>. 
 
 IN MEMORIAM 
 FLOR1AN CAJORI 
 
&*-, 
 
THE ELEMENTS 
 
 GRAPHICAL STATICS 
 
 AND THEIR APPLICATION TO 
 
 FRAMED STRUCTURES, 
 
 WITH NUMEROUS PRACTICAL EXAMPLES OF 
 
 A 
 
 CRANES BRIDGE, ROOF AND SUSPENSION TRUSSES BRACED 
 
 AND STONE ARCHES PIVOT AND DRAW SPANS- 
 
 CONTINUOUS GIRDERS, &c., 
 
 TOGETHER WITH THE BEST 
 
 METHODS OF CALCULATION, 
 
 AND CONTAINING ALSO 
 
 NEW AND PRACTICAL FORMULAE 
 
 FOR THE 
 
 Pivot or Draw Span Braced Arch Continuous Girder, Etc, 
 
 A. JAY r>U BQI^ r ,.;, p Ph.D. 
 
 PROFESSOR OF CIVIL AKP. iI 1 ECHA.'SI T rA <i J ENGrV-SFHiWC., LETI5GH' UNIVERSITY, PENNA. 
 
 "Peu a peu les connaissances aleg6briques d6viendront moins indispensables, et la science, re- 
 duite a ce qu'elle doit etre, a ce qu'elle devrait etre deja, sera ainsi mise a la port6e de cette classe 
 d'hommes, qui n'a que des moments fort rares a y consacrer." [Poncelet-Traite des proprietea 
 projectives des figures, Paris, 1828.] 
 
 With an Atlas of 32 Plates. 
 
 NEW YOKK : 
 
 JOHN WILEY AND SON, 
 15 ASTOR PLACE. 
 
 1875. 
 
COPYRIGHT, 1875, BY 
 JOHN WILEY & SON, 
 
 JOHN F. TROW & SON, 
 
 PRINTERS AND STEREOTYPERS, 
 
 205-213 Rast izth St., 
 
 MEW YORK. 
 

 ELEMENTS OF GRAPHICAL STATICS. 
 
 NOTE TO STUDENTS AND TEACHERS. 
 
 TEACHERS examining the work with a view to introduction will at 
 once perceive that it was not intended to be read through in sequence 
 by any class, but that it is required to be used intelligently with 
 reference to the degree of preparation of their students, with a knowl- 
 edge of the relative importance and relation of the various subjects 
 treated, and with a view to the end desired. Though the logical 
 order of presentation requires a certain order in the development of 
 the subject as a whole, it by no means follows that this order should 
 be preserved by the student, or that he should even be acquainted 
 with the whole. In fact some of the subjects treated of are best 
 taken up by the student at a later period, when better prepared for 
 their comprehension, others are best omitted at least in the first 
 reading, and others again may even be omitted entirely as of minor 
 importance, and the student left to pursue them for himself, as taste 
 or the exigencies of practice may demand. In this latter respect, as 
 well as in the completeness with which the several topics are dis- 
 cussed, the work is intended to serve as a book of reference. As a 
 Text Book, it is designed to teach those students possessing the 
 knowledge of mathematics and mechanics usual to the senior classes 
 in our technical schools and colleges, how to find the conditions of 
 stability in every kind of structure of common occurrence and this 
 not alone by graphical construction but also by calculation as well. 
 Structures of less common occurrence may or may not be then taken 
 up according to the ability of the class and the time at disposal. 
 With the above end in view, the teacher will find it not merely 
 desirable but even essential to depart from the plan of the work as 
 laid down, and to supplement it largely by various examples illustra- 
 tive of the general principles and bringing out as clearly and 
 repeatedly as may be necessary the various points noticed in the 
 text, 
 
 In the work four separate and distinct methods of solution are 
 given for such structures as bridge and roof trusses, and the student 
 
XIV ELEMENTS OF GRAPHICAL STATICS. 
 
 should become familiar with all. Thus there are two methods by dia- 
 gram, viz., by resolution of forces (Maxwell) and by the equilibrium 
 polygon ( Culmann) ; and two corresponding methods by calculation,, 
 viz., by composition and resolution of forces, and by moments (Hitter). 
 To give these in such manner and with such emphasis that the student 
 shall be conversant with all, and able to use them with discrimina- 
 tion where in any case they best apply, we recommend the following 
 order of perusal : 
 
 FIRST. Chap. I, which gives the first method by diagram (Max- 
 welPs) y and such of the Appendix as relates to this chapter, is read. 
 The class may then go into the drawing room, and under the super- 
 vision of the teacher actually solve a variety of roof trusses from 
 simple to most complex, both for dead load and ivind force in each 
 case. In each and every case also the results should be checked by 
 calculation by the method of moments (.Ritter's), at first thoroughly 
 and in detail, and afterwards only a few test pieces to check the 
 accuracy of the diagram. From Roof Trusses we then pass on ta 
 Bridges, and here also a series of selected examples of every class 
 used in practice is solved, and the method of tabulation of apex weights- 
 referred to in Art. 12 and Appendix to Chap I, brought out repeat- 
 edly until the student has thoroughly mastered it, and appreciates 
 fully the fact that for each form of truss the strains due to only two- 
 weights are really necessary to be found, and that the others may 
 then readily be found directly from these. Here also each example 
 should be checked by calculation by the method of moments. In 
 the case of curved flanges the various lever arms may first be 
 measured directly to scale from the frame, and then trigonometrically 
 computed. At this point the student is then already in possession 01* 
 two independent methods of solution for any kind of framed roof or 
 bridge which occurs in general practice. 
 
 There are in fact only two framed structures, the continuous 
 girder and the braced arch remaining, which he is not able to solve. 
 Should it be deemed undesirable to consider these, he can at once 
 pass to Chap. II and the stone arch. If, however, a knowledge of 
 these is desired, he is now ready to extend his principles and methods 
 to them also. Thus he has already recognized that provided only all 
 the outer forces are known he can both diagram and calculate any 
 framed structure. In such structures as he has hitherto had, these 
 outer forces are either given or are easily found. In the cases now 
 considered they are not all given, and must, therefore, first be found. 
 Once known, however, his way is clear. Kecognizing clearly, then, 
 what is aimed at, the supplements to Chaps. VII and XIII are first 
 
ELEMENTS OF GRAPHICAL STATICS. XV 
 
 taken and he is now able to find for the continuous girder, the outer 
 forces required. Chap. XIII will then give exercises in finding these 
 forces, and handling the formula he has just deduced. Finally, Chap. 
 XII resumes again in the light of his present knowledge, the same 
 old two methods with which he is already so familiar, of diagram 
 and calculation, and a few examples actually worked out by both 
 methods complete his mastery of the continuous girder and draw 
 span. He can now pass on to the braced arch, and in Chap. XIV 
 will find all that he needs. Here he must take at first the formulae 
 and constructions for finding the outer forces, on trust. Afterwards, 
 if deemed desirable, he can follow out the development of these 
 formulae as given in supplement to Chap. XI Y. 
 
 In the case of the parabolic arch, at least, the constructions are so 
 simple that it is well adapted to .class instruction. The draw span is 
 of such importance as to render some atttention to it, at any rate, 
 desirable in any full course. Thus the student is now able to solve 
 any case whatever of framed structure in two ways, by diagram and 
 calculation. The same simple principles have been applied throughout, 
 and formulae have been called in only in a subsidiary way to deter- 
 mine certain forces which are necessary to be first known before 
 these principles can be applied. 
 
 Thus, Chap. I, appendix to Chap. I, supplements to Chaps. 
 VII and XIII, then Chaps. XIII, XII, with appendix, and 
 XI Y with appendix, form by themselves and in the order a 
 complete and systematic course. 
 
 If, however, it is deemed undesirable to consider the continuous 
 girder and braced arch, the course indicated in the preceding para- 
 graph may be omitted, and then Chap. I, with appendix, constitutes 
 a course as thorough as can be desired, the student taking first those 
 examples given in the book, and then such others as the teacher 
 may select, and always solving in two ways, by diagram and calcula- 
 tion. He may then take Chaps. II- Y, and now possesses a second 
 method of diagram (Culmann's) by which he may check any or all of 
 the previous cases. In general, a few examples of application to 
 bridge girders, drawing the parabolas for total load (moments) and 
 moving load (shear) will be sufficient. He is now ready to pass on 
 to the Stone Arch, where again suitable problems should be proposed 
 by the teacher and solved under his supervision. The remainder of 
 the work, including moment of inertia and continuous girder treated 
 by the second method of diagram, will in general be found unneces- 
 sary and rather advanced, except in a very full course. 
 
Xvi ELEMENTS OF GRAPHICAL STATICS. 
 
 The course recommended is then as follows, in the order given, 
 chaps, in brackets being omitted or not at option of teacher : 
 
 Chap. I and appendix, supplements to Chaps. YII (and XIII, 
 Chaps. XIII, XII and appendix, XI V and appendix), Chaps. 
 II-Y, XY (XVI, VI, XI). It will be seen that the order in- 
 tended for class instruction is quite different from that of the 
 work itself. 
 
 We would ask teachers examining the book with a view to adop- 
 tion, to look it over in the order above given. 
 
PREFACE. 
 
 IT is now ten years since the appearance of the Graphical 
 Statics of Culmann,* during which time the method has been 
 greatly extended in its applications, and has met with such 
 acceptance that there is now scarcely a Polytechnikum in Ger- 
 many where it is not a prominent feature in the regular course 
 of instruction. 
 
 This rapid spread of a new discipline is the more remarkable 
 when we consider the obstacles which it encountered. Cul- 
 mann, with a boldness which we might almost term rash, based 
 his development upon the modern geometry of Yon Staudt, and 
 assumed in his re iers a familiarity with this very terse^)resen- 
 tation of a subject then, as indeed now, but little known, and 
 which, therefore, but few possessed. To practical engineers, 
 therefore, to whom his methods specially recommended them- 
 selves, his presentation of those methods was almost unintelli- 
 gible. 
 
 At a time when the students of the Zurich Polytechnic were 
 already overburdened, the new discipline was introduced ; while, 
 owing to want of familiarity with the fundamental principles 
 premised, they were unable to understand his lectures or read 
 his work. Yet such was the intrinsic value of the new method 
 that, notwithstanding these obstacles, even in spite of them, it 
 made rapid headway ; found friends everywhere ; crept into 
 other departments of the Polytechnic ; and finally the aim of 
 Culmann was completely attained when the modern geometry 
 was itself introduced, and a special lecturer in that branch ap- 
 pointed. Thus, as a direct result of the Graphical Statics of 
 Culmann, appeared the first and, till now, only complete text- 
 book upon the modern geometry, viz., Reye's " Geometric der 
 Lage" Hannover, 1868. Since then, hand in hand and with 
 remarkable rapidity, these two studies have made their way, 
 
 * Die Graphische Statik. Culmann. Zurich, 1866. Second Edition, 1st 
 vol., 1875. 
 
v PEEFACE. 
 
 until, as already remarked, they now form a notable feature in 
 the course of every technical institution in the land. 
 
 The acceptance which the method has found in France, and 
 the attention which it has there excited, is sufficiently indicated 
 by the work of Levy (La Statique Graphique et ses Applica- 
 tions, Paris, 1874), which contains a very clear and elegant 
 presentation of the principles, though the applications are of 
 the simplest character, while, as was perhaps not unnatural in 
 the author, the German origin of the system is very imper- 
 fectly indicated, and the special methods of Culmann but little 
 more than hinted at. 
 
 In Italy also the method has found an ardent expounder in 
 the distinguished mathematician Cremona (Le figure recip- 
 roche nelle statica grafica, Milan, 1872), and to his efforts and 
 labors its introduction and acceptance is due. 
 
 In England, Prof. Clerk Maxwell, in the Trans, of the Royal 
 Society of Edinburgh, 1869-70, has contributed a paper upon 
 i% Reciprocal Figures, Frames and Diagrams of Forces/' and, 
 among others, Jenkin, JRanken, Bow, and Unwin have contrib- 
 uted to the popularity and spread of " Maxwell's Method." 
 Maxwell and his followers give, however, only the very simplest 
 applications, based upon the resolution and composition of 
 forces, such as will be found in our first chapter. The entire 
 system developed by Culmann, the properties of the " equilib- 
 rium polygon," upon which the fruitfulness and value of the 
 graphical statics wholly depend, are unnoticed both by our 
 English and French authors. 
 
 The author feels, therefore, that no apologies are needed for 
 the present work. Whatever its shortcomings and defects, he 
 claims at least the honor of making the first attempt to intro- 
 duce among American Colleges and American Engineers a 
 knowledge of a subject of approved interest and practical value 
 to both, whether regarded as a geometrical discipline or as a 
 most efficient aid in investigations of stability. Nor is he with- 
 out hope that the next ten years may find the method as uni- 
 versally accepted at home as now abroad. 
 
 The same difficulties certainly have not here to be encoun- 
 tered. The subject as here presented requires only a knowl- 
 edge of the elements of geometry as universally taught, and 
 can thus be readily introduced into our schools as well as read 
 by those practical engineers for whose benefit the method 
 
 
PREFACE. V 
 
 seems so especially designed. A subject of such importance, 
 which has already endured successfully so severe a test, and 
 made headway against such obstacles, we cannot certainly af- 
 ford any longer to ignore, and it is hoped that the present 
 work may serve to excite a more general interest in the method. 
 
 For the practical engineer, the importance of graphical 
 methods needs, indeed, to-day no demonstration. Such methods 
 are everywhere in use. But a simple and general system 
 which shall include all special solutions the development of 
 the few principles upon which all such solutions are based, and 
 from which they all flow is at least in this country unknown. 
 Even in English literature there is to be found little more than 
 the very elementary deductions of our first chapter, so that it 
 may justly be said that the entire method owes its- existence 
 and development to the labors of German scholars and the en- 
 lightened appreciation of German engineers. How thorough 
 have been these labors, how widespread this appreciation, and 
 how various are the applications of the method itself, the reader 
 may gather from the Introduction to this work, and from the 
 appended list of literature upon the subject. A glance at this 
 list will also show that the selection of what was of most value, 
 and the omission of those applications of minor importance, 
 necessary to bring the present work within reasonable limits, 
 and at the same time preserve the logical unity and complete- 
 ness of the whole, was not the least difficult portion of our 
 task. It would, indeed, have been easy to have given the work 
 twice its present dimensions, though without a corresponding 
 increase in value sufficient to justify the additional cost. As 
 it is, no application of real and practical value to the engineer 
 strictly deducible from the graphical statics has been over- 
 looked, and discrimination has been chiefly exercised in those 
 departments where graphical and analytical processes are still 
 of necessity combined. Here we have selected only those cases 
 where such union shows itself most advantageous, and the 
 graphical constructions most simplify, illustrate, or interpret 
 the purely analytical process, and where such cases, moreover, 
 presented a useful, practical, and not merely theoretical value. 
 
 As to the plan of the work, a word of explanation is neces- 
 sary. We have endeavored to keep always in view the re- 
 quirements, of both students and practitioners, of technical 
 schools and practical engineers, and thus to combine a text- 
 
VI PREFACE. 
 
 book for school instruction, and a book of reference and manual 
 for practice as well. The attempt is a difficult, if not a dan- 
 gerous one, and one which, in other departments, has met with 
 more failure than success. If we venture to indulge a hope that 
 in this case at least partial success has been attained, and that 
 the attempt to occupy the two stools at once has not been dis- 
 astrous, our belief is due to the nature of the subject itself, 
 and not to any overweening estimate of our own abilities to 
 succeed where so many have failed. The subject seems, indeed, 
 especially suited to such a method of treatment. In fact, no 
 other would appear at this period to properly meet the necessi- 
 ties of the case. Its geometrical principles are simple, its ap- 
 plications eminently practical. To present the principles alone 
 would be to deprive the- study of its chief interest and attrac- 
 tion. To rest content with a few practical applications would 
 be to sacrifice, in a great measure, system and clearness of pre- 
 sentation. In the accomplishment of our double task we are 
 fortunate to have had at our disposal such works as those of 
 Bauschinger in the one, and Culmann in the other direction. 
 Our obligations to both authors are great, and are fully indi- 
 cated in the text. The same acknowledgment is due, in greater 
 or less degree, to Mohr and Winkler, Hitter and Reuleaux. In 
 every case where such assistance has been received, due ac- 
 knowledgment has been made. 
 
 For the historical and critical Introduction, we are indebted, 
 with few alterations, to the pen of Weyrauch* It will, we are 
 sure, prove of value to the student, and serve to awaken an in- 
 terest in those highly important developments which geometry 
 has within the last decade undergone. 
 
 Thus collecting in a connected form the scattered results and 
 researches of various authors, it has been a pleasurable duty to 
 recognize the labors of those men who have chiefly contributed 
 to' this new branch of geometrical statics, and to whom our own 
 obligations are so great. While thus crediting fully that which 
 others have done, we have felt the more justified in calling at- 
 tention to any deviations of our own. We have especially 
 sought to extend the application of the method by resolution of 
 'forces (known best, perhaps, as MaxweWs Method ) a method 
 
 * Veber die gra/phische Statik zur Orientirung. Yon Dr. phil. Jacob J. 
 Weyrauch, Privat docent an der polytechnischen schule zu Stuttgart. Leipzig, 
 1874. 
 
PREFACE. Vll 
 
 which bids fair to obtain widespread recognition, in direc- 
 tions in which it has hitherto been supposed of little service. 
 This often, indeed, by the aid of analytical results, we have 
 been enabled to do, and not, as we conceive, without a degree 
 of success. The formulae used are always simple and of ready 
 application, and this union of analytical results and graphical 
 processes the practical engineer will, we think, find of value. 
 Thus, in the braced arch (Chap. XIY.) and continuous girder 
 (Chap. XII.) new constructions will be found, and both these 
 important and difficult cases may thus be solved with an ease, 
 completeness and accuracy far : superior to that of the pure 
 graphical method itself. Those acquainted with the analytical 
 investigations of the " braced arch" as contained in Oapt. Eads> 
 Report to the III. and St. Louis Bridge Co., May, 1868 (App.), 
 will not, we feel sure, be slow to recognize the advantages of 
 the present method. The subject in its present state is thus 
 fairly brought within the reach of the practical Engineer and 
 Constructor. 
 
 To simple girders, contrary to usually received opinions, by 
 the means of apex loads, the above method applies directly, and 
 without the aid of analytical results a fact which has been too 
 generally passed over without sufficient notice by. writers upon 
 the subject. 
 
 We have devoted considerable space to the subject of the 
 continuous girder, but not, we feel sure, more than its impor- 
 tance demands. The subject deserves more attention at the 
 hands of the practical engineer and constructor than it has 
 hitherto received. That the present indifference upon the sub- 
 ject is due chiefly to lack of information can hardly be doubted, 
 when the opinion is current, and is even endorsed by those who 
 are considered as authorities, that the complete solution of the 
 /problem is " probably impossible by reason of its complexity," 
 and " too complex for mathematical investigation." * Opin- 
 ions like these are best met by the complete solutions of par- 
 ticular examples, and in Chapter XII. will be found the com- 
 plete calculation and tabulation of the strains in every piece 
 due to every apex load, for the central span of seven continu- 
 ous successive spans, and, as far as any inherent difficulties are 
 concerned, we might as well have taken 50 or 100 spans. 
 
 * OrapJiiccd Method for the Analysis of Bridge Trusses. Greene. 
 
Vlll PREFACE, 
 
 When engineers shall have become convinced of the fact that 
 there is in the continuous girder a saving of material amount- 
 ing usually to from 25 to 30 per cent, per truss, and in the ex- 
 treme case even reaching as high as 50 per cent., as compared 
 with the si^rle girder ; and that the only objection which can 
 be urged viz., the influence of small variations in level of the 
 supports has, when properly considered, no force whatever, 
 we shall probably hear less often of designs contemplating many 
 successive and independent spans of considerable length such 
 as, for instance, for a bridge over the Hudson at Poughkeep- 
 sie, consisting of five separate spans of 525 ft. each. Such a 
 design would find little favor in France or Germany, where 
 continuous girders are more favorably considered, possibly be- 
 cause the ability to calculate them is less rare, and reflects in 
 this respect little credit upon the American profession. About 
 the facts in the case there can now be no dispute ; the subject 
 has been too thoroughly investigated to admit of it, and we 
 refer the reader to the Appendix for the results. The mathe- 
 matician and theoretical engineer have done their part ; it 
 remains for the practical engineer and constructor to do theirs. 
 
 The present work contains the only complete graphical and 
 analytical presentation of this subject in English professional 
 literature, and should it succeed in causing a change of view 
 in the above respect alone, will not have been in vain. In this 
 connection the list of literature upon the continuous girder 
 appended to Chap. XIII. may also be of service. 
 
 We notice with pleasure in this direction the admirable little 
 treatise of Clemens Herschel, C.E., upon draw spans.* This 
 subject is at least of admitted practical value, and we have 
 treated it with a fullness which, in our opinion, leaves little to 
 be desired. We have borrowed from the above work the con- 
 ception of the " Tipper" or draw with secondary span, which 
 is both new and, as it would seem, most adequately represents 
 the true state of the case, and alluded to the idea, also original 
 with Mr. Herschel, of weighing off the reactions at the supports 
 of a continuous girder, instead of measuring the differences of 
 level. In this case, as in that of the continuous girder gene- 
 rally, we have clearly brought out the method of calculation ~by 
 
 * Continuous, Revolving Drawbridges. Little, Brown and Company, Bos- 
 ton, 1875. 
 
PREFACE. IX 
 
 apex weights, and here, indeed, lies the whole secret of thorough 
 practical solution. In fact, from this point of view* the com- 
 plete solution of a continuous girder for any number of spans, 
 equal or unequal, offers no more essential difficulty than the 
 calculation of so many separate simple girders. That this is 
 not exaggeration, but accurate statement of fact, a perusal of 
 Chaps. XII. and XIII. will suffice to prove. 
 
 We cannot leave this part of the subject without acknowledg- 
 ing our indebtedness to Mansfield Merriman, C.E., Assistant in 
 Engineering in the Sheffield Scientific School of Yale College, 
 for the formulae of the latter chapter. Mr. Merriman has 
 done for the practical solution of the continuous girder what 
 Weyrauch has for its theoretical discussion. We refer the 
 student to the Supplement to Chap. XIII. for a specimen of 
 his method of discussion. 
 
 By the proper use of " indeterminate multipliers," the whole 
 analytical discussion is most remarkably simplified. The only 
 one of the many writers upon the subject known to us who 
 seems to have hit upon this treatment is WinMer (Die Lehre 
 von der Elasticitdt und festigJceit). In Art. 144 of the above 
 work he gives formulae similar to Mr. Merriman's for the 
 moments at the supports of a continuous girder for all spans 
 equal only. He seems, however, to have failed to realize the 
 true significance of the method, or the important part played 
 by the Clapeyronian numbers. Independently of Winkler, 
 Mr. Merriman has reproduced these formulae in their true 
 light, and applied the method to any lengths and number of 
 spans, with any differences of level and any method of loading. 
 His formulae are simple, entirely free, even in general form, 
 from integrals, and are given in just the shape required in 
 practice. This compactness renders it possible for the engineer 
 to enter upon a couple of pages of his note-book all the for- 
 mulae required for the thorough calculation of a continuous 
 girder of any number of spans, equal or unequal ; and this cal- 
 culation in any particular case proceeds iu a manner precisely 
 similar to that of the simple girder, directly and without refer- 
 ence to authorities, tables, points of inflection, elastic line, 
 methods of loading, or any of the " other paraphernalia with 
 which the subject is usually encumbered." 
 
 It will be observed that here and throughout we have no- 
 where left out of sight analytical processes or methods. The 
 
X PREFACE. 
 
 reader who considers the present work as an attempt to super- 
 cede^ or even subordinate analytical investigation, misjudges 
 entirely our aim. So far from this, we indulge the hope that 
 its perusal cannot fail to render familiar the use of both 
 methods, to bring out their points of difference and relative 
 advantages, to illustrate the one by the other, to enable the 
 reader to check the results of the one by the other, and in any 
 case apply one or both, or a judicious combination of both, as 
 may in such case be most advantageous or desirable. This will 
 be especially noticed in the discussion of the simple and con- 
 tinuous girder and of the braced arch. (Chaps. XII., XIII., 
 XIY. and XVL, and Appendix.) 
 
 As to the use of the work, the practical engineer will find in 
 Chap. I., and that portion of the Appendix relating to this 
 chapter, an easy and simple method of solution applicable to 
 any framed structure having simple reactions, and including 
 thus all varieties of bridge and roof trusses of single span. In 
 the Appendix he will find detailed examples calculated to illus- 
 trate every practical point of importance, and also a full expo- 
 sition of Hitters "method of moments" The principles of 
 this chapter alone will enable him to solve readily, both by cal- 
 culation and diagram, every case usually arising in practice. 
 In problems involving the moment of inertia of areas, in the 
 case of the continuous girder, the braced arch and stone arch, 
 as also the suspension system, he will find Chaps. VI., XII., 
 XIII., XIV. and XVI. of value; and in the perusal of any or 
 all of these he will, it is hoped, find no trouble by reason of 
 logical connection with preceding principles. They are in this 
 respect, as far as possible, complete in themselves. We may 
 also call his attention to Chap. XV._, upon the stone arch, 
 though it is to be regretted that the practical importance of the 
 subject, in the present age of iron, renders the ease with which 
 it is graphically treated of less importance than formerly. For 
 his benefit also frequent practical examples are given in detail, 
 so' that in all important applications he can easily select a 
 parallel case, and follow it out, step by step, in the case in 
 hand, without studying up the whole process of development 
 in order to place himself in a condition to make use of the 
 methods employed. We would also refer him to THE NEW 
 METHOD or GRAPHICAL STATICS (Van Nostrand, 1875) a reprint 
 of a series of articles contributed by the author to Van Nos- 
 
PREFACE. XI 
 
 trand's Engineering Magazine during the present year, where 
 lie will find such a condensed presentation of the more essen- 
 tial principles of the subject as will enable him to follow the 
 practical examples in the present work without the perusal of 
 the more lengthy preparatory portion here given. 
 
 For the student much of the practical applications may well 
 l>e at first omitted. Notably Chaps. VII.-XIL, inclusive. 
 Chaps. I.-IY. and XIII.-XVL will put him in complete pos- 
 session of the method, and, moreover, enable him to solve with 
 ease any structure, including the continuous girder, braced 
 arch, suspension system, and stone arch, as well as all the more 
 ordinary forms of bridge and roof trusses, cranes, etc. Indeed, 
 if the first-named structures, which are of comparatively rare 
 occurrence, are at first omitted, Chaps. I.-IY. alone will con- 
 stitute a complete course upon framed structures so far as 
 usually taught in our schools at the present day. Afterwards, 
 in practice, and in the solution of the particular problems 
 treated of, he will, in common with the practical engineer, find 
 in the other portions of the work and in the Appendix just 
 such assistance as he needs. We would also call the attention 
 of the mathematician more especially to the investigation in 
 Chap. V., Arts. 47-51, of the effects of a given recurring system 
 of moving loads., the analytical treatment of which would be 
 almost impracticable by reason of the complexity of the for- 
 mulas obtained, and in this respect certainly worthless, even if 
 possible, but the geometrical treatment of which gives rise to 
 some of the most elegant constructions of the graphical statics; 
 also to the Supplements to Chaps. /IIL and XIV., in which 
 the analytical treatment of the continuous girder and braced 
 arch is given. 
 
 Finally, if our purpose in writing these pages is accom- 
 plished, the principles and methods here set forth will be found 
 easily acquired, accurate in their results, and amply sufficient 
 for the ready determination of the strains in the various pieces 
 of any framed structure which the civil engineer can legiti- 
 mately be called upon to design. 
 
 With this much of introduction and explanation, we present 
 our work to the engineering profession in America and to 
 American technical colleges, in the hope that the spirit which 
 has led to its production, if not the method of its execution, 
 may win for it a favorable reception. 
 
Xll PREFACE. 
 
 In this spirit and in this hope we may, we trust, be allowed 
 to appropriate the closing lines of Culmanrfs preface " Und 
 nunftlhre hin gem hdtte icJi dich zum Fundament einer auf 
 wissenschaftlicherer Basis gegrundeten IngenieurJciinde ge- 
 mac/itj allein Jcaum darf ich die Hcffnung hegen, so viel Kraft 
 in mir zu finden, um das Game dieses umfangreichen Fachen 
 umzuarbeiten : das ist ein Werk, das mir vor Augen sc/iwebt, 
 wie einer jener alien mittelaUerlicJien Dome sick vor dem 
 Kunstler erhob^ der ihn entwarf und der der Hoffnung sick 
 nicht hingeben konnte, ihnje in seiner Vollendung zu scJiauen. 
 
 " Dock es mogen dich A.ndere benutzen und weiter bauen" 
 nnd was ich nicht kann, werden nieine Nachganger voll- 
 bringen. 
 
 NEW HAVEN, April nth, 1875. 
 
GElsTEEAL CONTENTS. 
 
 PAOB 
 
 INTRODUCTION xxv 
 
 PART I. GENERAL PRINCIPLES. 
 CHAPTER I. 
 
 FORCES IN SAME PLANE COMMON POINT OF APPLICATION. ILLUS- 
 TRATED BY PRACTICAL EXAMPLES 1 
 
 CHAPTER II. 
 
 FORCES IN SAME PLANE DIFFERENT POINTS OF APPLICATION. PRO- 
 PERTIES OF FORCE AND EQUILIBRIUM POLYGON 16 
 
 CHAPTER III. 
 . CENTRE OF GRAVITY 29 
 
 CHAPTER IV. 
 "MOMENT OF ROTATION OF FORCES IN SAME PLANE IN GENERAL 33 
 
 CHAPTER V. 
 
 MOMENT OF ROTATION OF PARALLEL FORCES PRACTICAL APPLICA- 
 TIONS 36 
 
 CHAPTER VI. 
 MOMENT OF INERTIA OF PARALLEL FORCES PRACTICAL PROBLEMS. . 61 
 
 PART II. APPLICATION TO BRIDGES. 
 
 A. THE SIMPLE GIRDER. 
 
 CHAPTER VII. 
 
 THE SIMPLE GIRDER OR TRUSS SUPPORTED ONLY AT ENDS 84 
 
 SUPPLEMENT TO CHAPTER VII. 
 
 THE THEORY OF FLEXURE. 
 
 CHAP. I. Methods of calculation 98 
 
 CHAP. II. Principles of the calculus 102 
 
 CHAP. III. Theory of flexure , . . 110 
 
XIV GENERAL CONTENTS. 
 
 B. THE CONTINUOUS GIRDER OF CONSTANT 
 CROS&SECTION. 
 
 CHAPTER VIII. 
 
 PAGE 
 
 GENERAL PBINCIPLES 125 
 
 CHAPTER IX. 
 LOADED AND UNLOADED SPANS. 141 
 
 CHAPTER X. 
 SPECIAL CASES OF LOADING 147 
 
 CHAPTER XI. 
 METHODS OP LOADING CAUSING MAXIMUM STRAINS 155 
 
 CHAPTER XII. 
 COMBINATION OP GRAPHICAL AND ANALYTICAL METHODS 168 
 
 CHAPTER XIII. 
 ANALYTICAL FORMULAE 198 
 
 SUPPLEMENT TO CHAPTER ,XIIL 
 DEMONSTRATION OF ANALYTICAL FORMULAE. . . 239 
 
 PAET III. APPLICATION TO THE AKCH. 
 CHAPTER XIV. 
 
 THE BRACED ARCH. COMBINATION OF GRAPHICAL AND ANALYTICAL 
 
 METHODS 251 
 
 SUPPLEMENT TO CHAPTER XIV. 
 DEMONSTRATION OF ANALYTICAL FORMULAE 271 
 
 CHAP. I. General considerations and formulae 271 
 
 CHAP. II. Hinged arch in general 277 
 
 CHAP. III. Arch hinged at abutments only 283 
 
 CHAP. IV. Arch fixed at ends no hinges 287 
 
 CHAP. V. Influence of temperature 299 
 
 CHAP. VI. Partial uniform loading 304 
 
 CHAPTER XV. 
 
 THE STONE ARCH 31 1 
 
 CHAPTER XVI. 
 
 THE INVERTED ARCH SUSPENSION SYSTEM. . 327 
 
 APPENDIX.. 339 
 
TABLE OF CONTENTS. 
 
 INTRODUCTION. 
 
 ABT. PAGE 
 
 I. Upon mathematical investigations generally xxvi 
 
 II. Analytical and geometrical mechanics xxvii 
 
 III. Geometrical statics xxix 
 
 IV. The graphical calculus xxxi 
 
 V. Graphical representation xxxiv 
 
 VI. Graphical statics xxxv 
 
 VII. The methods and limits of the graphical statics xxxvii 
 
 VIII. The modern geometry xxxix 
 
 IX. The modern geometry in engineering practice xlii 
 
 X. Practical significance of the graphical statics xliii 
 
 XI. Literature upon the graphical statics xlv 
 
 XII. Graphical dynamics 
 
 PAET I. GENERAL PRINCIPLES. 
 
 CHAPTER I. 
 
 FORCES IN SAME PLANE COMMON POINT OF APPLICATION. 
 
 1. Notation Representation of forces by lines 1 
 
 2. Resultant of two forces 2 
 
 3. Resultant of any number of forces 2 
 
 4. Conditions of equilibrium 3 
 
 5. Properties of force polygon 3 
 
 6. Order of forces in force polygon a matter of indifference 4 
 
 7. Forces acting in same straight line 5 
 
 8. PRACTICAL APPLICATIONS 5 
 
 9. Braced semi-arch 6 
 
 10. Roof truss 8 
 
 11 . Diagram for wind force 8 
 
 12. Application of method to bridges 11 
 
 13. Braced arch 12 
 
 14. Ritter's u Method of Sections " 14 
 
 CHAPTER II. 
 FORCES IN SAME PLANE DIFFERENT POINTS OF APPLICATION. 
 
 16. Resultant of two forces 16 
 
 17. Case of forces parallel 18 
 
XVI TABLE OF CONTENTS/ 
 
 AKT. PAGE 
 
 18. Perpendiculars let fall upon components from intersection of outer 
 
 polygon sides, are inversely as the components 18 
 
 19. Equilibrium polygon 19 
 
 20. Case of a couple Conditions of equilibrium 20 
 
 21. Properties of a couple 21 
 
 22. Force and equilibrium polygons for any number of forces 22 
 
 23. Influence of a couple 24 
 
 24. Order of forces in force polygon, a matter of indifference 25 
 
 25. Pole upon the closing line Failing case for equilibrium.. 25 
 
 26. Relation between two equilibrium polygons with different poles .... 26 
 
 27. Mean polygon of equilibrium 27 
 
 28. Line of pressures in an arch 28 
 
 CHAPTER III. 
 
 CENTRE OF GRAVITY. 
 
 30. General method for determination of centre of gravity 29 
 
 31. Reduction of areas 30 
 
 32. Reduction of triangle to equivalent rectangle of given base 31 
 
 33. Reduction of trapezoid to equivalent rectangle 31 
 
 34. Reduction of quadrilateral generally 31 
 
 CHAPTER IT. 
 
 MOMENT OF ROTATION OF FORCES IN THE SAME PLANE 
 IN GENERAL. 
 
 35. Definition of moment of rotation 33 
 
 36. Culmann's principle 33 
 
 37. Application to equilibrium polygon 34 
 
 CHAPTER Y. 
 MOMENT OF RUPTURE PARALLEL FORCES. 
 
 38. Equilibrium polygon Ordinates to give the moments of rupture. . . 36 
 
 39. Beam with two equal and opposite forces beyond the supports 38 
 
 40. Beam with two equal and opposite forces between the supports 39 
 
 41. Special cases of importance 41 
 
 1st. Beam load inclined to axis. ... 41 
 
 2d. Force parallel to axis 42 
 
 3d. Forces, in different planes ' 42 
 
 4th. Combined twisting and bending moments 43 
 
 5th. Application to crank and axle 44 
 
 42. Loading continuous Load area 45 
 
 43. Beam uniformly loaded 46 
 
 44. Moment curve a parabola 46 
 
 45. Beam continuously loaded and also subjected to action of concen- 
 
 trated loads 48 
 
 46. Influence of moving load 50 
 
 47. Load systems 53 
 
TABLE OF CONTENTS. XV11 
 
 ABT. PAGE 
 
 48. Properties of the parabola included by the closing line 54 
 
 49. Application of the above principles 54 
 
 50. Most unfavorable position of load system upon a span of given 
 
 length 56 
 
 51. Greatest moment at a given cross-section 59 
 
 CHAPTER VI. 
 MOMENT OF INERTIA OF PARALLEL FORCES. 
 
 52. Application of moment of inertia in proportioning any cross-section. 61 
 
 53. Graphical determination of moment of inertia 62 
 
 54. Signification of the area of the equilibrium polygon 63 
 
 55. Radius of gyration 63 
 
 56. Inertia curves Ellipse and hyperbola of inertia 66 
 
 57. Construction of curve of inertia 69 
 
 58. Example 69 
 
 59. Central curve Central ellipse 71 
 
 60. Centre of action of the statical moments considered as forces 73 
 
 61. Cases where the direction of the conjugate axes of the inertia curve 
 
 can be at once determined 75 
 
 62. Practical applications 76 
 
 1st. The parallelogram 76 
 
 2d. The triangle 77 
 
 3d. The trapezoid 78 
 
 4th. The parabolic segment 80 
 
 C3. Compound or irregular areas 81 
 
 64. Cases where there is no axis of symmetry 83 
 
 PAET II. APPLICATION TO BRIDGES. 
 A. THE SIMPLE GIRDER. 
 
 CHAPTER VII. 
 
 THE SIMPLE GIRDER. 
 
 66. Forces which act upon a bridge 84 
 
 67. Bridge loading 85 
 
 68. Shearing force Moment of rupture, etc 85 
 
 69. Concentrated loads Invariable in position 87 
 
 70. Concentrated loads Variable in position 88 
 
 71. Position of a given system of loads causing maximum shear. 88 
 
 72. Construction of the maximum shear 89 
 
 73. Maximum moments 90 
 
 74. Construction of maximum moments 91 
 
 75. Absolute maximum of moments 92 
 
 76. Continuous loading 93 
 
 77. Total uniform load 93 
 
 78. Method of loading causing maximum shear 94 
 
 79. Live and dead loads. . 95 
 
XV111 TABLE OF CONTENTS. 
 
 SUPPLEMENT TO CHAPTER VII. 
 CHAPTER I. 
 
 METHODS OP CALCULATION. 
 
 ABT. PAGE 
 
 2. Hitter's method 99 
 
 3. Method by resolution of forces 100 
 
 CHAPTER II. 
 
 PRINCIPLES OP THE CALCULUS. 
 
 4. Differentiation and integration 102 
 
 5. Powers of a single variable 105 
 
 6. Other principles 106 
 
 7. Illustrations 107 
 
 a First differential coefficient 109 
 
 CHAPTER III. 
 
 THEORY OF FLEXURE. 
 
 9. Coefficient of elasticity 110 
 
 10. Moment of inertia 110 
 
 11. Change of shape of axis 112 
 
 12. Beam fixed at one end, load at other 113 
 
 13. Beam as above Uniform load 117 
 
 14. Beam supported at both ends Concentrated load 118 
 
 15. Beam as above Uniform load 119 
 
 16. Beam fixed at one end, supported at other Concentrated load 120 
 
 17. Beam as above Uniform load 121 
 
 18. Beam fixed at both ends Concentrated load 122 
 
 19. Inflection points, etc. 124 
 
 B. THE CONTINUOUS GIRDER OP CONSTANT 
 CROSS-SECTION. 
 
 CHAPTER VIII. 
 GENERAL PRINCIPLES. 
 
 80. Mohr's principle 125 
 
 81. Determination of tangents to the elastic curve 127 
 
 82. Effect of the moments at the supports 128 
 
 83. Division of the moment area 129 
 
 84. Properties of the equilibrium polygon 130 
 
 85. Polygon for the positive moment areas 131 
 
 86. Construction of the fixed points, and of the equilibrium polygon ... 132 
 
 87. Construction of the moments at the supports 135 
 
 88. The second equilibrium polygon 136 
 
 89. Determination of moments at the supports.. 138 
 
 90. Comparison with girder fixed horizontally at both ends 139 
 
TABLE OF CONTENTS. XIX 
 
 CHAPTER IX. 
 LOADED AND UNLOADED SPANS. 
 
 ABT. PAGE 
 
 91. Unloaded span 141 
 
 92. Two successive unloaded spans , 141 
 
 93. The fixed points 142 
 
 94. Shearing- force Reactions at the supports, and moments in the un- 
 
 loaded spans 143 
 
 95. Loaded spans 143 
 
 96. Two successive loaded spans 144 
 
 97. Arbitrary loading 145 
 
 CHAPTER X. 
 
 SPECIAL CASES OF LOADING. 
 
 98. Total uniform load 147 
 
 99. Practical examples Girder of four spans 148 
 
 100. Partial uniform load 149 
 
 101. Concentrated load 153 
 
 CHAPTER XI. 
 
 METHODS OF LOADING CAUSING MAXIMUM STRAINS. 
 
 102. Maximum shearing force 155 
 
 103. Maximum moments 157 
 
 104. Determination of the maximum shearing forces 159 
 
 105. Determination of the maximum moments 161 
 
 106. Practical simplifications of the method 162 
 
 107. Approximate practical constructions 163 
 
 1st. Beam of two spans Moments 163 
 
 Beam of two spans Shearing forces 164 
 
 2d. Beam of three or more spans Moments 164 
 
 Beam of three or more spans Shearing forces 165 
 
 108. Method by resolution of forces Draw spans 166 
 
 CHAPTER XII. 
 
 CONTINUOUS GIRDER CONTINUED COMBINATION OF 
 GRAPHICAL AND ANALYTICAL METHODS. 
 
 111. Method of finding shearing forces, when inflection points are 
 
 known 168 
 
 1st. Loaded span 168 
 
 2d. Unloaded span 169 
 
 112. Determination of inflection points Inflection verticals 170 
 
 113. Beam fixed horizontally at ends 171 
 
 114. Example 172 
 
 1 15. Counterbracing 174 
 
XX TABLE OF CONTENTS. 
 
 ART. PAGR 
 
 116. Beam fixed at one end, supported at the other 175 
 
 117. Practical construction 178 
 
 118. Beam continuous over three level supports 178 
 
 119. Practical construction 170 
 
 120. The pivot draw with secondary central span 181 
 
 121. Supports not on a level Reactions 182 
 
 122. Beam over four level supports 184 
 
 123. Practical construction 186 
 
 124. Pivot span Example 187 
 
 125. Method of passing in construction, from one span to another 190 
 
 126. Method of procedure for any number of spans 192 
 
 127. Example Central span of seven spans 193 
 
 128. Method of calculation by moments 196 
 
 CHAPTER XIII. 
 
 ANALYTICAL FORMULA FOR THE SOLUTION OF CON- 
 TINUOUS GIRDERS. 
 
 129. Introduction 198 
 
 130. Notation 200 
 
 131. Theorem of three moments 201 
 
 132. Example Total uniform load Moments 202 
 
 133. Triangle of moments 203 
 
 134. Total uniform load All spans equal Reactions 205 
 
 135. Triangle for reactions 205 
 
 136. Clapeyron's numbers 206 
 
 137. Uniform live load over any single span Moments at supports 207 
 
 138. Triangle for moments Loaded span 208 
 
 139. Unloaded spans Moments 209 
 
 140. Practical rule and table for finding the above 210 
 
 141. Reactions at supports Loaded span 212 
 
 142. Triangle for reactions 213 
 
 143. Reactions for unloaded spans Tables for 214 
 
 144. Concentrated load in any span Moments 216 
 
 145. Application of above formulae 216 
 
 1 46. Triangle and table for moments 218 
 
 1 47. Reactions at supports Concentrated load , 219 
 
 148. Shear at supports 219 
 
 149. Recapitulation of above formulae 221 
 
 150. Continuous girder with variable end spans 224 
 
 151. Application of formulae for 225 
 
 152. Continuous girder with fastened ends 22t'> 
 
 153. Beam of single span Fastened at both ends Fastened at one end, 
 
 etc. Examples 22f> 
 
 154. Tables for moments End spans variable 229 
 
 155. Continuous girder All spans different General formulae Exam- 
 
 ples 233 
 
 156. General method of calculation . . 237 
 
TABLE OF CONTENTS. XXI 
 
 SUPPLEMENT TO CHAPTER XIII. 
 
 DEMONSTRATION OF ANALYTICAL FORMULA FOR THE 
 CONTINUOUS GIRDER. 
 
 IRT. PAGfi 
 
 1. Conditions of equilibrium 239 
 
 2. Equation of elastic line 240 
 
 3. Theorem of three moments 241 
 
 4. Determination of moments Supports all on level 42 
 
 5. Uniform load w 244 
 
 0. Formulae for the " Tipper " [Art. 120] 244 
 
 7. Example of two span tipper 245 
 
 8. LITERATURE UPON THE CONTINUOUS GIRDER. . 247 
 
 PART III. APPLICATION OF THE GRAPHICAL 
 METHOD TO THE ARCH. 
 
 CHAPTER XIV. 
 
 THE BRACED ARCH. 
 
 157. Different kinds of braced arch 251 
 
 158. Arch hinged at both crown and abutments 251 
 
 159. Hinged at abutments only Continuous at crown 253 
 
 1st. Parabolic arch 254 
 
 2d. Circular arch, Tables for solution of 255 
 
 160. Arch fixed at abutments Continuous at crown 257 
 
 1st. Parabolic arch 258 
 
 2d. Circular arch 260 
 
 162. General method of solution 262 
 
 163. Analytical formulae for horizontal thrust, and vertical reactions 264 
 
 164. Arches with solid web 266 
 
 165. Strains due to temperature, Formulae for 267 
 
 166. Effects of temperature 269 
 
 SUPPLEMENT TO CHAPTER XIV. 
 
 DEMONSTRATION OF ANALYTICAL FORMULA FOR THE 
 BRACED ARCH. 
 
 CHAPTER I. 
 
 GENERAL, CONSIDERATIONS AND FORMULAE. 
 
 1. Fundamental equations ; 271 
 
 2. Displacement of any point 275 
 
 CHAPTER II. 
 
 HINGED ARCH IN GENERAL. 
 
 3. Notation The outer forces in general 277 
 
 4. Intersection line 278 
 
 5. Parabolic arch Concentrated load 278 
 
 6. Circular arch Concentrated load 280 
 
 7. Integrals used in above discussion 282 
 
XX11 TABLE OF CONTENTS. 
 
 CHAPTER III. 
 ARCH HINGED AT ABUTMENTS ONLY. 
 
 A. Parabolic arc. 
 
 ART. PAGK 
 
 8. Horizontal thrust 283 
 
 9. Intersection curve 283 
 
 B. Circular arc. 
 
 10.- Horizontal thrust 283 
 
 11. Intersection curve 285 
 
 CHAPTER IV. 
 
 ARCH FIXED AT ENDS. 
 
 12. Introduction 287 
 
 13. Concentrated load General formulae 287 
 
 A. Parabolic arc. 
 
 14. Determination of H, V and M 289 
 
 15. Intersection curve 291 
 
 16. Direction curve 291 
 
 B. Circular arc. 
 
 17. Fundamental equations 291 
 
 18. Determination of H, V and M 293 
 
 19. Intersection curve 297 
 
 20. Direction segments 297 
 
 20 (b). Transformation series 297 
 
 CHAPTER V. 
 
 INFLUENCE OF TEMPERATURE. 
 
 21. General considerations 299 
 
 22. Influence of temperature on the arch 299 
 
 23. Fundamental equations General 300 
 
 24. Arch with three hinges 301 
 
 25. Arch hinged at ends 501 
 
 26. Arch without hinges 301 
 
 CHAPTER VI. 
 
 PARTIAL UNIFORM LOADING. 
 
 27. Notation 304 
 
 A. Arch hinged at crown and ends. 
 
 28. Vertical reaction 304 
 
 39. Horizontal thrust 305 
 
 B. Arch hinged at ends only. 
 
 30. Vertical reaction 305 
 
 31. Horizontal thrust Parabolic arch 305 
 
 32. Horizontal thrust Circular arch 306 
 
 O. Arch without hinges continuous at crown. 
 
 33. Parabolic arch. Formulae for V, H and M 307 
 
 34. Circular arch. Formulae for V, H and M 308 
 
TABLE OF CONTENTS. XX111 
 
 CHAPTER XY. 
 
 THE STONE ARCH. 
 
 ART. PAGB 
 
 167. Definitions, etc ; 
 
 168. Pressure line 311 
 
 169 . Sliding of the joints 311 
 
 170. Forces acting upon a cross-section Neutral axis 312 
 
 171. Kernel of a cross-section 313 
 
 172. Position of kernel for different cross-sections 314 
 
 173. Proper position of resultant pressure 316 
 
 174. Pressure line True pressure line 317 
 
 175. Support line 318 
 
 176. Deviation of support from pressure line 319 
 
 177. Dimensions of the arch 320 
 
 178. Construction of the pressure line 322 
 
 179. Practical example " 324 
 
 180. Proper depth of arch at crown 324 
 
 181. Increase of depth due to change of form 326 
 
 CHAPTER XYI. 
 THE INVERTED ARCH SUSPENSION SYSTEM. 
 
 183. Methods of construction 327 
 
 184. Rear chains, and anchorages 327 
 
 185. Cable with auxiliary stiffening truss 329 
 
 186. Method of loading causing maximum strains 330 
 
 187. Example ,, 333 
 
 188. Analytical investigation of the forces acting upon the stiffening 
 
 ooo 
 
 truss 066 
 
 189. Concluding remarks 335 
 
 APPENDIX. 
 
 NOTE TO CHAP. VIII. OP THE INTRODUCTION UPON THE MODERN 
 GEOMETRY. 
 
 NOTE TO CHAPTER 1 346 
 
 2. Bent crane 346 
 
 3. Character of strains in the pieces as indicated by the strain diagram, 347 
 
 4. Pieces in equilibrium Points to be avoided in constructing the 
 
 strain diagram 347 
 
 5. Roof truss Method of checking the accuracy of the diagram 348 
 
 6. The French roof truss Solution apparently indeterminate Method 
 
 of solution Method of calculation by moments 348 
 
 7. Application to bridges Bowstring girder Method of tabulation 
 
 illustrated. 350 
 
 8. Strains in the flanges Table 353 
 
 9. Method of calculation by moments 354 
 
 10. Girder with straight flanges Howe or Murphy Whipple system of 
 
 . bracing 355 
 
XXIV TABLE OF CONTENTS. 
 
 ART. ,PAGB 
 
 11. Lenticular girder, or system of Pauli 358 
 
 12. Remarks upon above system The most economical system for long 
 
 spans 359 
 
 NOTE TO CHAPTER II. 
 
 14. Equilibrium polygon considered as a, frame 362 
 
 NOTE TO CHAPTER V., ART. 51. 
 
 15. Beam subjected to given force system Maximum moment at any 
 
 cross-section 364 
 
 NOTE TO CHAPTER XII. 
 
 16. Pivot or draw span Practical example Solution by diagram By 
 
 method of moments By resolution and composition of forces .... 365 
 
 17. Relative economy of the continuous girder 374 
 
 18. Continuous girder supports out of level 377 
 
 NOTE TO CHAPTER XIV. THE BRACED ARCH. 
 
 19. Practical example of the braced arch 382 
 
 20. Arch hinged at both abutments and crown 383 
 
 21. Arch hinged at abutments only 386 
 
 23. Temperature strains for above 391 
 
 26. Arch continuous at crown Fixed at ends 394 
 
 27. Temperature strains for above 398 
 
 28. Advantage of arch with fixed ends for % long spans Comparison with 
 
 the St. Louis arch. . 399 
 
INTRODUCTION. 
 
 HISTORICAL AND CRITICAL.* I ^/t? I 
 
 X^ ' 
 
 THE subject of Graphical Statics has, since the appearance of (/ulrnann's 
 work (Die grapliisclie Statik, Zurich, Meyer and Zeller, 18#t>), excited 
 considerable attention, but an accurate and just estimate of its methods 
 and practical value is still wanting. Thus there are some who oppose it; 
 others willingly accept it as an efficient and valuable aid in practical inves- 
 tigations of stability ; still others even profess to see in it a future rival of 
 Analytical Statics. This last somewhat remarkable claim seems apparently 
 justified by a passage in Culniann's preface, where it is asserted " that the 
 Graphical Statics will and must extend, as graphical methods find ever 
 wider acceptance but in such case, however, its treatment will soon escape 
 the hands of the practitioner, and it will then be built up by the geometer 
 and mechanic to a symmetrical whole, which shall hold the same relation 
 to the new geometry that analytical mechanics does to the higher analysis." 
 These various and conflicting opinions find their supporters in technical 
 schools and among engineers throughout Germany. 
 
 In the consideration of the subject, we shall endeavor especially to give 
 an objective presentation, but shall also feel at liberty to present our own 
 opinions as well, and generally to venture such reflections as seem suited 
 to throw light upon the matter. For both reasons it will sometimes be 
 necessary to make apparent deviations, in order to point out the various 
 fields in which these new investigations take root, to define their limits, and 
 to decide in what directions and to what extent impulse and sustenance 
 for further development may exist. In such a manner only can we satis- 
 factorily ascertain how far the graphical statics may safely count upon 
 more than a passing recognition and brief existence. 
 
 We have therefore to ask of the reader who wishes to obtain a just and 
 accurate estimate of this new and, as we venture to think, highly important 
 subject, patience for the following general considerations. 
 
 * Ueber die graphische Statik zur Orientirung. By J. I.Weyrauch. Leip- 
 zig, 1874. 
 
XX Vl MATHEMATICAL INVESTIGATIONS. [iNTIiOD. 
 
 L 
 
 y 
 
 UPON MATHEMATICAL INVESTIGATIONS IN GENERAL. 
 
 Mathematical truths may be attained in two essentially different methods 
 by synthesis or by analysis, by composition or by resolution. In synthe- 
 sis, we ascend from particular cases to general ones ; in analysis, we descend 
 from general cases to particulars. By synthesis we pass from the simplest 
 or admitted truths, by combination and comparison, to more complicated 
 phenomena. Analysis seeks to refer back such phenomena to their fun- 
 damental relations, or to deduce special properties from the general con- 
 ditions. 
 
 The analysis of a phenomena presupposes, then, an accurate comprehen- 
 sion of all its elements. So far as these last stand in relations of cause 
 and effect to the whole and its parts, or so far as such relations exist be- 
 tween the parts themselves, they may be expressed by equations. Thus 
 the operations which are necessary in analysis become independent of con- 
 crete phenomena, and are governed only by the laws of abstract quantities 
 as included by algebra in the widest sense of the word. Algebra, then, is 
 not analysis itself, but only its instrument, " instrument precieux et neces- 
 saire sans doute, parce qu'il assure et facilite notre marche,"mais qui ri>a par 
 lui meme aucune vertu propre ; qui ne dirige point I' * esprit, mais que V esprit 
 doit diriger comme tout autre instrument' 1 '' (Poinsot, Theorie nouvelle de la 
 rotation, pre"s a" T Acad., 1834). Ordinarily the higher branches of algebra, 
 with which numberless really analytical investigations are connected, are 
 designated as analysis. More properly, all investigations which rest upon 
 equations of condition may be termed analytical investigations. 
 
 Synthetic investigation rests mainly upon geometrical conceptions, and 
 attains to the knowledge of phenomena through concrete conditions, which 
 latter may be designated as space relations and processes. Hence the usual 
 division into analytical and geometrical methods, even in applied mathe- 
 matics. We have thus with equal appropriateness an analytical geometry 
 as also a geometrical analysis. When pure geometry (in distinction from 
 analytical) makes use of the symbols and operations of algebra, it is only 
 to express with corresponding generality and more concisely than in words 
 truths attained to by abstraction, and independent of the dimensions of the 
 auxiliary figure ; or so to formulate such truths that they may be applied 
 in analytical investigation. Accordingly, such use of algebraic formulae 
 has as little effect upon the synthetic process as from the above it would 
 seem essential to the analytic treatment. In either case, algebra is but the 
 instrument, the method lies back of and directs it. 
 
 If analytical formulae and operations are entirely excluded from the 
 more complicated geometrical investigations, we are at once restricted to 
 general laws of metrical relation. There remains only the faculty of 
 
INTROD.] ANALYTICAL AND GEOMETRICAL MECHANICS. XXV 11 
 
 abstraction and graphical construction. The power of abstraction alone 
 suffices, indeed, to comprehend in full generality metrical relations in ele- 
 mentary geometry and its simplest applications, but fails when the relations 
 sought must be attained step by step by the application of a number of 
 principles, or in the auxiliary figure by a number of constructions. If, 
 indeed, we take the relation sought directly from the auxiliary figure 
 itself, and even if it were possible to take out the required distances with 
 absolute accuracy, still this result obtained would stand to the general law 
 desired only in the same relation that the result of a particular numerical 
 computation does to the more general algebraic formula. 
 
 Investigations by the aid of graphical figures can, however, make known 
 general relations of form and position, and have in this respect their special 
 advantage. So far also as by them metrical relations are sought, then, by 
 the exclusion of algebraic formulae, only the process of deduction the 
 routine of construction remains of general significance. Sciences, then, 
 which proceed in this manner, furnish indeed, with respect to metrical 
 relations, no general laws, but for the deduction of these relations do give 
 general methods. In this category we may place descriptive geometry and 
 the more recent graphical statics. 
 
 -it 
 
 ANALYTICAL AND GEOMETRICAL MECHANICS. 
 
 It is hardly necessary in these days to call attention to the advantages 
 of a geometrical treatment of mechanical problems. This, however, was 
 not always the case, and the most important developments of geometrical 
 mechanics belong to the present century. It is to Poinsot, Chasles, Afobius, 
 etc., that these developments are due. 
 
 By the Calculus of Newton and Leibnitz (1646-1714), and its subsequent 
 development, analysis became such a powerful instrument that the activity 
 of mathematicians was for a long time solely directed towards analytical 
 investigations. The power of analysis was in mechanics carried to its 
 highest point by Lagrange (1736-1813), in his Mechanique analytique. He 
 undertook the problem of reducing mechanics to a series of analytical 
 operations : " On ne trouvera point de figures dans cet outrage. Les 
 methodes que fy expose ne demandent ni constructions ni raisonnement geo- 
 metrique ou mecanique, mais seulement des operations algebriques assujeties 
 a une marche reguliere et uniforme" (Mechanique analytique. Paris, 1788.) 
 The principle of virtual velocities formed his point of departure. A 
 number of text-books upon theoretical mechanics still follow the method 
 of Lagrange. 
 
 The revival of pure geometrical investigations by Monge (1746-1818), 
 the creator of descriptive geometry, and his followers, could not well have 
 been without its influence upon mechanics. In the year 1804 appeared 
 the Elements de Statique, by Poinsot, in which, in contrast to Lagrange, 
 
XXviii ANALYTICAL AND GEOMETRICAL MECHANICS. [iNTKOD. 
 
 we find : " que tons les theoremes de la Statique rationeUe ne sont plus au 
 fond que des theoremes de Geometric" This work was the beginning of a 
 series of treatises in which the advantages of the synthetic development 
 and geometrical treatment of mechanics were defended and, by most 
 important results, strikingly demonstrated. 
 
 At this time the views as to the best method of treating mathematical 
 problems were sharply opposed. Carnot (1753-1823), to whom, however, 
 the modern geometry itself owes no slight impulse, gives the preference 
 to analysis. For synthesis " est restreinte par la nature de ces procedes ; 
 elle ne peut jamais perdre de vue son objet, ilfaut que cet objet s^offre tou~ 
 jours d V esprit, reel et net, ainsi que tons les rapprochements et combinaisons 
 qu'on en fait" 1 (Oeometrie de position. Paris, 1803.) That which here Car- 
 not considers as a defect in the synthetic and geometrical method, Poinsot 
 claims as its special advantage : " On peut bien par ces calculs plus ou moins 
 longs et compliques parvenir d determiner le lieu ou se trouvera le corps au 
 bout d'un temps donne, mais ou le perd entierement de vue, tandis qu'on vou- 
 drait V observer et le suivre, pour ainsi dire, des yeux dans tout le cours de sa 
 rotation' 1 ' 1 (Theorie nouv. d. I. rot. d. corps}. 
 
 The example of Poinsot found numerous followers. In Germany, Mo- 
 bius followed with his " Lehrbuch der Statik." Mechanics as well as 
 geometry thus received enrichment. Mobius gives the preference always 
 to the synthetic method, and also endeavors to interpret geometrically, 
 analytically deduced formulae " because in investigations concerning 
 bodies in space the geometrical method is a treatment of the subject itself, 
 and is therefore the most natural, while by the analytical method the sub- 
 ject is concealed and more or less lost sight of under extraneous signs " 
 (Lehrb. d. StatiTc. Leipzig, 1837.) 
 
 Even in analytical operations, geometrical considerations came more and 
 more in the foreground. On all sides the development of Kinematics, the 
 theory of motion without reference to its cause, was prosecuted. But. 
 neglecting the cause of motion, there remains only its path ; that is, geo- 
 metry proper (Kinematical geometry, or the geometry of motion}. The in- 
 vestigations of Chasles, Mobius, Rodrigues, Jouquiere, and others, may yet 
 be still further pursued ; and when by the aid of geometry a certain com- 
 pleteness has been given to the theory of the motion of invariable systems, 
 the geometrical theory of regular variable systems (to which the flexible 
 and elastic belong) will be possible. For the discussion of such branches 
 of mathematics, the synthetic geometry is necessary ; for their foundation 
 lies in a theory of the relationship of systems. 
 
 The advantage of the synthetic method in mechanics is denied by no 
 one. Wherever it is possible, we obtain more comprehensive conclusions 
 as to the nature of the phenomena, while all the properties of the same fol- 
 low directly from the simple and known truths premised. In analytical 
 investigations it is necessary, even when definite equations are obtained, to 
 deduce the actual laws singly and in a supplementary manner, although 
 they are indeed all contained in the equations themselves. 
 
 It is not, however, always possible to preserve the synthetic process 
 throughout. From the first truth the ways diverge in all directions, and 
 
TNTROD.] GEOMETRICAL STATICS. Xxlx 
 
 a special ingenuity is often needed to reach the goal. Just here analysis 
 comes to our aid with its rich treasures of developed methods, and here it 
 is most certainly not for geometry to " undervalue the advantage afforded 
 by a well-established routine, that in a certain degree may even outrun the 
 thought itself " (F. Klclu : Vrrglaichemle JRetrachtungen aber neuere geome- 
 trische Forschung^en. _ Erlanggn-, 1 ft72 T p. 41). Algebraic operations are 
 thus, however, not the chief thing, but only the instrument a most excel- 
 lent instrument indeed, which can be almost universally applied, and 
 which, by reason of its connection with an extensive and independent 
 mechanism, often needs only to be set in action in order to work of itself. 
 
 Geometrical mechanics, moreover, can never entirely free itself from 
 analytical formulae and operations. For though it may be both interesting 
 and useful to follow, with Poinsot, the body during its entire rotation, yet 
 practically this is of minor interest, and the chief problem remains still, 
 " d determiner le lieu ou se trouvera le corps au T)out cTun temps donne" 
 
 In the present day all those familiar with both methods of treatment 
 hold fast the good in each ; they supplement each other. Often in the 
 course of the same investigation we must interrupt the general analytical 
 process with synthetic deductions, and inversely. Thus we may well close 
 these considerations with the sentence with which Schell begins his " Theo- 
 rie der Bewegung und der Krdfte " both methods, the analytic and the 
 synthetic, can only, when united, give to mechanics that sharpness and 
 clearness which at the present day ought to characterize all the mathemati- 
 cal sciences. 
 
 III. 
 
 GEOMETRICAL STATICS. 
 
 Statics is a special case of dynamics, though earlier treated as indepen- 
 dent of the latter. The principle of d'Alembert furnishes the means of 
 passing from one .to the other. In technical mechanics the distinction is 
 still preserved, and indeed, in view of the distinct branches in which the 
 applications on either side, are found, not without propriety. 
 
 After the mechanics of the ancients, as comprised in the mathematical 
 collections of Pappus; the first great step towards our present geometrical 
 statics was made by Simon Stemnus (1548-1603), when he represented the 
 intensity and direction of forces by straight lines. Stevinus himself gave 
 a proof of the importance of his method, in the principle, deduced from it, 
 that three forces acting upon a point are in equilibrium when they are pro- 
 portional and parallel to the three sides of a right-angled triangle. 
 
 A main discovery was the parallelogram of forces by Newton (1642- 
 1727). The composition of two velocities in special cases was long famil- 
 iar. Galileo made use of it for two velocities at right angles, and exam- 
 ples also occur in Descartes, Roberval, Mersenne, and Wallis, but the funda- 
 mental principle was first established when Newton replaced the theories 
 
XXX GEOMETRICAL STATICS. [iNTEOD. 
 
 of special by that of universal causation (Philosophic^ naturalis principia, 
 mathematica. London, 1687). 
 
 Varignon in his "Projet d'unenouvelle mecanique," in the same year (1687), 
 and independently of Newton, applied for the first time the general princi- 
 ple of the composition of motions. From this he passes, in the Nouvelle 
 mecanique ou Statique, dont le projet fat donne en 1687 (published after 
 his death, Paris, 1725), by means of the axiom that " les effets sont toujours 
 proportionnels d leurs causes ou, forces productrices " to the composition of 
 forces also. 
 
 The Statique of Varignon is purely geometrical. He postulates nothing 
 beyond books 1-6 and 11 of Euclid, and even explains the significance of 
 + and signs. In this work, the first founded upon the parallelogram of 
 motion and of forces, we find also the force and equilibrium polygons 
 (Funiculaire, Section II.), to the application and development of which 
 almost the whole of Graphical Statics is to be attributed. Varignon recog- 
 nized the value of the equilibrium polygon, and gave it as the seventh of 
 the simple machines. 
 
 After the great Interim of Geometry, Monge wrote a Traite elementaire 
 de Statique (Paris, 1786). The work claims to contain for the first time 
 everything in statics which can be synthetically deduced. In a later edi- 
 tion we learn that synthetical statics must be taken up as preliminary to 
 analytical, just as elementary geometry before analytical geometry. Thus 
 the work of Monge contains the necessary preparation for Poisson 1 s "Traite 
 de mecanique" (Paris, 1811). 
 
 The greatest influence upon the development of geometrical statics was 
 exercised by Poinsot. By the introduction of force pairs, he solved in the 
 most elegant manner the fundamental problem of any number of forces 
 acting upon a body (Elements de Statique, Paris, 1804, and Memoire sur 
 la composition des moments et des aires dans la mecanique). 
 
 Chasles completed the solution by the proof that the contents of the 
 tetrahedron, which is determined by the resultant forces, is constant, how- 
 ever the forces may be composed. 
 
 In the hands of Mobius, geometry and geometrical statics were most com- 
 pletely developed. 
 
 Of the greatest importance, for later applications, was the introduction 
 of the rule of signs. 
 
 The germ of this had existed already in the preceding century.* Mobius 
 recognized its significance, extended it to the expression of the contents of 
 triangles, polygons, and three-sided pyramids, and applied it systemati- 
 cally (Barycentrischer Calcul. Leipzig, 1827). 
 
 A new impulse, extended field of action, and numerous additions were 
 given to geometrical statics by the Graphical Statics of Culmann. 
 
 * Mobius alludes to this, and we find, for example, in Kastner ( Geometrische 
 Abhandlungen, I. Saml., 1790, p. 464), the equation A B + B A = o. 
 
INTROD.] THE GRAPHICAL CALCULUS. XXxi 
 
 IV. 
 
 THE GRAPHICAL CALCULUS. 
 
 The most extended applications of statics are in the Held of engineering. 
 Here, not only general properties of form and position are required, but in 
 a large number of cases numerical relations are also necessary. General 
 results of the latter character can, as we have seen, only be embraced by 
 algebraic formulae (I). The pure graphical theory of construction is there- 
 fore in this respect lacking in completeness, as it is unable to furnish gen- 
 eral metrical relations. 
 
 The practical engineer has almost always, however, to do with special 
 problems; dimensions and acting forces are numerically given. Geometry 
 in such cases could give no general relations, because the results desired 
 are the consequences of the special proportions of the figure. In any de- 
 terminate case, however, we may obtain a result holding good for that case, 
 and it only remains to show how generally to obtain such a result. The 
 graphical calculus treats of such methods, and so, although not exclusively, 
 does graphical statics. As soon now as practical use is made of the actual 
 proportions of the figure, everything depends upon the exactness of the 
 drawing. One condition for the application of the graphical method is, 
 therefore, skill in geometrical drawing a requisition, indeed, which the 
 practical engineer can most readily meet. 
 
 The idea at bottom of the graphical calculus is simple. The modifica- 
 tions of numbers in numerical calculations correspond always to similar 
 modifications of the quantities represented by these numbers. The measure 
 of a quantity can be as well given by a line as a number, by putting in 
 place of the numerical the linear unit. In order for a graphical calculus, 
 then, the modifications of lines answering to corresponding numerical 
 operations are necessary, and these are furnished by geometry. They con- 
 sist of graphical constructions, and rest upon the known properties of 
 geometrical figures. The scale furnishes the means of converting directly 
 any numerical quantity into its corresponding linear representation, and 
 inversely any graphically obtained result can be at once transformed into 
 numbers. 
 
 The graphical determination of desired or computable numbers is natu- 
 rally nothing new. From the " Traite de Q-nomonique " of de la Hire 
 (1682) to the " Geometric descriptive' 1 ' 1 of Monge (1788), many examples 
 are to be found. The graphical calculus, however, goes further than this. 
 It aims to found a method, a routine, which shall not only apply to bodies 
 in space, but which shall also, like the arithmetical or algebraic calculus, 
 be independent of concrete relations and of general application. It seeks 
 further to obtain its results (products and powers) in the shape of lines 
 convertible by scale into numbers. (Hence the important part which area 
 
XXXM THE GRAPHICAL CALCULUS. [iNTROIX 
 
 transformation plays in the graphical calculus.) Such was the problem 
 which Cousinery proposed, and whose solution he attempted in his " Calcul 
 par le trait" (Ses Elements et ses applications. Paris, 1839). 
 
 Cousinery applied the graphical calculus to powers, roots, proportion 
 and progression; to the measure of lines, surfaces, cubes, graphic inter- 
 polation, and the strength of retaining walls. The presentation is nat- 
 urally by no means complete, and labors also under a prolixity and 
 minuteness of detail to which the results obtained are by no means com- 
 mensurate. It sounds somewhat comic when Cousinery, in his " Calcul par 
 le trait," claims the then already-existing graphical solutions of Poncelet 
 (" Memoire sur la stdbilite des revetements, in Memorial de Toff du genie") 
 as an elegant example of the application of his graphical calculus. 
 
 While Cousinery thus sought to apply geometry in a direction where 
 until then analysis had held sway, he acted in entire accordance with the 
 spirit of his age, though without making use of those means for aid which 
 lay at his disposal. " Without effect upon him," says Culmann, " were 
 the researches of Steiner, already published in 1832, as well as those of 
 his predecessor; and instead of simply premising the elementary prin- 
 ciples of the modern geometry, he laboriously sought to deduce them in- 
 dependently by the aid of perspective." The works, at least, of the French 
 predecessors of Steiner were, at any rate, well known to Cousinery. In his 
 preface we read: " Peut-gtre mihne nos efforts eussent-ils 6t6 complete- 
 ment infructueux, sans les ressources que nous ont procurers et les annales 
 de M. Gergonne et les travaux de M. Brianchon, et ceux plus re"cents de 
 M. Poncelet. Nous avons envers M. Chasles une obligation encore plus 
 droite, car outre les pr6cieux documents que reufernie son ' Histoire des 
 methodes en geometric? nous avons a lui faire agre"er un tgmoignage par- 
 ticulier de reconnaissance pour la maniere dont il a bien voulu mentionner 
 nos premiers essais sur le systeme de projection polaire." 
 
 Why Cousinery made use of perspective and not of the modern geome- 
 try, is easily understood. The development of geometry at that time, as to- 
 day, proceeded in various almost independent directions, and Cousinery 
 himself had the pleasure of seeing his " Geometric perspective " (Paris, 
 1828) designated by the reporters for the Academy, Fremel and Matthien, 
 as new and ingenious, as well as favorably noticed by Chaxles.* He 
 sought, therefore, naturally to develop and render fruitful his own method, 
 so much the more as the true significance and value of the various growing 
 branches of geometry could not then, as now, be correctly estimated. Ac- 
 cordingly, the Inge"nieur-en-chef. B. E. Cousinery, wrote avowedly for his 
 colleagues, and did not feel justified in directly premising a knowledge of 
 the newest investigations, more especially of his own. 
 i We have noticed the above somewhat in detail, because it bears directly 
 
 * Its newness, at least, is not without doubt. According 1 to Fiedler, the 
 principles are completely given in Lambert's celebrated work, " Die freie 
 Perspective" (Zurich, 1759). Poncelet also takes issue with the estimation of 
 the " Geometrie perspective" by Chasles (" Traite des propr. proj.^ II. , ed. 
 1865, p. 412). 
 
INTKOD.] THE GRAPHICAL CALCULUS. XXXI 1 1 
 
 upon a point of our discussion ; for the introduction of the modern 
 geometry in the graphical method by Culmann, is still, thirty years after 
 Cousinery, a chief hindrance to its rapid spread.* 
 
 After Cousinery, no one occupied himself with the graphical calculus 
 till Culmann gave it a place in his GrapJiische Statih The presenta- 
 tion is here far better, and especially shorter. The rule of signs, which 
 was unknown to Cousinery, is at once brought out. Instead of such 
 long and tedious applications as the graphical interpolation, a few 
 examples from engineering practice are given, among which we may 
 especially notice earth-work calculations. In the extensive earth works of 
 roads, canals, and railways, the method shows not only most plainly the 
 extent and best arrangement of transport, but also allows, with the aid of 
 the planimetre, the cost of transport to be determined. 
 
 As to the rest, it would appear as if the graphical calculus should play 
 an important part in engineering practice. This circumstance, as well as 
 the interesting problems which present themselves in connection, has 
 gained for the Arithmograpliy many friends. Several publications 
 have since sought to win for it a wider recognition without furnishing 
 anything essentially new. [IT. Eggers : " Grundziige einer graphischen 
 Arithmetic," Schaffhausen, 1865. J. ScUesinger : " Ueber Potenzcurven," 
 Zeitschr. d. osterr. Arch. u. Ing. Ver., 1866. E. Jdger : " Das graphischen 
 Rechnen," Speier, 1867. K. von Ott : " Grundziige des graphischen Rech- 
 nens und der graphischen Statik," Prag, 1871.] 
 
 Recently the method of the graphical calculus has been applied to Dif- 
 ferentiation and Integration. A treatise by Solin show the first exact, so 
 far as possible in a construction, the last approximate only (" Ueber graph. 
 Integr. ein Beitrag z. Arithmographie, Abhand. d. konigl. bohm. Gesellsch. 
 d. Wissenbach." VI. Folge, 5 Bd. Separate reprint by Rivnac, Prag, 
 1871). It is to be remarked, also that examples of double integration and 
 differentiation were given by Mohr in 1868. The graphical construction 
 of the elastic line, and the determination of the moments at the supports 
 of a continuous girder, are essentially examples in point (Mohr: "Bei- 
 trag zur Theorie der Holz und Eisenconstructionen, 1 ' Zeitschr. d. Hannov. 
 Ing. und Arch. Ver., 1869 ; or W. Hitter : " Die elastische Linie," Zurich, 
 1871.) 
 
 As to the importance of the graphical calculus as an independent study 
 or discipline, it is, as we believe, often exaggerated. The theoretical value is 
 but little, and for graphical constructions, as given by the graphical calculus, 
 offer in no respect anything new. That which pertains to practical applica- 
 tions may be easily based directly upon geometry, and is nowhere found 
 as a consequence of the method itself. If it is considered advisable to call 
 special attention to a few general points before making such applications, 
 all that can be desired can be easily presented in ten or a dozen pages 
 octavo. 
 
 * See Preface ; also Chaps. VII. and VIII. of this Introduction. 
 c 
 
GRAPHICAL REPRESENTATION. [iNTROD. 
 
 V. 
 
 GRAPHICAL REPRESENTATION. 
 
 Graphical representation, in the widest sense of the word, includes every 
 visible result of writing or drawing. The written sentence is the graphi- 
 cal representation of a thought the drawn line the graphical indication of 
 an idea. In such generality we naturally do not here regard graphical 
 representation. In a narrower sense we understand the graphical represen- 
 tation of the diversity or dependence of numerical quantities. In this 
 sense we cannot speak of the graphical represention of pure geometry. 
 This last was introduced into analysis by Vieta (1540-1603). Here 'the 
 figure merely aids the conception, while the equation embraces the charac- 
 teristics of the phenomena (I.), and ensures the independent character of 
 the drawn lines. Thus the clearness of geometry is combined with the 
 fruitf ulness of analysis. 
 
 If the graphical representation is constructed frcm a number of suitably 
 chosen and calculated values, the intermediate values can be directly meas- 
 ured and. by means of the scale, reconverted into numbers. The graphical 
 representation, then, replaces numerical tables. Illustrative examples often 
 occur in practice. We instance, for example, the graphical representation 
 of maximum moments and shearing forces in the continuous girder. If 
 the several values are calculated from a formula, their graphical union gives 
 a simultaneous view a picture of the law which the formula represents. 
 If these values are merely known observed, for example their graphical 
 combination may enable us to deduce the law which connects them. Thus 
 the graphical representation is of assistance in the deduction of empirical 
 formulae, and indirectly in the discovery of exact relations. Illustrations 
 of such application occur frequently in applied mathematics, especially in 
 astronomy and meteorology. 
 
 In this connection we may also remark that graphical representation 
 plays also an important part in statistics. By its aid a comprehensive view 
 is obtained of a series of separate results. Or it may be applied to still 
 higher problems for example, from comparison of simultaneous but differ- 
 ent series of observations to determine an inner connection. 
 
 In engineering practice, graphical representations have in recent times 
 notably multiplied. All graphical constructions, so far as they do not de- 
 pend upon analytical formulae, and therefore are not directly given by 
 geometrical laws, are nothing more than consequences of graphical repre- 
 sentation. 
 
INTROD.] GRAPHICAL STATICS. XXXV 
 
 VI. 
 
 GRAPHICAL STATICS. 
 
 The few text-books upon graphical statics and the more numerous works 
 upon its applications, afford us no definition, and can afford none, because 
 neither the method nor scope of this new study are anywhere sufficiently 
 indicated. 
 
 If, following Culmann, we speak of it in contradistinction to the appli- 
 cations of a pure graphical statics, we may define it somewhat as follows : 
 Graphical Statics comprises the theory of those geometrical constructions which 
 occur in the graphical solution of statical engineering problems ; it treats 
 further of the general relations deducible from such constructions. This 
 limitation, so far as it does not follow from the preceding, we shall seek 
 in the course of these remarks still further to establish. 
 
 Graphical representations of analytically obtained results have, as has 
 been already noticed, long been used in engineering practice. They served 
 also the purposes noticed in the preceding chapter. Often also certain 
 values, whose analytical determination is somewhat complicated, have 
 been sought by graphical constructions. Examples of this may be found 
 in many text-books upon the theory of structures, and we notice only, as 
 one of the most notable of recent date, the construction of lever arms and 
 limits of loading in A. Ritter's "Theorie und Berechnung eiserner Dach 
 und Briickenconstructionen " (Hannover, 1862). Poncdet applied analy- 
 sis in general to practical investigations, but sought in several complicated 
 cases to elucidate the deductions of formulae by geometrical constructions, 
 and to deduce graphical solutions from analytical relations. This pro- 
 cedure found considerable acceptance, and the investigations of Poncelet 
 were afterwards resumed upon more general assumptions by Saint Guil- 
 liem (Memoir e sur la poussee des terres avec on, sans surcharge, aim. des 
 ponts et chauss., 1858, sem. 1, p. 319). 
 
 The first, however, to give pure geometrical determinations of stability 
 in structures was Cousinery. He gave a number of examples as applica- 
 tions of his graphical calculus, but his ideas appear to have found in 
 France little acceptance. On the other hand, the graphical construction of 
 the curve of pressure in the arch by Mery (Memoir e sur V equilibre des voutes 
 en bwceau ann. d. ponts et chauss., 1840, sem. 1, p. 50) was extensively 
 used, and has since been extended by Durand-Claye to iron arches also 
 (Ann. d. ponts et chauss., 1867, sem. 1, p. 63, and 1868, sem. 1, p. 109). 
 Special prominence was given to graphical investigations of stability by 
 Culrnann's " Graphische Statik " (first part, Zurich, 1864, entire work, 
 1866 ; second edition, 1st part, 1875.) 
 
 This work of Culmann must be considered as original in all those parts 
 relating to structures. Poncelet and Cousinery, beyond the general idea, 
 furnished only unessential contributions. Culmann recognized the fruit- 
 
XXXvi GRAPHICAL STATICS. [iNTROD. 
 
 fulness of the relations between the force and equilibrium polygon, upon 
 which most of the practical solutions depend. He developed these rela- 
 tions, applied them in the theory of moments by the introduction of the 
 closing line (Schluss Linie), and, accepting the rule of signs, obtained gen- 
 eral points of view for the discussion of the most diverse figures which 
 could arise in the same problem. In this and in many other respects even 
 geometrical statics can profit from Culmann' s work, as, for instance, in the 
 investigation of the projective relations between the force and equilibrium 
 polygon. 
 
 The fundamental importance of the force and equilibrium polygon was 
 also recognized by those who, after Culmann, occupied themselves with 
 the graphical method. Here we may notice two works of special influence 
 upon the development of the graphical statics those of MoTir and Cre- 
 mona. The idea of Molir, that the elastic line is an equilibrium polygon or 
 curve (" Beitrag zur Theorie der Holz und Eisenconstructionen." Zeitschr. 
 d. Hannov. Ing. und Arch. Ver., 1868) is of special significance forgraphi. 
 cal statics. 
 
 That from it MoJir obtained the graphical determination of the moments 
 at the supports of a continuous girder, is an example both useful as well 
 as interesting. Already it has been endeavored to utilize the same idea in 
 other cases (Frankel : " zur Theorie der Elastischen Bogentrager," Zeitschr. 
 d. Hannov. Ing. u. Arch. Ver., 1869, p. 115), and by it an impulse has been 
 given to similar investigations. 
 
 Cremona, has kept more especially in view the geometrical side of graphi 
 cal statics. Starting from the theory of reciprocal polyhedrons, he gave 
 the reciprocal relations between the force and equilibrium polygon with a 
 generality and elegance to be expected from this distinguished Italian 
 mathematician (Le figure reciproche nelle statica graficu,. Milan, Langer, 
 1872). By this investigation the theoretical development of the graphical 
 statics is essentially anticipated. 
 
 It was under the most unfavorable circumstances that Culmann intro- 
 duced his graphical statics in the engineering department of the Ziirich 
 Polytechnic in the year 1860. It was finally, indeed, admitted as a regular 
 study, but not the geometry of position which he premised. It was not 
 till 1864 that this last was given in a series of lectures by Reye, and then 
 the time at disposition for both courses was insufficient. Meanwhile the 
 method spread, crept into the construction department of the engineering 
 school, and wherever it came, even in the other departments of the Poly- 
 technic, gained friends. Finally, at the present time, it is to be found, to- 
 gether with the modern geometry of position, upon which it was based, in 
 every Polytechnic throughout Germany. 
 
 According to the above given definition of graphical statics, the methods 
 of the graphical calculus, as far as applied in statical investigations, may 
 also be regarded as belonging to graphical statics, and justly so ; for 
 these methods follow directly from geometrical principles, and can be ap- 
 plied by any one acquainted with geometry, without being collected under 
 the special name of the " graphical calculus." Thus, for instance, Bausch- 
 inger, in his " Elemente der graphischen Statik" (Miinchen, 1871), disre- 
 
INTROD.] METHODS AND LIMITS OF GRAPHICAL STATICS. XXXvil 
 
 gards entirely the graphical calculus, and also cuts loose from the modern 
 geometry ; he develops the elementary principles of the subject in a logi- 
 cal and easily comprehended, if not purely geometrical manner, and thus 
 brings the subject within the reach of those persons for whom it seems so 
 especially designed. The work is remarkable for clear presentation, but 
 expressly avoids all special investigations and practical applications, for 
 which it is merely intended to prepare the way. In the present work, also, 
 a similar plan is pursued, but all such applications as are of most value to 
 the engineer or mechanic find likewise a place. Thus, combining the 
 method of presentation of Bauschinger and the practical applications of 
 (Julmann, it has been endeavored to make it a practical manual, as well as 
 a text-book of elementary principles to serve the wants of the practical 
 engineer, and also meet the requirements of the engineering student. How 
 far this twofold design has been realized, the judgment of the reader 
 must decide. 
 
 VII. 
 
 THE METHODS AND LIMITS OF THE GRAPHICAL STATICS. 
 
 The most perfect method of the graphical statics is the synthetic or geo- 
 metric; since in geometrical statics the solution must always, when possi- 
 ble, rest upon pure mechanical or geometrical reasoning. Culmann pre- 
 sents his graphical statics to practitioners "as an attempt to solve by the 
 aid of the modern geometry such problems pertaining to engineering prac- 
 tice as are susceptible of geometrical treatment." 
 
 The graphical statics, however, is not in and of itself the product of 
 endeavors to make the modern geometry of service in applied mechanics ; 
 graphical solutions merely were required. How to obtain these, was 
 another question. Thus it is that Poncelet's solutions consist almost en- 
 tirely of graphical representations of analytical relations ; that Cousinery 
 avoided all use of formulae ; that Culmann made use of the new geometry 
 wherever it was possible ; that Bauschinger and others make use only of 
 the ancient geometry ; and that the latest graphical solutions in a certain 
 degree, those of Mohr also entirely in the spirit of Fencelet's, rest again 
 upon analysis. The pure geometric solution is, indeed, desirable, but is not 
 always attainable. 
 
 If now we review all the cases in which direct and exclusively geomet- 
 rical solutions are not possible, we see at once that this occurs when it is 
 required to make use of the physical properties of bodies, as elasticity, co- 
 hesion, etc. Why? The actual condition of a body after equilibrium 18 
 attained, is a consequence of the motion of a variable system of points. 
 The theory of the motion of variable systems has, however, by no means, as 
 yet, been brought to practical efficiency (II.). We are therefore obliged to 
 start from an hypothetical condition or state of the body (in the theory of 
 flexure, for instance, we rest upon the assumption that all plane cross-sec- 
 tions made before the action of the outer forces remain plane after their 
 
XXXVlii METHODS AND LIMITS OF GRAPHICAL STATICS. [iNTEOD. 
 
 action). To deduce now from this general condition the special relations 
 necessary for solution, demands an essentially analytical process (T) 
 Hence the dependence of the graphical solutions in such cases upon ana- 
 lytical relations relations which, when the body is assumed to be rigid, 
 as in the arch, in frame work, or the simple girder, no longer exist. 
 
 The sphere of action of an independent graphical statics is, then, con- 
 fined to those problems which, under the assumption of inflexibility, are 
 determined by a sufficient number of conditions. Beyond this point we 
 have chiefly graphical interpretations only. 
 
 It has been already noticed that graphical statics, without the application 
 of algebraic operations, can furnish no general laws (IV.). From relatively 
 simple figures, indeed, here and there, general formula of metrical relations 
 have been derived, as is, in fact, not theoretically impossible (I.), but such 
 formulae were always previously known. Such a result holds, in general, 
 immediately good only for that form of figure which has been discussed, 
 Or, according to the terminology of Carnot, only for the existing " primi- 
 tive figure," and must be proved or transformed for all " correlative 
 figures " which can occur in accordance with the conditions of the prob- 
 lem. When the graphical investigation is guided by analytical opera- 
 tions, it is these last which render possible the deduction of general metri- 
 cal relations. 
 
 Thus, in the theory of structures, there remains subject to pure graphical 
 treatment only the general relations of form and position. Here we have 
 the elegant deductions upon unfavorable loading, and here the graphical 
 method often attains its end in a more elegant manner than the analytical. 
 A complete exploration and development of such form and place relations, 
 without a geometry of position, would evidently be impossible (IX.). The 
 scientific future of the graphical statics, therefore, rests essentially upon 
 the influence of the modern geometry. To endeavor to separate the higher 
 geometry from the graphical method would be as unwise and fruitless as 
 the attempt to exclude the higher analysis from analytical investigations. 
 As, however, for certain purposes an elementary presentation of analytical 
 theories relating to engineering practice will ever be acceptable, so also an 
 elementary development of graphical methods is not without justification, 
 the more so as long as the modern geometry itself is not sufficiently well 
 known. 
 
 Culmann says of the graphical statics : " It includes, thus far, only the 
 general part which we need in the investigation of problems in construc- 
 tion, but it must and will extend, as graphical methods find ever wider 
 acceptance. Then, however, it will escape the hands of the practitioner, 
 and must be built up by the geometer and mechanic to a symmetrical 
 whole, which shall bear the same relation to the new geometry that analyti- 
 cal mechanics does to the higher analysis." Such an estimation does not 
 appear to be entirely correct. It is geometrical statics (or mechanics) for 
 which the above relation may subsist, and to this, indeed, Culmann's valu- 
 able work has itself greatly contributed. It was, moreover, developed 
 quite independently of and much earlier than graphical statics (III.). In 
 this respect, therefore, the spread of graphical methods is of less inipor- 
 
INTROD.] THE MODERN GEOMETRY. 
 
 tance than that of geometrical views and knowledge ; for when practical 
 calculations are disregarded, and the deduction of general truths alone 
 occupies us, then, first of all, we must exclude from the drawn figure all 
 special relations that is, strike out of graphical statics the essentially 
 graphical part. A truth comprehended only in the abstract holds good 
 for all figures which can be drawn in accordance with the given condi- 
 tions. 
 
 We place, then, in one line geometry and geometrical statics (mechanics). 
 From geometry we obtain a method of construction, or descriptive geome- 
 try, which finds its practical applications in architecture and machine 
 drawing. From geometrical statics we obtain also a construction method 
 or routina viz., graphical statics which finds its practical applications in 
 the graphical calculation of structures and machines. Both descriptive 
 geometry and graphical statics have still, with reference to these practical 
 ends, to develop and make use of the general relations which subsist be- 
 tween the geometrical constructions to which they give rise, and thus each, 
 according to its means, contribute to the discovery and spread of geo- 
 metrical and mechanical truths. 
 
 From this co-ordination of descriptive geometry and graphical statics 
 we must not, however, infer an equal importance ; for, while in geometri- 
 cal drawing we have always to represent an ideal image, and the graphical 
 method is therefore directly suggested, we have for statical calculations 
 the analytical process also at our disposal, and everything depends then 
 upon the relative advantages and disadvantages of the graphical and ana- 
 lytical methods. We have thus noticed all the most important points 
 which occur in a theoretical consideration, and there only, remains to make 
 a comparison from a practical standpoint (X.). 
 
 VIII. 
 
 THE MODERN GEOMETRY. 
 
 Geometry treats of figures or constructions in space. These figures and 
 their properties are not always regarded and treated in equal extent and 
 generality. 
 
 Geometrical knowledge found its origin in practical needs, and the 
 ancients confined themselves almost exclusively to special investigations 
 of individual figures and bodies of definite form, such as presented them- 
 selves to the eye. In the pTiorisms of Euclid (-285), according to Pappus 
 (end of the fourth century), the mutual relations of the circle and straight 
 lines were, indeed, given with a certain degree of completeness, but these 
 have not come down to us. 
 
 Properties thus determined had naturally only a limited significance, 
 and could neither count upon permanence nor give satisfactory conclu- 
 sions. Investigators sought, therefore, assistance where it was best afforded, 
 in analysis. This was, in the sixteenth century, by the algebra of Vieta 
 (1540-1603), notably enriched. 
 
Xl THE MODERN GEOMETRY. [iNTROD. 
 
 From this period geometry, for a long time, served merely as an aid to 
 analysis, interpreting graphically its results (V.). From this union the 
 greatest advantages were derived, as analysis led to the infinitesimal cal- 
 culus of Newton and Leibnitz, and geometry to the analytical geometry of 
 Descartes (1596-1650). 
 
 But the extension and generality which geometrical truths received by 
 this great creation of Descartes was essentially due to analysis. Desaryucts 
 (1593-1662) and Pascal (1623-1662) extended pure geometrical considera- 
 tions, and made the first step towards the modern geometry when they 
 regarded the conic sections as projections of the circle, and deduced the 
 properties of the first from those of the last. Then De la Hire (1640-1718), 
 Le Poivre (1704) and Hnygens (1629-1695) occupied themselves with geo- 
 metrical investigations. While the two first developed the methods of 
 Desargues and Pascal, Huygens and, later, Newton (1642-1727) applied 
 pure geometry in optics and mechanics. Soon, however, the Calculus of 
 Newton and Leibnitz (1684 and 1687) showed itself so wonderfully fertile 
 in analytical geometry, that geometry proper was put in the background. 
 Only a few, as Lambert (1728-1777), still regarded it with favor. 
 
 Then appeared Monge (1728-1777), and gave the impulse to a complete 
 revolution in geometrical views, and to the reconstruction of the science 
 upon a new basis. In his Lemons de Geometric descriptive (Paris, 1788), all 
 those problems previously treated in a special and uncertain manner in 
 stereotomy, perspective, gnomonics, etc., were referred back to a few gen- 
 eral principles, and, without the aid of analysis, the most important prop- 
 erties of lines and surfaces were deduced. While descriptive geometry 
 taught the relations between bodies in space and drawn figures, it strength- 
 ened the power of abstraction ; introducing into geometry the transforma- 
 tion of figures, it gave to its deductions an advantage till then possessed 
 only by analysis; and while, finally, it owed its comprehensive results to 
 the application of projections, it pointed the way for the further develop- 
 ment of geometry itself. 
 
 Meanwhile, in the field of analytical geometry, the conclusion had been 
 reached that the desired truths admitted of a still more general compre- 
 hension. All properties had been obtained only with respect to and by 
 means of a determinate co-ordinate system. But already Godin (1704- 
 1760) had announced " que I" art de decouvrir les proprietes dss courses est 
 a proprement parler, I 1 art de changer le systeme de co-ordonnees" (Traite den 
 proprietes communes a toutes les courses). This idea Car not seized upon 
 (1753-1823), and in the sixth chapter of his Geometric de position (Paris, 
 1803) he sought to obtain a more general comprehension of figures by 
 analysis, and to avoid the indeterminancy of this last by the introduction 
 of the idea of position, and by many solutions after the method already 
 pointed out by Liebnitz and d'Alembert. 
 
 Now began a veritable race in the condensation and promulgation of 
 geometrical truths, in which the pure geometrical method obtained the 
 palm. The scholars of Monge Brianchon, Servois, Chasles, Poncelet, Ger- 
 gonne working with him and in his spirit, filled the Annales des mathe- 
 matiyues and the Correspondance sur Vecole polytechnique with new re- 
 
INTBOD.] THE MODERN GEOMETKY. xli 
 
 suits the two last named discovering the general law of reciprocity or du- 
 ality. 
 
 The foundation proper of the modern geometry was 
 his Traite des proprietes projectiles des figures (Paris, l828fi"Aggrandir les 
 resources de la simple Geometrie, en generaliser les conceptions et le langage or- 
 dinairement assez restreints, lesrapproclier deceuxde la Geometrie analytique, 
 et surtout offrir des moyens generaux, propres a demontrer et a fair e decouvrir, 
 d'une maniere facile, cette classe de proprietes dont jouissent les figures quand 
 on les considere d'une maniere purem'ent abstraite et independamment d'au- 
 cune grandeur absolue et determinee, tel est Vobjet qu'on s'est speciolemejit 
 propose dans cet ouvrage." 1 " 1 
 
 The new ideas found in Germany especially fruitful soil. Mobius, 
 Pliicker, Steiner, Grassman, and many others, proceeding in part from 
 entirely different points of view, opened out an abundance of new direc- 
 tions which have not yet been thoroughly explored, and which, in union 
 with other investigations, have caused a thorough change in our concep- 
 tions of space relations, whose latest phases are indicated by the names of 
 Riemann, Helmlioltz and Lie-klein. 
 
 In this development period, also, still existed the two parties in analyti- 
 cal and synthetic, or pure geometry. Plucker held the analytical relations 
 as the most general, and which were with advantage to be illustrated and 
 interpreted geometrically ; while Steiner recognized in the space figure 
 itself the true object and most efficient aid of investigation. Both direc- 
 tions the modern analytic and synthetic lead naturally to the same results. 
 "With reference to the methods, however, they diverge the nearer the ideas 
 and transformations of geometry approach the generality and ease of the 
 alg3braic method, thus rendering possible an abandonment of this last. 
 Thus, while analytical geometry, through the theory of determinants of 
 Hesse, came into ever closer connection with analysis a direction in which 
 English and Italian investigators as Salmon, Cayley, Cremona brilliantly 
 assisted, the Erlangen Professor von Staudt cut loose from algebraic formu- 
 la and metrical relations, and gave us the geometry of position (Nilrnberg, 
 1847, Beitr. z. Geom. d. Lage). 
 
 After von Staudt, the strict geometry of position remained a long time 
 disregarded, while the synthetic geometry of Steiner has enjoyed, without 
 intermission till the present day, a special preference on the part of mathe- 
 maticians. One reason may indeed be that mathematicians take little in- 
 terest in an independence of geometry to which analysis can lay no claim ; 
 but another, still more potent, is the extremely condensed, almost schematic 
 presentation of von Staudt, which has not exactly an encouraging effect 
 upon every one. 
 
 Culmann gave the impulse to a change in this respect. In his graphical 
 statics he rests directly upon the work of von Staudt, and, with something 
 more than boldness, assumes a knowledge of the geometry of position 
 among all practical men. Such a course was not indispensable for the 
 foundation of his method, and impeded the spread of the graphical stat- 
 ics ; but by it the geometry of position gained. This last had next, of 
 necessity, to be introduced into the Zurich Polytechnic, and thus' arose the 
 
MODERN GEOMETRY IN ENGINEERING PRACTICE. [iNTROD. 
 
 first, until now, only complete text-book upon the subject, the " Geometric 
 der Lage," by Reye (Hannover, 1868), as the direct result of the graphical 
 statics of Culmaun. 
 
 Since then, the modern geometry has been introduced into all technical 
 institutions throughout Germany, and thus placed at the disposal of the 
 arts and sciences. 
 
 As, according to its founder, Poncelet, it reaches the highest range of 
 speculation, so also in the most practical relations it acts to simplify and 
 condense : " Pen d peu les connaissances algebriques deviendront mains in- 
 dispensdbles, et la science, reduite d ce qu'elle doit etre, d ce quSelle devrait etre 
 dejd, sera aimi mise d la portee de cette classe d' homines, qui n'a que des mo- 
 ments fort rares d y consacrer." 
 
 [For illustrations of the method of the modern geometry, the reader may 
 consult the Appendix to this chapter.] 
 
 IX. 
 
 THE MODERN GEOMETRY IN ENGINEERING PRACTICE. 
 
 One who should infer that a science created thus from its very inception 
 with reference to the needs of practice* must have found access, above all, 
 in technical circles, would be much mistaken. As Culmann sent out his 
 graphical statics, deep silence prevailed, and if the modern geometry ap- 
 peared here and there in the lecture plan of one and another polytechnic, it 
 was, without doubt, due to the zeal of some enthusiastic privat docent who 
 had undertaken the thankless task of holding forth to empty benches. 
 
 Whence came this indifference to a discipline proceeding from the Ecole 
 polytechnique ? It is hard, indeed, to find a sufficient reason. We often 
 hear it said that by reason of the colossal extension which engineering 
 sciences have experienced, students are already overburdened. Most true ! 
 and it is just here that the modern geometry comes to our assistance. It 
 is precisely to this that the learned critic of Monge, Dupin, alludes : " II 
 (terrible que dans Vetat actuel des sciences mathematiques le seul moyen d'ern- 
 pecher que leur domaine ne devienne trop vaste pour notre intelligence, c'est 
 de generaliser de plus en plus les theories que ces sciences embrassent, afin 
 qu'un petit nombre des verites generates et fecondes soit dans la tete des 
 Tiommes I 1 expression abregee de la plus grande variete des f aits particuliers" 
 
 The modern geometry in its present form starts with a small number of 
 elementary constructions whose properties are first set forth, and then, pro- 
 ceeding from these by combination and comparison, it covers the entire 
 department of space. The engineer, during and after his preparation, has 
 to do with space problems, with geometrical principles and constructions ; 
 
 * Poncelet himself set upon the title-page of his work : " Ouvrage utile d 
 ceux qui s^occupent des applications de la Geometric descriptive et d' operations 
 geometriques sur le terrain." 
 
INTBOD.] PRACTICAL SIGNIFICANCE OF GEAPHICAL STATICS. xliii 
 
 "how many superfluous definitions and demonstrations could not be 
 spared, if they were already completely comprehended and recognized by 
 the scholar as parts of a higher whole" (Gulmann "Die Graphische 
 Statik "). At no very distant day it will no* longer be possible to read a 
 scientific work upon applied mathematics without familiarity with- the 
 principles of the modern geometry.* Permitting pure graphical applica- 
 tions, without the aid of analytic symbols, it forms the common point of 
 view for descriptive geometry, practical geometry, and graphical statics. 
 
 Descriptive geometry existed before the modern, and this last has sprung 
 from it. Now, reversely, the geometry of position comes to the aid of 
 descriptive geometry, and offers in return its most fruitful principles and 
 efficient aid. Thus in descriptive geometry we may refer to the works of 
 Pohlke, tichlesinger, and Fiedler. The effect of the geometry of position in 
 this direction to simplify and condense may be seen from the work of 
 Staudigl ("Ueber die Identitat von Constructionen in perspective, schiefer 
 und orthogonaler Projection' 1 ), where it is proved that "all problems of 
 the descriptive geometry, in which neither linear nor angular measure are 
 considered therefore all problems which belong to the geometry of posi- 
 tion can in similar manner and by precisely similar constructions be solved 
 as well in perspective as in oblique and orthagonal projection." In shades 
 and shadows and in geometrical drawing, Burmeister and Paulus owe to 
 the modern geometry the simplicity of their constructions. 
 
 In the department of practical geometry also, in geodesy, perspective, 
 surveying, we mark the influence of the modern geometry in the works of 
 Mutter and Spangeriberg, of Franke and Baur. 
 
 In mechanics and physics, we see it again in the works of Lindemann, 
 Burmeiater and Zech. 
 
 PEACTICAL SIGNIFICANCE OF THE GBAPHICAL STATICS. 
 
 We have already remarked (VII.) that the importance of graphical 
 statics is in great part dependent upon its advantages as compared with 
 the analytical method, and have reserved for this place a comparison from 
 a practical point of view. 
 
 Here, first of all, we have to notice the independence of the graphical 
 construction of the regularity or irregularity of the given relations. 
 Whether the forces are equal or not, whether they act at equal or varying 
 distances, even their relative position, are matters of indifference. Centre 
 of gravity, central ellipse, kernel for all, even the most irregular figures, 
 are found in similar manner, with equal ease, even when exact analytical 
 solutions are hardly conceivable. Thus a process, a routine almost 
 mechanical is rendered possible in many investigations of stability, with- 
 out losing sight of interior relations ; for in the repeated and independent 
 compositions of the forces we always perceive the origin, connection and 
 
 * Well illustrated in Gillespie's Land Surveying. New York, 1870. 
 
PRACTICAL SIGNIFICANCE OF GRAPHICAL STATICS. [iNTKCD. 
 
 reason of the result obtained, which, in the substitution of numbers in a 
 formula, is not always the case. 
 
 With this advantage goes hand in hand a disadvantage. This very 
 regularity of the process is a consequence of its special, we might almost 
 say numerical, character (I.). In a numerical analytical example greater or 
 less regularity has also but little effect. This numerical character has also 
 for consequence that we can never attain to general laws and relations 
 (IV., VII). 
 
 The practical engineer becomes with time ever more familiar with the 
 dividers and rule, while facility in analytical operations gradually disap- 
 pears. A graphical construction once completed is not easily forgotten, or 
 a single glance at a similar figure suffices to recall the whole process. It is 
 indeed easy in clearly given formulae to substitute special numerical values ; 
 but formulas unfortunately are not always clearly given, in some cases can- 
 not be so given, without presuming upon the thorough familiarity of the 
 reader with the processes involved ; these and the very many and various 
 systems of notation in use leave to the constant, easily acquired and 
 remembered graphical solution many advantages. 
 
 But here we may remark that graphical solutions can only be easily 
 acquired, retained or quickly recovered when the constructions are based 
 upon methods purely geometric, and not when they are simply the interpre- 
 tation of previously obtained analytical results. In the latter case we 
 must recall the process of development of the formula as well as the 
 graphical construction, and the method is thus too often confusing instead 
 of simple. 
 
 Often it is desired to make visible the results of an investigation, as in 
 the case of the arch, where the graphical method is especially advan- 
 tageous, and has in France been long used (VII.). 
 
 Errors relating to the mutual relation of strains are more easily discov- 
 ered in graphical solutions than in analytical, as a certain law of regularity 
 is always visible, which breaks abruptly for an error in construction. By 
 calculation, on the other hand, we can more easily select any one place in 
 the structure, and determine the strain there independently of the others. 
 
 As to which of the two methods demands the least time is a matter of 
 minor importance. In a construction costing from thousands to millions, 
 it matters little whether the calculations require one or several days, more 
 or less, if only the results are clear and correct. It is a question also 
 which can hardly be decided in favor of one or the other, dependent as it 
 is upon elements other than those pertaining to the methods themselves 
 such as varying individual skill and capacity in either direction. The 
 declaration which is already sometimes encountered, that the numerical 
 calculation of a continuous girder requires about three times as much time 
 as the graphical solution, sounds questionable. Why not at once furnish 
 the statement with decimal places f In general, for ordinary cases, the ana- 
 lytical solution requires less time ; for irregular and more complicated cases, 
 the graphical. 
 
 The exactness of the graphical solution is sufficient, but it, too, depends 
 upon the care and skill of the draughtsman. The greater the forces and 
 
IXTEOD.] LITERATURE UPON GRAPHICAL STATICS. xlv 
 
 dimensions with which one works, the better the results obtained. The 
 scales should not, then, be taken too small. 
 
 It is hoped that these considerations, now drawing to a close, will suffice 
 to give the reader clear ideas upon the nature and origin, advantages and 
 disadvantages, of the graphical statics. The determination whether he will 
 enter more fully into the subject it may be, even take part in its develop- 
 ment (there is abundance of room for workers), and in this case the choice 
 of direction may thus be facilitated. 
 
 The graphical statics is certainly suited, especially in extended applica- 
 tions of the geometry of position, to furnish many new points of view, and 
 in a practical respect it can often greatly simplify. Whoever has really 
 studied the new methods must admit this. 
 
 On the other hand, the importance of the graphical statics is sometimes 
 exaggerated. It appears out of place when in works designed for practice 
 graphical solutions are given of problems which any reasoning being can 
 almost solve in his head. 
 
 Such solutions may find a place in special text-books upon the subject, 
 where they may, indeed, be desirable for completeness. 
 
 If it is desired to make two independent investigations of stability, as 
 for large and important constructions is always desirable, it will be found 
 of advantage, if a suitable graphical solution exists, to make the first deter- 
 mination graphically. Nothing more ensures a conviction of the correct- 
 ness of an investigation than a correspondence of the graphical and cal- 
 culated results. 
 
 XI. 
 
 LITERATURE UPON GRAPHICAL STATICS. 
 
 We have already referred in VI. to the most important contributions in 
 the branch of graphical statics, and now annex a list of the literature upon 
 the subject so far as known to us. 
 
 Where several works treat of the same subject, we have allowed ourselves 
 a brief critical notice. Opportunity is thus given to those who would take 
 part in the development of graphical statics to make themselves acquainted 
 with all existing works, and at the same time the practical man is enabled 
 in any case that may come up to inform himself as to where assistance 
 may best be sought. A short remark to specify the contents may in this 
 respect often help in the right direction. The succession is in each division 
 chronologically arranged 
 
 Although the literature of the subject would seem from the following 
 tolerably extensive, still the number of pure geometrical solutions in 
 which no analytical formulae appear is much less. Publications upon the 
 subject would, moreover, beyond doubt, be still more numerous were it 
 not for the difficulty and cost of production of lithograph plates. 
 
LITEKATUliE UPON GRAPHICAL STATICS. 
 
 I. TEXT-BOOKS UPON GRAPHICAL STATICS. 
 
 Culmann, K." Die graphische Statik." With Atlas of 36 Plates. Zurich, 
 Meyer and Zeller, 1866. [I. Part, 1864: Elements and Graphical 
 Investigations of Structures. Also, second edition, first volume, 
 1875, with 17 Plates. General Principles, second volume, to follow 
 shortly.] 
 
 Bauschinger "Elemente der graphischen Statik." With Atlas of 20 
 Plates. Miinchen, 1871. [Without the aid of modern geometry, and 
 without practical applications. Admirable exposition of the Princi- 
 ples.] 
 
 lleuleaux. An outline of the graphical statics is to be found in " Der Con- 
 structeur," by Reuleaux, third ed. Braunschweig, 1872. 
 
 Levy " La Statique Graphique et ses Applications." Paris, 1874. With 
 Atlas of 24 Plates. [Principles and several applications; clear and 
 elegant exposition of the subject.] 
 
 n. PAPERS UPON THE GRAPHICAL STATICS. 
 
 Most " Ueber eine allgemeine Methode, geometrisch den Schwerpunkt 
 beliebiger Polygone und Polyeder zu bestimmen." Archiv d. Math, 
 und Phys., IL. (1869), p. 355. [Also applicable to curve areas, with- 
 out equilibrium polygon.] 
 
 Culmann, K. "Ueber das Parallelogram und iiber die Znsammensetzung 
 der Krafte." Vierteljahrsschr. d. Naturforsch. Ges. zu Zurich, 1870. 
 [Correspondence of the graphical statics with the Statics of Pliicker.] 
 
 Mohr " Bsitrag zur Theorie der Holz- und Eisenconstructionen." Zeitschr. 
 d. Hannov. Arch. u. Ing. Ver., 1870, p. 41. [Relation between the 
 neutral axis and centre of strains.] 
 
 Grunert, J..A. "Ueber eine Graphische Methode zur Bestimmung des 
 Schwerpunktes eines beliebigen Vierecks." Arch. d. Math. u. Phys., 
 LII. (1871), p. 494. [Simple and brief. Compare also L., p. 212.] 
 
 Cremona, B. " Le figure reciproche nelle statica grafica." With 5 Plates. 
 Milan, 1872. German translation in Zeitschr. d. Ost. Arch. u. Ing. 
 Ver., 1873, p. 230. [Force and equilibrium polygon as reciprocal 
 figures.] 
 
 Du Bois, A. J. " The New Method of Graphical Statics." Van Nostrand's 
 "Engineering Magazine," Vol. XII., Nos. 74, 75, 76, 77, 78. [General 
 properties of force and equilibrium polygons, with practical applica- 
 tions to bending moments, and several important mechanical problems. 
 Also, Maxwell's Method applied to bridges, roof trusses, etc.] Sepa- 
 rate reprint, 1875. Van Nostrand, New York. 
 
LITERATURE UPON GRAPHICAL STATICS. 
 
 III. APPLICATION TO THE SIMPLE GIRDER. 
 
 Culmann, K. "Der Balken." Third chap, of d. graph. Statik, 1866. 
 [Contains also the construction of the inner forces.] I 
 
 Vojdcek " Graphische Bestimmung der Biegungsmomente an kurzen 
 Tragern." Zeitschr. d. Vereins Deutsch. Ing., 1868, p. 503. [Graphi- 
 cal interpretation of analytical relations.] 
 
 Cotterill, J. H. "On the Graphic Construction of Bending Moments." 
 "Engineering," 1869 (VII.), p. 32. [Equilibrium polygon for the sim- 
 ple truss, with references to Rouleaux and Culmann.] 
 
 Winkler, E. "Einfache Trager," "Theorie der Brticken," "Aeussere 
 Krafte gerade Trager." Wien, 1872. [Simultaneous presentation of 
 analytical and graphical methods.] 
 
 Ott, K. von "Wirkung paralleler Krafte auf einfache Trager mit Gerade 
 Langenachse." In die Grundziige d. graph. Reclmens u. d. graph. 
 Statik. Prag, 1872. p. 28. [The most elementary principles pertain- 
 ing to composition of forces in a plane are prefaced.] 
 
 IV. APPLICATION TO THE CONTINUOUS GIRDER, 
 
 Culmann, K. "Der continuirliche Balken." Fourth chap, of the Graph. 
 Statik, 1866. [With examples the moments at the supports are 
 analytically determined.] 
 
 MoJir " Beitrag zur Theorie der Holz- und Eisehconstructionen." Zeitschr. 
 d. hannov. Arch. u. Ing. Ver., 1868, p. 19. [Completion of Culmann' s 
 method the moments at the supports are graphically determined.] 
 
 Lippich "Theorie des continuirlichen Tragers Constanten Querschnitts." 
 Wien, 1871. Separate reprint from Forster's Bauzeit., 1871, p. 103. 
 [Graphical method, together with elementary analytical.] 
 
 Hitter, W. " Die elastische Linie und ihre Anwendung auf den continuir- 
 lichen Balken." Zurich, 1871. [Mohr's method given as a supple- 
 ment to the Graph. Statik of Culmann.] 
 
 Winkler, E. " Continuirliche Trager. Theorie der Bracken aeussere 
 Krafte gerade Trager." Wien, 1872. [The Mohr-Culmann method, 
 together with analytical.] 
 
 Solin, J. " Geometrische Theorie der continuirlichen Trager." Mitth. d. 
 Arch. u. Ing. Ver. in Bohmen, 1873. 
 
 Greene, Chas. E. " Graphical method for the analysis of Bridge Trusses ; 
 extended to Continuous Girders and Draw Spans." New York, 1875. 
 [Moments at supports found by successive approximation, or balancing 
 of moment areas.] 
 
 V. APPLICATION TO FRAME WORK. 
 
 Culmann, K." Das Fachwerk." Fifth chap. Graph. Statik, 1866. [Most 
 general form of parallel truss, suspension truss, Pauli's truss, roof 
 trusses.] 
 
LITERATURE UPON GRAPHICAL STATICS. 
 
 Keck, W. " Ueber die Erinittclung der Spannungen in Fachwerk tragern 
 mit Hulfe der graphischen Statik." Zeitsclir. d. hannov. Arch. u. Ing. 
 Ver., 1870, p. 153. Separate reprint, Hannover, 1872. [Presentation 
 of the method with reference to practice.] 
 
 Jenkin " On the Practical Application of Reciprocal Figures to the Cal- 
 culation of Strains in Frame-work. Transact, of the R. Soc. of Edin- 
 burgh, 1870, (XXV.) p. 441. 
 
 Maxwell, Prof. Cleric " Reciprocal Figures, Frames, and Diagrams of 
 Forces." Trans, of R. Soc. of Edinburgh, 1869-70. 
 
 Uhwin " Iron Bridges and Roofs.' 1 London, 1869. [Application to roof 
 trusses, wind force, etc.] 
 
 Banfon,F.A."The Strains in Trusses." New York, Appleton, 1872. 
 [Examples of simple trusses drawn to scale.] 
 
 Bow, Robert H. "Economics of Construction in Relation to Framed Struc- 
 tures." London, 1873. [Application of Maxwell's Method only to 
 roof trusses, etc.] 
 
 Ott, K. von " Das Fachwerk." In Grundzuge d. graph. Rechnens u. d. 
 graph. Statik. Prag, 1872. [Roof trusses, truss fixed at one end and 
 free at the other, bridge trusses.] 
 
 Reuleaux " Hilfslehren aus der Grapho statik." Second chap, of the Con- 
 structeur, third ed., 1872. [Compound truss, roof trusses, etc.] 
 
 Scliaffer "Graphische Ermittelung der Ordinaten des Schwedler'schen 
 Tragers." Zeitschr. fur Bauwesen, 1873, p. 237. [Proceeding from 
 the equation for the same.] 
 
 Heuser "Graphische Ermittelung der Ordinaten des Schwedler'schen 
 Tragers." Zeitschr. f. Bauwesen, 1873, p. 523. [Preceding method 
 simplified another by means of equilibrium polygon.] 
 
 VI. APPLICATION TO THE IRON ARCH. 
 
 Culmann, K." DerBogen." Sixth chap, der graph. Statik, 1866. [Con- 
 tains also the inverted or suspended arch. The arch as a rigid body.] 
 
 Durand-Claye, A. " Sur la verification de la stabilite des arcs m&alliqucs 
 et sur 1'emploi des courbes de pression." Ann. d. ponts et chauss., 
 1868, sem. 1, p. 109. [Mery-Durand pressure curves, but with refer- 
 ence to the absolute resistance of the material.] 
 
 Frankel, W. " Zur Theorie der elastischen Bogentrager." Zeitschr. d. han- 
 nov. Arch. u. Ing. Ver., 1869, p. 115. [Following out Mohr's idea of 
 the equilibrium polygon as elastic line. ] 
 
 MoJir " Beitrag zur Theorie der elastischen Bogentrager." Zeitschr. d. 
 hannov. Arch. u. Ing. Ver., 1870, p. 389. [Criticism of the preceding 
 method, and giving another.] 
 
 Beitrage zur graphischen Berechming elastischer Bogentrager mit 
 Kampfergelenken." Mitth. d. Arch. u. Ing. Ver., in Bohmen, 1873. 
 
LITERATURE UPON GRAPHICAL STATICS. 
 
 VII. APPLICATION TO THE ARCH. 
 
 Cousinery, E. B. " Application des proce'de's du calcul graphique a" divers 
 problemes de stability." Fourth chap, of Calcul par le Trait. Paris, 
 1839. [With especial reference to the strength of abutments pure 
 graphical treatment.] 
 
 M&ry " MSmoire sur I'Squilibre des voutes en berceau." Ann. d. ponts 
 et chauss., 1840, sem. 1, p. 50. [Geometrical determination of every 
 possible pressure curve.] 
 
 Culmann, K. " Der Bogen.". Sixth chap, of Graph. Statik, 1866. [Con- 
 taining also arch centerings ; exact discrimination of support and 
 pressure line.] 
 
 Durand-Claye, A. "Sur la verification de la stability des voutes en 
 maconnerie et sur 1'einploi des courbes de pression." Ann. d. ponts 
 et chauss., 1867, sem. 1, p. 63. [Reference to relative resistance of 
 material. ] 
 
 Harlacher, A. R. "Die Stiitzlinie im Gewolbe." Tech. Blatter, 1870, p 
 49. [Practical method by inscription of support line, according to 
 Culmann.] 
 
 Heuser "Zur Stabilitatsuntersuchung der Gewolbe." Deutsche Bauzeit. 
 1872, p. 365. [Also methods for unsymmetrical form and load.] 
 
 VHI. APPLICATION TO RETAINING WALLS. 
 
 Poncelet. J. V. " Me'moire sur la stabilite des reve'tements et leur fonda- 
 tion." Mem. de 1'off. du Ggnie, 1838 (XIII.) ; separate reprint, 
 Paris, 1840. [First analytical graphical theorie.] 
 
 Cousinery, E. B. " Application des precedes du calcul graphique & divers 
 problemes de stability." Hourth chap, of " Calcul par le Trait," 1839. 
 [Pure graphical, without formulae.] 
 
 ^aint-G-uilhem " MSmoire sur la poussSe des terres avec ou sans sur- 
 charge." Ann. d. ponts et chauss., 1858, sem. 1, p. 319. [Further 
 development and generalization of Poncelet's Theory.] 
 
 Rarikine "Manual of Civil Engineering." London, fourth ed., 1865. 
 [Containing graphical construction of pressure parallel to earth sur- 
 face upon vertical wall.] i 
 
 Culmann, K. " Theorie der Stutz- und Futter-Mauern." Eighth chap, of 
 Graph. Statik, 1866. [With use of equilibrium polygon, pressure 
 upon tunnel arches.] 
 
 JMzhey, E. " Beitrage zur Theorie des Erddrucks und graphische Bestim- 
 mung der Sta'rke von Futter-Mauern." Mitth. iiber Gegenst. d. Artill. 
 und Geniewesens; separate reprint, with two plates, Wien, 1871. 
 [Point of application of earth pressure for complicated contour.] 
 
 MoJir "Beitrage zur Theorie des Erddrucks." Zeitschr. des hannov. 
 Arch. u. Ing. Ver., 1871, p. 344. [Point of application of earth pres- 
 sure and new analytical theory.] 
 
1 GRAPHICAL DYNAMICS. 
 
 WinTder, K-" Neue Theorie des Erddrucks." Wien, 1872. [Containing 
 graphical methods according to the old theory.] 
 
 Haseler, 0. " Beitrage zur Theorie der Futter- tmd Stiitz-Mauern." Zeit- 
 schr. d. hannov. Arch. u. Ing. Ver., 1873, p. 36. [Graphical deter- 
 mination of earth pressure according to Culmann.] 
 
 MISCELLANEOUS APPLICATIONS. 
 
 Reuleaux" Die graphische Statik der Axen und Wellen." Published by 
 
 polytech. Ver. in Zurich, 1863. [Autograph copy of lectures.] 
 Culmann, K. " Der Werth der Constructionen." Seventh chap, of Graph. 
 
 Statik, 1866. [Best and cheapest systems under given conditions, 
 
 especially for bridges.] 
 Reuleaux " Graphostatische Berechnung verschiedener Axen, Kranpfosten, 
 
 Kurbeln," in the Constructeur, third ed., 1872. 
 
 Scattering graphostatical constructions are to be met with in many text- 
 books upon construction, especially since the appearance of Culmann's 
 work, a second edition of which is in course of preparation, and expected 
 soon to appear. 
 
 XII. 
 
 GRAPHICAL DYNAMICS. 
 
 The scientific or practical value of graphical solutions once recognized, 
 there remains no reason for limiting them to statical problems only, and 
 endeavors in the above direction are already forthcoming. We limit our- 
 selves to a passing notice. 
 
 First, we have an attempt to employ graphical constructions in the 
 theory of the overshot and breast- wheel (Qeeberger, " Arbeitung der Theo- 
 rie der oberschlachtigen Wasserrader auf graphischen Wege." Civil Ing. 
 1869, p. 398, and 1870, p. 339). We cannot here notice the value of the 
 solutions given, but the very sparing applications of geometry hardly jus- 
 tify the title of the work. 
 
 A short article, which gives the graphical determination of the force at 
 every position of a moving point, may also be noticed. (Rapp, " Zur 
 graphischen Phoronomie," in Zeitsch. f. Math. u. Phys., 1872, p. 19.) 
 
 The genuine foundation of a graphical dynamics has been first attempted 
 by Proll (" Begriindung graphischer Methoden zur Losung dynamische 
 Probleme," in Civil Ingenieur, 1873). From the fact that the effects of 
 forces in dynamics are measured by the changes of velocity of any point or 
 points of a machine system, Proll concluded that it must be possible to 
 represent these force effects by geometrical relations, such as kinematic 
 geometry teaches. 
 
 His investigations, since published in independent form ("Versuch 
 einer graphischen Dynamic," with 10 plates, 1874), fall into thr.ee parts. 
 The first part treats of the action of the " outer forces " in machines whose 
 
GRAPHICAL DYNAMICS. 11 
 
 motion is in a plane, the outer forces being also in this plane. In the sec- 
 ond part he subjects to graphical treatment the action of outer forces upon 
 a free movable material point. The third part, finally, considers the motion^ 
 of rigid- invariable systems acted upon by given forces. 
 
 In the course of the development extended use is made of analytical 
 formulae. The work is but the beginning of the future structure, but this 
 beginning will be thankfully received by all those with whom graphical 
 methods have found acceptance. 
 
PART I. 
 
 GENERAL PRINCIPLES. 
 
 CHAPTEE I. 
 
 FORCES IN THE SAME PLANE COMMON POINT OF APPLICATION. 
 
 1. Notation, etc. In order that a force may be " given " or 
 completely determined in its relations to other forces, we must 
 know not only its intensity, but also its direction, and the posi- 
 tion of its, point of application. These three being known, the 
 geometrical expression of our knowledge is very simple. We 
 have only to assume a certain length as the unit of force, and 
 then any force is at once given by the length, direction, and 
 position of a straight line. This method of force representa- 
 tion is so obvious, that it is in fact used in mechanics, even 
 where the treatment itself is essentially analytical. 
 
 Unless expressly stated, all the forces with which we have to 
 do, will be considered as lying and acting in the same plane. 
 Graphically then, any force is completely determined by a 
 straight line, the beginning of which represents the point of 
 application, and the length and direction of which give the in- 
 tensity and direction of the force. 
 
 We shall indicate a force in general by the letter P, its point 
 of application by A. When we have several forces we repre- 
 sent the points of application by A 1? A 2 , A 3 , etc., and the ends 
 of the corresponding lines by P 1? P 2 , P 3 , etc. The direction in 
 which a force is supposed to act is thus unmistakably indi- 
 cated. 
 
 When, however, lines representing several forces are laid off 
 one after another, the beginning of each at the end of the pre- 
 ceding, it will be sufficient to put at the beginning of the 
 first, and 1, 2, 3, etc., at the end of each. No confusion can 
 arise, as each force acts and reaches from the point indicated 
 by the figure which is one less than its index, to the point indi- 
 cated by that index. 
 
 When, finally, we designate a force by the two letters or fig- 
 ures which stand at the beginning and end, we shall always 
 indicate by the order in which the letters or figures are written, 
 
2 FORCES IN THE SAME PLANE. [CHAP. I. 
 
 the direction of action of the force, first naming the point of 
 application, and then the end. 
 
 A force due to the composition of several forces, as P l5 P 2 , P 3 , 
 we denote by Pi. 3 or R^. Thus R^ denotes the resultant of 
 the forces P l5 P 2 , and P 8 . 
 
 2. Parallelogram of Force. If two forces, P^ and P 2 , 
 given in direction and intensify by the lines OP X OP 2 [Fig. 1, 
 PL 1], have a common point of application O, the resultant 
 Ri.2 is found by the well known principle of the " parallelo- 
 gram of forces," by completing the parallelogram as indicated 
 by the dotted lines, and drawing the diagonal. OR then gives 
 the resultant of the forces P l and P 2 . If this resultant acts in 
 the direction from O to R, as indicated by the arrow, it replaces 
 P! and P 2 ; that is, it produces the same effect as both forces 
 acting together. If it were taken as acting in the opposite 
 direction i.e., from O outwards, away from R it would hold 
 the forces p t and P 2 in equilibrium. 
 
 Now, we see at once that it is unnecessary to complete the 
 parallelogram. It is sufficient to draw from the end of the 
 force P 2 the line P 2 R in the same direction that P l acts in, and 
 make it equal and parallel to P lf The point R thus found is 
 the end of the resultant R, or is a point upon the direction of 
 the resultant prolonged through O. 
 
 As to the direction of action of the resultant if we follow 
 round the triangle from O to P 2 and from P 2 to R and R to O 
 i.e., if we follow round in the direction of the forces the 
 direction for the resultant from R to O thus obtained is, as we 
 have already seen, the direction necessary for equilibrium. 
 
 3. If, instead of two forces, we have three or more, as P 1? P 2 , 
 P 3 , P 4 [Fig. 2] we still have the same construction. Thus com- 
 pleting the parallelogram for P x and P 2 we find R^. Complet- 
 ing the parallelogram for R^ and P 8 , we find R M , and again, 
 with this and P 4 we obtain R w . Again, we see it is unneces- 
 sary to complete all the parallelograms. We have only to draw 
 lines P! Rj.2, Ri_ 2 R^, Ri_ 3 R M , parallel to the forces P 2 P 3 and 
 P 4 respectively, and equal in length to the intensities of these 
 forces, and then, no matter what may be the number of forces, 
 the line drawn from the point of beginning to the end of the 
 last line laid off will give the intensity and position of the 
 resultant. As to direction, the same holds good as before. 
 
 If the end of the last line laid off as above, should coincide 
 
CHAP. I.] COMMON POINT OF APPLICATION. 3 
 
 with the point of beginning, there is, of course, no resultant, 
 and the forces themselves are in equilibrium. 
 
 4. The polygon formed by the successive laying off of the 
 lines parallel and equal to the forces, we call the "force poly- 
 gon" Hence we have the following principles established : 
 
 If any number of forces having a common point of appli- 
 cation and lying in the same plane, a,re in equilibrium, the 
 " force polygon" is closed. 
 
 If the "force polygon " is not closed, the forces themselves 
 are not in equilibrium, and the line necessary to close it gives 
 the resultant in intensity and direction. 
 
 This resultant, if considered as acting in the direction ob- 
 tained ~by following round the " force polygon " with the forces, 
 will produce equilibrium acting in the opposite direction, it 
 replaces the forces. 
 
 The resultant thus found in intensity arid direction can be 
 inserted in the force diagram at the common point of applica- 
 tion. 
 
 5. Thus, required the position, intensity, and direction of the 
 resultant of the forces P 1? P 2 , P 3 , P 4 , P 5 . 
 
 These forces are given in position, direction, and intensity 
 by the force diagram, Fig. 3 (a). The resultant of all these 
 forces must have of course the same point of application A as 
 the forces themselves it remains to find then its relative posi- 
 tion and the direction of its action, so that we may properly 
 insert it in the force diagram. 
 
 We have simply to draw the force polygon, Fig. 3, (b) by lay- 
 ing off successively O P 1? P t P 2 , etc., equal, parallel, and in the 
 same direction as the forces P 1? P 2 , etc., as given by Fig. 3 (a). 
 Then the line P 5 O necessary to close the force polygon gives 
 the intensity of the resultant, and in order to replace P^ it 
 must act in the direction from O to P 5 ; i.e., contrary to the 
 order of the forces. If then in Fig. 3 (a) we draw A R w equal 
 and parallel to O P 5 , we have the resultant applied at the com- 
 mon point of application A, and given in position, intensity 
 and direction. 
 
 Moreover, it is evident that any diagonal of the force poly- 
 gon as R^ [Fig. 3 (b)] is the resultant of P 8 ^, and acting in the 
 direction from P 4 to P 2 , it holds P^ in equilibrium. But it is 
 also the resultant of Pj, P 2 , P 5 , and R^, and acting in the same 
 direction as before, it replaces these forces. The force polygon 
 
FORCES IN THE SAME PLANE. [CHAP. I. 
 
 thus shows that the force which replaces P 1? P 2 , P 5 , and R w , at 
 the same time holds P 3 and P 4 in equilibrium, just as it should 
 do. 
 
 If, on the other hand, we had originally only P 1? P 2 , R^, P 5 , 
 and RLS forming a system of forces in equilibrium, we could 
 decompose Rg^ into two components by simply assuming any 
 point as P 3 [Fig. 3 ()] and drawing P 3 P 4 , P 3 P 2 . Then follow- 
 ing round this new polygon in. the direction of the forces, or, 
 what amounts to the same thing, taking the direction of the 
 components P 3 P 4 , opposed to the direction of R 34 for equilibri- 
 um, we obtain the direction of action of P 3 and P 4 as shown by 
 the arrows in Fig. 3 (b). These forces inserted in Fig. 3 (a), in 
 the place of R 3 . 4 and in these directions, will not disturb the 
 equilibrium. 
 
 Hence, any diagonal in the force polygon, is the resultant 
 of the forces on either side, holding in equilibrium those on 
 one side and replacing those on the other, according to the 
 direction in which it is conceived to act. 
 
 Also, any force or number of forces may be decomposed into 
 two others in any desired direction, ~by choosing a suitable 
 point in the plane of the force polygon and drawing lines 
 from this point to the beginning and end of the force or force 
 polygon. 
 
 6. It matter not in what Order we lay off the Forees in 
 the Contraction of the force Polygon. Thus, in Fig. 1, 
 whether we draw from the end of P 2 the line P 2 R^ equal and 
 parallel to P! or from the end of P x the line P! R^ equal and 
 parallel to P 2 , in either case we obtain the same resultant and 
 the same direction for the resultant. But by a similar change 
 of two and two, we can obtain any order we please. For exam- 
 ple, we lay off in Fig. 3 (c) the same forces in the order P 3 P 2 
 PI, P 5 P 4 , and obtain precisely the same resultant, in the same 
 direction as before. For, the resultant of P 3 and P 2 must be 
 the same as that of P 2 and P 8 in the first case. The resultant 
 of R 3 . 2 and P t must then be the same in both polygons, and so 
 on. 
 
 Generally, then, no matter what the order in which the 
 
 forces are laid off, the line necessary to close the force polygon 
 
 is the resultant of the forces, and the diagonals of the force 
 
 polygon give us the resultants of the forces on either side. 
 
 By assuming a point at pleasure, and drawing lines from this 
 
CHAP. I.] COMMON POINT OF APPLICATION. 5 
 
 point to the beginning and end of any side of the force poly- 
 gon, and taking the direction of these lines opposed to the 
 direction of that side, we can decompose any force in the force 
 polygon into its components. Thus the force polygon gives ns 
 complete information as to the action of the forces. 
 
 7. If the Force act in the same straight I-ine, the force 
 polygon of course becomes a straight line also, and the result- 
 ant is the sum or difference (algebraic sum) of the forces. 
 
 Thus, if we have P l5 P 2 , P 3 , all acting at the point A, as 
 shown by the force diagram Fig. 4 (a), we form the force poly- 
 gon by laying off from 0, Fig. 4 (&), the intensity of P 1? from 
 the end of this line P l P 2 equal to A P 2 and from P 2 , P 2 P 3 
 equal to A P 3 . Then the line necessary to close the polygon is 
 evidently P 3 P^ + P 2 P 8 . A single force acting then at A 
 in the direction of and having the intensity represented by the 
 line P 3 would replace P 1? P 2 , and P 3 . If acting from P 3 to 0, 
 it will produce equilibrium. 
 
 If we again choose an arbitrary point as C [we shall hereaf- 
 ter call this point the "pole" of the force polygon], and draw 
 lines S S 3 from this pole to the beginning and end of the force 
 polygon, we can decompose the resultant into two forces in any 
 required direction. If the resultant is supposed to act down, 
 then the arrows show the direction in which these components 
 must act in order to replace the resultant. If then at A we 
 draw lines parallel arid equal, we have these components in posi- 
 tion, direction, and applied at the common point of application. 
 
 . Practical Applications. Simple and even self-evident 
 as all the preceding may seem, we have already acquired all 
 the principles requisite for a rapid, accurate, and very elegant 
 method of finding by diagram the strains in the various mem- 
 bers of all kinds of framed structures, such as roof trusses, 
 bridge girders, cranes, etc., no matter how complicated the 
 structure, or what special assumptions are made as to the load- 
 ing, provided only, that all the exterior forces are known. A 
 complicated or unsymmetrical arrangement of parts increases 
 greatly the labor of calculation, but has no effect upon the ease 
 or accuracy of the graphical method. The method moreover 
 checks its own accuracy, does not accumulate errors, and shows 
 in one view the relation of the strains to each other, and the 
 variations which would be caused by a change in the manner 
 of load distribution, or in the form of construction. 
 
6 FORCES IN THE SAME P^ANE. [CHAP. I. 
 
 As this method is not as well known as it deserves to be, it 
 will perhaps be of advantage to pause for a moment in the 
 development of our subject, and make this direct application of 
 the principles already established. 
 
 BEACED SEMI-AKCH. 
 
 9. Stoney, in his " Theory of Strains," Yol. I., page 123, 
 gives the following example of a " braced semi-arch," repre- 
 sented by Fig. 5, PL 1. The dimensions are as follows: pro- 
 jecting portion, 40 ft. long, 10 ft. deep at wall. Lower flange, 
 circular, with a horizontal tangent 2 ft. below the extremity of 
 girder. Radius of lower flange, 104 ft. Load uniform and 
 equal to one ton per running foot supposed to be collected into 
 weights of 10 tons at each upper apex, except the end one, 
 which has only 5 tons. 
 
 Fig. 5 shows the arch drawn to a scale of 10 ft. to an inch. 
 
 This scale is too small in this case to ensure good results ; in 
 general the larger the scale to which the frame can be drawn, 
 the better; but for the purpose of illustration it will answer 
 well enough. With a large scale for the frame diagram, a 
 scale of 10 tons to an inch will in general be found to answer 
 well. Fig. 5 (a) gives the strains in the various members to a 
 scale of 10 tons to an inch and Fig. 5 (b) 20 tons to an inch ; 
 the first for the load at the extremity alone, the second for a 
 nniform load. 
 
 Fig. 5 (a) is thus obtained. We first lay off the weight, 5 
 tons, to scale, in the direction in which it acts ; i.e., down- 
 wards. Now this weight and the strains in diagonal 1, and 
 flange A, are in equilibrium ; therefore by article (4) the force 
 polygon must close. Drawing lines therefore from the ends of 
 the line representing the weight of 5 tons, parallel to these 
 pieces and prolonging them to their intersection, we obtain 
 the strains in A and 1. Commencing with the beginning of 
 the weight line and following down around the triangle thus 
 formed, we find that A acts from right to left, as shown by the 
 arrow. A acts then aioay from the apex ; it is therefore in 
 tension. Diagonal 1 acts towards the apex and is hence com- 
 pressed. 
 
 We pass now to apex a, of the frame. Here we have the 
 strains in E and diagonals 1 and 2, and these three strains hold 
 each other in equilibrium. The strain in 1 we have already, 
 
CHAP. I.] COMMON POINT OF APPLICATION. 7 
 
 and know it to be compressive. We have then simply to draw 
 lines from and b parallel to E and 2, and follow round the 
 triangle, to obtain the intensity and quality of the strains in E 
 and 2. We must remember that as 1 is in compression, and 
 we" are now considering apex #, we must follow round from o 
 * to b in Fig. 5 (#), and so round. We thus find 2 acting away 
 from apex a and therefore in tension, and E acting towards 
 this apex, and hence com/pressed. 
 
 Pass now to apex c. We have the strains in A and 2 in 
 equilibrium with B and 3. [No weights are supposed to act 
 except the one at the end.] But A and 2 we already have. 
 We draw 3 and B. Diagonal 2 has been found to be in ten- 
 sion. With reference to apex c it must therefore act away 
 fromc; i.e., from d to b in the force polygon. This is suffi- 
 cient to give us the hint how to follow round. We pass from 
 d to b for 2, from b to e for A, then from e to B and from B to 
 d for B and 3. B is therefore tension and 3 compression. 
 And so we proceed. For the next apex g, we have E and 3 in 
 equilibrium with F and 4. AVe draw parallels to F and 4 so 
 as to close the polygon of which we have already two sides, E 
 and 3, given, and remembering that as 3 is in compression, it 
 must therefore act towards g, we follow round the completed 
 polygon with this to guide us, and find 4 tension and F com- 
 pression. Thus we go through the figure, and when all is 
 ready we can scale off the strains. The strains in the lower 
 flanges it will be observed all radiate from o. The upper 
 flanges are all measured off on the horizontal e C, and the dia- 
 gonals are the traverses between. We see at once that however 
 irregular the structure, we can always easily and readily deter- 
 mine the strains at any apex, provided no more than two un- 
 known strain^ are to be found. If more than two pieces, the 
 strains in which are unknown, meet at an apex, we can evi- 
 dently form an indefinite number of closed polygons. The 
 problem is indeterminate, and the structure has unnecessary 
 or superfluous pieces. 
 
 Fig. 5 (b) gives the strains for a uniform load, taken, for con- 
 venience of size, to a scale of 20 tons to an inch. Here until 
 we arrived at apex c of the frame the strains are evidently the 
 same as before. Observe the influence of the weight at c. 
 Here we have the strains in A and 2 given in the diagram, in 
 equilibrium with B, 3 and the known weight acting at c\ viz., 
 
8 FORCES IN THE SAME PLANE. [CHAP. I. 
 
 10 tons. We lay off therefore 10 tons downward from e, Fig. 
 5 (), and follow down from e around the polygon. We thus 
 find B tension and 3 compression. Then 4 and F are found as 
 before for apex <7, 4 tension and F compression ; and then we 
 come to the next apex and the next weight. This is laid off 
 downwards from the end of the preceding, and then we follow 
 round, finding C tension and 5 compression ; and so on. 
 1O. As another example, let us take the 
 
 given in Fig. 6, PI. 2. This truss is given by Stoney, Yol. I., 
 page 128. Dimensions : span, 80 ft. : rise of top and bottom 
 flanges, 16 and 10 ft. respectively. Kadii, 58 and 85 ft. The 
 figure shows two different kinds of bracing. In the left-hand 
 part the extreme bay of the lower flange is half as long again 
 as the others. The upper flange is divided into 4 equal bays. 
 In the right-hand section, both flanges are divided into 4 equal 
 bays, and every alternate brace is therefore nearly radial. Each 
 upper apex in both cases is supposed to sustain a weight of 
 one ton. 
 
 The strains in the various pieces are given in Fig. 6 (a). 
 
 We form the force polygon by laying off the weights from 
 to Y and then laying off the reactions 3.5 apiece, upwards, we 
 come back to 0, and the force polygon is closed as it should be, 
 since the sum of the reactions must be equal and opposite to 
 the sum of the weights. Starting then with the reaction at the 
 left support A, we go through from apex to apex in a manner 
 precisely similar to the previous case. The operation is so 
 simple that it is hardly necessary to detail it again, but we 
 recommend the reader to go over it with the aid of Fig. 6 (#), 
 lettering the figure as he proceeds. The dotted part gives the 
 strains for the right-hand half. 
 
 DIAGRAM FOR WIND FORCE. 
 
 11. It is of considerable importance to investigate the influ- 
 ence of a partial load, such as that caused by the wind blowing 
 on one side of the roof, and this by the aid of our method we 
 can easily do. 
 
 From the experimental formulae of Hutton,* 
 
 * Iron Bridges and Roofs. Unwin. p. 120. 
 
CHAP. I.] COMMON POESTT OF APPLICATION. 
 
 P n =P sin. i 
 P h =:P sin. i 1 - 84cos - i 
 P v =Pcot. isin, i 1 - 84008 - 1 
 
 where P is the intensity of the wind pressure in Ibs. per sq. ft. 
 upon a surface perpendicular to its direction, i is the inclination 
 of any plane surface to this direction ; P n is the normal pres- 
 sure, P h the horizontal component of this normal pressure, and 
 P v its vertical component. 
 
 That is, if the wind blows horizontally, P h is the horizontal 
 and P v the vertical component of the pressure on the roof. If 
 we take P=40 Ibs., which probably allows sufficient margin 
 for the heaviest gales, we have the following values of the nor- 
 mal pressure and its components for various inclinations of 
 roof surface : 
 
 Angle of 
 Koof 
 
 Lbs. 
 
 per square foot 
 
 of surface. 
 
 
 Pa 
 
 Pv 
 
 Ph 
 
 5 
 
 5.0. ... 
 
 4.9. 
 
 0.4 
 
 10 
 
 9.7.... 
 
 9.6. 
 
 1.7 
 
 20 
 
 18.1 
 
 17.0. 
 
 6.2 
 
 30 
 
 26.4.... 
 
 22.8. 
 
 13.2 
 
 40 
 
 33.3.... 
 
 25.5. 
 
 21.4 
 
 50....... 
 
 38.1 
 
 24.5. 
 
 29.2 
 
 60 
 
 40.0.... 
 
 20.0. 
 
 34.0 
 
 70 
 
 41.0.... 
 
 14.0. 
 
 38.5 
 
 80 
 
 40.4.... 
 
 7.0. 
 
 39.8 
 
 90" 
 
 40.0.. 
 
 0.0. 
 
 40.0 
 
 The load at each joint may be taken as equal to the pressure 
 of the wind striking a surface whose area is equal to that por- 
 tion of the roof supported by one bay of the rafter, and inclined 
 at the same angle as the tangent to the rib at the joint. Thus 
 we can calculate P 1? P 2 , P 3 , P 4 , (Fig. 6), resolve these forces into 
 their horizontal and vertical components, and find the reactions 
 at the supports as well as the horizontal force at the left abut- 
 ment, which in our construction is supposed to be fixed. Should 
 the wind be supposed to blow from the right side, the strains 
 would be entirely different, and it would be necessary to form 
 a second diagram. Each piece must be proportioned to resist 
 the strains arising in either case. The forces P^ and their 
 horizontal and vertical components, as also the reactions, being 
 known, we can now form the force polygon. 
 
 Thus in Fig. 6 (&), we lay off the forces P 4 . l5 make a c equal 
 
10 FOKCES IN THE SAME PLANE. [CHAP. I. 
 
 to the vertical reaction at A, a b = the sum of the horizontal 
 components, or the horizontal force at A, and o 1) the vertical 
 reaction at the right snppprt. This last line should close the 
 force polygon and bring us back to o. 
 
 Now starting at the left support, we have the vertical reac- 
 tion a c, the horizontal force a b, and the wind force P l5 in 
 equilibrium with A and E. Closing the polygon by lines par- 
 allel to A and E, we obtain the strains in these pieces, E ten- 
 sion and A compression. At the next apex we have A and P 2 
 in equilibrium with 1 and B. Completing the parallelogram, 
 we find 1 compression and B compression. At the next apex 
 1 and E are in equilibrium with* 2 and F, and we find F and 2 
 tension and so on. The upper flanges are in compression and 
 start from the ends of the forces P l5 P 2 , etc. The lower flanges 
 radiate from b. If we were to carry out the construction for 
 the rest of the frame, the upper flanges after D would radiate 
 from o. 
 
 A comparison of Fig. 6 (a) and (b) shows that whereas under 
 uniform load the strain in 1 is tension, for wind force the same 
 brace is in compression. In fact in the first case all the braces 
 are in tension, while in the second 1, 3, and 5 are compressed, 
 and 3 and 5 quite severely. The strains in the bracing gener- 
 ally are much greater in the second case. 
 
 Were we to consider the wind as blowing from the other 
 side, or what is the same thing, suppose the right end fixed and 
 the left supported on rollers, then the horizontal reaction a b 
 will be applied at the right abutment. In this case the lower 
 flanges will radiate from a instead of &, and the first upper 
 flange will start from o. Supposing the first two lines of this 
 new diagram drawn, as indicated by the dotted lines, and fol- 
 lowing round from b to o, and so round to a and back to 5, it 
 may easily happen that the last upper flange is in tension and 
 the last lower flange in compression; that is, a complete reversal 
 of the ordinary condition of strain. 
 
 For an excellent presentation of the above method, we refer 
 the reader to Iron Bridges and Roofs, by W. C. Unwin, pp. 
 128-140. The above method is there referred to as "Prof. 
 Clerk MaxweWs Method" and as such is known and used in 
 England.* 
 
 * Phil. Mag., April, 1864, and a Paper read before the British Association for 
 the Advancement of Science, by Prof. Maxwell, in 1874. 
 
CHAP. I.] COMMON POINT OF APPLICATION. 11 
 
 BRIDGES. 
 
 12. For bridges the strains due to a uniform load are of 
 course easily found. In most cases a rolling load can be man- 
 aged also, without making a separate diagram for each position 
 of the load. Thus, if we diagram the strains for the load at 
 the first and last apex, the strains due to intermediate loads 
 will be multiples or submultiples of these. A calculation for a 
 simple Warren girder of small span, and a consideration of the 
 reaction for each position of the load, will at once illustrate 
 what is meant. [Compare Stoney, Theory of Strains. Pp. 
 99-111, Yol. I.] 
 
 Thus Stoney, in his Theory of Strains, Yol. I., p. 99, gives 
 the girder represented in Fig. 7, PL 2, span 80 ft., depth of 
 truss, 5 ft., 8 equal panels in upper flange, 7 in lower. 
 
 For the first weight of 10 tons, P 1? the strains are given by 
 Fig. 7 (a] to a scale of 10 tons to an inch. We form first the 
 force polygon by laying off from o, 10 tons, to P^ From the 
 end of this line we lay off upwards the reaction at right abut- 
 ment = $ of 10 tons, or 1.25 tons ; and then the reaction at the 
 left abutment = -J of 10 tons, back to 6>, thus closing the force 
 polygon. [Note. In any structure which holds in equilibrium 
 outer forces, the force polygon must close. If it does not, there 
 is no equilibrium, and motion ensues (see Art. 20).] Com- 
 mence now with the reaction at a in the frame diagram, Fig. 7, 
 because here we have a known reaction, a o (force polygon), 
 and only two unknown strains to be determined. Drawing 
 lines parallel to A and 1, we obtain the strains in A and 1. 
 Then pass on to apex b. With the now known strain in 1, we 
 can determine 2 and E. 
 
 Passing now to the next apex, we have A and 2 known, and 
 also the weight P t . Join therefore P l and E [Fig. 7 (a)~\ by 
 lines parallel to B and 3. B and 3 are both in compression. 
 We find diagonal 2 also in compression, and 1 in tension. That 
 is, ~both the diagonals under the weight are compressed, as evi- 
 dently should be the case. From 4 on we have tension and 
 compression alternately. 
 
 Fig. 7 (b) gives the strains due to the last position of the load 
 P 7 . The strains in the diagonals are evidently all equal, and 
 alternately tension and compression. 
 
 Now it is not necessary to construct more than these two dia- 
 grams. From these two alone we can determine the strains fur 
 
12 
 
 FORCES IN THE SAME PLANE. 
 
 [CHAP. I. 
 
 any intermediate weight. Thus scaling off the strains in Fig. 
 7 (a) and (&), we can tabulate them under P x and P 7 , as shown 
 by the table. 
 
 DIAGONALS. 
 
 Pi 
 
 P 2 
 
 P 3 
 
 P4 
 
 P 5 
 
 P 6 
 
 PT 
 
 C 
 
 + 
 
 T 
 
 1 
 
 12 4 
 
 10 6 
 
 89 
 
 71 
 
 53 
 
 35 
 
 1 8 
 
 
 49 6 
 
 2 
 
 + 12 4 
 
 + 10 6 
 
 +89 
 
 ' 
 
 + 71 
 
 + 53 
 
 + 35 
 
 + 18 
 
 + 49 6 
 
 
 3 
 
 + 1.8 
 
 10.6 
 
 8.9 
 
 71 
 
 53 
 
 3.5 
 
 18 
 
 + 18 
 
 37 2 
 
 4 
 
 18 
 
 + 10.6 
 
 + 89 
 
 + 7.1 
 
 + 53 
 
 + 35 
 
 + 18 
 
 + 37 2 
 
 18 
 
 5 .... 
 
 + 18 
 
 + 35 
 
 89 
 
 7 1 
 
 5.3 
 
 35 
 
 1.8 
 
 + 53 
 
 26 6 
 
 6 
 
 18 
 
 35 
 
 + 89 
 
 + 7.1 
 
 + 53 
 
 + 35 
 
 + 18 
 
 + 26 6 
 
 53 
 
 7 
 
 + 18 
 
 + 35 
 
 + 53 
 
 71 
 
 53 
 
 35 
 
 18 
 
 + 10 6 
 
 17 7 
 
 8 
 
 - 1.8 
 
 35 
 
 53 
 
 + 7.1 
 
 + 53 
 
 + 35 
 
 + 1.8 
 
 + 17.7 
 
 10 6 
 
 
 
 
 
 
 
 
 
 
 
 Now the reaction at the left abutment due to P 6 is twice 
 as great as that due to P 7 . Hence the values in the column 
 for P 6 will be twice as great ; in the column for P 5 three times 
 as great, and so on. For similar reasons the strain in 5 for 
 P 2 will be twice that for P^ In column P 2 , then, from 5 
 down we multiply the strains in P 1 by 2. In P 8 from 7 down 
 by 3. Thus we fill out the table of strains completely, and find 
 the maximum tension and compression. A similar procedure 
 will give the flanges.* 
 
 APPLICATION TO AN AECH. 
 
 13. For a " braced arch " (Stoney, p. 136) as represented in 
 Fig. 5 (c) PL 2, the strains in every piece due to any load are 
 in similar manner easily found by first finding the components 
 of the load acting at the abutments, and then proceeding as 
 above. Thus for a load P 2 , the left half of the arch is in equi- 
 librium with the forces acting upon it ; viz., a horizontal and a 
 downward force at a, and a horizontal and an upward force at 
 A. The resultant of the forces at a must then pass through 
 
 * The reader not familiar with the above method of tabulation will find it 
 further illustrated in Art. 7 of the Appendix. He cannot do better than to 
 refer to it here and now. 
 
CHAP. I.] COMMON POINT OF APPLICATION. 13 
 
 a and A, and be equal and opposite to the resultant at A. The 
 resultant at the right abutment must pass through that abutment, 
 and also through the intersection of P 2 with A a. So for any 
 other force, as P 6 , we have simply to draw B a to intersection 
 with P 6 , and then P 6 A. We can now decompose P 6 or P 2 along 
 the resultants through the abutments thus found. Thus resolv- 
 ing P 2 along A a and P 2 B, Fig. 5 (e\ we find the force acting at 
 apex a. This force resolved into A and 1 gives the strains on 
 these pieces both compressive. Passing then to the next apex, 
 we obtain the strains in 2 and E. Then to the next, and we 
 get 3 and B, compression and tension respectively, and so on, 
 as shown by diagram, Fig. 5 (<?), which, it will be seen at once, 
 is similar to Fig. 5 (a), already obtained for the " semi-arch," 
 except that the strain in A is less than for the semi-arch and 
 compressive, while B C and D are in tension. The reason is 
 obvious. At a [Fig. 5 (c)] the resultant lies between A and 1, 
 and therefore causes compression in both, while it passes out- 
 side of the arch entirely, to the right of the apex for diagonals 
 3 and 4, and hence causes tension in B C and D. Fig. 5 (d) 
 gives the strains due to P 6 . Here the resultant or reaction at 
 A is first found and resolved into 9 and H, and then we go 
 through the frame as before. We see that 4 and 5 under the 
 load are both compressed, that E and F are in tension and G- 
 and H, as also the entire upper chord, in compression. The 
 work checks from the fact that the line closing the polygon 
 formed by E and 2 should be exactly parallel to and give the 
 strain in diagonal 1, or A and 1 should be in equilibrium with 
 the resultant through a [see Fig. 5 (d)~\. 
 
 In every case of the kind we first, then, have to draw the 
 frame diagram. Then lay off \\\Q force polygon which should 
 close. Finally we construct the strain diagram. The frame 
 diagram should be taken to as large a scale as possible consist- 
 ent with reasonable size, and the scale for the force and strain 
 diagrams as small as possible, consistent with scaling off the 
 strains to the requisite degree of accuracy. A small frame 
 diagram does not give with the proper accuracy the relative 
 positions and inclinations of the various pieces, so as to ensure 
 the proper direction for the lines of the strain diagram. A 
 slight deviation from parallelism causes sometimes considerable 
 variation. Nevertheless with practice, care, and proper instru- 
 ments the accuracy of the method is surprising ; even in com- 
 
14 FORCES IN THE SAME PLANE. [CHAP. I. 
 
 plicated structures, the variation resulting from performing 
 the operation twice being inappreciable. Every symmetrical 
 frame gives also a symmetrical strain diagram, and the accu- 
 racy of the work is tested at every point by this double sym- 
 metry, and finally by the end or last point of the second half, 
 exactly coinciding with the last point of the first half. Thus 
 in Fig. 6 (a), if we had but one system of triangulation carried 
 through the frame, the strain diagram for the right half would 
 be precisely similar and symmetrical to that already found for 
 the first, and the end of the last line would fall, or should fall, 
 precisely upon the point b of the first. If it does not, and the 
 error is too great to be disregarded, then by checking corre- 
 sponding points in each half, we can find the point where the 
 error was committed. In any case errors do not accumulate. 
 Thus, armed with straight edge, scale, triangle, and dividers, 
 we can attack and solve the most intricate problems, without 
 calculation or tables, with ease, accuracy, and great saving of 
 time. 
 
 METHOD OF SECTIONS. 
 
 14. The results obtained by the above method are best 
 checked in general by Hitter's u method of sections," or the 
 use of moments.* This consists in supposing the structure 
 divided by a section cutting only three pieces. We can then 
 take the intersection of two of these pieces as a centre of mo- 
 ments, and the sum (algebraic) of the moments of all the 
 exterior forces, such as reaction, loads, etc., upon one of the 
 portions into which the structure is divided by the section, with 
 reference to this centre of moments, must be balanced by the 
 moment of the strain in the third piece, with reference to this 
 same point. Thus in Fig. 6, PI. 2, required the strain in D. 
 Take a section through D, 7 and H (right half of Fig.), and let 
 a be the centre of moments. The moments of the strains in 7 
 and H are then, of course, zero, since these pieces pass through a. 
 The moment of the strain in D with reference to a must then 
 be balanced by the sum of the moments of all the outer forces 
 acting upon the portion to the left (or right) of the section. 
 
 Thus, strain in D multiplied by its lever arm with respect to 
 #, is equal to moment of reaction at A, minus sum of the mo- 
 ments of loads between A and #, all with reference to a. If 
 
 * Dock- und Brucken- Constructional. Ritter. Hannover, 1873. 
 
CHAP. I.] COMMON POINT OF APPLICATION. 15 
 
 we take the direction of rotation of the forces on the left of the 
 section when in the direction of the hands of a watch as posi- 
 tive^ and find the moment of strain in D negative, it shows i 
 negative rotation about a, and the strain in D to resist this rota- 
 tion must act away from &, or be tensile. If the resultant 
 rotation of the outer forces is on the other hand positive, the 
 strain in D must act toward b, and D is therefore compressed. 
 
 This method of calculation, it will be observed, is both sim- 
 ple and general. It can be applied to any structure, when the 
 outer forces are completely known, and only three pieces are 
 cut by the ideal section. 
 
 15. It is unnecessary to give here further applications of our 
 graphical method. The reader can easily apply it for himself 
 to the " bowstring girder," bent crane, etc., and satisfy himself 
 as to its accuracy, and the ease with which the desired results 
 are obtained. 
 
 Enough has been said to indicate the many important appli- 
 cations which even at the very commencement of our develop- 
 ment of the graphical method we are enabled to make, and 
 here we shall close our discussion of forces lying in the same 
 plane and having a common point of application. As we pass 
 on to forces having different points of application, we shall 
 have occasion to develop new principles and relations not less 
 fruitful and useful in their practical results.* 
 
 * We refer the reader here to the Appendix to this chapter for further 
 illustrations of the application of the above principles, as well as for informa- 
 tion upon several points of considerable practical importance. We would also 
 remind him here once for all, that the Appendix to this work was NOT in- 
 tended to be disregarded, but has been thought desirable in order to avoid 
 encumbermg the general principles with too much of detail in the text. We 
 earnestly request him to neglect no reference to it which may be made in the 
 text. 
 
 He will do well in the present case, after first making himself familiar with 
 the above points, to solve for himself with scale and dividers a number of 
 similar problems, checking his results always by the method of moments. 
 He will thus in a very short time master the method, and be able to solve 
 readily and accurately every problem of usual occurrence in practice. 
 Though the method is very simple, actual practice with the drawing board is 
 here indispensable. 
 
16 FORCES IN THE SAME PLANE. [CHAP. II. 
 
 CHAPTER II. 
 
 FORCES IN THE SAME PLANE DIFFERENT POINTS OF APPLICATION. 
 
 16. Reultant of Two Forces in a Plane Different 
 Points of Application. Heretofore we have considered 
 forces having a common point of application, and have seen 
 that in any case the direction and intensity of the resultant is 
 easily found by closing the force polygon. 
 
 But suppose we have two forces P l P 2 having different 
 points of application A t A 2 ; required the position and direc- 
 tion of the resultant [PI. 3, Fig. 8]. 
 
 Any force acting in a plane may be considered as acting at 
 any point in its line of direction. 
 
 P! and P 2 may then be supposed to act at their common 
 point of intersection #, and through this point the resultant 
 should pass. The case reduces therefore to a common point 
 of application. The resultant is given in intensity and direc- 
 tion as before by the force polygon (&), and its position is deter- 
 mined by the point of intersection a. At this point, or at any 
 point in the line through a, parallel to 2, the resultant may 
 be supposed to act. 
 
 But the direction of the forces may not intersect within 
 reasonable limits, or the forces may be supposed parallel to 
 each other, so that they may not intersect at all. In any case 
 the force polygon will still give the intensity and direction of 
 action of the resultant, but its position in the plane of the 
 forces remains yet to be determined. Now we have se.en [Art. 
 5] that we can decompose a force into two components in any 
 desired directions, by choosing a "pole " and drawing lines to 
 the beginning and end of the force in the force polygon. Let 
 us choose then a pole C [Fig. 8 (5)] and decompose the result- 
 ant thus into two forces given in intensity by the lines C 
 and 2 C. The forces P x P 2 being supposed to act at the 
 points A! A 2 in the common plane, at what point in the plane 
 and in what direction must the resultant 2 be applied to keep 
 
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 17 
 
 this plane and hold the forces in equilibrium? The direction 
 of action of the resultant is given at once from the force poly- 
 gon [Art. 5 (&)]. It must act in a direction from 2 to 0, and 
 must be equal to 2 0, taken to the scale of force. Now at any 
 point in the line of direction of P 1? as for instance 1, let us 
 suppose the component given by C to act. What is then the 
 resultant of P! and CO? A glance at the force polygon gives 
 us 1C, because this line closes the polygon made by C 0, 1 
 and 1C. At I then, the three forces S (parallel and equal to 
 C 0) S t (parallel and equal to 1 C) and P! are in equilibrium, 
 and there is no tendency of the point 1 to move. But 1 C or 
 
 51 may be considered as acting in the plane at any point in its 
 line of direction ; therefore at 2 its intersection with P 2 pro- 
 longed. Suppose at 2, S 2 or 2 C to act. "We see at once from 
 the force polygon that 2 C, C 1 and P 2 are in equilibrium. 
 There is therefore no tendency of the point 2 to move, and the 
 two forces P l P 2 are then in equilibrium with C 0, 1 C, C 1 
 and 2 C. But since the resultant of C and 2 C or of S and 
 
 5 2 is also the resultant of the forces, and since it must there- 
 fore act through the point of intersection of S and S 2 : we 
 have only to prolo ng these lines to intersection b. Through 
 this point the resultants R^ 2 must pass and acting downwards 
 (from to 2) as indicated in the Fig., it replaces P l P 2 . Act- 
 ing upwards it would hold them in equilibrium. We thus 
 easily find the point 2 in the plane at which 2 C or S 2 must 
 be applied, when C or S acts at 1, and S ? 2 are thus found 
 in proper relative position. The position, intensity, and direc- 
 tion of the resultant are thus completely determined. 
 
 Had we taken ai*y other point than 1, as the point of applica- 
 tion of C 0, we should have found a different corresponding 
 point for application of 2 C, but in any case the prolongations 
 of 2 C and C would intersect upon the line a b, prolonged if 
 necessary. The same holds true for any position of the "pole " 
 C. This construction is evidently general whatever the posi- 
 tion or whatever the number of the forces. We may thus 
 obtain any number of points along the line a b ; that is, the 
 resultant also, may act at any point in its line of direction. 
 
 [NOTE. That b is a point in the resultant of P t andP 2 can 
 be proved in a method purely geometrical. In the two " com- 
 plete quadrilaterals " 1 2 C and 1 b 2 #, the Jive pairs of 
 corresponding sides 1 and a 1, 1 2 and a 2, 2 C and b 2, C 
 
18 FORCES IN THE SAME PLANE. [CHAP. II. 
 
 and b 1, C 1 and 1 2, are parallel each to each, therefore the 
 sixth pair 2 and a b must also be parallel / b is therefore a 
 point of the resultant passing through a, parallel to 2.] 
 
 17. The above Construction holds good equally well for 
 Parallel Forces. By means of it we find in PL 3, Fig. 9 (a) 
 and (b) and Fig. 10 (a) and (&), the resultant of a pair of paral- 
 lel forces, in the first case, both acting in the same direction ; 
 in the second, in opposite directions. 
 
 In both cases we have simply to choose a pole C, and draw 
 S Sj and S 2 . Then taking any point c in the line of direc- 
 tion of P l5 as a point of application for, S , draw through this 
 point S l5 thus finding d, the point of application for S 2 . S 
 and S 2 prolonged, intersect upon the resultant, whose intensity, 
 direction, and position thus become fully known. 
 
 18. Property of the Point b. It is plain that thus a point 
 of intersection J, through which the resultant must pass, can 
 always be found, provided S and S 2 do not fall together in 
 the force polygon, or intersect without the limits of the draw- 
 ing. By properly choosing the position of the pole C, this can 
 always be avoided if the points 2 and in the force polygon do 
 not themselves coincide, i.e., if the force polygon does not close. 
 
 The point &, Figs. 8, 9, and 10, which by reason of the arbi- 
 trary position of the pole may lie anywhere upon the resultant, 
 has a remarkable property. If we draw a line m n through 
 this point parallel to S^ and let fall from it perpendiculars p t 
 and p 2 upon P l and P 2 , then in all three cases, and therefore 
 generally, the triangle c m b is similar to C 1, and d b n is 
 similar to 1 C 2. Hence we have the proportions 
 
 1 : 1 C I ". c m : m b, and 
 
 1 C : 12 ; : n b : n d. 
 
 From these proportions we find 
 
 ~L : I 2 \\ c m x nb : mb x n d. 
 
 Now the triangles c m b and d n b have the same height 
 above the base m n ; the bases m b and b n are therefore pro- 
 portional to their areas. But their areas are equal to half their 
 sides cm and n d multiplied byj?i and^> 2 respectively. Hence 
 we have from the above proportion, since c m = n d, 
 1 : 1 2 ; : n d x p z : n d x p t or 
 01:12 ::p z :p,. 
 That is, the perpendiculars let fall from any point of th/>, 
 
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 19 
 
 resultant upon the components, are to each other inversely as 
 the components. Regarding any point of the resultant as a 
 centre of moments, the moments of the forces then are equal, 
 and of course the forces themselves are inversely as their lever 
 arms. 
 
 19. Equilibrium Polygon. If we consider the forces P t 
 P 2 , Figs. 8, 9, and 10, held in equilibrium by their components 
 C 0, 1 C, and 2 C, C 1, which act parallel to the lines S S x 
 and S 2 ; then regarding the line S t or c d as part of the mate- 
 rial plane in which the forces act, C 1 and 1 C balance one 
 another, and cause either tension or compression in c d. Sup- 
 pose the resultant R is to act so as to cause equilibrium, or 
 prevent the motion of the plane due to P t and P 2 . Then R 
 must act upwards in Figs. 8 and 9, and downwards from 2 to 
 in Fig. 10. In Figs. 8 and 9 then, S and S 2 act away from 
 c and d (Art. 4), and in Fig. 10 towards c and d. Following 
 round the force polygon, we find in the first two cases c d in 
 tension, in the last c d in compression. 
 
 In the first two cases, the points of application c and d of S 
 P! and S 2 P 2 if* connected by a string stretched between c 
 and d will be perfectly fixed and motionless ; while in the lat- 
 ter case, the string must be replaced by a strut. In case of 
 three or more forces the polygon or broken line which we thus 
 obtain, by choosing a pole, drawing lines to the beginning and 
 end of the forces in the force polygon, and then parallels to 
 these lines intersecting the lines of direction of the forces in the 
 force diagram, we call the " string " or "funicular polygon" 
 or the " strut polygon" according as the forces act to cause 
 tension or compression along these lines. We can apply to 
 both cases the general designation of polygon of equilibrium or 
 " equilibrium polygon" The perpendicular let fall from the 
 pole C upon the direction of the resultant in the force polygon, 
 we call the "pole distance "and shall always designate it by 
 H. The straight line joining the points c and d, or the begin- 
 ning and end of the equilibrium polygon, we call the "strut" 
 or " tie line " or generally the " closing line " and designate it 
 by L. The convenience and ' application of these terms and 
 conceptions will soon appear. In the present case of only two 
 forces, the equilibrium polygon becomes a straight line and 
 coincides with L, or c d. 
 
 [XoTE. We repeat that in order to determine the quality of 
 
20 FORCES IN THE SAME PLANE. [CHAP. II. 
 
 the strain in c d, we have only to follow round the force poly- 
 gon in the direction of the forces, and then refer to the force 
 diagram. Thus Fig. 9, at c, P x S and S x act, and are in equi- 
 librium. The corresponding closed figure is given in the force 
 polygon (a). S acts away from c, P x acts downwards from 1. 
 Continuing this direction we find S x acting from 1 towards C. 
 Reversing this direction (Art. 4), we find that the resultant 
 which replaces S and P t acts from C to 1. Referring now to 
 the force diagram (&), and transferring this direction to the 
 point c, we find this resultant acts to pull c away from d or 
 contrary to the direction of the force 1 C which replaces S 2 and 
 P 2 . The strain in c d is therefore tension. 
 
 A much better way of arriving at the same result is to con- 
 sider the triangle c b d as a jointed frame which holds in equi- 
 librium the forces P! P 2 and R^. Then the strains in any two 
 pieces c d, c ft, meeting at a point, are in equilibrium with the 
 force or forces acting at that point. 
 
 We have then the force Pj acting at apex c, decomposed into 
 strains along c b and c d (Art. 5). represented by C and 1 C in 
 the force polygon. All three are in equilibrium. P l acts 
 down. Follow down then from to 1 from 1 to C and C to 0. 
 Refer back now to apex c of the frame and transfer these 
 directions. The strain in c d acts away from the apex c and is 
 therefore in tension, while the piece c b would be in compres- 
 sion, since the direction of C is towards apex c. 
 
 See also " practical applications " of the preceding chapter 
 for illustrations of this. In the same way follow round 1 C 
 Fig. 10 (a) and refer to (b) and S is in compression^ 
 
 2O. Cae of a Couple. In Article 18 we remarked that the 
 pole can always be chosen in such a position as to give S and 
 S 2 intersecting within desired limits, provided that S and 83 or 
 the point and 2 do not coincide. This case however actually 
 happens, with a pair of equal and opposite forces that is, with 
 a couple. 
 
 Thus in Fig. 11, PI. 3, we have two equal and opposite force's 
 
 PI, P 2 - 
 
 The force polygon closes : therefore the resultant is zero. 
 S and 83 are parallel, hence their point of intersection in the 
 equilibrium polygon is infinitely distant. By changing the 
 position of the pole, we see that S and S 2 may take any posi- 
 tions in the plane. 
 
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 21 
 
 Two forces therefore which form a couple cannot be replaced 
 by a single force. Their resultant is an indefinitely small force 
 situated in any position in the plane of the forces, at an infinite 
 distance. 
 
 Conditions of Equilibrium. If then, similarly to Art. 4, 
 any number of forces lying in the same plane and having differ- 
 ent points of application, are in equilibrium, the force polygon 
 always closes. 
 
 For this reason, as already repeatedly seen in the practical 
 applications of our last chapter, the force polygon form'ed by 
 the exterior forces must always close. 
 
 But inversely, if the force polygon closes, it does not follow 
 that the forces are in equilibrium a couple may result. 
 
 To determine whether this is the case inspect the " equilibri- 
 um polygon." If this also closes [i.e., if S and S n intersect] 
 the forces are in equilibrium. If this does not close [i.e., if S 
 and S n are parallel] there is no single resultant, but the 
 forces can be replaced by a couple, and this couple, as we have 
 seen, may have any position in the plane. 
 
 21. Thus if we suppose in Fig. 11, PI. 3, P t and P 2 decom- 
 posed into their components S , S 1? and S 1? S 2 , the compressive 
 strains in S l at c and d are equal and opposite [see (a)]. We 
 have then S and S 2 remaining, which again form a couple 
 which must have the same action as the first. 
 
 Hence we see that one couple can be replaced by another with- 
 out changing the action of the forces. 
 
 It is easy to determine a simple relation between any two 
 couples. 
 
 If from c we lay off c a equal to o 1, and c o equal to Co, we 
 have o a parallel to C 1 or S 1? and therefore to c d. Join a d and 
 o d. The triangles c d a and c d o having a common base c d 
 and their vertices o and a in a line parallel to c d, are equal in 
 area. The side c a of one is known, and the opposite apex lies 
 in the line of the force P 2 . Its area is then ca.= P l multiplied 
 by half of the perpendicular distance of P t from P 2 , and is 
 therefore completely determined. So also for the other trian- 
 gle, one side of which o c is one force of the new couple, and 
 the opposite apex of which lies in the other force S 2 . 
 
 Hence a couple can be turned at will in its plane of action, 
 and the intensity and direction of its forces can be changed at 
 will if the area of the triangle the base of which is one of the 
 
22 FOSCES IN THE SAME PLANE. [CHAP. II. 
 
 new forces, and whose opposite apex lies in the other force, is 
 constant; or when the product of the intensity of the forces 
 into their perpendicular distance remains the same. The di- 
 rection *of rotation, of course, must also remain the same. 
 
 We shall see further on the significance of this area, or of 
 this product so much is clear, that a couple (or infinitely small, 
 infinitely distant force) is completely determined in its plane 
 when the direction of rotation is given, and the area of the tri- 
 angle or value of the product to which it is proportional, is 
 known. The couple itself can be replaced by any two parallel 
 equal and opposite forces whatever, if only the triangle having 
 one force as base, and the opposite apex in the other, has a given 
 constant area.* 
 
 22. Force and Equilibrium Polygons for any Number 
 of Force in a Plane. 
 
 In PL 3, Fig. 12 (b) we have the forces P w acting in various 
 directions and at different points of application. P 2 and P 3 
 form a couple ; that is, are equal, parallel, and opposite in di- 
 rection. Required the position, intensity and direction of action 
 of the resultant. 
 
 First, form the force polygon, Fig. 12 (a\ by laying off the 
 forces to scale one after the other in proper direction. Thus 
 we have 1, 1 2, 2 3, 3 4, 4 5 in Fig. 12 (a) parallel respec- 
 tively to P! P 2 P 3 , etc., in Fig. 12 (b). The line necessary to 
 close the polygon, 5, is the resultant in intensity and direc- 
 tion. In intensity because the length of 5 taken to the scale 
 of force, gives the intensity of the resultant ; in direction 
 because acting from 5 to it produces equilibrium, while act- 
 ing in the opposite direction, from to 5, it replaces the forces. 
 
 "We have, therefore, only to find the position of the resultant 
 in the plane of the given forces in Fig. 12 (b). Hence : 
 
 Second, choose anywhere a "pole " as C, and draw the lines 
 or rays, or " strings " S S t S 2 S 3 S 4 , etc. S and S 5 are evi- 
 dently components of the resultant, since they form with it a 
 closed figure in the force polygon. 
 
 Third, form the equilibrium polygon a b cde o', Fig. 12 (J), 
 as follows : 
 
 Draw a line parallel to S intersecting T? (produced if neces- 
 sary) at any point as a. From this point draw a line parallel 
 
 * Eltmente der Graphischen Statik. Bauschinger. Munchen. 1871. Pp. 
 11, 12. 
 
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 23 
 
 to Si to intersection with P 2 (also produced if necessary) at b. 
 From b parallel to S 2 to c, then parallel to S 3 to d, and finally 
 parallel to S 4 , to intersection e with P 5 . Through this last point 
 draw a line parallel to the last ray S 5 . Now S and S 5 are com- 
 ponents of the resultant 5 [Fig. 12 (a)~\ and are found in 
 proper relative position. Produce them, therefore, to intersec- 
 tion o '. Through this point the resultant must pass. Drawing 
 then through o', a line parallel to 5, we have the resultant in 
 proper position, and acting in the direction indicated in the fig- 
 ure, it produces equilibrium. 
 
 Any other point than a, upon the direction of P 1} assumed as 
 a starting point, would have given a different point o' ; so also 
 for any other assumed position of the pole C. But in every 
 case we shall obtain a point upon the line of direction of R^ 
 already found. The reader may easily convince himself of this 
 by making the construction for different poles, and points of 
 beginning. 
 
 ISTow the polygon or broken line, a b c d e, we call the equi- 
 librium polygon that is, it is the position which a system of 
 strings or struts, S Si S 2 , etc., would assume under the action 
 of the given forces at the assumed points of application. 
 
 Thus P t acting at a, is held in equilibrium by the forces along 
 S and S 1? P 2 acting at , by Si and S 2 and so on. If we join 
 any two points in the line of direction of S , and S 5 , as m n by 
 a line, we have then a jointed frame, which acted upon at the 
 apices a. . .e by the forces P!. . .P 5 , and at m and n by S and 
 S 5 is in equilibrium. 
 
 For S acting at m, we see from the force polygon may be 
 replaced by a force a parallel and opposed to the resultant R 
 and a force C a acting along the line L. In like manner S 5 may 
 be replaced by a C and 5 a parallel and opposed to the result- 
 ant. The two forces a C and C a being equal and opposed 
 balance each other through m n, while the sum of a and 5 a 
 is equal and opposed to the resultant 5. There is, therefore, 
 equilibrium, and m and n may be considered as the points of 
 support of the frame acted upon by the forces P t . . .P 5 at the 
 apices a. . .e, a Q and 5 a being the upward reactions at the 
 points of support. 
 
 As to the quality of the strains in the different pieces ; as 
 before the reaction at m, viz., a 0, is in equilibrium with the 
 strain in m n and m a. Following round, then, in the force 
 
24: FOKCES IN THE SAME PLANE. [CHAP. II. 
 
 polygon from a to 0, to C and C to a, and referring back to 
 the frame, we find strain in in n acting towards apex ra, there- 
 fore compressive ; strain in m a acting away from m, theref ore 
 tensile. In like manner Sx S 2 S 3 are in tension, while S 4 or d e 
 and S 5 or e n are compressed. 
 
 Hence we m&jfix any two points of the equilibrium polygon 
 by joining them by a line. The forces acting at these points 
 are at once found by drawing from C in the force polygon a 
 parallel to this line to intersection with resultant. Thus a C 
 (since we have taken m n parallel to S x ) is the force in m n and 
 a 0, 5 a, are the forces opposed to the resultant at m and n. 
 
 23. Influence of a Couple. Among the forces in Fig. 12 
 there are two, P 2 and P 3 which are equal, parallel and opposite, 
 the direction of rotation being as indicated by the arrow. Ex- 
 amining the equilibrium polygon, we see that the influence of 
 the couple is to shift S t through a certain distance parallel to 
 itself, to S 3 . Now suppose the forces composing the couple 
 were not given, but the value of the couple known, from the 
 direction of rotation and the area of the triangle A 2 P 2 P 3 , 
 which has its base equal to one of the forces and a height equal 
 to their perpendicular distance. In this case the lines 1 2, and 
 S 2 in the force polygon, would disappear, but we can none the 
 less find the point d, and from this point continue the polygon 
 by drawing S 4 and S 5 , and thus find the same points e and o' as 
 before. To do this we have simply to apply the principle 
 deduced in Art. 21, that one couple can be replaced by another 
 provided the area of the triangle is constant. 
 
 In the present case we must replace the given couple by 
 another whose forces are Si and S 3 , having the same direction 
 of rotation. 
 
 Lay off then from a, a i equal by scale to S 1 as given in the 
 force polygon. Describe upon S t the triangle a g h equal to 
 the given area A 2 P 2 P 3 . Draw g i, and then through h, h 7c 
 parallel to g i. The point Jc is upon the line of direction of 
 S 8 , or in other words the area of the triangle i J& a is equal to 
 a g h. The proof is easy. The two triangles ig hvxAigk 
 are equal, since they have the same base i g, and height. But 
 if from the triangle a i g we subtract i g h, we obtain a g h. 
 If from the same triangle a i g we subtract i g k, which is equal 
 to i g A, we obtain i k a. Equals subtracted from equals leave 
 equals. Hence ik a is equal to a g h. 
 
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 25 
 
 If then through 7c we draw a line parallel to S 3 and produce 
 it to d. we have the same point as before, and thus from d, can 
 continue the polygon. 
 
 \_Note that the direction of rotation shows the side of Sj upon 
 which the point 7c must fall. S t acts away from a [from 1 to 
 C in (a)] hence for rotation as shown by the arrow, g must fall 
 above S 1? and $! is shifted upwards. 
 
 24. Order of Forcc Immaterial. As in the case of a com- 
 mon point of application, so also here, the order in which the 
 forces are laid off is immaterial. To prove this for two forces 
 is sufficient, as by continued interchange of two and two, we 
 can obtain any desired order. 
 
 Let the two forces be P 4 and P 5 (Fig. 13, PL 3) existing either 
 alone, or in combination with others preceding and following. 
 
 Taking the forces first in the order P 4 P 5 , we have the equi- 
 librium polygon S 3 S 4 S 5 , (b) giving the point a in the result- 
 ant. Taking them now in reverse order, P 5 P 4 , we have the 
 polygon S 3 S' 5 S' 4 giving the same point a in the resultant. The 
 resultant in the force polygon (a), viz., 5, is of course un- 
 changed in intensity and direction in either case. It is required 
 to prove that in the second case the last string S' 4 is not only 
 parallel to S 5 in the first, but coincides with it. 
 
 This is easy. The resultant of P 4 P 5 goes through a, the in- 
 tersection of S 3 and S 5 . The same resultant in the second case 
 must also pass through the intersection of S 3 and S' 4 . But S 3 is 
 the same in position and direction in both cases. If the second 
 point of intersection does not coincide with a, still it must lie 
 somewhere upon S 3 . Hence as the resultant must pass through 
 both points, it must coincide with this last line ; viz., S 3 . But 
 this is not possible, as the resultant must also pass through d, 
 the point of intersection of the forces, or when these do not 
 intersect must be parallel to them. As therefore S' 4 must be 
 parallel to S 5 (shown by the force polygon), the intersections in 
 each case must coincide, as also the lines S' 4 , S 5 themselves, and 
 the polygon from e on has the same course in either case. 
 
 25. Pole taken upon closing line. We have seen (Art. 
 20) that when any number of forces are in equilibrium both 
 the force and equilibrium polygon must close. There is one 
 exception to this statement. Since the pole may be taken any- 
 where, suppose it taken somewhere upon the line closing the 
 force polygon. This line, as we know, is the resultant, and 
 
26 FORGES IN THE SAME PLANE. [CHAP. II. 
 
 holds the other forces in equilibrium. But now the equilibrium 
 polygon evidently will not dose. On the contrary the first and 
 last strings will be parallel. This position of the pole should 
 then be avoided. For any other position of the pole our rule 
 holds good ; viz., 
 
 If the force polygon closes as also the equilibrium polygon, 
 the forces are in equilibrium. If the equilibrium polygon 
 however does not close, the forces cannot be replaced by a single 
 force but only by a couple. The forces of this couple act in 
 the parallel end lines of the equilibrium polygon, and are given 
 in intensity and direction of action by the line from the pole 
 to the beginning of the force polygon [beginning and end coin- 
 ciding] . 
 
 26. Relation between two equilibrium polygons with 
 different poles. We may deduce an interesting relation be- 
 tween the two equilibrium polygons formed by choosing differ- 
 ent poles, with the same forces and force polygon. 
 
 Thus with the forces P! P 2 P 3 P 4 , we construct the force 
 polygon Fig. 14 (a), PL 4. Then choose a pole C and draw S M , 
 and thus obtain the corresponding equilibrium polygon S a b c d 
 S 4 Fig. 14 (b). Choose now a second pole C'. Draw S^ and 
 construct the corresponding polygon S' a! b' c' d' S' 4 . [In our 
 figure c and c' fall accidentally nearly together.] 
 
 Join the two poles by a line CO'. Then any two corre- 
 sponding strings of these two polygons intersect upon the same 
 straight line M N parallel to C C'. Thus S and S' intersect 
 at g, S'i and Sj at k, S' 2 and S 2 at I, S' 8 and S 3 at n, S' 4 and S 4 at 
 m and all these points g, k, I, n and in, lie in the same 
 straight line M N parallel to the line C C' connecting the 
 poles. 
 
 The proof is as follows.* If we decompose P t into the com- 
 ponents S Sj and S' S'^ these components are given in inten- 
 sity and direction by the corresponding lines in the force poly- 
 gon. If we take the two first as acting in opposite directions 
 from the two last, they hold these last in equilibrium. The 
 resultant therefore of any two as S and S' must be equal and 
 opposed to that of the remaining two, S t and S'^ and both re- 
 sultants must lie in the same straight line. This straight line 
 must evidently be the line gk joining the intersections of S S' 
 
 * Ekmente der Graphischen Statik. Bauschinger. Miinchen, 1871. Pp. 
 18-19. 
 
CHAP, n.] DIFFERENT POINTS OF APPLICATION. 27 
 
 and Si S\. But from* the force polygon we see at once that the 
 resultant of S and S' is given in direction and intensity by 
 C C', and this is also the resultant of Si and S\. The line join- 
 ing g and k must therefore be parallel to C C'. For the second 
 force P 2 we can show similarly that the line joining k and I is 
 parallel to C C'. But k is a common point of both lines hence 
 g k and I lie in the same straight line parallel to C C'. 
 
 [NOTE. The pure geometric proof is as follows : The two 
 complete quadrilaterals 1 C' C and g k a' a have five pairs of 
 corresponding sides parallel, viz., 1 and a a', a a' 1 C' and 
 a! k, C and a g, o C' and a' g, 1 C' and ak' hence the sixth 
 pair are also parallel, viz., C C' and g k. In like manner for 
 1 2 C C' and lkb r b and so on.~\ 
 
 We can make use of this principle in order from one given 
 equilibrium polygon S a b c d S 4 and pole, to construct another, 
 the direction of C C' being known. For this purpose, having 
 assumed the position of the first string S' we draw through its 
 intersection g with S a line M N parallel to C C'. The next 
 string must therefore pass through the intersection a' of S' and 
 P! and through the point k, of intersection of the second string 
 of the first polygon and the line M N. It is therefore deter- 
 mined. The next side must pass through V and Z, and 
 so on. 
 
 [Note. Observe that the intersections r and r' of the first and 
 last lines of both polygons must lie in a straight line parallel to 
 4, the direction of the resultant.] 
 
 27. Mean polygon of equilibrium. Since the pole may 
 have any position, let us suppose it situated in one of the angles 
 of the force polygon. It is evident that the first line of the 
 corresponding equilibrium polygon, then coincides with the first 
 force. If now the pole be taken at the beginning of the first 
 force in the force polygon, then the first side of the correspond- 
 ing equilibrium polygon will coincide with the first force, and 
 the last line will be the resultant itself in proper position. 
 
 Take for instance, the pole at o in the force polygon, Fig. 15 
 (a\ PL 4. The first side S reduces to zero. The next S t coin- 
 cides with 1. In (b) 'therefore P t is the first side of the equi- 
 librium polygon. The next side S 2 corresponds with S 2 in (a). 
 Thus we obtain the polygon a b c d e, the last side of which S 7 , 
 is the resultant itself. That is, S 2 is the resultant of P t and P 2 , 
 S : >, of P^, S 4 of P 14 and so on. Every line in the polygon then 
 
28 FORCES IN THE SAME PLANE. [CHAP. II. 
 
 is the resultant of the forces preceding, and we call such a 
 polygon the mean polygon of equilibrium. 
 
 If we wish to find the mean polygon for P^ we have only to 
 take the new pole C' at 2 in the force polygon (a). According 
 to the preceding Art., each side of the new polygon must pass 
 through the intersection of the corresponding side of the first 
 with the line S 2 which passes through a and is parallel to C C'. 
 Thus S' 4 must pass through ~b' and o. S' 5 through d and n, and 
 so on. S' 7 is the resultant of P S7 , and since S 2 is the resultant 
 of P^; S 7 , the resultant of P 1Jr , must pass through the intersec- 
 tion m of S' 7 and S 2 . 
 
 We observe here again the influence of the couple P 5 and P 6 . 
 S 4 and S 4 are simply shifted through certain distances, without 
 change of direction, to S 6 and S' 6 ; and as we have seen above, 
 knowing the direction of rotation, and the moment of the couple, 
 we might have omitted it in the force polygon and still obtained 
 S 7 and S' 7 as before. 
 
 2. Line of pressures in an arch. The practical applica- 
 tion of the above will be at once seen in the consideration of 
 an arch. Thus with the given horizontal thrust applied at a 
 given point of the arch, and the forces P w , we construct the 
 force polygon C o 5, and then the line of pressures abed. 
 [Fig. 16, PI. 4.] 
 
 Required with another thrust H' = o C' acting at another 
 point, and the same forces P w , to construct the corresponding 
 line of pressures. To do this we have only to lay off o C' equal 
 to the new horizontal thrust, then choose a point of the force 
 line, as 3, as a pole and draw the corresponding polygon, 
 Jc op Jc the point of intersection, &, is a point upon the line 
 m n parallel to o C, and upon this line will be found the inter- 
 section of corresponding sides of the two polygons. Thus from 
 the intersection of the side ap of the first polygon with m n, 
 draw a line to o and we have a'. From the intersection ~b of 
 the second line of the first polygon draw a line to a', and we 
 have b' #', and so on. 
 
 29. The preceding articles comprise all the most important 
 principles of the Graphical Method which can be deduced in- 
 dependently of its practical applications. Future principles 
 will be best demonstrated, and at the same time illustrated, by 
 considering the various special applications of the method, and 
 to these applications we shall therefore now proceed. 
 
CHAP. III.] CENTRE OF GRAVITY. 29 
 
 CHAPTER III. 
 
 CENTRE OF GRAVITY. 
 
 30. General Method. One of the most obvious applica- 
 tions of the new method as thus far developed, is to the deter- 
 mination of the centre of gravity of areas and solids. We 
 shall confine ourselves to areas only, merely observing that all 
 the principles hitherto developed apply equally well to forces 
 in space. The forces being given by their orthographic pro- 
 jections upon two planes after the manner of descriptive geo- 
 metry, the projections upon each plane may be dealt with as 
 forces lying in that plane, and thus the projections of the force 
 and equilibrium polygons, the resultant, etc., determined. 
 
 A body under the action of gravity may be considered as a 
 body acted upon by parallel forces. The resultant of these 
 forces being found for one position of the body [or the body 
 being considered as fixed, for one common direction of the 
 forces] may have its point of application anywhere in its line 
 of direction. 
 
 For a new position of the body [or another direction of the 
 forces] there is another position for the resultant. Among all 
 the points which may be considered as points of application of 
 these two resultants there is one which remains unchanged in 
 position, whatever the change in direction of the parallel forces. 
 This point must evidently lie upon all the resultants, and is 
 therefore given by the intersection of any two. 
 
 It is hardly necessary to give illustrations of the method of 
 procedure. 
 
 Generally, we divide up the given area into triangles, trapez- 
 oids, rectangles, etc., and reduce the area of each of these fig- 
 ures to a rectangle of assumed base. The heights of these 
 reduced rectangles will then be proportional to the areas, and 
 hence to the force of gravity acting upon them ; i.e., to their 
 weights. Consider then these heights as forces acting at the 
 centres of gravity of the partial areas. Construct the force 
 
30 CENTRE OF GRAVITY. [CHAP. III. 
 
 polygon by laying them off one after the other. Choose a pole 
 and draw lines from it to the beginning and end of each force. 
 These lines will give the sides of the funicular or equilibrium 
 polygon. Anywhere in the plane of the figure, draw a line 
 parallel to the first of these pole lines (S ). Produce it to inter- 
 section with the first force (P^, prolonged if necessary. From 
 this intersection draw a parallel to the second pole line (S^, and 
 produce to intersection with second force (P 2 ). So on to last 
 pole line, which produce to intersection with first pole line. 
 Through this point the resultant must pass, and of course it 
 must be parallel to the forces. 
 
 Now suppose the parallel forces all revolved say 90, the 
 points of application remaining the same. Evidently the new 
 force polygon will be at right angles to the first, as also the 
 new pole lines, each to each. It is unnecessary then to form 
 the new force polygon. The directions of the new pole lines 
 are given by the old, and this is all that is needed. 
 
 Anywhere then in the plane of the figure, draw a line (S' ) 
 perpendicular to the first pole line (S ) previously drawn, and 
 prolong to intersection with new direction of first force (P/). 
 Through this point draw a perpendicular (S/) to second pole 
 line, to intersection with new direction of second force (P 2 ') 
 and so on. We thus find a point for new resultant, parallel to 
 new force direction. Prolong this resultant to intersection 
 with first and the centre of gravity is determined. 
 
 [NOTE. If the area given has an axis of symmetry, that can 
 of course be taken as one resultant, and it is then only necessary 
 to make one construction in order to find the other.] 
 
 The given area of irregular outline must, as remarked above, 
 be divided by parallel sections into areas so small that the out- 
 lines of these areas may be considered as practically straight 
 lines. The forces are then taken as acting at the centres of: 
 gravity of these areas. This division will give us generally a 
 number of triangles and trapezoids. 
 
 It is therefore desirable to reduce graphically to a common 
 base the area of these triangles and trapezoids, and for this pur- 
 pose the following principles will prove of service : 
 
 32. Reduction of Triangle to equivalent Rectangle of 
 given Base. Let b be the base and h the height. Then area 
 
 = . Take a as the given reduction base, and let x represent 
 
CHAP. III.] CENTRE OF GRAVITY. 31 
 
 the height of the equivalent rectangle. Then 
 
 bfi k x 
 
 ax = or = -. 
 2, a %b 
 
 Now a, , and h being given, it is required to find x graphi- 
 cally. 
 
 Let A B C be the triangle, and D the middle of the base. 
 [Fig. 17, PL 5.] Lay off A E = h and A F = a. Draw F D, 
 and parallel to F D draw E x. Then A a? is the required 
 height. 
 
 -r, A a? A E x h 
 For: AD = AF r p = a 
 
 As to the centre of gravity of the triangle, it is evidently at 
 the intersection of the lines from each apex to the centre of 
 the opposite side ; since these lines are axes of symmetry. 
 
 33. Reduction of Trapezoid to equivalent Rectangle. 
 
 In the trapezoid A B C D, Fig. 18, PL 5, draw through the 
 middle points of A D and B C perpendiculars to D C, and pro- 
 duce to intersections E and F with A B produced. 
 
 Then lay off F g a the given reduction base, and draw 
 g E intersecting D C in a?. Then H x is the required height. 
 
 T, EF H x EF x 
 
 H Yyp (\Y 
 
 F^~HE a ~HE' 
 
 hence ##=:EFxHEi= area. 
 
 To find the centre of gravity , draw a line through the mid- 
 dle points of the parallel sides A B and D C. This line is an 
 axis of symmetry. Prolong A B and C D and make C a 
 A B and A b = C "D and join a and ~b. Then the intersection 
 of a ~b with the axis of symmetry gives the centre of gravity. 
 
 The construction for the reduction of a parallelogram is pre- 
 cisely similar. [Fig. 18 (5).] 
 
 The points F and E here coincide with A and B, and we 
 have 
 
 A x AB 
 
 T- = = , or # # h x A B =: area. 
 
 h, B g 
 
 The same construction also holds good, of course, for a rect- 
 angle or square. The centre of gravity in each case is at the 
 intersection of two diameters, since these are axes of symmetry. 
 
 31. Reduction off Quadrilaterals Generally. In general 
 
32 CENTRE OF GRAVITY. [CHAP. III. 
 
 any quadrilateral may be divided into two triangles which may 
 be reduced separately, or into a triangle and trapezoid. 
 
 It is also easy to reduce any quadrilateral to an equivalent 
 triangle, which may then be reduced by Art. 32 to an equiva- 
 lent rectangle of given base. 
 
 Thus we reduce the quadrilateral A B C D [Fig. 18 (c)] 
 to an equivalent triangle by drawing C C x parallel to D B to 
 intersection C t with A B, and joining C t and D. The triangle 
 D B Ci is then equal to D B C, and hence the area A D C x is 
 equal to A B C D. The triangle A D C^ can now be reduced 
 to an equivalent rectangle of given base by Art. 32. 
 
 The centre of gravity of the quadrilateral may be found as 
 follows : 
 
 Draw the diagonals A C and B D and mark the intersection 
 E. Make A E x = C E and B E 2 = D E, also find the centres 
 Q! and O 2 of the diagonals A C and B D. Join O 2 E t and O l 
 E 2 ; the intersection S of these two lines is the centre of gravi- 
 ty required. 
 
 The above is sufficient to enable us to find the centre of gravity 
 of any given area of regular or irregular outline. The method 
 may be applied to finding the centre of gravity of a loaded 
 water-wheel (as given in Der Constructeur, Reuleaux, Art. 47), 
 and many similar problems. The reader will have no difficul- 
 ty, following the general method indicated in Art. 30, in mak- 
 ing such applications for himself. The method itself is so sim- 
 ple that it is unnecessary to give here any practical examples 
 in illustration. We shall, moreover, have occasion to return to 
 the subject in the consideration of moment of inertia of areas. 
 
 "We pass on therefore to the moment of "rotation of forces in 
 a plane. 
 
CHAP. IV.] MOMENT OF ROTATION OF FORCES. 
 
 CHAPTEE IY. 
 
 MOMENT OF ROTATION OF FORCES IN THE SAME PLANE. 
 
 35. The "Moment" of a Force about any Point is the 
 
 product of the force into the perpendicular distance from that 
 point to the line of direction of the force. The importance 
 and application of the " moment " in the determination of the 
 strains in the various pieces of any structure will be evident by 
 referring to Art. 14, where Ritter's " method of sections " is 
 alluded to. In general, when the moments of all the exterior 
 forces acting upon a framed structure are known, the interior 
 forces, or the strains in the various pieces, can be easily ascer- 
 tained. 
 
 As we shall immediately see, these moments are given 
 directly in any case by the " equilibrium polygon" 
 
 36. Ciilmann's Principle. If a force P be resolved into 
 two components in any directions as b C, b C^ (Fig. 19, PL 5), 
 and these components be prolonged, it is evident that the 
 moment of P with reference to any point as a situated any- 
 where in the line c d parallel to P, is P x b a. But if from C 
 we draw the perpendicular H to P, then by similar triangles, 
 
 P : H ; : c d : I a ; 
 
 Pxb a = Hxc d. 
 
 That is, the moment ofP with respect to any point a is equal 
 to a certain constant H multiplied by the ordinate c d, paral- 
 lel to P and limited by the components prolonged. The con- 
 stant H we call the "pole distanced 
 
 This holds good for any point whatever, and we have only to 
 remember that if we assume the ordinates to the right of P as 
 positive, those to the left are negative. 
 
 "We can choose the pole C where we please, and thus obtain 
 various values for H, but for any one value the corresponding 
 ordinates are proportional to tlie moments. 
 
34 MOMENT OF KOTATION OF FOKOES. [CHAP. IV. 
 
 The above principle is due to Culmann, and will be referred 
 to hereafter as Culmann' s principle. 
 
 37. Application of the above to Equilibrium Polygon. 
 
 Let P 14 be a number of forces given in position as repre- 
 sented in Fig. 19 (a) PL 5. By forming foa force polygon Fig. 
 19 (b), choosing a pole C, and drawing S S 1? S 2 , etc., we form 
 the equilibrium polygon abed ef, Fig. 19 (a). 
 
 The resultant of the forces P 14 acts in the position and direc- 
 tion given in the Fig. How, as we have seen in Art. 22, 
 regarding the broken line a b c d e as a system of strings, we 
 may produce equilibrium by joining any two points as a and/" 
 by a line, and applying at a and f the forces S and S 4 . Let us 
 suppose this line a f perpendicular to the direction of the 
 resultant.- Since we can suppose the broken line or polygon 
 fastened at any two points we please, this is allowable, and 
 does not affect the generality of our conclusion. 
 
 Then the compression in the line a f is given by H, the 
 "pole distance" or the distance of the pole C from the result- 
 ant in the force polygon. We have therefore at a the force 
 H and Vi = H acting as indicated by the arrows. At a then 
 Y! acting up, H and S acting away from a, are in equilibrium, 
 or Vx is decomposed into H and S , as shown by the force 
 polygon. 
 
 According to Culmanrts principle then, the moment of V t 
 with reference to any point, as m or o, is equal to H x o m. 
 Therefore H being known, the ordinates between a f and S 
 are proportional to the moment of V t at any point. V x acting 
 upwards gives positive rotation (left to right) with respect 
 to m. 
 
 At the point b, P t may be replaced by a force K parallel to 
 R and a force K 1 along S t [see force polygon]. This we see 
 at once from the force polygon where K and K 1 make a 
 closed polygon with P 1? and taken as acting from to K and 
 K to 1, replace P lt But these two forces are in equilibrium 
 with Si and S , or 1 C and C [see force polygon], and since 
 K 1 and 1 K balance each other, all the forces acting at b may 
 be replaced by S , K and K C. We have then at b the force 
 K resolved into components in the directions S and S^ 
 
 By Culmanrfs principle, therefore, the moment of O K 
 about any point as m, is proportional to the ordinate n m, and 
 since K acts downward this moment is negative. Hence the 
 
CHAP. IV.] MOMENT OF ROTATION OF FOKCES. 35 
 
 resultant moment at m or o of the components at a and b par- 
 allel to R, is proportional to the ordinate o n. 
 
 So for any point, the ordinate included by the polygon ab c 
 d ef, and the dosing line af, to the scale of length multiplied 
 by the "pole distance " H to the scale of 'force ;, gives the mo- 
 ment at that point of the components parallel to the resultant. 
 
 The practical importance and application of this principle 
 will appear more clearly in the consideration of parallel forces 
 in the next Chapter. 
 
36 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V. 
 
 CHAPTEE Y. 
 
 MOMENT OF RUPTURE OF PARALLEL FORCES, 
 
 3. Equilibrium Polygon. Since the forces acting upon 
 structures are generally due to the action of gravity, these 
 forces may be considered as parallel and vertical, and in all 
 practical cases therefore, we have to do with a system of paral- 
 lel forces. 
 
 Given any number of parallel forces P^, PL 6, Fig. 20 ; 
 required to find the direction, intensity and position of the 
 resultant, and the moment of rotation at any point. 
 
 1st. Draw the force polygon, in this case it is, of course, a 
 straight lino. 
 
 2d. Choose a pole C, and draw the lines S . S 1? S 2 , etc. 
 
 3d. Draw the string or equilibrium polygon a b c d e f. 
 Considering this polygon as a system of strings, the forces will 
 be held in equilibrium if w r e join any two points, as a and g^ 
 by a strut or compression piece, and apply at a and g the up- 
 ward forces Vj and V 2 . 
 
 4th. Prolong a 1} and f g to their intersection o. Through 
 this point the resultant must pass. It is of course parallel and 
 equal to the sum of the forces. 
 
 Now, if a g is assumed horizontal, the perpendicular H to 
 the force line, or the "pole distance" divides the resultant 5 
 into the two reactions Y! and V 2 (Art. 22). 
 
 All the forces in the equilibrium polygon have the same 
 horizontal projection H, in the force polygon. 
 
 Let a g represent a beam resting upon supports at a and g. 
 We have then at once the vertical reactions V 1 and V 2 or ~k 
 and 5 &, which, in order to cause equilibrium, must act up- 
 
 For the moment at any point, as 0, due to V 1? we have, by 
 Culmann's principle, m o multiplied by H. The triangle formed 
 by a 5, a g, and P l5 gives then the moment of rupture at any 
 point of the beam as far as Pj. For a point o, beyond Pj, the 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 37 
 
 moment due to V 1? must be diminished by that due to P 1? since 
 these forces act in opposite directions, and rotation from left 
 to right upon the left of any point is considered positive. We 
 see at once from the force polygon that P t is resolved into S 
 and Sj or into a b and b c. Hence the moment at o due to P! 
 is m n multiplied by H. The total moment at o is then mo 
 m n = n o, multiplied by H. 
 
 Hence we see that the ordinates to the equilibrium polygon 
 from the closing line a g, are proportional to the total mo- 
 ments ; while the ordinate at any point between any two adja- 
 cent sides of this polygon, prolonged, represents the moment at 
 that point of a force acting in the vertical through the inter- 
 section of these two sides. 
 
 [The reader should make the construction, changing the order in which 
 the weights are taken, and thus satisfy himself that the order is a matter ' 
 of indifference. As to the direction of the reactions Vi, Va, it must be 
 remembered that a & is to be replaced by Vi and H, hence Vi must be op- 
 posed to C 0, the direction obtained by following round in the force poly- 
 gon the triangle 1 O. Force and distance scales should also be assumed. 
 Thus the ordiuates to the equilibrium polygon scaled off say in inches, and 
 multiplied by the number of tons to one inch, and then by the "pole dis- 
 tance " taken to the assumed scale of distance, will give the moments of 
 any point.] 
 
 The resultant of any two or more forces must pass through 
 the intersection of the outer sides of the equilibrium polygon 
 for those forces (Art. 16). Thus, the resultant of P t arid P 3 
 must pass through the intersection of a b and c d. Of V t and 
 P b through the intersection of a g and b c ; of P t P 2 and P 3 , 
 through intersection of a b and d e, and so on. In every case 
 the intensity and direction of action of the resultant is given 
 directly by simple inspection of the force polygon. 
 
 Thus from the force polygon we see that the resultants Tc 2 
 and Jc 3 of Vj. P! P 2 and V x P! P 2 P 3 , act in different directions. 
 Their points of application are at the intersection of c d and d e 
 respectively with a g, or upon either side of d in the equilibrium 
 polygon. At d the ordinate and hence the moment is greatest, 
 and at this point the tangent to the polygon is parallel to a g. 
 If we had a continuous succession of forces ; if a g, for in- 
 stance, were continuously or uniformly loaded ; the equilibri- 
 um polygon would become a curve, and the tangent at d would 
 then coincide with the very short polygon side at that point. 
 
38 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V. 
 
 The points of application of the resultants of all the forces 
 right and left of d are then at the intersection of this tangent 
 with a g, or at an infinite distance. 
 
 At d then we have a couple, the resultant of which is as we 
 have seen (Art. 20), an indefinitely small force acting at an 
 indefinitely great distance. That is, with reference to d, the 
 forces acting right and left cannot be replaced by a single 
 force. 
 
 Hence generally : at the point of maximum moment (" cross 
 section of rupture"), the resultant of the outer forces on either 
 side reduces to an indefinitely small and distant force, the 
 direction of which is reversed at this point, and the point of 
 application of which changes from one side to the other of the 
 equilibrium polygon.* 
 
 The " cross section of rupture " then, is that point where the 
 weight of that portion of the girder between it and the end is 
 equal to the reaction at that end, or where the resultant changes 
 sign. 
 
 The value of the moment at this point, is therefore equal to 
 the product of the reaction at one end into its distance from 
 the point of application of the equal resultant of all the loads 
 between that end and the point. 
 
 Thus for a beam uniformly loaded with w per unit of length, 
 
 the reaction at each end is -^-* From the above, the cross sec- 
 tion of rupture is then at the middle. The point of application 
 of the resultant of the forces acting between one end and the 
 
 A I , . w I I w Z 2 
 
 middle is at -r, hence the maximum moment is-^- x~: = -5-. 
 
 4: A 4: O 
 
 39. Beam with Two Equal and Opposite Forces beyond 
 the Supports. The ordinates to the equilibrium polygon thus 
 give, as it were, a picture or simultaneous view of the change 
 and relative amount of the moments at any point. The point 
 where the moment is greatest, i.e., where the beam is most 
 strained, is at once determined by simple inspection. 
 
 Let us take as an example a beam with two equal and oppo- 
 site forces leyond the supports. Thus, Fig. 21, PL 6, suppose 
 the beam has supports at A and B, the forces being taken in 
 the order as represented by P! P 2 . We first construct the force 
 
 * Die GrapMscJie Statik. Culmann, p. 127. 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 39 
 
 polygon from to 1, and 1 to 2 or 0. Next choose a pole C, 
 and draw S Si and S 2 . Draw then a parallel to S till intersec- 
 tion with first force, P t , then parallel with Sj to second force, 
 P 2 , then parallel to S 2 or S to intersection with vertical through 
 support B, and finally draw the closing line L. A line through 
 C, parallel to L, gives as before the vertical reactions. Follow- 
 ing round the force polygon, we find at A the reaction down- 
 wards, since S acts from C to and is to be replaced (Art. 4) 
 by L and V x ; at B reaction upwards, since P 2 acts up, and fol- 
 lowing round, S 2 acts from to C. .Both reactions are equal to 
 a 0. At A then the support must be above, and at B below the 
 beam. The shaded area gives the moments to pole distance H. 
 Had we taken the pole in the perpendicular through o, S would 
 have been parallel with the beam itself. This is, however, a 
 matter of indifference. The moment area may lie at any in- 
 clination to the beam. We also see here again the effect of a 
 couple (Art. 23). S is simply shifted through a certain distance 
 to S 2 , parallel to S , and therefore the moment at any point be- 
 tween P 2 and B is constant. This is generally true of any 
 couple, as we have already seen, Article 21, and may be proved 
 analytically as follows : 
 
 Let the distance between the forces be a = A B, Fig. 22. 
 Then for any point o, we have P X (& + B 6>) PxBo P [a + ^B o 
 B o] = P a. For o' between A and B, P x A o'+P x o' B= 
 P [A </ + <?' B] =Pa. 
 
 So also for any point to the left, the same holds true. 
 
 Graphically the proof is as follows : 
 
 Decompose both forces into parallel components, Fig. 23. 
 Then for any point, as o, we have the moment M = H x m n 
 Hxmp or M = H x n p. But n p is the constant ordinate 
 between the parallel components A n and A p. 
 
 We see, therefore, by simple inspection, that the distance of 
 P! and P 2 from the support B, Fig. 21, has no influence what- 
 ever upon the moment or strain in A B, provided the distance 
 between the points of application remains the same, and that 
 the moment at all points between P 2 and the support B is con- 
 stant and a maximum. From B and P 2 the moments decrease 
 left and right, and become zero at A and Pj. 
 
 1O. Beam with Two Equal and Opposite Forces be- 
 tween the Two Supports. Let the beam A B, Fig. 24, PL 6, 
 be acted upon by the two equal and opposite forces P 1 P 2 . 
 
4:0 MOMENT OF KUPTURE OF PARALLEL FORCES. [CHAP. V. 
 
 Construct the force polygon 012. Choose a pole C and draw 
 C 0, C 1, C 2. Parallel to C 0, draw the first side of the equi- 
 librium polygon to intersection with first force P t ; then paral- 
 lel to C 1 to second force P 2 ; then parallel to C 2 to d. Join d 
 and 0. Parallel to this draw C a in force polygon. Then a 
 is the vertical reaction at A, which acts upwards, since it niust 
 with C a replace C ; and C 0, when we follow round from o to 
 1 and 1 to C, acts from C to 0. 
 
 We have the same vertical reaction at B, but here, since we 
 must follow from 1 to 2 and 2 to C, C 2 acts from 2 to C, hence 
 following round, the reaction at B is downward. The shaded 
 area gives the moments to pole distance H, as before. 
 
 We see at once that at a certain point e the moment is zero. 
 Left and right of this point the moment is positive and nega- 
 tive. At the point itself we have a point of inflection, and 
 here, since the moment is zero, there is no longitudinal strain. 
 At jb and c the moments are greatest ; here the beam is most 
 strained, and at these points, therefore, are the " cross sections 
 of rupture." Here again, if we had taken the pole C in the 
 perpendicular through a, the closing line of the polygon o d 
 would have been horizontal. It is, however, indifferent at 
 what inclination a* d may lie, but we may if we wish make it 
 horizontal now, and then lay off from its new intersections with 
 P! and P 2 along the directions of these forces, the ordinates 
 already found at and c, and join the points thus obtained with 
 the ends of o d (i.e., with its intersections with the verticals 
 through the supports). The ordinates of the new polygon 
 thus found will be for any point the same as before, and will 
 also be perpendicular to the beam. 
 
 [NOTE. Had we taken the forces precisely as above but in reverse order, 
 the force line would be reversed, and we should have and 2 in place of 1, 
 and 1 in place of and 2 ; that is, in place of O 1 we should have O and 
 C 2. Constructing then the equilibrium polygon by drawing a line paral- 
 lel to new O to intersection with new Pi, then parallel to new C 1 to in- 
 tersection with new P 2 , then parallel with new O 2 to intersection with 
 vertical through B, and finally joining this last point with intersection of 
 the first line drawn (C 0) with vertical through A, we have at first sight a 
 very different equilibrium polygon. This new polygon will consist of two 
 parts. If the ordinates in one of these parts are considered positive, those 
 iii the other must be negative. The difference of the ordinates in these two 
 portions for any point, will give the same result as above. This, by mak- 
 ing the above construction, the reader can easily prove.] 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 41 
 
 41. Many other problems will readily occur, which may in a 
 similar manner be solved. The weights may have any position, 
 number and intensities desired ; in any and every case we have 
 only to construct with assumed pole distance the corresponding 
 equilibrium polygon, and we obtain at once the moments at 
 every point. By the use of convenient scales, numerical results 
 may be obtained which may be checked by calculation, and 
 the practical value and accuracy of the method thus demon- 
 strated. 
 
 The above principles will be sufficient for the solution of any 
 such problem w r hich may arise, and we shall therefore content 
 ourselves with the above general indication of the method of 
 procedure, and pass on to the consideration of a few cases 
 where the above needs slight modification, and which, from 
 their practical importance, and the ease with which they may 
 be treated graphically, seem worthy of special notice. 
 
 1ST. BEAM OR AXLE LOAD INCLINED TO AXIS.* [Fig. 25, PL 6.] 
 
 We have here simply to draw the " closing line "AC paral- 
 lel to the beam or axle. From d draw d B parallel to the force 
 P, then draw A B in any direction at. pleasure, and join B C. 
 We have thus the equilibrium polygon ABC, the ordinates to 
 which, as d B, parallel to the force P, will give the moments, 
 provided we know the corresponding pole distance. 
 
 But this can easily be found. As we have already seen, the 
 force polygon being given, the equilibrium polygon may be 
 easily constructed. Inversely, the equilibrium polygon being 
 given, the force polygon may be constructed. Thus from A 
 draw A c equal and parallel to P, and then draw c C^ parallel 
 to B C. A a and J c are the vertical reactions P! and P 2 ; a b 
 is the horizontal component of the force which must be resisted 
 at one or both of the ends ; and the moments at any point are 
 given by the ordinates parallel to P multiplied by the perpen- 
 dicular distance from C L to A c. If we suppose the force P, as 
 in the Fig., as causing two opposite vertical forces, instead of 
 acting directly upon the axis, we have only to prolong A B to 
 B! and join B! 62, and then the ordinates of A B! B 2 C parallel 
 to P or A c, multiplied by H (perpendicular distance from C t 
 to A c) will give the moments. 
 
 * See Der Constructeur, Reuleaux. 
 
4-2 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V. 
 2D. FORCE PARALLEL TO AXIS. [Fig. 26, PL 6.] 
 
 We have an example of this case in the " bayonet slide " of 
 the locomotive engine. 
 
 We have here two pairs of forces, the reactions V t and V 2 
 and the forces over B! and B 2 . The points of application of 
 these last change of course periodically, but for any assumed 
 position the moments are easily found. Thus draw A B! at 
 pleasure, and C B 2 parallel to it, and join B t B 2 arid A C, and 
 we have at once the equilibrium polygon. To find the corre- 
 sponding force polygon^ suppose P x applied at 5, and join ~b with 
 the other support. Make b c equal to P then c d = V 2 . Lay 
 off then A a = c d V 2 and draw a C l5 which is the pole dis- 
 tance. Draw GI e parallel to B! B 2 . Then A e and e A are the 
 forces acting over Bg and B 1? and A a is the reaction V^ The 
 case is, indeed, precisely similar to that in Art. 40. 
 
 [NOTE. The moment area should properly be turned over upon A C as 
 an axis, so that A a should be laid off and e fall lelow A. This can, how- 
 ever, cause no confusion.] 
 
 The application of the method to car axles,* crane standards, 
 and a large number of similar practical cases in Mechanics is 
 obvious. The formulae for many of these cases are too com- 
 plex for practical use ; in some, no attempt at investigation of 
 strain is ever made, the proportions being regulated simply by 
 " Engineering precedent " or rules of thumb. Those familiar 
 with the analytical discussion of such cases will readily recog- 
 nize the great practical advantages of the Graphical Method. 
 
 3D. BEAM OR AXLE ACTED UPON BY FORCES LYING IN DIFFERENT 
 
 PLANES. 
 
 The analytical calculation in such a case for instance is of 
 considerable intricacy, but by the graphical method, on the 
 contrary, the difficulty of investigation is scarcely greater than 
 before. 
 
 Thus, let Fig. 27, PL 7, represent a beam acted upon by two 
 forces P! and P 2 not in the same plane. 
 
 First, we draw the force polygons A C^ M and D O 2 2 for the 
 forces P! and P 2 , having both the same pole distance G O l = 
 O 2 H, the pole O 2 being so taken that the closing lines of the 
 
 * Der Constructeur, Reuleaux, pp. 215-222. 
 
CHAP. V.J MOMENT OF RUPTURE OF PARALLEL FORCES. 43 
 
 corresponding polygons A V D and A c" D coincide. This is 
 easily done, as if the closing line of the second polygon for any 
 assumed position of O 2 (O 2 H being equal to G OJ does not co- 
 incide with A D, the ordinate at c' can be laid off from C and 
 A G" D thus found in proper position, and then the pole O 2 can 
 be located. It will evidently be at the intersection of the ver- 
 tical O 2 O' 2 with G" D. 
 
 The two force polygons being thus formed, we construct the 
 polygon A C" D by drawing lines B B", E E", C C", etc., so 
 that their angles with the vertical shall be equal to the angle 
 between the planes of the forces, and making them equal to 
 the ordinates B 5", E e' 1 ', C c", etc., respectively. Join V B", 
 e' E", /' F", G' C", etc., "and lay off the ordinates B b, E e, F/, 
 C c, etc., respectively equal. The ordinates to the polygon 
 thus obtained, viz. : A 5 efc D multiplied by the pole distance 
 Oi G or O 2 H, give the moments at any point. A 1) and G D 
 are straight lines, ~b efc\& a curve (hyperbola). If we drop 
 verticals through C^ and O 2 , and draw the perpendiculars O/ M, 
 O' 2 K ; A M is the reaction R 1; and D K the reaction R 2 , both 
 measured to the scale of the force polygon. Their directions 
 are found by the composition of A G and H 2 and D H and 
 G M respectively, under the angle of the forces. 
 
 4TH. COMBINED TWISTING AND BENDING MOMENTS. 
 
 In many constructions pieces occur which are subjected at 
 the same time to both bending and twisting moments. Both 
 can be represented and given by moment areas. Thus., Fig. 
 28, PL 7, represents an axle turning upon supports at A and B 
 and having at C a wheel upon which the force P acts tangenti- 
 ally. We have then a moment of torsion M t = P R and reac- 
 
 * o ft 
 
 tions P t P - - and P 2 = P - ; s being the distance of P 
 
 a i S a -\~ S 
 
 from B, and a of P from A. 
 
 Let the bending moments be represented by the ordinates to 
 the polygon a C ~b ; then laying off a o equal to P and drawing 
 o O parallel to ~b c, we find the corresponding pole distance 
 O &, and the reactions P! and P 2 equal to Jc a and o Jc respec- 
 tively. 
 
 Now, in the force polygon O a o thus found, at a distance 
 from O equal to R, draw a line m n parallel to P. This line 
 m n evidently gives for the same pole distance the moment of 
 
44 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V. 
 
 torsion P x R. Laying off C'C^ = b &', in n, we have the 
 torsion rectangle C x b' b C'. 
 
 . Now the combined moment of torsion M t and bending M b is 
 | M b -f f v/Mg + Mp. We make then C' C equal to f C' C t = 
 m n and C C 2 equal to f C C' = M b , and draw C 2 b. Then 
 any segment of any ordinate, as/yj, is -f offf. Revolve now 
 C' C with C' as a centre, round to C' and join C' C 2 . Then 
 C 2 C' is equal to f V^b H-M?> an d therefore with C 2 as centre 
 revolving C 2 C' to C 3 , we find the point C 8 , C C 3 being equal to 
 f M b + f \/M^ 4- MjL In the same way we find any other 
 point as/3, by laying off//',, equal to// , joining/ 3 and/' and 
 making f 2 fs equal to f% f Q . The line C 3 f$ b thus found is a 
 hyperbola, and the ordinates between' it and b C give the com- 
 bined moments [for pole distance O Jc] at any point. 
 
 [NOTE. We suppose the axle to turn freely at A, and the working point 
 or resistance beyond B ; hence the moments left of the wheel are given by 
 the ordinates to a O.] 
 
 / 
 
 5TH. APPLICATION TO CRANK AND AXLE. 
 
 The above finds special and important application in the case 
 of the crank and axle. 
 
 Thus in PL 8, Fig. 29, let E D C B be the centre line of crank 
 and shaft. Lay off a P equal to the force P acting at A, choose 
 a pole o and draw o a o P and the parallels o a and a E. Join 
 E and d and draw o Pj parallel to E d. Then P P! is the 
 downward force at E and P 1 a the upward reaction at D. The 
 ordinates to E d a to pole distance o P, give the bending mo- 
 ments for the shaft. Make a F equal to the lever arm R, then 
 F G is the moment P R, and we unite this as above with the 
 bending moments and thus find the curves c' d' e' the ordinates 
 to which give the combined moments at every point of the 
 shaft [see 4th]. 
 
 For the arm B C, make the angle a Q B C equal to D a d, 
 and then the horizontal ordinates to a Q B give the bending mo- 
 ments for the arm. Make C c equal to C G and we have the 
 torsion rectangle C C Q b Q B, and as in the previous case we unite 
 the two and thus find the curve b h F, the horizontal ordinates 
 to which from B C give the required combined moments, to 
 
 * Der Constructeur, Reuleaux, p. 52, Art. 18. 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 45 
 
 polo distance o P. Thus h'h^ = f H A 0j H i f B 5 , and 
 H h = h, h' + ti i = f M b + | /MS + M?. 
 
 The application of the method when the crank is not at right 
 angles to the shaft, as also when thecrank is double, and gener- 
 ally in the most complicated cases, is equally simple and satis- 
 factory. Our space forbids any more extended notice of these 
 applications, and we must refer the reader to Der Constructeiir, 
 by F. Reuleaux, Braunschweig, 1872, for further illustrations 
 and applications of the method to the solution of various practi- 
 cal mechanical problems. 
 
 42. Continuous Loading Load Area. Thus far we have 
 considered only concentrated loads. But whatever may be the 
 law of load distribution, if this law is known, we can represent 
 it graphically by laying off ordinates at every point, equal by 
 scale to the load at that point. We thus obtain an area bounded 
 by a broken line, or for continuous loading, by a curve, the 
 ordinates to which give the load at any point. This load area 
 we can divide into portions so small that the entire area may 
 be considered as composed of the small trapezoids thus formed. 
 If, for instance, we divide the load area into a number of trape- 
 zoids of equal width, as one foot one yard, etc., as the case may 
 be, then the load upon each foot or yard will be given by the 
 area of each of these trapezoids. If the trapezoids are suffi- 
 ciently numerous, we may consider each as a rectangle whose 
 base is one foot or one yard, etc., as the case may be, and whose 
 height is the mean or centre height. The weight therefore for 
 each trapezoid acts along its centre line. We thus obtain a 
 system of parallel forces, each force being proportional to the 
 area of its corresponding trapezoid, and equal by scale to the 
 mean height or some convenient aliquot part of this height. 
 We can then form the force polygon / choose a pole ; draw 
 lines from the pole to the forces ; and then parallels to these 
 lines, thus forming the string or equilibrium polygon; and so 
 obtain the graphical representation of the moments at every 
 point. 
 
 Since, however, the polygon in this case approximates to a 
 curve, that is, is composed of a great number of short lines, the 
 above method is subject to considerable* inaccuracy, as errors 
 multiply in going along the polygon. 
 
 This difficulty can, however, be easily overcome. 
 
 Thus we may divide the load area into two portions only, and 
 
46 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V. 
 
 then draw the force and equilibrium polygon, considering each 
 portion to act at its centre of gravity, and so obtain an equili- 
 brium polygon composed of three lines only. These lines will 
 be tangents to the equilibrium curve. (Art. 76.) We thus have 
 three points of the curve, and its direction at these points. 
 In this manner we may determine as many points as may be 
 necessary, without having the sides of the polygon so short or 
 so numerous as to give rise to inaccuracy. 
 
 43. The above will appear more plainly by consideration 
 of a 
 
 BEAM UNIFORMLY LOADED. 
 
 The curve of load distribution becomes in this case a straight 
 line. The load area is then a rectangle, and hence the load per 
 unit of length is constant. Let us now divide this load area 
 [Fig. 30, PI. 8], into four equal parts, and considering each por- 
 tion as acting at its centre of gravity, assume a scale of force, 
 and draw the force polygon. Since in this case the reactions at 
 the supports must be equal, we take the pole C, in a perpendi- 
 cular to the force polygon at the middle point. This causes the 
 closing line of the equilibrium polygon to be parallel to the 
 beam itself, which is often convenient. We now draw C 0, C 1, 
 etc., and then form the polygon a c e g h. The lines a, a c, 
 c e, etc., of this polygon, are tangent to the moment curve at 
 the points 5, d,f, and A, where the lines of division prolonged 
 meet the sides. The curve can now be easily constructed, as 
 will appear from the next Art. 
 
 44. moment Curve a Parabola. Suppose we had divided 
 the load area into only two parts, of the length x and I x [Fig. 
 30, PL 8]. Then the moment polygon would be o a k h, and 
 the horizontal projection of the tangent a Jc would be % x 4- i 
 (txj-ll. 
 
 That is, the horizontal projection of any tangent to the mo- 
 ment curve is constant. But this is a property of the parabola. 
 The moment curve for a uniform load is therefore a parabola, 
 symmetrical with respect to the vertical through the centre of 
 the l>eam. 
 
 If, then, we divide o C and C h into equal parts, and join cor- 
 responding divisions above and below, we can construct any 
 number of tangents in any position. 
 
 ["NOTE. We may prove analytically that the moment curve is a parabola, 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 47 
 
 and hence that the line a Tc must le a tangent. Thus the moment at any 
 point is 
 
 p being the load per unit of length, I the length, and the reaction at sup- 
 
 f) 1 f) 
 
 port therefore ~-. Hence y = (Ix x 1 } for origin 0. 
 
 a a H 
 
 When the origin is at d, representing horizontal distances by y' and ver- 
 tical by x 1 , we have x = - y', and y = h #', k being the ordinate at 
 
 9 
 
 ^ 72 
 
 middle = 
 
 Hence by substitution 
 
 or reducing 
 
 2H 
 
 f = T x 
 
 which is the equation of a parabola having its vertex at <#.] 
 
 We may of course take the pole anywhere, and hence H may 
 have any value. It is in general advantageous in such cases 
 
 (i.e., for uniform load) to take H = * We have then 
 
 f = I so, 
 and for y %, or for the middle ordinate. we have x = j* 
 
 To draw the moment curve we have then simply to lay off 
 the middle ordinate equal to Jth the span. The curve can then 
 be constructed in the customary way for a parabola. Any 
 
 pi 
 ordkiate to this curve multiplied by H = ^ will then give the 
 
 moment at that point. 
 
 Enough has probably now been said to illustrate the applica- 
 tion of our method to the determination of the moment of rota- 
 tion, bending moment, or moment of rupture. The reader 
 will have no difficulty in applying the above principles to any 
 practical case that may occur. 
 
 It will be observed that the customary curve of moments in 
 the graphic methods at present in general use, comes out as 
 a particular case of the equilibrium polygon for uniform 
 load. 
 
 This polygon has other interesting properties, which we shall 
 notice hereafter. For instance, just as its ordinates [Fig. 30] 
 are proportional to the bending moments or moment of rotation, 
 
4:8 MOMENT OF RUPTURE OF PARALLEL FORCES. [dlAP. V. 
 
 so also its area is proportional to the moments of the moments, 
 or the moment of inertia of the load area. 
 
 As to the shearing force at any point of a beam submitted 
 to the action of parallel forces, the reactions at the ends being 
 easily found as above by a line parallel to the closing line in 
 the force polygon, we have only to remember that the shear at 
 any point is equal to the reaction at one end, minus all the 
 weights between that end and the point in question. 
 
 Thus for a uniformly distributed load we have simply to lay 
 off the reactions which are equal to one-half the load, above 
 and below the ends, and draw a straight line, which thus passes 
 through the centre of the span. The ordinates to this line are 
 evidently then the shearing forces. If we have a series of con- 
 centrated loads, we have a broken line similar to A\ 1' 1" 2', 
 etc., Fig. 32, PI 7, where each successive weight as we arrive at 
 it, is subtracted from the preceding shear. 
 
 44. Beam continuously Loaded and also Subjected to 
 the Action of Concentrated Loads. In practice we have 
 to consider not only a continuously distributed load, such as the 
 weight of the truss or beam itself, but also concentrated forces, 
 such as the weight of cars, locomotives, etc., standing upon or 
 passing over the truss. 
 
 In PL 8, Fig. 31, we have a continuous loading represented 
 by the load area A a b B, and in addition four forces P' w . 
 Now, since the total moment about any point is equal to the 
 sum of the several moments, we can treat each method of load- 
 ing separately and then combine the results. Thus with the 
 force polygon (b) we obtain the equilibrium polygon A' 1 2 3 
 
 B' for the continuous loading, and with the force polygon 
 
 (a) the equilibrium polygon A' I" 2" 3" B" for the con- 
 centrated loads. If now in (b) we draw C L parallel to the 
 closing line A' B', and in (a) C" L' parallel to the closing line 
 A' B", we obtain at once the reactions at the supports for each 
 case. 
 
 Thus for continuous loading we have L for reaction at A, 
 and 10 L for reaction at B ; for the concentrated loads, L' 0' 
 at A and 4' L' at B. These reactions hold the beam in equi- 
 librium. 
 
 For any cross-section ^, the shear to the right is composed of 
 the two components L 7 and I/ 3' (i.e., is equal to the reactions 
 minus the forces between cross-section and support). The mo- 
 
CHAP. V.] MOMENT OF KUPTUKE OF PARALLEL FORCES. 49 
 
 ment of L 7 is given by the ordinate o y to the corresponding 
 polygon, and we may consider L 7 as acting at the point of 
 intersection a of the side 7 8 with A! B' (Art. 38). In the same 
 way L' 3' acts at b. We may unite both these reactions and 
 find the point of application of their resultant , by laying off 
 in force polygon (b) 7 b equal to I/ 3', and then constructing 
 the corresponding equilibrium polygon e a d c. The resultant 
 R passes through c. This construction remains the same evi- 
 dently, even when the points a and b fall at different ends of 
 the beam, as may indeed happen. The components will then 
 have opposite directions, and must be subtracted in order to 
 obtain the resultant. 
 
 The total moment of rotation at y is proportional to the sum 
 of m n and o y. The greatest strain is where this sum is a 
 maximum. In order to perform this summation and ascertain 
 this point of maximum moment it is advantageous to construct 
 another polygon instead of A' 1" 2", etc., whose closing line 
 shall coincide with A' B'. This is easy to do, by drawing in 
 force polygon (a), I/ C' parallel to A' B', and taking a new pole 
 C' the same distance out as before, that is, keeping H constant, 
 and then constructing the corresponding polygon A.' V 2' 3', etc. 
 
 Thus the ordinate p y gives the total moment at y. We can 
 make use here also of the principle that the corresponding sides 
 of the two polygons must intersect upon the vertical through 
 A' (Art. 26). We have thus the total moment at any point, and 
 can easily determine the point of maximum moment or cross- 
 section of rupture. This point must necessarily lie between 
 the points of maximum moments for the two cases, or coincide 
 with one of them. In the Fig. this point coincides with the 
 point of application of P' 2 . 
 
 45. Case of Uniform Load. If the continuous load is uni- 
 formly distributed we can obtain the above result without 
 being obliged to draw the curve. As in this case we have a 
 very short construction for the determination of the point of 
 greatest moment, it may be well here briefly to notice it. 
 
 If we erect ordinates along the length of the beam as an axis 
 of abscissas, equal to the sum of the forces acting beyond any 
 cross-section, the line joining the end points of these ordinates 
 has a greater or less inclination to the axis according as the 
 uniform load is greater or smaller. At the points of applica- 
 tion of the concentrated loads this line is evidently shifted 
 4 
 
50 MOMENT OF RUPTTJKE OF PARALLEL FOKCES. [CHAP. V. 
 
 parallel to itself. Since at the point of maximum strain the 
 sum of the forces either side is zero, this point is given by the 
 intersection of the broken line thus found with the axis. 
 
 Thus in PL 7, Fig. 32, let A B be the beam sustaining a uni- 
 form load, and also the concentrated loads P t P 2 P 3 P 4 . The 
 reaction of the uniform load at the supports is equal to half 
 that load. To find the reactions for the concentrated loads we 
 draw the force polygon 01234, choose a pole C, then con- 
 struct the equilibrium polygon A' 1 2 3 4 B', and parallel to 
 A' B' draw C L. L and L 4 are the reactions at A and B. 
 ISTow through L draw A L horizontal, make it equal to the 
 length of the beam, and take it as axis of abscissas. [It is of 
 course advantageous here to lay off the forces along the verti- 
 cal through B, as done in the Fig. Then A falls in the vertical 
 through A and 1 2 3 4 are directly under the forces them- 
 selves.] 
 
 The ordinate to be laid off at A is equal to L + half the 
 uniform load. Between A and 1 the line A' x 1' is inclined to 
 the axis at an. angle depending upon the uniform load. Lay off 
 L U equal to this load and draw A U. A\ 1' must be parallel 
 to this line. At 1' the line A.\ I' is shifted to 1", so that 1 1 " 
 is the load P t . Then 1" 2' is parallel as before to A U, and 
 2' 2" is the load P 2 , and so on. The intersection 2 with A 1* 
 gives the point of maximum moment or cross-section of rup- 
 ture. The force P 2 at this point in our Fig. is divided, as shown 
 by L in the force polygon, into two portions, one of which is to 
 be added to the forces left, the other to the forces right. The 
 ordinate y y' at any point gives the shear or sum of the forces 
 acting at that point. This force acts up or down according as 
 the ordinate is above or below the axis. 
 
 Moreover, the area between the broken line and axis A L, 
 limited by this ordinate, gives the moment of rotation of the 
 forces beyond the section y, areas below the axis being nega- 
 tive. For a section at 2, therefore, we have area A A\ 1' 1" 
 2' 2 , minus 2 2" 3' 3" z' 2 , or what is the same thing, the area 
 ZQ z' 4' 4" B\ L, since the sum of the moments of all the forces 
 is zero. 
 
 46. Influence of a Concentrated Load, paiiig over the 
 Beam. If in addition to the already existing uniform and 
 concentrated loads, a new force operates, we have by (44) simply 
 to construct for this new force its force and equilibrium poly- 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 51 
 
 gon, and unite the forces and moments thus found with those 
 already existing. 
 
 In PL 7, Fig. 32, we have assumed a new force P\ near the 
 left support. The force polygon is 0' 1' C', the pole distance 
 being taken the same as before. For any one position of this 
 force we have then the equilibrium polygon A' 1' B ;/ , and 
 drawing a parallel C' L' to A' B" we obtain the reactions 0' I/ 
 and L' 1', which must be added to the reactions already ob- 
 tained. 
 
 If now we take a section y between P\ and the point of max- 
 imum moment 2 before found, the sum of the forces either 
 side of this section undergoes the following changes : Upon 
 the side where P^ lies, and the point 2 does not lie, where 
 therefore the sum was originally an upward force, we have the 
 downward force I/ V (equal to algebraic sum L' 0' 4- 0' 1'). 
 The sum of the forces at the section, or the shearing force, is 
 therefore diminished. 
 
 The total rotation moment is, however, increased by the 
 amount indicated by m n. Both changes, that of the sum of 
 the forces and the moment 'of rotation, increase as P\ ap- 
 proaches y, and are therefore greatest when P\ reaches y. 
 
 If P\ passes ?/, this point is in the same condition as z with 
 reference to the former position of P\ ; that is, the force and 
 point 2 are now both on the same side of the section. For z, 
 then, the original downward force to the left is increased by the 
 force I/ V. To the right the upward force is increased by 1' L'. 
 In like manner the moment of the forces beyond z is increased 
 by the amount indicated by op. This change is greatest when 
 P\ reaches z. 
 
 Therefore when a load passes over the beam the sum of the 
 shearing forces is diminished in all sections between it and the 
 original point of greatest moment, and increased m sections be- 
 yond this point, while the moment of rotation, or bending 
 moment, for all cross-sections is increased. These changes 
 moreover increase for any section as the load approaches that 
 section. The shear at any point is therefore least, and the mo- 
 ment greatest, when the load reaches that point. As soon, how- 
 ever, as the load passes this point, the shear passes suddenly 
 from its smallest to its greatest opposite value, and then dimin- 
 ishes as the load recedes, together with t'he moment of rotation. 
 On the other side of the point 2 of original greatest moment, 
 
52 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V. 
 
 the shear and moment increase as the load approaches, and 
 become greatest for any point when the load reaches that point. 
 At the moment of passing, these greatest values pass to their 
 smallest values, and increase afterwards, as the load recedes. 
 
 Since by the introduction of the load the shear for points 
 upon one side of 2 is diminished (between 2 and the load), and 
 on the other side increased, and the greatest moment is at the 
 point where the shear is zero, it follows that the point of greatest 
 moment moves in general towards the load. At a certain point, 
 then, both meet. As the load then advances this point accom- 
 panies it, passes with it the original position, and follows it up 
 to the point where it would have met the same load coming on 
 from the other side. From this point, as the load continues to 
 recede, it returns, and finally reaches its original position as the 
 load arrives at the further end. 
 
 It is evidently of interest to learn the position of these two 
 points, where the load meets and leaves the point of greatest 
 moment, or cross-section of rupture, and this in Fig. 32 we can 
 easily do. 
 
 When P\ arrives at 1', we have evidently the reactions by 
 laying off L E equal to P' 1? drawing A E, and through its 
 intersection with the vertical through the weight drawing the 
 horizontal A' B' . L B' is then the increase of reaction at B due 
 to P\. The entire reaction is B' B' 1? and the broken line A\ 
 1' 1", etc., holds good still, if we merely change the axis from 
 AO L to A' B' . The point of greatest moment, which is still 
 the intersection of the broken line with the new axis, in the 
 present case is not changed by reason of the overpowering in- 
 fluence of P 2 . It does not move to meet the load, but awaits it 
 until it reaches P 2 , and until, therefore, the new axis takes the 
 position A" B" . 
 
 If, however, the force P\ comes on from the right, we have 
 the reactions for any position as z, by laying off A E' equal to 
 P'u drawing L E', and then the horizontal A'" B'" through the 
 intersection of L E', with the vertical through z. Then A A'" 
 is the reaction at A, due to this position of the load. The in- 
 tersection #', corresponding to a?, shows the point to which the 
 point of greatest moment 2 moves to meet the load. As the 
 load passes towards the left, this point moves towards the right, 
 and both come together evidently at the point V^ correspond- 
 ing to the new axis A iv B iv . The point of greatest moments 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 53 
 
 passes then from 2 to V , and beyond these two limits it can 
 never pass. 
 
 Our construction, then, is simply to lay off the load in oppo- 
 site directions perpendicularly from each end of the axis A L, 
 and join the end points A E and L E'. The intersections of 
 these lines with the diagram of shear give the points 2 and 
 V required. 
 
 47. Load Systems.* Concentrated loads occur in general 
 in practice in a certain succession, as for instance the forces 
 acting at the points of contact of the wheels of a train of cars 
 passing over the beam, and it is necessary then to investigate 
 the influence of different positions of the train. It evidently 
 amounts to the same thing whether we suppose the weights to 
 move over the beam, or suppose the weights stationary and the 
 beam to move. In either case we obtain every possible posi- 
 tion of every weight relatively to the ends of the beam. 
 
 The severest load to which we can subject a railway truss, for 
 example, is when the span is filled with locomotives. If we 
 suppose, for illustration, in round numbers, the distance between 
 the three axles of the locomotive 3 ft. 6 in., between the 
 axles of the tender 5 ft. 6 in., between the foremost tender 
 and the back locomotive axle 4 ft., and the entire length of 
 locomotive and tender 34 ft. 6 in., and then suppose the weight 
 upon each locomotive axle 13 tons, and upon each tender axle 
 8 tons, we have a system of weights in fixed order and at fixed 
 distances, and the truss should be investigated for a series of 
 these systems, as many as can be placed upon the span, passing 
 over it from one end to the other. 
 
 In PL 9, Fig. 32 (a), we assume two such locomotives as 
 shown by P!_ IO , and construct the force and equilibrium poly- 
 gons. The forces are symmetrically arranged with respect to 
 a central point, and the pole in the force polygon is therefore 
 taken perpendicular from the middle of the force line. 
 
 Now the system of forces being as represented, suppose the 
 span to shift. Thus suppose the span of a given length repre- 
 sented by Sj Si in the Fig. Then 6 is the line closing the 
 polygon for this position of the span, and a parallel to 6 in 
 the force polygon, viz., C L gives the reactions at the ends. 
 Let now the span move from Si S x to s 5 s 5 ; we have a new po- 
 
 * Elemente der Gra/phitchen StatiJc, Bauschinger. 
 
54: MOMENT OF KUPTTJKE OF PAEALLEL FOKCES. [CHAP. V. 
 
 sition for the line closing the polygon and new reactions. As 
 the span continues to shift to the right, the lines closing the 
 polygon revolve, and as their projections are always constant, 
 viz., equal to the span, they are all tangent to & parabola, which 
 they therefore envelop. 
 
 48. Properties of tills Parabola. This parabola has sev- 
 eral important properties which will aid us in the investigation 
 of the case above proposed.* In PL 9, Fig. 32 (d), let XX be 
 the line along which the span is shifted ; a M and a N the 
 outer sides of the polygon, intersecting at a, along which the 
 closing lines slide as they revolve. For a given position s s of 
 the span, or cr is the corresponding line. S Q s is the position of 
 the span, for which the centre, C , lies in the vertical through a. 
 In this position cr cr is tangent to the parabola at o> , its middle 
 point, and upon this line lie the centres of all the other lines 
 (taken of course as reaching from a N to a M). Now the 
 point of tangency, /3, of any other line, as cr cr, with the parabola, 
 is as far from the centre of that line, 7, as the centre of that 
 line is itself from C Q . We have then only to make c b equal to 
 c CQ, and drop a perpendicular through 1} to find /3. Thus for 
 the position ^ s t and the line ^ ar^ to find the point of tangency 
 c\, make c d l equal to c C Q , and draw d c^ perpendicular to 
 intersection with cq ar^ 
 
 Inversely we may find that position for the span s s, for which 
 the vertical through a given point, b, shall pass through the 
 point of tangency. 
 
 We have only to move the span so that its middle point c 
 shall be as far from C Q as it is already from the given point, or 
 make c c equal to c b. (See Art. 75.) 
 
 If we shift now the span s s, and at the same time the point 
 ft through an equal distance, the intersections of the vertical 
 through b, with the corresponding closing lines of the polygon, 
 will all lie upon the same line a cr. 
 
 If therefore b\ is such an intersection, b has been moved from 
 b to b\, and hence the span from s s to s l s^. 
 
 49. Different Cases to be Investigated. We are now 
 ready to investigate the effect of a live load such as represented 
 in PI. 9, Fig. 32 (a). For the determination of the proportions 
 of the truss the following points are specially important : 
 
 * See Memente der Graphischen Statik, Bauschinger, pp. 108-114. Also, 
 Die GrapliiscJie Statik, Culmarm, pp. 136-141. 
 
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 55 
 
 1. When a certain number of wheels pass over the truss, but 
 without any passing off, or new ones coming on ; what position 
 of the system gives the maximum moment at any given cross- 
 section not covered by the system, and how great is this 
 moment? 
 
 2. Under the same supposition as above, what position of the 
 load gives the greatest moment for a given point covered by one 
 of the load systems ? 
 
 3. Among all the various points of the span, at which is found 
 the greatest maximum moment, for what position of the load 
 does it occur, and how great is it ? 
 
 4. If the number of wheels is indeterminate, how many must 
 pass on, and what position must they have to give at any point 
 the greatest maximum moment; where is the corresponding 
 cross-section, what position must the load have, and how great 
 is this maximum moment ? 
 
 The three first questions are easily solved by the aid of the 
 above properties of the parabola, enveloped by the closing lines 
 of the equilibrium polygon, corresponding to different positions 
 of the span. 
 
 Thus, as regards the first question, let the given cross-section 
 be , PL 9, Fig. 32 (df), and suppose the span s s in the position 
 where the vertical through ~b intersects a a at the point of tan- 
 gency 0. When now the span shifts, the intersection of the 
 ordinate through 5, with the corresponding tie line, will always 
 lie upon a cr. But this ordinate gives the reduced moments for 
 1) (reduced to pole distance H.) The greatest of these moments 
 will then be simply the greatest of the ordinates between a cr 
 and the polygon, and will always be found at an angle of the 
 same. When found, we have at once the position of t>, and of 
 course of the span with reference to the given loads. This is 
 always such that a wheel stands over the given section. 
 
 Thus in Fig. 32 (a\ supposing the * four wheels P 6 to P 9 to 
 pass over the span t ^, we seek the position of the load to give 
 the greatest moment at a point of the span from the left, 
 therefore -g-th from the middle. 
 
 We lay off the span in such a position, ^ t ly that its centre is 
 distant from the intersection a of the outer lines of the poly- 
 gon by Jth of the span. 
 
 The ordinate through the given point now passes through the 
 point of tangency of the tie line and parabola. We draw this 
 
56 MOMENT OF KUPTURE OF PARALLEL FORCES. [CHAP. Y. 
 
 tie line ^ 9, and seek the greatest ordinate between it and the 
 polygon. This we find at 7, and directly above 7 the given 
 point must lie, and hence we have the position of the span, viz., 
 1 1. If the scale of tons is ten tons to an inch, of distance 5 ft. 
 to an inch, and the pole distance H is assumed 12^ ft. = 2-J 
 inches, the scale of moments will be 10 x 2.5 x 5, 125 ft, tons 
 to an inch. 
 
 As to the second question; the position of the span required, 
 is that where the vertical through the given point of the system 
 S Fig. 32 (a), intersects the corresponding tie line at its point 
 of tangency with the parabola ; all other tie lines intersect this 
 vertical in a point between the tangent point and the polygon. 
 The middle of the span must then lie midway between the in- 
 tersection a of the outer polygon sides and the point s, where 
 the vertical through S meets the line X X. Thus the span has 
 the position t 2 t 2 . 
 
 The third question, finally, is easily solved if the parabola en- 
 veloped by the tie lines is drawn. The greatest ordinate be- 
 tween this parabola and the polygon gives the greatest moment, 
 and the point and the position of span required, since the 
 middle of the span must be half-w r ay between the point given 
 by this ordinate and a. 
 
 The greatest moment is always found upon an ordinate 
 through an angle of the polygon. 
 
 If, however, the parabola is not drawn, we find by trial at 
 several angles, drawing the tie lines and comparing the corre- 
 sponding ordinates, the ordinate required. Here the following 
 considerations may aid : 
 
 When the load is uniformly distributed, the maximum mo- 
 ment is in the middle of the span, and at the same time in the 
 vertical through the intersection a of the outer polygon sides. 
 The polygon itself becomes a parabola. The less uniform the 
 load is. the more this point approaches the heaviest loaded side, 
 as also the intersection a, though not in the same degree. For 
 loads not exceedingly unsymmetrical the point may be sought 
 for, then, in the neighborhood of a, i.e., near the resultant of the 
 forces acting upon the truss. Thus in our example we are jus- 
 tified in selecting the corner 7 of the polygon, nearest the point 
 of intersection a. 
 
 5O. Mot unfavorable Position of Load upon a Beam of 
 given Spun. The fourth question above requires a somewhat 
 
CHAP. V.] PARALLEL FOECES. 57 
 
 more extended consideration. The most unfavorable position 
 of a system of given concentrated forces is when it causes the 
 greatest moment at the cross-section of rupture. This position 
 is from the preceding, given by taking the centre of the beam 
 midway between the vertical through the point of intersection 
 of the outer sides of the equilibrium polygon and the nearest 
 angle of the same. If with this centre we increase the span, 
 the maximum moment increases until the span has the greatest 
 length possible without more wheels coming on. 
 
 Thus for the two wheels P 4 and P 3 , PL 9, Fig. 32 (a), a is 
 the intersection of the outer polygon sides, and 4 the nearest 
 polygon angle. The almost equally near angle 5 gives at any 
 rate no greater moment. In order then that these two weights 
 may cause the greatest maximum moment, the middle of the 
 beam must lie half-way between a^ and 4 ; and as the span 
 increases in length this moment increases, and is then greatest 
 when the span reaches to $i or P 3 . 
 
 If now the span still increases so as to also include P 3 , the 
 point of intersection of the outer polygon sides recedes to a 2 , 
 where in our Fig. it coincides almost exactly with the polygon 
 angle 4. Here then, approximately at 4, we must locate the 
 centre of the beam. If we take the same length of span a 
 before, that is, make the half span 0% s 2 equal to the distance 
 from Si to the point midway between a^ and 4, we see by draw- 
 ing the closing lines for these two positions of the span, that 
 the maximum moments measured upon the vertical through 4 
 are almost exactly equal in each case. For a smaller length of 
 span including the three weights, the maximum moment de- 
 creases, and is less therefore than the maximum moment already 
 caused by the t\vo wheels. The span ^ ^ may then be regarded 
 as the greatest for which the two wheels P 4 P 5 give the greatest 
 possible maximum moment. As the space s 2 &>, upon which we 
 have now three wheels, increases, the moment increases, and is 
 greatest when the span, its centre always remaining now at 2 
 reaches to s' 2 or to P 2 . 
 
 If now it still increases so as to also include P 2 , the intersec- 
 tion of the outer polygon sides retreats to a%. The nearest 
 polygon angle is still 4, and midway then between a$ and 4 we 
 must now locate the middle of the beam. If from this centre 
 we lay off the half span equal to a% s'%, to s 3 , and draw the clos- 
 ing line for this position of the span, we see as before that the 
 
58 MOMENT OF KUPTFRE. [CHAP. V. 
 
 moment given by the ordinate at 4 is for either case almost 
 exactly the same. Any less span including the four weights 
 would give a less moment ; less, therefore, than the moment 
 already caused by the three weights. The span s' 2 s' 2 then 
 precisely as before, is the extreme limit upon which the 
 three wheels P 3 to P 5 cause the greatest possible maximum 
 moment. 
 
 In a precisely similar manner we find that the span s' 3 s' s 
 with a centre midway between a 3 and 4 is the limiting span for 
 the four wheels P 2 to P 5 . 
 
 If now the span still increases so that P! comes on, the inter- 
 section of the outer polygon sides falls in our Fig. nearly at s 1? 
 and since this point also happens to correspond almost exactly 
 with the angle 3, we take the centre of the beam at s^. The 
 greater the span now becomes, the greater the maximum 
 moment. The greatest length, however, which the span can 
 have without including P 6 , is twice ^ 6, or twice the distance 
 between ^ and P 6 . If P also comes on, the intersection of the 
 polygon sides is found at 5 , and the nearest polygon angle is 4. 
 Midway then between a 5 and 4 is the new centre of the beam, 
 while before P 6 came on, it was nearly at s. But for centre 
 s t the half span was Si 6, while now it is somewhat less than 
 4 6 ; therefore considerably smaller. Since, however, we wish 
 to follow the span as it continues increasing, we must compare 
 those two spans which are equal before and after the coming 
 on of P 6 . The right-hand ends of these spans, viz., s' 4 and s 5 
 must evidently be distant each side of 6, by the half distance 
 of their centres s l and 4, or 0% (more accurately the point half- 
 way between a 5 and 4, but a 5 and 4 lie in our Fig. so nearly to- 
 gether that the centre cannot be indicated more exactly). We 
 make then ^ ' 4 = o^ = M x 6, provided that M L is taken half- 
 way between the centre ^ and 0%. 
 
 An exact construction shows that the maximum moments for 
 these two spans, the one given by the ordinate through 3, the 
 other by the ordinate through 4, are almost exactly equal, and 
 moreover, that the maximum moment for the span Si S l of equal 
 length whose centre is at M t is also almost exactly equal, when 
 measured upon the vertical through M!. We can therefore 
 take Si S t as the limit of those spans for which the five wheels 
 P t to P 5 cause the greatest maximum moment. 
 
 Taking on now the seventh wheel, the intersection of the 
 
CHAP. V.] PARALLEL FORCES. 59 
 
 outer polygon sides is at a 6 and the nearest polygon angle is 5. 
 Half-way between a 6 and 5 we must then take the centre, while 
 before it lay at a^ (nearly). If we take then M 2 half-way be- 
 tween 0-2 and this new centre, we find precisely as before the 
 span S 2 S 2 with centre M 2 , and right end at P 7 , as the limiting 
 span for the six wheels P l to P 6 . The same holds good for the 
 span S 3 S 3 with centre M 3 , for the seven wheels Pi to P 7 , and 
 so on. If, according to supposition, P 8 P 4 P 5 are 3 ft. 6 in. 
 apart, P 2 and P 3 4 ft., and P! and P 2 5 ft. 6 in. apart; then for 
 spans up to ^ S L = say 8 ft., the two wheels P 4 P 5 will give the 
 greatest maximum moment, and their place upon the beam is 
 given by the position of the centre (half-way between a^ and 4). 
 From about 8 ft. to 15 ft. span, or s 2 s 2 the three wheels P 3 to P 5 
 give the greatest maximum moment, and the centre of the span 
 is located at a%. For spans from 1 5 ft. to 19 ft. span, or s' z 8\, 
 the four wheels P 2 to P 5 give the maximum moment, and the 
 centre is at ^ ; and so on. Thus for a span of any given length 
 we have at once the weights and their position, in order to 
 cause the greatest maximum moment, as also the -place of this 
 moment, viz., the point vertically over that angle of the equi- 
 librium polygon nearest the centre of the span. The ordinate 
 through this point included by the equilibrium polygon, and 
 the closing line for the given span, taken to the moment scale 
 gives this moment at once ; or this ordinate taken to the scale 
 of force must be multiplied by the previously assumed jpale 
 distance. 
 
 51. Oreatet Moment of Rupture caused by a Sytem of 
 Moving ILoads at a given Cross-Section of a Beam of given 
 Span. For beams or trusses of long span, which are as a rule 
 caused to vary in cross-section, it is not sufficient merely to find 
 the greatest maximum moment which a given system of con- 
 centrated forces can cause ; we must also know for a number 
 of individual cross-sections, the maximum moments which can 
 ever occur. 
 
 For this purp6se the force and equilibrium polygons being 
 first constructed, we shift as above the given span along a 
 horizontal line, and draw for each successive position of the 
 span the corresponding closing line in the equilibrium polygon, 
 marking the point where each closing line is intersected by a 
 vertical through the given cross-section, which of course moves 
 with the span, keeping always the same position with reference 
 
60 MOMENT OF RUPTURE. [dlAP. V. 
 
 to the ends. The points thus obtained form a curve, and the 
 greatest ordinate between this curve and the polygon gives the 
 greatest moment which can act at the given cross-section. 
 This greatest ordinate will always be found at an angle of the 
 polygon, and hence a weight must always rest upon the cross- 
 section. Since the cross-section itself must lie upon this ordi- 
 riate, we have directly the position of the span with reference 
 to the given forces. The closing line for this position being 
 then drawn, a parallel to it in the force polygon gives the reac- 
 tions for this position. 
 
 The reader will do well to make the construction indicated 
 for an assumed span and system of weights, to convenient scales, 
 checking the results by computation.* 
 
 The above method applies more particularly to solid or 
 "plate " girders, beams, or trusses. It may of course be applied 
 to framed structures also, such as those illustrated in chapter 
 first. Thus the moment at any point, divided by the depth of 
 truss at that point, gives the strain in flanges. The more pre- 
 ferable, as perhaps also the simplest method of determining the 
 strains in such cases, however, is to find the reactions due to 
 each individual weight. Each reaction can then be followed 
 through the structure, as explained in that chapter, and the 
 strains in every member for every weight in every position can 
 thus be obtained and tabulated. An inspection of the table 
 \^11 then give at once the strains due to the united action of 
 any desired number of these weights. 
 
 We have thus two methods for the solution of such cases ; 
 first, by the composition and resolution of forces, and, second, 
 by the equilibrium polygon and moments of' rupture, and may, 
 if we choose, check the results obtained by one method by the 
 other. In most practical cases involving framed structures, 
 however, the first method is preferable as being simpler, quicker 
 of application, and of superior accuracy. 
 
 For solid-built beams or "plate girders," etc., the second 
 method comes more especially into play. The determination 
 of the strains in a structure of this kind from the known mo- 
 ment of rupture at any point, requires a knowledge of the 
 moment of inertia of the cross-section at that point, and this 
 may also be found by the Graphical method. 
 
 * This construction is given in Art. 15, Fig. VIII., of the Appendix. 
 
CHAP. VI.~| MOMENT OF INERTIA. 61 
 
 CHAPTER VI. 
 
 MOMENT OF INERTIA OF PARALLEL FORCES. 
 
 52. THUS far we have seen that by the graphic method we 
 can in any practical case determine the moment of the exterior 
 forces acting upon a piece at any cross-section of that piece. 
 But the exterior forces give rise to and are resisted by molecu- 
 lar or interior forces. Now the moment of the exterior forces 
 being found, the cross-section of the piece at any point being 
 known, and one of the dimensions of this cross-section being 
 assumed, it is required to find the other dimension, so that the 
 strain per unit of area of cross-section shall be less than the 
 recognized safe strain of the material as found by experiment. 
 
 The moment of the exterior forces at any cross-section we 
 call the moment of rupture ; and designate it by M. Let d = 
 the depth of cross-section.* 
 
 y = the variable distance of any fibre above or below the 
 neutral axis. 
 
 /3 =; 'the breadth of the section at the distance y from the 
 neutral axis, and consequently a variable, except in the case of 
 rectangular sections. 
 
 s = the horizontal unit strain exerted by fibres in the cross- 
 section at a given distance c from the neutral axis. 
 
 o 
 
 Then since the fibres exert forces which are proportional to 
 their distance from the neutral axis or to their change of length, 
 the unit strain in any fibre at a distance y from the neutral 
 
 o n i 
 
 axis will be . Let the depth of this fibre be d y, then, since 
 c 
 
 the breadth of section is /9, the total horizontal force exerted 
 
 o 
 
 by the fibres in the breadth /?, will be f3y dy. The moment 
 
 C 
 
 Q 
 
 of this force about the neutral axis will be /3 y* d y, and the 
 
 c 
 
 Thewy of Strains, Stoney, p. 43, Art. 07. 
 
62 MOMENT OF INERTIA. [CHAP. VI. 
 
 integral of this quantity will be the sum of the moments of all 
 the horizontal elastic forces in the cross-section round the neu- 
 tral axis, that is, equal to the moment of rupture of the section 
 in question. We have therefore 
 
 For a rectangular cross-section, for instance, (B is constant 
 and equal to the breadth b. Representing the depth by d we 
 
 have M = y^ , or if we make c the distance of the extreme 
 
 d 
 
 fibres - 
 
 from which M being known, as also s, if we assume b we can 
 find d or the reverse. 
 
 The integral //? y* dy is the moment of inertia of the cross- 
 
 section, and may be defined as the sum of the products obtained 
 ~by multiplying the mass of each elementary particle by the 
 square of its distance from the axis. [See Supplement to Chap- 
 terVIL.Art. 10.] 
 
 From the above, we see its importance in determining the 
 strain at any distance from the neutral axis, or in proportioning 
 the cross-section, so that the resulting strain shall be less than 
 a given quantity at any point. We see also that for a rectan- 
 
 ~b d z 
 gular cross -section the moment of inertia is -^-, where b is the 
 
 breadth and d the depth. 
 
 53. Graphical Determination. "We have already seen 
 that the moment of a force, as P t (PI. 6, Fig. 20) with reference 
 to any point, as 0, is given by the ordinate n m multiplied by 
 the constant H (Art. 38). The ordinate n m then represents 
 the product of P t multiplied by the horizontal distance of b 
 
 from n. But the area of the trianle ~b n m\$>mnK b n = 
 
 
 PI x -^ b n*, that is, the area of the triangle b n m represents 
 
 one-half the moment of inertia ofl?i with respect to o. Just 
 as the exterior ordinates of the equilibrium polygon have been 
 shown to have a certain significance, and to represent the mo- 
 
CHAP. VI.] MOMENT OF INEETIA. 63 
 
 ments of the forces, so the exterior areas of the equilibrium 
 polygon represent the moments of the moments, or the moments 
 of inertia. Thus in PI. 8, Fig. 30, the exterior parabolic area 
 oC/i should be one-half the moment of inertia of the rectangle 
 or load area o p r h. 
 
 Let us see if this is so. The area of the triangle o h C is 
 
 o h x the ordinate S C. This ordinate S C gives, as we have 
 seen, the moment, with respect to S, of the reaction. "We can 
 therefore find its value. Thus if p is the load per unit of length, 
 
 and I is the length, ^~ is the reaction, and ~ this moment. 
 
 I pi 2 p 1 B 
 The area of the triangle o c h is therefore -~ X r = -- . 
 
 The parabolic area odh is f of the circumscribing rectan- 
 gle. This rectangle is I x S d. The ordinate S d is equal to 
 S C dC. We have already found SC and dC is the sum of the 
 
 7} I I* 'D 1? T) ft 
 
 moments of P! and P 2 , or ~* x -. ^r~. Hence S d -- 
 
 A 4: O 4: 
 
 p p p p 
 
 ^ = Q~~~- The area f * ne circumscribing rectangle is then 
 
 7} 1? 2 V 1? T) 1? 
 
 p Two-thirds of this is -~ 9 which subtracted from &&- 
 gives for half the moment of inertia ~-j p Z 8 . Hence the 
 
 2i^c. 
 
 moment of inertia is p P, as should be. 
 
 54. We see therefore the significance of the area of the equi- 
 librium polygon. 
 
 If, when a number of forces are given, we form the force 
 polygon, and then the equilibrium polygon, the ordinates to 
 this last give the moments to the assumed pole distance. If 
 now we take these moments themselves as forces applied at the 
 same points, form a new force polygon with new pole distance, 
 and new equilibrium polygon, the ordinates to this new polygon 
 to the new pole distance will give the moments of the moments 
 or the moments of inertia of the forces. The same method is 
 applicable to moments of a higher order, but in practice we 
 have only to do with those of the second order alone. 
 
 55. Radius of Gyration. The moment of inertia of a 
 system of parallel forces P! P 2 etc., in a plane, with reference 
 
64 MOMENT OF INERTIA. [dlAP. VI. 
 
 to an axis from which the points of application are distant q l $>, 
 etc., is then 2 P <f. This is the product of three .quantities, 
 one of which is measured by the scale of force, and the other 
 two by the scale of length. We can therefore regard it as the 
 product of the square of a certain length by the sum of the 
 given forces, or 2 P <f = P 2 P. We call k the radius of 
 gyration. 
 
 In order to find the moment of inertia of a system of parallel 
 forces then, we must by the preceding Art. construct two force 
 and equilibrium polygons. If the pole distances are H and H', 
 and the segments into which the axis is divided by the produced 
 sides of the polygons are P\ P' 2 and P'^ P" 2 etc., respectively, 
 then 
 
 2 P (f = H H' 2 P" 
 and the radius of gyration is given by 
 
 HH' 2P" = L* 2P 
 
 , 
 or Js-. 
 
 This expression is easy to construct. Thus for example in 
 PI. 11, Fig 33, let o n C be the first force polygon, o n the force 
 line, containing the forces P ; C the pole, and H the pole dis- 
 tance. Make o b equal to the second pole distance H/ and draw 
 b c parallel to n c and c t parallel to H. Then 
 
 HH' 
 
 whence k = 
 
 If, therefore, in Fig. 33 (&), m" m" n is the segment of the 
 axis cut off by the outer sides of the second equilibrium poly- 
 gon, that is, if m" m" n -2* P", we have only to prolong m" 
 m" n to L, making m" L = A, and describe a semicircle upon 
 m" L, and erect the perpendicular m" 7t, which will be equal 
 to %. In general, the pole distance H and H' can be taken ar- 
 bitrarily, but it is often advantageous to take H (sometimes H' 
 also) equal to 2 P. Then 
 
 We should then have in Fig. 33 simply to increase m"'' m"^ 
 by the second pole distance H', and then proceed as above to 
 find L 
 
CHAP. VI.] MOMENT OF INEKTIA. 65 
 
 It is to be remembered that q q^ etc., the distances of the 
 points of application of the forces from the axis, may be meas- 
 ured in any direction, and H is parallel to this direction, and 
 is not therefore necessarily perpendicular to o n. 
 
 The above will be rendered plain by reference to Fig. 34-, 
 PI. 10. We suppose four forces applied at the points A t A 2 A 3 
 A 4 respectively, and acting parallel to XX. Required the mo- 
 ment of inertia of these forces and the radius of gyration, the 
 distances g l q^ etc., being measured parallel to Y Y. First we 
 form the force polygon by laying off along X X, 1, 1 2, 2 3, 
 3 4, parallel, and in the direction of action of the forces, cligos- 
 ing a pole C, and drawing C 0, C 1, C 2, etc. We now construct 
 the corresponding equilibrium polygon, CI, III, II III, in IV, 
 etc. The segments 01', 1'2', 2' 3', etc., represent the statical 
 moments of the forces with reference to X X. That is, these 
 segments to the scale of force multiplied by the pole distance C y 
 parallel to Y Y to the scale of distance, give the statical moments 
 of the forces. Now we take these segments themselves as forces, 
 and suppose them acting at the former points of application. 
 With the same pole as before we draw CO, C 1', C 2', etc., and 
 form the corresponding equilibrium polygon CI, III', II III'', etc. 
 The sum of the segments of XX cut off by the outer lines of this 
 polygon, or o y, to the scale of force multiplied by H H' or C if 
 gives the moment of inertia of the forces with respect to XX. 
 This moment then is M = y x Cy* 
 where Oy "= 2P" and"Cp = HH'. 
 
 The radius of gyration Jc is, as we have seen, given by 
 
 CT rrr ^ T)fr 
 
 , 5P being equal to 4 in the Fig. Hence 
 
 04 
 
 If, then, we lay off d = 4, and make c = C y, and make 
 the angle dee a right angle, we shall find a point e to the right 
 
 /"I ~ 2 TT TT/ 
 
 and Oe will be equal to -- = -y- Upon ey now describe 
 
 a semi-circle, the point of intersection y with the perpendicular 
 through will give (Art. 55) 
 
 /Oy x Cy 3 /HH'SP" 
 = V " Q4 c = V " ^P = ^ radius of gyration. 
 
 The square of this line, then, multiplied by 2 P or 4, will give 
 5 
 
66 MOMENT OF INERTIA. [CILAP. VI. 
 
 at once the moment of inertia of the given four forces with 
 reference to X X and Y Y as axes. If we were to suppose the 
 same forces with the same points of application to act parallel 
 to Y Y instead of X X, the distances q t q 2 being measured par- 
 allel to X X instead of Y Y, we should have the force polygon 
 G! O li %i 3 t 4 X instead of 001234, and a precisely similar 
 construction would give us o x multiplied by pole distance for 
 the moment of inertia, and a' for the radius of gyration. We 
 recommend the reader to follow through the construction as 
 shown by Fig. 34. 
 
 56. Curve of Inertia Ellipse and Hyperbola of Inertia. 
 II having found the radius of gyration as above, we lay it off 
 from the axis on either side, in a direction parallel to the direc- 
 tions in which <2i $2? etc., are supposed measured, and through 
 the points thus determined draw two parallels to the axis M' 
 and M" on either side, and then suppose the axis to revolve in 
 the plane of the forces about any point as O situated in the 
 axis ; the lines M' and M" also revolve and enclose a curve of 
 the second degree, whose centre coincides with O. Thus, if in 
 PL 10, Fig. 34, we lay off O I along Y Y on both sides of X X 
 equal to o V = k already found, and then let X X revolve 
 about O, K J and J K will also revolve, and enclose either an 
 ellipse or hyperbola. 
 
 In order to prove this, take O as an origin of co-ordinates. 
 Let the co-ordinates of the points of application of the forces A x 
 A 2 , etc., be x t y } , a? 2 y etc. From each of these points A draw 
 parallels to the axis of y, intersecting the axis of x in the points 
 C. Then O C = a?, A C = y. Now pass through the point O 
 an axis of moments M in any direction, and project for each 
 point OCA parallel to this axis upon the line q, which meas- 
 ures the distance of each point from the axis of moment (not 
 necessarily perpendicular distance). This projection is evi- 
 dently equal to q. Denote by a and /3 the ratios by which dis- 
 tances along X and Y must be multiplied, in order to obtain 
 their projections upon , by lines parallel to M. Then 
 
 q=a%+ @y 
 for each point of application, and hence 
 
 or since for one and the same axis M, and direction , a and 
 are constant, 
 
CHAP. VI.] MOMENT OF INERTIA. 67 
 
 In this expression a and /9 will vary with the position of M 
 and the direction of ^, but 2 P a? 2 , 2 P y 3 remain unchanged. 
 These last expressions are, however, nothing more than the mo- 
 ments of the second order (moments of inertia) of the given 
 force system with reference to the co-ordinate axis, the distances 
 of the points of application being measured in the direction of 
 the axis. They are known if the force system is given and the 
 co-ordinate system assumed. 
 
 If we put 2 P ar 1 = a 2 2 P, 2 P f = tf 2 P, 2 P x y = f 2 2 P, 
 and #, etc., are the radii of gyration of the moments of inertia 
 with reference to x and y, and the above eqiiation becomes 
 ^P (f = 2 P [a 2 a? + /3 2 & 2 + 2 a /9/ 2 ] 
 
 If we conceive for the assumed position of M, the radius of 
 gyration "k to be found, and M' and M" drawn on either side 
 at a distance &, measured parallel to , and indicate the dis- 
 tances cut oft' by these lines from the co-ordinate axes by x e 
 2/ e , and then project these distances parallel to M upon the 
 direction of q> or &, we have Jc = a x e = ft y^ whence 
 
 K Q K 
 
 a p = 
 
 e y* 
 
 and these values substituted in the above equation give 
 
 where ^ is essentially positive in the second term. 
 Hence, 
 
 If we suppose the axis M to change its position revolving about 
 O, the segments x e y & cut off from the axes of x and y by 
 M 7 and M" alone will change in this equation. It is therefore 
 the equation of the curve enclosed by M 7 M". If this curve is 
 known for a given force system, then the moment of inertia for 
 any axis passing through its centre is easily found. We have 
 only to draw parallel to the axis two tangents to this curve, one 
 on either side, and measure their distance from M, in the direc- 
 tion in which the distances q of the points of application from 
 the axis are taken. This distance is the radius of gyration, 
 and the moment of inertia is simply the product of its square 
 by the algebraic sum of the forces. 
 
68 MOMENT OF INERTIA. [CHAP. VI. 
 
 We call the curve represented by the above equation there- 
 fore, the curve of inertia. If we refer the curve to co-ordinate, 
 axes which coincide with the conjugate diameters, the equation 
 becomes 
 
 ^ + I 2 = i 
 
 o ' o - 1 * 
 
 a? y 2 
 
 where x and y are the new ordinates, and A, B, the conjugate 
 semi-axes of the curve. A and B are therefore the radii of 
 gyration of the force system, measured in the direction of the 
 co-ordinate axes, and hence 
 
 where x and y are the co-ordinates of the points of application 
 of the given forces. 
 
 . Since 2 P <f = Z? 2 P if the sign of -P <f is the same as 
 2 P, J is positive. When, on the other hand, these signs are 
 different, #* is negative. That is, when all the forces act in 
 the same direction W is positive, and we have 
 
 A 2 "R 2 
 + 1 
 
 X * + y*- 
 
 which is the equation of an ellipse. 
 
 If, however, the parallel forces act in different directions, k t 
 may be positive or negative. For cases where ^ is negative, 
 either A 2 or B 2 will be negative, and we shall have 
 
 ' 
 
 or 
 
 Both cases coincide. The double curve consists of two hyper- 
 bolas with common assymptotes, common centre, and equal 
 semi-axes. For every axis M passing through the common centre 
 O, we have a pair of parallel tangents either to one or the 
 other hyperbola. The corresponding ^ is positive for the one, 
 negative for the other. 
 
 If, then, in the method of construction to which we shall 
 presently refer, the square of the semi-axis B, which lies in the 
 axis of Y, is negative, that hyperbola whose imaginary axis lies 
 in Y gives ^ positive, the other gives ^ negative, and reversely 
 for the other case. If the axis of moments M coincides with 
 
CHAP. VI.] MOMENT OF INERTIA. 69 
 
 one of the common assymptotes, the radius of gyration and 
 moment of inertia with respect to it of the given force system 
 is zero. 
 
 57. Construction of the Curve of Inertia. The curve of 
 inertia for a given system of parallel forces and given centre 
 O, is determined by the direction of any two conjugate diame- 
 ters, since as we have seen in Art. 55, PL 10, Fig. 34, these 
 directions being assumed we can find the radii of gyration with 
 respect to XX and Y Y, and can thus determine O a and O 5, 
 the semi-diameters. We have then to develop a principle by 
 means of which these directions may be determined. 
 
 If we denote the distances of the points of application of the 
 forces from the axis of M measured in any direction by y, then 
 the statical moments of the forces, P y, are indeed dependent 
 upon the direction in which y is measured, but their relative 
 values remain the same. If then being found for any direction 
 of y, these statical forces are considered as being themselves 
 parallel forces acting at the points of application, and their 
 centre of action is found (for gravity centre of gravity) for 
 some other value of y, this centre of action remains unchanged. 
 For any axis passing through this centre of action the sum of 
 the moments of the forces is zero. If therefore we take a point 
 O in the axis M as origin of a system of co-ordinates, whose 
 axis O X may lie at will in the plane of the forces, while O Y 
 passes through the centre of action ; the sum of the moments 
 of the statical moments P y, considered as forces acting at the 
 points of application, with reference to O Y, will be zero. 
 These moments however, provided that the distances of the 
 points of application are measured along the co-ordinate axes, 
 are the moments of 'inertia, viz., 2 P y x. If these are zero we 
 see that the general equation of the curve of inertia (1) Art. 56, 
 becomes that of a hyperbola referred to its conjugate diameters 
 as axes. With the centre O therefore, the line joining O with 
 the centre of action, gives the direction of the conjugate di- 
 ameter of the curve. 
 
 This is the principle required. By means of it we can find 
 the conjugate diameters of the inertia curve, for a given centre 
 O, and thus construct it. 
 
 58. Contruetion of the Curve of Inertia for four paral- 
 lel force in a Plane. Example. As an example let us 
 take the four parallel forces, in PL 10, Fig. 34, supposed 
 
70 MOMENT OF INERTIA. [CHAP. VI. 
 
 to act in different directions, parallel to X X at the points 
 A! Ag, etc. 
 
 As before we have the force polygon C 1 2 3 4 for an arbi- 
 trary axis as X X, and from the corresponding equilibrium 
 polygon, we determine the statical moments with reference to 
 X X, 01' 1/2', etc., to the basis C 0. These moments we again 
 consider as parallel forces acting at A t A 2 , etc., for which we 
 have CO V 2' 3' 4' and corresponding equilibrium polygon 
 C I IT ffl', etc. We then determine the centre of action S, by 
 a second polygon 0" I II" III", etc., the sides of which are 
 respectively perpendicular to the first, according to the process 
 for finding the centre of gravity, Art. 30. The line joining O 
 with S gives th& direction ofYY, the diameter of the curve 
 conjugate to X X. To find the length of the semi-diameters 
 1} and O #, we must find the moments of inertia of the forces 
 with reference to X X and Y Y, taking the distances of the 
 points of application as measured parallel to these lines. 
 
 Therefore instead of C 0, we must take C y as basis or pole 
 distance, and then find the radii of gyration as already indi- 
 cated in Art. 55, viz., O b f and O a' '. These distances laid off 
 along Y Y and X X give the semi-conjugate diameters of the 
 curve of inertia. 
 
 From the Fig. we see that the force P l whose direction from 
 left to right we shall always consider positive, and 5* P 4 
 have the same sign. On the other hand the total moment of 
 inertia y and the moment of inertia of P l5 viz., l"_have dif- 
 ferent signs. The square of radius of gyration #* = ~ y p 
 
 is therefore negative, the radius itself or the semi-diameter O b 
 is imaginary. 
 
 In similar manner, we see that O a the radius of gyration for 
 Y Y is real, since the total moment of inertia O x and S P 4 1? 
 have the same signs. The curve is then a double hyperbola 
 with the conjugate semi-diameters O a and O b. 
 
 It is then easy to find the assymptotes K K and J J, and by 
 bisecting the angle which they make, the principal axes A A 
 and B B. In order to find the length of these axes, we have 
 the well-known principle that for any point as &, the product 
 of a k and k O (a ~k being parallel to the assymptote J J) is equal 
 
 to -r the sum of the squares of the semi-axes ( j ). If then 
 
CHAP. VI.] MOMENT OF INEKTIA. 71 
 
 we find Jc I, the mean proportional of O k and Jc a, and lay it off 
 twice from O to D along the assymptote O K, O D is the diag- 
 onal of a rectangle whose sides are the principal axes. We 
 thus find the vertices A, A, B, B. 
 
 We can thus construct the curves. Then for any position 
 of the axis X X as it revolves about O, we can find the cor- 
 responding radius of gyration and consequently the moment of 
 inertia, by simply drawing tangents to the curve above and 
 below the new position of X X w&& parallel to it. The radius 
 of gyration thus obtained measured to the scale of length and 
 multiplied by the algebraic sum of the forces, or 4 to the 
 scale of force, will give the moment of inertia required for the 
 assumed position of the axis. 
 
 59. Central Curve. Central Ellipse. If the point O about 
 which the axis turns coincides with the centre of action (or 
 gravity) of the forces, we call the curve enclosed by the paral- 
 lels M' M" at the distance k on either side, the central curve. 
 When the parallel forces all act in the same direction this curve 
 is always an ellipse. 
 
 For the central curve the principle proved in Art. 57 and 
 the method of construction given in Art. 58, are no longer 
 applicable, for the algebraic sum of the statical moments of 
 the given forces is zero for every axis through the centre of 
 gravity. We cannot therefore find the centre of gravity of the 
 moments of the forces, w r hen considered as forces themselves 
 and applied at the given points of application. 
 
 If we divide, however, these moments considered as forces 
 into two portions or groups, and find the centre of gravity of 
 each group, the line joining these two points has an important 
 property, viz., that for every moment axis parallel to it, the 
 algebraic sum of the moments of the statical moments consid- 
 ered as forces, that is, the algebraic sum of the moments of 
 inertia of the forces, is zero. In other words, 2 P e e' is zero, 
 e being the distances of the points of application from the first 
 axis, which passes through the centre of gravity of the forces, 
 and e' the distances from the axis parallel to the line joining 
 the two centres of gravity of the two groups of statical moments 
 considered as forces. If we draw then through the centre of 
 gravity of the forces themselves the moment axis X X, and 
 take it as the axis of abscissas of a co-ordinate system whose Y 
 axis passes also through the centre of gravity of the forces and 
 
72 MOMENT OF INERTIA. [CHAP. VI. 
 
 is parallel to the line joining the two centres of gravity of the 
 statical moments considered as forces, then the moments of 
 inertia 2 P y x are zero, and hence as in the preceding Art. 
 this axis ofYis conjugate to X X. 
 
 This holds good not only for the central curve, but also for 
 every inertia curve, whose centre O instead of coinciding with 
 the centre of gravity of the forces, lies in the axis passing 
 through that centre. In this case also the axis through the 
 centre O parallel to the line of union above, is a conjugate to 
 X X. Still more, the half length of this conjugate diameter is 
 in both cases the radius of gyration of the force system for the 
 axis X X and the direction of Y. 
 
 Hence in every inertia curve of a system of parallel forces, 
 whose centre lies in an axis passing through the centre of grav- 
 ity of the forces, the diameters conjugate to this axis are paral- 
 lel and equal. All these inertia curves are therefore touched 
 by two lines parallel to this axis and equally distant on. either 
 side. This distance is the radius of gyration for this axis. 
 
 For any such inertia curve, whose centre O is distant i from 
 the centre of gravity S of the forces, we call E and ( the par- 
 allel conjugate axes to S O for this curve, and the central curve 
 respectively; q and q the distances from them of any point of 
 application, these distances measured parallel to S O, and con- 
 sidered positive when the point of application lies on the same 
 side of E or (S respectively as the centre of gravity S from E. 
 Then *, the distance apart of E and ( is essentially positive, 
 and if we indicate by a and a the lengths of the semi-conj ngate 
 diameters for the inertia and central curve respectively, we 
 have 
 
 where q and q stand in the simple relation 
 
 q = q+i. 
 Hence 
 
 Since (: passes through the centre of gravity $P q = o, and 
 therefore 
 
 2 P q 2 = 5 P/ +* S P = a* 2 P. 
 Hence 
 
 # = a a +, 
 
 an equation which gives the relation between the lengths of 
 the semi-conjugate diameters of the central and. any inertia 
 
CHAP. VI.] MOMENT OF INERTIA. 73 
 
 curve, whose centre lies upon an axis through the centre of 
 gravity of the forces, at a distance \' from this centre. 
 
 Any two curves at equal distances either side of the centre of 
 gravity are therefore equal. If the semi-diameter of the central 
 curve a is real, and therefore a 2 positive, a 2 is also positive and 
 greater than a 2 . All the inertia curves are therefore of the 
 same kind as the central curve, and enclose the centre of grav- 
 ity. If, however, a 2 is negative, and the central curve there- 
 fore an hyperbola; all those inertia curves whose centres are 
 distant from the centre of gravity by a distance i less than a 
 are hyperbolas also. For a distance i equal to a, the curves re- 
 duce to straight lines equal and parallel to the conjugate diame- 
 ter of the central curve. For i greater than a, the curves be- 
 come ellipses. 
 
 6O. Centre of Action of tlie Statical Moments of the 
 Force.* We again suppose, through the centre of gravity of 
 the forces S [Fig. 35, PL 11] a line N N drawn which cuts the 
 central curve at A and A'. Two such points we have in every 
 case, except when the curve is an hyperbola, andN N coincides 
 with an assymptote. 
 
 Let (5 be the conjugate axis to N N in the central curve, E a 
 parallel to it through any point o distant i from S, and also 
 conjugate to N N in the inertia curve whose centre is o. Then 
 since the statical moments' of the forces with reference to N N 
 is zero, the centre of gravity of the statical moments with re- 
 spect to E ? considered as forces acting at the points of applica- 
 tion, will be somewhere upon N N. It is required to find 
 where. 
 
 We call q the distance of any point of application from E, 
 measured parallel to N N, and positive when upon the same 
 side of E as S, then i is essentially positive. 
 
 As before, q is the distance of the points of application from 
 (S, also measured parallel to N N, and positive in the same 
 direction as q. 
 
 Then we have always 
 
 q = q+i. 
 
 and for the moments of inertia of the forces with respect to E 
 and( 2P<? = ZP (q+i) 2 = 2Pf+#2P 
 
 or when a is the semi-diameter of the central curve, S A = S A' 
 and ^Pq 2 = a*+$ 2P 
 
 * See Supplement to Chap. VIL, Art. 10, latter part. 
 
74 MOMENT OF INERTIA. [CHAP. VI. 
 
 Let now m be the distance of the centre of gravity or action, 
 of the moments of the force's with respect to E, from E, and in 
 its distance from (5, positive the same as q and q. Then 
 
 m = tn + i 
 
 and since the sum of the moments is equal to the moment of 
 the resultant : 
 
 ^Pq 2 m^Pq. 
 
 But the sum of the moments P q of the forces with reference 
 to E, is equal to the product of the sum of the forces into the 
 distance i of the centre of gravity of the forces from E. Hence 
 
 and therefore 
 
 m i 2 P = 2P q 2 = (a 
 or, m. i a 2 + r. 
 
 Introducing the value for m 
 
 (m + i) i = a 3 -f ^ 2 
 or tit i a 2 . 
 
 If now a 2 is positive, which is always the case for an ellipse 
 as central curve, m is also positive, and is therefore to be laid 
 off from S along N N on the opposite side of Qj from o. If 
 then we conceive an axis E' drawn parallel to E, and symmet- 
 rical with reference to S, which axis we shall call for conven- 
 ience the symmetrical axis to E, we see from the above relation 
 that M is the pole of this axis in the central curve. 
 
 If, however, a 2 is negative, therefore a imaginary, nt is nega- 
 tive, and must be laid off from S towards 0, and the point M 
 thus found is therefore the pole of the axis E itself, or in the 
 case of an hyperbola is the pole of E' in that hyperbola which 
 is not cut by N N, and for which therefore A A! is imaginary. 
 
 Hence we have the principle 
 
 If we consider the statical moments of the forces with refer- 
 ence to any axis as E as themselves forces acting at the given 
 points of application, the centre of gravity of these moment 
 forces does not coincide with the centre of gravity of the origi- 
 nal forces, but is the pole * in the central curve of an axis E' 
 parallel and symmetrical to E. 
 
 In those cases where the central curve becomes an hyper- 
 
 * POLAR LINE OF A POINT, in the plane of a conic section, is a line such, 
 that if from any point of it two straight lines be drawn tangent to the conic 
 section, the straight line joining the points of contact will pass through the 
 given point, which is called a pole. 
 
CHAP. VI.] MOMENT OF INERTIA. 75 
 
 bola, we must observe whether the diameter conjugate to the 
 moment axis is real or imaginary. In either case the centre of 
 gravity is the pole of the line symmetrical to the moment axis 
 in that hyperbola for which that diameter is real or imaginary. 
 
 The construction is given in PL 11, Fig. 35. 
 
 Upon S o' = S o we describe a semi-circle. With S as cen- 
 tre, and S A' = a = semi-diameter of the central curve, describe 
 an arc, and from the intersection with the semi-circle drop a 
 perpendicular upon So'. The point M thus found is the centre 
 of gravity of the moments. For : a 2 = a M* + m 2 and a M 2 
 = ra (im) hence a 2 = m 2 +m i m 2 m i. The central curve 
 being known as also the distance i, the point M can be readily 
 found. 
 
 61. Cases where the Direction of the Conjugate Axis of 
 the Inertia Curve can be at once Determined. There are 
 certain special and practical cases in which the conjugate direc- 
 tions or axis of the inertia curve can be at sight determined, so 
 that only the length of the semi-diameters remains to be found. 
 The most important of such cases are as follows : 
 
 (1.) When in a system of parallel forces, these forces can be 
 so grouped in pairs, that the lines joining the points of appli- 
 cation of each pair are all parallel, and the centres of gravity 
 of each pair all lie in the same straight line. Then for the 
 central curve and all inertia curves whose centres lie upon this 
 straight line, the direction of the axis conjugate to this line is 
 the same as that of the lines joining the points of application 
 of each pair. 
 
 This is easy to prove. For, for each pair, the sum of the 
 moments with respect to the line joining their centres of gravity, 
 is zero. These moments regarded as forces and applied at the 
 points of application, give therefore for each pair two parallel 
 opposite and equal forces, the sum of the moments of which 
 for any line parallel to the line joining the points of applica- 
 tion, is zero. This is the case for all the pairs, and therefore 
 the direction of the lines joining the points of application is 
 that of the axis conjugate to the line joining the centres of 
 gravity, for the central curve as also all inertia curves whose 
 centres lie upon this last line. 
 
 (2.) When the forces can be so grouped that the points of ap- 
 plication of each group lie in parallel lines, and the centres of 
 gravity of the groups lie in the same straight line. Then this 
 
76 MOMENT OF INEBTIA. [CHAP. VI. 
 
 straight line gives the direction for the central curve and every 
 inertia curve whose centre lies upon it, of the diameter conju- 
 gate to an axis passing through the centre and parallel to the 
 lines joining the points of application. 
 
 For if we take any such axis, the points of application of the 
 forces in each group are equally distant. The statical moments 
 for each group are then proportional to these distances. If, 
 therefore, they are considered as forces, their centre of gravity 
 coincides with that of the forces themselves, and lies therefore 
 in the line joining the centres of gravity of the groups. The 
 centre of gravity of the whole force system lies then in this 
 line, which is therefore the direction of the axis conjugate to 
 the line parallel to the lines joining the points of application, 
 in the central curve, and also all curves whose centres lie upon 
 this line. 
 
 (3.) When the forces can be so grouped that the centres of the 
 central curves of each group lie in the same straight line, and 
 the diameters in each curve conjugate to this line, are parallel. 
 Then in the central curve of the entire system, the diameter 
 conjugate to this line is also parallel to these diameters. For, 
 for any axis parallel to these diameters, the centres of gravity 
 of the moments of the forces in each group lie upon the line 
 joining the centres of the curves. The centre of gravity of the 
 moments for the entire system lies then also upon this line, 
 which is therefore the direction of the axis conjugate to an axis 
 parallel to the diameters of the curves, for any inertia curve 
 whose centre lies upon this line. 
 
 In all these cases, if the directions thus found are perpendicu- 
 lar, we have to do with the principal axes. 
 
 62. Practical Applications. We can now apply the above 
 principles to practical cases, and as in the determination of the 
 moment of inertia of irregular figures, we have to deal with 
 triangles, parallelograms and trapezoids, we have first to con- 
 sider these three cases. 
 
 1st. The Parallelogram. PL 11, Fig. 36. 
 
 The moment of inertia of a parallelogram is, as is well known, 
 
 ^ == 12 a ^** a k e * n g tne breadth and 5 the depth. 
 
 ax 2 dx = ab* 
 
CHAP. VI.] MOMENT OF INERTIA. 77 
 
 Hence T& = b 2 = radius of gyration, or k= *J -bx b. 
 
 That is, the radius of gyration is a mean proportional between 
 
 1 , . 1 _ 
 -b and -b. 
 
 The centre of gravity of the parallelogram is at O the inter- 
 section of the diagonals, and this is therefore the centre of the 
 central curve. 
 
 If we suppose the parallelogram divided into laminae parallel 
 to D C, and suppose each lamina divided by G H parallel to 
 B C, the centres of gravity of each will lie upon G H. Right 
 and left of G H we then have a group of forces whose points of 
 application lie in lines parallel to G H, and the lines joining 
 any pair, one on each side of G H, are parallel. By (1) of the 
 preceding Art., therefore, GH and EF are conjugate axes of 
 the central curve. For the lengths of the half diameters, we 
 
 find the mean proportional between -^ b and -^ 5, -^ a and -^ #, 
 
 respectively, by the half circles B F and BH. We thus find k 
 and &', and can then construct the central ellipse directly, or 
 find the principal axes, and then construct it. The centre of 
 gravity of the moments of the parallelogram, with reference to 
 any axis parallel to A B, is as we have seen, Art. 60, the pole of 
 a line parallel and equally distant from O on the other side. If 
 we draw this line then, as D C, then from G draw two tangents 
 to the central ellipse, and unite the points of tangency by a 
 line ; the intersection of this line with O G is the centre of 
 gravity of the moments of the forces themselves considered as 
 forces, or area of the parallelogram, with reference to A B. 
 
 Zd. Triangle. PL 11, Fig. 37. 
 
 The moment of inertia of a triangle for the axis B C is 
 
 .H5 a A 3 * whence # = -~ A 2 , and for an axis E F distant i = 
 
 L O 
 
 o A, which passes through he centre of gravity, 
 a* = tf-# = tf- (Art. 59.) 
 
 I 
 
 a '=f tfdx = -^-ah 3 ,h being the line A D, a = B O. 
 
78 MOMENT OF IKERTIA. [CHAP. VI. 
 
 The conjugate axes of the central curve are by principle 1 or 
 2 of the preceding Art. E F and A D. 
 
 The above value of a is then the length of the semi- diameter 
 
 along A D, or a V -^ h x -= h. That is, a is a mean propor- 
 
 tional between -^ h and -~ h. This is found by the semi-circle 
 
 ODFig. 37. 
 
 The moment of inertia of the triangle with respect to A D is 
 
 g h (~o^) 3 - The radius of gyration then is y -^ {-^af = 
 
 ' "2 \% a ) x ^ \v a ) or a mean proportional between , and 
 |offorDC. 
 
 This is given by the semi-circle on D G = ^ D C, and we 
 
 thus have the four points 1 2 3 4 of the central ellipse, and the 
 semi-diameters 1 and 3, and can therefore construct it. 
 From the central ellipse as before, we can find the centre of 
 gravity of the moments considered as forces for any axis par- 
 allel to B C or A D, as also in either case, the radius of gyration 
 and therefore moment of inertia, for any axis passing through O. 
 
 3d. Trapezoid. PL 11, Fig. 38. 
 
 Here the lines E F joining the centres of the parallel sides, 
 and G H parallel to these sides, aird passing through the centre 
 of gravity 0, are the conjugate axes of the central ellipse. 
 
 For the axis A B and direction E F, the moment of inertia is 
 
 a and 5 being A B and C D, and h = E F. The square of 
 radius of gyration is then 
 
 1 . ,. , ~6 a+b 
 
 2 
 
 - 
 
CHAP. VI.] MOMENT OF INEKTIA. 79 
 
 For the radius of gyration for G- H, at a distance i = -r- h 
 
 - we nave 
 
 + H 
 
 9 
 
 la + Sb I , 2 ^ + 2 ^\ 2 
 
 0? = %? $ =- - T- A 2 rr A 2 I - --T- 1 = 
 
 6 a + b 9 V a + b / 
 
 1 /a 
 J *' 
 
 This radius a is half the diameter along E F. 
 To construct it, put (3 a) 2 = -^ A 2 + -/ T~T\2 ^ 2 - 
 
 Describe a semi-circle upon E F, and at the centre 1? arid 
 at the intersection of the diagonals &, erect perpendiculars 
 
 Ol J and K L. Then F"! 2 = A 2 and iTL 2 = 7-^ A 2 , since 
 
 ^ 
 
 E K = -- j h and K F = --- 7 A 
 a + b a + b 
 
 K L equal to J M from J, we have 
 
 E K = -- j h and K F = --- 7 A. If therefore we lay off 
 a + b a + b 
 
 
 and hence the half diameter sought is one-third F M. "We 
 thus find 1 and 2. 
 
 To find the other semi-diameter we have the moment of in- 
 
 ertia for E F and direction G H, jg- (a* -f- a* I + a 
 hence the square of the radius of gyration is 
 
 -^ 
 This last expression is easily constructed^ In the right-angled 
 
 triangle F B N, the hypothenuse F N = A/ (- a\ 2 + (^ b\*> 
 B N being made equal to C E. If we describe then a semi- 
 
 J / ' 7iy*dy+ I *h a -^-y^y 
 
80 MOMENT OF INERTIA. [CHAP. VI. 
 
 circle upon F U = -~ F N, and make F W ^ F N, F V is the 
 
 semi-diameter sought. We thus find 3 and 4, and can now 
 construct the central ellipse. This being constructed we can 
 find the centre of gravity of the moments with reference to any 
 axis parallel to A B or E F, according to Art. 60, or the moment 
 of inertia for any axis through 0, by drawing a parallel tangent 
 to the ellipse. The distance from to the point o tangency 
 gives then the radius of gyration for that axis. 
 
 Uh. Segment of Parabola. PL 11, Fig. 39. 
 
 Let the segment be limited by B C = 2 A, and A D = I. 
 Then it is evident that these two axes are conjugate (Art. 61), 
 and the centre of the central curve is 0, the ratio of A to 
 OD being as 3 to 2. Hence AD and E'F', parallel to CD 
 through 0, are conjugate axes of the central curve. To find the 
 length of the semi-diameters along these axes we find first the 
 moment of inertia of the segment with reference to an axis Y Y 
 parallel to E' F' and tangent to the parabola at A. We have 
 then for this moment of inertia 
 
 /' 
 
 Jo 
 
 where p is the parameter of the parabola, and I A D. Since 
 
 4 
 3 
 
 4 
 the area of the segment is -~ h I, we have for the square of the 
 
 radius of gyration 
 
 The square of the radius of gyration then for E' F' whose 
 
 distance from A is i = I is 
 o . 
 
 a being the semi-diameter along A D. It is easier here to com- 
 pute a, viz., a = 0.26186 I, and lay it off from O, thus finding 
 3 and 4. 
 
 For the other, semi-diameter we find the moment of inertia 
 for A D and the direction E' F'. Thus 
 
 
CHAP. VI.] MOMENT OF INERTIA. 81 
 
 The radius of gyration squared is, therefore, 
 
 and hence the radius of gyration is & = 0.44721 A. Laying 
 this off from 0, we obtain 1 and 2, and can therefore now draw 
 the central ellipse. 
 
 63. Compound or Irregular Cro-Section. Every cross- 
 section may be divided up into trapezoids, triangles, parallelo- 
 grams and parabolic segments, and the above cases will aid us, 
 therefore, in the application of the graphic method to compound 
 or irregular cross-sections. The engineer is often called upon 
 to determine the moment of inertia of such sections as the T, 
 double T, or different combinations of these in proportioning 
 the different pieces of bridges, such as chords, struts, floor-beams, 
 etc., as also in many other constructions. The calculation for 
 such cross-sections is sometimes very laborious. As an example 
 ' of the application of the graphical method best illustrating the 
 above principles, we take the cross-section shown in Fig. 40, 
 PL 12. 
 
 First we divide the cross-section into a series of trapezoids. 
 The first segment, bounded by a curve, we may consider a para- 
 bolic area. These trapezoids we reduce to equivalent rectangles 
 of common base a [Art. 32], and take the corresponding heights 
 as forces. These forces we lay off in the force polygon and 
 choose a pole C at distance H from force line, drawing CO, 01, 
 C 2, etc. Parallel to these lines we have the first equilibrium 
 
 polygon I II III VIII, the intersection of the two outer 
 
 sides of which gives the point of application of the resultant. 
 The intersection S of the resultant with the axis of symmetry 
 gives the centre of gravity of the cross-section [Art. 30]. The 
 segments o 1', 1'2', 2'3', etc., cut off from o S, give the statical 
 moments of the forces with reference to o S to the basis H. 
 We now choose another pole C' at distance H', and form another 
 force polygon, considering these moments as forces, and applied 
 at the centres of action of the moments of the separate areas 
 into which the whole cross- section has been divided. These 
 centres of action can be determined by forming the central 
 curve for each area according to Art. 62, and then applying the 
 6 
 
82 MOMENT OF INERTIA. [CHAP. VI. 
 
 principle of Art. 60. A little consideration will show that these 
 centres of gravity will coincide approximately with the centres 
 of gravity of the areas themselves, except for areas (3) (4) (5) 
 and (6). Finding then for these areas the centres of action of 
 the moments considered as forces, we construct the equilibrium 
 polygon O' I' IT. . . .VIII'. The distance 0" 8" cut off by the 
 first and last sides of this polygon gives the moment of inertia 
 to the pole distances H and H' and the reduction base a. Thus 
 0" 8" measured to scale of force and multiplied by a H H' is 
 the moment of inertia of the cross-section with reference to o S. 
 
 In TI H' (V'8" 
 
 The radius of gyration is then k=\J = . 
 
 a 8 
 
 The division will be performed if we take H' = 8 = 2 P. 
 This we can easily do now without drawing a new polygon, 
 since what is required is the intersection of the outer sides only. 
 Thus take a new pole C/ distant from o S, H' = 8. Now we 
 know that each side of the new polygon for this pole distance 
 will intersect the corresponding side of the first in a line paral- 
 lel to o GI [Art. 27]. Since the new polygon may start from 
 any point, we may take the first side to coincide with O VIII'. 
 Then the line of intersection of any two sides is O VnT 8". 
 Produce any side as IV' V 7 to intersection e with this line ; from 
 e draw e a/ parallel to C/ 4'. 
 
 Through a! the intersection of o' I' and V IV, the resultant 
 of (1) (2) (3) and (4), must pass. The change of pole cannot 
 affect this resultant, which must therefore pass through a/, the 
 intersection of <? a/ with the vertical through a' parallel to o S. 
 Hence 0/ a t ' is the direction of the last side of the new poly- 
 gon, and S f/ 0i" is the moment of inertia for the new pole dis- 
 tance o C/ = 8. The radius of gyration then is k = V H O/'S"! 
 In other words, Jc is a mean proportional between H and O/'S". 
 The construction of k is given by the semi-circle described upon 
 O/'S" + H. The ordinate to this semi-circle through O/ 7 per- 
 pendicular to <?S gives Jc. We thus find the semi-diameter 
 S a = S a' of the central ellipse. 
 
 In order to find the other semi-diameter S b = S ', we might 
 divide the cross-section into areas by lines parallel to S X, and 
 then proceed as above. This is, however, unnecessary. "With 
 the same areas as before, we can find the central curve for that 
 area on each side of X X, and then the centre of application of 
 
CHAP. VI.] MOMENT OF INERTIA. 83 
 
 the moment of each of these areas with respect to X X itself, 
 considered as a force. The method of procedure is then pre- 
 cisely as before. We draw a polygon the sides of which are 
 respectively perpendicular to those of the first polygon, and 
 thus find the statical moments 0"' I'" V" 2'", etc., to basis H. 
 
 Choosing then a pole C'" at distance H'" and drawing the 
 corresponding polygon, we have 8 1V for the moment of in- 
 
 ertia. The radius of gyration is then Jc = \ aHH "' Q8IV 
 
 a. 8 
 
 
 We have taken H'" = g 08, hence k = 1 H QQ- iv Hence k 
 
 1 _ 
 
 is a mean proportional between g 8 1V and H. The construc- 
 
 1 _ 
 
 tion is given in the Fig. by a semi-circle upon H+ ^ 8 IV . We 
 
 thus find the semi-axis S V S , and can now construct the 
 central ellipse. We have thus found graphically not only the 
 moments of inertia of the cross-section with respect to X X and 
 Y Y, but, by means of the central ellipse, for any other axis in 
 the plane of the Fig. passing through S. 
 
 64. The above method of procedure holds good generally 
 for any cross-section, except that, when there is no axis of sym- 
 metry, the centre of gravity must be found by a second equili- 
 brium polygon whose sides are respectively perpendicular to 
 those of the first. When the moment of inertia with reference 
 to a single axis only is required, the above method becomes 
 quite short and simple, as well as accurate. In our Fig. the 
 scale used as also the number of divisions taken make the pro- 
 cess appear more complicated than it really is. 
 
 With this we shall close our discussion of moment of inertia, 
 merely observing, that all the principles deduced in this chap- 
 ter for forces acting in a plane hold equally good for forces in 
 space. The central curve then becomes an area, we have a mo- 
 ment plane instead of moment axis M, and the ellipse and hyper- 
 bola of inertia become ellipsoid and hyperboloid respectively*. 
 
 For a much fuller discussion of the subject than is possible 
 here, we refer the reader to Culmanrfs Graphische Statik, pp. 
 160-206 ; also Bauschinger's Elemente der Graphischen Statik, 
 pp. 116-168. To the latter we are largely indebted in the 
 preparation of the present chapter ; Plates 10 and 12 are, with 
 slight alteration, reproduced from that work. 
 
84- APPLICATION TO BRIDGES. PART -II. 
 
 PART II. 
 
 APPLICATION TO BRIDGES. 
 
 65. Under the head of Parallel Forces we have already 
 given the general application of the graphical method to the 
 determination of the moments and shearing forces in beams 
 resting upon two supports only. We shall now take the sub- 
 ject up more in detail, and show the methods of determining 
 the maximum strains for all the possible conditions of loading 
 which may occur in Bridge Girders. In the following we shall 
 adhere closely to the development of the subject as given by 
 Winkler. {DerBruckenbau, Wien, 1872.] 
 
 66. Force wliicli act upon a Bridge. The forces which 
 act upon a bridge may be enumerated as follows : 
 
 1st. The weight of the bridge itself. This, previous to the 
 calculation of the strains, is unknown, since it depends upon the 
 intensity of the strains themselves. It is customary to assume 
 the weight to begin with, by comparison with existing struc- 
 tures of similar character, and then to find the resulting strains. 
 The weight answering to these strains can then be easily ascer- 
 tained ; the strength of the materials used being known, and 
 compared with the assumed weight. According as it is less or 
 greater, the weight was then assumed too great or the reverse, 
 A second approximation to the true weight may then be made, 
 and the strains proportionally diminished or increased. As 
 rules for estimating the weight of bridge girders under 200 feet 
 span, we have, for weight of girder G-, 
 
 where W = the assumed approximate total distributed load in 
 tons, including the weight of girder ; 
 
 I = length in feet ; 
 
 d = depth in feet ; 
 
PAKT II.] APPLICATION TO BRIDGES. 85 
 
 y= the working strain in tons per sq. foot of cross-section. 
 (See Stoney, Theory of Strains, vol. ii., p. 441.) 
 
 We have also the rule : " Multiply the distributed load in 
 tons by 4 ; the product is the weight of the main girders, end- 
 pillars and cross-bracing in pounds per running foot." Iron is 
 taken at 5 tons per sq. inch tension, and 4 tons per sq. inch 
 compression. 
 
 2d. The moving or live load; which is determined by the 
 purpose of the bridge. This load can take various positions 
 upon the bridge, and may even be divided into several por- 
 tions. It is therefore an important problem to determine that 
 distribution which shall cause the maximum strains. 
 
 The live load is, as the term implies, in motion, so that, in 
 combination with the deflection, there is a centrifugal force, 
 or increase of pressure. This is, however, in practice disre- 
 garded, while such a coefficient of safety is chosen in propor- 
 tioning the parts, that the increase of strain due to this cause is 
 fully covered. 
 
 3d. Horizontal forces, caused by the wind and the passage 
 of loads. 
 
 ^th. Pressures at the supports. The known forces cause re- 
 actions at the supports, which evidently must also be considered 
 as forces acting upon the bridge girder. For straight girders, 
 these reactions are vertical, while in suspension and arch sys- 
 tems they are inclined. 
 
 67. Bridge Loading. The heaviest load to which a railway 
 bridge can be subjected is when it is covered from end to end 
 with locomotives. " The standard locomotive is assumed to be 
 24 feet long, and to have six wheels with a 12-foot base ; to 
 have half its weight resting on the middle wheels, and one- 
 fourth on the leading and trailing pairs respectively, which are 
 supposed to be at equal distances on either side of the middle 
 wheels." (See Stoney, vol. ii., p. 405.) The standard engine 
 is assumed to weigh 24 tons. 30 tons and 32 tons, according to 
 the construction. This makes the standard load 1 ton, 1J ton, or 
 1^ ton per foot of single line. Short bridges of less than 40 
 feet span must be considered as subject to concentrated loads 
 from single engines. 
 
 The maximum load for public bridges is recommended by 
 Stoney at 100 Ibs. per sq. ft. 
 
 68. In the Straight Truss all the Outer Forces act in a 
 
86 APPLICATION TO BRIDGES. [PAET If. 
 
 Vertical Direction. The strain in any cross-section depends 
 upon, first, the resultant of all the outer forces acting either 
 side of the cross-section ; and second, the statical moment of 
 these forces with reference to the cross-section. The first, or 
 the algebraic sum of all the forces acting between the cross- 
 section and either end, we call the shearing force for this cross- 
 section, and indicate it by S. It is also designated as vertical 
 force, or transverse force. The moment of the resultant, or 
 the algebraic sum of the moments of all the exterior forces, 
 with reference to any cross-section, we call the moment for this 
 cross-section, and indicate it by M. It is also called bending 
 moment, or moment of rupture. For example, in a lattice 
 girder with horizontal flanges the strains in the web are pro- 
 portional to the shearing forces, those in the flanges to the 
 bending moments. 
 
 The shearing force is considered positive when it acts on the 
 left side upwards, or on the right side downwards. The mo- 
 ment M is positive, when on the left side the tendency of rota- 
 tion is to the left, on the right side to the right, or when it tends 
 to make the girder convex upwards, that is, causes compression 
 in the lower fibre or flange. 
 
CHAP. VH.] SIMPLE GIKDEKS. 87 
 
 CHAFTEE VII. 
 
 SIMPLE GIKDEKS. 
 
 69. Action of Concentrated Loads Invariable in Posi- 
 tion. By "simple girder" we understand a girder resting 
 upon two supports only, in opposition to a continuous girder 
 which rests upon more than two. 
 
 Suppose a number of forces P l . . . P 5 acting at various points. 
 [Fig. 41, PI. 13.] We form the force polygon by laying off the 
 forces to scale one after another ; then choose a pole O, and 
 draw O 0, 1, 2, etc., to the points of division. Parallel to these 
 lines we draw the lines of the equilibrium polygon between the 
 corresponding force lines prolonged. If now we dose the poly- 
 gon thus formed by the line A B, and draw through O the 
 parallel O L to A B, the segments L and L 5 of the force 
 line give the reactions Vj_ and V 2 . Further, the shearing force 
 between A and P t is S t *= V^ = L ; between P l and P 2 , S 2 = 
 V\ P! ; at P 3 , S 3 = V t P,, P 2 , etc. That is, the shearing forces 
 are the' distances of the points of the force polygon fro in~L>. It 
 is easy, then, to construct them, as shown in the lower shaded 
 area of the Fig. (See also Art. 46.) 
 
 If in the equilibrium polygon we let fall at any point a ver- 
 tical as I K, and from K draw K L perpendicular to A B, and 
 indicate by H the horizontal pull, by L the strain in A B, and 
 by M the sum of the moments of all forces left of I K, then, 
 for equilibrium about K, we have M^LxKL^LxIK cos 
 I K L, or, since the angle I K L = L O H in force polygon, 
 L x cos I K L = H, and hence M = H x I K, or representing 
 the variable ordinate I K by y : 
 
 M = Hy. 
 
 But H is the distance of the pole O from the force line ; 
 the moment at any point is therefore proportional to the verti- 
 cal height of the equilibrium polygon. (See also Art. 38.) If 
 we take H equal to the unit of force, we have 
 
88 SIMPLE GIRDERS. [CHAP. VII. 
 
 so that in this case the moment at any point is directly given 
 by the ordinate of the polygon at that point. It is this impor- 
 tant property of the equilibrium polygon which renders it espe- 
 cially serviceable in the graphical solution of this and similar 
 problems. 
 
 70. Concentrated Load Variable Position Shearing 
 Force. If the load lies to the right of any given cross-section, 
 then the shearing force at this cross-section will be S^V^ or, 
 since we regard a force to the left acting up as positive, S is 
 positive. As the load P moves towards the left, V t or S in- 
 creases. When the load is to the left of the cross-section, the 
 shearing force at the cross-section is S = V^ P, and since P 
 is always greater than V l5 S is negative. The nearer P ap- 
 proaches the cross-section, the smaller is S. 
 
 Hence : a concentrated load causes a positive or negative 
 shear, according as it is to the right or left of the cross-section 
 considered, and the shearing force is greater the nearer the load 
 is to the cross- section. 
 
 Moments. If the load lies to the right of the cross-section, 
 the moment is M = V\ x, x being the distance of the cross- 
 section from the left support. M is therefore negative and in- 
 creases with Y! ; that is, as the load .approaches the cross-sec- 
 tion. If the load is on the left of the cross-section, M V 2 
 (I #), V 2 being the reaction at the right support. Here also 
 M is negative and increases with V 2 ; that is, as the load ap- 
 proaches the cross-section. 
 
 Hence : a concentrated load wherever it lies causes in every 
 cross-section a negative moment, which for any cross-section is 
 a maximum, when the load is applied at that cross-section. 
 
 71. Position of a given System of Concentrated Loads 
 causing Maximum Shearing Force. If P! is the sum of all 
 the loads to the left of any cross-section, the shear at that cross- 
 section is S = Vj P x . As the system moves to the left with- 
 out any load passing off the girder or any load passing the cross- 
 section, Y! and therefore S increases as long as S is positive, or 
 as long as V x > P l8 If a load passes off the girder, then for the 
 remaining loads S increases anew as the system moves to the 
 left, until a load of the system passes the cross-section in ques- 
 tion. The same holds good for a system moving to the right, 
 where S is negative. 
 
 Hence : the shearing force is a maximum for any point, 
 
CHAP. VII.] SIMPLE GIRDERS. 89 
 
 when there is a load of the system at that point, and the maxi- 
 mum is positive or negative, according as the load lies just to 
 the right or left of the point. 
 
 Since for a single load (Art. 70) S is positive or negative, ac- 
 cording as the load is to the right or left, S will be in general 
 a positive or negative maximum when all the loads lie to the 
 right or left, and the heaviest nearest the cross-section. Only 
 in cases where a small load precedes, can S be greatest when 
 the second load lies upon the point in question. 
 
 If P is the resultant of all the loads and j3 its distance from 
 the right support, 
 
 V t = P and therefore S = P p le 
 L L 
 
 Now the position of the loads of the system or @ remaining 
 unchanged, P x will vary as the first power of x, the distance of 
 the cross-section from the left support. Therefore, between any 
 two cross-sections for which the load on the girder remains the 
 same, the shear S is represented by the ordinates to a straight 
 line. 
 
 72. Construction of the Maximum Shearing Forces. 
 Construct the force polygon with the given loads ; choose a 
 pole O [PL 13, Fig. 42 (a)~\ and draw the corresponding equi- 
 librium polygon. It is required to determine the shear S at a 
 crossrsection distant x from the left support, under the suppo- 
 sition that the first load P L of the system, moving towards the 
 left, acts at this cross-section. 
 
 Determine upon the outer side P! A of the polygon passing 
 through the point P l5 a point A distant from P lt by the distance 
 x, and then find the point B upon the polygon distant from A 
 by I, the length of span, and draw A B. Parallel to A B draw 
 O L in the force polygon, then A L = V t = S, the shear at P t . 
 Drop a vertical through B intersecting P! A produced, in M ; 
 then the triangles O A L and P l M B are similar, and there- 
 fore S = A L = B M , when a is the pole distance. If we 
 
 if 
 
 choose a = I, then S = B M. 
 
 Hence : the maximum shearing forces are proportional to 
 the vertical segments between the equilibrium polygon and the 
 prolongation of the outer side taken at the end of the system, 
 or are equal to these segments if the pole distance is taken equal 
 to the span ; provided that the last load is at the cross-section. 
 
90 SIMPLE GIKDEKS. [CHAP. VII. 
 
 We have, therefore, the simple construction given in PL 13, 
 Fig. 42 (b). The broken lines are parallel to the various posi- 
 tions of A B for corresponding positions of P t . The positive 
 and negative values of S equally distant from the right and left 
 supports are equal, so that it is only necessary to construct S 
 for one value. 
 
 If the second load is to be at the cross-section, and if e is the 
 distance between the first and second, we draw first a line 
 
 & _-_.-L- np 
 
 whose equation is 2/ PI -, , and construct, as above, a' 
 
 I/ 
 
 polygon, for which the second load lies on the right support B, 
 and whose second side (between second and third loads) coin- 
 cides with the above line. The ordinates to this line above the 
 axis of abscissas will give maximum of + S. 
 
 73. Maximum Moments. Since, according to Art. 70, a 
 concentrated load causes a negative moment at any point, 
 wherever it may lie, we must have evidently loads upon both 
 sides of any point, in order that the moment may be a maxi- 
 mum. Since a single load causes a greater moment at any 
 point the nearer it lies to that point, the greatest load must lie 
 nearest the cross-section in question. The method of loading, 
 causing maximum moments, can be best determined for a dis- 
 tributed load (not necessarily uniform). In this case the equi- 
 librium polygon becomes a curve [PL 13, Fig. 43]. If in this 
 curve we draw A B, and take C so that A C : C B \;x : I #, 
 then C D = M f or x. Suppose A B to take the position A! B', 
 the horizontal protection of C C' being indefinitely small, 
 then C' D' = M + d M. In order now that M may be a 
 maximum, C' D' must be equal to C D or C C' parallel to D D'. 
 If in the force polygon O A l is parallel to A A', O B x to B B'. 
 and O D t to D D', then A^ ^ and E>! B 1 are the loads upon 
 A C and B C. 
 
 Draw through C a vertical, and through A, A', B, B', paral- 
 lels to C C 7 or D D' intersecting this vertical in E, E', F, F'. 
 
 Then CE:CF::AC:BC;:a;:Z-tf, 
 
 C E' : C F' ; ; A' C' : B 7 C' : ; x : I - x; 
 therefore 
 
 C E : C F ; ; C E' : C F', or C E' : C E ; : C F' : C F ; 
 also CE' CE:CF'-CF;:CE:CF, 
 
 that is, E E 7 : F F' \\x\l- x. 
 
 If now we draw through A' and B' parallels to C C', or D D' to 
 
CHAP. VII.] SIMPLE GIRDERS. 91 
 
 intersections H and I, we have A H = E E', I B' F F'. Since 
 the triangle A A' H is similar to O A! ^ and B B' 1 to O B 1 T> ly 
 and since A 7 H = B I, we have 
 
 A! D t : B! D! ; ; A H : B' I ; ; E E' : F F 7 ; ; x : lx. 
 
 Since A 1 D x equals the load P t on A C, and B! "D^ the load P 2 
 on B C, we have P l : P 2 ; ; x : lx. 
 
 The same will hold true approximately for concentrated 
 loads. Hence, in order that the moment at any point may be 
 a maximum, the system of loads must have such a position 
 that the loads either side of this point are to each other as the 
 portions into which the span is divided. 
 
 In PI. 13, Fig. 44, let C D give the moment at C. If the line 
 A B moves so that the horizontal projections of A C and B C 
 remain equal to x and I x, then as long as the ends A and B 
 move on the same straight lines, the point C will also move in a 
 straight line. The point C describes, therefore, a broken line. 
 The verticals between this line and the polygon correspond to 
 the moments for various positions of the load and a given value 
 of x. Evidently the greatest ordinate will be over an angle of 
 the equilibrium polygon which is not under an angle of the 
 line described by C that is, for M maximum a load must lie 
 upon the cross-section. 
 
 For any cross-section, then, the moment is a maximum when 
 a load is applied at this cross-section. Which of the loads 
 must be so applied is determined by the preceding rule. 
 
 74. Contraction of Maximum Moments. After the 
 equilibrium polygon has been constructed, in order to find M 
 for a point C (PI. 13, Fig. 45), we determine two points P and 
 G upon the polygon which are distant horizontally from the 
 load on the given. cross-section corresponding to the angle E by 
 distances A C, B C. Then draw F G-, and the vertical K E is 
 equal to M when the pole distance is unity. We make C I = E K. 
 In this way we can construct the moments for different loads 
 of the load system at the given cross-section, and thus determine 
 that position of the load which gives the maximum moment at 
 the cross-section. 
 
 Generally when K E = y, and the pole distance is a, we have 
 JML = ay. The pole distance a is measured to the scale of 
 force, and then y is given by the scale of length. The unit for 
 
 M, in order that M may be equal to y, is evidently -th part of 
 
92 SIMPLE GIRDERS. [ CHAP - VIT - 
 
 the unit of length (when the pole distance is a force units), or, 
 what is the same thing, one unit of length is equal to a moment 
 units. The same equilibrium polygon can be used for any 
 number of girders of various spans, hence the method is of very 
 rapid application. 
 
 75. Abolute maximum of Moments. Since for any cross- 
 section M is a maximum when a load lies at that section, a load 
 must also lie upon the cross-section for which M is an absolute 
 maximum. , 
 
 If the line A B slides upon the equilibrium polygon, altering 
 its length so that its horizontal projection is constant and equal 
 to I, it will envelop a portion of a parabola so long as its ends 
 move in the same sides of the polygon. [PL 13, Fig. 46.] The 
 curve thus produced is therefore composed of portions of a 
 parabola. Let the ordinate D C correspond to the moment at 
 the point of application of the load P. DC will be evidently 
 greatest when A B is tangent to .the curve at C, so that the 
 maximum of the moments occurring at D is given by the dis- 
 tance C D between the polygon and curve enveloped by A B. 
 
 Let the prolongation of the sides upon which A B slides meet 
 in E, and F G be the tangent to the parabola at the point H in 
 the vertical through E, so that F H = H G-, and let I be the in- 
 tersection of A B and F G. Draw through A a parallel to E B, 
 intersecting F G in K. Then the horizontal projections of A F 
 and A K are equal, since those of E F and E G are equal. 
 
 Since, however, the projections of F G and A B as also of 
 A F and G B are equal, A K must be equal to G B. Hence 
 A I = B I. In a parabola the distances of the three diameters 
 passing through two points and the point of intersection of the 
 corresponding tangents are equal, hence the projections of H I 
 and C I are equal. 
 
 The middle point I of the tangent A B lies, then, half way 
 between the angle D vertically below the point of tangency and 
 the intersection E of the sides upon which it slides. 
 
 Since the projection of A B is I, its construction is easy. 
 The construction must, of course, be repeated for each angle, in 
 order to determine that for which M is an absolute maximum. 
 
 The above principle may, then, be thus expressed : The mo- 
 ment at any load is a maximum, when this load and the result- 
 ant of all the loads are equally distant from the centre of the 
 girder. (See also Art. 48.) 
 
CHAP. Vn.] SIMPLE GIRDERS. 93 
 
 76. In Arts. 46 to 50 the above principles have been already 
 deduced so far as relates to the moments alone, and a reference 
 to Art. 49 will show their application to the investigation of the 
 effect of a system of loads moving over the girder. We pass 
 on, therefore, to 
 
 CONTINUOUSLY DISTRIBUTED LOADING. 
 
 Suppose the load p per nnit of length laid off as ordinate. 
 The area thus obtained we call the load area. PI. 13, Fig. 
 46 {). 
 
 The equilibrium polygon becomes here a curve, for which 
 the same law holds good. If we draw tangents to the curve at 
 the points D / and E' corresponding to D and E, intersecting in 
 C', then the resultant of the load upon D E passes vertically 
 through C', or C' is vertically under the centre of gravity of the 
 area D D" E" E. 
 
 If we consider the load area divided into a number of parts, 
 the resultant for each will pass through the intersection of the 
 tangents at the points vertically under the lines of division. 
 Since these tangents are parallel to the lines in the force poly- 
 gon corresponding to these lines of division, they form the 
 equilibrium polygon for the concentrated loads, or resultants of 
 the portions into which the load area is divided. 
 
 Hence : if we divide the load area into portions, and replace 
 each ~by a single force, the sides of the corresponding polygon 
 are tangent to the equilibrium curve at the points correspond- 
 ing to the lines of division. (Art. 42.) 
 
 77. Total Uniform Load. In this case the reactions at the 
 
 supports are V t = V 2 = p I. Hence, for any cross-section dis- 
 tant x from the left support, the shearing force is 
 
 For a? = -~ Z ; S = 0. Sis greatest for x = and for x = I ; 
 
 that is, maximum S = -f -p Z, and S = ^ p I. 
 The moment at any cross-section is 
 
 M = V x + -px* = -^px (lx). 
 
 M will be greatest for x = -= l> and 
 
94 SIMPLE GIRDERS. [CHAP. VII. 
 
 Max. 
 
 s^ 
 
 The shearing forces are, then, given by a straight line inter- 
 secting the span in the middle, the ordinate at either end being 
 
 Ip I [PI. 14, Fig. 47.] 
 
 The moments, as we have already seen [Art. 44, Fig. 30], are 
 given by a parabola whose vertex is in the centre of the span 
 
 and whose middle ordinate is -p 1?. Since we have seen [Art. 
 
 o 
 
 70] that a load at any point causes at every point a negative 
 moment, the maximum moment at any point will be when the 
 whole span is loaded. 
 
 78. Method of Loading causing Maximum Shearing 
 Force. We have seen [Art. 70] that a single load causes at 
 any point a positive or negative shear, according as it lies upon 
 the right or left side of the cross-section at that point. Hence, 
 for a uniform load, 
 
 The shearing force will be a positive or negative maximum 
 according as the load reaches from the right or left support to 
 the cross-section in question. For the positive maximum we 
 
 have Y! p (l%) ^r-y ~^P - T~^- Therefore, max. + S = 
 
 a L 2i o 
 
 1 (l-xf 
 
 For the graphical determination we can apply the method 
 given in Art. 72, Fig. 42, by which we have for max. + S and 
 max. S two parabolas whose vertices are at the ends of the 
 
 span, and whose ordinates at these points are +~r and *- 
 
 2i 2i 
 
 Since, however, each point is found thus from the preceding, 
 the construction is not very exact. We may deduce a better 
 construction as follows. [PI. 14, Fig. 48.] Through any point 
 F of the curve drop a vertical intersecting A B in C and the 
 line B K parallel to the tangent at F in G. Let the tangent 
 
 at F intersect A B in H. Then C H = B H ; hence, C F = - C G-. 
 
 2 
 
 We have, then, A E = i A D =-1^ Z. Since C F = - C G-, 
 
 a A 2 
 
 we have also AI =- . AK ; therefore, A I : A E ; ; AK : A D. 
 
 a 
 
CHAP. VII.] SIMPLE GIRDERS. 95 
 
 Hence the following construction : 
 
 Make A E = \.p I. 
 
 \ 
 Divide A E and A B into an equal number of equal parts, and 
 
 draw lines from B to the points of division of A E, and verticals 
 through the points of division of A B. The curve passes 
 through the points of intersection of corresponding lines. 
 
 79. Live and Dead Loads. Let p be the load per unit of 
 length for dead, and m for live load. The maximum moment 
 for any point will be as before. 
 
 M -(j?-f ra) x (#); that is, will be 
 
 a 
 
 given by a parabola whose middle ordinate is - (p+m) P. 
 
 o 
 
 For the shearing force, we have 
 
 Max.+S = Ip (Z-2 
 2 
 
 Indicate A C [Fig. 49, PI. 14] by x^ for which max. S = 0. 
 then 
 
 =p I (I 2 Xi) m a?! 2 , 
 
 or 
 
 m 
 
 hence 
 
 I m m 2 m 
 
 For the point D for which max. +S = 0, B D = a^. The 
 shearing force within A C ia positive, within B D negative, 
 while within C D it is both positive and negative. 
 
 For Z = 5, 10, 20, 50, 75, 100, 150. 
 
 5=0.12 0.19 0.31 0.64 1.05 1.55 3.12 
 m 
 
 :0.24 0.29 0.33 0.38 0.42 0.44 0.46Z 
 CD =-.0.52 0.42 0.34 0.24 0.16 0.12 0.08Z; 
 that is, C D diminishes with increasing span. 
 
 O. Recapitulation. For girders of a length of about 100 
 feet or more, then, we may consider the live load as distributed 
 per unit of length. The maximum shearing force can then be 
 
96 SIMPLE GIRDERS. [CHAP. VII. 
 
 easily found according to the preceding Art., while the maxi- 
 mum moments will be given by the ordinates to the parabola 
 for full live and dead load [Fig. 30, Art. 44]. For a framed 
 structure, we have simply to multiply the shear at any point 
 by the secant of the angle which the brace at that point 
 makes with the vertical, in order to find the strain in that 
 brace. The moment, divided by the depth of truss at the point 
 in question, gives the strain in the flanges. For opiate girder, 
 the moment being found as above, and one dimension as the 
 depth given, we can, from Art. 52, so proportion the other di- 
 mension as that the strain in the outer fibre shall not exceed 
 the amount allowable in practice. The preceding Art. as also 
 Arts. 78 and 44 and 52 are all that we need to refer to for all 
 practical cases of parallel flange girders of large span. 
 
 The preceding will complete our discussion of the simple 
 girder. We have only to remark here that the strains due to 
 rolling load will, in general, be most satisfactorily found by the 
 method of resolution of forces, as illustrated in Art. 12. By 
 this method we first find the reactions at the supports for a sin- 
 gle apex load, either graphically or by a simple calculation 
 
 , and then follow this reaction through the 
 
 girder, arid find the resulting strains. We can thus find and 
 tabulate the strains in every piece due to a weight at each and 
 every apex. The maximum strains can, then, be easily taken 
 from the table thus formed. When the live load is supposed 
 thus concentrated at each apex, it is, as we have seen in Art. 12, 
 unnecessary to follow through every reaction. The reactions 
 due to the first and last weights are sufficient to fill out the 
 table. For solid-built beams or plate girders, the principles of 
 the present Chap., therefore, come more especially into play. 
 (See also remarks at close of Chap. Y.) 
 
 The preceding principles will, it is hoped, be found sufficient 
 to enable the reader to find the maximum moments and shear 
 at each and every cross-section of a beam of given span rest- 
 ing simply upon two supports, and acted upon by any given 
 forces or system of forces in any given position. The reader 
 will do well to take examples of simple trusses, and check the 
 results obtained by the method given in Chap. I. by the above 
 principles. The method of tabulation of single apex loads 
 
CHAP. VII.] SIMPLE GIRDERS. 97 
 
 upon which we lay so much stress is fully given by Stoney 
 [" Theory of Strains," vol. i.], and the examples there given will 
 be found of service. 
 
 Finally, then, the strains in upper and lower chords are great- 
 est for full load over whole span. We have, therefore, only to 
 erect upon the given span a parabola whose centre ordinate is 
 
 -~- , where p is the load per unit of length for dead, and 
 
 8 
 
 m for live load [Art. 44]. ,The ordinates to this parabola at 
 any point give at once the maximum moment at that point. 
 The depth of truss at this point, if a framed structure, or the 
 moment of inertia of the cross-section at this point, if it is a 
 solid beam [Art. 52], being known, the strain in the flanges or 
 outer fibres may be easily determined. The strain in the web is 
 given by the maximum shear. For dead load alone this is 
 given by the ordinates to a straight line passing through the 
 
 centre of span, whose extreme ordinates are^ - [Art. 77]. The 
 
 2i 
 
 maximum shear due to live load alone- (m I) will be given by 
 the ordinates to two semi-parabolas, convex to the span, having 
 
 their vertices at each end, and the extreme ordinates - - [Art. 
 
 A 
 
 78]. At any point, the greatest of the two ordinates to these para- 
 bolas is to be taken. For live and dead loads together, Art. 79 
 may also be useful. The shear being known, the strain in any 
 diagonal is equal to the shear multiplied by the secant of the 
 angle made by the diagonal with the vertical [Art. 10 of Ap- 
 pendix] i.m parallel flanges. For flanges not parallel, we must 
 find the resultant shear as given in Art. 16 (4) of Appendix, 
 or, better still, the flanges once known, the diagonals can be 
 diagrammed according to the principles of Chap. I. 
 
 For the investigation of load systems, the principles of 
 Arts. 70-75 will be found sufficient, and the application of 
 these principles we have already sufficiently illustrated in Arts. 
 49-51. 
 
 7 
 
$8 SUPPLEMENT TO CHAP. VII. [CHAP. \. 
 
 SUPPLEMENT TO CHAPTEK VII. 
 
 CHAPTER l' 
 
 METHODS OF CALCULATION. 
 
 1. In Chapter I. of the text we hare already obtained a method of dia- 
 gram which will be found both simple and general, and by which we can 
 readily determine the strains for any given loading in any framed struc- 
 ture, no matter how irregular in its shape or dimensions, provided only that 
 all the outer forces are Tcnown. 
 
 In Chap. VII. we have also been put in possession of another method of 
 diagram, by which we may for any structure of the above class, framed or 
 not, determine the moment at any point, and can then properly proportion 
 the cross-section. 
 
 Thus far, indeed, we are unable to apply these methods to the continuous 
 girder or braced arch, as in these cases there are not only upward reactions 
 but also end moments, and in the latter case a thrust also, which must first 
 be determined. The determination of these requires that the elasticity of 
 the material and cross-section of the structure be taken into account. But 
 with these exceptions, and they are of rare occurrence in practice, we can 
 already solve any case which may present itself. 
 
 In the Appendix, if he has attended to our numerous references to it, 
 the reader will have already become familiar with two corresponding meth- 
 ods of calculation, viz., that by resolution of forces and that by moments. 
 
 It is, however, in many cases desirable to know not only the strains in 
 every piece of a structure, but also the deflection of the structure, and this 
 also requires a knowledge of the theory of flexure or of elasticity. For the 
 sake of completeness, therefore, aiming as we do to put the reader in pos- 
 session of methods of calculation as well as of graphic determination, we 
 shall devote a few pages here to a brief notice of these two above-men- 
 tioned methods of calculation, and then pass on to the theory of elasticity 
 itself. This latter has been too generally considered by those unacquainted 
 with the methods of the calculus as difficult and abstruse. It is true that 
 the calculus must be called into requisition ; but so simple are the processes 
 for beams of single span and it is with these only we have at present to 
 do that we indulge the hope that by going back to first principles we may 
 enable even those at present unacquainted with the calculus to follow our 
 
CHAP, l'] SUPPLEMENT TO CHAP. VII. 99 
 
 I 
 
 demonstrations intelligently, and to comprehend perfectly and even apply 
 readily the method for themselves. 
 
 We cannot, indeed, make the reader familiar with all the principles of 
 the calculus, but all these principles are by no means needed. Its funda- 
 mental idea, a few of its terms and applications, are all that he need be 
 familiar with in order to perform the simple integrations we shall encoun- 
 c er, as readily as the most skilled mathematician. This portion of the 
 present Supplement may, perhaps, be considered by many as unnecessary and 
 superfluous. We are, indeed, justified in assuming such knowledge. But 
 as we believe our plan practicable, we cannot resist the desire of making 
 our development intelligible to all, and thus rendering our treatment of the 
 simple girder at least complete. 
 
 The practical man as well as the mathematician may thus have at his 
 disposal the powerful aid of the calculus, so far at least as his purposes 
 require it, and be able to deduce for himself the formulae which hitherto 
 he has accepted " upon faith." It may also not be improbable that here 
 . and there one may be found who, pleased with the simplicity of the prin- 
 ciples and the f ruitf ulness of their application, may be led to further prose- 
 cute the study for his own satisfaction. 
 
 We shall first, then, notice briefly the two methods of calculation above 
 referred to ; then devote a few pages to the development of those prin- 
 ciples and rules of the calculus of which we shall make use, and finally 
 apply these principles to the discussion of the curve of deflection of loaded 
 beams. 
 
 2. Hitter's Method. This method is referred to in Art, 14. It 
 rests simply upon the principle of the lever, or the law of statical moments ; 
 requires no previous knowledge, and converts the most difficult cases of 
 strain determination into the most elementary problems of mechanics. 
 Ritter, 'in his " Theorie eiserner Dach- und Briicken-Constructionen," has 
 applied this simple principle in such detail and fullness, and so clearly set 
 forth its elegance and simplicity, that it very generally, and justly, goes by 
 his name. 
 
 " Its results are clear and sharp as the results of Geometry, and of direct 
 practical application. There is hardly another branch of engineering 
 mechanics which, for such a small amount of previous study, offers such 
 satisfactory results, and which is so suited to engage the interest of the 
 beginner." 
 
 We have given in the Appendix to Chap. I. (Arts. 6, 9, 10) detailed ex- 
 amples of its application. Throughout this work similar illustrations of 
 its use will be met with, so that it is only necessary here to state more fully 
 than in the text its general principle. 
 
 If any structure holds in equilibrium outer forces, it does so by virtue of 
 the strains or inner forces which these outer forces produce. Now the 
 outer forces being always given, we wish to find the interior forces or 
 strains. If, then, the structure is framed, and we conceive it cut entirely 
 through, the strains in the pieces thus cut must hold in equilibrium all the 
 outer forces acting between the section and either end. Thus, in Fig. 6, 
 PL 2, a section cutting D, 7 and H completely severs the truss. Then the 
 
100 SUPPLEMENT TO CHAP. VII. [CHAP. 1.' 
 
 strains in these three pieces must hold in equilibrium the reaction at A and 
 all the forces between A and the section. 
 
 Now the principle of statical moments is simply that, when any number 
 of forces in a plane are in equilibrium, the algebraic sum of their moments 
 with respect to any point in that plane must be zero. 
 
 The application of this principle is simply so to choose this point of 
 moments as to get rid of all the unknown strains in the pieces cut, except one 
 only ; and then the other forces being known in intensity, position, and 
 direction of action, we can easily find this one ; since, when multiplied by 
 its known lever arm, it must be equal and opposite to the sum of the 
 moments of the known forces. 
 
 In a properly constructed frame it will, in general, always be possible to 
 pass a section cutting only three pieces. Then, by taking as a centre of 
 moments the intersection of any two, we can easily find the strain in the 
 third. 
 
 Even if any number of pieces are thus cut, if all but one meet at a com- 
 mon point, the strain in this one can be determined. 
 
 Thus, in Fig. IV., PL 1 of the Appendix, a section may be made cutting 
 2 3, d h, h c and c Y. But all these pieces, except the last, meet in 2, and 
 the strain in this last piece may, therefore, be easily determined. 
 
 The above is all that is necessary to be said as to this method. The ex- 
 amples already referred to will make all points of application and detail 
 plain as we proceed. We see no reason why the reader who has mastered 
 Chapter I. and diligently followed out the examples as given in the Appen- 
 dix, should not now be able to both calculate and diagram the strains in 
 any framed structure all of whose outer forces are known. 
 
 3. Method by Reolutioii of Forces. We have also yet another 
 method of calculation, based upon the principle that, if any number of 
 forces in a plane are in equilibrium, the sum of their vertical and hori- 
 zontal components are respectively zero. In structures all the forces acting 
 upon which are vertical, and such are all bridge and roof trusses, etc., of 
 single span, we have only to regard the vertical components. 
 
 In this connection we have to call attention to the following terms and 
 considerations. The shear or shearing force at any point is the algebraic 
 sum of all the outer forces acting between that point and one end. These 
 outer forces are the weights and reactions at the ends. At any apex of a 
 framed structure, where several pieces meet, the horizontal components of 
 the strains in these pieces must balance, or the structure would move ; and 
 for the same reason, the algebraic sum of the vertical components must be 
 equal and opposite to the shear. The shear being known, if the strains in 
 all the pieces but one are also known, that one can be easily found. Thus 
 the algebraic sum of all the vertical components of the strains in the other 
 pieces being found, and added or subtracted from the shear, as the case 
 may be, the resultant shear, multiplied by the secant of the angle made by 
 the piece in question with the vertical, gives at once its strain. 
 
 This method is also fully explained in the Appendix, Art. 16 (4), and a 
 practical rule is there given for properly adding the vertical components 
 and determining whether the result is to be added to or subtracted from 
 
J. 
 
 CHAP. I.] SUPPLEMENT TO CHAP. VII. 101 
 
 the shear. This rule we owe to Hiuriber* We have thus two methods of 
 calculation, which, for the sake of convenience, we may speak of as Bitter's 
 and Humberts. Corresponding to Humberts method we have also a graphic 
 solution, based upon the same principles precisely. This we have set forth 
 in Chapter I., and may call Prof. Maxwell's method. In Chapter II. and 
 the following we have also become acquainted with the graphic solution 
 corresponding to Hitter's method, or the method of moments, which we 
 may speak of as Culmanrit*. It is to this method, based upon the proper- 
 ties of the equilibrium polygon, that the graphical statics properly owes its 
 value and fruitfulness, and to it is due whatever pretensions it can claim 
 as a system. It will be seen hereafter that it alone can furnish a general 
 method applicable to all structures, whether framed or not ; whether all 
 the outer forces are known or not. By the same general method we are 
 enabled to find the centre of gravity and moment of inertia of areas, and 
 to solve thus a great variety of practical problems through which, how- 
 ever different, runs one universal method, one simple routine of construc- 
 tion. 
 
 * Strains in Girders, calculated ~by Formulas and Diagrams. 
 
102 SUPPLEMENT TO CHAP. VII . [CHAP. II. 
 
 CHAPTER II. 
 
 PRINCIPLES OF THE CALCULUS NEEDED IN OUR DISCUSSION. 
 
 4. Differentiation and Integration. We need but a very few 
 simple ideas and conclusions in order to have at our disposal the whole 
 theory of flexure for beams of single span. Those to whom these ideas are 
 not familiar already may find them indeed new, but will not find them 
 difficult or even abstruse, and with attention to the following will, we 
 venture to think, make a valuable acquisition. 
 
 The sign / is called the " sign of integration, 1 ' and integration 
 
 means 
 
 simply summation. It arises merely from the lengthening of the original 
 letter S, first used by Leibnitz for the purpose. The letter d is called the 
 "sign of differentiation-;'' in combination with a letter, as d x, it reads 
 " differential of #," and signifies simply the incitement which has been 
 given to the variable x. So much for terms. 
 
 Now suppose we have the equation 
 
 y = 5x\ . . ....... (1) 
 
 in which x and y, although varying in value, must always vary in such a 
 way that the above equation holds always true. This being the case, let 
 us give to y an increment that is, supposing it to have some definite value 
 for which, of course, x is also definite in value, increase this value by d y. 
 
 Then x will be increased by its corresponding amount d x, and as the 
 above relation must always hold true, we have 
 
 y + dy = 5 (x + dxy . . . ... . (2) 
 
 or y +dy = 5 (x*+Zxdx + d a; 2 ). 
 
 Inserting in this the value of y from (1), we have 
 
 dy = 5 (2xdx + dx*), ...... (3) 
 
 which is the value of the increment of y or d y, in terms of x and the in- 
 crement of x or d x. That is, the increments are not connected by the same 
 law as the variables. The variable y is always 5 times the square of the 
 variable x, but the increment of y is greater than 5 times the square of the 
 increment of x by an amount indicated by 5 x 2 x d x. From (3) we 
 have 
 
 (4) 
 
 which gives the value of the ratio of the two increments. Now, if we 
 assume a certain value for x, we find easily from (1) the corresponding 
 value of y. If we increase this value of x by a certain assumed increment, 
 d x, we find easily from (3) the corresponding increment of y, or d y. Then 
 (4) would give us the ratio of these two increments. 
 
CHAP. II.] SUPPLEMENT TO CHAP. VII. 103 
 
 Now we see at once from (4) that the smaller we consider d x to be, the 
 nearer this ratio approaches the limiting value 5 x 2 x. We may suppose 
 d x as small as we please, and then this ratio will differ as little as we 
 please from 5 x 2 a. This value, 5 x 2 x, forms, then, the limit towards which 
 
 the value of the ratio ~ approacfies as d x diminishes, but which limit evi- 
 
 Ou X 
 
 dently it can never actually reach or exactly equal. Because, in order that 
 this should bo the case, d x must be zero. But if d x is zero, that is, if # is 
 not increased, y also is not increased; d y is, therefore, zero, and there is no 
 ratio at all. 
 
 Now, just here comes in what we may regard as the central principle of 
 the calculus. 
 
 If two varying quantities are always equal and always approaching certain 
 limits, then those limits must themselves he equal. 
 
 The principle is too obvious to need demonstration. "Two quantities 
 always equal present but one value, and it seems useless to demonstrate 
 that one variable value cannot tend at the same time towards two constant 
 quantities different from one another. Let us suppose, indeed, that two 
 variables always equal have different limits, A and B ; A being, for ex- 
 ample, the greatest, and surpassing B by a determinate quantity A. 
 
 The first variable having A for a limit will end by remaining constantly 
 comprised between two values, one greater, the other less than A, and hav- 
 ing as little difference from A as you please; let us suppose this difference, 
 
 for instance, less than A . Likewise the second variable will end by re- 
 
 6 f 
 
 maining at a distance from B less than A. Now it is evident that, then, 
 
 the two values could no longer be equal, which they ought to be according 
 to the data of the question. These data are then incompatible with the 
 existence of any difference whatever between the limits of the variables. 
 Then these limits are equal." * 
 
 Now let us apply this principle to equation (4). In this equation -y is 
 a variable always equal to 5 (2 x + d x). But 5 (2 x+d a>), as we diminish 
 d x, approaches constantly the limit 5 x 2 x; and as -= is always equal to 
 
 5 (2 x + d x), it also constantly approaches the same limit. These limits, 
 
 dy 
 
 then, are equal, and the limit of -= = 5 x 2 x. 
 
 d x 
 
 Now, if we conceive, and such a conception is certainly possible, d x to 
 lie the difference between x and its consecutive or very next value, such that 
 between these two values there is no intermediate value of d x ; then d y 
 will be the difference between two consecutive values of y ; and regarding, 
 
 then, d x and d y in this light, -= will be the limit of the ratio of the in- 
 
 CL X 
 
 * The Philosophy of Mathematics. Bledsoe. 
 
104 SUPPLEMENT TO CHAP. VII. [CHAP. II. 
 
 crements, since the increments are then limiting increments, and can be no 
 smaller without disappearing. 
 We have thus 
 
 ^ = 5x2,, 
 d x 
 
 which is an exact relation between the increments upon this supposition. 
 From this we have d y = 5x2 x d x. 
 
 If now we sum up all the increments d y, then by virtue of the supposi- 
 tion we have made, / d y must equal y. We thus suppose y to flow, as it 
 
 were, unbrokenly along by the consecutive increments d y, just as the side 
 of a triangle moving always parallel to itself, and limited always by the 
 sides, describes the area of that triangle, while the change d y of its length 
 is the difference between two immediately contiguous positions. Upon 
 
 this supposition, we repeat, -= is the limit of the ratio of the increments, 
 a os 
 
 which limit is, as we see from (4), equal exactly to 5 x 2 x. We do not re- 
 ject or throw away d x from the right of that equation " because of its 
 small size with reference to 2 a,' 1 but simply pass to the limit, and then, ac- 
 cording to our fundamental principle above, equate those limits them- 
 selves. But if / d y = y, then the integral of 5x2xdx, or / 5x2 xdx=y 
 
 =5 & 2 . By " differentiating," as we say, equation (1) we get (5), and by 
 " integrating " (5) we obtain (1). 
 
 Hence we see the appropriateness of the term "fluent" given by New- 
 ton to the quantity d y or 2 x d x. So also we see the appropriateness of 
 
 the term "ultimate ratio " * for - itself. 
 
 d x 
 
 * Liebnitz undoubtedly discovered the calculus independently of Newton, 
 but he considered d x as a quanity so " infinitely" small that in comparison 
 with a finite quantity it could be disregarded " as a grain of sand in compari- 
 son with the sea." We see, indeed, from eq. (4) that if d x upon one side be 
 
 zero, we get the same value for - as before. But if d x is zero on one side, 
 
 it should be zero on the other side also. No matter how small we suppose d x 
 to be, we have no right to get rid of it by disregarding it. That Liebnitz rec- 
 ognized this cannot be doubted, and he was therefore inclined to consider his 
 method as approximate only. But to his surprise he found his results exact, 
 differing from the true by not even so much as a " grain of sand." There was 
 to him ever in his method this mystery, nor could he conceive what these 
 quantities could be which., though disregarded, gave true results. Bishop 
 Berkeley challenged the logic of the method, and adduced it as ac evidence of 
 k ' how error may bring forth truth, though it cannot bring forth science." 
 Strange to say, even the disciples of Newton were unable to answer Berkeley 
 without taking refuge in the undoubted truth of their results. And yet New- 
 ton in his Principia lays it down as the corner-stone of his method, that 
 44 quantities which during any finite time constantly approach each other, and 
 
CHAP. II.] SUPPLEMENT TO CHAP. VII. 105 
 
 The whole of the calculus is but the deduction of rules for finding from 
 given equations as (1) their " differential equations " as (5), or inversely 
 of finding from the differential equation by "integration," or summation, 
 the equation between the variables themselves. 
 
 Such of these rules as we need for our purpose we can now deduce. 
 
 5. Differentiation and integration of powers'of a single 
 
 variable. We have already seen that the / d y = y and / 2 xd x = a?, 
 
 hence d (x 2 ) = 2 x d x. 
 
 If we should take y = x ?> , we should have, in like manner, as before, 
 
 y + d y (x + d x) 3 = x* + 3 a 8 d x + 3 x d x z 4- d z 3 , 
 or dy= 3x*dx + 3xdx 2 + dx*, 
 
 or ~ = 3z 2 + 3xdx + dx' 2 , 
 
 dx 
 
 and passing to the limits, as before, 
 
 - = 3 x 2 , or d y=3 x 2 d x. Hence the differential of x 3 or d (a? 3 ) =3 x 2 dx, 
 
 it 3} 
 
 and reversely, the integral of 3 x z d x or / 3 x z d x = x 3 . In similar man- 
 ner, we might find 
 
 d (x 5 ) = 5 x* d x and A x*dx = x 5 . 
 
 Comparing these expressions, we may easily deduce general rules which 
 will enable us at once upon sight to " differentiate," that is, find the rela- 
 tion connecting the increments ; and " integrate " or sum up the successive 
 consecutive values of the variable; for any expression containing the 
 power of a single variable. 
 
 These rules are as follows : 
 
 To differentiate: 
 
 " Diminish the exponent of the power of the variable by unity, and then 
 multiply by the primitive exponent and ~by the increment of the variable." 
 
 Thus, d(x z ) = 2xdx, d (x*) = 3 x* dx, d (a? 7 ) = 7* 6 dx, d(x$) = ^X2dx, 
 
 d (#") n a; " 1 d x, etc. 
 
 To integrate : 
 
 "Multiply the variable with its primitive exponent increased "by unity, by 
 the constant factor, if there is any, and divide the result by the new exponent." 
 
 before the end of that time approach nearer than any given difference, an eguaV 
 There can be little doubt that Newton saw clearly that although the quantities 
 might never be able to actually reach their limits, yet that those limits them- 
 selves were equal, and hence the increment could be left out in the equation, 
 but not because by any means it was of insignificant size. His terms ' ' ultimate 
 ratio" and '''fluent' 1 ' 1 are alone sufficient to indicate that he understood the 
 true logic of the method he discovered ; while Liebnitz seems to have stood 
 gazing with wonder at the workings of the machine he had found, but whose 
 mechanism he did not understand. [See Philosophy of Mathematics. Bledsoe. 
 Lippincott & Co., 1868.] 
 
1U() SUPPLEMENT TO CHAP. VII. [CHAP. TI. 
 
 Thus A xdx= 2 -j-=x* f$afdx = -|" = *' /V dx = X ~ 
 
 /i as* 2 f / n x 
 x dx=- = 5-s / nx*-idx = = a, etc. 
 i a / 
 
 It is of this latter rule that we shall make especial use in what follows, 
 6. Other Principles Integration between limits, etc. 
 
 We may observe from (1) and (4) that a constant factor may be put out- 
 side the sign of integration. Thus / 5x2 x d x = 5 I 2 x d x = 5 x\ 
 
 It is also evident without demonstration that the integral of the sum of 
 any number of differential expressions is equal to the sum of the several 
 integrals. 
 
 Thus / he d a + 2 s d z + y*dy + x* 
 
 is the same as / x d x + I z 3 d z + I y* d y, etc. 
 
 If in (1) we had 
 
 y = 5 x*+a, 
 
 where a is a constant, we should have 
 
 y + d y = 5 (x + d x)* + a = 5 (x* + 2 x d x+d a?) + a, 
 
 or d y = 5 (2 x d x + d afy or ^| = 5 (2 x + d x) ; 
 
 whence 
 
 -^ = 5x2 a?, ordy = 5x2xdx, or 
 d x 
 
 just the same as before. 
 
 The integral of this will then be y = 5 x 1 as before, whereas it should be 
 y = 5 x* + a. 
 
 If two differential equations, then, are equal, it does not necessarily follow 
 that the quantities from which they were derived are equal. 
 
 We should, then, never forget when we integrate to annex a constant. The 
 value of this constant will in any given case be determined by the limits 
 between which the integration is to be performed. 
 
 We indicate these limits by placing them above and below the integral 
 sign. Thus the integral ofx*dx between the limits of x = + h and x= h is 
 
 r +h r x* 
 
 I x 3 dx. If we integrate a; 2 d x, we have, then, / a 2 d x = + C, 
 J-h J 
 
 where O is a constant whose value must be determined by the conditions 
 of the special case considered. If we introduce the value of x = h for one 
 
 limit, we have + O. For x = 2 h for another limit, we have - + O. 
 We have, then, two equations, viz. : 
 
 when x h, 
 
 c 
 
 J r ~h x ~ 3~ 
 
CHAP. H.] SUPPLEMENT TO CHAP. VII. 107 
 
 rx = 2h 
 
 JZ.J* 
 
 n the other, w 
 
 r h 7 
 
 x = h, I x'dx=- 
 *s h 
 
 and when x = 2h I x*dx= - + C ; 
 
 and by subtracting one from the other, we have for the integral between 
 
 r 2h 7 
 
 the limits x = 2 h and x li, I x' 2 dx 7* s , and O thus disappears. 
 
 We have, then, only to substitute in succession the values of the variable 
 which indicate the limits, and subtract the results. 
 If also there is but one limit, we could determine O if there were also a 
 
 condition, such, for instance, as that / x 1 d x should equal h when x = 2 h. 
 
 The ratio ^ is called the "first differential coefficient ; " if it were to 
 d x 
 
 be differentiated again, the next ratio, viz., that of the differential of the 
 
 differential of y to differential of or 2 , or -, is the " second differential co- 
 
 d x 
 
 efficient," and so on. 
 
 dy 
 Thus, y = x 5 ; dy = d (x 5 ) = 5 x*d x, or.-= = 5 x* ; differentiating again, 
 
 -i = 20 x 3 d x, or - 20 x 3 , and so on to third differential coefficient, etc. 
 dx dx* 
 
 7. Example. As an example of the application of our principles, let 
 it be required to determine the area of a triangle. Let the base be b and 
 the height h. Take the base as an axis, and at a distance of x above the 
 base draw a line parallel to &, and at a very small distance d x above this 
 line draw another, thus cutting out a very small strip. (Let the reader draw 
 the Fig.) Now for the base y of this strip we have the proportion h x : y 
 
 7) T ~h 'JC (L i/ 1 
 
 :: h : 5, or y = b , hence the area of the strip is b dx . But 
 
 n n 
 
 the area of this rectangular slip is not equal to the area of that portion of 
 it comprised within the triangle. It projects over at each end, and is, 
 therefore, somewhat greater. Thus for the small trapezoid actually within 
 the triangle we have for the upper side y', h(x+d x) : y'::h : b, or y' = 
 
 b - (x + d x). Hence yy' , and the area of the projecting portion 
 
 h, h 
 
 of the rectangle, that is, its excess over the trapezoid, is then (yy') d x, or 
 
 bdx" 1 __ , ,, bxdx b d x* , da , bx bdx 
 
 Therefore, bdx = d , or - = b , where 
 
 h h Ti d x h h 
 
 d a is the area of the small trapezoid itself. Now these latter two quanti- 
 ties are always equal for any value of d x. But as d x decreases, one side 
 
 of the equation approaches the limit & , and , therefore, approaches 
 
 h d x 
 
 this same limit. The rectangle itself is, then, the limit of the ratio of the 
 area of the small trapezoid to its height, and we can then equate the limits 
 themselves, remembering that in this case d a is the area passed over by the 
 
108 SUPPLEMENT TO CHAP. VII. [CHAP. II. 
 
 side y in passing from one position to the consecutive or very next. We 
 have, then, da = bdx , and if we integrate this expression, that is, 
 
 sum up all the d a's, we have the area of the triangle. Therefore, 
 
 x d x 5 x 1 
 
 =j i -IT- ) m .-ji+* 
 
 where O is the constant of integration, which we must never forget to annex. 
 Now, in the present case we wish to sum up all the areas d a, or " integrate," 
 between the limits x o and x = Ji. But for x o, A must be zero, and 
 hence we have O = o for the condition that x starts from the base. If in 
 addition to this condition we make x = h, we have the sum of all the areas 
 between x = o and x = h. 
 
 , , & h & h 
 A = & h . = , as should be. 
 
 The above reasoning is somewhat prolix. 
 
 If we thoroughly appreciate that d x is the difference between two con- 
 secutive values of x, we see at once that we obtain the limiting value of the 
 rectangle directly by multiplying its base by d x. The sum of all these 
 must be the area. This conception ofdx enables us to curtail much of our 
 reasoning. 
 
 Let us take the same problem again, but this time take the axis through 
 the centre of gravity of the triangle ; that is, at $h above the base. Then 
 for the base y at any distance x above this axis, we have 
 
 -A-av "h'l or = -l>-*J!L 
 3 3 h 
 
 Multiply this by d x upon the above conception of d x, and we have at 
 once not for the rectangle upon y, but for its limiting value, that is, for the 
 area of that portion of the rectangle included within the triangle, 
 , -. 2, b x d x 
 
 ~ y = 3 h ' 
 
 Integrating this, then, we have 
 
 2 , & x d x 2 , bx* 
 
 where O is a constant to be determined by the limits as before. For one 
 
 limit, x = 7i, and hence we have 
 3 
 
 A'=-l 
 
 2 
 
 For the other limit, x = + h, and hence we have 
 3 
 
 If we subtract the first from the second, O disappears, and we have A = 
 
 9 1 
 
 A" A' = & h = & A, as before. 
 18 2 
 
 o 
 
 We might also have integrated first between the limits x = and x h. 
 
 3 
 
CHAP. H.] SUPPLEMENT TO CHAP. VII. 101) 
 
 4 
 
 For x = 0, C = 0, and the area above the axis is then b Ji. For x = and 
 
 lo 
 
 x = h, we have for the area below the axis b h. This area has a dif- 
 
 3 18 
 
 ferent sign because below. If we give it the same sign as the other, and 
 then add it, we have the total area. If it also had been above, the total 
 area would have been the difference. Generally, then, we subtract accord- 
 ing to our rule. 
 
 8. Significance of the first differential coefficient. Any 
 equation between two variables of the first degree is the equation of a 
 straight line. If of the second degree, it represents one of the conic sec- 
 tions, an ellipse, circle, parabola, or hyperbola, Of a higher degree, a 
 aurve generally. If, then, we take the axis of x horizontal and y vertical, 
 and if d y and d x are the consecutive increments of y and x, that is, the dif- 
 ference between any value and the very next, the ratio ~ is evidently the 
 
 d x 
 
 tangent of the angle which a tangent to the curve at any point makes with the 
 horizontal. 
 
 If, then, we make = 0, and find the value of the variabb x corrc- 
 d x 
 
 spending to this condition, we find evidently the value of x for which the 
 tangent to the curve is "horizontal. If now the curve is concave towards the 
 axis, this value of x, substituted in the original equation, will give the maxi- 
 mum or greatest value of the ordinate y ; because for the point just one 
 side of this the tangent slopes one way, and for the point just the other 
 side it slopes the other. The point where the tangent is horizontal must 
 then be the highest. 
 
 If the curve is, on the other hand, convex to the axis, the value of x, which 
 
 makes - = 0, substituted in the original equation, will give y a minimum 
 d x 
 
 value for similar reasons. By setting the first differential coefficient, then, 
 equal to zero, we may find that value of x which corresponds to the maxi- 
 mum or minimum value of the ordinate, as the case may be. In the case 
 of the deflection of simple beams upon two supports, the curve is always 
 concave to the axis, and hence we obtain by this process always the maxi- 
 mum deflection. 
 
 The above comprises all the principles of which we shall make use in the 
 discussion of the theory of flexure. With a little study, we believe that 
 any one familiar with analytical operations, even although he may never 
 have studied the differential or integral calculus, can follow us intelligently 
 in what follows. Whatever points may still be a little obscure will clear 
 up as he sees more plainly than now their application. 
 
110 SUPPLEMENT TO CHAP. VH. [CHAP. III. 
 
 CHAPTER III. 
 
 THEORY OF FLEXURE. 
 
 9. Coefficient of EIaticity. Let us now take up the theory of 
 flexure, and see if it is not possible so to present the subject that, in the 
 light of the preceding principles, we may be able to solve all such prob- 
 lems as present themselves. 
 
 If a weight P acts upon a piece of area of cross-section A, and elongates 
 or compresses it by a small amount I, we know from experiment that, 
 within certain limits, twice, three times, or four times that weight will 
 produce a displacement of 21, 31, 4 I, etc. These limits are the limits of 
 elasticity. Within them practically, then, the displacement is directly as 
 the force. If we assume this law as strictly true for all values of the dis- 
 placement, and if we denote the original length by L, then, since the 
 
 force per unit of area is -r, and since this unit force causes a displacement 
 I, in order to cause a displacement L equal to the original length, this 
 
 unit force must be T time* as great, or equal to T . This force we call 
 I At 
 
 the modulus or coefficient of elasticity. It is always denoted by E. Hence 
 
 The coefficient of elasticity, then, is the unit force which would elongate a 
 perfectly elastic body BY ITS OWN LENGTH. It is a theoretical force then ; 
 but as the law upon which its value is based is true practically within cer- 
 tain limits, by experiments made within those limits, knowing P, A, and 
 L, and measuring I, we can find what the force would have to be if the law 
 were always true. Such experiments have been made, and the values of E 
 for different materials are to be found in any text-book upon the strength 
 of materials. 
 From (6) we have for the unit force of displacement 
 
 These expressions will be found useful as enabling us to replace often 
 expressions containing an unknown displacement by a definite or experi- 
 mentally known value. 
 
 1O. Moment of Inertia. This is also a convenient abbreviation, 
 and enables us to replace unknown expressions by a, in any given case, 
 perfectly determinate value. 
 
 The moment of inertia, with respect to any axis, is the algebraic sum of the 
 
CHAP. III.] SUPPLEMENT TO CHAP. VII. Ill 
 
 products obtained ~by multiplying the mass of every element of a given cross- 
 section T)y the square of its distance from that axis. 
 
 If a parallelogram stand on end, and then its support be suddenly pulled 
 away from under it, it will fall over backwards. But to knock it over 
 thus requires force. The force which, in this case overturns it is that of 
 inertia. At every point of the surface there is, then, a force acting, depend- 
 ing upon the mass of this point. But not alone upon the mass. A force 
 at the top acts evidently with more effect to turn the body over than one at 
 the bottom, which merely tends to make it slide. The moment of each ele- 
 ment of the area is. then, a measure of the force which at each point causes 
 rotation, and the sum of these moments is, then, the measure of the over- 
 turning action of the whole force of inertia upon the surface. The moment 
 of this latter force, or the sum of the moments of the moments, is, then, the 
 moment of inertia of the cross- section. Each element of the surface must 
 then be multiplied by the square of its lever arm, and the sum of all the 
 results thus obtained taken. In other words, the moment of each element 
 is itself considered as a force, and then its moment again taken. The sum 
 is denoted by I. ' For any given dimensions and axis it is a perfectly defi- 
 nite quantity, and may thus often replace expressions containing unknown 
 quantities. 
 
 The principles of the calculus just developed will enable us to deter- 
 mine it in some cases, at least, very readily. Its value for various forms of 
 cross-section, in terms of the given dimensions, is given in every text- book 
 upon the strength of materials. 
 
 Let us suppose a rectangular cross-section of breadth 5 and height h, and 
 take the bottom as axis. The area of any elementary strip is, then, b d x. 
 If its distance from the bottom is x, we have for its moment b x d x, and for 
 its moment of inertia, then, & # 2 d x. Integrating this expression, we have 
 
 _ &> 
 - 3 +0. 
 
 This integral is to be taken between the limits x = and x h. For x = 0, 
 
 1) x' d x = 0, and hence C 0. For x = h, then, we have -. If the axis 
 
 3 
 
 had been taken through the centre of gravity, we should have the above 
 
 integral between the limits + -and . For+ we have h O. For 
 
 2 2 2 24 
 
 , + O. Subtracting one from the other (Art. 6), we have 
 
 for the moment of inertia. For a triangle of height h and base Z>, we have 
 for axis through centre of gravity, from Art. 7, for the area of the very 
 
 small strip at distance a?, -Idx - x d x. Multiplying this by x*, we have 
 
 o fl 
 
 for its moment of inertia -lx l dx -x*dx. The integral of this is 
 o h 
 
 2 T) x 4 
 
 O A 
 
 For x = h, this becomes I 
 
 o 243 
 
112 SUPPLEMENT TO CHAP. VII. [CHAP. ITT. 
 
 For x =71, we have & 7i 3 + C. 
 
 27 1 
 
 Subtracting one from the other (Art. 0), we have -^ & A 3 , or _ & A 3 for 
 
 the moment of inertia. The moment of inertia of the rectangle I = 
 
 may be written = - x Ax , or the moinent of inertia of the half 
 
 parallelogram is equal to its area, into the distance of its centre of gravity 
 multiplied by f ds. of its height. We see at once that when we consider, 
 then, the statical moments as themselves forces, the centre of action, of these 
 moment forces does not coincide with the centre of gravity of the area. This 
 principle we have already noticed in Chap. VI., Art. 60. 
 
 I A 2 /I \ 2 1 
 
 We can also put --=_ = (_ - A ) . This value A is called the 
 
 A 12 \2V 3 ) 3y 
 
 radius of gyration. It is evidently the distance from the axis to that point 
 at which, if the mass were concentrated or sum of all the forces were con- 
 sidered as acting, their moment of inertia would be that of the cross-sec- 
 tion itself. The value of is, in general then, the square of the radius of 
 
 JA 
 
 gyration. We have already shown in Chap. VI. how to find it graphically 
 for various cross-sections. 
 
 We are now ready to take up the case of a deflected beam, and to find 
 the differential equation of its curve of deflection. 
 
 11. Change of SUiape of the Axis. In the Fig. given in the 
 Supplement to Chap. XIV., we have represented a beam deflected from its 
 original straight line by outer forces. Let the two sections A C, B D be 
 consecutive sections, parallel before flexure, and remaining plane after. Let 
 the length of the axis m a be s, then n a d s, and let d <p be the very 
 small angle between the sections after flexure. 
 
 If the deflection is small, s will be approximately equal to x, and d s to 
 d x. The elongation of any fibre at a distance from the centre is. then, 
 v d (f). The unit force corresponding to this elongation is from (7) T = 
 
 E - v. If d a is the cross-section of any fibre as d c, then the whole force 
 d x 
 
 of extension is 
 
 Ev da d $ 
 
 dx 
 
 The moment of this force is, then, = - -. The integral of this be- 
 tween the limits + -- and will give the entire moment of rupture. But 
 
 this is equal and opposite to the moment M of ail the outer forces ; hence 
 
CHAP. III.] SUPPLEMENT TO CHAP. VII. 113 
 
 But, as we have just seen, this integral is the moment of inertia I of the 
 cross-section with reference to the axis through the centre. Hence, 
 
 M = - . Since rf> is a very small angle, it may be taken equal to its 
 d x 
 
 tangent, or equal to ; hence - and M = B I . 
 
 dx d x d or d x* 
 
 But v d cf) : v :: d x : r, where r is the radius of curvature ; 
 
 v d & d x d <b 1 
 
 hence = or -r-^ = . 
 
 v r dx r 
 
 Therefore, M = Hli = EI^|= ;* v . * . (8) 
 
 r dx* v 
 
 and T = 5- B (9) 
 
 r 
 
 Equation (8) is our fundamental equation. 
 
 In any given case we have only to write down the expression M for the 
 
 d z y ! 
 moment of the outer forces at any point, and equate it with E I -^. 
 
 Cu <Xs , 
 
 Integrating once we shall then have for I constant, of course, E I - and, 
 
 d x 
 
 integrating again, E I y in terms of a?, or the equation of the deflection 
 curve itself. Making E 1~ = 0, we can then find the point of maximum 
 
 deflection, and inserting in the value for Ely the value of * thus found, 
 can find the maximum deflection itself. The discussion of any case reduces 
 thus to a simple routine, and every case is in many respects but a repetition 
 of the same processes. 
 
 12. Beam fixed at one end and loaded at the other 
 Constant cros-section. We shall always consider a moment positive 
 when it causes compression in the lower fibre ; negative when it causes ten- 
 sion in that fibre. Distances to the right of the origin are always positive, 
 to the left negative. Hence on the left of any section an upward force is 
 negative, a downward force positive ; while on the right of the section the 
 upward force is positive and the downward one negative. The reader 
 should always draw the Fig. for each case discussed, and in the beginning, 
 at least, review these conventions each time. 
 
 Now let a beam of length I have the weight P at the free end, and let it 
 be fixed horizontally or " walled in " at the right end. Then the moment 
 at any point distant x from the left or free end is M + P x. 
 
 (a) Change of shape. 
 
 From (8) we have now 
 
114 SUPPLEMENT TO CHAP. VII. [CHAP. III. 
 
 Integrating once (Art. 5) we have 
 
 BI^L^l 
 dx 2 
 
 where C is the constant of integration to be determined (Art. 6) by the 
 
 given conditions. Now by the condition in this case, when x = I, - 
 
 a x 
 
 must be zero, because the end is fixed, and the tangent there must therefore 
 
 P 1? 
 be horizontal (Art. 8). Hence O = - , and 
 
 E 2 
 
 _ 
 dx ~ 2 2' 
 
 We have thus introduced the condition that x cannot be greater than I. 
 Integrating again (Art. 5) 
 
 P a? PVx 
 
 Here again we have a constant to be determined, and here again we have 
 the condition that for x l,y must be zero, since at the fixed end there can 
 
 P 1? 
 
 be no deflection. Therefore, O = - and 
 
 o 
 
 P / \ P 
 
 E I y == [2 l z 3 Z 2 a; + a; 8 i = (21 + x) (I a) 2 . 
 6 y j 6 
 
 The deflection will evidently be greatest at the free end, and here, therefore, 
 for x = 0, we have 
 
 _ _ p * 3 . 
 
 If the cross-section is rectangular, I = 5 A 3 (Art. 10), and the maximum, 
 deflection A = - 
 
 (5) Breaking weight. 
 
 T I 
 
 We have also from equation (8) M = , where T is the tensile strain 
 
 t> 
 
 in any fibre distant from the centre. For v , T is the tensile strain in 
 
 2 
 
 O *p T Jj 
 
 the outer fibre, and M = . For v = we have the compressive 
 
 7i 2 
 
 2 C T 
 
 strain in the outer fibre upon the other side, or M = - . Theoretically 
 
 the two should be equal. Practically they are not. In fact, if we put for 
 
 .Pa = 
 Tbh 2 
 
 2 T I 
 
 M its value, we have P x = - , or for a rectangular cross-section P x = 
 
 h 
 
 i T & h 2 . This is greatest for x = I, hence the breaking weight P = . 
 
 ft TTJ 7 
 
 From this we have T = -r-rr' Now experimenting with beams of various 
 o fi 
 
CHAP. III.] SUPPLEMENT TO CHAP. VII. 115 
 
 materials, known dimensions and given weights, we may find experimen- 
 tally T. It would seem that this value thus found should equal either the 
 tenacity or crushing strength of the material, but the results of experiment 
 show that it never equals either, but is always intermediate between T and 
 C. Calling this intermediate value R, we have 
 
 The formula is based upon the condition of perfect elasticity, while R is 
 determined by experiments made at the breaking point when the condition 
 of perfect elasticity is no longer fulfilled. In the following table the tabu- 
 lated values of R are correct for solid rectangular beams, and sufficiently 
 exact for those which do not depart largely from that form. If instead of 
 
 we use the values of T or C, whichever is the smaller, we shall always 
 T)e on the safe side, since R is invariably intermediate between these. 
 
 In general we shall refer to the equation 
 
 when we have occasion to find the breaking strength. But it must be 
 always remembered that in any practical example we should replace T by 
 R for rectangular beams, or by T or O, whichever is the smaller, for others. 
 We give also the values of the coefficient of elasticity E. (Wood's Resist. 
 of Materials.) 
 
 TOR E 
 
 Cast-iron ....................... 16.000 96,000 36,000 17,000,000 
 
 Wrought-iron .................. 58^200 30,000 33.000 25,000,000 
 
 English Oak .................... 17,000 9,500 10,000 1,451,200 
 
 Ash ............................ 17,000 9,000 10,000 1,645,000 
 
 Pine ........................... 7,800 5,400 9,000 1,700,000 
 
 All in pounds per square inch. 
 
 2. Beam of uniform strength. 
 
 Suppose the cross-section or I is not constant, but varies so that at every 
 point the strain T is constant. From (11) we have 
 
 2 T I 
 
 M = P x ,' - for the outer fibre, whence 
 h 
 
 T = -. For a rectangular cross-section T = , ' . Now suppose the 
 21 o h 
 
 ' 
 2 
 
 breadth and height at the fixed end are &i and hi. Then at this end T = 
 n ~p 7 
 
 But this must be equal to T at any other point ; hence 
 
 6Pa QPl IV x 
 i A, 2 
 
 If we suppose the height constant, we have for the varying breadth at any 
 
 point b = li X . That is, the breadth must vary as the ordinates to a straight 
 I 
 
 line, and the plan of the beam is a triangle with the weight P at the apex. 
 
116 SUPPLEMENT TO CHAP. VII. [CHAP. III. 
 
 If the breadth is constant, h = hi./ ., or the elevation of the beam is a 
 parabola with the weight at apex. If the cross-section is always similar, 
 
 that is, if = -, we have 5 = -\, and substituting in the equation above 
 hi h hi 
 
 h = hi ? / . , which is a paraboloid of revolution. 
 
 (a) Change of shape 
 From (8) we have 
 
 d z y _ P x _ P x 
 
 where & and h are variable. If we suppose the height h constant and 
 always equal to hi, then, as we have seen, & = &i yj hence for rectangular 
 
 cross-section 
 
 d 2 y _ 12 PI 
 d x 2 E hi* 5i 
 
 /I nj 
 
 Integrating, since for x = I, = 0, we have 
 
 (L (K 
 
 dy _ 12 P Ix _ 12 F Z 2 
 d x E hi* &i E hi 3 &i* 
 Integrating again, since f or x = Z, y = 0. we have 
 _ 6P?a 2 _12PZ 2 a 6P 
 E hi 3 &i E hi z bi E h^ 
 
 For the maximum deflection x 0, and 
 
 . ^ ^^ 
 
 The above value of y can be written 
 
 but is , the deflection of a beam of constant cross-section &i A lt 
 
 as already found. Calling this deflection A , we have 
 
 Q 
 
 for the deflection at any point, or A = A for the maximum deflec- 
 
 tion. 
 
 In a similar manner, for constant breadth, we have 
 
CHAP. HI-.] SUPPLEMENT TO CHAP. VII. 117 
 
 For similar cross-sections, we have 
 
 9 r 5 x 3 I \5 ~ 9 36 P I 3 
 
 If we call the volume of the beam of constant cross-section V, then in 
 
 1 2 
 
 the first case the volume Vj = V ; in the second, V 2 = V ; in the third, 
 
 6 O 
 
 V 3= ?-V; or 
 
 V : V 2 : V 3 : V: = 30 : 20 : 18 : 15. 
 The maximum deflections, as we see above, are as 
 
 2 Ao, - Ao, - Ao, or as 20, 18, and 15. 
 
 o /* 
 
 That is, the deflections at the ends for a beam of uniform strength in the 
 three cases are as the volumes. 
 
 1.3. Beam a before fixed at one end Uniform load- 
 Constant cros-sectioii. If p is the load per unit of length, we have 
 for the moment at any point distant x from the free end, 
 
 x p x z p x z d z y 
 
 M = * x Sf = -a-.and hence =^ r . 
 
 ;.:.,. . p p 
 
 This moment is greatest for x = I, and hence Max. M = . 
 
 For the breaking weight, then, from (11) 
 
 pi 2 2TI 4 T I 
 
 T = or pl =-hT> 
 
 or twice as great as for an equal weight at the end. 
 For the change of shape, we integrate twice, precisely as before, the ex- 
 
 d 2 y p x z 
 pression -= j and obtain thus 
 
 * 24 B I 
 The maximum deflection, then, is 
 
 _ 4 p x 
 
 or 
 
 A = 8EI' 
 
 or only fths as great as for an equal load at the end. 
 2. ontant strength. 
 
 We have, as before, from (11) M = * = ^, whence T = ^f- 
 for rectangular cross-section, I = i I & and T = *-~. If h A, are the 
 
 breadth and heighth of the fixed end section, then, since T must be always 
 constant, 
 
 3 px 2 _ 3p I 2 I h* __ x 2 
 
 5 A 2 far*? or &7^7 = I s "" 
 
 For lieight constant, 5 = &j | X - \ 
 
118 SUPPLEMENT TO CHAP. VH. [CHAP. HI. 
 
 <y* 
 
 For breadth constant, h = hi -= . 
 
 , V 
 
 For similar cross -sections, h = fa 
 
 The first is in plan a parabola ; the second, in elevation a triangle ; the 
 third, a paraboloid of revolution. 
 
 For the change of shape, -we have, by proceeding in the same manner as in 
 Art. 12, A = 2 Ao, A = 4 Ao, and A = 3 A in the three cases, where A is the 
 deflection of a similar beam of constant cross-section &i hi. 
 
 14. Beam supported at both the ends Constant cross- 
 section Concentrated load. Let the weight P be distant from the 
 left end by a distance Zi and from the right end by Z a . Let the distance of 
 any point from the left end be x. For the upward reaction at the left end, 
 
 P k 
 
 T 
 
 The moment, then, at any point between the left end and F, for x less 
 than Zi, isM ""?"" ^ or any point to the right of P, or # greater 
 
 than l lt M' = . + P i x Zi I. Instead of this, however, we may 
 
 take the reaction at the other end, V a = P~; and then for x greater than Zi, 
 
 The moment is evidently greatest at the point of application of the load, 
 or for x = li Hence the maximum moment is T -. 
 
 (a) Breaking weight. 
 
 2 T I P I I 
 
 From (11) M = - = -- ' , or, for the breaking weight, P = 
 
 2 T I Z 1 T 5 A ? Z 
 
 -, 7 7 For rectangular cross-section, I = 77: 5 h* and P a -, 7 - 
 ti ti It t 12 o Li L-i 
 
 For a load in the middle, Zi = Z 2 = I and Max. M = -r P I, and P = 
 
 & 4: 
 
 r 4 times as great as for a beam of same length fixed at one end 
 and free at the other. 
 (5) Change of shape. 
 We have, then, from (8), for x less than Zi, 
 
 & y P Z 2 x , & y' P |j (7 x) ., , 
 
 d* = -HIT and 5/ "irr 2 for z greater than ' 
 
 Integrating, we have 
 
 d+y_ PZ,** dtf _ PI, r * f l, . 
 
 -~ = " " 
 
 For x = li t these two values of ^-are equal, and hence, since Z, = I l lt we 
 
 a a; 
 
CHAP. III.] SUPPLEMENT TO CHAP. VII. 119 
 
 P I 2 
 
 have C' = C H -^-. We have then the two equations 
 d y Pl*x 2 , d y' P I, 
 
 ^=-^TI +C and jf -i 
 
 containing both the same constant C. 
 Integrating these, we have 
 
 P Z 2 y? P Zi rZ a; 2 a*n P Z, 2 
 
 In the first of these, f or x 0, y ; hence d = 0. 
 
 P Zi 3 
 For x = Zi, y = y' ; and hence O 2 = . 
 
 D ! I 
 
 For x Z, y' ; and hence, finally, C = * ' (2 Z Zi] 
 We have, therefore, by substitution of these constants, 
 
 P Zi 2 Z 2 2 
 For x Zi, we have the deflection at the load y = 7. 
 
 O E I 6 
 
 >7 rti 
 
 Inserting the value of C in the value for above, and placing the 
 
 value of equal to 0, we have for the value of #, which makes y a maxi- 
 d x 
 
 mum, x =* / (2 Z Zj) Zi, an expression holding good only for x less than 
 
 Zj. Inserting this in the value for y, we have for the maximum deflection 
 itself 
 
 If the load is in the middle, we have for the curve of deflection 
 
 P Z 3 
 and for the deflection itself A = 
 
 48 El 
 
 The greatest deflection is not, then, at the weight, except when the load is 
 in the middle. When this is the case, the deflection is only ^-th of the 
 deflection for the same length of beam fixed at one end and loaded at the 
 other free end. 
 
 15. Beam as before supported at the end Uniform 
 load. For a load p per unit of length, the entire load is p Z. The reac- 
 
 tions at each end are -, and the moment at any point is 
 
 is evidently greatest at the centre, and hence 
 Mar. M = - 
 
120 SUPPLEMENT TO CHAP. VH. [CHAP. IH. 
 
 For the breaking weight, then, from (8) 
 
 16 T I 
 
 or 4 times as much as for a beam of same length loaded uniformly and 
 fixed at one end. 
 For the change of shape, we have 
 
 d z y _^px (lx} 
 
 dx*~ 2EI 
 The constants of integration are determined by the conditions that, for 
 
 X = W'dx = ' ^ ' y~'-> and x = l,y = 0. Integrating, then, twice 
 under these conditions, we have 
 
 This is greatest at" the centre, or for x = ; hence the maximum deflection is 
 
 2 
 
 A = - -, or only i&ths of a beam of the same length fixed at one 
 
 end and uniformly loaded. 
 
 16. Beam supported at one end and fixed at the other 
 Constant cross-section Concentrated load. Let the left 
 end be fixed horizontally so that the tangent to the deflected curve at that 
 
 point is always horizontal, and therefore = 0. 
 
 a x 
 
 Let the distance of the weight P from left be a, and the distance of any 
 point x. 
 Then, for x less than a, we have 
 
 M = V (l-x) + P (a-x) 
 for x greater than a, 
 
 M' = -V(Z-aO, 
 where V is the reaction at the free end, and is so far unknown. 
 
 If we put M = - and M' = '{, and integrate as usual, and remem- 
 d x d x* 
 
 ber that f or x = 0, ^ = 0, and for x = a, -^ ' = ~, we have 
 d x dx dx* 
 
 Integrating again and determining the constants by the conditions that, 
 for x = 0, y = 0, and for x = a, y y\ we have 
 
 V = ~~ [V (3 -aO-P (3 -*)] 
 
 y 1 = -^ [V * (3 Z_a)-P (3 x-a} a*]. 
 
CHAP. III.] SUPPLEMENT TO CHAP. VII. 121 
 
 Now, for x = I, y = ; hence V = P L_ - -. If the load is in the 
 
 middle, V = P. 
 
 , . V, or the reaction at the free end, is now known, and substituting it in 
 , the value of y' above, we have the equation of the deflection curve between 
 the weight and the free end. 
 
 y' --\ (3 x a) a 2 
 
 Substituting it also in the value of above, and placing then ^- equal 
 
 dx dx 
 
 to zero, we find for the value of x, which makes the deflection a maximum, 
 
 when x is greater than #, x = I 1\/ j . 
 
 ' 3 & o> 
 
 Substituting this value of x in the value of y' above, we have for the 
 maximum deflection itself 
 
 P Z 3 1 
 When the weight is at the middle, this becomes A = - x = , or 
 
 48 E I y/5 
 
 only = , as much as for a beam of same length fixed at end and with 
 tC y 5 
 
 load at other end, and only - as much as for same beam simply sup- 
 
 y5 
 ported at ends. 
 
 Breaking weight. 
 
 Having now V, we know M and M'. Rupture will occur where the 
 moment is greatest, that is, either at the fixed end or at the weight. Now 
 the moment at P is V (I a) V 1 + V a. The moment at the fixed 
 end is V l + P a. Now, as V is always less than P, we see at once that 
 for any value of a less than I, the moment at the weight is greatest. 
 
 We have for the moment at the weight from (8) 
 
 2 (3Z~a) /7 , 2TI ... 
 ij - (I- a) = , and hence 
 
 4 T 1 1 3 
 
 for the breaking weight P = - 
 
 ha* (3 I a) (l-o)* 
 
 64 T I 
 
 If the weight is in the middle, P -=-.-, 
 
 o fi l 
 
 or fths as much as for the same beam supported at the ends. 
 
 IT. Beam as before fixed at one end and supported at 
 the other Uniform load. In this case the moment at any point is 
 
 M V (l)-r-p (l-x)* = E I --Jj. Integrating twice and determin- 
 ing the constants by the conditions that, for x = 0, - = and y = 0, we 
 
 d x 
 
 easily obtain 
 
122 SUPPLEMENT TO CHAP. VH. [CHAP. III. 
 
 (B l -**-P (6 Z2 ~ 
 
 o 
 
 For x = 1, y = 0, and hence V = p 1. 
 
 8 
 
 Substituting this value of V 
 
 j r _ V'-iR 
 
 Putting this last equal to zero, we find x - - Z, or x = 0.5785 Z, 
 
 lo 
 
 for the value of a*, which makes the deflection a maximum, and this in- 
 serted in the value of y gives for this maximum deflection itself, 
 
 
 16 4 El El 
 
 For the "breaking weight, we have, since the greatest moment is at the fixed 
 
 1 72 __ 1 72 2 T I , . 16 T I 
 
 end and equal to p I 2 , M = p I 2 - ; hence p I = - - . 
 
 8 O fl hi 
 
 The strength is, then, f times as great as for the same load in the middle, 
 but no greater than for a beam of same length and load supported at both 
 ends. 
 
 18. Beam fixed at botli euds Coiitant cros-section 
 Concentrated load. Taking our notation as before (Art. 12), we 
 have in this case not only a reaction at the right end, but also a positive 
 'moment there as well, both of which must be found. If h be the distance 
 from left end to weight, and h from weight to right end, and if Vj and 
 V 2 are reactions, Mi and M 2 the moments' at left and right ends respec- 
 tively, then for equilibrium we must have Vi + V 2 = P, Mj + Vi ^ = 
 M 2 -V 3 Z a . 
 
 For x less than ^ we have M = Mj +V 1 x = EI |. Integrating 
 
 cL x 
 
 once, since the constant is zero, because, for = 0, ~ 0, we have 
 
 a x 
 
 d x 2 E 
 Integrating again, since, for x = 0, y = 0, and the constant is zero, 
 
 x\ 
 
 For the distance from the right end to the weight we may obtain similar 
 expressions, if we take that end as the origin, only we should have V 9 
 
 and M 2 in place of Vi and MI. At the weight itself - and y must in 
 
 (L x 
 
 each case be equal, but *=- of opposite sign. Therefore we have the equa- 
 
 CL $ 
 
 tions (2 MX + V, Zx) li = -(2 M 2 -V 2 Z 2 ) Z 2 , 
 
 (3 Mi + Vx Zx) Zx 2 = (3 M 2 -V 2 Z 2 ) Z a 2 . 
 
CHAP. III.] SUPPLEMENT TO CHAP. VII. 123 
 
 From these two equations, and the two equations above, viz., ~ Vi + V 2 = P 
 and Mi + Vi li = Mo V 2 Z 2 , we can determine Vi, V 2 , Mi and M 2 . 
 Thus from the last two we have 
 
 Vi Zi+V 2 ? 2 = M 2 -M: =V, k + PZa + Vi Z 2 =Vi Z + FZ 2 , 
 
 or Vi I = Ma-Mi-P Z 2 . 
 
 So also V 2 Z = M 2 -Mi +P Zi, and substituting these in the equations above, 
 
 we have 
 
 (Mi+Ma) ZrrFZj Z 2 , 
 
 M! Z (2 Z X -Z 2 )-M 2 Z (2 Za-ZO = P Zi Z 2 (k - Z 2 ) ; 
 and from these we have, finally, 
 
 and then from the values of Vi Z and V 2 Z above 
 
 Vl = -P" 2 v"^/ v ^ = 
 
 Z 3 Z 3 
 
 Change of shape. 
 Substituting these values, we can now find 
 
 077 
 
 Hence y is a maximum for x = ^-, and the maximum deflection 
 
 , 3 Zi + ly 
 
 itself is 
 
 2 P Z x 3 Z 2 2 
 
 A = 
 
 3 E I (3 
 
 This expression will be itself a maximum for Zi = Za or Zi = \ Z, that is, 
 the maximum deflection for a weight in the middle is at the weight and 
 equal to 
 
 PZ 3 
 
 A = 
 
 192 E I 
 
 This deflection is greater than the maximum deflection for any other 
 position of the weight, which in general is not found at the weight itself, 
 but at some other point between the weight and farthest end. 
 
 We see above that the deflection in this case for load in middle is only 
 one-fourth as much as for same beam and load when supported at the ends. 
 
 Breaking weight. 
 
 For the greatest moment, which we easily find to be at the end, we have 
 
 
 _PZ 1 Z 2 2 _ PZ, (Z-ZO 
 
 ~~ ~ 
 
 This is a maximum for Zi = Z. That is, the greatest moment at the end 
 occurs wlien the load is distant one-third of the length from that end. The 
 
124 SUPPLEMENT TO CHAP. VII. [CHAP. in. 
 
 value of this greatest moment is - P 1. Hence we have from (11) P I 
 
 A t 27 
 
 2 T I 27 T I 27 
 
 = 7 or P = , or as great as for the same beam supported at 
 
 /i d> fi I lo 
 
 p 7 
 
 the ends only. If the weight is in the middle, however, we have 
 
 8 
 
 2 T I 1 6 T I 
 
 - or P = , or twice as much as the same beam supported at the 
 tl ii L 
 
 ends. 
 
 19. The above is sufficient to introduce the reader to the theory of 
 flexure. He can now discuss for himself the above case for uniform load, 
 and prove that the maximum deflection is at the centre and equal to 
 
 pi 2 1 
 
 . That the greatest moment is at the end and equal to p l\ 
 
 24 T I 
 
 and that the breaking weight is p I = . We may also observe that 
 
 ft v 
 
 both in the beam fixed at one end and supported at the other, and fixed at 
 both ends, the moment at the fixed end is positive. From this end it de- 
 creases towards the weight, and finally reaches a point where the moment 
 is zero. Past this point the moment becomes negative, and in the case of 
 the beam, free at the other end, increases gradually to a maximum and then 
 decreases to zero. In the beam fixed at both ends, it increases to a maxi- 
 mum, then decreases to zero, then changes sign and becomes positive and 
 increases to the other end. These points at which the moments are zero 
 are points of inflection, because here the curvature changes from convex to 
 concave, or the reverse. 
 
 They can be easily found from the equations for the moments by finding 
 the value of x necessary to make the moments zero. 
 
 Thus, for a beam fixed at one end and supported at the other, uniform 
 
 load, the inflection point is at a distance from the fixed end x . For 
 
 both ends fixed, we make M p [I* 6 (I x) x] = 0, and find x = 
 
 12 
 
 V (3 T VW) I = 0.21131 I and 0.7887 Z. The reader will also do well to 
 b 
 
 discuss the curves of moments. He will find the moments represented by 
 the ordinates to parabolas, and limited by straight lines similarly to Figs. 
 73 and 75, PI. 18. 
 
 We shall give in the Supplement to Chap. XIV. much more general 
 formulae, from which, for one or both ends fixed or free, the moments and 
 reactions at the supports may be found, when any number of spans of vary- 
 ing length intervene, for single load anywhere upon any span, or uniformly 
 distributed over any span. 
 
CHAP. VIII.] CONTINUOUS GIRDERS. 125 
 
 CHAPTER VIII. 
 
 APPLICATION OF THE GRAPHICAL METHOD TO CONTINUOUS 
 GIRDKRS GENERAL PRINCIPLES. 
 
 8O. Mohr's Principle. Thus far, in addition to the general 
 principles of the Graphical method, we have noticed more or 
 less in detail its application to the composition and resolu- 
 tion of forces, and the corresponding determination of the 
 strains in the various pieces of such framed structures as Bridge 
 Girders, Roof Trusses, etc. We have also illustrated the 
 graphical determination of the centre of gravity and moment 
 of inertia of areas, as also of the bending moments and shear- 
 ing forces for simple girders, including several important cases 
 in practical mechanics. (See Art. 41.) Lastly, we have taken up 
 the subject of Bridge girders more in detail, and developed in 
 order the principles to be applied 'in the solution of any par- 
 ticular case. Although brief, it is hoped that this portion will 
 be found sufficient to illustrate fully the method of procedure 
 to be followed in practice. 
 
 As regards simple girders, the principles referred to are so 
 easy of application that the reader will find no difficulty in 
 diagraming the strains in any structure of the kind, as explained 
 in the " practical applications " of Arts. 8 to 13 ; or he can find 
 the maximum moment at any cross-section for given load- 
 ing according to the last chapter. In the case of beams or 
 girders continuoii,s over three or more supports, however, we 
 meet with difficulties which for some time were considered in- 
 superable. 
 
 Thus Cidinann, in the work which we have so often quoted, 
 says : * " The determination of the reactions at the supports for 
 a continuous beam, which depend upon the deflection, the law 
 of which is given by the theory of the elastic line, is impossi- 
 ble by the graphical method, at least so far as at present de- 
 veloped. The theory rests upon the principle that the radius of 
 
 *Culmann?8 GrapMsche8ta.Uk, p. 278. 
 
126 CONTINUOUS GIKDERS. [CHAP. VIII. 
 
 curvature of the deflected beam, for any cross-section, is in- 
 versely proportional to the moment of the exterior forces. 
 'Now the deflection at any point is so small, and the radius of 
 curvature so great, that its construction is impracticable, and 
 will so remain until Geometry furnishes us with simple rela- 
 tions between the corresponding radii of curvature of pro- 
 jected figures whose projection centre lies in the vertical to the 
 horizontal axis of the beam. If such relations were known, we 
 could by projection exaggerate the deflection of the beam 
 until the radius of curvature became measurable. Since we 
 are not yet able to do this, we must have recourse to calcula- 
 tion" He then enters into a somewhat abstruse analytic dis- 
 cussion of the continuous girder, and deduces formulae for the 
 reactions at the supports. These being thus known, the graph- 
 ical method is then applied. 
 
 Concerning this difficulty, Mohr* remarks that it has but 
 little weight, and may be easily overcome if the same simplifi- 
 cation of the graphical method is made which is considered 
 allowable in the analytical investigation, viz., when we take in- 
 
 stead of the exact value of the radius of curvative 
 
 as given by the calculus, the approximate value 
 
 
 Thus, let PL 14, Fig. 50 represent a perfectly flexible cord 
 A B D loaded by arbitrary successive forces. The variation of 
 these forces per unit of horizontal projection dvwe represent by 
 p. Take the origin of co-ordinates at the lowest point B. If 
 the cord is supposed cut at B and D, we have at B a horizontal 
 force H, and at D a strain S, which may be resolved into a 
 horizontal force Hj and a vertical force V. Since these forces 
 are in equilibrium with the external forces, the conditions of 
 equilibrium are 
 
 (1) ....... H = H! 
 
 and px 
 
 (2) ..... V= / pdx. 
 
 Jo 
 
 * Zeitsoh. fas kannov. Arch,-u. Ing. Vereim. Band xiv., Heft 1. 
 
CHAP. TUT.] CONTINUOUS GIRDERS. 127 
 
 Moreover, Cx 
 
 I p dx 
 
 (3) . . *La = = 
 
 dx H! H 
 
 Differentiating : 
 
 d 2 y p d x 
 
 1 ryp 
 
 ~7 
 
 6? X 
 
 Now, had we formed a force polygon by laying off the forces, 
 then taken a pole at distance H and drawn lines from pole to 
 ends of forces, the corresponding equilibrium polygon would, 
 as we have seen, Art. 43, be tangent to the curve A B D at the 
 points midway between the forces. The greater the number 
 of forces taken, the shorter, therefore, the sides of the polygon ; 
 the nearer it will approach the curve A B D. This curve is 
 therefore the equilibrium curve, found according to the graph- 
 ical method. Its equation is given above by (4). 
 
 But the equation of the elastic line is, as is well known, 
 
 (y) >'** 7 'o ~r~ ? 
 
 a (E> J- 
 
 where E is the modulus of elasticity of the material, M the 
 moment of the exterior forces, and I the moment of inertia of 
 the cross -section. 
 
 Comparing now this equation with equation (4) above, we 
 see that the elastic line is an equilibrium curve whose horizon- 
 tal force H is E, and whose vertical load per unit of length p 
 
 is represented by the variable quantity -- 
 
 This simple relation, first given by Mohr, renders possible 
 the graphical representation of the elastic line, and not only 
 solves graphically almost all problems connected with it, but in 
 many cases simplifies considerably the analytical discussion 
 also. 
 
 81. Elastic Curve. If we choose the pole distance H at -th 
 
 n 
 
 E instead of E, the ordinates of the elastic line will be n 
 times too great. If the scale of the figure is, however, - th the 
 
 * Stoney Theory of Strains, p. 146. Wood Resistance of Materials, p. 98. 
 Also Supplement to Chap. VII , Art. 11. 
 
128 CONTINUOUS GIRDERS. [CHAP. VIII. 
 
 real size, then in the diagram the ordinates of the elastic line 
 will l)e given in true size. 
 
 Equation (5) may also be written 
 
 that is, the elastic curve is an equilibrium curve, or catenary, 
 whose horizontal force H is E I or F, and whose corresponding 
 
 variable load per imit of length is M or respectively. 
 
 If we divide, then, the moment area by verticals into a number 
 of smaller areas, and consider these areas as forces acting at 
 their centres of gravity, these forces determine, as we have seen 
 (Art. 43), an equilibrium polygon which is tangent to the elas- 
 tic curve at the verticals which separate the areas. Thus we 
 can construct any number of tangents to the elastic curve ; areas, 
 which are positive or negative, must, of course, be laid off in the 
 force polygon in opposite directions. 
 
 If we divide the moment area by lines which are not vertical 
 [PL 14, Fig. 51], the directions of the outer polygon sides are 
 the same as for vertical divisions, because the vertical height 
 between the corresponding outer sides in the force polygon is 
 in any case always equal to the total load. 
 
 The two outer polygon sides for any method of division are, 
 therefore, tangents to the elastic curve at the- ends of the same. 
 Here also we can, of course, have negative areas. 
 
 82. Effect of End Moments. A beam or girder continuous 
 over three or more supports differs from a beam simply resting 
 upon its supports, in that, in addition to the outer forces, we 
 have acting at each intermediate support a moment or couple. 
 
 But, as we have seen, Art. 23, the effect of these moments or 
 couples will be simply to shift the closing line of the equilibrium 
 polygon through a certain distance. Thus [PL 14, Fig. 52 ()], 
 if the span ^ were uniformly loaded and simply supported at 
 the extremities A and B, the equilibrium curve, or curve of mo- 
 ments, would, as we know (Art. 44), be a parabola A D B. If, 
 however, the beam is continuous, we have at A and B moments 
 or couples acting, and the closing line A B is shifted to some 
 position as A' B'. If now w r e consider the moment area, we 
 see that by the shifting of the closing line the former moment 
 area, which we shall call \\\Q positive area, is diminished, while 
 to the right and left we have negative areas A A' C and B B' C.' 
 
CHAP. VIII.] CONTINUOUS GIRDERS. 129 
 
 It is evident that these areas have also a corresponding action 
 upon the elastic line. For a positive moment area this last is 
 concave upwards, while for negative areas it is convex upwards. 
 At the points of transition C and C' we have the inflection 
 points. This follows easily if we only hold fast the manner in 
 which the elastic line is constructed, viz., by dividing the mo- 
 ment area into laminae and regarding the area of each as a 
 force. The forces thus obtained must plainly act, some upwards 
 and some downwards, and the corresponding equilibrium poly- 
 gon or elastic line must be in part convex upwards and in part 
 convex downwards, and hence at the points of transition we 
 must hav r e points of inflection where the moment is zero. 
 
 83. Division of tne Moment Area. We shall assume the 
 cross-section of beam constant. Regarding the elastic line 
 simply as an equilibrium polygon, we can apply the principle 
 that the order in which the forces are taken is indifferent (Art. 
 6) when the resultant only is desired. Since in the considera- 
 tion of a single span only the first and last sides are of impor- 
 tance, we can, so long as we consider a single span only, take 
 then the laminae or divisions of the moment area in any order 
 we please. More than this, we can, as we have seen in Art. 81, 
 divide the moment area into laminae not vertical ; for example, 
 we may in any span distinguish three parts, one positive and 
 two negative, and consider each as a force acting at the centre 
 of gravity of the corresponding area. [This holds good only 
 for constant cross-section. For variable cross-section the hori- 
 zontal force E I is variable.] Still further, we can divide the 
 moment area for a single span into a positive area, which is pre- 
 cisely the same as for a non-continuous beam, and into a nega- 
 tive area, which will be evidently a trapezoid. 
 
 This is of great importance. To understand it fully we refer 
 to PL 14, Fig. 52. Here, in the second span, we see that the 
 real moment area consists of a positive part, viz., the parabola 
 C D C', and two negative parts A A' C and B B' C'. Instead of 
 these we may take the entire parabolic area A D B and the tra- 
 pezoid A A' B' B, or, finally, instead of this trapezoid, we may 
 take the two triangles A A' B' and B B' A'. The parabolic 
 area is positive, the triangular areas are negative. 
 
 If we assume the load as uniformly distributed, the first area 
 will be always parabolic, and we may, therefore, call it the 
 parabolic area. 
 
130 CONTINUOUS GIRDERS. [ciIAr. Vm. 
 
 By this division of the moment area we have obtained a great 
 advantage. While the three areas C D C', A A' C and B B' C' 
 are all three dependent upon the moments at the supports A A! 
 and B B', we have by his new division to do with three areas, 
 of which the first is entirely independent of the moments at the 
 supports, the second depends only upon that to the left, and the 
 third only upon that to the right. 
 
 84. Properties of the Equilibrium Polygon. Let us con- 
 sider now the case of a beam over four supports, that is, of 
 three spans Z , 4 and 4 the first and last being, as is usually the 
 case, equal, and the two first loaded with both live and dead 
 load, the last with dead load only. The parabolas for the ver- 
 tical loads [PL 14, Fig. 52] may be constructed by means of a 
 force polygon, or the ordinates at the centre calculated, and the 
 parabolas then drawn. The moments at the supports are A A' 
 and B B'. Although these are unknown, it is not necessary to 
 assume them at first. They may be directly constructed. 
 
 Thus, if we conceive the moment areas in each span divided 
 into positive parabolic areas and negative triangles, we have in 
 the first and last span one, in the middle two triangles. If we 
 consider these areas as forces acting at the corresponding centres 
 of gravity, we shall obtain an equilibrium polygon of the form 
 given in Fig. 52 (b). That is, this polygon must have eight 
 sides, and its angles must be somewhere on the verticals through 
 the centres of gravity of the parabolic and triangular areas. 
 The parabolic areas act downwards, the triangular areas up- 
 wards. The problem is, to make these last so great that this 
 polygon shall pass through all the points of support. 
 
 One of the properties of the polygon we have, therefore, just 
 noticed, viz. : its angles must lie in the verticals through the 
 middle points of the spans and through the points distant from 
 A and B one-third of the spans on each side (i.e., the centres of 
 gravity of the triangles). 
 
 If we prolong the second and fourth sides of the polygon, 
 they intersect in a point M, the point of application of the 
 resultant of the two contiguous triangular area forces (Art. 44). 
 
 The areas of these two triangles are - A A' Z and - A A' ^, that 
 
 is, the areas are as the spans 4 and ^. 
 
 Then by the principle of Art. 18 the resultant divides the 
 distance between the forces into two portions, which are to each 
 
CHAP. VHI.] CONTINUOUS GIRDERS. 131 
 
 other as ^ to 4> or inversely as the forces. Since the entire dis- 
 tance is - + 7: ^i> the distance of the resultant or of the point 
 o o 
 
 of intersection M from L is - ^ ; from N it is - 1 Q . The point 
 
 O t> 
 
 M, therefore, must lie somewhere in the vertical at - ^ from L, 
 
 o 
 
 the point of application of the triangular area force for the 
 span 4- The verticals through the centres of the parabolic 
 areas we call the parabolic or middle verticals / those through 
 the centres of gravity of the triangular areas, the third verti- 
 cals / those through a point as M, the point of application of 
 the resultant of two contiguous triangular area forces, the lim- 
 ited third verticals. Upon these verticals two sides must always 
 intersect. 
 
 85. Polygon for tlie Positive Moment Areas. It will be 
 
 found best to take as the reduction base for areas ^, i.e., half 
 
 2 
 
 the second span, and for pole distance l. Reducing the 
 
 3 
 
 areas of the parabolas to this basis, and considering the heights 
 thus obtained as forces, we can form a force polygon with pole 
 
 distance - \. It is not necessary to draw this polygon ; our ob- 
 3 
 
 ject is to find the corresponding equilibrium polygon. This 
 last, since we consider the entire parabolic area as a'force act- 
 ing at its centre of gravity, consists of two lines which inter- 
 sect in the vertical through the middle of the span. We pro- 
 long each of these lines and obtain two lines as shown in PL 14, 
 Fig. 52 (c). The segments cut off by these lines from the verti- 
 cals through the supports are the moments of the parabolic 
 areas with respect to the supports. These moments we can 
 easily find. 
 
 Thus, let the deflection of the parabola in the second span at 
 
 o 
 
 the centre be f, then its area is fl^. This reduced to the 
 
 3 
 
 1 4: 
 
 basis - 4 gives -/ as the force. The moment of this force 
 A o 
 
 4: 1 1 
 
 with reference to the supports is f x li = 2fxli. This 
 
 32 3 
 
 moment is equal to the segment sought multiplied by the pole 
 
132 CONTINUOUS GIRDERS. [CHAP. VIII. 
 
 distance. This last is - ^. The segment, therefore, is 2/! "We 
 
 o 
 
 do not need, therefore, to draw the force polygon, but have sim- 
 ply to take off with the dividers the middle or din ate of the para- 
 
 / D $ 
 
 bola f = ~rr , and lay it off twice on the verticals right and left 
 
 8 
 
 through the supports, and join the four points thus obtained by 
 lines crossing each other under the centre of the span. The 
 equilibrium polygon for the positive parabolic area is then 
 ready for the middle span. ["We advise the reader to construct 
 it for himself.] 
 
 For the two side spans the construction is different. Here 
 
 2 4 I 
 
 the area of the parabola is -f 1 0) the reduced area -f j, and 
 
 3 3 L\ 
 
 the moment x = 2' . 
 
 Dividing by - ^, we have for the segment required 
 3 
 
 Therefore, in the end spans, or generally in any span not equal 
 
 to the standard span, or that which furnishes the constants - l 
 
 2 
 
 and - Zj, we must multiply the middle ordinate of the parabola 
 o 
 
 by the square of the ratio of the " standard n span to the span 
 in question, and then lay the product off twice upon the verti- 
 cals through the supports. This multiplication is easily per- 
 formed graphically. If from the middle of span 4 we lay off 
 If horizontally and join the end with the end of y, then lay off 
 4 2 in same direction and draw a parallel to the first line, the 
 
 l z 
 segment on/" 7 will bejf'~& For we shall have 
 
 h 
 
 79 / / V O /*/ ''0 
 
 4 2 :/ :: tf:^ or os=f 'A. 
 
 Since these cross-lines depend upon given quantities, they 
 can be constructed for every span, and thus we have Fig. 52, c. 
 They give us not only the moments over the supports, but also 
 the moments for any point of the parabolic area forces. We 
 shall hereafter make use of them. 
 
 $6. Construction of the Fixed Points, and of the Equi- 
 librium Polygon. We have thus all the given and known 
 
CHAP. Vni.] CONTINUOUS GIRDERS. 133 
 
 quantities, have deduced the general properties of the equili- 
 brium polygon, and will now endeavor by their aid to draw the 
 polygon itself. We shall then be able to find the actual mo- 
 ments A A' and B B' at the supports. It is impossible at first 
 to draw any single side of the polygon in true position, and we 
 must, therefore, endeavor to find certain /points of the same suf- 
 ficient to determine it. 
 
 Lay off first the three spans, PL 14, Fig. 52 (d). Suppose 
 the second side of the polygon prolonged till it intersects the 
 vertical through the end support a, in a point K, Fig. 52 (). 
 This point is known. It is given by the moment of the para- 
 bolic area in the first span with respect to this end support. 
 This moment we have already by the cross-lines in Fig. 52 (c). 
 We have then simply to take it off in the dividers from (c) and 
 lay it off from d to K 7 in Fig. 52 (d). We have now in Fig. 52 
 (J), two points of the polygon known, namely, the end support 
 and K, which last must be in the second side prolonged. 
 
 The triangle L M N is now of special importance. What- 
 ever may be the position of KM and M N, we have already 
 seen that the intersection M must always lie somewhere in the 
 limited third vertical. The first side KM must, however ', 
 always pass through K, a known point. The second must pass 
 through the support, also a known point. The points L and N 
 must, moreover, always lie in the third verticals, distant from 
 
 A, 5- \ and ^ Z respectively. 
 
 If the line K M takes up various positions under these con- 
 ditions, the line M N will revolve about a fixed point which is 
 given ~by the intersection of a line through K and the support 
 A with M N. 
 
 If, then [Fig. 52 (d)], we draw a line in any arbitrary direction 
 through K 7 , and note the intersections L' and M 7 with the first 
 third vertical and the limited thtrd, then through L' and the 
 support draw a line to intersection N 7 with second third verti- 
 cal, and join M 7 N 7 , and finally through K' and the support 
 draw a line intersecting this last in I, the point I thus deter- 
 mined is & fixed point, and remains the same for any position 
 of K' M 7 . It is therefore a point on the fourth side M N of 
 the polygon. For the triangle L 7 M' N' may have any posi- 
 tion, yet so long as its angles lie in three parallel fixed lines, 
 ui id two of the sides- pass through two fixed points, the other 
 
134 CONTINUOUS GIRDERS. [CHAP. VIII. 
 
 side must also pass through a fixed point.* Out of all possible 
 positions of the triangle, one of these positions must coincide 
 \vith the polygon sides, and hence this fixed point is a third 
 known point, since we have already K' and the end support. 
 
 Although, then, we are as yet unable to draw any of the sides 
 of the polygon in true position and direction, still from the 
 hitherto known properties we have deduced a new one. We 
 know now a point through which the fourth side must pass. 
 But this is not all. We proceed still further. The fourth and 
 fifth sides must intersect upon the vertical through the centre 
 of the second span. These sides, moreover, cut off upon any 
 vertical the moment of the parabolic area with respect to any 
 point in that vertical. We know this moment thus for the 
 point I just found. It is found by taking the segment cut off, 
 from the vertical through I, by the cross-lines for the parabolic 
 areas found above in Fig. 52 (c). 
 
 Laying this segment off from I, we thus find I', a point in the 
 fifth side prolonged. From this point we proceed as before to 
 find the next fixed point I". We then lay off from 1" the mo- 
 ment of the parabolic area for this point and find F", a point 
 upon the eighth side. We can now draw the polygon itself. 
 
 Thus the eighth side passes, of course, through the last sup- 
 port and also I"'. It is therefore determined. Through the 
 intersection of this line with the vertical through the middle of 
 the span and the point I" the seventh side passes. The sev- 
 enth side is therefore determined. Through the intersection of 
 this with the third vertical and the support the sixth side 
 passes and continues till it intersects the third vertical on the 
 other side. Then from this point towards I' to intersection 
 with vertical through centre of middle span. From this last 
 point towards I to intersection with third vertical. From this 
 last point again through support to intersection with third verti- 
 cal on other side ; then towards K' to intersection with vertical 
 through centre of end span ; and lastly, from this last point of 
 intersection through the end support, and the polygon is com- 
 plete as given in Fig. 52 (b). 
 
 In this manner, however, inaccuracies may occur. To avoid 
 these we may start from the rig/it end support and also find 
 four fixed points as above. It is unnecessary to make the con- 
 
 * This proposition the reader can easily prove geometrically or analytically. 
 See Art. 112. 
 
CHAP. VIII.] CONTINUOUS GIRDEKS. 135 
 
 striiction. We see at once that we shall thus obtain in each end 
 span two points, in the middle span four points, which last, be- 
 ing joined by lines crossing each other, give in the middle span 
 two sides in proper position. It is also evident how the poly- 
 gon may then be completed. 
 
 87. Construction of the Moments at the Support.*. Thus 
 we are able to construct the equilibrium polygon, or rather the 
 extreme tangents to the elastic line for each span. "We have 
 now to determine the two moments over the supports. This is 
 very simple. The first moment to the left is cut off by the 
 fourth side, the second by the fifth side of the polygon, from 
 the verticals through the supports. We have therefore only to 
 prolong these two sides, take off the segments in the dividers, 
 and lay them off in Fig. 52 (a) in A A! and B B'. We have, 
 then, in PL 14, Fig. 52 (a) the moments for the given case and 
 loading at any point, as shown by the shaded area. 
 
 The proof is simple. The two lines N M and N L [Fig. 52, 
 Z*] evidently cut off upon the vertical at the support the moment 
 of the force acting at N. This force is the area of the triangle 
 
 A A.' B' equal to - A. A! ^. This reduced to the basis - l gives 
 
 A A!. If. we multiply this by the lever arm of the force, we 
 have its moment. This moment is, however, equal to the seg- 
 ment A A' multiplied by the pole distance, and since this pole 
 
 distance is itself ~ h the segment itself to the assumed pole 
 
 o 
 
 distance gives us the moment. 
 
 We see that it is not necessary to draw the line N L as it 
 passes through the support. We have simply to prolong the 
 side M N to intersection with the vertical through the support. 
 It is to be observed that the moments at the supports are cut off 
 at the supports only by those lines which pertain to the " stand- 
 ard " span, or that span from which we take our reduction basis 
 and pole distance. For lines in the other spans the above does 
 not hold good without modification. It is, however, always 
 possible, at least for from two to five symmetrical spans, to 
 observe the above conditions. In those cases where this is not 
 possible, an easy graphical multiplication of the segments by 
 the square of the ratio of the spans will give the moments. 
 We see also the reason why, for four symmetrical spans, the 
 and not the first must be taken as the standard span. 
 
136 CONTINUOUS GIBBERS. [CHAP. VIII. 
 
 If the construction of the moments over the supports is our 
 sole purpose, as is in practice the case, the polygon need not be 
 drawn. We have only to find our fixed points, and note the 
 intersection of the sides with the verticals through the supports, 
 without drawing the sides themselves. In the preceding Arts- 
 we have purposely considered only the particular case of uni- 
 form loading, and have taken only three spans, in order to 
 familiarize the reader with the nature of the problem and the 
 method of its solution. In oi'der to attain a clear understand- 
 ing of the subject as thus far developed, he would do well to 
 take some particular case, as, for instance, that of a girder of 
 two or three or four spans of given length, the end spans being 
 equal, and intermediate spans equal and say one-fourth longer 
 than the ends, and work out by diagram the moments at the 
 supports for a uniform load over the whole length of girder. 
 For two spans the moment at the centre support should be 
 
 g p P, I being the length of span,^> the load per unit of length. 
 
 For three spans the moment at the two inner supports is 
 
 1 -f- n 3 
 4 (3 i o \ P ft, where n I = the length of end spans. Thus, if 
 
 3 91 
 
 n = -, we have TT^Q & ^' "^ or ^ our s P ans ^ e momen t at the 
 
 1-4-2 w 3 
 second and fourth supports is . Q , rp P, and at the middle 
 
 4: \O ~r 4: 7l>\ 
 
 1 _l_ 2 n n z 
 
 support 4/014 r P P- By these formulae the graphical 
 
 results may be checked. 
 
 When the reader has thus become thoroughly familiar with 
 the principles of the preceding Arts, and their practical appli- 
 cation, he will be ready to resume at this point the more gen- 
 eral development which follows. 
 
 88. The Second Equilibrium Polygon. We see, there- 
 fore, that the actual form of the elastic line is not required to 
 be known. Only the outer forces and their moments are sought, 
 and to determine these it is sufficient to know the position of 
 the tangents to the elastic line at the supports. Thus the first 
 line of the equilibrium polygon [Fig. 52, &] being given in 
 position, by the aid of the middle, third, and limited third 
 verticals and the known point K, all the other sides may be 
 
CHAP. VIII.] CONTINUOUS GIKDEKS. 137 
 
 drawn, and the moments at the supports found. We conceive, 
 therefore, the moment area as the difference of the trapezoid 
 A' A" B" B' [PI. 15, Fig. 53] and the parabolic area A" C" B", 
 or equal to A" C" B" minus triang. A 7 A" B' 'minus triang. 
 B / B // A //_ rp} ie area A // c // B// we ca u t h e wrwpiQ or parabolic 
 
 moment area. 
 
 If we indicate the moments at the supports A 7 A." and B' B" 
 by M' and M", then, for a given span Z, 
 
 A' A"- B 7 = J M' Z and 
 A" B' B" = J M" I. 
 
 If we indicate further the height of a rectangle of base Z and 
 area A 7/ C" B", that is, the mean value of the 'moments of the 
 corresponding simple girder by 9)2, we have 
 
 area A" C" B" = 3tf Z. 
 
 The verticals through the centres of gravity of the triangles 
 divide the span into three equal parts. We call these the third 
 verticals. The load 9ft I acts at the centre of gravity of the 
 parabolic area A" B" C". 
 
 The four-sided equilibrium polygon A U S V B corresponding 
 to these forces we call the second equilibrium polygon. The 
 
 pole distance must be (Art. 81) -El. Instead of this, we take, 
 
 itf 
 
 which amounts to the same thing, the forces G F = - M' - , 
 EH- M"- and F E = 9ft -, and the pole distance 1} 
 
 2i A A 
 
 E I 
 
 -, where \ is any assumed length. For \ we may take the 
 n A/ 
 
 arithmetical mean of all the spans, or, as we have seen, one of 
 the actual spans. If the outer spans are both equal to 1 and the 
 other spans equal to 1 1? we should naturally choose \ = ? 1? since 
 
 then the forces would be - M', - M" and 9ft. 
 
 2t 2t 
 
 If the position of the tangents at the supports were known 
 or found, the equilibrium polygon could be easily drawn as 
 follows : Upon the two verticals distant each side of the centre 
 
 E I 
 
 S [Fig. 53] by the pole distance ~b - lay off the distance 
 
 n A 
 
138 CONTINUOUS GIBBERS. [CHAP. VIII. 
 
 A %)l - and join the points thus obtained by two lines cross- 
 
 ing each other. These cross-lines are the lines OP O E of the 
 force polygon. If now we make U IT and V V equal to the 
 ordinates of the cross-lines vertically under U and V, then the 
 sides of the equilibrium polygon U S and V Q prolonged, pass 
 through U' and V'. This will at once appear from an inspec- 
 tion of Fig. 53. 
 
 In this form the equilibrium polygon was first repre- 
 sented by Mohr. (Zeitschrift des Arch, und Ing. Ver. zu 
 Hannover, 1868.) 
 
 89. Determination of the Moments over tlie upports. 
 If we draw in the force polygon, lines parallel to the four 
 sides of the second equilibrium polygon, then the segments of 
 the force line between the lines parallel to A U, B V [PI. 15, 
 Fig. 53] and those parallel to S U, S V, are respectively F G = 
 
 - M' - and E H = - M" -.. If we prolong S U and S V to 
 A A/ 2t A 
 
 intersection^ M and N with verticals through the supports, and 
 represent A M and B N by y' and y", we have from the simi- 
 larity of the triangles U A M and V B N with O G F and O H E 
 
 hence 
 
 ' - ?^L " - M "^ 
 
 y ~ 6 I A y ~ 6 I A* 
 
 The segments A M and B N are, therefore, proportional to 
 the moments at the supports M x and M". 
 
 These moments themselves can now be determined in vari- 
 ous ways. 
 
 \st. It is in general best to choose the second pole distance 
 
 b = - \. We have then 
 o 
 
 M' = y' ( M" = y" (Art. 70). 
 
 If, then, at a distance from U and V either way equal to 
 
 2bl-\ = --\ we draw verticals, the segments A! M! and 
 \ LI 06 
 
 Bj Nj cut off from these verticals will evidently be equal to 
 the moments required, viz., 1VT and M". 
 
CHAP. VIII.]. CONTINUOUS GIRDERS. 139 
 
 %d. If we take X = I, we have at once M / = y' and M" = y" . 
 In the inner spans, therefore, we have directly, as we have al- 
 ready seen, for the special case (Art. 72), when \l, the mo- 
 ments at the supports. 
 
 3d. For a span adjoining a span whose length is X, we have 
 the moment for the intervening support directly from this last 
 span. If the inner spans have the same length I, and the two 
 outer the same length 1? we can accordingly, by making \ = I, 
 obtain directly the moments at the supports. 
 
 90. Comparison with Girder fixed horizontally at both 
 ends. If the ends are fixed horizontally, the lines in the force 
 polygon parallel to A U and B V coincide, i.e., H and G fall 
 together. Accordingly, if we designate the end moments now 
 by W W" we have (Fig. 53) 
 
 ") = W. 
 
 Therefore, in the first equilibrium polygon, the moment areas 
 on each side of the dosing line are equal. 
 
 Indicating the points for this case by the index [Fig. 54, 
 PL 15], we have the triangle A U M equal to U V V'o, 
 and therefore A M = V V V V, as also, in like manner, 
 
 B N = UolTo = U U', or, taking b = 1 X, 
 
 U U' = 3JT /-\ 2 V V = 
 
 Therefore, the ordinates between the cross-lines at the verti- 
 cals passing through U and V are, for girders fixed horizon- 
 tally, proportional to the end moments Wl' and 9ft". 
 
 For X = I, fi = - I, and these ordinates give the moments 
 6 
 
 directly. 
 
 If we draw through U 7 and V a straight line intersecting 
 the end verticals in Q and R, and prolong N U' and N V to 
 intersections S and T, then Q M 2 U U 7 , Q S = V V', and 
 
 hence M S = 2 U IF+V V' = (2 2U'+3B") (^Y; and in the 
 
 same way N T = (2 2tt" +2ft') {-j\ 
 
 Therefore, the segments cut off upon the end verticals ly the 
 cross-lines are proportional to2W+ W and 2 2ft" + W. 
 
140 CONTINUOUS GIEDEES. [CHAP. VIII. 
 
 The quantities SDc' and $31" being known, we can easily con- 
 struct the cross- lines. 
 
 If we draw a line through U and V to intersections O and P, 
 we have O M = V V, P N == U U'. Therefore, A O and B P 
 
 are equal to (>Itt'-M') ^ and (2JT-M") f -)*, where M' and 
 
 M" are the moments at the supports for a continuous girder, 
 and 9ft' W those for a girder horizontally fixed at its ends. 
 
CHAP. IX.] LOADED AND UNLOADED SPANS. 141 
 
 CHAPTER IX. 
 
 CONTINUOUS GIRDER LOADED AND UNLOADED SPANS. 
 
 91. Unloaded Span. If the span is unloaded, we have to 
 construct the second equilibrium polygon, only the two forces 
 
 - M 7 I and - M" I. If the position of the end tangents is known, 
 
 the polygon is completely determined. If we prolong the 
 middle side U V to intersections. M and N with the end ver- 
 ticals [PL 15, Fig. 55], then, by the preceding Art, A M == 
 
 MM , B N = M" - ) ; therefore, A M : B N ; : M / : M". 
 
 \A// \A// 
 
 If now w^e draw A B intersecting U V in I, the moment at this 
 point is zero. That is, the intersection I of the line joining the 
 supports with the middle side of the polygon is the point of in- 
 flection of the elastic line. 
 
 92. Two ucceive Unloaded Spans. Prolong the two. 
 middle sides U V and U t V t [PI. 15, Fig. 56] of the equilibrium 
 polygon for the two spans Z and ^. The point of intersection 
 W is a point in the resultant of the forces at V and U^ Since 
 
 these forces are - JV^ 1 G and - M : ^, we have W V : W U : : 
 
 li : 4- But the horizontal projection of V U is - (7 + k), 
 
 3 
 
 therefore that of V W is - ^ and of U W , - 4 ; while that of 
 
 o 3 
 
 B W is - (lilo). The vertical through W we have called the 
 o 
 
 limited third vertical. Its position is, as we see, easily found, 
 and depends simply upon the length of the spans. 
 
 Let us now consider more closely the intersections I and L of 
 the middle sides with the straight line joining the supports. 
 We have 
 
14:2 CONTINUOUS GIEDEK. [CHAP. IX. 
 
 But we have also 
 
 UoU^VoV-BUoiBVo:: l,:l^ 
 hence 
 
 L U : L W ; ; I V X \ : I W x 1 . 
 
 The ratio of the parts into which W U is divided by the 
 point L depends, therefore, for given spans only upon the ratio 
 of I V to I W , or upon the position of I alone. 
 
 If, therefore, we were to draw through the point I another 
 polygon, the point L would be unchanged, or still more gener- 
 ally, if the point I for different heights of the supports and 
 different polygons moves in the same vertical^ the point L will 
 also move in a vertical. 
 
 If the supports are in the same straight line, the points I and 
 L are the points of inflection of the elastic line. We have 
 therefore the principle, that if for different polygons the in- 
 flection point I remains the same, the inflection point L re- 
 mains also the same. 
 
 The point I being given, we can easily construct the point L. 
 We have only to draw through I at will any line intersecting 
 the third vertical through V and the limited third at, say, V 
 and W. Through V and the support B draw a line to inter- 
 section Ui with third vertical through U . Join now U^ with 
 W. The line U^ W cuts the line through the supports A and 
 B in the point L. (See also Art. 86, Fig. 52, d.) 
 
 93. The "Fixed Points." Suppose that, starting from the 
 left support A [PI. 15, Fig. 57], we have a number of unloaded 
 spans. The end A then is an inflection point, since the mo- 
 ment there is zero. Starting from this point, therefore, we can 
 construct, according to the preceding Art., the inflection point I 2 
 for the next span. Then starting from this we may construct 
 the point I 3 for the third span, and so on. Since these points, 
 under the assumption that the supports all lie in the same 
 straight line, do not change their position, whatever may be the 
 loading of the loaded spans, and whatever spans be loaded, we 
 call tlwm fixed points. 
 
 A second series of fixed points may be in similar manner 
 constructed, when a number of spans from the right are un- 
 loaded, so that there are two series of fixed points. In the 
 end spans the end supports are fixed points. 
 
 It follows directly from the construction that the fixed points 
 are always within the outer third of the span. 
 
CHAP. IX.] LOADED AND UNLOADED SPANS. 143 
 
 The construction of the fixed points is the first operation in 
 the graphical treatment of the continuous girder. 
 
 The above construction was first given by Mohr. 
 
 91. Shearing Force, Reactions at the Supports, and UIo- 
 meitts in the Unloaded Spans. The moments in the unloaded 
 spans are given, then, by the ordinates to a broken line whose 
 angles lie in the support verticals, and which, for the case of 
 supports on a level, passes through the corresponding fixed 
 points. 
 
 It follows directly that the moments at the supports are 
 alternately positive and negative, and increase from the 
 end, so that any one is more than three times the preceding. 
 (See Art. 111.) 
 
 Since now this polygon has alternately angles down and up, 
 the reactions at the supports must be alternately positive and 
 negative. From the corresponding force polygon it follows 
 that they must increase from the end. 
 
 The shearing forces are, therefore, also alternately positive 
 and negative, and increase from the end on. 
 
 95. Loaded Span. Let now the span A B [PL 15, Fig. 58] 
 be arbitrarily loaded. It can be proved here also, as in Art. 
 92, that the prolongation of the sides XT' V and S U, as also of 
 V" U" and S V, intersect in the limited third verticals. 
 
 When the supports are in a straight line, then, by the con- 
 struction of Art. 92, the fixed points I and K are the intersec- 
 tions of S V and S U with A B. We can, therefore, at once 
 assert, that the sides S U and S V of the second equilibrium 
 polygon pass through the fixed points I and- K, when the sup- 
 ports are on a level. ^ 
 
 For known position of the fixed points and for given load, it 
 is, therefore, easy to draw the second polygon by drawing ver- 
 ticals H! and KK t equal to the corresponding ordinates of the 
 cross-lines. S U and S V pass, then, through I K^ and K I t re- 
 spectively. Then, by Art. 89, the moments at the supports may 
 be determined. 
 
 Since A I < -Z, U must lie to the right of I, and the angle 
 3 
 
 AUS is concave downwards. Accordingly, the force at U, 
 viz., - M' I, acts upwards. The same holds good for V. Hence, 
 
144: CONTINUOUS GIRDER. [CHAP. IX 
 
 the moments M' M" for the loaded span are always posi- 
 tive* If we draw the lines M N and U V cutting the middle 
 
 vertical inP and Q, then, for b = \l, P O = 1 (M' + M") and 
 
 o 2 
 
 P Q = 2 (W + 2ft")- (See Arts. 89, 90.) Since now the points 
 U and V must lie under A B, 
 
 P O < P Q' or M' -f M'' < 3R'+ 3B". 
 If the supports are not upon a level, it follows from this Art, 
 and Art. 92, that the intersections of S U and S V prolonged, 
 with the prolongations of the lines A A' B B', joining the sup- 
 ports of two adjacent .spans, lie in the VERTICALS THROUGH THE 
 
 FIXED POINTS. 
 
 96. Two successive Loaded Spans. PL 15, Fig. 59. 
 
 1. Here also, as in Art. 92, we can prove that the prolongar 
 tions ofSV and S t Uj. intersect in the limited third vertical. 
 
 2. Draw through B a line which intersects S V and Sj U^ in 
 I' and r i? and the verticals through V, W and U t in V , W 
 and U . 
 
 Then 
 
 U U t : V V ; ; U B : V B ; : Z : Z x 
 
 v v : w w : : r v : r w . 
 
 Hence by composition 
 
 U U : W W 
 
 : r v x 4 : i' w x i ; 
 
 or since U 1^ : W W : U 1\ : W 1\ 
 
 UT W T ' T' V v 7 T "W v 7 
 01* 01 ~Q A Cfl * " o "*> *! 
 
 If, then, the point I' moves in a vertical, the ratio I' V to 
 I' W does not change, therefore the ratio of U 1/ to W I/ also 
 remains unchanged, and accordingly I\ must also move in a 
 vertical. If I' coincides with I, it follows from the construction 
 of Art. 93 that the point I\ becomes the fixed point 1^ Hence, 
 the intersections T and I\ of verticals through the fixed points 
 I and Ii with the sides S V and S t U\, or with the middle 
 sides of the two polygons adjacent to the support, lie always 
 in a straight line through that support, for any heights of 
 supports. 
 
 * A positive moment always indicates compression in lower flange. 
 
CHAP. IX.] LOADED AND UNLOADED SPANS. 14-5 
 
 This property of the second equilibrium polygon was first 
 made known by Culmann. 
 
 97. Arbitrary leading. According to the above properties 
 of the second equilibrium polygon, the general course of pro- 
 cedure for any given case of loading is then as follows [Fig. 60, 
 PI. 16] : 
 
 1. Construct all the fixed points A I 2 1 3 --- I^Ka, etc. (Art. 
 93), and draw verticals through them. 
 
 2. Construct the cross-lines for every span, Art. 88. 
 
 3. Make A C equal to O t Qi as given by the cross-lines, and 
 draw a line through C and A l to intersection D 2 , with vertical 
 through I 2 . Then make D 2 C 2 equal to O 2 Q 2 , and draw a line 
 through C 2 and A 2 to D 3 , and so on. Precisely the same con- 
 struction holds for the other way from the right end. Thus 
 A 4 E 4 is equal to R 4 P 4 , etc. 
 
 4. In this way we obtain for each of the middle sides of the 
 second equilibrium polygon two points, C and F l5 C 2 and F 2) 
 etc. ; A and E t , D 2 and E 2 , and so on ; so that now we can ac- 
 tually draw these middle sides. 
 
 The intersections of these lines with the support verticals 
 give, according to Art. 88, the moments at the supports. For 
 spans whose length 13 X these moments are given directly; for 
 other spans the ^ construction of Art. 74 must be applied. The 
 following simple construction may also be applied. Let 
 I K be the intersections of the verticals through the fixed 
 points, with the line A B joining the supports [Fig. 61, PL 16]. 
 
 Make I D' = I D *, K F' = KF Y, C' D' = O Q - 
 
 E'F' = RP (^Y, and draw C' F' and E' D'. These lines cut 
 
 \v / 
 
 the support verticals in M' and N', so that A M' and B N' are 
 the moments. 
 
 By the construction errors accumulate from one span to the 
 next, so that the diagram must be made with care. We have 
 also several checks, viz. : 1. The intersection of the middle sides 
 must lie in the vertical through the intersection of the cross- 
 lines. 2. The prolongation of the middle sides must intersect 
 in the limited third vertical. 3. The corresponding intersec- 
 tions of the middle sides with the third verticals must lie in a 
 straight line through the support. 
 10 
 
146 CONTINUOUS GIRDEK. [CHAP. IX. 
 
 If any span is unloaded, the cross-lines coincide. The above 
 method of construction holds good when the supports are not 
 upon a level. If the difference of height of the supports is 
 
 represented th of the real amount, the unit for the moment 
 m 
 
 scale is , as is easily seen by reference to Arts. 81 
 
 6 m E I 
 
 and 88. 
 
 The above method of construction of the moments at sup- 
 ports was first given by Culmann. 
 
CHAP. X.] SPECIAL CASES OF LOADING. 147 
 
 CHAPTEE X. 
 
 CONTINUOUS GIRDEK - SPECIAL CASES OF LOADING. 
 
 98. Total uniform Load. If a span is loaded with a uni- 
 formly distributed load of p pounds per unit of length, the 
 simple moment area is a parabolic segment whose vertical axis 
 passes through the centre of the span. [PI. 16, Fig. 62.] 
 
 1 -.-,.?, 
 
 The ordinate D C" is p P, and hence the area 
 
 It will be advantageous here to take p X 2 as the unit of the 
 moment scale ; and therefore 
 
 The vertical height of the cross-lines at the pole distance o 
 
 from the middle is 8ft -. If, then, we take b = - X, we have 
 X 6 
 
 and therefore 
 
 2. Moments. 
 
 If the moments A' A" and B' B" are known, we can find 
 the end tangents of the first equilibrium polygon by drawing 
 A" B", dropping a vertical through the middle D, and laying 
 
 off D E = 2 x \p P = \p P = \p X 2 (|y. The lines A" E 
 
 and B" E are, then, these end tangents. With the help of 
 these we may easily construct the parabola. 
 
 3. Shearing force. 
 
 If we draw in the first force polygon, lines parallel to the 
 
148 CONTINUOUS GIEDEE. [dlAP. X. 
 
 end tangents and the closing line A' B', the distances on the 
 force line are the reactions at the ends of the span. Instead 
 of this we may lay off from A and B, A G and B H equal to 
 the first pole distance #, and through G and H draw parallels 
 to the end tangents, intersecting the verticals through A and B 
 in A' and B^ We thus obtain the reactions A A 1? B B t . The 
 ordinates to A x B 1 then give the shearing force at any point. 
 
 If the line p X 2 representing the moment units is equal to m, 
 and that representing the force units p\ n^ then the first 
 
 pole distance must be a = r-ra X = X. 
 
 2 m 
 
 Accordingly, it is now easy, from the general construction 
 given in Art. 97, to construct the shearing force and moments 
 for uniform or dead load of girder in any case. Let us pass 
 on to an example illustrating more fully the above principles. 
 
 99. Example. As an example of the application of the 
 above principles, we take a girder of four spans, as given in 
 PL 17, Fig. 63. The two interior spans are each 96 ft., the 
 exterior spans 80 ft. each ; that is, l : I : : 5 : 6. Choose any 
 scale of length convenient, as, for instance, 50 ft. to an inch, 
 lay off the spans and construct first the fixed points. For this 
 purpose we draw the third and limited third verticals. These 
 last are easily found from the principle already deduced, that 
 they must divide the distance between the third verticals into 
 segments inversely as the corresponding spans [Art. 92]. Lay- 
 ing off. then, from the third vertical in the first span, -J I to the 
 right, or from the third vertical in the second span -J l to the 
 left, we have the first limited third vertical. The same at the 
 other end gives .the other. For the centre support, of course, 
 the limited third, since the adjacent spans are equal, passes 
 through the support itself. We can, therefore, now construct 
 i\\e fixed points according to Art. 93. 
 
 Let the load per unit of length p be -J ton per ft. Then 
 taking X [Art. 8] equal to Z, we have n =p\=pl = 48 tons 
 and m=j>\ 2 =pP= 4608 ft. tons [Art. 98 (3)]. It remains 
 to assume a scale of force. Let this be 20 tons per inch, then 
 our moment scale is 20 x 50 = 1000 ft. tons per inch. The 
 values of which we shall need to make use are, then, to scale 
 ^ = 1.6 inches, X = I 1.92 inches, 
 
 |) 2 = 0.6944 inches, (^ = 1.44 inches, (|)* = 0.4823 inches. 
 
CHAP. X.] SPECIAL CASES OF LOADING. 149 
 
 These values are repeated upon the PL for convenience of 
 reference. Also, p I = 48 tons 2.4 in. = n, p fi = 4608 ft. 
 tons = 4.608 in. = m. For the first pole distance [Art. 98 (3)] 
 
 n 2.4 25 
 
 we have a = X = . - A:> I -^ I = 1 in. Second pole dis- 
 m 4.608 48 r 
 
 tance [Art. 98] b = ^ X - 0.32 in. 
 
 According to Art. 98, we have now, for the cross lines, 
 
 :?X/ and O' P' = Q' P' = 
 
 Laying off these distances under the supports, we have thus the 
 cross-lines. 
 
 We have next to construct the second equilibrium polygon. 
 This, by the aid of the cross-lines and fixed points already con- 
 structed, we can easily do, as detailed in Art. 97 (3). Then 
 the moments at the supports are given directly to moment 
 scale in the interior spans, or we can find them from the end 
 
 spans by laying off j I [Art. 89]. 
 h 
 
 Finally, the moments thus found and laid off at the sup- 
 ports, we can construct the moment curve by making D' E' 
 
 / \2 fl\2 
 
 r and DE = i^X 2 - [Art. 98 (2)], and thus draw- 
 
 ing the end tangents and corresponding parabolas. 
 
 According to Art. 98 (3), we can then find the shearing 
 
 A-I 4 
 
 forces by laying off a = A and drawing parallels to the end 
 
 tangents to intersection with verticals through supports, as 
 shown in Fig. 
 
 We thus have both moments and shearing forces for uniform 
 load. By careful attention to the above, the reader will have 
 no difficulty in solving any case. We recommend him earn- 
 estly to perform the entire construction for himself, referring 
 to the proper Arts, at every step. [For convenience of size, 
 we have not observed our scales strictly in the Figs. TJie 
 reader should therefore ?wt attempt to check results with the 
 dividers.'} 
 
 1OO. Partial uniform Load. 1. When the girder is only 
 partially loaded, as, for instance, a certain portion of the span 
 id I from B, the simple moment area consists of a triangle ABC 
 
150 CONTINUOUS GIEDEK. [CHAP X. 
 
 and a parabolic segment C E B. [PI. 16, Fig. 64.] If in the 
 first force polygon B' D' is the total load upon the span, the 
 line O A' parallel to the end tangent A G divides B' D' in the 
 same ratio as the end of the load divides the span, or denoting 
 the length B C by /3 1. 
 
 B' A' :B'D ' :;/3l:l. 
 
 The intersection G of the end tangents lies in the vertical 
 G H, which halves B F. Since the triangle B C T is similar to 
 O A! B', we have 
 
 BT:B' A' :;:#; 
 
 a 
 
 or since B' D' =p I, B' A' =p j3 = <pl. 
 
 It is therefore easy to construct B T as in Fig. 64, where 
 
 /Q 
 
 B! D! p I, A! B t = B' A' = p I 7, and pole distance = i ; 
 
 then B 2 D 2 = B T. 
 
 If B T is thus found, we can easily, when A and B are given, 
 construct the end tangents, and then construct the first equili- 
 brium curve itself. 
 
 2. Make G K GI, then the triangle CKB is equal to 
 the parabolic area C E B. If, then, through K we draw a 
 parallel to C B, intersecting C G in L, and through L the ver- 
 tical LM, the triangle ALB is equal to the entire simple 
 moment area. This last is therefore proportional to L M, or 
 2ft I = $ I x L. M ; hence %tt = J L M. It is easily proved 
 also that F M = i F B. Thus, as GK is of GI, GL is 
 J- of G C ; hence M H is $ of F H, and therefore F M is f of 
 F H, or | of J- F B = ^ F B. L M can therefore be easily 
 drawn. 
 
 3. Let N be the middle of A B. Make N O equal to $ F N. 
 Then the centre of gravity of the triangle A C B is in the ver- 
 tical through O, while the centre of gravity of the parabola is 
 in H G. If through L we draw a parallel to AB, intersecting 
 the vertical through C in P, the two areas ^are to each other 
 as FC to C^P. If in the vertical through 6 we make OQ = 
 J CP, then, since IH = J CF, we have 
 
 IH:OQ::FC:CP. 
 
CHAP. X.] SPECIAL CASES OF" LOADING. 151 
 
 The intersection R of the line Q I with A B lies, then, in the 
 vertical through the centre of gravity of the simple moment 
 area. 
 
 Thus the construction of the cross-lines is now easy. 
 
 4. It is most convenient in the application of the above to 
 construct or calculate the distances of the cross-lines under the 
 supports once for all, for load over various parts of the span. 
 The necessary formulae can be directly deduced. Thus the 
 
 Triangle B AT = BT x l = ~ X 
 
 \$ 1=^ x 
 
 Triangle BC T = BT x $ 1= x 01. 
 
 Triangle BLT = BT X /3Z =xjS I 
 
 6 A (t, 6 
 
 The entire area is equal to the triangle ALB. 
 But A L B = B A T-B L T = 2& (I Z-l ft 
 
 2t d \^ O 
 
 The triangle A C B = B A T B C T ; hence 
 
 The parabolic area is equal to the entire area minus A C B, 
 
 7} BP 1 
 
 or parabolic area = TT *# P I- 
 
 2i Cb O 
 
 F is distant from B T by a distance ft I, N by a distance = 
 
 \l, NO is \ of N F ; hence O is distant from B T 
 2i o 
 
 Therefore, the moment of the triangular area, with reference 
 to the right support B, is 
 
 The moment of the parabolic area is 
 
 The total moment, with reference to the right support, is, 
 therefore, 
 
152 CONTINUOUS GIRDER. [CHAP. X. 
 
 In a similar manner we find for the left support 
 
 When ft = 1 the span is completely covered, and we have, 
 then, right and left 
 
 24 a 
 
 If we compare this value with those for partial loading, we 
 see that they differ only by certain coefficients, and that these 
 coefficients depend only upon the length of the loaded portion. 
 If, then, we have the distance between the cross-lines for total 
 load, we have only to multiply by certain factors to obtain the 
 distances for partial loading. For uniform or total load over 
 
 1 / A 4 
 the whole span, this distance is given by -p\ 2 I - ) (Art. 98). 
 
 4: \ A// 
 
 If we divide this distance in certain proportions we have at 
 once the distances for partial loading. These proportions are 
 given by (2 0*) $ for the right support, and (2-/3) 2 /3 for the 
 left, under the supposition that .the load comes on from the 
 right. The reverse is the case for load coming on from left. 
 
 ~| ~| Q 
 
 If we take /3 = -,-,- of the span, we can calculate these pro- 
 4 2i 4 
 
 portions once for all. We thus have the following table : 
 
 Support under load Support for wwloaded end 
 
 (2-02) /5. (2-p) p. 
 
 - span loaded = 0.1211 . . . 
 
 i span loaded | = 0.4375 .... = 0.5625. 
 
 3 207 225 
 
 - span loaded = 0.8086. ..:-&&** 0.8789. 
 
 The division of the distance between the cross-lines for uni- 
 form load over the whole span into these proportions is easily 
 accomplished graphically. Thus, from the end of the line to 
 be divided, draw a line in any direction, and lay off upon it the 
 six numbers above, to any convenient scale. Join the end of 
 the last division with the end of the line to be divided, and 
 then draw parallels through the other points. 
 
 It is important to observe some definite system of numera- 
 tion, otherwise, especially in the first attempt at construction, 
 confusion is apt to arise. 
 
CHAP. X.] SPECIAL CASES OF LOADING. 153 
 
 We can thus find the cross-lines for any position of the load, 
 and for each position can, if we wish, draw the equilibrium 
 polygon and determine the moments according to the general 
 method of Art. 97. 
 
 1O1. Concentrated Load. The simple moment area is in 
 
 this case a triangle [PL 16, Fig. 65] whose area is - I h, h be- 
 
 2 
 
 ing the height C D. Therefore 
 
 If from the centre of the span E we lay off E F = - E D, D 
 
 o 
 
 being the point of application, a vertical through F passes 
 through the centre of gravity. 
 
 As the height C D is proportional to 9Cft, we may take C D as 
 second force polygon. Since h =. 2 0?, the distance of the pole 
 
 N must be 2 x \ I = i Z, when X = I (Art. 89). Draw N P 
 6 3 
 
 parallel to A B, then is N P = - A B. 
 
 3 
 
 Parallel to N C and N D we may draw the cross-lines. A 
 simple construction may be given for them when they are made 
 to pass through A and B. Let A M and B L be the cross-lines. 
 Then the triangle S B M is similar to N C D, and 
 
 B M : C D ; ; B F : N P. But 
 
 and N P = A B; therefore, 
 
 B M : C D ; ; (2 A B-A D) : A B. 
 *Make D G = A B, then B G = 2 A B-A D. 
 
 The point M is accordingly found by drawing a straight line 
 through G and C, D G being equal to A B. In the same way 
 make D H = A B, and draw a line through H and C. We 
 thus obtain L. 
 
 The prolongations ofM.C and I* C, therefore, intersect A B 
 prolonged in the points G and H, distant from D ~by A B. 
 
 If, then, we have to investigate a concentrated load in various 
 positions, we draw the first equilibrium polygon C X and C Y 
 [PI. 16, Fig. 66], and lay off in the same the closing lines (for 
 
154 CONTINUOUS GIRDER. [CHAP. X. 
 
 the simple moment area) for the different positions of the load. 
 The distances cut off on the vertical through C by these lines 
 give, then, the various values of h or 2 %R. If we draw from. 
 
 these intersections lines to the pole N, at the distance - I from 
 
 3 
 
 D C, the cross-lines are parallel to these lines. It will be best 
 to keep for each pair the common line P Q parallel to N C. 
 "When the point of application of the load divides the span into 
 two equal parts, the point of intersection of the cross-lines di- 
 vides the middle third of P Q into equal parts. 
 
 The centre of gravity of the simple moment area cannot pass 
 beyond the middle third of the span. Since any load can be 
 considered as made up of a number of concentrated loads, it 
 follows generally that far any method of loading the centre of 
 gravity of the simple moment area lies between the third verti- 
 
CHAP. XI.] MAXIMUM STRAINS. 155 
 
 CHAPTEE XI. 
 
 METHODS OF LOADIN0 CAUSING MAXIMUM STRAINS. 
 
 1O2. Maximum Shearing Force -Uniformly distributed 
 Jloviiii? Load. Suppose, first, the span in question loaded 
 with a concentrated weight. The simple moment area is 
 A' C' B'. [PL 18, Fig. 67.] 
 
 In the force polygon let O A 1? O B x and O O t be respectively 
 parallel to C' A', C'B' and A 7 B 7 . Then ^ Aj and C^ are 
 the reactions at A and B. Since, according to Art. 80, the mo- 
 ments A A 7 and B B' are always positive, and the middle sides 
 A' S, B 7 S pass through the fixed points 1 and K, it follows 
 from the construction of the preceding Art. that the intersec- 
 tions O and P of the sides A' C' and B' V of the first equili- 
 brium polygon with the closing line A B must always lie within 
 A I and B K. That is, the points of inflection O and K are al- 
 ways between the fixed points and the ends. Therefore A', B' 
 and the point C' must lie on opposite sides of the closing line 
 A B, and consequently Cj. in the force polygon must lie between 
 A! and B^ 
 
 Accordingly the shear A t C^ at A is positive^ and the shear 
 G! B 1 at B is negative. 
 
 Let the distance of the load from the left support be /3, from 
 the right support &. The load itself is P, and the moment 
 A A' at the left support M', B B' at the right M". Kequired, 
 the shearing force S at a point distant x from the left support. 
 The partial reaction at the left is R'. Then 
 R' Z = M' - M" +P&, 
 M'^M" 
 
 ~~ ~ 
 
 M' and M" are always positive. If, therefore, M' > M", R' 
 is positive ; if M'< M", then M" - M'< M" + M 7 , or, since 
 by Arts. 75 and 80, M' + M"< 2R' + 3JT, M" - M^aft'+aR". 
 Now it can be easily proved analytically that for a girder hori- 
 
 zontally fixed,* m f = ~ and 2R"= 3 hence 3B' + 3R" 
 
 Supplement to Chap. VII., Art. 18. 
 
156 METHODS OF LOADING [CHAP. XI. 
 
 since + & = J. Therefore M" - M' 
 
 Since, however, /8 < Z, we have also M" M' < P $_. Hence, if 
 M' is < M", we have also R' positive. R' is therefore always 
 positive whatever may be the position of the load. In the same 
 way it may be shown that R" is always negative. 
 
 If now the load is to the right of the point distant x from 
 the left support, then for this point the shearing force S' R', 
 and is therefore positive. If the load is to the left of this 
 point, the shearing force S = R' P R", and is therefore 
 negative. S for any point is therefore positive or negative, 
 according as the load lies right or left of this point. Hence 
 for a uniform load we deduce directly 
 
 The shearing force at any point is a positive or negative 
 maximum when the load extends from this point to the right 
 or left support respectively. 
 
 The same principle holds good for the simple girder. 
 
 2. Thus far we have considered the load in the span itself. 
 Suppose now the load is in some other span, and the span in 
 question is unloaded, then 
 
 M' - M" M"-M'" 
 
 R =- -IT R = r 
 
 As we pass away from the loaded span the moments at the 
 supports are alternately positive and negative, and each is 
 greater than the one following (Art. 94). Since the moments 
 M' and M" are alternately positive and negative, R' will have 
 the same sign as M', and R" as M", and generally R^ as M m . 
 
 Adopting, then, the. notation shown in PL 18, Fig. 68, we 
 have for the span l m _ 
 
 . t - M m 
 
 where R ra has the same sign as M m . 
 In the same way 
 
 and therefore 
 
 1 _ 
 
CIIA1'. XI.] CAUSING MAXIMUM STRAINS. 157 
 
 M 
 
 But rc+ is negative and greater than 2 (Art. 94-). There- 
 fore in the preceding expression the numerator is positive and 
 
 ivi 1 
 
 > 3. Further, -^ L is negative and less than -, hence the de- 
 
 1V1 m 
 
 3 
 
 nominator of the above expression is negative and < -. There- 
 
 R I 
 
 fore R +1 is negative and > 2 -y 1 ^, that is, the shearing forces 
 
 at the supports are alternately positive and negative, and in- 
 crease (when 2 l m _ is not less than Z m ) towards the loaded span. 
 We have then 
 
 For any span, then, the shear at the left support R' will be 
 positive when the left adjacent span is loaded, the right adja- 
 cent span unloaded, and all the other spans each way alternately 
 loaded. The shear R' will be negative when the remaining 
 spans are loaded. Hence : 
 
 The shearing force is a maximum (positive) at any point 
 when the load extends from this point to the right support, and 
 the other spans are alternately loaded, the adjacent span to the 
 right ~being unloaded, that to the left, loaded. The negative 
 maximum, on the contrary, occurs when the load extends from, 
 the point to the left support, when the right adjacent span is 
 loaded and the left unloaded ; the other spans alternately 
 loaded. 
 
 PL 18, Fig. 69, gives these two cases. 
 
 In practice such a loading can never occur. If we suppose 
 the rolling load divided into two portions only, the above rule 
 reads as follows : 
 
 The shearing force at any point will be a positive maximum 
 when the load reaches from the right support to this point, and 
 when the left adjacent span is covered. The negative maximum 
 occurs when the load reaches from left support to the point, 
 and the right adjacent span is covered,. 
 
 1O3. Maximum Moments. 1st. Loaded Span. Let a weight 
 act at the point D, PL 18, Fig. 67. Then A U S V B is the 
 second. A' C' B' the first equilibrium polygon, and A A', B B' 
 
158 METHODS OF LOADING- [CHAP. XI. 
 
 are the moments at the supports. We have already seen that 
 the inflection points O and P (for which the moment is zero) 
 lie outside of the fixed points. We can therefore assert that 
 WITHIN the fixed points the 'moments are negative wherever the 
 weight may Replaced. From the Fig. we see at once that the 
 inflection points O and P move to the right or left as the weight 
 moves to the right or left. Accordingly when for the weight 
 at D the moment at O is zero, the moment at this point will be 
 positive when the load moves to the right of D, negative when 
 it moves to the left of D. 
 
 Hence for the maximum moment we have at once the follow- 
 ing principle : 
 
 For any point O outside the fixed points the moment will ~be 
 a positive or negative maximum when the load reaches from 
 the point D, where a load must Replaced to cause the moment 
 at O to ~be zero, to the right or left support respectively. For 
 the negative maximum, therefore, the load reaches from A to D; 
 for the positive, from D to B. 
 
 If the point O is given, it is indeed possible to determine by 
 construction the point D to which the load must reach. It is, 
 however, simpler to assume D and then construct O. 
 
 If we choose for the different positions of D an arbitrary 
 length for C' D' (Fig. 67), so that the point C' falls in a parallel 
 Q R to A' B' (Fig. 70), and, moreover, take D at equal intervals, 
 then the points L and M will be at equal distances (Fig. 67), 
 and hence the points I and K (Fig. 70), in which the verticals 
 through the fixed points are intersected by the lines A' M and 
 B' L (Fig. 67), will be at equal distances. We have, then, the 
 following simple construction [Fig. 70, PL 18] : 
 
 Between the verticals through the supports draw two parallel 
 lines A B and Q R at any convenient distance apart, and divide 
 Q R into a number of equal parts ; four or five are sufficient. 
 Draw lines from A to R and the middle S intersecting the ver- 
 tical through the fixed point I in I x and I 2 . In the same way find 
 K! and K 2 . Divide I t I 2 and K t K 2 into the same number of 
 equal parts as Q R has been divided into, and join these points 
 in reverse order by lines. The intersection of these lines with 
 the lines drawn from A and B to the points upon Q R give the 
 points O, for which the moment is a maximum when the load 
 is limited by the corresponding point upon Q R. 
 
 This construction was first given by Mohr. 
 
CHAP. XI.] CAUSING MAXIMUM STRAINS. 159 
 
 IO 1. Determination of the maximum Shearing Forces. 
 
 According to the general method of construction given in 
 Art. 97, we can now determine by reference to Arts. 98 and 99, 
 which treat of total and partial distributed loading, the shear- 
 ing forces corresponding to the methods of loading which cause 
 maximum strains. 
 
 As a review of the preceding principles, we take the same 
 example as before, as given in PL 19, Fig. 71. Here again the 
 reader should construct the Figs, for himself. The scales are 
 as before, Art. 99. 
 
 Fig. a shows the method of loading for positive maximum 
 shear in first span ; and the second Fig. below, the same for 
 the second span. [Art. 102.] , 
 
 We first find, precisely as in Fig. 63, the shearing forces in 
 the third and fourth spans for the total loads over those spans, 
 and lay off the shear thus obtained in the first and second spans, 
 as indicated by the broken lines in Fig. b in those spans. Thus 
 having first found the fixed points, which we may here take 
 directly from Fig. 63, we construct as in that Fig. also the cross- 
 lines for total load in third and fourth spans. Thus laying off 
 54 equal to Oj. P 1? and drawing a line through 4 and support to 
 intersection with vertical through fixed point in second span 
 [Fig. 60], we determine D', and then from the cross-lines find 
 at once D". In like, manner, supposing for the moment the 
 load on the other two spans, we have e a, a D, D a and a F', 
 and then at once F''. F" D' cuts off then the moment at the 
 right support, and joining 1, 2 and 5, we find, according to Art. 
 98 (2), precisely as in Fig. 63, the end tangents ; and then from 
 these, with the first pole distance a, find the shear. This is 
 given by the broken lines in third and fourth spans. Lay off 
 these lines in first and second spans, remembering, since the 
 shears at the supports alternate, that the positive shear at left 
 of fourth span must be laid off as negative (down) at right end 
 of first, etc. [See Fig. 63.] 
 
 Now for the positive shears in first and second spans for the 
 different methods of loading, we have only to determine the 
 direction of the tangents through end of load [Fig. 64, Art. 
 100]. The lengths of the segments cut off upon the verticals 
 through the supports by these tangents are given for first and 
 second spans by Figs, /and e, for each position of load. [Art. 
 100 (1).] 
 
160 METHODS OF LOADING [dlAP. XI. 
 
 We have first, then, to construct the cross-lines [Art. 100 (4)] 
 as shown in Fig. d. 
 
 . Take first the second span. For this span, as shown in Fig. 
 dj the first span is fully loaded, as also the last. Make, there- 
 fore, e a = Oi P 1? draw a D, and we thus find the point D, com- 
 mon to all the middle sides of the second equilibrium polygon 
 for different positions of load. So also make on right 54 
 O t P! and draw 4 D', and then, since third span is empty, D' F, 
 and we thus have P. Now from D and F lay off the cross- 
 line distances, as "D a, D 5, D <?, equal to e a, e b, e <?, etc. Draw 
 lines through these points and D and F respectively, and note 
 their points of intersection abed with the verticals through 
 the supports. [NOTE. Be careful to preserve an orderly nota- 
 tion.] These points give the moments for each position of 
 load in second span. Take the length a a from Fig. e and lay 
 it off from a on the right support vertical, and join the end 
 with a on left. This is the tangent for full load in second 
 span. A parallel to it at distance a from left gives the shear 
 in Fig. b. Then lay off b b taken from Fig. e on right, and join 
 with b on left. This is tangent for load over three-fourths 
 second span from right. A parallel to it at distance a from 
 foot of perpendicular one-fourth of span from left cuts off 
 shear for this position of load. So for tangents c c, d d. We 
 thus obtain the curve for positive shears in second span. The 
 negative shears are obtained by subtracting these from the 
 shear already found for full load. We thus have the lower 
 curve, and the shear diagram for second span is complete. 
 
 For first span, only the third is loaded. We lay off, then, 
 78 equal to the distance between cross-lines corresponding, 
 draw 8 F 1? and thus find F t . Lay off now at left end ed, ec, 
 e by draw lines from these points through second support, and 
 note intersections with vertical through D. Through each of 
 these intersections draw a line through F 1? and produce to 
 intersections abed with vertical through second support. It 
 is from these last points that the distances a a, bb, etc., taken 
 from Fig. f must be laid off respectively in order to find the 
 tangents e a, e b, e c, e d, e e. 
 
 Parallels to these tangents above in Fig. b give, as before, 
 the positive shear for each position of load. The negative 
 shear is, as before, found by subtracting the positive from total 
 load shear. Thus shear diagram for first span is complete. 
 
CHAP. XI.] CAUSING MAXIMUM STRAINS. 161 
 
 Of course, the same circumstances can hold good for third 
 and fourth spans as for first and second, except that positive 
 shears on one side of centre support are the corresponding 
 negative shears upon the other, and vice versa. 
 
 Following carefully the above with the aid of the Fig., the 
 reader cannot fail to grasp the method. An independent con- 
 struction for a similar case will make both principles and de- 
 tails familiar. Once thoroughly understood, the method is 
 rapid, accurate, simple, and of general application. 
 
 1O5. ]>etcr 111 illation of the Maximum Moments. In 
 like manner, it is easy, according to the general construction 
 given in Art. 97, and referring to Arts. 98 and 99, to determine 
 the maximum moments. In Fig. 72, PL 20, we have the same 
 example as before, concerning which we have but little addi- 
 tional to remark. Fig. 64, Art. 100, shows that the end tan- 
 gents give the moments within the unloaded portion of the 
 girder. These tangents are constructed precisely as before in 
 the several spans, except it will be noticed that in the first 
 span we have made use of the construction given in Art. 97, 
 Fig. 61. Thus the point F' is determined so that K F' = 
 
 (A 2 
 j-] , and thus the moments are measured directly at the 
 
 end vertical. Also upon the left support vertical we have laid 
 off the distances between the cross-lines in the second span 
 
 (7 \2 
 7/ ' 
 
 The only thing new in the PL is Fig. c, which, as we have 
 seen in Art. 103, Fig. 70, gives the points at which the positive 
 moment is a maximum for each position of load.. The positive 
 moments can then be taken directty off upon the verticals 
 through these points, and are limited by the horizontal through 
 the supports and the tangents, as above. 
 
 Thus, at second support the vertical distance to a, for first 
 span, gives the moment at the support. La} r it off in Fig. a 
 from the support line. For a load over f span, w r e see at once 
 the point for which the positive moment is a maximum from 
 Fig. c. Follow up the vertical through this point. The dis- 
 tance on this vertical in Fig. J, between the support line and 
 tangent b b^ gives the moment to be laid off in Fig. a upon 
 this vertical. So, for load over span, we have next vertical 
 
 and tangent cc^ and so on. We thus obtain the curve for 
 11 
 
162 METHODS OF LOADING [CHAP. XI. 
 
 
 
 positive moments at the right end of first span, aticde. From 
 e draw a line to left support. At the other supports, in like 
 manner, we determine the 'positive moments, and join the 
 points ee in second and third spans, and e and right support 
 in fourth. 
 
 We have already seen (Art. 103) that within the fixed points 
 the moments are negative wherever a load may be placed. 
 The maximum, therefore, occurs for full load. We have, there- 
 fore, found for 3d and 4th spans the parabola for full load, 
 precisely as in Fig. 63. These parabolas are given in broken 
 lines in the Fig. a. By subtraction of the positive moments 
 outside the fixed points from the positive moments at the same 
 points for total load given by these parabolas, we obtain directly 
 the lower curves as far as the points e in each span. 
 
 The second parabola (partly full, partly broken) is all that is 
 needed to complete our Fig. To obtain this we have simply, 
 in 2d and 3d spans, to make the vertical through centre of 
 line e e equal to its length already laid off for total load, viz. : 
 
 Z\ 2 
 
 -J, Art. 98 (2), produce e e to intersections with sup- 
 port verticals, and join these intersections with the extremities 
 of the first verticals above. We can then construct the para- 
 bola, which completes our diagram, and gives us Fig. a. 
 
 1O6. Practical simplifications. In practice, the construc- 
 tions given in Figs. 63, 71 and 72, admit of many simplifica- 
 tions, which, in order to avoid confusion at first, have been di& 
 regarded. The whole solution, given for the sake of clearness 
 in three separate Plates, can be performed upon a single sheet, 
 since the Fig. for the second equilibrium polygon in Figs. 63, 
 71 and 72 may be combined in one. Indeed, the lines neces- 
 sary for the construction of the maximum shearing forces can 
 be applied directly to the determination of the maximum 
 moments. It is therefore unnecessary to divide the construc- 
 tion into separate sheets. 
 
 2. The cross-lines in the end spans can be omitted, since 
 all that is required are the distances to be laid off upon the 
 end verticals, and these when found can be laid off at once. 
 
 3. We can apply the second equilibrium polygon directly 
 in order to find the moments for the dead and moving load. 
 Thus the transferring of ordinates from one Fig. to another is 
 avoided. 
 
CHAP. XI.] CAUSING MAXIMUM STRAINS. 163 
 
 4. It is evidently unnecessary to actually draw all the vari- 
 ous lines. We need only to mark the different points of inter- 
 section. 
 
 5. The construction for dead and live loads can be per- 
 formed at once, thus avoiding the necessity of a subsequent 
 addition. 
 
 1O7. Approximate Practical Constructions If the suc- 
 cessive steps of the preceding development are carefully fol- 
 lowed, the method will be found simple and easy of appli- 
 cation. Indeed, the complete and accurate solution of the 
 difficult problem of the continuous girder by a method purely 
 graphical, is the most important extension of the system since 
 the date of Culmann's treatise, and well illustrates the power 
 and practical value of the Graphical Method. 
 
 Humber gives the following constructions, " which may be 
 relied upon for safety without extravagance." * As rapid 
 means of obtaining approximate results, they may not be with- 
 out value to the practical engineer, and we therefore append 
 them here. It must be remembered that the constructions 
 hold good ONLY for end spans three-fourths the length of the 
 others. 
 
 I. Beam of Uniform Strength, continuous over one 
 Pier, forming two equal Spans, subject to a fixed Load 
 uniformly distributed, and also to a moving 'Load. 
 Maximum Moments.* PI. 18, Fig. 73. 
 
 The greatest moment at the pier (positive) will be when both 
 spans are fully loaded. 
 
 The greatest negative moment will obtain in the loaded span 
 when the other span bears only the fixed load. 
 
 (A moment is positive when the upper fibres or flanges are 
 extended, negative when the upper flange is compressed.) 
 
 Construction. Let A B C be the beam. On A B draw the 
 
 parabola whose centre ordinate D E is (p + m) ^-, and on 
 
 p 1? 
 B C the parabola whose centre ordinate G F is "-. 
 
 i * 
 
 At the pier B erect the perpendicular B H = ^ -., 
 
 and make B L = ^ - . Join AH, A L, and L C. 
 \A 
 
 * " Strains in Girders, calculated by Formulas and Diagrams." Humber. 
 New York : D. Van Nostrand, publisher. 
 
164: METHODS OF LOADING [CHAP. XI. 
 
 Then the vertical distances between the parabolic arc A E B 
 and the lines A H and A L, the greatest being taken, will give 
 the maximum moments, positive in the first case and negative 
 in the last. The points of inflection approach as near the pier 
 as K and recede as far as M. 
 
 If 4r~ is less than , the beam must be latched 
 
 2i \.2i 
 
 down at the abutments. The load comes on from the left; 
 p and m are the loads per unit of length of the permanent or 
 fixed and the moving or live loads. 
 
 Shearing Forces (PI. 18, Fig. 74). The maximum shearing 
 force at either abutment will obtain when its span only sus- 
 tains the moving load. The maximum shear at the centre pier 
 will obtain when both spans are fully loaded. 
 
 Construction. Lay off AC = (4:p + 5m) and A D = 
 
 o (p + m). At B lay off B P = twice A D. Take a point 
 
 M distant -J I from A, and join D and F to M. Draw C N 
 parallel to D M. Sketch in a curve similar to that dotted in 
 the figure, giving an additional depth to the ordinates at the 
 
 point of minimum shear of m -. Then the vertical ordinates 
 
 between AB and COPF may be considered to give the 
 maximum shearing force for either span. 
 
 II. Beam as above continuous over three or more 
 Pier. li = end spans. I = the other spans. 
 
 Moments. 
 
 The maximum moment (positive) will obtain when only the 
 two adjacent spans, and every alternate span from them, are 
 simultaneously loaded with the total load the remaining spans 
 sustaining only the fixed load. 
 
 The maximum moment at the centre of any span will obtain 
 when it and the alternate spans from it are fully loaded the 
 remaining spans sustaining only the fixed load. 
 
 Construction (PL 18, Fig. 75). Let A B C be part of the 
 beam. On B C draw the parabola whose centre ordinate E F 
 
 = ~ , and on A B the parabola whose centre ordi- 
 
 nate C D = (P + W t At B and c make B H _ c L = 
 
 8 
 
CHAP. XI.] CAUSING MAXIMUM STRAINS. 165 
 
 Join A Hand I, Make B G = C S = *l. 
 
 o 
 
 Join A G and S. The maximum vertical ordinates between 
 the two parabolas and the lines A H, A G, H L, and G S, as 
 shown in the figure, give the maximum moments. 
 
 The points I, K and O, P or M, N show the limits of devia- 
 tion of the points of inflection. 
 
 If -^ is less than - - , the beam will require to be 
 
 2i oA 
 
 held down at the abutments. 
 
 If the beam be continuous for three spans only, I in the 
 
 expression for B H = ( =- + ) must have a value given to 
 
 Shearing Forces. 
 
 The maximum shear at any pier (B or C) will obtain simul- 
 taneously with the maximum moment over that pier. 
 Construction. PL 18, Fig. 76. 
 Let A B C be part of the beam. First, for any inner span as 
 
 Z__. At B and C erect B G = C H = i&^J&A. Make B D 
 
 and C E each = - (p + m). Join D and E to midspan F, and 
 
 2 
 
 draw G K and H K parallel to D F and F E respectively. 
 Second, for either end span as \ . At B erect a perpen- 
 
 2 7 3 
 
 dicular = - (p+m), which, if ^ = - I, will coincide with B D. 
 
 3 4: 
 
 At A make AL = - B D. Join D and L to M distant - \ from A. 
 2 o 
 
 Make A O =| (p + m)- ^- 2 , and draw ON parallel to L M. 
 
 2i ' 32 1- 
 
 Sketch in curves as shown by the dotted curves in the figure, 
 giving additional depth to the ordinates there of ^ and --ir- 
 
 respectively. Then the vertical distances between O a b D and 
 A B give the maximum shearing forces for either end span, and 
 those between G c d H and B C, the shearing forces for the re- 
 maining spans. 
 
166 METHODS OF LOADING [CHAP. XI. 
 
 If the beam be continuous for three spans only, B G and C H 
 
 must be made equal to (18 * + 16 "*>*+" ( 2 -# + ?), where 
 
 oz o I \ 7 2i ' 
 
 L = ~. Further, the value given to B D for the inner span 
 
 3 
 1O8. JUethod by Resolution of Forces Draw Spans. 
 
 The most usual cases of continuous girders which occur in prac- 
 tice are draw or pivot spans, which when shut must be consid- 
 ered as continuous girders of two spans. The graphical method 
 becomes for such cases short and easy of application. In the 
 case of framed structures of this character, it may, however, be 
 more satisfactory to first find the maximum shearing forces 
 (Art. 104), and then follow the reactions thus obtained through 
 the structure from end to end by the method of Arts. 8-13. 
 As a check upon the accuracy of the work, we may apply the 
 " method of sections " referred to in Art. 14. In either case 
 we must, of course, start from an end support where only two 
 pieces intersect and the moment is zero. 
 
 Still again, we may find the reactions by calculation, and 
 then apply the method of Arts. 8-13. In the case of two spans 
 only, the formulae for the reactions are sufficiently simple, and 
 the ready and accurate determination of the strains offers, there- 
 fore, no difficulty. 
 
 We shall give here, therefore, the analytical formulae requi- 
 site for our purpose, referring the reader to treatises upon the 
 subject for their demonstration.* 
 
 Formulae for Reactions. Continuous Girder of two un- 
 equal /Spans, I and n I. 1st. Concentrated weight P, in first 
 span Z, distant /3 from left end support. Reaction at left end 
 support : 
 
 P 
 
 Reaction at middle support : 
 
 * Bresse La Flexion et la Resistance, and Cours de Mecanique Appliquee. 
 Weyrauch Theorie der Trdger. CoUignon Theorie Elementaire des Poutres 
 Droites, etc. Also Supplement to Chap. XIII. 
 
CHAP. XI.] CAUSING MAXIMUM STRAINS. 167 
 
 Reaction at right support : 
 
 0= _P f-^4-^1 
 
 2tt(l + 7l) L J "*>_! 
 
 2d. Uniformly distributed load extending to a distance /3 
 from left support. Load per unit of length =p. 
 
 o= 
 
 For two equal spans we have only to make n = 1 in the 
 above equations. For a uniform load over whole span /3 = L 
 
 From the above formulae we can find the reactions for any 
 case, and then proceed as indicated above. 
 
 109. By means of the graphical method, as we have now 
 seen, we are enabled to solve completely the problem of the 
 continuous girder, and that too without the aid of analytical 
 formulae, tables, or tedious computation. The method can also 
 be applied to continuous girders of variable cross-section, or of 
 uniform strength. We shall not, however, proceed further 
 with the development of the method in this direction. The 
 preceding will, we think, be found to contain all that is practi- 
 cally serviceable. For the application of the method to girders 
 of variable cross-section, we refer the reader to Winlder " Der 
 Bruckenbau" Wien, 1873 where will be found a thorough 
 presentation of the subject, both analytically and graphically, 
 to which we are greatly indebted in the preparation of the 
 preceding pages. Plates 17, 19 and 20, are, with but few alter- 
 ations, reproduced from that work. 
 
 * These formal EG are demonstrated in Van Nostranffs Eng. Mag., July, 
 1875. 
 
168 GRAPHIC AND ANALYTIC [CHAP. XH. 
 
 CHAPTEE XII. 
 
 CONTINUOUS GIRDER (CONTINUED) COMBINATION OF GRAPHICAL 
 
 AND ANALYTICAL METHODS. 
 
 1 1O. In the present chapter we shall develop a method for 
 the solution of continuous girders not purely graphical, but 
 based upon the method of resolution of forces illustrated in 
 Arts. 8-13, together with well-known analytical results, which 
 method for accuracy, simplicity, and ease of application will, 
 we think, be found superior to any hitherto proposed. The 
 method is, of course, applicable only to framed structures, but 
 for such cases is the most satisfactory of any with which we are 
 acquainted. 
 
 111. The Inflection Points being known, the Shearing 
 Forces and Moments at the Support* can, by a simple 
 construction, be easily determined. ~Lst. Loaded Span 
 Fig. 77, PL 21. Thus in the span B C = I, let the distance of 
 the weight P from the left support be a, and let i and i' be the 
 distances of the inflection points from B and C respectively. 
 Then if through any point P of the weight we draw lines, as 
 P D, P E, through i and i', intersecting the verticals at B and C 
 in the points D and E, the vertical ordinates between these lines 
 and B C will be proportional to the moments. For, as we see 
 from the force polygon, the equilibrium polygon must consist 
 of two lines as D P, P E, parallel to O and O 1, and because 
 of the moments at the ends, the closing line D E is shifted to 
 B C (Art. 23). Since the moments at the points of inflection 
 are zero^ the ordinates to P D and P E to pole distance H will 
 give the moments. Now the points of inflection being known, 
 and P J3 and P E drawn, we can easily find the pole distance 
 H and the shearing forces L and 1 L by laying off P verti- 
 cally, and drawing from its extremities lines parallel to P D 
 and P E intersecting in O. A perpendicular through O upon 
 1 gives H and the reactions L and 1 L. In other words, 
 we have simply to decompose P along P D and P E. 
 
CHAP. XII.] METHODS COMBINED. 169 
 
 The construction, then, is simply as follows : Take any point 
 on the direction of P, and draw P D, P E through the points 
 of inflection. Lay oft P to the scale of force as A P, and draw 
 A O parallel to P E or P D. We have thus the pole distance 
 H, and the shearing forces P H and H A at B and C. 
 
 B D or C E to the scale of distance, multiplied by H to the 
 scale of force, give the moments at B and C. That is, D and 
 E may be regarded as the points of application for H. The 
 forces along P D and P E considered as acting at these points 
 are held in equilibrium by the reactions P H and H A = L 
 and 1 L and H. Since H, acting as indicated in the figure 
 with the lever arm B D or C E, causes tension in the upper 
 fibres, the moments at B and C are positive. 
 
 2d. Unloaded SpanFig. 78, PL 21. 
 
 As we have already seen in Art. 93, the inflection points in 
 the unloaded spans are independent of the load, and are found 
 by the simple construction there given for the "fixed points" 
 Since, each fixed point lies within the outer third of the span, 
 we have in Fig. 78 the broken line a b c, referred to in Art. 94, 
 where the moments are alternately positive and negative, and 
 increase from the end, so that any one is more than twice the 
 preceding. Lines drawn parallel to these lines in the force 
 polygon, cut off from the force line the reactions at the sup- 
 ports. Thus, in Fig. 78, c b in the force polygon gives the 
 reaction at D, a b the reaction at C, and if B were an end sup- 
 port that is, if b a went through B a H would be the re 
 action at B. For the resultant shear at D, we should then have 
 aH ab+-cb = Hc. So for any number of spans ; the in- 
 flection points in the loaded span being known, we can easily 
 find the fixed or inflection points in the other spans, which are 
 independent of the load, and depend only on the length of 
 these spans. Then draw the broken line a b c P d. Then find 
 the pole distance H by laying off c P = P to scale, and draw- 
 ing c O parallel to c P, and through the point O thus deter- 
 mined drawing H O. Then find the reactions at the other 
 supports, or the shear at any support, by lines in the force 
 polygon parallel to a b, b c, etc. Thus the shear at B is the 
 distance H a cut off by H and O a parallel to a b. Since the 
 shear at D is plus and alternates from D, we have at B the 
 shear + H a. The shear at C is H b ; at D, H- H c, etc. O c 
 being parallel to PC; O a, to b a ; O b, to c b, etc. 
 
170 GRAPHIC AND ANALYTIC [CHAP. XII. * 
 
 112. Inflection Verticals, Draw a line from P through 
 the support D, and through its intersection with cb draw a 
 vertical 3 (Fig. 78). This vertical we call the inflection ver- 
 tical. 
 
 The equation of the line c ft is 
 
 where m l = D C, n I = C D, i = Ci'. The origin being at D. 
 For the line C, 
 
 where ^ = D i. 
 
 If in this last equation we make x = a, we have for the or- 
 dinate at P, 
 
 1ft! fa a) 
 
 h 
 and hence for the line P D, 
 
 m t (*, a) 
 y = *; - -x. 
 
 For the intersection of P D with b c then 
 
 m m (^ a) 
 
 ^ - . x + m = ^ - - x. 
 nl v i^a 
 
 Hence 
 
 - /. \ / 7 -\ ... 
 
 (^ a) (n I i) \ a 
 
 We see at once that the value of x is independent of m l or 
 B C, hence the intersection of P D and c b lies always in the 
 same vertical, whatever be the position ofPC. In other words, 
 if the three sides of a triangle pass always through three fixed 
 points (a 1 , D, ^), and two of the angles (P and c) be always in 
 the same verticals, the third angle must also always lie in the 
 same vertical. 
 
 For the distance of the inflection vertical on the other side 
 of the loaded span (beyond E), we have similarly 
 
 (I a) (B + 4) 
 
 where I is the loaded span and i B the distance of the inflection 
 point to the right of E. 
 
 Equations (1) and (2) give the distances of the inflection ver- 
 ticals from the supports D and E. 
 
CHAP. XII.] METHODS COMBINED. 171 
 
 113. Beam fixed horizontally at both ends Supports 
 011 level. Consider the span D E (Fig. 78) as fixed at the sup- 
 ports so that the tangent to the deflection curves at D and E is 
 always horizontal. Conceive the span prolonged right and left 
 beyond the supports a distance equal to the span I. It is re- 
 quired to find the position of the inflection verticals. 
 
 From equation (1) of the preceding Art. we have, since 
 n = 1, i = 0, 
 
 ii a \ 
 
 t/U 
 
 "" 7 -w 
 
 tj a) l ^ l a/ 
 and from equation (2), since i s I, 
 
 i% I (I a) 
 
 x = 
 
 Now for a beam fixed at the ends the distances of the points 
 of inflection are 
 
 a I I (I a) 
 
 Substituting these values in the equations above, we have 
 
 x = - and x = 4- o- That is, the position of the inflection 
 6 o 
 
 verticals is in this case independent of the load, and always 
 
 equal to ~ from the supports* 
 o 
 
 This remarkable property of the beam fixed at both ends 
 enables us to find the inflection points by a construction similar 
 to that for the fixed points in the unloaded spans, as given in 
 Art. 86. 
 
 Thus we have simply to draw from C distant I from A (PI. 
 21, Fig. 79) a line in any convenient direction, as C ~b intersect- 
 
 ing the inflection vertical I, which is distant from A, - Z, at a. 
 
 Through a and the fixed end support A draw a line to inter- 
 section with the weight P. Then draw P b. The intersection 
 i t of this last line with A B is the inflection point. A similar 
 construction gives i%. 
 
 We can now find the reactions and moments. Thus H O to 
 
 * This important result, which renders possible a complete graphical solution 
 of this case, has, so far as we are aware, never before been published. 
 
172 GRAPHIC AND ANALYTIC [CHAP. XII. 
 
 the scale of force, multiplied by A b to the scale of distance, 
 gives the moment at A, while H G is the reaction at A (Fig. 79). 
 114. Beam fixed at both ends Example. Since when 
 the points of inflection are once determined, we may draw P b 
 or P c at any inclination (Fig. 79), provided we afterwards find 
 the corresponding pole distance HO; if A b or B c be made 
 equal to the height of the truss, H O will ~be the strain in the 
 upper or lower flange at the wall (the flange in question being 
 always that for which there is no diagonal at its union with the 
 wall). Thus in PL 21, Fig. 80, we lay off D E = I, draw the 
 
 vertical I at ~ I from D, and for the given position of the load 
 
 o 
 
 P find the inflection point 4 by the preceding Art. A similar 
 construction on the other side gives %. Now laying off P M 
 equal by scale to the weight P, and decomposing it along P D 
 and P C, we find O H the pole distance which to the scale of 
 force will give directly the strain in the lower flange B m at 
 the wall, provided P D is made to pass through the intersection 
 of the upper flange with the wall. If the triangulation were 
 reversed, O H would be the strain in the upper flange at the 
 wall. In any case it is the strain in that flange at whose junc- 
 tion with the wall there is no diagonal. 
 
 The reaction at D is also H M, at C it is P H. Lay off then 
 B B' in Fig. 80 (a) equal to P M, and make B A = H M and 
 A B' = P H. Now draw m B parallel to O H and m A paral- 
 lel to O M, and produce both lines to intersection at m. Then 
 m B to scale of force is evidently the strain in the lower end 
 flange at the wall. We assume the following notation.* 
 
 Let A represent all the space above the girder, B all the space 
 below, and a h c d 7 etc., the spaces within the girder included 
 by the flanges and diagonals. Then, for instance, A b is the first 
 upper flange at the left, B a the first below ; a b the first diag- 
 onal at the left, and so on. 
 
 Now draw in Fig. 80 (#), m I and A I parallel to the corre- 
 sponding lines in the frame, and we have at once the strains in 
 these pieces to scale. Following round the triangle according 
 to our rule (Arts. 8-13) from m to A, A to I and I to m, we 
 
 * See an excellent little treatise on " Economics of Construction in Relation 
 to Framed Structures," by R. H. Bow, to whom this method of notation is 
 due. 
 
CHAP. XII.] METHODS COMBINED. 173 
 
 find A I tension and I m tension. [The strains in the upper 
 flanges must always be tension, since the moments at the sup- 
 ports for loaded span are always positive.] Moreover, the Fig. 
 thus far shows that A I, I m and m B are in equilibrium with 
 the shear B A = H M, as evidently should be the case ; hence 
 the strain in B m is compressive. 
 
 We have thus the strains in the three pieces at the right, and 
 can proceed from these to find all the others. Thus the strains 
 in B k and k I are in equilibrium with m I and B m. Lines 
 parallel to B It. and k I, therefore, which close the polygon com- 
 menced by B m and m I, give us the strains in B k and k I. 
 Observe that the line I k crosses A B, thereby making B k op- 
 posite in direction, consequently in strain from B m. This may 
 also be seen by following round the triangle m 1 k B, remem- 
 bering that, as m I is always found to be in tension, it must act 
 away from the new apex, that is, from m to I. We thus find 
 k I in compression, and k B acting away from this apex, or in 
 tension, therefore of opposite strain from the preceding flange 
 B m, which, as we have seen, is in compression. The reason is 
 obvious. The inflection point i% falls in the flange B k. If the 
 beam w r ere solid, the strain at i 2 would be zero ; to the right of 
 4 we should have compression, to the left, tension. In the 
 framed structure the strains can only change at the vertices. 
 The crossing of A B by I k indicates such change, and B k gives 
 its amount by scale. 
 
 Now taking the upper apex, we have here A I and I k in 
 equilibrium with k h and A h. As we already know, k I is in 
 compression. We must, therefore, now take it acting from k to 
 I, and following round the triangle we find A h compression, 
 and h k tension. From h on, the traverses between A h and 
 B k produced towards the right [Fig. 80 (&)] will give the di- 
 agonals, while the upper and lower flanges will be given by the 
 distances to them from A and B respectively, until we arrive 
 at the weight P. Observe the influence of the weight. We 
 have k h and B k in equilibrium with h g and g B, and also the 
 weight P = B' B. We must take, therefore, A B' = P H, and 
 then draw h g and B' g. Distances to the right of B' along 
 B'<7 are compressive lower flanges, to the left, tensile; while to 
 the right of A we have compressive upper, and to the left of A 
 tensile upper flanges. The two diagonals at the weight k h and 
 h g are in tension. From h g on, the diagonals are alternately 
 
174 GRAPHIC AND ANALYTIC [CHAP XII. 
 
 tension and compression. Moreover, the diagonal e d passes 
 through A, that is A d is zero. The weight P causes no strain 
 in A d, and for this one position of P, A d might be omitted 
 from the structure. The reason is again obvious. The point 
 of inflection i coincides with the apex under A d. Since at 
 ^ the moment of rupture is zero, if the flange A.~d were cut 
 there would be no tendency to motion. We have at ?' t the 
 shearing force only, A B' giving the strains in the diagonal 
 e d and d c. The upper flange A ', we see again, is in tension, 
 which is also shown by its lying to the left of A. 
 
 Thus we have the strains in every piece by a very simple 
 construction for any position of P, without any calculation what- 
 ever. The method in this case is purely graphical. We have 
 only to find the points of inflection and then proceed as above. 
 
 Fig. 80 (b) gives the strains for the same girder and position 
 of weight P, merely supported at the ends. For this case P D 
 in Fig. 80 not only passes through D, but P O also passes through 
 the upper left-hand corner at C. Hence A B will be less than 
 H M, and A B' greater than P H. Moreover, the end lower 
 flanges B a and B m no longer act, and must be removed. 
 Starting now with the reaction B' A [Fig. 80 ()], we go along 
 to the weight, from which point at h k we go ~back towards the 
 force line, and the reactions are such that the last diagonal 
 must pass exactly through B, just as in Fig. (a) e d passed 
 through A, because the points of inflection or zero moments are 
 now at the ends C and D. A careful comparison and study of 
 the two cases and their points of difference will be advanta- 
 geous to the reader. 
 
 115. Counterbracing. The objection may arise that the 
 above method applies only to a system of bracing such as rep- 
 resented in the Fig., where the diagonals take both compres- 
 si ve and tensile strains. In case, as in the Howe or Pratt Truss, 
 for instance, we had vertical pieces as also two diagonals in 
 each panel, then the strain in any diagonal as m I and flange as 
 B m, even if found, are apparently in equilibrium with three 
 pieces, viz., Jcl,~Bk and a vertical strut or tie at the intersec- 
 tion of these pieces. Hence, having only two known strains 
 and three to be determined, the method would seem to fail, as 
 any number of polygons may be constructed with sides par- 
 allel to the forces, and hence the problem is indeterminate 
 (Art. 9). 
 
CHAP. XII.] METHODS COMBINED. 175 
 
 jSTow in any framed structure of the above kind, the counter 
 ties are inserted to prevent the deforming action of the Tolling 
 load only. For the dead load but one system of triangulation 
 is required, and the strains in every piece due to this dead load 
 can therefore easily be determined. 
 
 We have then only to determine the strains in the same pieces 
 due to the rolling load also. If now in any diagonal the strain 
 due to this rolling load exceeds the constant strain due to the 
 
 O 
 
 dead load, and is of opposite character, and if the diagonal is 
 to be so constructed as to take biit one kind of strain, then a 
 counter diagonal must be inserted in that panel, and propor- 
 tioned to this excess of strain only. For instance, if a diagonal 
 takes only the compressive strain (a condition which is easily 
 secured in practice) due to the dead load, and the live load 
 would cause in that diagonal a tensile strain, then the excess of 
 this tensile strain over the constant compressive strain due to 
 the dead load must be resisted by a counter diagonal, which 
 also takes compressive strain only. The method is precisely 
 the same as by calculation (see Stoney and other authors 011 
 the subject), and we only notice the point here, as in all our 
 examples we have taken a single system of triangulation only 
 a system which, we may here remark in passing, has many ad- 
 vantages, and is worthy of more general attention * than it has 
 hitherto obtained. 
 
 [See also on this point Art. 10 of Appendix.] 
 116. Beam fixed horizontally at one end, supported at 
 the other Supports on Level. In this case, equation (1), 
 Art. 112, becomes for left end fixed, since n = 1, i% = 0, i 0, 
 
 i-i a I 
 
 (^ a) l ^\ a 
 
 But for this case the distance of the point of inflection from 
 the fixed end is 
 
 . _ (2 I a) la , 
 
 * See " A Treatise on Bracing." By R. H. Bow. D. Van Nostrand, pub- 
 lisher. 
 
 f The values of the distance of the inflection points which we assume above 
 as known, may easily be deduced by the theory of elasticity. See Supplement 
 to Chap. VTL, Arts. 16 and 19. See Wood, Strength of Materials ; Bresse, 
 Mecanique Appliquce ; or other treatises on the subject. 
 
176 GRAPHIC AND ANALYTIC [CHAP. XII. 
 
 Inserting this value of ^ in the value for x above, we have 
 
 x - _ l ( 2 t-a) 
 5 I a 
 
 for the distance of the inflection vertical to the left of the left- 
 hand support, which is supposed fixed. 
 
 Now this is the equation of an hyperbola, as shown in PL 21, 
 Fig. 81, whose vertex is at/J the distance Ay being 2 I, whose 
 assymptotes are respectively parallel and perpendicular to the 
 span, the perpendicular distance of E above the span A B be- 
 
 5 9 
 
 ing 1 I, and which intersects A B at ^ I from A. The ordinate 
 2i 5 
 
 d e, A d being equal to Z, is - 1. The diameter passes through 
 
 3 
 
 E and f, and E f is, therefore, the semi-transverse axis. The 
 hyperbola can, therefore, be easily constructed. We need only 
 to construct that portion between A B and the point e. 
 
 The construction for the point of inflection i is, therefore, 
 simply as follows : 
 
 Lay off A k vertically upwards and equal to the distance of 
 the weight P from A, and draw the horizontal h b to intersection 
 b with the curve. Now make A a = I and draw a b to inter- 
 section c. Draw b A to intersection P with weight, and then 
 P c intersects A B at the point of inflection i lm Decomposing 
 P along P B and P c, as in Art. 114, we have at once the re- 
 actions at A and B. Here also we see that, by a construction 
 purely graphic and abundantly exact, we can find the inflection 
 point and the reactions. 
 
 The method detailed in Art. 114 can then be applied to de- 
 termine the various strains in the different pieces. It is un- 
 necessary to give an example, as the process is precisely similar. 
 We have simply in this case to start with the reaction at the 
 free end B and follow it through. Observe only that, as this 
 reaction must be less than for a girder with free ends for the 
 same position of P, the point h will lie nearer the force line 
 B' AB (Fig. SO, &), hence Im will not pass exactly through B, 
 but will lie to the right of it, giving thus a reversal of strain in 
 the flanges, as by reason of the inflection point should be the 
 case. 
 
 Instead of constructing tho hyperbola, we may calculate its 
 ordinates from the equation for x above, for different values 
 of a. 
 
CHAP. XII.] METHODS COMBINED. 177 
 
 Thus, for 
 
 x=OAl %=-0.3SSl x=-0.375l o?= 0.35YZ aj= 0.333 J 
 
 This will be sufficient to construct the curve in any given 
 case. The inflection vertical moves, therefore, between the 
 narrow limits of x = f I and % = % I, or within J^th of the span, 
 as the load passes from A to B. 
 
 Inasmuch as all that is needed for the determination of the 
 strains in the various pieces are the reactions at the supports, 
 and (for girder fixed at both ends) the moments at the supports 
 also, and as the formulae for the two cases above are very sim- 
 ple, we may determine these quantities at once by interpolation 
 of the given distance of the weight P in the formulae, and then 
 apply the graphical method for the strains, as illustrated in 
 Art. 114. 
 
 Thus, for a horizontal beam fixed at both ends, we have for 
 the moment at the left support A, 
 
 - V 
 
 At the right support B, .. 
 
 M B = ^ (I - a). 
 For the reaction at the left, 
 
 For the reaction at the right, 
 
 R B = ? (3 a 2 l - 2 a 3 ). 
 
 In the case of a horizontal beam fixed at left end and merely 
 Besting upon the right support, we have 
 
 M A = (3 a^l - 2 a 1? - a*\ M B = 0, 
 
 a being always the distance of the weight P from the left. 
 These formulae are simple, and easily applied to any case. 
 
 We may also observe that in Figs. 79 and 81 the ordinates to 
 the lines P 5, P c, and P c, P B, from A B, are proportional to 
 the moments (Art. 110). These ordinates to the scale of dis- 
 
178 GRAPHIC AND ANALYTIC [cilAP. XTT. 
 
 tance, multiplied by the pole distance to scale of force, give the 
 moments at any point. ' Onr construction, therefore, gives the 
 moments also at every point, and we may thus check the re- 
 sults obtained by Art. 114 by the results obtained by the 
 method of moments. 
 
 117. Approximate Construction. It will be readily seen 
 that the portion of the hyperbola in Fig. 81, PI. 21, needed for 
 our construction, is nearly straight. In most cases it will be 
 practically exact enough to lay off f I to the left of A, and -J I 
 also to the left of A at a vertical distance equal to Z, and join 
 the two points thus obtained by a straight line. This line can 
 be taken instead of the curve, and the construction is then the 
 same as above. The error due to thus considering the curve 
 as a straight line is greatest for a weight in the middle of the 
 span, where it does not exceed To^th of the span for the posi- 
 tion of the inflection vertical, and diminishes from the centre 
 both ways. 
 
 118. Girder continuous over three Level Supports- 
 Draw Spans. This case is perhaps of the most frequent 
 practical occurrence, and an accurate and simple method of 
 solution is therefore very desirable. 
 
 In the first place, the formulae for the reactions are very 
 simple and easy of application. Thus, for left end support A, 
 the load being in the second span, or to the right of the middle 
 support B, 
 
 R A = j?(3^-2?- 3 ); 
 
 * ^ 
 
 for the reaction at middle support, 
 
 for reaction at right end C, 
 
 where a is always the distance of the weight P from the mid- 
 dle support.* We are therefore already in a position to solve 
 completely the case under consideration. We have only to 
 
 * As already remarked, the development of the formulas assumed in this 
 chapter must be sought for in special treatises on the subject. We assume 
 them as known, and then apply them graphically as above. 
 
 See also Supplement to Chap. XIII. 
 
CHAP. Xn.] METHODS COMBINED. 179 
 
 find the reactions and follow them through by the method of 
 Art. 114. 
 
 From the above reactions we can, however, easily determine 
 the distance of the inflection point. This will, of course, be 
 found only in the loaded span, at a distance from the middle 
 support. 
 
 VU - ~^ TQ \ ^T 7 O ' -" v 
 
 4&&+ 2 al a 2 
 
 We can find the values of x corresponding to different values 
 of a, and thus plot the curve for the inflection points. Thus, 
 for 
 
 a :0 a = ^l a = $l a = %l a = I 
 x = cc -fa I x = -f^l x = ^l x = \l. 
 
 This curve being drawn for any particular case, we can 
 easily find the position of the inflection point for any given 
 value of #, and hence the reactions, and then find the strains in 
 the various pieces. 
 
 Thus, in PL 21, Fig. 82, the curve B e d being drawn, we 
 can at once find the inflection point i for any position a of 
 the weight P. We have simply to make B J a and draw 5 e. 
 b e is the distance of the point of inflection from B. We can 
 now, as explained above, draw any line as P i, and then P C 
 and A h. The ordinates to the broken line A h P C from A C, 
 to the scale of distance, multiplied by the pole distance H to 
 scale of force, will give the moments at any point. Moreover, 
 H E is the shear at B. E a is the reaction at B, H a the re- 
 action at A, and H P the reaction at C. The reactions at B 
 and C are, of course, positive or upwards, that at A negative 
 or downwards. Hence E H # -f- H P = P, as should be. 
 
 The value of x for the inflection vertical is by Art. 112 
 
 i a I 
 
 X = -T-. - r-j - ; j 
 
 (i a) I ^ a 
 or, substituting the value of i above, 
 
 a I (2 I - a) 
 
 Since, therefore, in this case the value of x is no simpler than 
 that for i given above, it will be preferable to plot the first 
 curve directly as represented in Fig. 82. 
 
 119. Approximate Construction. In practice it will be 
 
180 GRAPHIC AND ANALYTIC [CHAP. XII. 
 
 found abundantly accurate to assume the curve for i between 
 the required limits, as a parabola whose equation is i = x = 
 
 a ^ -. The greatest error for a = - will then be about 
 
 5 t 4: 
 
 of the span, and decreases both ways to a = o and a = -. 
 100 * 2 
 
 From a = to a = I the parabola coincides closely with the 
 
 2 
 
 q -i 
 
 true curve. The difference for a = I is only -^^-l, and we 
 
 4 500 
 
 have, therefore, a very simple practical construction for both 
 reactions and moments. We have only (Fig. 82) to erect a 
 vertical at the centre support B and make it equal to I, and 
 then construct a parabola passing through B whose ordinate 
 
 c d = - I. The horizontal ordinates to this parabola for any 
 5 
 
 vertical value of &, give the distance out from B of the inflection 
 points. For the load in first span A B, of course this parabola 
 lies on the other side of B c, and b e is laid off to the left. The 
 remainder of the construction is as in Art. 118 for the reactions 
 and moments. When great accuracy is required, we can find 
 the reactions from the equations of Art. 118. In any case, the 
 reactions being given, we can follow them through the structure 
 by the method of Art. 114, and thus determine the strains in 
 every piece due to every position of each apex load. A tabu- 
 lation of these strains will then give by inspection the maximum 
 strain in any piece due to the live load. All the weights taken 
 as acting simultaneously will then give the strains due to uni- 
 form total live load. The strains due to dead load will be 
 multiples or sub-multiples of these. Thus if total live load 
 causes, say, 100 tons compression in a certain piece, and if the 
 
 Q O 
 
 dead load is - of the live load, then we shall have - of 100 == 
 
 2 2 
 
 150 tons compression in the same piece due to the dead load 
 alone. If now the live load causes a maximum tension in the 
 same piece of 200 tons, then the piece must be made to resist 
 both tensile strain of 200 150 50 tons and compressive 
 strain of 150 + 100 250 tons. If a diagonal, the counter tic 
 is strained 50 tons, while the maximum strain on the diagonal 
 
CHAP. XII.] METHODS COMBINED. 181 
 
 is 250 tons compression. It is only necessary, therefore, to find 
 the strains due to each weight of the live load. From the 
 tabulation we can, then, by means of the ratio of the dead to 
 live load, find the strains due to dead load alone, and then by 
 a comparison of the two find the maximum compressive and 
 tensile strains. If the maximum strains due to live load are of 
 opposite kind, but less than the constant strains clue to dead 
 load, we shall need no counterbracing. The resultant strains 
 will then always be of the character given by the dead load. 
 If greater, we must counterbrace accordingly. The process is 
 the same as by the methods of calculation, and the reader may 
 refer to Stoney Theory of Strains for illustrations. 
 
 12O. The " Tipper," or Pivot Draw, with secondary cen- 
 tral Span. We have said that a pivot draw may be considered 
 as a beam continuous over three supports. In practical con- 
 struction this statement needs some modifications which deserve 
 special notice. Thus practically that portion of the beam over 
 the central support forms a short secondary span D D [Fig. 83, 
 PL 22] the reactions at the supports D and D being always 
 equal and of the same character. If a weight acts, say, on the 
 first span A B, and the beam itself is considered without weight, 
 the end C must be held down, that is, the reaction there is neg- 
 ative. ISTow as the weight P deflects the span A B (Fig. 83), it 
 causes one secondary support D to sink, and the other to rise 
 an equal amount. In practice D and D may be the extremities 
 of the turn-table, and the reactions are then evidently different 
 from those given by the formulas of Art. 118. 
 
 If in this case we take a as the distance of the weight P from 
 the left support A, the reaction for load in A B will be given 
 by the following formulae : 
 
 Where the ratio -^ = &, a being the distance of the weight P 
 
 from the left support A (for load in the span A D), I = span, 
 A D = D C, and n 1= span D D, and where the constant 
 (4 + 8 n + 3 n 2 ) is put for convenience = H, then 
 
 \ 2 H - (10 + 15 n + 3 n*) & + (2 + n) & I 
 |_ J 
 
 R A = 2 H - (10 + 15 n + 3 n* 
 2 H 
 
 * See Supplement to Chap. XIII., Art. 6. 
 
182 GRAPHIC AND ANALYTIC [CHAP. XH. 
 
 R = 2^ r (2 + n \ ~~ (2 + 3 n + 3 ^ ^ \ 
 
 These reactions, it will be observed, when added together 
 R A 4- 2 R D + R c are equal to P, as should be the case. 
 
 By the application of these formulae, which are for any par- 
 ticular case by no means intricate, we can find the reactions at 
 A and C as also at D or D ; and then starting, say, from A, can 
 follow the reaction there through the frame by the method of 
 Art. 114. A negative reaction indicates that the support tends 
 to rise, and unless more than counterbalanced by the positive 
 reaction due to uniform load, the end where this negative reac- 
 tion occurs must be latched down. 
 
 121. Supports in Pivot Span are not on a level Reac- 
 tions for live load, however, are the same as for level sup- 
 ports. The three supports of a pivot span should not be on a 
 level. It is evident that if this were the case, the first time the 
 draw is opened the two cantilevers deflect and it would be diffi- 
 cult to shut it again. The centre support should therefore be 
 raised until the reactions at the end supports are zero, that is, 
 until they just bear. The centre support is then raised by an 
 amount equal to the deflection of the learn when open, due to 
 the dead load. Even when shut, then, there are no reactions at 
 the end supports except when the moving load comes on. Now 
 this being the condition of things, it may seem, strange to assert 
 that these reactions %XQ precisely the same as for three level sup- 
 ports, and yet such is the fact. If the beam, originally straight 
 were held down at the lower ends by negative reactions, then 
 the reactions would have to be investigated for supports out of 
 level, and a load would diminish these negative reactions, or 
 might even cause them to become positive. But such is not 
 the state of things. The end reactions are in the beginning 
 zero, and any load gives, therefore, at once positive reaction at 
 its end support. This positive reaction is just what it ivoulo, 
 be for the same beam over three level supports. 
 
 An analytical discussion of the case would be out of place 
 here, but assuming the expression to which such a discussion 
 would lead us, we may show that this is so. 
 
 Thus, for a beam over three supports A, B, and C, not on a 
 level. GI being the distance of A below B, and c% the distance 
 of C below B, the modulus of elasticity being E and the mo- 
 
CHAP. XII.] METHODS COMBINED. 183 
 
 inent of inertia I, we have for the moment at the centre sup- 
 port B due to any number of weights in both spans, 
 
 a being always measured from the left support. 
 
 Now in this expression the last two terms are precisely the 
 same as for supports on a level ; the influence of the different 
 levels is contained in the first term on the right only. Now by 
 the supposition, c t and c 2 must be taken equal to the deflection 
 due to the dead load, and the value of this term will therefore 
 be entirely independent of the live load, which enters only in 
 the last two terms. 
 
 A particular case may perhaps render this plainer. If a 
 girder of two equal spans over three level supports is uni- 
 formly loaded, the reaction at an end support is, as is well 
 known, |ths of the load on one span. 
 
 Now let us take the girder over three supports not on a level, 
 and from our formula above find the reaction at one end due 
 to uniform load when ^ and c 2 have the proper values given 
 to them. First the dead load p I over each span causes a de- 
 
 flection at each end of the two cantilevers ii&s-i This, 
 
 then, is the value for ^ and <? 2 m the formula. Now let us 
 take an additional moving load of ml over the whole beam, 
 and with this value of c 1? C 2 find the reaction. We have from 
 our formula 
 
 4 M B I = - lp 1? - \ (p + m) P, 
 or M B = ->Z 2 - 
 
 Now we have by moments, 
 
 I 2 
 RAX Z-(^ + m)-= 
 
 hence, inserting the value of M B above, 
 
 * Theorie der Trdger: Weyranch. Also Supplement to Chap. XIII., Art. 3. 
 f Supplement to Chap. VII., Art. 13. 
 
184 GEAPHIC AND ANALYTIC [CHAP. XII. 
 
 That is, the reaction at A is due to the moving load alone, 
 as evidently should be the case, and is, moreover, just what it 
 should ~be for a girder with level supports; viz., \ml. (See 
 also Appendix, Art. 18, Ex. 5.) 
 
 The raising of the centre support, then, will not affect our 
 construction for the reactions as given in Figs. 81 and 82, pro- 
 vided there are only three supports. 
 
 We have deemed it well thus to call special attention to the 
 considerations of the last two articles, both on account of their 
 practical importance and because they are not brought out 
 clearly, nor indeed, so far as we are aware, ever alluded to in 
 any treatise upon the subject.* 
 
 122. Bctiiii continuous over four Level Supports. We 
 thus see" that a draw or pivot span is more properly considered 
 as a beam of three spans instead of two, of which the centre 
 span is very small compared to the end spans ; it may be only 
 two or three panels long. Moreover, we must often in practice 
 consider the beam as a " tipper," and therefore apply the formulae 
 for reactions of Art. 120. If, however, by reason of the method 
 of construction, as often happens, for instance, by the under 
 portion of the beam coming in contact with the frame below, 
 this tipping of D D (Fig. 83) is confined between certain limits, 
 beyond which the supports must be considered fixed, it will be 
 necessary to find the reactions as for a beam over four fixed 
 supports, and determine the corresponding strains in this case- 
 also. 
 
 Comparing, then, the strains obtained each way, we take only 
 the maximum strains from each. 
 
 The formulae for the reactions at \h.Q fixed supports A B C D 
 are as follows (PI. 22, Fig. 84) : 
 
 1st. Load P in left end span A B at a distance a from left: 
 support A, the end spans being n I and the centre span B C=L 
 
 We put Jc = ~ and H = 3 + 8 n + 4 n\ Then 
 n L 
 
 R A =g| H-(H + 2^-h2^)&-f(2rc+2rc, 2 )F 
 
 * Clemens Herschel, in his treatise upon " Continuous, Revolving Draw- 
 bridges" (Little, Brown & Co., Boston, 1875), notices this fact for the first 
 time. 
 
CHAP. XII.] METHODS COMBINED. 185 
 
 R B =j|| (3 + 10^+9 2 +2^ 3 )&-(27i + 5^ 2 +27i 3 )& 3 
 R c r=-| -(n+3n* + 2n*)fc+(n + Sn*+%n*)& 
 
 RD= H| njc ~ n ^ 
 
 These reactions add up, as they should, equal to P. 
 
 In practical cases of pivot spans, we have only to consider the 
 the outer spans ; as a load in the middle span B C = I rests 
 directly upon the turn-table. The above formulae are then all 
 we need. For a load in the right end' span the same formulae 
 hold good, only remembering to put now R D in place of R^ 
 R c in place of R B , R B in place of R c , and R A in place of R D . 
 
 If, however, neglecting the particular case of pivot spans, we 
 suppose the middle span B C = I loaded, we have a being 
 
 now the distance of P from B, and & being now -j instead 
 
 C 
 
 of ^, as above, H remaining the same. 
 2d. Load in B C. 
 
 1 
 
 LH 3 4n-M-4n* *- 6 15 6 2 H 1 
 
 [si 
 J 
 
 r ^ ' _ ^ ^ i 
 L J 
 
 These reactions should also add up to P, as is the case. The 
 number n may be taken at pleasure, so that the end spans may 
 be as much larger or less than the centre spans as is desired. 
 
 H, P and the quantities in the parentheses, it will be observed, 
 are for any given case, constants which may be determined and 
 inserted once for all. 
 
 "We have, then, only to insert the values of 7c for different 
 positions of the load P. Thus the equations for any particular 
 'ase'are very simple and easy of application. 
 
186 GRAPHIC AND ANALYTIC [CHAP. XII. 
 
 123. Construction. We may, if desired, apply our method 
 of construction to the determination of the reactions. Thus 
 from the above reactions we may easily determine general ex- 
 pressions for the inflection points. For the case of a load in 
 C D = n I (PI. 22, Fig. 85), we have, when i is the distance of 
 the inflection point from C, 
 
 -R D x (nl-i) +P(a-i) = 0; 
 
 P a - R D n I 
 
 whence ^ = = - = . 
 
 P KJJ 
 
 For the inflection point distant i from B in the unloaded 
 span, 
 
 hence 
 
 For the second case of load in B C = I, we have for the in- 
 flection point between B and P 
 
 R A (n l+i) + Rji i = 0, or 
 
 i= Knl_ 
 
 For the point between P and C 
 
 R D (n I + i) + R c i = 0, or 
 
 The insertion of the proper values of the reactions for each 
 case, as given above, will easily give general expressions for the 
 inflection points, which the reader may, if desired, deduce for 
 himself. 
 
 Our construction is, then, as follows [PL 22, Fig. 84] : 
 
 \st Case. Load in C D. -. 
 
 Having found %, draw a line at any inclination, as ^ d through 
 & 1? intersecting P at d, and the vertical through C at c^ Then 
 lay off B i and draw d D, c b and ~b A. 
 
 Make d c = P by scale, and c D drawn parallel to ^ d then 
 gives the pole distance H. The ordinates, then, to the broken 
 line A b c : d D taken to scale of distance, multiplied by H to 
 scale of force, give the moments at every point. Moreover, H d 
 is the reaction at D. Draw D b parallel to c 5, then c b is the 
 reaction at C. In like manner a b is the reaction at B, and H a 
 the reaction at A. The moment at C, and reactions at C and 
 
CHAP. XII.] METHODS COMBINED. 187 
 
 D, are positive. Reaction and moment at B negative / reaction 
 at A positive ; as a little consideration of what the curve of the 
 deflected beam must be, will show. The shear at C is, there- 
 fore, + H a a b + b c + H c. The shear at B is H b or 
 4- H a a b, and so on. The shear being always given by the 
 segment between H and lines parallel to A b, b c l} c { d and d D. 
 
 2d Case. PL 22, Fig. 85. Load in B C. 
 
 Having found the distance B i from the equation for this dis- 
 tance of the point of inflection above, we lay off B c\ = B -& t 
 and thus draw c E at an angle of 45. Finding then the value 
 of C i 2 from its equation above, we can draw E c% and then 
 c 2 D and Ci A. The construction is then the same as before. 
 Thus H is the pole distance, H c the negative reaction at D, 
 H b the negative reaction at A, and c l b, c E the positive re- 
 actions at B and C. The shear at B is H c ly etc. Thus the outer 
 forces are completely known for a weight at any point. It will, 
 however, in general, in practice, be found more satisfactory to 
 use the formulae for the reactions which we have given than to 
 find these reactions by the above construction. 
 
 We shall now illustrate the preceding principles by an exam- 
 ple taken from actual practice. 
 
 124. raw Span Example. In PL 22, Fig. 86, we have 
 given to a scale of 20 ft. to an inch the elevation of one of the 
 trusses of the pivot draw over the Quinnipiac Eiver at Fair 
 Haven, Conn.* 
 
 Length of span A B = 89.88 ft. B C = 21.666 ft, divided 
 into seven panels of 12.84 ft. and two of 10.833 ft. respectively. 
 
 Height at B and C, 16 ft. ; at A and D, 12.1 ft. Diagonal 
 bracing as shown in Fig. Line load 9 tons per panel. 
 
 21 ( 
 In this case n = ' - or n = 0.24106, hence the equations 
 
 of Art. 120 become 
 
 A = P (l-1.1298 ?+0.1836 ^ 
 
 B 
 
 = C = P /0.6S36 2-0.1836 
 
 * Designed and erected by Clemens Herschel, C.E., and probably the only 
 structure of the kind in this country for which the strains have been accu- 
 rately and thoroughly determined. For the above data I am indebted to M. 
 Merriman, assistant engineer in charge. 
 
188 GRAPHIC AND ANALYTIC * [CHAP. XII. 
 
 D = -P fo.2374 0.1836 -V 
 
 \ \ I 1? I 
 
 Now -' is -, -, -, -ths, etc., according to the position of the 
 
 L 7 i 7 * 
 
 weight at lst % 2d, 3d apex from end. So also is r , , 
 
 / Otto 343 343, 
 
 etc. The above equations for the reactions, then, may be 
 written A = P (1-0.1614 + 0.000535 3 ), 
 
 B = C = P (0.09766 -0.000535 3 ), 
 D = -P (0.03391 -0.000535 3 ), 
 
 where has the values 1, 2, 3, 4, etc., for P!, P 2 , P 8 , P 4 . 
 
 Thus, if we wish the reactions due to a weight P 4 of 9 tons at 
 the fourth apex, as shown in Fig., we have only to make P 9 
 and 4, and we find at once A = 3.498 tons, B = C = 
 3.207 tons, D = 0.912 tons. The sum of all these reactions 
 exactly equals P, as should be. 
 
 The middle supports are supposed raised by an amount equal 
 to the end deflections of the open draw, therefore the strains 
 due to dead load are easily found, as in the " braced semi-arch," 
 Art. 9. 
 
 The reactions due to live load, according to Art. 121, will not 
 be affected by this raising of the supports. 
 
 To find the strains due to P 4 , we draw the force line F 2 F 
 [Fig. 86 (a)] by laying off P 4 9 tons down from F to F l3 then 
 F! E 2 downwards equal to the negative reaction at L , viz., 
 0.912 tons. Then from E 2 lay off upwards E 2 E t = to positive 
 reaction at C +3.207 tons. Then ^ E = reaction at B = 
 + 3.207 tons, and finally E F equal to reaction at A = +3.498 
 tons, which should bring us back exactly to point of beginning 
 F, since the reactions and the weight P must be in equilibrium. 
 \JN~ote. When we wish to begin at the left end of the frame, 
 it is best, as in this case, to lay off the reactions in order, com- 
 mencing at the right.'] We have taken the scale of force 4 
 tons per inch. 
 
 The weight P 4 acts upon the triangulation drawn full in the 
 figure. Using now the notation of Art. 114, and representing 
 all the space above the truss by E, all lelow by F, we have at 
 A the reaction E F [Fig. 86 (a)] in equilibrium with E 1 and 
 F 1, and drawing parallels to these lines from E and F, we find 
 the strain in F 1 = 3. 54 tons tension, and E 1 = 5.1 compression. 
 
CHAP. XII.] METHODS COMBINED. ISO 
 
 So we go through the truss and find the strains in every 
 piece. Heavy lines in the strain diagram denote compression. 
 We see at once that for this position of the weight, all the 
 upper flanges in span A B are compressed, the last lower flange 
 P 7 is also compressed, and all the other lower flanges are in 
 tension. At the point of application of the weight P 4 , the two 
 diagonals 3 4 and 4 5 are in tension, and either side they alter- 
 nate in strain as far as C or diagonal 8 9. Diagonals 8 9 and 9 10 
 are both tension, and then the strains alternate to support D. 
 All the upper flanges of the right half are tension and increase 
 towards the middle. All the lower are compression and like- 
 wise increase towards the middle. 
 
 If we go through the whole truss from A to D, the last diago- 
 nal 15, 16 should evidently pass exactly through E 2 , thus check- 
 ing the accuracy of the construction. The diagonal 6 7 crosses 
 the force line, thus causing the strain in the lower flange to 
 change from tension in F t 5 to compression in F! 7. The point 
 of inflection, therefore, falls to the right of diagonal 5 6. 
 
 The reaction at B diminishes greatly the strain which would 
 otherwise take effect in 7 8 and E 8 ; while the reaction at O 
 reverses the strain which would otherwise take effect in 9 10 
 and diminishes E 10. We recommend the reader to follow 
 through carefully the strain diagram, Fig. 86 (a). 
 
 A series of figures similar to Fig. 86 (a) (in the present case 
 seven separate figures) will give completely the strains due to 
 the rolling load. A table may then be drawn up containing 
 the strains due to dead load, and the maximum strains due to 
 live load in every piece, and the total maximum tension and 
 compression in every piece may then be found. [Compare 
 Art. 12, Fig. 7.] 
 
 For the supports fixed, instead of B and C tipping, the pro- 
 cess is precisely similar, except that we have to make use of 
 the formulae of Art. 12i4. The reaction at A will then be 
 somewhat less than in the present case ; the inflection point is 
 therefore found further from the right support B ; it may be 
 even to the left of diagonal 5 6, in which case (see Fig. 86, a) 
 we should have tension in upper flange E 6. The reaction at 
 B would then be still positive, but greater than E E 1? while C 
 would be negative and no longer equal to B, and D would be 
 positive. We should thus have 7 8 tension and E 8 tension ; 
 F 7, as before, compression, 8 9 compression, and 9 10 com- 
 
190 GRAPHIC AXD ANALYTIC [CHAP. XII. 
 
 pression, and E 10 compression ; while F 9 would be tension. 
 From 9 10 to the right the diagonals would alternate in strain, 
 the compressed upper flanges, as also the tensile lower flanges, 
 would diminish towards D, and the last diagonal should pass 
 exactly through new position of F 2 , thus closing the strain 
 diagram and checking the work. The reader will do well to 
 construct the diagram. 
 
 The strains should be found for both cases, and the maximum 
 strains taken from each, which, compared with the permanent 
 strains due to the dead load, will give the total maximum 
 strains. 
 
 "We have taken for convenience of size too small a scale for 
 the frame to ensure good results. With a large and accurately 
 constructed ymw e diagram, dealing as we do with only single 
 weights, and consequently small strains, the above force scale 
 of 4 tons per inch would give very accurate results. 
 
 If the strains due to uniform load (no end reactions) are 
 found by addition of the strains for each apex load diagramed 
 separately, the same scale may be employed ; but if all the 
 loads are taken as acting together (Fig. 5, b\ a smaller scale 
 for strains will have to be adopted, as the force line will other- 
 wise be too long. [See Art. 16 of Appendix for the method of 
 calculation.] 
 
 125. Method of passing direct from one Spun to next. 
 
 By inspection of Fig. 86 we see that we might find the strains 
 in the intermediate span B C without first going through the 
 whole of A B or C D. Thus, if we knew the moment at B, 
 this moment^ divided by depth of truss at B, would give the 
 strain in flange F 7 for the system of triangulation indicated 
 by the full lines. If then we knew also the shear at B = 
 P A B = E! F! (Fig. 86, a), we could at once lay off F l 7 
 and E! F t (Fig. 86, a\ and then proceed to find E 8 and 7 8, 
 just as before. In the same way the moment at C, divided by 
 height of truss at C, would give us strain in F 9, and with 
 shear at C = P - A - B - C = E 2 F l = D, we could find E 10 
 and 9 10, as before. As we know already, a load anywhere 
 upon a beam causes positive moments at a fixed end i.e., 
 makes upper flange over support tension and lower flange com- 
 pression. But as we see from the last case, owing to the tri- 
 angulation, the last upper flange may also be compression (see 
 E 6 in Fig. 86) if the inflection point lies between diagonal 5 6 
 
CHAP. XH.J METHODS COMBINED. 191 
 
 and the support. The known moment gives, then, the charac- 
 ter of the 'strain only for that flange which does not meet a 
 diagonal at the support. The moment at B, therefore, being 
 positive, gives us compression here in lower flange, because, for 
 the system of triangulation corresponding to the weight, that 
 flange does not meet a diagonal at B. For a weight upon the 
 other system of triangulation (dotted in Fig.), the same moment 
 would give us the tension in E 7. The construction assumes 
 equilibrium between F 7, 7 8, and E 8, and the shear at B ; 
 that is, between the pieces cut by an ideal section to the right 
 of B through the truss and the shear at that section. That this 
 is so is shown by the strain diagram, since there we see that the 
 strains in these pieces form a closed polygon with the shear at 
 B = E! F x . This must evidently be so if these are the only 
 pieces cut by such a section, since then the horizontal com- 
 ponents of the strains in these pieces must balance, and the 
 resultant vertical component must be equal and opposite to the 
 shear. 
 
 It is important to know which side of E t Pj_ to lay off F! 7, 
 since, if we had laid it off in this case to the right, we would 
 have obtained a very different value for E t 8. For this pur- 
 pose we have only to suppose the strain in the flange (either 
 upper or lower, as the case may be) to be applied at the point 
 of junction or apex of the other two pieces, and then lay it off 
 in the direction with reference to that apex corresponding to 
 the known character of its strain. The direction of the shear 
 is always known from the reactions. 
 
 Thus in our Fig. the shear between B and C acts down from 
 E! to F l5 because P 4 , which also acts down, is greater than the 
 sum of the upward reactions at A and B. The strain in F 7 
 is also known to be compressive, and therefore, in following 
 round the strain polygon commencing from E x to F 1? it must 
 act towards apex at 7. We must, therefore, lay it off to the 
 left of E! Fj. In similar manner, for the other triangulation, 
 the strain in flange E 7 is, in span B C, in equilibrium with 7 8 
 (dotted diagonal) and P 8 and shear E t F 1? and is, moreover, 
 known to be tension. Consider it acting then at B ; and then, 
 since it is tension, we go round the polygon from E t to P t , and 
 then to the right of E x F 1? or away from B, the point at which 
 it is supposed to act. 
 
 Now for the case of the " tipper : " the reaction at D, and 
 
192 GKA.PHIC AND ANALYTIC [CHAP. XII. 
 
 therefore the moment at C, is also positive. The lower flange 
 F 9 is therefore compression, or for the dotted system of trian- 
 gulation E 9 is tension. The shear to the left of C, E t F! acts 
 down, since P + A + B = ^ F lt Consider F 9 acting at 
 apex 9, and then, since it is compression, it must act towards 9 
 (from right to left), and passing down then from E x to F 1? we 
 must lay off F 9 to the left of E x F^ For similar reasons, for 
 the other system, E 9 must be laid off to the right. 
 
 For fixed supports B and C, the moments alternate from B, 
 and the moment at C is therefore negative that is, gives com- 
 pression above and tension below. Flange F 9, for the system 
 of triangulation of P, would then be tension instead of com- 
 pression, as above ; P will, however, still be greater than A + B, 
 and hence the shear is to be laid off down, and F 9 must be 
 laid off to the right. 
 
 If, then, it were required to find the strains in the span B C, 
 preceded and followed, it may be, by many others, it is suffi- 
 cient to know the moment and shear at one support. We can 
 then commence and continue the strain diagram, without being 
 obliged to go off to a distant free end and trace all the strains 
 through till we arrive at the span in question. 
 
 126. Method of procedure for any number of Spans. 
 Let us take, then, any number of spans, say seven [PL 22, Fig. 
 87], and let it be required to find the maximum strains in the 
 span D E. It is not, as we have just seen, necessary to com- 
 mence at the extreme end A or H, and follow the reaction there 
 through, from span to span, till we arrive at D. As we have 
 seen from the preceding Art., we may start directly from D, 
 provided we know the moment and shear there. Now, since a 
 load in any span causes positive moments and reactions at the 
 two ends of that span, and since either way from these ends the 
 moments and shear at the other supports alternate in character 
 [Art. 102], any and all loads in A B cause positive moments 
 and reactions at D. So also for loads in C D and in F G. 
 Loads in B C, E F and G H, on the other hand, cause negative 
 moments and reactions at D. [See Fig. 87.] 
 
 To find the maximum positive moment and shear at D due 
 to the other spans, we must then suppose the method of 
 loading shown in Fig. 87 (a). For the maximum negative 
 moment and shear at D, we have the system of loading shown 
 in Fig. 87 (b). 
 
CHAP. XII.] METHODS COMBINED. 193 
 
 these two moments and shears being once known, we 
 can find by diagram and tabulate the respective strains in every 
 piece of the span D E. Thus dividing the moment at D for 
 either case by the height of truss, we have at once the strain in 
 either upper or lower flange at D depending upon the system 
 of triangulation as explained in Art. 125. With this strain and 
 the shear at D properly laid off to scale, we can commence the. 
 strain diagram precisely as though we had traced all the loads 
 through from the extreme end A or H to D or E. 
 
 We must next find and tabulate the strains in D E due 
 to each apex load in the span itself, and for this we must 
 know to begin with the moments and shears for each separate 
 load. 
 
 [Note. Distinguish carefully between shear and reaction at 
 a support. The shear at D, or at a point just to right of D, is 
 the algebraic sum of all the reactions and weights between that 
 point and A. See also Fig. 84 (Art. 123), where the reaction 
 at B is l> a, but the shear atB is b a + H. a = H b. 
 So also the reaction at C is -f ~b c, but the shear at C is 
 + I c ba + Ha HC, etc.] 
 
 Conceiving n-ow that we have found and tabulated the strains 
 due to the first and second systems of loading as shown in Fig. 
 87, and also the strains for each load P in D E, the sum of these 
 strains will give the strains due to live load o-ver the whole 
 length of girder, and taking the proper proportion of these, we 
 shall have the strains due to the dead load. Combining then 
 these strains with those first found, we can easily find the total 
 maximum strains which can ever occur in D E. 
 
 127. Example. Let us take, as an illustration of the pre- 
 ceding, the girder shown in Fig. 87, of seven equal spans, and 
 seek the maximum strains which can ever occur in the middle 
 span D E. Let Fig. 88, PI. 23, represent the span D E length 
 80 feet, divided into 4 panels ; and let the live load at each 
 apex be 40 tons,* the uniform load being half as much, or 20 
 tons per apex. Height of truss = 10 feet. 
 
 Now the quantities which for the present we must suppose 
 known or already found are as follows : 
 
 * A very great load : half the resulting strains would give more nearly the 
 strains in a single truss. 
 13 
 
194 GRAPHIC AND ANALYTIC [CHAP. XII. 
 
 Positive moment at D (1st system of loading), 
 
 as shown by Fig. 87 (a) .... + 788.2 feet tons. 
 
 Corresponding shear at D = -1- 14.63 tons. 
 
 Negative moment at D 
 
 (2d system of loading) = - 382.54. 
 
 Corresponding shear ~ 14.63. 
 
 Also for the loads in D E : 
 
 For the first load P 1? moment -f 158.92, shear = + 36.17. 
 P, = + 271.96 = + 25.88. 
 P 3 " = + 203.36 = + 14.16. 
 P 4 " = + 62.88 " = + 3.82. 
 
 In Fig. 88 we have found by diagram the strains due to P 3 . 
 [For notation, see Art. 114.] 
 
 We l&y off to scale the shear 14.16 upwards, since it is posi- 
 tive, and then, since the moment 203.36 at D is positive, and 
 hence the strain in A a must be tension, we lay off A a = 
 
 ?H^- = 20.3 tons to the right of B A (Art. 125). With B A 
 
 and A a thus given, we can rapidly and accurately find all the 
 other strains. Thus from our diagram we have, representing 
 tension by minus and compression by plus, 
 
 A a = 20.3 A c + 8.0 A e = + 36.4 A g = + 24.4 
 A ]c = - 27.2 
 
 B6= + 6.0 B<Z=:-22 B/=-50.8 B A = + 1.2 tons ; 
 
 and for the diagonals : 
 
 a &=+19.6 be =-19.6 cd = + 19.6 de=-l9.6 
 
 ef = + 19.6 fg =4- 36.4 g h =-36.4 h Jc = + 36.4 tons. 
 
 Heavy lines in the diagram represent compression. 
 
 In a manner precisely similar we can find the strains due to 
 the other weights, as also to the two systems of loading shown 
 in Fig. 87. Suppose all these strains thus found. Then the 
 method of tabulation is as follows : 
 
CHAP. XII.] 
 
 123 
 
 METHODS 
 56 
 
 10 
 
 11 
 
 195 
 
 12 
 
 DIAGONALS. FLANGES. 
 
 'l 
 
 Live Load in D E - 
 
 Interior Loading 
 Maximum 
 Strains. 
 
 Exterior Load- 
 ing. 
 
 Dead 
 Load 
 =# 
 Live. 
 
 Total 
 Maximum 
 Strains. 
 
 P : 
 
 * 2 
 
 p 3 
 
 ?4 
 
 Tens. 
 
 Comp. 
 
 + 
 
 1st 
 Case. 
 
 2d 
 Case. 
 
 Tens. 
 
 Comp. 
 
 + 
 
 Aa 
 
 - 15.6 
 
 - 27.2 
 
 _ 20.4 
 
 - 6.4 
 
 - 69.6 
 
 
 -78.8 
 
 + 38.2 
 
 -55.1 
 
 - 203.5 
 
 + 63.6 
 
 Ac 
 
 + 16.4 
 
 + 24.4 
 
 + 8.0 
 
 + 1.2 
 
 
 + 50.0 
 
 - 49.5 
 
 + 8.9 
 
 + 4.7 
 
 -44.8 
 
 Ae 
 
 4- 8.8 
 
 + 36.4 
 
 + 36.4 
 
 + 8.8 
 
 .... 
 
 + 90.4 
 
 -20.3 
 
 - 20.3 
 
 + 24.9 
 
 - 15.7 
 
 + 115.3 
 
 Afir 
 A* 
 
 + 1.2 
 
 + 8.0 
 
 + 24.4 
 
 + 16.4 
 
 
 + 50.0 
 
 + 8.9 
 
 - 49.5 
 
 + 4.7 
 
 -44.8 
 
 + 63.6 
 
 - 6.4 
 
 - 20.4 
 
 -27.2 
 
 - 15.6 
 
 - 69.6 
 
 
 + 38.2 
 
 - 78.8 
 
 - 55.1 
 
 _ 203.5 
 
 
 B6 
 
 -20.4 
 
 + 1.2 
 
 + 6.0 
 
 + 2.8 
 
 -20.4 
 
 + 10.0 
 
 + 64.2 
 
 - 23.6 
 
 + 15.1 
 
 -28.9 
 
 + 89.3 
 
 B<1 
 
 -12.8 
 
 -50.8 
 
 -22.0 
 
 - 5.2 
 
 -90.8 
 
 
 + 34.9 
 
 + 5.6 
 
 -25.1 
 
 -115.9 
 
 + 15.4 
 
 B/ 
 
 - 5.2 
 
 - 22.0 
 
 - 50.8 
 
 -12.8 
 
 -90.8 
 
 
 + 5.6 
 
 + 34.9 
 
 - 25.1 
 
 - 115.9 
 
 + 15.4 
 
 B/, 
 
 + 2.8 
 
 + 6.0 
 
 + 1.2 
 
 - 20.4 
 
 -20.4 
 
 + 10.0 
 
 - 23.6 
 
 + 64". 2 
 
 + 15.1 
 
 -28.9 
 
 + 89.3 
 
 a 6 
 
 + 51.2 
 
 + 36.4 
 
 + 19.6 
 
 + 5.2 
 
 
 
 + 112.4 
 
 + 20.7 
 
 - 20.7 
 
 + 56.2 
 
 
 + 189.3 
 
 fo c 
 
 + 5.2 
 
 - 36.4 
 
 - 19.6 
 
 - 5.2 
 
 -61.2 
 
 + 5.2 
 
 -20.7 
 
 + 20.7 
 
 - 28.0 
 
 _ 109.9 
 
 
 <: d 
 
 - 5.2 
 
 + 36.4 
 
 + 19.6 
 
 + 5.2 
 
 - 5.2 
 
 + 61.2 
 
 + 20.7 
 
 - 20.7 
 
 + 28.0 
 
 
 + 109.9 
 
 d e 
 
 + 5.2 
 
 + 19.6 
 
 - 19.6 
 
 - 5.2 
 
 -24.8 
 
 + 24.8 
 
 -20.7 
 
 + 20.7 
 
 0.0 
 
 - 45.5 
 
 + 45.5 
 
 ef 
 
 5.2 
 
 -19.6 
 
 + 19.6 
 
 + 5.2 
 
 -24.8 
 
 + 24.8 
 
 + 20.7 
 
 - 20.7 
 
 0.0 
 
 - 45.5 
 
 + 45.5 
 
 f(J 
 
 + 5.2 
 
 + 19.6 
 
 + 86.4 
 
 - 5.2 
 
 - 5.2 
 
 + 61.2 
 
 -20.7 
 
 + 20.7 
 
 + 28.0 
 
 .... 
 
 + 109.9 
 
 gh 
 
 - 5.2 
 
 -19.6 
 
 -36.4 
 
 + 5.2 
 
 - 61.2 
 
 + 5.2 
 
 + 20.7 
 
 - 20.7 
 
 -28.0 
 
 - 109.9 
 
 
 hk 
 
 - 5.2 
 
 + 19.6 
 
 + 36.4 
 
 + 51.2 
 
 
 + 112.4 
 
 -20.7 
 
 + 20.7 
 
 + 56.2 
 
 
 + 189.3 
 
 Having found and tabulated the strains clue to each weight 
 as shown by the first five colums of the table, add all the ten- 
 sions and compressions for each piece and place the results in 
 columns 6 and 7.* ^Ve thus have the maximum strains of each 
 kind which can be caused by the weights in span D E alone. 
 In the next two columns, 8 and 9, place the strains due to the 
 two cases of loading of Fig. 87. Now if the uniform or dead 
 load is taken at one half the live, we have simply to take the 
 algebraic sum of the strains in columns 6, 7, 8, and 9 horizon- 
 tally, and divide by 2. We thus find column 10. Finally, 
 from columns 6, 7, 8, 9 and 10, we can find the total maximum 
 
 * A more convenient form of table, perhaps, is obtained by putting the 
 weights Pj_ 4 in the vertical left-hand column, and the pieces A a, Ac, 
 etc., in the top horizontal line. The numbers are thus more conveniently 
 placed for add tion. 
 
196 GKAPHIC AND ANALYTIC [CHAP. XII. 
 
 strains as given in the last two columns. Thus take the piece 
 A c. In this piece there is a constant compression due to dead 
 load of 4.7 tons. The second system of loading adds to this 
 for live load 8.9 tons, and the loading of D E itself causes 50 
 tons compression. Since all three cases may exist together, we 
 have 4.7 + S. 9 + 50 63.6 tons compression. Again the first 
 system of loading, which may act also alone, causes 49.5 tons 
 tension in A c. Diminishing this by the 4.7 tons compression 
 dne to dead load already existing, we have 49.5+4.7 =44.8 
 tons tension. These two strains are the greatest which can ever 
 occur in A c. 
 
 For A a we see that the greatest compression is due to the 
 second system of loading acting alone, but this compression = 
 '38.2 tons, is less than the tension always existing in A a from 
 the dead load, which is 55.1 tons. Compression, then, can never 
 occur in A a. So also for A Jc. In like manner diagonals a , 
 b c, c d, y <7, g A, and h k do not need counterbracing, as the 
 constant dead load strain overbalances any strain of opposite 
 character which can ever come upon them. 
 
 As we have in the present example taken a middle span, 
 observe that the strains of P t and P 4 , P 3 and P 2 are similar. Thus, 
 strain in A a due to P 4 is the same as in A k due to P l5 and so on . 
 
 1. Method of Moments. We can very easily check our 
 results by the method of moments of Art. 14. 
 
 Thus, for the first system of loading, we have the moment at 
 D = +788.2 ft. tons and the shear =14.63. For any upper 
 flange, then, as A g, since the positive moment causes tension 
 in upper fiange, and positive shear causes compression, we have, 
 taking apex g as centre of moments, 788.2 + 14.63 x 60 = 
 - 788.2 + 877.8 = + 89.6. Dividing this by depth of truss 
 = 10 ft., we have 8.96 tons compression in A g. 
 
 For the strain in By due to the weight P 2 , since the moment 
 at D for this case is +271.96 and shear =+25.88, and the 
 positive moment and shear cause compression and tension re- 
 spectively in lower flanges, while the weight P 2 =40 causes 
 compression in By we have, taking f as centre of moments, 
 + 271.96-25.88x50+40x20 = -222.0:1:, rind dividing this by 
 10, we obtain 22.2 tension in By, and so on. 
 
 For the diagonals, the shear at any apex, multiplied by the 
 secant of the angle with the vertical, gives at once the strain. 
 In this case the angle is 45; therefore the secant is 1.414. 
 
CHAP. XII.] METHODS COMBINED. 197 
 
 Hence, for strain in c d due to P 2 , we have 
 25.88x1.414:= +36.6 tons. 
 
 The calculation, then, of the strains in any span of a continu- 
 ous girder, as also the diagraming of these strains, is simple and 
 easy, and offers no more difficulty than in the case of a simple 
 truss, provided we know or can find the moments and shearing 
 forces at the supports for the various cases of loading. The 
 method of finding these necessary quantities will form the sub- 
 ject of the next chapter. 
 
 The reader will do well to compare the strains in the above 
 table with those for the same simple girder similarly loaded.* 
 
 It will be found that there is a saving of material in the 
 flanges of about eleven per cent, over the corresponding simple 
 girder. Some of the flanges are, to be sure, subjected to both 
 tensile and compressive strains, instead of being always of the 
 same character ; but in wrought-iron girders this is of little con- 
 sequence. 
 
 It is worthy of remark that a slight relative difference of level 
 of the supports of a continuous girder may cause very great 
 changes in the strains, and hence, in structures of the kind, the 
 foundations must be secure from settling, and the pier sup- 
 ports accurately on level. Under these circumstances, .where 
 long spans are desirable, the continuous girder is to be preferred 
 to a succession of single spans. 
 
 If the greatest negative reaction at any support, as E (Fig. 87), 
 is greater tnan the constant positive reaction at that point due 
 to the dead load, the girder will require to be latched or held 
 down at that support. 
 
 * See Appendix, Art. 16, where this comparison is made. 
 
198 CONTINUOUS GIEDEE. [CHAP. XIII. 
 
 CHAPTER XIII. 
 
 ANALYTICAL FORMULAE FOE THE SOLUTION OF CONTINOUS GIEDEES. 
 
 129. Introduction. As we have seen in the preceding 
 chapter, the complete and accurate determination of the strains 
 in the continuous girder, both for uniform and moving loads, is 
 easy, provided we can find the moments and shearing forces at 
 the supports for the various states of loading,, and for each apex 
 load. Now this we are able to do with mathematical accuracy, 
 and without much labor. The formulae necessary for the pur- 
 pose, when put into proper shape for use, are neither difficult of 
 application nor more complicated than many which the practi- 
 cal engineer is often called upon to manipulate. Since the 
 publication of Clapeyron's paper * in 185 7, in which, for the 
 first time, his well-known method was developed, and his cele- 
 brated "theorem of three momenta" made known, the subject 
 has engaged the attention of many mathematicians. In 1862 
 WinJder f first developed a general theory, and gave general 
 rules fo? the determination of the methods of loading causing 
 greatest strains, together with tables for the maximum values 
 of the moments, shearing forces, etc., for various numbers of 
 spans of varying length. In the same year Bresse \ followed 
 with a similar work. In 1867 WinHer gave a general ana- 
 lytical theory, and, finally, in 1873 Weyrauch \ has treated the 
 subject with a degree of completeness and thoroughness which 
 leaves but little to be desired. He discusses the subject in its 
 most general form, for any number of spans of varying length, 
 
 * Clwpeyron Calcul d'une poutre elastique reposant librement sur des ap- 
 puis inegalement especes. Compte rendus, 1857. 
 
 f Beitrage zur Theorie der continuirlichen Briickentrager Civil Ingenieur, 
 1862. 
 
 \ Bresse Cours mechanique appliquee. Paris, 1862. 
 
 Winkler Die Lehre von der Elasticitat und Festigkeit. Prag. 1867. 
 
 H Weyrauch Allgemeine Theorie und Berechnung der continuirlichen und 
 einfachen Trager. Leipzig, 1873. 
 
CHAP. XT!!.] ANALYTICAL FORMULA. 199 
 
 and for all kinds of regularly and irregularly distributed and 
 concentrated loads both for constant and varying cross-section 
 of girder. His formulae are mathematically exact, and for 
 given loading are free from integrals. 
 
 The above is but a very imperfect sketch, and we have named 
 but a few of the many writers who have been occupied with 
 the subject. Clapeyrorfs Theorem above alluded to, as origi- 
 nally given by him, applied only to uniform load over whole 
 length of girder, or over an entire span. But as early as JBresse's 
 Treatise, it had been extended to include concentrated and 
 local loads as well, and Winkler has also given a very complete 
 and practical discussion of the subject. 
 
 Notwithstanding the labors of these and many other mathe- 
 maticians, there seems to be a wide-spread idea, even among 
 those who are supposed to have considerable familiarity with 
 mathematical literature, that the results deduced are unpracti- 
 cal. It is not uncommon to meet with even recent publica- 
 tions* in which it is stated that the authorities pass over such 
 problems with "judicious silence;" that the mathematical in- 
 vestigations are intricate, and the formulae deduced trouble- 
 some in application ; that even a a partial solution of the prob- 
 lem by mathematical calculation is attended with considerable 
 difficulty, and that a complete solution for the bending moment 
 and shearing force at every section, under moving partial and 
 irregular loads, taxes the powers of the best mathematicians, 
 and is well-nigh impossible, so far as any practical application 
 of them by the engineer is concerned." How far such ideas 
 are justified may be seen from the following pages. That the 
 authors and works above referred to can only be read by good 
 mathematicians is not to be denied. It may also be admitted 
 that the subject is an intricate one, and when treated mathe- 
 matically in its most general form the results are naturally in an 
 unpractical shape. But that these results are, therefore, worth- 
 less, or that the formulae, when applied to any particular case, 
 are " too intricate for practical use," by no means follows. 
 
 The desirability of formulae for the application of our graph- 
 ical method as developed in the preceding chapter ; the erro- 
 neous ideas prevalent on the subject which we have just noticed ; 
 
 * Graphical Method for the Analysis of Bridge Trusses : Greene. D. Van 
 Nostrand, publisher, New York. 1875. 
 
200 CONTINUOUS GIRDER [CHAP. XIII. 
 
 and the deplorable fact that the " authorities " do but too often 
 treat the subject with "judicious silence," and that, therefore, 
 there exists in our engineering literature no collection of prac- 
 tical and useful formulae for this important class of bridges, 
 though such formulae are, and have been for years, free to all 
 for the asking, all these facts may serve as apology for the in- 
 troduction of the present chapter, in a work which professedly 
 treats only of Graphical methods. The apologies of those who 
 professedly treat the subject analytically, and have yet omitted 
 such formulae, are not so numerous. 
 
 "We propose to give the analytical results necessary for com- 
 plete solution of a girder of uniform cross-section over any 
 number of level supports, with end spans any desired ratio of 
 intermediate ; for uniform load over whole length from end to 
 end of girder, for uniform load over any single span, and for 
 concentrated load in any single span at any point of that span. 
 These three cases, as we have seen in the preceding chapter, 
 are sufficient for the complete solution of framed Bridge 
 Trusses.* 
 
 Many of these results are here given for the first time, at least 
 in their present shape, in any published treatise, though, as re- 
 marked, some of them in more or less practical form have 
 long been common property for all who may have desired to 
 make use of them. 
 
 The formulae only will be given, in such shape and with 
 such illustrations of their application that, we trust, they will 
 be found free from complexity, and of considerable practical 
 importance. In the Supplement to this chapter a demonstra- 
 tion of the formulae is presented. 
 
 13O. Notation. The notation which we shall adopt is as fol- 
 lows [see Fig. 89, PL 23] : 
 
 * The formulae for concentrated loads are alone all that is really necessary. 
 Their addition gives, as we have seen in our tabulation, Art. 127, the strains 
 for uniform load also. In fact, for strict accuracy, only single isolated loads 
 should be considered, as the results given by the formulae for uniform load are 
 not perfectly accurate. This may be seen from the well-known fact that, for 
 a girder fixed at one end, and supported at the other, the reaction at the free 
 
 5 
 end for a load in the middle is -TO. of the load, while if the same load were uni- 
 
 3 6 
 
 formly distributed, the reaction is ^ths, or ^ of the load. The difference, 
 
 however, for any practical case, where there are a number of panels, is very 
 Blight. 
 
CHAP. XIII.] ANALYTICAL FOKMULJE. 201 
 
 Whole number of spans is indicated by s ; 
 
 Hence, whole number of supports is s + 1 numbered from 
 left to right. 
 
 Number of any support in general, always from left is m. 
 
 The supports adjacent to a loaded span left and right are 
 indicated by r and y + 1. 
 
 When extreme end spans vary in length from the interme- 
 diate, they are always denoted by n I, where n is a given fraction 
 or ratio for any particular case. Thus, if intermediate spans 
 are all 70 feet and end spans 50 feet, I = 70, n I = 50, and n 
 
 When spans next to ends also vary, they are similarly denoted 
 
 All other spans are of equal length and denoted by I. 
 
 The length of the span in which the load is supposed to be 
 is in general 4> where the value of r for any particular case in- 
 dicates the number of the loaded span from left. 
 
 A concentrated load is indicated by P. 
 
 Its distance from nearest left-hand support, by a. 
 
 The ratio of a to length of loaded span Z r , is It = . 
 
 4 
 
 Moment at any support in general is M mJ where m may be 
 1, 2, 3, r, r 4- 1, s, etc., indicating in every case the moment at 
 corresponding support from left. 
 
 In same way reaction at any support is Rjn, shear S m . 
 
 At supports adjacent to loaded span, then, we have M r , M r+1 , 
 R r) R^, S r , S r+1 , for the moments, reactions, and shears at those 
 supports. 
 
 A dead uniform load is u per unit of length. 
 
 A uniform live load, w per unit of length. 
 
 w 4x, then, indicates a uniform live load over any span. 
 
 These comprise all the symbols we shall have occasion to use* 
 By reference to Fig. 89, the reader can familiarize himself with 
 their signification, and will then find no difficulty in under- 
 standing and using the following formulae. Certain symbols 
 which we shall use for expressions of frequent occurrence, will 
 be best explained as we have occasion to introduce them. 
 
 131. <4 Theorem of Three Moments.'' This remarkable 
 Theorem, due to Clapeyron, expresses a relation between the 
 moments at any three consecutive supports, both for uniform 
 
202 CONTINUOUS GIRDER. [CHAP. XIII. 
 
 load over whole length of girder from end to end, and for uni- 
 form load over the whole of any single span. It may be writ- 
 ten as follows : 
 
 M m l m + 2 M m+1 (l m+ l m+1 ) + M m+2 Z m+1 = [ + + J. 
 
 If we suppose only one of the two adjacent spans as l m to 
 contain the full live load -10, while all the spans are of course 
 covered with the dead load u, the above equation becomes 
 
 +1 
 
 m ' * - " A m -t- 1 i_-iu -1U + J.J in +4 w ia + i 
 
 If both spans bear the same uniform load u alone, 
 
 ]\I m 1+2 ]VT m i\l + I i"] + Hff I = \f-\-f "I 
 
 If the spans are equal, the above two equations become re- 
 spectively 
 
 n .. l O . 
 
 M m Z + 4M m+1 1 + 
 and 
 
 M m I + 4 M m+1 I + M m+2 I r&?~* 
 
 Now in every continuous beam, whose extreme ends are not 
 fixed, two moments are always known, viz., those at the ex- 
 treme supports, which are always zero. Hence, by the applica- 
 tion of this theorem, we can" form in any given case as many 
 equations as there are unknown moments, and then, by solving 
 these equations, can determine the moments themselves. 
 
 132. Elx&mple Total uniform Load all Spans equal. 
 Thus let it be required to find the moments at the supports for 
 a beam of seven equal spans, uniformly loaded over its whole 
 length. The moments at the end supports M t and M y are zero. 
 We have then, by the application of the last equation above, the 
 following equations : 
 
 Fdr the first three supports 1, 2 and 3, m 1, and 
 
 4 Mg I 4- M s I ^ , or 4 3YT 2 -f- M 3 = . 
 For supports 2, 3 and 4, ny, = 2, and 
 
 M 2 + 4 M 3 + M 4 = " 
 
CHAP. XIII.] ANALYTICAL FORMULAE. 203 
 
 For supports 3, 4 and 5, m 3, and 
 
 or since in this case the moments equally distant each way 
 from the middle are equal, this last equation becomes 
 
 M 3 + 4 M 4 + M 4 = ?tf. 
 
 a 
 
 We have therefore three equations between three unknown 
 moments, M 2 , 3YT 3 and M 4 , and by elimination and substitution 
 can easily find 
 
 1 9 
 
 If, as in our example of Art. 127 in the preceding chapter, 
 we take u = 1 ton per ft., I = 8'0 ft., then u 1? 6400, and the 
 moment at the fourth support becomes 540.8. If the height of 
 truss is ten feet, this gives [Fig. 88] 54.1 tons strain in the 
 upper flange A a. By reference to our tabulation, Art. 12T, 
 we see that this agrees closely with strain in A 1 due to uni- 
 form load, found in a manner entirely different, viz., by sum- 
 mation of the strains due to first case of loading, and the several 
 loads in -the span itself, and serves therefore as a check upon 
 our results. 
 
 133. Triangle of Momciit. For the benefit of the practi- 
 cal engineer, who may object to the algebraic work involved in 
 elimination of the unknown moments from the equations above, 
 when the number of spans is great, we offer the following tabu- 
 lation, from which he may easily and directly determine the 
 moments at the supports for any "desired number of spans 
 without formulce or calculation. 
 
 Thus, if we were in the above manner to find the moments 
 for a number of spans, and tabulate our results as given in the 
 annexed table, an inspection of the table will show us that we 
 can produce it to any extent desired without further calcula- 
 tion. 
 
204 
 
 CONTINUOUS GIRDER. 
 
 [CHAP. XIIL 
 
 MOMENTS AT SUPPORTS TOTAL UNIFORM LOAD ALL SPANS EQUAL. 
 
 Coefficients of u I 2 given in triangle. 
 
 The Roman numerals along the sides of the triangle indicate 
 tfye number of spans, and the horizontal line to which they be- 
 long give the moments. Thus, for our example of seven spans 
 just worked out, we have the extreme moments M! and M 8 = 0, 
 
 15 
 
 M 2 and M 7 = u Z 2 , etc. 
 
 Now, a simple inspection of this table will show us that for 
 any even number of spans, as VIII., for example, the numbers in 
 the horizontal line are obtained by multiplying the fraction 
 above in any diagonal column, both numerator and denomina- 
 tor, by 2, and adding the numerator and denominator of the 
 fraction preceding that. 
 
 Thus, 
 
 15 x2 + 11 
 
 41 
 
 142 x 2 + 104 388 ' 142 x 2 + 104 
 
 30 _ 15 x 2 
 388 "" 142x2 
 
 12 x 2 + 9 33 
 m the other diagonal column ; j^~2 + 104 == 388 r 
 
 11 x 2 + 11 . 
 
 = :r77r~ T777 m the other diagonal column ; and so on. 
 142 x 2 + 104 
 
 For any odd number of spans, as IX., we have simply to add 
 numerator to numerator and denominator to denominator, the 
 two preceding fractions in the same diagonal column. 
 
CHAP. XIII.] ANALYTICAL FORMULAE. 205 
 
 41 H- 15 56 33 + 12 30 + 15 45 
 
 I n n Q __ Ol* _ _ _ 
 
 ' 388 + 142 7 530' 388 + 142' " 388 + 142 530' 
 and so on. 
 
 We can, therefore, independently of the theorem and analyt- 
 ical method by which the above results were deduced, produce 
 the table to any required number of spans.* 
 
 134. Total uniform L,oad all Span equal Reactions. 
 The moments being known, the reactions at the supports can 
 be very easily found. 
 
 Thus, the reaction at the first or last support is 
 
 _ul M s 
 
 ' : T~ T ; 
 
 at any other support 
 
 Thus, in our example in Art. 127, we find 
 
 56 161 _ 137 143 
 
 Hence, the shear at the fourth support is 
 
 56 161 137 143 . . 71 
 
 71 
 or when u I = 80 tons, - ul = 4:0 tons. 
 
 Multiplying this shear by 1.414 (the secant of the angle with 
 vertical), we find for the strain in diagonal a b (Fig. 88) due to 
 uniform load + 56.5 tons, the same nearly as already found in 
 our tabulation. 
 
 135. Triangle for Reactions. The reactions for a number 
 of spans being found, and tabulated, as above, in the case of the 
 moments, we sjiall have a triangular table precisely similar to 
 the one above, in which the same rule holds good for odd and 
 even numbers of spans. 
 
 * The above relations between the moments can be shown analytically to 
 be a result of the properties of the well-known " Clapeyronian numbers." 
 For the table above, as also the others which follow, we are indebted to the 
 kindness of Mr. Mansfield Merriman, Instructor in Civil Engineering in the 
 Sheffield Sci. School of Yale College. They are given, so far as we are aware, 
 in no treatise upon the subject yet published. 
 
206 
 
 CONTINUOUS GIRDER. 
 
 [CHAP. xm. 
 
 REACTIONS AT SUPPORTS TOTAL UNIFORM LOAD ALL SPANS 
 
 EQUAL. 
 
 Coefficients of u I given in triangle. 
 
 vni. 
 
 VII. 
 
 VII. 
 
 VIII. 
 
 "We are thus able to find both moments and reactions at the 
 supports for any number of spans, so far as uniform loading is 
 considered, and may then either diagram the strains in the 
 various pieces or calculate them as explained in Arts. 127 and 
 128. No formulae are required. Any one who understands 
 the method of moments as applied to simple girders can, by 
 the aid of the two tables above, find accurately the strains in 
 every piece of a girder, continuous over as many equal spans as 
 is desired, and uniformly loaded over its entire length, all sup- 
 ports being on the same straight line. 
 
 As we have seen, Art. 127, this is one of the cases which 
 must be considered in order to find the maximum strains in 
 any span,* and the results above given for its solution will, we 
 trust, be found by the practical engineer to be neither " com- 
 plex " nor " difficult of application." 
 
 136. Clapeyroiiian Numbers. In the analytical discussion 
 of continuous girders, certain numbers having many remarka- 
 ble properties play a very important role. 
 
 We have seen that the theorem of three moments furnishes 
 us with as many equations between the moments as there are 
 moments to be determined. For a small number of supports, 
 
 * See note to Art. 129. 
 
CHAP. XIII.] ANALYTICAL FORMULAE. 207 
 
 these equations can, be solved by the ordinary rules of algebra ; 
 but for a great number, or in the general analytic discussion of 
 any number, we must have recourse to a special artifice. Tims 
 we multiply our equations, beginning with the last, by numbers 
 indicated by <? 1? <? 2 , c 3 , . . . . c^_i, and then choose these num- 
 bers such that, by the addition of all the equations, all the JVC's, 
 with the exception of M l5 disappear. We thus easily determine 
 M! without the tedious process of substituting from one equa- 
 tion to the other, through the entire list. 
 
 The following relations must then evidently hold between 
 these numbers, as is evident from the theorem of three mo- 
 ments of Art. 131 : 
 
 2< 
 
 \ (4-i + 4) + <? 2 i 
 4-i + 2 C2 (4_ 2 + 
 
 Li = o. 
 
 4-2 = o. 
 
 
 
 
 
 tf s -3 4 + 2 c s ^ (4 + 4) + c s _, 4 = o. 
 
 If the first number is chosen at will, say 1 3 the other num- 
 bers can be found from these equations. 
 
 Now in the present case of all spans equal, we have between 
 any three of these numbers the relation : 
 
 Cm-l + 4 C m + G m + l = 0. 
 
 If we take the first, c t 0, and the next, c% = 1, we have 
 for the others the following values : 
 
 Gl = c, = + 15 c, = - 780 CM = + 40545 
 
 c 2 = -f 1 c 5 = - 56 <% = + 2911 c u = - 151316 
 
 c 3 zn 4 c 6 = + 200 c 9 = - 10864 c 12 = + 564719 
 
 These are the so-called Clapeyronian numbers. They alter 
 nate, as we see, in sign, and each is numerically 4 times the 
 preceding minus the one preceding that. We shall always indi- 
 cate these numbers by the letter <?, the index denoting the par- 
 ticular one. Thus, c 7 is the seventh number, counting and 
 1 as the two first. 
 
 No table of these numbers is needed. The index being 
 given, any one can write down the series for himself, till he 
 arrives at the desired number. 
 
 137. Uniform Live Load over any single Span Moments 
 at Supports of Loaded Span. These numbers being pre- 
 mised, we can now give the following formulse for the moments 
 at the supports r and r + 1 of the uniformly loaded span : 
 
208 CONTINUOUS GIRDER. [CHAP. XII'. 
 
 For tl.e left support, 
 
 M --- W P | g r g S -r+2 + 4. 6%_ r + 1 | 
 
 L c +i J 
 
 For the right support, 
 
 M r+1 = - -W P pr g s-r + l 
 
 These formulae, it will be seen, are very simple an<} easy of 
 application. 
 
 Thus, for seven spans, load over the fourth from left, we 
 have s = 7, r = 4, and hence 
 
 1 72 F^ <"5 + <4 '4! 
 
 --wP ~ 
 
 Both moments are equal, as should be the case for a middle 
 span. Inserting now the proper values for the Clapeyronian 
 numbers from the preceding Art., we have 
 
 x 56 4-15 2 "! 615 
 
 So for any desired number of spans, the values of r and s 
 being known, the corresponding Clapeyronian numbers can be 
 easily found, and, inserted in our formulae, give us at once the 
 moments at the supports. 
 
 Turning again to our example, Art. 127, and making w = 2 
 tons, and I = 80 ft., we have w Z 2 = 12800, and therefore M 4 = 
 676, and dividing by depth of truss 10 ft., we find the strain 
 in A a (Fig. 88) 67.6 tons, nearly what we have found by the 
 summation of the strains due to the loads P t _ 4 in our tabulation. 
 
 138. Triangle of Moments Uniform Live Load over any 
 single Span. If from the above formulae we find the moments 
 at supports for a number of spans, and tabulate as before, we 
 shall have a triangle of moments similar to those already given, 
 which may be produced to include any desired number of 
 spans. We have only to observe that the numerator or de- 
 nominator of any fraction in the table follows the law of the 
 Clapeyronian numbers that is, is four times the preceding in 
 the same diagonal column minus the one preceding that. 
 
CHAP. XIII.] ANALYTICAL FORMULAE. 209 
 
 MOMENTS AT SUPPORTS OF LOADED SPAN. UNIFORM LIVE LOAD 
 
 OVER ANY SINGLE SPAN. 
 Coefficients of w I' 2 given in triangle. 
 
 m, - 153 x 4- 41 571 , 
 
 Thus for seven spans, - _ = , and so on. 
 
 The triangle above gives the moments for uniform load over any 
 span, both right and left. For left supports we have simply 
 to count the span from right to left. Thus for seven spans for 
 load in the sixth span from left, we have moment at left-hand 
 
 627 
 support = - w Z 2 , counting the spans from left to right in 
 
 triangle. For the moment at right support of same loaded 
 span, we count six the other way from right to lejct, and find 
 
 571 
 
 13d. Moments at Supports of Unloaded Spans. The 
 
 triangle and formulae above give the moments at the supports 
 of the loaded span only, both positive that is, always tending 
 to cause tension in upper flange and compression in lower. 
 
 If m represents the number of any support counting from 
 the left, the moments at any support generally may be found 
 by the following formulae : 
 
 When m<r + l, M m == \w V p m Gs ~ r+ * + ^ m Cg - r+1 1 
 
 g_l + 4: C B 
 
 When m>r M^l^Z 2 r c s - m+2 + <Vn M +i1 
 
 <?s_i + 4 <? a J 
 
 If we make in these formulas m = r in the first, and m = 
 14 
 
210 CONTINUOUS GIKDEK. [CHAP. XIII. 
 
 r 4- 1 in the second, we obtain the formulse of Art. 137. For 
 any other support left of r, or right of r + 1, we have only to 
 give the proper values to m, s and r for any given case, and 
 find the corresponding Clapeyronian numbers. 
 
 14O. Practical Rule, and Table. The moments at the 
 supports of the loaded span having been found by the formulse 
 of Art. 137, or the triangle of moments of Art. 138, instead of 
 using the above formulse, we may find the moments at the 
 other supports as follows : 
 
 For all supports left of the loaded span: Commencing at 
 the left end support, place over each support the Clapeyronian 
 numbers 
 
 1 4 15 56 209 780, etc. 
 Take the last number thus obtained, before reaching the left 
 support of the loaded span, as a common denominator. Then 
 the moment at the left end is of course zero. At the second 
 support 1, at the third 4, at the fourth 15, at the fifth 56, etc., 
 all divided by this common denominator, will express the frac- 
 tional part of the moment at the left support of the loaded 
 span, which the moment at the support in question is. For the 
 moments at the supports right of the loaded span, proceed simi- 
 larly, only count from the right end. 
 
 Thus, for a girder of ten spans, sixth span from left loaded : 
 The moments M 6 and M 7 due to load being found, suppose we 
 wish the moments left of M 6 . Commencing at left end, num- 
 ber the supports 1, 4, 15, 56, 209. (Let the reader draw a figure 
 representing the case.) The number 209 is the last before 
 reaching the sixth support. We take this, therefore, for a^com- 
 
 mon denominator. Then we have M^ = M 2 = M 6 
 
 209 
 
 So for supports to the right of support 7, we have 
 
 Remembering that M 6 and M 7 are both positive, and that 
 the moments alternate either way from these supports, we find 
 easily the proper signs for the moments right and left. 
 
 We can now, therefore, find the moments at D and E due to 
 the first and second cases of loading of Fig. 87 (Art. 126). 
 
CHAP. XIII.] 
 
 ANALYTICAL FORMULAE. 
 
 Let us take the first case. For load on A B, we have from 
 
 /-rorv 
 
 our triangle or formulae the moment at B = w P. At D, 
 
 , 56 780 
 then, we have * 
 
 56 
 
 wl?. 
 
 For load on C D, we have at once from triangle the moment 
 
 Finally, for load on F G, we have for moment at F, from 
 
 * 1 627 
 
 triangle = ^r^-rr w 
 
 11644 
 
 therefore at D, ^ x ^ P = 
 
 45 
 
 11644 
 
 All these moments at D are positive ; we have therefore, for 
 
 the first case of loading, the total moment at D 4- 
 
 717 
 
 w 
 
 If we make I = 80 ft. and w = 2 tons, we find the moment 
 at D = 788, and dividing by 10 we obtain 78.8 tons as the 
 strain in A a, Fig. 88, corresponding with our tabulation, Art. 
 127. 
 
 Table for all the Moments. 
 
 All the moments may be found from a simple table similar 
 to the following, which will be found perhaps preferable to the 
 triangle of Art. 138. 
 
 TABLE FOE MOMENTS. UNIFORM LOAD IN SINGLE SPAN. 
 
 Support counted from Left. Denominator A 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 1 
 
 
 
 1 
 
 4 
 
 15 
 
 56 
 
 209 
 
 I. 
 
 
 
 3 
 
 12 
 
 45 
 
 168 
 
 627 
 
 II. 
 III. 
 
 
 
 11 
 
 44 
 
 165 
 
 616 
 
 
 
 
 41 
 
 164 
 
 615 
 
 
 
 IV. 
 
 
 
 153 
 
 612 
 
 
 
 
 V. 
 
 
 
 571 
 
 
 
 
 
 VI. 
 
 I 
 
 Spans. 
 
 iA. 
 
 1 
 
 1 
 
 2 
 
 4 
 
 3 
 
 15 
 
 4 - 
 
 56 
 
 5 
 
 209 
 
 6 
 
 780 
 
 7 
 
 2911 
 
212 CONTINUOUS GIEDEK. [CHAP. XIII. 
 
 This table, it will be observed, can be produced to include 
 any number of supports desired. The law of the Clapeyroniaii 
 numbers runs both horizontally and vertically. The smaller 
 table gives the denominator, the larger the numerator of the 
 coefficient of w ft for any case. Thus, for seven spans we have 
 four times 2911 = 11644 for the common denominator. For 
 load on second span from right, moment at sixth support from 
 
 627 
 left, we have then directly w P ; for fourth support from 
 
 45 
 left, w P, the same as above. 
 
 For load in fifth span from right, the table gives us at once 
 
 612 
 and + ^ -r-r w P, for supports 1, 2 and 3. 
 
 For the other supports, since if now we were to continue count- 
 ing from left we should have to pass a loaded support, we 
 must count the loaded span from left, and count the supports 
 in reverse order. For fifth support from right, then, the num- 
 ber required is at intersection of III. (instead of Y.) and 5, or 
 
 (\~\ A 
 
 5-j-j w P, as found above. Thus the tables above cover all 
 11644 
 
 cases, giving supports at loaded span itself, as also right and 
 left of this span. We have only to remember to count sup- 
 ports from left, and loaded span from right, for all supports 
 left of load, and inversely for all supports right of load. [The 
 reader should always, when using Table, make a sketch of the 
 given number of spans, indicate the loaded span, and number 
 the supportsl\ 
 
 141. Reaction* at Supports Live Load over single Span. 
 For the reactions at ends of loaded span, we have 
 
 For reactions at extreme ends, when end spans are loaded, or 
 
 w 
 
 When any other spans are loaded, or 
 
 when r > 1 and < s, R, = - ??? R s + 1== _^?!. 
 
CHAP. XIII.] ANALYTICAL FORMULA. 213 
 
 For all other reactions, 
 
 Thus for load covering the first span of seven spans, we find 
 from the known moment, given in preceding Art., for the fourth 
 support, 
 
 6 x 56 336 
 
 R ^n^ wl = u^ wL 
 
 For load over third span from left, 
 
 6 x 616 1 , 6607 
 
 For load on sixth span, 
 
 Hence, total reaction at fourth support for first case of loading 
 
 7213 
 is jpj-rr; w & I n the same way we can find the reactions at the 
 
 first, second, and third supports, for the second case of loading, 
 as shown in Fig. 87, and then can easily find the shear at any 
 support, as D, by taking the algebraic sum of all the reactions 
 and loads between that support and the end. 
 
 We can now, therefore, find the shear and moment at D, and 
 thus determine the strains in the span D E for both cases of 
 loading, as given in our tabulation, Art. 127. 
 
 142. Triangle for Reactions Single Span loaded. If 
 we calculate from our formulae the reactions at supports of 
 loaded span, for a number of spans, we can tabulate the results, 
 as on next page, in a triangle, where each number is four 
 times the preceding minus the one preceding that, all in the 
 same diagonal column. 
 
214 CONTINUOUS GIRDER. [CHAP. XIII. 
 
 REACTIONS AT SUPPORTS LIVE LOAD OVER SINGLE SPAN. 
 Coefficients of w I given in triangle. 
 
 This triangle, similar to the preceding one for moments, 
 gives the moments at the left support of the loaded span, when 
 we count from left to right. Counting the other way, we have 
 the reactions at the right support of the loaded span. 
 
 Thus for six spans, fourth span from left loaded, we count 
 
 1770 
 
 four from left in horizontal line for VI., and find w I for 
 
 3120 
 
 reaction at left support. For reaction at right, we count four 
 also from right end, and find w I. 
 
 143. Reaction in unloaded Spans Load over one Span 
 only Table. The formulae of Art. 141 for the reaction at 
 any unloaded support are sufficiently simple and easy to apply ; 
 still we may, if thought preferable, also draw up tables for 
 these, to be used in connection with the triangle of the pre- 
 ceding Art. The following tables give the coefficients of w I 
 for the reactions not adjacent to the loaded span. The denomi- 
 nator of the fraction is to be taken from the triangle above ; the 
 tables referred to give only the numerators. 
 
CHAP. XIH.] 
 
 ANALYTICAL FORMULAE. 
 
 215 
 
 REACTIONS AT UNLOADED SPANS. 
 Supports counted from left. Supports counted from right. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 
 1 
 
 6 
 
 24 
 
 90 
 
 336 
 
 1254 
 
 r. 
 
 3 
 
 11 
 
 18 
 
 72 
 264 
 
 270 
 990 
 
 1008 
 
 
 ii'. 
 
 66 
 
 
 
 
 in 1 . 
 
 41 
 153 
 571 
 
 246 
 
 918 
 
 984 
 
 
 
 IV. 
 
 V. 
 
 
 
 
 
 
 
 
 VI'. 
 
 2131| 
 
 
 
 
 
 VII'. 
 
 B -g 
 
 
 6' 
 
 5' 
 
 4' 
 
 a- 
 
 2' 
 
 1' 
 
 I. 
 
 II. 
 III. 
 
 1254 
 
 336 
 1008 
 
 90 
 
 270 
 
 24 
 
 72 
 
 6 
 
 18 
 
 1 
 ~3~ 
 11 
 
 990 
 
 264 
 
 66 
 
 IV. 
 V. 
 VI. 
 VII. 
 
 
 
 
 
 
 984 
 
 246 
 918 
 
 41 
 153 
 
 
 
 571 
 2131 
 
 
 Tables give the numerators of the coefficients of w L 
 triangle on page opposite. 
 
 Denominators from 
 
 These tables may be carried out to any desired extent by the 
 law of the Clapeyronian numbers in the vertical columns. 
 
 As an example of their use, take seven spans load in fifth 
 from left, that is, in third from right. (Make sketch.) From 
 the triangle we take the common denominator 11644. Then 
 from first table in the horizontal column of III/ we have for 
 left end 
 
 Ri- 
 
 wl, Ra = . 
 
 66 
 
 264_ 
 11644' 
 
 990 
 
 11644 w 11644 ~ 11644 ~ 11644' 
 
 For supports right of loaded span, we must take the second 
 table, and look in horizontal column for Y. We thus obtain 
 
 153 918 
 
 ^-11644^ *- ~ 11644** 
 
 We can now, therefore, either by our tables or formulae, or 
 both, find the moments, reactions, and shearing force at any 
 support for both cases of loading given in Fig. 87. The reader 
 will do well to take the example of Art. 127, and find the mo- 
 ment and shear at D for both cases, and thus check our results 
 as giren in Art. 127, viz., + 788.2 ft. tons and 382.5 ft. tons 
 for the moments, and 14.63 tons for shear. 
 
216 CONTINUOUS GLBDEE. [CHAP. XIII. 
 
 144. Concentrated Load in any Spun Moments at Sup- 
 ports. It only remains to consider a concentrated load at any 
 point. If the formulae for this case do not prove to be too com- 
 plex or intricate for practice, we may consider the case, so far 
 as equal spans are concerned, as fully solved. 
 
 We have seen that the " theorem of three moments," so far 
 as uniform loads are concerned, enables us to solve the case 
 thoroughly. It is more especially as regards concentrated or 
 partial loads that the opinion widely prevails as to the impossi- 
 bility of obtaining practically useful formulae ; and this, not- 
 withstanding that it has been shown by Bresse, WinJder, Wey- 
 rauch, and many others, that the theorem of three moments 
 can be extended to include concentrated loads also. 
 
 The Theorem as thus extended is as follows: 
 
 ^-1 + 2 M m /^ + Z m + M m+1 l m = 
 
 Pm-l a' 
 
 where, by our notation (Fig. 89, Art. 130), a! and a are the dis- 
 tances of P m _i, P m , from the nearest left supports.* 
 
 By the aid of this theorem, we are able to deduce the follow- 
 ing formulas : 
 
 For moments left of r, and including support r, that is 
 
 when m <* + !, M m = - c m A Cs ~ r+2 + A ' CS ~ T+I . 
 
 C &+1 
 
 For moments right of r + 1, including support r + 1, or 
 
 u TUT A Cp + A' c I+l 
 
 when m > r, M m = c a _ m+2 - - - *&* 
 
 C B + 1 
 
 In these formulae, c represents, as above, the Clapeyronian 
 number, and A A' stand for the following expressions : 
 
 A = PZ(2-3# J + P) A' = PZ (&-#), 
 k being the fraction -, or the ratio of the distance of the weight 
 
 L 
 
 P from the left support, to the length of span. 
 
 145. Illustration of Application of above Formulae. 
 
 These formulae are by no means difficult of application. Let 
 
 * For demonstration of this Theorem, see Supplement to this chapter. 
 
CHAP. XIII.] ANALYTICAL FORMULA. 217 
 
 us take the example of Art. 127 (Fig. 88), where P = 40 tons, 
 1=80 ft., and a becomes 10, 30, 50 and 70 ft. respectively. 
 First, as regards the expressions A A' : 
 
 These become in the present case 3200 (2 & 3 ^ + ^) and 
 3200 (k %?) respectively, where Jc has the values -J-, f , -|, and -J 
 successively. Now as the denominator is in each term always 
 the same, in the first 8, in the second 64, in the third 512, and 
 only the numerators of the values of Jc vary for the different 
 positions of P, we may put these values of A and A' in the 
 forms 
 
 or 
 
 A = 800 h 150 A 2 4- 6.2305 A 3 , 
 A' = 400 h - 6.2305 A 3 , 
 
 where A has successively the values 1, 3, 5 and 7, for P 15 P 2 , 
 P 3 and P 4 respectively. These are then the practical formulae 
 for substitution in the present case. 
 
 We can now apply the formulae for M above. Thus, sup- 
 pose for seven spans we have P 3 in the fourth, as shown by 
 Fig. 88, and wish the moment due to P 3 at the fourth support 
 D. Then s = 7, r = 4, and m = 4, and we have 
 
 M 4= -o 4 
 
 or, referring to Art. 136 for the Clapeyronian numbers, 
 -56 A + 15 A' 840 A - 225 A' 
 
 M 4 = - 15 
 
 2911 2911 
 
 Now for P 3 we have k = -, or A = 5, and therefore 
 
 o 
 
 A = 1028.81 A' = 1221.2. 
 Hence 
 
 = 202.4 ft. tons. 
 
 This divided by 10 = height of truss gives tension in A a = 
 20.2 tons, nearly what we have already found in our tabulation, 
 Art. 127. 
 
 In like manner we may easily find the moment at D due to 
 
218 CONTINUOUS GERDEE. [CHAP. XIII. 
 
 every weight, or by giving the proper value to m in our for- 
 mulae, we may find the moment at any support we please. 
 
 The moments at the supports of the loaded span being found, 
 the moments at the other supports may be obtained according 
 to the rule given in Art. 140 for uniform live load over single 
 span. 
 
 146. Triangle of Moments. The reader may also by the 
 aid of the formulae above form a triangle similar to those al- 
 ready given, containing the coefficients of P I for the moments 
 at the supports of the loaded span. 
 
 Thus for two spans, for moment at left support, we should 
 obtain and J [2 Jc - 3 Z? + F] P I, and this last value will 
 run down the right diagonal column without change, except in 
 its coefficient J, which will become successively T 4 ^, -J|-, $, 
 for three, four and five spans respectively. For three spans 
 we shall have, 0, -^ [7 k 12 I 2 + 5 %?] P Z, and, as above, 
 TS- t 2 ^ 3 ^ + ^] P I. The second of these will run down 
 the second diagonal column from the right without change, 
 except in its coefficient, which will be -fa, ^V 5 ?? etc., for five 
 and six spans. 
 
 So, for four spans we have 
 
 0, .fa [26 Jc - 45 & -f 19 %?] P /, ^ [7& - 12 W + 5 ^] PI, 
 ^ [2 Jc - 3 J<? + #] P I, 
 
 for moments at left, for load in 1st, 2d, 3d and 4th span re. 
 spectively. The second of these runs down the third diagonal 
 column from the right, changing coefficient as above. 
 
 If the triangle be now drawn, and these expressions properly 
 inserted, we shall observe that along the diagonal columns 
 sloping down and to the left, the values of Jc in the parenthesis, 
 as also the denominators of the outside fractions, follow the law 
 of the Clapeyronian numbers. The numerators of these out- 
 side fractions in these columns remain unchanged. The outer 
 left column is of course always zero. 
 
 Another triangle must be found for moments to the right of 
 load, and then the moments at the unloaded supports -may be 
 found by the rule of Art. 140. 
 
 All the moments may also be found from a couple of tables 
 formed similarly to those of that Art. It is unnecessary to 
 give such tables here. From the above the reader can form 
 them for himself, if desired. The formulae for moments given 
 
CHAP. Xin.] ANALYTICAL FORMULA. 219 
 
 above are so simple, and with a little practice so readily worked, 
 that tables are scarcely needed. 
 
 147. Reactions at Support for concentrated Load in 
 ingle Span. For the reactions we have the following formulae : 
 
 1st. Abutment reactions. 
 
 When the end span contains the load, that is, when r = 1 or 
 
 When the load is not in the end spans, i.e., when r > 1 and 
 r <s, 
 
 R . M 2 M s 
 
 Rl- -J-, Rs+l = . 
 
 2d. Reactions at supports of the loaded span itself (not end 
 span), 
 
 3d. For all other reactions. 
 
 
 
 The above formulae, in view of what has been said in Art. 
 145, are sufficiently simple to need no illustration. 
 
 For load in fourth span of seven spans we find easily for the 
 reaction at left support, 
 
 R 4 = g [2911 - 3 Jc - 6387 & + 3479 #]. 
 
 This can be put in working order as explained in Art. 145, 
 and the reader can check the results which we have given in 
 the example of Art. 127 for himself. 
 
 A triangle and two subsidiary tables for the reactions at the 
 supports of loaded spans may be formed similarly to the tri. 
 angle and tables of Arts. 142 and 143. We leave this for the 
 reader to accomplish for himself, if thought desirable. 
 
 148. Shear at Supports of loaded Span. We are now in 
 possession of all the formulae necessary for the complete solution 
 
220 CONTINUOUS GIRDER. [CHAP. XIII. 
 
 of a girder over any number of supports, all spans equal. For 
 any desired span, we can find the maximum positive and nega- 
 tive moments by the cases of Fig. 80, as also the moments due 
 to various positions of the weight P. We can also find the re- 
 actions at all the supports due to these cases. From the reac- 
 tions and known forces, we can then easily find the algebraic 
 sum, or shear , at any support. The moment and shear at any 
 support due to any case of loading are, as we have seen, the 
 quantities required for calculation. 
 
 Now it is not necessary to find all the reactions in order to 
 obtain the shear. The moments at the supports being known ? 
 we can find the shear directly. 
 
 Thus, for concentrated load in a span 4 .(Fig. 89) we have for 
 any point x 
 
 M r - S r x + P (4 - a) - m = 0, 
 
 where S r is the shear at the left of the loaded span, and m is 
 the moment at any point. We see at once that, to determine 
 this moment, it is the shear that we wish, and not the reaction. 
 For a uniform load we have similarly, 
 
 M r - S r x + ~ - m = 0. 
 2i 
 
 If in both these equations we make x = 1 T) m becomes M^, 
 and we have 
 
 where q = P (1 Jc) for a concentrated load, and q = -g for 
 
 uniform load. 
 
 In an unloaded span at the left support, or when m < r, 
 <] disappears, and we have 
 
 For the shear at the right support of the loaded span we have 
 simply S r P or S r w I, and hence 
 
 
CHAP. XIII.] ANALYTICAL FORMULA. 221 
 
 where q r P k for concentrated load, and q' = ^-^- for uni- 
 
 2i 
 
 form load. For any other span at the right support 
 
 S' M m - M m-l 
 
 ^m-1 
 
 Thus, S m and S' m are the shears at any support just to right and 
 left of that support respectively. The reaction at any support 
 is then R m = S m + S' m . 
 
 The moments, then, at two successive supports being known, 
 we can readily find the shear at any support, and these two, 
 moment and shear, we repeat, are the quantities required for 
 calculation. The reactions, and the tables for the reactions 
 above, are only useful as enabling us to find the shear. It is 
 this last, together with the moment at the support, which gives 
 us the moment m at any point of the span in question, as is 
 evident from the above equations. It is only in the case of the 
 simple girder that the reactions at the ends are the same as the 
 shears. In the continuous girder only the latter should ~be 
 used, except for ends of end spans, where the two are identical. 
 We have only to remember, then, that the shear at any support 
 is the algebraic sum of all the reactions and loads from that 
 support to the nearest extreme end, and then, knowing these 
 reactions and loads, the determination of the shear is easy. 
 
 We might give tables for shears directly, as above, for reac- 
 tions; but this is unnecessary. Having taken the reactions 
 from our tables already given, and found the moments either by 
 our formulae or tables, we can then find the shears both by means 
 of the reactions and also directly from the moments them- 
 selves, and thus check at once the accuracy of our determina- 
 tion of both. From what has already been given, the reader 
 can easily construct tables of shears similar to those alreadj^ 
 given for reactions for himself, if desired. 
 
 149. Recapitulation of Formulae Continuous Girder 
 over any Number of Level Supports, all Spans equal.* 
 For notation, see Art. 130, Mg. 89. 
 
 * These formulae are given in similar form in Winkler's "Der Lehre von 
 der Elasticitaet und Festigkeit," Art. 144, p. 122. They were also indepen- 
 dently deduced by Mr. Merriman, to whose kindness we are indebted for much 
 of this chapter. The above methods of tabulation were communicated by 
 him, and are given in no treatise upon the subject. 
 
222 CONTINUOUS GIRDER. [CHAP. XIII. 
 
 1st. Moments at supports. 
 When m<r + l, M = - C A gg ~ r+2 + A/ GS -*+ I 
 
 when m > r, M m = - 
 
 in which & = * and A = P J (2& - 3/P-f /fc 3 ), A' = P J (^- #) 
 
 for concentrated loads, and A = A' = J w Z 2 for a uniform 
 load over any one span. 
 
 2d. Shear at the supports. 
 
 In the loaded span, to the right of the left support, 
 
 To the left of the right support, 
 
 S' r+1 = ! 
 r 
 
 In the wMloaded spans, to the right of the left support, 
 
 8 _ 
 
 To the left of the right support, 
 
 ' Mm ~ Mm - 1 
 
 For the reaction at any support, Rm = S' m + S m . 
 
 3d. Reactions. 
 
 (a) Abutment reactions : 
 
 when r = 1, R 1= -?^ + ^ ; w hen r = s, R s+1 = - s +/; 
 
 when y > 1 and < s, R t =: ^ R s+1 = ???. 
 
 ^ ^ 
 
 (5) Eeactions at supports adjacent to loaded span (when this 
 span is not an end span) : 
 
CHAP. XIII.] ANALYTICAL FORMULAE. 223 
 
 (<?) All other reactions : 
 
 Where for concentrated load, 
 
 A = P I (2 Jc- 3 & + IF) A' = Pl(%-X*) 
 q = P (1 Jc) q f = P Jc. 
 
 For uniform load in single span, 
 
 A = A' = J w P, q = c[ = \ w I, Jc being always y, and 
 
 Cl = c, = 15 c 7 = - 780 c w = 40545 
 
 c 2 1 c 5 = - 56 c 8 == 2911 cu = - 151316 
 
 c 3 - 4 e 6 = 209 c 9 = - 10864 c 12 564719, etc. 
 
 We give also, for sake of completeness, although not needed 
 for calculation, the formulae 
 
 FOR UNIFORM LOAD OVER ENTIRE LENGTH OF GIRDER.* 
 
 Moment at any support, 
 Reactions at abutments, 
 
 Reactions at other supports, 
 
 where, as before, A = ^ wP, q \ w I, 
 and the numbers indicated by 5 are as follows : 
 
 &! = & 4 = - 3 5 8 = - 41 &a = - 571 
 
 5a = 1 5 = - 4 J 9 = - 56 5 18 = - 780 
 
 J 3 = 1 5 6 = + 11 J 10 = 4- 153 etc. , 
 
 J 7 = + 15 l n = + 209 
 
 * The above equations were first given by Mr. Merriman, in the Jour. 
 Franklin Institute, April, 1875. 
 
224 CONTINUOUS GIRDER. [CHAP. XIII. 
 
 These numbers change signs by pairs alternately, and every 
 other one follows the law of the Clapeyronian numbers. In 
 fact those with the odd indices are those numbers, and the even 
 ones, commencing with 0, 1, 3, follow the same law. ' 
 
 The above comprises all the formulae thus far given, in a 
 shape very convenient for reference. The reader who has fol- 
 lowed attentively our explanation of their use, needs nothing 
 more to solve the case of equal spans completely. The expres- 
 sions, however, for the reactions are unnecessary. As we have 
 seen from Art. 148, we need only the moments and shears at 
 any support in practical calculations. The practical formulae 
 necessary and sufficient for any case will be found in the next 
 Art. 
 
 15O. Girder continuous over any Number of Level Sup- 
 ports ; Symmetrical with respect to the Centre, and with 
 two variable end Spans n I and p I on each side. [Fig. 
 89.]* 
 
 Moments at Supports : 
 
 M m = 
 
 when m> r, IV^ = %* - A <* + . 
 
 I p C B _! + 2 (n + p) c a 
 
 Shear at Supports loaded span, 
 
 4 
 s Wi = - ~ V~ ~"~ "*" ^' ' 
 
 wwloaded spans, 
 
 B ^ = M,-M^t 
 
 M m - m^^ 
 
 For uniform load, 
 
 A = A' = J w f r ; q q' = \ w l r 
 
 * Jour. Franklin Institute, March and April, 1875. 
 
CHAP. XIII.] ANALYTICAL FORMULA. 225 
 
 For concentrated load, 
 
 A = P g (2 k - 3 & + #), A 7 = P (& - If), 
 
 and ^ = P (1 &), ' = P k, k being always =-. 
 
 ^r 
 
 The quantities denoted by c are also as follows : 
 
 _ 
 
 + 12 j9) - n (14 + 
 
 j> (52 + 45 j>) + ^(52 + 60 p) 
 
 T 
 
 _ -p (194 + 168 p) - n (194 + 224 p) 
 -J~ 
 
 p (724 ^ 617^) + w (724 + 836^) 
 c 8 - , etc., 
 
 following the law of the Clapeyronian numbers. 
 
 151. Application of the above Formulae. The for- 
 mulse of the preceding Art. comprise in a most compact form all 
 the formulae hitherto given, and are all that is necessary for the 
 complete solution of any practical case. 
 
 Thus, by making p unity and retaining only 71, we have 
 the case of a girder with variable end spans n I, of different 
 length from the others, which latter are all equal and repre- 
 sented by I. The reader will find no difficulty in using the 
 above. For any particular case, when w or P and Z, k, n and 
 p are given, A, A 7 , q and q' can be easily found, and the prob- 
 lem is solved. If n and^) be both unity, we have the formulae 
 for all spans equal. The expressions for M m will then reduce 
 to those already given in Art. 144. Thus, in Art. 145 we have 
 already found for seven equal spans, I = 80 ft., load P = 40 
 distant 50 ft. from left ; the moment M 4 = 202.4. Now, from 
 our formulae above, we find for MS making m 5, 5 = 7, 
 r = 4; M 5 = 272.8. 
 
 Then by our formulae for shear, S 4 = + 14.12, or nearly 
 15 
 
226 CONTINUOUS GIKDER. [CHAP. XIII. 
 
 what we have assumed in Art. 145. We may also find the 
 same shear by finding the algebraic sum of the reactions at 
 ABC and D from the formulae of Art. 147. This is more 
 tedious, and, as we see, unnecessary. The moments can be 
 easily found, and then the shear obtained directly from these. 
 
 We must bear in mind that l r always denotes the span the 
 load is upon, whether nl,pl, or I, while l m is any span in gen- 
 eral, according to the value of m. 
 
 152. Continuous Girder with fixed ends. It is worthy of 
 remark that if n be made zero in the formulas of Art. 149, we 
 have a girder with fastened ends and variable end spans p I. 
 If in addition^? is unity, then all the spans become equal. We 
 must, however, remember that when we thus make n = 0, the 
 number of spans is s 2 instead of s, as before/ and the end 
 spans are p I ; the end supports are also 2 and s instead of 1 
 and s + 1. 
 
 153. Examples. As illustrations of the use of the formulae 
 of Art. 150, we give a few examples. 
 
 Ex. 1. A beam of one span is fixed horizontally at the ends. 
 What are the end moments and reactions for a concentrated 
 weight distant k \from the left end? 
 
 Here the two outer spans of three spans are supposed zero. 
 Therefore, 5 2 = 1, and s = 3. The left end is 2 instead of 
 1, and the right end 3. Hence, r 2,^> = 0, and n = 1 in the 
 formulae of Art. 150. We have, then, 
 
 d = 0, c 2 = 1, CQ = 2, 4 = 4> and hence, 
 
 or, inserting the values of c above, 
 
 M 2 = 1(2A-A'), M 3 =-1(A-2A'). 
 
 For a concentrated load, A = P Z 2 (2 k - 3 # + #), 
 and A' = P P (& - ;&). Hence, M 2 = P I (k - 2Z? + .#), 
 and M 3 = P I (1$ - V s ). 
 
 For the reaction at the left end, which is in this case the 
 same as the shear, we have 
 
CHAP. Xm.] ANALYTICAL FORMULAE. 227 
 
 5 2 = M 2- M s + P (1 _ k) or S 2 = P (1 - 3 # + 2 #), 
 
 f 
 
 5 3 = M 3- M s + P k or S 3 - P (3 # - 2 #). 
 
 ^ 
 
 For a load anywhere, we have simply to give the proper value 
 to k, and we have at once the reactions and moments. Thus, 
 for a load at \ the span from the left, Tc, = , and 
 
 s 2 = fjp, S S = AP ; M 2 = &Pi,m i = &Pi 
 
 For a load in centre, k J, and 
 
 [Compare Supplement to Chap. VII., Arts. 16 and IT.] 
 Ex. 2. J^r # uniform load over the same beam, what are the 
 
 end moments and reactions f 
 
 We have simply to introduce the proper values of A and A' 
 
 for this case, and we have at once 
 
 Mg T ^- w Z 2 = M 3 and S 2 = S 3 = w I 
 
 Ex. 3. A girder of three equal span's is " walled in " at the 
 ends, and has a concentrated load in the first span. What are 
 the moments, shears, and reactions at the ends and intermediate 
 
 In this case, s 2 = 3, and hence s = 5, r = 2, n 0, 
 j9 = 1, and therefore 
 
 C 2 A i 5 + A' c 4 
 = T * 4 + 2* ' 
 
 frA^ + A'fr ^AC^ + A^ 
 
 2 c 4 4- 2 c? 5 ^ c 4 + 2 c 5 
 
 also, c t = , c 2 = 1, <k 2, c 4 = 7, c 5 = 26, etc. 
 
 Inserting these values and the values of A and A' for con- 
 centrated load, we have 
 
 P 7 91 
 
 = - (3 & - 3 #), M 5 = (3 ^ - 3 
 
 4bO 4:0 
 
228 CONTINUOUS GIEDEE. [CHAP. XIII. 
 
 Observe that the moments are positive at each end of the 
 loaded span and alternate in sign from that span, varying as 
 the numbers 1, 2, 7, or as the numbers c [Art. 140]. A positive 
 moment always denotes compression in lower fibre. For the 
 shears we have, then, from Art. 150, 
 
 S 2 = -?- (45 - 99 # + 54 #% S' 8 = -^ (99 & - 54 Jf). 
 4o 45 
 
 S s = ^ (27 ^ - 27 #), S' 4 = ^ (- 27 /fc 2 + 27 #), 
 
 4:0 4:5 
 
 S 4 = ^ (- 9 # + 9 #), S' 5 = ^ (9 #> - 9 #). 
 
 4-0 4:0 
 
 For the reactions, then, 
 
 Ra = S 2 , R 3 = S' 8 + S 8 = ^ (126 # -81.& 3 ), 
 
 4-O 
 
 R 4 = S' 4 + S 4 = ?- (- 36 & + 36 #), R 5 = S' 5 . 
 
 4:0 
 
 Observe that the reactions as also the shears are positive at 
 the supports of the loaded span, and alternate in sign from 
 those supports. A positive shear or reaction acts always up- 
 wards. Disregarding, then, for the present, the weight of the 
 beam itself, it would have to be held down at first pier from 
 right end. 
 
 If the weight is in the centre of first span from left, Jc J, and 
 
 36 _ 9 
 
 360 ' - 360 ' 360 ' ^ ~ 860 
 
 .. _p B R _ 
 
 ^ ~ ' * - ' 4 ' ^ ~ * 
 
 The reactions add up to P, as they should. 
 If P = 100 tons, and 1 = 16 ft., we have 
 
 M 2 = 237.5 ft. tons, M 3 = 87.5, M 4 = - 50, M 5 = 25 ft. tons. 
 R 2 = 60 tons RS = 47.5, R 4 = - 10, R 5 = 2.5 tons ; 
 
 S 2 = 60 tons, S' 8 = 40, S 8 = 7.5, S' 4 = - 7.5, 
 
 S 4 = -2.5, S' 5 = 2.5. 
 
 Ex. 4. A beam of five spans, free cut ends ; centre and adja- 
 cent spans 100 ft., end spans each 75 ft., has a uniform load 
 extending over the whole of the second span from left. What 
 are the moments at the ends and supports f 
 
CHAP. Xin.] ANALYTICAL FORMULAE. 229 
 
 Here s = 5, n = f , p 1, r = -2 ; therefore, from Art. 150, 
 . 2 _A. 5 + A'* 4 ^A. 2 + A'. 3 
 
 * ^4 + 1 4, Z C 4 + J C 5 
 
 and <?! 0, 2 = 1, <? 3 = f, c 4 13, 5 = 48.5. 
 
 w Z 3 
 Since, then, A = A' = for uniform load, we have 
 
 ! = 0, M 2 = -g-VV P> M 3 - T f I* w ?, M 4 = - y 35^ w 
 
 M ^ 0. 
 
 If the load is two tons per ft., w 1? = 20,000, and 
 M t = 0, M 2 = 414.7, M 3 = 1036.6, M 4 = - 279.5, M 5 = 79.7. 
 
 Find the shears and reactions at each support. 
 
 Ex. 5. A beam of four equal spans, has the second span from 
 left covered with full load. What is the moment and shear at 
 left of load f 
 
 Ans. M 2 = Tg^j- w I*, S 2 = JJJ- w I. 
 
 What at right of load ? 
 
 Ans. M 3 = 3^- w I, S' 3 = % Jf w I. 
 
 What are the formulae for concentrated load ? 
 
 5o 
 
 TV/r r'r Z> 1 Z2 i K L3~| ~p 7 
 
 J-TXo f /* I * tv ~~" -L^ 'v ~i *-^ *v I ^ v* 
 
 OD 
 
 S 2 = ^ [56 - 58 ^ + 3 t? - ^ 3 ], 
 
 S' 8 = [58^-3^ + ^] ^>- 
 
 Examples might be multiplied indefinitely. 
 
 The above is sufficient to show the comprehensiveness of our 
 formulae, and the ease with which results may be obtained, 
 which, by the usual methods, would require long and intricate 
 mathematical discussions. The points of inflection and the 
 deflection may also in any case be easily determined, and gen- 
 eral equations similar to the above deduced, but, as we have 
 seen, the above are sufficient for full and complete calculation. 
 
 154. Table* for Moment*. From the formulae of Art. 150 
 we can easily find the moments for both uniform and concen- 
 trated load in a single span for various numbers of spans. If 
 these results are tabulated we, shall obtain tables from which 
 
230 
 
 CONTINUOUS GIEDEE. 
 
 [CHAP,, xm. 
 
 the moments may be at once taken. The formulae for the 
 shears are so easy when for any case the moments are known, 
 that it is unnecessary to give tables for these. 
 
 The reader will do well to make himself perfectly familiar 
 with the formulae by calculating the moments for various cases, 
 and comparing with the following tables. We give the prac- 
 tical case of variable end spans n I and equal intermediate 
 spans I. 
 
 TABLE FOB MOMENTS UNIFORM LOAD OVER ANY SINGLE SPAN. 
 Coefficients 'of w I 2 from talk. End spcuns n 1. 
 
 Supports counted from left. 
 
 1 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 
 n* 
 
 (2 + 3n)n3 
 
 (7 8) 
 
 (26 + 3071)^3 
 
 (97 + 11271)7*3 
 
 I. 
 
 1+271 
 
 (l + 2ra) (2 + 27i) 
 
 (1+270(7+871) 
 
 (l+27i)(26 + 80n) 
 
 etc. 
 
 II. 
 
 5 + n 
 
 (5 + 6ra)(2 + 3/i) 
 
 (5 + 6 n) (7 + 8 n) 
 
 etc. 
 
 
 III. 
 
 19 + 2271 
 
 (19 + 22)(2 + 2) 
 
 (19 + 227i)(7 + 87i) 
 
 
 
 IV. 
 
 71 + 8271 
 
 (71+83n)(2 + 8n) 
 
 etc. 
 
 
 
 V. 
 
 365+ 304 n 
 
 etc. 
 
 
 
 
 VI. 
 
 Number of 
 
 spans. 
 
 One-fourth of Denominator. 
 
 1 
 
 
 2 
 
 
 3 
 
 3 + 8 Ti + 4 7i a 
 
 4 
 
 12 + 2871 + 16 n 
 
 5 
 
 45 + 104 71 + 60 7i 2 
 
 6 
 
 168 + 3837* + 224 r 
 
 I* 
 
 The above table can be easily extended to include any num- 
 ber of spans. It is precisely the same as the table of Art. 140, 
 and, in fact, includes that table. We have only to make n = 1 
 and we have at once the table for equal spans. Suppose we 
 take five spans, load in second from right. From the smaller 
 table we have at once for the denominator of the coefficient 
 of w Z 2 , 4 (45 + 104 n + 60 n 2 ). Then from the other table 
 we have at support 1 from left M x = 0, 
 
 (l+2n)wP 
 
 at support 2, 
 
 4 (45 + 104 ?i 
 
ANALYTICAL FORMULA. 
 
 CHAP, xm.] 
 , at support 3, M 3 (1 4- 2 n) (2 + 2 n) w 
 
 231 
 
 , etc. 
 
 4 (45 4- 104 n + 60 
 
 If the load were in second span from left, and supports to 
 right of load were required, we have simply to count the sup- 
 ports the other way in the table. 
 
 Thus, 3YE 6 = 0, 3YC 3 = - - - ., or same as 3VT 2 
 
 in first case, etc. 
 
 TABLE FOR MOMENTS CONCENTRATED LOAD IN ANY SPAN, k = . 
 
 End spans n 1. Coefficients of P \ T from table. 
 
 
 a = a' 
 
 (P 
 
 y 
 
 0=0' 
 
 <i 
 
 V' 
 
 
 11. 
 
 3 + 271 
 
 1 
 
 2n 
 
 1 
 
 2 + 271 
 
 3 + 4 
 
 II'. 
 
 III. 
 
 9 + 10/1 
 
 2 + 2 n 
 
 3 + 471 
 
 2 + 271 
 
 7 + 871 
 
 12 + 1471 
 
 III'. 
 
 IV. 
 
 33 + 38% 
 
 7 + 871 
 
 12 + 14w 
 
 7 + 87i 
 
 26 + 307* 
 
 45 + 5271 
 
 IV. 
 
 v. 
 
 123 + 142 n 
 
 26 + 3071 
 
 45 + 52n 
 
 26+3071 
 
 97 + 11271 
 
 168 + 19471 
 
 V. 
 
 VI. 
 
 459 + 53071 
 
 97 + 11271 
 
 168 + 194 n 
 
 97+11271 
 
 362 + 41871 
 
 627 + 72471 
 
 VI'. 
 
 VII. 
 
 etc. 
 
 etc. 
 
 etc. 
 
 etc. 
 
 etc. 
 
 etc. 
 
 VII'. 
 
 Number of 
 spans. 
 
 Denominator A. 
 
 2 
 
 
 3 
 
 3 + 87i + 4n 2 
 
 4 
 
 12 + 2S?t + 167i2 
 
 6 
 
 45 + 10471+60712 
 
 6 
 
 168 + 388 + 224712 
 
 7 
 
 etc. 
 
 The above tables give only the moments at the supports of 
 the loaded span. The Roman numerals L, II., III., etc., denote 
 the number of this span from left, and I'., II'., III'., the num- 
 ber of the loaded span from right. The expression for the 
 moment at left support is 
 
 M 
 
 = P l r ^ [7' & - 3 ff 
 
CONTINUOUS GIRDER. [CHAP. XIII. 
 
 For the right support, 
 
 where the expressions for 0, 0', A, 7 , 7 ' ? ', a, a', are to be 
 taken from the tables and inserted. 
 
 Thus, for five spans load in second span from right, or third 
 from left, we have at once A = 46 + 104 n + 60 n\ For the 
 moment at the left support, to find 0, we must take horizontal 
 line for III., and thus find (2 + 2 *). For 7', /3' and a' we 
 must take II'., and find, therefore, 3 + 4 n, 2 + 2 n and 3 + 2 n. 
 Hence, moment at left support is 
 
 For moment at right support, we must take line II'. for 6' and 
 line III. for 7, ft and a, and hence 
 
 
 Since the span in question is not an end span, 1^ = 1 and 
 
 
 For a load in an end span, use the formulae of Art. 150. 
 For a load in middle span of five spans, i.e., third span from 
 each end, we have 
 
 +- n n) *~ 3 (7+8 n) *' + 
 
 M '== 
 
 When Jc = -=, both these moments become, as they should, equal 
 
 2i 
 
 for any assumed value of n, as the reader may readily prove by 
 insertion. 
 
 For the other moments not adjacent to the loaded span, the 
 rule of Art. 140 holds good. 
 
 Thus, M t = 0, M, = I M,, M 8 = J M,, M 4 = -J M,, etc., 
 
 P " " 
 
CHAP. Xin.] ANALYTICAL FOKMUL^EJ. 233 
 
 and similarly on the other side, 
 
 M 8 = 0, M^ = * M r+1 , M s _ 2 = Jz? M;^, etc. 
 C/ r+1 U I+l 
 
 We must remember always to give the proper signs to the 
 moments, viz., positive for extremities of loaded span, and 
 alternating each way from these for the others. From the 
 formulae of Art. 140 we can then easily find the shear at any 
 support. 
 
 155. Continuous Girder I^evel Supports Spans all dif- 
 ferent Oeneral Formulae.* The preceding formulae com- 
 prise the case of one or, at most, two variable end spans. We 
 give below the general formulas for all spans different. These 
 formulas include all the others as special cases. Thus, if we 
 make all spans equal, we have the formulae of Art. 149. If end 
 spans \ and l s are made zero, and we take the number of spans 
 equal to s 2, and first support 2, we have the continuous gir- 
 der with fixed ends, in which the intermediate spans may or 
 may not be equal, as we choose. If we make ^ or Z 8 = 
 alone, and s 1 = No. of spans, we have a continuous girder 
 fixed at one end only. In short, the formulae comprise the 
 entire case of level supports. They are as follows : 
 
 Let s = number of spans, 4=length of loaded span, 7c = , 
 
 Ly 
 
 a being distance of load from left support ; l^ 4> 4 4-i> fc 
 the lengths of the various spans counting from left. 
 Then, when m < r + 1, 
 
 TVT / A 
 
 m 4<-i 
 when m > r, 
 
 M A A e r + B c r+l 
 
 A -m "'s m+27 ; o (1 i 7 - \ * 
 
 V_i tfs-i + 2 (4 + ^.j c a 
 
 For the shear at supports of loaded span, 
 
 gr = M r -M r+1 + g SWi ^M -M r + 
 
 6 r l t 
 
 For -zmloaded spans, 
 
 * These formulas were first given by Mr. Merriman, and may be found In 
 the London Phil. Magazine, Sept., 1875. 
 
234 CONTINUOUS GIEDEE. [CHAP. XIII. 
 
 For the reaction at any support, R m = S' m + S m . In which 
 
 we have always Jc = - , q = P (1 #), q' = P Jc. 
 i> r 
 
 A = P l r 2 (2 k - 3 & -f If) and B = P I* (k - F) for concen- 
 trated load ; and q = q' = ^ wl iy and A = B = % wl* for uni- 
 form load entirely covering any one span. 
 
 Also for G and d we have the following values : 
 
 = 1, 
 
 2 ft + 
 " 
 
 _ 4 ft 4- 4) (4 + 4) - V 
 
 ""' 
 
 or, generally, 
 
 _ 4: ft + d) (4-1 + 4-2) ~ 
 
 74-7 7 
 
 "s m+3 i ^s m+2 J ^s m+3 
 
 d 5 = 2 d 
 
 or, generally, 
 
 d m = 2d 
 
 As an illustration of the use of the above formulae, let us 
 take three unequal spans, load in the first. Then s = 3,r = ~L, 
 
 Tc, = y. For moment at second support, m 2, or m > ^ ; hence 
 ^i 
 
 M -- /7 A ^i + B g 2 
 
 ^ * 
 
CHAP. Xm.] ANALYTICAL FOKMUL^E. 235 
 
 But Ci = 0, C 2 = 1, 
 
 hence, since B = P If (k If), 
 
 _ 2 P y (* - Jf) (4 + 4) 
 * (4 + 4) (k + 4) -V 
 
 If in this we make \ = 4? we have the extreme spans equal, 
 and then 
 
 M _ a P y (* - #) ft + 4) 
 
 * (4 + 4) 2 - V 
 
 If we make in this, again, \ = 4, we have for all spans equal 
 
 just what we should have from Art. 149. 
 For the reaction at the end support, we have 
 
 or, since M t = 0, 
 
 For all spans equal, or 1^=1^=1^ = 1^ this reduces to 
 
 S! = Z (15 - 19 Jc + 4 #), 
 15 
 
 as we should have found from Art. 149. 
 
 Ex. 1. A beam of one span is fixed horizontally at the 
 right end ; what are the reactions and the moments for concen- 
 trated load ? 
 
 Here 5 1=1 or s = 2, r = 1, 4 = 0, and from the for- 
 mulae of Art. 155, c l = 0, c 2 = 1, and <h = 0, d% = 1, d$ = 2, 
 
 , A GI + B c 2 B c% 
 
 
 .= - 2 + P (!-&) = (2 - 
 ' a = ? (3 *->). 
 
236 CONTINUOUS GIRDER. [CHAP. XIII. 
 
 Ex. 2. A 'beam of three spans of 25, 50 and 40 feet respec- 
 tively is fixed horizontally at the right end, and has a concen- 
 trated load of 10 tons at 12 feet from the third support from 
 left. What are the moments at the supports f 
 
 Here I, = 25, ^ = 50, 4 = 40, 4 = 0, P = 10, k 4 = 12, 
 Jc = 0.3, 5-1 = 3. s = 4, 4 = and r = 3. Also, c l = 0, 
 c 2 = 1, GS = 3, c 4 = 12.25 and ^ = 0, ^ = 1, 4 = 2, 
 <Z 4 = 6.4, d 5 = - 32.4. 
 
 When, then, m < 4, 
 
 Inserting & = 0.3 and the values of <?, 
 
 for m = 1, M! = ; m = 2, M 2 =- 8.20 ; m = 3, M 3 = 24.62 ; 
 for n = 4, 
 
 M 4 = (_ 3 A + 12.25 B) = (6.25 + 9#- 15.25 #), 
 
 or M 4 = 42.29 ft. tons. 
 
 Find the shears. Also moments and shears for uniform load 
 over third span. 
 
 Ex. 3. A beam of four spans \ = 80, 1 3 = 100, 1 3 = 50, 
 1 4 = 40 ft., free at the ends, has a load of 10 tons in the sec- 
 ond span at 40 ft. from left. What are the moments? 
 
 Here s = 4, ^ = 0, c* = 1, <? 3 = 3.6, c 4 = 19.6, c 5 = -83.7, 
 dj. = 0, 4 = 1, <& = - 3.6, d, = 10.3, 4 = - 41.85, r = 2. 
 
 For m < 3, M m = - ^- (A ^ 4 + B 4), 
 
 m>3, 
 Hence, Mj = 0, 
 
 Us < 17 * - 30 - 9 * + 13 - 9 *> = 
 
 M 5 = 0. 
 
 Find the shears. Also find the moments and shear for uni 
 form load over second span. 
 
CHAP. XIII.] ANALYTICAL FORMULA. 237 
 
 156. Thus we see that, as in Art. 150, a few short and sim- 
 ple formulae, which may be written on a piece of paper the size 
 of one's hand, are all that we need for the complete solution of 
 any case of level supports whether the spans be all equal or 
 the end ones only different, or all different ; whether the girder 
 merely rest on the end supports or be fastened horizontally at 
 one or both ends. We have only to remember that a positive 
 moment causes tension in upper flange at support, and there- 
 fore compression in lower; inversely for negative moment. 
 Also, that a positive shear acts upwards, and a negative shear 
 downwards. Also, that both moment and shear are positive at 
 supports of loaded span, and alternate in sign both ways. This 
 is all that we need to form properly the equation of moments 
 at any apex, and determine the quality of the strains in flanges 
 and diagonals. We can thus solve any practical case of framed 
 continuous girder which can ever occur with little more diffi- 
 culty than in the case of a simple girder. 
 
 Thus, for the span DE (Fig. 87) we have only to find the 
 moments at D and E due to every position of P in the span 
 D E, and the corresponding shears at D. These once known, 
 and, as we have seen, they can be easily obtained from our 
 formulae, we can find and tabulate the strains in every piece 
 due to each weight, as shown in Art. 127. An addition of these 
 strains gives, then, the maxima of each kind due to interior 
 loading. 
 
 We have, then, to find, in like manner, the strains due to the 
 two cases of exterior loading as represented in Fig. 87. From 
 the four columns thus obtained, we can deduce the dead load 
 strains, and then finally the total maximum strains of each kind 
 for every piece. [See, for illustration of the above, Art. 127.] 
 
 Thus, the whole subject is solved with the aid of but four 
 simple formulas, and for a problem generally considered impos- 
 sible by reason of its " complexity," our results will, we trust, 
 be found sufficiently simple and practical. 
 
 In view of the fact that the necessary formulae for practical 
 computations have been often given in the later works of 
 French and German authors, although perhaps never before 
 in so compact and available a shape as above, it is indeed sur- 
 prising that they should have been so completely ignored by 
 English and American writers. 
 
 The tables and formulae which we have given will, we trust, 
 
238 CONTINUOUS GIRDER. [CHAP. XIII. 
 
 bring the subject fairly within the reach of the practical engi- 
 neer, and should they be the means of calling more general at- 
 tention to this important class of structures, will not, we hope, 
 be considered as out of place in the present treatise. 
 
 For the influence of difference of level of the supports, as well 
 as for variable cross-section and the relative economy of the 
 continuous girder, see Arts. 17 and 18 of the Appendix. 
 
ART. 1.] SUPPLEMENT TO CHAP. XIH. 239 
 
 SUPPLEMENT TO CHAPTER XIII. 
 
 DEMONSTRATION OF ANALYTICAL FORMULAE GIVEN IN TEXT. 
 
 IN the following we shall give the complete development of the general 
 formulae of Art. 155. As these formulae include, as we have seen, all the 
 others as special cases, it is sufficient to show how they are obtained in 
 order to enable the reader to deduce all the others. 
 
 1. Conditions of Equilibrium. In the rth span of a continuous 
 girder, whose length is IT (see Fig.), take a point o vertically above the rih 
 
 support as the origin of co-ordinates, and the horizontal line o I as the axis 
 of abscissas. At a distance x from the left support pass a vertical section, 
 and between the support and this section let there be a single load P r 
 whose distance from the support is a. 
 
 Now all the exterior forces which act on the girder to the left of the 
 support r we consider as replaced, without disturbing the equilibrium, by 
 a resultant moment M r and a resultant vertical shearing force S r . This 
 moment is equal and opposite to the moment of the internal forces at the 
 section through the support r ; while the vertical force is equal and oppo- 
 site to the shear. 
 
 Not only over the support, but also at every section, the interior forces 
 must hold the exterior ones in equilibrium, and therefore we have the con- 
 ditions : 
 
 1st. The sum (algebraic) of all the horizontal forces must be zero. 
 
 2d. The sum (algebraic) of all the vertical forces must be zero. 
 
 3d. The sum (algebraic) of the moments of all the forces must be zero. 
 
 Thus, for the section #, we have from the third condition 
 
 2 M = M r - Sr x + P r (x - a) - m = . . . . (1) 
 where m is the moment at the section. From this we have 
 
 m = Mr S r x + P r (x a) . (2) 
 
 If in this we make x = 1& m becomes M r+ i, and we thus have for the shear 
 just to the right of the left support of the loaded span 
 
24:0 SUPPLEMENT TO CHAP. XIII. [ART. 2. 
 
 For an unloaded span the weight P disappears, and 
 
 For the shear just to the left of the right support of loaded span, 
 
 For unloaded span, the weight P disappears, and 
 
 S' m is then the shear to the left of any support m, and S m that to the 
 right. The reaction at any support is therefore 
 
 Rm = S'm + Sm. 
 
 These are the formulae already given in Art. 148. 
 
 2. Equation of tlie Elastic Line. We can now easily make 
 out the equation of the elastic line for the continuous girder of constant 
 cross-section, or constant moment of inertia. 
 
 The differential equation of the elastic line is,* 
 
 where E is the coefficient of elasticity, and I the moment of inertia. 
 If now we insert in (3) the value of m, as given in (2), we have 
 
 d? y _ M r S r x + P T (x a) 
 dH?~ El 
 
 Integrating f this between the limits x = and .?, and upon the condition 
 
 that x cannot be less than a, the constant of integration ~~ = t r = the tan- 
 gent of the angle, which the tangent to the deflected curve makes with the 
 horizontal at r ; and we have, since we must take the / P r (x a) simul- 
 taneously between the limits x a and z for x and x ; 
 
 dy_ 2M r g-S t a? + P r (g-a) 
 
 cTx- tr+ 2EI ' (3 '^ 
 
 If we take the origin at a distance h T (see Fig.) above the support r, 
 then integrating again, the constant is h T , and we have 
 
 which is the general equation of the elastic curve. If in this we make 
 
 * See Supplement to Chapter VII., Art. 12. 
 
 f Notice that when x = 0, a 0, and hence (x a) also. 
 
ART. 3.] 
 
 SUPPLEMENT TO CHAP. XIII. 
 
 x lr, y becomes h r+ i. If also we put - = &, or a = k Z r , and insert also 
 
 for S r its value as given in (2 a), we find for 
 
 _ _ 2 M r Z r -f M r+1 Z r - P r Z r 2 (2 * - 3 
 
 ..(5) 
 
 We see, then, that the equation of the curve is completely determined, 
 when we know M r and M r+1 , the moments at the supports. These, as we 
 shall see in the next Art., are readily found by the remarkable "theorem 
 of three moments," already alluded to in Art. 144. 
 
 3. Theorem of Tliree moments. 
 
 In the Fig. we have represented a portion of a continuous girder, the 
 spans being li Z 2 . . . / r ? etc., and Ihe supports 1, 2 ... r, etc. Upon the 
 spans IT-I and l r are the loads P r _i and Pr, whose distances from the near- 
 est left-hand supports are k / r _i and k l r ; k being any fraction expressing 
 the ratio of the distance to the length of span. 
 
 The equation of the elastic line between P r and the r + 1 th support is 
 given by (4), and the tangent of the angle which the curve makes with the 
 axis of abscissas is given by (3 a). If in (3 ) we substitute for ST its 
 value from (2 a), and for t r its value from (5), and make at the same time 
 
 x l r , then - becomes 
 a x 
 
 1 , the tangent at r + 1, and we have 
 
 
 2 M 
 
 r+1 
 
 - p r 
 
 Remove now the origin from o to ft, and we may derive an expression 
 for t r by simply diminishing each of the indices above by unity ; therefore 
 
 7, T, ^ I- -I 
 
 -- 
 
 ^ Z r _i + 2 M r ^-Pr,! Z r 2 _! (*- 
 
 r 1 - 
 
 Now, comparing these two equations, we may eliminate the tangents, 
 and thus obtain 
 
 M r _i ? r _! + 2M r (7 r _! + Z r ) + M r+l t = 
 
 _ 6 E 
 
 (2 yfc- 
 
 which is the most general form of the theorem of three moments for a 
 girder of constant cross-section. 
 
 When the ends of the girder are merely supported, the end moments are, 
 of course, zero. Then, for each of the piers, we may write an equation of 
 the above form, and thus have as many equations as there are unknown 
 moments. 
 
 16 
 
242 
 
 SUPPLEMENT TO CHAP. XIII. 
 
 [ART. 4. 
 
 4. Determination of the Moments Support all on 
 
 level. When all the supports are in the same horizontal, the ordinates 
 hij hi, #r, etc., are equal; and hence the term involving E I disappears, 
 and we have simply 
 
 M r _! Z r _! + 2 M? (Z r _i +lr) + M r _! Z r = 
 
 P r _i IJ-i (Tc - F) + P r V (2 k - 3 # + & 3 ), 
 as already given in Art. 144. 
 
 Now let s = number of spans, and let a single load P be placed on the 
 rth span. [PL 23, Fig. 89.] 
 
 From the above theorem, since M! and M s+1 are zero, we may write the 
 following equations : 
 
 2M 2 (I, + Z 2 ) + M 3 Z 2 = 0; 
 
 M 2 Z 2 + 2 M 3 (Z 3 + 1 3 ) + M 4 Z 3 = 0. 
 
 Z r ) 
 
 M r+1 ^ = 
 
 = A 
 
 M r _! Zr-l + 2 M r (Z r _ 
 PZ r 2 (2&- 3 
 
 M r Z,. + 2 M r+1 (Z,. + Zr+i) + M r+2 Zr+i = 
 
 P Z,- 2 (& - P) = B. 
 
 M s _ 2 Z s _ 
 
 M s ^ = 0; 
 
 =, 0. 
 
 (6) 
 
 2 M s _! (7 S _ 2 
 M s _i Z s _! + 2 M s (Z s _! + 
 
 The solution of these equations can be best effected by the method of 
 indeterminate coefficients, as referred to in Art. 136. 
 
 Thus we multiply the first equation by a number c 2 , whose value we 
 shall hereafter determine, so as to satisfy desired conditions. The second 
 we multiply by c 3j the third by c 4 , the rth by Cj. + i, etc., the index of c cor- 
 responding always to that of M in the middle term. Having performed 
 these multiplications, add the equations, and arrange according to the co- 
 efficients of M 2 , M 3 , etc. We thus have the equation 
 
 [2 C 2 (li + Z 2 ) + C 3 Z a ] M 2 + [c a Z 2 + 2 C a (l a + Z 3 ) + C 4 Z 3 ] M 3 + . . . 
 + [Cr_i Zr_i + 2 Cr (Z r _i + ZT) + CT+I Z r ] M r + . . . . 
 
 + [c s _ 2 Z g _ 2 + 2 c s _i (Zs_ 2 + Z s _i) + C B Z s _i] M s _! 
 
 + [Cg_i Z s _i + 2 C 8 (Z s _! -f ?)] M s = A Cj. + B Cr+1. 
 
 Now suppose we wish to determine M s . We have only to require that 
 such relations shall exist among the multipliers c that all the terms in the 
 first member of the above equation, except the last, shall disappear. We 
 have then evidently, for the conditions which these multipliers must 
 satisfy, 
 
 2 1 c 2 (Zi + Z 2 ) + Z 3 Z 2 = ; 
 
 c 2 Z 2 + 2 c 3 (Z 2 +Z 3 ) + c 4 Z 8 = ; 
 
 Cr_l Zr_! + 2 C,. (? r _! + Zr) + CT+I Z,. = J 
 Cs_2 Z 8 _ 2 + 2 C 8 _! (Z s _ 2 + Z 8 _!) + C B Z?_i = Oj 
 
ART. 4.] SUPPLEMENT TO CHAP. XIH. 24:3 
 
 while for M s we have at once, 
 
 __Ac r 
 
 - _ 
 
 Cs_i Z s _i + 2 C s (l a _-L + Zg ) C B +i I* 
 
 If, in like manner, we should multiply the last of equations (6) by the 
 number d 2 , the last but one by d 3 , the rth by <4_ r+1 , etc. ; then add, and 
 make all terms, except that containing M 2 , equal to zero ; we should have 
 the conditions : 
 
 2ds(k + Zs_i) + <ZsZB-i=0; 
 
 d* 4-1 + 2 d 3 (4_i + 4-2) + d t 4_ 2 = ; 
 
 2 <4-r+2 (4 + 4-l) + ^s-r+3 4-1 = ; 
 
 4_i Z 3 + 2 ds_i (l t + Z a ) + <4 Z 2 = ; 
 while for the moment we have 
 
 The values of M 2 and M s are thus given in terms of the quantities 
 A and B and c and d. 
 
 A and B depend simply upon the load and its position in the rth span. 
 Thus A=PZ r 2 (2-3& 2 + P), B = PZ r 2 (fc-F). 
 
 As for the multipliers c and d, they depend only upon the lengths of the 
 spans, and need only satisfy the conditions above. Hence, assuming 
 d = 0, ca = 1, and d l = 0, d* 1, we can deduce the proper values for 
 all the others. Thus, 
 
 d = 0, d, = 0, 
 
 ci --1, d* = 1, 
 
 Za + ?3 Za ^ T Zg_i -f Z s _2 , Z s _ 
 
 C 4 = 2 C 3 - -- C 2 -, 4 = 2 rfs - 5 - rf a ^ 
 Zs Z 3 Z 8 -2 4- 
 
 Zs + Z 4 Zs , n J 4-2 + 4-3 * 4- 
 
 C 6 = 2 C 4 = - C 3 , a*, = 4 Ci4 - - - 3 j 
 
 O ^ ^ 4 + ^ 5 ^ ^ 4 
 
 Ce = 2 Cs = - C 4 , 
 
 65 Z B 
 
 4- 
 
 etc., etc. etc., etc. 
 
 Now from equations (6) we see at once that M 3 = c 3 M 2 , M 4 = c 4 M 2 , 
 etc., or, universally, when n < r + 1, 
 
 . . (7) 
 
 Also taking the same equations in reverse order, M^ = d* M 8 , M s _2 = 
 d* M B , etc., or, universally, when n > r, 
 
 ... (8) 
 
SUPPLEMENT TO CHAP. XIII. [ARTS. 5, 6. 
 
 Equations (7) and (8) are the general equations given in Art. 155, which, 
 as we have seen, include the whole case of level supports. 
 
 5. Uniform Load. For uniform load the same equations hold 
 good. We have only to give a different value to A and B. 
 
 Thus, for several concentrated loads we should have 
 
 A = S P Z r 2 (2 lc - 3 F + & 3 ). 
 
 For a uniform load over the whole span I r, let w be the load per unit 
 of length, then 
 
 ri ri 
 
 2P = I wd a; or since a = lc Z r , 2 P = / wl T dlc. 
 
 Jo Jo 
 
 Inserting this in place of 2 P above, and integrating, we have 
 
 A = B = i w Z r 3 . 
 
 Thus the equations of Art. 155 hold good for concentrated and uniform 
 load in any span, for any number and any lengths of spans. 
 
 The above formulae were first published in an article on the Flexure of 
 Continuous Girders, by Mansfield Merriman, C.E., in the London Phil, 
 Magazine, Sept., 1875. 
 
 6. Formulae for the Tipper. The expressions for the reactions 
 in this case, already given in Art. 120, may be easily deduced. The solu- 
 tion is tedious by reason of lengthy reductions, but the process of deduc- 
 tion is simple. - 
 
 The construction in this- case is indicated in Fig. 83, PL 22. We sup- 
 pose, as shown there, a weight upon the first span only. Under the action 
 of this weight the beam deflects, and one centre support falls and the 
 other rises an equal amount. Thus, if we take the level line as reference, 
 A 2 = ha. Moreover, the reactions at these two supports must always be 
 equal. 
 
 We have, then, as representing this state' of things, A 2 = A 3 , and calling 
 the supports 1, 2, 3 and 4, we have from Art. 1, since Mj M 4 = 0, and 
 
 Zi = Za, 
 
 R, = Si = S, = - ^ + P (1 - jfe), 
 
 B a' ms 
 
 **- St = --!7' 
 
 These reactions will evidently be known, if we can determine the mo- 
 ments. 
 
 Let Y r = 6 E I F^r - ftr-l + ftr-ftr + ri Then the gen _ 
 
 L 4r-l TT J 
 
 eral equation of three moments of Art. 3 becomes, when we neglect P r , that 
 is, suppose only the first span loaded : 
 
 M r _i Z r _i 4- 2 M r (7 r _! + Zr) + M r + 1 Zr = - T r 4 P r -l V-l (^ ~ *") 
 
 This expresses a relation between the moments at three consecutive 
 
AET. 7.] SUPPLEMENT TO CHAP. XIII. 245 
 
 supports for load between the first two. Let r 1 = 1, or r = 2. Then, 
 since Mj = M 4 = 0, we have 
 
 2M a (fa + fa)+M 3 fa = -Y a + Pfa a (;&-A; 3 )=R . . (10) 
 where R stands for convenience equal to the expression on right. 
 
 Let r 1 = 2, or r = 3. Then the weight disappears, and since fa = fa, 
 M 3 fa + 2M 3 (fa + fa) = -Y 3 ...... (11) 
 
 From (11) we have 
 
 But since R 2 must always equal R 3 , we have from (9) 
 
 M.-M. _.-- 8 M. = _ Pit 
 
 fri ta 
 
 Substituting (12) in (10), we have 
 
 
 _ R k a Y, (fa + fa) 
 
 3fa'+8fafa + 4fa' 
 Substituting (12) in (13), 
 
 _fafaPft-Y a (fa + 2fa) 
 
 8fa'+8fafa + 4fa' 
 From (14) and (15), we have then 
 
 - Y 3 - R = I, fa P k. 
 Insert in this the value of R from (10), and 
 
 Y a - Y 3 = P I? (k - If) + P I, fa T& = P (IS k - fa 2 + fa fa A). 
 Now in the present case hi = 0, 7i t = 0, and h 2 A 3 , and since also 
 fa = fa, 
 
 and Y 2 = 6Eir 3 --+-l- That is, Y 2 = - Y 3 . 
 
 Hence, from our equation above, 
 
 Substituting these values of y 2 and y 3 in (11) and (13), we can obtain at 
 once M 2 and M 3 , which finally substituted in eq. (9), will give us the re- 
 actions as already given in Art. 120, when we put n I in place of fa. 
 
 7. In similar manner we can solve other problems. Thus what are the 
 reactions for a girder continuous over three supports, the two right-hand ones 
 resting upon an inflexible body which is pivoted at the centre ? 
 
 This is the case of the tipper when raised at the centre so that the ends 
 just touch, and then subjected to a load at any point of first span the 
 other end not being latched down, so that it rises freely, as though without 
 weight of its own. 
 
246 SUPPLEMENT TO CHAP. XTTT. [ABT. 7. 
 
 In this case we have from (9), since now M 3 = 0, 
 
 By the conditions R a must equal R 3 , hence 
 
 2 I, M 2 + Z 2 M 3 = - l l I, P TQ . . . * . . (16) 
 From the equation of three moments above, we have, making r 1 = 2, 
 or r = 3, since then P disappears, 
 
 M 2 Z 2 = - Y 3 . . ,> , .... . . (17) 
 
 ivr Ts 
 
 or M '=-T 
 
 Substituting this value of M 2 in (16), we find 
 
 Y, = ^*; hence M. = - ^ 
 
 2 ii + ta < 61 + ta 
 
 and therefore, at once, 
 
 Putting w Z in place of Z 2 , we have 
 
 R! + 2 R 2 , it will be observed, equals P, as should be. 
 
 The conception of a beam tipping, as in the last two Arts., is due to Clem- 
 ens Herschel (Continuous, Revolving Drawbridges, Boston, 1875), and the 
 above formulae were first deduced by him in the above work. 
 
LITERATURE UPON THE CONTINUOUS GIEDEE. 247 
 
 LITERATURE UPON THE CONTINUOUS GIRDER. 
 
 "We give below, for the benefit of students and those interested in the 
 subject, a list, chronologically arranged, of works upon the continuous 
 girder. A glance at this list will convince the reader as to the thorough- 
 ness with which the problem has been treated. 
 
 1. REBHANN. "Theorie der Holz-und Eisenconstructionen." Wien, 
 1856. [Treats the continuous girder of constant cross-section and equal 
 spans according to the old method ; first determining the reactions at 
 the supports. A load in any single span only is considered, either total 
 uniformly distributed, or concentrated and acting at the centre.] 
 
 2. KOPKE. " Ueber die Dimensionen von Balkenlagen, besonders in 
 Lagerhausern." Zeitschr. des Hannov. Arch. u. Ing. Ver., 1856. [The 
 simple and continuous girder. Attention is here first called to the advan- 
 tage gained from sinking the supports.] 
 
 3. SCHEFFLER. " Theorie der Gewolbe, Futtermauern und Eisernen 
 Bracken." Braunschweig, 1857. [Continuous girder with total uniformly 
 distributed load, and invariable concentrated loading. Advantage of 
 sinking the supports.] 
 
 4. CLAPEYRON. Calcul d'une poutre elastique reposant librement gui- 
 des appuis inegalement especes." Comptes rend us, 1857. [Here, for the 
 first time, the well-known Clapeyronian method is developed, by which a 
 series of equations between the moments at the supports is first obtained. 
 Application to total distributed loads, but varying in different spans.] 
 
 5. MOLLINOS ET PRONIER. " Traite theoretique et practique de la con- 
 struction des ponts metalliques." Paris, 1857. [Treatment of the con- 
 tinuous girder of constant cross-section, according to Clapeyron.] 
 
 6. GRASHOF. "Ueber die relative Festigkeit mit Riicksicht auf deren 
 moglichste Vergrosserung durch angemessene Unterstiitzung und Einmau- 
 erung der Trager bei constantern Querschnitte." Zeitschr. des Deutsch. 
 Ing. Ver., 1857, 1858, 1859. 
 
 7. MOHR. " Beitrag zur Theorie der Holz-und Eisenconstructionen." 
 Zeitschr. des Hannov. Arch. u. Ing. Ver., 1860. [Theory of continuous 
 girder, with reference to relative height of supports. Application to gird- 
 ers of two and three spans. Best sinking of supports for constant cross- 
 section. Disadvantage of accidental changes of height of supports. In- 
 fluence of breadth of piers.] 
 
 8. II. " Continuirliche Briickentrager^" Bornemann's Civil-Ingenieur, 
 1860. [Continuous girder of constant cross-section of three spans. Best 
 ratio of spans, and sinking of supports.] 
 
 9. WINKLER. "Beitrage zur Theorie der continuirlichen Briicken- 
 triiger." Civil-Ingenieur, 1862. [General Theory. Determination of 
 methods of loading causing maximum strains; and, for the first time, 
 general rules for the same given. Best ratio of end spans.] 
 
248 LITERATURE UPON THE CONTINUOUS GIRDER. 
 
 10. BUESSE. "Cours mecanique applique*e professe a" Fecole imperiale 
 des ponts et chaussees." Seconde Partie. Paris, 1862. [Analytical treat- 
 ment of the continuous girder of constant cross-section. The transverse 
 forces are not considered. The exact determination of the most dangerous 
 methods of loading, with reference to the moments in the neighborhood of 
 the supports, is also wanting.] 
 
 11. ALBARET. "Etude des ponts metalliques a poutres droits reposant 
 sur plus de deux appuis." Ann. des ponts et chaussees, 1866. [Continu- 
 ous girder of constant cross-section, treated after Clapeyron.] 
 
 12. RENAUDOT. "Memoire sur le calcul et le control e de la resistance 
 des poutres droites a" plusiers travees." Ann. des ponts et chaussees, 1866. 
 [Continuous girder, treated according to Clapeyron.] 
 
 13. CULMANN. "Die Graphische Statik." Zurich, 1866. [Graphical 
 treatment of simple and continuous girder of constant and variable cross- 
 section. Moments at the supports are determined analytically.] 
 
 14. H. SCHMIDT." Ueber die Bestimmung der ausseren auf em Briick- 
 ensystem wirkenden Krafte." Forster's Bauz., 1866. [Data for the amount 
 of live load for Railroad and Way Bridges. Determination of the equiva- 
 lent uniformly distributed load. Data for dead weight and wind pres- 
 sure. ] 
 
 15. GRASHOF. "Die Festigkeits Lehre." 1866. [General analytical 
 treatment of the girder without special reference to bridges. Continuous 
 girder of uniform strength.] 
 
 16. WINKLER. " Die Lehre von der Elasticitaet mid Festigkeit." Prag., 
 1867. [General analytical theory of the continuous girder of constant and 
 variable cross-section. Application to total uniformly distributed loading. 
 Influence of difference of height of supports.] 
 
 17. FRANKEL. "Ueber die ungiinstigste Stellung eines Systems von 
 Einzellasten auf Tragern iiber eine und iiber zwei Oeffnungen, speciell auf 
 Tragern von Drehscheiben." Bornemann's Civil-Ingenieur, 1868. 
 
 18. MOHR. "Beitrag zur Theorie der Holz- und Eisenconstructionen." 
 Zeitschr. des Hannov. Arch. u. Ing. Vcr., 1868. [Here, for the first time, 
 the elastic line is regarded as an equilibrium curve, and the graphical 
 treatment of the continuous girder founded.] 
 
 19. H. SCHMIDT. " Betrachtungen iiber Briickentrager, welche auf zAvei 
 und mehr Stiitzpunkte frei aufliegen, sowie iiber den Einfluss der unglei- 
 chen Hohenlage der Stiitzpunkte." Forster's Bauz., 1868. 
 
 20. COLLIGNON. " Cours de mgcanique applique*e aux constructions." 
 Paris, 1869. [Continuous girder of constant cross-section and uniform 
 load.] 
 
 21. LAISSLE and SCHUBLER. "Der Bau der Briickentrager init beson- 
 derer Riicksicht auf Eisenconstructionen." III. Ami., I. Theil. Stutt- 
 gart, 1869. [Treatment of the continuous girder, according to Clapeyron.] 
 
 22. LEYGUE. " Etude sur les surcharges a conside"rer dans les calculs 
 des tabliers metallique d'apprs les conditions generales d'exploitation 
 des chemins de fer." Paris, 1871. 
 
 23. LIPPICH. " Theorie des continuirlichen Triigers constanten Quer- 
 sehnittes. Elementare Darstellung der von Clapeyron und Mohr begriin- 
 
LITEKATUKE UI>ON THE CONTINUOUS GIKDER. 249 
 
 deten Analytischen und Graphischen Methoden und ihres Zusammen- 
 hanges." Forster's Bauz., 1871, also separate reprint. [The geometrical 
 constructions are deduced from the analytical formulae.] 
 
 24. SEEFEHLNER, G. " A tobbnyngpontu vasr&cstartokrdl A magyar 
 m6rrok es Spitesz egylet kozlonye," 1871 [Hungarian]. 
 
 25. HITTER, W. " Die elastische Linie und ihre Anwendung auf den 
 continuirlichen Balken. Ein Beitrag zur graphischen Statik." Zurich. 
 1871. [This and the preceding work treat the continuous girder after the 
 Culmann-Mohr method.] 
 
 26. OTT. " Vortrage uber Baumechanik," II. Theil. Prag, 1872. 
 [Analytical determination of the shearing forces and moments for the sim- 
 ple and continuous girder of constant cross-section and level supports.] 
 
 27. WEYRAUCH, J. I. " Allgemeine Theorie und Berechnung der con- 
 tinuirlichen und einfachen Trager." Leipzig, 1873. [A work well deserv- 
 ing to close the list. Gives the general theory for constant and variable 
 cross-section for any number of spans from 1 to oo, and for all kinds of 
 regular or irregularly distributed and concentrated loads. The formulae 
 are general, and for given loading free from integrals. Difference of level 
 of supports ; most unfavorable position of load ; exact theory of the fixed 
 and movable inflexion and influence points, etc. Examples illustrating 
 use of formulae, and complete calculations of girders.] 
 
 This last work leaves but little to be desired in thoroughness and com- 
 prehensiveness. 
 
 It will be observed that England and America have contributed but lit- 
 tle to the literature of the subject. Indeed the standard works of both 
 countries show scarcely a trace of the influence of the labors of French and 
 German mathematicians in this field. The only works which, so far as we 
 are aware, can be mentioned in this connection are as follows : 
 
 RANKINE, W. J. M. " Civil Engineering." 1870. [Very brief and in- 
 complete.] 
 
 HUMBER. " Strains in Girders." American Ed. New York : Van 
 Nostrand. 1870. [Graphical constructions, holding good only under the 
 supposition that the end spans are so proportioned that the girder may be 
 considered as fixed at the intermediate supports, for full load.] 
 
 STONEY, B. " Theory of Strains." London, 1873. [Very brief notice 
 of the subject. Points of inflection are found for full load, and the flanges 
 then cut at these points.] 
 
 HEPPEL, J. M. Phil. M^. (London), Vol. 40, p. 446. 
 
 Also Minutes of the Proceed, of the Inst. Civ. Eng. [Excellent papers, 
 which might well have been followed up.] 
 
 In the latter publication also : 
 
 BELL, W. Vol. 32, p. 171. 
 
 STONEY, E. W. Vol. 29, p. 382. 
 
 BARTON, JAMES. Vol. 14, p. 443. 
 
 In American literature : 
 
 FRIZELL. "Theory of Continuous Beams." Jour. Frank. Inst., 1872. 
 [Davelopment of the subject according to Scheffler. See 3.] 
 
250 LITEEATUEE UPON THE CONTINUOUS GIEDEE. 
 
 GREENE, CHAS. E. "Graphical Method for the Analysis of Bridge 
 Trusses." Van Nostrand. 1875. [Force and equilibrium polygons are 
 used, but the moments at the supports are found by an original method of 
 approximation, or balancing of moment areas. ,] 
 
 HERSCHEL, CLEMENS. " Continuous, Revolving Drawbridges." Boston, 
 1875. [The formulae of Wey ranch are made use of. The case of " second- 
 ary central span " is for the first time investigated, and the appropriate 
 formulas given. The fact that the live load reactions for supports out of 
 level are unchanged, provided the dead load reactions are zero, is also for 
 the first time clearly stated. The draw span is thoroughly treated, and the 
 idea of weighing off the reactions at the piers of a continuous girder sug- 
 gested.] 
 
 MERRIMAN, MANSFIELD. " Upon the Moments and Reactions of the 
 Continuous Girder" Journal of the Franklin Inst..for March and April, 
 1875; Van Nostrand's Eng. Mag., July, 1875; also the London Phil. 
 Mag., Sept., 1875, as well as the formulae contained in Chapter XII. of this 
 work. [By the aid of the properties of the Clapeyronian numbers, Mr. 
 Merriman has deduced new and general formulae eminently suited for 
 practical use. Also relations are deduced from which tables for moments 
 and reactions may be drawn up to any desired extent by simple additions 
 and subtractions, independently of the general formulae. (See Chap. XII.) 
 The simple girder appears as a special case of the continuous girder. The 
 formulae are, in respect to simplicity and ease of application, superior to 
 any heretofore given.] 
 
CHAP. XIV.] THE BEACED AECH. 251 
 
 PART III. 
 
 APPLICATION OF THE GRAPHICAL METHOD TO THE ARCH. 
 
 CHAPTEK XIV. 
 
 THE BKACED AECH. 
 
 157. Different kinds of Braced Arches. Just as in gird- 
 ers, we may distinguish between the solid beam, or " plate gird- 
 er," and the open work, or framed girder ; so, regarding the arch 
 as a bent beam, we may distinguish the braced arch and the 
 solid arch, or arch proper. The strains in the various pieces 
 composing the braced arch may be easily found by the method 
 of Arts. 8-15, or by calculation by the method of moments of 
 Art. 14 for any loading, if only all the outer forces acting 
 upon the arch are known : that is, so soon as, in addition to the 
 load, we know also the reactions at the abutments, or the hori- 
 zontal thrust and vertical reactions at the points of support, and 
 the moments, if any, which exist at these points. 
 
 We may distinguish three classes of braced arches : viz., 1st. 
 Arch hinged at both crown and springing ; 2d. Arch hinged 
 at spring line only continuous at crown ; 3d. Arch continu- 
 ous at crown &&& fixed at abutments. 
 
 158. Arch hinged at both Crown and Abutments. This 
 form of construction [PL 23, Fig. 90], owing to the hinges at 
 crown and abutments, affords for live load but little of the 
 advantage of a true arch. It is, in fact, an arch only in form, 
 but in principle is more nearly analogous to a simple triangular 
 truss of two rafters, these rafters being curved and braced ; the 
 thrust being taken by the abutments, instead of resisted by a 
 tie line A B. 
 
 The case presents no especial difficulty, and may be easily 
 calculated or diagramed, provided that not more than two 
 pieces, the strains in which are unknown, meet at any apex. 
 Thus, in PI. 23, Fig. 90, the resultant at the abutment due to 
 any weight P being known, it may be directly resolved into tnc 
 
252 THE BRACED AKCH. [CHAP. XIV. 
 
 two pieces which meet there. The strains in these two pieces 
 being thus found, those in two others in equilibrium with each 
 of them may be obtained. In Art. 13 we have already illus- 
 trated the method of procedure for such a case, as also the 
 method of finding graphically the resultant at crown and abut- 
 ments due to any position of the weight. 
 
 Thus the resultant at the crown for the unloaded half must, 
 for equilibrium, pass through the hinge at B also. Its direc- 
 tion is thus constant for all positions of P upon the other half. 
 The resultant for the other half must then pass through a and 
 the hinge at A (Fig. 90). 
 
 We have then simply to draw a B, prolong P to intersection 
 a, and draw a A. A a and B a are the directions of the resul- 
 tant at A and B, and by resolving P along these lines, we may 
 find the vertical reaction V = d l> and the horizontal thrust 
 H = cb. 
 
 We can thus easily find the reactions at the abutments in 
 intensity and direction, and following these reactions through 
 the structure, as illustrated in Arts. 8-13, Chap. I., can deter- 
 mine the strains upon all the pieces for any position of the 
 weight. A tabulation of the strains for each weight will then 
 give us the strains for uniform load as well as live load, as al- 
 ready explained in the preceding chapter, Art. 156. 
 
 There must be only two pieces meeting at the abutments. 
 Thus the pieces in Fig. 90, represented by broken lines, can 
 serve only to support a superstructure, or transmit load to the 
 arch, and have no influence upon the strains in the other pieces. 
 
 If the span A B = 2 , the rise of the arch is A, and the dis- 
 tance of the weight P from the crown is x, positive to the left ; 
 then taking moments about the end B, we have 
 
 2 V0 = P (a + a), or V = P ( + ^- 
 
 A Cb 
 
 Similarly, taking moments about the crown, 
 
 , n V a Px P (a x) 
 
 v # + H A = P #, or H= = - ^ ^ '-. 
 
 fi 2t fi 
 
 The same formulae apply for a weight upon the other half, for 
 V and* H at the other end. 
 
 The values of V and H can easily be found from these for- 
 mulae, and the strains then calculated by moments, thus check- 
 ing the diagrams. If these reactions are found for the given 
 
CIIAr. XIV.] THE BRACED ARCH. 253 
 
 dimensions of the centre line, we may, if we choose, suppose 
 the depth of the arch to vary above and below the centre line 
 equally, from the crown to ends. The lever arms of the pieces, 
 and hence their strains, will be different, but V and H are the 
 same as before. Thus, whatever the shape of the arch, we can 
 easily find the strains both by diagram and calculation. If we 
 draw a line through A and the hinge at crown, we may easily 
 prove that the greatest vertical ordinate between this line and 
 the arch is 
 
 y = 
 
 O /v 
 
 where r is the radius. 
 
 Now if the depth d of the arch is made greater than this or- 
 dinate, it may be shown that both flanges will always be in 
 compression. This condition serves, then, to determine the 
 proper depth of circular arch, which should not be less than 
 
 It is unnecessary to give here an example.* The method is 
 so simple that the reader will find no difficulty in applying the 
 principles above to any case. He will do well to calculate or 
 diagram the strains in an arch similar to that shown in the Fig. 
 for comparison with the two cases which follow. 
 
 159. Arcli hinged at Abutments continuous at Crown. 
 If we suppose the hinge at the crown removed those at the 
 abutments being, however, retained then, for any position of 
 the weight, the resultant at each end must for equilibrium 
 pass, as before, through the end hinges. In the preceding 
 case, a, for load on left half, was always to be found at intersec- 
 tion of weight with the line through B and hinge at crown, and 
 was therefore fully determined. Now, however, &, the com- 
 mon intersection of weight and resultant abutment pressures, 
 has a different position, and hence the resultants and horizon- 
 tal and vertical reactions are different. 
 
 If we can find or know the locus or curve in which this point 
 a must always lie, we can easily find, as before, the resultants 
 or reactions by simply prolonging the line of direction of the 
 weight till it meets this locus, and then drawing from the point 
 
 * See Note to this Chap, in Appendix. 
 
254: 
 
 THE BRACED ARCH. 
 
 [CHAP. xiv. 
 
 of intersection lines to A and B, and resolving P in these direc- 
 tions. 
 
 The equation of this locus can be found analytically without 
 much difficulty. 
 
 1st. PARABOLIC ARC. Thus, for & parabolic arc, we have* 
 
 32 a? h 
 y ~ 5 (5 ^ - ar 8 )' 
 
 Where [PL 23, Fig. 91] a is the half span, and h the rise of 
 the arc ; x the distance of the weight from the crown, and y 
 the ordinate N dot the locus cd eiJc. 
 
 For a given arc, then that is, a and h given we have only to 
 substitute different values for a?, as x 0, 0.1, 0.2, etc., of the 
 span, and we can easily find the corresponding ordinates y, and 
 thus construct the locus cdeiJc. It is then easy to find the 
 reactions at A and B for any position of P, as above indicated. 
 
 The vertical reaction at the abutment may also be easily 
 found by moments thus, 
 
 V t x 2 a = P (a + a?), or V t = (a + x). 
 
 2i a 
 
 The horizontal thrust is 
 5 
 
 (5 a 9 - a?) (a 2 - a?) 
 a 3 A 
 
 These values, though not needed for the construction above, 
 may be of use, and are therefore given. In the following 
 tables we give the values of H and y for different values of x : 
 
 X 
 
 H 
 
 y 
 
 x 
 
 H 
 
 y 
 
 
 
 0.3906 
 
 1.280 
 
 0.5 
 
 0.2783 
 
 1.347 
 
 0.1 
 
 0.3859 
 
 1.283 
 
 0.6 
 
 0.2320 
 
 1.379 
 
 0.2 
 
 0.3706 
 
 1.290 
 
 0.7 
 
 0.1797 
 
 1.415 
 
 0.3 
 
 0.3490 
 
 1.304 
 
 0.8 
 
 0.1226 
 
 1.468 
 
 0.4 
 
 0.3176 
 
 1.322 
 
 0.9 
 
 0.0622 
 
 1.527 
 
 0.5 
 
 0.2783 
 
 1.347 
 
 1.0 
 
 
 
 1.620 
 
 .a 
 
 
 
 .h 
 
 . 
 
 ft 
 
 .A 
 
 * For the demonstration of the analytical results made use of in this chap- 
 ter, we refer the reader to Die Lehre von der Elasticitat und Festigkeit, by E. 
 Winkler. Prag, 1867. See also the Supplement to this chapter. 
 
CHAP. XIV.] THE BRACED ARCH. 255 
 
 From the table, a and h and P being known, H and y can be 
 found for the successive positions of P at 0.1, 0.2, etc., of #, or 
 
 the half span, by multiplying P by the tabular number for 
 
 H, and h by the tabular number for y. 
 
 2d. CIRCULAR ARC. For a circular arc we have for the equa- 
 tion of the locus cdeik [Fig. 91], 
 
 where K = -r 3, I being the moment of inertia of the constant 
 
 cross-section, A its area, and r the radius of the circle : also 
 where 
 
 (sin 2 a sin 2 j3) (a 3 sin a cos a + 2 a cos 2 a) 
 
 sin a [sin 2 a sin' 2 /3 + 2 cos a (cos /3 cos a) 2 cos a (a sin a sin 0)]' 
 
 a being the angle subtended at centre by the half span, and 
 . __ 2 cos a (a sin a /3 sin /3) 
 
 B = 
 
 sin 2 a - sin 2 /3 + 2 cos a (cos @ cos a a sin a H- /3 sin 
 2 a cos 2 a 
 
 2 (a 3 sin a cos a + 2 a cos 2 a)' 
 or, approximately, 
 
 24 6 a a 
 
 5 A 2 
 
 = IT^ 4 = 64 P 
 where $ is the angle from crown to weight. 
 
 the square of the radius of gyration, or, approximately, 
 the square of the half depth, hence K = -_ approximately. 
 
256 THE BRACED ARCH. [CHAP. XIV. 
 
 For the exact values of A and B, we have the following table : 
 
 
 fi 
 
 a= 
 
 a=10 
 
 a= 20 
 
 a- 30 - 
 
 a-=40 
 
 a=50 
 
 a=W 
 
 a=90 
 
 
 
 
 
 1.20 
 
 1.19 
 
 1.17 
 
 1.14 
 
 1.08 
 
 1.00 
 
 0.88 
 
 
 
 
 
 0.2 
 
 1.21 
 
 1.20 
 
 1.18 
 
 1.15 
 
 1.10 
 
 1.01 
 
 0.90 
 
 
 
 
 A 
 
 0.4 
 0.6 
 
 1.24 
 1.29 
 
 1.24 
 1.29 
 
 1.21 
 1.27 
 
 1.18 
 1.24 
 
 1.13 
 1.20 
 
 1.05 
 1.13 
 
 0.94 
 1.02 
 
 
 
 
 - 
 
 
 0.8 
 
 1.88 
 
 1.38 
 
 1.36 
 
 1.34 
 
 1.30 
 
 1.24 
 
 1.18 
 
 
 
 
 
 1.0 
 
 1.50 
 
 1.50 
 
 1.49 
 
 1.47 
 
 1.45 
 
 1.41 
 
 1.36 
 
 
 
 
 B 
 
 a 
 
 0.234 
 
 0.233 
 
 0.221 
 
 0.203 
 
 0.178 0.146 
 
 1 
 
 0.107 
 
 
 
 a 4 
 
 For the values of y Q we have the following table 
 
 
 
 a=0 
 
 =10- 
 
 a=20 
 
 a=30 
 
 = 
 
 a=50 
 
 .=60- 
 
 a=W 
 
 
 
 1.280 
 
 1.282 
 
 1.288 
 
 1.300 
 
 1.316 
 
 1.340 
 
 1.375 
 
 1.571 
 
 0.2 
 
 1.290 
 
 1.292 
 
 1.298 
 
 1.309 
 
 1.327 
 
 1.348 
 
 1.380 
 
 1.571 
 
 0.4 
 
 1.322 
 
 1.324 
 
 1.320 
 
 1.340 
 
 1.354 
 
 1.374 
 
 1.403 
 
 1.571 
 
 0.6 
 
 1.379 
 
 1.380 
 
 1.385 
 
 1.393 
 
 1.405 
 
 1.421 
 
 1.443 
 
 1.571 
 
 0.8 
 
 1.468 
 
 1.4G9 
 
 1.471 
 
 1.476 
 
 1.483 
 
 1.490 
 
 1.504 
 
 1.571 
 
 1.0 
 
 1.600 
 
 1.600 
 
 1.599 
 
 1.597 
 
 1.594 
 
 1.591 
 
 1.588 
 
 1.571 
 
 .a 
 
 .h 
 
 It will be seen that for the semi-circle the locus is a straight 
 line, for which y = ^irr = 1.5708r. Thus, for any given case 
 that is, I, A and r given we can easily calculate K. Then 
 from our tables, for given value of a, we can find A, B and y Q for 
 values of ft of 0, -j^ths, T \ths, etc., of a. These values inserted 
 in the equation for y above, will enable us to plot the curve or 
 locus cdeik) which being once known, the rest is easy. We 
 have thus by a union of analytical results with our graphical 
 method a very easy and practical solution of this important 
 case. We may, jf we choose, only use our method to determine 
 the horizontal thrust and vertical reaction as shown by the Fig. 
 91, and then calculate the strains by the method of moments. 
 The availability and ease of the method here given, as com- 
 pared with calculation, will be seen from a consideration of the 
 
CHA.P, XIV.] THE BRACED ARCH. 257 
 
 analytical formulae for the horizontal thrust and vertical reac- 
 tion at A. Thus, for the vertical reaction, we have, as before, 
 simply 
 
 For the horizontal thrust, however, we have the following 
 very clumsy formula : 
 
 sin* a sin 2 p + 2 cos a (cos {I cos a) 2 (l-f-ic)cosa (a sin a sin/9) 
 2 [a 3 sin a cos a + 2 (1 + *) a cos 2 a] 
 
 For the semi-circle, this reduces to 
 
 K being, as before, = -r ^ ; where A is the area and I the 
 
 moment of inertia of the cross -section, r the radius of the arch, 
 and the angles a and /3, as represented in Fi^. 91, viz., the 
 angle of the half span, and the angle to the load, subtended by 
 x. The first of the above formulae is sufficiently simple, and 
 by it we may check the accuracy of our construction. Thus 
 having plotted the curve cdeiJc by the aid of our expression 
 for y and the tables above for any position of P required, we 
 draw d A d B, and resolve P along these lines, thus finding V 
 and H [Fig. 91]. We can then calculate V from the formulae 
 
 p 
 
 above, viz., V = (a + x). If this calculated value agrees 
 
 2 Ob 
 
 with that found by diagram, we may have confidence that the 
 curve is properly plotted, and hence that the value of H is also 
 correct. Thus, with very little calculation and great ease, 
 rapidity and accuracy, we can find the reactions at the end A 
 for any given position of P in any given case. These reactions 
 once known, we can easily find the strains either by diagram, 
 as illustrated in Chap. 1., or by calculation by the method of 
 moments of Art. 14. 
 
 16O. Arch fixed at Abutments continuous at Crown. 
 This is by far the most important case of braced arch, as by the 
 continuity of the crown and fixity of ends we obtain all the 
 advantage possible due to the combined strength and elasticity 
 of the arch. It is also the most difficult case of solution, as the 
 
 formulae obtained by a mathematical investigation are complex, 
 17 
 
258 THE BE AGED ARCH. [CHAP. XJV. 
 
 and give rise to tedious and laborious computations in practice. 
 A method combining simple analytical results with graphical 
 construction similar to the preceding, will, however, obviate 
 these difficulties, and bring the subject fairly within the reach 
 of the practical engineer. 
 
 In the present case, as before, the common intersection of the 
 weight and the reactions lies in a curve, the equation of which 
 may be found, and the curve itself thus plotted for any given 
 case. 
 
 But this curve, or locus, ILK [PI. 24, Fig. 92] being con- 
 structed, in order to find the directions of the reactions which 
 now no longer pass through the ends of the arc A and B, it is 
 necessary to find and construct also the curve enveloped ~by these 
 reactions for every position of P ; that is, the curve to which 
 these reactions are tangent. If, then, these two curves are con- 
 structed, we have only to draw through L [Fig. 92] lines tan- 
 gent to this enveloped curve, and we have at once the reactions 
 in proper direction, and by resolving P along these lines, can 
 easily find their intensities, and therefore V and H, as before. 
 
 \st. PARABOLIC ARC. 
 
 For a parabolic arc we have for the locus ILK, y = ^ h ; 
 that is, the locus is a straight line at \th the rise of the arch 
 above the crown since we now take y as the ordinate to the 
 locus measured above the horizontal tangent at the crown. The 
 origin is, therefore, at the crown instead of at the centre of the 
 half span, as in the previous case. 
 
 For the second curve, or curve enveloped by the reactions, we 
 have,* taking v as the abscissa and w as the ordiuate of any 
 
 4. n?- oon 2 <# (23 a 2 + 20 ax + 5 a?) k 
 
 point [Fig. 92], v = , w = ^ , . ,- f- , 
 
 3 a + x ' 15 (a + x) (3 a + x) 
 
 where, as before, a is the half span, h the rise, and x the dis- 
 tance of the weight from crown. For x = 0, v = f a, and 
 w = ff h. For x = a, v = a, and w = f h. For x = a, 
 v = a, and w = oo. Eliminating x from both equations, we 
 
 5 a 2 5 av + 2v* t 
 
 have ^ K - h. 
 
 15 a (a v) 
 
 Hence the curve enveloped by the reactions is on each side an 
 
 * For the proof of all the expressions assumed, see the Supplement to this 
 chapter. 
 
CHAP. XIV.] 
 
 THE BEACED ARCH. 
 
 259 
 
 hyperbola, which has for asymptotes the vertical through the 
 abutment and a straight line which cuts the axis of symmetry 
 of the arch at the point b [PL 24, Fig. 93], ^ h under the crown, 
 the tangent at the crown at f a from the crown, and the chord 
 of the arc at 6 a from the centre. The centre of the hyper- 
 bola is at e, yV h below the horizontal through the crown. The 
 two hyperbolas osculate at the point -5- a vertically below the 
 crown. [See Fig. 93.] 
 
 As an aid to the construction of these hyperbolas, we give the 
 following ^table : 
 
 oc 
 
 V 
 
 w 
 
 X 
 
 V 
 
 w 
 
 1 
 
 1.0000 
 
 00 
 
 
 
 0.6667 
 
 0.5111 
 
 0.9 
 
 0.9524 
 
 2.7721 
 
 0.1 
 
 0.6452 
 
 0.4897 
 
 0.8 
 
 0.9091 
 
 1.5455 
 
 0.2 
 
 O.B249 
 
 0.4722 
 
 0.7 
 
 0.8695 
 
 1.1065 
 
 0.3 
 
 0.6061 
 
 0.4577 
 
 0.6 
 
 0.8334 
 
 0.9999 
 
 0.4 
 
 0.5882 . 
 
 0.4463 
 
 0.5 
 
 0.8000 
 
 0.7600 
 
 0.5 
 
 0.5714 
 
 0.4349 
 
 0.4 
 
 0.7693 
 
 0.6756 
 
 0.6 
 
 0.5555 
 
 0.4258 
 
 0.3 
 
 0.7407 
 
 0.6155 
 
 0.7 
 
 0.5405 
 
 0.4160 
 
 0.2 
 
 0.7144 
 
 0.5714 
 
 0.8 
 
 0.5263 
 
 0.4102 
 
 0.1 
 
 0.6897 
 
 0.5377 
 
 0.9 
 
 0.5128 
 
 0.4053 
 
 
 
 0.6667 
 
 0.5111 
 
 1.0 
 
 0.5000 
 
 0.4000 
 
 .a 
 
 .a 
 
 .h 
 
 .a 
 
 .a 
 
 .h 
 
 From the table it is easy to construct the hyperbola for any 
 given case. We have,, of course, a perfectly similar hyperbola 
 for the other half, its centre e being similarly situated with 
 respect to the crown, to the right of c. "We have then simply 
 to draw a line through the intersection m of the weight P 
 [Fig. 93] with the line i k, at -g- h above <?, tangent to the hyper- 
 bola, and we have at once the direction of the resultant. This 
 tangent may be drawn by eye, or geometrically constructed if 
 
THE BRACED ARCH. [CHAP. XIV. 
 
 desired.* A similar tangent to the hyperbola on the other side 
 determines the direction of the other reaction. We can then 
 resolve P in these two directions, and find at once V and H. 
 The problem, then, so far as a parabolic arc is concerned, is 
 sufficiently simple and easy of solution. We have only to draw 
 a straight line and two easily constructed curves. The formulae 
 for V and H and moment at crown M are for this case also 
 simple, and may be used for checking our results. They are : 
 
 TVr - IP ( a ~ X Y ( 3 ^ ~ 10 a x ~ 5 ^ 
 ~ * ~> 
 
 H- !*p- V - 
 
 " J - 
 
 ( a ~ 
 
 where, as before, a is the half span, h the rise, and x the dis- 
 tance of weight P from crown. A negative moment always 
 indicates tension in lower or inner flange. 
 
 2d. CIRCULAR ARC. 
 
 In this case we 'have for the locus ILK [Fig. 92], for small 
 central angles a, the equation : 
 
 #, A, and x being as above, and <f r the square of radius 
 
 of gyration ; A being the area and I the moment of inertia of 
 cross- section. 
 
 [Note. In all the cases hitherto considered, or which we 
 shall consider hereafter, the cross-section is assumed constant.'] 
 
 According to the exact formula, which is too complicated to 
 make it desirable to be given here, we have the following 
 
 * For the construction of a tangent to a conic section, see Appendix, Fig. 4. 
 
CHAP. XIV.] 
 
 THE BRACED ARCH. 
 TABLE, FOR VALUE OF y. 
 
 261 
 
 
 
 a = 
 
 a = 30 
 
 a = 60 
 
 a =90 
 
 
 
 0.200 
 
 0.211 
 
 0.252 
 
 0.329 
 
 0.2 
 
 0.200 
 
 0.210 
 
 0.246 
 
 0.312 
 
 0.4 
 
 0.200 
 
 0.307 
 
 0.236 
 
 0.280 
 
 0.6 
 
 0.200 
 
 0.202 
 
 0.217 
 
 0.228 
 
 0.8 
 
 0.200 
 
 0.197 
 
 0.200 
 
 0.158 
 
 1.0 
 
 0.200 
 
 0.188 
 
 0.151 
 
 0.082 
 
 
 
 
 
 
 a 
 
 h 
 
 Instead of determining the curves enveloped by the reac- 
 tions, the expressions for which are in this case somewhat com- 
 plicated, it will be found preferable to find the distances c c 2 
 of the intersections <p and ty of the reactions [see Fig. 92] with 
 the verticals through the centres of gravity of the end cross- 
 sections. For small central angles a, we have 
 
 C, 
 
 15 (a + x) I A A 2 
 
 ___2A_r 
 15 (a + x) I 
 
 where 0, A, A and a?, have the same signification as above. 
 
 Since I divided by A equals the square of the radius of gyra- 
 tion = ^, we have 
 
 2A 
 
 4- 
 
 15 (a + a?) 
 
 [ 
 
 For braced arches when the material is nearly all in the 
 flanges, the material in the bracing being very small, we may 
 call the radius of gyration half the depth of the arch measured 
 upon the radius from centre to centre of flanges ; or represent- 
 ing this depth by d, 
 
262 THE BRACED AECH. [CHAP. XIV. 
 
 2A . 
 
 5a? 
 
 [A negative result indicates that the distance is to be laid off 
 below the centre of cross- section.] These formulae are easy of 
 application, and sufficiently exact for arches whose rise is small 
 
 compared to the span ; when - is, say, not greater than -^. 
 
 2 a> 
 
 All the above formulae are for constant cross-sections. Exact 
 formulae for variable cross-section give results but little less, 
 and are much more complicated. The effect of using the above 
 formulae is therefore, merely, to increase slightly the coefficient 
 of safety. 
 
 161. We are now able to determine readily and accurately 
 the strains in the various pieces of braced arches hinged at 
 crown and abutments, and hinged at abutments only. We 
 have only to construct in each case the reactions at the abut- 
 ments, as explained in Arts. 158 and 159, Figs. 90 and 91, and 
 then, by the method already detailed in Arts. 8-13, we can fol- 
 low these reactions through the structure, and thus find the 
 strains in each piece due to every position of the load. We 
 may also, having found the reactions for given position of 
 weight, calculate the strain in each piece by moments. 
 
 For the case of the arch continuous at the crown w&& fixed at 
 the abutments, we must remember that we have also a moment 
 at each end tending to cause either tension or compression in 
 the inner flanges according as it is negative or positive. The 
 case is precisely analogous to the continuous girder, or girder 
 fixed at ends. As in that case [see Fig. 77, Art. Ill] the 
 moment at one end, as B, was the product of H into the vertical 
 distance B D, so here the moment at A (Figs. 92 and 93) is the 
 product of H into c t , found by the formulae above. This 
 moment can, then, be easily found when c l and H are known. 
 We can then lay it off, according to the directions of Art. 125, 
 for " passing from one span to another of a continuous girder," 
 and thus commence our diagram of strains; or we can cal- 
 culate the strains by the method of moments. 
 
 162. Illustration of method of Solution. As an illustra- 
 
CHAP. XIV.] THE BRACED AECH. 263 
 
 tion, take a portion of a braced arch, as represented in PI. 24-, 
 Fig. 94. We have first to plot the upper curve or locus of m 
 for the given dimensions of the centre line of the arch. This 
 curve once plotted, then, for any position of the weight, we have 
 only to prolong P to m, and draw a line from m to the end of 
 centre line if the arch is hinged at ends, or to < at a distance 
 <?!, above or below the end of centre line if the arch is fixed at 
 ends ; c l being easily found from our formulae above. In sim- 
 ilar manner, we draw a line from m to the other end, or C 2 . 
 Now these two lines are the resultants of the outer forces P, 
 and by simply resolving P in these directions, we have at once 
 V and H, while the moment at the end M t = H c 1? positive 
 if it tends to cause compression in lower flange, or since c is 
 negative down, if it acts below the end. 
 
 We can now easily find the strain in any flange, as D, 
 whether the arch vary in depth or not, provided only it is sym- 
 metrical with respect to its centre line. Thus for D, take the 
 opposite apex a as the centre of moments. The moment of H 
 with reference to #, as shown in the Fig., tends to cause tension 
 in D, while that of V causes compression. We have then, 
 representing tension by minus, 
 
 moment of V moment of H 
 
 strain in D = = , 
 
 lever arm or D 
 
 all with reference to a. If the result is minus, it indicates thus 
 tension, if plus, compression ; if it is zero, the two moments are 
 equal, and at , therefore, no moment exists ; hence a must be 
 a point of inflection. Note that H and V must be taken as act- 
 ing at $, Fig. 94 We can also evidently take them as acting 
 at the centre of the end cross-section, if we take into account 
 the moment H c^ 
 
 In similar manner, for C we take 5 as centre of moments, and 
 then, since H now causes compression in C and V tension, we 
 have for V and H, acting at <, 
 
 . . ^ moment of H moment of V 
 
 strain in C = - . 
 
 lever arm or C 
 
 For V and H, considered as acting at the end of centre line, we 
 
 moment of H + H^ moment of V 
 have C = - = , . 
 
 lever arm or 
 
 taking <\ without regard to its sign, but simply to the kind of 
 
264: THE BRACED ARCH. [CHAP. XIV. 
 
 strain it tends to cause in the piece in question. Properly, 
 since when H is belo\v ^ is negative, we should have H ^ 
 for moment causing compression in C. 
 
 Thus we may proceed till we pass P, and then the moment 
 of P, with its proper sign, as producing tension or compression 
 in the piece in question, must also be taken into account, or we 
 may instead take the moments of V and H at the other end, 
 that is, the same side of the weight as the piece itself. 
 
 The diagonals may be similarly found by moments. It will, 
 however, be best to determine them by diagram, one of the 
 flanges being first calculated (in this case the first upper flange), 
 as explained in Art. 125. They may also be calculated from 
 the resultant shear at any apex. Thus, for diagonal 3 find the 
 vertical components of the previously determined strains in D 
 and C. These vertical components, together with the vertical 
 component of the strain in 3, must for equilibrium be equal and 
 opposite to the total shear at ft. 
 
 Calling this shear F, and a, fi and y the inclinations of D 
 and 3, we have for the strain in 3, 
 
 S 8 = (F S l sin a S 2 sin ft) cos 7. 
 
 If either of the vertical components of the strains in D or C 
 acts opposite to the shear F, it must, of course, be subtracted ; if 
 in the same direction, added to F. For the ready determina- 
 tion of the proper signs, see Appendix, Art. 16 (4). 
 
 The moment H ^ is the moment at the fixed end, and is con- 
 stant throughout the arch for any one position of the load. It 
 causes tension in outer and compression in inner flanges, pro- 
 vided, as in the Fig., <f> fall below the centre of the end section. 
 This moment is increased (or diminished if <j> is above) by the 
 varying moment of H for each apex. 
 
 The above method of determining the strains in the braced 
 arch, though not strictly graphical, but rather a combination of 
 analytical and graphical methods, offers such a ready solution 
 of this important and difficult case, that we have not thought it 
 out of place to notice it somewhat in detail. We consider it by 
 far the simplest and easiest method which has yet appeared. 
 
 163. Analytical Formulae for V and H. A comparison 
 of our method with the long and involved analytical expres- 
 sions to which the theory of flexure conducts us, will render its 
 advantages still more apparent. 
 
CHAP. XIV.] THE BRACED AKCH. 265 
 
 Thus, for a load of w per unit of horizontal length, reaching 
 from left end to a point whose angle from vertical through 
 crown is ft (Fig. 92), a being the angle subtended by the half 
 span, we have * 
 
 H = & sin (B-Jci +.k + sin 
 
 where R is radius of arch, and 
 
 7 . 2 sin a , sin a cos a . sin 2 a 
 
 fc = a -r sm a cos a . , \ = -f s , 
 
 a a 2 
 
 7 sin a 7 sin a r n -, 
 
 #2 = , #3 = a sin a cos a , and 
 
 /z, = /3 + 2 yS sin 2 /3 + 3 sin ^Q cos j3. 
 
 For V we have 
 
 v _ R F cos a sin 2 8 sin/3 K 3 cos a cos 3 a . "I 
 
 L.2 (a sin a cos a) 2 a sin a cos a 6 (a sin a cos a) J' 
 
 , cosyS 8 sin 8 cos s /3 
 Where -^ + '__J yZ t 
 
 For a concentrated load P for any point [Fig. 92], we havef 
 
 or, more correctly, 
 
 __ _ p a /3 sin a cos a sin /3 cos /3 + 2 cos a sin ft 
 2 (a sin a cos a) 
 
 For the semi-circle, this becomes 
 
 v _ p TT 2 /3 2 sin /3 cog ft 
 
 2 TT 
 For H we have 
 
 H 
 
 p 2 s i p [ C03 3 cos a + (1 + ) /3 sin 3] (1 + ) a (sin 2 a + sin 2 ff) 
 2 [(1 + K) (t (a + sin a cos a) 2 sin 2 a] 
 
 * Taken from Capt . Bads' Report to the lUinois and St. Louis Bridge Co. , 
 May, 1868. 
 f Die Lehre von der Eiasticitdt und Festigkeit. Winkler. Prag. 1867. 
 
266 THE BKACED AKCH. [CHAP. XIV. 
 
 where K = g ; I being the moment of inertia, and A area of 
 
 the cross-section, and r the radius of circle. 
 
 These formulae, it will be observed, involve much labor in 
 any particular case. Where the number of weights is large, 
 the computation is tedious in the extreme. A method which 
 shall give accurate results and avoid such formulae as the above 
 is certainly very desirable, and such we believe to be the 
 method which we have given. 
 
 For the analytical investigation of arches, and the demon- 
 stration of the formulae for the curves of which we have made 
 use, the reader may consult Die Lehre von der Elasticitaet und 
 Festigkeit, by Dr. E. WinJder, to which we have already re- 
 ferred, and which contains a thorough discussion of the whole 
 subject. The tables which we have given, as well as the for- 
 mulae for /, G! and <? 2 , will, it is hoped, give the method here 
 presented a practical value, and render the solution of any par- 
 ticular case easy and rapid. 
 
 164. For a solid or plate girder arch of given cross-section, 
 we may also determine the proper proportions by finding, as 
 above, the moment M of the exterior forces at any point. 
 
 TI 
 
 Then M = _, 
 
 t 
 
 where T is the strain per unit of area in any fibre distant t 
 from. the axis, and I the moment of inertia of the cross-section. 
 Thus, for a rectangular cross-section I = ^ b d 3 , where b is the 
 breadth and d the depth. 
 
 Hence M = % T I d\ 
 
 if we take t = -. 
 
 The strain, then, per unit of outer fibre will be 
 
 T _6M 
 
 " bd 2 ' 
 
 The safe working strain should not exceed for iron 5 tons 
 per sq. inch for tension and 4 tons for compression, and there- 
 fore d being assumed, we can easily proportion b so as to sat- 
 isfy this condition. 
 
 As examples of braced arches, such as we have considered, 
 viz., continuous at crown and fixed at abutments, we may men- 
 
CHAP. XIV.] THE BRACED AKCH. 267 
 
 tion the Bridge over the Mississippi River at St. Louis, ~by 
 Copt. Eads / one over the Elbe near Hamburg on the Paris- 
 Hamburg R. R., in which, however, the outward thrust of the 
 arch is balanced by a precisely similar inverted braced arch, 
 or suspension system. Thus the piers have to support a verti- 
 cal reaction only, and the necessity of large and expensive abut- 
 ments of masonry for resisting the horizontal thrust is obviated. 
 
 The strains in the inverted arch of this character are found 
 in a precisely similar manner. The only difference is that the 
 reactions, and therefore the vertical and horizontal components, 
 act now in a direction opposite to the direction for the upright 
 arch, and the strains, though the same in amount, are of re- 
 verse character in each piece. 
 
 The bridge over the Rhine at Coblenz is an illustration of 
 the braced arch pivoted at the abutments only. 
 
 Examples of the solid or cast-iron arch of all kinds are 
 common. 
 
 165. Strains due to Temperature. In the first class of 
 braced arches, viz., pivoted at both abutments and crown, there 
 are evidently no strains due to changes of temperature. The 
 arch can accommodate itself to any change of length by rising 
 at the crown and turning at the abutments, and no strains are 
 induced. 
 
 We represent by e the coefficient of linear expansion for one 
 degree [about 0.000012 for iron, for every degree centigrade], 
 and by t, the temperature above or below the mean tempera- 
 ture , for which no strain exists. 
 
 Then for arch pivoted at abutments only, we have for the in- 
 crease of thrust,* 
 
 __ 2 El et sin a 
 ~ 7* (a 3 sin a cos a + 2 a cos 2 a) + 2 K r* a cos 3 a, 
 where E is modulus of elasticity, I moment of inertia of cross- 
 
 section, and K = ; A being area of cross-section, r radius of 
 arch, a the angle of half span, or, approximately, 
 
 8 A A 2 + 15 I' 
 where h is the rise of arch. 
 
 *Lehre von der Elasticitdt. Winkler. Also Supplement to this chapter, 
 Art. 26. 
 
268 THE BKACED ARCH. [CHAP. XIV. 
 
 For the moment at any point, then, due to change of temper- 
 ature, we have 
 
 M = H r (cos /3 cos a), 
 
 ft being the angle from vertical to that point. 
 
 This moment, if positive, causes tension in outer and com- 
 pression in inner flanges, and we can, as before, easily find the 
 corresponding strains either by diagram or calculation. 
 
 For an arch fixed at ends and continuous at crown, we have 
 
 TT 
 
 2 Ele t (1 + K) a sin a 
 
 [(1 + K) (a 2 -4- a sin a cos a) 2 sin 2 a] ' 
 
 or, approximately, 
 
 45 E I e t 45 E I A e t 
 
 ~~ r 2 (a 4 + 45 K) ~ 4= A A 2 + 45 I' 
 
 But this thrust does not act at the abutment, since, if it did, 
 there would be no moment there. It must be considered as 
 acting at a distance for rise of temperature, below the crown of 
 
 (1 + K) a sin a 
 e '--= (l + )a -*' 
 
 or, at a distance above the end abutment of h e Q . 
 Approximately, we have 
 
 _ a 2 + 6 K _ ( A a 2 + 6 I) h 
 6 3Aa? ' 
 
 a being the half span. 
 
 This thrust and its point of application being known, we can 
 easily find the moment, and hence the strains at any point. 
 We see that the horizontal thrust is about six times as great as 
 for the case of an arch pivoted at both ends. 
 
 The constant moment acting at the abutment, which may be 
 considered as acting at every point, is 
 
 _ /sin a \ 
 
 M A = 1 -- cos a t H r ; 
 
 it acts to cause compression in outer and tension in inner 
 flanges at abutment. If we find this moment, we can then con- 
 sider H as acting at the end, and then we have for the moment 
 at any point 
 
 * Capt. Eads' Report to the Illinois and St. Louis Bridge Co., May, 1868. 
 
CHAP. XIV.] THE BRACED ARCH. 269 
 
 a positive result, giving tension, a negative, compression in the 
 outer flanges. 
 
 166. Effects of Temperature. We are now able to solve 
 accurately and thoroughly any class of braced arch, both for 
 variable loading and changes of temperature, and here the fol- 
 lowing remarks upon the latter subject may not be without 
 interest. We quote from Culmann Die Graphische Statik, 
 p. 487 : 
 
 " The question arises whether the fears which the additional 
 strains, due to variations of temperature, have given rise to, are 
 well founded. Before the construction of the Arcole Bridge 
 in Paris the Engineer Oudry made various experiments with 
 a rib of about the same span as the bridge itself, of which the 
 following seems decisive as regards the present question. By 
 driving in the wedges upon which the rib rested above and be- 
 low, he could raise and lower the crown much more than the 
 distance due to variation of temperature without diminishing 
 
 its supporting capacity Oudry, having thus assured 
 
 himself of the harmlessness of temperature variations, decided 
 upon broad and firm bearing surfaces. 
 
 " Interesting observations have also been made upon the 
 changes of form of the cast-iron arch of 60 metres span over 
 the Rhone at Tarascon, published in the Annales des Ponts et 
 Chemins, 1854, from which, however, it only appeared that the 
 changes of form followed slowly the temperature; that they 
 were less than the received coefficients would have led us to 
 expect, and were nowhere found to be prejudicial. 
 
 " Since, then, this question appears to have been settled more 
 than ten years ago, may we not fear that those who still wish to 
 pivot iron may some day seize upon the idea of pivoting stone 
 arches also ! 
 
 " Stone, as is well known, expands not much less than iron 
 for equal changes of temperature, and, moreover, its modulus of 
 elasticity is much less. The expanded stone arch cannot accom- 
 modate itself to the given span, therefore, as easily as the iron 
 arch, and it would then be clearly more advantageous to pivot 
 the stone arch ! As, however, such a clumsy contrivance would 
 give no great impression of stability, we feel justified in recom- 
 mending a broad and solid bearing surface for all arches." 
 
 As the opinion of an eminent engineer, the above may not be 
 without interest. We would only add that, according to the 
 
270 THE BRACED AKCH. [CHAP. XIV. 
 
 accepted formulae for temperature strains already given, the 
 results are of more importance than the above remarks would 
 indicate. As will be seen in the Appendix, the temperature 
 strains in the braced arch, fixed at ends and continuous at 
 crown, are very considerable, and, if the formulae are accepted 
 as correct, can by no means be disregarded. By comparison of 
 our numerical results for the three cases of braced arch there 
 given, it appears that the one hinged at crown, and springing, 
 is by far the best form of construction, but it must be remem- 
 bered that a different proportion of span to height and depth 
 may considerably affect this conclusion. Upon this point we 
 refer the reader to Art. 28 of the Appendix. 
 
 With the above, we conclude our discussion of braced arches, 
 or arches whose weight is not so great that the effect of the live 
 load can be disregarded, and pass on to the stone arch, or arch 
 proper.* 
 
 * See Appendix, Art. 17, for a practical application of the principles of this 
 chapter. 
 
CHAP. I.] SUPPLEMENT TO CHAP. XIV. 271 
 
 SUPPLEMENT TO CHAPTEE XIY. 
 
 DEMONSTRATION OF ANALYTICAL FORMULA GIVEN IN TEXT. 
 
 In order to complete our discussion of the braced arch, we shall now 
 give the analytical development of the formulae of which we have made 
 use in the preceding chapter. We do this the more readily, as in no book 
 of easy access to the student are these formulae made out. In the work of 
 WinUer, already referred to in the text, will be found a very thorough 
 discussion of the subject. We shall confine ourselves at present to the 
 case of a single concentrated load. 
 
 CHAPTER I. 
 
 GENERAL CONSIDERATIONS AND FORMULAE FOR FLEXURE. 
 
 1. Fundamental Equations The resultant of all the forces act- 
 ing upon a curved piece in a common plane may be decomposed into a 
 force normal to the piece N, and into a compressive or tensile force in the 
 direction of the axis or of the tangent to the axis G ; and this latter force, 
 if taking effect above or below the axis, acts to bend the piece, and gives 
 rise to a moment as well as to a compressive or tensile force G. These 
 forces cause corresponding strains. Thus, if P is the tangential strain per 
 unit of area d a, then 
 
 (1) 
 
 while, if v is the distance of any fibre from the axis, 
 
 M (2) 
 
 (a) Coefficient of elasticity. 
 
 Let the length of a piece be s, its area of cross-section A, and, as above, 
 
 G 
 
 the force acting upon this area G. Then ^ will be the force per unit of 
 
 area. Let the displacement [elongation or compression] produced by this 
 G 
 A 
 
 force be As; the sign A indicating and reading "elongation. 1 " Now 
 
272 
 
 SUPPLEMENT TO CHAP. XIV. 
 
 [CHAP. i. 
 
 experiment shows that within narrow limits, i.e., within the elastic limits, 
 the elongation or compression is directly as the force of extension or com- 
 pression. Supposing that this held true always for all values of A , then, 
 
 since a force produces a displacement A s, the force necessary to produce 
 
 a displacement *, will be 
 
 A s 
 
 E = 
 
 as great. Calling this force E, we have 
 Gs 
 
 The force E, then, is the force which would le necessary to produce a dis- 
 placement s equal to the original length, if the law of proportionality of the 
 displacement to the force always held good for all values of A s. This 
 
 value 
 
 E = 
 
 _ 
 
 AA 
 
 (3) 
 
 we call the coefficient of elasticity. 
 
 From (3) we easily obtain for the force in the direction of the tangent 
 to. the axis 
 
 = EA 
 
 A s 
 
 and for the relative displacement 
 
 A a 
 
 G 
 
 E A 
 
 (4) 
 (5) 
 
 (&) Fibre strain P, and moment M. 
 
 As seen in eq. (1), the longitudinal strain upon an element of cross-sec- 
 tion da is called P. In a curved piece conceive two cross-sections, as 
 
CHAP. I.] SUPPLEMENT TO CHAP. XIV, 273 
 
 shown in the Fig., as A O B D perpendicular to the axis of the piece m n. 
 Let these sections be infinitely near; then, the distance J>a upon the axis is 
 d s. Let d * v be the length of any fibre, as d c, before the change of form. 
 Then, after deformation, its length is = d s v + A d s v . But if d (f> is the 
 small angle between the normals, dsv = ds + vd(f), where v is the distance 
 a c of any fibre from the centre of gravity of the cross-section. After 
 deformation, d s becomes ds + Ads, and d <j> becomes d $ + Ad<f>, and 
 d s v becomes d s v + A d s v . Hence the length of any fibre after deforma- 
 tion is d s v + A d s y = d s + A d s + v (d < + A d <). 
 Subtracting this from the eq. for d s v above, we have 
 
 Therefore the ratio of the change of length to the original length of fibre 
 
 Ads 
 
 is 
 
 ds v 
 
 If r is the radius of curvature, then rdd> = ds, = ; hence 
 
 d s r 
 
 A d s v r A d s A d 0-i r ... 
 
 -^ = h^ +B ir^i (0) 
 
 G 
 
 From eq. (1) we have the strain on a fibre P = -r . 
 
 From eq. (4), G = E A . Hence P = B . In the present 
 
 case 
 
 
 is given by (6) ; therefore 
 
 ds d s 
 
 Since now from (1) G = / Pda, we have from (7) 
 
 G__ Ads r da Ad<f) f*v d a 
 
 E = r ~d7J 7^o + ' ~~d~J r~TV 
 
 Since from (2) M = / P v d a, we have again from (7) 
 
 M_ Ads rvda Ad<f> /V d a 
 B = "~d~s~J 7T^ + "TT'J 7+~V 
 
 But / - is, when 9 is very small compared to r, equal to 
 
 / 2 d a, which is the moment of inertia of the cross-section I. Also, 
 
 da 
 
 7^ = 
 
 r 
 18 
 
274: SUPPLEMENT TO CHAP. XIV. [CHAP. I. 
 
 or T I = I da - I vda + - I = A -f 3, because since 
 
 1 J r + v J rj rjr + v r* 
 
 v is measured from the centre of gravity, / v d a = 0. 
 
 rvda f* fv z da I 
 
 Again, r I = I v a a I = -. 
 
 / T + 1) I I T + V T 
 
 Therefore the insertion of these values of the integrals in the equations 
 
 for _ and _ above gives 
 E E 
 
 Gr Ads r~Ad(f> AflJsnl 
 
 E ~" ds ~ [_ d s ~ r ds\ r 
 
 M rA d ^ A d ~j I 
 E r~ L d s rdsj r' 
 
 M 
 
 G- A a s 
 Inserting the second in the first of the above equations, ^- = v- A = , 
 
 and hence 
 
 _- - + - . (8) 
 
 ds EA + EAr 
 
 Inserting this in the second equation above, 
 
 M M G 
 
 E I + E~A~^ + E~A~7 
 
 (c) Change of length and position of axis. 
 
 From (8) we have at once for the elongation of axis, 
 
 or, when v is very small compared to r, 
 
 1 
 
 From (9) we have for the change of direction of the tangent to the axis 
 
 A ^ = E/ > (T + A~7* + A^r) d8 ' or from ( 10 ) for r 
 constant, that is, for a circle, 
 
 When D is very small compared to r, we have 
 
 ds ....... (13) 
 
CHAP. I.] SUPPLEMENT TO CHAP. XIV. 275 
 
 If the piece had been originally straight, d A (j> would be equal to d <, and 
 
 dA(b dd> 1 El 
 
 3-^ = = -; hence from (13) we have M = - . 
 
 as rd r r 
 
 From the calculus we have the radius of curvature 
 
 r -- * ' or> a PP roximatelv > r = 
 
 hence M = B I ~ . . . . - . . . (13 &) 
 
 This is the equation assumed in the Supplement to Chap. XIII. , Art. 1. 
 2. Displacement of any point. We indicate the horizontal dis- 
 placement of any point of the axis along the axis of a by A x, along y by 
 A y. The corresponding changes of d x, d y, and d s are A d x, Ady, Ads. 
 The total horizontal displacement is then dx + Adx = (ds + Ads) cos 
 A <). The total vertical displacement is d y + A d y = (d s + A d s) sin 
 + A $). Hence, since Adx = d A z, 
 
 d Ax (d s + Ads) cos (< + A$) dx, 
 n (0 + A </>) d y. 
 
 By Trigonometry, cos (0 + A $) = cos </> cos A < sin < sin A 0, 
 sin (0 -f A 0) = sin cos A < + cos sin A <, or if cos A < = 1, 
 
 <? ^ <w 
 
 sin A = A <, cos </> = -r- , sin < = -= . 
 
 Substituting these in the equations above, 
 
 . / Ads\ 
 
 dAx = (dx A(pdy) II +5 I d x, 
 \ I 
 
 (Ads\ 
 1 + j I d y, 
 as / 
 
 or, removing the parentheses, and neglecting quantities of the second order 
 with respect to A (f> and = , 
 
 dAx= A<f>dtM ^ dx 
 
 -di~ dy 
 
276 SUPPLEMENT TO CHAP. XIV. [CHAP. T. 
 
 When the radius of curvature is very great with reference to the thick- 
 ness of the beam, and the relative change of length , is disregarded, 
 
 CL 8 
 
 we have simply 
 
 dAy = A<}>dx. 
 But from (12) A < is equal to / _ _ 
 
 Mds Ads 
 
 H 
 
 /WLds 
 E j , hence, for very small with 
 
 reference to r, 
 
 We shall have frequent occasion to refer to formulae (8), (12), (13), (14) 
 and (15) in the following discussion. 
 
CHAP. H.] SUPPLEMENT TO CHAP. XTV. 277 
 
 CHAPTER II. 
 
 HINGED ARCH IN GENERAL. 
 
 3. Notation Tlie outer fbrce in general. We suppose the 
 ends of the arch to be hinged at the abutments at the centre of gravity of 
 the end cross-sections. Then the end reactions must pass through these 
 points. These end reactions and the loads constitute, then, the outer forces. 
 For equilibrium, then, the horizontal components of these reactions must 
 be equal. Each of these components we call the horizontal thrust. 
 
 We use the following notation [PL 23, Fig. 91] : 
 
 R and R', the reactions at ends A and B. 
 
 V and V, their vertical components. 
 
 H, their horizontal component, or the thrust. 
 
 <z, the half span. 
 
 A, the rise of arch. 
 
 a, the half central angle, if arch is circular. 
 
 The origin of co-ordinates we take at crown, x horizontal, y vertical. The 
 angle of radius of curvature at any point with y, or of tangent to curve at 
 any point with #, we call 0. 
 
 THE OUTER FORCES IN GENERAL. 
 
 Suppose a single load P to act at any point E. Let its horizontal dis- 
 tance from crown be 2, the corresponding central angle E O C be /3. 
 Then the conditions of equilibrium are : 
 
 V + V ' = P, 
 
 Va V a Pz = Q. 
 
 From these last, we have 
 
 (16) 
 
 For a circular arc, since a = r sin a, z = r sin 0, 
 
 sma + sinff sma-sin/3 
 
 2sina ' I ~~ 
 
 We distinguish three segments in the arch [Fig. 91], viz., A E, or from end 
 to the load ; E O, or from load to crown ; C B, or from crown to right 
 end. Quantities referring to the second we indicate by primes, to the 
 third by double primes. Thus for the tangential component of the result- 
 ant at any point within A E, we put G ; for the normal component, N. 
 In E O, then, we have G' and N' ; in O B, G" and N*. We have then 
 
278 SUPPLEMENT TO CHAP. XIV. [CHAP. II. 
 
 G = H cos (f> V sin <, G' = H cos + V sin $ ) ^ 
 N = H sin + V cos $, N' = H sin V cos $ ) 
 
 M = H(fc-y)-V(0-aO, M' = H(fc-y)- V ( + *) ( 19 ) 
 In the case of a circular arc, a = r sin a, h = r (1 cos a), # = / sin <, 
 y = r (1 cos (/>), and hence 
 
 M = 
 JVT 
 
 = H r (cos cos a) V r (sin a sin <) ) 
 = H r (cos cos a) W (sin a + sin <) j 
 
 4. Intersection Line. We call the locus of d [PL 23, Fig. 91], or 
 the curve cdeik, the intersection line. 
 
 If now there are three hinges, one at crown and one at each abutment, 
 then the resultant for each half must pass through the crown O. If, there- 
 
 fore, for the crown (*=0, y=0), we make in (19) M'=0, we have H=V -, 
 or inserting the value of V from (16), 
 
 H = ^|^ ......... (21) 
 
 If the load lies to the left of O, then only the resultant If acts upon the 
 right half, and must pass, as above, through the crown O. The point d 
 lies then always upon B C or A O prolonged. Hence, the intersection lines 
 are two straight lines, which pass through the crown and ends. 
 
 5. Parabolic Arc concentrated Load. For a parabola we 
 
 have y x z ; hence, d y x d x, and, approximately, ds =d x. 
 a' a 
 
 (a) Change of direction of tangents. 
 
 Inserting this value of y in equations (18) for M and M', we have from 
 equation (13), Art. 1, since r d = d s = d x for the change of direction of 
 the tangent at any point before and after flexure, 
 
 Integrating this, we have for the three segments A E, E C and O B, 
 
 . . (22) 
 
 where A, A', A" are constants of integration, to be determined by the as- 
 signment of the proper limits. Thus, if we make x = z, the two first of 
 equations (22) are equal; hence, 
 
CHAP. II.] SUPPLEMENT TO CHAP. XIV. 279 
 
 and accordingly A - A' = (V - V) a z - \ (V + V) z. Since V + V = P, 
 and from (16) V V = P , we have, 
 
 a 
 
 A-A'=lPz* ....... (I.) 
 
 (&) Horizontal displacement. 
 
 Inserting the value of dy for the parabola, viz., ay = xdx, and the 
 
 value of M from (19), and inserting in this last the value of y, viz., 
 
 y = x*, we have from equation (15), Art. 2, 
 2 
 
 Integrating this, we have for the three divisions of the arch, as before, 
 
 For the point B, or x = z, we have from the two first of these equations, 
 B - B' = 1 (V - V) a z* - i (V + V) z 4 - (A - A') z*, that is, 
 
 B-B = - 2 \-Ps* ....... (II.) 
 
 For the crown, x = 0, and the second and third equations are equal, 
 hence 
 
 B' = B* ........ (III.) 
 
 For the left end, that is, for x = a, since the end of the arch must not slip, 
 we must have x x = 0. So also for the right end, for x = a. There- 
 fore, from the first and third equations, putting B' for B", we have 
 
 &Ua 3 h--*sVa 4 ' + iAa 2 + B = 0, 
 - A- H a 3 h + -fr V a 4 + i A" a 2 + B' = 0. 
 
 By the addition and subtraction of these equations, we have, since 
 V + V = P, and (V - V') a = Pz, 
 
 A") 2 -(B + B') = .... (IV.) 
 ^ + 2 4 ) + i(A-A")a 2 =0 . . (V.) 
 
 We might, in a precisely similar manner, form three equations similar to 
 (23) for the vertical displacement A y. This would introduce three more 
 constants and four more equations of conditioh between them. By the 
 nine equations I. to IX. thus obtained, these constants may be then deter- 
 mined in terms of the known quantities H, A, P, a and z, and thus the 
 change of shape at any point may be fully determined. 
 
 The complete discussion, as indicated, is unnecessary for the purpose we 
 
2SO SUPPLEMENT TO CHAP. XIV. [OHA.P. II. 
 
 have in view, and we shall not, therefore, pursue it further. We have 
 already all the general formulae of which we shall need to make use in the 
 discussion of the parabolic arch. 
 
 6. Circular Arc concentrated Load. In a perfectly similar 
 manner we may make out analogous formulae for the circular arch. Thus, 
 referring to equation (8), Art. 1, and inserting for G and M their values as 
 given in (18) and (19), Art. 3, we have for the force in the direction of 
 the axis (see eq. 4), 
 
 . 
 E A - = H cos a V sin a 
 
 <E A - = H cos a V sin a 
 
 ds 
 
 Putting for H r Vand V their values from (21), Art. 4, and (17), Art. 3, we 
 have, 
 
 A d s _ _ sin a + sin /3 2 cos a sin $ 
 
 = 
 
 . . (25) 
 A d s' _ sin a sin /3 
 
 ~dT' 2(1 -COS a) 
 
 (a) Change of direction of tangents. 
 
 Referring to equation (12), Art. 1, we have, since rd<f) = ds and r </> = , 
 
 /M Ads 
 
 EI r ** + -dT*- 
 
 The value of M is given in (20), of ^U-J in (24). Inserting these values, 
 we have 
 
 A < = ^j-j / |~H (cos $ cos a) V (sin a sin 
 
 ==-r- (H cos a -H V sin a) $. 
 
 Performing the integration,* and putting, for brevity, * = -, we have for 
 the three segments of the arc, as before, 
 
 ElA = r2 pK (sin < - < cos a) - V Casino + cos <f>)~^ - K T* (H cos a + V sin a) $ + A "| 
 El A<' =r a [~H (sin</> cosa) V(<sina cos </)"! K r 2 (H cos a + Vsina)0 + A' t 26 
 E I A $" = r* [H (sin <- cos a) - V (< sin a - cos<)]- K r 2 (Hcos a + V sin a)<f + A" J 
 
 For the point E, < = 3, and the first and second equations become simul- 
 taneous. Hence, after reduction, 
 
 A-A' = (V- V) r 2 /3sina + (V + V) r* cos + K (V-V) r z /3sin. 
 
 But V + V = P, and from (6) V - V = P , hence 
 
 sin a 
 
 . . . (I.) 
 
 . * See Art. 7, f allowing. 
 
CHAP. II.] SUPPLEMENT TO CHAP. XIV. 
 
 (b) Horizontal displacement. 
 
 According to eq. (14), Art. 2, since x = r sin 0, y r (1 cos 0), 
 dx = r cos d 0, d y = r sin d 0, we have 
 
 E I A x = r I [H r' 2 (sin cos a) W (0 sin a + cos 0), 
 
 K r* (H cos a + V sin a) + A J sin <Z 0, 
 
 K r 2 / (H cos a + V sin a) cos d 0. 
 The integration gives us the three equations, 
 
 E I A x = r* H ( \ sin cos $ cos a sin + cos a cos 0) 
 
 V (sin a sin sin a cos + sin 2 0)1 
 
 K r 3 (H cos a + V sin a) cos A r (1 cos 0) + B. 
 
 I A #' r 3 1 H ( -J sin cos cos a sin 04-0 cos a cos 0) 
 
 V (sin a sin <^> sin a cos (f> i sin 2 0) I 
 
 K r 3 (H oosa + V sin a) cos A' r (1 cos 0) + B'. 
 
 E I A x" = r* I H (f | sin cos cos a sin + <j> cos a cos 0) 
 
 V (sin a sin sin a cos \ sin 2 <^>) 
 
 K r 3 (H cos a + V sin a) cos A" r (1 cos 0) + B". 
 
 For = , that is at the load, A a? must equal A x'. 
 Hence, after reduction, 
 
 B - B' = -$Pr 3 (2 + sin a /3- 2003/3-2)8 sin /3) + * P r 3 /3 sin /3 . . (II.) 
 For the crown = 0, and A x' A x" ; hence 
 
 B' = B" (III.) 
 
 For the left end, 0=a and A = 0; for the right end, = a and 
 A a" = 0; that is, 
 
 iHr 3 (a 3 sin a COS a -f 2aCOS 2 a) 4- i V r 3 (3 sin 2 a 2 a sin a COS a) 
 K r 3 (H cos a + V sin a) a cos a A r (1 cos a) + B = 0, and 
 
 4- i H r 3 (a 3 sin a cos a + 2 a cos 2 a) | V r 3 (3 sin 2 a 2 a sin a cos a) 
 f K y 3 (H cos a + V sin a) a cos a A" r (1 cos a) + B" = 0. 
 
 The subtraction and addition of these equation gives, after reduction, 
 
 H r 2 (a 3 sin a cos a + 2 a cos 2 a) 
 iPr 2 (3 sin 2 a 2 a sin a cos a 2 sin 2 3 + 2cos/3+ 2/3sin/3) 
 
 + K r z 12 H a cos 2 a + P (a sin a cos a /3 sin /3) I 
 
 + (A-A")(l-cosa) = (IV.) 
 
 and -J Pr 3 sin 3 (3 sin a 2 a cos a) 
 
 - (A + A") r (1 - cos a) - K Pr 3 a cosa sin/3 + B + B' = . . (V.) 
 
282 SUPPLEMENT TO CHAP. XIV. [CHAP. II. 
 
 Here, as, before, we shall leave the discussion, as we have already all the 
 equations of which we shall make use. 
 
 7. Integrals used in the above Discussion. For convenience 
 of reference, we here group together the known integrals employed in the 
 preceding discussion. 
 
 / sin x d x = cos x, / cos x d x sin x. 
 I sin 2 ado: = $x isinxcosx, I cos 2 xdx = ix + sin x cos x. 
 
 I sin x cos x dx i sin 2 x. 
 
 I sin 3 x dx \ cos x (2 + sin 2 x), I cos 3 x d x = $ sin x (2 -+- cos 2 x). 
 I sin 2 x cos x dx = $ sin 3 x, I sin x cos 2 x d x = $ cos 3 x. 
 I xsmxdx = smx x cos x, I x cos x d x = cos x + x sin x. 
 I x sin 2 xdx = x*+i sin 2 x | x sin x cos x. 
 I x cos 2 xdx = x* i sin 2 x + %x sin # cos . 
 I xsinx cos # tZ a? = J (2 sin 2 x # + sin x cos #). 
 
CHAP. III.] SUPPLEMENT TO CHAP. XIV. 283 
 
 CHAPTER III. 
 
 AEOH HINGED AT ABUTMENTS ONLY - CONTINUOUS AT CROWN. 
 A. PARABOLIC ARC CONSTANT CROSS-SECTION CONCENTRATED LOAD. 
 
 8. Horizontal Thrut. We can apply here directly the results of 
 Art 5. Thus, in equations (22) for x = 0, A $ = 0, and A 0" = 0, hence 
 A' = A". If then in eq. (V.) of that Art. we put A' for A", and then for 
 A A' its value from (I.), we have at once 
 
 H _ 5 p5a*-6aV + g*_ (5 a* - z*) (a* - z>) 
 
 H=r * P ~ a* h ~ 64P ~^h~ " (27) 
 
 This is the formula which we have given in Art. 159 of the text, without 
 demonstration. The thrust is greatest when the load is at the crown. We 
 
 have then z and H = |f P % The value of V is given in Art. 3. 
 
 n 
 
 9. Interection Curve. Denote the ordinate of the curve cdei/k 
 (PL 23, Fig. 91), taken above the line A B by y. Then we see from the 
 
 y 
 
 Fig. that y = A N tang, d A N (a z) =. The value of H is given 
 
 above, that of V has already been found in Art. 3, eq. (16). viz., 
 V = P - . Hence we have, after reduction, 
 
 
 (28) 
 
 which is the equation already given in Art. 159, and from (18) and (19) 
 the values of the table in that Art. have been calculated. 
 
 The above values of H and V are simple and of easy application, not in- 
 volving much calculation in any special case. Hence we can readily com- 
 pute H and V, and thus check the accuracy of our method of construction 
 given in Chap. XIV. 
 
 B. CIRCULAR ARC CONSTANT CROSS-SECTION CONCENTRATED LOAD. 
 
 1O. Horizontal Thrut. Here we can apply directly the results of 
 Art. 6. Thus, inserting in eq. (IV.) of that Art. A A' for A A", and 
 taking the value of A A' from (I.), we have an equation for the deter- 
 mination of H. This, after reduction, becomes 
 
 _ sln2 a sin- ft + 2 cos a (cos /3 cos a) 2 (1 + K) cos a (a sin a )3 sin g) 
 2 [a 3 sin a cos a + 2 (1 + *) a cos2 a ] 
 
 which is the equation given in Art. 159 (2) of the text. 
 
 For the semi-circle, a = 90, sin a 1, and cos a = 0, and this becomes 
 
284: SUPPLEMENT TO CHAP. XIV. [CHAP. III. 
 
 If we put 
 
 A! = sin 2 a sin 2 + 2 cos a (cos /3 cos a a sin a + /3 sin j3), 
 A 2 = 2 cos a (a sin a /3 sin $), 
 B! 2 (a 3 sin a COS a + 2 a COS 2 a), 
 B 2 = 2 a COS 2 a, 
 we have 
 
 A 1 -A 2 ^_ *~A/ /A, 
 Bx+B, *V B, \B, 
 
 H* 
 
 A TJ A 
 
 or, if we put A = -^, B= g 2 , and H = P ~, we have 
 
 , 1 A K 
 
 But Ho = P .g- 1 is the value of H from the formula above, when the terms 
 
 containing Ic are disregarded. 
 
 We have also, by series (see Art. 20, following), 
 
 A! = -,V ( 2 2 ) [ (5 a 2 - /3 2 ) - ^o (49 a 4 + 34 .a 2 /3 2 - 11 *) + . . .] 
 A 2 = 2 (a 2 - /3 2 ) [1 - i (4 a 2 + ^ 2 ) + . . .] 
 B : = ^ a 6 (1 - -/ r a 2 + . . .) B 2 = 2 a (1 - a 2 + . . .) 
 Approximately, therefore, since for h small with respect to a, the tan- 
 gent may be taken for the arc, and hence = , or a = , we have 
 
 rah 2 h 
 
 24 6 a 2 15 _ 15 a* 
 
 ~~- 
 
 Hence, when rise is small compared with span, we have the approximate 
 expression 
 
 . 24 - 
 
 5 A 2 1-- - 
 
 15 15 
 
 By means of a table calculated for H , for various values of a and /3 = 0, 
 0.2, 0.4, etc., of a, the thrust can be readily found in any case from the 
 above formula. 
 
 We give in the following Tables the values of H, , A and B, calculated 
 from the exact formulae. The formula for H above is thus made of easy 
 practical application, without tedious calculation, and the results given by 
 the method of Chap. XIV. may easily be checked. 
 
 The value of V is given in Art. 3. 
 
CHAP. HI.] 
 
 SUPPLEMENT TO CHAP. XIV. 
 
 TABLE FOB H . 
 
 285 
 
 a 
 
 a = 
 
 a = 10 
 
 a = 20 
 
 o = 30 
 
 a = 40 
 
 a = 50 
 
 a = 60 
 
 a = 90 
 
 
 
 0.391 
 
 0.391 
 
 0.388 
 
 0.385 
 
 0.380 
 
 0.373 
 
 0.364 
 
 0.318 
 
 0.2 
 
 0.372 
 
 0.372 
 
 0.369 
 
 0.365 0.359 
 
 0.352 
 
 0.342 
 
 0.288 
 
 0.4 
 
 0.318 
 
 0.317 
 
 0.315 
 
 0.309 
 
 0.301 
 
 0.292 
 
 0.278 
 
 0.208 
 
 0.6 
 
 0.232 
 
 0:231 
 
 0.228 
 
 0.222 
 
 0.213 
 
 0.202 
 
 0.187 
 
 0.110 
 
 0.8 
 
 0.123 
 
 0.122 
 
 0.119 
 
 0.115 0.108 
 
 0.099 
 
 0.086 
 
 0.030 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 a 
 
 Coefficients of P^. 
 h 
 
 Thus, for /3 = 0, 
 
 , etc., of a, the numbers in the table give the co- 
 
 efficients of P for a = 0, 10, 20, etc. 
 h 
 
 For the values of A and B, we have the following 
 TABLE FOR A AND B. 
 
 
 a 
 
 gj 
 
 a = 10 
 
 a =20" 
 
 a =30 
 
 a = 40 
 
 a =50 
 
 a = 60 
 
 a =90 
 
 i 
 
 
 
 
 1.20 
 
 1.19 
 
 1.17 
 
 1.14 
 
 1.08 
 
 1.00 
 
 0.88 
 
 
 
 Coeffi- 
 
 
 0.2 
 
 1.21 
 
 1.20 
 
 1.18 
 
 1.15 
 
 1.10 
 
 1.01 
 
 0.90 
 
 
 
 of 
 
 A 
 
 0.4 
 
 1.24 
 
 1.24 
 
 1.21 
 
 1.18 
 
 1.13 
 
 1.05 
 
 0.94 
 
 
 
 a* 
 
 
 0.6 
 
 1.29 
 
 1.29 
 
 1.27 
 
 1.24 
 
 1.20 
 
 4.13 
 
 1.02 
 
 
 
 
 
 0.8 
 
 1.38 
 
 1.38 
 
 1.36 
 
 1.34 
 
 1.30 
 
 1.24 
 
 1.18 
 
 
 
 
 
 1' 
 
 1.50 
 
 1.50 
 
 1.49 
 
 1.47 
 
 1.45 
 
 1.41 
 
 1.36 
 
 
 
 
 B 
 
 
 0.234 
 
 0.233 
 
 0.221 
 
 0.203 
 
 0.178 
 
 0.146 
 
 0.107 
 
 
 
 a 4 
 
 ,0 2 
 
 Thus, for various values of , we have the coefficients of ^, which give 
 
 A for a = 10, 20, etc., and for the values of a have the coefficients of 
 
 a 4 
 
 , which give B. 
 
 11. Intersection Curve. Indicating, as before, by y the ordinate 
 
286 SUPPLEMENT TO CHAP. XIV. [CHAP. III. 
 
 Nd of the curve cdeilc [Fig. 91], we have, as before, y=ANtan^. 
 
 d A N = (a - z) ^, or since a = r sin a, z = r sin fl. V = P sm + sin ff 
 
 1=1 2 sin a 
 
 from eq. (17), 
 
 2 sin a H 
 Inserting the value of H above, we have 
 
 sin 2 a sin 2 /3 ~1 + B B 4 
 
 
 2 sin a l - A AC A 
 
 /sin 2 a sin'/SXBi 
 or if yo = r I -- g-jjT^ - 1^- ; that is, if y is the value of y when & i 
 
 neglected, 
 
 which is the value of y given in Art. 159 (2) of the text. In that Art. we 
 have already tabulated the values of A and B, as also of y for various 
 values of a and /3. 
 
 For /3 = a, that is, for the end ordinate, our expression for y reduces to 
 
 In this case, by differentiating numerator and denominator, we have 
 
 a 3 sin a cos a + 2 (1 + K) a cos" a 
 sin a a COS a K (sin a + a COS a)' 
 
 For the semi-circle, a = 90 = , sin a = 1, cos a = 0, and hence 
 
 2 
 
 y = \ nr = 1.5708 r. Hence, for the semi-circle the intersection curve be- 
 comes a horizontal straight line at 0.5708 r above the crown. In all cases 
 for small central angle c, K may be disregarded. 
 
 The above results are sufficient to enable us to either diagram or calcu- 
 late the strains in every piece for any given position of load. 
 
CHAJP. IV.] SUPPLEMENT TO CHAP. XIY. 287 
 
 CHAPTER IV. 
 
 ARCH FIXED AT ENDS. 
 
 12. Introduction. In the previous case, the end reactions pass 
 always through the ends. If, however, the ends are " walled in," so that 
 the end cross-sections remain unchanged in position, and cannot turn, these 
 reactions pass then no longer through the centres of the end cross-sections. 
 In the first case, the moments at the ends are zero ; now, however, we have 
 end moments to be determined, viz., Mj and M 2 , left and right. For their 
 determination we have the condition that the tangents to the curve at the 
 ends must always remain invariable in direction, or for the ends, A ^> = 0. 
 
 In the arch above with hinges at ends, we have always considered a por- 
 tion lying between the end and any point. In the present case, however, 
 we shall consider the portion between the crown and any point. Both 
 methods lead, of course, to the same results, but the latter, in the present 
 case, is somewhat simpler. 
 
 Accordingly, we conceive the arch cut through at the crown [PI. 24, 
 Fig. 93]. The total resultant force exerted upon the one-half by the other, 
 we decompose into a vertical force V at the crown, and a horizontal force 
 H. The distance c & of this last from the centre of gravity of the section 
 at crown is e , and hence the moment at crown is M H e . 
 
 13. Concentrated Load General Formulae. Let a weight 
 P act at any point; then representing, as before, by primes, quantities 
 relating to the portion between the load and right end, we have, as in (18), 
 
 G = H cos <j> (P V) sin 0, G' = H cos + V sin < | 
 
 N ; = - H sin + (P - V) cos 0, N' = - H sin $ - V sin $ \" ^ 
 
 Also, M = - H (e + y) - Vx + P (x - z), M' = - H (e + y} - V x, 
 
 or, since H e M = moment at crown, 
 
 M = Mo - H y - V x + P(x - ), M' = M - H y - V x . . (32) 
 
 (a) Intersection curve. 
 
 The two reactions, R and R', intersect, as before, in a point L (Fig. 92), 
 which must lie upon P prolonged, as otherwise R, R' and P could not be 
 in equilibrium. The locus of the point L we call, as before, the intersection 
 curve. The equation of this curve can be easily found when V, H and M 
 are known. 
 
 The force acting upon the portion B E (Fig. 92) is the resultant of V 
 and H. The component H acts at the point of intersection o of this 
 resultant L ^ with the vertical through C. The vertical distance of this 
 point o from c is, as above, e ; its horizontal distance from P is z. Then 
 z cot L ^ Tc is the vertical distance of this point from L, and e 4- z cot 
 
288 
 
 SUPPLEMENT TO CHAP. XIV. 
 
 [CHAP. iv. 
 
 = e + z ~ = y, where, as in the Fig., e is negative. In any case, e 
 
 IB 
 
 jyr 
 
 is given by e = ^F 5 hence, for the intersection curve, 
 
 H 
 
 (33) 
 
 (&) Direction curves and segments. 
 
 The direction of the resultants R and R' can be determined in two ways 
 First, by the points of intersection $ and ^ with the verticals through the 
 centres of the end cross-sections ; second, by means of the curves enveloped 
 by these resultants for every position of P. We call the first distances 
 A (f) = Ci, B \|r = c 2 , the direction segments, and the enveloped curves the 
 direction curves. 
 
 Taking Ci and c 2 as positive when laid off upwards above the ends, we 
 have MI = H c\ M 2 = H c 2 ; therefore 
 
 (34) 
 
 We may also easily determine the equation of the direction curves. Let 
 the co-ordinates with reference to the crown of any point be v and w (Fig. 
 92). If the load P is now moved through an indefinitely small distance, 
 the new resultant cuts the former in a point of the curve required. These 
 two resultants intersect the vertical through C in two points. Let the dis- 
 tances of these points from O be c and c + d c, and let y and y + d y be 
 the angles of the resultants with the vertical. Then v (w + c) tan y, 
 v = (w + c + d c) tan (y + d y). 
 
 Eliminating v, 
 
 (c +dc) tan (y + d y) c tan y _ d (c tan y) 
 tan (y + d y) tan y d tan y 
 
 From the first of the equations above we have then 
 
 dc 
 
 v = -= tan 2 y. 
 
 d tany 
 
 But tan y = -, c = - -, c tan y = -, hence 
 
 . . (35) 
 
 w = 
 
 d 
 
 Thus we see that in any case we have only to determine H, V and M , and 
 we can then from (33) and (34) or (35) determine at once the intersection 
 
CHAP. IV.] SUPPLEMENT TO CHAP. XIV. 289 
 
 curve, and the direction segments or curves. These are all we need for our 
 method of construction as given in Chap. XIV. ; once given, we can then 
 easily construct H, V and M 9 or Mi for any position of weight. 
 
 A. PARABOLIC ARC CONSTANT CROSS-SECTION CONCENTRATED LOAD. 
 
 # 2 
 
 14. Determination of H, V and Mo. We put y = h , 
 
 dy = 2 -jdx, ds = dx, as before. Then from the values of M given in 
 
 (32) we have, according to equation (13), Art. 1, after inserting the values 
 of y and d s above, and integrating, 
 
 and 
 
 r- T3T 7. -i 
 
 + A'. 
 
 For x = z, A $ = A $', hence | P (z 2 z) z + A = A', or 
 
 A-A' = iPs 2 ....... (L) 
 
 For x = a, A $ 0, and for x = a, A 0' = 0, hence 
 
 = a 
 
 = -ajMo -iHA + *Vo + A', 
 
 and by addition and subtraction, 
 
 A + A' = Va' t -lP(a-2z)a .... (II.) 
 
 2M a-fHaA + iP(a-2) 2 = . . . (IE.) 
 From I. and II. we have 
 
 A = |V 2 -iP(a 2 - 2as-2 2 ) ) 
 
 A'^-LVa 2 ~iF(a 2 -2as + s 2 ) j 
 
 For the horizontal and vertical displacement of any point, we have from 
 ( 5), Art. 2, after integration, 
 
 E I A x =- i Mo z 3 - rf-i Vz<+iP Q^- | sz 3 ) + i A 2 + B 
 
 ], 
 J 
 
 El Art =- ^-[i Mo z 3 -^ V-* V 4 +i AV+B'l 
 and 
 
 E I A y = iM 2 - ?^ 2 ^ 4 -| (V-P) z'- 
 i a 
 
 B L A y= i Mo x 9 - ? s a; 4 -i V aj+ A' 
 19 
 
290 
 
 SUPPLEMENT TO CHAP. XIY. 
 
 [CHAP. rv. 
 
 For x = z, A x = A a?', and A y = A y', hence 
 
 B-B' = / 4 -Ps 4 -i(A-A> 2 , and C - C' = $P z s - (A - A')z, or 
 
 (IV.) 
 (V.) 
 
 For a = a, A a? = 0, A y = 0, and f or 35 = a, A *' = 0, A y' = 0. 
 hence, 
 
 = + I Mo a 2 - nfeH Ji a* - i (V - F)a 3 - $ Pzo? + Aa + O, 
 
 = + i Mo a 2 ff H h a 2 + % V a 3 A' a + O'. 
 
 The addition and subtraction of the two first and two last of these equa- 
 tions gives, when we put for A +A', A A', B B', O C', their 
 values above : 
 
 2 M a 3 | H Ji a 3 + P (3 a 4 8 a 3 z 4- 6 a 2 s 2 2 4 ) = . . (VII.) 
 
 O + C' = Mo a? + | H h a? P a (a 8 3 a z + 3 s 2 ) . . (VIII.) 
 
 4 f V a 9 i P (2 a 9 3 a 8 s + z*) = (IX) 
 
 Equations HE. and VII. contain only H and M unknown. Their solu- 
 tion gives 
 
 (36) 
 
 a 3 h 
 
 M = - 
 
 From IX. we find directly 
 
 . . . (37) 
 
 (38) 
 
 These are the equations given in Art. 160. From them we have the fol- 
 lowing table : 
 
 z 
 
 H 
 
 V 
 
 Mo 
 
 z 
 
 H 
 
 V 
 
 Mo 
 
 
 
 0.4688 
 
 0.5000 
 
 - 0.09375 
 
 0.5 
 
 0.2637 
 
 0.1563 
 
 + 0.02539 
 
 0.1 
 
 0.4594 
 
 0.4252 
 
 - 0.04936 
 
 0.6 
 
 0.1920 
 
 0.1040 
 
 + 0.02400 
 
 0.2 
 
 0.4320 
 
 0.3520 
 
 - 0.01606 
 
 0.7 
 
 0.1219 
 
 0.0607 
 
 + 0.01814 
 
 0,3 
 
 0.3882 
 
 0.2818 
 
 + 0.00689 
 
 0.8 
 
 0.0607 
 
 0.0280 
 
 + 0.01025 
 
 0.4 
 
 0.3308 
 
 0.2160 
 
 + 0.02025 
 
 0.9 
 
 0.0169 
 
 0.0073 
 
 + 0.00314 
 
 0.5 
 
 0.2637 
 
 0.1562 
 
 + 0.02539 
 
 1.0 
 
 0. 
 
 0. 
 
 0. 
 
 .a 
 
 *f 
 
 .P 
 
 .Pa 
 
 .a 
 
 -: 
 
 .P 
 
 .Pa 
 
CHAP. IV.] SUPPLEMENT TO CHAP. XIV. 291 
 
 From (34) and (35) we can now find the intersection and direction curves. 
 The preceding table gives us sufficient data for complete calculation by 
 moments according to Art. 162. The intersection and direction curves 
 will, as already explained, enable us to find the above quantities graphi- 
 cally. 
 
 15. Intersection Curve. From (33) we have y = , or m- 
 
 H 
 
 serting the values of H, V and M above, and reducing, 
 
 Hy = & P (a ~ ^ (a + z? = iHft, hence y = *fc 
 a 
 
 For the parabolic arch with fixed ends, then, the intersection curve becomes 
 a straight horizontal line, i h above the crown. 
 
 16. Direction Curve. From (36), (37) and (38) we have 
 
 -2 2 g dV SP'-s 8 
 
 dz ~ 8a*h ' dz 4 a 3 
 
 P (a - z) (4a a - 5az - 5g 2 ) 
 
 dz Sa* 
 
 Inserting these in (35), as also the values of H, V and M themselves, and 
 reducing, we have 
 
 (39) 
 
 15 (a + z) (3 a + 2) 
 
 For z = 0, = | a, w = f h. For z = a, v = %a, w h. 
 For z a, v = a, w = oo . Eliminating s, we have 
 
 x 
 15 a (a v) 
 
 This is the equation of an hyperbola. Hence, for the parabolic arc with 
 fixed ends, the direction curve is upon each side of the crown an hyperbola. 
 This hyperbola is described in Art. 160 of the text, (Fig. 93), and a table 
 to facilitate its construction is there given. 
 
 B. CIRCULAR ARC CONSTANT CROSS-SECTION CONCENTRATED LOAD. 
 
 17. Fundamental Equations. From eq. (32) we have, since 
 
 x = r sin $, y = r (1 cos <), z = r sin 0, 
 
 M = Mo Hr(l cos$) + (P V)rsin$ Ffsin/3 
 
 ' = M - Hr(l - ' 
 
 The expressions for G, Art. 13, eq. (12), apply here directly. 
 Therefore, from eq. (8), Art. 1, we have 
 
 
292 SUPPLEMENT TO CHAP. XIV. [CHAP. IV. 
 
 Hence from (12), Art. 1, since ds = rdfa 
 
 ^rM --Hr(l-cos0) + (P-V)rsin(/)-P7'sm/3]^0-f d<p, 
 
 -^rMo-Hr(l-cos<)-Vrsind>l dd) + Sds ' 0. 
 E I L J d s 
 
 Substituting the values of above, integrating, and putting, as be- 
 fore, for brevity, K = ^ we have 
 
 E I A < = r [MO H r (<f> sin <) (P V) r cos $ P r $> sin /3 J 
 
 + K r (M - H> - Gr sin /3) $ + A. 
 
 E I A <' ' = r [M O <j>- Hr ($ sin <) + Vr cos $J + K r (M H>) $+A'. 
 For = 0, A ^ = A 0', and we obtain 
 
 A - A' - Pr 2 [cos ft + (1 + K) j3 sin J . . . (I.) 
 
 For <p = a, A < = 0, and for <J> = a, A #' = 0. Adding and subtracting 
 the equations thus obtained, and eliminating A A', we have 
 
 A + A' = P r 2 1 cos a + (1 + K) a sin /3~| - 2 V r 8 cos a . . (II.) 
 2 Mo a - 2Hr(a- sina)-Pr [cos a - cos 4- (a - 0) sin /3 J 
 
 + K [2 Mo a 2H r a - P r (a j3) sin /3j = . . (HI.) 
 
 From eq. (14), Art. 2, we have, as before, after integrating, for the hori- 
 zontal displacement, 
 
 E I A a? Mo ?* 2 (sin <j> <p cos 4>)+ H> 3 (2 sin <p 2 $ cos </> <j> + sin <j> cos 0) 
 i V r 3 sin a <p + i P r 3 (sin 2 <j> + 2 sin /3 sin ^> 2 ^ sin /3 cos <J>) 
 
 + K T* 2 (M Hr P r sin /3) $ cos + A r cos + B. 
 E I A x' = Mo r 2 (sin <J> <j> cos 0) 4- H r 3 (2 sin <f> 2 < cos <+ sin 4> cos ^>) 
 
 i V r 3 sin 2 + AC r 2 (M H r) <p cos ^ 4- A' r cos + B'. 
 For * = /3, A a; = A #', hence 
 
 B-B' = -iPr(2 + sin 2 /3) .... (IV.) 
 
 Further, for $ = a, A* = 0, and for </> = a, A x' = 0. Hence, by add- 
 ing and subtracting, 
 
 B + B' = Vr (2 - sin 2 a) - Pr (2 - sin 2 a + 2 sin a sin /3) . . (V.) 
 2 Mo (sin a a COS a) HT* (2 sin a 2 a COS a a + sin a COS a) 
 
 + iPr [2 - sin 2 a + sin 2 3 - 2 cos (a - /3) + 2 (a - 0) cos a sin /3j 
 - K f"2(M - Hr a cos a - Pr(a - 0) cos a sin /3J = 0. .(VI) 
 
CHAP. IV.] 
 
 SUPPLEMENT TO CHAP. XIV. 
 
 293 
 
 Multiplying HI. by cos a, and adding to VI., we haye 
 2 Mo sin a H r (2 sin a sin a cos a a) + | P r (sin a sin /3) 2 =0. 
 In similar manner, we have from eq. (14), Art. 2, for the vertical dis- 
 placement 
 E I A y = MO r 2 (cos 4> + </> sin #) i H r 3 (2 cos <f> + 2 $ sin <f> sin 2 0) 
 
 ^P? >3 (4> + sin4>cos4>4-2sin/3cos4> + 2 < sin /3 sin <?>) 
 (Mo H r Pr sin /3) < sin 4- Ar sin <J> + O. 
 2 (cos # + 4> sin 0) |Hr s (2 cos<|> -+- 2 < sin < sin 8 <j>) 
 + V r 3 (< + sin < cos <p) + K r 2 (M H r) < sin ^> + A' r sin # + C'. 
 For $ = 0, A y = A y', hence 
 
 C-C'^Pr'O -f sin /3 cos 0) . . . . (VIH.) 
 Finally, f or ^ = a, Ay = Q. For $ = a, Ay' = 0, and hence 
 
 C + O' = SMor 2 (cos a -I- a sina) + Hr 3 (2 cos a + 2a sin a sin 2 a) 
 
 + P r 3 1 a + sin a cos a 2 sin (a - - j8) + 2 (a 3) sin a sin 3| 
 
 - 2Hr 2 (M -Hr)asina-|-KPr J (a-/3)sinasin/3 . . .(IX.) 
 Vr (a sinaCOSa) = Pr(a 13 sina cos a sin /3 cos /3 4- 2 cos a sin /3). .(X.) 
 18. Determination of H, V and M . 
 (a) Vertical ^Reaction. 
 The vertical force V is given directly by eq. X. Thus 
 
 _a /3 sina cos a sin /3 cos /3 + 2 cos a sin /3 
 
 2 (a sin a cos a) 
 an expression independent of AC. 
 Transforming by means of series, we have, approximately, 
 
 For the semi-circle 
 
 TT 2 3 2 sin (3 cos 3 
 
 27T 
 
 (43) 
 
 (44) 
 (45) 
 
 From the exact formula (43) we have the following table : 
 
 3 
 
 a = 
 
 a = 10 
 
 a = 20 
 
 a = 30 
 
 a = 40 
 
 a = 50 
 
 o = 60 
 
 a = 90 
 
 
 
 0.5 
 
 0.5 
 
 0.5 
 
 0.5 
 
 0.5 
 
 0.5 
 
 0.5 
 
 0.5 
 
 0.2 
 
 0.3520 
 
 0.3515 
 
 0.3500 
 
 0.3475 
 
 0.3439 
 
 0.3392 
 
 0.3332 
 
 0.3065 
 
 0.4 
 
 0.2160 
 
 0.2152 
 
 0.2130 
 
 0.2092 
 
 0.20B7 
 
 0.1966 
 
 0.1876 
 
 0.1486 
 
 0.6 
 
 0.1040 
 
 0.1033 
 
 0.1014 
 
 0.0981 
 
 0.0934 
 
 0.0874 
 
 0.0799 
 
 0.0486 
 
 0.8 
 
 0.0280 
 
 0.0277 
 
 0.0269 
 
 0.0255 
 
 0.0238 
 
 0.0211 
 
 0.0182 
 
 0.0065 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 .a 
 
 .P 
 
 
 . . 
 
294 
 
 SUPPLEMENT TO CHAP. XIV. 
 
 [CHAP. iv. 
 
 (5) Horizontal thrust. 
 
 Eliminating M from HI. and VI. we obtain, after reduction, 
 
 2 sin a Fcos ft - cos a + (1 + *) /3 sin /3~| - (1 + *) a (sin 2 a + sin2 /3) 
 
 H = P L= J . . (46) 
 
 2 I (1 + K) a (a + sin a cos a) 2 sin 2 al 
 
 If we put Ai = 2 sin a (cos /3 cos a + /3 sin /3) a (sin 2 a + sin 2 /3), 
 A 2 = a (sin 2 a + sin 2 /3) 2 /3 sin a sin /3, , 
 
 Bi = 2 a (a + sin a cos a) 4 sin 2 a, 
 B 2 = 2 a (a + sin a COS a), 
 
 A ip A 
 
 and let A = -^- 2 -, B = =*-, and H = =* P, we have 
 
 AI J3l JD! 
 
 . 1- AK 
 
 1 + B *' 
 
 where H is the value of H from the above formula, when terms containing 
 K are disregarded. 
 Transforming by series, we have 
 
 H = 
 
 From the exact formula (46) above, we have the following tables : 
 TABLE FOB H . 
 
 /B 
 
 a=0 
 
 a = 10 
 
 a =20 
 
 a = 30 
 
 n = 40 
 
 a =50 
 
 a = 60 
 
 a =90 
 
 
 
 0.4688 
 
 0.4687 
 
 0.4683 
 
 0.4678 
 
 0.4671 
 
 0.4661 
 
 0.4610 
 
 0.4592 
 
 0.2 
 
 0.4320 
 
 0.4317 
 
 0.4309 
 
 0.4291 
 
 0.4272 
 
 0.4243 
 
 0.4173 
 
 0.4017 
 
 0.4 
 
 0.3308 
 
 0.3301 
 
 0.3281 
 
 0.3244 
 
 0.3196 
 
 0.3128 
 
 0.3012 
 
 0.2601 
 
 0.6 
 
 0.1920 
 
 0.1912 
 
 0.1887 
 
 0.1845 
 
 0.1784 
 
 0.1703 
 
 0.1578 
 
 0.1087 
 
 0.8 
 
 0.0608 
 
 0.0603 
 
 0.0590 
 
 0.0566 
 
 0.0534 
 
 0.0490 
 
 0.0421 
 
 0.0181 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 .a 
 
 | 
 
CHAP. IV.] 
 
 SUPPLEMENT TO CHAP. XIV. 
 
 295 
 
 For the values of the quantities A and B we have the f ollowing table : 
 VALUES OF A AND B. 
 
 
 * 
 
 . = 
 
 a =10 
 
 a =20 
 
 =30 
 
 a =40 
 
 a=50 
 
 a=60 
 
 o=90 
 
 
 
 
 
 3.00 
 
 3.01 
 
 3.02 
 
 3.06 
 
 3.10 
 
 3.16 
 
 3.25 
 
 3.66 
 
 
 
 0.2 
 
 3.12 
 
 3.15 
 
 3.16 
 
 3.19 
 
 3.26 
 
 3.34 
 
 3.45 
 
 4.07 
 
 
 
 0.4 
 
 3.57 
 
 3.58 
 
 3.63 
 
 3.70 
 
 3.81 
 
 3.79 
 
 4.20 
 
 5.65 
 
 a* 
 
 A 
 
 
 
 
 
 
 
 
 
 
 
 
 0.6 
 
 4.69 
 
 4.70 
 
 4.81 
 
 4.96 
 
 5.20 
 
 5.57 
 
 6.14 
 
 10.57 
 
 h* 
 
 
 0.8 
 
 8.33 
 
 8.42 
 
 8.67 
 
 9.10 
 
 9.78 
 
 10.87 
 
 12.93 
 
 35.62 
 
 
 
 1 
 
 CO 
 
 CO 
 
 CO 
 
 CO 
 
 CO 
 
 CO 
 
 CO 
 
 CO 
 
 
 B 
 
 .a 
 
 2.813 
 
 2.825 
 
 2.861 
 
 2.931 
 
 3.039 
 
 3.198 
 
 3.417 
 
 5.279 
 
 a* 
 
 From the above tables it is easy to find the thrust for any given position of 
 load, and any given span and rise. The preceding table gives the reac- 
 tion ; it only remains to determine the moment M at crown. 
 
 (c) Moment at crown. 
 
 From VII. we have 
 
 (sin a sin B) 2 
 v 
 
 /~ a \ 
 
 Mo = JHr [2 cos a- - - )- 
 
 \ sin a/ 
 
 sin a 
 
 Substituting the value of H, already given, eq. (46), we obtain 
 2 Mo f"(l + K) a (a + sin a cos a) 2 sin 2 a~| 
 
 = Pr I sina sina cos (a )+2 sina (cos # cosa) sina (sina sin /3) sin ,8 
 
 a (cos /3 cos a) (1 + K) { a (sin 2 a -t- sin 2 [3) 2 ^ sin a sin /3 J 
 
 + (1 + K) (a /3) (a + sin a cos a) sin /3 | . 
 
 In similar manner, as before, for H we have 
 
 0* 
 
 MO=M O 
 
 l+B/c' 
 
 where M 00 is the value of M when terms containing Tc are disregarded, 
 and B has the same value as above. By series we have 
 
 [~3a 2 - 
 L 
 
 10a/3-3/3 2 
 
 and 
 
 ^4 (3a 4 + 6a* 0+ 45 a 2 /3 2 + 308a/3 3 + 154/3 4 )!, 
 
 = 
 
 360 
 
 a 2 (3a 2 -10a/3-5/3 2 ) 
 
296 SUPPLEMENT TO CHAP. XIV. [CHAP. IV. 
 
 From the exact formula above we have the following tables. 
 VALUE OF M 00 . 
 
 & 
 
 a = 
 
 0=10 
 
 a = 20 
 
 o = 30 
 
 a =40 
 
 a =50 
 
 a = 60 
 
 a = 90 
 
 
 
 0.09375 
 
 0.09426 
 
 0.09582 
 
 0.09849 
 
 0.10241 
 
 0.10777 
 
 0.11613 
 
 0.15108 
 
 0.2 
 0.4 
 
 0.01606 
 0.02025 
 
 0.01615 
 0.02021 
 
 0.01658 
 0.02008 
 
 0.01742 
 0.01973 
 
 0.01846 
 0.01942 
 
 0.02063 
 0.01888 
 
 0.02280 
 + 
 0.01719 
 
 0.03083 
 + 
 0.01441 
 
 0.6 
 
 0.02400 
 
 0.02393 
 
 0.02369 
 
 0.02326 
 
 0.02267 
 
 0.02181 
 
 0.02001 
 
 0.01454 
 
 0.8 
 
 0.01025 
 
 0.01019 
 
 0.00999 
 
 0.00962 
 
 0.00917 
 
 0.00850 
 
 0.00716 
 
 0.00332 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 .a 
 
 .Pa 
 
 VALUES OF B AND C. 
 
 
 
 
 a=0 
 
 a = 10 
 
 a = 20 
 
 a = 30 
 
 a =40 
 
 a = 50 
 
 a =60 
 
 o = 90 
 
 
 
 
 
 7.50 
 
 7.52 
 
 7.57 
 
 7.78 
 
 7.94 
 
 8.52 
 
 8.61 
 
 11.12 
 
 
 
 0.2 
 
 28.13 
 
 28.09 
 
 27.99 
 
 28.14 
 
 28.43 
 
 27.99 
 
 27.76 
 
 31.81 
 
 
 
 
 0.4 
 
 12.50 
 
 12.61 
 
 13.05 
 
 14*08 
 
 15.40 
 
 17.35 
 
 21.11 
 
 37.47 
 
 a 4 
 
 
 0.6 
 
 4.69 
 
 4.77 
 
 5.00 
 
 5.51 
 
 6.24 
 
 7.31 
 
 9.10 
 
 20.27 
 
 
 
 0.8 
 
 2.74 
 
 2.85 
 
 3.20 
 
 3.89 
 
 4.94 
 
 6.61 
 
 9.77 
 
 42.53 
 
 
 
 1 
 
 1.88 
 
 00 
 
 00 
 
 00 
 
 00 
 
 00 
 
 00 
 
 00 
 
 
 B 
 
 .a 
 
 +2.81 
 
 +2.83 
 
 +2.86 
 
 +2.93 
 
 +3.04 
 
 +3.20 
 
 +3.42 
 
 +5.28 
 
 a 4 
 
 Here, as always, a negative moment denotes tension in lower or inner 
 flange. We see at once from the table that the maximum compression in 
 this flange at crown does not occur for full load, but for load extending 
 from both ends towards the crown as far as about gths of the span or fths 
 of the half span. "Within the middle half of the arch, then, a load any- 
 where causes tension in lower flange at crown outside of this middle half 
 a load anywhere causes compression in the lower flange at crown. For 
 large central angles, K may be disregarded, and we have simply M = M 00 . 
 
CHAP. IV.] SUPPLEMENT TO CHAP. XIV. 297 
 
 19. Intersection Curve. From Art. 13 we have 
 V r sin 3 - M 
 
 y= ^ 
 
 Hence, by substitution of the values of V, H and M , 
 
 - r j l~l - 12 ( a * -2afl-/3 2 ) 1 
 y "" 10 L " a 4 (a + ft) 2 *J 
 
 which is the equation given in the text, for which a table is there given. 
 
 2O. Direction Segments. It will in the present case be found 
 most convenient to determine the directions of the resultants by d and e 2 
 equation (34). 
 
 M! M 2 - 
 
 Thus, Ci = - -|p c 2 = -jj-. 
 
 But M! = Mo - H h + (P V) a P z, M 2 = M H h + V a. 
 We have, by series, then the approximate formulae, 
 
 45Ififl 2h T 451 
 
 15 (+) L A7i 2 J 15 (-+- 
 
 where positive values of Ci and c 2 are laid off upward above, negative 
 values downward below, the centres of gravity of the end cross- sections. 
 
 From the preceding tables we can calculate easily in any case H and V 
 an^. Mo, and thus check the results obtained by the method of Chap. XIV. 
 The formulae above for d and c 2 do not admit of tables, nor, in fact, are 
 such needed. They are sufficiently simple for ready insertion. 
 
 Thus, by the aid of our tables, having computed V and H, and, if neces- 
 sary, Mo and 0o, we can by the method of moments, as explained in Chap. 
 XIV., Art. 162, readily calculate the strains in the braced arch, whether 
 continuous at crown and fixed or hinged at the ends, or hinged at both ends 
 and crown. 
 
 2O (ft). Transformation Series. We have in the preceding repeat- 
 edly made use of series in the transformation of angular functions, such as 
 sin, cos, etc., into functions of the arc itself. We group here, for conve- 
 nience of reference, the series thus used : 
 
 sin x = x (I - i 2 + 7^0 aj 4 - <ro L 4 -u 6 + Wainr & - uinrunn) o * 
 
 COS X = 1 - i X 2 + & X* - Ti b ? + Toku " ~ TSjtv ^ 
 
 sin 2 x = x 2 (I - -U 2 
 cos 2 0=1- x 
 sin x cos x = x (I I x 2 -t- -ft x* 3-73- x + dW 9 ITT&TS a? 1 + ...) 
 
298 SUPPLEMENT TO CHAP. XIV. [CHAP. IV. 
 
 COS 3 X = 1 - t X 9 + U 4 - #jj X s + Tfffcj X 8 - TTWoo Z 1 + ... 
 
 tan a; = x (1 + 1 x 2 + -fr a* + & x + 7 f| F z s + rHfle 10 4- ...) 
 
 1 # 
 
 cot * = - - (1 + -iV x 2 + ^ tf + -nky a; 6 + ^if ? 8 + ...) 
 
 sin x siny = xy [l 
 + 7 3 2 
 
 cos a; cos y=l- 
 +15 
 
 sin cos y=a [l- 
 +35 
 
CHAP. V.] SUPPLEMENT TO CHAP. XIV. 299 
 
 CHAPTER V. 
 
 INFLUENCE OF TEMPERATURE. 
 
 21. Oeneral considerations. When the temperature of a per- 
 fectly free body, which possesses in every direction the same coefficient of 
 elasticity and expansion, changes equally at all points, there can be no 
 strains in the body. For were there such strains, then, as there are no 
 outer forces, there could be no equilibrium. 
 
 If, however, the change of temperature is not the same at all points ; or 
 if the body is not free, so that it is possible for outer forces to act, there 
 are strains. 
 
 In the following we assume the change of temperature to be everywhere 
 the same, but that the body is not free. 
 
 We assume that at a certain temperature t no strain exists in the body, 
 and call this the mean temperature. The deviation above or below the 
 mean temperature we call + t or t, and denote the coefficient of expan- 
 sion for one degree by e. 
 
 The determination of the strain in a straight beam held at both ends, is 
 very simple. If the length is I, its relative change of length is e t. Since, 
 however, it cannot expand, the strain S per unit of area is precisely as great 
 as the force which would be required to produce this relative elongation, 
 
 or from eq. (4) s = + E e t ........ (48) 
 
 If the area of cross-section is A, then the strain at each end is 
 
 In equation (48) it is assumed that a compressive strain, due to + t, is 
 positive, a tensile strain, due to t, is then negative. 
 
 22. Influence of Temperature on the Arch. Since by a 
 change of temperature the length of the arch varies, while the span remains 
 always the same, the shape or curvature must change, which naturally must 
 give rise to strains and outer forces. In the following we have only to 
 determine these outer forces, since, as shown in Chap. XIV., these are all we 
 need to determine the strains themselves. 
 
 *f / TVT\ 
 
 The relative change of length is from eq. (8), Art. I., ^-^ JG + J. 
 
 This change is caused by the outer forces. The relative change of length 
 due to temperature alone is e t. Hence the total relative change of length is 
 
 Gr + M 
 
300 SUPPLEMENT TO CHAP. XIV. [CHAP. V. 
 
 Hence the change of length of the axis is 
 
 / d s f t s (50) 
 
 The change of the angle between two infinitely near cross-sections, and the 
 actual turning of a cross-section is from (9) and (12), Art. 1, given by 
 
 M G r + M 
 
 d - El EA?> 
 
 Gr 
 
 1 /M 1 /^ 
 
 l - (j> = Ej T ds+< B~rJ 
 
 A 
 
 Finally, from (14) we have 
 
 d s 
 
 ds 
 
 /(*&d s 
 &<!>dy + I -j^ 
 
 / 
 ^t X 
 
 Substitute in these last two equations for A <j> and - their values from 
 
 (.1 8 
 
 (49) and (52). The double integral thus arising can be resolved by par- 
 tial integration. 
 
 Thus /* //(*)<& 
 
 Applying this, we obtain 
 
 r r& 
 
 Ax yA0+ / yd&+ I dx 
 
 J J *' .... (5S) 
 
 /r&ds 
 zdA<f>+ I -jdy 
 
 We shall assume in the following the axis always circular. 
 
 23. Fundamental Equations General. Upon this assump- 
 tion of a circular axis we have generally 
 
 G = H cos <, N = H sin <f> "J 
 
 M = M -f Hr(l-cos^) I ..... (54) 
 
 Gr + M = M + Hr j 
 
 Hence, from the preceding Art, 
 
 H r + Mo ' /** 
 A s = -Id + Ttt$ ..... (55) 
 
 E A Jo 
 
 AEy . 
 
 
 A a? = r A (1 cos <j>) H / M (1 cos $>) d $ 
 
 r + Mo r _ 
 
 EA J * *?*+ 
 
CHAP. V.] SUPPLEMENT TO CHAP. XIV. 301 
 
 24. Arch with three Hinges. If there are three hinges, the mo 
 ment M at the crown must be zero, and therefore M = H?*(l cos <). 
 But for = a, M must also be zero, hence H r (1 cos a) = 0, and therefore 
 H is zero. Then for any point G = 0, and M = 0, and N = 0. That is, 
 for the arch with three hinges there are for a change of temperature no outer 
 forces, and hence no strains. 
 
 25. Arch hinged at Ends. Here, since f or = a, M = 0, we 
 have from. (54) 
 
 Mo = H r (1 cos a) \ 
 
 M = - H r (cos - cos a) I (58) 
 
 G r ~\- M = + H r cos a J 
 . From (56), since for 9 = 0, A = 0, 
 
 A ^ = / (cos cos a) d + cos a / d $ . . (59) 
 
 J o Jo 
 
 From (57), since for = 0, A * = 0, 
 
 Hr'f f* C* 1 
 
 A 02= I (1 COS 0) / (COS COSa)^0 / (1 COS 0) (COS COS a) d I 
 
 L Jo Jo J 
 
 H r cos a cos 61, 
 
 I ^r~a I d ^ r e t sin (j>. 
 
 E A / 
 
 Jo 
 
 For ^ = a, this becomes zero, and we have for the horizontal thrust 
 _, _ E c t sin o 
 
 r 2 C a >cos /"" cos 2 a /*" ' 
 
 I cos <4 (cos cos a) d <h I (cos0 cos a)d6-\ i I a$ 
 
 I / I / ^ A / 
 
 /0 i/0 t/0 
 
 .... (60) 
 Performing the integrations indicated (Art. 7), and putting, for brevity, 
 
 K = , we have 
 
 Ar 2 
 
 _ 2EIfsina 
 
 r 2 (a 3 sin a cos a + 2 a cos' 2 a) + 2 x?* 2 a cos 2 a 
 
 By series (Art. 20), we obtain the approximate formula 
 
 r 2 (2 4 + 15/c) 8AA 2 + 151 
 
 The above are the expressions given in Art. 165 without proof. The less 
 h, the greater for equal dimensions is H. For h = 0, we have 
 H = E A e t, as we should have for a straight beam. 
 
 26. Arch without Hinges. In this case we have the general equa- 
 tions (54) and (55), which apply directly without change. 
 
 From (56), since for = 0, A = 0, we have 
 
302 SUPPLEMENT TO CHAP. XIV. [CHAP. V. 
 
 From (57), since for $ = 0, A x 0, we have, inserting the value of A <, 
 above, 
 
 ."r A r , i 
 
 A x = 1 cos & I dfy I cos a I 
 
 H :J F A A 
 
 ~^F I C 1 ~ cos <?) / (1 - cos 0) <Z0 - / (1 - 
 EI L Jo Jo 
 
 r* 
 
 / d $ 
 Jo 
 
 -+- =-. - - cos / d $ r 1 1 sin 
 E A 
 
 For = a, A ^ = 0, hence from the first of these expressions 
 r 2 C" 1 C a 
 
 = 
 
 r* c* i r 
 
 I dQ + j Ic 
 
 l jo A Jo 
 
 If the distance at which the horizontal thrust H acts from the crown is 
 o, we have M = H e , whence we see at once that e is the fraction in 
 (63) multiplied by r. For </> = 0, A a must also be zero, and we thus 
 obtain another relation between Mo and H which does not contain A. If 
 we multiply the expression thus obtained by r cos a, and then subtract the 
 result from that previously obtained for = 0, A $ = 0, we have 
 
 
 ^ /" a /* a 
 
 | / (l-cos0)d0- / (l- 
 
 .... (64) 
 Performing the integrations indicated in (63), we have (Art. 7) 
 
 where, as before, K = - . 
 Ar 2 
 
 From (64) we obtain 
 
 M r sin a iHr 2 (a 2 sin a + sin a cos a) = E I f t sin a. 
 Inserting the value of M above, we have 
 
 H = 8BI,<(l + ,).rin. 
 
 2 [(1 + K ) (a 2 + a sin a COS a) 2 sin 2 a] 
 
 r 
 and hence 
 
 M - 
 
 ""[(! + f) (a 2 + a sin a cos a) 2 sin 2 a] 
 
 From these two we obtain for the point of application of H 
 
 = -igt = + ( 1 +')''- rin " r .... (88) 
 (1 + K) a 
 
CHAP. V.] SUPPLEMENT TO CHAP. XIV. 303 
 
 By series, we hare (Art. 20) , 
 
 without reference to K, e = $ h. 
 
 For small central angles, then, for which < may be disregarded, the 
 thrust given above by (66) acts at } Ji below the crown for a rise of tem- 
 perature of t degrees above the mean. For a decrease of temperature be- 
 low the mean it acts above, M is negative, and the strain in the lower 
 flange tensile. 
 
 Further, we have, by series, the approximate formulae 
 
 r 2 (a 4 + 45*) 4 A A 2 +451 
 
 These are the expressions given in Art. 165 without proof. 
 
304 SUPPLEMENT TO CHAP. XIV. [CHAP. VI. 
 
 CHAPTER VI. 
 
 PAETIAL UNIFORM LOADING. 
 
 27. Notation. In the preceding discussion of the -arch we have con- 
 sidered the influence of a > single concentrated load only, and this, as we 
 have repeatedly seen in the case of the simple and continuous girder, etc., 
 is sufficient for full and accurate solution. When once we are able to find 
 and tabulate the strains in every piece due to a single load in any position, 
 the thorough solution becomes simply a question of time. 
 
 It may often happen, however, that we may wish to determine the strains 
 for a full load only, or for a uniformly distributed load extending from 
 one end to some given point. In such case it would be unnecessarily tedi- 
 ous to obtain our result by the successive determination and addition of all 
 the intermediate apex loads. We may easily deduce from the preceding 
 the general formulae for partial loading also. 
 
 As before, we shall let a = the half span, h = the rise, I the moment of 
 inertia, and A the area of the cross-section. But we shall represent by p 
 the load per unit of length of horizontal projection, and by z the distance 
 of the end of the load extending from the left, from the crown. This dis- 
 tance z, from the crown to the end of load, is then positive towards the 
 left. In the circular arch the angle subtended by this distance -z we call /3. 
 The angle /3 is then positive to the left of the vertical. The angle sub- 
 tended by the half span is, as before, a. For /3 = a, then, or for z a, 
 there is no load upon the span. For /3 = 0, or z = 0, the load extends 
 from the left to the centre. For /3.= a, or z = a, the load covers the 
 whole span. Pis. 23 and 24, Figs. 91 and 92, still hold good, therefore, for 
 our notation. We have only to conceive, instead of the concentrated load 
 P, a uniformly distributed load, per horizontal unit, extending from left end 
 as far as the position of P. This much being premised as to notation, we 
 shall treat, as before, the three cases of arch hinged at crown and ends, 
 hinged at ends only, and without hinges. 
 
 A. ARCH HINGED AT CROWN AND ENDS. 
 
 28. Reaction. This case is too simple to demand any extended 
 notice, in view of what has already been said. We have from eq. (16), Art. 
 3, for the reaction at the left or loaded end, for concentrated load, 
 
 2a 
 If now we put P = p d z, and integrate, we have 
 
 r ., a + z 2az + z 
 V = / p d z = p -. -- h C, 
 J * 2a a 
 
 where O is the constant of integration. By taking the proper limits, we 
 can eliminate this constant, and thus obtain the reaction for load covering 
 
CHAP. VI.] SUPPLEMENT TO CHAP. XIV. 305 
 
 any desired portion of the span. As we shall in every case suppose the 
 load to extend from the left end up to any point, we shall take the limits 
 of z = a and 2, and therefore obtain 
 
 For z a, this is zero, as it should be, since then the load has not come 
 on. For z = a, the load extends over the whole span, and V = p a, or 
 half the whole load, as it should. We might have obtained this result at 
 once by moments. Thus, 
 
 V x 3a = a - 
 
 but have preferred the above method as showing how uniform loading is 
 deduced directly from concentrated by inserting pdz for P and inte- 
 grating. 
 29. Horizontal Thrut. In precisely similar manner we have 
 
 from (21), Art. 4, for the thrust due to concentrated load P, H = P ^ ~ e \ 
 
 2 h 
 
 Put P p d 3 and inte-grate between the limits z a, and 2, and we have 
 
 For z = a, this is zero, as should be. For z = a, or for full load over 
 
 'D QJ^" 
 
 whole span, H = ^T. We may also deduce the above equation directly 
 
 tii fi 
 
 by moments. 
 
 The above formulse (71) and (72) are all that we need either for calcula- 
 tion or diagram. They apply evidently equally well, whether the arch be 
 circular or parabolic, or, in general, whatever its shape may be. The form 
 has no influence upon either the thrust or the reaction. 
 
 For the moment at any point whatever, whose distance horizontally 
 from crown is x and vertically below crown y, we have at once 
 
 M = H (h - y) - V (a - x) + 2- (a - a) 2 . 
 
 2 
 
 If this point is an apex, then the moment divided by depth of arch at 
 this point is the strain in flange opposite that apex. A positive moment 
 throughout this work always indicates compression in the inner or lower 
 
 B. AKCH HINGED AT ENDS ONLY. 
 
 30. Reaction. The vertical reaction at the end is precisely the same 
 as before for three hinges, and is given by equation (71). This reaction is 
 evidently independent of the shape of the arch, and the above formulae 
 holds good generally. 
 
 31. Horizontal Thrt Parabolic Arch. We must here 
 distinguish the shape of the arch, and treat first the parabola. We have 
 already from eq. (27), Chapter III., Art. 8, for a single load, 
 
 5 
 
 ~ 
 
 20 
 
306 SUPPLEMENT TO CHAP. XIV. [CHAP. VI. 
 
 We have, as before, simply to make P = p d z, and then integrate between 
 the limits z = a and z indeterminate. 
 We thus find at once 
 
 n r -i 
 
 . . (73) 
 
 For z = a, this reduces to zero, as it should. For z = a, the load covers 
 
 the whole span, and we have H ^-. For z = o, the load reaches from 
 
 2 h 
 
 2 
 
 the left as far as the crown, and H -r-^-. The formulae is simple, and re- 
 
 4ft 
 
 quires no table. Numerical values may be easily inserted. 
 
 32. Horizontal Thrust Circular Arch. As already noticed, 
 the vertical end reaction for this case has been given in eq. (71). It re- 
 mains to determine the thrust. We have, as before, simply to insert pdx = 
 p r cos /3 d 3 in place of P in the expression for the thrust for concentrated 
 load of Art. 10, and then integrate between the limits /3 = a and /3 inde- 
 terminate. 
 
 We have thus similarly to that Art. 
 
 where H is the value of H when terms containing K are neglected, or 
 
 H.=?;|- ; andA = ^ B = B 1 
 12 BI AI BI 
 
 The quantities Ai, Bi, A 2 and B 2 are as follows : 
 
 A, = 7 sin 3 a + 3 a cos a 3 sin a 6 a cos a sin 2 a 6 sin 2 a sin /3 
 + 2 sin 3 /3 3 /3 cos a 9 cos a sin cos /3 + 12 cos 2 a sin /3 
 + 12 a cos a sin a sin 6 cos a sin 2 /3. 
 A 2 = 3 [2 a cos a sin 2 a + a cos a sin a cos 2 a 4 a cos a sin a sin j3 
 
 + 2 |3 cos a sin 2 /3 cos a + cos a sm cos 3]. 
 Bj = a 3 sin a cos a + 2 a cos 2 a. B 2 = 2 a cos 2 a. 
 
 These expressions can be tabulated as in Art. 10, or developed into 
 series as in that Art., and the formula thus made practically available. 
 
 For = a, we have H zero, as should be. For ]3 = a, we have the 
 load covering the entire span. 
 For this case we have 
 
 H _l f sin3 q - 3 (12 sin2 a) (sin a -acos a) - 3 K cos a (a + 2 a sin a sin a coa a) 
 ' ~ 6 "a + 2 a cos'-i a 3 sm a cos a + 2 K a cos 1 ^ a 
 
 For the SEMI-CIRCLE, this reduces simply to 
 
 H = pr = 0.424 pr. 
 
 3 7T 
 
 In any case where exact results are desired, eq. (74) must be used, and a 
 table calculated for the central angle a. We have approximately by 
 series also, more especially for small central angles, or for a large in respect 
 to hj for total load over whole span : 
 
 j>a 2 8 Aft* _pa* 8ft 2 
 
 2 ft 15 I + 8 A ft 2 ~ 2 ft 15 g* + 8 ft 2 
 
CHAP. VI.] SUPPLEMENT TO CHAP. XIV. 307 
 
 where A is the area and I the moment of inertia of cross-section, and g 
 the radius of gyration. In framed arches this may be taken as approxi- 
 mately equal to the half depth from centre to centre of flanges. 
 
 C. ARCH WITHOUT HINGES FIXED AT ENDS, CONTINUOUS AT GROWTH. 
 
 33. Parabolic Arch Formulae for V, H and M. In this 
 case the reactions no longer follow the law of the lever, and eq. (71), there- 
 fore, no longer holds good. 
 
 (a) Vertical reaction at unloaded end. 
 
 We have from eq. (28), Art. 14, for the reaction at the right end for a 
 single load, 
 
 1 (a - zY (2 a + e) 
 ~4 P ~~^~~ 
 
 Making P =pdz, and integrating between the limits z = a and a, we 
 find the reaction for a load coming on from left, 
 
 V = 
 
 L. 3 a 4 - 8 a :J z + 6 a 2 z 2 - z*\ . . . (76) 
 
 for a full load z = a and V=pa, as should be. 
 
 (5) Horizontal thrust. 
 
 In like manner we have for the horizontal thrust at end from (36), Art. 
 14, 
 
 15 p g* - 2 a 2 z* + g* 
 H = 32 a*h 
 
 Replacing P by p d z, and integrating as before, we obtain directly 
 
 H= - -15a2 + Wa*z*-3z* . . (77) 
 
 in $2 
 
 for a full load z = a, and H = . 
 
 to h 
 
 (c) Moment at unloaded end. 
 
 In precisely similar manner we have from (37), Art. 14, 
 
 _ 1 (a zY (3 a 2 - 10 a z - 5 g 2 ) 
 ~32 ~~^~~ 
 
 Putting P = p d z, and integrating, we harve for the moment, always at 
 the right end, or for load not extending past the centre, at crown 
 
 M 
 
 = -- [3a 4 z-8a 3 z* +6a*z* - g 8 ] . . (78) 
 32 a L J 
 
 For z = a, this is zero, as should be. For z 0, or for load extending as 
 
 far as crown, it is also zero. For z = a, the moment at the end is ^ . 
 
 2 
 
 A negative moment, as always, denotes tension in lower flange. 
 
 Just as for concentrated load, as shown in Art. 14, as the load comes on, 
 the moment at crown is positive, and increases with increasing load up to 
 a certain point, beyond which any load causes a negative moment, and be- 
 yond which the moment at crown, therefore, decreases, until, when the load 
 
303 
 
 SUPPLEMENT TO CHAP. XIV. 
 
 [CHAP. vi. 
 
 reaches the crown, it becomes zero. This point, which gives M , the mo- 
 ment at crown, a positive maximum, is at a distance z = a + aVf 
 0.264911 a, or nearly i a from the crown. 
 
 The values of V 2 , H and M a (M 2 and V? being always the moment and 
 reaction at unloaded end), for various values of z, are given in the following 
 table: 
 
 z 
 
 V 2 
 
 H 
 
 M 2 
 
 z 
 
 V 2 
 
 H 
 
 M 2 
 
 1 
 
 
 
 
 
 + 
 
 0.1 
 
 0.2349 
 
 0.29656 
 
 -0.01206 
 
 0.9 
 
 0.0002437 
 
 0.000579 
 
 0.0001096 
 
 -0.2 
 
 0.3024 
 
 0.34128 
 
 -0.03024 
 
 0.8 
 
 0.0014 
 
 0.00421 
 
 0.00076 
 
 -0.3 
 
 0.3707 
 
 0.38242 
 
 -0.055611 
 
 0.7 
 
 0.00624 
 
 0.01643 
 
 0.002185 
 
 -0.4 
 
 0.4459 
 
 0.41846 
 
 -0.08918 
 
 0.6 
 
 0.0144 
 
 0.02889 
 
 0.00432 
 
 -0.5 
 
 0.5273 
 
 0.44824 
 
 '-0.131836 
 
 0.5 
 
 0.0273 
 
 0.051757 
 
 0.006836 i 
 
 -0.6 
 
 0.6144 
 
 0.47104 
 
 -0.18432 
 
 
 
 
 1 
 
 
 
 
 
 0.4 
 
 0.0459 
 
 0.08346 
 
 0.00918 
 
 -0.7 
 
 0.7062 
 
 0.48667 
 
 -0.24718 
 
 0.3 
 
 0.07074 
 
 0.11777 
 
 0.010611 
 
 -0.8 
 
 0.8014 
 
 0.49884 
 
 -0.32076 
 
 0.2 
 
 0.1024 
 
 0.15872 
 
 0.01024 
 
 -0.9 
 
 0.90024 
 
 0.49942 
 
 -0.405109 
 
 0.1 
 
 0.1412 
 
 0.20315 
 
 0.007062 
 
 -1 
 
 1.0000 
 
 0.5000 
 
 -0.5000 
 
 
 
 0.1875 
 
 0.25 
 
 
 
 
 
 
 
 a 
 
 pa 
 
 pa* 
 
 p a 2 
 
 a 
 
 pa 
 
 p a z 
 
 p a 2 
 
 
 
 h 
 
 
 
 
 k 
 
 
 It will be seen that the moment at the unloaded end, which, as long as 
 the load is left of crown, is the moment at crown also ; increases as the 
 load passes on, is positive and increases up to about z = .25 a. Then it 
 diminishes, becomes zero when the load reaches the crown, changes to 
 negative as the load passes the crown, and this negative value increases up 
 to full load when it is i p a 2 . For full load, then, the lower end flanges 
 are in tension. At the crown the moment is zero, and the compression 
 there in both flanges is due to H only. 
 
 34. Circular Arch Formulae for V, H and H. 
 
 (a) Vertical Reaction. 
 
 Here we have r sin /3 = x, r cos j3 d /3 = d $, and P=pdx = prcosfidp. 
 Inserting this in place of P in eq. (43), Art. 18, and integrating between 
 the limits /3 = a and /3 indeterminate, we have for the reaction at w?aloaded 
 end, or for reaction at crown when load does not extend past the crown, 
 
 V = ^L | COS a cos a a sin /3 + cos ft + /3 sin |8 
 
 2 (a sin a cos a) L 3 
 
 + sin a cos a sin - 22jl5 _ cos a sin 2 01. . . .(79) 
 
 J 
 
CHAP. VI.] SUPPLEMENT TO CHAP. XIV. 309 
 
 For = a, this is zero, as should be, since then the load is not upon the 
 span. For = a, V = p r sin a, as should be, for full load over whole 
 
 span. For the semi-circle, a = 90 = , sin a 1, cos a = 0, and 
 
 2 
 
 TT nrH 3 R 
 
 - sin + cos ft + sin 
 
 V p r . 
 
 7T 
 
 If the semi-circle is uniformly loaded over whole span, = a = 90 
 
 = , sin = 1, cos 0, and V pr, as should be. The for- 
 2 
 
 mula (79) above is precisely the same as that given by Capt. Eads in his 
 Report to the Illinois and St. Louis Bridge Co., May, 1868. 
 
 (5) Horizontal Thrust. 
 
 In similar manner, from eq. (46), Art. 18, by inserting p d x = p r cos dj3 
 in place of P, and integrating between = a and 8, we have similarly to 
 Art. 32 
 
 H = Ho^A_* (80 ) 
 
 and 
 
 , , 
 
 12 -ti AI Uj 
 
 A! =3 a 3 sin a cos a 2 a sin 2 a 3 9 sin cos 0+12 COS a sin 
 
 6 sin 2 0+6 a sin a sin 0+2 a ^? -, 
 
 sin a 
 
 A 2 =3a 3 sinacosa+2a sin 2 a+60sin 2 3 0+3 sin/3 cos0 
 
 6 a sin a sin 02 a - , 
 sin a 
 
 Bi=a (a+sin a COS a) 2 sin 2 a, B 2 =a (a+sin a COS a). 
 Formula (80) agrees exactly with that given by Capt. Eads in the Report 
 above quoted, if terms containing K are neglected. Since K = - -, where I 
 
 is the moment of inertia and A is the area of cross-section ; r being the 
 
 radius ; for small central angle r is very large in proportion to , or the 
 
 A 
 
 square of the half depth. In such case, then, K. may be neglected. For 
 == a , we have H = 0, as should be. For = a, we have load over 
 entire span, and 
 
 H _i. 3 a 2 n sin- 3 sin a cos a K (3 a+2 a sin g a 3 sin a cos a) 
 (1+K.) a (a+sin a cos a) 2 sin 2 a 
 
 Approximately we have, by series, for full load, from (80) : 
 
 _ 
 2h 2 22 
 
310 SUPPLEMENT TO CHAP. XIV. [CHAP. VI. 
 
 For o = 90, or for semi-circle, we have from (80) 
 
 t [37r-6/3+18sin/3cos/3-12/3sin 2 +67rsin/3 
 
 hK (-3 7T-12 ft sin 2 /3+6 -6 sin ft cos ft+Q ir sin /3+2 TT sin 3 /3j. 
 For /3 = 90, or for full load upon semi-circle, 
 
 PL- V,- 9 . 
 
 (c) Moment at unloaded end. 
 
 From Art. 18 (c) we have, for concentrated load, 
 
 (sin a sin /3) 2 
 
 /2- 
 \ 
 
 cosa- -- 
 
 sin a/ Sill a 
 
 The value of H we have already given in (80). 
 
 Inserting in the second term p d x = p r cos ft d ft for P, and then inte- 
 grating, we have 
 
 M = AH-B ....... (82) 
 
 where A = | r 1 2 cos a -- ^ ) = JT-? (2 cos a -^ \ 
 y sin a/ 2 sin a \ sin a/ 
 
 and B = P f (sin a sin ft } . 
 
 12 sin 3 a \ / 
 
 For a uniform load over whole span, ft = a, and 
 
 ^ . (83) 
 
 sm / 3 
 
 "We have from (83), by series, the approximate formula for moment at 
 crown 
 
 175y a* . 
 
 where A is the area and I the moment of inertia of cross-section, g the 
 radius of gyration, or, approximately, the half depth for framed arch as 
 always a negative moment indicates tension in lower or inner flange. 
 
 Equations (79), (80) and (82) may, if desired, be tabulated as in Art. 18. 
 For small central angles, or for h small with respect to a, K may be disre- 
 garded, and the results already given for parabola (Art 33) may be taken 
 as sufficiently exact. 
 
CHAP. XV.] THE STONE AECH. 311 
 
 CHAPTEE XY. 
 
 THE STONE ARCH. 
 
 167. Definitions, etc. In the stone arch we have a system 
 of bodies in contact with each other, and so supported between 
 certain fixed points, that they are not only in equilibrium 
 among themselves, but also with the exterior forces. The sur- 
 faces of contact we call the bed-joints ; the fixed points are 
 the abutments ; the central or highest arch stone is called the 
 key -stone, and those resting upon the abutments, the imposts. 
 The inner and outer limiting surfaces of the arch, generally 
 curved, are designated as the intrados and extrados, and the 
 arch stones generally are called voussoirs. 
 
 16. L.IIIC of Presure in Arch. We have already indi- 
 cated (Art. 28, Fig. 16) the manner in which a number of suc- 
 cessive forces are resisted by an arch. We see from the force 
 polygon in that Fig. that the horizontal pressure is the same at 
 every point, and that the vertical pressure is equal to the sum 
 of the weights between the crown and any point. The pres- 
 sure line is then an equilibrium polygon formed by laying off 
 the weights of the arch stones, choosing a pole, and drawing 
 lines from this pole, etc., as described in our second chapter. 
 
 If the weights are very small, and their number very great, 
 the equilibrium polygon becomes a curve. This curve for 
 equilibrium should never pass outside the limits of the arch. 
 
 169. Sliding of the Arch Johns. The arch is properly, 
 then, nothing but a curved wall. Upon a vertical wall, which 
 may also support loads, but which has no horizontal thrust, only 
 vertical forces act, and the resultant is known in position and 
 direction. We may, then, investigate the stability of an ordi- 
 nary wall, and apply the results directly to the arch. 
 
 We assume the wall divided by plane bed-joints extending 
 through its entire breadth, whose distances apart depend upon 
 the dimensions of the stones. These joints are the weak places 
 of the wall, since separation here is not resisted by the greatest 
 strength of the stone. Neglecting the influence of the mortar, 
 
312 THE STONE ARCH. [CHAP. XV. 
 
 we assume that any section along a bed-joint resists only a per- 
 pendicular pressure due to the parts above, and a force paral- 
 lel to the joint which must not exceed the resistance to sliding 
 due to friction. If this parallel force is greater than the resist- 
 ance of friction, the upper part will slide upon the joint. 
 
 If we represent the greatest angle of repose by <, then the 
 resultant of the vertical forces, acting upon the joint in ques- 
 tion, must make an angle with the normal to the joint less than 
 the angle <. Thus at the joint A (Ph 24, Fig. 95), this angle is 
 greater than </>, and the upper part will slide along this joint. 
 At B this angle is less than </>, and no sliding can take place. 
 
 The ratio of the force of friction due to the component of P 
 normal to the joint, to the component of P parallel to the joint, 
 we call the coefficient of safety against sliding. It is evidently 
 
 equal to , or to the distance G-N divided by PN. 
 
 tan P N 
 
 Since we can dispose the bed-joints at pleasure, we may, 
 always make them perpendicular to the direction of the pres- 
 sure, for instance in Fig. 95 horizontal ; or at least so place 
 them that their normals vary from the direction of the resultant 
 of the outer forces, at most by an allowable angle P N. 
 
 The sliding of the joints can then always be prevented by the 
 position of the bed-joints. 
 
 17O. Force acting upon a Cros-ection Neutral Axis. 
 Let us consider what happens when the resultant of the outer 
 forces acting upon a joint, instead of acting at the centre of 
 gravity, approaches the edge of a joint, under the assumption 
 that sliding cannot take place, or that the direction of this re- 
 sultant is perpendicular to the joint. There is no reason for 
 assuming the distribution of pressure upon the joint surface 
 any different from the case of a beam. The stone, as well as 
 the mortar, is elastic, though in a less degree than wood or iron, 
 and accordingly the pressure at any portion of the joint is pro- 
 portional to the approach of the limiting surfaces of the upper 
 and lower portions of the wall. If, then, we assume that these 
 surfaces are plane before and after loading, if the resultant 
 pressure does not act at the centre of gravity, but near to one 
 edge, the pressure at different points will vary, and there will 
 be a neutral axis, or line of no pressure, either within or wholly 
 t without the joint surface. 
 
 Every cross-section is therefore acted upon by a system of 
 
CHAP. XV.] THE STONE ARCH. 313 
 
 parallel forces whose intensities are directly as their distances 
 from a certain axis. 
 
 Now, neglecting the influence of the mortar, the wall can 
 resist compression only. No tension can exist at any point of 
 the joint surface. 
 
 Clearly, then, the neutral axis should lie wholly without the 
 cross-section, or at most only touch it. It should never be 
 found within the cross-section, as in that case all the material 
 on the other side is useless, and might be removed entirely 
 without affecting the pressure upon the actual bearing surface. 
 
 The neutral axis, then, should always lie without the cross- 
 section of the joint. 
 
 IT I. System of Parallel Forces wluse Intensities are 
 proportional to their Distances from a certain Axis The 
 Kernel of a Cross-section. If in a system of equal and paral- 
 lel forces we find the moment of each of these forces with 
 reference to a certain axis, and then consider these moments as 
 themselves forces, we shall have a system of the kind referred 
 to, since each moment force will be directly proportional to its 
 distance from a given axis. 
 
 Now, as we have seen in Art. 60, Chapter VI., the centre of 
 action of such a system of moment forces does not coincide 
 with the centre of gravity of the original simple forces, but for 
 any given axis is found from the central curve of the cross-sec- 
 tion. In PL 11, Fig. 35, we have already given the construc- 
 tion for finding; this centre of action, the semi-diameter of the 
 
 O ' 
 
 central curve being known, for any given axis. 
 
 Suppose now this axis to envelop in all its different posi- 
 tions the outline of the given cross-section, and find the corre- 
 sponding positions of the centre of action of the moment forces. 
 These different points lie in a closed figure which we may call 
 the kernel of the cross-section. Then, in order that we may 
 always have compression in every part of the joint surface of 
 our wall, the resultant of the forces acting upon it should 
 always act within the kernel. 
 
 In Plates 11 and 12, Figs. 36, 37, 38 and 40, we have con- 
 structed the kernels of the various cross-sections represented. 
 
 Thus in Fig. 36, according to the construction of Art. 62, for 
 an axis at A, we describe upon O C a semi-circle. Then with O 
 as a centre and radius equal to semi-diameter of the central 
 ellipse on A C, describe an arc intersecting the semi-circle in a. 
 
314 THE STONE AKCH. [dlAP. XV. 
 
 From a drop a perpendicular upon A C, and we obtain the 
 centre of action for axis at A. A similar construction for other 
 axes, as A B, B C, etc., give us other points, and we thus find 
 the small central parallelogram, which is the kernel or locus 
 of the centres of action of the moment forces for all positions 
 of the axis enveloping the parallelogram A B, C D. A similar 
 construction gives us the kernel for the other figures. 
 We have from Art. 60 
 
 a' 
 
 m = > 
 
 ^ 
 
 where m = the distance of the resultant P of the forces acting 
 upon the cross-section from the centre of gravity, and a = the 
 semi-diameter of the central curve, and i = the distance of the 
 neutral axis from the parallel diameter of the central curve. 
 
 If we call c the distance of an outer fibre from this diameter 
 measured on the side of P, its distance from the neutral axis 
 is i + c. If the strain in this fibre is S, we have 
 
 P 
 
 s -i + c ; ; -r- : i, 
 
 -A. 
 
 where A is the area of the cross-section. Hence 
 
 If P acts at the centre of gravity of the cross-section, i= oo 
 
 p 
 
 (Art. 60), the neutral axis is infinitely distant, and S = . If 
 
 A. 
 
 P moves away from the centre of gravity, the neutral axis 
 approaches, and is always parallel to the conjugate diameter 
 in the central ellipse. When P reaches the perimeter of the 
 kernel, the neutral axis touches the perimeter of the cross- 
 section, and at least, then, in one point of this perimeter, the 
 pressure is zero. If P passes beyond the kernel, the neutral 
 axis enters the cross-section, and tensile strains enter on one 
 side to balance the compressive strains on the other. The ker- 
 nel then forms a limit beyond which the resultant P must 'not 
 act. 
 
 172. Position of Kernel for different Cross-sections. 
 If the cross-section is symmetrical with reference to the cen- 
 
 c P 
 
 tre of gravity, we have . = 1, and therefore S = 2 ; that is, 
 
 1 A. 
 
CHAP. XV.] THE STONE AKCH. 315 
 
 when the neutral axis touches the cross-section, or P acts in 
 the kernel, the strain S is twice as great as when P passes 
 through the centre of gravity of the joint surface and is uni- 
 formly distributed. 
 
 As P passes beyond the kernel, the neutral axis, as we have 
 seen, enters the joint area, and on the side away from P occa- 
 sions, or would occasion in a beam, tensile strains. But as the 
 assumption is that the joint (neglecting mortar) cannot resist 
 tensile strains, we may remove all that portion on the opposite 
 side of the neutral axis without increasing the pressure on the 
 other side. 
 
 In this case, then, the central ellipse is not that for the whole 
 joint area, but only for that portion 011 the same side as P, and 
 P is upon the kernel for that portion. 
 
 This portion can be determined directly for a certain posi- 
 tion of P only in a few individual cases ; generally, it must be 
 found by trial. We must first find 'for the central ellipse of 
 the entire joint area the neutral axis corresponding to given 
 position of P, and then draw a parallel cutting off somewhat 
 more of the area. Then determine the central ellipse of the 
 cut-off portion, and see if the pole lies symmetrically to the 
 pole of the cutting line. 
 
 The parallelogram is one of the areas in which we can de- 
 termine directly the amount cut off when P acts at a point 
 upon the line joining the centres of two opposite sides. For if 
 we cut off by a parallel to these sides a portion so that P is at 
 d of the line joining the centres of the opposite sides of the 
 new parallelogram, then P lies upon the kernel for this new 
 area. The proof is easy. The moment of inertia of the 
 parallelogram is -% b A 3 , with reference to the diameter b. The 
 square of the radius of gyration a a is then y 1 ^ A 2 . The distance 
 of the point of application of P from one of the sides is 
 
 a" 
 
 ^ -h m = ^ + . Hence 
 
 The half height of the kernel is, then, -J-th the height of the 
 parallelogram, or the kernel occupies the inner third. (See Fig. 
 36 ; also Woodbury : Theory of the Arch, p. 328, Art. 3.) 
 
 For any given position of P, then, three times its distance 
 
316 THE STONE ARCH. [CHAP. XV*. 
 
 from the nearest side on a line parallel to the other two, gives 
 the position of the fourth side of the parallelogram for which 
 P is upon the kernel. 
 
 173. The resultant pressure should therefore act 
 within the middle third of the joint area. As this prin- 
 ciple is most important, and the demonstrations of Chapter VI., 
 upon which the above result is based, may appear to some too 
 purely mathematical, we give here the demonstration of the 
 same principle as given by Woodbury, in the work above 
 cited. 
 
 " Suppose the pressure to be nothing at the in trad os #, and 
 to increase uniformly from that point to the extrados b (PL 
 24, Fig. 96). It is plain that the pressure at any point along a ~b 
 will be represented by the brdinate of a certain triangle. The 
 whole pressure will be represented by the surface of that tri- 
 angle ; and the point of application of the resultant of all the 
 pressures will be at c opposite the centre of gravity of that 
 triangle. We have then c b = ^ a ft. Vice versa, if the point 
 of application be at c, c I % a 5, we know that the pressure 
 is nothing at a. 
 
 " If the point of application be at c, c 7> being less than -J a J, 
 c being still opposite the centre of gravity of the triangle 
 whose ordinates represent the pressure, we know that the ver- 
 tex of that triangle and point of no pressure are at e } be = 3 
 xbc. 
 
 " In this case, the joint a b will open at a as far as e ; the 
 adjacent joints will also open until we come to one where the 
 curve of pressure passes within the prescribed limit. 
 
 " This reasoning is, of course, applicable to all the joints ; 
 and we readily conclude that the curves of pressure should lie 
 entirely between two other curves which divide the joint into 
 three equal parts." 
 
 Thus, in PL 24, Fig. 97, suppose .the resultant P of the 
 upper part of the wall to have the position as represented, so 
 that it intersects the joint B D in C outside of the middle third 
 of the cross-section. The entire pressure is distributed over 
 3 C B = A B, and the area D A does not act at all. Moreover, 
 the pressure at B is twice as great as when P passes through 
 the centre of gravity and is uniformly distributed over A B, or 
 is f ds of the uniformly distributed pressure of 'P upon C B. 
 
 Beyond A the pressure is zero, and the conditions of load 
 
CHAP. XV.] THE STONE AKCH. 317 
 
 and equilibrium would not be changed if the stone beyond A 
 were removed. 
 
 If C approaches still nearer B, so that the pressure is distrib- 
 uted upon an ever-decreasing area, the resistance of the mortar 
 will be finally overcome ; it will be forced out, and stone will 
 come in contact with stone, and there will be rotation about the 
 edge at B. This rotation can never occur if the pressure P is 
 distributed over the whole joint area. If, then, we consider 
 rotation to commence at the moment when P is no longer dis- 
 tributed over the entire area when, therefore, the neutral axis 
 just enters the joint then, in order that no rotation may occur, y 
 P must pierce the joint area inside the kernel. 
 
 174. LJne of Pressures in tlie Arch. When the dimen- 
 sions and form of a wall are given, we can determine directly 
 the resultant P of the outer forces acting upon a joint, and then 
 by the two preceding Arts, can determine the condition of sta- 
 bility of the wall. In the arch, however, we cannot determine 
 P directly for a given cross-section, but must first make certain 
 assumptions. 
 
 In the first place, it is clear that an arch is stable when it is 
 possible in two joints to take two reactions P 1 and P 2 (PI. 24, 
 Fig. 98) such that, with the weight of the intervening portion 
 of the arch and its load, the resulting line of pressure shall lie 
 so far in the interior of the arch that rotation about a joint edge 
 cannot take place. If the arch is so feeble and the resistance 
 of the material so slight that only one such assumption of P^ and 
 P 2 can be made, and only one such pressure line drawn, this is 
 plainly the true pressure line for stability, and by it P t and P 2 , 
 as also the pressure at every joint, are determined. 
 
 If, however, the arch is so deep and the resistance of the 
 material so great that by variation of P 1 and P 2 several such 
 pressure lines may be drawn, none of which causes rotation 
 about a joint edge, which of all these possible pressure lines is 
 the true pressure line of the arch ? 
 
 We assert : That is the true pressure line which approaches 
 nearest the axis, so that the pressure in the most compressed 
 joint edge is a minimum. 
 
 If we assume the material so s}ft that the pressure line ap- 
 proaches the axis so near that only one assumption of P, and 
 P, is possible, then this would evidently be the true pressure 
 line. If now the material hardens without altering any of its 
 
318 THE STONE ARCH. [CHAP. XV. 
 
 other properties, such as its specific weight or modulus of elas- 
 ticity, then the position of the pressure line is not altered. As 
 there is no reason for supposing the pressure line different in 
 an arch built of hard material from that in one originally soft 
 which has afterwards gradually hardened, it follows that the 
 pressure line in all arches of same form and loading has the 
 same position which it would have had if the arch had been 
 originally of the softest material ; that is, that position which 
 makes the pressure in the most compressed joint edge a mini- 
 mum. 
 
 In order to draw the pressure line in an arch, we may then 
 seek by means of the formula 
 
 o * 
 
 this pressure in the joint, where the pressure line approaches 
 nearest the edge, and ascertain whether it can be still further 
 diminished by change of position of the pressure line. This is, 
 however, not necessary. We have only to ascertain whether it 
 is possible to draw a pressure line whose sides cut the corre- 
 sponding joint a.rea, within the kernel, for then, since we know 
 that there can be a still more favorable position, there is no 
 danger of rotation. 
 
 175. The Line of Support. The curve formed by joining 
 the intersections of the sides of the pressure line with the joint 
 areas we call the support line, or line of support. 
 
 If the joints of an arch answer to the condition of Art. 169, 
 so that sliding of the joints cannot occur, we see at once from 
 the position of the support line on what side and where rota- 
 tion will take place. If at any point this line passes beyond 
 the kernel, we have theoretical beginning of rotation ; if it 
 passes outside of the arch, there is actual rotation, and if it lies 
 within the kernel, there is no rotation. 
 
 The manner of determining from the position of the support 
 line all the possible motions of an arch is illustrated in the fol- 
 lowing Figs. 
 
 In PI. 24:, Fig. 99, we have a possible support line touching 
 the extrados at crown and springing, and the intrados between 
 these points. We have accordingly rotation at crown, and at 
 the points between crown and springing, so that the joints at 
 these points open on the sides of the arcli opposite the support 
 line. The crown will sink, and as at the crown and flanks the 
 
CHAP. XV.] THE STONE AECH. 319 
 
 support line is approximately parallel to the extrados and intra- 
 dos, there will be several joints in the same condition, and 
 several will open, as indicated in the Fig. 
 
 In PI. 25, Fig. 100, we have the condition of stability of a 
 pointed arch, not loaded at the crown. The support line ia 
 horizontal at crown, and there is no angle there, as in the arch 
 itself. The rotation at various points is indicated in the Fig. 
 We shall soon see that the support line deviates but very little 
 from the pressure line. From the direction of the tangent to 
 the support line at any point, therefore, we may conclude as to 
 the conditions of sliding. 
 
 From Fig. 101 we may conclude that the arch will slide out- 
 wards upon the right abutment. The rotation at various points 
 is given by the Fig. It is sufficient, as we see, to make the 
 abutment surface more nearly perpendicular to the support 
 line, as shown in the left abutment, to prevent this sliding, and 
 at the same time a more favorable support line can be drawn. 
 Since, as we have seen in Art. 100, sliding can and must in 
 this manner be always prevented, we shall give no more exam- 
 ples of arches unstable in this particular. 
 
 The arches of Figs. 99 and 100 can be made stable by suffi- 
 ciently increasing their thickness, or conforming their shape 
 more nearly to that of the support line. 
 
 176. Deviation of the Support from the Presure Line. 
 This deviation is not great. In order to make it apparent, we 
 must draw a pressure line for slight pressure in the lower part 
 of an arch with very long and inclined voussoirs [PL 25, Fig. 
 102]. Thus, if we combine the weights of the voussoirs 1, 2, 3, 
 4, etc., acting at their centres of gravity, with the pressure^Q in 
 the first joint, we have the pressure line shown by the broken 
 line 1, 2, 3, 4, 5, 6, 7, 8, whose sides 1 2, 2 3, 3 4, etc., give the 
 direction of the pressure in the corresponding joints between 
 the voussoirs 1 and 2, 2 and 3, etc. Thus 5 6 is the direction 
 of the pressure upon the joint between voussoirs 5 and 6. This 
 direction cuts the joint at 5', which is therefore the point of 
 application of the pressure, or a point upon the line of support. 
 Thus we find 3', 4', 5', 6', and the line joining these points is 
 the support Iww. In general, then, the support and pressure 
 lines coincide when the vertical through the centre of gravity 
 of any very small element coincides with the joint, and they 
 deviate when this vertical does not coincide with the joint. 
 
320 THE STONE ARCH. [CTIAP. XV. 
 
 In the ordinary form of joint, as shown in Fig. 103, the sup- 
 port line varies from the pressure line, since the vertical through 
 the centre of gravity S does not coincide with the joint under 
 S. If, however, we should conceive the arch divided into ver- 
 tical laminae, then the support and pressure lines fall together. 
 This is precisely the assumption always made in the analytical 
 discussion of the theory of the arch. 
 
 Thus we take the area A = / y dx, and this expression sup- 
 poses the arch divided into vertical laminse. 
 
 The first to make clearly this distinction between the lines of 
 pressure and support, was Mosely (Civil Eng.). Other authors 
 have after him adopted this distinction, and then proved that 
 the two lines always coincide, without remarking that this coin- 
 cidence is only because of the adoption of the above integral. 
 The same assumption simplifies greatly the graphical construc- 
 tion also (the analytical treatment is without it well-nigh im- 
 possible). We shall therefore assume vertical lam i 1133 where 
 it is at all permissible. This is always permissible at the crown 
 of arches with horizontal tangent, because there the joints are 
 vertical, and over all, when the pressure line lies below the 
 axis of the arch ; for the support line lies always above the 
 pressure line, and therefore, in this case, the conditions of sta- 
 bility are more favorable for it than for the pressure line itself, 
 when considered as the line of support. 
 
 Moreover, it is easy at any point of the pressure line con- 
 structed with vertical laminae to pass to that line for another 
 form of joint, and to the corresponding support line. Thus, if 
 for the point A (PL 25, Fig. 104) we have found the pressure 
 Q, and if now we wish to pass to the joint ABC, we prolong 
 Q till it meets P, the weight of the voussoir A B C D, and re- 
 solve P and Q at this point into Q'. Then Q' is a side of the 
 new pressure line, and it cuts A B in a point of the support 
 line. 
 
 In this way we can easily determine whether the error com- 
 mitted when we substitute the pressure line for vertical laminae 
 for that for the actual joints, which is given by the segment of 
 the joint A B between QJ. and the pressure line, can be disre- 
 garded. 
 
 177. Dimension* of the Arcli. The object of the con- 
 struction of the pressure or support line in the arch is to deter- 
 
CHAP. XV.] THE STONE ARCH. 321 
 
 mine the stability and the joints of the abutments. When, the 
 live load of the arch can be neglected with respect to its own 
 weight, and when the material of the arch possesses the usual 
 strength, and the pressure line lies within the inner third, then 
 the lower point of rupture lies so low that the back masonry 
 reaching from this point beyond the pressure line completely 
 encloses it. 
 
 There is, therefore, nothing arbitrary, when the form of the 
 arch is given, except the depth. Since in an arch of less depth 
 than is allowable in practice a support line can still be in- 
 scribed, the graphical method is unable to determine the 
 proper depth. We must then leave to theory the development 
 of formulae by which this can be determined, and assume that 
 not only the form of the arch is given, but also its proper 
 depth and the lower joint of rupture. It is required to deter- 
 mine the stability of the abutments. 
 
 The stability of the abutments can be regarded from two 
 points of view. We may consider it as a continu|tion of the 
 arch, as in many English and French bridges, in which the 
 arch is continued as such, clear to the foundation ; or we may 
 regard it as a wall whose moment about the joint of rupture 
 resists the rotation about this joint due to the thrust. Both 
 views are identical, as the entire theory of the support line rests 
 upon the investigation of the rotation. They differ only in the 
 method of expressing the safety of the abutment. 
 
 If the arch is continued to the foundation, and the space be- 
 tween it and the road line filled up by spandrels ; or if the 
 thickness of the abutment increases from above as the support 
 line requires ; or, as is often the case in England, the abutment 
 consists of walls parallel to the crown, separated by hollow 
 spaces; still, in every case the abutment is not to be distin- 
 guished from the arch proper it is stable when the support 
 line lies in the interior. If the prolonged arch is separated 
 entirely from the adjacent masonry, there is no reason for not 
 giving the axis of the prolongation the form of the support 
 line itself. 
 
 If, on the other hand, there is no separation of the arch and 
 abutment, as in the English hollow abutments, it is sufficient 
 that the support line lie in the inner third, and the abutment 
 will be certainly stable. 
 
 The supposition that the resistance of the rnortar is suffi- 
 21 
 
322 THE STONE ARCH. [CTIAP. XV. 
 
 cientlj great to unite the whole abutment as a single block 
 which turns about its under edge, gives too small dimensions. 
 To ensure safety it is assumed that equilibrium exists with 
 reference to rotation about the lower edge, when the thrust of 
 the arch is 1.5 greater than the actual. Investigations of 
 French engineers have shown that this coefficient of safety for 
 very light arches is not less than 1.4. The old tables of Petit 
 give 1.9. We assume it, therefore, = 2. 
 
 If, therefore, the double thrust of the arch at the lower point 
 of rupture is united with the weight of the abutment, the re- 
 sultant should still fall within the base. Since it is indiiferent 
 in what order the elements of the abutment are resolved, it is 
 best to divide it into vertical laminae, and unite these with the 
 double thrust. The equilibrium polygon thus obtained should 
 cut the foundation base within the edge of the abutment. 
 
 When the thickness of the abutment is thus determined, we 
 must construct the actual pressure line for the simple thrust 
 in order to Determine the joints. In drawing this second pres- 
 sure line, we should properly take the divisions of the arch by 
 the joints themselves. If, however, we take the division in 
 vertical laminae, the deviation, as we have seen, is insignificant. 
 The normals to the actual joints must, then, not deviate from 
 the sides of this pressure line by more than the angle of 
 repose. 
 
 178. Construction of the Pressure Line. In PL 25, Fig. 
 105, we give the method of construction of the proper width of 
 abutment for an arch. We first divide the arch into vertical 
 laminse, and determine their weight. If the surcharge has 
 vacant spaces, or is generally of different specific weight from 
 the material of the arch itself, it must first be reduced. Thus, 
 if the surcharge (spandrel filling, etc.) weighs, for instance, only 
 f ds as much as an equal area of masonry in the arch, we have 
 simply to diminish the vertical height above the arch by -Jd. 
 We thus obtain the dotted line given in the Fig., which forms 
 the limit of the reduced laminae, and we can treat the areas 
 bounded by this line the vertical lines of division and the 
 intrados as homogeneous. We have then only to determine 
 the centres of gravity of the various laminae according to the 
 construction for finding the centre of gravity of a trapezoid 
 (Art. 33), and suppose at these points the weights, which are 
 proportional to the reduced areas of the trapezoids to act. 
 
CHAP. XV.] THE STONE ARCH. . 323 
 
 Laying off these weights in their order, we have the force 
 line (Fig. to left). The weights of the abutment laminae 9, 10 
 and 11 are laid off to same scale one-half 'of their proper in- 
 tensities. The reason will soon appear. 
 
 1st. To determine the thrust H, and also the joint of rup- 
 ture. 
 
 We first inscribe a pressure line by eye, and assume the 
 point of the intrados to which this line most nearly approaches 
 as the edge of the joint of rupture. Draw next from the cor- 
 responding point of the force line a line parallel to the assumed 
 pressure line at this point. This line will cut off from the 
 horizontal through the beginning of the force line our first 
 approximate value of H. 
 
 Thus, suppose we have inscribed by eye the pressure line 1, 
 
 2, 3, 4, etc., which gives us the point a for the position of the 
 edge of the joint of rupture. Then a line drawn from 5 on 
 the force line, parallel to the side 4 5 of the pressure line, gives 
 us our first value for H. 
 
 Now assuming this value of H, we erase the first assumed 
 pressure line, and proceed to construct the pressure line cor- 
 responding to this value of H, and the force line divisions 1, 2, 
 
 3, 4, etc. If this pressure line lies always within the middle 
 third of the arch, it may be taken as the proper pressure line, 
 and H as the true thrust. In general, however, this will not 
 be the case. The pressure line thus determined may even pass 
 without the arch entirely. We then determine the new point 
 of rupture, as given by the point of exit of this pressure line, 
 and produce the side at this point back to intersection with H 
 prolonged. From this point of intersection draw a line which 
 does lie within the middle third of the arch at the lamina of 
 rupture, and then in the force polygon from the corresponding 
 point of the force line draw a parallel to this line, thus cutting 
 off a new value for H. Erasing now the preceding pressure 
 line, we construct a third with this new value of H, which will 
 in general give us a pressure line lying everywhere within the 
 middle third of the arch. If not, another approximation may 
 be made. We thus find by successive approximation the true 
 joint of rupture and the corresponding thrust. 
 
 2d. Width of abutment. Since we have laid off the arch 
 weights to scale in their true value, the pressure line thus ob- 
 tained is the true pressure line for the arch. But we have laid 
 
324 THE STONE ARCH. [CHAP. XV. 
 
 off the abutment laminse 9, 10 and 11, one-half \\\G\Y true value, 
 and the pressure line thus obtained with the same thrust and 
 pole O is the same as if we had taken their true value and twice 
 H. Its intersection with the foundation gives us, then, the 
 proper width of the abutment for stability, according to our 
 assumption of 2 for the coefficient of stability (Art. 177). 
 
 179. Thus we can easily determine for any given case of 
 arch and surcharge the horizontal thrust and the proper width 
 of abutment, and then from the pressure line can easily so dis- 
 pose the joints as to prevent sliding. If the dimensions of the 
 arch as given are not such as to be stable, it will be found im- 
 possible to inscribe, as above, a pressure line which shall lie 
 within the middle third, and the curve of 'extrados or intrados 
 will have to be altered so that this shall be the case. The pres- 
 sure line thus obtained, it is true, does not exactly correspond 
 with the true one, as it is still possible to inscribe another which 
 shall deviate still less from the true line. We have also taken 
 the double thrust for the abutment laminse alone, instead of 
 for all laminse from the joint of rupture of the arch. Both 
 deviations are made on account of the far greater ease and 
 rapidity of construction. It would be found very tedious to 
 take first the force polygon up to somewhere about the section 
 of rupture, then by long trial find the innermost support line, 
 and finally, after the section of rupture is by this line deter- 
 mined, to lay off the remainder of the force polygon, and pro- 
 long the pressure line through the abutments. 
 
 It is far simpler to 'proceed, as above, by assuming the point 
 of application of the horizontal thrust, as also temporarily the 
 section of rupture. We obtain thus a somewhat smaller value 
 for the width of abutment, but, on the other hand, we have 
 taken the coefficient of stability at 2 instead of 1.9, as assumed 
 in Petitfs tables. 
 
 Moreover, the widths of abutment thus obtained are greater 
 than those obtained by these tables, as it is assumed in them 
 that the point of application of the horizontal thrust is at the 
 upper edge of the abutment. Thus in every respect the con- 
 struction gives results reliable and even more accurate than the 
 tables, as we take the arch as it really is in any given case, 
 while in the tables suppositions are made with reference to 
 spandrel filling, etc., which do not hold good for every case. 
 
 1O. Proper Thickness of Arch at Crown. The proper 
 
CHAP. XV.] THE STONE AKCH. 325 
 
 depth of the arch at the key depends not only upon the rise 
 and span, but also upon the load. The pressure at the extrados 
 at the key, which, is in general, the most exposed part of joint, 
 should not, according to the best authorities, exceed -^th the 
 ultimate resisting power of the material. If P is the pressure 
 per unit of surface, H the thrust, and d the depth of key-stone 
 
 2H 
 
 joint, then P -y, 
 
 since, on the assumption that the curve of pressure does not pass 
 outside the kernel, the maximum pressure is twice the mean 
 
 TT 
 
 pressure -j. This mean pressure, then, should not exceed ^th 
 
 CL 
 
 the ultimate resistance of the material. In the best works of 
 Rennie and Stevenson the thickness at key varies from -^-th to 
 gJg-d the span, and from ^-th to ^th tne radius of the intrados. 
 The augmentation of thickness at the springing line is made by 
 the Stevenson's from 20 to 40 per cent., by the Rennie's at 
 about 100 per cent. 
 
 Perronet gives for the depth at crown the empirical formula 
 
 d= 0.0694 /* + 0.325 meters, 
 
 in which r is the greatest length in meters of the radius of cur- 
 vature of the intrados. 
 
 For arches with radius exceeding 15 meters, this gives too 
 great a thickness. According to Rankine, 
 
 d = 0.346 Vr 
 for circular arches, and 
 
 d = 0.412 Vr, 
 
 where r is the radius of curvature of the intrados at the crown 
 in feet. 
 
 " The London Bridge is in its plan and workmanship per- 
 haps the most perfect work of its kind. The intrados is an 
 ellipse, the span 152 ft., the rise th as much, the depth of key 
 -gig-th the span. The crown settled only two inches upon remov- 
 al of centres." [ Woodbury : Theory of the ArchJ] 
 
 In general, we must first assume the depth at key in view of 
 the strength of the material, the character of the workmanship, 
 the load, etc. Then the thrust being found, we find the mean 
 pressure per unit of area as above. If this mean pressure 
 exceeds ^th the ultimate resisting power of the material, make 
 
326 THE STONE ARCH. [CHAP. XV. 
 
 a new supposition, increase the thickness, find the thrust and 
 pressure anew, and so on, till the results are satisfactory. 
 
 The ultimate resisting power of granite may be taken at 
 6,000 Ibs., brick 1,200, sandstone 4,000, limestone 5,000 Ibs. 
 per square foot. These values are, of course, very general, 
 and subject to considerable variations, according to the kind 
 and quality of the stone. The strength of the material to be 
 used must, for any particular case, be determined by actual 
 experiment. 
 
 The weight of a cubic foot of stone may, in general, be as- 
 sumed at 160 Ibs., brick masonry at 125 Ibs. 
 
 IN!. Increase of thickness due to change of form. 
 Having obtained a thickness which satisfies all the conditions, 
 we must, if the arch be very light, make some further provi- 
 sion for the change of form which is sure to take place af ter 
 the removal of the centres. By this change of form the pres- 
 sure line is altered, and the thickness may need to be increased. 
 In general, we need only to increase the depth from the key to 
 the springing. This increase need not exceed fifty per cent, at 
 the joint of rupture and weakest intermediate joint. [ Wood- 
 l>ury : Theory of the Arch.] 
 
 182. Thus, by a simple and rapid construction, we can de- 
 termine, for any particular case, the thrust, joint of rupture, 
 and proper thickness of the abutments, without the use of 
 tables or the intricate formulae usually employed. There is no 
 difficulty in laying down on paper and verifying all the ele- 
 ments of the most complex case. The method is entirely in- 
 dependent of all particular assumptions, and is therefore 
 especially valuable when irregularities of outline or construc- 
 tion place the arch almost beyond the reach of calculation. It 
 is general, and may be applied with equal ease to loaded and 
 unloaded, full circle, segmental, or elliptical arches with any 
 form of surcharge. 
 
CHAP. XVI.] THE INVESTED AKCH. 327 
 
 CHAPTEK XYI. 
 
 THE INVERTED AKCH SUSPENSION SYSTEM. 
 
 183. The inverted arch forms the supporting member of 
 chain or cable suspension bridges. Whether the cable be com- 
 posed of chains, links, or wires, we suppose them so flexible 
 that they can perfectly assume the curve of equilibrium. As, 
 therefore, disregarding the dead weight, any partial load would 
 cause a change of shape, the cables must be stiffened in order 
 to prevent the motion which would otherwise take place. 
 
 We may stiffen the chains, as shown in PL 26, Fig. 106, by trian- 
 gular bracing, thus making a rigid system ; or we may have two 
 parallel chains and brace them to each other, as shown by Fig. 
 90 inverted ; or we may introduce an auxiliary truss, the office 
 of which is not to add in any degree to the supporting power 
 of the combination, but simply to distribute a partial load over 
 the whole span, so as to cause it to take effect as a distributed 
 load, and thus prevent change of shape. 
 
 As in the first and last cases the structure is commonly 
 hinged at the centre in order to eliminate the effects of tem- 
 perature, the method of resolution of forces explained in Arts. 
 8-13 will, in general, be applicable for the determination of 
 the strains. 
 
 In the second case, we can apply the principles of Arts. 158- 
 161. ! 
 
 The rear chains, anchorages, and stiffening truss deserve, 
 however, special notice. 
 
 14. Rear Chain* and Anchorages. The greatest ten- 
 sion in the main chains occurs, of course, for full load. To 
 find the tension^at top of tower, as also the horizontal pull, we 
 have simply to lay off half the whole load vertically from o to 
 d [PI. 26, Fig. 106], and then draw O o horizontal and O d 
 parallel to the last side at tower. Then O d is the tension in 
 that side, and o O the horizontal pull. This pull is neutralized 
 by the opposite and equal pull of the rear chain leading to the 
 anchorage ; provided, as should always be the case, it makes an 
 
328 THE INVERTED ARCH. [CHAP. XVI. 
 
 equal angle with the vertical. We have thus acting upon the 
 tower simply the half load ; and the tension in the rear chain 
 is equal to that in the last link, O d. 
 
 If from O we draw parallels to the other links, we have at 
 once the strains in these links, O <?, O 5, O a, etc. 
 
 Now, if the anchorage is a solid block of masonry, its con- 
 dition of stability is, of course, very easily determined. The 
 moment of the tension in the rear cable, with reference to the 
 edge of rotation, must be more than balanced by the moment 
 of the weight of the block acting at its centre of gravity, with 
 reference to this edge. The case is too simple to need further 
 notice. 
 
 It is, however, more economical to make the anchorage hol- 
 low that is, in the form of an arch. The preceding method 
 for determining the stability of the arch has then here direct 
 application. 
 
 Thus, laying off along the vertical through the centre of the 
 tower the weights of segments of the arch, we form with these 
 segment weights and the double tension in the chain an equi- 
 librium polygon. For this we have the pole O n A O 1 being 
 double the tension Od already found. We then draw O t l, (^2, 
 Oj3, etc., and then from A parallels to these to the segment 
 verticals 1, 2, 3, etc. We thus have the polygon A 1, 2, 3, 4, 5. 
 \_N~ote. We take the double tension, as before, for the arch, we 
 took 2 H instead of H, in order to ensure stability.] 
 
 The last line of this polygon 4 5 prolonged must, for sta- 
 bility, pass within the pier abutment, and its resultant, when it 
 is combined with the weight of the pier and pier abutment, 
 must pass within the abutment foundation. Through its inter- 
 section with the vertical line through the axis of the tower the 
 curve of pressure for the arch must pass. 
 
 Drawing now O a 4 parallel to the rear chain, and making it 
 also equal to the double tension, or twice O d, we find the pole 
 O 2 , and from it draw O 2 1, O 2 2, O 2 3, etc., and then construct 
 the pressure line for the arch. It must, for sfeaiulity, lie within 
 the middle third. 
 
 To ensure stability when the tension in the rear chain dimin- 
 ishes, or when the bridge is unloaded, the arch must be stable 
 lyy itself. We must, therefore, construct the curve of pressure 
 f < >r the arch alone, neglecting the tension of the rear chains, 
 as explained in the preceding chapter. 
 
CHAP. XYI.] THE INVERTED AKCH. 329 
 
 If this also passes within the middle third of the arch, as 
 represented by the dotted line, the arch is, under all circum- 
 stances, stable, and can fully resist the tension of the rear 
 chains. 
 
 We can now, finally, so dispose the joints as to prevent 
 sliding. 
 
 185. Stiffened Suspension System. We have already re- 
 ferred to the methods of stiffening the cable or chain so as to 
 prevent the changes of shape due to partial loading. Of these 
 methods, it only remains to notice particularly the last, viz., by 
 means of an auxiliary truss. The office of this truss is to dis- 
 tribute a partial load over the whole length. We have now to 
 investigate the forces which act upon the truss. 
 
 In PL 27, Fig. 107, let the chain be acted upon by the truss 
 represented by A B, which is called into action only by a par- 
 tial load, and not at all by a total uniform load. We can neg- 
 lect then the weight of the truss itself, as this is borne by the 
 cable. At the apices 3, 4, 5, 6, 7 let us suppose partial loads 
 indicated by the small arrows pointing down. Then, at every 
 point of connection with the chain, we have the reactions 1', 2', 
 3', etc., acting upwards.- Now the truss must prevent deforma- 
 tion, and hence these forces are dependent upon the form of 
 the cable itself. Indeed, if we take any point, as O, as a pole, 
 and draw lines parallel to the respective sides of the cable, 
 these lines will cut off upon a vertical P' these forces. -The 
 absolute value of these forces will, it is true, vary according to 
 the position of the pole assumed, but their relative proportions 
 remain always the same. The resultant P 7 of all these forces 
 passes then through the intersection of the two outer sides of 
 the catenary. 
 
 Since the truss distributes its load P upon the cable, the reac- 
 tion B at the right support is here zero. The reaction, however, 
 at A cannot be zero unless P and P' coincide, as is the case for 
 total uniform load. These, then, are all the forces which are 
 kept in equilibrium by the truss. If P is given, P' and the re- 
 action at A can be easily found, and if we then divide P', ac- 
 cording to the form of the chain, into the portions I/, 2', 3', 
 etc., we have the forces at each apex. 
 
 Thus we lay off to scale the given forces 3, 4, 5, 6, 7 P, and 
 with a pole distance any convenient multiple of the height of 
 truss draw lines to these points of division, and then construct 
 
330 THE INVEKTED ARCH. [CHAP. XVI. 
 
 the corresponding equilibrium polygon A 3, 4, 5, 6, 7, B. Pro- 
 long then the outer side B 7 to intersection with P', and draw 
 the closing line A P'. A parallel through O to this line cuts 
 off from the force line P the reaction at A and the cable reac- 
 tion P'. 
 
 Now P' being thus found and the form of cable given, we 
 have only to lay off P' vertically, draw from its extremities 
 lines parallel to the two outer sides of the given cable arc, and 
 from the pole thus determined, lines parallel to the other sides 
 will give us the forces 1', 2', 3', etc. These when thus found 
 we lay off on our force line for the pole O, as shown in the 
 Fig., and then construct the corresponding equilibrium poly- 
 gon A 1', 2' 9', 10', B. 
 
 Thus the vertical ordinates between A P', P' B and this 
 polygon give us the moment at any point for a truss acted 
 upon by the forces I/, 2', 3', etc., alone. The ordinates between 
 A P', P' B and the polygon A 3, 4, 5, 6, 7, give, in like manner, 
 the moments for a truss acted upon by the forces 3, 4, 5, 6, 7, 
 whose reactions are A and P'. The ordinates, then, included 
 between both polygons give us the moment at any point of the 
 stiffening truss. Thus the ordinate y, multiplied by the pole 
 distance, gives us the moment in the truss at the point o. If 
 we had taken the pole distance O equal to the height of the 
 truss, then these ordinates would give us at once the strain in 
 the flanges. We can thus easily find the strains in the stiffen- 
 ing truss for any weight or system of weights in any position. 
 
 186. Most unfavorable method of Loading. Let us in- 
 vestigate the action of a single weight P at any point. In PI. 
 27, Fig. 109, we have a single weight P acting between A and 
 P'. 
 
 The Fig. is nothing more than a repetition of Fig. 108, only 
 we have a single load P instead of a system of four loads, and 
 therefore the polygon for P consists only of two straight lines 
 instead of having as many angles 3, 4, 5, etc., as there are apex 
 loads in the first case. All lines have the same position as in 
 Fig. 108, and hence the construction needs no further explana- 
 tion. 
 
 We see at once from the Fig. that any load between A and 
 P' increases the moment at every point of the span A B, and 
 therefore at the point of rupture or of maximum moment also. 
 So also for the shearing force. When, therefore, the moment 
 
CHAP. XVI.] THE INVERTED ARCH. 331 
 
 at any point, and the sum of the forces between that point and 
 A, is a maximum, at least the entire distance from A to P' 
 must be covered with the load. 
 
 In Fig. 110 we have the weight P on the other side of the but 
 centre P'. The construction is identical with Figs. 109 and 108, 
 the position and the direction of action of the forces is now 
 different. Since the resultants A and P' now lie on the same 
 side of P, A and P' act in opposite directions, and since P' 
 must still act upwards, A must act downwards. In the neigh- 
 borhood of 5' the moment is zero. Between this point and B 
 the moments have the same signification as in Fig. 109 ; on 
 the other side the moments have then a different siom. In 
 
 O 
 
 order, then, that the moment at 5 ' shall be a maximum, the load 
 must cover the length from A to P, this last point being the 
 point at which a load causes no moment in 5'; for if any 
 point between A and P were not loaded, as we have seen, a 
 load at that point would increase the moment at 5'. A load 
 beyond P, however, would diminish the moment at 5'. 
 
 The above holds good for every point between A and P', and 
 therefore for the point of rupture or of maximum moment it- 
 self. In order that this maximum moment can be no more 
 increased, the load must extend from A beyond the centre to 
 that point at which a load being placed causes no moment at 
 the cross-section of rupture. 
 
 As for the shearing force, at the end A it will evidently be 
 greatest for load from A to P', or over the half span, since 
 every load on the other side of P' diminishes this reaction. 
 Hence we have the following principles established : 
 
 The moment cut any Gross-section of the stiffening truss is a 
 maximum, when the load reaches from the nearest end beyond 
 the centre to a point for which the moment at this cross-section 
 is zero. 
 
 The above condition holds good, therefore, for the maximum 
 of all the maximum moments, or for the cross-section of rup- 
 ture itself. 
 
 The maximum shearing force is at one end of the truss 
 when the adjacent half span is loaded. 
 
 If the arc is unsymmetrical, we must understand by " half 
 span " the distance between the end and vertical through the 
 intersection of the outer arc ends produced. 
 
 1ST. Example. As an illustration of the above principles, 
 
332 THE INVERTED ARCH. [CHAP. XVI. 
 
 let us take the structure represented in PL 28, Fig. 111. Span, 
 60 ft. ; depth of truss and panel length, 5 ft. Scale, 10 ft. to an 
 inch. We suppose the live load to be 2 tons per ft., giving 
 thus 10 tons for each lower apex, and take the scale of force 50 
 tons per inch. 
 
 On the left we have laid off the force lines for the loads 2, 3, 
 4, 5, 6 and 7 to 11, and have taken the poles 'for each, so that 
 the first lines are all parallel to each other and to the first link 
 of the cable ; the common pole distance being 2 j- times the 
 height of trass, or 1.5 inches. The moment scale is then 
 1.5 x 10 x 50 = 750 ft. tons per inch. Since the full load is 
 entirely supported by the cable, we have only to investigate the 
 effect of the live load upon the truss. 
 
 Precisely as in Fig. 108, we construct the polygons for forces 
 2-11, 3-11, 4-11, etc., and draw the closing lines as indicated 
 by the broken lines radiating from the centre O. Parallels to 
 these from the poles cut off from the force lines the end and 
 chain reactions. The upper portions are the chain reactions, 
 the lower the reactions at the right end for the loads 2-11, 
 3-11, etc. 
 
 Now we have to divide these chain reactions into as many 
 parts as there are load apices by lines parallel to the sides of 
 the chain. This we have done by drawing two lines parallel to 
 the two chain ends, inserting the chain reactions between these 
 lines, and then drawing parallels to the chain sides. If, as in 
 this case, the curve of the chain is a parabola, these reactions 
 are divided into 11 equal parts. If the chain has any other 
 form, the parallels to the chain sides determine the relative 
 lengths of these portions. 
 
 It will only be found necessary to construct the moment 
 polygons for 4, 5 and 6-11 ; the other polygons already drawn 
 are necessary for the determination of the shearing forces only. 
 
 Thus, on the force line for loads 4 to 11 we can now lay off 
 the 11 equal parts just found, into which the chain reaction is 
 divided. So for 5-11 and 6-11. These portions we have indi- 
 cated by Roman numerals. We can now draw the correspond- 
 ing polygons precisely as in Fig. 108, which are indicated also 
 by Roman numerals. 
 
 It is then easy with the dividers to pick out the maximum 
 moment at any apex. These moments, laid off as below, give 
 the curve of moments for the truss, which being scaled off and 
 
CHAP. XVI.] THE INVERTED ARCH. 333 
 
 divided by the depth of truss, give at once the strains in the 
 flanges. Since the moment scale is 750 tons per inch and the 
 depth of truss 5 ft., the moment ordinates scaled off at 150 tons 
 per inch will give at once the strains in the flanges, without 
 division. 
 
 For the shearing forces, we know from the preceding that 
 the maximum, reaction at right end is for loads 6-11. This 
 reaction we have already found in the corresponding force line 
 by means of the closing line already drawn. We lay it off then 
 right and left, half-way between the ends and first apex, that 
 being the effective length of load, the two half-end panels rest- 
 ing directly upon the abutments. 
 
 The maximum shear at any point is evidently when the load 
 reaches from right support to that point, and is equal to the 
 sum of the chain reactions at the unloaded apices. Thus, max- 
 imum shear at 3 is equal to the interval I II for the line 
 3-11; at 4, I III for line 4-11; at 5, I IV for line 5-11; 
 and at 6, I V for line 6-11. Laying off the shear at 6, we 
 can draw the line 6-11, as indicated in the diagram, and thus 
 determine the shear at 2. This we cannot find, as above, for 3, 
 4, etc., as for the load 2-11 ; owing to the shape of the chain 
 as represented, there is no upward reaction at 1, as there is no 
 angle of the chain at 6. 
 
 The shear diagram is, of course, symmetrical on each side of 
 the centre. We can therefore construct it as represented, and 
 then the determination of the strain in the diagonals is easy. 
 We have only to multiply the shear at any apex by the secant 
 of the angle which the diagonals make with the vertical. This 
 we may do by properly changing the scale at once, and thus 
 scale off the strains directly. 
 
 188. Analytical Determination of the Forces acting 
 upon the stiffening Trus. Assuming that the truss distrib- 
 utes the partial loading uniformly over the whole arc, we may 
 deduce very simple formulae for the forces acting upon the 
 truss. As we have already seen, for a maximum moment at 
 any point, the load must always extend out from one end. 
 
 Let us represent, then, the ratio of the loaded part from left 
 to the whole span by k. 
 
 Let the entire span be 2 Z, then the loaded portion is 2 k I. 
 Let m be the load per unit of length ; then the whole load 
 [Fig. 108]. 
 
334 THE INVERTED ARCH. [CHAP. XVI. 
 
 The distance of P from the left is then half the loaded por- 
 tion, or k I. Its distance from P', which acts at the centre of 
 the span, is I (1 &). 
 
 Hence we have for the left reaction A 
 Axl = Pl(l-k) or A = -p(l-fy 
 Also P'l = P x Tel or P' = 2k*l 
 
 The chain reaction per unit of length is then 
 
 Now let x be the distance to the point of maximum moment. 
 Now since at this point the shear must be zero, the weight 
 of the portion x must be equal to A (Art. 38). 
 We have then 
 
 Ax - ~ = P (kl - x) = 2Jclm(Jcl - x), 
 
 Zi 
 
 whence, by substituting the value of A, 
 
 x = ^. 
 
 1 + k 
 
 But the maximum moment is Ax - = - , and there- 
 fore, substituting the value of x above, 
 
 _ _ fC \ X "^ fC ) /-\ 70 
 
 M max. = j = . 2 1 m. 
 
 1 + K 
 
 This becomes a maximum for 1 k %? = 0, or for 
 Jc = i 1/5 - i = 0.618034. 
 
 Therefore, the greatest moment occurs when 0.62 of the span 
 is covered with the load. 
 
 We have then the 
 
 Length of the loaded portion, = 2 k I = 0.61803 x 2 .1 
 Reaction, A = 2 & Z ra (1 - &) = ( 4/5 2) 2Zm = 0.23607. 
 Chain reaction, P' = *2&?lm = (J- ^ 1/5)2 lm = 0.38196. 
 Load per unit in loaded portion, or the difference between 
 the load m and the chain reaction m 1 per unit of length 
 = m (1 - &') = i ( 1/5 - 1) m ?= 0.61803 m. 
 
 The distance of the point of maximum moment is 
 
 A 
 
 x = - = (t - 4 ^5) 2 ^ = 0.38196 . 2 Z = 
 
CHAP. XVI.] THE INVESTED AECH. 335 
 
 The maximum moment itself is 
 
 - TO\ -4 7 u- . * v \ v ' " --- *. F v * * v 
 
 2 m (1 F) 1 + & 
 
 For a simple girder uniformly loaded, the maximum moment! 
 is \p 1?. The maximum moment is then reduced from \ to 
 0.18, or to about d, or is -f^ths the maximum moment for a 
 simple girder of same span and load* 
 
 If we represent the dead load by p, then, since the stiffening 
 truss sustains only the moving load, we have 
 
 0.18 ml? _ 36 m \ m (0.6 If 
 
 That is, the maximum moment in the stiffening truss is the 
 same as for a simple girder of -f^ths the span, loaded only with 
 the moving load. 
 
 189. Summary, The reaction at the end abutment and the 
 chain reaction at each apex having been found, as above, for 
 any given load, we might have found the strains in every 
 piece by the method of Arts. 8-13. This would, however, in 
 this case have proved long and tedious. The construction of 
 the curve of maximum moments and shear is preferable. 
 
 We can therefore readily determine the strains in such a 
 combination as that represented in Fig. 111. We have already, 
 Arts. 90-94, given practical and simple methods for the deter- 
 mination of the strains in braced arches of the usual forms of 
 construction. 
 
 It will be observed that it is by no means necessary that the 
 arrangement of bracing and flanges should be the same as that 
 shown in Figs. 90 and 94. 
 
 Thus we may treat the arch represented in Fig. 5 (c) accord- 
 ing to Art. 158. as hinged at both abutments and crown, or, 
 making the lower flange continuous at the crown, we may find 
 the resultant pressures at the abutments by Art. 159, and then 
 follow these pressures through precisely as shown in the Fig. 
 
 The combination of Fig. Ill being of considerable impor- 
 tance, as the more usual form of construction of suspension 
 bridges, and not falling under our classification of " braced 
 arches," we have considered it desirable to discuss it somewhat 
 
 * Rankine gives -/ 7 -ths for a girder whose ends are fixed, the greatest mo- 
 ment occurring for a load over Jds the span. 
 
336 THE INVERTED ARCH [CHAP. XIV. 
 
 fully. A better form of construction is that shown in Fig. 
 106, which is perfectly rigid, and the strains in which are 
 easily found by Art. 158 or 159, according as we hinge it in 
 the centre or not. 
 
 Reviewing now the preceding, we see that the graphical 
 method, as here developed, furnishes us with a simple, accurate 
 and practical solution of nearly every class of structure occur- 
 ring in the practice of the engineer or builder. In our first 
 chapter we have a method by the resolution of forces applica- 
 ble to any framed structure, however irregular or unsymmetri- 
 cal, provided only there are no moments at the ends to be 
 determined. 
 
 In Art. 125 we have explained fully the application of the 
 method for this case also, when these moments are known, and 
 in Chaps. VIII. to XIV. inclusive we have given practical con- 
 structions for the determination of these moments for all the 
 important classes of structures in which this condition occurs, 
 such as the continuous girder, braced arches, etc. 
 
 When the structure is not framed, or composed of pieces the 
 strains in which can be definitely determined, we have the 
 method of moments of Chap. V., which, as we have seen, may 
 be extended so as to completely solve the difficult case of the 
 continuous girder, and which may, of course, be applied to 
 framed structures also, as illustrated in Fig. Ill (Art. 187) in 
 the case just discussed. Thus we have two distinct graphical 
 methods by which our results may be checked. The first- 
 method includes a great variety of the most important and 
 usual structures, such as bridge girders, roof trusses, cranes, 
 etc., and in view of its ease and accuracy will undoubtedly be 
 found of great service by the engineer and architect. The 
 second method has important mechanical applications, as no- 
 ticed in Art. 41 ; and aside from these, and its application to 
 structures having end moments, such as the continuous girder, 
 etc., furnishes us with ready determinations of the centre of 
 gravity of areas (Chap. III.), the moment of inertia of areas 
 (Chap. VI.), and also gives us a very complete solution of the 
 stone arch (Chap. XV.). 
 
 We have also the analogous methods of calculation, viz., 
 
CHAP. XVI.] THE INVEKTED ARCH. 337 
 
 both by resolution of forces and by moments (Arts. 9 and 16 
 of Appendix). The latter being so general and simple in its 
 application, we have not felt justified in leaving it entirely out 
 of sight, and in those cases where it seemed of especial service, 
 or assisted the graphical solution, we have illustrated it more or 
 less fully (Chap. XII.). In this latter chap, we have also given 
 constructions as well as formulae, and developed principles 
 which, it is believed, render possible, for the first time, the com- 
 plete and accurate solution of the important .case of the " draw 
 xpan." (Arts. 118-121.) 
 
 The formulae of Chap. XIII. in connection with the method 
 of calculation by moments, render the calculation of the con- 
 tinuous girder generally as simple, and but little more tedious 
 than for the simple girder itself. Whatever may be thought 
 of the advantages or disadvantages of this class of structures 
 by engineers generally, it is at least time that such structures 
 as draws or pivot spans should be calculated under suppositions 
 which approach somewhat more nearly the actual case than is 
 at present the practice. As to the relative economy of con- 
 tinuous girders, we have endeavored to enforce the fact that 
 the saving over the simple girder is from 15 to 20 and even 50 
 per cent. We give in the Appendix a tabular comparison of a 
 few cases sufficient to show the point beyond dispute, and any 
 one may easily add to the list, or verify the calculations. 
 
 The " graphical arithmetic," as it might be called, such as 
 graphical addition, subtraction, multiplication, division, extrac- 
 tion of roots, determination and transformation of areas, etc., 
 we have entirely omitted in the present work, judging it of but 
 little practical value, except in rare cases, when we have ex- 
 plained the necessary constructions as they occur, and unneces- 
 sary for the development of the graphical method proper. [See 
 Chap. IV. of Introduction.] 
 22 
 
APPENDIX. 
 
 GRAPHICAL STATICS. 
 
 A. JAY DUBOIS. 
 
APPENDIX. 
 
 NOTE TO CHAPTER VIII. OF THE INTRODUCTION - UPON THE 
 MODERN GEOMETRY.* 
 
 IT is to be regretted that, notwithstanding its beauty of form, 
 simplicity, and many happy applications in the technical and 
 natural sciences, the Modern Geometry is yet hardly known, 
 scarcely by name even, in our schools and colleges. 
 
 The work of Gillespie upon Land Surveying, already cited in 
 the Introduction, and a treatise on Elementary Geometry by 
 William Chauvenet (Phil., 1871), are the only ones which 
 occur to us in this connection. 
 
 It has already been stated that the modern orj)ure geometry 
 of space differs essentially from the ancient, and from analytical 
 geometry, in that it makes no use of the idea of measure ny r>f 
 We find in it no mention of the bisection 
 
 of lines, of right angles and perpendiculars, of areas, etc., any 
 more than of trigonometrical quantities, or of the analytical 
 equations of lines. We have nothing to do with right-angled, 
 equilateral, or equiangular triangles, with the rectangle, regular 
 polygon, or circle, pxcept in a supplementary manner. So also 
 for the centre, axes, and foci of the so-called curves of the second 
 order, or the conic sections. 
 
 On the contrary, we obtain much more general and compre- 
 hensive properties of these curves than those to which most 
 text-books upon analytical geometry are limited. 
 
 A new path is thus opened to the conic sections, without the 
 aid of the circular cone, after the manner of the ancients, or of 
 the equations of analytical geometry. 
 
 As a direct consequence, the principles and problems of the 
 modern geometry are of great generality and comprehensive- 
 
 * The following- remarks and illustrations are taken from the Geometrie der 
 Lage, by Reye. Hannover, 1866. 
 
342 NOTE TO CHAP. VIH. OF THE INTKODUCTION. [APPENDIX. 
 
 ness. Thus the most important of those properties of the conic 
 sections which are proved in text-books of analytical geometry 
 are but special cases of its principles. A few particular ex- 
 amples taken from the Geometrie der Lage, by Reye, which 
 could not well have been inserted in the Introduction to this 
 work, will best explain and illustrate our general remarks the 
 more so as these examples are of special interest and value to 
 the engineer. 
 
 It is a problem of frequent occurrence in surveying to pass a 
 line through the inaccessible and invisible point of intersection 
 of two given lines. The Geometry of Measure, or ancient 
 geometry, gives us any required number of points upon this line 
 by the aid of the principle, that the distances cut off from par- 
 allel lines by any three lines meeting in a common point are 
 proportional. The__Geometry of Position furnishes us with a 
 simpler solution. 
 
 FIG. 1. 
 
 Thus the two lines , 5 being given [Fig. 1.], we have sim- 
 ply to choose any point we please, as P. From this point draw 
 any number ;of lines desired, in any direction intersecting the 
 given lines. Now, in any quadrilateral which any two of these 
 lines form with the two given lines a and &, we have simply to 
 draw the diagonals. The intersections of all these diagonals 
 lie in the same straight line passing through the intersection A 
 of the two given lines, and therefore determine the line re- 
 quired. Observe that the construction is entirely independent 
 of all metrical relations, and depends solely upon the relative 
 position of the two given lines. 
 
 Again : If we take upon any straight line three points, A, B 
 and C [Fig. 2.], and construct any quadrilateral, two opposite 
 sides of which pass through A, one diagonal through B, and the 
 
APPENDIX.] 
 
 THE MODERN GEOMETRY. 
 
 343 
 
 other two opposite sides through C, then will the other diagonal 
 intersect the line in a point D, which for the same three points, 
 
 FIG. 2. 
 
 A B C D 
 
 A, B and C, is always the same for every possible construction. 
 Moreover, these four points, A, B, C and D, are always harmonic 
 points, so that D is harmonically separated from B by the 
 points A and C. Thus, A B : B C ; ; A D : C D. This construc- 
 tion may also be applied in surveying, as in passing around an 
 obstruction, as a wood, etc., into the same line again. 
 
 Again : We may notice the following principle concerning 
 the triangle [Fig. 3] ; 
 
 FIG. 3. 
 
 u. 
 
 If two triangles, ABC and A! B t C 1? are so situated that the 
 lines joining corresponding angles, as A A 1? B B 1? C C 1; meet in 
 a common point S, then will the intersections of corresponding 
 sides, as A C and A x G l9 A B and A t B 1? B C and B! C 1? meet in a 
 common line, as u u. The inverse also, of course, holds good : 
 that if the sides intersect on a line, the lines through the angles 
 intersect in a point. 
 
 Another series of principles are connected with the curves of 
 the second order, or conic sections. From analytical geometry, 
 
344 NOTE TO CHAP. VIII. OF THE INTRODUCTION. [APPENDIX. 
 
 as is well known, a curve of the second order is completely de- 
 termined by five points or five tangents. But the length of the 
 calculation or construction of a curve thus determined is also 
 well known. The geometry of position, however, proves two 
 very important principles, which render it easy to construct to 
 the five given points or tangents any number of new points or 
 tangents, and thus quickly draw the curve itself. The reader 
 already acquainted with these principles will also probably re- 
 member how much auxiliary demonstration their proof in the 
 analytical geometry requires. The first of these, due to Pascal, 
 is, that the three pairs of opposite sides of a hexagon inscribed 
 within a conic section intersect upon a straight line. The 
 second, due to Brianchon^ is, that the three principal diagonals 
 of the circumscribing hexagon, which unite every pair of oppo- 
 site angles, intersect in one and the same point. Both prin- 
 ciples are easily deduced from the circle. It will be observed 
 that they are independent of the relative dimensions, centre, 
 axes, and foci of the curves. For this very reason they are of 
 the greatest generality and significance, so that an entire theory 
 of the conic sections can be based upon them. Thus Pascal's 
 principle solves the important problem of tangent construction 
 from a given point, even when the curve is given by five points 
 only, without completely constructing it. 
 
 This problem of tangent construction to curves of the second 
 order can in many cases be solved by the aid of a principle 
 which expresses one of the most important properties of the 
 conic sections, but which, nevertheless, is seldom found in text- 
 books upon analytical geometry, because its analytical proof is 
 somewhat complicated, and little suited to set forth the property 
 in its proper light. 
 
 For example : If through a point A [Fig. 4] in the plane of 
 but not lying upon a curve of the second order, we draw se- 
 cants, every two secants determine four points, as K, L, M, N, 
 upon the curve. Any two lines joining these four points, as 
 L M and K N or K M and Li l\ r , intersect in a point of a 
 straight line a a, which is the polar of the given point A ; that 
 is, which intersects the curve in the two points of tangency 
 G G. Thus the lines through A and the intersections of a a 
 with the curve are the tangents to the curve through A. If 
 the point A were within the curve, this line a a would not in- 
 tersect it. This construction can be used in order to draw 
 
APPENDIX.] THE MODEEN GEOMETRY. 345 
 
 through a given point tangents to a conic section by the sim- 
 ple application of straight lines. Upon every secant through 
 
 FIG. 4. 
 \L 
 
 A, moreover, there are four remarkable points, viz. : the point 
 A itself, the first intersection B with the curve, the intersection 
 with the polar, and, finally, the second intersection D with 
 the curve. These four points are harmonic points, and the 
 polar a a contains, then, every point which is harmonically 
 separated from A by the two curve points. The important, 
 principles relating to the centre and conjugate diameter of 
 conic sections are merely special cases of the above important 
 principles. These last can be easily extended to surfaces of 
 the second order, as the intersection of these by a plane is, in 
 general, a curve of the second order. 
 
 From these few examples, which might be indefinitely multi- 
 plied, it may easily be seen how very different, but not less im- 
 portant than those of analytical geometry, are the theorems of the 
 geometry of position. Thus the latter are generally proved by 
 aid of the angle which the tangents make with the line through 
 the focus, or by the distances cut off from the axes that is, by 
 metrical relations. We refer, of course, to the elements of 
 analytical geometry as contained in most text-books, and not to 
 those most fruitful and later methods .whose existence are 
 chiefly due to the sagacity of Plucker (Introduction, VIII.). 
 
346 NOTE TO CHAP. I. [APPENDIX. 
 
 NOTE TO CHAPTEE I. 
 
 1. The method by the resolution of forces developed in 
 Chapter I. is so simple and easy of application, and its principles 
 are so few and self-evident, that we have not considered it ad- 
 visable to tax the patience of the reader by any great variety 
 of practical applications. A large number of such applica- 
 tions are to be found in a most excellent little treatise by 
 Robert H. Bow, entitled The Economics of Construction in 
 Relation to Framed Structures. There are, however, a few 
 important practical points- of detail, and a few general consid- 
 erations, which we think it well to notice here, and to which, 
 in illustration of the remarks in Chap. I., the reader will do 
 well to attend. 
 
 2. In PL 1, Fig. I. (Appendix), we have represented the 
 "Bent Crane " given by Stoney in his Theory of Strains, p. 
 121, Art. 200. 
 
 We assume the following method of notation. Let all that 
 space above the Fig. be indicated by X, and all that space 
 below by Y, and the triangular spaces enclosed by the flanges 
 and diagonals by the numbers 1, 2, 3, 4, etc. The first upper 
 flange is then denoted by X 2, the second by X 4, and so on. 
 So also the first lower flange is Y 1, the next Y 3, etc. The 
 first diagonal is then X 1, the next 1 2, the next 2 3, etc.* 
 
 The flanges are equidistant, forming quadrants of two cir- 
 cles whose radii are respectively 20 and 24 feet. The inner 
 flange is divided into four equal bays, on which stand isosceles 
 triangles, and a weight of 10 tons is suspended from the peak. 
 The scale for this and all the Figs, of PI. I. is 20 tons to an incji 
 and 10 feet to an inch. Laying off, then, the weight X Y = 10 
 tons, we form, according to the method of Chapter I., the strain 
 diagram. It will be seen at once that all the lower flanges, Y 1, 
 Y 3, etc., radiate from Y, all the upper flanges, X 2, X 4, etc., 
 from X, and everywhere the letters in the one diagram indi- 
 cate the corresponding pieces in the other. 
 
 * For this very elegant niethod of notation, we are indebted to the work of 
 B. H. Bow, above alluded to. 
 
APPENDIX.] NOTE TO CHAP. I. 347 
 
 "We can now at once take off the strains to scale in the vari- 
 ous pieces. 
 
 A comparison of our method with that given by Stoney for 
 the same case will be instructive, as illustrating the compara- 
 tive merits of the two. 
 
 3. Character of the Strain* in the Pieces. One of the 
 most important points of our method is the ease and certainty 
 with which the character of the strains in the pieces may be 
 determined. We have only, as detailed at length in Chapter 1., 
 to follow round any closed polygon in the direction of the 
 forces, and then refer back to that apex of the frame where for 
 the moment we may happen to be. 
 
 Thus for the peak, since we know that the weight acts down, 
 we follow down from X to Y, and then from Y to 1, and 1 
 back to X. Referring back now to the frame, and remember- 
 ing that a force acting away from the apex means tension, and 
 towards, compression, we have at once Y 1 compression and 
 X 1 tension. 
 
 Now for apex a, since X 1 is tension, with respect to this new 
 apex, it must act away. We go round then from X to 1, 1 to 
 2, and 2 back to X, and then, referring these directions to the 
 corresponding pieces meeting at &, we have 1 2 compression 
 and X 2 tension. 
 
 We find thus all the outer flanges in tension, as evidently 
 should by simple inspection be the case. Also all the inner 
 flanges compression. As for the diagonals, they alternate, the 
 first being tension, the next compression, until we arrive* at 4 5, 
 which we find to be also compression. 
 
 A glance at the strain diagram shows how this comes about. 
 The line X 4 crosses Y 5, and thus gives us a reverse direction 
 for 4 5. 
 
 In such a simple structure as the present, the character of the 
 strains would present no especial difficulty in any case ; but in 
 more complicated ones, the aid of such a simple and sure crite- 
 rion as the above is indispensable, and we have been thus even 
 prolix upon this point, the more so as it is not so much as 
 alluded to, as far as we are aware, in those few works which 
 notice the above method at all. 
 
 4. There are other points which we may here illustrate by 
 our Fig. 
 
 According to our first principle (Art. 3, Chapter I.), when 
 
348 NOTE TO CHAP. I. [APPENDIX. 
 
 any number of forces are in equilibrium, the force polygon is 
 closed. Inversely, then, a closed force polygon indicates forces 
 which, if applied at a common point, would hold each other in 
 equilibrium. 
 
 Thus Y 3, 34, X 4, and the weight are, or would be, if all 
 applied at a common point, in equilibrium. This we see 
 directly from the Fig. Thus we know that when any num- 
 ber of forces are in equilibrium, the algebraic sums of their 
 vertical and horizontal components must be zero, otherwise 
 there must, of course, be motion. Now the vertical component 
 of Y 3 plus that of 3 4 minus that of X 4 is exactly equal and 
 opposed to the weight, while the horizontal component of Y 3 
 plus that of 3 4 is equal and opposed to that of X 4, and there 
 is then equilibrium. 
 
 Again, according to the principle of Art. 5, Chap. I., any 
 line, as the one joining 2 and 6 (broken line in Fig.), is the 
 resultant of X 2 and X 6, as also of 2 3, 3 4, 4 5 and 5 6. 
 
 The Fig. also well illustrates the points to be avoided in 
 making a strain diagram, already alluded to in Art. 13, Chap. 
 I. The scale to which the frame is taken is here altogether 
 out of proportion to the scale of force. The first should be 
 increased or the second diminished, or both. The present 
 length of the diagonals and flanges is inadequate to give 
 with sufficient accuracy the directions of strain lines of such 
 length. 
 
 Nevertheless we have experienced no difficulty in checking 
 to tenths of a ton the results given by Stoney for this structure. 
 
 5. In PI. 1, Fig. II., we have represented a roof truss, span 
 30 ft., rise 8 ft., camber 1 ft. ; and the strain diagram illustrates 
 in its two symmetrical halves (one full, the other dotted) the 
 remarks of Art. 13, Chap. I., upon the check which in such cases 
 our method furnishes of its accuracy. 
 
 We lay off the weights 1, 2, 3, 4, 5, and then the reactions at 
 A and B, which should bring us back to the point of beginning, 
 and thus complete the force polygon. The strains are then easily 
 found, and the two halves should be perfectly symmetrical, and 
 give the same results. 
 
 In Fig. III. we have given another form of truss with strain 
 diagram, the other half of which the reader can complete and 
 letter for himself. 
 
 6. In Fig. IY. we have a form called the French roof truss 
 
APPENDIX.] NOTE TO OH A P. I. 349 
 
 and two strain diagrams the larger for vertical reactions, the 
 smaller for inclined reactions. 
 
 This last brings out the force polygon in perhaps a clearer 
 shape than before. The weights 1 to Y being laid off down- 
 wards, the two reactions must always bring us back to the 
 starting-point, and thus close the polygon in this case a trian- 
 gle, in the preceding case a straight line, and in the case of Fig. 
 6, Art. 10., Chap. I., a true polygon. Both strain diagrams 
 illustrate the check we have upon the accuracy of the work. 
 The second half should be perfectly symmetrical with the first, 
 and the lines Y k and points Jc in each should coincide. 
 
 We have here also to notice a point which in roof trusses is 
 of frequent occurrence, and may, if not noticed, cause diffi- 
 culty. 
 
 We have already observed in Art. 9, Chap. I., that we can 
 always find the strains in the pieces which meet at an apex, 
 provided only two are unknown. Now in the strain diagram 
 to Fig. IY., we readily determine the strains in X , Y a, X b, 
 a b, Y c and b c successively, and arrive finally at apex 2, where 
 we have the two known strains in X 5 and b , and wish to find 
 the strains in three pieces, viz., X cZ, d Ji and c h. At first sight 
 this seems impossible. If, however, we assume that the pieces 
 of the frame can take only strains of a certain kind, as, for 
 instance, hd only tension, and not compression, the problem is 
 perfectly determinate. This assumption is easily realized in 
 practice. Thus if h d is a rod of small diameter, it cannot 
 act as a compression member at all. Moreover, the strain 
 of tension in h d must evidently be precisely equal to that in 1 1 
 b c, already found. We have then to form a closed polygon u 
 with the weight at 2 and the known strains in X b and b c, whose * 
 other three sides shall be parallel to X d, h d and c h respec- 
 tively, and in which, moreover, the strain in h d shall be equal 
 to that in b c, and where both these strains must be, when the 
 polygon is followed round according to rule, tensile. We have 
 evidently, then, in accordance with these conditions, only tlio 
 polygon 2 X d h c b X, thus finding the point d, from which 
 we can now proceed to find e, etc. The points a, b, d and e are 
 evidently in the same straight line parallel to c h. This point 
 is one of importance, and the reader should carefully follow 
 the above remarks with the aid of the Fig. 
 
 The strain diagram thus constructed shows us many facts 
 
350 NOTE TO CHAP. I. [APPENDIX. 
 
 about the system not otherwise apparent. Tims ~b c, c h and h d 
 are in equilibrium with the load at 2. Again, a b and b c are 
 in equilibrium with Y a minus Y <?, as also are hd and de with 
 Jc e minus kh. Also Teh, h c, Y c and Y Jc are in equilibrium, 
 and Yc, cb and X b are in equilibrium with the reaction minus 
 the weight at 1, or with the shear to the right of 1. This last 
 principle is general. When a section can ~be made entirely 
 through a structure, the strains in the pieces cut are in equi- 
 librium with the shear at the section. If only three pieces are 
 cut, then, by taking as a centre of moments the point of inter- 
 section of any two, we can easily find, knowing the moment of 
 the shear, the strain in the third. 
 
 Thus we have the general and easy method of calculation 
 given in Art. 14, Chap. I. The moment of the shear is, of 
 course, the sum of the moments of all the exterior forces be- 
 tween the section and one end. 
 
 We have then two methods, one graphic and one by calcula- 
 tion, by which we can find the strains in every kind of simple 
 truss which can ever occur in practice. By li simple " we mean 
 merely resting at the supports, or not acted upon at the ends by 
 a couple or moment, as is the case, for instance, in the continuous 
 girder. 
 
 When the structure is unsymmetrical, or complex, the deter- 
 mination of the different lever arms is often very tedious, involv- 
 ing a good deal of trigonometrical computation. On the other 
 hand, the frame can always from its known proportions be 
 easily and accurately drawn to scale, and then the exterior 
 forces, whatever their relative intensity or directions, can be 
 laid off, and the strains at once determined. Here we see, then, 
 one of the great advantages of our graphical method. An 
 unsymmetrical frame and different directions of the forces 
 requires no more time or labor than a more simple case. 
 
 7. Application to Bridges Bow-string Girder. In Art. 
 12, Chap. I., we have alluded to this application, and shown 
 how by two strain diagrams only we can completely calculate a 
 bridge of any length. As this application is so important, and 
 as the method is stated by several authors to be inapplicable to 
 bridges,* or, at best, to be unsatisfactory, we will here call more 
 
 * Iron Bridges and T&w/s Unwin p. 143. Economics of Construction 
 Bow- -p. 61. 
 
APPENDIX.] 
 
 NOTE TO CHAP. I. 
 
 351 
 
 special attention to the points to be observed in the tabulation 
 of the strains. There is, indeed, no more satisfactory, complete 
 and rapid method for the solution of bridge girders generally 
 than that afforded by the graphic method. 
 
 As an example, let us take the Bow-string Girder given by 
 Stoney, p. 131. Span, 80 ft., divided into 8 panels ; rise of 
 bow, 10 ft. Load, 10 tons at each lower apex. 
 
 We construct the two strain diagrams* given in Fig. V., 
 PL 2, viz., one for the load P 7 at the first apex, and one for the 
 load at the last apex, P,. Referring, if necessary, to Art. 12, 
 Chap. I., the reader can easily follow out these diagrams. We 
 then scale off the strains, and obtain, for the strains in the 
 diagonals 
 
 
 ab 
 
 be 
 
 cd 
 
 de 
 
 ef 
 
 fff 
 
 ffh 
 
 FT 
 
 -2.7 
 
 -11.4 
 
 + 4.8 
 
 - 4.3 
 
 + 2.4 
 
 -2.3 
 
 + 1.4 
 
 Pi 
 
 -0.4 
 
 + 0.23 
 
 -0.56 
 
 + 0.51 
 
 -0.9 
 
 + 0.88 
 
 - 1.4 
 
 Now from the strains thus obtained for these two weights 
 we can easily obtain all the others. 
 
 Thus, as the end reactions are inversely as the distances of 
 the weight from the ends, the reaction at the left end due to 
 P 2 will be twice that due to P a . For P s three times that due 
 to P,. The strains will therefore be twice and three times 
 those due to P 1? until we arrive at the weights P 1 and P a re- 
 spectively. So also for P 6 the reaction at the right is twice 
 that due to P 7 , and the strains are therefore double up to the 
 weight P 6 . To the right, then, of P 6 the strains are twice 
 those due to P 7 , and to the left of P 6 they are six times those 
 due to P!. Take, for instance, P 5 . The right reaction is f ths 
 of the apex load, and the right reaction of P 7 is -|th of that 
 load. For P 5 , then, the strains in all pieces to the right of that 
 weight are 3 times those due to P 7 . Again, the left reaction 
 is for P 5 |ths the apex load. But the left reaction for P l 
 is -Jth the same load. The strains then in all the pieces to the 
 left of P 5 are 5 times those due to P!. So for any other load. 
 We can therefore form at once the following table : 
 
 * Strain diagrams in Fig. V., and also in Fig. VL, are, for obvious reasona, 
 drawn to different scales. 
 
352 
 
 NOTE TO CHAP. I. 
 
 [APPENDIX. 
 
 
 
 
 PI 
 
 P 2 
 
 P 3 
 
 P 4 
 
 5 
 
 P 6 
 
 P 7 
 
 Uniform 
 Load. 
 
 Max. 
 Comp. 
 + 
 
 Max. 
 Tens. 
 
 Total 
 Strains. 
 
 ab 
 
 -0.4 
 
 - 0.8 
 
 - 1.2 - 1.6 
 
 -2.0 
 
 -2.3 
 
 -2.7 
 
 - 8.25 
 
 1-n.o 
 
 - 19.25 
 
 be 
 cd 
 
 +0.23 
 ^.56 
 
 + C.5 
 1.1 
 
 + 0.7 
 1.7 
 
 + 0.9 
 - 2.2 
 + 2.0 
 
 + 1.1 
 
 + 1.4 
 
 -11.4 
 +4.8 
 
 -4.9 
 
 + 4.8 - 11.4 
 
 + 4.bj - 11.8 
 
 - 16.3 
 
 - 2.8 
 
 -3.4 
 
 -5.2 
 
 - 17.0 
 
 de 
 ef 
 
 +0.51 
 ^0^ 
 
 + 1.0 
 - 1.8 
 
 + 1.5 
 
 + 2.6-8.6 
 
 -4.3 
 + 2.4 
 
 -3.97 
 
 -4.8 
 
 + 7.6 
 + 7.1 
 
 - 12.9 
 
 - 16.9 
 - 18.3 
 
 - 2.7 - 3.6 
 
 -4.5 
 
 + 4.7 
 
 - 13.5 
 
 fa\ 
 
 d 
 
 +0.88 
 
 + 1.8 
 
 + 2.6 + 3.5 
 
 -6.9 
 
 -4.6 
 
 -2.3 
 
 0.55 
 - 4.2 
 
 + 8.8 
 
 - 13.8 
 
 -20.5 
 
 -1.4 
 
 -2.8 
 
 -4.2 
 
 -5.6 
 
 + ,2 
 
 + ,8 
 
 +1.4 
 
 + 8.4 
 
 -14.0 
 
 - 18.2 
 
 In the columns for P t and P 7 we put the strains already 
 found by diagram. The strains for P 2 on the entire left half 
 will be double those for P x ; for P 3 three times, and for P 4 four 
 times those for P x . We have therefore at once the columns 
 for P 1? P 2 , P 3 and P 4 . Now for P 5 we see from the Fig. that 
 the strains in diagonals ef and fg must both be tension. 
 From the left end, then, as far as ef, the strains are 4 times 
 those due to P l5 and from the right, as far asfg, 3 times those 
 due to P 7 . We thus obtain the column for P 5 . In the same 
 way for P 6 , all above or to left of cd are 6 times P l5 all below 
 or to right of de twice P 7 . Thus we fill out the whole table. 
 Adding now all the tensions and compressions in each piece, we 
 obtain the maximum strains of each kind due to the live load, 
 as given in the last two columns but one. Suppose now the 
 dead load or weight of the girder itself to be fths of the roll- 
 ing or live load. We have only to take, then, f ths the sum of 
 these last two columns and we have the strains due to uniform 
 or dead load, as given in the fourth column from the right. 
 
 We can now easily obtain the total strains. Thus the ten- 
 sion in a 1) due to the live load only is 11 tons. The tension 
 due to the dead load is 8.25 tons. Total greatest strain which 
 can ever come upon a 5, then, is 19.25 tons tension. No com- 
 pression can ever come on this |)iece ; it does not need, there- 
 fore, to be counterbraced. On the other hand, all the other 
 diagonals, except perhaps cd, must be counterbraced, as the 
 maximum compression due to the live load overbalances the 
 constant tension of the dead. Had the dead load been taken 
 much greater than the live, the diagonals might always have 
 been in tension. Hence the appropriateness of this class of 
 girder for long spans. 
 
APPENDIX.] 
 
 NOTE TO CHAP. I. 
 
 353 
 
 We see also from the table just what weights, and where 
 placed, give the greatest strain of eacli kind- in any piece. 
 
 . Strains in tlie Flanges. The method is precisely simi- 
 lar for the flanges. Thus we scale off from our diagrams 
 
 
 Xa 
 
 Xd 
 
 JLd 
 
 */ 
 
 Th 
 
 P! 
 
 + 2.82 
 
 + 3.08 
 
 + 3.47 
 
 + 4.11 
 
 + 5.11 
 
 PT 
 
 + 19.7 
 
 + 21.6 
 
 + 10.4 
 
 +6.8 
 
 + 5.1 
 
 
 Ya 
 
 Yc 
 
 Ye 
 
 ?9 
 
 Pi 
 
 - 2.52 
 
 -3.01 
 
 -3.62 
 
 - 4.46 
 
 P 7 
 
 - 17.6 
 
 - 13.1 
 
 -7.9 
 
 -5.7 
 
 Tabulating these, we obtain the following table : 
 
 Flanges. 
 
 
 PI 
 
 ! 
 
 P 2 
 
 P 3 
 
 P 4 
 
 P 5 
 
 P 6 
 
 P 7 
 
 Uniform 
 live 
 load. 
 
 Uniform 
 dead 
 load. 
 
 Total 
 
 Strains. 
 
 Xa 
 
 +2.82 
 
 + 5.6 
 
 + 8.5 
 
 +11.3 
 
 + 14.1 
 
 + 16.9 
 
 +19.7 
 
 + 78.9 
 
 + 59.1 
 
 + 138 
 
 X6 
 
 +3,08 
 
 + 6.2 
 
 + 9.2 
 
 +12 3 
 
 + 15.4 
 
 +18.5 
 
 +21.6 
 
 + 86.3 
 
 + 64.7 
 
 + 151 
 
 Xrf 
 
 +3.47 
 
 + 6.1) 
 
 + 10.4 
 
 +13.9 
 
 + 17.3 
 
 +20.8 
 
 +10.4 
 1^.8 
 +5.1 
 
 + 83.2 
 
 + 62.4 + 145.6 
 
 x/ 
 
 ~XA 
 
 +4.11 
 
 + 8.2 
 
 + 12.3 
 
 +16.4J+20.5 
 
 + 13.7 
 + 10.2 
 
 + 82.0 
 "+81.6 
 
 + 61.5 
 + 61.2 
 
 + 143.5 
 
 +5.11 
 
 + 10.2 
 
 + 15.3 
 
 +20.4 
 
 +15.3 
 
 + 142.8 
 
 Ya 
 
 -2.52 
 -3.01 
 
 -5.0, 
 
 -7.G 
 
 -9.0 
 
 -10.1 
 -12.0 
 
 -12.6 
 
 -15.0 
 
 -15.1 
 
 ^-IsT 
 
 -17.6 
 
 -70.5 
 -76.2 
 
 -52.8 
 
 -128.3 
 
 Yc 
 
 -6.0, 
 
 -13.1 
 
 -57.1 
 
 - 133.3 
 
 Ye 
 Y7 
 
 -3.62 
 ^A6 
 
 -7. 2, -10.9 
 
 -14.5 
 
 -18.1 
 
 -15.9 
 
 -7.9 
 
 -78.1 
 
 -58.5 
 
 - 136.6 
 
 -8.9 -13.4 
 
 -17.8 
 
 -17.1 
 
 -11.4 
 
 -5.7 
 
 - 78.8 
 
 -59.1 
 
 137.9 
 
 This table is obtained precisely as before. Thus for P 6 the 
 strains in Xa and Xb are multiples of P 1? while those in the 
 other flanges are multiples of P 7 . So also for Y c and Y e. We 
 f see at once that the greatest strains are for full load, since for 
 all loads the upper flanges are always compressed and the 
 lower extended.* The above is sufficient to illustrate fully the 
 application of our method to bridges. It is evidently appli- 
 
 * A more convenient form of tabulation is to put the weights in the left 
 vertical column and the pieces in the top horizontal line. The numbers can 
 then be more easily added. 
 23 
 
354 NOTE TO CHAP. I. [APPENDIX. 
 
 cable to any structure where the reactions are inversely as the 
 distances from the end. The strains due to the first and last 
 weights are all that we need in order to thoroughly solve any 
 case of the kind. It is advisable, however, to construct a 
 third diagram for an intermediate weight, in order to serve as 
 a check upon the others.* 
 
 9. Method of Calculation by Moments. AVe may illus- 
 trate here the method of calculation by moments referred to in 
 Art. 14, Chap. L, a little more fully. Thus in the example 
 above, Fig. V., suppose we wish the strains due to P { . Reac- 
 tion at left end is evidently th of 10 tons = 1.25 tons. Con- 
 ceive the lower flange Y a cut. Rotation would evidently take 
 place about apex a, and we have, therefore, strain in Y a x its 
 lever arm from apex a = 1.25 x 5. The depth of truss, or 
 lever arm of Y$, from apex a, is 2.58 feet. Hence we have 
 
 ' , K0 = strain in Y a 2.42 tons. 
 a.Do 
 
 This strain is evidently, by reason of the direction in which 
 the two portions of the truss would rotate about a, tension. In 
 like manner, for upper flange X &, if we know the lever arm of 
 this flange from the opposite apex, we can easily find the strain ; 
 for the diagram shows that X 5, b c and Y c are in equilibrium 
 with the reaction, and hence, if we take the point of moments 
 at the intersection of the two pieces b c and Y c, the moments 
 of these pieces are zero, and we have remaining only the mo- 
 ment of the strain in X b balanced by the moment of the reac- 
 tion. 
 
 Again, if X b and Y<? are thus found, and if these two strains, 
 together with the reaction, are in equilibrium with the diago- 
 nal b c, we can find the strain in this diagonal by taking the 
 apex d as a centre of moments. The moment of Y c then is 
 
 * It may also be well to notice here that the practice of deducing in the 
 tabulation the dead load from the live load strains is not strictly accurate, as 
 the live load acts at the lower apices only [or at the upper apices only, if the 
 bridge is under grade],, while the dead load is distributed along both flanges, 
 and acts at both upper and lower apices. 
 
 In every case, however, the greater portion of the dead load, say, for in- 
 stance, fds of the whole, owing to the track, platform, cross-girders, etc., 
 acts at the same apices as the live load itself ; and the error is in any case 
 very slight, and practically of no account. 
 
APPENDIX.] NOTE TO ClfrAP. I. 355 
 
 zero, and we have the moment of the strain in ~b c balanced 
 by the moment of the reaction and the moment of X 5 ; the 
 first causing compression, the second tension, and the difference 
 then giving the resultant moment strain, which, divided by the 
 lever arm of & <?, gives the strain itself. 
 
 The method is easy of application, but, as we have already 
 remarked, the determination of the lever arms for each case is 
 frequently tedious. " These may, however, be scaled off from 
 the frame diagram with sufficient accuracy in practice. 
 
 As before, we need only the strains due to the first and last 
 weight, and can then form our tabulation as above. This 
 tabulation we can also check by finding the strains due to uni- 
 form load independently, and seeing whether it agrees with 
 the sum of the separate apex weight strains. 
 
 Thus, for all the weights acting, suppose we wish the strain 
 in Y g. The lever arm of Y g is 9.85 feet. We have then re- 
 action = 35 tons, multiplied by 35 feet = 1225. This must be 
 diminished by P 7 X 25 = 250, P 6 x 15 = 150, and P 5 x 5 = 
 50. We have then 1225 - 450 = 775, which, divided by 9.85, 
 gives 78.7 tons tension in Y </, agreeing with our tabulation 
 above. 
 
 For the method of calculation by resolution of forces, see 
 Art. 16 of this Appendix. 
 
 10. Girder with Straight Flanges. In such a case, as the 
 lever arms are at once known and are constant, the above 
 method is of very easy application. In this case the strains in 
 the diagonals are best found by multiplying the shear by the 
 secant of the inclination of the diagonal with the vertical.* 
 Thus, if this angle is 45, we have simply to multiply the shear 
 at any point by 1.4142, and we have at once the strain in the 
 
 * This is but a particular result of the general method of moments. Thus, 
 for any diagonal, as ab (Fig. VII.), according to our rule, we take the centre of 
 moments at the intersection of the two other sides cut by a section through 
 the truss, viz. , the flanges. But these two sides are here parallel, hence their 
 intersection is at an infinite distance. The lever arm of a b is then GO x cos </-, 
 6 being the angle with the vertical. If the weight PI acts, we have then, 
 calling the reaction R, R x GO P GO = S cos </> x GO, where S is the strain in 
 
 a b. This can be put (R P) GO = S cos x co, hence S = - . But 
 
 GO 1 
 
 R P is the shear at 5, C08 > x 00= T = sec ; hence we have only to 
 
 multiply the shear by the secant. 
 
356 NOTE -TO ciiAr. i. [APPENDIX. 
 
 diagonal at this point. The shear is always in such cases the 
 reaction at the end minus the weights between that end and 
 the apex in question. 
 
 The flanges are easily obtained by moments, as above. 
 
 The following points need attention, however. First, if 
 there are two or more systems of diagonals, as represented in 
 PL 2, Fig. YIL, by the full and dotted diagonals (omitting the 
 upright lines), we must find the strains for each system sepa- 
 rately, and then add them together. Thus, if the strains found 
 in a c and c e* etc., for one system, are 50 and 60 tons, and 
 those in df and fg, for the other, are 40 and TO tons, we have, 
 when the two systems are combined, d c = a c + df = 50 + 40 
 = 90, cf= df+ce = + 60 = 100, fe = ce +.fg = 
 60 + 70 = 130, and so on. This holds true, of course, whether 
 the strains are obtained by calculation or diagram. Thus, for 
 a lattice girder, we calculate or diagram each system ~by itself, 
 and then the strain in any flange, when the two are combined, 
 is equal to the sum of the strains on that flange due to each 
 system of triaiigulation which includes it. 
 
 There is another point to be observed in connection with 
 the system known as the Howe or Pratt Truss. Inserting the 
 dotted verticals into our Fig., we have this system of square pan- 
 elling. Let us suppose that the diagonals take tension only, 
 and the verticals compression only. 
 
 Now for a weight at apex 9 of 10 tons, we have a right re- 
 action of 1 ton, which, running through the system, causes strain 
 in the diagonal of f P 4 . For the flange D, then, our point of mo- 
 ments is at/, and if the height of truss is equal to panel length, 
 
 1 x 50 
 viz., 10 feet, we have the strain in D = = 5 tons, for 
 
 P 9 . In the same way for P 8 , we have for D 10 tons ; for P 7 , 15 
 tons ; for P 6 , 20 tons ; for P 5 , 25 tons. For P 4 , on the other 
 hand, we have a left reaction of 4 tons, which causes strain in 
 diagonal e &, and for this weight and all succeeding weights 
 our point of moments for D is then at e. We have then P 4 
 
 4 x = 24 tons ; for P 3 , 18 tons ; for P 2 , 12 tons ; and for 
 
 P!, 6 tons. 
 
 For all these weights, then, acting together, we have 135 tons 
 strain in D. 
 
APPENDIX.] NOTE TO CHAP. I. .357 
 
 But for all tlie weights acting together, it is evident that 
 only all the braces sloping each way from the centre are 
 strained. Hence e Tc is not strained, and our point of moments 
 is for D always at/! Thus for total load we have strain in 
 
 45 x 50 - 10 x 40 - 10 x 30 - 10 x 20 - 10 x 10 ' 
 D = - 1Q - = 125 tons, 
 
 whereas we found by addition of the several weights 135 tons. 
 
 There is thus an ambiguity in this class of bracing as to the 
 way in which the strains may go. Two symmetrical weights, 
 as 9 and 1, may either go left and right directly to the abut- 
 ments or a portion of each go towards the centre. The inter- 
 mediate diagonals may be either all strained or not strained at 
 all. The strains may go partly in one way or partly in the 
 other. We should then not rely on our summation of the sepa- 
 rate weights, but always check them by calculation or diagram 
 for the total load also, and take the greatest strain*. Practi- 
 cally, for long spans, it is very rare that the difference is of any 
 importance. 
 
 In diagraming by our method such a system of bracing as 
 the above, we should consider but one series of braces, viz., 
 those strained by the uniform load alone. Thus, for our Fig. 
 and loads on the lower apices, we should take only the diago- 
 nals parallel to fh on the left of centre, and yP 4 on the right. 
 If, on the other hand, the verticals are ties and the diagonals 
 struts, we should retain only those parallel to c & on the left, 
 and those parallel to k e on the right of centre. The others 
 are to be omitted. Then, the tabulation being formed, if in 
 any diagonal a strain may occur of reverse character to that 
 which it is intended to resist, a counterbrace must be inserted 
 in this panel to take this reverse strain. 
 
 As in our examples we have taken always a triangular system 
 of bracing, it is important that the reader clearly understand 
 the method to be pursued in other forms. For the rectangular 
 system of bracing generally, the point where for uniform load 
 the shear is zero is the point from which the braces must slope 
 both ways. The other diagonals, or the count erbr aces, are then 
 omitted in both calculation and diagram, and replaced from the 
 tabulation when necessary to replace a strain of the reverse 
 character to that which the braces are intended to sustain. 
 
 Attention to the above points will enable us to both calculate 
 
353 
 
 NOTE TO CHAP. I. 
 
 [APPENDIX. 
 
 and diagram with ease and accuracy any form of truss which 
 occurs in engineering practice. 
 
 11. In the bow-string girder represented ir Fig, Y. it is 
 evident that the bottom flange serves merely to resist the thrust 
 of the bow and keep it from spreading. It adds nothing to the 
 supporting power of the combination. We might remove it 
 entirely and replace it by abutments which would equally well 
 sustain this thrust, and if we then introduced a horizontal flange 
 at crown, and inserted diagonals between for stiffness, we should 
 have the form of braced arch given in Chap. I., Fig. 5 (c). If, 
 however, we should resist the thrust of the bow by an inverted 
 arc, it would answer the same purpose as the bottom flange, 
 and we should, in addition, double the supporting power. 
 
 We have illustrated this in Fig. YI. 
 
 The span is the same as before. The lower apices only are 
 supposed to be loaded, for comparison. [Properly, we should 
 have distributed the load over both upper and lower apices.] 
 The rise of each arc is one-half as great as before, or 5 f t % only, 
 thus making the total depth the same as in the preceding 
 case. 
 
 By means of two strain diagrams, we find the strains due to 
 P ia ndP 7 . Thus: 
 
 
 xa 
 
 Xb 
 
 Xd 
 
 x/ 
 
 Xh 
 
 Ya 
 
 Y c 
 
 Y e 
 
 Y<7 
 
 Pi 
 
 +2.3 
 
 +2.6 
 
 +3.2 
 
 +3.7 
 
 +5.0 
 
 -2.25 
 
 -2.87 
 
 -3.48 
 
 -4.32 
 
 P 7 
 
 +16.1 
 
 +17.8 
 
 +9.1 
 
 +5.7 
 
 +4.5 
 
 -15.75 
 
 -12.35 
 
 -7.3 
 
 -4.8 
 
 Then, precisely as in the preceding Art., we can fill out our 
 table of strains. This the reader can now easily do for himself. 
 We thus find, for a uniform dead load f ths the live load, the 
 total maximum strains below. 
 
 Xa 
 
 Xb 
 
 Xd 
 
 x/ 
 
 Xh 
 
 Y a 
 
 Yc 
 
 Y<5 
 
 Y 
 
 +112.7 
 
 +126.7 
 
 +133.5 
 
 +127. 
 
 +134.7 
 
 -110.1 
 
 -126.7 
 
 -129.3 
 
 -125.6 
 
 Comparing these with the corresponding strains for the bow- 
 
APPENDIX.] NOTE TO CHAP. I. 359 
 
 string, we find that they are very much less in every piece. In 
 fact, there is a total gain of over 10 per cent., and that, too, not- 
 withstanding that the rise of each arc is only half that in the 
 first case. Had we taken a double depth, the saving would 
 have been very great, and as in this case also, for a long span 
 and relatively large dead load, the diagonals would always be in 
 tension, the increased length of these last would be no dis- 
 advantage. 
 
 12. The above construction is worthy of the careful consid- 
 eration of the bridge builder. It peculiarly recommends itself 
 for long spans, and has several important advantages possessed 
 by no other form of truss. For long spans the strains in the 
 flanges are nearly uniform. The diagonals are less strained 
 than in any other system, and are always in tension. Every 
 member acts to support, as well as to strengthen. The height* 
 is everywhere proportional to the maximum moment of the ex- 
 terior forces. The load is distributed along the neutral axis, 
 thus securing the maximum of rigidity ; while the neutral axis 
 itself passes through the points of support. 
 
 This construction is known in Germany, from the name of its 
 inventor, as Paulas Truss. Upon this system are the double 
 track bridge over the Isar at Grossheselohe, 2 spans of 170.6 
 ft. ; a large number of smaller bridges, such as one over the 
 Rodach, 109 ft. span ; over the Main in Schweinfurt, 116.4ft. 
 span ; and especially one over the Rhine at Mayence, of 32 
 spans, 4 of 345 ft., 6 of 116 ft., 20 of 50 ft., and 2 of 82 ft. ; all 
 upon the same system. 
 
 In England, we might notice the famous bridge over the 
 Tamar at Saltash, near Plymouth, whose two principal spans 
 are 455 ft., which is also constructed upon this system. 
 Finally, we may mention the bridge over the Elbe, near 
 Hamburg, the three principal spans of which are 325 ft. each. 
 
 In this latter structure both the upper and lower members 
 are braced or ribbed arches, of 'a constant depth of about 10 ft., 
 a combination which, for long spans, seems most excellent. A 
 single arch alone, similar, for example, to the steel arch over the 
 Mississippi, by Capt. Eads, would have required heavy abut- 
 ments. 
 
 The same arch inverted would have required equally heavy 
 anchorages. The combination does away with both. The 
 
360 NOTE TO CHAP. I. [APPENDIX. 
 
 thrust of the upright arch is opposed by the pull of the inverted 
 one, all the advantages of Pauli's system are obtained, and 
 there are no temperature strains such as occur in the single 
 arch, while the bracing is reduced to a minimum. At the same 
 time all the rigidity due to the arch is obtained.* 
 
 13. In the construction of the diagrams, care should be exer- 
 cised in the selection of the scales, that the frame diagram may 
 be large enough to secure the desired accuracy. Lines should 
 be drawn very fine with a hard, sharp-pointed pencil, so as to be 
 scarcely discernable, and their intersections accurately marked 
 by needle point. 
 
 With an accurate scale and good instruments, strains can be 
 taken off in nearly every practical case to hundredths of a ton 
 
 * Compare Long and Short Span Railway Bridges, by John A. Roebling, C.E. 
 In this work, Mr. Roebling proposes a system in principle essentially the 
 same as the above, to which he gives the name of "Parabolic Truss." He, 
 however, constructs the arch of channel irons bolted to the sides of a straight 
 truss, the sole office of which is to give rigidity to the system. Also, claiming 
 that iron in the shape of wire will safely sustain three times as much as in the 
 shape . of bars or rods, he introduces a wire cable in place of the inverted 
 braced arch. 
 
 It will thus be seen that for rigidity the system is wholly dependent upon 
 extraneous members, such as the auxiliary truss and the tower stays, which are 
 liberally introduced. By dividing the material composing the upright arch 
 into two portions, bracing between them, and thus forming a braced arch sim- 
 ilar to Capt. Eads, the stays and stiffening truss might be entirely dispensed 
 with, the construction greatly simplified in the number of its members, and 
 the bracing reduced to a minimum. If, also, as claimed by Capt. Eads, the 
 conditions for cast steel are just the reverse of iron, and it is most advantageous 
 to use it in compression, then it seems that such a modification of Mr. Roeb- 
 ling' s design with wire cable and a cast-steel braced arch would better sustain 
 the thesis with which his work, above quoted, opens, viz. : that "the greatest 
 economy in bridging is only to be obtained by a judicious application of the Para- 
 bolic Truss." 
 
 Such a combination of the suspension and upright arch would seem to avoid 
 the principal objections urged against each separately. ' The anchorages and 
 abutments are dispensed with, the greatest rigidity is secured with the mini- 
 mum of bracing, and the material is used in the most advantageous way. In 
 addition to the advantages of Paulas system-being secured, we have the ease 
 of erection of the suspension system combined with the rigidity of the arch. 
 The system is self -balancing, and practically unaffected by changes of temper- 
 ature. 
 
 For the practical details of construction of such a system, the reader can 
 with profit consult Mr. Roebling' s work, above quoted. They will be found 
 to be neither expensive nor difficult of execution. 
 
APPENDIX.] NOTE TO CHAP. I. 361 
 
 with perfect accuracy. The use of parallel rulers is not to be 
 recommended. The T square, triangle and drawing-board are 
 far preferable. It should be remembered, finally, that careful 
 habits of manipulation, while they give constantly increased 
 skill and more accurate results, affect in no degree the rapidity 
 and ease with which those results are obtained. 
 
362 NOTE TO CHAP. H. [APPENDIX. 
 
 NOTE TO CHAPTER II. 
 
 14. The reader will observe that in Chapter I. we had given 
 forces acting at certain points of a given frame, and we found 
 by simple resolution of forces the strains in the pieces of that 
 frame. In Chapter II. we have given forces acting in certain 
 directions, and having assumed the strains, we find the equi- 
 librium, polygon or frame, which, having its angles on these 
 force directions, and having these strains, will hold the given 
 forces in equilibrium. Thus in Figs. 12 (b) and (c), PL III., of 
 the text, by choosing a pole and drawing lines to the forces in 
 the force polygon (a), we virtually assume the strains which 
 are to act upon our frame. Then lines parallel to these strains 
 in (b), forming a polygon whose angles are upon the forces, 
 must give us the frame which holds these forces in equilibrium, 
 provided we close the polygon by a line and apply at the ends 
 forces which balance each other horizontally, and whose com- 
 ponents parallel to the resultant of the forces balance the 
 forces. 
 
 Thus the polygon maficdenm is a frame along whose 
 sides the forces S S l5 etc., act, and whose reactions at the sup- 
 ports m and n must then be a o and 5 a, as given in (a). 
 
 This frame keeping the same pole, that is, the same strains 
 we may put anywhere in the plane, its angles being always on 
 the forces, and its sides always respectively parallel, though 
 varying in length according to the position assumed. 
 
 We might also have assumed different strains, that is, taken 
 a different pole, and constructed a different frame; but evi- 
 dently the end reactions will not be altered, and will be always 
 equal to a and 5 a, as given in (a). 
 
 The peculiarities of the frame thus obtained are, as we see 
 further on, that its end sides always intersect upon the result- 
 ant of the forces ; its depth is always proportional (for paral- 
 lel forces) to the moment at any point ; its area to the moment 
 
APPENDIX.] NOTE TO CHAP. II. 3G3 
 
 of inertia of the forces ; while, finally, in a loaded beam the de- 
 flection curve itself is but a polygon or frame of this character, 
 when the curve of loading follows the law of the moments in 
 the beam. 
 
 It is upon this polygon and its properties that the entire 
 system of Graphical Statics is based. 
 
364 NOTE TO CHAP. V. [APPENDIX. 
 
 NOTE TO CHAPTEE V., AET. 51. 
 
 15. In Fig. YIIL (Appendix) we have given the construction 
 referred to in Art. 51 of the text for a system of loads of given 
 intensities. The span s s is supposed to shift to s x s 1? s% s 2 , etc., 
 and a certain cross-section & to shift with it to Tc^ &%, etc. The 
 intersections of the respective closing lines with verticals 
 through & , &u &2> etc-? gives us a curve between which and the 
 polygon the greatest ordinate gives the maximum moment for 
 the assumed cross-section. The place of this ordinate is the 
 position of the cross-section from which we determine the ends 
 of the span, and thus have its position with reference to the 
 loading when the moment in k is the greatest possible. 
 
 Thus if this greatest ordinate is- at the angle YIIL in the Fig., 
 the weight P 8 must rest upon the cross-section. The distance 
 then from P 8 to the left end of span s, is the distance from S Q 
 to &o on left, and to right end of span s, is the distance from 
 #o to S Q on right. 
 
 The ends s and s being thus found, perpendiculars through 
 them determine the closing line L, and the parallel to this in 
 the force polygon gives the end reactions L and 20 L for the 
 position of span which makes moment at Jc a maximum. 
 
 
APPENDIX.] PIVOT SPAN. 365 
 
 NOTE TO CHAPTEE XII., ART. 124. 
 
 16 In Arts. 120 and 121 we have given the formulae and 
 principles necessary for the complete solution of the pivot span. 
 We propose here to illustrate more fully their application by a 
 simple example. 
 
 Fig. IX. represents such a structure. The two outer spans 
 A B C D = 40 f t. The central or turn-table span, B C = 
 20 ft. Centre height at B and C = 10 ft. End height = 6 ft. 
 Panel length, 10 ft. ; each apex live load, 10 tons, or 1 ton per 
 foot. Dead load, half 'as much. Two systems of triangulation, 
 as shown in the Fiff. 
 
 O 
 
 Our proportions are taken for the sake of illustration merely, 
 and not as an example of actual practice. All the points to be 
 observed are, however, illustrated as well as by a much longer 
 span, and more usual proportions. 
 
 It is to be observed that the end verticals are compression 
 members only, and cannot take tension. This is necessary -to 
 prevent ambiguity as to the way in which the strains go. A 
 negative reaction might otherwise cause tension in 1 2, and 
 compression in F, or tension in 1 5, compression in 5 6, and 
 tension in A. If 1 5 cannot take tension, we have but one 
 course for the strains, and the problem is determinate. 
 
 We also, for similar reasons, construct the centre span so 
 that the diagonals take tension only, and the verticals compres- 
 sion only. These points as to construction being settled, let us 
 proceed, first, to determine the reactions. 
 
 1st. REACTIONS. 
 
 We shall consider the case of the " Tipper" or secondary 
 central span only [Art. 120], as this case most nearly ap- 
 proaches the true state of things. The method of procedure 
 for fovrjixed supports is precisely similar, only taking the for- 
 mulse for that case from Art. 122. 
 
 The less the span B C, the nearer the case approaches to three 
 fixed supports ; and when the distance B C is zero, n is zero, and 
 our formulas are the same as for beam over three supports. 
 
 For a load in the left span distant a from A, these formulae 
 are as follows [Art. 120] : 
 
366 NOTE TO ART. 124. [APPENDIX. 
 
 R A = j 2H - (10 + 15 w, + 3 n>) Jc + (2 + n)ji\, 
 
 C ^g |( 6 + 9 
 " 
 
 RD =2H 
 
 1 
 J 
 
 I (2 + ri)k* (2 -f 3 n + 3 ft 2 ) k I, 
 
 in which Jc=^ I =A B=C D, n I = B C and H = 4 -f 8 7* + 3 ft 2 . 
 
 
 We have first to put these formulse into the most convenient 
 shape for use in the particular case under consideration. Thus 
 
 "1 Q K 
 
 in this case I = 40, n I = 20 ; hence n = - and H = - , and 
 
 2 . 4 
 
 Jc = , where a has the successive values of 10, 20, 30, 40 for P 15 
 
 P a , P 3 , P 4 . J& is therefore successively -, -, - and -. 
 Our equations for reactions are then, after reducing, 
 
 J 45 -10 I 
 
 [1 
 1A7i 8 17^" I 
 107, 17^J 
 
 Now, as we may notice, the denominator of & is always 4, of 
 always 64 ; the numerator only changing according to the 
 position of the weight. These equations can then be written 
 
 RA - 224 
 
 ^J2240-584a+5a 3 I 
 
 A |~72 -</], 
 L -I 
 
 R D = 
 
 where a has the values 1, 2, 3 for P 1? P 2 , P 3 , etc. 
 
 These, then, are the practical formulas for this case, and from 
 
APPENDIX.] PIVOT SPAN. 367 
 
 them we can easily find the reactions for the apex loads of 10 
 tons each. 
 
 Thus, for P! make a = 1, and we have 
 
 R A = 7.415, R B = RC = 1.58, R D = 0.584 
 
 For P a make a 2, and 
 
 R A = 4.964, R B = R c = 3.035, R D = 1.035. 
 
 For P 3 make a = 3, and 
 
 R A = 2.78, R B = R c - 4.22, R D = - 1.22. 
 
 For P 4 make a = 4, and 
 
 RA = lj -^B RC == 5, RD ! 
 
 Loads upon the centre of the span B C acting, that is, at 
 apex 10, give no reactions, but are supported directly by the 
 turn-table. Hence, for P 5 ; R A , R B , R c and RD are zero. For 
 the first load, P 7 to the right of C, the reactions at A and B are 
 the same as for P 3 at D and C, already found. For the next 
 load, P 8 , the reactions at A and B are the same as for P 2 at D 
 and C, already found. For P 9 , the same as for P t . For P 6 , as 
 for P 4 , etc. 
 
 We thus have the reactions at A and B due to every indi- 
 vidual apex load, and can now proceed to find the strains. 
 
 Our formulae, it will be observed, thus become very simple 
 and easy of application for any particular case. 
 
 2c7. FLANGES BRIDGE SHUT. 
 
 Let us first find the strains in the flanges. We have only to 
 apply the method of moments, and the work is so simple that 
 an example or two will suffice. 
 
 We repeat again the rule. Conceive a section cutting only 
 three strained pieces. Take the intersection of two of these as 
 the centre of moments for finding the strain in the third. The 
 moment of the strain in this last about this point must be equal 
 to the algebraic sum of the moments of all the forces acting 
 between the section and one end. Take P x for example. Its 
 upward reaction at A is 7.415. [A negative reaction acts 
 down. Thus, for P 7 above, the reaction at A is, from our for- 
 mulae, 1.22. The minus sign indicates that the reaction is 
 down, and that, neglecting the dead load, the girder must be 
 held down to the support A. If the reader will draw roughly 
 the curve of deflection, he will see that this is so.] 
 
368 NOTE TO AKT. 124:. [APPENDIX. 
 
 Conceive a section through the girder at, say, the centre of 
 flange A. It cuts 4 pieces, but, since the weight P t acts only 
 through its own system of diagonals, only three are strained. 
 The point of moment for A is then at 6, the intersection of the 
 other two strained pieces. The strain, then, in A x by its lever 
 arm = 7.415 x 10. The lever arm of A is 6.965 ; hence 
 
 A x 6.965 = 7.415 x 10, 
 or A = + 10.64 tons compression, 
 
 because the upward reaction acting with 6 as a centre of rota- 
 tion tends to compress A. 
 
 This strain evidently acts through both A and B, since both 
 these flanges are included by the two diagonals of the system 
 for P! ; hence also, B = + 10.64 tons. 
 
 For flanges C and D, since 78 is the strained diagonal, 8 is 
 the centre of moments. The same reaction acts now with the 
 lever arm 30 to cause compression, and P x acts with the lever 
 arm 20 to cause tension. We have then 
 
 C x 8.955 = + 7.415 x 30 - 10 x 20, 
 or . C = D = 2.5 tons compression. 
 
 Now we come to the centre span, and must carefully observe 
 the following points. Since D has been found to be compres- 
 sion for P t , we see at once that the whole upper flange for the 
 span A B is for this weight in compression. Diagonal 8 9 is 
 therefore in tension. Were there no vertical strut at B, this 
 would cause compression in 910. But brace 910 cannot by 
 construction take compression. The strained pieces cut by a 
 section through E are then E, B 11 and K, which give us the 
 centre of moments at B for strain in E. Observe, that were 
 it not for the vertical, we should have had 10 for the centre 
 of moments; or, with the vertical, had D been found tension, 
 8 9 would have been compression ; there would then have been 
 no strain in the vertical, that being incapable of tension, and 
 diagonal 9 10 would have been strained, thus giving us also 10 
 for the centre of moments. Attention to the above is necessary 
 in order to properly pass from the span A B into the middle 
 span. 
 
 We have then for strain in E 
 
 E x 10 = 7.415 x 40 - 10 x 30, or E = - 0.34, 
 
APPENDIX.] 
 
 PIVOT SPAN. 
 
 369 
 
 or in tension, as indicated by the sign, since the moment of P, 
 overbalances that of the reaction. 
 
 \Note. The different lever arms are easily obtained from 
 the known dimensions of the truss. We have considered it 
 unnecessary to detail how they are to be found. They may 
 either be measured to scale from the frame or computed trigo- 
 nometrically.] 
 
 The lower flanges are found in similar manner. 
 
 Thus, strain in F is zero, since it passes through the point of 
 moments. 
 
 For G and H, we have 
 
 G x 8 = - 7.415 x 20 + 10 x 10, or G - 6.04 tension. 
 In like manner, for I, 
 
 I x 10 = - 7.415 x 40 + 10 x 30 = + 0.34. 
 
 For K, for similar reasons as above for E, we have centre at 
 11, and therefore the reaction at B also enters into the equa- 
 tion of moments, and 
 
 K x 10 = - 7.415 x 50 + 10 x 40 1.58 x 10, or K = + 1.34. 
 We have then, finally, for the strains in the flanges due to P t 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 H 
 
 I 
 
 K 
 
 Pi 
 
 +10.64 
 
 +10.64 
 
 +2.5 
 
 +2.5 
 
 -0.34 
 
 
 
 -6.04 
 
 -6.04 
 
 +0.34 
 
 +1.34 
 
 In a precisely similar manner we find the strains due to P 2 , 
 P 3 and P 4 . 
 
 We have only to observe that for P 7 , the first weight to the 
 right of C in the other span, the reaction at A is negative and 
 equal to the reaction of P 3 at D, already found, or 1.22. 
 
 Now as we suppose the end A bolted down, this reaction acts 
 as a weight of 1.22 tons suspended from the end. So for the 
 reactions of P 8 and P 9 , viz., 1.035 and 0.584. These reac- 
 tions, moreover, must all take effect through diagonal 1 2 and 
 flange F, as the end vertical cannot take tension. 
 
 Finding then the strains due to each of the other weights, 
 we can, finally, tabulate our results as on next page : 
 24 
 
370 
 
 NOTE TO ART. 124. 
 
 [APPENDIX. 
 
 STRAINS IN FLANGES LIVE LOAD BRIDGE SHUT. 
 
 
 A 
 
 B 
 
 
 
 D 
 
 E 
 
 F 
 
 G 
 
 H 
 
 I 
 
 K 
 
 PI 
 
 +10.64 
 
 + 10.64 
 
 + 2.5 
 
 + 2.5 
 
 0.34 
 
 
 
 ?.r 
 
 6.04 
 
 6.04 
 
 + 0.34 
 
 -1.34 
 
 p a 
 
 
 
 + 12.47 
 
 + 12.47 
 
 0.14 
 
 0.14 
 
 7.1 j 5.32 
 
 5.82 +2.14! 
 
 P 3 
 
 + 3.9 
 
 + 3.9 
 + 2.5~ 
 
 + 9.31 
 
 + 9.31 
 
 + 1.12 
 
 
 
 6.95 6.95 
 
 1.12 
 
 +1.88 
 
 P 4 
 P 6 
 
 
 
 + 2.5 
 
 + 4.0 
 
 + 4.0 
 
 1.42 
 
 1.42 3.;i 
 
 3.33 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 P 6 
 
 
 
 2.5 
 
 2.5 
 
 4.0 
 
 4.88 
 
 4.0 
 
 + 1.42 
 
 + 1.42 + 3.33 
 
 + 3.33 
 
 
 
 PT 
 
 
 ~~0 
 
 3.1 | 3.1 
 
 4.88 
 
 + 1.74 
 
 + 1.74 
 
 + 4.U6 
 
 + 4.U6 
 
 +l.bb 
 +2.14 
 
 P 8 
 
 -2.6 
 
 -2.6 
 
 4.4 
 
 4.14 
 
 + 1.46 
 
 + 1.46 
 
 + 3.45 
 
 + 3.45 
 
 P 
 
 
 
 1.4 
 
 1.4 
 
 2.34 
 
 2.34 
 
 + o.as 
 
 + 0.83 
 
 + 1.95 
 
 + 1.95 
 
 +1.34 
 
 Total 
 Strains 
 
 +14.54 
 
 + 29.51 
 
 + 26.78 
 
 + 15.81 
 
 + 5.12 
 
 + 5.45 
 
 + 5.45 
 
 + 12.79 
 
 + 13.13 
 
 +9.38 
 
 ; 
 
 -9.6 
 
 - 9.6 
 
 15.76 
 
 15.84 
 
 - 8.52 
 
 21.61 1 21.64 
 
 9.77 
 
 -1.34 
 
 In the two horizontal lines at bottom, we have the total 
 strains of each kind caused by the live load. 
 
 3d. FLANGES BRIDGE OPEN DEAD LOAD. 
 
 We have next to find the strains due to the dead load when 
 the span is open. 
 
 We have then 5 tons at each apex, except the ends, where 
 we have P 2.5 tons. 
 
 These strains are easily found by moments as above, and we 
 have then the following table : 
 
 
 A 
 
 B 
 
 C 
 
 D 
 
 E 
 
 F 
 
 G 
 
 H 
 
 I 
 
 K 
 
 PO 
 
 
 
 
 6.2 
 
 6.2 
 
 10.1 
 
 10.1 
 
 + 3.57 
 
 + 3.57 
 
 + 8.3 
 
 + 8.3 
 
 + 10 
 
 P> 
 
 
 
 11.1 
 
 11.1 
 
 15.0 
 
 
 
 + 6.1 
 
 + 6.1 
 
 + 15.0 
 
 + 15.0 
 
 P 2 
 
 
 
 
 
 
 
 10.1 
 
 10.0 
 
 
 
 
 
 + 5.5 
 
 + 5.5 
 
 + 10.0 
 
 P 3 
 
 
 
 
 
 
 
 
 
 5.0 
 
 
 
 
 
 
 
 + 5.0 
 
 + 5.0 
 
 Total 
 Strains 
 
 
 
 6.2 
 
 17.4 
 
 31.2 
 
 40.0 
 
 + 3.6 
 
 + 9.7 
 
 + 20.1 
 
 + 33.8. 
 
 + 40.0 
 
 
 
 
 
 
 
 
 
 
 If now, as should be the case, we suppose the centre sup- 
 ports raised above the level of the ends, so that the ends just 
 bear, then these strains above act even when the bridge is shut. 
 As we have already seen in Art. 131, our formulae for the 
 reactions are not affected by this state of things. The strains 
 due to live load will then be increased by those above, and we 
 thus have for the total maximum strains which can ever occur, 
 
APPENDIX.] 
 
 PIVOT SPAN. 
 
 371 
 
 A 
 
 B 
 
 
 
 D 
 
 * E 
 
 F 
 
 a 
 
 H 
 
 I 
 
 K 
 
 + 14.5-1 
 
 + 29.51 
 
 + 26.78 
 
 + 15.81 
 
 + 5.12 
 
 + 9.05 
 
 + 15.15 
 
 + 32.89 
 
 + 46.93 
 
 + 49. S8 
 
 
 15.8 
 
 27.0 
 
 46.96 
 
 55.84 
 
 8.52 
 
 21.51 
 
 21.64 
 
 -9.77 
 
 1.34 
 
 Of course, for this condition of things the ends must always 
 be bolted down. 
 
 It is sometimes customary to raise the ends by an apparatus 
 for that purpose, after closing the draw, until the proper pro- 
 portion of the dead load takes effect also as a positive reaction. 
 
 We can easily find the strains in this case also by adding 
 the numbers in the last horizontal line of our table for bridge 
 shut, with their proper signs, and taking half the results for a 
 new line for dead load strains. The resulting strains can 
 then be found precisely as in the table of Art. 8 (Appendix). 
 We must also find the strains for bridge open as above, and then 
 take the greatest strains of each kind from these two tables. 
 
 In this case 'the strains would be differently distributed. 
 Flange E will be always in tension, A and K always in com- 
 pression ; the compression in B C and D will be somewhat 
 greater than above, and the tension in the same flanges less. 
 The reader can easily deduce the strains for this case from the 
 two preceding tables. 
 
 If the truss may act as a girder over four fixed supports, 
 we should, in order to be certain of the maximum strains, 
 make the calculation for this case also, using the formulae of 
 Art. 122. This is unnecessary, however, if the supports B and 
 C can never sink far enough to strike the turn-table, or be im- 
 peded in their motion. 
 
 4cth. STRAINS IN THE DIAGONALS. 
 
 We may find the strains in the diagonals also for each 
 weight separately, both for bridge open and shut ; and a pre- 
 cisely similar method of tabulation will give the strains. 
 
 It will here be found preferable to make a series of dia- 
 grams, as illustrated in Fig. 86, Art. 124, for each weight and 
 its own system of triangulation. We obtain thus the diagonal 
 strains, and at the same time check the results obtained for the 
 flanges above. 
 
 If we wish to calculate the diagonals, it will be better to find 
 the resultant shear acting upon the diagonal, and multiply it by 
 the secant of the angle the diagonal makes with the vertical. 
 
372 NOTE TO ART. 124. [APPENDIX. 
 
 We can also, if we wish, apply, the method of moments. 
 Thus, if we determine the point of intersection in the present 
 case of the inclined upper flange with the horizontal lower 
 flange, this point will be a common centre of moments for the 
 diagonals. The lever arms of the diagonals with reference to 
 this point must next be determined, and then we are ready. 
 
 This point above for centre of moments is easily found ; thus 
 4 : 40 : ; 10 : 100. 
 
 It is therefore 60 ft. to the left of A, or 100 ft. left of B. 
 Take now any diagonal, as 3 4. Its angle with the horizontal 
 is very nearly 42, and with the vertical 48. Its lever arm is 
 then 80 sin 42 = 53.5, and sec. of angle with .vertical is 1,49. 
 
 Now take the weight P 3 . Its upward reaction at A is 4.964, 
 P 2 being 10. 
 
 We have then 
 
 [str. in 3 4] x 53.5 = 10 x 80 4.964 x 60 = + 502.16. 
 
 The resultant rotation is then positive, or from left to right. 
 The point P 2 then sinks and 4 rises, and 34 is in tension and 
 502.16 
 
 "KO~K~ = ~~ 9-38 tons. 
 oo.o 
 
 This is sufficient to illustrate the method. 
 
 For the first method referred to above, viz., that by resultant 
 shear, the following points are to be observed : 
 
 When a piece slopes towards the nearest support, we say it is 
 sloped as a strut, whatever the real strain in it may be. 
 
 When it slopes away from the nearest support, it is sloped as 
 a tie. 
 
 The simple shear is the reaction at the support minus the 
 weights between any point and that support. 
 
 If any three strained pieces are cut by a section through the 
 structure, the strains in these pieces are in equilibrium with the 
 simple shear at this section. Hence the algebraic sum of the 
 vertical components of these pieces must be equal and opposite 
 to the shear itself. 
 
 In order to add these vertical components with proper signs, 
 we must remember that if a flange is in tension and sloped as a 
 strut, or in compression and sloped as a tie, we add the vertical 
 component of the strain in it to the simple shear already 
 obtained. If in compression and sloped as a strut, or tension 
 and sloped as a tie, we subtract. 
 
APPENDIX.] PIVOT SPAN. 373 
 
 The resultant shear thus obtained then, multiplied by the 
 secant of the angle with vertical, gives the strain in diagonal. 
 
 If the sign of the result is negative ( ), it shows that the 
 strain on the diagonal is contrary to that indicated by its slope. 
 
 To illustrate, let us again take the weight P 2 and consider 
 diagonal 34. 
 
 The simple shear at apex 4 is 4.964 10 = 5.036. The 
 strain in C for P a we have found to be compression, and equal 
 to -h 12.47. It is sloped as a strut, and its vertical component ia 
 therefore to be subtracted from the shear above. Since its 
 angle is nearly 5 43' with the horizontal, this vertical com- 
 ponent is 
 
 12.47 x sin 5 43' = 1.24. 
 
 Since H is in this case horizontal, it has no vertical com- 
 ponent. 
 
 The resultant shear is then 
 
 - 5.036 - 1.24 := - 6.276. 
 
 As the secant of the angle of 3 4 with the vertical is 1.49, 
 we have for the strain in 34, - 6.276 x 1.49 = 9.35. 
 
 This result being minus, and 3 4 being sloped as a strut, 
 the strain is 9.35 tons tension, agreeing closely with the value 
 found above by moments. 
 
 The above method is preferable to the method by moments 
 for the diagonals, as we have only to determine the secants for 
 the verticals and the sines for the flanges, which is in most 
 cases easier than to find the lever arms for the diagonals and 
 the points of intersection of the upper and lower flanges in each 
 panel. It is, like the method of moments, of general applica- 
 tion to any framed structure whose outer forces are known. 
 
 The method of diagram in Art. 124 will be found preferable 
 to both. 
 
 It is unnecessary to pursue our example further. With the 
 mutual checks of the two methods of calculation explained 
 above, as well as the diagrams, correct results cannot fail to be 
 obtained. The diagrams should always be made first, as they 
 settle by mere inspection many points which may at first cause 
 trouble such as whether the shear in a piece is subtract! ve or 
 not according to our rule, the character of the strains in dif- 
 
 O ' 
 
 ferent pieces, etc. It is well to indicate on the diagrams com- 
 pressive strains by double or heavy lines. 
 
374 
 
 NOTE TO ART. 128. 
 
 [APPENDIX. 
 
 NOTE TO AKT. 128, CHAPTER XII. 
 
 17. We wish here to call more particular attention to the 
 relative economy of the continuous as compared with the 
 simple girder. This, we think, is greater than is generally sup- 
 posed. It may reach from 18 to 25, and even as high as 50 
 per cent. 
 
 Take the example worked out in Art. 128, Fig. 88. We 
 have obtained the maximum strains in that Art. upon every 
 piece. 
 
 We give them below, compared with the strains in the same 
 pieces for a simple girder of same dimensions anb load : 
 
 Aa Ac Ae Ag A* B6 Ed Bf BA 
 Continuous.. 203.5 +63.6 +115.3 +63.6 203.5 +89.3 115.9 115.9 +89.3 
 
 Simple +180 +240 +180 90 210 210 90 
 
 a& be cd de ef fg gh hk 
 
 Continuous.. +189.3 109.9 +109.9 +45.5 +45.5 +109.9 +109.9 +189.3 
 Simple +127.3 127.3 +56.5 +56.5 56.5 +56.5 127.3 +127.3 
 
 It will be seen at once that there is a saving in the flanges 
 about 11 per cent, in all but the bracing is heavier, giving lit- 
 tle or no saving. The span is too short to properly represent 
 the relative economy of the two systems. 
 
 If we take a truss such as represented in PI. 2, Fig. VII., 
 Appendix, by the full lines only, omitting the dotted verticals and 
 diagonals height 6 ft., span 50 ft., panel length 10 ft,, dead 
 load 5 tons per panel, live load 7 tons per panel and calculate 
 the strains in the pieces for a simple girder, and then as a con- 
 tinuous girder of two spans and three spans, we have the fol- 
 lowing results : 
 
 
 1 span 
 strain in tons. 
 
 2 spans 
 
 3 spans 
 end. 
 
 3 spans 
 middle. 
 
 
 205 2 
 
 231.6 . 
 
 229.8 
 
 239.4 
 
 
 
 
 
 
 Lower Chord . 
 
 200. 
 
 168.7 
 
 152.9 
 
 122.6 
 
 Upper Chord .... 
 
 200. 
 
 ^158.7 
 
 159.9 
 
 135.2 
 
 Total 
 
 605 2 
 
 559 
 
 542 6 
 
 497 2 
 
 
 
 
 
 
 Per cent, saving. . 
 
 
 8 per cent. 
 
 10 per cent. 
 
 18 per cent. 
 
APPENDIX.] THE CONTINUOUS GIRDEK. 375 
 
 We have then, in the first case, a saving of 8 per cent., in the 
 second, 10, and in the third, 18 per cent, over a simple girder. 
 Quite a notable saving, although the spans are very short, and 
 although, in the first two cases (the spans being end spans), we do 
 not obtain the full advantages of continuity. If, then, instead 
 of three simple girders of the above dimensions, we should con- 
 struct the girder continuous over the piers, we should save in 
 strain, and hence in material, 10 per cent, in each end span, and 
 18 per cent., or nearly twice as much, in the centre span. 
 
 The advantage of continuity is rendered still more apparent 
 by taking a longer span. Thus for a girder of 200 ft., height 
 20 ft. 10 panels, and double system of triangulation, similar to 
 Fig. YII. for a live load of 20 tons per panel, and dead load 
 of 10 tons we have the following results : 
 
 One span. Two spans. Five spans. 
 
 r centre. 
 
 Bracing 1398.6 1428.2 1596.2 
 
 Lower Chord., 2400. 1793.2 1395.7 
 
 Upper Chord . . 2550. 1981.6 1622.6 
 
 Total 6348.6 5203.0 4614.5 
 
 Per cent, saving 18 per cent. 27 per cent. 
 
 That is, we have a saving of 18 per cent, instead of only 8 
 per cent, as before, for two spans, and of 27 per cent, for the 
 centre span of five spans. For three spans, then, of this length 
 we should save 18 per cent, on the end, and at least 30 per cent, 
 on the centre span. 
 
 If we suppose the same girder as above fastened or fixed 
 horizontally at the ends, we shall have the case of a middle span 
 in a very great number, and may expect to find the greatest 
 saving possible for this length. 
 
 The formulae of Chapter XIII., as also the simple graphical 
 method for this case, given in Chapter XII., Art. 114 [Fig. 80], 
 enable us to solve this case easily. The reader will find, on mak- 
 ing the calculation, the following strains : 
 
 Strain in tons. 
 
 Diagonals 1279.2 
 
 Upper Chord 940.2 
 
 Lower Chord 965.4 
 
 Total 3184.8 
 
 Per cent, saving 49.8 
 
376 NOTE TO ART. 128. [APPENDIX. 
 
 The above will serve to illustrate the point in question quite 
 as well, perhaps, as an extended theoretical discussion. We see 
 that the saving increases rapidly with the length of the span, 
 and may easily rise as high as 30 or 40 per cent., while in some 
 cases even 50 per cent, may be realized. 
 
 THE DISADVANTAGES OF THE CONTINUOUS GIRDER ABE I 
 
 1st. The fact that the various pieces, especially the chords, 
 undergo strains of opposite character. 
 
 This, in wrought- iron structures, we venture to think of little 
 importance. The extra work and cost of chords and chord con- 
 nections necessary to secure the flanges against both compressive 
 and tensile strain, can hardly amount to 10, 18, 30, or even 50 
 per cent, of the cost of girder ! 
 
 %d. Difficulty of calculation. 
 
 We have, we trust, in what precedes, and in Chapter XIII., 
 succeeded in removing this objection. 
 
 The opinion is widespread among engineers that the deter- 
 mination of strains in the continuous girder is impracticable 
 and involved in mystery. No opinion could well be more un- 
 founded. The accurate and complete calculation for all pos- 
 sible loading, live or dead, is precisely similar to and offers no 
 more difficulty than the simple girder itself. 
 
 The formulae for moments and shears are, as we have seen, 
 simple and easy of application. 
 
 The graphic method here developed offers also a thorough 
 solution. In view of both, and of the extensive literature upon 
 the subject (which seems, by the way, to have been so generally 
 ignored), we can finally pronounce the problem to be fully 
 solved. 
 
 3d. The changes of strain, unforeseen and often considerable, 
 which a small settling of the piers or change of level of the 
 supports may occasion. 
 
 This, be it observed, is only of importance when the piers 
 settle after the erection of the superstructure. If piers are to 
 be considered as settling indefinitely, or continuously during a 
 succession of seasons, continuous girders are not to be thought 
 of. If, however, as is generally the fact, the piers take their 
 permanent set during the first season,, and afterwards are im- 
 movable, the above objection has no weight. It is not necessary 
 
APPENDIX.] SUPPOETS OUT OF LEVEL. 377 
 
 that the piers should be exactly on level or even on line, or even 
 that the differences of level be known. 
 
 As shown in Art. 121, these differences produce no effect, 
 provided the girder be built to the profile of the supporting 
 points. 
 
 If in any case these differences are required, and it is con- 
 sidered difficult to determine them over water with sufficient 
 accuracy, then the proper reactions at the several piers may be 
 weighed off* and the girder thus left in position under pre- 
 cisely the circumstances for which it has been calculated. 
 
 THE PRINCIPAL ADVANTAGES OF THE CONTINUOUS GIRDER ARE : 
 
 1st. Ease of erection, where false works are difficult or ex- 
 pensive. The girder may be built on shore, and then pushed 
 out over the piers. 
 
 2d. Saving in width of piers, as compared with width re- 
 quired for separate successive spans. The girder may be placed 
 upon 'knife edges at the piers. In fact, such a construction is 
 preferable, as better ensuring the calculated strains. Width of 
 piers is undesirable. 
 
 3d. Saving in jnaterial usually from 25 to 30 per cent. 
 
 18. Continuous Crirder Supports not on a level. In 
 Chapter XIII. we have all the formulae required for the solution 
 of the continuous girder for supports on a level, or all on line, 
 when the deviation from level is small, whatever may be the 
 number or relative length of the spans. If for a continuous girder 
 of constant cross-section, the supports are properly lowered after 
 the girder is placed upon them, we may obtain a saving of 23 
 per cent., or more in material over the same girder with sup- 
 ports all on level. If, however, the cross-section varies accord- 
 ing to the strain in other words, if the girder is of constant 
 strength no advantage is gained from thus lowering interme- 
 diate supports. Such disposition of the supports may even act 
 injuriously. 
 
 The formulae for shear and moments which we have given 
 are, indeed, based upon the hypothesis of constant cross-section, 
 but the strains in every piece of the girder being found for the 
 shears and moments thus obtained, each piece is proportioned 
 
 * An idea first suggested by Clemens Herschel, C.E.: Continuous, Revolving 
 Draw Spans. Little, Brown & Co. , Boston, 1875. 
 
378 NOTE TO ART. 128. [APPENDIX. 
 
 to its strain, and the actual girder erected is not of constant 
 cross-section, but more nearly one of uniform strength. Formu- 
 lae for the case of supports out of level, as well as determina- 
 tions of the best differences of level, are hence of but little prac- 
 tical importance, and have not been given. If, however, it be 
 required to find the effect due to the sinking of any one pier, 
 the following may be found of service. 
 
 Let the n ih support be depressed below the level of the others 
 by the distance A. Then the moments at all the supports are 
 changed. The moments at n and at each alternate support 
 from n are diminished, and at the others increased. 
 
 Let H = ?=i 
 
 Then, when all the spans are equal, the following formulae 
 give the moment at any support : 
 
 1st. All spans equal, n number of lowered support, from 
 left, 
 
 when m<n, M m = 
 
 when m=n, M m = - 5 - n ~ l C *-+ 2 + n G *- +1 H, 
 
 when m > n, M m = HLtl H, 
 
 C B+l 
 
 where ^ = 0, c z = + 1, c 3 4:, c 4 = + 15, c 5 = 56, 
 c 6 = + 209, etc. 
 
 From the moments at the supports the shears can be readily 
 determined from the formula of Art. 148, viz. : 
 
 M m+l a, _M m -M m _ 1 
 
 - 
 
 where q = P (1^) for concentrated load and r ; for uniform, 
 
 2d. Spans att unequal. 
 
 when m < n, c m \ ^-+i ~ ^-n+a + s_ J+ 8 -n+2 I 
 L 4 ^-' J 
 
APPENDIX.] SUPPORTS OUT OF LEVEL. 379 
 
 when m = 
 
 + *- 
 
 4 4-1 ^ (41 + 4) 
 
 [ Cn - 1 ~ gn + ^ +1 ~ Cn 1 6A n EI, 
 4-i 4 
 
 s m+2 
 
 when m > n* M = 7 n /7 
 
 i_i <v-i + 2 ( 
 
 in which expressions 
 
 - ^ - , 4 - 7 7 
 
 =_2o 4 ^-c^,etc. 
 
 64 64 
 
 _ A /71 /7 - 9 ^s + 4-1 
 
 = U, <^ 2 = Ij ^3 ^ - 7 -- > 
 
 ^a-1 
 
 _ 4: ft + Cl) (Cl + ^-^s-l 
 
 -- 7 - 7 J 
 
 V-l ^&-2 
 
 = _ 2 <Z 4 ^ + ^ - d. etc. 
 
 The reader who has learned the use of the formulae of Chap- 
 ter XIII. will have no difficulty in applying the above to any 
 particular case. In the same way as there explained, by mak- 
 ing 4 and 4 zero, we may fix the girder at the ends, etc. The 
 formulae for shear at any support are, of course, the same as 
 before (Art. 150). 
 
 Ex. 1. Let a beam df two equal spans be uniformly loaded 
 throughout its whole length, and let the centre support be low- 
 ered by an amount h% = * . What are the moments and 
 
 reactions f 
 
 The moments due to the full load alone before the support is 
 
 lowered are M! = 0, M 2 = , M 3 (Art. 150). For 
 
 8 
 
 the moment due to the lowering of the support alone, we have 
 from the above formulae, since 
 
 J " J y 
 
 Hence the total moment is 
 
 .__ w Z 2 w P wfi 
 
 IyL _ _ _ 
 
 8 16 " " 16 ' 
 
380 NOTE TO ART. 128. [APPENDIX. 
 
 For the shears, then, we have 
 
 M 2 wl 7 , 9 
 
 7 18 7 
 
 Hence, R 1 = ~ w I, R 2 = w I, R B = --w I. 
 ID lo lo 
 
 Ex. 2. Zfow mw?A m^stf w<2 lower the second support in the 
 above example, in order that the reaction at the centre support 
 may he just zero f 
 
 In this case we have 
 
 _wP_ H__ <z0P 3 
 
 ~~ ~ Z: ~ 
 
 g/ _ 2 M 2 , _ wl __ 6 A 2 E I 5 7 6 AS E I 
 
 T~ " T Z 3 ~ 4 ^ ~~p " 
 
 If this is to be zero, we have 
 
 _ 5wl* and M = _ 1 ^ ? 
 
 and R! = w Z, Rg = 0, R 8 = w I, 
 
 or precisely as for a beam of single span ^and length 2 I. 
 
 Ex. 3. J. beam of four equal spans is unloaded, and the 
 third support is lowered by an amount h% = _ . 
 the reactions f 
 
 Ans. Ri = 
 
 Ex. 4. ^i J^^m of five equal spans rests as a continuous gir- 
 der over six supports. Having given the dimensions of the 
 beam, length of spans, and coefficient of elasticity ; to deter- 
 mine the reactions due to a sinking of the third support one- 
 eighth of an inch. 
 
 Let the beam be of wood, 1 foot wide, 1.5 deep, 
 I = 20 feet, 5 = 5, r = 3, E = 288,000,000 Ibs. per sq. ft., 
 A, = in. = 0.010417 ft. 
 
APPLNDIX.] SUPPORTS OUT OF LEVEL. 381 
 
 Then, c, = 1, c 3 = - 4, c, = 15, 5 = - 56, c 6 = 209, 
 
 540EI&, __ 906EI^ _576EU 3 
 
 and M2 = 209 P ' M3= 209 /* ' M4 -~~~~ 
 
 1 3 375 
 
 or, inserting the constants above, and I = b d B = ' , 
 
 M! = M 6 1= 0, M 2 = 5448, M 3 = - 9142, M 4 = 5812, 
 M 5 = - 1453 ft. Ibs. 
 
 If all the spans are unloaded. For the reactions necessary 
 to bend it and keep it down to the supports, R t = 272 Ibs., 
 112 = 1002, R 3 =-1477, R 4 = llll, R 5 = - 436, R, = 73. 
 
 If the beam weigh 75 Ibs. per foot^ what deflection of third 
 support will raise the left end from the abutment f 
 
 Ans. R t = w I = ? or 7^ = 0.0226 f t.=0.2712 in. 
 
 It will be observed that a small difference of level has then 
 considerable effect. 
 
 Ex. 5. Two equal spans are uniformly loaded. How high 
 must the centre be raised in order that the ends may just 
 touch f 
 
 This is the case of the pivot span with centre support raised. 
 (See Art. 121.) 
 
 The reactions at the ends are zero. At pier, then, Rg = 2 w I, 
 hence moment at pier JYT 2 = \ w P. But the moment when the 
 supports are on level is M 2 = -J- w~P, hence f wff must be due 
 to the elevation of the support. Then from our formulae, 
 
 3 E I ^ , w P 
 
 - , or A,= - 
 
 which is precisely the same as the deflection of an horizontal 
 beam, fastened at one end, and free at the other (Supplement 
 to Chap. VII., Art. 13).* 
 
 * The calculations and formulae given in this Note (pp. 374-381) are taken, 
 by permission, from a manuscript work on the Theory and Calculation of Con- 
 tinuous Bridges, by Mansfield Merriman, C.B., which, we hope, will be soon 
 given to the public. 
 
382 NOTE TO CHAP. XIV. [APPENDIX. 
 
 NOTE TO CHAPTEE XIV. 
 
 THE BRACED AKCH. 
 
 19. The subject of braced arches is an important one, and 
 is treated in no work with the fulness and completeness it 
 deserves. The methods and formulae of Chapter XIY. will, 
 we believe, render the determination of the strains in this class 
 of structure easy, and we propose in the following to illustrate 
 their use, so far as may be necessary to render their application 
 clear. 
 
 In PI. 4, Fig. X., we have represented a braced circular 
 arch with parallel flanges. Span of centre line = 175 ft. ; 
 radius, 201.4 ft. ; rise, 20 ft. In practice, the panels would be 
 taken of equal length ; for convenience of calculation, however, 
 we suppose the panel length to vary so that the horizontal pro- 
 jection is constant, and equal to 25 ft. Depth of arch, 10 ft. 
 Hence, span of lower flange = 170.6 ft. ; rise, 19.5 ft. ; radius, 
 196.4 ft. Span of upper flange, 179.34 ft. ; rise, 20.5 ft. ; radius, 
 206.4 ft. 
 
 Since the flanges are, in practice, broken lines, and not true 
 curves, the depth or lever arm for upper flanges is 9.43 ft., for 
 lower flanges, 10.4 ft. 
 
 The determination of the other dimensions required is then 
 easy, and a simple question of trigonometry. 
 
 Thus we have for the half central angle a = 25 45', and for 
 the distances of the apices from the chord of the centre line : 
 
 For 1... 4.5 3.... 4.7 5. ...11.3 7 14.6ft. 
 
 " 2.... 10.8 4.... 18.5 6.... 23.4 8.... 25 " 
 We suppose the load at each apex 10 tons, and shall consider 
 
 ~Lst. Arch hinged at crown or ap6x 8, and at the ends of 
 the lower flange the flanges H and A being removed. 
 
 2d. Arch hinged at apex 8, and at the ends of the centre 
 line the flanges A and E butting against a skew ~back or pivoted 
 plate, and the flange H only being removed. 
 
 3d. Arch continuous at crown the flange H being retained, 
 and hinged at ends of lower flanges. 
 
APPENDIX.] THE BRACED ARCH. 383 
 
 kth. Arch, as in 3d, but pivoted at ends of centre line. 
 
 5th. Arch without hinges, or continuous at crown and fixed 
 at abutments. 
 
 These cases will illustrate all the principles of Chapter XIY., 
 and a comparison of the results obtained in each case may 
 prove instructive. 
 
 2O. Arch hinged at apex 8, and at the extremities of the 
 lower flange flanges H and A being removed. 
 
 From Art. 158 we can easily find the reaction and horizon- 
 tal thrust at left end either by construction or formula for 
 every weight. Thus 
 
 
 For the first weight P,, then, 
 
 For the weight P 1 
 
 In similar manner, we find 
 
 P! = 0.603, P 2 = 1.33, P 3 = 2.06, 
 
 H! = 2.74, H 2 = 3.84, H 3 = 5.9, 
 
 P 4 = 2.8, P 5 = 3.53, P 6 = 4.2, 
 
 H 4 = 8.1, H 5 = 10.2, H 6 = 12.1, 
 
 P 7 = 5.0, P 8 = 5.7, P 9 = 6.47, P 10 =7.2, 
 
 H 7 = 14.4, He = 12.1, H 9 = 10.2, H 10 = 8.1, 
 
 P u =7.94r, Pu=S.6, PIS- 9.4, 
 
 H u = 5.9, H 12 = 3.84, H 13 = 1.74. 
 
 It will be at once seen that the reaction of P 8 at A is the 
 same as of P 6 at B, or equal to 10 P 6 ; while the horizontal 
 thrust is the same for both. We need them only to find P and 
 H for weights 1 to 7, and can then at once write down the 
 others. We are now ready either to calculate or diagram the 
 strains. 
 
384: NOTE TO CHAP. XIV. [APPENDIX. 
 
 Thus, for instance, for P 10 (see Fig. X.), we lay off the reaction 
 at A upwards to scale from C to D, then the horizontal thrust 
 at A from D to 1, then the equal thrust at B from 1 back to D, 
 then the reaction at B from D to 8, and finally the weight 
 down from 8 back to C, thus closing the polygon for the exterior 
 forces. Lines parallel to the pieces then give the strains. 
 Thus the thrust and reaction at A are in equilibrium with E 
 and 1 2. Then 1 2 is in equilibrium with B and 2 3, and so on. 
 Observe that the diagram checks itself. Thus the last diagonal 
 
 7 8 must be in equilibrium with 6 7 and G (flange H being re- 
 moved), and that this is so is shown by the strain in 7 8 passing 
 exactly through 8, thus making the strain in H zero. We can 
 also check the work by calculating the strain in the last flange 
 D by moments. Thus for P 10 
 
 D x 9.43 = 7.2 x 72.8 - 8.1 x 19.1 - 10 x 25 = 119.45, 
 or D = + 12.6. 
 
 If this agrees with D as found by diagram, and if the diagram 
 also checks, we may have confidence in the accuracy of the 
 work, and at once scale off the strains. Observe that diagonals 
 45 and 56 are both tension ; also that F and G are tension. 
 
 We have given also the diagram for P n , which the reader 
 can easily follow through for himself. F and G are both ten- 
 sion, 3 4 and 4 5 both compression. The horizontal thrust is 
 
 8 &, and the reaction at A = b 1. 
 
 We thus make a diagram for each separate weight, and then 
 taking the dead load at -J the live, we can form the following 
 table of strains. Since we wish only the maximum strains on 
 one-half the arch, those on the other half being precisely simi- 
 lar, we can diagram the strains due to all the weights upon the 
 right half at once by taking the sum of their reactions and 
 thrust at A. We have then the following table : 
 
APPENDIX.] 
 
 THE BRACED ARCH. 
 
 385 
 
 s 
 
 
 . 
 
 . 
 
 . 
 
 10 
 
 "* 
 
 . 
 
 o 
 
 
 o 
 
 "<* 
 
 05 
 
 co 
 
 R 
 
 
 
 
 
 
 
 
 ; 
 
 7 
 
 7 
 
 1 
 
 Th 
 
 1 
 
 1 
 
 1 
 
 <M 
 
 1 
 
 3 
 
 1 
 
 n 
 
 \ 
 
 Maximuir 
 
 CO 
 
 
 
 + 
 
 *> 
 
 1 
 
 + 
 
 C5 
 
 
 
 TH 
 
 + 
 
 J> 
 
 
 
 + 
 
 
 
 
 
 f 
 
 rH 
 
 oof 
 
 * 
 
 + 
 
 10 
 
 + 
 
 
 00 
 
 TH 
 
 1O 
 
 + 
 
 
 t- 
 
 9 
 
 + 
 
 00 
 
 TH 
 
 + 
 
 05 
 
 X 
 
 + 
 
 Jl-* 
 
 * 
 
 i> 
 
 05 
 
 O5 
 
 + 
 
 OS 
 
 5 
 
 + 
 
 <* 
 
 
 10 
 
 + 
 
 CO 
 
 cs' 
 
 CO 
 
 + 
 
 *. 
 
 3 
 
 c^ 
 
 ! 
 
 CO 
 
 5* 
 
 + 
 
 q 
 
 ! 1 
 
 1 
 
 CO 
 idj 
 
 TH 
 
 + 
 
 00 
 
 as 
 
 1 
 
 TH 
 
 o 
 
 TH 
 
 + 
 
 J> 
 
 10' 
 
 1 
 
 TH 
 
 05 
 
 1 
 
 
 T- 
 
 
 
 i> 
 
 TH 
 
 o 
 
 o 
 
 ^ 
 
 o 
 
 GO 
 
 10 
 
 10 
 
 1O 
 
 11 
 
 M g 
 
 00 
 
 
 
 
 o 
 
 TH 
 
 1 
 
 * 
 
 : 
 
 10 
 
 1 
 
 g 
 
 1 
 
 OS 
 
 I 
 
 ^ 
 
 1 
 
 05 
 
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 $ 
 
 
 
 \ 
 
 
 (5 
 
 00 
 
 00 
 
 <* 
 
 ^* 
 
 os 
 
 03 
 
 
 
 10 
 
 TH 
 
 
 
 o 
 
 CO 
 
 11 
 
 g 
 
 + 
 
 g 
 
 + 
 
 8 
 
 + 
 
 + 
 
 s 
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 CO 
 
 + 
 
 $ 
 
 + 
 
 o 
 
 + 
 
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 CO 
 
 + 
 
 J> 
 
 + 
 
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 s 
 
 + 
 
 
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 00 
 
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 * 
 
 c? 
 
 T^ 
 
 
 
 TH 
 
 
 
 rH 
 
 O5 
 
 CS 
 
 ,_, 
 
 M 
 
 p7 
 
 00 
 
 + 
 
 ^h 
 
 05 
 
 + 
 
 J> 
 
 1 
 
 Tt? 
 
 1 
 
 ~ 
 
 rt< 
 
 TH 
 
 + 
 
 TH 
 
 + 
 
 CO 
 
 1 
 
 ; 
 
 " 
 
 ; 
 
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 Pk 
 
 *>' 
 
 1 1 
 
 10 
 
 
 
 T 1 
 
 + 
 
 T-l 
 
 co' 
 
 + 
 
 CO 
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 1 
 
 t^* 
 
 o 
 
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 CO* 
 
 1 
 
 
 05 
 
 + 
 
 CO 
 
 o 
 
 TH 
 
 1 
 
 
 
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 s 
 
 + 
 
 
 
 ^ - 
 
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 oq' 
 
 + 
 
 
 05' 
 
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 I> 
 
 00 
 
 o 
 
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 w 
 
 10 
 
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 CO 
 
 to 
 
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 10 
 
 10 
 
 to 
 
 00 
 
 T-t 
 
 + 
 
 CD 
 
 + 
 
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 + 
 
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 ^ 
 
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 OS 
 
 + 
 
 00 
 
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 as 
 
 + 
 
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 + 
 
 CD 
 
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 + 
 
 ^ 
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 PU 
 
 z> 
 
 00 
 
 + 
 
 IO 
 O5' 
 
 05 
 
 + 
 
 o 
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 + 
 
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 ^ 
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 z> 
 
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 GO 
 
 l> 
 
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 d 
 
 + 
 
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 : 
 
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 10 
 
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 e 
 
 rH 
 
 o 
 
 t- 
 
 TH 
 
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 Ol 
 
 Pi 
 
 so 
 
 + 
 
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 t 1 
 
 + 
 
 10 
 
 II 
 
 + 
 
 00 
 
 + 
 
 
 
 + 
 
 o 
 
 TH 
 
 J 
 
 rJH 
 
 + 
 
 CO 
 
 1 
 
 
 
 + 
 
 
 
 1 
 
 CO 
 
 + 
 
 CO 
 
 + 
 
 t- 
 
 1 
 
 00 
 
 fk 
 
 T 1 
 
 05 
 
 + 
 
 o 
 
 OS 
 
 + 
 
 JO 
 
 as' 
 
 + 
 
 OJ 
 
 w 
 
 + 
 
 <N 
 J> 
 
 + 
 
 CO 
 
 i-T 
 
 1 
 
 ^ 
 
 T-H 
 
 + 
 
 CO 
 
 1 
 
 to 
 
 ^ 
 + 
 
 TH 
 -4 
 
 I 
 
 Y 1 
 
 CO* 
 
 + 
 
 - 
 
 I> 
 
 \ 
 
 
 
 as 
 
 1 
 
 
 J> 
 
 to 
 
 CO 
 
 o 
 
 w 
 
 i> 
 
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 TH 
 
 CO 
 
 rH 
 
 CO 
 
 as 
 
 TH 
 
 1 
 PH" 
 
 <^ 
 
 1 
 
 C5 
 
 + 
 
 i 
 
 + 
 
 o 
 
 CO 
 
 + 
 
 "* 
 
 CO 
 
 + 
 
 TH 
 
 <M 
 
 + 
 
 CO 
 
 1 
 
 CO 
 
 + 
 
 1O 
 
 + 
 
 TH 
 
 1 
 
 T- 
 
 fi 
 
 + 
 
 TH 
 TH 
 
 1 
 
 10' 
 
 + 
 
 CO 
 94 
 
 ft 
 
 ^ 
 
 CO 
 
 1 
 
 2 
 
 4- 
 
 CO 
 
 id 
 
 + 
 
 CO 
 
 ;' 
 
 + 
 
 10 
 
 & 
 
 + 
 
 o* 
 
 
 
 TH 
 
 + 
 
 CO 
 
 oi 
 
 1 
 
 o 
 
 05 
 
 + 
 
 CO 
 CO 
 
 + 
 
 TH 
 
 co' 
 
 1 
 
 1O 
 00 
 
 + 
 
 as 
 
 CO' 
 
 1 
 
 05 
 
 TH 
 TH 
 
 + 
 
 
 w 
 
 
 
 Q 
 
 H 
 
 h 
 
 
 
 05 
 
 TH 
 
 CO 
 <M 
 
 "* 
 
 CO 
 
 10 
 
 "tf 
 
 CO 
 10 
 
 t- 
 
 CO 
 
 GO 
 
 t- 
 
NOTE TO CHAP. XIV. [APPENDIX. 
 
 21 . Arch hinged at Apex and at the Extremities of the 
 centre Une ; FIange A being retained, and only H re- 
 moved. The method of solution is precisely the same as 
 before, the only difference being that the span is now 175 ft. 
 instead of 170.6, and the rise 25 ft. instead of 29.5. The reac- 
 tions and thrust will then be somewhat different. Thus, for 
 the left abutment A, 
 
 P 1 =0.71, P a =1.42, P 8 =2.14, P 4 =2.85, P 5 = 3.57, 
 H!=2.48, H 2 = 4.97, H 8 = 7.5, H 4 = 9.97, H 5 = 12.5, 
 
 P 6 = 4.28, P 7 = 5.00, P 8 = 5.72, P 9 = 6.43, P lo = 7.15, etc. 
 H 6 = 14.98, H 7 = 17.5, H 8 = 14.98, H 9 = 12.5, H 10 = 9.97, etc. 
 
 We have therefore the following table : 
 
APPENDIX.] 
 
 THE BRACED ARCH. 
 
 387 
 
 i Strains. 
 
 
 
 
 
 
 O5 
 
 TH 
 
 1 
 
 
 
 o 
 
 T 
 
 D 
 }" 
 
 h 
 
 
 10 
 
 1 
 
 j 
 
 ^ 
 C 
 
 T- 
 
 H 
 
 :> 
 
 ( 
 
 10 
 
 os* 
 
 1 
 
 CO 
 
 ol 
 
 1 
 
 c 
 
 c 
 
 5 
 H 
 
 5 
 
 00 
 CO 
 
 1 
 
 3. 
 a 
 
 (O 
 
 ,_, 
 
 05 
 
 05 
 
 05 
 
 
 
 1. 
 
 5 
 
 
 
 05 
 
 C 
 
 ^> 
 
 * 
 
 o 
 
 c 
 
 6 
 
 
 
 1 
 
 1 
 
 o 
 
 rp 
 C- 
 
 TH 
 
 GO 
 O5 
 
 TH 
 
 + 
 
 ** 
 
 I 
 e 
 
 D 
 h 
 
 5 
 
 TH 
 
 C 
 C 
 
 j 
 
 h 
 
 o 
 
 + 
 
 a 
 
 T- 
 
 3 
 
 4 
 
 h 
 
 + 
 
 Ii 
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 10 
 
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 0* 
 
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 00* 
 
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 9 
 
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 3 
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 10 
 
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 tt 
 
 ^ 
 
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 5 
 
 t- 
 
 pi 
 
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 3 
 
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 10 
 
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 te. 
 
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 5 
 
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 TH 
 
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 c: 
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 5 
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 10' 
 
 <y 
 i> 
 
 j 
 
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 fc 
 
 10 
 
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 05 
 
 
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 cc 
 
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 i 
 
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 co 
 
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 c 
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 i 
 
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 1 
 
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 i 
 
 TH 
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 05 
 
 ^ 
 
 05 
 
 10 
 
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 <* 
 
 T- 
 
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 t- 
 
 00 
 
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 10 
 
 00 
 
 ^ 
 
 
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 j^ 
 
 05 
 
 o 
 
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 00 
 05 
 
 
 
 g 
 
 J 
 
 H 
 
 4 
 } 
 
 00 
 
 1 
 
 + 
 
 c 
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 5 
 
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 oo 
 
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 1 
 
 
 10 
 
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 1 
 
 M 
 
 O 
 
 P 
 
 W 
 
 fc 
 
 d 
 
 J 
 
 TH 
 
 CO 
 
 Tj 
 
 o: 
 
 
 10 
 
 CO 
 10 
 
 i> 
 
 CC 
 
 
 00 
 
388 NOTE TO CHAP. XIV. [APPEKDIX. 
 
 A comparison of this table with the preceding shows that we 
 have gained nothing by introducing two end flanges at A at 
 each end, and pivoting the arch at the extremities of the centre 
 line. We have indeed slightly diminished the strains in the 
 lower flanges E and F, as also in the bracing, but the other 
 strains are much greater than before a result which might 
 have been anticipated, since the effect of hinging at the centre, 
 instead of at the extremities of the lower flange, is simply to 
 reduce the effective height or rise from 29.5 ft. to 25 ft. In our 
 example, since the depth of arch is half the whole rise of the 
 centre line, this reduction is considerable. 
 
 For a much longer span and smaller proportional depth the 
 difference would not be so marked, but it would seem that the 
 strains in the second case must always be greater than in the 
 first. The best construction, then, seems to require the hinge 
 in the upper flange at the crown, and at the extremities of the 
 lower flange at the abutments. By this means, the greatest ef- 
 fective rise is obtained, and both ribs aid in supporting the 
 load. Were the hinges all three in the same rib, then, for 
 uniform load, that rib alone is the sole supporting member, and 
 is unassisted by the other. This should then be avoided. 
 
 22. Arcli continuous at Crown, and hinged at End of 
 Lower Rib. For this case, referring to Art. 159, we have sim- 
 ply to interpolate from our table there given the values of A, 
 B and y i n the equation, 
 
 1 +B/c 
 y = r=ST*y 
 
 and thus plot the curve cdeik. Fig. 91. The construction of 
 the reactions and horizontal thrust for each weight is then easy. 
 These once known, we proceed as above, in order to find the 
 strains. 
 
 Now, in the formulae above K = -5, and since we can put 
 
 A T" 
 
 for the square of the radius of gyration, and this radius is 
 A. 
 
 approximately the half depth of the arch, we have 
 
 25 25 1 
 
 1 (170.6) 2 - 2910436 ~~ 1164' 
 
 Now B and A are, as we see from the table, small, and he^nce 
 in our present example the terms containing K can be disre- 
 
APPENDIX.] THE BEACED ABCH. 389 
 
 garded, and the value of y can be taken directly from the 
 table for y , given in Art. 159. For a 25 45', then we have 
 at once, since h = 19.5, 
 
 for = 0, y = 1.295 h = 25.25 ft., 
 
 = 0.2 a, y = 1.304 A = 25.42 ft., 
 ft = 0.3 a, y = 1-335 A = 26.03 ft., etc. 
 
 The corresponding value of x is R cos ft. 
 
 Having thus plotted the curve, and constructed the reaction, 
 and thrust for each weight, the diagram for strains proceeds as 
 before. We thus form the following table : 
 
390 
 
 NOTE TO CHAP. XIV. 
 
 [APPENDIX. 
 
 
 os 
 
 t> 
 
 CO 
 
 
 
 
 
 o 
 
 00 
 
 OS 
 
 CO 
 
 t> 
 
 
 
 ,_, 
 
 OD 
 
 00 
 CO 
 
 1 
 
 % 
 \ 
 
 9 
 
 1 
 
 : 
 
 
 
 ! 
 
 O3 
 
 Oi 
 
 CO 
 
 10 
 
 1 
 
 
 
 TH 
 
 1 
 
 CO 
 O3 
 
 CO 
 TH 
 
 1 
 
 oo' 
 
 TH 
 
 1 
 
 s 
 
 p 
 
 OS 
 
 i> 
 
 II 
 
 os 
 
 Tj< 
 
 o 
 
 > 
 
 CO 
 
 J> 
 
 OS 
 
 03 
 
 OS 
 
 *> 
 
 GO 
 
 
 03 
 
 CO 
 
 + 
 
 5 
 
 + 
 
 to 
 
 T 1 
 
 + 
 
 i 
 
 + 
 
 g 
 
 w 
 
 + 
 
 os' 
 
 TH 
 
 w 
 
 + 
 
 10 
 O3 
 
 + 
 
 s 
 
 + 
 
 oo' 
 
 03 
 
 + 
 
 
 
 + 
 
 O3 
 
 + 
 
 00 
 
 + 
 
 g 
 
 + 
 
 s 
 
 + 
 
 g 
 
 to 
 
 o 
 
 00 
 
 T _ l 
 
 TH 
 
 03 
 
 os 
 
 ^ 
 
 *> 
 
 10 
 
 i> 
 
 03 
 
 TH 
 
 TH 
 
 U H " 
 
 P . 
 
 TH 
 1 
 
 TH 
 
 4- 
 
 
 
 1 
 
 os 
 l> 
 
 + 
 
 g 
 
 + 
 
 
 
 + 
 
 > 
 
 + 
 
 CO 
 
 + 
 
 co' 
 
 + 
 
 to 
 
 + 
 
 03 
 
 4_ 
 
 TH 
 
 CO 
 
 + 
 
 TH 
 
 -f 
 
 1 
 
 a 3 
 
 *a o 
 
 * 
 *>' 
 
 CO 
 
 1 
 
 t> 
 
 >' 
 
 CO 
 
 1 
 
 
 
 3 
 
 1 
 
 * 
 
 * 
 
 1 
 
 
 
 to 
 
 '. 
 
 j 
 
 p 
 
 CD 
 OJ 
 
 1 
 
 to 
 J> 
 
 1 
 
 * 
 
 TH 
 TH 
 
 1 
 
 
 
 CO 
 
 1 
 
 to 
 
 O3 
 O3 
 
 1 
 
 TH 
 
 co' 
 
 TH 
 
 1 
 
 10 
 
 co' 
 
 O3 
 
 1 
 
 s p ~ i 
 
 35 g 
 
 TH 
 
 t- 
 
 OS 
 
 00 
 
 CO 
 
 
 
 00 
 
 os 
 
 o 
 
 TH 
 
 to 
 
 TH 
 
 CO 
 
 ^H 
 
 2^ 
 
 H 
 
 TH 
 
 CO 
 
 + 
 
 g 
 
 + 
 
 0* 
 
 + 
 
 (M 
 
 o 
 
 TH 
 
 1 
 
 s 
 
 TH 
 
 + 
 
 5 
 
 
 
 2 
 
 + 
 
 03 
 
 CO 
 
 + 
 
 to' 
 
 O3 
 
 + 
 
 g 
 
 + 
 
 CO 
 
 + 
 
 
 
 + 
 
 3 
 
 + 
 
 & 
 
 + 
 
 
 05 
 
 
 
 so 
 
 1O 
 
 CO 
 
 O3 
 
 o 
 
 CO 
 
 CO 
 
 ?> 
 
 03 
 
 
 
 1O 
 
 I> 
 
 oi 
 
 ; 
 
 TH 
 
 + 
 
 
 r^ 
 
 1 
 
 
 
 + 
 
 CO 
 
 + 
 
 to 
 
 + 
 
 CO 
 
 TH 
 
 + 
 
 03 
 
 + 
 
 CO 
 
 1 
 
 03 
 
 + 
 
 03 
 
 I 
 
 TH 
 
 + 
 
 
 
 1 
 
 ci 
 
 h 
 
 o 
 to 
 
 + 
 
 p 
 
 J> 
 
 + 
 
 00 
 
 
 4- 
 
 CO 
 
 TH 
 
 + 
 
 !> 
 
 <M' 
 
 1 
 
 l> 
 
 co' 
 
 + 
 
 CO 
 
 oo' 
 
 + 
 
 CO 
 
 o 
 
 TH 
 
 + 
 
 03 
 
 os 
 
 1 
 
 GO 
 CO 
 
 1 
 
 to 
 
 CO 
 
 + 
 
 CO 
 
 TH 
 
 1 
 
 10 
 CO* 
 
 + 
 
 o 
 
 03 
 
 1 
 
 Pk 
 
 TH 
 
 CO 
 
 + 
 
 SO 
 
 i> 
 
 + 
 
 co 
 w 
 
 1 
 
 00 
 
 + 
 
 00 
 058 
 
 ! 
 
 to 
 
 00 
 1 
 
 to 
 
 to 
 
 TH 
 
 + 
 
 - 
 
 10* 
 
 + 
 
 o 
 to' 
 
 1 
 
 00 
 !>' 
 
 + 
 
 CO 
 
 
 
 + 
 
 TH 
 l> 
 
 I> 
 
 to" 
 
 + 
 
 CO 
 
 co' 
 
 1 
 
 fc 
 
 tc 
 
 TH 
 
 + 
 
 CO 
 
 TH 
 
 TH 
 
 + 
 
 
 
 <tt' 
 
 + 
 
 CO 
 0? 
 
 + 
 
 00 
 
 "*' 
 
 + 
 
 ^ 
 
 to 
 
 1 
 
 I 
 
 o 
 
 co' 
 
 + 
 
 l> 
 
 03* 
 
 CO 
 
 o 
 
 + 
 
 <* 
 
 co- 
 
 00 
 09 
 
 | 
 
 TH 
 
 co 
 
 + 
 
 co 
 
 1 
 
 tf 
 
 00 
 
 o 
 
 + 
 
 o 
 J> 
 
 + 
 
 Tf< 
 I> 
 
 + 
 
 CO 
 
 o 
 
 rH 
 
 + 
 
 oi 
 
 + 
 
 o 
 
 co' 
 
 + 
 
 OS 
 
 TH 
 
 TH 
 
 + 
 
 !> 
 
 o 
 
 + 
 
 o 
 o 
 
 10 
 
 ; 
 
 co 
 
 TH 
 
 O 
 i> 
 
 + 
 
 03 
 
 co' 
 
 + 
 
 TH 
 
 10* 
 
 1 
 
 & 
 
 CO 
 O3' 
 
 t> 
 
 TH 
 
 + 
 
 1 1 
 
 TH 
 
 + 
 
 CO 
 00 
 
 + 
 
 CO 
 I> 
 
 TH 
 
 + 
 
 00 
 
 
 TH 
 
 + 
 
 5 
 
 Tjj 
 TH 
 
 1 
 
 CO 
 
 TH 
 
 + 
 
 . 
 
 03 
 
 + 
 
 CO 
 03 
 
 1 
 
 
 
 ; 
 
 CO 
 
 CO 
 
 1 
 
 oo 
 t> 
 
 1 
 
 
 00 
 
 oo 
 
 1C 
 
 0> 
 
 -1 
 
 o 
 
 TH 
 
 rH 
 
 OS 
 
 o 
 
 CO 
 
 > 
 
 CO 
 
 t- 
 
 (Kc 
 
 J> 
 
 1 1 
 
 1 
 
 
 
 T 1 
 
 1 
 
 oo 
 
 TH 
 
 1 
 
 oo 
 to 
 
 + 
 
 to 
 o 
 
 _j_ 
 
 
 
 + 
 
 l> 
 
 CO 
 
 -j- 
 
 O3 
 
 TH 
 
 1 
 
 
 
 TH 
 
 + 
 
 TH 
 
 + 
 
 03 
 + 
 
 O 
 
 TH 
 
 + 
 
 z> 
 
 1 
 
 co 
 
 TH 
 
 + 
 
 <o 
 1 
 
 (C 
 
 CO 
 
 J> 
 
 TH 
 
 1 
 
 OS 
 
 i 
 
 1 
 
 -* 
 
 CO 
 
 TH 
 
 00 
 
 55 
 
 + 
 
 p 
 
 :o 
 
 10 
 
 + 
 
 to 
 
 s 
 
 + 
 
 05 
 
 
 Tfl 
 
 + 
 
 03 
 03 
 
 1 
 
 to 
 
 o 
 
 1 
 
 f 
 
 03 
 "* 
 
 + 
 
 CO 
 
 co' 
 
 - 
 
 CO 
 03 
 
 I 
 
 f> 
 
 CO 
 
 + 
 
 
 B 
 
 D 
 
 p 
 
 w 
 
 h 
 
 C5 
 
 w 
 
 03 
 
 s 
 
 ^ 
 
 CO 
 
 10 
 
 Tt< 
 
 to 
 
 10 
 
 r^ 
 
 CD 
 
 00 
 
 J> 
 
APPENDIX.] THE BRACED ARCH. 391 
 
 Comparing these resiflts with those in our table above, Art. 
 19, for the same case not continuous at the crown, we see that 
 the strains in the upper flanges are much less, and are, more- 
 over, of opposite character ; while the strains in the lower 
 flanges are greatly increased, and nothing is gained. This re- 
 sult might also have been anticipated, since the effect of insert- 
 ing the flange H is to reduce the effective height from 29.5 to 
 19.5 ft., and, moreover, for total dead and live load, nearly the 
 whole weight comes directly upon the continuous lower rib, and 
 the upper aids but very little. 
 
 23. Strain* due to Temperature. We have, in addition, 
 strains due to change of temperature to be taken into account 
 in determining the total maximum strains. 
 
 For the present case we have, from Art. 165, for the thrust 
 due to change of temperature, 
 
 15 El A et 
 ~8AA 2 + 15T 
 
 or, substituting in the place of the square of the radius of 
 
 A. 
 
 gyration = <f, we have 
 
 15 in A ^2 */ 
 H = 
 
 Now g is approximately the half depth of arch ; hence cf 1 
 25 sq. ft. = 3600 sq. inches. Taking 5 tons to the square inch 
 as our unit strain, we may take the area of our flanges, as de- 
 termined from the above table of strains at about 25 square 
 inches. Hence A = 50. Taking E = 14,000 tons per sq. inch, 
 A 2 19. 5 2 = 54,756 sq. inches, and supposing the temperature 
 to vary 25 (Centigrade) on each side of the mean, we have, 
 assuming e at 0.000012, the thrust H about 25 tons. 
 
 It is easy to find either by moments or diagram, or both, the 
 strains due to this thrust. Since the temperature varies between 
 25 on both sides of the mean, this thrust can be both positive 
 and negative, and the corresponding strains have, therefore, 
 double sign. We find, therefore, 
 
 B = =F 24.5 C = T= 40.4 D = =f 49.1 E = 37.5 
 
 F = i 55.9 G- = 66.9 H = j_ 69.6 
 
 12==Fl?.l 23 = 15.0 34=^8.6 45 =12.5 
 
 56 = ^2.2 67 = i 9.7 7 8 = =p 5.7 
 
392 NOTE TO CHAP. XIV. [APPKNDIX. 
 
 Hence we have, for the total maximum strains for the case of 
 the preceding article, 
 
 B = + 57.4-63.4 C= + 81.1-77.1 D= + 64.2-91.4 
 E=+ 279.4 F = + 286.3 G= + 286.5 
 
 H=+ 285.3 12 = + 53.4-39.723=+ 43.7-28.8 
 
 34= + 37.322.445=+ 40.4-28.456=+ 23.4-12.5 
 67=+ 36.4-22.778=+ 42.5-23.8 
 
 24. Arch continuous at the Crown, pivoted at the 
 extremities of the Centre Line. The method of solution is 
 precisely similar to the preceding. We have only to take the 
 rise of the centre line, or h 20, instead of h =19.5, as before, 
 and the radius and span of centre line, instead of the radius and 
 span of the lower rib. 
 
 One point only needs to be noticed. Having found the reac- 
 tion and thrust for any weight, these forces now act at the ex- 
 tremity of the centre line. "We can therefore form the strain 
 diagram as follows : 
 
 First calculate by moments the strain in A and E. Then, in 
 diagram (c), Fig. X., having laid off the thrust o C and the reac- 
 tion C 5, draw from o and b lines parallel to A and E, and lay 
 off o A equal to the strain in A, and b d to the strain in E. 
 Then, if these strains have been correctly found, the line A d 
 must be parallel to and give the strain in diagonal 1 2. The 
 diagram thus commenced, can then be continued as shown, and 
 the strains in all the pieces determined. 
 
 We may also form the strain diagram without calculating A 
 and E. Thus o b is the resultant acting at the end of the 
 centre line. Since it acts then half way between A and E, 
 bisect it in 0, and draw a A perpendicular to flange A. Then 
 o A is the strain in A, and drawing A d parallel to diagonal 1 2, 
 we have at once the strain in E and 1 2. 
 
 Performing the operations indicated, we obtain the following 
 table of strains : 
 
APPENDIX.] 
 
 THE BRACED ARCH. 
 
 393 
 
 
 
 
 
 
 
 
 
 
 
 O 1 
 
 3 
 
 rH 
 
 co 
 
 1 
 
 3 
 
 b 
 
 
 cr 
 
 5 
 
 ^ 
 
 1 1 
 
 ; 
 
 
 1 
 
 1 
 
 j 
 
 
 
 I 
 
 
 1 
 
 5 
 
 < 
 
 OS* 
 TH 
 
 IO 
 
 1t 
 
 ^ 
 
 T- 
 
 fl 
 H 
 
 ir 
 
 7 
 
 i 
 
 g 
 
 i 
 
 C^ 
 
 O2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 
 
 
 
 
 
 1 
 
 a 
 
 1 
 
 IO 
 
 o 
 o> 
 
 05 
 
 05 
 
 OS 
 
 TH 
 
 CO 
 
 co 
 
 1-' 
 b 
 
 T- 
 
 5 
 
 H 
 
 1-H 
 
 CO 
 
 c 
 c 
 
 } 
 
 H 
 
 3 
 
 10 
 
 Os' 
 rH 
 
 GO* 
 
 TH 
 
 If 
 
 a 
 
 T- 
 
 5 
 
 2 
 H 
 
 C 
 li 
 
 5 
 i 
 
 o 1 
 
 T- 
 
 
 
 i 
 
 H 
 
 i 
 
 i 
 
 1 
 
 + 
 
 + 
 
 * 
 
 + 
 
 4 
 
 + 
 
 
 - 
 
 * 
 
 
 h 
 
 + 
 
 + 
 
 
 - 
 
 
 K 
 
 - 
 
 h 
 
 * 
 
 fl- 
 
 IO 
 
 4- 
 
 CO 
 
 oo" 
 
 CO 
 
 CO 
 
 OS 
 
 oo 
 
 CO 
 
 h 
 
 CO 
 
 o 
 
 r . 
 
 ? 
 
 3 
 
 {- 
 
 CO 
 
 c 
 
 r- 
 
 J 
 
 H 
 
 h 
 
 rH 
 
 
 00 
 
 
 8 
 
 d 
 
 3 
 
 5 
 1 
 
 c 
 c 
 
 5 
 5 
 
 Of 
 
 C 
 
 3 
 3 
 
 
 
 1 
 
 02 
 
 
 
 
 
 05 
 1 
 
 CD* 
 
 1-H 
 
 1 
 
 -y 
 
 L' 
 
 D 
 
 5 
 
 
 If 
 
 b 
 
 O 
 
 ) 
 
 j 
 
 05 
 
 OS 
 
 1 
 
 co' 
 
 1 
 
 T- 
 T" 
 
 1 
 
 H 
 
 1 
 
 3 
 
 3 
 / 
 
 a 
 
 , 
 C 
 
 *' 
 J 
 
 o 
 
 
 
 CO 
 
 1 
 
 Cfi 
 
 PH 
 
 
 
 co 
 
 
 
 co 
 
 
 
 05 
 
 cr 
 
 D 
 
 5 
 
 c 
 
 5 
 
 Tt< 
 
 OS 
 
 C/ 
 
 3 
 
 c 
 
 5 
 
 ^ 
 
 3 
 
 rH 
 
 1 
 
 EH 
 
 8 
 
 4 
 
 TH 
 
 00 
 
 CO 
 
 00 
 
 
 
 os 
 
 4 
 
 O5 
 
 i 
 
 H 
 
 I 
 
 - 
 
 
 c 
 
 -1 
 
 i 
 
 5 
 
 h 
 
 05 
 
 t- 
 
 ? 
 
 r~ 
 
 S 
 
 H 
 
 h 
 
 ! 
 
 5 
 h 
 
 T- 
 
 C 
 
 J 
 
 F 
 
 4 
 
 
 
 
 rH 
 
 05 
 
 
 
 TH 
 
 00 
 
 c 
 
 5 
 
 tw 
 
 c: 
 
 5 
 
 05 
 
 O5 
 
 r- 
 
 -1 
 
 t 
 
 
 c 
 
 > 
 
 CO 
 
 
 
 CO 
 
 4 
 
 05 
 
 + 
 
 o 
 
 1 
 
 1 
 
 
 
 i 
 r 
 
 T 
 
 
 
 5 
 
 H 
 
 H 
 
 05 
 
 1 
 
 c 
 
 5 
 
 t- 
 
 c 
 
 4 
 
 1 
 
 c 
 
 ~i 
 1- 
 
 rH 
 1 
 
 
 CO 
 
 Z> 
 
 OS 
 
 o 
 
 CO 
 
 o . 
 
 If 
 
 5 
 
 00 
 
 T- 
 
 H 
 
 b- 
 
 05 
 
 o 
 
 5 
 
 o 
 
 3 
 
 ? 
 
 3 
 
 00 
 
 fc 
 
 IO 
 
 00 
 
 TH 
 
 00 
 
 05 
 
 
 1 
 
 o 
 - 
 
 i 
 " 
 
 + 
 
 C 
 
 H 
 
 s 
 
 . 
 
 00 
 
 1 
 
 00 
 1 
 
 T 
 
 H 
 h 
 
 li 
 
 3 
 
 V 
 
 3 
 
 h 
 
 05 
 
 I 
 
 
 CO 
 
 
 10 
 
 05 
 
 
 
 * 
 
 a 
 
 D 
 
 CO 
 
 T 
 
 5 
 
 CO 
 
 CO 
 
 i- 
 
 3 
 
 * 
 
 H 
 
 G 
 
 5 
 
 05 
 
 i 
 
 co 
 
 3 
 
 ; 
 
 + 
 
 1 
 
 00 
 
 i 
 
 
 H 
 
 f 
 
 00 
 
 4 
 
 1^ 
 
 5 
 f 
 
 IO 
 
 i 
 
 + 
 
 n 
 
 3 
 h 
 
 I 
 
 
 T 
 
 j 
 h 
 
 CO 
 
 1 
 
 * s 
 
 b-' 
 
 1 
 
 7 
 
 GO 
 
 OS 
 
 fc- 
 
 s 
 
 o 
 
 TI 
 
 O 1 
 
 C 
 
 D 
 5 
 
 h 
 
 00 
 
 o 
 
 rH 
 
 T 
 
 
 
 h 
 h 
 
 05 
 
 co' 
 
 1 
 
 TH 
 
 10' 
 
 L: 
 
 c: 
 
 3 
 
 5 
 
 -T- 
 
 c 
 
 T 
 
 H 
 3 
 
 H 
 
 a 
 
 c; 
 
 3 
 S 
 
 h 
 
 CO 
 
 10' 
 
 1 
 
 
 CO 
 
 TH 
 
 GO 
 
 Tj< 
 
 00 
 
 rH 
 
 d 
 
 5 
 
 05 
 
 If 
 
 2 
 
 t. 
 
 t. 
 
 T- 
 
 H 
 
 T- 
 
 -i 
 
 c: 
 
 3 
 
 05 
 
 OS 
 
 GO 
 
 o 
 
 co 
 
 7 
 
 + 
 
 CO 
 
 ^ 
 
 5 
 1 
 
 10 
 
 c 
 
 5 
 h 
 
 TH 
 
 1 
 
 00 
 
 If 
 
 D 
 
 a 
 
 5 
 h 
 
 c; 
 
 3 
 h 
 
 1 
 
 
 co 
 
 CO 
 
 00 
 
 OS 
 
 CO 
 
 TH 
 
 c 
 
 i 
 
 CO 
 
 c 
 
 5 
 
 10 
 
 oo 
 
 C 
 
 5 
 
 c 
 
 B 
 
 b 
 
 
 05 
 
 * 
 
 00 
 
 CO 
 
 05 
 
 00 
 rH 
 
 o 
 
 TH 
 
 OS 
 
 
 
 5 
 h 
 
 + 
 
 o 
 
 5 
 
 o 
 
 rH 
 
 c 
 
 5 
 
 -r 
 
 H 
 h 
 
 c: 
 
 > 
 
 OS 
 
 1 
 
 
 GO 
 
 b- 
 
 !> 
 
 05 
 
 CO 
 
 T}< 
 
 C< 
 
 D 
 
 10 
 
 a 
 
 3 
 
 OS 
 
 TH 
 
 b 
 
 
 C 
 
 5 
 
 Tl 
 
 $ 
 
 T* 
 
 fc 
 
 CO 
 O5 
 
 CO 
 
 4 
 
 ' 
 
 oo' 
 
 s 
 
 58 
 
 S 
 
 5 
 ^ 
 
 CO 
 
 TH 
 
 a 
 
 T- 
 
 2 
 H 
 
 00 
 
 05 
 
 I 
 
 C 
 
 5 
 h 
 
 a 
 
 D 
 - 
 
 C 
 
 1 
 
 co 
 
 rH 
 
 * : 
 
 00 
 00 
 
 TH 
 
 IO 
 CO* 
 
 o 
 
 rH 
 
 
 rH 
 
 00 
 
 o 
 
 t 
 
 C 
 
 o 
 
 L 
 
 b-' 
 
 
 
 o 
 
 T- 
 
 ? 
 
 5* 
 
 H 
 
 00 
 
 OS 
 rH 
 
 | 
 
 a 
 
 r 
 
 5 
 
 H 
 
 h 
 
 c 
 c: 
 
 5 
 5 
 
 h 
 
 o 
 c: 
 
 > 
 j 
 
 05* 
 
 rH 
 
 
 < 
 
 w 
 
 D 
 
 Q 
 
 H 
 
 CM 
 
 c 
 
 5 
 
 
 
 C 
 
 T- 
 
 5 
 
 H 
 
 CO 
 05 
 
 CO 
 
 If 
 
 *: 
 
 3 
 
 r> 
 
 Ci 
 
 u 
 
 5 
 3 
 
 b 
 
 c: 
 
 5 
 
 00 
 
394: NOTE TO CHAP. XIV. [APPENDIX. 
 
 A comparison of the above with the same case hinged at 
 the ends of the lower rib, shows a decided gain. The effective 
 height is increased, being now 20 ft., in place of 19.5 ; in 
 addition to which both ribs under total load bear their proper 
 proportion. If, then, we wish the arch continuous at crown, 
 both ribs should butt against an end plate, pivoted in the 
 centre. This is preferable to hinging the lower rib at its ex- 
 tremities, and removing the end flanges A. 
 
 25. Temperature Strains. For the strains due to tem- 
 perature, we may take, as before, the thrust H = 25 tons, and 
 rind thus 
 
 A=11.2 B=q=13.2 C=q=29,4 D = qp 37.6 
 
 E = 27.0 P = 45.5 G= 56.4 H == 59.2 
 
 12 =T 17.8 2 3 =13.9 3 4 = =F 9.8 4 5 = 11.3 
 
 5 6 = =F 2.8 6 7 = 8.1 7 8 = 4.3 
 
 all somewhat less than in the previous case, as they should be, 
 since the point of application is at the centre between the 
 flanges. 
 
 We have then from the preceding table the total maximum 
 strains : 
 
 A = 
 
 + 135.7 
 
 B = + 135.2 C = + 144.3 
 
 D = + 155.5 
 
 E = + 158.2 
 
 +176.7 r +173.9 
 - 7.2 - - 19.2 
 
 H + 172.3 
 ti ~ - 25.5 
 
 12 = 
 
 + 49.0 
 -44.1 
 
 o o + 33.4 o A + 28.5 
 - _ 33.0 - _ 25.4 
 
 45 _ +24.8 
 ~ -26.1 
 
 56 = 
 
 + 27.8 
 28.5 
 
 Rf7 + 29.4 ^ ft + 33.0 
 67 -_30.7 78: "-34.7 
 
 
 26. Arch continuou at Crown and fixed at Ilie Ends 
 
 From our table, Art. 160, we have directly for a = 25 45', 
 and Ti 20, y being now measured above the horizontal tangent 
 at crown of centre line, 
 
 for ft = 0, y = 0.209 h = 4.18 ft. 
 
 /3 = 0.2a, y = 0.208 h = 4.16 
 
 ft = 0.4a, y = 0.206 h = 4.12 
 
 = 0.6a, y = 0.201 h = 4.02 
 
 ft = O.So, y = 0.198 h = 3.96 
 
 ft = l.Oa, y = 0.189 h = 3.78 
 
APPENDIX.] THE BRACED ARCH. 395 
 
 Also from our formulae of Art. 160, we have 
 40 
 
 Cl ~ 
 
 15 (87.5 + x) 
 40 
 
 I 87.5 5 x 246.1 I 
 I 87.5 + 5 x - 246.1 I 
 
 15(87.5+0)1 
 
 Hence for 
 
 P! x = 75 d = + 8.7 c 2 = - 3.5 
 
 P 2 x = 62.5 o t = + 8.3 c 2 = - 2.7 
 
 P 3 a = 50 ^ = + 7.9 es, = - 1.7 
 
 P 4 cc = 37.5 d = + 7.3 <fe = - 0.61 
 
 P 5 z = 25 ^ = + 6.7 c 2 = + 0.8 
 
 P 6 ,7? = 12.5 d = + 5.8 e 2 = + 2.5 
 
 P 7 x = c l = 4- 4.8 a* = + 4.8 
 
 P 8 #= 12.5 G! = + 2.5 <? 2 = + 5.8 
 
 P 9 rrz_25 ^=+0.8 c 2 = + 6.7, etc., 
 
 negative values of c being laid off above the ends of centre 
 line. 
 
 We can therefore easily construct the left reactions, as 
 explained in Art. 160, Fig. 92. We thus obtain 
 
 P t = 0.5, P 2 = 0.9, P 3 = 1.2, 
 
 U, = 4.4, H 2 = 8.6, H 3 = 13.1, 
 
 M! = + 38.3, M 2 = + 71.4, M 3 = + 103.5, 
 
 P 4 = 2.3, P 5 == 3.1, P 6 = 4.4, 
 
 H 4 = 15.9. H 5 = 18.9, H 6 = 21.1. 
 
 M 4 = + 116.1, M 5 = + 126.6, M 6 = + 122.4, 
 
 P 7 = 5.0, P 8 = 5.6, P 9 = 6.9, 
 
 H 7 = 22.4, H 8 = 21.1, H 9 = 18.9, 
 
 M 7 = + 107.5, M 8 = + 52.7, M 9 = + 15.1, 
 
 P_77 p=88 P=91 P=95 
 
 H 10 =15.9, H n = 13.1, H M = 8.6, H 1S = 4.4, 
 
 Mj =-9.7, Mu=-22.3, M^ = 23.2, M 13 = - 15.3. 
 
 A positive moment indicates tension in the lower flange at 
 abutment, and compression in upper. 
 
396 NOTE TO CHAP. XIV. [APPENDIX. 
 
 We can therefore easily calculate the strains in. flanges A 
 and E for each weight. 
 
 Thus, for u 
 
 A x 9.7 = 8.8 x 2.2 + 13.1 x 4.5 - 22.3, 
 or A = + 5.7. 
 
 E x 10.4 = - 8.8 x 12.5 + 13.1 x 10.8 + 22.3, 
 or E + 5.1. 
 
 The strain diagram can then be commenced, as shown in dia- 
 gram (<?), Fig. X., and explained above. We can find by cal- 
 culation the flange H, and thus check our diagram. 
 
 Proceeding thus, we obtain the following table of strains : 
 
APPENDIX.] 
 
 THE BRACED AECH. 
 
 1 
 
 
 
 
 
 
 
 
 
 rH 
 
 rH 
 
 
 
 CO 
 
 00 
 
 CO 
 
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598 NOTE TO CHAP. XIV. [APPENDIX. 
 
 The strains in the present case are, we see, much greater 
 than for any of the others. Unless the maximum of stiffness 
 is essential, it would appear, then, undesirable to fix the arch at 
 the ends for an arch of the above dimensions. 
 
 27. Temperature Strains. The strains due to tempera- 
 ture are also very great. Thus, from Art. 165, we have 
 
 45 EIA et 
 
 H = 
 
 4 A A 2 + 45 I 
 
 and for the distance of the point of action of this thrust, below 
 the crown of the centre line, 
 
 _ (A a 2 + 6 I) h 
 ~ 3 A a* > 
 
 or since - = g 2 = 25 ft., 
 
 A 
 
 _ _ 
 
 ~ 
 
 _ h(a? 
 
 For A = 60 square in., a = 87.5 ft., h = 20 ft. = 240 in., 
 g = 60 in. = 5 ft., e = 0.000012, E = 14,000 tons, t = 30, 
 we have 
 
 H = 125 tons and e = 6.7 ft. 
 
 Hence we have the strains 
 
 A = 228, C = d= 29.0, E = =F 30.0, G = 118.5, 
 B = 112.5, D = =p 13.0, F = =F 65.0, H = 137.5, 
 
 12=^96.0, 34=^67.0, 5 6 =' T 29.5, 
 
 2 3 = d= 51.0, 4 5 = 39.5, 6 7 = 25.0, 7 8 
 Therefore the total strains are 
 
 A + 48o.4 -D + 318.3* /-, , -ic\-i K T\ , i K-I c\ 
 
 XI. 
 
 142.9 
 
 ~ - 43.9 
 
 
 " ' 
 
 E = 
 
 + 108.2 
 - 5.7 
 
 p + 185.4 
 ~ - 28.1 
 
 r + 242.1 
 
 ~ - 77.4 
 
 w + 284.7 
 ~- 88.4 
 
 12 = 
 
 + 103.8 
 - 167.4 
 
 2 o + 86.3 
 - - 63.4 
 
 34- + 72 ' 3 
 - 116.0 
 
 A K + 67.9 
 -45.3 
 
 5 6 = 
 
 + 46.2 
 -46.3 
 
 ft + 51.6 
 6 7 = _ 35.3 
 
 7 8 - + 22 ' 1 
 7 8 - _ 31.7 
 
 
 With the above we close our discussion of the braced arch. 
 Our design has been to illustrate the application of the for- 
 mulae and methods of Chapter XIY., and to show that by their 
 aid such a structure can be calculated with ease and certainty. 
 
APPENDIX.] THE BRACED AKCH. 399 
 
 In short, the difficulty is but little if any greater than for a 
 simple girder, only for a long span and many panels the work 
 becomes tedious and wearisome. 
 
 In such a case, perhaps the method of moments will be found 
 preferable to diagrams. Thus, for any condition of loading, we 
 can easily find the strains at certain given intervals or portions 
 of the span, as T Vth, T 2 0-ths, etc - These strains being plotted to 
 scale along the span, we have a curve, from which we can 
 readily determine the strain at other points. 
 
 The strains in the flanges being thus known, we can readily 
 determine the transverse force, or force at right angles to the 
 rib, at any point. This force causes strain in the diagonals, and 
 has simply to be multiplied by the secant of the angle made 
 with it by any diagonal. 
 
 As to the effects of temperature, the remarks of Art. 166 do 
 not seem to be substantiated by our results. It would seem 
 that, according to the received formulae, the strains due to tem- 
 perature are very great, and that by far the best form of con- 
 struction for short spans is that in which the arch is hinged at 
 both abutments and crown. 
 
 28. Advantage of Arch with fixed Ends for long Spans. 
 We cannot conclude from our results above anything as to the 
 comparative advantages or disadvantages of the arch with fixed 
 ends. Different proportions will give altogether different re- 
 sults. We can only say that for small spans the arch with 
 three hinges is undoubtedly the best construction. The ad- 
 vantages of continuity will be apparent only for long spans 
 where the point of inflexion is distant from the ends by a 
 greater proportion of the span. We have already seen the 
 same to be true of the continuous girder. If we were to judge 
 from comparisons of short spans only, we should be inclined 
 to discredit any great advantage for continuity. If, however, 
 we take longer spans, so as to bring the points of inflection well 
 out, we find a marked saving.* 
 
 We had intended to give here a comparison of the strains in 
 a hinged arch with those in the central span of the St. Louis 
 bridge, as given in the Report of Capt. Eads to the Illinois and 
 St. Louis Bridge Co. for May, 1868. 
 
 As this goes to press, however, our attention has been called 
 
 * Art. 17 of this Appendix. 
 
400 NOTE TO CHAP. XIV. [APPENDIX. 
 
 to an article in the Trans, of the Am. Soc. of Civil Eng. for 
 May, 1875, by Mr. S. H. Shreve, which, although written with 
 precisely the opposite intention, seems to prove so clearly the 
 superiority for long spans of the arch without hinges, that it is 
 unnecessary to give a comparison here. We have only to take 
 Mr. Shreve's results and properly interpret^ them. 
 
 Thus, while ostensibly investigating the strains in the centre 
 arch of the St. Louis bridge an arch which is continuous at 
 crown and fixed at the end Mr. Shreve uses the formula given 
 
 Q 
 
 in Art. 27 of the Supplement to Chap. XIV., viz., H = TTT' 
 
 That is, he considers the arch as having hinges at both crown 
 and ends. 
 
 Then, supposing the arch to be affected by temperature, he 
 applies the above formula to an arch hinged at crown in lower 
 chord and at ends in upper chord of the same dimensions as 
 the St. Louis bridge. It is hardly necessary to point out here 
 that if the arch is really thus hinged, or can be supposed thus 
 hinged, there can be no temperature strains. If, however, it is 
 not hinged, then the above formula does not apply. The one 
 
 assumption contradicts the other. The formula H = ~. can 
 
 2i fi 
 
 be applied to no arch which is strained by temperature. Such 
 a treatment would seem justified on Mr. Shreve's part in view 
 of the statement of Capt. Eads, that for the greatest rise of 
 temperature above the mean, the lower arch does all the duty 
 at crown, and the upper at the ends. If this were accurately 
 so, then Mr. Shreve's results would give the true strains. All 
 that Capt. Eads evidently intended to imply was, that a rise of 
 temperature relieved the upper chord at crown of a great part 
 of its compression and increased that of the lower. It does not 
 by any means follow that the upper chord is entirely relieved, 
 under which supposition only can the lower chord be supposed 
 hinged. On the contrary, for an equal fall of temperature 
 below the mean, the lower chord is relieved and extra strain 
 brought in the upper chord at crown. If the adjustment were 
 just such that the previous compression in the lower chord 
 should be exactly neutralized, then the arch might be consid- 
 ered as hinged at the upper flange and lower ends, and thus 
 Mr. Shreve should increase the rise of his arch by the depth, 
 
APPENDIX.] THE BRACED ARCH. 401 
 
 which would decrease greatly his strains. The one supposition 
 is as much justified by the remarks of Capt. Eads, which he 
 quotes, as the other and neither are correct. Apart, however, 
 from the merits of the controversy, with which we have nothing 
 to do, Mr. Shreve's results are undoubtedly correct for an arch 
 of the same dimensions as the St. Louis uniformly loaded and 
 hinged at the ends in upper flange and at the crown in lower. 
 If, then, a comparison of these results with those given by Capt. 
 Eads shows them all too large, then, since Capt. leads' formulae 
 are, as we have seen, undoubtedly correct, it clearly shows the 
 superiority of the arch without hinges. This is the only legiti- 
 mate deduction which can be made. 
 
 Mr. Shreve's formulae are undoubtedly as " true as the prin- 
 ciples of the lever," and apply, beyond question, to an arch 
 hinged as he supposes. Our formulae in Art. 27 of the Sup- 
 plement to Chap. XIY. are also as true as these principles ; but 
 to apply correctly even so simple a principle as that of the 
 lever, demands a knowledge of all the forces and their points 
 of application. From our formulae, as we have shown in Art. 
 34 of the above Supplement, we may easily deduce Capt. Eads', 
 thus proving the accuracy of both. Though the " calculus will 
 not determine the strains affecting a truss, whether arched or 
 horizontal," it may nevertheless be exceedingly serviceable in 
 determining the forces which act upon the truss without an 
 accurate knowledge of which the '"'principle of the lever" can 
 only mislead. This principle, upon which Mr. Shreve lays such 
 stress, is precisely that which we have employed so often in 
 this work, and s'hown to be of universal application. In Art. 
 36 of this Appendix we have made use of it, just as Mr. Shreve 
 does, in the calculation of an arch similar to the St. Louis. 
 Our results differ from those he would obtain, simply because 
 we take into account a force and lever arm whose existence he 
 ignores. Mr. Shreve assumes that V and H and the load are 
 all the forces which act, and these are all of which his formula 
 takes account. In common with Capt. Eads, we take in addition 
 a moment due to the continuity of the ends, while V and H them- 
 selves, by reason of this continuity, have very different values. 
 
 Thus, for full load, we have from eq. (81), Art. 34 of Sup- 
 plement to Chap. XIY., 
 
 u __pa? 4 A 2 
 
 ~ 2A 45 
 26 
 
402 NOTE TO CHAP. XIV. [APPENDIX. 
 
 and from eq. (84) 
 
 M fl - - 
 
 70 
 
 Thus, instead of H = ^- . , as given by Mr. Shreve, we have 
 
 this into a certain coefficient which is less than unity. 
 
 Taking pa = 936,000 Ibs., g = 6.025 ft., a = 257.88 ft, 
 and h 46.65 ft. for centre line, we have H = 2,178,317 for 
 thrust at crown, instead of 2,586,184.9 Ibs., as given by Mr. 
 Shreve. This thrust alone would cause, then, 1,089,158 Ibs. 
 compression in each flange. But due to continuity of ends and 
 crown, we have also a moment at crown M 6,587,335, 
 which being negative causes tension in lower flange at crown. 
 Dividing by 12.05, the depth of arch, we have 546,666 Ibs. 
 tension, and therefore only 1,089,158546,666 or 542,492 Ibs. 
 resulting compression. This at 27,500 Ibs. per square inch, 
 requires 19.72 square inches area, while Mr. Shreve requires 
 in his arch 126.42 square inches area. It is, however, but just 
 to notice, that while this loading (uniform) causes the maxi- 
 mum compression in lower flange at crown for Mr. Shreve's 
 arch, it does not for the arch fixed at ends and continuous at 
 crown. 
 
 In this latter case, as we may see at once from the table for 
 M of Art. 18, Supplement to Chap. XIV., a load within the 
 centre half anywhere causes tension in the lower flange, and 
 the maximum compression is when the flanks are loaded and 
 this portion is empty. It is with the maxima that the com- 
 parison must be made, and as Capt. Eads has, very properly, 
 taken the rolling load into account, it is with these maxima 
 that the comparison has been made. From such comparison 
 Mr. Shreve h'nds that " every member of the two tubes is 
 deficient in area, many containing much less than half the 
 material that is necessary." As his results are correctly calcu- 
 lated for a hinged arch, and Capt. Eads r results are also correct 
 for an arch without hinges, we can only conclude not that 
 " the great importance of immediately strengthening the ribs 
 of the St. Louis bridge can no longer be ignored," but rather 
 that, for long spans of small relative rise, the arch without 
 hinges is much preferable and more economical. The case 
 
APPENDIX.] THE BRACED ARCH. 403 
 
 is, indeed, perfectly analogous to that of the continuous girder. 
 Here also we have end moments, and here also for long spans 
 the advantage over the simple girder is marked. 
 
 In Mr. Shreve's arch it is, indeed, perfectly true that, " when 
 one segment is loaded, any weight whatever in any other 
 position on the other segment will lessen the tension on the 
 lower arc of the loaded segment." In the arch without hinges 
 the case is altogether different, owing to the influence of the 
 end moments, which Mr. Shreve so persistently ignores. 
 
 The two cases have, indeed, nothing whatever in common, 
 and from the strains in one no conclusion whatever can be 
 drawn as to what should be the strains in the other. ' With the 
 same propriety might one comparing the strains in the same 
 girder fixed at ends and free at ends, as given in Art. 17 of 
 this Appendix, infer that the strains in the first were unduly 
 small. The only legitimate conclusion from such comparison 
 is the one there drawn, viz., that the one in which the strains 
 are least is the one most economical of material. In this 
 respect, and in this only, Mr. Shreve's results are valuable, and 
 we can only thank him for having saved us the labor of 
 making the comparison for ourselves. 
 
 As a case in point bearing out our conclusion above, we 
 may instance the ' Coblentz bridge, which, as originally con- 
 structed, was continuous at the crown, but pivoted at the ends 
 of the centre line, as in our example, Art. 20. But unlike 
 that example, owing to the length of span being much greater, 
 and the rise and depth much less in proportion, it was found 
 advantageous to block up the ends after erection, and thus fix 
 it at the ends. 
 
 If Mr. Shreve's deductions are to be believed, this was a 
 very dangerous thing to do; but, as experience has proved, 
 greater rigidity has thereby been secured, and no evil effects 
 have as yet been perceptible. It is, however, quite possible 
 that before thus blocking the ends, the effect of the end 
 moments thus brought into play was duly considered; and in 
 view of the result, it would appear as if they really had some 
 influence upon the character and distribution of the strains. 
 
 It would seem, therefore, that, for the present at least, the 
 " strengthening " of the arches of the St. Louis bridge by hinging 
 them (!) at crown and ends may be safely postponed until it 
 can be satisfactorily shown in what manner, for rise of tern- 
 
4:04 NOTE TO, CHAP. XIV. [APPENDIX. 
 
 perature, the end moments mysteriously disappear, and the 
 previously existing compression, due to load in upper flange at 
 crown and lower at ends, is exactly and entirely neutralized. 
 
 Meanwhile it would seem that the St. Louis arch, as con- 
 structed, is far superior to the same arch hinged, more eco- 
 nomical of material and more rigid, and sanctioned alike by 
 theory and precedent. 
 
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