y = circumference of a great circle multiplied by the altitude = (ztrr) (2rsmO l ); x l = rsinfl, sm0 z r- 236 APPLIED MECHANICS. 2. Semicircular Arc (Fig. 139). Length of arc = nr, surface of sphere described = 4?rr 2 ; 2r .'. 2Trx (Trr) = 4?rr 2 .*. x = 7T 3. Trapezoid (Fig. 147). Let AD = b, BC b ; let it revolve around AD : it generates two cones and a cylinder. AD + BC Y Area of trapezoid = - BG, B Volume = ~ -(AG + HD) + 7r(G) 2 . BC \-HD+ 3 BC) FlG - J 47. = ^ ^(^Z) + BC + ^C) GBI BC \ GBI = KL 4. Circular Sector AGO (Fig. 146). Area of sector = r*0 lt volume described = -Jr (surface of zone) = \r(2-rrr} (?r sin 0,) = sin 0! 165. Centre of Gravity of Solid Bodies. The general formulae furnish, in most cases, a very complicated solution, and hence we generally have recourse to some simpler method. A few examples will be given in this and the next section. CENTRE OF GRAVITY OF SYMMETRICAL BODIES. 237 Tetrahedron ABCD (Fig. 148). The plane ABE, containing the edge AB and the middle point E of the edge CD, bisects all lines drawn parallel to CD, and terminating in the faces A ABD and ABC : hence a similar reasoning to that used in the case of the triangle will show that the cen- tre of gravity of the pyramid must be in the plane ABE ; in the same way it may be shown that it must lie in the plane ACF. Hence it must lie in their intersection, or in the line AG joining the vertex A with the centre of gravity (intersection of the medians) of the opposite face. FIG. 148. In the same way it can be shown that the centre of gravity of the triangular pyramid must lie in the line drawn from the vertex B to the centre of gravity of the face A CD. Hence the centre, of gravity of the tetrahedron will be found on the line AG at a distance from G equal to \A G. 1 66. Centre of Gravity of Bodies which are Symmet- rical with Respect to an Axis. Such solids may be gener- ated by the motion of a plane figure, as ABCD (Fig. 149), of variable dimensions, and of any form whose centre G remains upon the axis OX ; its plane being always perpendicular to OX, and its variable area X being a function of x, its distance from the origin. Here the centre of gravity will evidently FIG. i 49 . jj e on tne ax j s QX^ an d the elementary vol- ume will be the volume of a thin plate whose area is X and thickness A;r ; hence the elementary volume will be Take moments about OY, and we shall have or x fXdx = fXxdx and Volume = fXdx, fXxdx ** = 7xJ*" F =/^ APPLIED MECHANICS. EXAMPLES. x 2 y 2 z 2 I. Ellipsoid -f ^- + = i (Fig. 150). Find centre of gravity a D c of the half to the right of the x plane. Let OK = x. Now if, in the equation of the ellipsoid, v X 2 Z 2 we make y = o, we have H = I ; where z = Make z = o in the equation of the ellipsoid, and + ij = I > where ^ = .-. EK are the semi-axes of the variable ellipse EGFH, which, by moving along OX, generates the ellipsoid. Hence hence irbc Area EGFH = Tr(EK . GK) = (a* - x 2 ) = X; Elementary volume = (a 2 irbc (* a . ( a 2 x 2 x 4 ) I (a 2 x x*)dx < > a 2 Jo _ ( 2 4 ). trbc ^ J ( x*} a a 2 -x*)dx \a 2 x--\ 3)0 V = - - I a (a 2 - x 2 )dx = \irabc. d 2 t/0 a. Hemisphere. Make a = b = c, and x = f a, V CENTRE OF GRAVITY OF SYMMETRICAL BODIES. 239 If the section X were oblique to OX t making an angle 0. with it, the elementary volume would not be Xdx, but Xdx sin 0, and we should have 3. Oblique Cone (Fig. 151). Let OA = h; let area of base be and let the angle made by OX with the base be 6; X x> A FIG. 151. r h sin0 / ^^ ** o 4. Truncated Cone (Fig. 151). Let height of entire cone be h = OA ; let height of portion cut off be h l ; AT* h*- h* I x*dx 4 ,/^-A 4 ^TTSi 240 APPLIED MECHANICS. CHAPTER VI. STRENGTH OF MATERIALS. 167. Stress, Strain, and Modulus of Elasticity When a body is subjected to the action of external forces, if we imagine a plane section dividing the body into two parts, the force with which one part of the body acts upon the other at this plane is called the stress on the plane ; it may be a tensile, a compressive, or a shearing stress, or it may be a com- bination of either of the two first with the last. In order to know the stress completely, we must know its distribution and its direction at each point of the plane. If we consider a small area lying in this plane, including the point O, and represent the stress on this area by /, whereas the area itself is repre- sented by a, then will the limit of <- as a approaches zero be the a intensity of the stress on the plane under consideration at the point 0. When a body is subjected to the action of external forces, and, in consequence of this, undergoes a change of form, it will be found that lines drawn within the body are changed, by the action of these external forces, in length, in direction, or in both ; and the entire change of form of the body may be correctly described by describing a sufficient number of these changes. If we join two points, A and B, of a body before the external forces are applied, and find, that, after the application of the external forces, the line joining the same two points of the body has undergone a change of length &(AB), then is the STRESS, STRAIN, AND MODULUS OF ELASTICITY. 241 limit of the ratio ' as AB approaches zero called the strain of the body at the point A in the direction AB. If AB 4- &(AB) > AB, the strain is one of tension. If AB + A (^4-5) < ^4-#, the strain is one of compression. Suppose a straight rod of uniform section A to be subjected to a pull P in the direction of its length, and that this pull is uniformly distributed over the cross-section : then will the in- tensity of the stress on the cross-section be If P be measured in pounds, and A in square inches, then will / be measured in pounds per square inch. If the length of the rod before the load is applied be /, and its length after the load is applied be I ~\- e, then is e the elongation of the rod ; and if this elongation is uniform through- x> out the length of the rod, then is - the elongation of the rod per unit of length, or the strain. Hence, if a represent the strain due to the stress / per unit of area, we shall have The Modulus of Elasticity is commonly defined as the ratio of the stress per unit of area to the strain, or *-*-; a and this is expressed in units of weight per unit of area, as in pounds per square inch. This definition is true, however, only for stresses for which Hooke's law " The stress is proportional to the strain " holds. 242 APPLIED MECHANICS. For greater stresses the permanent set must first be deducted from the strain, and the remainder be used as divisor. The limit of elasticity of any material is the stress above which the stresses are no longer proportional to the strains. The modulus of elasticity was formerly defined as the weight that would stretch a rod one square inch in section to double its length, if Hooke's law held up to that point, and the rod did not break. EXAMPLES. 1. A wrought-iron rod 10 feet long and i inch in diameter is loaded in the direction of its length with 8000 Ibs. ; find (i) the intensity of the stress, (2) the elongation of the rod ; assuming the modulus of the iron to be 28000000 Ibs. per square inch. 2. What would be the elongation of a similar rod of cast-iron under the same load, assuming the modulus of elasticity of cast-iron to be 1 7000000 Ibs. per square inch ? 3. Given a steel bar, area of section being 4 square inches, the length of a certain portion under a load of 25000 Ibs. being 10 feet, and its length under a load of 100000 Ibs. being 10' o".o75 ; find the modulus of elasticity of the material. 4. What load will be required to stretch the rod in the first example Y 1 ^ inch ? 1 68. Resistance to Stretching and Tearing. The most- used criterion of safety against injury for a loaded piece is, that the greatest intensity of the stress to which any part of it is subjected shall nowhere exceed a certain fixed amount, called the working-strength of the material ; this working-strength being a certain fraction of the breaking-strength determined by practical considerations. The more correct but less used criterion is, that the great- est strain in any part of the structure shall nowhere exceed the working-strain ; the greatest allowable amount of strain being a fixed quantity determined by practical considerations. RESISTANCE TO STRETCHING AND TEARING. 243 This is equivalent to limiting the allowable elongation or compression to a certain fraction of its length, or the deflection of a beam to a certain fraction of the span. If the stress on a plane surface be uniformly distributed, its resultant will evidently act at the centre of gravity of the surface, as has been already shown in 42 to be the case with any uniformly distributed force. If a straight rod of uniform section and material be sub- jected to a pull in the direction of its length, and if the result- ant of the pull acts along a line passing through the centres of gravity of the sections of the rod, it is assumed in practice that the stress is uniformly distributed throughout the rod, and hence that for any section we shall obtain the stress per square inch by dividing the total pull by the number of square inches in the section. If, on the other hand, the resultant of the pull does not act through the centres of gravity of the sections, the pull is not uniformly distributed ; and while will express the mean stress per square inch, the actual inten- sity of the stress will vary at different points of the section, p being greater than at some points and less at others. How A to determine its greatest intensity in such cases will be shown later. With good workmanship and well-fitting joints, the first case, or that of a uniformly distributed stress, can be practi- cally realized ; but with ill-fitting joints or poor workmanship, or with a material that is not homogeneous, the resultant of the pull is liable to be thrown to one side of the line passing through the centres of gravity of the sections, and thus there 244 APPLIED MECHANICS. is set up a bending-action in addition to the direct tension, and therefore an unevenly distributed stress. It is of the greatest importance in practice to take cogni- zance of any such irregularities, and determine the greatest intensity of the stress to which the piece is subjected : though it is too often taken account of merely by means of a factor of safety ; in other words, by guess. Leaving, then, this latter case until we have studied the stresses due to bending, we will confine ourselves to the case of the uniformly distributed stress. If the total pull on the rod in the direction of its length be P, and the area of its cross-section A, we shall have, for the intensity of the pull, P On the other hand, if the working-strength of the material per unit of area be /, we shall have, for the greatest admissible load to be applied, P = fA. If / be the working-strength of the material per square inch, and E the modulus of elasticity, then is the greatest admissible strain equal to Thus, assuming 12000 Ibs. per square inch as the working tensile strength of wrought-iron, and 28000000 Ibs. per square inch as its modulus of elasticity, its working-strain would be 1 2000 28000000 7000 Hence the greatest safe elongation of the bar would be of its length. Hence a rod 10 feet long could safely be stretched ^ of a foot = 0.05 14". VALUES OF BREAKING AND WORKING STRENGTH. 245 169. Approximate Values of Breaking Strength, and of Modulus of Elasticity. In a later part of this book the attempt will be made to give an account of the experiments that have been made to determine the strength and elas- ticity of the materials ordinarily used in construction, in such a way as to enable the student to decide for himself, in any- special case, upon the proper values of the constants that he ought to use. For the present, however, the following will be given as a rough approximation to some of these quantities, which we may make use of in our work until we reach the above-mentioned account. (a) Cast-Iron. Breaking tensile strength per square inch, of common quali- ties, 14000 to 20000 Ibs. ; of gun iron, 30000 to 33000 Ibs. Modulus of elasticity for tension and for compression, about 17000000 Ibs. per square inch. (b) Wrought-Iron. Breaking tensile strength per square inch, from 40000 to* 60000 Ibs. Modulus of elasticity for tension and for compression, about 28000000. (c) Mild Steel. Breaking tensile strength per square inch, 55000 to 70000 Ibs. Modulus of elasticity for tension and for compression, from 28000000 to 30000000 Ibs. per square inch. (d) Wood. Breaking compressive strength per square inch : Oak, green 3000 Ibs. Oak, dry 3000 to 6000 Ibs. Yellow pine, green 3000 to 4000 Ibs. Yellow pine, dry 4000 to 7000 Ibs. 246 APPLIED MECHANICS. Modulus of elasticity for compression (average values) : Oak 1300000 Ibs. per square inch. Yellow pine 1600000 Ibs. per square inch. 170. Sudden Application of the Load. If a wrought- iron rod 10 feet long and I square inch in section be loaded with 12000 pounds in the direction of its length, and if the modulus of elasticity of the iron be 28000000, it will stretch 0.05 14" provided the load be gradually applied : thus, the rod begins to stretch as soon as a small load is applied ; and, as the load gradually increases, the stretch increases, until it reaches 0.05 14". If, on the other hand, the load of 12000 Ibs. be suddenly applied (i.e., put on all at once) without being allowed to fall through any height beforehand, it would cause a greater stretch at first, the rod undergoing a series of oscillations, finally settling down to an elongation of 0.05 14". To ascertain what suddenly applied load will produce at most the elongation 0.05 14", observe, that, in the case of the gradually applied load, we have a load gradually increasing from o to 12000 Ibs. Its mean value is, therefore, ^(12000) = 6000 Ibs. ; and this force descends through a distance of 0.05 14". Hence the amount of mechanical work done on the rod by the gradually applied load in producing this elongation is (6000) (0.0514) = 308.4 inch-lbs. Hence, if we are to perform upon the rod 308.4 inch-lbs. of work with a constant force, and if the stretch is to be 0.05 14", the magnitude of the force must be 308.4 0-0514" = 6000 Ibs. RESILIENCE OF A TENSION-BAR. 247 Hence a suddenly applied load will produce double the strain that would be produced by the same load gradually applied ; and, moreover, a suddenly applied load should be only half as great as one gradually applied if it is to produce the same strain. 171. Resilience of a Tension-Bar. The resilience of a tension-rod is the mechanical work done in stretching it to the same amount that it would stretch under the greatest allowable gradually applied load, and is found by multiplying the greatest allowable load by half the corresponding elongation. Thus, suppose a load of 100 Ibs. to be dropped upon the rod described above in such a way as to cause an elongation not greater than 0.05 14", it would be necessary to drop it from a height not greater than 3.08". EXAMPLES. 1 . A wrought-iron rod is 1 2 feet long and i inch in diameter, and is loaded in the direction of its length; the working-strength of the iron being 12000 Ibs. per square inch, and the modulus of elasticity 28000000 Ibs. per square inch. Find the working-strain. Find the working-load. Find the working-elongation. Find the working-resilience. From what height can a 5o-pound weight be dropped so as to produce tension, without stretching it more than the working- elongation? 2. Do the same fora cast-iron rod, where the working-strength is 5000 pounds per square inch, and the modulus of elasticity 17000000; the dimensions of the rod being the same. 172. Results of Wohler's Experiments on Tensile Strength. According to the experiments of Wohler, of which an account will be given later, the breaking-strength of a piece 248 APPLIED MECHANICS. depends, not only on whether the load is gradually or suddenly applied, but also on the extreme variations of load that the piece is called upon to undergo, and the number of changes to which it is to be submitted during its life. For a piece which is always in tension, he determines the following two constants ; viz., /, the carrying-strength per square inch, or the greatest quiescent stress . that the piece will bear, and u, the primitive safe strength, or the greatest stress per square inch of which the piece will bear an indefinite number of repetitions, the stress being entirely removed in the inter- vals. This primitive safe strength, u y is used as the breaking- strength when the stress varies from o to u every time. Then, by means of Launhardt's formula, we are able to determine the ultimate strength per square inch for any different limits of stress, as for a piece that is to be alternately subjected to 80000 and 6000 pounds. Thus, for Phoenix Company's axle iron, Wohler finds / = 3290 kil. per sq. cent. = 46800 Ibs. per sq. in., u 2100 kil. per sq. cent. = 30000 Ibs. per sq. in. Launhardt's formula for the ultimate strength per unit of area is t u least stress ) a = u{\ + u greatest stress)' Hence, with these values of t and u y we should have, for the ultimate strength per square inch, ( i least stress ) a 2100% i -h - ->kil. per sq. cent., ( 2 greatest stress) or ii least stress ) i -f- - -\ Ibs. per sq. in. 2 greatest stress) WOH LEWS EXPERIMENTS ON TENSILE STRENGTH. 249 Thus, if least stress = 6000, and greatest = 80000, we should have a = 30000$ i -f . -fo\ = 30000^1 + isul = 3 II2 5> if least stress = 60000, and greatest = 80000, a = 30000 | i -f i . f I = 30000^1 + fj = 41250; if least stress = greatest stress = 80000, a 30000^1 -j- \\ = 45000 = carrying-strength. Hence, instead of using, as breaking-strength per square inch in all cases, 45000, we should use a set of values varying from 45000 down to 30000, according to the variation of stress which the piece is to undergo. For working-strength, Weyrauch divides this by 3 : thus obtaining, for working-strength per square inch, ( i least stress ) b 10000 < i H [ Ibs. per sq. in. ; ( 2 greatest stress ) lor Krupp's cast-steel, notwithstanding the fact that Wohler finds / = 7340 kil. per sq. cent. = 104400 Ibs. per sq. in., u = 33 kil. per sq. cent. = 46900 Ibs. per sq. in., Weyrauch recommends ( q least stress ) a 3300 \ i + - -Vkil. per sq. cent., ( ii greatest stress) ( o least stress ) a = 46000 < i 4- >lbs. per sq. in., ( ii greatest stress) = I5633J o least stress ) i 4- - ; - - > Ibs. per sq. in. 1 1 greatest stress j EXAMPLES. Find the breaking-strength per square inch for a wrought-iron tension rod. 1. Extreme loads are 75000 and 6000 Ibs. 2. Extreme loads are 120000 and 100000 Ibs. 3. Extreme loads are 300000 and 10000 Ibs. Find the safe section for the rod in each case. 250 APPLIED MECHANICS. 173. Suspension-Rod of Uniform Strength. In the case of a long suspension-rod, the weight of the rod itself some- times becomes an important item. The upper section must, of course, be large enough to bear the weight that is hung from the rod plus the weight of the rod itself ; but it is sometimes desirable to diminish the sections as they descend. This is often accomplished in mines by making the rod in sections, each section being calculated to bear the weight below it plus its own weight. Were the sections gradually diminished, so that each section would be just large enough to support the weight below it, we should, of course, have a curvilinear form ; and the equation of this curve could be found as follows, or, rather, the area of any section at a distance from the bottom of the rod. Let W = weight hung at O (Fig. 152), Let w = weight per unit of volume of the rod, Let x = distance AO, Let 5 = area of section A, Let x + dx = distance BO, Let 5 + dS = area of section at B t Let f = working-strength of the mate- rial per square inch. i. The section at O must be just large enough to sustain the load W; *ff ' W FIG. 152. * S Q = f- 2. The area in dS must be just enough to sustain the weight of the portion of the rod between A and B. The weight of this portion is wSdx ; _ wSdx .'. <>= dS w iv .*. -= = -fdx /. log, S = -7X H- a constant > / / CYLINDERS SUBJECTED TO INTERNAL PRESSURE. .2$ I W When x = o, 5 = -^ ; W IW\ w .-. log* Y = the constant .% log,.S log,( -y- ) = -^ This gives us the means of determining the area at any dis- tance x from O. EXAMPLES. 1. A wrought-iron tension-rod 200 feet long is to sustain a load of 2000 Ibs. with a factor of safety of 4, and is to be made in 4 sections, each 50 feet long; find the diameter of each section, the weight of the wrought iron being 480 Ibs. per cubic foot. 2. Find the diameter needed if the rod were made of uniform section, also the weight of the extra iron necessary to use in this case. 3. Find the equation of the longitudinal section of the rod, assum- ing a square cross-section, if it were one of uniform strength, instead of being made in 4 sections. 174. Thin Hollow Cylinders subjected to an Internal Normal Pressure. Let/ denote the uniform intensity of the pressure exerted by a fluid which is confined within a hollow cylinder of radius r and of thickness / (Fig, 153), the thickness being small compared with the radius. Let us consider a unit of length of the cylinder, and c ( let us also consider the forces acting on the upper half-ring CED. PIG. 153. The total upward force acting on this half-ring, in conse- quence of the internal normal pressure, will be the same as that acting on a section of the cylinder made by a plane pass- ing through its axis, and the diameter CD. The area of this 252 APPLIED MECHANICS. section will be 2r X I = 2r : hence the total upward force will be 2r X / = 2pr; and the tendency of this upward force is to cause the cylinder to give way at A and B, the upper part separating from the lower. This tendency is resisted by the tension in the metal at the sections AC and BD ; hence at each of these sections, there has to be resisted a tensile stress equal to \(2pr) pr. This stress is really not distributed uniformly throughout the cross-section of the metal ; but, inasmuch as the metal is thin, no serious error will be made if it be accounted as distributed uniformly. The area of each section, however, is t X i = / / therefore, if T denote the intensity of the tension in the metal in a tangential direction (i.e., the intensity of the hoop tension), we shall have Hence, to insure safety, T must not be greater than f, the working-strength of the material for tension ; hence, putting f-pr /- / , we shall have / = 7 as the proper thickness, when/ = normal pressure per square inch, and radius = r. The above are the formulas in common use for the deter- mination of the thickness of the shell of a steam-boiler ; for in that case the steam-pressure is so great that the tension induced by any shocks that are likely to occur, or by the weight of the boiler, is very small in comparison with that induced by the steam-pressure. On the other hand, in the case of an ordinary water-pipe, the reverse is the case. RESISTANCE TO DIRECT COMPRESSION. To provide for this case, Weisbach directs us to add to the thickness we should obtain by the above formulae, a constant minimum thickness. The following -are his formulae, d being the diameter in inches, / the internal normal pressure in atmospheres, and / the thickness in inches. For tubes made of Sheet-iron ......... /* = 0.00086 pd -f 0.12 Cast-iron ......... t = 0.00238^ -f- 0.34 Copper ....... . . . / = o.ooi48/^/ + 0.16 Lead .......... /= 0.00507^+ 0.21 Zinc ........... / = 0.00242 pd -+- 0.16 Wood .......... / = 0.03230^ -f- 1.07 Natural stone ........ /= 0.03690/^4- 1.18 Artificial stone ....... t = 0.05380/^4- 1.58 175. Resistance to Direct Compression. When a piece is subjected to compression, the distribution of the compressive stress on any cross-section depends, first, upon whether the resultant of the pressure acts along the line containing the cen- tres of gravity of the sections, and, secondly, upon the dimen- sions of the piece ; thus determining whether it will bend or not. In the case of an eccentric load, or of a piece of such length that it yields by bending, the stress is not uniformly distributed ; and, in order to proportion the piece, we must determine the greatest intensity of the stress upon it, and so proportion it that this shall be kept within the working-strength of the ma- terial for compression. Either of these cases is not a case of direct compres- sion. In the case of direct compression (i.e., where the stress over each section is uniformly distributed), the intensity of the stress is found by dividing the total compression by the area of the 254 APPLIED MECHANICS. section ; so that, if P be the total compression, and A the area of the section, and / the intensity of the compressive stress, On the other hand, if f is the compressive working-strength oi the material per square inch, and A the area of the section in square inches, then the greatest allowable load on the piece subjected to compression is The same remarks as were made in regard to a suddenly applied load and resilience, in the case of direct tension, apply in the case of direct compression. 176. Results of Wohler's Experiments on Compressive Strength. Wohler also made experiments in regard to pieces subjected to alternate tension and compression, taking, in the experiments themselves, the case where the metal is subjected to alternate tensions and compressions of equal amount. The greatest stress of which the piece would bear an indefi- nite number of changes under these conditions, is called the vibration safe strength, and is denoted by s. Weyrauch deduces a formula similar to that of Launhardt for the greatest allowable stress per unit of area on the piece when it is subjected to alternate tensions and compressions of different amounts. Thus, for Phoenix Company's axle iron, Wohler deduces / = 3290 kil. per sq. cent. = 46800 Ibs. per sq. in., u = 2100 kil. per sq. cent. = 30000 Ibs. per sq. in., s = 1 1 70 kil. per sq. cent. = 1 6600 Ibs. per sq. in. EXPERIMENTS ON COMPRESSIVE STRENGTH. 2$$ Weyrauch's formula for the ultimate strength per unit of area is {u s least maximum stress | u greatest maximum stress \ ' and, with these values of u and s, it gives least maximum stress a = 2100 !i least maximum stress | 1 ~ 2 greatest maximum stress j ' per S( ** cent * 11 least maximum stress ) i - = - / Ibs. per sq. in. 2 greatest maximum stress ) With a factor of safety of 3, we should have, for the greatest admissible stress per square inch, ( i least maximum stress ) b = i oooo < i : Jibs. ( 2 greatest maximum stress) For Krupp's cast-steel, / = 7340 kil. per sq. cent. = 104400 Ibs. per sq. in., u = 3300 kil. per sq. cent. = 46900 Ibs. per sq. in. approximately, s = 2050 kil. per sq. cent. = 29150 Ibs. per sq. in. approximately. We have, therefore, for the breaking-strength per unit of area, according to Weyrauch's formula, least maximum stress a or a !c least maximum stress ) - il greatest maximum stress } kiL per Sq ' Cent ' ( c least maximum stress ) - 4 6 9 oo| , - fi greatestmaximumstress [lbs. per sq. .; 256 APPLIED MECHANICS. and, using a factor of safety of 3, we have, for the greatest admis- sible stress per square inch, !r least maximum stress ) , i f: - : - > Ibs. per sq. in. 1 1 greatest maximum stress j b 15630 The principles respecting an eccentric compressive load, and those respecting the giving-way of long columns so far as they are known, can only be treated after we have studied the resist- ance of beams to bending; hence this subject will be deferred until that time. EXAMPLES. Find the proper working and breaking strength per square inch to be used for a wrought- iron rod, the extreme stresses being 1. 80000 Ibs. tension and 6000 Ibs. compression. 2. i ooooo Ibs. tension and 100000 Ibs. compression. 3. 70000 Ibs. tension and 60000 Ibs. compression. Do the same for a steel rod. 177. Resistance to Shearing. One of the principal cases where the resistance to shearing comes into practical use is that where the members of a structure, which are themselves subjected to direct tension or compression or bending, are united by such pieces as bolts, rivets, pins, or keys, which are sub- jected to shearing. Sometimes the shearing is combined with tension or with bending ; and whenever this is the case, it is necessary to take account of this fact in designing the pieces. It is important that the pins, keys, etc., should be equally strong with the pieces they connect. Probably one of the most important modes of connection is by means of rivets. In order that there may be only a shearing action, with but little bending of the rivets, the latter must fit very tightly. The manner in which the riveting is done will necessarily affect very essentially the strength of the joints; RESISTANCE TO SHEARING. 257 hence the only way to discuss fully the strength of riveted joints is to take into account the manner of effecting the rivet- ing, and hence the results of experiments. These will be spoken of later ; but the ordinary theories by which the strength and proportions of .some of the simplest forms of riveted joints are determined will be given, which theories are necessary also in discussing the results of experiments thereon. The principle on which the theory is based, in these simple cases, is that of making the resistance of the joint to yielding equal in the first three, and also in either the fourth or the fifth of the ways in which it is possible for it to yield, as enumerated on pages 258 and 259. A single-riveted lap-joint is one with a single row of rivets, as shown in Fig. 154. A single-riveted butt-joint with covering plate is shown in T C C one Fig. 155 A single-riveted butt-joint with two covering, plates is shown in Fig. 156. FIG. 154. FIG. 155. FIG. 156. 25 8 APPLIED MECHANICS. FIG. 157. FIG. 158. A double-riveted lap-joint with the rivets staggered is shown in Fig. 157; one with chain riveting, in Fig. 158. Taking the case of the single-riveted lap-joint shown in Fig. 1 54, it may yield in one of five ways : i. By the crushing of the plate in front of the rivet (Fig. 159). FIG. 159. FIG. 160. 2. By the shearing of tne nvet (Fig. 160). RESISTANCE TO SHEARING. 259 3. By the tearing of the plate between the rivet-holes (Fig. 161). 1 FIG. 161. 4. By the rivet breaking through the plate (Fig. 162). 5. By the rivet shearing out the plate in front of it. Let us call d the diameter of a rivet. / the pitch of the rivets ; i.e., FlG - l62 - their distance apart from centre to centre. t the thickness of the plate. / the lap of the plate ; i e., the distance from the centre of a rivet-hole to the outer edge of the plate. f t the ultimate tensile strength of the iron. /, the ultimate shearing-strength of the rivet-iron. f s > the ultimate shearing-strength of the plate. f c the ultimate crushing-strength of the iron. We shall then have i. Resistance of plate in front of rivet to crushing =r f c td. 2. Resistance of one rivet to shearing = //- Y 3. Resistance of plate between two rivet-holes to tearing = /'(/ - d). 4. Resistance of plate to being broken through = a~ , d where a is a constant depending on the material, taken as empirical for the present. A reasonable value of this constant is /". This may be 260< APPLIED MECHANICS. 5. Resistance of plate in front of the rivet to shearing = 2/,7/. Assuming that we know the thickness of the plate to start with, we obtain, by equating the first two resistances, which determines the diameter of the rivet. Equating 3 and 2, we obtain which gives the pitch of the rivets in terms of the diameter of the rivet, and the thickness of the plate. Equating, next, 4 and i, we have which gives the lap of the plate needed in order that it may not break through. By equating 5 and i, we find the lap needed that it may not shear out in front of the rivet. A similar method of reasoning would enable us to determine the corresponding quantities in the cases of double-riveted joints, etc. There are a number of practical considerations which modify more or less the results of such calculations, and which can only be determined experimentally. A fuller account of this subject from an experimental point of view will be given later. 178. Intensity of Stress. Whenever the stress over a plane area is uniformly distributed, we obtain its intensity at each point by dividing the total stress by the area over which it acts, thus obtaining the amount per unit of area. When, how- ever, the stress is not uniformly distributed, or when its inten- INTENSITY OF STRESS. 26l sity varies at different points, we must adopt a somewhat differ- ent definition of its intensity at a given point. In that case, if we assume a small area containing that point, and divide the stress which acts on that area by the area, we shall have, in the quotient, an approximation to the intensity required, which will approach nearer and nearer to the true value of the intensity at that point, the smaller the area is taken. Hence the intensity of a variable stress at a given point is, -- The limit of the ratio of the stress acting on a small area containing that point, to the area, as the latter grows smaller and smaller. By dividing the total stress acting on a certain area by the entire area, we obtain the mean intensity of the stress for the entire area. 179. Graphical Representation of Stress A conven- ient mode of representing stress graphically is the following: Let AB (Fig. 163) be the plane surface upon which the stress acts ; let the axes OX and OY be taken in this plane, the axis OZ being at right angles to the plane. Conceive a portion of a cylinder whose elements are all parallel to OZ, bounded at one end by the given plane surface, and at the other by a surface whose ordinate many units of length as there are units of force in the intensity of the stress at that point of the given plane surface where the ordinate cuts it. The volume of such a figure will evidently be V = ffzdxdy = ffpdxdy, where z / = intensity of the stress at the given point. FIG. 163. at any point contains as 262 APPLIED MECHANICS. Hence the volume of the cylindrical figure will contain as many units of volume as the total stress contains units of force ; or, in other words, the total stress will be correctly repre- sented by the volume of the body. If the stress on the plane figure is partly tension and partly compression, the sur- face whose ordinates repre- sent the intensity of the stress will lie partly on one side of the given plane sur- face and partly on the other ; this surface and the plane surface on which the stress acts, cutting each other in some line, straight or curved, as shown in Fig. 164. In that FlG - case, the magnitude of the resultant stress P V will be equal to the difference of the wedge-shaped volumes shown in the figure. It will be observed that the above method of representing stress graphically represents, i, the intensity at each point of the surface to which it is applied ; and, 2, the total amount of the stress on the surface. It does not, however, represent its direction, except in the case when the stress is normal to the surface on which it acts. In this latter case, however, this is a complete representa- tion of the stress. The two most common cases of stress are, i, uniform stress, and, 2, uniformly varying stress. These two cases are repre- sented respectively in Figs. 165 and 166; the direction also being correctly represented when, as is most frequently the case, the stress is normal to the surface of action. In Fig. 165, AB is supposed to be the surface on which the stress GRAPHICAL REPRESENTATION OF STRESS. 263 acts ; the stress is supposed to be uniform, and normal to the surface on which it acts ; the bound- ing surface in this case becomes a plane parallel to AB ; the intensity of the stress at any point, as P, will be represented by PQ; while the whole cylinder will contain as many units of volume as there are units of force in the whole stress. Fig. i(56 represents a uniformly varying stress. Here, again, AB is the surface of action, and the stress is supposed to vary at a uniform rate FlG> l65 ' from the axis O Y. The upper bounding surface of the cylin- drical figure which represents the stress becomes a plane inclined to the XOY plane, and containing the axis O Y. In this case, if a represent the in- tensity of the stress at a unit's distance from O Y, the stress at a distance x from OY will be/ = ax, and the total amount of the stress will be FIG. 166. P = ffpdxdy = affxdxdy. When a stress is oblique to the surface of action, it may be represented correctly in all particulars, except in direction, in the above-stated way. 1 80. Centre of Stress. The centre of stress, or the point of the surface at which the resultant of the stress acts, often becomes a matter of practical importance. If, for con- venience, we employ a system of rectangular co-ordinate axes, of which the axes OX and OY are taken in the plane of the surface on which the stress acts, and if we let p = $(x, y) be 264 APPLIED MECHANICS. the intensity of the stress at the point (x, y), we shall have, for the co-ordinates of the centre of stress, ffxpdxdy J'Sypdxdy : (see 42), where the denominator, or ffpdxdy, represents the total amount of the stress. When the stress is positive and negative at different parts of the surface, as in Fig. 164, the case may arise when the posi- tive and negative parts balance each other, and hence the stress on the surface constitutes a statical couple. In that case Sfpdxdy = o. 181. Uniform Stress. In the case of uniform stress, we have i. The intensity of the stress is constant, or / = a con- stant. 2. The volume which represents it graphically becomes a cylinder with parallel and equal bases, as in Fig. 165. 3. The centre of stress is at the centre of gravity of the surface of action ; for the formulae become, when / is constant, _ pffxdxdy _ ffxdxdy _ Xl ~~~ pffdxdy ~~~' ~~ pffydxdy = pffdxdy ~' ffdxdy ' "where x , y , are the co-ordinates of the centre of gravity of the surface. Examples of uniform stress have already been given in the cases of direct tension, direct compression, and, in the case of riveted joints, for the shearing-force on the rivet. UNIFORMLY VARYING STRESS. 26$ 182. Uniformly Varying Stress. Uniformly varying stress has already been denned as a stress whose intensity varies uniformly from a given line in its own plane ; and this line will be called the Neutral Axis. Thus, if the plane be taken as the XOY plane (Fig. 166), and the given line be taken as OY, we shall have, if a denotes the intensity of the stress at a unit's distance from OY, and x the distance of any special point from O Y, that the intensity of the stress at the point will be p = ax. The total amount of the stress will be P= affxdxdy. The total moment of the stress about O Y will be found by multiplying each elementary stress by its leverage. This lever- age is, in the case of normal stress, x ; hence in that case the moment of any single elementary force will be and the total moment of the stress will be M - affx^dxdy = al. In the case of oblique stres x s, this result has to be modified, as the leverage is no longer x. Confining ourselves to stress normal to the plane of action, we have, for the co-ordinates of the centre of stress, _ ffpxdxdy _ affx*dxdy _ ffx^dxdy _ ffx*dxdy_ I ffpdxdy ~~ P = ffxdxdy '' x Q A ~ x A _ ffpydxdy _ affxydxdy _ ffxydxdy _ ffxydxdy ~~ ffpdxdy ~~ P ~~ ffxdxdy = x*A since P = affxdxdy = aXoA, where x m y m are the co-ordinates of the centre of gravity, and A is the area of the surface of action. 266 APPLIED MECHANICS. 183. Case of a Uniformly Varying Stress which amounts to a Statical Couple. Whenever P = o, we have affxdxdy = o /. ffxdxdy = o .'. x^A = o .*. x = o. In this case, therefore, we have i. There is no resultant stress, and hence the whole stress amounts to a statical couple. 2. Since X Q = o, the centre of gravity of the surface of action is on the axis OY, which is the neutral axis. Hence follows the proposition : When a uniformly varying stress amounts to a statical couple, the neutral axis contains (passes through) the centre of gravity of the surface of action. In this case there is no .single resultant of the stress ; but the moment of the couple will be, as has been already shown, M = affx 2 dxdy. 184. Example of Uniformly Varying Stress. One of the most common examples of uniformly varying stress is that of the pressure of water upon the sides of the vessel contain- ing it. Thus, let Fig. 167 represent the vertical cross-section of a reservoir wall, the water pressing against the vertical face AB. It is a fact established by experiment, that the intensity of the pressure of any body of water at any point is propor- tional to the depth of the point below the free upper level of the water, and normal to the surface pressed upon. Hence, if we sup- pose the free upper level of the water to be even with the top of the wall, the intensity of the pressure there will be zero ; and if we represent by CB the intensity of the pressure at the bottom, then, joining^ and STRESSES IN BEAMS UNDER TRANSVERSE LOAD. 267 C we shall have the intensity of the pressure at any point, as D t represented by ED, where ED : CB = AD : AB. Here, then, we have a case of uniformly varying stress nor- mal to the surface on which it acts. 185. Fundamental Principles of the Common Theory of the Stresses in Beams under a Transverse Load. Fig. 168 shows a beam fixed at one end and loaded at the other, while Fig. 169 shows a beam supported at the ends and loaded at the middle. Let, in each case, the plane of the paper contain a vertical longi- tudinal section of the beam. In Fig. 1 68, it is evi- dent that the upper fibres are lengthened, while the lower ones are shortened, and vice versa in Fig. 169. In either case, there is, somewhere between the upper and lower fibres, a fibre which is neither elongated nor com- pressed. Let CN repre- sent that fibre, Fig. .168, and CP, Fig. 169. This line may be called the neutral FIG. i6g. line of the longitu- dinal section ; and, if a section be made at any point at right 268 APPLIED MECHANICS. angles to this line, the horizontal line which lies in the cross- section, and cuts the neutral lines of all the longitudinal sec- tions, or, in other words, the locus of the points where the neutral lines of the longitudinal sections cut the cross-section, is called the Neutral Axis of the cross-section. In the ordinary theory of the stresses in beams, a number of assumptions are made, which will now be enumerated. ASSUMPTIONS MADE IN THE COMMON THEORY OF BEAMS. ASSUMPTION No. I. If, when a beam is not loaded, a plane cross-section be made, this cross-section will still be a plane after the load is put on, and bending takes place. From this assumption, we deduce, as a consequence, that, if a certain cross-section be assumed, the elongation or shortening per unit of length of any fibre at the point where it cuts this cross-sec- tion, is proportional to the distance of the fibre from the neutral axis of the cross-section. Proof. Imagine two originally parallel cross-sections so near to each other that the curve in which that part of the neutral line between them bends may, without appreciable error, be accounted circular. Let ED and GH (Fig. 168 or Fig. 169) be the lines in which these cross-sections cut the plane of the paper, and let O be the point of intersection of the lines ED and GH. Let OF = r, FL = y, FK = /, LM = / + a/, in which a is the strain or elongation per unit of length of a fibre at a distance y from the neutral line, y being a variable ; then, because FK and LM are concentric arcs subtending the same angle at the centre, we shall have the proportion r + y I -\- ol y ^ = -y- or i + a = i + y .'.a = - Or a = r ASSUMPTIONS IN THE COMMON THEORY OF BEAMS. 269 but as y varies for different points in any given cross-section, while r remains the same for the same section, it follows, that, if a certain cross-section be assumed, the strain of any fibre at the point where it cuts this cross-section is proportional directly to the distance of this fibre from the neutral axis of the cross- section. ASSUMPTION No. 2. This assumption is that commonly known as Hooke s Law. It is as follows : " Ut tensio sic vis ; " i.e., The stress is proportional to the strain, or to the elonga- tion or compression per unit of length. As to the evidence in favor of this law, experiment shows, that, as long as the mate- rial is not strained beyond safe limits, this law holds. Hence, making these two assumptions, we shall have : At a given cross-section of a loaded beam, the direct stress on any fibre varies directly as the distance of the fibre from the neutral axis. Hence it is a uniformly varying stress, and we may repre- sent it graphically as follows : Let ABCD, Fig. 170, be the cross-sec- tion of a beam, and KL the neutral axis. Assume this for axis OY, and draw the other two axes, as in the figure. If, now, EA be drawn to represent the intensity of the direct (normal) stress at A, then will the pair of wedges AEFBKL and DCHGKL represent the stress graphically, since it is uni- formly varying. POSITION OF NEUTRAL AXIS. ASSUMPTION No. 3. It will next be shown that, on the two assumptions made above, and from the further assumption that the deformation of each fibre of the beam parallel to its* longitudinal axis is due to the forces acting on its ends FIG. 170. 2~0 APPLIED MECHANICS. and that it suffers no traction from neighboring fibres, it fol- lows that the neutral axis must pass through the centre of gravity of the cross-section. D N 1 1 |I C E 1 A A iff * ""vi B ' FIG. 171. FIG. 172. Since the curvatures in Figs. 168 and 169 are exaggerated in order to render them visible, Figs. 171 and 172 have been drawn. If, now, we assume a section DE, such that AD = x (Fig. 171) and NE = x (Fig. 172), and consider all the forces acting on that part of the beam which lies to the right of DE (i.e., both the external forces and the stresses which the other parts of the beam exert on this part), we must find them in equilibrium. The external forces are, in Fig. 172, i. The loads acting between B and E ; in this case there are none. 2. The supporting force at B ; in this case it is equal to W , and acts vertically upwards. In Fig. 171 they are, The loads between D and N ' ; in this case there is only the one, W at N. The internal forces are merely the stresses exerted by the other parts of the beam on this part : they are, i. The resistance to shearing at the section, which is a vertical stress. 2. The direct stresses, which are horizontal. Now, since the part of the beam to the right of DE is at rest, the forces acting on it must be in equilibrium ; and, since POSITION OF NEUTRAL AXIS. 2 7 l they are all parallel to the plane of the paper, we must have the three following conditions ; viz., i. The algebraic sum of the vertical forces must be zero. 2. The algebraic sum of the horizontal forces must be zero. 3. The algebraic sum of the moments of the forces about any axis perpendicular to the plane of the paper must be zero. But, on the above assumptions, the only horizontal forces are the direct stresses : hence the algebraic sum of these direct stresses must be zero ; or, in other words, the direct stresses must be equivalent to a statical couple. Now, it has already been shown, that, whenever a uniformly varying stress amounts to a statical couple, the neutral axis must pass through the centre of gravity of the surface acted upon. Hence in a loaded beam, if the three preceding assump- tions be made, it follows that the neutral axis of any cross- section must contain the centre of gravity of that section. By way of experimental proof of this conclusion, Barlow has shown by experiment, that, in a cast-iron beam of rectangu- lar section, the neutral axis does pass through the centre of gravity of the section. RESUME. The conclusions arrived at from the foregoing are as fol- lows : i. That at any section of a loaded beam, if a horizontal line be drawn through the centre of gravity of the section, then the fibres lying along this line will be subjected neither to tension nor to compression ; in other words, this line will be the neutral axis of the section. 2. The fibres on one side of this line will be subjected to tension, those on the other side being subjected to compres- sion ; the tension or compression of any one fibre being proper tionai to its distance from the neutral axis. 2/2 APPLIED MECHANICS. The first of the three assumptions of the common theory was not accepted by St. Venant, who developed by means of the methods of the Theory of Elasticity a theory of beams based upon the second and third assumptions only. A study of. St. Venant's theory involves, however, far more complica- tion, and requires a good previous knowledge of the Theory of Elasticity. Moreover the results of the two theories as far as the determination of the outside fibre-stresses and of the de- flections are practically in agreement, while, on the other hand, the intensities of the shearing-forces as computed by the two theories are not in agreement. The St. Venant theory may be found in several treatises upon the Theory of Elasticity. 1 86. Shearing-Force and Bending-Moment. In deter- mining the strength of a beam, or the proper dimensions of a beam to bear a certain load, when we assume the neutral axis to pass through the centre of gravity of the cross-section, we have imposed the second of the three last-mentioned conditions of equilibrium. The remaining two conditions may otherwise be stated as follows : i. The total force tending to cause that part of the beam that lies to one side of the section to slide by the other part, must be balanced by the resistance of the beam to shearing at the section. 2. The resultant moment of the external forces acting on that part of the beam that lies to one side of the section, about a horizontal axis in the plane of the section, must be balanced by the moment of the couple formed by the resisting stresses. The shearing-force at any section is the force with which the part of the beam on one side of the section tends to slide by the part on the other side. In a beam free at one end, it is equal to the sum of the loads between the section and the free end. In a beam supported at both ends, it is equal in magnitude to the difference between the supporting force at either end, 'and the sum of the loads between the section and that support. SHEARING-FORCE AND BEND ING-MOMENT. 2? 3 The bending-moment at any section is the resultant moment of the external forces acting on the part of the beam to one side of the section, these moments being taken about a hori- zontal axis in the section. In a beam free at one end, it is equal to the sum of the moments of the loads between the section and the free end, about a horizontal axis in the section. In a beam supported at both ends, it is the difference be- tween the moment of either supporting force, and the sum of the moments of the loads between the section and that sup- port ; all the moments being taken about a horizontal axis in the section. Hence the two conditions of equilibrium may be more briefly stated as follows : i. The shearing-force at the section must be balanced by the resistance opposed by the beam to shearing at the section. 2. The bending-moment at the section must be balanced by the moment of the couple formed by the resisting stresses. It is necessary, therefore, in determining the strength of a beam, to be able to determine the shearing-force and bending- moment at any point, and also the greatest shearing-force and the greatest bending-moment, whatever be the loads. A table of these values for a number of ordinary cases will now be given ; but I should recommend that the table be merely considered as a set of examples, and that the rules already given for finding them be followed in each individual case. Let, in each case, the length of the beam be /, and the total load W. When the beam is fixed at one end and free at the other, let the origin be taken at the fixed end ; when it is supported at both ends, let it be taken directly over one support. Let x be the distance of any section from the origin. Then we shall have the results given in the following table : 274 APPLIED MECHANICS. At Dista from O ifeh. ft Distance m Origin. ' " pq Si I T? & g '-3 >, IS 11 4'g ! It W iJ .g r- 4> u PQ MOMENTS OF INERTIA OF SECTIONS. 2/5 In a beam fixed at one end and free at the other, the great- est shearing-force, and also the greatest bending-moment, are at the fixed end. In a beam supported at both ends, and loaded at the middle, or with a uniformly distributed load, the greatest shearing-force is at either support, the greatest bending-moment being at the middle. In the last case (i.e., that of a beam sup- ported at the ends,- and having a single load not at the middle),. the greatest bending-moment is at the load ; the greatest shear- ing-force being at that support where the supporting force is greatest. 187. Moments of Inertia of Sections. In the usual methods of determining the strength of a beam or column, it is necessary to know, i, the distance from the neutral axis of the section to the most strained fibres ; 2, the moment of in- ertia of the section about the neutral axis. The manner of finding the moments of inertia has been explained in Chap. II. In the following table are given the areas of a large number of sections, and also their moments of inertia about the neutral axis, which is the axis YY in each case. These results should be deduced by the student. 276 APPLIED MECHANICS. Distance of YY from Extreme Fibres. S. HP | 1! II MOMENTS OF INERTIA OF SECTIONS. 277 fa ^r S- SS 5ia 5lx 13 w rt -o CQ 278 APPLIED MECHANICS. o a 1 rt X w W Q vO f I I -S -8 S 3 ^ ^H ^ > MH 55 "5 g -s ^ bfl tfl ^ bo !* MOMENTS OF INERTIA OF SECTIONS. '279 + + 4- ^ IN *?! II- ^ I H ^ c/T > | -S'S 4 s,-^ 's ^ s N 2 " s * * K if fi ii + S o < H o '^3 rt oS rt Vj (U (L) o i- >_ i- JA, ^ U 280 APPLIED MECHANICS. o g 4> c SJIN u II ** rj- fp 1 8" V V. II II i2 x a J5 131 ' 11 I '~ rt i MOMENTS OF INERTIA OF SECTIONS. 28l 1 -^1 ^ 8 8 u 3 1 V* - 1 . - 1* * ^ "5 rt ^-t-> ^3'^g 1 ~ .2 " '?:! '-= ^ ^ | ^ f X rt> G O "-> N ro bJO C JU 2" 2" 2 s 2" ^ O C4 jS>- < crj *"O o>- \ ^ 'o ^ c w <---i \ m ? ^' s - r >. fo 282 APPLIED MECHANICS. .5 + C T3 rt X! fi it!! ^ P S < : i 286 APPLIED MECHANICS. 188. Cross-Sections of Phoenix Columns considered as made of Lines. It is to be observed that the moments of inertia are the same for all axes passing through the centre. Thickness /, radius of round ones = r, area of each flange = a, length of each flange = /. Figure. Description. A. Y 2 II Y2I2 Four flanges 2-nrt + 20. Eight flanges 2irrt + 8a (-0 Square, four flanges, r radius of cir- cumscribed circle Six flanges 6a REPRESENTATION OF B ENDING-MOMENTS. 287 189. Graphical Representation of Bending-Moments. The bending-moment at each point of a loaded beam may be represented graphically by lines laid off to scale, as will be shown by examples. I. Suppose we have the cantilever shown in Fig. 215, loaded at D with a load W ': then will the bending-moment at any section, as at F t be obtained by multiplying W by FD ; that at AC being W X (AB). If, now, we lay off CE to scale to represent this, i.e., having as many units of length FIG. 215. as there are units of moment in the product W X (AB), and join E with D, then will the ordinate FG of any point, as G, represent (to the same scale) the bending-moment at a section through F. II. If we have a uniformly distributed load, we should have, for the line corresponding to CE in Fig. 215, a curve. This is shown in Fig. 216, where we have the uniformly distributed load EIGF. If we take the origin at D, as before, we have, for the bending-moment, at a distance ^ from the origin, as has been W shown, -(/ xY ; and by giving x dif- ferent values, and laying off the corresponding value of the bending-moment, we obtain the curve CA, any ordinate of which will represent the bending-moment at the corresponding point of the beam. When we have more than one load on a beam, we must draw the curve of bending-moments for each load separately, and then find the actual bending-moment at any point of the beam FIG. 216. 283 APPLIED MECHANICS. by taking the sum of the ordinates (drawn from that point) of each of these separate curves or straight lines. If we then draw a new curve, whose ordinates are these sums, we shall have the actual curve of bending-moments for the beam as loaded. Some examples will now be given, which will explain them- selves. III. Fig. 217 shows a cantilever with three concentrated loads. The line of bending-moments for the load at C is CE, that for the load at O is OF, and for the load at P is PG. They are combined above the beam by laying off AH DE, HK = DF, and KL = DG, and thus obtaining the broken line LMNB, which is the line of bending-moments of the beam loaded with all three loads. FIG. 217. IV. Fig. 218 shows the case of a beam supported at both ends, and loaded at a single . point D; ALB is the line of bending- moments when the weight of the beam is disregarded, so that xy = bending-moment at x. FIG. 218 V. Fig. 219 shows the case of a beam supported at the ends, and loaded with three concentrated loads at the points B, C, and D re- spectively; the lines of bending-mo- ments for each individual load being respectively AFE, AGE, and AHE, FIG. 219. and the actual line of bending-mo- ments being AKLME. REPRESENTATION OF BEND ING-MOMENTS. 289 VI. Fig. 220 shows the case of a beam supported at the ends, and loaded with a uniformly dis- A E F B tributed load ; the line of bending- moments being a curve, ACDB, as shown in the figure. FlG . 220> VII. In Fig. 221 we have the case of a beam, over a part of which, viz., EF, there is a distributed load ; the rest of the beam being unloaded. The line of bending-moments is curvilinear be- tween E and F, and straight outside Xjs^/tt y^ of these limits. It isAGSHB; and, when the curve is plotted, we can N/ find the greatest bending-moment graphically by finding its greatest ordinate. We can also determine it analytically by first determining the bending- moment at a distance x from the origin, and on the side towards the resultant of the load, and then differentiating. This process is shown in the following: Let A (Fig. 222) be the point where the resultant of the load acts, and O the middle of the beam, and let w be the c OA B load per unit of length ; let OA = a, AB = AC = b, and ED = 2^, so that the whole load = 2wb : there- a A- c wb(a 4- c] fore supporting force at D = 2ivb = - -. If we take a section at a distance x from O to the right, we shall have, for the bending-moment at that section, wb(a -+ c) w (c x) (a -f b x) 2 = a maximum. Differentiate, and we have wb(a -\- c\ a(c &) I tioi ( /T j h */\ /~\ * c T V -r ; ... - c 290 APPLIED MECHANICS. hence the greatest bending-moment will be ^ \ C ) 2 \ C ) 7 ( a ac VIII. In Figs. 223 and 224 we have the case of a beam supported at the ends, and 1 J J <.!_ loaded with a uniformly dis- tributed load, and also with a c o n c e n- trated load. In the first Fi G .2 24 . FlG - 22 3- figure, the greatest bending-moment is at/?, and in the second at C. IX. In Fig. 225 we have a beam supported at A and B, and loaded at C and D with equal weights; the lengths of AC and BD being equal. We have, con- sequently, between A and , a uniform bending-moment ; while on the left of A and on the right The line of bending- FlG - 225 ' of B we have a varying bending-moment. moments is, in this case, CabD. We may, in a similar way, derive curves of bending-moment for all cases of loading and supporting beams. AT DIFFERENT PARTS OF A BEAM. 2QT 190. Mode of Procedure for Ascertaining the Stresses^ at Different Parts of a Beam when the Loads and the Di- mensions are given, and when no Fibre at the Cross- section under Consideration is Strained beyond the Elastic Limit. When the dimensions of a beam, the load and its distribution, and the manner of supporting are given, and it is desired to find the actual intensity of the stress on any particular fibre at any given cross-section, we must pro- ceed as follows :-- i. Find the actual bending-moment (M) at that cross-sec- tion. 2. Find the moment of inertia (/) of the section about its neutral axis. 3. Observe, that, from what has already been shown, the moment of the couple formed by the tensions and compressions is al, where a = intensity of stress of a fibre whose distance from the neutral axis is unity, and that this moment must equal the bending-moment at the section in order to secure equilib- rium. Hence we must have Moreover, if / denote the (unknown) intensity of the stress of the fibre where the* stress is desired, and if y denote the distance of this fibre from the neutral axis, we shall have from which equation we can determine/. EXAMPLES. i. Given a beam 18 feet span, supported at both ends, and loaded uniformly (its own weight included) with 1000 Ibs. per foot of length. The cross- section is a T, where area of flange = 3 square inches, area of web = 4 square inches, height = 10 inches. Find (a) the 292 APPLIED MECHANICS. bending-moment at 3 feet from one end ; (b) the greatest bending- moment; (c) the greatest intensity of the tension at each of the above sections ; (d) the greatest intensity of the compression at each >of these sections. 2. Given an I-beam with equal flanges, area of each flange = 3 square inches, area of web = 3 square inches, height = 10 inches; the beam is 1 2 feet long, supported at the ends, and loaded uniformly (its own weight included) with a load of 2000 Ibs. per foot of length. Find J(a) the bending-moment at a section one foot from the end ; (<) the greatest bending-moment ; ( a, wx* To find the value of x for the section of greatest bending-moment, differentiate each, and put the first differential co-efficient = zero. We shall thus have, in the first case, W l W(l -a) I W(l - a) -- 1 -- ~ - - w# = o, or x = - H -- - -. - : 2 / 2 wl and in the second case, wl W(l -a) I W(l -a) W - + , - - -wx- W** o, or x = - + v . - -- . 2 / 2 Wl W Now, whenever the first is < #, or the second is > a, we shall have in that one the value of x corresponding to the section of greatest bending-moment. But if the first is > a, and the second < 0, then the greatest bending-moment is at the concentrated load. These conclusions will be evident on drawing a diagram representing the bending-moments graphically, as in Figs. 223 and 224; and the greatest bending-moment may then be found by substituting, in the cor- responding expression for the bending-moment, the deduced value of x. 296 APPLIED MECHANICS. 2. Given an I-beam, 10 feet long, supported at both ends, and loaded, at a distance 2 feet to the left of the middle, with 20000 pounds. Find the bending-moment at the middle, the greatest bending-moment, also the greatest intensity of the tension, and that of the compression at each of these sections. Given Area of upper flange = 8 sq. in. Area of lower flange = 5 sq. in. Area of web = 7 sq. in. Total depth = 14 in. 193. Beams of Uniform Strength. Abeam of uniform strength (technically so called) is one in which the dimensions of the cross-section are varied in such a manner, that, at each cross-section, the greatest intensity of the tension shall be the same, and so also the greatest intensity of the com- pression. -Such beams are very rarely used ; and, as the cross-section varies at different points, it would be decidedly bad engineering to make them of wood, for it would be necessary to cut the wood across the grain, and this would develop a tendency to split. In making them of iron, also, the saving of iron would gen- erally be more than offset by the extra cost of rolling such a beam. Nevertheless, we will discuss the form of such beams in the case wlien the section is rectangular. In all cases we have the general equation y applying at each cross-section, where M = bending-moment ''section at distance x from origin), / = moment of inertia of same section, ' y = distance from neutral axis to most strained fibre, and p intensity of stress on most strained fibre ; the condition for this case being that / is a constant for all values of x (i.e., for all positions of the section), while M, /, and y are functions of x. BEAMS OF UNIFORM STRENGTH. 297 As we are limiting ourselves to rectangular sections, if we let b = breadth and h = depth of rectangle (one or both vary- ing with x\ we shall have as the condition for such a beam, with/ a constant for all values of x, when the same load remains on the beam. We must, therefore, have bk 2 proportional to M. Hence, assuming the origin as before, i. Fixed at one end, load at the other, bh* =(-) W(l *). 2. Fixed at one end, uniformly loaded, bh 2 = ( - ) (/ x) 2 . \ 21' , 3. Supported at ends, loaded at I 2 \/ 2 middle - | 2 p 2 4. Supported at ends, uniformly loaded, bh 2 = ( --- }(lx x 2 ). \j> 2l ' Now, this variation of section may be accomplished in one of two ways: ist, by making h constant, and letting b vary; and 2d, by making b constant, and letting h vary. Thus, in the first case above mentioned, if h is constant, we have, for the plan of the beam, and if one side be taken parallel to the axis of the beam, this will be the equation of the other side ; and, as this is the equa- tion of a straight line, the plan will be a triangle. APPLIED MECHANICS. If, on the other hand, b be constant, and h vary, we shall nave, for the vertical longitudinal section of the beam, and, if one side be taken as a straight line in the direction of the axis, the other will be a parabola. A similar reasoning will give the plan or elevation respect- ively in each case ; and these can be readily plotted from their equations. CROSS-SECTION OF EQUAL STRENGTH. A cross-section of equal strength (technically so called) is one so proportioned that the greatest intensity of the tension shall bear the same ratio to the breaking tensile strength of the material as the greatest intensity of the compression bears to the breaking compressive strength of the material. This is accomplished, as will be shown directly, by so arranging the form and dimensions of the section that the distance of the neutral axis from the most stretched fibre shall bear to its distance from the most compressed fibre the same ratio that the tensile bears to the compressive strength of the material. Let f c breaking-strength per square inch for compression, f t = breaking-strength per square inch for tension, y c = distance of neutral axis from most compressed fibre, y t = distance of neutral axis from most stretched fibre. If p c = actual greatest intensity of compression, and p t = actual greatest intensity of tension, then, for a cross-section of equal strength, we must have, according to the definition, <=; but we have = = intensity of stress at a unit's pt ft yc yt CROSS-SECTION OF EQUAL STRENGTH. 2Q9 distance from the neutral axis. Hence, combining these two, we obtain y - = 7 y t ft EXAMPLE. Suppose we have^ = 80000 Ibs. per square inch, and/ = 20000 Ibs. per square inch. : find the proper proportion between the flange A t and the web A 2 of a T-section whose depth is h. 194. Deflection of Beams. We have already seen ( 185), that, in the case of a beam which is bent by a transverse load, we have -'oo, a r where (having assumed a certain cross-section whose distance from the origin is x) a = the strain of a fibre whose distance from the neutral axis is y, and r = radius of curvature of the neutral lamina at the section in question. Hence follows the equation but from the definition of E t the modulus of elasticity, we shall have V* where / = intensity of the stress, at a distance y from the neutral axis. Hence it follows, assuming Hooke's law, that r Ey E y We have already seen, that, disregarding signs, M = - / 3^0 APPLIED MECHANICS. (making, of course, the two assumptions already spoken of when this formula was deduced), where M = bending-moment at, and / = moment of inertia of, the section in question ; i.e., of that section whose distance from the origin is x. This gives - = , if, denoting tension by the + sig n > and taking y positive upwards, we call M positive when it tends to cause tension on the lower, and compression on the upper, side; these being the conventions in regard to signs which we shall adopt in future. Hence, by substitution, we have 1 - p - M (i\ ~r~Ey- El Now, if we assume the axis of x coincident with the neutral line of the central longitudinal section of the beam, and the axis of v at right angles to this, and v positive upwards, no matter where the origin is taken, we shall always have, as is shown in the Differential Calculus, ' (+(1)7 Hence equation (i) becomes (*) M and / being functions of x : and, when we can integrate this equation, we can obtain v in terms of x, thus having the equation of the elastic curve of the neutral line ; and, by com- puting the value of v corresponding to any assumed value of x, we can obtain the deflection at that point of the beam. FORMULA FOR SLOPE AND DEFLECTION. 3OI The above equation (2) is, as a rule, too complicated to be integrated, except by approximation ; and the approximation usually made is the following : Since in a beam not too heavily loaded, the slope, and con- sequently the tangent of the slope (or angle the neutral line makes with the horizontal at any point), is necessarily small, it follows that is very small, and hence (--) is also very small, dx \dxl and i + ( ) is nearly equal to unity. Making this substitu- tion, we obtain, in place of equation (2), -. d* ~ EI* and this is the equation with which we always start in com- puting the slope and deflection of a loaded beam, or in finding the equation of the elastic line. By one integration (suitably determining the arbitrary con- stant) we obtain the slope whose tangent is , and by a second dx integration we obtain the deflection v at a distance x from the origin ; and thus, by substituting any desired value for x, we can obtain the deflection at any point. 195. Ordinary Formulae for Slope and Deflection. We may therefore write, if i is the circular measure of the slope at a distance x from the origin, since i = tan i = -j- dx nearly, -*< ?- To determine c, observe that when x = o, i = o ; c = o is the slope at a distance x from the origin. The deflection at the same point will be , . f,** = - y& = -^- 2 - ^w J EIJ \ 2 ) EI\ 2 6 ) but when x = o, v = o .*. ^ = o /. the deflection at a distance x from the origin will be The equations (i) and (2) give us the means of finding the slope and deflection at any point of the beam. To find the greatest slope and deflection, we have that both expressions are greatest when x = I. Hence, if * and V Q rep- resent the greatest slope and deflection respectively, Wl 2 SPECIAL CASES. 33 2. Next take the case of a beam supported at both ends and loaded uniformly, the load per unit of length being w. Assume the origin at the left-hand end ; then wl wx 2 w M x --- = Ux x 2 ) and W= wl 2 2 2 V w /w /lx 2 x*\ (lx - x ^ x = _(___) + ,. 2EI ^ To determine c, we have that when x = -, then i = o; W // 3 / 3 \ 2// 3 V 8 " V + - } - W ^ W (6/x 2 - /3^ f ff '/ ' V -5 "?/ r> A 7? T Therefore, for the slope up to the middle, we have w r w x 2 i = ^r, I xdx = =- T h c. 2EIJ 2EI 2 When x ~ , then i = o ; wr W . and w /v , r\. w (x> rx\ v - I \x lax = I AJcl'J \ 4/ 4.fi/\3 4/ But when x o, v = o ; c = o. i 4 The slope is greatest when x o ; .'. z' = The deflection is greatest when x = -; 4. In the following table 7 denotes the moment of inertia of the largest section : SPECIAL CASES. 305 Uniform Cross-Section. Greatest Slope. Greatest Deflection. Fixed at one end, loaded at the other Fixed at one end, loaded uniformly . . Supported at ends, load at middle . . . Supported at ends, uniformly loaded . . i Wl 2 2 7?7 i Wl 2 i Wl* ZEI i Wl* *>EJ i Wl 2 *EI i Wl* '6 El i Wl* ^EI 48^7 5 #73 384^/ Uniform Strength and Uniform Depth, Rectangular Section. Fixed at one end, load at the other . . Fixed at one end, uniformly loaded . . Supported at both ends, load at middle . Supported at both ends, uniformly loaded, Wl 2 ~EI i Wl 2 2 El i Wl* *~EI i Wl 2 i Wl* ~*~EI iWZ* 4^7 i Wl* V~EI i m* & EI 64^7 Uniform Strength and Uniform Breadth, Rectangular Section. Fixed at one end, loaded at the other, Supported at both ends, load at middle, Supported at both ends, uniformly loaded wr 2 ~EI i wr 2 wr 3 EI i wr 4 El o wr 24 EI 0018 Wr 0-098 EI 0.010 EJ 306 APPLIED MECHANICS. 197. Deflection with Uniform Bending-Moment If the bending-moment is uniform, then M is constant ; and, if / is also constant, we have _ ^L f - Mx but when x = -, then i = o; Ml 2EI Ml l\ dv " t = T^rl x -- ) = -r EI\ 2] dx lx the constant disappearing because v = o when x = o. Hence, for a beam where the bending-moment is uniform, we have _ J ^\ M Y* 2 ^ and for greatest slope and deflection, we have -Ml Mil* / 2 \ i Ml 2 1 .71 ^O * ~~ r\ T * V<~\ -r- 198. Resilience of a Beam. 7^ resilience of a beam is the mechanical work performed in deflecting it to the amount it would deflect under its greatest allowable gradually applied load. In the case of a concentrated load, if W is the greatest allowable gradually applied load, and v l the corresponding deflection at the point of application of the load, then will the W mean value of the load that produces this deflection be W and the resilience of the beam will be z/,. 2 SLOPE AND DEFLECTION OF A BEAM. 3O/ i99- Slope and Deflection of a Beam with a Con- centrated Load not at the Middle. Take, as the next A a case, a beam (Fig. 228). Let < a the load at A be W, and dis- tance OA = a y and let a > -. 2 FIG. 228. W(l - a) x < a M = ^ '- x, ^f - a) - x we have that when x /, v = o ; Substituting this value of i in the equations (i), (2), (3), and (4), we obtain for SLOPE AND DEFLECTION OF A BEAM. 309 Wa , r Wa (4) v = To find the greatest deflection, differentiate (2), and place the first differential co-efficient equal to zero : or, which is the same thing, place i = o in (i), and find the value of x ; then substitute this value in (2), and we shall have the greatest deflection. We thus obtain (/-*)*> = ^( 3 / -*/-*>) .-. = "-(* ~ & + a -\ 3 3\ ' ' / or ' *=' ' a ~'> and the greatest deflection becomes Wa(l - a)(2l - a) _ 2OO. EXAMPLES. 1. In example i, p. 294, find the greatest deflection of the beam when it is loaded with \ of its breaking-load, assuming E = 1200000. 2. In the same case, find what load will cause it to deflect ^J^ of its span. 3. What will be the stress at the most strained fibre when this occurs- 4. In example 3, p. 294, find the load the beam will bear without deflecting more than ^J^ of its span, assuming E = 24000000. 5. Find the stress at the most strained fibre when this occurs. 6. In example 6, p. 295, find the greatest deflection under a load J the breaking-load. 3io APPLIED MECHANICS. 2Oi. Deflection and Slope under Working-Load. If we take the four cases of deflection given in the first part of the table on p. 305, and calling/ the working strength of the material, and y the distance of the most strained fibre from the neutral axis, and if we make the applied load the working- load, we shall have respectively ! m =*L W=^ y ty Wl // 2/7 2 . = /. W 2 y ly m = fj 4 '" y wi fl 4- T -T *y : W= ^ And the values of slope and deflection will become respectively, Slope. Deflection. Slope. Deflection. / 2. From these values, and those given on p. 305, we derive the following two propositions : i. If we have a series of beams differing only in length; and we apply the same load in the same manner to each, their greatest slopes will vary as the squares of their lengths, and their greatest deflections as the cubes of their lengths. SLOPE AND DEFLECTION OF RECTANGULAR BEAMS. 31! 2. If, however, we load the same beams, not with the same load, but each one with its working-load, as determined by allowing a given greatest fibre stress, then will their greatest slopes vary as the lengths, and their greatest deflections as the squares of their lengths. 202. Slope and Deflection of Rectangular Beams bfc h If the beams are rectangular, so that / = and y = -, the values of slope and deflection above referred to become further simplified, and we have the following tables : Given Load W. Working-Load. Greatest Fibre Stress =/. * Slope. Deflection. Slope. Deflection. 1. 2. 3. 4- 6W1 2 4 #73 I 2 / 2 Ebfc 2W* Ebte 3 Wl* Eh 2_fl 3 Eh i Ebte 3 Wl* 46fc i Wl* *Ebh* i Wl* 3 Eh i fl 2 Eh i/' lEbfc 5 wr* *Eh 2ft 1 Eh (>Eh sfi* ^Eh * Ebfc 32 Ebh* So that, in the case of rectangular beams similarly loaded and supported, we may say that Under a given load W, the slopes vary as the squares of the lengths, and inversely as the breadths and the cubes of the depths ; while the deflections vary as the cubes of the lengths, and inversely as the breadths and the cubes of the depths. 312 APPLIED MECHANICS. On the other hand, under their working-loads, the slopes vary directly as the lengths, and inversely as the depths ; while the deflections vary as the squares of the lengths, and inversely as the depths. 203. Beams Fixed at the Ends. The only cases which we shall discuss here are the two following ; viz., i. Uniform section loaded at the middle. 2. Uniform section, load uniformly distributed. CASE I. Uniform Section loaded at the Middle. The fixing at the ends may be effected by building the beam for some distance into the wall, as shown in Fig. 229. The same result, as far as the effect on |w the beam is concerned, might be effected as follows : Hav- ing merely supported it, and placed upon it the loads it has to bear, load the ends outside of the supports just enough to make the tangents at the sup- ports horizontal. These loads on the ends would, if the other load was re- moved, cause the beam to be convex upwards : and, moreover, the bending-moment due to this load would be of the same amount at all points between the supports ; i.e., a uniform bending-moment. Moreover, since the effect of the central load and the loads on the ends is to make the tangents over the supports horizontal, it follows that the upward slope at the support due to the uniform bending-moment above described must be just equal in amount to the downward slope due to the load at the middle, which occurs when the beam is only sup- ported. Hence the proper method of proceeding is as follows : i. Calculate the slope at the support as though the beam were supported, and not fixed, at the ends ; and we shall if we represent this slope by i u the equation BEAMS FIXED AT THE ENDS. 313 w* 2. Determine the uniform bending-moment which would produce this slope. To do this, we have, if we represent this uniform bending- moment by M lf that the slope which it would produce would be and, since this is equal to * we shall have the equation _^/_ ;w_. M TEI -*"- V3) .: Jf, = ~ (4) This is the actual bending-moment at either fixed end ; and the bending-moment at any special section at a distance x from the origin will be where M is the bending-moment we should have at that sec- tion if the beam were merely supported, and not fixed. Hence, when it is fixed at the ends, we shall have, for the bending- moment at a distance x from O, where O is at the left-hand support, W W, M=oc-l. (5) When x = -, we obtain, as bending-moment at the middle, *-?; (6) o and, since M l = M , it follows that the greatest bending- moment is W 8 ; 3H APPLIED MECHANICS. this being the magnitude of the bending-moment at the middle and also at the support. POINTS OF INFLECTION. The value of M becomes zero when x = - and when x = ; 4 4' hence it follows that at these points the beam is not bent, and that we thus have two points of inflection half-way between the middle and the supports. SLOPE AND DEFLECTION UNDER A GIVEN LOAD. We shall have, as before, W& Wlx . /M , EI dX = and since, when x = o, i = o, .. c ' o """ * = ~dx = W I 2X l IX 2 v I 3 2 the constant vanishing because v = o when x = o. The slope becomes greatest when x = -, and the deflection when x = -. 4 Hence for greatest slope and deflection, we have Wl 2 f . 64^7' BEAMS FIXED AT THE ENDS. 315 SLOPE AND DEFLECTION UNDER THE WORKING-LOAD. If f represent the working-strength of the material per square inch, and if W represent the centre working-load, we shall have fiP7 = /7 8 " y CASE II. Uniform Section, Load uniformly Distributed. Pursuing a method entirely similar to that adopted in the former case, we have i. Slope at end, on the supposition of supported ends, is Wl* 2 4 ^/ 2. Slope at end under uniform bending-moment M t is () Hence, since their sum equals zero, Wl 12 ' which is the bending-moment over either support. The bending-moment at distance x from one end is W Wl M - \lx x 2 ) . ^4) 2/ 12 Wl This is greatest when x = o, and is then . Hence great- 1 2 est bending-moment is, in magnitude, (5) 12 316 APPLIED MECHANICS. POINTS OF INFLECTION. M becomes zero when x = - =. (6) 2 2 ^3 Hence the two points of inflection are situated at a distance / on either side of the middle. SLOPE AND DEFLECTION. , M 7 t s= I -~dx = the constant vanishing because z = o when x = Oi W v = the constant vanishing because ^ = o when .r = o. Hence for greatest slope and deflection we have, t is greatest when x = -f i zt -y=\ and z; is greatest when * = - ; SLOPE AND DEFLECTION UNDER WORKING-LOAD. For working-load we have Wl fl 77 = 7 B ENDING-MOMENT AND SHEARING-FORCE. 3'7 EXAMPLES. 1. Given a 4-inch by 12 -inch yellow-pine beam, span 20 feet, fixed at the ends ; find its safe centre load, its safe uniformly distributed load, and its deflection under each load. Assume a modulus of rupture 5000 Ibs. per square inch, and factor of safety 4. Modulus of elasticity, 1200000. 2. Find the depth necessary that a 4-inch wide yellow-pine beam, 20 feet span, fixed at the ends, may not deflect more than one four-hun- dredth of the span under a load of 5000 Ibs. centre load. 204. Variation of Bending-Moment with Shearing- Force. If, in any loaded beam whatever, M represent the b ending-moment, and F the shearing-force at a distance x from the origin, then will *-<* > Proof (a). In the case of a cantilever (Fig. 230), assume the origin at the fixed end ; then, if M represent the bending-moment at a distance x from the origin, and M '+ ^M that at a distance x + tx from the origin, we shall have the following equations : x = l M= S W(a- x), X = X x = l M + AJ/ = -2 W(a- x A#) nearly. X = X a being the co-ordinate of the point of application of W, x = l AJ/= AjeS W nearly ====- = 2 W: 3 I 8 APPLIED MECHANICS. and, if we pass to the limit, and observe that we shall obtain (b) In the case of a beam supported at the ends (Fig. 231), , A.,*. assume the origin at the left-hand /|^ I ^j,j 7\ end, and let the left-hand support- ing-force be S ; then, if a represent FIG. 231. the distance from the origin to the point of application of W, we shall have the equations M = Sx - 2 W(x - a), M 4- bM = S(* + A#) - S ^(tf d5 4- A*) nearly. Hence, by subtraction, X- X = S . tec 2 WA* nearly JT = o = o 2 ^nearly; mt-' x = o and, if we pass to the limit, and observe that p ^ g 2 iff Jf = O we shall obtain as before. LONGITUDINAL SHEARING OF BEAMS. 319 205. Longitudinal Shearing of Beams. The resistance of a beam to longitudinal shearing sometimes becomes a mat- ter of importance, especially in timber, where the resistance to shearing along the grain is very small. We will therefore pro- ceed to ascertain how to compute the intensity of the longi- tudinal shear at any point of the beam, under any given load ; as this should not be allowed to exceed a certain safe limit, to be determined experimentally. Assume a A section AC (Fig. 232) at a distance x from V the origin, and let the bending-moment at that section be M. Let the section BD be at a distance x + &>* from the origin, and let the bending-moment at that section be FIG. 232 . M + kM. Let y be the distance of the outside fibre from the neutral axis ; and let ca y^ be the distance of a, the point at which the shearing-force is required, from the neutral axis. Consider the forces acting on the portion ABba, and we shall have i. Intensity of direct stress at A = -j^. 2. Intensity of direct stress at a unit's distance from neu- M tral axis = -j. My 3. Intensity of direct stress at ^, where ce = y, is =-. (M + So, likewise, intensity of direct stress at / is Therefore, if z represent the width of the beam at the point ?, we shall have M (*y Total stress on face Aa = -j- I yzdy, 1 Jyi M + ^M r?o Total stress on face Bb = - j - I yzdy ; l J yi 320 APPLIED MECHANICS. ^.rr bM C y ,*. Difference I yzdy : / J yi and this is the total horizontal force tending to slide the piece AabB'on the face ab. Area of face ab, if #, is its width, is therefore intensity of shear at a is approximately &M C? -r I yzJy or exactly (by passing to the limit) /dM\ dM And, observing that F = -T-, this intensity reduces to (0 We may reduce this expression to another form by observ- ing, that, if y z represent the distance from c to the centre of gravity of area Aa, and A represent its area, we have /Vo J yzdy=y 2 A; therefore intensity of shear (at distance j/j from neutral axis) at point a = ( -M } - (a) This may be expressed as follows : LONGITUDINAL SHEARING OF BEAMS. 321 Divide the shearing-force at the section of the beam under consideration, by the product of the moment of inertia of the section and its width at the point where the intensity of the shearing-force is desired, and multiply the quotient by the statical moment of the portion of the cross-section between the point in question and the outer fibre ; this moment being taken about the neutral axis. The result is the required intensity of shear. The last factor is evidently greatest at the neutral axis ; hence the intensity of the shearing-force is greatest at the neutral axis. LONGITUDINAL SHEARING OF RECTANGULAR BEAMS. For rectangular beams, we have th* /=-, *, = *. Hence formula (2) becomes ^)- (3) For the intensity at the neutral axis, we shall have, therefore, I2F /h bh\ 3 F b*h?> \4 2 / 2 bh? since for the neutral axis we have h bh v a = - and A = . 4 2 EXAMPLES. i . What is the intensity of the tendency to shear at the neutral axis of a rectangular 4-inch by 1 2-inch beam, of 14 feet span, loaded at the middle with 5000 (bs. 3 22 APPLIED MECHANICS. 2. What is that of the same beam at the neutral axis of the cross- section at the support, when the beam has a uniformly distributed load of T 2000 Ibs. 3. What is that of a 9-inch by 14-inch beam, 20 feet span, loaded with 15000 Ibs. at the middle. 206. Strength of Hooks. The following is the method to be pursued in determining the stresses in a hook due to a given load ; or, vice versa, the proper dimensions to use for a given load. Suppose (Fig. 233) a load hung at E; the load being P, and the distances AB n\ OF=y y O being the centre of gravity of this section, conceive two equal and opposite forces, each equal and parallel to P, acting at O. Let A = area of section, and let 7 = its moment of inertia about CD (BCDF represents the section revolved into the plane of the paper) ; then i . The downward force at O causes a uniformly distributed stress over the section, whose intensity is 2. The downward force at E and the upward force at O constitute a couple, whose moment is and this is resisted, just as the bending-moment in a beam, by a uniformly varying stress, producing tension on the left, and compression on the right, of CD. COLUMNS. 323 If we call p^ the greatest intensity of the tension due to this bending-moment, viz., that at B, we have and if / 3 denote the greatest intensity of the compression due to the bending moment, viz., that at F, we have therefore the actual greatest intensity of the tension is and this must be kept within the working strength if the load is to be a safe one ; and so also the actual greatest intensity of the compression, viz., that at F, is, when/, >/,, ,-, _, -A*+*)y, ^ A -A A- 7 ^, which must be kept within the working strength for compression. 207. Strength of Columns. The formulae most commonly employed for the breaking-strength of columns subjected to a load whose resultant acts along the axis have been, until recently, the Gordon formulae with Rankine's modifications, the so-called Euler formulae, and the avowedly empirical formulae of Hodg- kinson. These formulae do not give results which agree with those obtained from tests made upon such full-size columns as are used in practice. The deductions of the first two are not logical, certain assump- tions being made which are not borne out by the facts. When a column is subjected to a load which strains any fibre beyond the elastic limit, the stresses are not proportional to the strains, and hence there can be no rational formula for the break- ing-load. Hence, all formulae for the breaking-load are, of necessity 3 2 4 APPLIED MECHANICS. empirical, and depend for their accuracy upon their agreement with the results of experiments upon the breaking-strength of such full-size columns as are used in practice. Nevertheless, the ordinary so-called deductions of the Gor- don, and of the so-called Euler formula? will be given first. 208. Gordon's Formulae for Columns. (a) Column fixed in fc Direction at Both Ends. Let CAD be the central axis of the column, P the breaking-load, and v the greatest deflection, AB. Conceive at A two equal and opposite forces, each equal to P; then i. The downward force at A causes a uniformly distributed stress over the section, of intensity, ^P_ [ D 2. The downward force at C and the upward force at A Fio.234. constitute a bending couple whose moment is M=Pv. If p 2 = the greatest intensity of the compression due to this bending, where 7= distance from the neutral axis to the most strained fibre of the section at A. Then will the greatest intensity of stress at A be and, since P is the breaking-load, p must be equal to the breaking- strength for compression per square inch=/'. (i) where p= smallest radius of gyration of section at A. Thus far the reasoning appears sound; but in the next step it is assumed that GORDON'S FORMULA FOR COLUMNS. 325 where c is a constant to be determined by experiment. Hence, sub- stituting this, and solving for P, P = fA , 9 , (2) which is the formula for a column fixed in direction at both ends. (b) Column hinged at the Ends. It is assumed that the points of inflection are half-way between the middle and the ends, and -jr~ hence that, by taking the middle half, we have the case of bending of a column hinged at the ends (Fig. 235). Hence, to obtain the formula suitable for this case, substitute, in (2), 2/ for /, and we obtain * FIG. 235. " M <3) (c) Column fixed at One End and hinged at the Other (Fig. 236). ~~r In this case we should, in accordance with these assumptions, take J of the column fixed in direction at both ends; hence, to obtain the formula for this case, substitute, in (2), J/ for /, and we thus obtain j- 1 (4) i6/ 2 ' gcp 2 FIG. 236. >. 236. Rankine gives, for values of / and c, the following, based upon Hodgkinson's experiments: f (Ibs. per sq. in.). c. Wrought-iron 36000 36000 Cast-iron 80000 Dry timber . 72OO 3OOO 326 APPLIED MECHANICS. 2o8a. So-called Euler Formulae for the Strength of Columns. (a) Column fixed in Direction at One End only, which bends, as shown in the Figure, i. Calculate the breaking-load on the assumption that the column will give way by direct compression. This will be PI=/A, CO where /"= crushing-strength per square inch, and A = area of cross- section in square inches. 2. Calculate the load that would break the column if it were to give way by bending, by means of the following formula : f, = where E= modulus of elasticity of the material, 7= smallest moment of inertia of the cross-section, and /= length of column. Then will the actual breaking-strength, according to Euler, be the smaller of these two results. To deduce the latter formula, assume the origin at the upper end, and take x vertical and y horizontal. Let p= radius of curvature at point (x, y), and let 3/=bending-moment at the same point. Then we have, with compression plus and tension minus, PIG. as?. M_ El Py El' (3) But d*y P "dx* = "El*' dy d*y , P f dy, JL . -J-dx I y-~dx dx dx* EIJ J dx dx ^\ = --^y 2 4- c; and, since for y Er ~dx EULER FORMULA FOR STRENGTH OF COLUMNS. 327 dy J y ' Sm - a = And since, when x=o, y=> " c=o, we have When y=a, x=l\ hence, substituting in (5), and solving for P, >-(=) (b) Column hinged at Both Ends (Fig. 235). i. For the crushing-load, P~/A. 2. For the breaking-load by bending, put 1/2 for I in (6) ; hence (7) (c) Column fixed in Direction at One End, and hinged at the other 90, is p A * = 46000-125-. A p On the other hand, those recommended by Prof. J. Sonde- ricker, of which the first was devised by Mr. Theodore Cooper, are as follows: (a) For Phoenix columns with flat ends l/p > So, P_ 36000 A 18000 For lattice columns with pin-ends and l/p>6o, P_ = 340QO A 12000 (7-) For solid web, square, or box columns with flat ends, and l/P>8o, P 33000 Z = ~(//fl-8o)2' 10000 ($) For solid web, square, or box columns with pin-ends, and l/P>6o, P _ 31000 A (l/p -60)* 6000 The number of tests that have been made upon full-size steel columns is very small, hence no formulae will be given here, but the subject will be discussed in Chapter VII. The number of tests that have been made upon full-size timber columns is con- siderable, but this subject will also be discussed in Chapter VII. 211. Columns subjected to Loads which do not Strain any Fibre beyond the Elastic Limit. Under this head will be discussed, first, the mode of determining the greatest fibre THEORY OF COLUMNS. 331 stress in a straight column subjected to an eccentric load, and, secondly, the general theory of columns. (a) Straight column, under eccentric load. Let O' be the centre of gravity of the lower section, and let A'O' = x , where A' is the point of application of the resultant of the eccentric load. Conceive two equal and opposite forces at O', each equal and parallel to P. Then we have: i. Downward force along OO r causes uniform P stress of intensity p\ = -r . 2. The other two form a couple whose moment is Px , and the greatest intensity of the stress due to this couple is p2= - r > where a = O'B'. Hence, FIG. 238. 1 the greatest intensity of the stress is P Px a and this should be kept within the limits of the working-strength. (b) Theory of columns. The theory of columns is that of the Inflectional Elastica, and is explained in several treatises, among which is that of A. E. H. Love on the Theory of Elasticity. It is as follows: Let the curve OP be an elastic line, on which O is a point of inflection. It follows that there is no bending-moment at this point, and hence we may assume that at O a single force R acts. Take Y| the origin at O, and axis of X along the line of action of the force R. Let E\ = modulus of elasticity of the material, 7 = moment of inertia of section about an axis through its centre of gravity, and perpen- 33 2 APPLIED MECHANICS. dicular to the plane of the curve, (> = angle between OX and the tangent at any point P whose coordinates are x and y, a = value of -4= area of section, p = -*rr> R = ' Then we have for any such elastic line, when compressions are plus and tensions minus, i M_ p ET Moreover, since = 3 and M = Ry, we have, for a column p ds d6 R of the same cross-section throughout its length, ~y~ = ~~7>'> ID where the quantity j^j is a constant. By differentiation we obtain R dy R Integrating, and observing that at O, ~^J =O an ^ =a > obtain The integration of this equation requires the use of elliptic integrals, hence only the results will be given here. THEORY OF COLUMNS. 332* They are : (2) and (3) (4) where E denotes the elliptic integral of the second kind, and K the complete elliptic integral of the first kind. Moreover, for the determination of the load R, we obtain from equation (4) K=- and hence 4K2 (6) From these equations, we can, by using a table of elliptic functions, deduce the following results for the coordinates of points on the inflectional elastica, for various values of a : a 5 T X / y I 10 o.oo o . oooo . OOOO 0.25 0.50 0.2476 o . 4962 o .0392 -554 20 o.oo o . oooo . OOOO 0.25 o. 50 0.2376 o . 4849 0.0773 o. 1079 30 o.oo . OOOO . OOOO 0.25 0.50 0.2224 o . 4662 0.1135 o. 1620 Moreover, these results agree with those which we obtain by APPLIED MECHANICS. experiment, and thus we can, by making use of our calculations, compute the load required to produce a given elastica, determined by the slope at the points of inflection, which, in the case of pin- ended columns, are at the ends, and, in the case of columns fixed in direction at the ends, are half-way between -the middle and the ends. All this can be done, and can be verified by experiment, provided that the load is not so great that any fibre is strained beyond the elastic limit of the material, and provided the value of l/p is not so small that the curvilinear form is unstable. For smaller values of l/p the only stable form is a straight line, and the column does not bend. To ascertain the least value of l/p for which a curved form is stable, observe that K cannot be less than 71/2, and since this cor- responds to one bay, and hence to the case of a pin-ended column, we have in that case, by substituting n/2 for K in equation (6) , 7T 2 n and, since I=AfP and -7=<7, we have for the line of demarcation between the straight and curved form in a pin-ended column -; (7) and for that in the case of a column fixed in direction at the ends As an example, if a= 10,000 and 1=30,000,000 we should find that a pin-ended column would not bend unless l/p were greater than 172, and that a column fixed in direction at the ends STRENGTH OF SHAFTING. 333 would not bend unless l/p were greater than 344. Columns with smaller values of l/p would remain straight when the resultant of the load acts along the axis, and no fibre is strained beyond the elastic limit. 212. Strength of Shafting. The usual criterion for the strength of shafting is, that it shall be sufficiently strong to resist the twisting to which it is exposed in the transmission of power. Proceeding- in this way, let EF (Fig. 239) be a shaft, AB the driving, and CD the following, pulley. Then, if two cross-sections be taken between these two pulleys, the por- tion of the shaft between these two cross-sections will, during the trans- mission of power, be in a twisted con- F riG. 239. dition ; and if, when the shaft is at rest, a pair of vertical parallel diameters be drawn in these sec- tions, they will, after it is set in motion, no longer be parallel, but will be inclined to each other at an angle depending upon the power applied. Let GH be a section at a distance x from O, and let KI be another section at a distance x -f- dx from O. Then, if di represent the angle at which the originally parallel diameters of these sections diverge from each other, and if r = the radius of the shaft, we shall have, for the length of an arc passed over by a point on the outside, rdi; and for the length of an arc that would be passed over if the sections were a unit's distance apart, instead of dx apart, rdi _ di dx dx This is called the strain of the outer fibres of the shaft, as it is the distortion per unit of length of the shaft. 334 APPLIED MECHANICS. In all cases where the shaft is homogeneous and symmet- rical, if i is the angle of divergence of two originally parallel diameters whose distance apart is x, we shall have the strain, di i v = r = r-. dx x This also is the tangent of the angle of the helix. A fibre whose distance from the axis of the shaft is unity, will have, for its strain, dt_ = / dx x A fibre whose distance from the axis of the shaft is p, will have, for its strain, di i v = p - = p-. dx x Fixing, now, our attention upon one cross-section, GH, we have that the strain of a fibre at a distance p from the axis (p varying, and being the radius of any point whatever) is where - is a constant for all points of this cross-section. X Hence, assuming Hooke's law, " Ut tensio sic vis" we shall have, if C represent the shearing modulus of elasticity, that the stress of a fibre whose distance from the axis is p, is which quantity is proportional to p, or varies uniformly from the centre of the shaft. The intensity at a unit's distance from the axis is 0- STRENGTH OF SHAFTING. 335 and if we represent this by a, we shall have for that at a dis- tance p from the axis, Hence we shall have (Fig, 240), that, on a small area, V J dA = dp( P dB) _ pdpdO, ^^ the stress will be pdA = apdA = ap 2 dpd9. The moment of this stress about the axis of the shaft is ppdA = ap z dA = ap^dpdO, and the entire moment of the stress at a cross-section is afp*dA = affpidpdO = al, where / = fp 2 dA is the moment of inertia of the section about the axis of the shaft. This moment of the stress is evidently caused by, and hence must be balanced by, the twisting-moment due to the pull of the belt. Hence, if M represent the greatest allowable twisting- moment, and a the greatest allowable intensity of the stress at a unit's distance from the axis, we shall have M = al = - /. P If / is the safe working shearing-strength of the material per square inch, we shall have / as the greatest safe stress per square inch at the outside fibre, and hence M=- I r will be the greatest allowable twisting-moment. 33^ APPLIED MECHANICS. For a circle, radius r t 2 " ~ * ~~2~ ~ J ~i6~* For a hollow circle, outside radius r v inside radius r M Moreover, if the dimensions of a shaft are given, and the actual twisting-moment to which it is subjected, the stress at a fibre at a distance p from the axis will be found by means of the formula The more usual data are the horse-power transmitted and the speed, rather than the twisting-moment. If we let P = force applied in pounds and R = its leverage in inches, as, for instance, when P = difference of tensions of belt, and R = radius of pulley, we have and if HP number of horses-power transmitted, and N = number of turns per minute, then TT T) _ \ *' *_ /_ . 12 X 33000 ' 12 X -l^OOoIfP ^r^r. ^ Jyi 271 N EXAMPLE. Given working-strength for shearing of wrought-iron as 10000 Ibs. per square inch ; find proper diameter of shaft to transmit 2o-horse power, making 100 turns per minute. TRANSVERSE DEFLECTION OF SHAFT. 337 Mp Angle of Torsion. From the formula, page 336, p~ =- % combined with we have = ap = Cp-, oc . _ MX " ~' which gives the circular measure of the angle of divergence of two originally parallel diameters whose distance apart is x ; the twisting-moment being M, and the modulus of shearing elas- ticity of the material, C. EXAMPLES. 1. Find the angle of twist of the shaft given in example i, 212, when the length is 10 feet, and C = 8500000. 2. What must be the diameter of a shaft to carry 80 horses-power, with a speed of 300 revolutions per minute, and factor of safety 6, break- ing shearing-strength of the iron per square inch being 50000 Ibs. 213. Transverse Deflection of Shafts. In determining the proper diameter of shaft to be used in any given case, we ought not merely to consider the -resistance to twisting, but also the deflection under the transverse load of the belt-pulls, weights of pulleys, etc. This deflection should not be allowed to exceed y^- of an inch per foot of length. Hence the de- flection should be determined in each case. The formulae for computing this deflection will not be given here, as the methods to be pursued are just the same as in the case of a beam, and can be obtained from the discussions on that subject. APPLIED MECHANICS. 214. Combined Twisting and Bending. The most com- mon case of a shaft is for it to be subjected to combined twisting and bending. The discussion of this case involves the theory of elasticity, and will not be treated here ; but the formulae com- monly given will be stated, without attempt to prove them until a later period. These formulae are as follows : Let M l = greatest bending-moment, M 2 = greatest twisting-moment, r = external radius of shaft, / = moment of inertia of section about a diameter, TTf 4 for a solid shaft / = , 4 f = working-strength of the material = greatest al- lowable stress at outside fibre ; then i. According to Grashof, /= LjfJ/i + fVJ/x 2 + M*\. (i) 2. According to Rankine, / = j M, + \!M* + M; j . (2) 215. Springs. The object of this discussion is to enable us to answer the following three questions : (a) Given a spring, to determine the load that.it can bear without producing in the metal a maximum fibre stress greater than a given amount. (&) Given a spring, to determine its displacement (elongation, compression, or deflection) under any given load, (c) Given a load P and a displacement & t ; a spring is to be made of a given material such that the load P shall produce the displace- ment 6 I , and that the metal shall not, in that case, be subjected to more than a given maximum fibre stress. Determine the proper dimensions of the spring. SPRINGS. 339 There are practically only two cases to be considered as far as the manner of resisting the load is concerned. In the first, the spring is subjected to transverse stress, and is to be calcu- lated by the ordinary rules for beams. In the second, the spring is subjected to torsion, and the ordinary rules for re- sistance to torsion apply. It is true that in most cases where the spring is subjected to torsion there is also a small amount of transverse stress in addition to the torsion ; but in a well- made spring this transverse stress is of very small amount, and we may neglect it without much error. We will begin with those cases where the spring is subjected to torsion, and for all cases we shall adopt the following nota tion : P = load on spring producing maximum fibre stress/; f = greatest allowable maximum fibre stress for shearing ; C = shearing modulus of elasticity ; x = length of wire forming the spring ; M l = greatest twisting moment under load P\ L = any load less than the limit of elasticity ; M = twisting moment under this load ; p = maximum fibre stress under load L ; p distance from axis of wire to most strained fibre ; / = moment of inertia of section about axis of wire ; z'j = angle of twist of wire under load P- i = angle of twist of wire under load L ; V = volume of spring ; #j = displacement of point where load is applied when load isP; d displacement of point where load is applied when load isZ. Then from pages 335 and 337 we obtain the following four formulae : *=/, (i) 34O APPLIED MECHANICS. MX '=C7' (3) These four formulae will enable us to solve all the cases of springs subjected to torsion only. Moreover, in the cases which we shall discuss under this head, the wire will have either a circular or a rectangular section : in the former case we will denote its diameter by d, and we shall then have net* d /= -- and p = ; 32 2 while in the latter case we will denote the two dimensions of the rectangle by b and h, respectively, and we shall then have We will now proceed to determine the values of P, #, S l , and V in each of the following four cases, all of which are cases of torsion : CASE i. Simple round torsion wire. Let AB, the leverage of the load about the axis, be R ; then we shall have M = LR, M, = PR ; and we readily obtain from the formulae (i), (2), (3), and (4) ^\ ,f C- <" (7) SPRINGS. 341 and from these we readily obtain (8) CASE 2.' Simple rectangular torsion wire. In this case we readily obtain (9) , D . , . = =ri ' (IC >=*'> = "' CASES 3 and 4. Helical springs made of round and of rec- tangular wire respectively. A helical spring may be used either in tension or in compression. In either case it is important that the ends should be so guided that the pair of equal and opposite forces acting at the ends of the spring should act ex- actly along the axis of the spring. This is of especial importance when the spring is used for making accurate measurements of forces, as in the steam-en- gine indicator, in spring balances, etc. Moreover, it is generally safer, as far as accuracy is con- cerned, to use a helical spring in tension rather than in com- pression, as it is easier to make sure that the forces act along 34 2 APPLIED MECHANICS. the axis in the case of tension than in the case of compres- sion. Whichever way the spring is used, however, provided only the two opposing forces act along the axis of the spring, the resistance to which the spring is subjected is mainly torsion, inasmuch as the amount of bending is very slight. This bending, however, we will neglect, and will compute the spring as a case of pure torsion, the same notation being used as before, except that we will now denote by R the radius of the spring, and we shall have M = LR, M, = and now formulae (5), (6), (7), and (8) become applicable to a spring made of round wire, and formulae (9) and (10), (n) and (12), to one made of rectangular wire. We must bear in mind, however, that x denotes the length of the wire composing the spring, and not the length of the spring, d and d l now denote the elongations or compressions of the spring. GENERAL REMARKS. By comparing equations (8) and (12), it will be seen that if a spring is required for a given service, its volume and hence its weight must be 50 per cent greater if made of rectangular than if made of round wire. Again, it is evident that when the kind of spring required is given. SPRINGS. 343 and the values of C and f for the material of which it is to be made are known, the volume and hence the weight of the spring depends only on the product Pd lt and that as soon as P and d\ are given, the weight of the spring is fixed inde- pendently of its special dimensions. If, however, we fix any one dimension arbitrarily, the others must be so fixed as to satisfy the equations already given. Next, as to the values to be used for /and C, these will depend upon the nature of the special material of which the spring is made, and these can only be determined by experiment. Confining ourselves now to the case of steel springs, it is plain that /and 7 should be values corresponding to tempered steel. As an example, suppose we require the weight of a helical spring, which is to bear a safe load of 10000 Ibs. with a deflec- tion of one inch, assuming C= 12600000 and/= 80000 Ibs. per sq. in., and as the weight of the steel 0.28 Ib. per cubic inch. From formula (8) we obtain _, 2 X 12600000 X 10000 X i =39.4cu.m. Hence the weight of the spring must be (39.4) (0.28) = II Ibs. We may use either a single spring weighing 1 1 Ibs., or else two or more springs either side by side or in a nest, whose com- bined weight is 1 1 Ibs. Of course in the latter case they must all deflect the same amount under the portion of the load which each one is expected to bear, and this fact must be taken into account in proportioning the separate springs that compose the nest. FLAT SPRINGS. Let P, L, V, d, and d^ have the same meanings as before, and let 344 APPLIED MECHANICS, f= greatest allowable fibre stress for tension or compres- sion : R = modulus of elasticity for tension or compression ; /= length of spring; M l = maximum bending-moment under load P ; M= maximum bending-moment under load L. Moreover, the sections to be considered are all rectangular, and we will let b = breadth and h = depth at the section where the greatest bending-moment acts, the depth being measured parallel to the load. Then if / denote the moment of inertia of the section of greatest bending-moment about its neutral axis, we shall have f= M 12 We will now consider six cases of flat springs, and will de- termine P, tf, tf z , and V for each case, and for this purpose we only need to apply the ordinary rules for the strength and de- flection of beams. CASE i. Simple rectangular spring, fixed at one end and loaded at the other. I 3 L i*> I* f (24) E E (26) SPRINGS. 345 CASE 2. Spring- of uniform depth and uniform strength, tri- in plan , fixed at one end and loaded at the other. (27) (28) (29) (30) CASE 3. Spring of uniform breadth and uniform strength, parabolic in elevation, fixed at one end and loaded at the other. (31) (33) (34) CASE 4. Compound wagon spring, made up of n simple rec- tangular springs laid one above the other, fixed at one end and loaded at the other. 346 APPLIED MECHANICS. Let the breadth be b, and the depth of each separate layer be h. Then '- n bh* 6 / /' (35) i N^ 6 = 4 /* L nbh* E* (30 i =3=^ /'/ (37) i \ i y^ ^fi 1 * (38) CASE 5. Compound spring composed of n triangular springs laid one above the other, fixed at one end and loaded at the other. *-\r=r ^ 6 = -nW L E> < 4 > *=ji ; (4I) CASE 6. This case differs from the last in that in order to economize material we superpose springs of different lengths, SPRINGS. 347 and make them of such a shape that by the action of a single force at the free end they are bent in arcs of circles of nearly or exactly the same radius. The force P bends the lowest triangular piece AA in the arc of a circle. length of this piece is -. In order that the re- maining parallelopipedical portion may bend into an arc of the same circle it is necessary that it should have acting on it a uniform bending-moment throughout, and this is attained if it exerts a pressure at A l upon the succeeding spring equal to the force P, and following this out we should find that the entire spring would bend in an arc of a circle. The values of P, 6, d z , and Fare correctly expressed for this case by (39), (40), (41), and (42). For any flat springs which are supported at the ends and loaded at the middle, or where two springs are fastened to- gether, it is easy to compute, by means of the formulae already developed, by making the necessary alterations, the quantities P. 3 d z , and V, and this will be left to the student. COILED SPRINGS SUBJECTED TO TRANSVERSE STRESS. Three cases of coiled springs will now be given as shown in the figures, and the values of P, 3, d lt and Fwill be deter- mined for each. In each of these cases let R be the leverage, of the load, and let GO = angle turned through under the load. Then we may observe that all the three cases are cases of beams sub- jected to a uniform bending-moment throughout their length, this bending-moment being LR for load L and PR for load P. 348 APPLIED MECHANICS. CASES I and 2. Coiled spring, rectangular in section. f b>i i \ ^ = i/-^> (43) UP L , , (44) (45) f Rl CASE 3. Coiled spring, cir- cular in section. f = -^f^> (47) 64 l^_L . . ~* j* z^> \4w (49) E TIME OF OSCILLATION OF A SPRING. (46) (So) Since in any spring the load producing any displacement is proportional to the displacement, it follows that when a spring oscillates, its motion is harmonious. SPRINGS. 349 Suppose the load on the spring to be P t and hence its nor- mal displacement to be , so that the actual displacement varies from # x -f- tf to <$, 1'X X 3 3 X 3 X c g2 c 1 JfS S 1C oS^o-S "S-e-c cx CX CX ex a. a. .0^ -5 . w . _^tc ^ ^ " S "3 T3 - - 1 -o a 3 '3 j5' 3 'rt js '3 *o 2f 'O .2 c/5 -i- 8 in 6 m en 6 6 m oo O 'S a s Q 2 B D o 6 in N O co en oo M 6 CO en " -S r rt tuo a . 5 w ^ O O Q O O co m O "^ r^ oo en m co en co en en o O en ^ en w en C-O cr- C) co CO o in CO en Cl CO o en en R 2 m r^ 8 lip p w 1 1 1 1 N *, rf in w M en O M o in in ^- en en 352OOO in in 1 i CO 00 m en 8 O CM *! O in r^* m rt o CO en ,, vn co m en co H CO o 8 Cl 8 -f 4- -t r^ Tf to ^ t*** QO ! f|f s 5 tJ < cr i^ .be