Pe hig Se eee eat oe ote: SS ae 2 i Si Gs oe NM 3 1924 080 782 216 In compliance with current copyright law, Cormell University Library produced this replacement volume on paper that meets the ANSI Standard Z39.48-1992 to replace the irreparably deteriorated original. 1997 Theory of Arches and Suspension Bridges BY J. MELAN (Kaiserlicher-Koniglicher Hofrat) Professor of Bridge Design at the German Technical Schoo] at Prague & AUTHORIZED TRANSLATION BY D. B. STEINMAN, C. E., Ph. D. Professor of Civil Engineering at the University of Idaho. Author of ‘‘Suspension Bridges and Cantilevers’’ CHICAGO THE MYRON C. CLARK PUBLISHING CO. LONDON r.& FN! Spon, Lrpv., 57 Haymarket 1913 w COPYRIGHT, 1913 BY THE Myron C. CLARK PUBLISHING Co, TRANSLATOR’S PREFACE. TIMULATED by a profound admiration for the work of Professor Melan and prompted by a desire to fill a recog- nized gap in American engineering literature, the writer has undertaken the translation of this book on the Theory of Arches and Suspension Bridges. No other work of the same scope was to be found in any language, at least none that could compare with Melan’s for thoroughness of treatment and masterly style. The place it has been given in the offices of consulting engineers and bridge departments, the frequent references to it in our technical literature and the employment of its formule or methods in the design of some of our largest structures indi- eate that, even before translation, there has been a real de- mand for Melan’s book among American engineers. The work has been enthusiastically received in Europe where it has al- ready gone through three editions and the highest honors have been awarded the author. In order to widen the sphere of use- fulness of the book and to render it accessible to the entire pro- fession in this country, the writer has been encouraged to give the time and effort required for this translation. Professor Melan’s book is here faithfully reproduced with- out any omissions. An appendix devoted especially to the de- sign of masonry and reinforced concrete arches and a very ex- haustive bibliography on arches and suspension bridges are included. The translator has carefully checked the derivation of all the formule and has corrected the few typographical errors found in the original. In some of the examples and tables the quantities have been converted from metric to English units. The notation has been altered wherever necessary to eliminate the German characters and abbreviations and to conform more closely to our standard symbols, and corresponding changes have been made in the figures and plates. In a few places brief explanatory notes have been inserted to help the American reader who, on account of the difference in training, may not be as familiar with the analytical and graphical devices em- ployed or as agile in passing from formula to formula as the German student. With the exception of the above minor changes the original work has been kept intact. In addition to its use as a work of reference by those en- gaged in the design of higher structures, the writer believes that this volume will be found admirably adapted as a text- book for advanced or post-graduate students in structural engi- Ill Iv ARCHES AND SUSPENSION BRIDGES. neering or applied mathematics and, from personal experience, he strongly recommends it for this purpose. Used in this way, the book will prove a liberal education in itself, rendering clear the fundamental principles of the analysis of structures, familiarizing the student with many helpful analytical and graphical devices of general application, training him in the independent derivation of formule and preparing him to do original work in structural theory and design. The writer wishes to thank his friends in the engineering profession who have encouraged him to undertake this task, particularly Professor W. H. Burr and Dr. Myron 8. Falk. D. B.S. University of Idaho, November 1, 1912. FOREWORD BY Horrat J. MEuan. HE present work, produced in English with the consent of the author, comprises the Theory of Arches and Suspension Bridges as it appears in the third enlarged edition of the “Handbuch der Ingenieurwissenschaften,’’ Volume 2, Part 5, 1906. The design of arches and suspension structures is here thor- oughly and exhaustively developed, both the analytical and graphic procedures being given in conformity with modern prac- tice. All types of arches and suspension bridges, of any practical importance, are considered. The principles of the exact theory, taking into account the deformations produced by the loading in the case of lightly stiffened structures, are also developed (§ 9 and § 28). The author can but express his hearty satisfaction and pleasure at this undertaking by a professional colleague to translate his work into English and thus render it accessible to the engineers of America. He hopes that the work will be found adapted for use in teaching this branch of bridge design and that the Ameri- can bridge engineers, famous for the magnitude and skill of their achievements, will judge it favorably and find it useful for refer- ence in theoretical problems. Prague, Sept. 20, 1912. Cotta ee ee k, k. Hofrat, Professor des Briickenbaues an der deutschen technischen Hochschule in Prag, on CONTENTS PAGE TmtTOdUCon ice scene sine cmadcnsecie ius au See Watnee ae Mleds ew Godan 1 General Method of Design........... 00. cece eee e eee cece eens 8 A. The Flexible Arch and the Unstiffened Cable. The Funicular Polygon for Given Vertical Loading............ 10 The Unstiffened Suspension Bridge................. 00.0000. 14 1. Form of the Cable and Value of the Horizontal Tension. 14 2. Maximum Attainable Span...................0....0.. 17 3. Economic Ratio of Rise to Span............. 0... e eee 18 4, Deformations: scnecceces os nex ter sessetats Gear seen 19 5: Secondary Stresses: 2 wicsc. eal ecn sie wee sree ween sas we cou 2S B. The Stiffened Suspension Bridge. Approximate Theory sasccids ovederescas ees evrweea ved. veux 25 1. Conditions for Equilibrium of the Funicular Polygon... 25 2. Forces Acting on the Stiffening Truss ................ 26 3. Determination of the Horizontal Tension H........... 29 Values of H for Special Cases of Loading.................0.. 45 1. Stiffening Truss with Central Hinge ................ 45 2. Stiffening Truss without Central Hinge ............. 45 Stresses in the Stiffening Truss ..............00. 00000 e eee 47 J. General Principles: ccc iiscceeneeeen eae eireccegs was 47 2. Reaction Locus. Critical Loading ..............-.. 48 3. Determination of Moments for Full Loading and for the Criti@al, LO@dinge oo iesitiieedeeisseccaiae a Give eurieaace aie see 53 4. Determination of the Maximum Shears .............. 58 5. Effect of Temperature Variations .................. 63 6. Secondary Stresses 24 sss cadeneessersatetecoawnn waa 67 Computation of the Deflections ....................-..55 000s 69 1. Deflections due to Loading ..................... 0058- 69 2. Deflections Produced by Temperature Variations or by Displacements of the Cable Supports ................ 75 The More Exact Theory for the Stiffened Suspension Bridge... 76 1. Truss without a Middle Hinge. Single ve eaieaes tase 16 2. Truss with a Middle Hinge........... Icetdemeses: OF VII Vill § 11. § 13. tn Or § 17. 15. 16. ARCHES AND SUSPENSION BRIDGES. C. The Arched Rib. PAGE Internal Stresses in Curved Ribs................22.004. cea 80 1. Determination of the Normal Stress (c).............. 88 2. Determination of the Radial Shearing Stress (7,)...... 92 Conditions for Stability. Line of Resistance. Core-Points. Graphie Determination of Fiber Stresses............. .... 95 Determination of the Deformations..................4.. 05. 98 1. Elongation of the Axis........... ccc cece cee ees cee 98 2. Variation of the Angle @ ..........-... ese -s ose 98 3. Variation of the Radius of Curvature............. .... 98 4. Variation of the Codrdinates.................5-. 000: 99 External Forces of the Arched Rib...................0. 0005 102 Reaction Locus and Tangent Curves. Critical Loading... .... 105 1. Reaction Locus and Tangent Curves.............. 0... 105 2. Laws of Loading for the Normal Stresses......... .... 107 3. Laws of Loading for the Shears (S)............5. 004. 109 1. The Three-Hinged Arch. Pixtertal BOrces: s s.ecsae sineceiels oes Gardin) Row apheaed Suara dene ale Go Dee 110 DGLORNIB TIONS: 24inive vaaains canara Me bioun a aun wine ecnteon s Matetea a thar 114 1. Effect of Temperature Variation and of a Horizontal Displacement of the Abutments................ 0... 114 2. Deformations due to Loading................0.08 c0e 114 2. Arched Rib with End Hinges. Determination of the Horizontal Thrust................0.... 120 We General Case: cz, cexices cise quate aie uae Se a Bee IN ee Rte 120 2. Flat Parabolic Arch with Parallel Flanges.............. 127 3. Plate Arch with Parabolic Intrados............... 2... 129 4. Segmental Arch with Uniform Cross-Section........ .... 130 Maximum Moments and Shears.............. 00200022 e cee 133 1. General Case. Graphic Method ................. 0... 133 2. Parabolic Arch with Variable Moment of Inertia... .... 134 3. Flat Parabolie Arch with Parallel Flanges............ 135 4. Segmental Arch with Uniform Cross-Section........... 137 5. Temperature Stresses ........ 0.0... c ccc e vee 139 Deformations: 16-s.58gc aera nea nei wes aeeeeriaads ee% 141 3. Arched Rib without Hinges. Determination of the Reactions..................00. 000.00 e ee 147 1. General Case. Concentrated Load .................... 147 2. Simplification with Constant Moment of Inertia, I’ ..... 154 3. TMlat Parabolic Arch with Constant Moment of Inertia... 157 4. Effect of Temperature Variation and of a Displacement Of. ‘the Abmtiients:. 2 cs. owas 2 caces ea cax dears ews eau oo 159 5. Arch Connected to Elastic Piers...................... 161 CONTENTS. IX PAGE 6. Segmental Arch with Constant Moment of Inertia....... 163 § 21. Maximum Moments and Shears........... 00.0002 e eee ees 168 1. General Case. Constant Moment of Inertia............ 169 2. lat: Parabolie Areh. a4 5 gnc acgaie co gate elas sae eee te 170 30 “Segmental Areh .. 2, but will be subject to deformation under moving loads. There are s + 4 unknown quantities, consisting of the stresses in the bars and the components of the end reactions; while the requirements of static equilibrium at the k==s-+1 panel points furnish 2 (s +1) equations of condition. Consequently, for s > 2, i. e., for a poly- gonal frame containing more than two bars, the number of con- ditional equations exceeds the number of unknowns. The super- fluous equations can not, in general, be satisfied except by special relations between the constants, corresponding to a definite adjustment of the geometric form of the system to each partic- ular disposition of the loading. This adjustment of form to Fig. 2. [>] TITATATON COOLIO TTT ECO sta=174+5=22, 38+2kh=3x14+2x9=21, varying loads can actually occur only in the case of the sus- pended polygon; while the arch-polygon can merely assume a position of unstable equilibrium, and will collapse upon the first change in the loading on account of the altered configura- tion required for equilibrium. Arch-polygons, by themselves, are therefore unsuitable for use as bridges; they require, in addition, some type of stiffening construction to preserve their form, thereby converting them into rigid systems. A similar arrangement also becomes neces- sary in suspension bridges if it is desired to reduce the deformations of these structures. In the unstiffened suspension Fig. 3. ff | ee FT TTT NY AY sta=114+3=14, 39+2h=3x142x5=13, bridges and arches, with loads applied at the panel points of the funicular polygon, the only stresses will be those of pure tension or compression, respectively; in the stiffened struc- tures, however, subjected to changing load, bending stresses will also appear. The stiffening construction for flexible arches or suspension systems may consist in their connection with a straight truss (Figs. 2 and 3) whose elastic deflections will limit, in magnitude, the static deformations of the funicular polygon. INTRODUCTION. 3 . Instead of using this construction, the arch or suspension poly- gon may be made rigid in itself and thereby rendered capable of supporting any loading, even if the resultant forces do not coincide with the axes of the structural parts. Such a rigid system may be conceived as derived from the funicular polygon Fig. 4. sta=154+7=22, 3§+2h=3x14+2x9=21, by replacing the hinged joints between the successive bars by rigid connections, and giving these bars such sections as will enable them to resist bending moments. The funicular polygon (Fig. 1) is thus replaced by a single curved rib, which may con- sist of either a girder or truss construction (Fig. 4). Since the polygonal frame with hinged ends, hence with four reaction unknowns, constitutes a non-deformable construc- tion when the number of bars is reduced to two, the union of two ribs through an intermediate hinge likewise produces a stable bridge structure (Fig. 5). To this system other parts may be linked, according to the number of new reaction un- knowns thereby contributed. We thus derive various types of stable bridges (Figs. 5 to 8) which are classed as suspension bridges or arches whenever, as specified at the outset, they produce oblique reactions at the ends of the principal curved or polygonal portion of the struc- ture; these reactions may be taken up either by the abutments or by a tie-rod construction (Fig. 3). For determining the stresses in these structures, the equa- tions of static equilibrium may or may not suffice. In the former case we have statically determinate systems, in the latter statically indeterminate. This distinction is to be drawn par- ticularly with reference to the unknown external forces; any 4 ARCHES AND SUSPENSION BRIDGES. bridge type can easily be classified as to its external deter- . minateness or indeterminateness if we conceive it as built up of bars, or of bars and ribs, and compare the number of un- known forces with the number of independent equations of condition available. The pivoted connection of two ribs is called a hinge, and the similar connection of two or more bars is called a panel-point. Each rib furnishes three equations of equilibrium for its applied forces, and each panel-point fur- nishes two such equations. The unknowns include first the reac- tion components whose number is given by the rule: one unknown for each free end, i. e., each end capable of free motion on sliding plates or rollers; two for each hinged end, i. e., each end capable of rotation but not of translation; and three for each fixed end, rigidly anchored. Similarly each intermediate hinge represents two unknowns and each intermediate sliding joint one; and the stresses-in the bars are additional unknowns. Therefore, if a structure consists of S ribs with G hinges and of s bars with k panel-points, and if a is the number of reaction unknowns, the total number of unknowns is a + s + 2 G and the number of corresponding equations of equilibrium is Fig. 7. at2@=64+2x3=12, 39=3x4=12, 3 8 + 2k. For all combinations in which the external forces are statically determinate, a +s+2G=3 8 + 2 k; and for those statically indeterminate,a+s+2G>3S8 42k. We thus find that such systems as shown in Figures 2 and 3 belong to the statically indeterminate class, but that they may be rendered statically determinate with reference to the forces affecting the component structures (polygon and stiffening truss) by introducing an intermediate hinge in the stiffening truss. The same applies to the arch rib with end-hinges (Fig. 4), which also becomes statically determinate when a center-hinge is introduced, making it a three-hinged arch (Fig. 5). In this case, however, the same result may be obtained by using some INTRODUCTION. 5 weighting device to keep the horizontal component of the end- reactions at a constant value; or by changing one of the end- hinges to a free end (sliding on an inclined plane) so as to diminish the number of reaction unknowns by one (Fig. 9). The corresponding reaction will then act constantly normal to the plane of the rollers, so that the reaction at the other end may be determined by simple resolution of forces. This arrange- at+s+2@=648+4+2=16,39+2k=3x4+2x2=16, ment, however, is not classed among true arches, but is consid- ered simply a girder whose expansion end slides on inclined bearings. In general, subdividing any rib by the insertion of a hinge will diminish the relative number of unknowns by one. Thus the arrangement in Fig. 6, which is three-fold statically indeter- minate, may be rendered determinate by introducing three hinges. On the other hand, the statically determinate structure in Fig. 7 may be rendered doubly indeterminate by omitting the hinges in the two side-spans. Not all combinations which may thus be formed, however, are practical for construction. Notwithstanding the static Fig. 9. determinateness or even indeterminateness of the entire com- bination, there may be so great a degree of mobility in the individual parts or in the structure as a whole that its rigidity may be inadequate for practical requirements. This will be the case in the three-hinged arch (Fig. 5), for example, if the two end-hinges and the center-hinge lie nearly or exactly in a straight line; or, in general, whenever any one of the com- ponent parts of the structure is unrestrained or imperfectly restrained from rotation (or translation = rotation about an infinitely distant point). To make this clear, it is necessary to consider the structure as a kinematic chain and to enlarge the concept of the’ hinge. 6 ARCHES AND SUSPENSION BRIDGES. Ina plane rib abc (Fig. 10), conceived as a kinematic element, let two points a and 0 be constrained to move along definite paths. The intersection o of the normals to these paths consti- tutes the instantaneous center of rotation, having the effect of a fixed pivot. The rib will not be restrained from rotating about this point o unless a third point c in the rib is either fixed or constrained to move along a path whose normal does not pass Fig. 10. through o. This requirement is not satisfied, for example, in the case of the middle rib of Fig. 11, since it has but a single instantaneous center 0; this arrangement, therefore, although satisfying the general criterion for static determinateness, is not stable and cannot be used for a bridge. It is similarly apparent that the arrangement in Fig. 12 is not stable if the center-hinge c lies in the line joining the two centers 0 0,; the Fig. 12. closer c comes to the line 0 0,, the less is the rigidity of the structure and the greater will be the deformations. In accordance with the generalized concept just developed, the hinge-joint between two ribs may be replaced by a pair of pin-ended links forming a ‘‘hinged quadrilateral.’’ The inter- section of the axes of the links determines the center of rotation, i. e., the virtual pivot taking the place of the fixed hinge. Thus the arrangement in Fig. 13 represents a statically determinate three-hinged arch; but it will not be a stable system if the two links ad and 6c intersect in the line joining the two end-hinges or are parallel to this line. By the insertion of another bar in the ‘‘hinged quadrilateral’ (e. g., ac in Fig. 13), the hinge effect is suspended and the structure Fig. 13 is changed to a singly indeterminate two-hinged arch. INTRODUCTION. re The end-hinges, also, may be replaced in their action by pairs of links (Fig. 14). By adding a third link, rotation is prevented and the ends become anchored. Thus with the bars bc and fg provided, Fig. 14 represents a hingeless arch and the degree of indetermination is raised to three. The individual ribs composing the structure may either be made solid, i. e., of plate-girder form, or framed of separate members like a truss. If the external forces acting on each rib are known, then the internal forces, i. e., the stresses in the dif- ferent members of the rib, can be determined. For this pur- pose the methods of pure statics will suffice in the case of statically determinate truss systems, but for indeterminate forms, including solid plate girders, the elastic deformations must be taken into account. For criteria of static determinate- ness or indeterminateness of trusses, the reader is referred to other works on the subject. 8 ARCHES AND SUSPENSION BRIDGES. § 2. General Method of Design. The fundamental prin- ciples of design of ordinary trusses or girders also apply to arches and suspension bridges. As in the simpler structures, it is necessary to first determine the external forces or reactions produced by a specified loading. Knowing these, we can proceed to evaluate the stresses within the structure. For the solid-web portions, we apply the general theory of flexure of curved ribs which, as an approximation, may be reduced to the simpler theory of straight beams; for the open-web or framed portions, the ordinary rules and methods of truss design are applicable. These methods, as is well known, assume an ideal framework in which the external forces are applied and the members meet one another only at their end-points, and all necessary rotation about these points may take place without any resistance,—assump- tions which, as a rule, are not fulfilled in structures as built, rendering it necessary to estimate the magnitude of the devia- tions or so-called secondary stresses. It should also be noted here that the results of the theory deduced for solid ribs may often be extended to structures with latticed or trussed webs. If the structure is statically determinate with reference to the external forces, the ordinary rules for the composition and resolution of coplanar forces suffice to definitely determine the end reactions and, in determinate truss systems, the stresses in the members as well. In statically indeterminate arrangements, the missing equations of condition must be deduced from the displacements of the points of application of the external forces, i. e., from the elastic deformations of the structure. To estab- lish these equations, we may use the ‘‘Theorem of Virtual Dis- placements,’’ first applied to the design of indeterminate struc- tures by Mohr. The same equations may also be derived, and sometimes more conveniently, by applying the ‘‘Theorem of Least Work’’ established by Castigliano and Frankel: In any elastic system in a condition of equilibrium, such stresses must appear as will make the total work of deformation a minimum. Consequently, if this work of deformation is expressed as a func- tion of the indeterminate stresses or reactions, the differential coefficient of the function with respect to these unknowns, equated to zero as required for a minimum, will give the equa- tions of condition necessary for the determination of these unknowns. In place of the analytical treatment, most of the problems of this class may also be solved by graphic methods. For statically indeterminate structures, the graphic process consists in drawing one or more deformation polygons (Williot’s dis- GENERAL MEtTHop oF DESIGN. : 9 placement diagram or deflection polygon), enabling the indeter- minate quantities to be found. But even if the latter are deter- mined analytically, graphic methods may still be used to advan- tage for arches and suspension bridges, especially in determin- ing the internal stresses. There remains to be noted that in evaluating the work of deformation or in drawing the displacement diagram, the opera- tions may be simplified within the limits of permissible approx- imation by neglecting those elements, as the web members of a truss or the shears in a girder, which have a negligible influ- ence on the total work of deformation. 10 ARCHES AND SUSPENSION BRIDGES, A. The Flexible Arch and the Unstiffened Cable. §3. The Funicular Polygon for Given Vertical Loading. If concentrated vertical loads are applied on a cord, fastened at its ends and considered weightless, it will assume a definite polygonal form dependent upon the relations between the loads. If A denotes the vertical component of the tension at one of the supports, and if P,...P, are the applied loads as far as the m™ side of the polygon, the vertical component of the stress in this side will be Vn=A— P, —P,...—Prm, while the horizontal component H will be the same for all the Fig. 15. sides. The angle 7, between the m™ side and the horizontal is given by A — Py —P....— Pm Vm tan ™ = FP oa, Se a gts salad (1. and the tension in the cord by Pin V Vint B= FA sO tin icc ec ccc cece eee ee (2. If gm, Ym and 2m.,, Ymi, Genote the coordinates of the vertices adjoining the m™ side, referred to the point A (Fig. 15) as origin, we also have A—P,—P,...—Pm 6 i i) eo FunicuLar PouyGon ror VERTICAL LOADING. 11 A —P,...— Pm-1— Pm or A Ym = A, In like manner we obtain for the side preceding —— ia ay po ex AYn-1= a a a 7 ALm-1. If we write AYn— Ay m-1—— Ym+1 — 2Ym + Yyn- 1=A’Yn and if Atm—= Atm..— 4, i. e, if the horizontal spacing of the vertices of the polygon is uniform, then we obtain AS ileion Eid atau acaaenea yeatetes (3° In general we have (tan tm — tan tm-,) = — ce ide ceawaeea tn (3° The funicular polygon for the loads P and the horizontal tension H may also be determined graphically in the well-known manner (Fig. 15). If the points of suspension of the cord lie in a horizontal line, the force A is the same as the reaction of a simply supported beam and is found from the moment of the external loads: a ee (4. In this case, if f is the versine of the funicular polygon and r the lowest vertex, we have, from the equation of moments, H—™ __ Aar—P, (ar —2,)—P, (ae) a Pra (er @r= 1) re (5. If the versine f is not given, but the length of the cord in- stead; and if, in addition, the points of attachment of the suspended loads 1, 2... and, thereby, the lengths e,¢,e, .. of the individual sides of the polygon are fixed, then the horizontal tension is given by the relation: e e; I €o 1 2 Vrpat V H+ (A—P,)? + ‘Vira aba be gee It is readily seen that the stresses T in the successive members of the polygon increase toward the points of support and attain their maximum values in the first and last members of the system; also that for downward acting loads the stresses in a polygon convex downward will be purely those of tension, and in a polygon convex upward they will be purely those of com- pression. If the loads are continuously distributed, the funicular polygon becomes a continuous curve. If gq is the load per 12 ARCHES AND SUSPENSION BRIDGES. horizontal linear unit at any point having the abscissa x, equ. (3) becomes ay —4= Sade gy from which (by differentiation) we obtain the following as the differential equation of the equilibrium curve: ay = A Gah eet eee eee eee (7. If + is the radius of curvature of the equilibrium curve, 2, ot == = sec*r. Substituting this in the last equation (7), we obtain, with the aid of equ. (2), EE SQ COST a ba eis ah AUR es RRA Save (8. and, for a load q, and radius of curvature r, at the crown of the curve, For a uniformly distributed load, 1. e., for a constant q, if we take the origin of coordinates at the crown, the integration of equation (7) will give Hence the equilibrium curve will, in this case, be a parabola. If lis the span and f the rise (versine), then The largest stress in the cord will be Tmax = Vi + (Fai) ey 1+(4).. «as. If the load is not constant per horizontal unit, but per unit length of the cord, then the equilibrium curve takes the form of the common catenary. In that case, with the origin of co. ordinates at the crown of the curve, equ. (7) becomes HF4——g secr——g (1 +(34))*. 5 ag 1 From this, with os — as the parameter of the catenary, we obtain the following equation for the curve: Funicutar PoutyGon ror VERTICAL LOADING. 13 mgm (OP O-F HD) cyst eset eens (14. The cord length s of the catenary is given by = + Vicy poy Es Aihara aa (15, The foregoing equations for H apply also to an obliquely suspended cord if the ordinates y are measured from the arch- chord or line joining the points of support. If the funicular polygon is carried over an intermediate sup- porting pier (Fig. 15a) in such a manner as to leave it free either Fig. 15a. --- oe tee gence een - eee to slide itself or to displace the point of support, then the cord will assume a position of equilibrium yielding the same value of H on both sides of the pier. If the load is uniformly distributed per horizontal unit of length, with the intensity q in one span and q’ in the other, the equalizing of the horizontal tensions given by equ. (11) will impose the following relation between the versines of the two parabolas: ft 14 ARCHES AND SUSPENSION BrinGES. §4. The Unstiffened Suspension Bridge. 1. Form of the Cable and Value of the Horizontal Ten- sion. Let q, denote the weight per horizontal linear unit of the roadway suspended from the cable together with the assumed uniformly distributed live load. a.) Chain-cable bridges. We make the assumption, near enough to the truth for all practical conditions, that the weight of the chain is so distributed as if its curve were a parabola and its cross-section at every point proportional to the stresses under maximum loading. Let g, be the weight per unit length of the chain at the crown; then, at a distance x from the crown, the weight of the chain per horizontal linear unit will be Gx = Go Sect. Substituting the following value for the parabola of equ. (12), Bes yt dy __ 64 fa? sectr—=1 + (4% —14 90, there results 64 Pr 2 g.= go(1+ a ). Let the weight of the suspension rods per horizontal linear unit ‘ a fe Pe Then, transferring the origin of coordinates to the crown of the curve, equ. (7) becomes, Gy 647 | . 4f HS = d+ G+ (90 Go + IGF) 2*. Introducing the abbreviations ‘ Jo + Jo= Y 2 f f : a4 (69. +i)=", then the double integration of the above differential equation will yield Se eee nee (17. Substituting in this equation of the curve the coordinates of one of its points, namely z= + and y = f, we obtain Tue UNSTIFFENED SUSPENSION BRIDGE. 15 vc ? H=s(a+rz deacctaccmnelss erations (18. Replacing the value of r from equ. (16), we have the fol- lowing expression for the horizontal tension: H=g(1+- 5 )g +(o+ti)e as If A, is the cross-section of the chain at the crown in sq. in., y its specific weight in lbs. per cu. ft., s its stress under severest loading (in lbs. per sq. in.), then gy==C.Ap. lbs. p. 1. f.) v eg | and A, =. Here C is a coefficient representing the increase in weight allowed for the details of construction such as eye-bar heads, pins, etc., and may be assumed equal to 1.15 to 1.20. Substituting the above relations in equ. (187)and solving for A,, we find tes p got ei A, == tet Lye yeae CIS: S~ yaa (arp 7 3 2 The weight of the semi-cable will be G = f gx. dx, or = Ao-y 1 16 f? G=C. 4750454 aiid (20. The vertical component of the end-reaction will then be 1 ees Vay Gy tag de fabs Gesceeisaverei shes (21. and the greatest tension in the chain is given by Tax = V V? + H? aid vrs ho ieee aay Sc Reh Ee eee (22. The cross-section of the chain at the points of support must therefore be At these points, the angle which the cable makes with the horizontal is determined by tan 7 = ae On approximately, epee ie. 8 tant =— G+ if Bo Gr xn e ecw news (28. b.) Wire-cable bridges. A wire cable has a uniform eross- 16 ARCHES AND SUSPENSION BrinGEs. section throughout. If g is the weight of the cable per linear unit (p. 1. f.), the horizontal projection of this weight will give gx =9.SeCT or, assuming a parabolic curve for the cable, oe rz 9x = 9 . 8€C T= av 1+ The weight of the suspension rods ii be assumed as above. On integrating the differential equation of the equilibrium curve, we obtain xv Ysa (G+ Got £0?) nee cinisw ew deweeee ne (24. where pao EO Sat fee eenacncate ones (25. The horizontal tension will be 4 Noe ae GQ+ 3 3 +) a7 t+ (a+ si)ay er (26. The weight of the half-cable will be approximately @=g¢(14+44) bestta Mush actobee leaahitetrise (27. The values of V and Jmax are given by formulae (21) and (22). The angle which the cable at the tower makes with the horizontal is obtained from tant =~, or approximately, 4f 47 tan r =—! f(a ae aoa) eesoetiges (28. The constant cross-section of the cable is determined by A= — = 2 sect. With the value of H from equ. (26) and with g = a it , there results PA a 6 j) as 7 (29. $s is (4 +F =f) secr The length of the chain or cable will be given by re —2{ 2+ wy G [= (34)+.-J}s 0 Tue UNSTIFFENED SUSPENSION BRIDGE. 1? where oe Sub- stituting their ae we obtain ss i (g +r 3) te ee ae) Or (380. [+ J] For small values of r, i. e., fo a a ratio of rise to span, we may write the following expression for the cable-length, ap- plicable also to flat parabolic curves: de arabe snide: (31. 2. Maximum Attainable Span. The theoretical limit of span attainable with a suspension bridge is determined by the condition that the cable shall have a finite cross-section, hence that the denominator of equ. (19) or (29) must remain positive. This criterion becomes for chain-cables, 1 ay max < H 8 f ee oe for wire-cables, wenger G32: 1152+. ¥ (F434) Vitis We thus obtain the following theoretic values of the maximum spans for the given working stresses and rise-ratios: Mare Pr ul a2 jy 4 1 as. 7 8 10 2 7a 16 lbs./ft.* — lbs./in.? tt. ft. ft. ft. ft. . = s = 12000] 2960| 2403] 2020] 1740] 1527 Chain-Cables....| Cy = 560) § _ yg000 | 4440] 3605| 3030] 2610] 2290 ; s = 45000 | 11580) 9700! 8290] 7220] 6380 Wire-Cables ....| y= 490) 5 __ g0000 15440] 12933] 11053] 9627 | 8507 In the above spans, if the cable is to have a finite cross-section, it must carry no other load than its own dead weight. The above solution, consequently, does not give the maximum practicable span for suspension bridges which have to carry a definite use- ful load in addition to the weight of the cable. We may, how- ever, calculate the span lJ for a cable of the maximum practicable cross-section when the rise-ratio, working stress and applied load 18 ARCHES AND SUSPENSION’ BRIDGES. are specified. If the weight of the cable per unit length —g, and if we put g’=g (1 + = + , also if g, denotes the suspended load p. 1. f., then we have ea. g Ue Dans apne cant eas eietoni areas (33. It is important to note here that q, is not quite independent of the span-length, but increases with it on account of the weight of the stiffening construction and the wind-bracing. Note:—The problem of the maximum practicable span for suspension bridges, arising in connection with the projects for bridging the North River, was investigated by a special commission of U. 8. army officers. Assuming 16 cables, each containing 6000 steel wires, No, 3,B. W. G., with a total cross-section of 5058 sq. in. and weighing g=17,200 lbs, p. 1. f.; also adopting a working stress of 60,000 Ibs, per sq. in., a live load consisting 27,540,000 l 27,764,726 + 3.24906 1+ 0,00055335 7? + 0,000000003 7, there is obtained, for a rise-ratio of 1:8, the value of 14,335 ft. for the limiting span.— (Report of Board of Engineer Officers as to “the maximum length of span practicable for suspension bridges’: Major Chas, W. Ray- mond, Captain W. H. Bixby, Captain Edward Burr; Washington, Govern- ment Printing Office, 1894.) of a six-track railway giving p= total load of go = 13605 + Ibs, p. 1, f, and a corresponding 3. Economic Ratio of Rise to Span. The rise (or versine) of the cable affects, on the one hand, its own weight and that of the suspension rods and, on the other hand, the cost of the towers. If we overlook the effect on the backstays and on the masonry of the anchorage, we may determine approximately the rise-ratio which will make the total cost a minimum. De- noting this ratio by I =n, the ratio of the cost of the towers per foot of height to the price per pound of steel construction by P, and using the abbreviation a. =e, then, with the aid of equ. (20) and a small admissible simplification, there is obtained the equation of condition 16 (14+ v) : oO ot +7 en Ql? + Pnl=min. Differentiating and solving for n, and, for abbreviation, put- ting the very small quantity 2eql+ 3P el eer Tg awe obtain, c—usn eek UE a ic lelabela ta: (34. eql+tLV Tuer UNSTIFFENED SUSPENSION BRIDGE. 19 Neglecting the value of a, we have, mess 1 Eqol ON et cece cette ene eeee es (35. For example, if P =1650, «= 0.0003, q, = 2000 Ibs. p. 1. f. and J = 300 ft., then n= 011= +. In order, however, to reduce the deflections as far as practi- cable (see following paragraph), it may be desirable to reduce this value of the best rise-ratio. 4. Deformations. In order to simplify the following investi- gations, we will assume the cables when unloaded, i. e., subjected merely to their own weight, to conform to parabolic curves; this is very closely true with small ratios of rise to span. We will also neglect the resistance to deformation afforded by the friction at the pins of the chain or by the stiffness of the wire- cable. Let g = the total dead weight of the bridge, p. 1. f. p = the uniformly distributed live load, p. 1. f. a.) Maximum crown-deflection produced by deformation under load. This will occur when a certain central portion of length 2é is loaded. For the present we will not consider the possible displacements of the cable-saddles on the towers, but will assume the cables as fixed at the ends. Let f’ denote the versine of the cable when loaded, so that by equ. (5), to4: fo wo. To 9.1 + 0.025 7 i 2438: Af,= (0.007 + 0.046 a —0.0075 a) f b.) Crown-deflection due to elongation of cable. The dif- ferentiation of equ. (31) yields, on putting the rise-ratio - =n, 15 AL eis, sa Gao) DD cecion ts a ete ace a (39. The elongation of the cable (AZ) may be due to its elastic strain, to temperature variation, or to a yielding of the anchorages. If s is the intensity of stress in the cable under maximum live load (—p), then the elastic stretch due to dead load (g) will be =e Ee AL, a gtp Uitte es ce rere oiel ety (40. and the increase in stretch due to live load will be soe PS AL, wba ice era eat lel safeopteh cheats (41. The elongation produced by a temperature variation of + ¢°, with a coefficient of expansion of » = .0000069, is If the cable over the Pig. 17, main span is continued on each side as a_ backstay (Fig. al A i to the anchorage = - Mo 17), and if the cable “k-+~4 AT is capable of slipping over the fixed saddles, then the value Tur UNSTIFFENED SUSPENSION BRIDGE. 21 of Z in equs. (39) to (42) must consist of the total length of cable between anchorages; consequently L=Wit FS wt) $ 2h sec ayers. (43. If, however, a displacement of the saddle will occur before the cable will slip, then with A ZL = the elongation of the main cable and AL, the elongation of one of the backstays, we should have 15 2 15—8 (5n?—36n') Af= 16 (5n—24n’) AL+ seca, 16(5n—24n’) Aly. Substituting the values AL=c.L=c.(1 +30 = nt)l, AL, =c.L,=c. seca, .h, where c is given by the coefficients of Z in equs. (40) to (42), we obtain 15 96 n* dfa= {$l— alt ape (42h) f 0. (44. ce.) Crown-deflection produced by displacement of the saddles. If a displacement of the saddles reduces the effective span of the cable by an amount Al, without a simultaneous change in the total length of the cable, there will result a sag or deflection at the crown amounting to 15 — 8 (5 n?— 36 nt) Af,= =“ WGa=te sec www ewe ewes If the saddle-displacement is accompanied by a slipping of the cable, so that the total length of the latter between anchor- ages (Fig. 17) remains unchanged, then the crown deflection becomes 15 — 8 (5 n? — 36 n*)— 15. seca, Af,= 16 Gn— Dany AL-+ ee. d.) Maximum horizontal displacement of the crown. For the lowest point of the cable curve, we have a =0Q; accord- Fig. 18. ingly, with the notation of Fig. 18, 7 ; so long as the crown (S) is to the XY es Uy fo left of the head of the load (E), x : i dM 1 1 ¢ pen ee Ge 79 I-22) +5 p45 =0. ee Consequently x will have its maximum value when ¢ has its 22 ARCHES AND SUSPENSION BRIDGES. maximum value, hence when =1— xz. Substituting this value in the above equation, we obtain — g V po — Dp rae p + ye Hence the maximum deviation of the crown from the center of the span will be Sa ie =p rs aaa pk we eee ween (47. The total rise of the cable may be assumed invariable for all ordinary values of g:p. Consequently, the uplift of the cable at the center of the span will amount to 2e a Af, ($s f ee ee (48. We thus obtain the following values: g 1 1 for a 3 > 1 2 3 e= 0.167 0.134 0.086 90.051 0.0361 Af = 0.0625 0.0445 0.0214 0.0084 0.0045 f Example. It is required to design a suspension bridge with steel-wire cables to carry a roadway 12 meters wide. Span?—100m= (328 ft.), rise f= 8m, Two cables are used, The loads per cable per linear meter are: Dead weight of roadway = 1400 kg. (= 940 Ibs. p.1. f.), live load = 400 X 6 = 2400kg, (= 82 X 19.7 = 1615 lbs, p.1.f.), hence g, = 3800 kg.; also the weight of the suspension rods, j = “. ee x ae 3.7 kg. s 10 800 (= Go ¥ _ 2555 yx 188 s | 144 11,390 “ 144 is taken at s = 2000 kg. per sq. em, (= 28,460 Ibs. per sq. in.), Using = 7.6 lbs. per ft, p, 1. t.) The working stress tant = . = 0,32, sec r = 1,050, the value of the uniform cable cross-section is given by equ. (29) as A = 333.7 em? (= 51.7 sq.in.). The weight of a cable is therefore g = 0.78: 4 = 260.3kg. per m, (= 175 lbs, per ft.), and the horizontal tension is, by equ. (26), H = 635.6 tonnes ( = 1,402,000 Ibs.) The angle of suspension is now more accurately given by equ. (28) as tan T= 0.32024; and the length of the cable is given by equ, (30) as L = 101,672 m, If there is no stiffening construction, the deflections are calculated as follows: The maximum crown deflection under partial (symmetrical) load- 2400 ere 3 p __ 2400 ing is given by equ, (38) (tor F rai) as A f,= 0,057 - f = 0.456 m. If the horizontal lengths of the backstays are taken as J, 25 m., then the crown deflection due to elastic stretch under full load is given by equs. (44) and (41), A f, =0,208 mm, The deflection due to a temperature variation of + 30° C. is, by equs. (44) and (42), Afi = + 0,144 m, Tue UNSTIFFENED SUSPENSION BRIDGE. 23 Finally, by equ. (47), the largest horizonva: displacement of the crown will be ¢ = 0.110 1 = 11m.,, and the corresponding uplift at the center of the span is, by equ. (48), A f, = 0.0325 f = 0.260 m, The total vertical displacement at the center of the span caused by the crossing of the maximum live load amounts therefore, in the unstiffened suspension bridge, approximately to +0.46 m, and — 0,26 m., i. e., 0.72 m., to which must be added the tempera- ture deflections of + 0,144m, 5. Secondary Stresses. The preceding theory of the flex- ible cable is based on the assumption that the cable (or chain) is constantly free to assume the curve of equilibrium correspond- ing to the momentary loading; this neglects the resistance to deformation offered by the friction in the hinges of the chain or the stiffness of the wire of the cable. Although the latter effect may be so small as to have no appreciable influence on the stresses in the cable, it will generally be otherwise with the frictional resistances in the chain-hinges; experiments made on existing bridges by Steiner and Frankel with the aid of Frankel’s Extensometer, have demonstrated the occurrence of considerable bending strains in the individual links of the chain. If 7 denotes the axial tension in one of the links, d the diameter of the pin at the hinge, and ¢ the coefficient of fric- tion, then the bending moment transmitted to the link by friction may reach the value M=—=¢.T a Approximately we may take ¢.7T—=¢’.H, where, in the most severely stressed link, ‘=¢ (1 + 8 c ). Then, to obtain the greatest possible value of the bending moment, H must be determined for that intensity of loading which, applied asymmetrically, will produce moments of deformation equal to the friction-moment ¢’.H. = : For this purpose, we may consider the live load of p per linear unit resolved into two parts p’ and p”, where the former covers the span only partially so as to produce the bending moment given by equ. (104*) as Mf = 0.0165 p’l?; while the second part p’”’ is supposed to cover the entire span. Then, neglecting the effect of the load p’ upon the value of H, we have / , 1 ” i? d whence Oa, ee Hence, the maximum bending Aca will be y—t.__* (ptg)->. 0.264 4 oe 24 ARCHES AND SUSPENSION BRIDGES. Thus, knowing the sectional dimensions of the members of the chain, we may readily determine the bending stresses. For the friction coefficient ¢, particularly in old chain bridges where rust may occur, a high value should be adopted, at least 60.20. Example. In an existing chain-bridge of 275 ft. span and 18.6 ft, rise there is carried by each chain its own weight of g= 650 lbs, p. 1, f, together with an applied load of p= 820 lbs. p.1. f. For these values Hg,»=746,000 lbs, The diameter of the pins amounts to d=0.164 ft. Using ¢' = 0.23, 0.264 fete J = 0,25 2 46.0 i = 0264494 0,99 and M = 0.25 X 0.082 x 746,000 x 0.99 15,140 ft. lbs. The chain consists of 5 eye-bars, each 5.2 inches deep and 1,22 inches thick; hence each bar must take a bending moment of yee = 36,330 in. Ibs,, and with a section modulus of 5.5 in’, the 5 : 36,330 resulting maximum fiber stress will be : 5 we find = 6,600 lbs. per sq.in, THE STIFFENED SUSPENSION BRIDGE. 25 B. The Stiffened Suspension Bridge. §5. Approximate Theory. In order to reduce the static distortions of the funicular polygon or flexible cable discussed in §4, there is introduced a straight truss connected to the cable by suspension rods. This stiffening truss may either extend over a single span, i. e., simply rest on two supports, or it may be built continuous over several spans. In the latter form, the only cases of practical application are those of two equal spans or of three spans. As already mentioned in the Introduction, this type of structure is statically indeterminate; but, by intro- ducing a hinge in the stiffening truss, we may either secure determinateness or at least reduce the degree of indeterminate- ness. We first adopt the assumption that the truss is sufficiently stiff to render the deformations of the cable due to moving load practically negligible; in other words, we assume, as in all other rigid structures, that the lever-arms of the applied forces are not altered by the deformations of the system. This approximate theory will usually be sufficiently accurate for all practical pur- poses; but, in order to determine the limits of its applicability, a method for a more exact design will be appended. 1. Conditions for Equilibrium of the Funicular Polygon. Equations (1) to (5), (§3), will also apply here if the forces P in those equations are put Fie. 19 equal to the stresses of the an suspension rods 8 ine yA. creased by the panel-loads Tat aaa K corresponding to the f dead weight of the cable. ane If the rods are located at uniform intervals a, then Weta eer ca te entaneaP saat nu cueede tee te - we have by equ. (3°) ; wT Re II iad pairs etek GoM (49. If the vertices of the funicular polygon lie along a parabola with a vertical axis, and if the adjacent rods are distant a and a’ from the m™ rod, then equ. ‘(3°) gives, HAGE) gt Kee cece e ees (50. Hence, if the cable is parabolic, and if the panel-points are uniformly spaced (horizontally), the suspender-forces must be 26 ARCHES AND SUSPENSION BRIDGES. uniform throughout. It thus becomes the function of the stif- fening-truss to distribute any arbitrary live load in whatever manner may be demanded by the particular form of the cable- curve. If the cable is continuously curved, then the applied loads per horizontal linear unit are given by = ap ee Sm k= — Hig eee ee ee eee (51. and if the curve is a parabola, the uniform loading becomes Sk SL sic ane eee (52. The approximate theory assumes that the above equations (49) to (52) hold true even after deformation. This assumption is all the more admissible the stiffer the truss, for the less will then be the deflections transmitted by the rods to the cable. Fig. 20. 2. Forces Acting on the Stiffening Truss. These com- prise the dead weight of the truss and any construction it may earry, the live load, and the stresses in the suspenders which are, in effect, vertical loads directed upward. If we imagine the last forces removed, then the bending moment M and the shear S at any section of the truss distant x from the left end may be determined exactly as for an ordinary beam (simple or con- tinuous according as the truss rests on two or more supports). This moment and shear would be produced if the cable did not exist and the entire load were carried by the truss alone. If — M, represents the bending moment of the suspender-forces at the section considered, then the total moment in the stiffening truss will be M = M — M,. Let the straight line passing through A’ and B’, the points of the cable directly above the ends of the truss (Fig. 20), be adopted as a coordinate axis from -which to measure the ordi- nates y of the funicular polygon. Also consider the dead weight TuE STIFFENED SUSPENSION BRIDGE. 27 of the cable (k p. 1. f.) to be uniformly distributed. We then have the following cases: a.) For a truss simply resting on two supports, with span A B=l, by a familiar property of the funicular polygon, M, + ke (I—2)=H.y; consequently M—=M+4 he (l—2)—Hiy..eeeeee, (53, Considering the cable-weight as included in the load carried by the stiffening truss, also representing M by the ordinates y of an equilibrium polygon or curve constructed for the applied loading with a pole distance = H, we have M=M — HH .y=H.(y—y) ....e eee cea (54, Hence the bending moment at any section of the stiffening truss is proportional to the vertical intercept between the axis of the cable and the equilibrium polygon for the applicd loads drawn through the points A’ B’ (Fig. 20). 8.) For a stiffening truss continuous over several spans, MeHg Me MAZE (yam wi He) hence M=M—H(y—m” + —m' >"), ele seied (55. Here M’, and M”,, denote the end bending moments produced by the suspender forces acting on the continuous truss, intro- duced with their appropriate negative signs. These moments may readily be represented by the lengths m’ and m”, with a pole distance H = 1, for any given form of cable. In the case represented in Fig. 21, where the truss extends 28 ARCHES AND SUSPENSION BRIDGES. symmetrically over two side spans but without connection to the backstays, we have in the main span, M —M — HZ (y—m) (56 in the side span, M— M+ H.m. ae With a uniform distribution of the suspender forces, i. e., for a parabolic eable-curve or chain-polygon, if the ratio of span- 4 is the ratio between the mo- lengths is r= and if i= i ments of inertia of the cross-sections of the stiffening truss in the main and side spans respectively, the theorem of three moments gives = aap ot BAS 6 RRS Gere ee SS Saw Se Sais Siete Denoting the coefficient of f in the above equation by the symbol « (a constant for any given structure), m—e.f, hence in the main span, M—M — H (y~—«.f) in the side span, M=M-+H.-~.ef wee (OT. If the truss is connected to the cable by suspension rods in the side spans also (Fig. 22), then, with a parabolic cable curve, B’ Fig.22, © strap trac rr erens As before, let us designate the coefficient of f in this equation by the constant ¢, so that m—e.f. We then obtain, for the main span, M= M — H (y—«.f) | .. (58%, for the side spans, M= M — ZH (y, — ae e.f) where y, represents the ordinates of the side-cable below the connecting chord A’ B’. The transverse shears 8 are given by the following equations: a.) In the single-span stiffening truss S=S — A (tanr — tanc) THe STIFFENED SUSPENSION BRIDGE. 29 8.) In the continuous stiffening truss, S=S—H (tanz—tano+">",) Sees (60. Here + denotes the inclination to the horizontal of the tangent to the cable curve at the given section, and o is the inclination to the horizontal of the cable-chord joining the points of suspen- sion; both of these angles are reckoned positive when directed downwards to the right. m’ and m” have the significance pre- ay assigned. For the case represented by Fig. 21, the shears will be : Z j 2 in the side spans, S=S+4H.- tis’ in the main span, S = S — H tanr (61 If the truss is suspended in the side spans also, the second of equs. (61) is replaced by 2+4+2ir7 f S=S+H (F570. 4 + tano — tans) .... (62. 3. Determination of the Horizontal Tension H. In the bridge systems considered above, the horizontal tension is statically indeterminate unless another condition is furnished by some special construction so as to eliminate the indeterminateness in this respect. For instance, the cable might be given a definite, invariable H by replacing one of the anchorages by an attached weight. The stiffening truss would then be subjected to an un- changing upward loading.—Another arrangement that has been proposed, but never yet executed, is to balance the end-reactions or to fix the ratio of the cable and truss reactions by supporting both of these members on the ends of a common lever. If a Fig. 23. r= is the ratio of the lever a2 wecepirees aes arm of the truss to that of the «]/Né \ cable-tower, then, with the ar- tr rangement and loading shown in Fig. 23, GM nee [sscxavexsouastcteecns : aa? In actual practice, however, the only cases of importance are the following two arrangements: a.) The stiffening truss is provided with a central hinge. This furnishes a condition which enables H to be directly de- termined; viz., at the section through the hinge the moment Af must equal zero. . Consequently, if the bending moment at the same section of a simple beam is denoted by M,, and if f is the 30 ARCHES AND SUSPENSION BRIDGES. ordinate of the corresponding point of the cable, then for a single-span bridge, by equ. (54) and for a continuous truss having a hinge in the central span, by equ. (56) r= Mo f—™ b.) The stiffening truss has no central hinge, and is not pro- vided with any of the devices, described above, for making it statically determinate. The required equation for the determi- nation of the horizontal tension must therefore be deduced from the elastic deformations of the system. Various procedures may be adopted for this purpose. We may, for example, equate the variations in the cable-ordinates, increased by the elongations of the suspension rods, to the deflections of the stiffening truss, and thus develop an expression for H. A simpler method, how- ever, consists in applying the ‘‘Theorem of Least Work.’’ If N denotes the axial stress and M the bending moment at any section of an individual member of the system, and if s is the length of the member, A its cross-section, J its moment of inertia about the neutral axis, and F its coefficient of elasticity, then the familiar expression for the work of deformation is w= faue + (supa min pier ee ies oop (66. and the resulting equation of condition for the case under con- sideration is aw — N dN Ny. adit BA di” oe To apply this equation, it is simply necessary to express the axial stress and bending moment for each part of the system in terms of the external forces and the unknown horizontal tension I, a.) Let us first consider the type of construction shown in Fig. 19, consisting of a suspension bridge with stiffening truss extending over but one span. Let A = cross-section of the cable at any point. A, = cross-section of the cable at the crown. d= length of cable between consecutive rods. a = horizontal distance between consecutive rods. 2 A, = cross-section of the stiffening truss. (= cross-section of the two chords.) I = moment of inertia of stiffening truss. THE STIFFENED SUSPENSION Brince. 31 A, = cross-section of a suspension rod. y’ = length of a suspension rod. y = ordinate to cable measured below the closing chord. h = effective depth of the stiffening truss. ( = distance between the neutral axes of the two chords.) f = the versine (or rise) of the cable. f’ = height of the end posts or towers. A, =cross-section of end posts or towers. l, = horizontal projection of the backstays. a, = inclination to horizontal of backstays. a = angle of suspension of the cable. G = total load on the stiffening truss. A,B = resulting vertical end reactions. K = weight of cable per panel point. k = weight of cable per linear foot. Remembering that the stress in the cable (or chain) is Ps 8S Aa ie awa 0 Waa Bin ha OR Chea (68. and assuming that the cross-section at every point is proportional to the stress, then we have A = A, - = and for the backstays = A, ~ SEC ay. ‘We must now write out the expression of equ. (67) for each member of the system, for which purpose the following tabula- tion will be used. In this table, the stress in the suspension rods is taken from equ. (49) ; and it is assumed either a.) that the stiffening truss is not independently supported but is hung, at the ends, from the points of suspension of the cable, or b.) that the stiffening truss has independent supports, one fixed and the other movable horizontally, which, on account of the possible negative reactions, must be considered as anchored down. ARCHES AND SUSPENSION BrinGEs. 0 0 0 Ay—(@—1)" I 3 I Pee) lem et) el rae | ie ari Ce RS ssniy, op ——-~__. ooo Hp _ (#—1) wh , : T e awd xp as “| mf I # vy = +W Surweyys % 8 'p uD} (ip up} ure +,('o up7 + vUn2) f° e + a) + vun}) W SIOMOT, (9 J av 8 8 yun} H —@ oud) f° = — vd} yf fiz | ov UDI — nud? H — at ““sisog pug (0 Oz 0 Vy 2» vb D AwAg a —(f.v) Ag H fi °y fy MS ieee |e spoy 1M c : av worsuadsng 0 Fr *p,008 | * Fel woos 4g ‘yp 008° °V peas ‘9 098" HH |**- ‘sfkeqsyoeg 9 0 FD D D D XZ ee ee! lnaerenggs eT oH x Ae as a eH arqeg HP f | : : HP? . HP i Ss yao 7 —— 2 WI0N equa wip 7 P a ee quay 32 2 THE STIFFENED SUSPENSION BRIDGE. 33 l l 1 ! Noting thats *7= 3 a?+ 3 (Ay)?=l.a— SyA*y, and 0 0 0 0 that for an approximately parabolic curve of cable we may —_— 2 2 . write fee= «da =F a ae , we obtain from the rela- aw tion cara =0 the following expressions for the horizontal tension : For the case a): ~{e = def Ae tees a+ (4230 ty +42 0 H= fe) l aty + 2h sectat 3S H(A A*y)?+2 f’ 4 tana 2 1 l— > 0 ale tah eet teense nena as ok (69. For the case Wye My ‘ Al’? cy’dz af av + (4° ayaty +See SU) H= 1 1—xZaty +21, secta, + aes (Aty )? 0 @ A, oa 1 ,Ao # y* ax +2f a gr (tan at tan a)?+ Ao f een uhudt i tydaaae jae Ouse (70. Here £;, is the elastic coefficient for compression of the towers. Fig. 24. Fig. 25. foxx aa ee Ym + Ym-1) | 6 ao (tm41 6 tee (2 Ym + Ym); Ting1 and substituting in equ. (717), we obtain a final expression for the horizontal tension due to the external loading: l = Ma va BS a eer (73. ’ oT SO a aa Ye oe do Ao If the loading consists merely of a concentration G, distant é from the left support, then the two summations in the above expression may be given a static significance, enabling them to be determined graphically. Thus, if zm is the abscissa of any panel point, : i—s oc > Mn va—@[ 7 3am Ua + 7% (l — £m) va|—G.m 0 0 é & 1 and 3 Ym Um = By where mé signifies the moment, at the load-point (é), produc- ible in a simply supported beam loaded at every panel point (m) with a vertical force of magnitude v,. Similarly, » is the static moment, about the chord A B, of the forces v conceived as acting horizontally at the respective panel points. These mo- ments may readily be constructed as the ordinates of the proper funicular polygons. If the panel-length a is constant, then Io Io 5 Um = Gye (2 Ya + Yaa) +e (2 Ym + Yar); mat and if the moment of inertia, J, is also constant, i : Um = | Yart 4Ya+ Yn), or, with sufficient accuracy, Um=Yn. The graphic construction is executed exactly as in Fig. 54, (see §17). We first divide the area between the cable and its chord into strips of width a, which may be taken equal to the spacing of the suspension rods; and the load-elements v are THE STIFFENED SUSPENSION BRIDGE. 37 either computed by equ. (72) or assumed approximately equal to the ordinates y. With the pole distance p, we next construct the two funicular polygons b and c for the forces v acting vertically and horizontally, respectively. The ordinate of the polygon ), directly under the point of application of the load, gives the value of H, if the concentration G is represented by a length composed of the intercept non the polygon c on the chord A B plus the length nn, = c = = z . Thus, mM SS, Gi ten tren vee ese eravenete a ags 74, e+ep ( In an actual design, the data were: 1=50 m.,, f=65 m., % % aty = 4.506 m., 1, = 15m. —2— = 2.049; the effect of the: suspension rods was neglected. Choosing a = 2.5 m., p= 35 m., there was obtained c= 2.167 m. For a load at the conten of the span, the construction yielded H = a = 1.286 G4. If the stiffening beam is a truss with parallel chords, then in equs. (71) to (74) we may write] — A,: ae Again I may be assumed constant either throughout cach panel a or, for a first approximation, throughout the entire span. If we desire, furthermore, to include the effect of the web members, if A, is the section of any web member and y its inclination to the ver- tical, then, by equ. (59), we must correct equs. (69) and (70) by adding to the numerators of the expressions for H the term, 1 : - secty - 3S Bo ap. and to the denominators the term - A a, : The following values, therefore, 1 + tt: sey + = a 0 must be substituted in equ. (71) and the subsequent formulae, for the case (b) : Ao Ce.” oM=~ , 2 h 3 : Ay yo Ay + pore easy q=l— os va A?y + 21, sec? a +435 (Az y)? A, , Ao. 2 he 3 tA 2 +2f 7. geet Ga) ge they Bgl By Ys However, in the total work of deformation, the influence of the web members of the stiffening truss, as well as that of the suspension rods, is so slight that the corresponding terms may, 38 ARCHES AND SUSPENSION BRIDGES. without appreciable error, be neglected. Under the usual con- ditions of practice, their effect on the value of H cannot exceed 0.3 to 0.5%. ‘If the cable or arch is parabolic in form, then, taking the origin of coordinates at the left point of support, the equation of the curve will be 4 y= sf (la) «2. If, furthermore, the distances a between suspension rods are assumed vanishingly small, the summations in the expression for H may be replaced by the corresponding integrations, thus yielding the following values: so fet) aeni(n4 38) 3% (4 wf v) (4 ‘da = BE (pe ) 1 2 wa, 18 fs, fae ful U i 2 ¥ pw =f ty), dx =8 (f" Ae - ci ae if- sf) 0 l YMy= (Fy F¥ de—UL. pe _ fu dx? 7 3 1 0 Finally, if the moment of inertia of the stiffening truss is assumed constant (— J) for the entire span, and if the loading consists of a concentration G located at a distance é from the left end of the span, then we have t 1 {M.y.d2—G = ae 0 =sé (@-Me +E) 4, Substituting the above values for the summations of equ. (70), we obtain the following expression for the horizontal ten- sion: Tue STIFFENED SUSPENSION BRIDGE. 39 a LG G+ te [5 or—2n gp a] a+ sa: + Ppt 2 sec" a) +64 (81 —2/) 90 it EE (44t0 ana, ) In this expression, A, denotes the sectional area of the sus- 3 : : A pension rods per linear foot of span, 1. e., rae Neglecting the effect of the elongation of the rods, as well as that of the compression of the towers, and omitting the cable- weight from consideration, we have aes & l =(,-2%+1).64 16 5 ae 3 As (1 eS 4 s+ 2" sot =) Sea eae 3 (753, The last equation will usually be accurate enough for a pre- liminary design; on the basis of the results thus obtained, an estimate can be made of the probable variation of the moments of inertia (J), and a re-design may then be executed by the more exact equation (71) or (73). Of course it will be neces- sary, in the first design, to guess the probable value of the ratio I/A,, but the range of this quantity in practice lies between such narrow limits that the corresponding variation in H is quite i=; by the symbol n, inappreciable. If we denote the ratio re : ik AS 5 5 A and if An is its variation, then approximately af = —n: An. As may be shown by a study of actual designs, the ratio » always lies between the limits 0.01 and 0.02, so that, at the mean value, aE = 3.55 . An; consequently, if the varia- tion of n is 0.01, the resulting error in H will not be more than 3.6%. If the horizontal span of the cable l’ differs from the span of the truss 1, we must substitute l’ for | in the first two terms of the denominator in equ. (69) or (70), also multiply the terms containing *. in the denominator of equ. (75) by <. In this case, f is the versine of the cable for the span I. 8.) By the same analysis as was applied in deriving equ. (69), we may also obtain the expression for H for a three-span suspension bridge (lig. 27). It will here suffice to derive the approximate equations based on the assumptions of a parabolic =! or 40 ARCHES AND SUSPENSION BRIDGES. weightless cable, and a stiffening truss with moment of inertia constant throughout each span. In the following, let f =the versine of the cable in the main span (J), f, =the versine of the cable in the side spans (l,), measured vertically,* l, =the horizontal distance from tower to anchorage, inclination to the horizontal of the cable-chord in the side span, = moments of inertia of the stiffening truss in main and l l, side spans respectively. Fig, 27, necesd Wik : te | ete ‘ ‘ 4 ~f aaa Feige 4G Resta eae hisssseced > Weta jnjstetcts shes > i a Then, for a concentration G in the main span, distant £ from either tower, (aot) [ibea(4—e) Ht ( :) = (5 41-5 2° eG i) sy eal +3? “4a 72 sua 2 efte ize 1 ee Ga ee) (76. Eg t If the concentration is applied in the side span at a distance from the end-support, the ‘horizontal tension in the cable will g be: [()2(2)'n]-4 O-)) nics (#8 G)) [i—5 +(G-- 3} el ape | Stren! 8 Ih 8 fe f goes Te eae e P+ e*) 167 oh 16 f? +5 35424 (1+ 1 | If the stiffening truss is suspended from the cable in the side spans also, then, in the above equations: *In Fig. 27, f, should be measured vertically, not obliquely as wrongly H indicated by the drawing. THE STIFFENED SUSPENSION BRIDGE. 41 I ex Sa) eaenueie. Pio teno aus (17. P (31+2 aa 1) But if the stiffening truss, although continuous, is not con- nected with the cable in the side spans, then, in the above equa- tions, we must put f; = 0 and 21 eS — ———_. 31421, ae ee eee ee ee ee If, in the latter case, it is desired to include the effect of the elongation of the rods in the main span and the compression of the towers (of height/’), then there must be added to the denominators of the above expressions for H the terms: 31 atlas (45 + tana) + (7 -F AP] The last term in the numerators of the above expressions, i. e., il, 311,+ 2,1’ cable curves in the main and side spans are parts of like para- bolas, when there are no suspenders in the side spans, or when the stiffening truss is interrupted (i. e., hinged) at the towers. In the last case, viz., the two-hinged suspension bridge, we must also put the factor of continuity «—0, thus obtaining the fol- lowing simplified equations: For a load G in the main span, — (7) -2G)+1 | sC42ri Be) tala et eit settee) ] | co, Sg f the term containing the factor vanishes when the For a load G in the side span, “7 [@-2@‘+ JE 5 (142757) +244 7o+24 (REueaemy b. (79). é 78 To investigate the effect of a moving load, it is advisable to apply the Method of Influence Lines. 42 ARCHES AND SUSPENSION BRIDGES. On account of their frequent application in the present and succeeding parts of this book, the general properties of influence lines will here be reviewed. To obtain the influence line for any given function for which we wish to investigate the law of loading, we erect at each position of the load an ordinate equal to the corresponding value of the function. The resulting curve then indicates definitely the positions at which the influence of the load is maximum or minimum, and also assists in deter- mining the most unfavorable position for a train load, If the influence line is constructed for a moving load equal to unity, then the function in the case of a series of concentrations will be equal to the sum of the products of the loads by the ordinates (Y) of the influence line corresponding to their points of application. If, instead of a unit load, any force G be taken as the basis of the influence line, then for the 1 concentrations P we must take the sum € = PY. The influence line is a@ continuous curve if the loads are applied directly to the main truss (or other structure). If, on the contrary, the loading may act only on isolated points of the truss, as in the case of principal trusses sup- porting transverse floor-beams, then the curve is replaced by an inscribed polygon of straight lines with vertices located in the vertical lines through the points of direct loading. (Proof: Let P be the load in any panel a between two transverse beams at which the ordinates for a unit-load influence line are Y, and Y, 3 let e be the distance of the load from the first of these panel points; then the ordinate Y of the influence line at the point of application of the load must satisfy the equation ee P(a—e) P.e 2 ae, showing that Y is the ordinate of the straight line joining Y, and Y,,). If a train of concentrations is placed arbitrarily upon the structure, it will in general be necessary to shift the entire system of loads one way or the other before they will yield a maximum value of the given function. In order to establish a criterion for the proper direction of this shift or displacement, let a,,a2,a;,..., be the angles of inclination of the sides of the influence polygon and let P,, P., P; ... be the loads lying within the projections of the respective sides. Then the combined influence of these loads is represented by the sum P, Y, + P, Y. + P; Y; + ete.; and the variation of this function producible by a displacement of the entire train through the distance Ag (it being assumed that during this displacement no load enters or leaves the span or crosses a panel point) will be given by AE (P, tan a, + P, tan a, + P; tan a3 + ...) since, by this displacement Aé, the ordinate Y corresponding to any load will be increased by Aé,tan u. If the truss is directly loaded, so that the influence line is a curve instead of a polygon, the same expression applies provided that a,, a, ... are the inclinations to the horizontal of the tangents to the curve at the load-points and Aé is assumed to be an infinitesimal displacement. We thus find that to secure the most unfavorable loading (yielding the maximum value of the function) there must be a displacement either to the right or to the left according as: the expression in the parenthesis is positive or negative in value; also that, in general, to secure the above condition one of the concentrations must rest at a panel point. The above expression may be represented graphically as follows: Lay off the loads upon a vertical line; commencing at the top of this load-line construct a polygon whose sides make the angles vz, @, ... With the vertical and have for their vertical projections the respective forces P,, P, .... The position of the terminal point of this THE STIFFENED SUSPENSION BRIDGE. 43 polygonal series of lines, to the right or left of the load-line, determines the direction of the required displacement of the series of concentrations. Finally, the influence line also shows the effect of a continuous uniform load. This effect will be represented by the area included between the influence line and the axis of abscissae within the limits of the loading. To obtain the H-influence line or H-curve, we must cal- culate the values of H by the preceding formulae for different possible positions of a unit load and plot these values as ordi- nates at the respective load-points. We may then find the values of H for any loading consisting either of a train of concentra- tions or of a continuous load. Thus, if we denote by y,,7,....the ordinates of the H-curve at the points of application of the loads P,, P,..... , then the horizontal force is given by H=P, i Pg tah ees If, instead, we have a continuous load of intensity q per unit length distributed uniformly over the distance x,— %,, then the horizontal force will be Xa H—=qf{ 1de—q.8, xy where ® is the area of the H-curve included between the ordi- nates which mark the limits of the loading. If a hinge is intro- duced in the middle of the span, then, by equs. (64) and (65), the H-curve will be identical with the influence line for moments at the central section of the truss. With a simple (non-contin- uous) stiffening truss, the H-curve thus becomes a triangle having a maximum ordinate of = G Fat the crown. Example. a) In a suspension bridge with a continuous stiffening truss, as in Fig. 27, let = 041, f = 0.11, fi = 0.041, 7 = 5 1, and I 1 Ad? a; also let = 0.5 1, and tan a, = 0.25. Then, by equ. (77), 0, = 60 2 (14(0.4)*.5) e="3+4+2(0.4) .5 ing values of the horizontal tension: -= 0.377, and equs.(76) and (76") yield the follow- Load in a side span: 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 £ 4 Hf = —0.0174 —0.0406 —0.0721 —0.1109 —0.1525 —0.1890 —0.2087 —0.19680 —0.1345 @ Load in the main span: =0.1 0.2 0.3 0.4 0.5 oon H = 0.7061 1.8832 1.9354 2.2944 2.4185. G. 44 ARCHES AND SUSPENSION BRIDGES. Fig. 28. The accompanying Fig. 28 gives the H-curve constructed from the above values. Example. tb) If the stiff- ening truss is hinged at the towers, then, for the same data as in the preceding example, we obtain the following values of H by equs (79) and (79°): For a load in a side span: = =0.1 0.2 0.3 0.4 0.5 L, H= 0.0671 0.1270 0.1739 0.2036 0.2139. G. For a load in the main span: + =0.1 0.2 0.3 0.4 0.5 H = 0.5245 0.9922 1.3587 1.5912 1.6710. G. According to the expression for H, the value of the horizontal force . But, as previously depends upon the assumed value of the ratio ‘0 noted, this ratio in all practical cases will fall between sufficiently close limiting values, and any error in estimating it will not affect the value of H appreciably. Thus, in the above example (a), if I 1 1 0 Af? 20’ 80’ 0.85, 1.021, 1.093 times the values calculated above. then H will be , SPECIAL Cases or LOADING. 45 §6. Values of H for Special Cases of Loading. The following formulae are derived by integrating the expressions derived above for the action of concentrated loads. 1. Stiffening Truss with Central Hinge. Single Span. If the truss is completely loaded with p per unit length, by (64), 1 U H=->.+-pl seme merce ccc rer eesaee eee (80. If the truss is uniformly loaded for a distance \ from one end, by (64), 1 7 for A, H = (21a—4- i 2. Stiffening Truss without Central Hinge; constant mo- ment of inertia (1), parabolic cable. a.) Single Span. If the truss is completely loaded with p per unit length, by (75), 1 I H=+sy 7 P! Se) oie Wie aero be elie Sia de Stace Gee (82. If the truss is uniformly loaded for a distance A from one end, Pare LTTIsA\® 1fA\2, LT A\Z 1 H=5[5G)-2G) + 2] G) 7-2-8. Here N,, represents the denominator of equ. (75) or (75*). b.) Three Spans. Stiffening Truss Continuous. (Fig. 27.) Main span completely loaded: By (76), qt eas : Sia : (84 1h fk yt Sake : — 2 BSIL+ 241 Sete e+ Jal Both side spans completely loaded: es ( 76"), 2f1f, 1 1 ih 3 a ase Pele 8 4 31,4241 2 tae rrr ee | (85. Ut mY re 2 oe | Re F Ph 46 ARCHES AND SUSPENSION BripGEs. Main span loaded for a distance A from either tower: By (76), 1 1 r»\3 Beis G) eG ae Ge 1 lly Tl; +437 [i-3 a a ee (86. (2-2) FO) 4m Side span loaded for a distance A from the free end: By (76°), ae wilsG)- 3(7) + 13 fe —(je or Ta Le” Ahi—se im rid) cet co C-G)) hie Here N,, represents the denominator of equs. (76) and (76#). ec.) Three Spans. Stiffening Truss Interrupted (Hinged) at the Towers. (Fig. 29.) Fig. 29. ae et eg ecco For loading in the main span, equs. (82) and (83) apply if there is substituted for N the denominator of equ. (79). For a complete loading of both side spans, by (79?) : = 8 ego a ee ee (88. For a continuous load in either side span covering a distance A from the free end, by (79*): i vel3 Ge 2) t3]G)- 4. a “ply .. (89. STRESSES IN THE STIFFENING TRUSS. 4y § 7. Stresses in the Stiffening Truss. 1. To Find the Maximum Stresses in the individual members of the stiffening truss, it is necessary to first determine the max- imum bending moments and shears in the structure. If it con- sists of a truss with parallel chords, the chord stresses are obtained from the bending moments and the web stresses from the shears. The moments M, according to equs. (54) to (58), may be represented by the products H .z, where z denotes the vertical intercepts between the equilibrium polygon of the external loading and the axis of the cable. These moment inter- Fig. 30. r. ee 3 Nia 5 #8 p= cepts must be measured in the vertical lines passing through the panel-points lying opposite the given chord members, and con- sequently (see Fig. 30), U.-—@a-*. 7 ee Soe. op anieite gates shee (90. eg If, as in the system of Fig. 32, the horizontal thrust of the arch is taken up by the stiffening truss, then the resulting axial stress must naturally be taken into consideration in proportion- ing the members. The chord stresses will then be _ Ms h\ = 2 +3— —(4a-3 _ i Wy Bop Reet ess (91. =a h + —+ (e+ os) h In general, therefore, we may put U,= ee 7 -H BO pet en ak gare weariness (92. Ly=+77-+H 48 ARCHES AND SUSPENSION BRIDGES. where e, and e, are the intercepts between the equilibrium polygon and the axis of the arch-curve, or (in the structures of Figs. 25 and 32) between the equilibrium polygon and two lines obtained by shifting the arch curve upward or downward by an amount 4. 2. Reaction Locus. Loading for Maximum Stress. In order to determine the positions of the moving load for maximum stresses in the stiffening truss, it is necessary first to investigate the influence of a moving concentration. If the graphic method is selected for this purpose, the problem reduces to finding the equilibrium polygon for any position of the load. This involves Fig. 31. monies la! no special difficulty after the value of the horizontal thrust for the construction of the equilibrium polygon is obtained from the H-curve. a.) For a Simple Non-Continuous Stiffening Truss, the equilibrium polygon must pass through the points of suspen- sion (A’B’) of the cable. In this case the direction of the sides of the equilibrium polygon (or triangle) is simply found from the resultant of the horizontal tension H and the vertical end- reaction; this reaction, for a load placed at a distance é from L—é the end A, is defined by @ ; , and hence is given by the ordinates of the straight line bc (Fig. 31). If we construct the equilibrium polygon, as indicated in the figure, for other posi- STRESSES IN THE STIFFENING TRUSS. 49 tions of the load, the geometric locus of the point FE in which the two sides of the polygon intersect each other and the load- line will yield a curve K EF K, which we name the Reaction Locus. The ordinates of this curve, measured from the chord A’B’, are analytically defined by G(l— £)é y= Tt SEE ist wtnatuecih cous (93. Substituting the approximate expression of (75*) for H, we have Nw. 0 "= pEu—e fF From the latter equation, by putting é=0, we-obtain the ordinate of the point K; to determine the same point by con- ws Fig. 32. Tension ) Upper Compn = =) chord Tension {t— moms} LOWEr Compn NB oe Chord. Loken t tio a pene Mi, yeh “pA T rt \ P= Sis, SOs 4, x ; 5 x 5, A ul s | | oy t 1 gP | Since, by equ. (90), the stress in the upper chord member rt (Fig. 31), or in the lower chord member su, is determined by the intercept between the equilibrium polygon and the point M of the cable, then the point J (Fig. 31), in which the con- necting line A’M intersects the reaction locus, will be the critical point for the above members, i. e., the position at which the moving concentration produces a change of stress in the given chord members. For all loads placed to the right of J, the equilibrium polygon will be above the cable-curve at M, hence z will be negative; while for all loads to the left of J (up to some point J, where the connecting line B’ M may cut the reaction locus), the equilibrium polygon will fall below the eable-point M, so that z will be positive. Consequently, a load covering the distance J B will produce the maximum tension in the upper chord rt and the maximum compression in the lower chord su; while a load covering the remainder of the span A J (or J J,) will produce the maximum compression in the upper chord and the maximum tension in the lower chord. For those systems in which the horizontal force H is taken up by the chords of the stiffening truss (Fig. 25), it is necessary, as deduced above from equ. (91), to replace the cable-curve by two similar curves distant + 4 from it, the upper one govern- ing the design of the upper chord and the lower one governing the design of the lower chord (Fig. 32). Analytically, the abscissa of the load-division point or critical point J is given by =. pits Elec cae g dekh orentaere sca aes (96. where x and y denote the coordinates of the point M referred to the closing chord A,B,. Substituting for 7, its value from (94), we obtain the following equation for the bridge without a central hinge: EMEP Ng PFS net ee eas (97. For a center-hinged stiffening truss, by equ. 95, the critical point will be given by & 2fa TLyt+2fe Soh de lave d0 8 Ta Oba cat atte erie ati veces eras Ske sive (98. and for a parabolic cable-curve, & Bg) tiuaneeeeran ue (98%, STRESSES IN THE STIFFENING TRUSS. 51 b.) In the Three-Span Suspension Bridge Having a Contin- uous Stiffening Truss, there is an added difficulty ; as the bend- ing moments at the intermediate supports are no longer zero, the equilibrium polygon does not pass through the suspension-points of the cable. If M,—H.m, and M,—H.m, are the bending moments produced at the intermediate supports by the vertical loads acting on the continuous truss, then (by Fig. 21) the equilibrium polygon must pass through two points respectively (m,— m) and (m, — m) above the cable suspension-points, where m is determined by equs. (57) and (58). (This is equivalent to expressing the resultant bending moments at these points as the difference between the moments due to the downward-acting loads and those due to the upward-acting suspender forces.) The quantities M, and M,, or m, and m,, must be determined by the theory of the continuous beam (Theorem of Three Moments) either by computation or by construction: In Fig. 33, the graphic solution is indicated. FF’, are two fixed points located in the main span by laying off B,, F=C,,F, =1+(38+4+ 2+). Then lay off C,,E=B,,C,,—3F,C,, PP, 1 & (l—£) . and EG=BC - ByF ? also ca -G- =a The last dis- tance (GH) is equal to the intercept M T, at the point of load- ing, of the ordinary equilibrium polygon constructed with the pole distance H. The connecting line # H intercepts on the load- vertical a distance M L; this is m,. (The corresponding analyt- ical relation, derived from the Theorem of Three Moments, is m —|< ic) | 2irt (1-4) (3 + 2ir) 1 LA 1 (1+2ir) +4 2ir) where r is the span ratio, J, : 1, and 7 is the ratio of the moments of inertia, J :J,.) Similarly, the distance intercepted by # H on the symmetrically located vertical will be m,. Projecting these two intercepts (m, and m,) upon the respective end-verticals, the resulting connecting line N O will be the closing side of the equilibrium polygon; and transferring the altitude MT to P K, we obtain N K O as the true equilibrium polygon in its proper position relative to the cable-curve. Going through this construction of the equilibrium polygon for different positions of a concentrated load, there is finally obtained the Reaction Locus K, K,, as the geometric locus of the point K in which the sides of the equilibrium polygon, repre- senting the end-reactions, intersect the load vertical. Further- more, the different directions of the sides of the equilibrium polygon envelop a curve UU, which we call the Reaction Tangent 52 ARCHES AND SUSPENSION BRIDGES. Curve. We also construct the curve A,, @ whose ordinates are the end-moments per unit H( —m,), for different positions of the concentration in the side span. With a load at Q, if the point Q is projected horizontally to Qi, the line drawn from Q, through the fixed point F, represents the line of the equilibrium polygon in the main span. (The corresponding analytical expressions are oo _fe@, é(—é) 2ir (1+ir) £ 8, Q=—m=[F S|] apr asm Ot7, and 01,9, = FM =| lammaemOt+) where i and r have the significations given above.) It now becomes a simple matter to determine the manner Fig. 33. of loading for producing the maximum bending moments at any section of the stiffening truss. If S, is the point of the cable at the given vertical section, then the tangent from S, to the Reaction Tangent Curve U will intersect the Reaction Locus at some point J which will be the ‘‘load-division point’? (= eritical point) in the main span; while the critical point in the left side span will be given by the point Q. In Fig. 33 are shown the diagrams of loading for three different cross-sections, S,, S., S53. If the three-span bridge is hinged at the towers, the equi- librium polygons for a concentrated load and the resulting reac- tion locus are determined for each span exactly as in Fig. 31. STRESSES IN THE STIFFENING TRUSS. 53 For producing maximum positive (or negative) moments at any section, the corresponding span must be loaded up to the critical point, and the other two spans must be free from (or full of) load. 3. Determination of the Moments for Full Loading and for the Severest (Critical) Loading. a.) Method of Influence Lines. In a single-span bridge (Fig. 31), the moment produced at any cross-section M is expressed by M=M—Hy—y|~—H].. ee ates dane 190 : z M ‘ For a moving concentration, a represents the moment influence line of a simple beam, constructed with the pole distance y. Hence the moment M is proportional to the difference between the ordinates of this influence line and those of the H-curve. The influence line for the simple-beam moments is familiarly obtained as a triangle whose altitude at the given a (l—«) ley the point J (Figs. 31 and 32), which is obtained by the aid of the reaction locus as previously described, and projected down to the point 7 in the H-curve. If ®, and , are the areas respect- ively above and below the H-curve, included between it and the above-mentioned influence line (areas 7b h and a mi, in Fig. 31), then, for a uniformly distributed load of p per unit length, section = It may be constructed, however, by using For finding the areas ©, proceed as follows: First determine the areas of the H-curve by analytic or graphic integration, divide by 4, and plot the resulting lengths as the ordinates of a curve bk 1 ;s0 that, for instance, the area bi’ ih =i’'k + (Fig. 31). Then the ordinate 7’k gives the value of H for a uni- form load covering the length BJ if, at the same time, ac=G= B . Similarly, dividing the area of the triangle av’b by + gives the ordinate 7’n. Consequently we have 54 ARCHES AND SUSPENSION BRIDGES. b,=kn- + and mar (—M) =kn.y. In the same manner we find the effect of a full loading to be Mit=tl.y, and therefore the maximum positive moment will be max (+M)=(kn+tl).y. The distances i n and tl are to be measured on a force-scale fixed pl =o The above treatment may also be applied to the continuous stiffening truss (Fig. 33), but in that case the influence line for the moments M is no longer simply a triangle but must be specially derived from the theory of the continuous beam. For this purpose we may utilize the equilibrium polygons which were required in constructing the reaction locus; the ordinates of these polygons measured below the horizontal line B,, C,, and divided by (y—m) give the corresponding ordinates of the required influence curve. At the sections near the points V and W, where the values of 747—-m are very small, this method is not conven- iently applicable; but, for these sections, the moment Jf is very nearly (and for the points V and W exactly) equal to the moments M of an ordinary continuous beam. b.) Determination of the Maximum Moments from the Equilibrium Polygon of the Loading. If z denotes the intercept between the equilibrium polygon of the loading and the axis of the cable or arch, measured in the vertical line through the section M, and if H is the horizontal force, then, by the prin- ciple established in § 5, the moment in the stiffening truss at the section M is given by Mia A 2B. oe aa Dea ar a secs noes weet aed Sa (101. The quantities H and z may be determined either by computa- tion or by construction. In the latter case, we must first find the value of H for the given loading from the ordinates or areas of the H-influence line and, with this as a pole distance, con- struct the equilibrium polygon of the loading in its proper posi- tion relative to the cable-curve. For uniformly distributed loads, the expressions for the horizontal force H have been deduced in § 6. These expressions may he plotted to give an H-curve which, as previously men- tioned, may also be constructed by graphic integration of the by the relation ac = aa STRESSES IN THE STIFFENING Truss. 55 ordinary H-curve. If, in addition, we construct, for a single- span stiffening truss, the parabola b vc as the integrated influ- ence line for vertical reactions (Fig. 31), then the two inter- cepts 7’y and 7’k will represent respectively the horizontal and vertical components of the left end-reaction for a load covering the section J B of the truss; and the vector resultant of these two forces will give the direction of the side of the equilibrium polygon passing through the suspension-point A’. The distance Mz intercepted by this line on the vertical through M is the required factor z. The determination of this quantity may be simplified by drawing the locus of the point LZ in which the polygon-side A’ intersects the center-line of the loaded seg- ment; the resulting curve is called the Second Reaction Locus.— We then obtain the maximum negative moment at the section M by multiplying together the two distances i’k and Mz, meas- uring the first by the scale of forces (ac— 4) and the second by the scale of lengths. Assuming a parabolic cable-curve, we may establish the following analytical expressions with the aid of equations (80) to (89): 1. Three-Hinged Stiffening Truss. (Single Span with Cen- tral Hinge). Under complete loading, the equilibrium curve co- incides throughout with the ecable-curve. Hence the load is carried entirely by the cable and the stiffening truss remains unstressed, i. e., Miotr = 0. Under the critical loading defined by equ. 98*, we find = pxe(l—z) (l—22) . Mmax = + “337 —8ay (102. From this expression we find that the absolute maximum moment will occur at «= 0.2341 and will have the value Absolute max. M = + 0.01883 p?......... (1022. If one-half of the span is loaded, there will be produced positive moments in the loaded half and negative moments in the unloaded half amounting to Mf = = px (4-—): and the maximum moment, occurring at the quarter-point, will be Mac. M= j pl’ =0.01562 pr’ 2. Stiffening Truss without Central Hinge; Parabolic Cable, Constant I for Truss. a. Single Span. With the span completely loaded, the bend- 8: : ; 1 ing moment at any section will be M=-> px (l—z) — H.y; 56 ARCHES AND SUSPENSION BRIDGES. substituting the value of H from equ. (82), we obtain Moot =] P © (1-2) A eee eee (103. For the critical loading, applying equs. (83) and (97), we obtain Mau —2840=2 [3 (8) —4(8)+3 8) ](0—$)"c08 From this value the maximum moment may be derived by the relation In equ. (104) the value of £, must be obtained from equ. (97). N,, denotes the denominator of the expression for H in equ. (75). The absolute maximum moment is affected by the value = corresponding to a very weak stiffening truss, we obtain the following value of the absolute maximum moment, occurring at « = 0.2501: Absol. maz. M = + 0.01652 pl?............ (104°. With the half-span loaded, we find in this case, as in the ‘ 1 ‘ preceding, max. M— | pl*; this moment occurs, as before, at of N,,; substituting its smallest value, N;,; = the quarter points. As the quantity N, representing the stiffness of the truss, increases in value, the maximum negative moments diminish, whereas the full-load moments and the maximum positive mo- ments are increased with respect to the above values. Equation (104) applies to all sections from «—0 to 2, = 4 : N,,. For the minimum moments at the sections near the cen- ter, from x, to 1—2,, it is necessary to bring on some load also from the left end of the span; so that the expression for these moments consists of two parts which may be obtained by apply- ing equ. (104) to the two symmetrically located points x and 1— gz. In Figs. 34a and 34b are plotted the graphs representing the maximum moments in suspension bridges with and without a central hinge. For the second graph there was selected the example given on page 37. The data were 1—50 meters, f—6.5 m., 1, sec? a, = 19.05 m, —— = 2.049, 4,=0.03 sq. meters; hence NV = 1.8692, giving a moment at the center of the span, for a full-span load, equal to 0.0180 pil’. STRESSES IN THE STIFFENING TRUSS. 57 B. Three Spans. Stiffening Truss Continuous. (Fig. 27.) With all three spans loaded, the moment at any section of the main span is given i Moa=[s p—HZ]e (—x) —e (jpt- Hf).. (105. and at any section of fe side span distant « from the free end 1 4f, 1 2 o 4 = (sp—HF)2(—2) —e (sn! —H/)#.. (106. where the value of « must be taken from equ. (77) and that of H from the combination of equs. (84) and (85). The maximum moments, produced by the critical loading, must be caleulated by the general equs. (58+). Fig. 34a, Maximum Moments in Stiffening Truss with Central Hinge, Fig. 34b. Maximum Moments in Stiffening Truss without Central Hinge. + =0.13 N=1.8692 Hi a +++ 1H a Got 402 402 gov gos pe? . Three Spans. Stiffening Truss Hinged at the Towers. (Fig. 29.) With all three spans loaded, we have =a 8 fy 1%; Me >p a (l—2) fp—s- (424 ae | -- (107. or, since N,, = N,, (i+ 2 = 7 apply equ. (103) to this a the moments agreeing very closely with those of a single-span bridge, completely loaded. The moments in the side spans are also given by equ. (103) if 1, is written instead of I. The maximum negative moment at any section of the main span is obtained by loading a length 1— é, in that span and completely loading both side spans. Then, by (83) and (88), Haw — “5 {[2—G) 4G) (108 +3) | G-i) 44a} J ) very nearly, we may also 58 ARCHES AND SUSPENSION BripGEs. The maximum negative moment at any section of a side span is obtained by loading a length J, —&, in that span and com- pletely loading both the other spans. Then, by (82), (88), (89), pat oS 2 sree —s— & Moe — 5 {[2—f 46) 4064 OT} ce G-%) - Fiz (14: Li rf For the main span, in equ. (108), the value of é, is defined by the relation similar to equ. (97), — &3 +16? +P E= Nl f= y? and for the side span, in equ. (108), the value of &, is defined by the following relation obtained from equs. (797), (93) and (96), c IT q, c (= é°+ hé’+ &1). ie Ng bh fy Pe The maximum positive moments are given by the relation Minax = Mot ae Main? 4. Determination of the Maximum Shears. The critical loading for shears at any section may be obtained from a consideration of equ. (59). Accordingly, the load-point yielding a change in sign of the transverse shear is found by locating the point J (Fig. 35) in which the reaction locus is cut by a polygon-ray drawn at an angle 7, i. e., parallel to the cable- tangent at the given section. In addition, the sign of S changes at the section itself, so that the critical loading is as indicated in Fig. 35. The same rule applies to the main span of the continuous form of suspension bridge but, at the same time, the side span nearer the section must be completely loaded and the other side span free from load or partially loaded for a maximum positive shear. To determine the maximum shears we may again employ the two methods previously introduced: a.) Jethod of Influence Lines. The shear at any section of the stiffening truss is given by equ. (59) as S=S—H (tan + — tan o) = S = (tan7 — tano) |! H\ — tang Here o denotes the inclination of the connecting line A’ B’ below the horizontal (-+ if A’ is higher than B’) (cf. Fig. 20). If the stiffening truss is one with parallel chords, and if 8 is the STRESSES IN THE STIFFENING TRUSS. 59 inclination to the vertical of the web member at the given section, then the stress in this web member will be Ss tant—tane P=S . sec8=sec8. (tan+— tan) [ = H]..(108. The values assumed by the bracketed expression for different positions of a concentrated load may be easily represented. They are, namely, the differences between the ordinates of the H-curve and those of the influence line for the shears S, the 1 latter being reduced in the ratio of ——————— or, if the tan? — lane Fig. 35. a Bel + ension-"7 at Fi B A a . suspension points are at the same level, in the ratio of cotr :1. The latter influence line is familiarly obtained by drawing the two parallel lines a ¢, and bs, (Fig. 35), their direction being fixed by the intercept ad= G eae =a laid off on the tant — tango end vertical. The vertices ¢, and s, lie on the verticals passing through the panel-points of the given web member. The inter- section 7 with the H-curve, which should fall directly under the critical point J, affords a check on the construction. The maximum shears or shearing stresses produced by a uniformly 60 ARCHES AND SUSPENSION BRIDGES. distributed load are determined by the areas included between the H and §S influence lines; all areas below the H-curve are to be considered positive and all above negative. These areas must be multiplied by the coefficient + sec § (tan + — tan o) to obtain the greatest stresses in the web member. Equ. (109) is applicable also to the stiffening truss con- tinuous over four supports, if o = 0 and the structure is sym- metrical about the center line; but in this case the S-influence line no longer consists of straight lines. At the sections near the center line, where r= 0, the shears are very nearly the same as in an ordinary truss, i.e, S=S very nearly. b.) Determination of the Shears by Means of the Equtli- brium Polygon for Partial Loading. With the aid of the second reaction locus, we may again construct the equilibrium polygon and the corresponding force polygon for partial loading. (Fig. 350.) In the latter diagram, 0 a B is the force polygon for a load covering the length J B: 0a—AH, o B is parallel to the polygon-side at A’, hence a8 —§S, and if oy is drawn at an angle r, y@—S for the load J B; similarly 0’ a’ B’ is the force polygon for the load M B, so that y B’ is the corresponding value of 8; the difference between the two shears S, represented by 8’, is the maximum positive shear (+ Smax) for the sec- tion M. On the assumption of a parabolic cable-curve, we may develop the following expressions for the maximum shears: 1. Stiffening Truss with Central Hinge. Single Span. With the span completely loaded, the equilibrium curve for the load- ing coincides with the curve of the cable; hence the shears will be Stot = 0. For the maximum positive shear, the truss must be loaded ; * : . 4 P from the given section to a point whose abscissa is £, = 37-4, - For «> 0.251, the load must extend from the section to the end of the truss. We thus obtain, for c= 0 to 0.251, __ _ p(P—3lx + 427)? Max. (xt 8) == 5 ~ 8F (31 — 4x) Sep aeeeetee ace (110. x (31— 4 max.(+S)=H+ pe (Ol fe) The absolute maximum shears occur at the supports and at STRESSES IN THE STIFFENING TRUSS. 61 the center of the span and amount to maz.( +S) = - pl and 1 1 . “gpl respectively. 2. Stiffening Truss without a Central Hinge. Constant Moment of Inertia. a.) Single Span. With complete loading, the shear at a distance x from the end of the span will be, by (59) and (82), Sto = vp (I-20) RF, Loading the truss from the given section to the end of the span, the maximum positive shear will be, by (59) and (83), Sea 3 (1-7) 1-9 — FBO 4) -30-8) +4} For the sections near the ends of the span, from «=O to oat (a— =); the loads must not extend to the end of the 4 span to give the maximum positive shears, but must extend only to a point whose abscissa, é,, is determined by the following equation, deduced from (94) : &(2 +l — 4) =A eee cee (113. For these sections, the shears given by equ. (112) must be increased by an amount C39) CDEC a) | ae i ene (112°. -3(:-9)' +33} We have also: Simin aaa Srot =o, Smaz- The largest shear occurs at the abutments; with the approxi- ..(112. mation N=; its value is Absolute max. ( +8 ) =0.1528 pl. At the center of the span, max. (+8)=+ipl In Figs. (86a) and (36b) are shown the graphs of the max- imum shears. 62 ARCHES AND Suspension Bringrs. B.) Three Spans. Stiffening Truss Hinged at the Towers. (Fig. 29.) With the three spans completely loaded, the shear at any section in the main span will be, by (82), (88) and (59), a oe ha ' Sis —3p (1—22)f1 sy +2735)] pace (IES and in the side spans, 1 Thy’ fy Stoe =F (,— 22) fa - ee be Ae (a + 2h )] aca. The maximum positive shears in the main span are ealcu- lated, exactly as for a single span, by equs. (112) and (112+) Fig. 36a. Maximum Shears for Stiffening Truss with Central Hinge, (tinea Lea] Fig. 36b. Maximum Shears for Stiffening Truss without Central Hinge, F013 N=18692 Lend a mel ee 3 Tess 02336 l- ° os o2 q3 pe but substituting for N,, the denomination of equ. (79). In the side spans, Sm tpl, (1—£) {1 — 8. Bh ees (a6. a0 —£) +3]} and for the sections near the ends of these spans, namely from r=(0 to rab (1-4 Nu, ie and from x7 = i ( 1 aa oN tl :) to x —1,, the value given by equ. (115) for the maximum shear must be increased by the amount STRESSES IN THE STIFFENING TRUSS. 63 1 £5 2° 8 If? 2a 1 _ & 2 Beh ¢ - i) te Lf? ( i dL: ( I (116 1 GV EL t : —a( a Te } where &, the abscissa of the second critical point, must be obtained from 1 hiv if be (Ht ibs —@) = GN ne Spgs pgs (117. The above equations (115), (116) and (117) for the side spans are analogous in form and derivation to equs. (112), (112#) and (113) for the main span. 5. Effect of Temperature Variations.—If ¢ is the variation of temperature from the condition of no stress and w is the coefficient of linear expansion, then the equation of condition for the horizontal force H to replace equ. (67) will be SGA tot) eas + fo SY ds—o...(118. If the denominators of the expressions in (69), (70), ete., are denoted by No, Nz, ete, and if & ~ 421, sec? u, is assumed approximately equal to the cable-length ZL, then the development of the above equation in the same manner as on page (32) will yield the following expressions for H, for the several cases : a.) Single Span. Stiffening truss without a central hinge, cases a) and b), (p. 31): A? [ 7 - — 2f' tan a| Ay. B.w.t A, — la x < . (119. [:. AoE .w.t oe Nr and, neglecting the effect of the suspension rods, a ee eee (120. 71 In the type of structure shown in Fig. 25, a uniform change of temperature in all the members will produce no stresses in the structure. But, in that construction, if the arch under- goes a temperature change of ¢ and the stiffening truss and rods are subjected to a temperature variation of t’, the horizontal stress will be 64 ARCHES AND SUSPENSION BRIDGES. y 1—=Z- A y)AD.o(t—t’) pees! ot aah 2 ae (121. Neo +94, l For determining the moments and shears we have the equa- tions =—Hy.y, S=—=—AH,.tanr. In a similar manner we may obtain the effect of a displace- ment of the points of support of the cable produced either by a yielding of the anchorages or by a stretching of the backstays. If the span-length is thus diminished by an amount 61, then the variation in the horizontal tension will be Am cece (122. With the approximate assumption of a continuously curved parabolic cable, the denominators of the expressions (119), (120) and (122) may be replaced by © gE - N,; (where N,, repre- sents the denominator of expression (75) or (75*)). Neglecting the effect of the suspension rods there results, 15 I L 1 ha Si hee 8 ad 1 FF LE .... (123. i 13 ee ( or, substituting the value of Z, (+254 ee sec” a) .E.o.t.1 A.= oF ey HDS 8 I A 2 rie (at 3 pt 7 sec 7) In the example, for which the maximum live-load moments are plotted in Fig. 34b, the horizontal tension due to a tempera- ture variation of + 30° is computed by equ. (1237) as follows: (1.8521) (20,000,000) (0.0000118) (30) (0.06147) A. = 24.3854 = = 33.05 tonnes. The resulting moment at the center of the stiffening truss attains a value of + 214.8 t. m. b.) In the Three-Span, Continuous Suspension Bridge, the horizontal tension due to a uniform variation of temperature throughout the structure will be: 3EIw [(2 + et)+2 ot (1+ na + tan?a,) | f?. Nie If the stiffening truss has a central hinge, it might be sup- iS sas, CDE, STRESSES IN THE STIFFENING TRUSS. 65° posed that no temperature stresses can occur, since, in a static- ally determinate structure, no change can be produced in the horizontal foree except by the application of external forces to the system. Actually, however, there will be a change of stress since, in general, the deformation of the cable accompanying the sag or rise at the central hinge cannot be neglected from con- sideration; so that the original assumption underlying the approximate theory can no longer be retained. Engineer G. Lin- denthal was the first to call attention to this point:* but his approximate method, too inaccurate for a very stiff truss, results in overestimating the effect of temperature in the structure with a central hinge. When the cable is lengthened by a temperature rise of ¢° (or by a displacement of the cable supports), the crown or cen- tral hinge will sag through some distance Af. If we assume that the axes of the two halves of the truss remain straight during this sag, there will result a break in the cable-curve at the mid- point; hence the cable will no longer conform to the parabolic equilibrium curve corresponding to a uniformly distributed load. There must therefore result a change in the distribution of the suspender-forces: the suspension rods near the ends A and B, as well as those near the central hinge, C, will increase in stress while the intermediate rods will be relieved of load. The effect of this is to give the stiffening truss an increased share of the load at the intermediate points, causing it to deflect downward. The reverse occurs with a rise of the cable crown: the rods at the points specified above are subjected to increased stress and the truss is bent upward. A rigorous treatment of the problem,t along the main lines developed in § 9, leads to the following results: Let H be the horizontal tension in the cable for that condition in which the eable coincides with the equilibrium curve of the external load- ing, so that no forces are acting on the stiffening truss. As a result of temperature variations or other cause, let the cable- crown deflect downward through the distance Af. Let the accompanying deflection of the truss at a distance x from the abutment be y. Then, neglecting the elongation of the sus- penders and denoting the half-span by a, the cable ordinate is changed into y+ 2 -Afty and the horizontal tension becomes , f H’=4. 7a * Report of Board of Engineer Officers, Washington, 1894. Appendix D: Temperature Strains in Three-Hinged Arches. t Melan, Die Ermittelung der Spannungen im Dreigelenkbogen und in dem durch einen Balken mit Mittelgelenk versteiften Hangetriiger mit Beriicksichtigung seiner Formanderung. Osterr. Wochenschr. f. d. 6ffentl. Baudienst. 1903, No. 28. 5 66 ARCHES AND SUSPENSION BRIDGES. The bending moment in the stiffening truss will then be M=(H—H’) y—W' (2 af+n) or M=[(y—2f)af—at| + - If the cable is initially a parabola, then ip ee Gea a hence, uM [2° af—a] H’. The differential equation of the elastic curve of the truss, 2 EI aa ==—M, on introducing the abbreviation H’ H — “EL Ske Fl ee ey (125 becomes iy 2 4 .2 @(a—e) da? Cn“ e @ -Af=0. The integration of this equation finally yields the following expression for the moments in ‘the stiffening truss: at — ec% + ec 8® Af u—[1 “te [eer Sansui (126. The maximum value, at c= will be (1—e%"*)? 1 + ect The bending moments are therefore directly proportional to the crown deflection Af; but not exactly to the quantity J, because this is also involved in the coefficient c. Only for small values of I is M proportional to the stiffness of the truss, since, at the Af, a At the other limit (with an infinitely rigid truss), I= o, (c=0), 7»=0, hence, Max = -QHI- “f aa (127. limiting value, with I= 0, (c= ©), we obtain M,,,., = 2 EI The last expression gives pretty close results even for trusses of moderate stiffness. STRESSES IN THE STIFFENING TRUSS. 67 The sag Af at the crown, caused by an elongation of the cable amounting to A D = w. t. l, is given with sufficient accuracy by Af = (+554 esha foes na anh (128. The equations (127) and (1277) apply also in the case where the crown deflection Af is caused by a uniform load covering the entire span, if H is the total horizontal tension resulting from this load. Consequently it cannot be asserted that the introduction of a hinge in the stiffening truss will wholly eliminate the second- ary stresses due to temperature variation. 6. Secondary Stresses. Up to this point in the theory of the stiffened suspension bridge it has been assumed that there is no friction either at the pins of the chain-cable or at the panel-points of the truss. The effect of riveted connections at the latter points is the same as ordinarily given in discussions of secondary stresses in framed structures;* so that we need to consider only the effect of friction at the joints of the chain upon the values of the horizontal tension and of the external forces acting on the stiffening truss. In wire-cable suspension bridges, the resistance of the stiffness of the cable takes the place of frictional resistance. If d denotes the diameter of the pins, ¢ the coefficient of friction, 7 the cable tension, then the moment taken up by the joint-friction cannot exceed My =¢.T. a or approx- imately My = $.H. q; the moment transmitted to the stiffen- ing truss will then be M—=M—H.y—M,=M—H(y+¢-¢)- The expressions for H established in § 5 (equs. 69 to 73) need not be altered, therefore, except to increase the arch ordinates y by¢. 4. The latter quantities, however, will always be very insignificant compared to y; even with the full effect of friction, assuming ¢ = 0.2, the value of ¢ 4 will never exceed a few centimeters or millimeters; consequently this correction may be completely neglected in the expressions for H. Thus the value of the horizontal tension is not sensibly affected by friction in * Handbuch des Briickenbaues, Chap, JX; Konstruktion der eisernen Balkenbriicken. Leipzig, 1906. 68 ARCHES AND SUSPENSION BRIDGES. the chain or stiffness in the cable. The external forces acting on the stiffening truss likewise remain unaffected; at the most, the bending moments may be diminished by the friction-moment d $ H.+. If the members of the chain are not pin-connected, but riv- eted instead, the stiffness of the chain must be added to that of the stiffening truss; and the moment M—M—Hy must be shared by both of these structural elements in the relative pro- portion of their moments of inertia. If 7, is the moment of inertia of the cross-section of the chain, and 7, that of the stiff- ening truss, then the chain will carry a bending moment of M’ = Sa . M, and the truss will carry M” = M— M’. In pin-jointed chains, M’ cannot exceed the value ¢. HT . a 2 From the value of M’ we may compute the resulting secondary fiber stresses in the chain. DEFLECTIONS OF THE STIFFENING TRUSS. 69 §8. Computation of the Deflections. 1. Deflections Due to Loading. By Castigliano’s Theorem, the deflection at any point of an elastic system may be obtained as the differential coefficient of the work of deformation with respect to a force supposed to be acting at the given point in the direction of the deflection. Fig- 37. On the truss A B, Fig. 37, let a concentration G be ap- plied at a distance é from the abutment. Let the resulting AB axial forces and moments in an é--4 Ve i the structure be denoted by t ‘e N,andM;. Then the deflec- tion at any section having the abscissa x will be, aw Ne dNx M d Mx eam Tm (FE SN dst Be. oH ds oo... (129. where N, and M, denote the axial force and moment producible by a load G, applied at x. Writing for the horizontal tensions corresponding to the load-positions é anda, H;=h;.G@ and H, =h, . G, respectively, where the coefficients hz and hx, repre- senting the horizontal tensions per unit load, may be obtained from the previously established equations (69) to (75*) or from the H-curve and may therefore be considered as known quanti- ties, there results dNx_dNz 1 Mx (aMx__ eye iG dh "* 2a \eee 9 ™ we thus obtain for a single-span suspension bridge without a central hinge: Cable: aqw_! a a a6. ea BA Mee 1 iy? ea nee “G Backstays: @W __ 2h, seca. eo x Hy . seca, Dacre hy 8€C a4 2 70 ARCHES AND SUSPENSION BRIDGES. Suspension Rods:dW __ SH, vy oy A*y dGx a EA, Bes a 1 aul Ay a Y =A. hg hy. dy (=") G Truss: a F inf (M t —Hyy). (<& —y hx)dx — mx + he. hx set ana]. G Here m, denotes the moment at the point M of a simple beam produced by a loading composed of the ordinates of the moment-curve for a unit load at C. For this we obtain the expressions a (l—€) oy ...(130. for a > & m,=[2le—x?—@ £2) fora < & m,—([2lé—f—x’| Furthermore, mg and mz represent the moments producible at the sections ( and M, respectively, in a simple beam, when the loading consists of the area between the cable-curve and its chord. y» is the doubled static moment of the same area about the chord (. —f y? dz). Since, by equ. (74), hg= me and e+ep hy ee therefore m;z.d:=mx-h;. Adding together the several values of oe, we obtain d Gx Ax = Cs A?y +2], sec? ta + Sy’ (=4 “)’) . (131. + or (mx thy. hé.w~— 2 he ; ms) | GC Noting that the first bracketed expression in the above equation, with the notation previously introduced, == cp, and that m,=h, (u + cp), we finally obtain the following simple formula for the deflection of the truss at M: 1 Uae [mx —h, 2 Mx). G...eee, diseawetne- ae (131° which could also have been derived directly from the differen- tial equation of the elastic curve oe -—.. In the above expression for y, m, is determined by equ. (130), hg by equs. (71) to (75%), and mx by the relation DEFLECTIONS OF THE STIFFENING TRUSS. V1 a ! mM: = — feydet+Ffy (l—ax) du. .. (182) 0 wo or, for a parabolic cable, my = FS (@—21 aH D) ee eeeeeee es (132°. The ordinates of the elastic curve may also be obtained by a graphic construction. This is essentially the same as the method used for the two-hinged arch in Figs. 12 to 1‘ (Plate I). Thus, if we write the expression for 7 in the form then that quantity may be represented as the difference between the ordinates of two funicular polygons, one of which (I) coin- cides with the H-curve, while the other (III) is constructed for an imaginary loading consisting of the simple moment- diagram (Fig. 14). The latter is drawn by means of the force polygon (Fig. 1°) having H; for its pole distance. If 2, and 2, are the ordinates of the two funicular polygons, measured to the seale of lengths, and if a is the width of the strips into which the load-diagrams are divided, and p is the pole distance of the force polygons (Figs. 1° and 1‘) which were used for the construction of the funicular polygons I and III, then we have i oe (23—41) ° I, ET He.a-p so that the scale for measuring the deflection ordinates is times the scale of lengths. Jn the example on page 37, IJ = 0.0625 m4, w = 2.5 m., p = 35 im., 11.45 He = q59 ¢ = 0.764 G, E = 20,000,000 t./sq. m.; if the scale of lengths adopted is 1 m. = 2 mm., then the scale for the platted deflection ordinates will be 1 mm. (actual) = 37.4 mm. on the drawing. With the aid of the deflection diagram drawn for a unit load at C, it becomes a simple matter to determine the deflec- tion producible at C by any system of concentrations or by any continuous loading. By the principle of the reciprocity of deflections (Maxwell’s Theorem), the deflection at C caused by any load at M is equal to the deflection producible at Jf by an equal load at C. Consequently, the curve of deflections (7) for a unit load at C is at the same time an influence line for deflections at the point C; hence to find the total deflection at 72. ARCHES AND SUSPENSION BrRIDGEs. C, multiply each load by the corresponding ordinate of the deflection curve, and add the products thus obtained. The above construction, based on the assumption of a con- stant moment of inertia of the stiffening truss, may also be extended to the case of a variable moment of inertia. If J is the moment of inertia at any section and J, a fixed moment of inertia, then it is simply necessary to reduce the ordinates of the load-diagrams, i. e., the area between the cable-curve and its chord and the area of the simple moment diagram, in the ratio I, : I, and proceed with these reduced areas exactly as in the above construction. In the example of the arch in Plate I, the variation of the moments of inertia was thus taken into account. For any specified loading, the deflections of the stiffening truss may be computed in another manner. On the truss there are acting, in addition to the external loads which produce the 2 moments M, the upward-directed suspender-forces, H ca or Hy at a ane x of Wide — : = ee 2M —z)dz ate (@ m,) while the latter loads, the suspender-forces, in the case of a parabolic cable, are uniformly distributed and = =f es they produce a deflection — 7” which, with a constant J in the stiffening truss, is determined by the equation 1 we 1 3 2 3 f ag f= = «x (1 21x? + x*) = =e 3EI The resulting deflection of the truss is therefore 717’ — 7”, which coincides with equ. (1317). We thus find the- deflection at the mid-point produced by a uniform load of p per unit length covering the entire we to be given by the equation pl ; =a (1— ar FE esas i ee where N denotes the denominator of equ. (75) or (75*). If merely the half-span is loaded with p per unit length, then the deflection at the quarter point will be, by (133) and (134), DEFLECTIONS OF THE STIFFENING TRUSS. "3 . eet a 57 8 pl tha in the loaded half, = y=—qy (31 — sey) h; 1 578 a it in the unloaded half, »=— sig (= ° by 23 i . (136. By equ. (75), N will always be greater than = ; substituting this minimum value in (136) we obtain the upward or down- ward deflections at the quarter points: al 5 Pp L\*. 1S" 384 BT (=) For the stiffening truss with central hingc, we must substi- tute in the deflection formula, equ. (131), the valucs hg = + and hy where it is presupposed that é and x < i pa 2f ~~ BF? Denoting the first bracketed expression, which represents the effect of the axial forces, by the symbol 3, equ. (131) takes the form: _ _Gat 3 EP BAe sens (19% 1 +35 [mit ro 37 (mt + mee) | The symbols my, », my and mg have the significations pre- viously specified. For « > 4, __ad(l—a) G ne 1 Vx *s+5™ m,;-+ 4p? ~ BF [mx 4 PE Ao +m, (l—«)]] Substituting the values of m;, » and mg which obtain for a parabolic cable, there result the following expressions: .. (138. Porz<é5>é: 7é(l—a2) 2 i =i ne a + Sopp EU 8) +52 (I—2)*—P] (141, Here, again, we may apply the graphic method for deter- mining the deflection-ordinates. Thus, equ. (137) may be written = Om (m,—h gmx) +3" a icin Pa euea Cale (142. where b= [ae (nt Ca)—m JE -S sa. The bracketed expression in equ. (142) may be represented, as before, by the intercepts between the two funicular polygons I and III. (Figs. 1* to 1t, Pl. I), where the simple moment dia- gram (Fig 1°) is obtained from the force rales (Fig. 1°) constructed with the pole distance H t=G. ar To obtain the actual deflections, the above intercepts must be increased by the ordinates of a triangle whose altitude, at the center of EI Hg.ap © If, as before, 7’ and 7” denote the deflections specified by equs. (133) and (134) for the hingeless structure, produced by the external loads and by the suspender-forces respectively ; if H is the horizontal tension produced in the hinged structure by the given loading and A, is the horizontal tension producible i in an identical structure without the hinge; and if NW is the denom- inator of the expression for H, in equ. (75), then by a trans- formation of equ. (137) we obtain, , r N 2 ta Pal! Be) > a From this we find the deflection at the mid-point produced by a uniform full-span load of p per unit length to be the span, is 8° pl* ag (WE) cece oxen 145, and the deflections at the quarter-points produced by a half- span load are, in the loaded half, 1 5 13 * 384 Hs ) + 384 “a (N—2 5 no in the unloaded half, .... (146. on 1 1 2° 384° Es) +3n(" 3 5 z DEFLECTIONS OF THE STIFFENING TRUSS. "5 2. Deflections Produced by Temperature Variations or by Displacements of the Cable Supports. Equ. (1312), which defines the deflection-ordinate at a distance x from. the end of the span, applies also to the case of deflections produced not by the loading but by temperature effects or by a yielding of the towers or backstays. In this case, m,—0; and G.h must be equal to the horizontal tension H, caused by the given influence. Consequently, Here H, must be determined by equs. (119) to (123) or, if the deflection is caused by a yielding of the cable supports, by equ. (122), while m, is given by equ. (132) or (1327). Since the quantity m, is proportional to the horizontal tension pro- ducible by a concentration applied at the point x, the deflec- tions of the stiffening truss are given by the ordinates of the H-curve, i.e., the funicular polygon I (Fig. 1*, Pl. I), where, RI lta with the previous notation, the scale unit to be adopted is times that of the scale of lengths. In the stiffening truss interrupted by a central hinge, any stretch in the cable will cause a sag at the point « (< +): if the bending of the truss is neglected, this sag will amount to 2 | Af : = , where Af is determined by equs. (39) to (44). 76 ARCHES AND SUSPENSION BRIDGEs. § 9. The More Exact Theory for the Stiffened Sus- pension Bridge. The theory developed in the preceding paragraphs (§§ 5-8) gives satisfactory results only for those systems in which the elastic deformations are limited in amount. In the systems treated above, the admissibility of that approximate method of design depends upon the degree of thoroughness with which the stiffening truss performs its function of limiting the mobility of the system. If the truss is provided with but small chord- sections or has but small depth, that is, if it has a small moment of inertia and consequently great flexibility, or if it is provided with an intermediate hinge, then, under certain cases of loading, the deformation of the system ought no longer to be neglected in determining the stresses. 1. Truss without a Middle Hinge. Single Span. Let H, denote the horizontal tension in the cable in its initial position of equilibrium (when it is carrying no live load), in which con- dition it is assumed to form a parabolic curve. The stiffening truss is then without stress. Let the suspender-forces, under this loading, be denoted by s, (per unit length), and the dead. weight of the cable by k; then by equ. (51) we obtain the follow- ing differential equation of the equilibrium curve: On bringing a live load upon the structure, of intensity g per unit length, let the suspender-forces change to s,, the hor- izontal tension to (H,-+ H), the ordinates of the equilibrium curve to (y+ 7); we may again write the relation between these quantities: (Hot H) Yt ag th (B. By the addition of the live load q there is also produced a deflection of the stiffening truss. Neglecting the elongation of the suspension rods, the deflection of the truss at any section may be equated to the sag (7) of the cable at the same section; so that 7 is the ordinate of the elastic curve of the truss for which we may write the differential equation d BE OG) oad ae eas (y. Exact THEORY FOR THE SUSPENSION BripGE. UY The addition of equs. (a), ses and (y) — on — (A, = ¢. Introducing the ots ga + Hy — 72 rr. OO eS aR a toe ea ee a LS See ee Sere (148 and assuming that J and q are constant within the limits of integration, the integration of the above differential equation yields an Sp Act + Bet a (+2 oat. -); or since, on account of the initial parabolic form of the cable, eh ay dz? ~2 and Tet Os En 4 cx -x_ «=f BE dg 4e +Be a+. (a 7 (8. The forces g — (s, —s,) acting on the truss produce a bend- ing moment M—M — (H+ 4H,) (ytyn) +H. y,i-e., M=M—(74-) p Sy nsec iecavcank (149. where M denotes the moment producible by the loading g in a simply supported beam. Consequently, EE dx ng Oe ere (M— Hy). aa he Eiquating the expressions (8) and (e), substituting y= i x.(l—g) and replacing the arbitrary constants A and B by C,= ant a nd C, = aie we finally obtain aoe q H+ 4H, (150. (Cre + Cne% +26 Ff a (la) +B — 54 The arbitrary constants C, and C, are determined by the eondition that at c—0 and x=1, we must have »=—0. If the load qg covers but a fraction of the span, different values must be given to the constants in the equations of the elastic curve for the loaded and the unloaded segments. The necessary equa- tions are then given by the condition that, at the division point, the - adjacent segments must have identical values of y and —— at 78 ARCHES AND SUSPENSION BRIDGES. Substituting the value of » in equ. (149), we obtain an expression for the moment M=—H[Cye™+ C,e%+2(f —4 )| aitiat From equ. (149) we observe that the moment M is no longer proportional to the intensity of the load q which produces it, but depends in amount also upon the horizontal tension H, which originally existed in the structure before the application of the load g. In distinction from the approximate theory, the term —(H-+H,)7 represents the effects of the deformation and the initial tension H, of the cable. This correction is in a desirable direction in the case of the stiffened suspension bridge, since, for the loading which produces a maximum posi- tive moment, 7 is positive, so that the bending moment will be diminished; and the same is true with respect to the numerical value of the negative moment. In the arch construction, how- ever, the reverse obtains, for there the deflections of the struc- ture tend to increase the bending moments. The ratio of the missing term — (H+ A.) 7 to the approximate value of the moment (M — Hy) may, under certain conditions, become very large even with a rigid stiffening truss, particularly at those sections where (M— Hy) is very small or equal to zero. On the other hand, the effect of this deformation will not be sensible for those cases of loading which produce maximum bending moments unless the coefficient ¢ is very large, corresponding to a very small value of J. (Compare the examples calculated below.) The bending moments produced in the stiffening truss by any given loading may be determined exactly by equ. (151), if the value of the horizontal tension H is known. Since the deformations of the cable will in general be but very small, we may state in anticipation that these values of H will differ but slightly from the values given by the foregoing approximate equations (§§5 and 6). It will therefore usually suffice to use the approximate values of H in computing the bending moments by equ. (151). To determine H more accurately, we make use of the prin- ciple that, for small deformations, the work done by the forces (s, +) acting on the cable must equal the work of deformation of the internal stresses in the cable; hence, with the previously adopted notation, 1 H+4H, ? f cls ian: dea Ah) (=~ +21, sec?a;) 0 From equ. (8) we may write, approximately, Exact THEORY FOR THE SUSPENSION BRIDGE. 79 =a ay Sf or (s, +k) =—(H + Ho) ar = (H+ Hy) ; 2 also, adopting the abbreviation & « + 21,sec? a,=L, so that, for a parabolic cable, L=U(1t+ 2 F) 42h sectay, and substituting the value of 7 from equ. (150), the above equation of virtual work becomes 1 1 8 4 = [Jeet ont (mg) 4 tf tetas : ag TE Ae: BF Solving this equation for the horizontal tension, we obtain M—-4)d H= f¢ w)as 2. (152 2 8 a ft EG —f Cert Ce) det size t og, 5 In order to apply this formula, it is necessary first to find the approximate value of H by the preceding method, namely 1 sMyd x by the equation H = aa ae , and to use this value in | SPdot |. computing the quantities c, C, and C, appearing in the above expression, Equations (151) and (152) hold good for any condition of loading. Let us first apply them to the case of a single concen- tration (P) located at the distances € and |—£— &" from the respective abutments. For this loading, the following values are obtained for the arbitrary constants: For the segment from 0 to é: ; 1 i i 8 3 O.=a2—r{ Fe (er? —ert y—sh- (le ve = 2cH 1 P Cc Ef, -ct 8 C= aren {7G f’—e e)+ a (1—e") } For the segment from é to 1: oan C183, 1 P F 8 if C= see {(sG (e°§—e jit (1—e“) } C= poetry (fot) tap—e) — f ei—e*! |2cH 80 ARCHES AND SUSPENSION BRIDGES. We thus obtain: 1 pee 72. = 2 = ee eee (154, a2 a where, for abbreviation, K=5 = sy so (ete) (f°) (ee) | eas 426 (ett et—ay bP The bending moments in the stiffening truss are then to be computed by equ. (151). With this more exact theory, as previously observed, there is no longer any proportionality between the applied loads and the internal stresses. The method of influence lines is therefore inapplicable. If the loading consists of a load g’l uniformly distributed over the entire span and a similar load pd covering a distance A from the left abutment, then the arbitrary constants of equ. (150) are defined by the following expressions: For the segment from zs =—0 to r=A pp 2—etan— et Py) 8 Ff 1 HEC 3 ~~ eal Cri e aa He? C,=—He?C,—( H—p+g’) os. (155. For the segment from s—~ to c=, He?C,=He®C,— 5 pe He?C,=He?C,— spe and equ. (151) yields the following expressions for the moments: For s=0 to x= A, 1 2 cx 2 zi 8 f u=—i | He C,e* + He? Ce + FA (p+ 9’) l . (156. For <= to r=—1, M=—%[He*Cye*+ Hot Ce*+ 7 H—g'| Here H is the horizontal tension produced by the loading gi+tpa, ?= Hy Mle , and H, denotes the horizontal tension cor- responding to the initial parabolic curve of the cable and the Exact THEORY FOR THE SUSPENSION BRIDGE. 81 unstressed condition of the stiffening truss. In determining the value of c, the value of H given by the approximate theory may be used. If only the uniform load g’, covering the entire span, is applied, the constants reduce to Om Bota) eas eee (157. = 1 4 8f ec! ( C= qa —F BE) ia: and the moment becomes ees 8f i er + ec (l-x) u=+(9¢ —Ya)[1-“E J. (158. Example 1. Double-Track Railway Bridge. Span 7 — 150 meters. Steel-wire cable: “ Rise = 20 m., horizontal length of back-stays = 75 m., tan a, = 0.4, cable-section 40 = 328 cm’. Stiffening truss: parallel- chords, depth = 7 m., mean moment of inertia per truss J = 69,473,000 em*. = 0.69473 m*. Dead load per truss g = 2.4 tonnes, live load p = 4.0 tonnes per linear meter. It is assumed that the total dead load is carried by the cable alone, and hence there are no bending moments in the stiffening truss when the bridge is unloaded. To attain this condition, the stiffening truss is provided with a central hinge during erection and not until the construction is completed and the suspension rods adjusted is this hinge to be replaced by a rigid connection. We have 21, sec? a, “16 f ne L=1 (14 3 Pp + aes) = 2,255 1. Hence the denominator of equ. (75°) kecomes 8 3 X 0.69473 By the aLove assumption, =e 1 150 Ho= = x 50 X 2.4 = 337.5 tonnes. With the half-span loaded, by the approximate theory (equ. 82), 4 = = 5 = 114.90 & aor = 229.8 t. With the same half-load, the. tending moments computed by the approzi- mate thcory are as follows: (Af =M— Hy.) 1 1 At c= 40 = 45 pl — Wy, = 5625 — 229.8 x 15 = 2178 t. m. ey 1 1 Ata=-yl M= 4g pl’ — ll. f = 5625 — 229.8 x 20 = 1029 t. m, 3 1 Ate= 4, Ma= 39° pl? — My, = 2812.5 — 229.8 x 15 —— 634.5t.m, 6 82 ARCHES AND SUSPENSION BRIDGES. In applying the more exact theory, we have: Hot H 567.3 = ay = 730,000,000 x0.69473 — 9-000040835 1 8f y= 24,488.88 = 0.00639 ol = 0.9585 2-H = 1.6348 1 Using these values and putting g’ =—0,A = a 1, in equs. (155) and (156), we obtain the bending moments: 1 1 Ate= 71 = ae X 0.0847565 = 2075.6 t. m. 4 1 1 At c= ob l= an X 0.038268 = 937.1 t. m. 3 1 Ate= zl, Ms=— aE X 0.027238 = — 667.0 t. m. Comparing the two methods, it is seen that the exact design, which allows for the deformation of the cable under load, yields results for the positive moments M, and M, smaller by 4.7 and 8.8% respectively and for the negative moment M, larger by 4.9% than the corresponding results of the approximate theory. If the stiffening truss is more flexible, as in the following example, larger differences will appear, If the entire span is loaded with p= 4.0 tonnes per linear meter, we find by the approximate theory, HH =114.9 X 4= 459.6 t. and the moment at the mid-point 1 The more exact method yields: ; 797.1 2 ?= 39,000,000 0.69473 — 0-000050173 1 = 19,931.0 cl= 1.06249 and by equ. (158), substituting for g’ the value p = 4.0 t., 1 1 at g= > l, M=0.92195 = 1837.6 t. m. This is about 10% less than the result of the approximate design. Example 2. Highway Bridge. Span and rise as above. Stiffening truss, 2.5 meters high; mean moment of inertia J = 7,812,000 em‘ = 0.07812 m*. Cable-section do 200 em*. Dead load per truss, g = 2.40 tonnes, live load p= 1.4 t. per linear meter. It is again assumed that the total dead weight of the bridge is carried by the cable. With L = 2.255 1, we find 8 3X 0.07812 N= 5 + pon 400” X 2-255 = 1.66606, 1 (150)? mag hoy RIE, Exact THEORY FOR THE SUSPENSION BRIDGE. 83 and, with the half-span loaded, by equ. (82), H = 135.048 x =94.53 t. For this half-load, the approximate theory yields the following values of the moments: 1 Ats= 7 |, M,= 1968.75 — 94.53 & 15 = 550.8 t. m. 1 At a= oh M, = 1968.75 — 94.53 K 20 = 78.15 t. m. 3 At g= ah M,; = 984.37 — 94.53 XK 15 = —433.57 t. m. Applying the exact theory, we have: 2 eT a 0002766 c= "EI 20,000,000x.07812. =" , 1 8 —y =3,615.4, cl = 2.49466, =/- H = 0.672213, : 1 Using these values and A= lin equ. (156), we compute the following moments: 1 1 1 Ty ie - At r= > l, M,= oe 0.01304 = 47.15 t. m. 3 1 At = ty M.= — —@ 9.10701 = —386.89 t.m. The deviation from the approximate values amounts to about 17% in M,, and about 40% of the approximate value in the small bending moment at the mid-point; and in every case the approximate theory mais the bending moments too large. With the entire span loaded with p = 1.4 t. per linear meter, H = 135.048 X 1.4 = 189.06 t. and the moment at the mid-point is given by the approximate method as 1 M = 5 pl — H. f = 3937.5 — 3781.2 = 156.3 t.m. For the more exact design we have: 526.56 1 e= 1562400 = 0.0003370, —z =2,967.2, cl=2.7537 c l : and by equ. (158), for = e we obtain M = 0.029203 - = 86.6 t. m. e Hence the approximate design in this case yields a value too great by more than 40%. 84 ARCHES AND SUSPENSION BRIDGES. 2. Truss with a Middle Hinge. Let the stiffening truss have a constant moment of inertia 7; and, in the unloaded con- dition, with only the dead load of g units per linear meter act- ing, at a definite temperature, let the form of the cable coincide with the equilibrium curve of the dead load, so that there is then no stress in the stiffening truss. Let H, be the horizontal tension of the cable when this condition obtains. Upon applying a live load p, which increases the horizontal tension by H, or for any other cause, let the cable sag at the crown by an amount Af, producing a bending in the stiffening truss. Writing for the half-span ia, and denoting by 7 the deflection due to bending of the stiffening truss at a distance x from the abutment, then the total deflection at that point will ben+ = .Af, and the same amount will represent the increase in the ordinate y of the cable-curve if the stretching of the suspenders is neglected. The horizontal forces (H,-+ H) cor- responding to the initial position of the cable-crown will now change to (H’,+H’) = 57 (H,+ H). If M denotes the moment of the loads g + p in a simply supported beam, then the bending moment in the stiffening truss will be M=M— (H’,+H’) (y+ Af4n)....--- bei and the differential equation of the elastic curve of the structure will be an = 42 2 M Zz ge O18 (Ge AN) (B. when, for abbreviation, we put eta Hot 2 Hot Hd El ~ & If we assume Af independent of y, which is very nearly the ease, then, for a parabolic initial form of the cable and for the cases of loading occurring in practice, the bracketed quantity in (8) will be a homogeneous algebraic expression of the second degree; let this expression be denoted by F(x) and its second derivative with respect to x by F’”’(x). Then the integration of (8) yields n= Act + Bet+F(x) +4P" (2) 1.000... te Cc é The constants A and B are given by the conditions that at z=0 and =a, 70. Substituting the expression (y) in equation (a) gives the bending moment Exact THEORY FOR THE SUSPENSION BRIDGE. 85 M=-—(A'.+H’) (Ae*+ Be) —EI.F’(2)..(8 The design for any given loading now becomes a simple matter. With a ee load of p per unit length we obtain P(e) 2S" 4, go wine and the maximum moment at the quarter point (e%** — 1)? a = api QETs — asedewesacets (159. When the half-span is loaded with p per unit length, z__ 1 (@g+p)e’ El 4f Mnax= and the maximum moments at the quarter-points in the loaded and unloaded segments will be M __ (e%**—1)* 2EI 2(g+p)Af+ pf 1max et*+1 a 2g+p ie (c#e*—1)?_ 2H 2gAf—pf .. ee. (160. 2max~_ eet a? 2g+p . For a very stiff truss (J = , ae we have wea _ 7 2 2EI , s Er a? =} (Ao +H’ ) so that the extra moment at the quarter points due to the sag Af at the crown will be given with sufficient accuracy for most cases by the formula Me TD AP erwin Stanenciedens (161. The crown sag A f, due to an elongation of the cable amount- ing to AD, is expressed with sufficient accuracy by the equation (128), _AL ar—(44+55)- 4 Example. If the bridge of the above example 1, is constructed with a 1 6.4 (150)? 2. Het H naa El = 0.0000648, ca = 0.6037; hence, by equ. (159), 900 Mmax = 216.83 A f or approximately, by equ. (161), Mmax = as Af= 225 .Af. The elastic stretch of the cable will be 86 ARCHES AND SUSPENSION BRIDGES. H pee 6 AL= BA = 2000 > X 328 = 0.290 m,, hence 75 1 20 and Mmex = 216.83 X 0.563 = 122.1 t. m. . 1 8.8 (150)? With only the half-span loaded, H + H,= ié— 20 = 618.75 t., c? = 0.0000445, ca = 0.5002, and we obtain, by equ. (160), My max = (1369.3 + 219.1 Af) t. m. Mz max = (—1369.3 + 82.2 Af) t. m. The approximate formula for the rigid truss gives, 1 1 My, max = + 6] pU + 7 (Hot H). Af= (#1406 + 154.7A/) tra Here, neglecting the effect of temperature variations, the value of A f must be taken as one-half the value for full load, or A f = 0.281 m. STRESSES IN CuRVED Riss. , 87 C. The Arched Rib. (Arch with Solid Web.) §10. Internal Stresses in Curved Ribs. We conceive an arched rib as generated by a plane figure whose center of gravity moves along a line of single curvature while the plane of the figure remains continually perpendicular to this directrix. The generating figure is called the cross- section of the arch and the directrix is called its axis. It is presupposed that the lines of action of all the external forces acting on the arch (loads and end-reactions) lie in a single plane—the force-plane, which coincides with the plane of the arch-axis and contains a principal axis of each cross- section. If the external forces are known, then the resultants of the internal stresses at any section are also determined, since these must be in equilibrium with all external forces acting on the rib between the given section and the end of the span. If the resultant of these forces at every section passes through the center of gravity of the cross-section, that is if the loads are so distributed that their equilibrium polygon coincides through- out with the arch-axis, then the stresses will be uniformly dis- tributed over each cross-section and there will be none but normal stresses (pure tension or compression) throughout the structure. Under any other loading, for which the line of resultant pressures departs from the arch-axis, bending stresses will appear in the structure. Upon applying any load, the axis of the rib assumes a new form called the elastic curve for the given loading. To be pre- cise, the quantities employed in the following analysis ought to relate to the condition of the structure after deformation (ef. § 28). Nevertheless let us presuppose — what appears to be admissible in all practical cases—that the deformations are so slight that all relations based on the original coordinates of the axis will hold good, even after deformation. We locate the origin of coordinates at the left end-point of the arch-axis and measure the positive abscissae toward the right and the positive ordinates upward. Let the inclination to the horizontal of the tangent to the arch-axis at the point (2, y), or the inclination to the vertical of the plane of the cross-section, be denoted by ¢. Then, for a positive dz, if the curvature is convex upward, df must be taken as negative. Also let R denote the 88, ARCHES AND SUSPENSION BRIDGES. resultant of the external forces acting on the portion of the arch to the left of the section (z,y). Let its components paral- lel to the coordinate axes be Fig. 38 H and V, and the components z a parallel respectively to the er F fr tangent and normal to the arch-axis at the point (2, y) Be be P and S (Fig. 38). Let M Kn. be denote the moment of the ex- en a ternal forces about an axis - on perpendicular to the force- \ 2 ly, plane at the point (z, y), con- eo ie Ms i sidered positive if directed \ hogr clockwise. The external 2. 4 ae forces may then be replaced by a single force R, or its components (H,V) or (P, S) applied at the center of gravity of the given section together with a couple whose moment is M. With these external forces, ze the internal stresses at the sec- t tion (z, y) must be in equilib- rium. 2 Consequently, if oc, 7, and 7, ig denote the normal and shearing u " stresses on any elementary area dA = du. dv of the cross-sec- tion (Fig. 39), we may write the following equations of equi- librium: P+fodA =0 S+ fr,.dA =( T- dA =0 (162 M+ few.aA St ‘ Sry.u.dA—Sry.0.dA=0 1. Determination of the Normal Stress ¢. It is assumed that the axis will remain a plane curve after deformation and that all bending takes place in the force-plane. The radius of curvature of the arch-axis at the point (z, y) will be denoted by r (positive if the curve is convex upward). If ds, denotes the length of a fiber between two cross-sections infinitely close together before deformation, and A ds, is its change in length, then the length of the fiber after deformation will be d sy+Ads, STRESSES IN CURVED RIBS. 89 We adopt the same assumption as in the common theory of flex- ure of straight beams, namely that the material cross-sections remain plane and perpendicular to the axis after deformation. By this assumption, if—d¢ denotes the angle between the two consecutive cross-sections, and if this changes upon deformation into — (d¢+Ad¢), (d¢ and Ad ¢ being considered negative when the curve is convex upward and when the curvature is increased by the deformation), we have dsy=ds—vdd and dsytAdsy=ds+Ads—v (d¢+Ad¢). Subtracting, the first of these equations from the second, we obtain : Adsy=Ads—v.-Ad¢. The proportional elongation or strain of the fiber is, therefore, Adsy ___—Ads—vAd@ dsw = ds—vd¢ or, since ds==—r d¢, Adsv Ads Ad@d r a US ropes (163. If, in addition to the normal stress, o, there is also a tem- perature variation ¢ contributing to the fiber-strain A ds,, then, with a coefficient of expansion o, Adsy oC i a Combining this with equ. (163) and solving for o, Ads P Add r ds ds rip. ee Substituting this expression in the first and fourth of the equa- tions of equilibrium (162), we obtain P=—BAB(TIALE Ad¢ rp —dA+EBot.A r+v ds ads rv sae rv M==—B - 92° ( 7° da }.. (165. +EHotfv.dAa Observing that fdA=A, fv.dA=0, 90 ARCHES AND SUSPENSION BRIDGES. spac fuae fit ae Wen a and hence that rdA ieee =e and vdA __ 1 rv rfo =| dA, and oe the abbreviation C ore = fauasas, sctatedheucaieemoateeaes (166. then the equations (165) become P Ads Add 1 Ads\ J gan Pe) 4—Ca tae A 1a Me ( Aae s: eg, and E ds Solving these equations, we find Ads P M a ee pp ie Ad@ _ P se ea eee Hoe ( ‘ ds BAr BAr? ¢ EJ Substituting these values in equ. (164), we obtain the normal fiber stress Mrvo == +44 Tiree Steer eese This equation (168) shows that o is not a linear function of v but, on the contrary, is represented by the ordinates of an equilateral hyperbola whose asymptote normal to the given section passes through the center of curvature. The fiber stress o will be zero at Pr M a ae oe Tr a+H4+5 Hence. if P—0, the neutral axis does not pass through the center of gravity of the cross-section (v0), unless r= o, i. e., unless the axis of the rib is originally straight. The quantity .J may be expanded into a series: J=f Boda forda—+ (1-24 45-...)aa. STRESSES IN CurveD Riss. 91 But if v?dA represents the moment of inertia J of the cross- section about the U-axis. If, furthermore, the cross-section is symmetrical about this axis, all the terms in v’, v*... will vanish, so that we obtain JaItafo(l+3+54+...)aa. If the radius of curvature r is very large in proportion to the depth of cross-section, all the other terms in the expression for J may be neglected in comparison with the term J, so that we may substitute the moment of inertia J for the quantity J in the formulae established above. For example, in a rectangular cross-section of width b and depth h, we find: fon we atic 08 2-005 0.10 0.20 = 41,0004 1,0015 1.0060 a[s 3 This approximation appears to be admissible in all prac- tical applications of the formulae to arch-bridges. In most cases it even suffices to replace equs. (167) and (168) by the simpler equations applying to straight beams: Ads P ——-=,, —et ds nA ; Add _ M a (167 ds BJ oF 4" S0e toe: Wa ee mths Se eee eich! Bi ites (168° The fundamental relations employed in the preceding analysis impose one more test: to determine under what con- ditions, if any, the fifth of the equs. (162), also involving the quantity o, will be satisfied. Substituting the expression of (164) and making a slight omission, we obtain ECS —ot) fu.dA — EA (rue dA—0. or, since fru .dA=0, uwv.dA r farts — 0. This equation will be satisfied by the condition, among others, that the cross-section be symmetrically disposed about the force-plane; if r is very large compared with v, the equation reduces to 92 ARCHES AND SUSPENSION BRIDGES. f uv.dA=0O, which shows that one of the principal axes of the section must lie in the force-plane. 2. Determination of the Radial Shearing Stress 7,. Asa rule, the shearing stresses in an arch may be neglected; never- theless a knowledge of them is sometimes important, and the value of +, should therefore be derived. We shall adopt the assumption, however, that the radius of curvature is suf- ficiently large compared with the depth of cross-section to per- mit the use of equ. (168*) in determining the normal stresses. If, from the rib-segment included between the sections at (2, y) and (x+dz, y+dy), we cut off an upper portion Fig. 40. by means of a section perpendicular to the force-plane and parallel to the arch-curve, there will be acting upon this elementary body the forces and stresses represented in Fig. 40. Here d G denotes the external load acting on the arch segment and qds the dead weight of the latter. It is assumed that the shearing stresses 7, are the same for any two mutually perpendic- ular planes at any point and that they are uniform in intensity throughout the elementary tangential section. Distinguishing the stresses at the neighboring section by affixing an index (’) and presupposing but a slight variation of cross-section, we have, by (168*), A I et 4f—— VPrePie (M+dM)z : sacuie’ inamncaliiaeey aia STRESSES IN CuRVED RIBS. 93 Here a denotes the area of cross-section included above the tangential cutting plane, and % denotes the static movement of this area about the U-axis. Now (by Fig. 38) we have P=H cost V sin o,..-.... ccc erent eee (169. hence e=(— H sing +Vcosg) ae + e© sing + cos ¢. But, since (by Fig. 38), S=—H.sind+Vecos¢.......... meow (170. and dV=dG@.sina—qds, dH=dGeos a, “4 =, we obtain S , d@ 8 PF =— 24S cos (a—$) —9 sing=—=+ ga, -(A7L. where gn denotes the intensity (per unit length of axis) of the normal component of the total load on the elementary solid. In like manner we obtain dM=—H.ds.siné+V.ds.cos¢+dG@.m.cos (a—¢); hence, dM dG oo S+m7, cos(a—¢), aaah eee .. (172, where m is the radial lever arm of the force dG about an axis through the point (2, y). Tinally, the condition that the algebraic sum of all the force- components parallel to the tangent must be zero, yields the fol- lowing equation adapted for determining 7, : ae (ds—vdg)—d¢ (rd A+fo'dA—fo.dd + dG cos (a—)—q.ds. © ‘sind =0. Substituting ds—vdd—=— dd (r+) or, by the above assumptions, =—d¢.r—ds, putting fred4=fn.d A, and substituting the value of f o dA—fo.dA, also the expressions of (171) and (172), we obtain 2 0 fee dA =s (+ _— =a) a cos (a—¢) GQ—4—"™? < 94 ARCHES AND SUSPENSION BRIDGES. dd 1 If we put — aes = 0, the ahove reduces to red =— 4 00s (a—¢) (1a—4—#).. ... (173. If, in addition, 44 = 0 or cos (a—$) =0, the above becomes ET Neptiiniwcssineatnianienss (1738. The above results (equs. 168? and 173*) show that the for- mulae for straight beams may be applied, with close approz- imation, to the arched rib, provided the radius of curvature is comparatively large; in all further stress-analysis, determina- tion of principal stresses, etc., the same principle will be applied and flat arches will be treated approximately as straight beams. In bridge-arches for the usual descriptions of loading, the shear S will always be very small compared with the axial-force P, so that, as a rule, the shearing stresses in arched ribs may be onitted from consideration. NorMat FIBER STRESSES IN THE ARCHED RI. 95 §11. Conditions for Stability. Line of Resistance. Core-Points. Graphic Determination of Normal Fiber Stresses. Neglecting the shearing stresses, as just suggested, the prin- cipal stresses become identical with the normal fiber stresses. These, by equ. (168), attain their greatest values at the extreme fibers; if these fibers are distant v, and — v, from the gravity- axis, the respective stresses will be: a os re M Mr vy A Ar J (r+n,) (174 _ : ‘sf Aes Wt aes 3 On A Ar J (r—) Neither of these values should exceed the safe working intensity of stress. If + is sufficiently large relative to v, the above expressions become pP AP 1% AieteP i : 1 Pde os eaten, ae oe eee (1749. P Mov a ae which values may also be obtained directly from equ. (168*). . If we substitute M—P.p, then it appears that the dis- tribution of fiber stresses in the cross-section is especially dependent upon the value of p, 1. e., upon the distance of the piercing point D (Fig. 38) of the external resultant & from the gravity-axis. It is therefore desirable to know the position of this point at every cross-section; and, for any given loading, the curve constituting the locus of these piercing points is named the Line of Resistance. The curve enveloping the successive resultants of the ex- ternal forces (generally differing but little from the line of resistance), i. e., the equilibrium curve of the external loading, is called the Line of Pressures or, more briefly, the Pressure Line. A line drawn from any point of the line of resistance tangent to the pressure line will constitute the line of action of the corresponding resultant pressure R. The name, line of resistance, is adopted because that line determines at each sec- tion the point at which the internal resisting stresses must balance the external forces to produce stable equilibrium. An- other important property of the line of resistance is that its dis- 96 ARCHES AND SUSPENSION .BRIDGES. tance from the gravity axis controls the magnitude and distri- bution of the stresses in the cross-section. If 7 denotes the radius of gyration of the cross-section in the force-plane, then J = A . 7”, and by equ. (168+), Equ. (175) shows that the normal stresses vary along the section as the ordinates of a straight line; specifically, atv=4,, o=—t(14+244) joe — sid LEG, atv——v., Equations (176) show that the extreme stresses co, and a2, and hence all the stresses in the section, will have the same sign provided p> £ and, at the same time, < + sites (177. Hence, in each cross-section, if we fix two points at dis- tances k, and — k, from the gravity axis, determined by =, cet a ee (78. or by the corresponding graphic construction of Fig. (41), they will so divide the section that all the normal stresses will be of Fie. 41 the same sign so long as the line of resistance, g. 41. . : representing the resultant pressure R, passcs between the two points. These points, whose position depends only upon the form of the cross-section, are called the Core-Points, theic continuous loci are the two Core-Lines, and the portion of the force-plane intercepted between these lines is called the Core. All the normal fiber stresses at any section will therefore have the same sign or direction, so long as the line of resistance cuts the section within the core. Introducing the quantities k, and k, into equations (176), we obtain oe —_— =e pee Mer, : I = —2 (2 Hs?) — _ ran ) yy + Mute .. (179. NorMaL Fisper STRESSES IN THE ARCHED Rip. 97 where M,, and M,, represent the moments of the external forces about the two core-points of the cross-section. These equations, or the equivalent forms ai oe —F0-£)] form the basis for the graphic method of Fig. 42 for determining the normal stresses. N,, N., are the core- points lying in the foree- plane; and the ordinate SC, at the center of gravity, is made equal to =. The lines drawn from the core-points N,, N,, to the point C will intercept on the resultant normal pressure P, two lengths representing extreme fiber stresses o, and o, respectively. From this construction it is evident that the stress at one of the extreme fibers will be zero when the line of resistance passes through the corresponding core-point; that the stresses o, and o, will have like signs when the resultant force passes inside the core, and unlike signs when it passes outside the core; and the neutral axis will come the nearer to the eenter of gravity the farther the line of resistance is removed therefrom, With the aid of analytical geometry it may also be shown that the neutral axis is the anti-polar of the central ellipse with respect to the point of application of the resultant pressure as pole. If the cross-section consists of two flanges of area A, and A,, and if h is the effective depth or distance between the centers of gravity of the flanges, then approximately, neglecting the effect of the web, we may write A,v,==A,.v2, A, v,?+ A,v2= (A,+4, )?, and k 1V2=V,7,k20,=7; hence we find : k=, Ey, 1. e., the core-points coincide with the flanges of the girder. Consequently, by equ. (179), _ P(pths) 1 A (ka ln) SOAR (180 Fie) gee . 2 Aatia) | deh Here M, and M, represent the moments of the external forces taken about the centers of gravity of the upper and lower chords respectively. 98 ARCHES AND SUSPENSION BRIDGES. §12. Determination of the Deformations. 1. Elongation of the Axis. If (x,y) and (2, y.) are the co- ordinates of two points on the arch axis and if s and s, are their respective distances measured along the axis from any assumed mitial point, the change in the axial distance between the two points is obtained by integrating the first of equs. (167); there results: s As—ASy~=wo t (s—S,) ~—f (4 + So and approximately, if r is very large, i VW i) ds.. (181. As—A s5=wt (s—Sp) = ( aa ) OU Siescsies (181°. So 2. Variation of the Angle ¢. The variation of the angle between the normals or tangents at two points whose initial coordinates are (x, y) and (2, y,) is given by the integration of the second of equs. (167): 8 s P M M ds sotto Gar t gap ta) demo! fF § 0 —fpess eee If r is very large compared with the dimensions of the cross- section, then approximately, eaten fF Mds -# ft 3 .. .. (1828, =f To + ot (s—de) 3 . (182. 3. Variation of the Radius of Curvature. If +r, is the ra- dius of curvature at any point after deformation, then 1 db4Aad TY ds+-Ads ” DEFORMATIONS OF THE ARCHED Rip. 99 so that the variation of curvature is 11 dg+Aag ae _ ds ag ae "1 rr ds+Ads Ads I+ ds On substituting the values from equs. (167), the above becomes 1 BF ce a eee: BS (183. TY, r ES (: P M ) BA Bart? , 3 ‘ Ads. As an approximation, since ee always very small com- pared with unity and as r is assumed very large relative to the sectional dimensions, we may write the o_o a Fo pe BE teers (1833. If the arch is made to conform to an equilibrium curve, then, as the line of resistance coincides with the axis, Af —0, and, if the ends are 1 1 free to undergo the necessary displacements, by equ. (183), 7,7’ 1 or the radius of curvature remains constant at every point. The deforma- tion then consists merely of a shortening of the axis by the distance Ads. Consequently there must be, on the whole, a shortening of the span. As this, however, is impossible with an arch whose ends are immovable, it is seen that a perfect coincidence of arch-axis with line of resistance cannot be attained, or may be assumed approximately only when the deformations are negligible or the material is considered incompressible. 4. Variation of the Coordinates. Let Ax, Ay denote the displacements of the point B (x,y) in the directions of the respective coordinate axes (Fig. 43); and let Ada, Ady, Ads represent the varia- tions in the distances dx, dy, ds between the two consecutive points, B, B’. Let the section at B’ undergo a rotation A ¢ which we assume equal to that at B. The point B of the arch-axis is thus displaced to B’’; and the projections of this displacement BB’ upon the coordinate axes may be determined from the triangle BB” C which is similar to the triangle B B’ D. We thus have Fig. 43. BC=—A¢.ds. v——ag . dy, and da BY C=Ad¢d.ds. ais =A¢.dz. If the arch-axis suffers, in addition, a change in length, in particular if the arch element BB’ = ds is lengthened by Ads, 100 ARCHES AND SUSPENSION Uripces. then the point B” receives a displacement whose rectangular components are A ds 42 and A ds 42, so that the total dis- placement of B relative to B’ is given by the expressions, Adz=—Ad¢.dy+Ads. “ Ady=A¢.dz+Ads. ay Considering such displacements in all the arch elements begin- ning at some initial point A, and summing these up to get the total displacement of B, we obtain, Ac=—fag.dy+ fae -d sy=faedz+ foe -d If we indicate the values of the coordinate and angular varia- tions pertaining to the initial point A with the subscript 0, and the values pertaining to the point B with the subscript 1, then, by partial integration of the above equations and substituting the proper limits, we obtain Xi do Ad AX, —AMM=—AP Yi FA foYot fos Se yds f a dx z, a (184. Ad Ad ieee nae : rds+f 1d So Lo ae =i Substituting for 42% ana 4® ; the expressions (167), but intro- ducing the abbreviation Bee 0 $1 A : f at . ds, we obtain, oi s a =P’, and writing A¢d, =A do+ SL Jt AL,— AXp=— Avo =| EJ (y.—y) ds 8 : -. «(185 —J ga (4ptastaz) tut f (Uce | DEFORMATIONS OF THE ARCHED Rip. 101 ) Si M Ay, — Ayo =A $5 ( 2:—2o ) +f4 (z,— x) ds sy A Si (2 uae) otf es ds—dy) | 8, 8, J The above equations may be simplified if the radius of curvature is very large relative to the sectional dimensions. We may then put P’—P, J=T, and, in the temperature terms, omit the parts containing =. There results: $1 M AL,—ALp=—A $0 (Y1—Yo) — if EI (¥i—y)ds a L.. (185% —S ga (OFE totae) peta), 81 Ay,—AYo Ado (%:—2o) + {2 (2,—«) ds * r.. (1868. Sti +S EEF ds—dy) + ote) . 102 ARCHES AND SUSPENSION BRIDGES. § 13. External Forces of the Arched Rib. We here consider plate arches or curved ribs whose ends are connected to rigid supports. The connections may be either such as to keep the ends of the rib completely or practically immovable, although permitting free rotation of the axis— Arches with Hinged Ends—or such as to prevent rotation also, so that the ends of the axis remain fixed in direction—Arches with Fixed Ends (or Hingeless Arches). Both of these types of arches, as explained in the Introduction (§ 1), belong to the statically indeterminate class of structures, 1. e., the conditions of static equilibrium do not suffice to determine the external reactions; it thus becomes necessary to establish additional equations of condition on the basis of the elastic deformations. For this purpose we may employ the deformation equations established in the preceding section (§12), or, as will later be shown, we may apply the Theorem of Least Work. Static determinateness, however, may be attained in the arch with hinged ends by introducing a third hinge, thus obtaining the Three-Hinged Arch. In the general case of the fixed-ended, hingeless arch, there exist the following relations between the external forces: Let A and B (Fig. 44) be the two end points, assumed to be at different elevations. Let G@ represent one of the forces acting in any direction upon the arch. Let us, for the mo- ment, imagine the end A free to slide horizontally while the end B, at the same time, has a hinged connection; and, in the freely supported girder thus obtained, let us deter- mine the bending moment M at any point (2, y), the ordinate y being measured Fig. 44. from the closing chord A B. If g and y are the lever arms of the external forces about the end B and the point (z,y), respectively, then ee l M= oM~s x CIE FY ce eeerersceeeeens (187. The fixedness of the ends of the rib, preventing either rota- EXTERNAL Forces oF THE ARCHED Rip. 103 tion or horizontal displacement, may be replaced by the effect of a force with horizontal component H,, applied at A and acting in the direction of the chord A B, and a moment M, applied at A together with another M, applied at B; these reactions must be added to those of a freely supported girder. The bending moment at the point (z,y) then becomes: M=M + Mie) tie iy oe, (188. Using the notation indicated in Fig. 44, we may also write the following general equations: t ae =o Gg eB, tei 0 1 1 —V.=V— 2G sing ..-- (189 1 H,=H,+%3G@cos p 0 J At any section .r, the horizontal and vertical components will be x H=I/,+ 3G .cosy 0 a V=V,— 3G. siny 0 Jf ¢ is the slope of the tangent to the arch-axis at the point (x, y), the axial and shearing components are given by oo ee ee ee (191. S=V cosp— Hsing. ....cccccccc ceeeecees (192, With M and P known, we may determine the internal stresses in the arch by equ. (168) or (1687), and the maximum fiber stresses by equ. (174) or (174*), so that the entire problem is solved. But in order to find M and P, in the general case, it is necessary to first determine three statically indeterminate quan- tities: H,, M, and M,. In the two-hinged arch the end moments M, and M, vanish; hence the number of indeterminate quantities reduces to one, viz., the horizontal thrust H,. Finally, if a third hinge is put into the arch, it provides another static condition, namely that the moment of the external forces about the center of the hinge must vanish; this condition determines the value of H,. If the two ends are at the same level, and if all the applied loads are vertical in direction, the above general equations reduce to ql H,=H=H, , Vi=72@.g+ om sous les: 104 ARCHES AND SUSPENSION BrinGEs. If the loading is vertical, M may be represented by the ordi- nates of a funicular polygon. If this polygon is constructed with the pole distance — H,, and placed in such a position rela- tive to the arch as to intersect the end-verticals at distances Mi ~~ Ay nates of the polygon measured above the arch-chord and multi- Mi(l—w#)+Ma-@ . U 2 ey and eal above the points of support, then the ordi- 1 . plied by H, will represent the moments M + consequently, by equ. (188), the vertical intercepts between the equilibrium polygon of the loading and the axis of the arch, mul- tiplied by H,, will give the bending moments at the correspond- ing points of the axis. This principle was first established by Winkler. Reaction Locus aND TANGENT CURVES. 105 § 14. Reaction Locus and Tangent Curves. Critical Loading. 1. Reaction Locus and Tangent Curves. When the end reactions, i. e., the resultants of (H,, V,) and (H,, V,), and the end moments, M, and M,, are known for any given loading, we may construct the force polygon and equilibrium polygon and the line of resistance for that loading. The two end reactions interseet each other in a point on the resultant of the applied loads, which changes as the positions of the loads change; when this load variation follows some definite rule, the curve described by the intersection of the reactions is named the Reaction Inter- section Locus or, more simply, the Reaction Locus. With the shifting of the load resultant, the end reactions alter their position; one way in which these reactions may be determined is to have for each point of the reaction locus the Fig. 45. two points in which the reactions intersect the vertical lines pass- ing through the ends of the arch-axis. Another way of fixing the positions of the end reactions is by means of the two curves which are enveloped by them when the load resultant passes through the successive points of the reaction locus. The reactions are thus determined in posi- tion and direction, for any load, by the two tangents drawn from the corresponding point of the reaction locus to the two nappes of the enveloping curve. The magnitudes of the reac- tions are then found from the force triangle composed of the load-resultant and lines parallel to the two tangent directions. The two curves enveloping the reactions are called the Reaction Envelope Curves or,-more simply, the Tangent Curves. In Fig. 45, let (a, yx) be the coordinates of any point of one 106 ARCHES AND SUSPENSION BsinGEs. of the reaction-lines, (,, 72) the distances intercepted by the reac- tion-lines on the left and right end-verticals respectively, and ( é, n¢ ) the coordinates of the point of intersection of the two reaction-lines; then, in the general case, _ Mh MSP eee eee e eer e eee ees (194 and 2— Yo= a or m=Yet Fe a noe d wocteiehe ee (195. Also, for the left reaction, eee = 1 + 7 os ore, ee (196. and for the right reaction, —— Mit Hi y+ V2 l—a@) ee ee (197. Putting «= é in equ. (196) or (197), we obtain the ordinate of the reaction intersection point, or, with variable & the equation of the reaction locus: — M+Vie Mo+ Hs 2 + Ve (I—&) gg tT ee (198. V2 Se a (l—z) To obtain the equation of the tangent curve, the value of € must be considered as the variable parameter in equations (196) and (197). The equation of the tangent curve for the left reactions is derived by eliminating é from equ. (196) M,.+ Vie ( Nee — Ay ) ferentiating this with respect to é. Similarly, we derive the equation of the tangent curve of the and the equation obtained by dif- right reactions by eliminating é from equ. (197) ( m>~E = Athy th: Game) ) and the equation obtained by dif- ferentiating this with respect to é. If end hinges are considered, M,—M,—0, and hence "1 = 0, 7. = Y2, or the reactions always pass through the centers of gravity of the end-sections; consequently the tangent curves in this case reduce to the two end-points. The reaction locus in this case will be Hz y2+ V2 (lL—&) 2 Az In the case of the three-hinged arch, the above formulae eas Laws or LOADING ror THE ArcuED Ris. 107 apply in a general way. We may, however, make use of the relation that the line of resistance must, in this case, always pass through the three hinges, so that it is readily determined for any given loading. For a moving concentration, the reac- tion locus will then consist of two straight lines, meeting at the middle hinge, which are the prolongations of the lines joining the end hinges to the middle hinge. On the basis of the above principles it is readily seen how the reactions may be found graphically as soon as the reaction locus and tangent curves are known. It should be observed that with the aid of the curves drawn for a single concentration we may construct not only the reactions for different positions of this load, but also those for a train of concentrations resting on the structure. These procedures become particularly simple when the ends are hinged since, in that case, all the reactions for the individual loads pass through the same point so that the position of the resultant reaction, determined in magnitude and direction by the cor- responding force polygon, is at once known; whereas in hingeless arches this line of action must be determined by a special funicular polygon. The reaction corresponding to a train of concentrations may be deter- mined either as the resultant of the sums of the horizontal and the vertical components of the reactions for the individual loads, or as the closing side of a force polygon, whose sides represent the reactions for the individual loads. Next, combining the reaction determined by either of the above methods with the total loads acting on the portion of the rib separated from the rest by a given section, or by constructing the cor- responding equilibrium polygon, we obtain the magnitude and position of the resultant pressure R at the section considered and with these data, as previously indicated, we may calculate the values of the extreme fiber stresses. With the aid of the reaction locus and tangent curves drawn for a single concentration, the critical loading for any stress, i. e., the manner of loading for making the stress a maximum, may be determined. 2. Laws of Loading for Normal Stresses. As previously shown (§11), the normal fiber stresses attain their maximum values in the fibers farthest from the axis. Hence it will gen- erally suffice to determine the critical loads just for these extreme fiber stresses; and it should here be recalled that the stress o, in the uppermost fiber will have a sign the same as or opposite to that of the normal pressure P according ‘as the resultant pressure R intersects the section above or below the lower core-point. On the other hand, the stress o, in the lowest fiber will have its sign the same as or opposite to that of P, according as R cuts the section below or above the upper core- point. Furthermore, for a concentration G, the resultant R coincides with the first reaction R, if the load is located in the segment (1 — x), but with the equilibrant of the second reac- tion R, if the load is located in the segment (x). The sign of the normal pressure P for different positions of the load is thus dependent upon the signs of the individual values of R,, R,, and upon their position relative to the cross-section. For the 108 ARCHES AND SUSPENSION BRIDGES. cases under present consideration, it may be stated that P will have the same sign for all sections and all load-positions; in fact, for vertical loading, P will be positive if the axis is con- cave downward (arch), and negative if the latter is convex downward (suspension cable). To investigate the effect of the loading on the normal stresses in the uppermost fibers of a cross-section (Fig. 46), we draw two lines from the lower core-point n, of the section tangent to the tangent curves. The points J and G, in which these tangent rays cut the reaction locus, determine the limits of loading or ‘‘critical points.’’ Any load applied between C and J yields a resultant pressure cutting the section below n,, hence producing tension (if P is positive) ; any load between J and EF (the intersection of a vertical through the uppermost fiber with’ the reaction locus) yields a resultant pressure acting above J H, and every load between EF and @ yields a resultant above F G, Fig. 47. > ry - - PZO0F. Pz0e: P20F; SSE. SEE so that, in either case, the resultant will pass above n,, producing compression in the upper fibers; every load between G and D yields a resultant pressure acting below FG or n,, producing tension in the given fibers. For the normal stresses in the lowest fibers of a cross-section (Fig. 47), the tangents J, H, and F, G, must be drawn through the upper core-point »,; the points J, G, are then the critical points, so that all loads lying between C and J,, or G, and D wi Laws or LoapInG For THE ARCHED Rip, 109 . produce compression if P > 0, and all loads between J, and G, will produce the reverse stresses. If, from the points C and D, lines are drawn tangent to the tangent curves, their intersections with the core-lines will divide the rib into portions having alternately one or two critical points, according as the intersections of the tangents from the core-points with the reaction locus fall within or without the limits of the span. In the two-hinged arch, the end points take the place of the tangent curves; and, in the three-hinged arch, the reaction locus consists of two straight lines meeting at the middle hinge. 3. Laws of Loading for Shears 8. The shears S determine the shearing stresses in the plate-arch or the stresses in the web members in the framed arch with parallel chords It is there- fore important to determine the critical values of the shears. Fig. 48. Drawing the vertical line H E (Fig. 48) through the upper fiber H of the section and, parallel to the arch-tangent F G another line F, G, tangent to the tangent curve, the intersec- tions EF and G, with the reaction locus give the division points of the loading or ‘‘critical points’’ for shearing stress. For a section to the left of the crown, all loads between C and £, or G, and D, produce a negative shear, and all loads between F and G, produce a positive shear. For sections to the right of the crown, the reverse rule obtains; and, in general, the loading beginning at any section will have a positive or negative effect according as it extends toward the right or the left, and the limits of the loading are given either by the end of the rib or by the intersection of the reaction locus with the parallel tangent. Drawing tangents from the points C and D to the tangent curves, the points at which the arch axis is parallel to these tangents divide the rib into portions having alternately one or two critical points. Naming’ the points of contact of the tangents under consideration M and N, then for all sections between M and WN the vertical through the uppermost fiber gives the sole critical point, since the intersection G,, used above, here falls outside the span. The extreme values of the shear § will then be produced by loads extending from the given section to the left or right ends of the span, exactly as in the case of maximum shears in a simple truss. 110 ARCHES AND SUSPENSION BripGEs. 1. THE THREE-HINGED ARCH. § 15. External Forces. By equ. (188), if the ends of an arched rib are hinged, so that 11, = M, = 0, we have in general Mae Wal Gan a hacen lao ees (199. The insertion of a third hinge at the point (c—g, y =f) of the arch-axis supplies an additional equation of condition enabling H, to be determined directly. Namely, at the hinge point the moment of the external forces must vanish, and hence M,—H,f=0 where M, is the simple beam bending moment at the center of the hinge. Consequently H,= FCT tees eet eee eee ee eee (200 If a vertical load @ is applied at a distance é ( g, Fis i Mabie (2018. l.f The influence line for horizontal thrust (H7) in a three-hinged arch thus proves to be a triangle with base 1, whose apex lies on the vertical line through the middle hinge and whose altitude g(l—g) Lae If the middle hinge lies at the crown, then g = » so that the is equal to G - horizontal thrust for a unit load is given by equs. (201) and (2014), as —a_s —n U—— H=Gyp or H=G DF 4 With the value of H known, we may determine the moment M for any load applied at « = é, by equ. (199): Tre Tnrri-ll:scen Arcu. 111 Mxcp= Gr —Hey Mx>g= G4 (ln) Hey We also obtain the following expressions for the axial and transverse forces from equs. (191) and (192): Prep =G is sind +H cos ¢ abauae nave (203. Prsp=— G+ sing + H cos ee les x>g—=G ; cos¢—H sing (204 Sxcp=— @ F-cosp —H sing It is advantageous to take the moments MJ about the core- points of the cross-section, substituting for y the ordinates of these points, since, by equ. (179), the extreme fiber stresses are proportional to these moments; it then becomes unnecessary to proceed with the determination of the axial forces. The moment and shear influence lines are formed of broken straight lines whose vertices lie in the verticals of the crown and of the given section, and whose equations are given by (202) and (204) if € is considered the variable. These influ- ence lines may also be obtained by a simple construction which, for the moments, coincides with that of Fig. 32 if My and Ih occurring there, are understood to represent the core-points of the given section. The shear influence line is drawn in Fig. 49. The two parallel directions at and bs are determined by the intercept ad=G.cot¢, on the end-verticals; furthermore the point 7 must lie vertically below J, the point at which the reac- Pig. 49, tion locus is intersected by a line drawn from the end hinge parallel to the section tangent. The intercepts of the figure included between atsb and acb give the val- ues of the shear for a unit G load = a The horizontal thrust for a uniform load covering the entire span, with an intensity of p per unit length, is 112 ARCHES AND SUSPENSION BrIDGES. a If the load extends only for a distance A, from one end, then, by equ. (200), pry a 4f for Ms H= 2% M >: H=— (44.1—2n—1*) By the general rules given in § 14 it is a simple matter to determine, for any given form of arch, the greatest moments and shears producible by a moving load. The influence lines may be used for this purpose, this method being advantageous whenever the load consists of a train of concentrations. For a continuous uniform load, the critical loading for moments is determined by: AL 2 fax : TS Bfeplge erences (207. where a, and y, are the coordinates of the appropriate core- point in the given section (cf. equ. 98). If the distance A, < + from the left end is covered with the uniform load p, we have Ay. ax (21—A, k Mg PCI) _ pee — pr i * Ux (208. 1 =F Pte (4y— tr) which is identical with the moment in a simple beam of span A,. If the distance (1—A.) is covered with the load p, we obtain Min P Tx (l— ax) ms Yn — Minax ts ... (209. =o Pr (l—2,) a ee "Yx The critical point for maximum shears is given by Ae 2f T= Btliang ee (210. We then have for any section for which A,< a Snax = $j [ (1-2)? — (1A, )*] cos 6 — Pp (Mat) sin g _Or—a)? (211 2h =p.cos¢. ‘THE ''HREE-HINGED ARCH. 113 and 1 pt - Smin—=-> p (L—2 x) cos 6— Se : sind — Smax ; e (212 =p.cosp. 3I—2d2—2e°—P as ; 4r, For all sections for which equ. (210) gives A, > 4 we must substitute A, = 1 in equ. (211). If the axis of the arch is of parabolic form, equs. (110) also yield the values of S.sec ¢. Figs. (50) and (51) show the plotted graphs of the maximum moments and shears in a segmental arch having a constant distance between core- points. The curves acmdb and egnhf in Fig. 50 represent the moments about the upper and lower core-points when the entire span is loaded; the curves apgqrsb and p1q1,71 5; give the maximum negative and positive moments about the upper core-points, and the curves ¢,%,,v,w, and etunvuwf give the same moments about the lower core-points, In Fig. 51, abcd is the curve of shears for a full-span load, aefgh and kimnd are the curves for loads extending from the section to the right or left ends of the span, respectively, while pe or ql represents the effect of a load extending from the point J (see Fig. 49) to the end-point B. If the half-span is loaded, and if the arch-axis is a parabola, the moments in the loaded or unloaded segments, respectively, will be M=+ 2 px (3-2 ), and at the quarter - points M= + 3; vl’. S 114 ARCHES AND SUSPENSION BRIDGES. § 16. Deformations. 1. Effect of Temperature Variation and a Horizontal Dis- placement of the Abutments. We consider a three-hinged arch whose ends are at the same level. Let the span J be increased by the amount AJ through a yielding of the abutments; also let there be a simultaneous temperature variation of + t°. Setting M0 and P= 0 in the general deformation equation (185), extending the integration over the half-arch, denoting the total length of axis by b and the length from the end to the point (xz, y) by s, and assuming a segmental form of axis for the arch, we find: Al b—1 l “FA do ftot| 3 +3): hence Al b Ate — op ota ee ee ey (213. which formula may also be apphed to non-segmental arches of flat pitch. Substituting the expression (213) in (185) and (186), we obtain the following values for the displacements of any point of the arch-axis: Ac=—5 i (1- Ce Re Ay=— . 2 tot(s. sin @ +> + or approximately, ee eae bigs cee (214° Ay=— “pS ot (y+ The upward deflection at the crown is given by the second of equs. (214) as l bl ging ah - af 4f 2. Deformations Due to Loading. In equs. (185) and (186), which serve to determine the deformations, there remains to be determined the quantity A dp, 1. e., the rotation at the left Af=— DEFORMATIONS OF THE TuREE-HINGED ARCH. 115 end A. The following values may be established for this quan- tity in three-hinged arches: If 6 is the total length of axis, and A ¢’, the rotation of the right segment of the arch at the crown-section, then, applying equs. (185) and (186) to the two half-arches with the admissible approximations J—I and P’=P, we obtain the following ex- pressions for the displacement of the crown: ey 2B: Am—=—A do f—f fy) ds ~ 0 %, = a (44 -ds+az) 0 —Am=+A¢'° + fh (y) ds 5 b . (216. J Adding the first and second of the above equations, also the third and fourth, and eliminating A ¢’, from the resulting equa- tions, we finally obtain Ago= — {7 G= a *)ds flog (218. 116 ARCHES AND SUSPENSION BRIDGES. Since M and P may now be considered as known for any given loading, the above equation will, in general, enable A ¢, to be determined and hence, by equs. (185) and (186), we may find the deflections at any point of the arch and, by equs. (216) and (217), the deflections at the crown. a.) Flat Parabolic Arch, Loaded with a Concentration. The equation of the axis of the rib is y = “t a(l—a); also, for a flat arch, we may put ds=da2,r= ae and P, constant for all sections, = H. TI and A will likewise be assumed constant. We then obtain for a concentration G, lying at a distance (< >) from the left end, by equ. (218), é 1 soa [Fife tag)art £ fom ® é (3+ AF )ae— $s fa (09 HF aa] 1 l ~artlef O-F+ = pat r=) | Hence, beens @.é(P—5 (—£)*] GE (87 4+3P 0 380 EL. P 12E A fl Similarly we find the rotation at the right end B, —vAg — &-blb—5? I—4)] G.E(8f+31) A¢,= 30 BIE a EAP? l oe (220. ) (219. The components of the deflection at any point (z,y) of the axis of the rib are found by equs. (185) and (186) to have the following values: For x < and KE Ago. 45[2¢{5 (2-1 } (l—2x) Ss {52 (2132) —£(301°—551a+ 16 x?) ba] .. (221, + ye ee {16 (22) P30} (I-22) DEFORMATIONS OF THE THREE-HINGED ARCH. 117 Ay=—- sor be {e—é (31—2) } +e{e—s5 (i—e)} ] .. (222. G 2 + sp ar [16 (82 —21) P—3P)] For «<< + but >é: dome [ser (21) +221 (4+ 151e—5e) —227 (19 +15 215) +702 2®—55la*+160°] $ (223. rae [16 (1-22) P30] (122) au= 57° 3p (EE — es ee ts 1501+ 5at} + - 5%, [16 (8a—21) P—30" Finally, for « >> i (and hence > €): aay ep [P+ 201 (6P—5E1+ 58) — 22? (21P—5f1458) 4+ 702°? — 55241 + 162°] (225. — £2 {16 (22—1) fP+3e b (22-2), ay= GE) [p52 (I) —50(l—2)*] 4. S0=2) 1161132) P—3)) ne 12BAPE The expressions for the vertical deflections agree in the first parts, viz., the terms depending on the moments, with the cor- responding terms given in equs. (139) to (141) for the system of arches or cables stiffened with a three-hinged truss. We may therefore apply here also the graphic construction there given for determining the deflections. The displacements at the crown are found to be: _ _@fé Le= ae + 7 7 eel) eSeoaaeserusele ese (227. _ Ge(8F +32) Yo= =a (bets —5£*l) mae eee (228. The rotations at the crown are found to be, for the left arch- er 118 ARCHES AND SUSPENSION BRIDGES. and for the right arch-segment, Ad’ == (2+ 2021—208 (8—16f?) .. (230. 120. oATF ) +a The preceding formulae apply only if é< ae ; the effect of a load in the right half of the arch is obtainable by the princi- ple of symmetry. If the load is at the crown, there results Ax.=0 _ GE G8 430) AYce = — 79 BI “48 BAP ofa Yes, a RE te, Ee sale (231. = pe, Hee pene Pa on Ad.=—Ad’.=— ison | apap +++: (282. By equ. (227), the horizontal displacement of the crown will be a maximum for a load at r=, furthermore, every load on the left half of the rib produces a displacement of the crown toward the right. Also, if we consider only the first term in equ. (228), we find Aye =O at about = 0.3361, so that, approximately, all loads in the outer thirds of the span produce an uplift, and all loads in the middle third produce a depression of the crown. b.) Flat Parabolic Arch; Continuous Load. The deforma- tions produced by any loading, whether a train of concentra- tions or a continuous load, are readily evaluated by means of the influence lines which may be constructed in accordance with equs. (219) to (230). Let us consider here only the case of a uniformly distributed load of intensity p extending for a distance A from the left end, where A < a For such loading we obtain the following formulae: [81?—2002a-+ 1512402] — PXBPF3P) ogg, A¢o=— 0 FTF a ? 248A fl (fP— heat] eNO (234, —Ad = ah a Pe 24 BA Pi and, at the ae [92?—301A? + 8a] — pd (32—16/*) So cust 7 E cece eee , pn (3 7—16 f?) A$’ o= sgh (+ LOLA 808] + PE. (236. At. = ave (od? 20: EA? 4216?) eat eewe cues eases (237; DEFORMATIONS OF THE TuREE-HINGED ARCH. 119 x n?7(8 f? 37? A ¥e= seh yj [BE —20 1A? + 16a8] — PEED) taaned (238. If the loaded length A extends from the left end to some point beyond the crown, i. e., if A > + we have the following formulae: Ago — 2G [2185 1(1—a) 2+. 4 (Ia) 9] 120 EIF p (4IN—L— 2?) (8774317) -+ + + (239. = 48 BA f?l —A¢, = BUA [BP Lea + BIA + 408] p(4IX—P—2) (87 +392) .. +. (240. ey 48 BA f7l __ p(l—r)? 2 A $e= SiO RTE (4? + 101(l— a)? —8 (I—A) ?] (241 p (41X— VP? —2)’) (83 — 167?) oP . _ 48 BA fl A ge =2C=™" [99—-301 (1A)? + 8(I—A)?] BI p (4IN—F— 20) (3 —16 72) +++ + (242. t 48 BAS? l (l—A)? An —= Pt [518—201(1— A)? +16 (I—A)?] ...... (243. __ p(I—)? 2 AYo= O60BTI [3 2—201(1— A)? 16 (I—A) *] (244 _ p(4tA—P—2) (8748/7) Eee 96 BA f The horizontal deflection of the crown is found to be a max- imum for A= 4; we then obtain —s pre 9 MAX.(ALe)= ggg gp crrrt ttre ttt eee (245. The maximum uplift of the crown occurs approximately when the first and last thirds of the span are loaded; we then obtain _— 37 opt ___ ph BE +8f?) The maximum downward deflection at the crown, which occurs approximately when the middle third of the span is loaded, is found to be 37_ pit 4 5 phBU+8P) 864 max.(—Ay,.)= 116,640 Er Tage ee 120 ARCHES AND SUSPENSION BRIDGES. 2. ARCHED RIB WITH END HINGES. §17%7. Determination of the Horizontal Thrust. 1. General Case. By equations (187) to (192), putting both of the end moments M, and M, equal to zero, the external forces may be determined as soon as the horizontal force H is known. To determine the latter it is necessary to consider the elastic deformations. The appropriate equation of condition may be developed from the Principle of Virtual Work, or it may be derived directly from equ. (185). From the latter, with the approximation J —J, we obtain the displacement of the end B toward A; ej vast fe Fz (4 ds—ar) aot f (0 —dz)..(248. 0 By the Principle of Virtual Work, equ. (248) may be established as follows: The general expression for the work of deformation of curved ribs may be written, P?ds M’ds w=\-tea + a x (1-2) Va—% (h—y) val : 1! : ; The expression T x (l1— x) Um gives the reaction at A for 0 the forces vm considered as acting vertically. If we imagine this : ay s ‘ if ; reaction, which is easily determined and — 5% vm in sym- 0 124 ARCHES AND SUSPENSION BRIDGES. metrical arches, to be applied horizontally at A, also the forces Um to be acting in the opposite direction at the respective panel- points, then the above expression in brackets denotes the static moment of those forces about the line of action of W and may be represented by the ordinates of a funicular polygon in the well-known manner. Fig. 54, 1 1 oul f----+4 1 vt ‘ Sut Seeu ps 4 Zl nice Sid [ass renee elec ce ie ew cid 4 ' feaereocanca ‘ ro ' hae sas ' i ' weerfocnnnnnpeccengqebena rad The construction for the horizontal thrust produced by a vertical load G, also by a horizontal force W, is shown in Fig. 54. Fig. 54? is the force polygon constructed with the quanti- ties vn computed by equ. (255) ; Fig. 54° is the funicular polygon of these forces considered as acting vertically; Fig. 54° is the funicular polygon for the same forces acting horizontally. These funicular polygons give the intercepts pp, and qq, on the lines of action of G and W respectively. If we add to the latter inter- ape then the length pp, and q,q, will represent the horizontal thrusts for the forces G and W respectively, provided the latter forces are represented by the distance n,,. The latter length is composed of cept the small distance qq, = 1 b.cosB. 1’ No R= ZYnVm and igi BIO Pads 0 Ao.@o-p where p is the pole distance of the force polygon (Fig. 54*).* With an, and I’ constant, i. e., with the moment of inertia increasing toward the abutments in the same ratio as sec ¢, we obtain approximately Un = Yn. *That the H-curve for concentrated loads on arches with end hinges may be represented by a funicular polygon, was first demonstrated by Mohr. HorizontaL TuHrust In THE Two-Hingep ARCH. 125 In plate-girder arches, in which the depth of web is usually fairly constant, the approximation vm = Ym will always give sufficiently accurate results. It appears that the variation in the moments of inertia must be very marked before it will exert any appreciable influence upon the value of Z. If the two ends lie in a line inclined at an angle a to the horizontal, the preceding method for determining the horizontal thrust H must be modified as shown in Fig. 55. The quantities Fig, 55, v are given here, as above, by equ. (255), the ordinates y being measured from the closing chord; and, in l’—IJcos¢, ¢ denotes the inclination of the arch elements to the horizontal. The funicular polygon giving the quantity % yv must be con- structed for the forces v considered as acting parallel to the arch- chord, using a force polygon whose vertical pole distance is p. b.cosB.secta. I" Ao. @-p Equ. (250), likewise (251*) and (251°), may be extended to the case of an arch having its two ends connected by an elastic tie. If A, is the cross-section of this tie-rod, the lengthening of The correction length is, in this case, nn, = 126 ARCHES AND SUSPENSION BrIDGEs. the span becomes AL= —%—. 1+ tl, so that equ. (250) assumes the form: The problem becomes somewhat more difficult if the tie-rod itself is arched and is connected to the main arch by suspension rods. The equation for H may then be established by applying the Principle of Least Work. If the suspension rods are distributed at uniform inter- vals of length a (Fig. 56), and accepting the approx- imation, admissible in flat arches, of equating the axial forces in the arch to H, we have M—=M—Hy+Hy=M-Ty’'; and if A, =a mean cross-section of the rib, A, = the cross-section of the tie-rod, A, =the cross-section of a suspension rod, f, =the rise of the tie-rod, and = =1(1+5 #6 yo) then we obtain, b f M4 ds+ Eot(d cosp—l) fo .. +. (260. b $88 bcos cos B m A’n Se ds tite By (ey This formula leads to the same graphic method for deter- mining H as has been deduced above from equ. (250) et seq., but in place of the arch-ordinates y there must be introduced the intercepts y’ between arch and tension-rod. If the tie-rod does not connect the ends of the axis but two higher points thereof (CD, Fig. 57), its stress is to be com- puted by the formula: a HorizontaL Turust in THE Two-Hincep ARCH. 127 D y bi cos B hi J ds + OSE A Here y is the arch-ordinate measured from the tie-rod, M is the moment of the given loading in a simple beam of span AB=l, 1, and A, are the length and cross-section of the ten- sion-rod, and b, and A, are the length and mean section of the arch between the points C and D. The influence line for H is given by the funicular polygon of the forces v (computed from the values of y); this polygon is prolonged outward, above the points c and d, to the closing side a b. If the arch is connected through hinges to elastic piers, and if h is their height, 7, their constant moment of inertia, and E, the coefficient of elasticity of their material, then the outward deflection of each pier by the force H must equal = , so that F é ‘ 2 H.W the increase in span will be Al = + 3 Eh equ. (250) and solving for H, we find 1 My f 7 .as 0 H= ; Lo. (2598, 2 b cosB 2E i ye +o Be fs vas Ao 3A i 0 Substituting this in 2. Arch with Flat Parabolic Axis and Parallel Flanges. Concentrated Load. Adopting the same simplifying assump- 128 ARCHES AND SUSPENSION BRIDGES. tions as above, equ. (250) for determining the horizontal thrust is also applicable to the present case. Assuming A cos ¢ = Ao, = =TI,, constant, and as that, for flat parabolic arches, b cos pat ae r) (1 — gf constant, likewise I.cos ¢= ua— = c fr), then for a parabola of rise f with a load G at a dis- tance é from the end, we have (cf. the derivation of equ. 75) : 5§(8—21° +0) Ff ~ [set we Pe)|e ee racist (261. ae L gf +151 -(A- i a5) For the effect of a concentration alone, we obtain from the first term of the above expression 5E g ‘ H= oP —25+1)4.6 Lezucierean ae (2614. +15EI, where, for abbreviation, f=(p+eeO-F5)] (262. The substitution of f’ =f would be an approximation corre- sponding to the assumption that the axial force contributes but very little to the deformation of the arch. The equation of the reaction locus is derived from equ. (198) : —e 8f’ Z y= B(FLUE—FP) Se ee emer eee reece er weererns (263. A much simplified, approximate expression for the hor- izontal thrust in a flat, parabolic, two-hinged arch of uniform section has been established by Engesser and Miiller-Breslau; it reads: é (i—é) . a = iG. 7 (2633. With this approximation, the influence line for H becomes a parabola, and the reaction locus a straight line parallel to the arch-chord. The error in comparison with the more exact equa- tion (2617) amounts to — 4% for a load at the crown, but, in the small values of the horizontal thrust for loads near the abut- ments, the error becomes relatively greater up to about + 10%. HorizONTAL THRUST IN THE Two-HINGED ARCH. 129 3. Plate-Girder Arch with Parabolic Intrados and Straight Extrados. Although the general formulae [equ. (250) together with (253) to (255)] are sufficient for the treatment of this ease, let us establish a direct, approximate formula for the horizontal thrust producible by a concentrated load. If A, B, is the gravity axis of the rib (Fig. 58), f’ its rise, f the height of its crown above the line connecting the end hinges, h, and h, the effective depths (or distances center to center of flanges) of the cross-sections at the crown and abutment respectively, and I, and I, the moments of inertia of the same cross-sections ; also introducing the ratio-symbols = ff f him ho iC Slept wae wien (264 = Ith _ 2h x Io ho and the abbreviations 2¢ LE and, approximately, I’, =1,(14+8«% + 16x 5) where x is the distance of any section from the crown; then equ. (250) yields the following formula for the horizontal thrust producible by a load placed at a distance é’ from the crown: H= 1 1 jE. {5 (1-97) +. a (1-11) - gb 1-7") t age (1-9) | 4 they td cos B 271 2 1 4 Ip 6 cos B 1-30tt¥ +8 7G Fret zp tedt (42-x) ome (26 =x) ay wees (265 9 130 ARCHES AND SUSPENSION BrinGEs. where f a= ve —1)—26 b=srf E+ 2Q0ve (875 ;—1) —42+y este #EE 42=0 nee 1 2 1 d=5—srtertsevg :(+ —tv) 1 2 f° e=st+7r(4r5 —1) The akove equation may also be used as an approximate, general formula for calculating the horizontal thrust in any arch-rib having a parabolic gravity-axis and any continuous variation of depth of girder. Thus, if the cross-section is uniform, we may obtain the horizontal thrust from the above formula by putting e-—=y—0 and, if the end hinges are Fig. 59. in the axis, y= 1. For the crescent arch (Fig. 59), we must write y =1, e=—1, x =+1; we then obtain for this case, H= 1 l EI Lev) + [ie 19) 499-1) +e -1)] ) fap ta- #0 008 8 8 72 Inbcos B 43 et—aPl (266. Another formula for crescent arches, given by Schaffer, is as follows: | +7) loge + (1-7) loge j= Z\@- gyt pp wtb cosB H= cis . (266" 1 +554 ob cos B eT The reaction locus for the above types of arches is deter- inined by the general equation (1987). 4. Segmental Arch of Uniform Cross-Section. Substitut- ing P’=P+ ~ y=r (cos ¢—cos B), You yi—0, 4 ds—da—=—ds cos B and P=Hcos¢d+Vsind élso putting the small quantity af 8 eee eens Pewee ewer ere scene wee ec cee (267, HorizontaL Trust In THE Two-Hincep ARci. 131 where J is supposed to be constant, equ. (185) yields the fol- lowing relation : +B EI.Al=r es conten ae—ot fone -2H ef (cos $¢-cos 8) “assanesf (cos¢-cosB) cosBdd (268. ~2meaf oe cos¢.d¢—sr fy cosB.sind.dd -B +2EIwtcosp.r.g If two concentrations G are applied at two symmetrically located points of the arch, and if the included central angle = 2y, we find ee PN Aa oa o=y7 _anl¢= : M=Gr (sing siny) ]5— 0 v—o]$- — fle (oF Set o=B ie ; M=Gr (sing sing) |o— V=G ae Substituting these expressions in equ. (268) and solving for H, we derive the following formula for the horizontal thrust produced by a single concentration (together with a tempera- ture change ¢ and an end-deflection A 1) : H= [sin?B — sin?y + 2cosB (cosy —cosB) —2(1+5) cos B (BsinB—vysiny) ]G¢ +22f (20 tr 8 cosB—Al) 2[B—3sinBcosB+2 (1+.3)8 .cos?B] ... (269. In the semicircular arch, 8 = 3 consequently, Fa PE A eri nas (270. wv ra 132 ARCHES AND SUSPENSION BriIpGEs. or, with Al=O and if x is the distance of the load from the abutment, a astoceedan nating (2708. Par The H-influence line for a semicircular arch is therefore a parabola. The equation of the reaction locus is derived from equ. (198?) : __ Gr (sin? B —sin?y) = 2H sin B (271. For the semicircular arch, this reduces to Hence the locus is, in this case, a horizontal line. Flat semicircular arches may, with little error, be designed by the formulae for parabolic arches, MoMENTs AND SHEARS IN THE Two-HIncEep ARCH. 133 § 18. Maximum Moments and Shears. 1. General Case. Graphic Method. There is nothing essen- tial to be added to the general rules given in § 14, for the con- struction of the reaction locus and the determination of the critical loading, in applying them to the two-hinged arch. If the H-curve (or H-polygon, if the loads cannot be transferred to the arch except at distinct points) is plotted according to the values of-H calculated for different positions of a moving load, or obtained graphically as described in § 17:1, then the reac- tion locus may be simply constructed by drawing the successive end-reactions, through the end hinges, as the resultants of the Fig. 61. ose eee nt corresponding values of H and V,. With the aid of this locus, the critical loads may be determined according to §14. In finding the maximum moments and shears, we may apply the principles given in § 7 :3,4. As ‘before, we may either use the method of influence lines or construct the equilibrium poly- gon of the loading. As in the case of the three-hinged arch, it is advisable to take the moments about the core-points of the cross-sections since, by equ. (179), the extreme fiber stresses are directly determinable from these moments. Fig. 61 shows the application of the method of influence lines. K,K, and K,K, are the two core-lines. Since, in general, M=M—Hy=y (= = H), the moment for a moving load will be represented by the difference between the ordinates of 134 ARCHES AND SUSPENSION BRIDGES. the H-curve and those of the ordinary moment influence line for a simple beam, the latter figure being constructed with the distance y as the unit load. Since J, is the critical point (or load-position giving zero moment) for the upper core-point, and J, is the critical point for the lower core-point of the section M, M,, the figures a m, 7, 6 and a m, i, b will be the moment influence lines for the respective core-points; and if v, and v, are the distances of the upper and lower extreme fibers from the gravity axis, then, for a uniformly distributed live load of p per unit length, we find: Maximum tension in the upper _ “¥: ene Pe ore : Ee a 4 : area inh b (Suen Gi) fiber I Maximum compression in the __%¥%: area aM, ( pl ) upper fiber I 2area adb Maximum tension in the =2% . grea am,i, (. Pp lower fiber I 2 areaadb i i i a) sake l Maximum compression in the __ t% area i,hb ( p ) lower fiber I 2areaadb The evaluation of the areas of the influence lines may be accomplished graphically, as already outlined in § 7:3 (a): The influence line for shears is represented by Fig. 49 if the line abc is replaced by the H-curve. The method of influence lines is particularly advantageous when the live load consists of a train of concentrations. In that case the position of loading for maximum stress may be found by the rules given on page 42, It should also be noted that, usually, even continuous loading is transmitted to the main arch at distinct points. The influence lines are then no longer continuous curves but assume the form of polygonal figures with the vertices on the vertical lines through the points receiving the load. The second graphic method, for determining the moments and shears by means of the equilibrium polygon for partial load- ing, is to be carried out exactly as described in § 7 :3 (b). 2. Arch with Parabolic Axis and Variable Moment of Inertia. The horizontal thrust for a uniform load p covering the entire span is found by integrating equ. (265). If N de- notes the denominator of that equation, we obtain 1 1 1 1 a5 Tan a ap bb tase 2 12 60 140 252 U ee ¥ ppt (272. For a partial load, extending from one end to a point distant = €’, from the crown, if we put + ae =+y, (where the MOMENTS AND SHEARS IN THE Two-HinGep ArcH. 135 + sign applies if the load extends ovet the crown and the — sign if the load falls short of the crown), we find the horizon- tal thrust, (24+3n—8) +f bb 5— 1") a 1 Sea 1 we gt a C2(Ss _ iy CHW) b+ yg BHONT-W OP ( = 48 NV f If x, yx are the coordinates of the core-point of any cross- section, the limits of loading for maximum stress are determined by the conditions: 2 H on = Vv, and ya V, ‘s Substituting the value for H from equ. (265), also V, =G1+yv 2 , we obtain, 1 a—w) fst 2 a4y a—t G4yi+y) b 4 1 6 . 15 t 1 1 a 4 6 tl _ ax T tog Qtwtatty) oe] b=S.a Yk 1... (274, The value of y, given by solving this equation is to be used in calculating H by equ. (273); we may then find the greatest negative moment by the relation 1 Main=V. oe —H y= = (1 +y,)?pla— H.yx eee eee (275. 8 This expression holds good for all sections from the end to a point whose abscissa is given by l—2c,.= INF z “UY For the middle portion from x’, to 1—2’,, for which an interrupted load extending on each side to the end of the span must be taken, there is to be added to the value of Af given by equ. (275) the value given by the same formula for the sym- metrically located point of the arch. The general formulae may be used for calculating the shears. 38. Arch with Flat Parabolic Axis and Parallel Flanges. This is merely a special case of the preceding one; if the moment of inertia is constant throughout, the corresponding formulae may be obtained by simply substituting «= x¥—0 and v—=1 in equs. (272) to (276). If the moment of inertia increases toward the abutments with the secant of the slope, i. e., if J, = 136 ARCHES AND SUSPENSION BRIDGES. I, sec $, we obtain the following expressions directly by integrat- ing equ. (261?) : For a full-span load [cf. (82)], 1 2 Hiw= 5 ‘rie. ieiaile “eeiter ale aria dis ealer cleo @ ani Aa vars (277 For a load covering a length A, from one end [cf. equ. (83)], 5 a\2 MY 1 2a Tp? H=3(¥)[1-G)'+3G@) |]@ (278. For a crown-load of length a, 1 a-, a\yl H = 7595 (25-10G +5) 5 - pa... (279. For the greatest negative moment, A, is determined by i=) (PHD) =F ES © acebocas (280. and the value of the moment is given by Moin=V .%—H. yr BM ..—H Yu eeeee (281. For the central portion included between the abscissae x’, and 1—x’,, determined by the equation bal p Fey's cadet sees cia nein nense (282. we must add to the above values of M the values given by the same formula for the symmetrically located points of the arch. For a full-span load, the moment about any point of the parabolic axis is given by ra(l—a) , f'—f Big ge eee vaioasteen eee (283, or, if the moments are taken about the core-points, Mico PO BE vaeeeeee eee (283°. From the above-values, the maximum positive moments may readily be deduced. If f’, defined by equ. (262) or approximately by 1 7 f'=f (14 “S AP? ; is put equal to f, which is equivalent to neglecting the effect of the 2 axial force upon the deformations, then we should have Hioe = + or =H'tor and the moment about any point of the parabolic axis —0; hence, MoMENTS AND SHEARS IN THE Two-HINGED ARCH. 137 under this assumption, the line of pressures coincides with the axis of the arch. The error thas introduced into the determination of the stresses, especially near the crown, may sometimes become considerable. For example, if the cross-section at the crown consists of two flanges having a combined area Ao, then, with the above approximation, the intensity H’ of stress in each flange will ke ao = — ae But if we substitute oO ; (1 15 I? a the value f’ = f + 35 7? ), we nd: The stress in the upper flange, 14 5 oh ce ALF SPA i fs tet Cdr ge oe 32 f? The stress in the lower flange, ,_ 13k oz mG 2 (f/—1) j= Lf ee ESC. gee h ne oe 32 fF? The former attains its maximum valce with + = 0.742, wher h 1 ou=1,35 o’ ando,=0.242 0’. With p= yw find for the upper flange, ou= 1,24 0’, and for the lower flange, 7, = 0.65 o’, The critical loading for shears is determined by (PLA, AF) = fF cbt ds cadences (284. and, if H,,, denotes the horizontal thrust calculated by equ. (278) for a load extending from z to d,, the maximum + shear will be: Smax = 37 [ (t—2x)?— (L—d,)*] cos? — Hy r, . sing. . (285. and, similarly, Sain=—z P (1— 22) cos — BE. sing — Smax oe. (286. 4. Segmental Arch of Uniform Cross-Section. The hori- zontal thrust for a continuous, uniformly distributed load p may be obtained by integrating equ. (269) between the ap- propriate limits, after substituting G = p.dx——prcosy.dy. If the load extends from one end to a point departing from the crown by the central angle y, the limits to be used are y = 8 and y, so that we obtain: 138 ARCHES AND SUSPENSION BripGEs. 1 J, sinB.A—siny.B+ 9 cos B. (B—vY) rod . cosB.G 2(B—3 sin Boos B42 (1+ 5) B cos) RUNG Here, ae if 1).33 : A= 3 sin’ B — = cos” B— B sin B cos B, ee ee ‘ ; 3 B= sin’p— x sin y— cos B(2cosB+2B sin B—ysiny— + cosy) ‘i C=sinB (2B sinB —4Bsiny —cosB) +siny (Qy siny+cos y) Tp r If the entire span is loaded, y==— £, and we have, sin? B —3 (1—2sin?B) (sinB—BcosB) He= me AC a a Weems (288. 6(B—3sinBcosB+2(1+5) Bcos*B) and, if the arch is a complete semicircle, Fg aed enc hens: (289. With the aid of the reaction locus given by equ. (271) we may determine the critical loadings to be used in finding the maximum moments and shears by equs. (281) and (285) re- spectively. These are plotted in Figs. (62) and (63) for a STRESSES IN THE Two-HINGED Akc.:. 139 segmental arch of uniform core-depth. The same notation is here used as for the three-hinged arch in Figs. (50) and (51), so that we may refer to the explanation there given (p. 113). 5. Temperature Stresses. The horizontal thrust, H;, pro- duced by a uniform change of temperature, ¢, is given by equs. (250) to (269), if we substitute G=0, M—0, and Al—0O. For structural steel we may take Ho =196 lbs. / sq. in. / °F. (= 248 tonnes / sq. m. / °C.) and, for the variation of tem- perature, assume t= + 55°F. Equ. (250) then becomes: | ate, es (290. I,b cos B Go 2 ym. Vm - ———— ¢ Ao or, for flat parabolic arches of uniform cross-section, equ. (261) gives approximately, Io a a H.=+10,780 —— = * oe ee (2908, In the crescent rib of parabolic form, equ. (266) shows that the value of H is reduced to about half the above value. The moments due to temperature change, referred to the core-points, are given by Nh Bee Ye ered aad SS (291. These moments are thus proportional to the distances of the core-points from the closing chord, and hence they attain their greatest value at the crown. The shear due to temperature variation is, by equ. (192), SiS Bigs SUD. a cee ee ei pe a aes (292. In an arched rib consisting of two equal flanges, if A is the total cross-section and h the effective depth, the extreme fiber stresses due to temperature variation are, by (179), h h a ve) 3 2a (yey os =+ => A.h and, introducing the value of H: from equ. (290"), we obtain, for the parabolic arch with parallel flanges, a tye D 2 rit ie =) Accordingly, with Ewt= 10,780 lbs. per sq. in., the extreme fiber stresses at the crown will be: o=tEFot 140 ARCHES AND SUSPENSION BRIDGES. lh sojedo 24 i ‘i f 3 15 he Pag is h The stress in the upper flange is a maximum with e = 0.742; the following values then obtain: ou = + 0.371 X 10,000 = + 3,710 lbs, / sq, in. o, = = 0.808 X 10,000 == = 8,080 lbs, / sq. in. : : nol With a ratio of — = “35 f ou = + 2,640 Ibs, / sq. in. o1 = = 3,700 lbs, / sq. in, In the case of the parabolic crescent arch, the temperature stresses are constant throughout each flange, and their approximate value is 1 lh o= + 5,820 2 is i TF Hence, with the same ratio of crown-depth to rise, the temperature stresses in the crescent arch are less than in the arch with parallel flanges. In an arch having its ends connected with a tie-rod of the same material as the rib itself, no stresses are produced by uniform changes of temperature. DEFORMATIONS OF THE Two-HINGED ARCH. 141 § 19. Deformations. To determine the elastic deflections of any point (2,, y,) of the axis of the arch, equations (185) and (186) may be employed. Let us first apply equ. (186) to the end-points; simplifying by writing P’—P and J=—I, and assuming a symmetrical arch-form, if 6 = the total length of axis, we find for the rotation at the end A, sae” ( a7 l—2).ds— 2 (hat) fa—2) 42a] 0 0 Again substituting 1’ =—Icos¢=I 2, and writing, for abbreviation, [ wit (4 —Fot) |] =z ....... (293, the above equation assumes the form: t Ely. Ado — 5 f2(l—2)da— 1 "(EB ot)ay.. (294. 0 c= Substituting this relation in ee (185) and (186), we obtain the following expressions for the variations of the co- ordinates : —ET,.dy=% fea—mae— fea arte, - (295. El,.d0—% fea) da fet y) dx +Cy.... (296. where gal 142 ARCHES AND SUSPENSION BRIDGES. represent the contribution of the axial compression to the total deformation. If we adopt the usually admissible approximation i = = = constant, where A, is a mean area of section, and if ¢ is supposed uniform throughout the span, then the second definite integral of equ. (294) vanishes and we have of putting Cat (3 —Eot)y, C.= —1,( —Eot)z, The definite integral containing the quantity 2 may readily be given a static interpretation: If the quantities z are con- sidered as forces acting on the horizontal projection of the arch-axis, then the second member of equ. (294) represents the end-reaction Z, for these forces. Let M,, and M,, denote the z. moments of the forces Z, and 3 2 about the arch point (2, y;), 0 when these forces are applied vertically and horizontally, re- spectively, at their respective points. Then we have, PT ou peace alee ut hetasa’ 294". SBI ky Si oso hime acoees (295%, PTW PO co eee near Gece. (2962. If the radius of curvature of the arch axis is very large 6 pM hence, neglecting the term C, which depends upon the axial compression, we may obtain the vertical deflections (—A y) of the arch as the ordinates of a funicular polygon, constructed with the pole distance E I,, for a loading consisting of the relative to the sectional dimensions, we may write z= area of the curve of reduced moments, M ae With the above assumptions, the deflections may be determined by rules which are practically the same as those for a simple beam. For finding the deflections graphically, a treatment suggests itself similar to that applied to suspension bridges in § 8. If we write, with the usual notation, / —MW— Hy, there re- sults e—H[(¥-y)* +e], be atattiadon ke (298. es DEFORMATIONS oF THE Two-Hincren ARCH. 143 where shy x BAe Io —Q-43"" SF tte (299. The moments, M,, and M,,, may then be represented by the differences between the ordinates of two funicular polygons: M one constructed for the supposed loads eae oO + c, and the other for the supposed loads y. + The former polygon is obtained by reducing the ordinates of the simple moment curve, Ih 7 thereto the small and nearly constant quantities c; while the constructed with pole distance H, in the ratio and adding funicular polygon for the loads y . a is identical with the H-influence line described in § 17:1 and constructed as shown in Fig. 54. Little need here be added to make clear the graphic determination of the deflections of a crescent-shaped arched rib as executed in Plate I, Figs. 1 to 1%. In Fig. 1, the ordinates of the arch-axis are multiplied aa od by the ratio a and from the resulting values, with the aid of the force polygon Fig. 1°, there is constructed the funicular polygon I, i. e., the H-curve for a moving concentration G. The value of this unit load G is determined in the familiar manner by the funicular polygon JI; in the example selected, with 1=160 meters, Io = 2.80m.*, do = 0.112 bcos B. I, Ao. @o.7p =1.13 m. We proceed to find the vertical deflections of the points of the arch produced by a load G=1 applied at the point C (a,y,). The load values zg are derived from the simple moment diagram (Fig. 1¢) constructed from the force polygon Fig. 1° with the pole distance H. The sq. m., a= 16 m., p = 180 m., the necessary correction nn, = I ordinates of this moment diagram are multiplied by a and should 6 Aofo are vanishingly small on the scale of the drawing. From the resulting ordinates there are constructed, with the aid of the force polygon Fig. 1*, the funicular polygon III and, for horizontal action of the forces, the funicular polygon IV. With the provisional neglect of the effects of the axial compression, the vertical intercepts between the funicular polygons I and III and the horizontal intercepts between the funicular polygons II and IV, or rather between the corresponding funicular curves, are propor- tional respectively to the vertical and horizontal deflections of the points of the arch under the given loading; and the scale for these deflections is Ely H.a.p of lengths is 0.5 mm.=1m.,G=—1t., H = 0.454 t., H = 20,000,000 t. per i ; __ __ 20,000,000 x2.8 05 sq. m.; hence the scale for the deflections is 1 m. = 0.454x16x180 * be increased by the quantities c= = 0.33 m.; but these corrections times the scale of lengths. (In the example on Plate I, the scale e 144 ARCIIES AND SUSPENSION BripDGEs. mm.,or 1 mm, in the structure = 21,4 mm, on the drawing: Seale I.) The and : C. C. : corrections . — —~, representing the coordinate components El, El, of the deflections due to the axial forces, may be assumed proportional to EA, H 20,000,000 x 0.112 0.454 X 0.5 mm. or, with G=1 t,, 1mm, in the structure = 2,467 mm. on the drawing. It thus appears that the effect of the axial compression, com- pared with the deflections first investigated, is an absolutely negligible quantity. The intercepts between the two funicular polygons I and III serve not only to determine the deflections at all points produced by a unit load (G) at C but also, by Maxwell’s Theorem of Reciprocal Deflections, the intercept at. any point M will give the deflection at C produced by a unit load at M. Consequently the above intercepts constitute the influence line for the vertical deflections at the point C; and, with this, the total deflection at C produced by any given loading may be obtained by the usual rules. Similarly, the influence line for the horizontal deflections of the point C may be represented by the intercepts between the funicular polygons I and V (Fig. 1"), if we apply the principle that the horizontal deflection of C produced by a unit vertical load at any point M is equal to the vertical deflection of Mf produced by a unit horizontal load at C. The latter deflection, however, may be constructed, as before, by finding the simple moments of the horizontal load and treating the resulting values the coordinates x and y; the corresponding scale is times the scale of lengths. Thus, in the example, the scale is 1 m, = of Mw’ a as vertically applied forces. The moments Mw for VW =—1, CSL, eal defined by Mv =y —a@ a and Mw= (I—«) a =, 7 2re - l z=0 mS easily constructed; and from the ordinates of this diagram, multiplied by a there are derived the force polygon (Fig. 1*), whose pole distance is Hy, a and the funicular polygon V. The differences between the ordinates of this polygon and those of the H-curve (Funicular Polygon I) determine the horizontal deflection of the point @ for any position of the mov- Io Hyia p (In the example, Hs, = 0.405, hence the scale for the horizontal deflections is 1 mm. in the structure = 23.9 mm. on the drawing. Scale II.) The influence lines for the quantities C, and C., i. e., the deflections due to axial compression which should be added to the above deflection values, may be represented, according to equ. (297°), by the H-curve; and the scales for measuring this curve, to obtain the vertical and horizontal deflections of the point C, are == and at 1 a: force-unit, respectively. Thus, in the example, the scale for vertical 20,000 X 0.112 "37. * 37.5 = 3082 mm. on the drawing (Scale III); and the scale for horizontal deflections ing load; the corresponding scale is times the scale of lengths. times the deflections is 1 mm. in the structure = DEFORMATIONS OF THE T'wo-HINGED ARCH. 145 20,000 x 0.112 is 1 mm, in the structure = : a — >< 37.5 = 2625 mm. on the drawing (Scale IV). If the arch has a uniform moment of inertia, there may be deduced from equ. (295*) a direct expression for the deflection producible at the point (z,y) by a concentration @ placed at a distance é from the end of the span. If, as an approximation, we write —___ _ 1 equ. (2957) yields the following: rcos @ To (cf. equ. 1312.) ? — Ely. by=G@m,—H m+ He(P OS™ + rey)... (800. Here, as in equs. (130), m,—= (2lé—& — > for a<é A ee « (801, m= (21 a—#?— &) ————_,, for x >é i is the horizontal thrust to be determined as in § 17; and, as in (132), x I pee Jeude+tfo—s)yas saaae (302. 0 In general, with variable moment of inertia, m, is io to the ordinates of the H-curve. With a parabolic arch-axis, tem EE (a9 21a LP) eee cette (303. and —EIAy=Gm,—HAm+Hexr (l—2)...... (804. The vertical deflection of the arch-point (z,y) due to tem- perature variation may be obtained from equ. (300) by putting G—0O and H = the value of H; given by equ. (290). Hence, El,Ay=Hi.m,+2(Lot—*)Ioy ounces (305. In the example executed in Plate I, assuming Zw = 248 tonnes / sq. 2.8 X 128 m. / °C., we have Ht = 248. t 6x 180 X75 7 is the ordinate of the H-curve (Funicular Polygon I) measured to the scale of lengths, then mz =7.p.a—180 X 16 X 4; consequently EI, Ay = (1184.67 + 1368.3 y)t or, finally, Ay (in mm.) = (0.021167 + 0.02444 y)¢, where 7 ad y, measured to the scale of lengths, must be 10 = 0.411. ¢ tonnes. If 146 ARCHES AND SUSPENSION BripGEs. given in meters. For t= + 30°C., the vertical deflection at the crown amounts to A ye = + (0.02116 X 51.75 4+ 0.02444 XK 42.5) 30 = + 64mm. Similarly, in the case of a displacement of the abutments increasing the length of span by Al, the vertical deflection of the arch-point (x,y), is given by where i. e., with reference to the graphic representation, N =n n,.a. p. REACTIONS FoR THE HIINGELESS AROH. 147 3. ARCHED RIB WITHOUT HINGES. § 20. Determination of the Reactions. 1. General Case. Concentrated Load. We assume an arch so rigidly supported that its end sections cannot undergo rotation; hence the ends are to be considered as fixed. In general, the ends may lie at different elevations; but the form of the axis must be such that the vertical center line bisects all chords parallel to the closing chord. Fig. 64. ca! The fundamental equations are obtained from equs. (185) and (186) by extending their limits of integration over the entire arch and introducing the condition A ¢, 0 which must be satisfied on account of the fixedness of the ends. The hori- zontal and vertical coordinates (x, y) of the points of the arch are to be measured from a pair of axes so located that the Y-axis coincides with the vertical center-line of the arch while the X-axis is parallel to the closing-chord at a vertical distance t, above that chord. With the admissible simplifications J = J, P’ = P, and provisional disregard of the effect of temperature, also assuming an unstressed initial condition of the rib, the general equations become: b b b6,—0= f gdst f a . ds 0 6 b b M P —an,—0= f Bydst+ { H(4as—ar) 0 0 b b Ay, —0— f # ads + fF CEas+ay) 0 0 148 ARCHES AND SUSPENSION BRIDGES. Observing the smallness of the terms containing P in com- parison with those in which M occurs, it appears fully per- missible to put P= H and to consider A — constant — an average cross-section A,. We then have, either exactly or very nearly, Py os, aR ga (fds—de ~~ BAS b P (x Sia Gaetay)=0 so that, upon replacing I by I’ =I cos ¢ =I ae. (where x is not measured parallel to the axis A, but horizontally), we obtain : l ve M Hb § Gea a me 2 l Fo M Hil T°: eS a + te ee eee eee (307 _t 2 l PEs M yp: rdz—=0 aah, 2 J Ags the hingeless arch is triply indeterminate with respect to the external forces, the above three equations of condition will suffice for the evaluation of the unknown reaction elements. As proved in § 13, the moment M is given by the vertical intercept between the equilibrium polygon of the loading and the axis of the arch, multiplied by H. Consequently, if DFE (Fig. 64) is the funicular polygon for a concentration applied at any point C, constructed with H as the pole distance, we have: M=H .MN=H(QN—PM—PQ)=M—4H(y+2) REACTIONS FoR THE HINGELESS ARCH. 149 where M, as before, is the moment which would be produced in a freely supported beam of span J. With a= 2+ 2 ibis there results: M=M— Hy—Hz,—H?°—* 2 As unknowns whose values are necessary and sufficient to fully determine the equilibrium polygon, let us choose* HH ee, Srl Se eo cess (308. U The expression for the bending moment then becomes: M = M— Hy — X,.0—Xp.... eee (309. Upon substituting this value in equs. (307) and solving them for the three unknowns, remembering that, on account of the assumed (oblique) symmetry of the arch about the Y-axis, 1 1 +> +3 ada xcyde f i 0, also f 7 = 0, ae! ou 2 = 9 we obtain the following expressions: +} M se) | te] w]e] ~ = oo — a as yr daz 1 f I’ a Ao t 2 * We here follow, in the principal features, the method of design of F. B. Miiller-Breslau. (See Zeitschr. d. Arch. u. Ing. Ver. zu Hannover, 1884.) 150 ARCHES AND SUSPENSION BrinGEs. 1 + M edz eal Xx, = 1. an See hait ate fotos +> wide ‘a 1 aaa +d Mda Hd a tie ays ee a ee (313 tee The definite integrals appearing in the above expressions may be replaced by summations in a manner similar to that deduced for the integrals occurring in the expressions for H in § 5 and § 17. For this purpose we divide the arch into horizontal panel-lengths @n_,, @m, my ------ , within which the moments of inertia of the cross-sections may be assumed constant and equal to Im_,, Im, Im .------ We introduce an arbitrary constant panel-length a,, also a mean value for the moment of inertia, 7,, and adopt the following notation: Ua Be pe (2Ymt Ya) + Ge Gee (2Yat- Yr) -. (B14. +1 am JI, ams I Um Ga, Tq (22m + Gm) GE (2tmt sm) (315. 1 an I 1 am I oF peas A 5 tie Oi fot +1 0 vu= 2 Qo I'm ae 2 ao my oe or, with sufficient approximation, WO nore ocirteg iiaret autos (317. Substituting the above values in equs. (311) to (318), [cf. equs. (253) and (254)], we obtain: H ieee OR i ae amuses (318. Zymvm + ane ZMmv'm LSS ewe een niaea wees (319 z Mnv’m+ H fob ies ERTS ov esnn ese ven (320. REACTIONS FoR THE HINGELESS ARCH. 151 The summations appearing in these expressions should ex- tend over the entire arch, hence from — 4 to + 4. The position of the axis of abscissae, from which the ordinates y are measured, is fixed by equ. (310) or, if y’ denotes the arch ordinates measured from the closing chord, the axis is determined by q In, sf Sage “YmvUm one 0” (321 by tt ; ; -dx Dae I’ 0 For a constant value of I’, the Y-axis becomes the rectifying line for the arch-curve; so that ¢, represents the altitude of a parallelogram erected upon the arch-chord with an area equal to that included between the chord and the arch-curve. We proceed to determine the influence lines for the quan- tities H, X, and XY, for a moving concentration. The graphic method may here be applied, since the summations appearing in the above expressions are readily represented by the ordinates of funicular polygons. Thus, if the load consists of a unit concentration applied at a distance é from one end, we have, as shown in § 5, p. 36, the following relation: 1—¢ é see = Mnva= pe oC mVm + a (l—2’m) m= My 0 é where M, is the moment producible at the section é of a beam freely supported at A and B by a loading consisting of the vertical ‘‘forces’’ vm. In similar manner we obtain the quantities 1 1 SMnv’m and= Mnv’m as the moments M, and My of the 0 0 “forces’’ v'm and vm, respectively. Since the values of vm, vm and v’m may be calculated directly by equs. (314) to (316), there is no difficulty in constructing the ahove funicular polygons and, hence, the influence lines for H, X, and X,. In Figs. (657) to (65*), this construction is carried out. We first have to find the rectifying axis A,, for which we use the force polygon for the quantities v” (Fig. 65°) and the re- sulting funicular polygon constructed upon the arch-rise (Fig. 65°). The calculated quantities v give the force polygon (Fig. 657) and, corresponding to the vertical and horizontal directions of application of these forces, we obtain the funicular polygons (Fig. 65°) and (Fig. 65'). The intercept of the latter polygon Iyl Aep? gives the o A on the X-axis, augmented by the quantity 152 ARCHES AND SUSPENSION BRIDGES. magnitude of the unit force G; while the ordinates of the former polygon give the thrusts H. Constructing, next, the funicular polygon for the loads v’” considered as acting vertically on the straight beam AB, we obtain (in Fig. 65%) the influence line for the quantity X,. Similarly the funicular polygon of the forces v’ (Fig. 65*), constructed with the aid of the correspond- ing force polygon (Fig. 65"), represents the influence line for the quantity X,. [The correction to be applied to X, by equ. (320), an . H=c.H, can be introduced; in general, however, this quantity will be so very small that it may safely Fig. 65, F Ay a ¢ f. a Bi So Se Ww &G A Force Polygon d Force-pol f V. ‘orce-polygon of V. ig s be neglected.] If the pole distance for the force polygon (Fig. 1 ; 65°) be chosen equal to + = v”, then X, is to be measured 0 on a scale whose unit — n force-units. The scale for X, is ‘ u 3 . ae . tory is given by = v'm 4m, i.e. by the distance p q which is inter- 0 cepted on the Y-axis between the first and last sides of the funicular polygon. From the influence lines for X, and X, we may easily derive the curves for (e,—e,) and 2 which are to be used directly in the construction of the end reactions. To do this, we simply multiply the ordinates of the X,-curve by the ratios Hy ae, and those of the X,-curve by the ratios H pq REACTIONS FoR THE HINGELESS ARCH. 153 ae nH’ accomplished graphically. The construction of the reaction lines no longer offers any difficulties, since the distances z, and (e, —e,) fix the closing side of the equilibrium polygon D E F, which is to be constructed with H as pole distance. If this polygon be drawn for different positions of the load, the curve generated by the vertex-point F will be the reaction locus and the enveloping curves of the reaction lines will be the tangent curves. Through X, we also obtain the vertical end-reactions V, and V,. These will be for supports at the same elevation: the corresponding figures indicate how this may be V.-V,—2, . (322. where V, and V, denote the reactions for a simple beam. Since X, is influenced only by the variation of the moment of inertia and not by the form of the arch, it follows that V, and V, are also independent of the arch-curve. The end moments are an by the formulae: M,=H .t.—- —X,=—H.e, ate awe oes ee (309% M,=—H. a ae - 2 b.) Horizontal Loading. If a concentration W is supposed to act on the arch horizontally, or parallel to the arch-chord (Fig. 65*), at a point whose horizontal distance from the crown is w, there may be derived from the fundamental equations (307) the following equations of condition for the unknowns Hy, Xiw and X., which have the same significance as the similar quantities in the preceding analysis: t sMo+w.? 252 w= — Thl err 1W W Xywl(1— aac) — Xe nn (323. — SMe 1+Wt=0 t in 1W Aiw gq, — Xow (1— 2 Ue op 2M —7Wt+O=0. | 154 ARCHES AND SUSPENSION BRIDGES. Here M denotes the moments producible about the points of the arch-axis by the load W if the right end of the rib were free to slide horizontally ; v, v’ and v” are the qnanh Hes defined L 1 by equs. (314) to (316) ; and Cm [Het —W al = is a correction term which is so small that it may safely be neglected. H,, is the horizontal (inward) thrust at the end B, and (H,,— W) is the corresponding thrust at the end A. The summations }Mv, 3Mv’ and 3Mv” may be deter- mined graphically as the moments producible about the line of action of the load, W, by forces v,v’,v” conceived as acting horizontally, together with opposing horizontal forces at the ends A and B equal in magnitude to the vertical reactions which would be produced if the forces v, v’, v” were applied vertically. Hence the summations are given by the intercepts of the funicular polygons Fig. 65', Fig. 65! and Fig. 65°, being rep- resented by the lengths af, y8, and 4, respectively, measured by the same scale units as were used in the preceding construc- tion for H, X, and X, for vertical loading. The end moments for this case are given by: mame (ME —BVO~B) | gop M,—=H,¢,=Ht,+X,4 —X, 2. Simplification with Constant Moment of Inertia, J’. As a rule the variation of the moment of inertia in hingeless arches is so slight, that we may assume l’ — 1. cos ¢ = constant. Then, with a uniform panel-length a, we have 1 Um = 6 (Ym1+4Ymt Ym ) VU m==XImn v m=1. Frequently, with greater subdivision into panels, we may also write Um = Yn- But, for a constant v”, the influence line for XY, becomes a parabola; and, for a load G at a distance € from the end, equ. (313) becomes: X.—; f Mazt+c—eG +e Ib Here C denotes the small quantity H = ; if this is neglected, we obtain: REACTIONS FOR THE HINGELESS ARCH. 155 1G &(l—é) ae Ea i. @., 2 1s equal to one-half the altitude of the simple moment triangle D E F (Fig. 65°). Similarly, with the same assumptions as above, we find, from (312): + lew Madz 4 nee ft H—8) gg 128) l 7 aXe v an ae "da or, referring back to equ. (308), Ca Cu 1—2€é Accordingly z. may be derived from (e,—e,) by a simple construction. Furthermore, by equ. (322), Vi=V, 6 Se icin (326. Vi Vi— X= GO ees. (327. The vertical reactions are thus identical with those occurring in a straight beam having fixed ends and a constant moment of inertia. The ordinates of the reaction locus, measured from the axis A,, are given by equ. (198). after substituting the values from (308) and (309): GMO 858 (Le); substituting the values for z, and (e,—e,), this equation reduces to: _@ | 288 _ gad 1 B — l@d 156 ARCHES AND SUSPENSION BRIDGES. The ordinate of the point D (Fig. 66) is also found: 22+ 25% a2 kz, 2 l or, 1—t)? @ Lt gM SN eee cee ee ec eens (328. Hence the end moments will be given by: M,——H(2,—t) -— G25" 4 Ht...... (329. and M,= —H(2,—t) —-@*49 + Ht, ...... (330. Fig. 66. £ . ———— Se ns DD Sere or a i : R 3 HN. Gee |P e ia In these expressions, the terms not containing H are identical with the end moments of a straight beam fixed at the ends. On the basis of the foregoing results, the construction given in Figs. (657) to (65*) for the general case may be simplified as follows for the special case of Z’ = constant: First, by the method described previously (equ. 318) construct the H-curve ahb (Fig. 66); also, with H as pole distance, for the given position of the load, construct the moment triangle dce. Now lay of ML = 2z= : cf, (equ. 324), draw the connecting line KL and the line ND parallel to the arch-chord. Then DL fixes the closing side of the equilibrium polygon (equ. 328); REACTIONS FOR THE HINGELESS ARCH. 157 and the vertex thereof may be obtained by joining D with M and drawing JF parallel to DM (equ. 328). Asa check, the altitude of the point F above the closing line must equal the length cf. We thus obtain the successive points F of the reaction locus as well as the tangents D F and F EF to the tangent curve. Assuming a constant moment of inertia I’ and neglecting the effect of axial compression, the equations of condition (307) may be interpreted as follows: 1.) The total area between the axis of the arch and the line of pressures must vanish. ( f Maz =0.) 2.) If we conceive the elements of the arch-axis to be invested with weights proportional to the elements of the above-described area, affixing different signs to the areas on the two sides of the arch-axis, then the static moment about any two axes must vanish. (fuyao =0 and f{ Moda = 0); in other words, the two areas, included between any horizontal line and the arch-axis and pressure-line respectively, must have a common center of gravity. Winkler* has combined the above conditions in a single theorem: ‘‘In an arch of uniform cross-section, that line of pressures is approximately the right one for which the sum of the squares of the deviations from the arch-axis is a minimum.’’ This theorem may be derived from the uM? Pp? F principle of least work, whereby W = f Fr det Sf HA ds =min., or, with the above approximations, f IP da=min. 3. Flat Parabolic Arch of Constant Moment of Inertia. The rectifying line of the arch-curve is, in this case, located at 7 the rise of the arch. For a load @ placed at a distance ¢ : 4 from the end, equ. (311), with the values y= hy (l— 2) — — f and J’ = constant, will yield the following expression : 15 #(1=8)? i — a pikeenese 331. Sa a a vel f taes2 4 Af? For the vertical reactions V, and V,, the expressions of (326) and (327) remain valid. The equation of the reaction locus becomes, 8 45 | : n= f(t eye): acne an ees (332, hence, in the present case, the reaction locus is a straight line * Deutsche Bauzeitung, 1879, p. 128. 158 ARCHES AND SUSPENSION BRIDGES. parallel to the axis of abscissae. The directions of the reaction lines are then fixed by the values, from (328), of z, = fe +4 1 i “ and z, = =i - 4; they are most conveniently obtained by the construction shown in Fig. 66. In addition, we may also establish the equations of the tangent curves which, for this case, are hyperbolas. This is done with the aid of the following diagram (Fig. 67). If at and yr are the coordinates of any point of the line of action of the left reaction referred to the point EF as origin, then the equation of this reaction line is ye tt an (AS oe). (333 or 2Pyr=En (2ar—l) Haar... eee (a. Differentiating this with respect to &, Leys F (28r— 1). dscaaae aor sgeegecds saeadend angered (B. Eliminating & from (a) and (8), we obtain the following equation of the tangent curve: (2ar—1)?qt BaryrI=O.. ee (3332, Fig. 67. AE=BEL=+f BC=E,D=n= =f (2 45 Io ui Tan) 1 BJ=JC= 31 Of the two asymptotes of the hyperbola represented by the above equation (333"), one coincides with the vertical line through the end of the arch-axis, and the other is an inclined line (J E,) passing through the points (*=0,y= + a) and (e=1,y=0). Ate= 4, or the center of the span, the hyperbola is tangent to the axis of abscissae. The end moments are = 2G abi =@2f = a = i i PR 2 2.0 REACTIONS FOR THE HINGELESS ARCH. 159 MBE gig 41) aes ase Ue eb Re 335. Zog2 far é) | ( +2 DP Neglecting the effect of the axial compression, we may sub- stitute in all the above equations 1+ Ee “ha =1 and y= 2 f. 4. Effect of a Change of Temperature of the Arch and of a Displacement of the Abutments. Let the arch, supposed weightless and without load, alter its temperature uniformly in all parts by ¢°; also let the span be increased, through a displacement of the abutments, by an amount Al. Then, re- taining the assumptions introduced at the beginning of this article, the fundamental equations derived from equs. (185) and (186) take the following form: +e f ALyde—(H—BAgot) LEAL tn (336. 1 aa 2 uM f : czdxz=0 a Also, with M= 0, equ. (309) becomes M = — Hy— X,2¢4—X, +t : Choosing the axis of abscissae, as before, so that sf - 1” =0, 1 «| the solution of the above equations will yield the following expressions : 160 ARCHES AND SUSPENSION BrIDGES. Be tka (337, da ae 0 X,= eS naan (338 AoTo — Introducing the quantities v, v’, v’”, defined by equs. (314) to (316), we obtain = EI, (wtl—Al) = en a Gere (339. to (2 ym ont pe) Ib(H—EA, et X,=0, X,— I re ees (340. Apdo? 2 UV" m 0 The denominators of the above expressions are determined by the previously described graphic construction. X, will always be a vanishing quantity and may, without sensible error, be equated to zero. The line of action of the horizontal thrust thus coincides with the X-axis, i. e., if J’ = constant, with the rectifying line of the arch-curve Fig, 68, (Fig. 68). os With parabolic arch curve and 1 constant J’, since f ydr= oO a 4 gg PL A 45 BI,(wt— 5") Hi Botte (341 La ae or, introducing the constant ordinate, y, of the reaction locus as given by equ. (332), 1 econ: (342. Comparing this with the corresponding (approximate) expression for the parabolic arch with hinged ends, as given by equ. (290°): REACTIONS FOR THE HINGELESS ARCH. 161 it appears that in the hingeless arch the horizontal thrust due to tempera- ture change or displacement of the abutments is nearly six times as great as the corresponding thrust produced in the two-hinged arch under the same conditions. 5. Arch Connected to Elastic Piers. Let the arch-axis AB (Fig. 69) be continued in the vertical pier-axes A A, and BB,. We assume that the two piers, throughout their height h, have constant moments of inertia J, and are formed of a material whose coefficient of elasticity is #,=«.K. The base- sections, A, and B,, are assumed incapable of rotation or dis- placement. Let the arch-rib have a variable moment of inertia I and a mean cross-section A,. In calculating the deformations, we will consider the axial force in the arch constant and — H, and neglect the effect of the corresponding force in the piers. Fig, 69, With the notation indicated in Fig. 69 or previously intro- duced, we have, for a vertical loading on the arch, V, = V, + Y,, at any section of the left pier, M = M, — Hz at any section of the right pier, M— M,— Hz+ X,1 at any section of the arch, M—M,+2X,x—H(h+y)+M Substituting the following symbols for certain definite integrals: q 1 i 1 _ (ds __ (yds __( xds —f x'ds af, 0- (4, co- fH, a-f2e..... (343, 0 0 0 0 and equating to zero the differential coefficients of the work of deformation with respect to the three unknowns H, X,, M,, there are obtained the following three equations of condition: 11 162 ARCHES AND SUSPENSION BRIDGES. aw a —EaY =, 2 ag 2 1 a va Mds . fi Hat +p tu, (c +2h)— EC.) L (344. +X,(a+ 2) 14 (Mano 0 +E ST =m, (a+2")Z—H(ahto++)Z 1 +X,(4+ 2) + (4 as—0 0 J Solving these equations, we obtain py = a Ih ah hyvit ; aot = ( Fi ayE and 1 1 f Ma as—4+ fas Xie : sl ahead auld ody see esat (346. " 1, 2h and M, may be found by substituting these values in any one of equations (344). If h = 0, the above equations reduce to 1 1 a f Mt as—v fas a i ag saMais (347. ue ba 1 Mea 1 M pe aSpas ga ee aha Lees (348. 1 2 ane a=d REACTIONS For THE TINGELESS ARCII. 163 and M,a—Hb+51aX,+ {7 ds—0 Seaman (349. This constitutes another solution for the general case (1.) as previously given by equs. (311) to (813). 6. Segmental Arch with Constant Moment of Inertia. It is usually advisable to adhere to the general formulae established in paragraph (1) of this article for any form of arch. Neverthe- less, for segmental arches, it is desirable to present here a more exact analysis* to be applied whenever the sectional dimensions cannot, as above, be considered negligibly small in comparison with the radius of curvature. With the notation of Fig. 70, the fundamental equations derived from equs. (182), (185) and (186) may be written as follows: g sean f Gir tait lel .. (350, Ge F J Av-A ay—=—y.A gor cosh (A b—Ago) —r f Meeete (351, : Ay-AYo=x-Ado-r sing (Ad—-Ago)-1? ween mneee (352. 0 Assuming a constant area of cross-section and writing J E ae Pe ee ee er ee eee ee ee ee a ee ee eee (353. * Schaffer— “Handbuch des Briickenbaues”—Chapter XI—Ist edition. 164 ARCHES AND SuSPENSION BripGEs. and observing that, for vertical loads, x . P=(V,—3G)sing+ Hcosg...-...e-e (354. 0 and M=M,+ Vyr (sin d9—sin¢) —Hr (cos ¢ — cos $0) —3Gr (sin dy — sing) . RP: where M,, H and V, refer to the left end, we obtain: ¢ g 6 146 : A p-Ado=- 1? (Fe 4 EE? ot (b-$0) do Go =—Z; [MA +8) (6-40) + Var (cos $ — 008 $0) + (1 +8) r(# — do) sin go} +H {-r (sin d-sin do) + (148) r (h- $0) cos bo} + . @{-r (cos ¢—cos ¢,)-(1+8) r (6-41) sings} | + wt (b— 40) = — 2, [1.1 48) (6-40) + Vif y+ +8) a (6- do) } +H{x+ (1+8) (r—f) (¢— 40) } L .. (356. ~3a{y—h+ (1+8) (0-€) (4s) HF t(4-40) g At—A fo—=—y A orcs p (A b-Age)—12 f ATE Do =~ YA $o-1 008 ¢ (AGA bo) - > [M, (sing—singo) _ Vr (sin & sin fo)? -ar{?; ete and (cos d- Cos o) | (357 cos COs bo (sin P-sings) \ + 32 = a (sin $,—sin ¢)] =-y Ago-r cos $ (Ad- eee M,2- Ee -H (Gp), oy | ey 3G Gao") 0 REACTIONS FOR TUE HINGELESS ARCH. 165 where (A ¢d—A 4q,) is to be found by equ. (356). og Ay-Ayo= xr. Ado-rsing (AG-A gy)—12 f ASR Se po =. Ago—r sind (Ag-A do) - l- M, (cos 6 cos $9) — Vyr{ > # (sin p—sin $0) (cos ¢o-Cos $) \ H . -b. , sings +7 (cos-coso)? +3 Gr{ 2% 4 208 (cos¢-cos¢,) }.. (358. + (sing,-sing) } 24 60- (one) Adan a.) _ 7 J. r(bd-d) , w(ytr-f) ay|\, Hy al M,y+Vi{ : 2 + “yh, a. = r(o-¢:) , (a-§) (y-h) | (ytr-f) (w-8) fee erect | Here h is the ordinate of the arch-curve corresponding to the angle ¢,. If we substitute s=l1, y= y,, 6 = ¢,,, the above equations (856), (357), (358) will constitute the three required equations of condition; these need not be rewritten since they do not differ from the foregoing except in the affixing of the subscripts and the substitution of J for z. If we have a symmetrical segmental arch, then t a=3, Yi=0; $+, = — $03 and if we write, for the ends of the span, A¢é—Ad =<, Av — Atyp== Cy Ay — Ayo=—c;, we obtain the following cx- pressions for the equations of condition: A $-Abo= = [-2 My (1+8) go- Vi (148) Uo +2 {3 - (148) (r-f) do} (359. 1 (148) (1-26) (betes) VT] 1 $ofas tem este tas8) Foci V1 A L-Atyp= Cy == — C, 1 COS ho- a [- Ile +H{ r*$0- Lop | +3 gue .. (360. 166 ARCHES AND SUSPENSION BripDGEs. 1b 2 l(r-f) eee Adot a eae 361. a hs -+ ~ 32 {8 (Hyg) + OOO -pa-ol] | Equ. (361) yields: Ca= EJ (ce; —lAg,—cairsin bo) = 7? (do— sin bo cos bo) ) +36 (ho + ¢1— sin gy cos 6) — 2 sin 1 cos dy + sin dicos 1) | 2 (bo — sin bp 00s ho) (362. _ BI (2%—214 b — esl) ‘ia ; r[(2r°d—l(r—f)] +36 2r? (doth) th (L—2£) —2(r—f) Ul —8) 27r bd —l(r—f) From equs. (359) and (360) we obtain Fat [= G+ 4) di (eater cos bo) + ar sin po] r [bo (1 +4) (bo + sin by cos bo) — 2 sin? o] 2ZE J wt dosin bo 1 FF Te CLF 8) Cha + ein Ge 008 Ga) — Dein ds] 1 2 sin bo [cos d1.— cos bot go: (1 + 5) sin ds) +34 — A+5) $0 (sin? bo+sin’® 1) : 0 = 2 [bo (1 +5) (do+ sin cos b.) —2 sin? bo] L (363. — EJ[=2 148) ho (a tea (r—f) +00) 7 [bo(1 +8) 2r oo +1 (r—f))—P] 2ETwt gol + 716 +8) @F7641G—P)—FI 1[2h+¢, (1+ 5) (1—2t)] — 1+5) & (P— 2lé+ 2%) 2[G (1+8) (2°d.4+1(r—f)) —?] +36 With these values, M, may be calculated by either equ. (359) or (360) ; the latter yields: _ BJ (a +arcos $0) = : M. =~ trsnk Virsinds H (¢.— sin $. cos $o) r l Gr (sin bo + sin 1)? at 2 sin do i 4 sin bo . (364, — Eilatatr—f)) _ VA ns rl ae ~ 21 G (L—$)* +345 REACTIONS FoR THE HINGELESS ARCH. 167 M, is most readily found in this manner ane V, and H have been determined. Then we also find 1 Ma My + Vib 3G (Lb) oes eeeecceees (366. The preceding equations express not only the effect of a concentrated load, but also that of a change of temperature; the latter effect, by itself, -is obtained by putting G—0 in the various formulae. Similarly, these formulae give the effects of a rotation or displacement of the abutments, if the magnitude of such displacement (c,, c, or c,) be known. If the ends are rigidly fixed, we must take c, —c,—c,—0. For a flat segmental arch, the formulae of the parabolic arch may be applied without sensible error. 168 ' ARCHES AND SUSPENSION BripGEs. § 21. Maximum Moments and Shears. In determining the critical loads for maximum moments and shears, we need merely to apply the general principles presented in § 14. The same general rules apply to the hingeless arch as to the arch with hinged ends except that the reaction lines do not pass through the end-hinges but are to be drawn tangent to the tangent curves. Again, the maximum moments and shears may be determined either by the method of influence lines or by means of the equilibrium polygon of the loading. The influence lines for moments and shears, however, are not so simply found as for the hinged arch, but may be derived from the influence lines for H, X,, X., according to the formulae (309), (192), (322), which may be rewritten: M=M— Hy—X,24—X, ...........6... (367. S=(V,+2X,) cos¢—Hsind ........... (368. where ua—@ US) OS G sta Pg dese and x is measured from the center of the span. To find the extreme fiber stresses in the sections, the moments should be taken about the core-points. The method of influence lines is advantageous when the load consists of a train of concen- trations. To obtain the maximum moments and shears directly from the equilibrium polygon of the critical loading, it is necessary to find the values of the horizontal thrust, end reaction and end moment for that loading. It is a simple matter to obtain these quantities by summation of the values found for the different positions of a concentrated load; this may also be accomplished graphically. If, with the resulting values of I, V, and M,, there is constructed the equilibrium polygon of the loading (or line of pressures) in its proper position with respect to the axis of the arch, then the moment at the section under investigation will be M— Hy’, where y’ is the distance of the equilibrium polygon above the arch-axis or, rather, above the core-point of the section. If the load is taken as uniformly distributed, we may use, as before, a second reaction locus and tangent curve to facilitate the construction of the equilibrium polygon. For special arch forms, the influence of a partial uniform load may be calculated as follows: MoMENTS AND SHEARS IN THE HINGELESS ARCH. 169 1. Any Arch-Curve, with /’ — Constant. If the load of p per unit length extends for a distance X from the left end, the resulting vertical end reactions may be derived from ecus. (326) and (327) by integrating between the limits 0 and ), giving pr (218 — 21d? +d?) rr) sew ew te mw ew we we pr (21—)d) Qe tet t serene V.= y= If H is the horizontal thrust produced by this loading, the integration of the expressions of (329) and (830) yields the following values of the end moments: 6 —BAIF3N u,——[p 2 a —H to] aelet (371, 1—3n E Ma—— [pve - Ha]... (372. where ¢, is defined by equ. (321). For a full-span load, if H,., denotes the corresponding horizontal thrust, we have 1 M,— = — [dot — Hew te] oe (374. If A, is the distance to be covered with load, as determined by the rules of § 14, in order to produce the greatest positive moment at the core-point (ax, yx) (referred to the end A as origin), that moment will be given by Mmax== M,+V, (l— ax) = Hy, — p SS pn [ s’-3a,-6 (21-1, 5 = | H(yicte) —p Be): (375. ae Mynax= 12E For a full load, there results 1 Mie = Gp P [6% (l— zx) —P]-He (Yi tg) hae ae a als (376. with which we may calculate Minin= Mot — Maax. If A, is the abscissa of the critical point for shears, and if H,-», is the horizontal thrust for a load extending from 170 ARCHES AND SUSPENSION BRIDGES. x to d,, we obtain, by the use of equs. (192) and (369), the following expressions for the greatest shears: Smar— fo [ (I-a)? (I2~ac®) — (1-Ag)? (Tag? | cos¢-H,.x, sing. . (377. Sain = = p (L—2r) cosd— Htor sind — Smax------+-- +0 (378. 2. Flat Parabolic Arch. If a uniformly distributed load p extends for a distance 4 from one end of the span, the integra- tion of equ. (331) yields ___ pa?(107-151A+60?7) 5 ANB Br , BMY pw? ace Fe ae ie az) (l-s5 tz) + OM. fe Le eth at in 45 1, where, as an abbreviation, f’ =f (1 a ae : For a full-span load, pt 1 __ : Hit = Gr <5 ape wees (380. “4 Af? The vertical reactions and the end moments M, and M, are to be determined by substituting the above values of H in the formulae previously developed; similarly, the maximum moments and shears may be found by equs. (375) and (377), when the distances A, and d,, corresponding to the critical loads, are known. In the case of the parabolic arch, those distances may be calculated as follows: For the maximum moment at the core-point (a, yx), (re- ferring to the end A as origin), we have, by equ. (333), De FIR OK i \8 45 Yu~ 3 f= (ye ey, Jef Q4+2 Ao or (u-— Ff) P+ EP (1 2a)r— se fleece (381. The critical length, A,, will equal 1? for any arch-point whose coordinates (xx, y’,) satisfy the equation: , 2 4 , 2 77 pt 4 (yu sf) +e ira plate cece. (382. If this yields x’, > q, then, for all points in the middle portion between the abscissae (U—a’,) and (x’,), the critical loading will be an interrupted one (with two critical points); and, for any of these points, we must add to the value of Mnax given MOMENTS AND SHEARS IN THE HINGELESS ARCH. 171 by equ. (375) the value given by the same formula for the symmetrically located point of the arch. The critical point for shears is determined by r+ 1 8. 4f - 5 Ba? Tr f’=tan 6 =r (l—2z) seer -. (383. For a full-span load, the end moments will be, by equs. (374) and (380), 1 F f and the crown moment will be, by (376) and (380), If we put f’=/f, which is equivalent to neglecting the effect of the axial force upon the deformation, we obtain, for a full load, M, = M, = M. =0 so that the pressure line coincides with the axis of‘the arch. With respect to the error resulting from this approximation, practically the same remarks apply as in the case of the two-hinged arch (see page 137). If the arch-rib consists of two parallel flanges of combined area A with an effective depth between them of h, and if @ is the fiber stress for the pressure line coinciding with the arch-axis, then the extreme fiber stresses at the crown in the case of the eccentric location of the pressure line will be by equ. (179), in the upper flange, 15 oh aT oy = FC — — : 45 ha lta, and, in the lower flange, 15 Oh vit —_ 45 hh? A ae The maximum value of the upper fiker stress occurs with ee = a and amounts to o, = 1.250; while the lowcr fiber stress changes its sign, hence becomes a tensile stress, with ae, = ie 5 Similarly we find, for the end sections, 1 oh ee ee ou=o.———,- sand ai=c, = 14-2 2 7 ee 6 f? 16 f? Here the lower flange stress will be a maximum with f = 0.387, giving : h 4 o, =1.72¢. With = 5 TR the upper flange at the end section receives 2 ; Hailas 1 a tensile stress. For illustration, if — = 3; the stresses at the crown will be oy= 1.24 o, o:=0,28 o; 172 ARCHES AND SUSPENSION BRIDGES. and the stresses at the end section will be ou=—0.19 a, o=1%6. If the half-span is loaded with p per unit length, we obtain, for the 2 parabolic arch of constant section, H = a os and the moments at 2 the end-points of the axis are MM, = — pt et spt and P 8 7’ 24 2 M.= (f-3 z. The greatest positive moment occurs at x2 = BE . - : 4 and is given by: ae 3/13 ff y?_ (tlt pa tS je te 7) i 7) (2 gr) pv max ; 9 24 - ¥ Neglecting the axial compression, i. v., with f’ =f, we obtain g g Pp 1 9 M,=——p?=— 3] ax = 57 pl’. eg Bt Ms, and Mm Tosa ?! 3. Segmental Arch. For a uniform load p extending for a distance 4 from the left end, the end reactions V, H and M are obtained from equs. (362), (363) and (364) by substituting G= pdx=——prcos ¢,d¢, and integrating be- tween the limits 0 and 2 (or ¢,‘and ¢,). There will be given here only the formulae for the special ease of the load p covering the entire span. By symmetry, we obtain directly ViVi OL = pr sindy. sce ccee cree eens (384. We also have, by equ. (363), __ prsinds (_ 3 (1-8) (bo-singocos bo) —2 (1+) dosin® do Bea ( bo (148) (Go+sin boC08 bo) —2 sin? dy ). (385. With this value, we obtain from equ. (364) : 3(1-5) (bo-sin bo cos bo)?—2 sin*d, [ (1+5) __ pr (3 bo’t+ Po sin Go COs Ho) —4 sin” Go] i 12 Go (148) (hot sin bocos bo) —2 sin? po = M, .. (386. For the moment at the crown, we obtain, by substituting the above values in (355), 8 (1-8) (sin?.—$.7—2 sin? .cos +2 dosin bo) oat —4 (145) dosin®do+sin' (1438) a 12° Go (148) dotsin bocos go) —2 sin? ho -. (387. MoMENTS AND SHEARS IN THE HINGELESS ARCH. 173 The maximum moments and shears are determined by the general rules. In the accompanying graphs, these values are plotted for a segmental arch having a uniform core-depth; Fig. 71 gives the maximum moments and Fig. 72 the maximum shears. The same notation is used as in the corresponding diagrams for the two-hinged and three-hinged arches; and since arches of the same rise and depth of rib have been con- sidered, there is thus provided an opportunity for comparison of the three classes of arches with respect to the resultant stresses at the different sections. (cf. Figs. 50, 51; 62, 63.) Fig. 71. s* é 4, Temperature Stresses. The horizontal thrust produced by a variation of temperature has been derived in § 20:4; its value for the general case is given by equ. (337), for the para- bolic arch by equ. (341), for the cireular arch by equ. (363). With regard to the magnitude of the temperature stresses, the following notes are given: In a parabolic arch-rib having two equal flanges of combined area A and effective depth h, the intensity of stress in the flanges at the crown is approximately om (iiss). A.h Come tH similarly, at the ends, 174 ARCHES AND SUSPENSION BRIDGES. 2H. (Zr+4) ee Ach Substituting Ht from equ. (341), we obtain 3 hy h 3 hy h 15 CFrzs sg osleee ata ve= FoF ot ———75 47 > an aS ag ee i + 6p For steel, we may use Fwt =10,780 pounds per square inch, thus giving, in round numbers, lis 3h 1+ 3h bo eee hoot ay oo= + 20,000 —>; » ands, = = 40,000 — =>: P14, 2H P4q 5H 16 f? 16 f? The upper flange stress at the crown attains its greatest value with a ratio of = = ee we then have ocu = + 2,670 lbs, / sq, in, and oc: = — 6,220 lbs. / sq. in.; similarly the lower flange stress at the end section attains its maximum value with + = 0.39, when we find ou =— 14,040 Ibs. / sq. in. and o,,; = + 7,710 lbs. / sq. in. For a ratio of a= = we find gu=+ 2,540, oc1 = — 7,620 Iks. / sq. in. ow = — 12,700, on = + 7,620 Ibs. / sq, in. It thus appears that the temperature stresses may become very considerable and that, in this respect, the hingeless arch is at a disadvantage in comparison with the arch having hinged ends. DEFORMATIONS OF TITE HINGELESS ARCH. 175 § 22. Deformations. The elastic deflections of any point (x,y,) of the arch, referred to the end A as origin of coordinates, are determined by equs. (185) and (186). Again employing the abbreviation Ty P In “ Mo + (4 Bet) ty =? SaieteeOeeetoe (388. [ef. equ. (293)], and conceiving these quantities 2 as forces acting on the arch, we obtain, exactly as for the two-hinged arch in § 19, the following expressions for the coordinate deflections : — El,Ay=Ma +, ElAr=Mn+C, Here M.y and M,, denote the static moments about the point (x, y,) of the forces 2 applied at the points of the arch-axis together with a force — Z, —— E'I, A ¢, applied at the end A, these forces being applied vertically and horizontally for the respective moments. If the ends of the arch are perfectly im- movable, then A¢d=0O, and hence Z,—0. In general, Z, is the end reaction produced by loading the horizontal projection of the arch with the forces 2. Furthermore, we may write C,—1,f (2 — Bot) ay Yo i= =f —Eut) de 0 or, with sufficient approximation, G15 t —Eot) Yi C= —1,(4 —EFot) xy Hence exactly the same rules apply to the determination of deflections in the hingeless arch as in the two-hinged type; and the contribution of the bending moments alone may be 176 ARCHES AND SUSPENSION BRIDGES. obtained, exactly as in the preceding case, as the ordinates of a funicular polygon constructed with E J, as its pole distance for a loading consisting of the reduced moment-diagram of M i. For the graphic determination, proceed as follows: : For the given loading, construct the equilibrium polygon or pressure line. The vertical intercepts m between this polygon 0 and the arch-curve, multiplied by the ratio a and increased by the small and nearly constant correction c=(1- Betas) In = 7 Fay mong 2S will give the load-quantities Some (me) cecceeerereeeeseee eee: (392. These have to be laid off in a force polygon, with proper observance of the algebraic signs (m is positive if the pressure line lies above the arch-axis), and the resulting funicular polygons for vertical and horizontal action of the forces con- 1 rP structed. If the pole distance is selected = —,* ue ae being the assumed uniform horizontal spacing of the ¢-loads, then the ordinates of the two funicular polygons, measured by n times the scale of lengths, will give the vertical and horizontal deflections, respectively, of the points of the arch under the given loading. To be precise, there should be added the cor- er ee El, El, too small to warrant consideration. We may also represent the influence lines for the horizontal and vertical deflections of any arch-point (z, y,) by construct- ing the curves of vertical deflections producible by a unit load acting vertically and horizontally, respectively, at the point (z,y,). Compare § 19. In the above formulae, the effect of temperature is included. If it is desired to treat this effect separately, an expression may be established for it which is identical with that for a two- hinged arch, equ. (305). This gives the upward deflection of any arch-point due to temperature variation by the formula: EIAy=Him.+2 (Lot— =) Tice Basha (393. Here H, denotes the horizontal thrust due to the change of temperature and m,; the static moment, about a vertical line through the given point, of the area included between the arch- curve and its rectifying line or line of action of the horizontal rections ; but, as a rule, these quantities are DEFORMATIONS OF THE HINGELESS ARCH. 1%? thrust. This quantity is proportional to the ordinate (y) of the H-curve (Fig. 65°) ; to be specific, m,=17.p.a, where p is the pole distance and a the panel-length used in constructing the H-curve, and y is measured to the scale of lengths. We thus obtain for the temperature effect, in general, Ais (erg 82) te cscxseeen! pened (394. For the example executed in Fig. 65, in which 7= 160 m., f= 42.5 m., A, =0.112 sq.m, and Jo= 0.70 m.‘, we find by the approximate formula (341), assuming E w = 248 tonnes per sq. m. per degree C. and Al=—0, Hy. = 1.040 X ¢ tonnes, The H-curve was constructed with a= 16 m. and p=25 m., so that Hemz =16 X 25 X 1.04 yn. ¢and EI, Ay = (416 + 334.2 y) ¢; or, finally, Ay (in mm.) = (0.0297 4 + 0.0238 y) 4. 178 ARCHES AND SUSPENSION BRIDGES. 4. THE CANTILEVER ARCH. § 23. Arch With Hinged Ends, Having Cantilever Extensions Beyond the Points of Support. The general equations (200) and (250) for determining the horizontal thrust hold good for loads on either cantilever arm if we substitute for M in those formulae the moments which would be produced in the span J of a simple beam. In the three-hinged arch (Fig. 73), a and is pro- Fig. 73. GUA B65 = portional to the moment M, at the crown-hinge; hence the influence line for H is formed by prolonging the sides of the triangle, ca and cb. The reaction lines are given by the rectilinear prolongations of the reaction locus. For maximum moments, only one of the cantilever arms has to be fully loaded, THE CANTILEVER ARCH. 179 namely that one adjacent to the unloaded segment of the interior span. If the right arm is completely loaded with p per unit length, then, with the notation of Fig. 73, we obtain: ’ perx “ “ In the two-hinged arch, the horizontal thrust may be deter- mined by equ. (250); introducing the abbreviations N for the eS I ing on the external loading, equ. (250) may be written denominator and v, and retaining only the terms depend- 180 ARCHES AND SUSPENSION BriDGEs. 2Mv H=—~ For a load G placed on the cantilever arm, M = —G $2 ; hence 1 Ha ¢@+ re ee (397 to oN : In the H-influence line, Fig. 54, obtained as the funicular polygon of the v-forces according to § 17, the intercept of the prolongation of the last side upon the vertical through A gives i the quantity = vz; hence, by equ. (397), this line also consti- 0 tutes the continuation of the H-influence line on the side of the cantilever arm (Fig. 74). If the arch possesses a vertical axis of symmetry, then 1 1 ve =134, hence 0 0 1 1 hs 0 a Sa ae Beau aires eee a ween a (3978. The vertical component of the reaction at A is Gs V,.=—-G 7 hence the direction of the left reaction is determined by Vi N war=> = UL oMea and is therefore independent of the position of the load G. Con- sequently the reaction locus is continued over the cantilever arm as a straight line consisting of the end-tangent from the point A. If we substitute 2 f, 1 tan-+, then I xv N fo= a Sh ae eee ea tees (398. Zve Zv 0 0 consequently, ee a a 2 cl ee 8 Gk ace (399. THE CANTILEVER ARCH. 181 Thus, for loading on the cantilever arms, the two-hinged arch acts exactly like a three-hinged arch having its crown-hinge at a height of f,; H and M may also be calculated by the same formulae (395) and (396), provided f, be substituted for f. If a load p completely covers both cantilever arms, assumed equal in length, the moments at the core-points of the arch-sections are easily derived. They are, in fact, proportional to the distances of the core-lines from a horizontal line drawn at a height of fo above the end-points, and J 1 2 are given by these intercepts multiplied by H = — zr Es [cf. equ. 0 (396)]. In Fig. 75, these moments are represented in the graphs mno Fig. 75, . p and pqr. Adding to these the moments due to a load p’ covering the entire span l of the arch, we obtain the combined moments for the total loading as indicated by the shaded ordinates of Fig. 735; there is thus shown graphically the effect of the cantilever arms upon the: distribution of stresses in the arch. 182 ARCHES AND SUSPENSION BRIDGES. § 24. General Case of the Cantilever Arch. Let an arch with fixed ends have a cantilever arm anchored to it at any point (Fig. 76). The general'equations (311) to Fig. 76a, oot tle a 1 cer cscr alt pee ii si fe? aes Figs, 76b and 76c. F (313) apply here also, and may be written in an abbreviated form as follows: THE CANTILEVER ARCH. 183 where M, as usually, represents the moments within the span 1 of a simply supported beam. For the concentration G, mao ft! (L—2)]%? l 2 w=Cc =-0-$(h49 2 consequently l d ia! fi ésnt aml =e a) =F30+ x) rea: The summations appearing in the above expression are rep- resented as static moments of the v-forces by the intercepts of the corresponding sides of the funicular polygon upon the ver- ticals through the ends A and B; i. e., the two summations are represented respectively by the end ordinates of a line drawn tangent to the H-curve at the point c (Fig. 76%). Hence, according to the equation l i) 452 73 (-*2 -4) or (kv) =>. 2] 2=0 [ @ a’=a in constructing the deflection curve directly. If, instead of the moment of inertia 7, we substitute (where J, is any assumed constant moment of inertia), then the seale unit for the ordinates of the deflection curve will be El, Z ‘ a a ee times the unit of the scale of lengths. Here H, = Dy denotes the horizontal thrust for a unit load G@ at the end of the cantilever arm, A x is the horizontal spacing of the k and v ordinates, and p is the pole distance used in constructing the funicular polygon. The expression for the deflections (— Ay) of the end of the cantilever arm, due to a load G@ at any point z, as given by equ. (300), may be written as follows if we neglect the term depend- ing on the axial thrusts: —EI.Ay=Gm,— H,y.m,x. With a constant moment of inertia, we obtain: —[2 Geeta?) fm] 2 (for 2 EI.ay=[z2(t a 0’) mz] @ st (for r=0 tol) (401. EI.sy——[3 ws fe 25 ; (a—x') | Ga (for x’=0 toa) mx denotes the moment of the quantities v; i. e., if wu is the ordinate of the H-curve drawn with a pole distance, p, m= u.p.Ax. Again, N is the denominator of the expression for H. The vertical displacement of the end of the cantilever due to a change of temperature, if we neglect the expansion of the cantilever itself, is given by EI. Ay—=—Hi ge 4, or, substituting the value of H, from equ. (290), Ay=—otb.cosB. 3 Ooo Sila teen icthalerafSiGrh es (402. If the arch has a crown-hinge, we may first neglect the inter- ruption of continuity at the crown and, taking k — v = 0 at this point, proceed as above to obtain a curve of deflections as the funicular polygon of the forces k — v. To the deflections thus determined we must add the effect of the rotation at the crown. This produces a crown sag, — 8; and, by equ. (143), neglecting the term referring to the axial thrusts and with the altered 186 ARCHES AND SUSPENSION BripGEs. a notation p= vy = N, =a, me= ap we obtain E1,8—G@.9-(1-L)q.N ee AL eae, (403. Using the same scale as for the ordinates of the deflection El, El curve, namely Hapae eae the deflection ordinates due to the crown-hinge rotation are given by the moment ordinates for a crown-load of (a—-£) ks which, since N—=p.Ax.2n.n and h=s> j a times the scale of lengths, = -™2-P , ig represented by a distance 2 (4 nn — v4 Us) Vos =-— 2rv, (Fig. 73b). The deflection lines d, c, e, are therefore fixed by drawing d,c,|| Or, and the resulting (full-drawn) inter- cepts of Fig. 736 give, by the above scale, the deflections of the cantilever end D for a moving concentration. (Compare equ. (148) et seq.) The effect of temperature, as before, is giver. by Ay. —=—otb.cos B.z- iOa Sx egos cetentels (402°. THE Continuous ARCH. 187 5. THE CONTINUOUS ARCH. §26. Continuous Arch with Hinged Connections Be- tween Successive Ribs. Determination of the Horizontal Thrust. By continuous arches we understand a form of construction in which several arch-ribs have common intermediate piers on which their ends are free to slide together, but with the outer- Fig. 78. most ends definitely connected to the abutments. Let us first consider the ease of such a bridge composed of n successive arch- ribs with hinged ends. Again allowing the approximation of writing P’—H and of substituting a mean cross-section A,, A,, ...., for each rib, also writing for the mth span Ty a. __ __ _bm.cos Bm yds i. fq (4 4s—az)= cos and f = 344% and, finally, assuming that the displacements on the piers may occur without friction and that the changes of temperature from the unstressed condition amount to ¢,, ¢t., t,.. for the respective arches, we obtain for any loading on the mth span the following expression for the increase in length of that span: Al—sMY gs— Hs tte _q. Pnceebn ts tb cos Bn. Adding together all such expressions for the individual spans, we obtain: 188 ARCHES AND SUSPENSION BRIDGES. sap—pos tds m bi cos By db. cos B, bacos Bn nf tat EA, a EA, aE seen =P EB An ] + o[t, 0, cos B,+t,b,c0s Bp +....+ trbacos Ba] Hence, > = .ds+ EB» (t,b,cos By + teb.cos B+... + tn ba cos Ba) ae a 4+ yds +..42 yds +4 bi cos By 422008 By ae 1 TI 2 I I Ay Aa Si... ts fn.cos Be Comparing this with the case - the simple two-hinged arch we find the following: The horizontal thrust due to any load is essentially less in the continuous arch; this reduction of thrust increases with the number of arches linked together. If all the arches have the same dimensions, then for n arches for which only one is loaded, eee aus as te acetindeie aren (4048, where H’ denotes the thrust in the single loaded arch if its ends were immovable. For constructing the H-influence line, the procedure given in §17 may be directly applied. In obtaining the maximum stresses, the method of influence lines recommends itself; and it is thus made apparent that the maximum positive moment at any section will occur when the corresponding span alone is partially or entirely loaded, whereas, for the greatest negative moment, all the remaining spans must be completely loaded. If the frictional resistance at the piers is considered, the calculation of the horizontal thrust is changed as follows: Let H, be the horizontal thrust transmitted to the left abut- ment; also let it be assumed that the vertical pressures on the intermediate piers due to the dead weight of the arch-ribs are practically uniform and equal to A, and that the coefficient of friction = f. Let the loading in the mth span produce on the (m—1)th and mth piers the vertical reactions V, and V, respectively. Then the horizontal forces acting in the first to the nth spans are: H,, H,+fA, H,+2fA.. 9 H,+(m—2)fA, H,+(m—1)fA+fV,, By (m—2) {AF V1 — Ve), Bat CS) FATT VV gliccey Hy=- (2m—n—1) f4-Lf (V.—V 2). Hence, retaining the same simplifications as above and omit- THE Contunuovs ARCH. 189 ting the effect of temperature as given by equ. (404), we obtain, from the equation of work, the following expression for H,: 3 Mus _ 0, f.4—C2fV,—Csf (Vi-Va) H,= .. (405. yds yds yds bi cos B, bz cos By ope pee 1, 8a bn cos Bn ee gee Here, = yds - yds ee yds c-E4 422284. (m—1) 34 +(m—2) 3 28 +... (2m—n—1)3 OM 4 Bee 42 be.cos + (m-1) bm 008 Pw + (m2) Pascoe Bree ; o ee ow (406 +...(2m—n—1) ete bm cos Bm Am ad C,=% - m ids ‘ds bm m+ bn Ba O—3 & +... 3 288 4 Deecooe Amn 4 | bcos Pa m+1 n m41 D If, as before, we assume all the arches to have the same cross- section, then C7, = . [2m (2n—m-+1) —n?—3n] ( yas + eosb as ae ome .. (406%, t= (n—m) (: ve fe Pos | and we obtain 2m (2n—m +1) —n?—3n 2n Tees 1 H.=—H'— f.A——fV, 2 —A—™. f(V,—V;) It should be observed, however, that the horizontal thrust due to loading can never be negative in any span, i. e., the horizontal thrust, already existing on account of dead load cannot become diminished. If the contrary is indicated by the above formula, then we must consider as included in the system only those arches on each side of the loaded span for which the formula gives a positive value of the horizontal thrust; H, must 190 ARCHES AND SUSPENSION BRIDGES. be positive in the outermost spans. The horizontal force in the loaded span is H=H4 (1) pA+7.V. Fig. 79. A special form of the continuous arch, designated by Haber- kalt* as the ‘‘Balanced Arch,’’ is shown in Fig. 79. The end spans consist of half-arches whose free ends are connected to- gether through a straight tension rod AH. This tie-rod has no connection with the intermediate arches. Only one of the sup- ports is fixed, the others are free to slide horizontally; in the terminal supports, A and £, this freedom of movement may be attained through the use of rocker-piers. To calculate the force H acting in the tie-rod, we may again employ equ. (404), if we denote by y the arch-ordinates measured from the respective chords and add to the denominator a term -. representing the effect of the tensile strain in the tie-rod. (£—Jength and A = cross-section of the tie-rod AE.) If the ends A and E rest on rocker-piers of section A, and height h, and if 1, and h are the horizontal and vertical distances between the end-points of each terminal arch, there enters into the denominator of equ. (404) 2h, h? Ail? * end span at a distance é from A or E£, the numerator of the ‘expression for H must be increased by the small term (I—£) hits C= as an additional term: For a load & located in either The effect of a temperature change of ¢° is represented by adding to the numerator of equ. (404) the term —Eot (L+ ahh), 4 This system of arches has found application, in the form shown in Fig. 80, in a foot-bridge over the Seine built at Paris in 1900. In this case, however, the supports on the two inter- mediate piers are made fixed; so that the middle arch, acting as * Osterr. Wochenschr, f. d. dffentl. Baudienst, 1901. Tue Continuous ARCH. 191 an ordinary two-hinged arch, is excluded from the continuous structure. The tension in the tie-rod is determined by MY ds ee ia a ie 4 pees +4 a “hi Fig, 80- In the above expression, the summations cover a single side-span and the quantities A, A,,h,, etc., have the denotation previously assigned. In Fig. 80 the shaded ordinates give the influence of a con- centration G upon the moment at the arch-point M measured to tar ae a scale whose unit is a" 192 | ARCHES AND SUSPENSION Bripees. §27. The Continuous Arch with Hingeless Connec- tions between the Ribs. If the successive arches are rigidly connected together at the intermediate piers, so that any change in the angle between the arch-axes at the points of support is prevented, although a horizontal displacement at these points is permitted, the struc- ture thus formed of n successive spans without any hinges is of n-fold static indeterminateness. It is here assumed that the connection to the abutments possesses freedom of rotation. 1. A statically determinate structure is obtained if a hinge is inserted in each span. However, in general, such a system will be practicable only with an odd number of spans; for, if the number of spans is even, the system lacks stability. The Fig. 81. system ceases to be a rigid one, but enters into unstable equilib- rium, when it is possible to connect the end-hinges by a chain of straight lines passing through all the intermediate hinges (the chain of lines A B’ C, D’ E, F’ in Fig. 81). If the last point (F’) does not coincide with the end-support (/"), the stability of the system is assured; but the capability of deforma- tion is even then considerable, if F’ approaches close to the point F. If only one span, as BC in Fig. 81, is loaded, the reactions in the unloaded spans are given by the chains of lines from the abutments, A B’ and FE’ D, C’, while the loaded span acts as a three-hinged arch supported at B’ and C’. In the second, fourth, and all the even spans, these imaginary points of support lie above the crown-hinge; in the odd-numbered spans these points fall below the crown-hinge; and, in such case, they will coincide with the actual pier-supports B, C, D.., if all the crown-hinges are located at the same height above the closing chords ABC D.., i. e., if all the arches have the same rise. Loading in the odd spans will thus produce compression at the end abutments, and loading in the even spans will produce Tur Continuous ARCH. 193 tension. The individual spans act alternately as erect or in- verted arches of equal virtual rise f, throughout. All parts of the structure are thus subjected to reversals of stress. For a moving concentration, the reactions vary as a linear function of the position of the load; and, for a structure of three spans of equal rise, the influence lines for the horizontal thrust Fig. 82. a Horizontal Thvust 1’, ‘ L, + b.Vertical Reaction atA * Var ¢ ! Pee = 6G + cVertical Reaction at B at the end-supports and for the vertical reactions at A and B are shown in Fig. 82, a, b,c. The moment, referred to the point (x, y) in the first span or to the point (#’, y’) in the second span, is given by M—=M— Hy or M=M + Hy’, respectively, where M denotes the moment of the vertical forces for a single arch. Hence the influence quantity for M is obtained in Fig. 82, d and e, as the difference between the ordinates of the H-curve and those of 13 194 ARCHES AND SUSPENSION BRIDGES. the or 7 lines. These intercepts must be multiplied by the factors y or y’; and these ordinates y must be measured, in the odd spans, above the true points of support A B, and, in the even spans, from the virtual points of support B’ C’. There thus remains no difficulty in establishing the analytical expressions for the resultant forces and moments. If the friction at the movable intermediate supports is con- sidered, then the reaction at those points must no longer be taken as vertical but as inclined at the angle of friction. The reactions of the loaded span tend to be displaced outward, but the direc- tion of displacement changes alternately for the other spans. Since the reactions in these spans also alternate regularly be- tween tension and compression, it appears that the directions of the reactions at all the piers on each side of the loaded span must be inclined toward that span (Fig. 83). The intermediate Vig. 83, supports of the successive spans are replaced by the imaginary points B, C,.., B’ C’..; and the changes in the reactions from those of the frirtionless condition are easily obtainable. 2. If no crown-hinges are provided, the structure with n distinct spans is n-fold statically indeterminate; and, to deter- mine the reactions, » equations of elasticity must be established.* To establish these equations of condition, it is simplest to make use of the principle of least work. For flat arches, with the depth of section small in comparison with the radius of curva- ture, the work of deformation may be written Ie H?1 w= (popdat+ fe —oH ftdz,.... (407. *The theory ofthis type of structure, which may be considered as a more general case of the continuous beam, was first presented by H. Miiller-Breslau (Woehenbl. f. Arch. u. Ing., 1884). In the following treatment, this work has been utilized. THE Continuous ARCH. 195 if we adopt the approximations of taking the axial-force = H and assuming a mean cross-section A, and denote by J the moment of inertia multiplied by cos é = at. The integration is to cover the entire length of the structure and 1 denotes this total length of span between end abutments. If V, V,.. denote the vertical reactions at the Ist, 2nd, .., intermediate piers, and if y, y,.. are the moments at any point of a straight beam of span | loaded with a unit force success- ively at the different points corresponding to the intermediate supports; if, furthermore, M is the bending moment producible in this beam by the external !oads acting on the continuous arch and if y is the arch-ordinate measured from the line joining the end-points A B; then we have M=M — Hy — V,y,— Voy2— Vays — If, as a general case, we assume that, under the loading, the end-points are displaced outward by an amount AJ and the intermediate supports are depressed by the small amounts 8, 8,.., then Castigliano’s Theorem, concerning the derivatives of the work of deformation, yields the following relations: aw aw aw ht Ses Oy pe? gone ae But, by equ. (408), aM au aM aH oY, Yu RS Hence, if J, denotes an arbitrary moment of inertia, equ. (407) yields the following equations of condition: EI.Al—oEI,tl={ My dx—HI E1,8,—{ Mf y.de E1,8,—{ M$ y.dz. If we substitute here the value of Af from equ. (408) and intro- duce the abbreviations comfy pdetle tm fy ym de wi sat dsb See rene eat A ee (409. L MM Cmr =fy0 Yr - dx 196 ARCHES AND SUSPENSION BRIDGES. we finally obtain the equations: -oBIgtl+EIAl=f M 4! yda-Heo~Vitr~ Vite Vaca~ B1.81= { M7? ysdar-H c1-Vieu-Voea- Vacer~ +... (410. B Iy8.= [ M 72 yoda—H c2— Vien VaCu- Vater +--+ If the load consists of a single concentration G in the mth sas 45 ° I, span, the definite integrals f My d x, f Nie Y,AL..--, Fig. 84, iii Mm. 4) URAL, |e, Si | ii HAT Ss 1 nee ATT aa)s Cc ees may be represented by the ordinates of respective funicular polygons. They are in fact equal (cf. §17) to @ times the moments, about the load point, in a simple beam of span J loaded successively with the quantities y 4s, Yi -, Yo i ...(Fig. 84). Let us denote these moments, which may readily be determined by construction, by m, m,, m2... THE CONTINUOUS ArcH, 197 The quantities c,c,..Cm and the general term cm, are also given by these funicular polygons, as follows: Since y;—= fe x (from s<=0 to e=1,) and y;= = (l—«) (from c=—l, to r=1), we have lr 1 Uy Io lr To mr Ff (unt) ede + { (ym) (2) dz, 0 lr i. e. = the moment about the rth point of support of the area of the graph of yn *, or, since by equ. (409) cm: Crm, it 1s also equal to the moment about the mth point of support of the area of the graph of y, is The equations (410) thus take the form —voLE I, tlt EI, Al=G m—Hc.— V,c,—V2e2.— V3C3— - - - EI,8,=G m,- H ¢,-V,01,-VoC1-Vala, — - - - - ET,8.= G m2—H ¢2— Vy Cy2— V2C22- V3C32— - - - . (411. There remains to be determined the definite integral cy = 1 fv Bart ite. 0 the moment, about the chord A B, of the load-quantities y + dz This quantity is nothing else than applied horizontally at the respective points of the arch. Hence this quantity may, in a familiar manner, also be obtained by graphic construction; and the solution of equations (411) will yield the values of the reactions H, V,, V.... If the effects of temperature and of yielding of the piers are, for the sake of simplicity in computation, omitted for separate consideration, then the left-hand members of the above equations must be replaced by 0, and the solution gives: H=G (am+ Bm,+ym,-+ bike) V,=G4 (a,m+Bhim,+yim.+..) Vi=G (agm+ Bom, +y2M2+ --) where aPy...-, 4, Bi¥i-+-) 42 B2y2-+-- represent coefficients which are independent of the position of the load G; so that it 198 ARCHES AND SUSPENSION BRIDGES. is necessary to solve the equations but once in order to obtain the influence of any position of the load. Equations (411), with the omission of the first, also serve to determine the reactions for a continuous beam on putting H = 9. For a more exact analysis, the friction at the supports should also be considered. More Exacr Turory or THE ARCH. 199 § 28. More Exact Theory of the Arch, Including the Effect of the Deformations Under Loading. As previously stated, in § 10, the foregoing theory of the arch-rib is founded on the approximation of assuming that the elastic curve under loading is identical with the initial form of the arch-axis; hence, in determining the moments, the deforma- tion of the arch caused by the loading has not been considered. This is the same approximation as was made in the treatment of stiffened suspension bridges in §§ 4 to 8; and, for a more accurate analysis, we must follow the same general procedure developed in outline in § 9. However, since the error of the approximate theory diminishes as the deformations of the struc- ture become less, it may be generally stated that in arches, which as a rule are more rigidly constructed than suspension bridges (especially the earlier examples), it will less often be necessary to apply the rigorous theory. We will therefore take up the exact analysis only so far as may be necessary in order to obtain an estimate of the admissibility of the approximate treatment and of the magnitude of its error. Let us consider a two-hinged arch of uniform cross-section, and let (x, y) denote the coordinates of any point of the arch- axis, measured from the left point of support, in the unstressed condition of the arch-axis; and let the vertical deflection of the point, under loading, be y. We will neglect the horizontal deflec- tions of the points of the arch and, furthermore, assume that the end-reactions are not affected by the deformations. "With the usual notation, the moment at the given point of the arch may be written: M=M—H(y— ); and, for a flat arch of approximately constant curvature, the change of curvature due to deflection will be, by equ. (183), (22 ) Sea EA da? EI’ or, substituting the above expression for M, x I an M_ = (zI-~7) aed y) H Hy. Substituting the abbreviations C= - i Mite . (413, 3 I EI-7ZH 200 ARCHES AND SUSPENSION BRIDGES. M a F (2) =U—-y, the differential equation of the deflection-ordinates cf the arch becomes <7 +07q+0°F (2) =0. x For the cases of loading considered in practice and with a segmental or parabolic arch-axis, (x) is an algebraic function of no higher than the second degree, so that its second differen- tial coefficient will be constant and its higher derivatives will vanish. Then the integral of the above differential equation is n=A sine +Bcosca—F (#) +P” (a) .. (414. with which the expression for the moment becomes M=H (Asinca+Beosca+F” (n)) .. (415. The constants A and B are determined by the condition that at x—0O and x1, y»—0; and, at any interruption of contin- uity, equal values must obtain for y and at. To determine the horizontal thrust H, the following method may be used. Let us consider an arch loaded with a unit con- tinuous load, and assume the resulting pressure-line to coincide with the axis of the arch; then, if we thus disregard the bending- moments, the axial thrust in the arch = 1.r, where ra is the radius of curvature of the arch at the crown. The actual distributed load q, in the place of the unit loading, produces the deflections 7 and, with the length of the axis = b, a shortening H ‘EA ciple of virtual work to the assumed load 1, on the basis of the deflections 7, the equating of the external and the internal work of the arch-axis by the amount . 6. If we apply the prin- 1 Hb gives Ll gels ae Substituting in this equation the above expression for 7 (equ. 414) and observing that, for a parabolie arch uniformly loaded, A 8 F” (2) =—£4-1, we obtain, with the aid of equ. (415), More Exacr Tueory or THE ARCH. 201 f (M+ G)ae ote H= . (416. 1 f (A sin ca + Boosca) age 44 + + fl—- ort 0 To apply this formula, an approximate value for H must first be caleulated by the formulae previously established. [Com- pare this expression with that of equ. (152) for the case of the suspension bridge.] ~ For a uniformly distributed full-span load, the constants A and B are found, by the condition that »—0 for xO -and a= 1, to have the values _ il gq 8f elie=> 2473 3) B=3(q- : Hence, by equ. (415), 1 cos ¢ ( —a) M=-;(¢-4 =F) p- a aa and, by equ. (416), 1 1 =e 1 qd Bf) 2 cl, 8f H , 5 ftt a [($—-F) geft+f] ea? For very small values of c, hence for very rigid arches, we thus find, in agreement with the-approximate theory, 1 8 M= 342 ( f and approximately (assuming J = ~), gt H= 2 a At the limiting value of S cl = +, we find M—o, i.e. the arch must fail under the bending-stress. This yields the relation CP=r= ees EI—-~H mI mET or SS BL +43) Se eee (419. _ The limiting value thus fixed for the horizontal thrust at which the arch will fail by bending is nearly identical with the 202 ARCHES AND SUSPENSION BRIDGES. buckling load given by Euler’s formula for straight columns whose unsupported length is J. If the loading consists of a single concentration G, applied at distances € and é’ from the respective ends of the span, the moment at any point to the left or right of the load may be written H 8 bas or =H (A’sincxz+B’coscr)+ = i ; The constants are determined by the conditions that, for «—0 and z=1, y= 0, and, at the point of application of the load, y and at must have the same values for the left and right segments of the rib. Substituting & —1l—é and «’=—Il—z we obtain w= since’ sincar g—_ sinca+sinca’ —1) -. (420. c sincl Cr sincl which expression applies for all points from c=—0 to c—é. For the right segment of the arch, i. e., from c= é to r=—1, we must interchange é’ with € and z’ with z in the above expres- sion. For the horizontal thrust, equ. (416) yields the expression - @.4.¢ Fee (421. = i sincE+ sinc&—sincl 3 silt sincl -<+ ef @ tan ~ o-))- aE With c=0, i. e, for the infinitely es — the above expression gives H —0. But if we assume c only so small that the are-functions sin and tan may be replaced by the corre- sponding arcs, we obtain - Sate and, if we neglect the relatively very small second term in the denominator, this formula becomes identical with that of Engesser and Miiller-Breslau : w=" Goa - (see p. 128). For this case, namely with a fe iia of equ. (420) take the indeterminate form of a whose evaluation by the usual More Exact THeEory oF THE ARCH. 203 methods gives the following expression in ‘agreement with the approximate theory: ag 2h M=G ; —H. “ta (i—2). If the arch is uniformly loaded over the entire span with g per unit length, and, in addition, with a load p extending for a distance A from the left end, the following expressions for the moments at the arch-points are derived from equ. (415): From z—0tor—=), —_ 7 [(9_ 8f\,.. i ae Mal [2-5 (sin catsin cx’—sincl) + (sincx’+sincxcoscrA—sincl) | From z~iX tor—l, .. (422, M= or [i — “t) (sin ex + sin cxr’—sincl) pee (1—cos cd) sinex’| Here, «’ =1—~2zx and ’ =!1— .. For the horizontal thrusts in this case, equ. (416) yields the expression: 1 l » == [oet+px (31—2a) | 4+ te = gg aq ee eee ees (423. 1 8 2 K+ 54+ $n- Ae where K= : [2 (£ _— +) (1-cosel)+ 2 (1+cosed—coscl-coser) | csincl H vr H For comparison of these formulae with those of the approxi- mate theory, the following example is worked out: For a parabolic arch of constant section, let 1120 m., f= 12.13 m., A = 0.06696 m?,, J = 0.5549 m*. Let the loading consist of a dead load of g =2 t. per meter covering the entire span and a live load of p= 2 t. i per meter covering the left half of the span. Hence \=2’ =. By the approximate theory, equs. (277) and (278), we obtain 1 P q=% (o+ - any where 157 =f (1+ =e ap) = 12,13 x 1,180 = 13.44 m, 204 ARCHES AND SUSPENSION BRIDGES. Hence 1 120? H=-% X3X 34 = 401,79 t. The moments are given by M =M— Hy as follows: 1 1 3 For ¢= ql gl 4! M = 844.72 526.29 — 55,28 tonne-meters. For the exact design, we first assume the above value for H and thus obtain Bea I a a ae = 17,098,000—8,287 401.8 — 9-000 i Hence c = 0.0060176 and cl = 0.722112 = 41° 22’ 16.3, With \ = 4, equ. (423) yields 432,000+9,941,538 H = 9591344 29,332,014+97040—5,34 = 401,79 t. which is precisely identical with the result of the approximate theory. U Furthermore, with A =, equ. (422) yields, 1 for ¢ = a4 3 = wna (o— H -t) (ma q ol+ sing + cl—sinct) +p (sin. 3 4 cht sing : cl, cos yet —sinel) | 1 forg = ol w= seal (0+ 3 g Hp “t) (2 sina x el—sinct) |. fore = {7h 1 8f ce Jl 3 M = FF sinol (o-a P ) (sin get sin {ol —sincl) . i 1 + p.sin | el (1— cos 5 cl) | Substituting the numerical values in these expressions, é 1 1 3 or a= ql gl al Af == 87241 556.38 — 39,99 tonne-meters, The variation from the approximate theory thus an.ounts to More Exact Turory or tHe ARCH. 205 + 27.69 + 30.09 + 15.29 t.-m. or 3.17% 5.41% 38.2%, The error involved in the approximate theory, with an arch of adequate stiffness, is thus found to be zero with respect to the horizontal thrust and but a few per cent in the ruling positive moments. At any rate, it should be noted that the accurate theory always yields increased values for the positive moments and that, in a very flexible arch, the error in this respect may be very considerable. However, in all cases of arches applied to bridge construction it will suffice to conduct the design by the approximate theory, especially as the exact theory: becomes very complicated when the moment of inertia is variable. If it is desired to calculate the additional moments due to the deformation, an approximate method may advan- tageously be applied if the arch is not too flexible. This con- sists in determining the deflections of the arch under the given loading by the rules developed in § § 16, 19 and 25, and then referring the moments to the altered form of the axis. If Aw and Ay are the displacements of an arch-point due to the loading, H the horizontal thrust and V the sum of all the vertical forces on one side of any given section, then the correction to be added to the bending moment on account of the deformation of the axis is given by AMS VAP = FAG soiccnvica stoma (424. 206 ARCHES AND SUSPENSION BRIDGES. §29. Proportioning of Section in Plate Arch Ribs. It is difficult to directly determine the exact sectional areas in plate arch ribs, even if the dead load is known or assumed, since the maximum moments cannot be determined without the core-lines and the position of the latter is in turn dependent upon the sectional dimensions and areas. The best we can do is to adopt an approximate method which consists in first estimating the position of the core-points, taking them at about the outer leg of the flange angles. Fig, 85, Let, the fundamental cross-section consist of a web plate and four equal flange angles (Fig. 85); let h be the depth of this section, a its area, and J, its moment of inertia about its gravity axis. Also, let a, and a, be the required areas of the flange plates, and J the moment of inertia of the combined cross-section A= a,-+a,-+ a, whose center of gravity is at the distances d, and d, from the centers of the flanges. Then we have, very nearly, @,d,— ,d.—ag(d—%)—0 and d,+d,—h. Hence g=(-“4°)h G4-014+45%)4 a or, if we write the ratio Puiate Arci Riss. 207 then, d—(1—¢)4, G=(4¢) $e. (426. Using these relations, the following approximate expressions for the moment of inertia may be written: Imad ita,ds+Iota( 5 —h) =A (1-$*) Gth—a(F) - If Af, and M, are the maximum moments about the two core-points of the cross-section, (which we have assumed to lie at the flange surfaces of the flange angles), and d’, and d’, are the distances of the center of gravity of the section from the upper and lower extreme fiber-planes, then the maximum fiber stresses which should equal the permissible intensities of compressive stress are Mad’, Mid’, pos se) UE. , dy’ Vy Consequently, ee or, with sufficient approximation, dy 1—¢ Mi, dy l+o 0 WM,’ from which we obtain M.—M, oe ee saieleslaoias (427. But, since lo = MM, d’, or, approximately, [4a—¢#) = tI ao Jo= Me (1—#) + we have 2M. ah’?—4], Asap rayne tee (428. or, substituting the value of ¢ from equ. (427), M.+M, (ah? —41g) (M+ A.)? Aes 4 ON sy eee (429. The areas of the flange plates may then be calculated by the formulae: mp AeA apap 3 | (430 2 2 M,+ Mf, 1—¢ Lh tt a= A-— Fo ay, A 3% | This approximate design of the flange sections thus requires that the maximum moments M, and Af, about the outer edges 208 ARCHES AND SUSPENSION BRIDGES. of the flange angles should be determined. In calculating these moments, however, it will usually suffice to find the critical loading with reference to the center of gravity of the section instead of for the separate core-points. The error introduced by this inaccurate determination of the loading will never be material, since the influence of the loads in the vicinity of the critical points is always comparatively insignificant.* At the same time the design is thus greatly simplified since the number of different cases of loading is reduced by one-half. With the aid of the cross-sections determined by this first approximation, a more accurate design may then be made in which the true core-lines are located, the maximum moments referred to these, and the effect of temperature taken into con- sideration. For calculating a, and a,, the above formulae may again be used. , *In Weyrauch’s design of the Neckar Bridge at Cannstaat, it is shown that the error in the maximum stresses due to the simplified assumption of the loading does not exceed 1.7%. (See Weyrauch: “Die Elastischen Bogentrager,” 2nd edition, p. 41. Munich, 1897.) STRESSES IN FRAMED ARCHES. 209 D. Arch and Suspension Systems with Open Web. (Framed Arches.) § 30. Determination of the Stresses in the Members, when the External Forces Are Known. 1. General. According to the explanation given at the beginning of this book, the class of bridges under immediate discussion includes all those framed structures which, on ac- count of the occurrence of oblique reactions, may be designated as Arches or Suspension Bridges. In construction, these bridges consist of two chords, at least one of which is curved or poly- gonal in outline and is connected to the other by a framework or system of web members. In the design, the same funda- mental assumptions are made as in the general theory of Trusses, namely that the individual members of the framework are connected at the panel-points with frictionless hinges, so that only axial stresses can appear in the members. The secondary stresses resulting from rigid joints are to be separately con- sidered but, with proper construction, these stresses will as a rule be slighter in arches than in ordinary trusses on account of the smaller deformations; hence it will generally be un- necessary to investigate these stresses in designing an arch. It is further assumed that the deformations are so slight that the initial form of the structure may be made the basis for the computation of the stresses. (Cf. § 28.) If the external forces, i. e., the applied loads and the result- ing reactions, are known, then the calculation of the stresses v in the individual members may be executed according to the generally applied rules. If the truss-work itself contains no superfluous members, then the stresses can be simply derived from the equations for static equilibrium; and this may be accomplished : 1. By applying the conditions of equilibrium to the forces acting at each panel-point, which may be done graphically by the method of Cremona’s force-polygon; or 2. By the so-called Method of Sections. The latter method is applicable whenever it is possible to pass a section through the structure so as to cut not more than three non-concurrent members. The stresses in any one of these members may be determined either analytically, by 14 210 ARCHES AND SUSPENSION BRIDGES. equating the moment of the external forces with that of the unknown internal stress, taking the point of intersection of the other two members as the center of moments; or graphically, by resolving the resultant of the external forces along the direc- tions of the three members in the section. (Fig. 86.) If the web-bracing contains redundant members, then an exact design requires the application of a procedure based upon a consideration of the elastic deformations of the individual members. There are thus to be obtained as many equations of condition as there are redundant members. In many cases, however, the procedure may be simplified and the number of statically indeterminate quantities reduced Fig, 86. by adopting a certain approximation which will here be con- sidered in connection with several systems of web-bracing. a.) The framework consists of three sets of web members, all of rigid construction (Fig. 87). Let the panel-length be relatively small so that the break in direction of consecutive chord members may be inappreciable. Each panel contains a redundant member, viz., one of the two diagonals. If, as is sufficiently accurate, we assume the external reactions to remain unaffected, then the stress in the diagonal will affect no members of the contiguous panel except the radial posts common to both. Introducing the notation indicated in Fig. 87 for the lengths and sections of the members composing any panel, then the stress S, of the diagonal d, is determined by the following expression of the Theorem of Least Work: BrP SSP 8 we cess secseer anes (431. in which the coefficient of elasticity is assumed uniform through- out, 7 represents the ratio of the length to the cross-section STRESSES IN FRAMED ARCHES 211 of each member, and u denotes the stress produced in each member by two opposing unit forces acting in the direction of the memberd,. If Z represents the stresses which would he produced in the members by the given loading upon the omission of the redundant diagonal, then the actual stress will be SSL 8 Ab ice eh a thd eb ASsw tale (432. and only the posts c, and c, require additional terms, S’ w’ and 8S” uv” respectively, to represent the influence of the adjacent diagonals. Accordingly there results from equ. (431), SrZuts8,srwtr, 8’ uz,w +r,S8” uw’ =0........ (433. With approximately parallel directions of c, and c,, the stresses uw are easily expressible with the aid of the force polygon, Fig. 87, by the following values: (from similar triangles), U= se eee t= — ‘i= ge 1 ds s 2 ds s 3 d, , 4 dy a %S§ dy 2 Furthermore, ce cc” y= —_ oe ) uu” Ty ; or, with close approximation, , C1 uw” eu ut? USS a rat si as Introducing in the above equation the further approximation S’ = S” = 8,, and solving for S,, there will result the expression (2.642: 542,908 +250 —Z, ue eh Pontus + A)p ay No appreciable error will be introduced by using a mean depth of panel c, putting C,C,= C7, 6, Cy (C4, + C2) = 20%, and, in addition, assuming an average cross-section A, for the posts c, and c, there will result (6a, t a eA ot) ds a a3 OE Hee eae gw . (434, So= Sa » (434, For a final simplification, introduce the closely approximate values Z3=2Z,= a Zs in the above equation, thus obtaining: 212 ARCHES AND SUSPENSION BrinpGEs. a b? a d? dy (2. + 2g) 60 +a) a 434" a v? é de d? as : ae ae a de S.— By equ.(432), the value of S;, fixes the stresses in all the other members in the panel. Accordingly, the stress in the diagonal d, will be d. S,=Z,+ ae S. or, on substitution, e vu? a bv? é (4G +27) at% +a ea + a b? 3 dé de “Aa Ta ata As aa 2 6 d,* Ae ) Ss= The ratio between the stresses of the two intersecting diagonals will be oe v? a ue é d\ dz s, (ag tea) at2(et+et2o+4)4 a, 5 (2,4 +2,~) d,—Z, (22 4+#) & a An approximate value, sufficiently accurate in many cases, may be obtained by neglecting the terms containing A, or A, as a denominator. There will then result the simple relation . (435, 3 d; Oe A oo cnn RE S, 9 de dy, A As If the radial posts ¢ are of relatively heavy section when compared with the diagonals, then a final simplification will give, Ss; Aaa The expressions (436) and (437) are dependent only upon the lengths and cross-sections of the web members; consequently, if the latter quantities or their ratios be assumed at the outsct, the resultant of the two stresses S, and S, may easily be obtained (Fig. 88). For this purpose, lay off upon the two diagonals from their point of intersection the distances K-1 and K-2 proportional respectively to the numerator and denominator of the above expressions. The diagonal of the resulting parallelo- gram gives the direction of the resultant of 8, and S, and its intersections o and wv with the two chords will be the moment- centers for the determination of the chord stresses. If M, Ss cashes assoc irles (437. STRESSES IN FRamMED ARCHES. 213 and M, are the moments of the external forces about these two points, p, and py the perpendiculars from 0 and uw upon the lower and upper chords respectively, then the chord stresses will be Spats Sy — eee (438. b.) The framework consists of doubly intersecting diagonals, without radial posts. (Fig. 89.) This system of bracing, in the consideration of its internal stresses, is also statically in- determinate. An approximate determination, sufficient in all cases, may be made by putting A, —0 in the equations derived above. There is thus obtained, from equation (435), the following ratio of stresses in two intersecting diagonals: Fig. 88, Fig, 89. So dy Be nts eens (439. Laying off the distances d, and d, from the point of intersection K, the diagonal of the resulting parallelogram will cut the chords in the points 0 and u, about which the moments of the chord stresses are to be taken. The stress in the diagonal will be, by the above equation, Rae 2) eceubanmensoneeemuens (440. from which may be derived the well-known rule, that the analysis of such a multiple web system may be effected by resolving it into its simple component systems. ec.) The framework is constructed as in (a.) but without rigid diagonals. The approximate formulae (436) and (487) show that under any given loading the stresses in the two diagonals of any panel will be of opposite character. The exact formula (435), however, may yield a like sign for the two stresses; but, in such case, at least one of these stresses will be very small. It may therefore be assumed that one of the two diagonals in each panel must always remain without 214 ARCHES AND SUSPENSION BripGEs. stress. There remains to be determined, under any given load. ing, merely which of the two diagonals reccives a positive Z, or a tensile stress; this member is then to be considered as acting alone in its panel, so that the web system is reduced 19 a statically determinate one. 2. Severest Loading and Determination of Maximu Stresses. If we can find the reactions for any arbitrary position of a concentrated load, then no difficulty will be involved in determining the manner of loading for maximum stress in each member. This may generally be accomplished by the aid of influence lines for the individual stresses, as explained on page 42 of this book. There may also be applied the re- action locus and tangent curves for Pig. 90. determining the critical point or load- division point for each stress. In arches with pin-ends, the tangent curves are replaced by the two end- points: and in three-hinged arches, the reaction locus consists of the two lines joining the end-points with the crown- hinge. Arches with double end-con- nections (Fig. 90), corresponding to arched ribs with fixed ends, may be considered as exerting two reactions at each abutment, the resultant of which coincides with a tan- gent to the Tangent Curve. The position of continuous loading for producing maximum stress is determined by rules similar to those applying to arched ribs. It is merely necessary to find the center of moments for the member under investigation, and to pass through that point the tangents to the Tangent Curves. For a chord member, the center of moments will be the opposite panel-point or some point determined as in Fig. 88 or 89; for a web member, it will be the intersection of the two chords of the corresponding panel. The intersections of the above tangents with the Reaction Locus constitute the division-points of the loading. The stresses produced by the loading thus determined may be evaluated by the method of sections or moments, either graphically or analytically, provided the corresponding reactions can be ascertained. Except for the signs of the stresses, there will be no difference in the procedure whether arches or sus- pevsion structures are considered. If the loading consists of a train of concentrations, the maximum stresses are best determined by the aid of influence lines. This procedure, in the case of hinged arches, is to be STRESSES IN FRAMED ARCIIES. 215 especially recommended for its simplicity and will therefore be given a detailed consideration. 3. Method of Influence Lines for Framed Arches with Pin-Ends. a.) Determination of the Chord Stresses. According to equa- “tion (438), the upper-chord stress rs (Fig. 91) is determined by Ma Mu S=— ——-=— ——. seco (rs) Du ho where M/, is the moment of the external forces about the point w (to be determined as above), hy is the vertical depth of truss at the point u, and o is the inclination of the upper flange above the horizontal. The moment Jf,, however, may be expressed in the case of the two-hinged arch as the moment M, of a freely supported heam diminished by the moment of the horizontal thrust, or My=M,—Z y,, sc that the chord-stress will he — — I s0¢ = — (2 =) .seco.. (441. ho Yu Ito Fig.91. If the influence line for the horizontal thrust H, or the ‘“Hf-curve,’’ is constructed or computed according to methods to be later explained for a moving load = 1, there remains merely to draw an influence line for the expression — ,i. e., Yu 216 ARCHES AND SUSPENSION BRIDGES. for the moments of a freely supported beam divided by the constant ordinate yy. This moment influence line, in the case of vertical loads, is known to be a triangle with a maximum Guna ordinate at w= Us Yu Lu Accordingly, if we make the distance AN = - and draw the connecting line N B (Fig. 91), the latter will intercept on Lu aa Ls Yu so that A UB will be the Moment-Triangle. Within the panel, however, all loads must first be transferred to the panel-points r and s, so that the influence line between R and S must be a straight line; consequently the entire moment-triangle will not be effective but the influence line will be represented by ' RS B. Subtracting from the ordinates of this line those of the H-curve, the differences, multiplied by the numerical factor the vertical line through the point wu a distance U U’ = = sec o, will give the stresses in the chord member rs ac- cording to equation (441). For a system of concentrations, the severest position of load- ing and the corresponding stress may be determined by the general principles of influence lines. For a uniformly dis- tributed load of intensity p per unit length, the largest stresses in the chord rs will be, Snin—=— Area ARS J. i: .SeCa.D, Smax = + Area JHB .+~.seco.p, in which the areas are to be determined by measuring the abscissae to the scale of lengths and the ordinates to the scale of forces. If the panel-points of the lower chord lie along a parabola, then it should he noted, for simplifying the construction, that the locus of the points U corresponding to the panel-points of the lower chord will be a horizontal line and its constant ordinate will be U U’ =a b.) Determination of the Web Stresses. (Fig. 92.) It will be sufficient to analyze the stresses for a simple web system; and then to make an approximate determination of the stresses in a multiple system either by separating it into simple systems or by eliminating the redundant members with the aid of equations (434) to (435). The stress in the member rv (Fig. 92) is obtained by taking the moment of the external forces about the point of inter- STRESSES IN FRAMED ARCHES. 217 section z of the chord members of the corresponding panel, so that i dz S—+ (rv) where dz represents the perpendicular let fall upon the member from the point z. Again, we may write M4,=M,— UH. y,, thus obtaining sate t- at (= Ms —H)¥ (442, M, is again the moment which would exist if the end-points were Fig, 92. ee - ee Aaa ne , BA go geal N J Se NAN A Leet a pote it er 2...>> AW ; “ ie ii ae : u Xt io > wi J il) 3 |W Fe “Ss * P horizontally movable, or the moment produced in a simply supported beam. For all positions of a unit load (@=—1) to the left of the point r, M, =+ x’,, or equal to the moment cf the right reaction about the ae z; for the load positions to the right of the point s, the moment of the left reaction will give M, =£. 2; this affords a simple construction for the influence line of =. For this purpose erect at .! a > vertical line A P = ao and at Ba vertical line BQ= = - the connecting lines AQ and BP, intersecting in the vertical 218 ARCHES AND SUSPENSION BRIDGES. line through z, determine the polygon A R S B which constitutes the influence line for the function ae Deducting the H-ordi- nates, the shaded area in Fig. 92 will be the influence area for the required web stress; the intercepts of this area, multiplied Yu dz member rv for the different positions of the load. In the case represented, there are two division-points for the loading (critical points) ; all loading within the limits J, J, produces a tensile stress, loading in the remaining portions of the span will produce compression in the member. In some cases there may be but one critical point, i. e., but one intersection of the moment polygon with the H-polygon; this will occur when the vertex Z lies outside of the H-polygon. In the case of a uniformly distributed load p per unit length, the largest stresses in the member ¢ v will be Smax= + Area J,SJ_. 4p : Sin = — (Area ARJ,+ ArcaJ,H B) “ —p- by the constant factor , represent the stress in the web For an approximate solution in the case of a double web system (Fig. 91), resolve it into its component systems, consider one-half of p acting on each, and apply the method of influence lines as above. Two-HINGeD FRAMED ARCH. 219 § 31. The Simple* Framed Arch with Hinged Ends. Under this head we consider all those framed structures of a single span, which have their ends A and B capable of rotating but not of sliding freely. If, in addition, such a framework is interrupted by a central hinge, there results the statically determinate a.) Three-Hinged Framed Arch. The equations of static equilibrium suffice to determine the reactions in this type of structure. The reaction locus in this case consists of the two lines passing through the end-points and the crown-hinge; and the reactions are given by equations (200) and (201). The H-polygon for vertical loads reduces to a triangle whose altitude, if the crown hinge is at the middle of the span and if the corresponding rise is denoted by f, is H = Te To the foregoing discussions, nothing need be added con- cerning the determination of the stresses in the Three-Hinged Arch. These may be found, according to the previous general remarks, either analytically or graphically. The influence lines for chord and web members are constructed as in Fig. 91 and 92, except that the H-polygon is replaced by the triangle already mentioned. Fig, 93, The suspension system represented in Fig. 93 is to be con- sidered as an inverted three-hinged arch and is to be designed in exactly the same manner. In such a system, if the curve of the cable or chain forming the lower chord is made to coincide with the equilibrium polygon of the dead load, then all the dead load and the main part of all live load that may be uni- formly distributed over the entire span will be carried by the lower chord, while the upper chord and the bracing will be stressed only by partial or non-uniform loading. b.) Two-Hinged Framed Arch. As previously stated, this system is statically of single indetermination with reference to *i, e., having a single span; non-continuous. 220 ARCHES AND SUSPENSION BRIDGES. the external forces, so that the elastic deformations must be considered in determining the unknown reaction. Various methods of procedure are here applicable. 1. General Equation of Condition for the Horizontal Thrust (H). If one of the ends of the framed arch, e. g., A, were free to slide horizontally, then the system would act as a simple truss for which the internal stresses could be determined in the usual manner. Let these stresses be represented by Z and let « denote the stresses which would be pro- duced by a pair of forces of magnitude = 1. seca applied at the end-points and directed toward each other. The stresses uw are also easily determined. Tf, upon anchoring the ends, there is produced a horizontal thrust = H, so that the actual force acting along the direction of the arch-chord = H seca, then the stresses in the arched framework will be SS Ae A. ei ey eee os pee (443. These stresses will produce elastic deformations in the in- dividual members amounting to s As=;,8 where s is the length and A the cross-section of the member, and E is the coefficient of elasticity. The work of deformation of the internal stresses of the system will be _il ee! S?s 1 2 W=55 S-As= 5 5g =9527, Fig. 94. where r= for abbreviation. Assuming that the abutment is displaced horizontally under loading by an amount Al, the span | being increased by this amount, then, by a familiar principle concerning the work of deformation (Castigliano’s Theorem), aw : d (H seca) 1 ds or —Al=F =F it5- aR: Substituting the value of S from equ. (443) and putting oF =u, — E.Al=SrSu=SrZu+Asru’? —Al.cosa= dw whence ___ ErZu+F#.Al H=— Spae ttt eee (444, A different analysis gives the value of H in terms of the Two-Hincep FRAMED ARCH. 221 virtual deflections at the abutment. Considering the end B as fixed, let the horizontal deflections of the free end A produced by a unit load at M, be 8ma, produced by a load Py at M, be Pm 8ma, produced by a force 1. sec a acting along the chord AB, be 8ua, produced by a similar force H. seca, be A . 84a. The actual horizontal displacement of the point A with reference to the point B, caused by a yielding of the abutments, is denoted by —Al. This must agree with the resultant effect of all the applied loads P and of the force H.seca, so that —AL=(3 Pm bma +H 822) whence eae as eke (445. aa Comparing this with equation (444) shows that SrZu =EH.S Padma and Sru?—F.8,.; so that the terms of equ. (444) represent the horizontal deflections of the freely supported end of a simple truss, caused respectively by the external loading and a force 1. seca directed along AB, multiplied by the coefficient E. The above determination of the horizontal thrust tacitly assumes an unstressed initial condition of the system; in other words it is presupposed that upon the removal of the external loading the system will remain free from stress. In general, however, this unstressed condition will not obtain except when every member is at that temperature at which it may have been fitted into the structure without initial strain. If the actual state of the individual member differs from this temperature by 7°, then, if » represents the coefficient of expansion, the equation of condition for H becomes —E .Al=3(rS+E# .ot.s)u. If it is desired to find the effect of the temperature alone, the external loading may be omitted from consideration; accord- ingly, putting Z— 0, —E.AIL=SE.o.t.s.ut+ HH snw? whence E.Al+ 2 Eetsu Ee ay ceacteier ds (446. It should be observed here that it is impracticable to consider separately the temperature range (t) of each individual mem- ber from its unstressed initial condition. It must suffice to assume certain uniform temperature limits for all the members 222 ARCHES AND SUSPENSION BrinGEs. or at least for groups of members, although this makes it un- certain whether. the severest combination of conditions is pro- vided for. It is therefore important that the range of tempera- ture, at any rate, should not be assumed too small. Under the conditions of our climate, there should be assumed a value of t— + 30°C. (= + 54°F.) ; furthermore, it is easily practicable to apply the above formula to the case of the unequal heating of the two chords occurring when one of these is shielded from the sun. Assuming of to be the same for all the members, then Swotsu—wtSsu. There is a theorem, first demonstrated by Mohr*, as follows: If, in a free, unloaded structure, two unit forces directed toward each other are applied at any two panel- points A B separated by a distance s,, the resulting stresses w’ produced in the various members s will satisfy the condition Su’sts,=0. In the present case, the distance between the panel-points AB=l1.seca, and w represents the stresses produced in the framework by the forces 1 . sec a acting along the chord AB; consequently wu’ —=wcosa and Suws—=—l.sec?a, thus giving Al— 2 i SO saiapabavsonees (447. TU For steel we may use # w» = 250 tonnes per square meter per degree C. (= 197.5 pounds per square inch per degree F.), so that for £= 30°C. (—54°F.), Hot 7,500 tonnes per square meter (= 10,660 pounds per square inch). It may be observed here that the above derivation of H is not exact inasmuch as it assumes, in determining the stresses, that the initial geometric form of the system remains unchanged. Only under this assump- tion can the theorem of virtual work be applied. This is, however, the same approximation as was made for the theory of the Arched Rib, and the same remarks apply concerning its permissibility as have been made in § 28. This method of analysis becomes better applicable to the framed arch as its radial depth increases and as its deflections, in consequence, diminish. 2. Analytical Determination of the H-Influence Line. In order to obtain the influence line for the horizontal thrust ZH, it is necessary to consider a unit load placed successively at the individual panel-points and to compute the corresponding values of Sr Zu. In a symmetrical arch, this computation is most easily accomplished by finding the stresses wu and z produced in each of the members of the framework by a hori- zontal and vertical unit reaction applied at the end A. The stresses wu and z may be determined either analytically or graphically, the force polygon being used in the latter procedure. * Zeitschrift d. Arch. u. Ing. Ver. zu. Hannover, 1874. Two-Hincep FRAMED ARCH. 223 If, now, a load of unity be placed at any panel-point whose horizontal distances from the ends A and B are x and 2’ re- spectively, then the horizontal thrust is given by the formula x a’ LZrzutaz rzu 0 x Fe ee Hawa g elew ne Hy (448. lsru* 0 x The summation _% refers to all the members between the end A 0 , x and the load, and the summation % refers to all the members x between the point of application of the load and the panel-point symmetrical thereto. If the panel lengths are all equal, the distances x, «’ and I may be replaced by the corresponding numbers of panels. In an asymmetrical arch, an additional force polygon, 2’, is to be constructed, viz., for a vertical load of unity applied at the end B; and the numerator of the expression for H must x ax be changed to #’ 3 rzutaedr zu. 0 0 3. Graphical Determination of the H-Influence Line. The effect of a concentration P is given by equation (445) as Pébma La the projections upon the chord A B of the horizontal deflections of A produced respectively by the load P= 1 and by the force (1.seca) acting along the line A B. According to Maxwell’s Theorem, however, the deflection of the point A along the direction of the chord caused by the vertical unit load, must be equal to the vertical deflection of the loaded panel-point that would be produced by a unit force acting along the chord. Consequently the influence line for H is given by the vertical deflections of the panel-points (elastic curve of the loaded chord) produced by two opposing unit forces acting along the chord AB. The scale unit for H is the relative displacement of the points A and B produced by two opposing forces of (1.seca) acting along the same chord AB. For evaluating these displacements, various exact or simplified procedures may be followed. a.) Method of the Williot Deflection Diagram. (Plate 1, Figs. 2-2°.) We first construct a Cremona Force Polygon for the stresses u, produced by two unit forces acting along AB (Fig. 22). From these values and the known lengths and cross- sections of the members, are computed the elongations multiplied The expressions $n, and 8, may also represent 224 ARCHES AND SUSPENSION BRIDGES. by E: EAs=—.u,. The latter quantities are used for the construction of a Williot Diagram (Fig. 2°). For this purpose we consider any arbitrary member (kd in Fig. 2°) as fixed and lay off its elongation (here—0O) from a pole k, to the point d,. From k, d, are drawn the elongations of the contiguous members kz and di in their respective directions (a lengthening being drawn in the direction of the member, a shortening in the opposite direction) and at the ends of these two lines are erected perpendiculars thereto, representing the angular dis- placements of the members. The intersection 7, of these per- pendiculars represents the change in position of the panel-point a relative to kd. Proceeding thence in the same manner, the elongations of the members ic and dc fix the position of c,, and so on to A,. Proceeding similarly on the other side of kd, we obtain the point J,, then e,, and finally the point B,, thus determining the relative displacement of the points A and B. Since A, however, is constrained to move in a horizontal plane, there must be effected such a rotation of the entire framework as will eliminate the difference of elevation of the points A, and B,. In this rotation about B, A will move in a direction perpendicular to AB through the distance B, A,, and the diagram of displacements corresponding to this rotation consists of the figure A,b,c, ..., geometrically similar to the actual framework; finally, therefore, the distance between correspond- ingly designated points of the two displacement diagrams will give the actual deflections of the respective points b, c, ete. If the moving concentration G is applied only at the panel- points g, h, i, ete., of the upper chord, then the vertical dis- placements of these points, scaled from the Williot diagram, will represent the ordinates of the H-influence line, all upward deflections corresponding to positive values of H. These values are plotted in Fig. 2° to one-half the scale of the displacement diagram. The scale unit for H or the value of the correspond- ing concentration G is therefore 8aa, Where 6,, is the displace- ment of A in the direction A B produced by a force 1. seca acting in the same direction, or, what is equivalent, the dis- placement produced by a unit load multiplied by seca. This quantity is represented by the horizontal displacement A, A, to be scaled from the Williot Diagram. b.) Construction of the Deflection Curve (H-Influence Line) as the Funicular Polygon of the Deformations of the Angles. According to the above discussion, the influence line for H coincides with the deflection curve (elastic curve) of the loaded chord produced by the action of a unit force directed from A toward B. This deflection curve may be derived from the Two-Hincep Framep Arco. 225 angular and linear deformations of a chain of members con- necting the panel-points in question, which chain may consist either of chord members or web members (as indicated by the Fig, 95, a. b, heavy lines in Fig. 95a and Fig. 95b respectively). Fig. 96. The changes in the angles of this linkwork will produce vertical deflec- tions of the panel-points; these deflec- tions may be obtained as the ordinates of a funicular polygon constructed by treating the angular deformations as loads concentrated at the respective panel-points. The angular distortions may be de- termined either analytically or graphic- ally. If MNO (Fig. 96) is one of the triangular frames connected with the point M of the linkwork, and if An, An, Ao are the longitudinal strains.in the bars composing it, then the angular distortion at M will be A (0) = (Am—An) cotan (mn) + (Am — Ao) cotan (mo). The same quantity may be found graphically by constructing, to any convenient scale the vector summation MO, N, M, of the strains in the triangular frame MNO, (all elongations being plotted in the direction of the cyclical succession of the sides of the triangle M-N-O, and all compressions being plotted in the reverse direction) ; the change in any angle will then be represented by the projection of this chain of lines upon the opposite side of the triangle. Adopting an arbitrary radius k for the angular scale-unit, then the distance w, measured to the linear scale of the plotted strains and divided by k, will represent the actual change in the angle in radians. Note that w is positive if it is directed from the perpendicular line through M in the positive cyclical direction, MNO. In the link-chain of Fig. 95a, the changes in all the angles meeting at a panel- 15 226 ARCHES AND SUSPENSION BRIDGES. point must be added algebraically to get the total angular distortion at the point; in the arrangement of Fig. 95b, it is necessary to consider only those angles formed by the web members themselves. Construct the force polygon of the distances w with a pole distance p; then, considering these quantities as vertically applied loads, construct the corresponding funicular polygon; the ordinates of this polygon, measured from the closing side, will represent, to a scale of times that of the plotted elonga- tions, the deflections due to the angle-changes. The polygon will therefore be an influence line for H. In similar manner we may represent the denominator 6, of the expression for H, i. e., the displacement of A in the direction A B produced by a force 1.seca. The part of this displacement due to the angular deformation will be given by the intercept upon the chord A B of a funicular polygon constructed for the quantities w considered as forces acting parallel to AB, with a pole- distance p. cos a (since the quantities w were determined for a force of unity). In Plate I, Figs. 3 to 3°, this procedure is shown applied to the same structure as that used in Fig. 2. The linkwork was formed of the web members according to the scheme of Fig. 95b. The quantities w were obtained graphically from the elongations in the table (column 5) which were plotted to the same scale as that of the Williot Diagram (Fig. 2°). All the values of w were of the same sign (negative), indicating that all the displacements were upward; this corresponds to positive values for H. Furthermore, having assumed k = 5 meters and the pole distance p10 meters, the funicular polygons Figs. 3° and 3¢ give the displacements to a scale of a times that of the elongations, i. e., to the same scale as that of Fig. 2. To the deflections caused by the angular deformations must be added those produced by the pure elongations of the mem- bers in the chain. These are scaled from a simple Williot Diagram (Fig. 3°), obtained by the vector addition of the elongations of the members, to which is attached a small repro- duction of the structure to provide for the rotation of the system back to its horizontally constraining abutment. The vertical deflections of the panel-points of the framework are taken from this diagram and (reduced to > scale) are annexed to the Deflection Curve (Funicular Polygon, Fig. 3¢); similarly, the corresponding horizontal displacement 8”,, of the point A is added to the displacement 8’,, obtained in Fig. 3° from Two-Hincrep FramMep ARCH. | 227 the angular distortions. We thus obtain the quantity - Saa =5 (8’aa + 8’nn) as the magnitude of the load G which produces the values of H given by the ordinates of Fig. 34; G thus constitutes the scale or unit-load of the H-influence line. The two procedures a) and b) yield identical results; in general, method b) permits of securing a somewhat greater degree of graphic precision. ce.) Simplified Method for the Determination of H. The determination of the horizontal thrust HW will be substantially simplified if the elongations of the web members be neglected and only the effect of the chord members considered. The web members have relatively but a small share in the total deformation of the system, so that this simplification will usually be within the permissible limits of accuracy. It can be intro- duced in the analytical determination of H (paragraph 2.), as well as in the graphical procedures (a) and (b). It is ihus necessary to find the stresses uw and the resulting elongation terms u=r.u merely for the chord members. The chord stresses wu (produced by the force 1 . sec a) may be obtained either from a force polygon or by computation. Consider the chord member lying opposite to the panel-point m (Fig. 97); let om be its inclination from the horizontal, y,, the vertical height of the point m above the arch-chord, and hm the vertical depth of truss; then the stress in the member will be Um = e secom Be ge ah a Aa Faeairatto et as GS lal aa lata Bae (449. Again applying method (b) on the basis of a chain of links consisting of the web members, the weight w, to be applied at m and representing the angular distortion, is given by w= rw. = where d is the perpendicular distance of m from the chord member and k is an arbitrary constant. Putting k=1 and d =hy COS om, We have Sm U m Ww yw m he Tim 30 Om or, with a, as the length of the horizontal projection of the chord member, __ am Um 2 Am hm Wn Introducing an arbitrary standard section A,, the quantities 228 ARCHES AND SUSPENSION BripGEs. w may be replaced by the new numbers v = A,.w, whence Ac m Ac m m ta 5 Um $06? Om—= 4 * ao ; j (451. We can now obtain the ordinates for the H-influence line as the moments M, producible in the framework when acting as a simply supported beam and loaded vertically at the panel- points with the weights vm; while the unit load G will be given by vm Ym; accordingly, My, Fe ete eeeeeeeeeeee scenes (452. This relation could also have been obtained from equ. (448). Fig. 97, The signs of the quantities v need no further consideration as we have already seen that the angular deformations w will have the same sign for all the panel-points of the upper and lower chords. The graphic method again involves the construction of two funicular polygons, one for the forces v acting vertically and a horizontal pole distance p’, the other for the forces v directed parallel to A B with a vertical pole distance p’ (Fig. 97). In Plate I, Figs. 4*-4°, this simplified determination of H is carried out for the framed-arch treated in the preceding illustrations and, for comparison, the resulting influence line for H is in- troduced in Fig. 3¢ in dash and dot lines. For this purpose, the pole distance p’ was so chosen as to make the funicular polygon Fig. 4° yield an intercept identical with the distance 1 ‘ > S22 of Fig. 3, so that the same scale unit @ will apply to the H-polygon. The discrepancy with the exact determination of H amounts to 15% in the case considered. Two-HinGep FRAMED ARCH. 229 If the loads in the arch act only at the panel-points of one of the chords, then the H-influence line will have in common with the funicular polygon of Um only those vertices which lie under the panel-points of the loaded chord (Fig. 97). If the framework contains vertical web members, then the values of vm belonging to panel-points in the same vertical line must of course be added together (Plate I, Fig. 4). If the framework consists of radial posts and intersecting diagonal struts (Fig. 98), then instead of the panel-points we must use the points 0 and u determined by the method of Fig. 88; for these points, if the chord-sections are - and A, respectively, the quantities v must be calculated as ollows : Ac @uYfo Vo = - - seca a ae : (453. a ore Lee ease Sareiees aa dee ae The same applies to structures consisting of simply inter- secting diagonals. If h, the depth of the framed arch, is small compared to the rise, and if the cross-sections of the upper and lower chords are put equal to each other (compare sub- sequent remarks on this matter), then it will generally be sufficient to use the approximation vot w= 244 SEY" sectam .......... (454 Vm where Ym, Nm, @m, S€C om refer to the axis of the arch and to the point of intersection of the two diagonals. This load, Vo + Vu, is then to be considered as acting in the vertical line through this crossing-point. The horizontal thrust produced by a displacement of the abutments of Al and a uniform change in temperature will be, by equ. (447), d Fi aS OAS ae cence cee fe In the braced arch with straight upper chord, the brac- ing near the crown is commonly replaced by a full plate web (Fig. 99). For this central portion, equ. (250) of the theory of the Plate-Arch-Rib may be used for calculating the hori- zontal thrust; if A, is the 230 ARCHES AND SUSPENSION BRIDGES. average cross-section of the arch-rib and s the length of its axis, the resulting expression for the horizontal thrust will be = Me Gots (456. Syv + 3YmUm+$- Here Syv refers to all the panel-points of the framed portion, the values of v being calculated by formula (451); 3 yma, on the other hand, refers to successive sections of the plate-rib, Ym Yrepresenting the ordinate of the axis of the rib and Um being figured by equation (255), viz., m Ae m1 Un=F- (2ym+Ym1) 5 +3 (2Ym+ Ym) F— »- + (457. M, is again the moment produced by loads of aa v acting on a simple beam of span /; so that, as before, the influence line for H will appear as the funicular polygon of the v-forces. Fig, 100. d.) Correction of the Preceding Method of Determining H by Introduc- ing the Effect of the Web Members. Pass a vertical section through the diagonal dm (Fig. 100); then, for a vertical end-reaction of unity, the con- dition of static equilibrium requires the following relation between the hori- zontal components of the stresses in the three rnaembers cut by the section: Zq Stn B= — Zm_1 COS Ou — Zm COS Oo, or e (- aes Mn dm a Lm-1 ae ‘1 Ga hm 1 dm-1 oS hm-1 hm am.’ for a horizontal end-reaction of unity, e , a= — (1+ tim-1 C08 ou + ttm C08 G0) gst 9) oi am_1 ait Ama hm dm-1 If Aa is its cross-section, then the contribution of the diagonal d, to the summations Srzu and Sru? of equ. (448) will be given by 1 d'm [ Bm Symi ue) ym Zu=-—: = = ee ia ak bin, iy ies os is —*)] r= 1 @ = [ (42) m Yoon yma Ym ynt Aa a’ m4 hin 1 i) hm_ 1 hm_1 hm hm Noting that exactly similar expressions may be obtained for the diagonal dn,, (ascending toward the right), whose cross- Two-Hincep FramMep ARCH. 231 section may be denoted by Aj, a comparison with equ. (451) will show that the influence of the clongations of the web- members may be taken into consideration by augmenting each panel load, vm, at the mth panel-point by an amount: e ben m me c Bing m ae ee ee ym jae 1 (9m yon “).. (458. Aa @mihm \hn hm, Ad a@m.im \hm hima and at the (m—1)th panel-point: Ac dn yma Ym AUVyqn-}=— - = 2 un) ml Aa @ma-hnaNhma hm (459 ++ Ac , Bins Umar _ yas) : 7 Aa’ a’my hm \ hms Nim_2 In addition, the denominator & vy must be increased hy A vm. Ym and AUm,-Y’m41, respectively. These formulae are easily modified for the case in which one set of web members are vertical. (dn. =m, = .... =0). 4. Tentative Estimate of Cross-Sections. In evaluating H it is necessary to know the areas of cross-sections of the indi- vidual members. As these, however, remain to be determined by the design, we are compelled to provide for a recomputation ; and, for the first determination of H, we must make a pre- liminary estimate of the cross-sections of the members. As shown above, the chords have a considerable effect upon the magnitude of the deformations while the influence of the web members is much less. It will therefore be advisable first to determine H by the simplified method in which the elongation of the web members is neglected; or else, if it is desired to include the effect of these members, a uniform section may he assumed for all of them. The assumption of uniform sections may even be admissible for the chord members, in a preliminary computation. Designs actually carried through have shown that the error of such assumptions is negligible, particularly in arches having the hinges in their neutral axis and with chords more or less parallel, as well as in arches of crescent shape. Arches with straight upper chord or those having their end-hinges in the intrados, will of course display a larger variation in chord- section. In order to obtain the closest values of H by the first computation in such eases, it is advisable first to determine the chord-sections as those of a three-hinged arch whose crown- hinge is somewhat above the mid-point between the two chords. If we assume a constant cross-section for the upper chord and another for the lower, then only the ratio between these values will be required in the expression for H,; the mean cross-section, however, must be known or assumed in evaluating H;. Observ- ing, furthermore, that the elongations in the vicinity of the crown of the arch are most effective in influencing the value 232 ARCHES AND SUSPENSION BrIDGES of H, we should therefore use that ratio of cross-sections which will actually obtain near the crown. Various values may, of course, be adopted for this ratio, but only a certain particular value will yield equal intensities of stress in both chords. Actual designs have shown that in framed arches with either parallel or non-parallel chords, subjected to a moving load, the most effective value for the ratio between the chord-sections at the crown is approximately unity. It is therefore advisable, for a first design, to assume equal sections for the two chords at the crown; and, in arches having their end-hinges in the neutral axis, the chord-sections may be assumed uniform throughout. In an arched truss subjected to a moving load, as in all other statically indeterminate structures, it is possible to determine the position of loading .for maximum stress in each member and thus obtain directly, without trial, the required cross-sections so that all parts of the structure shall be proportioned for equal intensities of stress. Such a design, however, would not only be impractically laborious, but would also be of doubtful value since there are many other sources of error such as non-uniform elastic coefficients, erection stresses and uncertainties in the temperature stresses. We will therefore always employ the above indicated approximate method based on a provisional assumption of cross-sections and, if necessary, go through a recomputation based on the results of the preliminary design. It is a different matter, however, with an arched truss to be designed for a fixed position of loading. In such a case, as demonstrated in the paper cited below,* it will not, in general, be possible to secure equal intensities of stress in all parts of the structure, except with such perfect erection as is very rarely attained. Such structures may be designed 8 A for the individual members as to satisfy the equation of condition 8 dS A dH cross-sections will therefore be superfluous; however, as the value of the statically indeterminate reaction or of the stress in one of the redundant members may here be assumed arbitrarily within certain limits, namely so as to exclude any consequent change in the sign of the stresses, it will be possible to obtain a series of differently proportioned structures which shall all be stressed to satisfy the above condition. (Compare the previous citation.) by a simpler method which consists in so choosing the stresses + o = ‘s=2 (tous) =—EAI. The provisional assumption of 5. Computation of Deflections. In the framed arch, let a load P be applied at any panel-point C whose horizontal distances from the two abutments are é and &. Let H ¢ be the horizontal thrust producible by a load of unity at C. In order to calculate the consequent deflection A y of a panel-point D, distant x and #’ from the two abutments, let us consider a unit load applied at D and determine the resulting stresses zx in the various members of the statically determinate, i. e., freely supported, truss. If the stresses in the framework pro- *«Beitrag zur LBerechnung statisch unbestimmter Stabsysteme.” Zeitschr. d. dsterr. Ing. und Arch. Ver., 1884, No. 3. Two-HiIncep FRAMED ARCH. 233 duced by the load P at C are denoted hy S=P (4; + H;.u), so that a .S represents the elongations of the individual mem- bers, then the principle of equality between external and internal virtual work will give the equation Ay. 1=358.2= 5 [Srz-2x+Hesruzx] which may be rewritten either as I 1 Ay=— [-2r2 ZetSruzx| Hy .....6.. 460. I=F H, £ | é ( or, since & ruwz,—=-— H, & ru’, as Aye [Srzg2x— Hg HsSrw] oo... (460°. The latter form will be suitable for computation if the thrusts H, and H,, which are produced by unit-loads at C and D, have already been determined by equ. (448) or in any other way. The summation %rz;zxz may likewise be computed in terms of the series of stresses z, which are produced in the members of the framework by an upward unit reaction at the end A, by means of the formulae: wb ® é wiz wet 1 for Srz'+ pw r2?4+ FR rz, a oe = .. (461. for >a, Srigax = TES rz? Spare are 0 x In a completely loaded span, the crown deflection may be found with sufficient accuracy from the average of all the panel deflections. If the panels are uniform in length, if P is the load per panel-point of which there are n, and if S denotes the stresses produced in the system by the application of a unit load at every panel-point, then, on the assumption of a parabolic deflection polygon, the crown displacement will be given by equating the external and internal work as follows: (Here S and z are identical.) 3 Pir? Ap=F MB CUT t ttre erate eee The horizontal deflection A x of the panel-point D (positive if inwards) produced by a vertical load P applied at C is given by the principle of virtual work as L.Ar=35- 8,w, of As=> [Srz,u’ +H, Xruu’]....... wee. (463. “Yere wu’ denotes the stresses produced in the members of the 234 ARCHES AND SUSPENSION BRIDGES. simply supported structure by a horizontal force of unity, directed inwards, applied at D. These stresses, again, may be determined either analytically or by means of a force polygon. The following equations will serve to determine the dis- placements produced by temperature changes and the simul- taneous yielding of the abutments. The vertical deflecticn of the panel-point D will be Ay=32x(ots+ H.u)—ot3szx +4 Sruzy. Introducing the value of H from equ. (446), Al+twtZtsu Zru? Ay: =ot3sz,— “SruZy=otSd $z,— (wtSsu+Al) Ay. Here s denotes the length of the individual members, and H, is the horizontal thrust produced by a unit load acting at D. According to Mohr’s Theorem, cited on page 222, we have ZS su——lsec?a, also 3 57z,—0, so that the formula for tem- perature deflections becomes Ay = 2 3ruzr= (wtlsecta—Al).Hx .... (464. The horizontal displacement of the panel-point D on account of temperature changes will be, An=3u'(ots +5 A, w) =otSu's+% Sruu’ or Ag—= tot dw’ s+(wtlsec?a—Al) aru ear aee (465. In the graphic procedure, our design will be to determine the influence lines for the horizontal and vertical displacements of an arbitrary panel-point D. These influence lines (by Maxwell’s Theorem) represent the sag-diagrams or elastic curves of the framework for a vertical or horizontal load applied at D. Each of these elastic curves may be obtained: a.) By drawing the Williot Displacement Diagram. For this purpose it is necessary to determine the stresses in the members, produced by a vertical or horizontal unit load applied at D, and to ecaleulate the resulting elongations. b.) From the funicular polygon of the angle-changes con- sidered as vertical loads. Here the angle-changes may either be determined directly from the strains in the members of the structure or else they may be taken as the resultant of the angle-changes in the freely supported truss and those produced by the force H,.seca acting along the chord AB. The latter quantities yield an elastic curve which coincides with the H- influence line; and, in fact, if the construction described in Two-Hincep Framep Arc. 235 method 3.b) is followed, these (negative) deflections will be given by the ordinates of the H-curve if they are measured k p.Hx seca to which the strains were plotted. It therefore suffices merely to compute the elastic curve for the simple truss loaded at D. For this purpose, we must compute the stresses z, and the resulting changes in length and angle of the chain of members; laying off the angle-changes upon a radius k, the resulting lengths w are to be treated as vertical forces applied at the panel-points. For these, construct a funicular polygon with a pole distance of p . H.. seca and correct the resulting ordinates by the small vertical deflections which correspond to the elonga- 1 . seca ? to a scale whose unit is times that of the scale tions of the members multiplied by — i. e., to YZx. and which are obtained by means of a Williot Hx seca? diagram (similar to Fig. 3°, Plate I). The differences between these corrected ordinates of the funicular polygon and the ordinates of the H-influence line will then give the actual E times that to which p.Hxseca vertical deflections to a scale of the strains were plotted. If the simplified method 3.c) has been used for obtaining the H-curve, the latter being constructed as the funicular polygon of the panel loads vm (equ. 451), then in determining the elastic curve of the simple truss there again need be con- sidered the strains in the chord members only. The loads v’m, proportional to the angle-changes in the linkwork, are then to be determined by the formula — Ac AmZx 2 Um =F SCOT sa eee e eee e eens (466. where Zx is the stress in the member m of the simple truss loaded with P=1 at D. With these panel loads and a pole distance p. H,, construct a funicular polygon for vertical load- ing; as before, the differences between the ordinates of this polygon and those of the H-curve, will give the deflections of AcE times that of Ax Pp the framed arch to a scale whose unit is the scale of lengths. The horizontal deflections are similarly obtained, if for z, in the above construction are used the stresses which are pro- duced in the framework, considered as a simple truss, by a unit horizontal force applied at D. If the structure has a central hinge, we again first find the elastic curve for any point D from the stresses and strains in 236 ARCHES AND SUSPENSION BRIDGES. the three-hinged arch, assuming, however, an unyielding mem- ber (D E, Fig. 101) to be connected across the crown-hinge so that its angle-change will be o =O. The stresses in the mem- bers will be S=z,-+H,.u, where H, is the horizontal thrust of the three-hinged arch produced by the unit load applied at D. With wo, = 0, these stresses will induce a horizontal displacement of the end A relative to B which may be calculated by the formula, $= SrSu—Srzu+H,. Srv? Fig. 101. or, since Srzu=— H, . Srv’, 3. = (Hy — Az) Sru?=(Hx— Ax) 8. Here H, is the horizontal thrust of the two-hinged arch formed by introducing the rigid crown-strut, and 8, is the corresponding horizontal deflection of A, produced by a unit Two-HIncep FrAMED ARCH. 237 horizontal force acting at A. The quantities H, and 8, are to he determined by the graphic procedure described under 3.b}) and 3.¢). On account of the ends being anchored, the displacement 3b cannot take place; there must therefore occur, at the crown, an angular rotation amounting to This will produce deflections which may be represented by the ordinates of a triangle constituting the funicular polygon for a load w. applied at the hinge. In Fig. 101d, there is first drawn the deflection curve of the three-hinged arch for a unit load at D. This curve is obtained, by the method described in 3.b), as the funicular polygon of the panel loads w’; these are obtained from the strains produced by the stresses S of the three-hinged arch, using a unit k—20 meters and assuming w’,=0. According to equ. (467), the load now to be assumed at the crown-hinge is Wem k.we=k. (Ha— He) =F (Yes) nies (467°. where y; and y, are the ordinates of the two H-curves measured to the seale of 6, == 1. Combining the funicular triangle of the Joad w, with the previously constructed deflection polygon, there will be obtained the total deflections of the three-hinged arch. (ig. 101d.) 238 ARCHES AND SUSPENSION BriIDGES. §32. The Framed Arch with Tie-Rod. If the ends of a two-hinged arch are joined by a tension-rod to take up the horizontal thrust (Fig. 102), then, in calculating Fig, 102, Ae ee ee the value of H, the effect of the stretching of this tie-rod must be considered. If A, is its cross-section and l its length, so that Al= = 1 is its elongation, then equs. (444) and (445) will give for an arch with ends at the same level, ZrZ TPSma Hoa en oe (468, SHOE AT ae San + slo The H-influence line, constructed according to the preceding paragraphs, requires no modification, but the unit load deter- mined by the displacement 5,, must be increased by 1. ae Compared with the arch without a tie-rod, the value of H is : 3 1 r reduced in the proportion of 1: (a Tee . ai). A uniform temperature change in both arch and tie-rod will, in this case, produce no stresses. Tue Framep ArcH witH TrE-Rop. 239 If the tie-rod does not connect the end hinge-points, but joins the two panel-points #F (Fig. 103) instead, then the tension produced in it will be = : ZPS8me oe ree has P ‘ Zrutt — he si ae oT where uw’ represents the stresses in the framework produced by a unit tension along £ F, 8 represents the accompanying horizontal displacement of HF relative to #, and 8mne denotes the similar horizontal displacement produced by a unit load applied at M. The latter symbol also represents the vertical deflection of M produced by a unit horizontal force in EJ’. As this ten- sion H produces stresses and strains only in those members lying between # and F’, the angle-changes of the linkwork and the panel loads w or v derived from these are also limited to the panel-points between H and F/. The H-influence line obtained as the funicular polygon of the loads w will be straight in the portions A, #, and F, B, (Fig, 103), 240 -ARCHES AND SUSPENSION BRIDGES. § 33. Other Types of Framed Arches with Hinged Abutments. 1. The Free-Ended Cantilever Arch. The horizontal thrust at the abutments will produce stresses and strains only in the members situated between the abutments. Accordingly, the loads w or v, employed for the construction of the H-influence line, will be limited to the panel-points between A and B and the H-curve will continue beyond these points in straight lines. (Compare the case of the Arched Rib, Fig. 74.) If the free end of the cantilever arm supports a simple truss (suspended span), so that the arch forms part (anchor span) of a ‘‘Gerber’’ Bridge, Fig, 104, then the course of the H-curve is easily specified (Fig. 104). The deflections of the panel-points of the arch, as well as those of the cantilever arm, are easily obtainable by the procedure presented in § 31:5. 2. The Cantilever Arch with Ends Fixed to Horizontal Rollers. If the ends of the cantilever arms of an arch are constrained to move in horizontal planes, there results a three- fold indeterminate system. We will choose as our unknowns the horizontal thrust of the arch H, and the vertical reactions, D and HE, at the end-supports. By introducing a central hinge in the arch, we may reduce the number of unknowns to two. The equations of condition are obtained by putting the hori- zontal displacement of A, also the vertical displacements of D and E, equal to zero. In a free-ended cantilever arch, horizontally movable at A, let there be denoted by 8ma, the horizontal displacement of A relative to B | Produced by 8ma, the vertical deflection of D a unit load at 8ne, the vertical deflection of EF any point Af. Tue FRAMED CANTILEVER ARC-I. 241 8a = 8a, the deflection of D acting horizontally -in- Sao = Sea, the deflection of L ward at A. Saa, the deflection of D Ana : 3 hia, the deduational i \ Produced hy a unit load at D. See, the deflection of H, produced by a unit load at HL. 8a, the horizontal displacement of A feet by a unit force Then the equations of condition for H, D and E may be written, SP .8na tA .dsat D.8aa +E .8ea= SP .8me t+ A. de +D.8ae+ LE .8e50e=0 For the analytical treatment we will determine the series of stresses wu, wu, and u,, which are produced in the members of the statically determinate structure when there act upon it SP See i Bh et siag CAOD: Fig. 105. the forees H—1, D=1 and E=1, respectively. If the external loading produces stresses Z in the determinate struc- ture, then the accession of the forces H, D and EF will render S=Z+Hu+Du,4+EL£u, and the theorem of virtual work will therefore yield the follow- ing equations of condition, equivalent to equs. (469), SrZutAsrwtDsruuteHsruu =9 SrZu, +H SruuitDsru2+ESru,u,=07-- (470. SrZu,tHsruwtDdDrsruw.+HSru’?=0 The coefficients in these equations may be evaluated either analytically (equs.470) or by means of a graphical treatment. 16 242 ARCHES AND SUSPENSION BRIDGES. In the latter method, it is necessary to first construct the curve of deflections produced in the framework by a unit force acting along the chord A B (Fig. 105). In this curve, the ordinates under M, D and E will give the quantities 8ma, daa, Sea; then, by use of the graphic method described above (§ 31:3), there is to be found the horizontal displacement 8,2. In a similar manner, construct the deflection curve of the freely supported cantilever arch for a vertical load of unity applied at D, to obtain the deflections 8a, 8aa, and Sae. Finally, a deflection curve constructed for a unit load applied at KF will give, at the points M and £, the deflections 8me and Sec. If the structure is symmetrical in form, it is seen that the third deflection curve need not be drawn. There are thus obtained all the coefficients in equations (469), and solving these will give the influence values of the unknowns H, D and EL. These may then be used in the equation S=Z4+Hu+Du,+E£u, to determine the influence values of the stresses in the individual members of the structure. If the structure, instead of being anchored at the abutments A and B, is provided with a tie-rod joining these two points (Fig. 106), then the first of the equations (469), representing the horizontal displacement of A relative to B, is to be modified l BA, where I is the length and A, the cross-section of the tie-rod. by replacing the zero in the second member by — H Fig. 106. pe aaemmaae jae EN B YSRBBWSE f-7 ak De cm If the structure has a central hinge, the problem becomes simplified since the mere omission of the end anchorages will make the structure statically determinate so that there are but two unknowns, D and E, to be determined. If the bridge is symmetrical, a single deflection curve will in this case be sufficient, namely, the deflection curve of the three-hinged arch for a unit load applied at D or EF. This type of bridge is treated in greater detail in the paper cited below.* 3. The Framed Arch with Anchored Ends. This arrangement sketched in Fig. 107, proposed by Engineer A. * Melan, Bogentrager mit vermindertem MHorizontalschube. Osterr. Monatsschr. f. d. offentl. Baudienst, 1897. Tue FramMep ArRcH witH ANCHORED ENDs. 243 Schnirch,* consists of an arched structure, with the ends A and B on rollers having the framework extending outward beyond these abutments and anchored by the ties C A’ and DB’ to the deepest practicable points of the picr-masonry. This anchorage corresponds, in effect, to a lowering of the end-hinges; the structure may be treated as an ordinary two-hinged arch with its abutments at A’ and B’, the portions 4 A’ and BB’ heing considered as members of inelastic and unyielding material. This arrange- : ment, by increasing the arch-rise, Fig. 107. affords the advantage of a redue- tion in the horizontal thrust, es- pecially in that due to changes of temperature; in consequence, there 9 may be secured a marked saving of material in the masonry of the abutments, and, under certain con- ditions, in the superstructure as well. This anchoring will be distinctly advantageous in flat arches. The horizontal thrust may be calculated by equ. (448). The stresses z remain the same as in the arch supported at A and B, whereas the stresses u must be determined for a horizontal unit load applied at A’. Graphically, the influence line for H may be determined as above by means of the deflection curve for a unit force at A’. Similarly, the displacements of the movable ends, A and B, may be found either by the graphic method developed above, i. e., by constructing the deflection curve for a horizontal load at A or by the analytic method of equ. (463). If any loading is placed on the structure, producing the stresses S = Z + Hu, the resulting combined displacement of the two rolling endg will be, by equ. (463), aaa Ke. d Re hee ee --5 f Ala [Er Zw tH Eruu’] see RY MOR RE 4 SR (471, The stresses w’, i. e., the stresses in the members producible by a unit force acting along AB, may be determined either by a force polygon or by the relation wu’ = .u where y is the ordinate of the moment- y yth center for any member measured above the arch-chord. The roller-dis- placement producible by temperature change is obtained by equ. (465), with Su’s—=—lI and Al=O, Zrw— Zruw A ; a An=op— Se (Zrw—Zruu’) .... (472, * Zeitschr. d. sterr. Ing.-u, Arch.-Ver., 1884, p. 184. 244 ARCHES AND SUSPENSION BRIDGES. § 84. The Continuous Framed Arch and the Braced Sus- pension Bridge of Multiple Span. This group of bridges corresponds to that of the arched ribs treated in § 26 and is found exemplified particularly in suspen- sion structures. The defining condition for this type of con- struction is that the arches (or suspension spans) shall be hinged together at the intermediate piers and capable of hori- zontal displacement at these points, while the ends must be securely anchored. Such framed structures, with reference to their external reactions, are of single static indetermination. Fig. 108, They may be rendered determinate by introducing a central hinge in one of the spans. 1. We will first treat the latter case, particularly in the form of a three-span Suspension Bridge with a Central Hinge. The continuous tension chord is to be conceived as hinged at A and B, while the bracing is interrupted at these points of support. - The method of support at these intermediate piers is to be either by means of rollers or rocker-arms; the ends of the bridge are held fixed by the roller-supports and the oblique anchorage. a.) Central Span. The stresses in the main or central span are not affected by any loads in the sidespans. Loads THe Bracep SUSPENSION Bripce or MULTIPLE Span. 245 in the central span will produce stresses according to the principles of the three-hinged arch. In Fig. 108a, B’ 0’ J, O A’ represents the influence line for the lower chord member r w. The factor by which this influence area is to be multiplied to give the stress in the member r wu produced by a loading of p tnits per unit length, is p.—4. With a parabolic upper chord, The P O would lie on the horizontal line through C’; the two triangles A’ C’ B’ and A’ O B’ would then be of equal areas, and the positive area A’ O J, would thus be equal to the negative area B’ C’ J,.. The lower chord, under this assumption, would there- fore receive no stress from a uniform load covering the entire span, and its greatest stresses would be Limax—= — Dmin—=p. (Area A’ OJd,) — In Fig. 108@ are also drawn the influence lines for the upper chord member os and the web member mn. With the aid of these we may determine the limiting stresses producible in the upper chord by a uniformly distributed dead load of g per unit length and a similar live load of intensity p per unit length, by the formulae Unax—=[(p +g) Area B’C’ J,—g.Aread,RA’ = Seco Umin= [g- Area B’ 0’ J,— (p-+g) Aread,R.A’| hs seca and the corresponding stresses of the web member mn, by the formulae Daax={(p+g) (Area B’NJ,+AreaJ,C’ A’)-g. Area J, M J,] Dain=[g (Area B’NJ,+ Aread,C’ A’) - (ptg) Aread,M J,] 8 Simple expressions may be written for the chord stresses in the main ‘pan. First, for any form of cable, if Ug and Le, represent the stresses due to dead load, we have Li=[$ [Cee 2fl —1)-£. 4 4]2 nn 8 lyt+2fe fy he Fl 1 Ties = De HS (sae 1) a _ly—2fa yy 8 lyt2fa fF ho =f a? 2f1 _ thai 2fa | nya v.=[- 2 Walt 2a —1)+5 lie ote Ff hie Umax = ee Lmin= Le —— . a2 fe ky 1 lat ofe Ff be QFL 1 min = aa —1 eos . Umin=Us — Pe ~Gfoe- - ce seco 246 ARCHES AND SUSPENSION BRIDGES. With a parabolic upper chord, the above expressions for the lower chord stresses reduce to Le=0 pa(l—ax) (I-20) 2(31—2a) ho For the web members, the expressions for the limiting stresses become very complicated in this case; it will ke more convenient to apply equ. (442), using the values of Mz and H corresponding to the critical loading. Dmax = — Lmin= b.) Side Span. The construction in the side span is to be treated as a simply supported truss subjected to the full hori- zontal tension arising in the main span. A load in the side span will produce the same stresses as in a framework simply resting on two supports. The upper chord will therefore receive its maximum compression when the side span is completely loaded and the main span carries no load at all; and the maximum tension is produced when the main span is completely loaded and the side span is free from load. The opposite law of loading holds true for the lower chord. If H is the horizontal tension and if M is the bending moment for the loading in the side span, then the stress produced by any loading of main and side span in the upper chord member 0’s’ (Fig. 108) will be x U=—(M—H.y) $= — (5 and the stresses in the lower chord oe r’ w’ will be = (w= joe (a ee L=(M H.ys') hs’ -(— i hs’ By these formulae, the stress influence lines may be readily constructed. These are represented in Fig. 108), to one-half the seale of Fig. 108a, for the indicated chord members of the left-hand span; they are obtained by making ae te on -l. Here v’~ and x’, are the horizontal distances of the panel-points ‘7’ and s’ from the intermediate pier. The largest stresses producible by an assumed uniform dead load of g in the main span and of g, in the side spans together with a uniform live load of p per unit length will be, Umax=[ (p-+g) Area A’ C’ B’—g,. Area D’ R’ A’) seca. the vertical line of the pier = = 1 and ys’ Umin= [g. Area A’ C’ B’— (p+ g,) Area D’ R’ A) - seco. Imax [ (p+ 91). Area D’ 8’ A’ —g. Area A’ C’ B’] we. Lmin= (91. Area D’ 8’ A’ — (p+g).Area A’ C’ B’] te THE BraceD SUSPENSION BrincE oF MULTIPLE SPan. 247 The stresses in a web member, e. 5 wn’ with. have the ’ g, , general value —H) . 7 ’ where M,’ and y,’ refer to the point of cia 2’, of the two chord members belonging to the panel containing the diagonal. If 2’ falls within the span, then the same law of loading obtains for the web member as for the chords, viz., the limiting stresses arise when either the main or side span alone is completely loaded. The influence lines for this case are shown in Fig. 1080, in the right-hand span; the resulting stresses in the diagonal m/’n’ will be DS= (My= Ay += Dmax=[(p +g) Area A’ C’ B’ —g,. Area B’ N’ M’ E’| aa . Dmin= [g.AreaA’ C’ B’ — (p+g,) Area B’ N’ M’ E’| ai The analytical expressions for the chord stresses in the side spans may be written: 1 P s’ 1 Lg= [> 91 as’ (ly — as’) =9 “Ee hee 1 : pi Lmax = Le+ @ pats’ (—as') ae Pys’ 1 Lmin= Lg — p “3F_ he? 1 yr | seco ve=—[ 5 2 (uae) =o Be I he ae sec o Umax = Ug + Pop he? 7 1 seco Umin= Us — 9 par’ (i—ar') ae The web stresses are to ke calculated by means of the general formula (442). 2. Continuous Braced Suspension Bridge without a Central Hinge. If none of the spans is provided with a hinge, then the framework in each span acts as a two-hinged arch whose horizontal thrust H is conditioned by the displacements of the end-points. Let us first consider each of these arches as independent, and let P,P’, P” ... = loads in the 1°, 2°, 3°, ..., spans. P81ay P’8’ may P’8 ma, -.. = the relative horizontal displacement of the ends of each span when acting as a simple truss under the above loads; considered positive if outward. 248 ARCHES AND SUSPENSION BrIDGEs. 8aay 8’an, Oa, --- == the horizontal displacements of the ends of each span produced by unit forces acting horizontally outward at these points; accordingly FP bma + ve 82a, P’ oma + H’ 8’ aa, Pp” Oma + H” O" aa ... = total horizontal displacements of the ends of each span. Then the total horizontal displacement of the anchored ends of the system must be — Ab= P8ma+P! 8m P83 ma -- +» HH (8004-8 cat 800 +--+) whence the following value of the horizontal thrust may be written : PSme+ P'8'mat P1d"mot.... Al H== — fin COG (473. The deflections § are to be found either analytically or graphically according to § 31; for the analytical treatment we use the following expression corresponding to equation (444) : H=— ZrZutlrZ’w+lrZu"+... +All a Zrwt+trut?+2ru” +..... ere Coes Here r= represents the ratio of length to cross-section of each member; Z, Z’, Z”,. ... are the stresses, produced by the actual loading, in the members of the 1°, 2°, 3°, ... spans considered as simple trusses; u, u’, wu’, ... are the correspond- ing stresses producible by a load of H=—1, or a force of 1, sec a acting along the chord of each span where a is the angle of inclination. A load in any span will thus contribute a positive horizontal force, viz., a cable-tension in the case of a suspension bridge. Having constructed the influence line for H, the influence lines ‘for the stresses in the members are obtainable in the same manner as with the center-hinged construction. Example. (Includes Plates II and III.) In the following we give, as an example, the complete design of G. Lindenthal’s project for a railway suspension bridge over the St. Lawrence River at Quebec. The main span is 548.02 meters ( = 1,800 ft.), and each of the two side spans -is 207.87 m. (= 682 ft.). The bridge consists of inverted two-hinged arches swung from rocker arms and anchored at the ends. The bracing consists of vertical posts with a double set of diagonal tension members. DESIGN OF A BRACED SUSPENSION BRIDGE. 249 TABLE I, » 2 % 8 Q ” ~ a 5 u y ly 1 az | *'s o 1 A |seec/100—| y h — |= seca|— seco : Aa 5 A h |h h . ~I = al l ttn | 19.07) 2579 |1.0093) 0.739 | 3.5853 | 0.0533 (0.1410/ 9.4966 XX: 67.24| 17.68 3.8045 Ug | 18.99 | 2579 |1.0051) 0.737 | 4.1223 | 0.0514 |0.1867]12.6456 XxX 67.1 | 15.24'4.3902 tag | 18.93 | 2579 |1 0934) 0.735 4.6300 | 0.0698 |0.2348 15. 7746 XXII 67.08] 13.87|4.8384 um | 18.91] 2579 |1.0014] 0.734 5.0463 | 0.0752 |0.2782/18. 6710 XXIV 67.04] 12.80.5.2400 ttos | 18.90 | 2579 |1.0003] 0.733 | 5.3689 | 0.0800 10.3146) 21.1120 XXV 67.04] 12.19 15.4945 tse | 18.90 | 2579 [1.0000] 0.733 5.4945 | 0.0820 10. 3300|422.1281 XXVI (11.0641) la | 29.43} 953 [1.5568] 3.080 0.7463 | 0.1023 |0.2347| 1. 7163 7.31) 15.2410.4794 hs | 20.96] 953 [1.1087] 2.195 0.7076 | 0.0676 |0.1051] 1.1248 14.19] 17.68'0.7970 la | 20.96} 953 11.1087] 2.195 | 0.9943 | 0.0581 |0.1276| 2.1977 5 20.36 20 42 0.961 ns 20.96 | 1018 |1.1087] 2.057 1.1610 | 0.0504 |0.1204] 2.7948 ve 26.50] 23.77 1.0976 le | 20.96 | 1018 |1.1087] 2.057 | 1.2412 | 0.0435 19.1110] 3.1851 eee. 31.30] 27.43 1.1414 .96 | 1018 |1.1087| 2.057 1.2594 | 0.0376 |0.0973| 3.2753 = ea) 2-087) 95 04 81.85 1.1306 .90 | 1277 |1.0000} 1. 1.2978 | 0.0339 |. 0651] 2.4812 ae 1.480) 49 9 | 27.43'1.4050 .90 | 1277 {1.0000} 1. 1.6545 | 0. .0962| 4.0349 a . 1.480) 9 a7] 23.7/1.8440| » DES? (0062 | 42038 18.90} 1450 [1.0000] 1. 2.0720 | 0. 1297] 5.5613 - ne 1.303} 16 95] 20.42'2.3000 coo .90 | 1625 11.0000) 1.16 2.6522 | 0.0528 0.1565] 7.5540 : reat 1.0000) 1.163) 44 50] 17.6812. 8015 DASGe Olde .90 | 1798 |1. . 3.0973 | 0. 1986] 10. 0545 i ou 0000) 1.051} - 65| 15.2413.3902 DASE, [ONE .90 | 1971 |1.0000] 0.958 3.6143 | 0.0389 |0. 2383] 12. 4950 * ae - 0-958) -3.29| 13.87|3.8384 eo vee x | 18.90] 1971 [1.0000] 0.958 4.0392 | 0. 2965] 15.6135 lass 54.25] 12.80/4.2400 pate 90 | 2146 11. : 4.3872 | 0. .307¢| 16.7 os 1.0000] 0.880] «| 5 saly gags} 42572 | 0-0800 |0. 307°] 16.7804 la {18.90 | 2146 |1.0000| 0. 880 4.4945 | 0.0820 /0.3115] 8. $853 1 Anchorage] 48.75 | 7426 {1.11901 0.655 : ela bet | Anchorage! 48.75 0.655 1.1130 a ¥%Main Span 13.9491 219.8551 931.6250 1. Influence Line for H — By equ. (473°), t= Zrw+2zru” ZrZut=rZ’'w+eE.Al Here Z and wu, with the usual significance, refer to the members of the main span, Z’ and wu’ refer to the members of a side span. For determining H, we will apply the approximate method (§ 31: 3. ¢), in which the chord strains alone are considered and the H-influence line is constructed as Design or A BraAcEpD SUSPENSION BRIDGE. 251 the funicular polygon of the panel loads v which are obtained, according to equ. (451), from v=r. He sec" o. In this expression, y represents the ordinate of any panel-point measured from the line joining the points of suspension of the corresponding span; h is the vertical depth of truss; o is the inclination from the horizontal of the chord member situated opposite the panel-point; and r= the ratio, length: cross-section of the member. In the arrangement under consideration, consisting of inter- secting tension-diagonals, of the two such members in each panel only that one should be considered as acting which is put in tension by the given loading. This would make the determination of the H-influence line very complicated, so that we adopt the admissible approximation of calculating each value of v from the mean of the values for the two diagonals, and considering it as applied at the mid-points of the chord- members. We thus obtain My SS es es Loyt 220'y' + 2 secta The extra term in the denominator takes care of the elongation in the anchor-chains (of length 1, section A and inclination a). The quantities entering into the calculation are given in Table I. In Plate II, Figs, 1—4, the H-curveisconstructed as the funicular polygon of the v and v’ forces and the quantities Dvy and Zv’y’ are obtained as the proper intercepts of funicular polygons. In the last column of Table I, the products v y are also computed and their sum is found to be 1 1 100 (= Zoytz vy’ + 7 secta) = 234.62. The pole distance of the v and wv’ force polygons having been taken as p = 3.875 times the unit to which the 100-fold v's were plotted, therefore in the expression H = = . G, where € is any ordinate of the H-curve, the 2X 234.62 3.875 the construction. The ordinates of the H-curve are compiled in Table II. value of N must be N = = 121.1 meters, which agrees with Panel- baie | Point TABLE II. ae OG aS I |u|] ivi{|v VI | Vil | VIII) Ix x XI > a, £ 1.95 | 3.20 | 3.99] 4.42] 4.79] 4.79 | 4.42 3.96 | 3.35 2.13 0 | 26 gt XII | XIII] XIV | XV | XVI/XVII| XVII) XIX| XX |XXI| XXII | XXIII] XXIV|XXV) = | | | | 22,10) 41.76] 60.66] 78.18] 95,21]111.10) 126.55 141,18 154.31 166 53} 177.32 | 185.86 | 191.47 |194.46, 1746.6 252 ARCHES AND SUSPENSION BRIDGES. Hence, if P is the load per panel-point, for a sidespan completely loaded, 36.91 Sige H=Ti7 P=°. P; for the main span completely loaded, 1746.69 os . H=2 -pit P= 798 P. The horizontal thrust produced by temperature is Ewt(L+2L,seca) . A= Sra ; here [= 548.0 m., L,= 207.87 m., sec? a = 1.0752, F = 2040 tonnes/em.’, 40 2 wot= “30,000” Ewt=1.02 tonnes/cem,’, so that 1,02 x 994.2 ‘= 330 934.62 216,16 tonnes, 2. Chord Stresses. The panel-points are numbered consecutively from 0 to Xlinthesidespan and from XI to XXV to the mid-point of the center span. The lower chord members will be denoted by L, the upper chord members by U, the diagonals inclined downward to the right by D, those slanting downward to the left by D’, and the posts by V. As an index to each member will be attached the number of the corresponding panel (1-11 in the side span and 12-26 in the center span), the posts being designated by the number of the panel-point. In Plate II, Fig. 4, are drawn the influence lines for the chord stresses in the main span. Since the upper chord conforms to a parabola, all its panel-points will have the same value of a the lower chord influence lines are therefore easily eonstructed, all of them being represented by triangles of the same altitude, viz. Le 1 548.0 Ep O= | ag X 121.1 = 302.75 m, The altitudes of the influence triangles for the upper chord stresses are given by — ==") 4, where z and y; are the coordinates of 1 Yi a lower chord panel-point referred to the panel-point XI. We thus obtain the following values: There were thus constructed the upper chord influence lines in Plate II, Fig. 4. ead as XI | xir | x11] XIV | XV | XVI ae XIX) XX |XKI) XXII | XxIII]XXIV|XxV AA als, | | 0 ee eee en Oe 180,03 ]196 69'211,14]223, 30] 233.07 | 240.31 | 245.18 |247.55 A . ~ DESIGN OF A BRACED SUSPENSION BRIDGE. 253 The limiting stresses in the members produced by a uniform load (p) will now be found from the areas of the influence lines. With u as the panel-length, let & = area of H-curve in the main span = 3493.38 ¢#, = area of H-curve in the side span = 36.91l@ F, =the negative portion of the influence area for a chord member in the main span, few eceepee eee LIM. Wee “ = area of the < triangle = M~ 29a ge Then, : Fmax=F+F,— ®?, Frin= F, + 2,, Smex=Paax.g 8000. Be, Smin=— Fria. 4.800 eae For all the lower chord members, F = 302.75 X 14.5.4 = 4389.88.a (constant). ‘For any member and any loading, that influence line should be used which refers to the panel-point belonging to the positively stressed diagonal in the panel. In Table ITI are calculated the chord stresses produced by the moving load. In the actual design, the moving load was p = 4464 kilograms per meter (= 3000 pounds per linear foot), or a panel load of pa = 4464 X 18.9 = 84,370 kg. Woe thus have —P# oe 696.7, so that the quantities > Fmax and = Fmin in the table must be multiplied by the factor 0.6967. % seco in order to obtain the chord stresses in tonnes. It was not necessary to draw the chord influence lines for the side spans, since just two different eases of loading need be considered in order to obtain the extreme stresses in all the members: one of the side spans completely loaded, or the main span and the other side span covered with load. The data for computation are arranged in Table IV. Remark on Tables III and IV: The oblique lines in columns 6 and 10, and 7 and 11, respectively, indicate which panel-point is to be the cen- ter of moments for each chord stress according to the diagonal considered acting. With reference to the dead load, it is assumed that this is carried directly by the tension-chord acting as a cable and consequently the bracing receives no stress from this source. This condition may be realized by making the diagonals adjustable in length and leaving them loose during erection, so that only the posts serve to transmit the load to the cable. Not until the entire construction, including the roadway, is completed, are the diagonals to be adjusted; they should be given a slight initial tension at a mean temperature of about 10°C. (= 50°F). An approximately uniform .distribution of the dead load corresponds to the parabolic curve adopted for the tension-chord and to the ratio of rises or versines in the main and side spans of fi ASL Ly, The stresses in the cable produced by the dead load of 8900 kg. per meter ( = 6,000 lbs. p.1.f.) are computed by the formula 2 U0, = — 8900, 21 seco 6096.5. seca (tonnes). These values are given in Table V. 254 ARCHES AND SUSPENSION BRIDGES. Taste III. Maximum Moving-Load Chord Stresses in the Main Span. a G an n = & als au n[s = OO & leslie fail] #] Bs | 8 2 ie | es |e] 2 1 ° i 8 oD o 8 E on o € “— |2& elle o> % 2 be 3 a 2 loo| & Sd Re Dg 5 s A als z als ry (m.) | (m) |] (m) (t.) {| (m) -(t) Ly 0.7463| 4805.9 _-— |0.7463] —378.0 xl 4389 ,9| 654 .5||1551.0] Dy 728.3) Da Lis ~~ |0.5315| 573.9 ~ 19,8836] —411.1 XIII 4389.9] 695.2)/1491.7) DisDig 669.0] D’13 D'14 hea “* J0.8836] 917.7 {1.1050} —488.3 xIVv 4389 ,9| 536.7||1433.2| D.,Dac 610.5] DDis Lis “XY |1.1050] 1102.7 “LY |1.1050] —488.3 xv 4389 ,9| 477.1||1373.6] Dy, D'ie 550.9] DisDie Lis — |1,2655| 1157.4 A {1.2169} —466.3 XVI 4389,9| 417.0||1313.5| D’y_ D'1 490.8] Die Diz Ly DOD, |b 2584] 1091.7 “\" {..2655] —431.9 XVII 4389 ,9| 354.2||1250.7] D7 Die 428,0| Diz D'e Lis —* [1.1305] 984.6 pop,,|t 4950] —374.0 XVII 43899] 293.3||1189.8] Dis Dio 367.1] De D's Lio SZ [1.4650] 1213.9 11,8440} —391.9 XIX 4389,9| 231,8]|1128.3| Dip Dap 305.6] Die D 20 Lao - |1.8440] 1449.0 _—~ 12,3000] —395.8 XX 4389.9] 173.6]|1070.1] Dao Dax 247.4] D’q Day La {2.3000 1714.3 ~* |2. 3000] —395.8 XXI1 4389 ,9| 117.4]|1013.9] Do, Dos 191.2] Do: Dz Le “T= |2.8045] 1980.8 n 6 ® o © bo b £ e ‘eto b " Bee) v| Mm eel 7 | be | ete | Bee | elle oe. [es s bi % R + wo % R e& |Os at 1 rile r S4 nls a (m) | (t.m.)|(t-m.) | (tem.) | (t.) s A (t.) ‘ Middle and Right Spans Left Span Completely Loaded. Completely Loaded. L 0.08051) +636.4 0.08051) —512.2 1} ~* | 2.59} z971.6} 66.8} 7994.8 a 6361.2) D's Lo | 0.06765] 584.8 —— | 0.05591] —640.1 II 4.66] 14349.0] 119.7] 14229.3 DaDe yes cee D's; D's L 05591 i 0.04646] —709.2 111 ee 6.22 19132.0] 159.7] 18972.3} DsDs | | aa olf 15260.7 WD, é : 0.04646] —709,2 Iv 2 7.25} 2289.7] 186.2] 22184.5) DD's | ne] ay off 17SILA Di Ds 032 0.03894] —693.6 vi? | 7.77} 23915.0! 199 5] 23715.5] sD’. ooosis! esz.el| 120822 Dede B : : 0.03297] —629.2 vi|-“° | 7.77] 23915.0] 199.5] 23715.5 HD, ooaso7] ap. qlf 1208 DoD5 L .03297| 782. 0.03953] —704.2 vit| "| 7,25} 22390. 7] 186.2] 22184.5 DD ocnosal sap all Lett] D’22's ra : : 0.04813) —724.8 vull| ““° | 6.29} 19132.0| 159.7] 18972.3] De D's o.osag7l s69 all 2220% 7] ZePe 58 ; ; ~ | 0.04813] —734.8 TX ie 4.66] 14349.0] 119.7] 1429.3] D’eD'ro| || gan gif 1450.0 DoDra 07 : 0.05997] —686.7 x|%° | osol suz.c} 66.s| 79048] 0% | 6361.21 Dy Vike {| 0.40338) 817.2 | 0.10338] —657.6 U 0.06827, 513.8 0.06827] +2912.5 1) ~* | 17.37] 7971.6) 446.0] 7525.6 an o.osacal 780 ol{ 225° ° ae U. : —780. 0.06857] 2925.1 11] ~* | 29.55] 14349.0| 579.0] 1370.0] Da Dz ooratl —ar.s\| 2279-2] ZeP”s | U : 871. ~ | 0.05692] 3152.4 ut} —° | 97.73] 19132.0] 712.1] 18419.9 Ded’, o.c4sr| 276 al O15] P’aDs | og U ~~ | 0.04757] —876. 0.03986] 3221.9 Iv | ~* | 32.93] 223207] 845.1] 21575.6 Die o.oaoral —s61.7|| SO2"| BP U, ” | 0.04013] —861. 0.03396] 3176.6 vy} —° | 38.10] 23915.0] 978.1] 2936.9] D’s D’o 93545.8| Is Do Us ~~" | 0.03419) —784.1 _— | 0.02920] 3103,1 v1 43.25] 299 5.0]1111.1] 22803.9] D’sDz | |. |/100268.0 ID's v 03520] —752. 0.02940} 3124.0 vit] —” | 36.91] 22320.7] 947.6] 21378,1 DiDs. salsa Diode ; U 04324] —793, 03550) 3217.2 vit} ° | 30.57] 19132.0] 784.9] 18347.1 DD o-o1360) —a00 oll 21 3} BeDe | U i —800, 05433) 3228.5 1x} ° | 24.20] 14349. 0] 621.3] 18727.7 DoD" on Dadi a U 05482) —752. .07828] 3076.7 x | —”° | 17.83] 7971.6] 457.7] 7513.9] D’1o 43779 5] Dio Un | 0.07097) —583.2 ~~ | 0.07097] 3106.9 The temperature changes, which give rise to the horizontal force computed above, produce stresses in the chords of the structure amounting to m., seco. For a range of temperature of + 40°C. (=72°F.), we found above that Ht = + 216.16 tonnes; the resulting chord stresses are listed in Table V. This Table V contains in addition, in the last two columns, the extreme values of the chord stresses producible by the combined action of dead load, live load and temperature variation. 256 ARCHES AND SUSPENSION BRIDGES. TABLE V. Summary of Maximum Chord Stresses. pena Live Load Stress Temperature Effect Total Stresses Ea Load —-— ress ’ ; E és Max. | Min. ||—#0°c |P188° |+4000.) P1828] Max. | Min. Acting Acting (t) (t) (t) (t) (t) (t) (t) U, |46153.1]} +2912.5 | —513.8 | 956.4 — 256.4 +9322.0 |+5382.9 U2 | 6178.2|| 2925.1 | —780.2 || 257.4] D’e |— 276.2} Do || +9360.7 |+5121.8 3 | 6207.5|| 3152.4] —871.5 |! 277.6] D’s |~ 282.4] D3 || +9637.5 |+5053.6 Us, | 6241.0]| 3221.9 | —876.3 |] 283.6] D, |— 283.9] D%, || +9746.5 |+5080.8 Us | 6279.4|] 3176.6 | —861.7 || 279.7] Ds |—285.6| D’, || +9735.7 |+5132.1 o «6| 6322.1/| 3103.1 | —784:1 || 273.2] De |— 281.5] D’s |} -+9698.4 |45256.5 1 | 6366.7|| 3124.0 | —752.4 || 275.0] D’, |— 281.0] Dz || +9765.7 |+5333.3 Ug | 6417.3]| 3217.2 | —793.4 || 283.2] D’s |—285.7| Ds || +9917.7 |+5496.2 Us | 6471.6|| 3223.5 | —800.0 || 294.2] Dp, {|— 288.1! D%») || +9984°3 |+5383.5 Uio | 6529.5|| 3976.7 | —%52.6 || 270.9} Dio |— 286.8) Dio || +9877.1 |+5490.1 Ux | 6591.7|| 2103.9 | 533-2 |] 273.5 — 273.5 +9972 .2 |+-5784.0 by 312.2 1836-4 — 45.2 o [tf #4 4 eres 681.6 Ty “Wo 2 | fons ene! B Beal Sl ania It Roos Ta —709.2 362.0 || 62.4| 2 61.1 2 |[— 771.6 |+ 923.1 Ls —693.6 782.0 ||— 61.1] ©® 55.3] © || — 754.7 |+ 837.3 Le 629.2 667.6 ||— 55.3] 2 47.3] 3 — 684.5 |+ 714.9 or ae 782.0 ||— 62.0] 4 55.4) || — 766.2 | 837.4 a i 2 = up a Tts| wa mee7|] & | o's] & ||— fo0:8 [4 size zs Sera) Beglegs) # | eg 4 [are oes —557. -2 ||— 57. i — 715.5 | 815. Tia |46536.1|| +2615.2 | —250.5 | 4349.9 — 342.9 9494.2 |+5942.7 Urs | 6476.9]| 2579.6 | —396.6 || 339.7] Dis |— 412.7] D'is 9395.6 |15694.6 nu | 6492.7]) 2464.6 | —153.9 || 409:2] Du |— 454.7] D's 9296.5 |-+5514.1 Uys 6370.8]| 2381.9 | —450.2 || 451.1] Dis |— 473.5] D’is5 9203.8 |15446.8 re | 6324.5]| 2280.2 | —449-9 || 470.4] Die |— 480.2! D’se 9075.1 |-15401.4 az | 6281.2|| 2168.0 | —383.5 || 474.5) D's |— 476.9] Diz || +8943.7 |15420.8 Uya | 6242.8)| 2057.1 | —387.5 |} 471.6] Dis |—~ 545.6] D’is || +8791.5 |15309.7 Ure | 620).3]| 1943.2 | —502.4 || 542.7] Dio |—~ 626.1] D’io |] +8695.2 |15080.8 20 | 6178.8]} 1802.6 | —626.6 || 623.1] Dzo |— 723.0] D’2o 8604.5 |1 4829.2 Ua 6153.2|| 1610.6 | —730.6 || 720.0] Da |— 830.0] D’a 8483.8 | 1 4592.6 Usa | 6133.7|| 1405.1 | —829°6 || 827.4] D2 |— 954.8] D’oo || +8367.2 |+4349.3 U23 6117.2}| 1118.3 | —869.3 |} 952.2 | Dos |—1049.3) D’os || +8917.7 |14258.6 Us | G0s8-3| 79:8 | erica || dasco | Des (ale S) Dect |] 8068-4 eames Us | 6096.5|| 79.6 | —843.6 |] 1187°8 | Doo rier os Tinh 4065.1 Ina —378.0 +805.9 ||—161.3 + 161.3 — 539.3 |t 967.2 i =i) Gerices| | ae) cee thes ‘14 . * i; A oe . Lis —488:3]} 110277 |/263:0] 2 238.9] = BL 11341.6 Lie —46613] 1157.4 ||-973: 5 %3.0| & ||— 73919 1420.4 Li —431/9| 1091/7 ||-273.6 | 2 270.9} 2 — 105.5 {1362.6 Lis —374'0 | 984.6 ||-316'7] 244.4) — 690.7 | 1229.0 Lio —391.9 | 1213/9 |]-39816 | 316.7, 8 — 790.5 |11530.6 Lao —395.8| 1449.0 ||-497.2] ©& 398.6, © — 3930 |1 1847.6 La —395.8 | 1714.3 ||606.2 2 497.9, @ —1002 6 | 2211.5 Lan 32-0) 1980.8 | ~7a2-8 5 606.2) & | —1005.g |12587.0 La 357.8} 248820 |Loss | go, | Tyee 8 aie Los 252.0} 268217 ||-971:6 916.5 —1293.6 |+3599.2 Lg 337.6 2807.4 ||—971.6 971.6 1209.2 |+-3779.0 3. Web Stresses. . d Employing the notation indicated in the adjoining diagram, Fig. 110, the stress in any diagonal in the main span will be pent) 2 (En). Here Z is the stress for a simply supported truss, and its influence line DESIGN OF A BRACED SUSPENSION BRIDGE. 257 may be constructed in the well known manner. Using an intercept =G. 23 laid off on the left reaction vertical, we obtain the influence line for Zz _M., this is represented by A,miniBy (Fig. 111), from which aa iediiontes of the H-ourve are to be subtracted. The maximum tension in the diagonal, D, is represented by the intercepted area (shaded) F'max, and its value will be Dmax= we = 7 + Pmax. The maximum tension in the opposite diagonal D’ is similarly determined by the negative portion (F,) of the influence area augmented by twice the area ($) of the H-curve in the side span so that ape? 4 (RP Le D Ge (Fi +2¢) In Plate III, Fig. 1, are constructed the influence lines for the diagonals, and the data for their computation are arranged in Table VI. The ordinates of the H-curve are those of Plate II reduced to == the scale, co that the unit load G= = X 121.1 = 30.3 meters. Column 4 of Table VI gives the intercepts (« =) used in constructing the influence lines; y and the areas in column 5, when multiplied by the factor ae a 84.37 2 ; ; : a3 a 2.786 +, give the maximum tensions in the diagonals & producible by the moving load. The temperature stresses, with Ht = 216.16 t:, are given by Dt = 216.16 Soe these are listed in column 9 in the table. 17 258 ARCHES AND SUSPENSION BRIDGES. Taste VI. ; Stresses in the Diagonals of the Main Span. i x 1 L. L.|tem,|| & 2 = y y .G — Fmax & y + E (@=30.3) | ¢ ze [Piet De item, & (m) (m) qm) (m) (m) (t) | (t) (t) Dys | —106.52 | —37.79 85.47 371.22] —72.23 | 0.5992 |} 540.3) 113.1] 653.4 Dy | —79.85 | —24.%5 97.85 391.93] —73:30 | 0.3376 |] 368.1) 73.0] 441-1 Dis | —57.36 | —14.05] 123.79 534.03 | 77.42 | 0.1815 || 259.6] 39.2'| 308.8 Di, | 39.62 | 5.52] 217.80 1371.81] —83.82 | 0.0058 || 251-1] 1412/] 235.3 Diz | —24.32] 41.95] 318.17 [—$ 498-93) 92.05 | 0.0212 |} 252.4) 4.6|] 257.0 Dy, | +251.15 | 466.69] 114.20 259.45 | 4102.26 | 0.6522 |] 470.7] 141.0)| 611.7 iw | 269.59] 66.75] 122.48 254.19 | 97.23 | 0.6865 || 485.4] 148. 4!| 693.8 so | 6288.64} 66.75] 131.13 252.29 92.96 | 0.7180 || 503.9] 155.2'| 659.1 | 312.71 | 66.75| 142.07 264.36 90:52 | 0.7374 |] 542.3) 143-1|| 685.4 Des | 345.02 | 66.63| 157.03 297.78 93.94 | 0.7092 || 587-5] 153.3'| 740.8 o, | 391.23] 66.63; 177.74 351.65] 104.39 | 0.6394 || 625.5] 138.51| 763.7 Du | 481.87] 66.57| 219.83 436.25] 139.11 | 0.4779 |! 646.4] 103.3]| 749.7 Do, | 767.46 | 66.44] 350.27 936.851 281.72 | 0.2358 |] 614/6| 51.0|| 665.6 Dox | =3.5p a. sec B =3.5X84.37X1.838= 540.6] 0 || 540.6 Py =147.26 D's —37.79 lo gh 48-45 ¢ = 165-71] —107.35 | 0.3527 || 162.6] 76.3]] 238.9 Bis —24.75| 182-261 _ 380.71 | —109.72 | 0.2256 || 113.4] 48.8l| 162.2 D'is —14..05 293-811 _ 312.26 | —114.30 | 0.1229 || 106.8] 99.6|| 133.4 D'v 5.52} 100-431 _ 118.86 | —120.70 | 0.0457 || 142.2] 9.91] 152.1 D'y 1.95 — 4404.21] 123.16 |—0.0152); 186.5] 3.3]] 189.8 Drs 66.69 is ae}= 57-01] 113.08 | 0.589s|] 93.5] 127.5]] 221.0 D's 66.75 #2-891 _ 46.34] 105.52 | 0.6266|] 80.8] 135.4]| 216.2 9 Zi%5p 66.75 18:4 = 46.02] 100.31 | 0.6655]| 85.2} 143.9|| 229.1 : S 41:05) _ D'n 66.75 qe ae t= 89.50 96.86 | 0.6891]} 114.1] 149.0]| 263.1 D'x 66.63 ets BLot 98.63 | 0.6755]| 164.4) 146.0]| 280.4 D'o 66.63 172-001 — 141.05 | 107.90 | 0.61861] 242.7] 133.7]| 375.4 D'x 66.57 245-60} 264.05 | 141.73 0.4590 344.5] 101.4]| 445.9 D's 66.44 676.401 — 694.95 | 282.85 | 0.2349 | 454.1) 50.8|] 504.9 . | D'x asfor Dos i| 540.6] 0 || 540.6 To find the stresses in the posts of the main Span, the same method of treatment will be followed. If z is the distance of the mth post from the intersection of the mth upper chord member with the (m + 1) th lower chord member, and if y is the ordinate of this intersection point, then Pp y Vinax= — Gr . "2. + Fmax. Fmax is the positive portion of the influence line for the post constructed by using the intercept G—~ on the reaction vertical. With the exception of V,,, the influence lines for the posts are easily obtained from those of the diagonals; they are drawn in Plate III, Fig. 1, as dash and dot lines. The minimum stress in the posts occurs when the system of diagonals D’ is acting; for this case, py Vain — ~~“ (CF + 2%) where 7’ and y’ refer to the point of intersection of the (m-+1) th upper chord, with the mth lower chord member, and #” represents the negative portion of the corresponding influence area. The quantities for computa- DEsIGN oF A Bracep SUSPENSION Bripar. 259 tton are arranged in Table VII; it should be observed that the stresses in columns 10 and 11 are obtained by multiplying the areas in columns é 2 and 6 by the factors 2.786 + and 2.786 % respectively. z& ca The stresses Vmax are found to be larger throughout than Pmin; only the former need therefore be considered, and it consequently suffices to calculate the temperature stresses for rising temperature only (which brings the diagonals D into action); these temperature stresses are given by the expression 216.16 2 z ; Tasre VII. Stresses in the Posts in the Main Span. | x eS - 3 2 Moving Load . 2 x ' Temp. S a| 8 is Vmax | Vmio Ve f (m) | (m) | (m) (m) | (m) | (m) (t) (t) (t) (t) Vig [310.73] 56.39 183, 480.3073 —265.6 -- 66.4]} —332.0 | is \ 103.22 5 * ; Via [296.99 ie 320.2019] "72-42 | | 24. 75/117.05]0.2104|| —216.4] — 71.2] — 66.6|] —273.0 Vis [317.35 ee 0,1812/144-83 1) 14 05/114.05]0,1292/} —161.1] — 55.8) — 39.2!| —200.3 \ Vie 440.26) 14.05 192.950.1057) °19-42 } | 5.52]116.21/0.0426]) —131.2] — 78,1] — 22.9] —151 1 Vyo [1216.17] 5.52,134.11 9.4114]2892.44 | 1.£5)118,81.0.01642)| 139.2] —132.1] — 8.9|] 148.1 Var [227.92] 66,69[124.05|0.5376 928-23 || 8.53192. 99.0, o688|| —340.8] —177.7] —125.9|] —496.7 Vis |205.43] 66.69/118.87|0.5610| 26-87 |) oo.75)137.31/0, 4861|| —820.6| — 61.0) —121.2|] —441.8 Vio |200.89] 66,751118.41|0.5038] 28-87 | | 66.75]137, 46]0, 4856|} —315.1] — 58.2} —121.9]/ —437.0 Vo 200.50} 66, 75]118.56|0.5630] 93-93 | | 66.75|142.64 0. 4680]] —314.0) — 70.0) —a21. |] —435.7 Vo, [193.65] 66.75|123.74]0.5394] 99-27 + | 66. 63]156.05'0. 4269] —290,6] — ¢2.3) —116.c]] —407.2 Veo 245.14) 66,63]137.16)0, 4858/1034 | | 66.63 183. 3610.3633| 331.3, 125.6) —105.0|] 436.3 Vg 274.26} 66,63]164,46)0.4051]°14- 7 { | 66.57/253.11 0.2609)] —331.5] —167.1] — 87.6|] —419.1 Vag \412.74] 06,57 236.21 0, 2818)°99 fg § | 66.44)621 .£0,0.1273|| —828.6) 212.2] — 60.9|| —B84.5 Vis |805.44| 66.44 502.90)0,1321)2.069 pa —296.0] —253.9| — 28.5]) —324.5 | The stresses in the side span are given by D=Z+H.u. The Z influence lines for the diagonals of the system D are constructed in Fig. 3, Plate ITI, with the aid of the intercepts D and D’ laid off on the two reaction verticals. The quantities D and D’ represent the stresses in any member producible by a unit force acting vertically at the left or right end of the span; they are obtained from a force polygon, Fig. 4, Similarly, the stresses wu producible by a foree of 1. sec a acting alone the line joining the two points of support are obtained by a force polygon, Fig. 2. The maximum tension in any diagonal D is obtained when the load covers the portion of the side span to the right of the panel con- taining the member. With the diagonals D,,D;,D.,D, and Dy, there must be added to the above a load covering the center span together with the other (right) side span. In Table VIII, the quantities of column 6 are obtained from the influence line for Z, and those of column 5 from the ordinates of the H-curve in the loaded section. From these, the stresses given in the remaining columns are obtained, using the values pa = 84.37 tonnes and . = 0.6967. The temperature stresses are given by Dy = 216.16 u. 260 ARCHES AND SUSPENSION BRIDGES. If the other diagonal D’, in any panel, is brought into action, then its stresses will be D’ = — D =. where d and d’ are the lengths of the respective diagonals. of these influence areas and multiply the resulting stresses by =. We can therefore obtain the maximum stresses in the diagonals of the system D’ by using the influence lines drawn for the system D; for this purpose we simply measure the negative portions ’ The stresses D’ in Table VIII are thusobtained from D’ = (Z+ Hu) a a TaB_e VIII. Diagonal Stresses in the Side Spans. Lo} o . Qa a & a a A ae £S é a| oD D’ u H a tort | Be g Z sid a E+/ 83 || + : eS eal By 4 (t) (t) (t) (ty Lary Ha pa Da |+1.063 — 0.100) 25.28——[4.1757pa=352.9) —2.4 349.9] 21.6|| 371.5 Da |+0.710 — 0.037] 31.50 [2 1820 “= 184.1] —0.8 183.3] 8.0|| 191.3 Dz |+0.470| 2.826) + 0.015)°327:8 11} .osg0 «= 91.9, +37.1 129.0} 3.3/| 132.3 Ds |+0.283] 2.388] 0.015513 «. {0.4748 «= 40.1] 4114.8 154.9} 10.1|| 165.0 Do |+0-145 — 2.084) 0.0705) 6 67 « 0.1669 “= 14.1] 4173.9 188.0) 15.2|] 203.2 Dz |+2-040| — 0.185, — 0.040) 15.66“ [1.so78 « =152.5] —0.4 152.1] 8.7] 160.8 De |-+2-432|— 0/305|— 0.014] 19.67“ Ir9517 « =105.6) —0.1 105.5| 3.0|| 108.5 Do |+3-056|— 0.485] + 0.020/°3%3 <1} lo.7040 « = C11] +49.2 110.3) 4.31} 114.6 Dao |+4.080| — 0.760 + 0.07199343 ° I lo.o511 «© = 21.2) 4171.4 ne 195.6| 15.31] 210.9 D's POS v | [0.2616 “* = 21.2) +245.6-—— =1.0840}] 289.3) 28.5]| 312.8 w 28.64 D's peal «| [0.8128 = 68.5 +91.0-——=1 1006|| 175.6] 8.8}) 184.4 87 D' 10.04 }1.4397 “ =190.8] —0.1-———=1. 1128]] 134.2} 3.6]] 137.8 i D's 16.14“ fo.s062 “ =177.6 —0.5 1.1212] 198.6] 11.3] 209.9 40.2 D's 20.34" fp.60s9 =220.] —1.0———=1.1966]) 246.6] 17.1] 269.7 ee 45.23 Dr BL og « } 40.3001 = 26.1) +98.§———=1.4852]| 185.5] 12.9] 198.4 30.52 / 3530.3 aie D's Diad « } [07012 ** = 69.2] -F34.¢-——=1 5497] 145.3) 4.1] 150.0 36.37 D's 30.96" |14804 «* =124.0] —9,4-——— = 1.5033] 198.3] 6.9] 205.2 22, 22, D'xo 34.97 |2,9937 “ =252.5) —1.7/—_=1 .5974)| 400.7] 23.9]| 424.6 DESIGN OF A BrAcED SUSPENSION BRIDGE. 261 The stresses in the posts can also be determined from the influence lines constructed for the system of diagonals D. If, with this system acting, Vin and V’m are the stresses produced in the mth post by a unit force applied vertically at the left or right end of the span, then we have Vin eo Dg for any loading which would stress the diagonals D in the adjoining panels, and V'n Va= “Day. . Dros m+ for any loading which would stress the diagonals DU’. Vm and V'm are obtained from the force polygon Fig. 4; Dm and Dm are calculated by the expression Z-+ Hu from the appropriate positive or negative portions of the influence areas for the mth and (m+ 1)th diag- onals. These cases are marked in Fig. 3, Plate III, by dash and dot lines. For the members V’,, V, and V’,,, separate influence lines are drawn. We thus find, for each post, two stress-values (compression), the larger of which is retained. The quantities involved in the calculation are given in Table IX. 7 TasLe IX. ; Stresses in the Posts in the Side Spans. a ts Vim @ |Tem|/L. L. a Dm and aoe] E u A Z 2+Hu Vim 4 —40°!/Tem. a Dm+i ea (t) () | ce) |} pa | — Vi |+0.110 35.217t |e set pa, 274.94 2.722.211 00. —272.2-23.8]| 296.0 Va|+0.100 s1.61 «+ [3.2524 “« |-274.44 2.2=—272.2[- = =0.6164 —167.5'~-13.3||-180.8 Ve|+0.087| 27.19 + {1.6650 * |-140.5+ 0.7=—139.8]?°°-0.0901||— 96.5 — 6.9]| 103.4 3680.3 « : se 9 |0-350_ | V.|—0.015| “33-00 « { [0.8073 |— 68.1—37.1=—105.2[--=0.7447||— 78.3— 9.5 Vz |-0.. 0465] 2929-3 (10.3208 « |— 27.8-114.7=—142.5 em =0.7986 —113.8|15.3 Vol-+0.075| 16.09 ** |1.8643 « |-157.34 0.8=—155.5)1.00 —156.5|_28.9 V,|4+0.040 12.22 [1.0081 “ |- 89.74 0.3=— 89. 4| 2 =1.0631 —104.3'_ 9.21 |_143.5 ii = 1.380_ fe Ve/+0.014) 6.19 « |0.5914 — 49.94 0.1=— 49.8) o7g7 11350 56.6/— 6.9 Vo |— 0.020) 85392 1 | [0.1982 «+ | 16.7—49.2=— 65.9] ?-1.0602||— 69.8 —11.4 Vio F 22.0} Vi — 6.9 V2 |—0.037/8530.3 1 + 18.9—90.9-— 0.588 _ 2 | 0.0378580.3 1 } 19 ove 18.2—90.9= isla ae — 90.4 66 Vs|-+0.015| 5.61 [0.6682 “ |_ 56.44 0.1=— 56.3] ==0.8900/|— 50.1] 9 Va |+0.0465] 10.30 “ 4.2085 «* |-103.6-+ 0.3=—103.3 par=0.047 — 97.81 9.4||—100.2 Ve |+0.0705] 15.27 ** 11.7989 “* |—-151.8-+ 0.8=—151.0/1-010_9 go4 |! ss0.11_ ¢ ol|_a5s.1 Loe” I 8.0]/—158. Ve | +0.095) 20.57 “* = 12.4530 ** |—-207.04 1.4=—205.6/1.00 —205.6|_ ¢,9]/—212.5 ce ‘ 1.131 V2 | 0. 0147938043 ©. lo.5083. «+ |— 42.9-24.5=— 77.4] 7 =0.930 |] 72.0 9.8 1216 1.295 Vo|+0.020| 26.45“ |1.1238 “| 94.84 0.4=— 94.4 Tappn 0-848 — 80.1} 0 |/— S01 1.500 Vo|+0.071] 28.83 * [2.8152 “ |—195.3-+ 1.4=—193.9]5773=0.7444||—144.31_ 4 gl] 448.9 00 248.0 6.9]] 254.9 ne 35.08 ‘ 2.9690 ‘* |—250.5-+ 2.5=—248.0 262 ARCHES AND SUSPENSION BRIDGES. §35. The Framed Arch with Fixed Ends or with Double Anchorage. If a framed arch is anchored or supported in two panel- points at each end, then it requires at least 6 conditional equations to fix the reactions. For ex- Pig. Re. ample, in Fig. 112, if A, B, are the panel- x. points in which one end of the structure ps4 is connected to the abutment, there are ae required two conditions to fix the reaction at A; while the panel-point B is fixed by a the member A B and by the plane of the rollers on which it may slide, so that but one additional unknown is introduced at this point. There are thus a total of 6 reaction unknowns for the two ends of the structure; against these we have only the three equations of static equilibrium, so that three more equa- tions of condition must be established from a consideration of the elastic deformations of the system. For the unknowns, to be found from these equations of strain, we may use either the horizontal thrust (1. e., the resultant of the horizontal forces at A and B), the vertical shear and the moment at any cross- section of the arch (e. g., at the crown), or we may consider three of the horizontal reaction-components at the four points of support as our unknowns. The latter scheme will first be applied. 1. General Equations of Condition for the Horizontal Reactions. If we consider the two vertical members A B and C D inserted at A and D (Fig. 113), we shall obtain the four Fig, 113, 2 B 4 “ Cb, Ay _ 79 CS "y ey na of oo D panel-points i, B, C, D, which (upon neglecting the slight strains in the members B B, and C C, may be substituted for the actual points of anchorage; at these points, therefore, the horizontal reactions H,, H., H, and H,=— (H,+ H,+ H,) may be assumed to act. Again let Z denote the stresses which would be produced by the external loading if all the horizontal reactions were equal to zero; in other words, if the structure were held fast at D but free to slide horizontally at A; also, let u, — stresses due to a unit horizontal force at A u, == stresses due to a unit horizontal force at B z = stresses due to a unit vertical force at A. Tue FRAMED ARCH WITH FIXED ENps. 263 Then the stresses in any member of the framed arch would be S=Z+ Hyw,+H,u,— (A. 4 +H, +H,)z.. (474. \ Ifr= e is the ratio of length to cross-section of a member, then the internal work of deformation for the entire structure i 1 “ : is expressed by W =o xr 8?; hence the horizontal displace- ments A, A, A, A, of the end-points A, B, C, D (considered positive if directed inward) will be determined by the following well-known law (Principle of Virtual Work) : ds ads BOS ane Te E, ars dis = (A,+A,) E, .. (AT5, srs ——(a,—A,)E addas Substituting here the value of S from equ. (474), also putting dS dH; ds se SF Un ey Uy > Z2Z=uU, C2 FT -2=U, a oe dH Ee where u, and u,, denote the stresses produced in the members of the framework by forces of unit horizontal component applied at A and B and acting in the directions A D and BD respec- tively, there are obtained the following equations of condition for the three horizontal reactions: SrZu,+H,> ruy?+H,Xru,u,—H, + Srzu,=(A,+A,)E SrZu,t+H,3ru,ust Hy 3ru,?—Hy-F Xrzuz=(A,+A,)E }.. (476. SrZz+H,3ru,z+H,> ru,z~H,-% rzt==(As-A,)E For abbreviation, let us put sru’z—a, Sru,u,—b) Sru’ =a, Sru,z—e, ar2 =a Tru, 4 —C, E (A, i) —irZ isa. pga esi (477. E (A, +4A,) —3rZu,—d, LE (A,—A,) —3r Zz =4d, 8 264 ARCHES AND SUSPENSION BRIDGES. Then the solution of the above equations (476) for the unknowns (H) will give H.= d; (a203— ¢27) + dz (¢1¢2— bas) + d; (b¢2— a2¢1) z G1 A2 3 + 2b cy C2 — a; C2? — a2 C1? — a3 b? a dy (¢1¢2— bas) +d, (a,a3— ¢:’) +d; (be,— ai) H,= 4 Oz 3 + 2B 0, C2 — Ay C2? — 2 Cy? — a b? (478, Ll d, (bc: —a2¢;) + ds (b¢:— 102) +d, (a1a2— b”) €3 Q1 G2 a3 + 2b C102 — G1 C2” — a2 €,? — a; ? The series of stresses, u, u, Zz, which are needed in making up the quantities a,, a,, etc., may be obtained graphically by means of force diagrams. We will again construct influence lines for H, but for this purpose we must first determine the stresses Z produced by different positions of a concentration P=1. With a symmetrical form of arch, it will suffice to obtain merely the stresses z; since, for a load situated at any panel-point distant x and x’ from the ends A and D respectively, h== x a Oe SrZu=Xrzut—srzu 0 x x e x. Ae 28 ele ee where 3 ree to all the members between the reaction A and the joad, and 3 refers to all the members between the load and the symmetrically located panel-point. Having found the horizontal reactions, the stresses in the members of the framework are determined by equ. (474). The total horizontal thrust transferred to the abutment is A ys san Sey ve a eee ee (479. while its position above the point A is Lats Az (e2.—e1) @ mm nt teeter terre tree ees (480. and its position above the point D is ,__ _Hs.e CO a ttre tresses eee (481. The vertical reaction at the left end of the span will be ey €2 3 V=V—-2,- — 8 BS Etisncee ae (482. when V denotes the reaction for a freely supported beam. Equations (478) determine the horizontal reactions for the general case, assuming that certain displacements A are produced at the abutments. If the abutments are unyielding, then we THe Framep Arcu with Fixrep Enps. 265 must put the displacements A—0O in the expressions for d. The equations are, however, also applicable to the case of yielding points of support such as are involved in the braced suspension bridge outlined in Fig. 114. The elastic displace- Fig. 114, ments of the points of support are here governed by the eclonga- tions of the anchor chains and these are determined by the following equations in which s, is the common length, A, the So : common cross-section and r, = Ao” Aa A, A,=—-+H,, A= +A, Ap = ~ (H+ H+ B,) Equations (478) may be applied directly to this case if the coefficients @,, ds, ..., are replaced by the following values: a,’=a, +27, b’=b+r, d,=—XrZu, dy’ = a, + 215 ery dy —3rZu, | ¢4gs, ay =a, +2515 oy =O — 1o dz=—3rZz To find the stresses produced by temperature variation, we must go back to equation (475) which may now be written ds H 3 (r$+Eots) — -=E (A, +4,) or SrSu,—E (A, +A,—ot Ssu,) From this it follows that the stresses accompanying a uni- form temperature change may be obtained by equ. (478) if, retaining the other coefficients, we merely put d,= — EotXsu,= Eotl’ d= — BatSsvz— Bott” | se ccasas (484. g—— Botts sz=0 Here Il’ and 1” denote the lengths of the connecting lines A D and B D; and the relations Ssu,——lU, Ssu,=——l”, =sz=—0, are obtained from Mohr’s Theorem, cited on page 222. 266 ARCHES AND SUSPENSION BRIDGE. In Plate I are constructed the force polygons of u,;=%,, u. and z for the arch represented in Fig. 5, and the resulting influence lines for H, and H., are drawn in Fig. 5a, Assuming a uniform cross-section Ac = 250 sq. cm for all the chord members and 4a = 25 sq. em. for all the diagonals, there are obtained the expressions 149,951 d, — 163,026 d, + 641d, a 3,407,067 __ — 163,062 d, + 239,439 d, + 6,873 ds a 3,407,067 From these are obtained the following values for the horizontal reactions: fo 1 : Braye eee Cas aes cpt Eg ton OP lap "aes ee. ae 26 26 26 25 H,= 0.157 0.372 0.565 0.666 0.658 0.612 0.481 H, = 0.100 =0.187 0.272 0.407 0.594 0.734 0.922 — ii; = 0.364 0.782 1.030 1.183 1.187 1.056 0.922 H,= — 0.107 —0.223 —0.193 —0.110 0.065 0.290 0.481 Using the values l= 80 m., Ewt= + 9600, and v= Aes = 0.004 for a chord member near the crown, the temperature effects are represented by 13111 x 0,004 Aiyt=— —3.407,067 Botl= += 11.8 tonnes 76377 X 0.004 at = 3,407,067 Ewtl==+68.7 tonnes, 2. A Simplified Method of Design for Braced Arches with Fixed Ends. This consists in establishing the required equa- tions of condition and thence devising a graphic treatment, exactly as in the case of the plate arch rib with fixed ends. Adopting an arbitrary pair of coordinate axes, with the Y-axis vertical, and defining all the panel-points of the structure by their distances x and y from these axes (where x should be measured horizontally even though the axis of abscissae be inclined), then the bending moment about any panel-point (xz, y) may be expressed, as in equation (309). by MM — Fy — Xe —X,, where M is again the moment for a simple beam, H the resulting horizontal thrust, and, by equ. (308), X,=H. X,=H.2z, With less error than in the case of the two-hinged arches, we may here neglect the strains in the web members. The stress in the chord member lying opposite the panel-point (z, y) will be S== © sec o, where the upper sign applies to the upper chord and the lower sign to the lower chord. Considering the system as unstrained and free from temperature stress, the Tue Framep ArcH witH FIxep Enns. 267 Principle of the Virtual Work of Deformation yields the follow- ing equations of condition Sd 8 4s Soa ae Aya ea ae - Observing that dS y ads x dS 1 MER oe at ES — Ae wy aH = séca, ax, =F SE€CO, aX. aS S€Co and using the abbreviation p= 2h° sec? o = AY se8o eee es (485. where A, is an arbitrary mean cross-section, then, with a con- stant E, the above equations of condition become: SMpy—HArpy?—X,Spxry—Xspy—0 SMpx—HAHipyx—X,3px?>—X,Spx—0 =Mp—Hipy—X,spxr—X,3p—0 As the location of the coordinate axes is not governed by any condition, we may so choose these as to simplify the above equations as much as possible. This may be accomplished by making Spxry=—0, Spxr—0, Spy =0,.-..0-. (486. which three conditions together with that of a vertical Y-axis, suffice to fix completely the coordinate axes. There is then obtained =Mpy __ =Mpz =Mp H= Zpy? ’ i= "soe? A,= =p . If the loading consists of a unit concentration, the numerators of the above expressions assume the familiar form whereby they may be represented as the static bending moment produced at the point of application of the load by the following quan- tities considered as forces acting at the individual panel-points: Ao ; =p. oe . Sea Ao 0 jmp AE ES Nance | ised ats (487. Ao vw’ = p= q rs Seeha We thus obtain the following formulae, identical with equations (318) to (320) of the theory of the Plate-Arch: My H— Zvy - Mv 5 ee (488, a My" Se 268 ARCHES AND SUSPENSION BRIDGES. There now remains no difficulty in constructing the influence lines for the quantities H, X, and X., these lines being obtained as the funicular polygons of the panel-loads v, v’ and v”. [See Fig. 65, which applies also to the framed arch if the loads v, v’, v” are calculated by the above equations (487) and are considered as acting at the different panel-points. | The approximation of considering the elongations in the chord members only is generally admissible. Nevertheless we may include the effect of elongations of the web members, as was demonstrated in the case of the two-hinged arch, by apply- ing certain corrections to the panel-loads v. It is evident, however, that the design will thus be rendered more com- plicated; besides, for this degree of precision, it is preferable to use the more direct method described in part 1 of this section. In order to calculate the panel-loads v, v’, v’”, we must first locate the coordinate axes so as to satisfy the conditions ex- pressed by equations (486). Let us first choose any coordinate system, having its origin at A, and let x’ y’ denote the corres- ponding coordinates of any panel-point, a and b the coordinates of the origin of the required system of axes, and a the angle between the two X-axes (both Y-axes being vertical) ; we will then have eect ee cay y=y —b+4+ (a—z’) tana and, substituting these expressions in equ. (486), we obtain the following values for a. b and tana: Zpa’ Zpy’ geen aa Zp Zp = api ey tana = a= pa’ — 2px” These expressions will evidently become much simplified when the arch has a vertical axis of symmetry. In such easc, tana—O (i. e., the X-axis is parallel to the arch-chord), 1 Zpy’ =-, and = eke naetaa te Ontos (490. To find the effect of temperature variation, we must again start with the fundamental equations S ds S . dS S(grtet)stg—% 2(grtet)sg,—4, Ss dS : (aa +wt)s wo Assuming the coordinate axes as defined by equ. (486), we obtain o EB Aowt 2s 25 x EAowtds $8 spa = — 1 t lpi ’ it pa? ; ds BAowt zs i= =p Tue FramMep ArRcH wiItH FrxEp ENps. 269 By applying Mohr’s Theorem (see page 222), the summations appearing in the numerators of the above expressions may be determined by a simple geometric construction.* If the arch as is symmetrical about a vertical center-line, then 3s 7 ie 0 dS as and, very nearly, 3s ae =—I1, Xs ax, —% so that the horizontal thrust due to temperature effect will be EAowtl A. = 7 (491. and its line of action will coincide with the X-axis defined by equ. (490). 3. Calculation of the Deflections. The vertical deflection of any panel-point D caused by any loading which produces the stresses S in the individual members of the framework, will be where Z, denotes the stresses producible in the structure, were it freely supported, by a unit load applied at D. If the external loading consists merely of a concentration P at the point &, then S=Plzg t+ Hie w+ Hog ts — Hyg 2] Similarly, the horizontal displacement of D is given by At=3——.8. wx ok eu eeee narnia aeees (493. where w, represents the stresses producible in the structure, were it freely supported, by a unit load applied horizontally at D. *See the article (previously cited) by Miiller-Breslaw in the Zeitschr. d. Arch. u. Ing. Ver. zu Hannover, 1884. 270 ARCHES AND SUSPENSION BRIDGES. E. COMBINED SYSTEMS. §36. Combination of Arched Rib with Straight Truss. 1. Exact Method. Let a rigidly constructed arch rib, A B (Fig. 115), be connected with a straight truss, CD, by means of the verticals 0,1, 2, ... n. Let the loading consist of a concentration P bearing directly upon the truss; this load will be partially transmitted to the arch by the verticals. The compressions in these verticals, designated by V,, Vi, Vz... Vas Fig, 115. ert VY ‘I will be the only loads acting upon the arch; while the truss will be loaded by the concentration P and the forces —V,, —V,, —V,., ete. There is no difficulty, now, in determining the vertical deflection of the arch-rib at the mth vertical. In general, let 8, represent the deflection at the mth panel- point produced by a unit load at panel-point r. By Maxwell’s Theorem, this is known to be equal to the deflection producible at the rth panel-point by a unit load acting at panel-point m. The influence number (8) may be calculated for a plate- arch by the following expression based on equations (295) and (304) : 1 Sar = qr [Mm — Ay tim + Hye a (l—a)]... (494. and, for a framed arch, by the following expression derived from equ. (4607) - Bar = [Sr 2mZr— Hm Ap Sr u?] ..ccceees (494", With given or assumed sectional areas of the arch, these deflections may at once be determined, since the reactions H produced by the unit load may be computed exactly as for CoMBINATION or ARCHED RIB WITH Srraicut Truss. 271 an independent arch. The total deflection at the mth panel- point caused by the loads V will then be AYm= V, Sim V.8em+ soe +VinSmm Se9) + Vn 8 n_aym- In like manner, we may compute the deflection at the mth panel-point for the straight truss, considering its ends as simply supported. If em, denotes this deflection for a unit load at the rth panel-point, its value may be calculated for a plate arch, or approximately for a framed arch, by the formula A, €mr = rr” 596 re (495. and more accurately for a framed arch by 1 émr = = =P Ds ie ey Spans daly Mare See ea CN (4952. where 2’, and z’, again represent’ the stresses in the truss pro- duced by unit loads placed at m and r respectively. The total deflection of the truss at panel-point m, produced by the loads —V and + P, will then be A re =— (Vy emt Veet. +++ Vnémmt ++ ++ Vn €ea-1ym) + Pe gm Also, let ¢m Vm represent the shortening of the vertical m due to its compressive stress V,,, where ¢, may be calculated in the familiar manner when the length and cross-section of the mem- ber are given. We then have Ay+¢nVn—=Ay’; applying this relation to each separate vertical, and introducing the abbreviated notation Sep eee eo eee ees (496. we obtain the following system of equations: Vi (Q,, +61) + VeG.2+ Vedigt--- tVntimt+... + Vai Gyn) =f &1.é V, QrotV, (@y2-+ $2) + V 3423+ «++ bt Vin@oam+... 4- Vay Geen) =P. €2.¢ Pe V, @,m+ V2. @emt V; “ Qgm+ ah ee + Van (@m.m— om) + Slane + Va-3 @mn-1) = P. €mg a V, Quay tVe- Qinryot Va@uagt- - ++ Vin Gam: -- + Va [Gray ny Hon] =P. €n-ayet J These (1 —1) equations suffice to determine the (n—1) 272 ARCHES AND SUSPENSION BRIDGES. forces V. We can then obtain the horizontal thrust of the arch by means of the expression H= HV, 48, V ee ox 4+ Ba Veo cnt Bg Vege (498 Using the above equations, we will construct the influence lines for V,, V,, ete., and for H; these will give, for any loading, the values of the external forces for the arch rib as well as for the truss. If M, denotes the bending moment at any section x of a simple straight beam loaded with the forces V, then the bending moment in the arch will be a lions oe esis hala aues (499 and the bending moment in the straight truss will be Mp =P. HB! yee eeees (500. 2. Simplified, Approximate Method of Design. By the results of § 19, if we neglect the effect of the axial compression, the deflection of the arch at any section distant x, from the abutment may be calculated by the following expression (applicable also to framed arches), sn Li fs where M’, is the bending moment in the arch-axis, and I’ de- notes the moment of inertia of the arch cross-section multiplied ie (tte (- I (2, —2) da] 0 by cos r= ae Similarly we obtain the deflection of the truss: ois 4[sftza-nar— fee 2) dz], where M” and I” are the bending moment and moment of inertia, respectively, for the given section of the truss. Neglect- es further, the shortening of the verticals, we have Ay=Ay’, sf = a) )a—ayae— f(a —43)(#,—2) de=0. The left member of the above equation, however, represents the bending moment at the section x, of a simply supported beam which is loaded at each point with a force of magnitude Ute, 7 a). This bending moment must therefore vanish for all the sections z,, i. e., for all the points at which there is a connection between arch and truss. If we imagine these CoMBINATION OF ARCHED Ris with Straicut Truss. 273 connecting verticals to become infinitely close together, then the above condition must hold true for every point of the span; this cannot possibly occur unless all the loads upon the structure reduce to zero, in other words we must have M’x =o M"x SF teens Hence, the bending moments in the arch and the truss will bear the same proportion as the respective moments of inertia. If these moments of inertia are constant, then the two bending- moments will also have a constant ratio. If this ratio is r . 7% if M is the moment producible in a simple beam by the external loading, and if M,, as above, is the moment producible by the loads V, then, by equs. (500) and (499), M”,=M—M,, and MM’, =1M", = M, — Hy. Hence wo M—H yy ee a 1+i To evaluate Z/, neglecting the effect of the axial compression, we use the equation l i M's M alm fe yee (ny fates 0 whenee M'. = 47 (M—Hy).. (502. 1 pe farude -Al Vf oe eee eee (503. yo. pr oF Omitting the end-displacements from consideration, the hori- zontal thrust assumes exactly the same value as in the directly- loaded arch without a stiffening truss. Also, the critical loads and maximum moments are to be found in the same manner as for the simple structure; but the actual ao in the arch and the truss will be respectively tag and -—— times the i corresponding moments for the directly loaded arch. 18 274 ARCHES AND SUSPENSION BRIDGES. § 37. Combination of Arched Rib and Cable. Let the cable, supported at C and D on rollers or rocker- arms, be connected by vertical rods with the arch A B (Fig. 116). Let the horizontal thrust of the arch be denoted by A,, and Fig, 116. the horizontal tension of the cable by H,. Finally, let y, rep- regent the ordinates of the arch-axis referred to the line joining its end points (A—B), and y, the ordinates of the cable- polygon referred to the connecting line EF. In developing the following theory we assume, as with the systems previously considered, that the deformations are so small as to have but a negligible effect upon the bending moments. Adopting the notation \} = length of cable between panel-points, a= distance between two suspension rods, A.= cable cross-section at the crown, A, = average cross-section of the arch, A, = cross-section of a suspension rod, I,=Icos¢=moment of inertia of an arch-section multi- plied by the cosine of the inclination of the arch-axis, b = total length of the arch-axis, then the stress in the cable section will be n i= Hye and the tension in a suspension rod will be R= 7 a Noting that the moment of the suspension forces at the section x y, of the arch — H, y,, and if M represents the moment CoMBINATION OF ARCHED Rip with Srraicutr Truss. 275 of the applied loading in a simple beam of span J, then the bending moment in the arch is found to be M=M—H,.y,—H,.y, .............. (504. The axial thrust in the arch may, with sufficient accuracy, be assumed equal to H,. Assuming, furthermore, that the anchored ends of the cable are displaced through 8, and 4, in the direction of the backstays, that the supports at C and D are forced together through 8, and 6, respectively, and that, by a yielding of the abutments, the span of the arch is increased by Al; then, if the temperature differs from that of the un- stressed condition by 1°, the Principle of Virtual Work (Castigliano’s Theorem) yields the following two equations: 1 aly I an t “42 M M dat as eo ahh Hy El, OH, SAL = — 8, seca, — 8, sec a,— 8, (tana, + tana’,) — 8, (tana, + tan a’g) aw M aM Hi y™ | A: 5 Ay ae pion, 8° TBA at Bag (i sec?a Fl, 800? ay) Fn AP y.\? x 2 2 s Af ye Se = (=*) stots @ +1, sec?a,+l,sec?a,+ e 8) or, using the abbreviation 8, seca, +8,8¢C a. +8, (tana, + tana’,)+8,(tana,+ tana’,)=Ak, and substituting the vaiue given by equ. (504) for M, ¢ l 1 m[ f yd o oe | +4: Jae ae l «fa ae 0 1 Ll F Be 1 a nfm aes m | fee 2 hae Hee Nae Bas (2 )|- fae dz—Eak ~Bwt(s~ +1, sec* a, +1,sec? a. +3") (505. These two equations serve to determine the two unknowns, - I, and Hi. If the external loading consists merely of a concentration, then, as rte in an earlier oe i. this book, the detidts integrals f dx and fA dx may be conceived as 276 ARCHES AND SUSPENSION BRIDGES. the moments in a simple beam loaded with the quantities : = and T , and may be thus represented by means of a funicular polygon. If we make y,—O in the first of the equations (505), it reduces to the corresponding formula for a simple arch with hinged-ends. Making y,—0O in the second equation, there results the approximate formula for the horizontal tension in a cable connected with a straight stiffening truss. If both arch and cable have an initial parabolic form, with versines f, and f, respectively, then, for a constant moment of inertia in the arch, 1 l yr da = 8f71 fe#2- 8 f.71 wae dz — Bhifet ih 157, ? I, 157,’ i “isi,” Furthermore, for a concentration P acting at a distance é from the end A, 1 Myida __ fig I—~£) (P+1éE—#) P, I, 371, 0 l My:dax fzE(1—E) (V+1E— B®) P. I; 38 I, « 0 Putting ne z~ +1, sec? a, +12 860? ag+ 1° Ss (= Yo neglecting the term representing the elongations in the rods, and assuming a continuous curvature of the cable, there results = +34) +1, secta, 1,sec?ay.... (506. Using this value, also putting AL=Ak—=O and t—0O, the solution of equations (505) will give z P.€ ee) Woe 1 3 Ac mp tg {i+ fF Ae +eral ey , ut P.é(l—£) (P+1E—2#) - (907, is kf? Aa 15 k I fat [2+ bf? Act 8 1 Le The two horizontal forces therefore have a constant ratio, namely, Ay i: f;Aa Fe 1 Kenai enna (508. This ratio will hold true for any arbitrary loading. CoMBINATION OF ARCHED R1B AND CABLE. 277 The effect of temperature, together with a yielding of the abutments and anchorages, will produce horizontal forces given by the following formulae: 7 BPG nar Ge at oF 15 by ke (a+ ate cee a ae) ipa feo 8 : (509. ee [(a+ coe) ea »| Of: (+ ae ate eames If we put H,=C. H,, so that, by equ. (508), C= bf2Ac fide’ the bending moment at the section zy, of the arch may be calculated by the relation SMF a cere oureratabin (510. From this we derive the following rule for finding the maxi- mum moments and critical loads for the arch: Increase the BAe. f Kk fy Aa a and take the resulting parabola as the axis of a new arch for which the maximum moments, ete., are to be determined in the usual manner, using the value of H, given by equ. (507). In calculating the temperature effect, we must keep in mind that a negative value of H,, with flexible cable and suspenders, cannot be transmitted to the arch. If, upon an increase of temperature, H, becomes negative, then the cable becomes in- effective and the entire structure acts merely as a simple two- hinged arch. It can be easily demonstrated that this condition may occur with a quite moderate rise of temperature and that, even with the structure completely loaded, the cable may become slack unless there is a simultaneous yielding of the arch abut- ments or unless the cable has been very skilfully mounted. The horizontal tension in the cable due to temperature will exceed that due to the loading (of intensity q), when rise (versine) of the arch by the amount Cf, = Aa Eb LOK 15 Lk 1 “or, on substituting J, = A a and o’ A, = = uF , when Bet ih - he ot [a+ tg oe 3 cele Here o’ represents, approximately, the mean stress at the crown of the arch produced by a full-span load. This stress, in a 278 ARCHES AND SUSPENSION BRIDGES. wrought iron arch, need not be taken higher than 500 kg. per sq. em. (= 7120 pounds per square inch), bearing in mind that this does not include the stresses due to temperature, etc. If we put the expression within the brackets approximately =2, the temperature rise need not exceed a 500 ee A ‘— 350 = ry oe = 10° C.(=18° F.) before the cable will become slack. In reality, however, we may expect much greater fluctuations of temperature than this, whence we may conclude that the cable will not be effective at all times and that, consequently, this combination cannot be considered an advantageous form of construction. It should also be observed that the versines, f, and f,, of the cable and of the arch, must have a certain ratio in order that the maximum intensities of stress in these two parts of the structure may be approximately equal to each other or to assigned eae Introducing the abbreviation N—=1-+ bP ae > b oh “Rh fP Aa =F = “he Aap? the maximum stress in the cable becomes eg Gi = Vis CoB FN Rafi da” and neglecting temperature stresses, Similarly, the maximum stress in the crown-section of the arch will be, when the cable is slack, 2qP4+15EwtAa.h + off 4e\ 16 45 f.(¥— a) We therefore have 1S Poet & N k fi Cae ) eS eee 4 k f*, Aa or approximately oa 15 960 A k fs z= (+5 G00 fd? 5 If arch and cable are to be equally stressed, we must accordingly have a 1 fa 2k tit 3" tt or the rise of the arch must be less than = to ’ the versine of the cable. APPENDIX. The Elastic Theory Applied to Masonry and Concrete Arches. The theory of arches developed above, in §§ 10 to 24, based on the elastic deformations, makes no other assumption in re- gard to the material of construction than that it is to be con- sidered perfectly elastic within the limits of stress occurring in the structure. If this assumption is equally applicable to masonry as to metallic structures, the above theory of the solid, elastic arch rib may be as appropriately applied to masonry arches; and it need only be observed that, on account of the small tensile strength of the material of a masonry arch, no large tensile stresses may occur, so that wherever such stresses are indicated by the computations the entire cross-section of the arch is not to be considered effective. The elastic behavior of masonry, previously maintained by Tredgold and Bevan, appears to be established beyond ques- tion by the tests of Bauschinger*, Foeppl and others. All natural and artificial building stones, within the practical limits of stress, are to be regarded as more or less perfectly elastic bodies; and the conclusion is therefore justified that the distribution of normal pressures in any smooth, plane joint in masonry is determined by the elastic law or, with the simplify- ing assumptions of Navier’s hypothesis, by the rules expressed by equations (174) in § 11. An arch composed entirely of squared stone, i. e., without any mortar, will therefore act as an elastic arch, at least so long as no tensile stresses occur in the planes of the joints and the deformations do not exceed certain limits. Masonry composed of stone and mortar, how- ever, on account of its heterogeneity, cannot be treated as per- fectly elastic and will be affected not only by the manner of construction but also by the age of the masonry. The’ more recent the work, the larger will be the ratio of the permanent set to the total deformation. Furthermore, there does not ap- pear to be a definite elastic limit for such masonry. Neverthe- less laboratory investigations as well as observations on actual structures’ indicate that within certain limits of stress even *Mitteilungen aus dem mechanisch-technischen Laboratorium zu Mun- chen, No. X, tKépeke, ‘‘Die Messung von Bewegungen an Bauwerken mit der Libelle.’’ WProtokolle des sachs. Ing. Ver., 1877.—‘‘Are d’experience,’’ Report of M. de Perrodil. Ann. des ponts et chaussées, 1882.—‘‘ Gewdlbe- Versuche des Osterreichischen Ingenieur- und Architekten—Vereins.’’ Zeit- schrift of this Society, 18954 also published as a separate reprint. 279 280 ARCHES AND SUSPENSION BRIDGES. mortar masonry acts as an elastic body. Furthermore, even with conditions of imperfect elasticity where permanent com- pression accompanies the temporary strains, the distribution of pressure in plane joints is really given by the same laws as in perfectly elastic materials, provided we may only assume that the conditions are uniform throughout the mass of masonry and that the permanent strains are proportional to the normal pressures. In fact, as Foepplt has shown, we arrive at results practically identical with those of an elastic arch if we treat the voussoirs as incompressible bodies and ascribe elastic com- pressibility only to the mortar. Actually, however, the rela- tions are not so simple. The masonry at different points may have varying elasticity and, with uneven hardness, varying compressibility ; furthermore, in masonry of unsquared stones, the joints are not true planes. The perfect elastic behavior may therefore best be expected in the case of monolithic con- erete arches; nevertheless even for this material, according to the experiments of Hartig, Bauschinger, von Bach, and others, Hooke’s law of proportionality is but approximately true and only within low limits of stress, since for larger stresses the compression increases more rapidly than the pressure. On ac- count of these considerations, the elastic theory is not as accurate a treatment for determining the stresses in masonry arches as in metallic arches; nevertheless, on the basis of the above-men- tioned investigations, especially the tests on arches by the Aus- trian Society of Engineers and Architects, the results of the elastic theory may also be applied to masonry arches thereby arriving at the actual relations in accord with all preceding theories for the masonry arch. The perception that a correct theory for masonry arches should be founded on the laws of elasticity had been expressed by Poncelet (1852), but later writers (Winkler, Culmann, Schwedler and others) were the first to develop this idea and to apply it to the actual construc- tion of such a theory. Now the theory of the elastic arch is generally recognized as the basis for a correct analysis of the masonry arch, and the methods of design developed in the pre- ceding chapters may therefore be directly applied to masonry arches. The static analysis of a masonry arch involves the finding of the lines of resistance for specified cases of loading and, ultimately, the determination of the maximum stresses in the joints, which may be found either directly or by means of the line of resistance belonging to the corresponding severest con- dition of loading. For the rigid solution of this problem, there- fore, two cases of loading must be considered for each joint, one for the upper and lower core-point respectively; neverthe- less these may be replaced for each joint, without great error, tFoeppl, ‘‘Theorie der Gewolbe.’’ Leipzig, 1881. APPENDIX. MASONRY AND CONCRETE ARCHES. 281 by a single case of loading corresponding to the middle point of the joint. (Compare the remark in § 29, p. 208.) If F is the resultant force acting at any section (or joint) of the arch, N its component normal to the section and 4M its moment referred to the center of gravity of the section (arch- axis), then, with A as the area and Z as the section modulus of the cross-section, the extreme fiber stresses will be = a aie a If N and M are taken for a unit width of the arch, then, with d as the thickness of the arch, v 6M Pg eae It is advisable to determine the stresses due to the dead and live loads separately, combine these and then, in the case of hingeless arches, add the stresses due to temperature variation. If the arch is provided with hinges at the ends and at the crown, so that it reduces to the statically determinate case of the three-hinged arch, there is no special difficulty in determin- ing the forces acting; for each section to be investigated the critical loading may readily be determined and used as a basis for evaluating the maximum live load stresses in the extreme fibers. For this purpose either the analytical or graphic pro- cedures given in § 15 may be used. If p is the live load per linear foot and if the length of loading is determined for the middle point of each section, then, for a parabolic arch, also approximately for a flat segmental arch, we have the expressions __ pu(l—a) (l—2@)_ M— 2(31—22@) E48/l (I-22) (51-42) | N= 4fl (31—2a@ )* p- cosh, Here ¢ is the angle of inclination of the section from the vertical and a the distance of its middle point from the end of the span. If we are dealing with concentrated loads, it is best to em- ploy the method of influence lines. The effect of the dead load is obtained by drawing the corresponding line of resistance or, for greater accuracy, analytically from the relations M=M — H.y and N= H.cos ¢+V. sind where H is found from the crown moment M, of the dead load Mc as H = — Far more detailed and tedious is the correct design of thé masonry arch constructed without hinges, which is, of course, 282 ARCHES AND SUSPENSION BRIDGES. the common arrangement. In such an arch, if the applied load- ing and the accompanying secondary effects produce no tensile stresses in the end and intermediate joints exceeding the tensile strength of the masonry, the arch will act exactly as a hinge- less elastic rib and the theory given in §§ 20 and 21 may be directly applied. Here also, to be exact, it would be necessary to find the most unfavorable loading for each section to he in- vestigated and then determine the corresponding stresses. On account of the tediousness of such a design, however, it is fre- quently deemed sufficient to consider just one or two special cases of loading, usually a loading over the entire arch and one covering the half span. The former gives the maximum stresses in the joints near the crown of the arch, the latter corresponds approximately to the maximum stresses at the quarter points and ends of the span. Accordingly the external forces together with the pressures in the joints, for each of the three conditions of loading, namely, a.) dead load, b.) full live load and ce.) half-span live load, Fig. 1 may be determined directly without the necessity of first find- ing the influence lines. The dead load is composed of the weight of the arch and the pressure of the construction resting upon it. The latter, in the ease of arches with spandrel filling, may be taken as equal at each point to the weight of the superincumbent filling; but in arches with large rise and large depth of filling this pressure is to be determined in accordance with the theory of earth pressures. In the case of secondary arches or piers rest- ing upon the main arch, the weight they carry is applied in the form of concentrated loads. Continuous weights and loading, however, will also be replaced by concentrations since we will divide the arch into segments and consider their weights applied at their respective centers of gravity. a.) Analytical treatment for a given case of loading. We first have to determine the resultant force at any one joint; with this known, the pressures at all the remaining joints are readily found. To completely determine such a force, however, three APPENDIX. MASONRY AND CONCRETE ARCHES. 283 things must be known: the components normal and parallel to the section and the moment about the axis of the arch. To establish the corresponding equations of condition, the three derivatives of the work of deformation will be employed. In the following, we assume a masonry arch symmetrical about the crown and adopt as our unknowns the data for the thrust at the crown of the arch, namely, its horizontal and vertical components H and V and its moment M, about the center of gravity of the section. In addition, we divide up the arch ring by sections normal to the axis of the arch and uni- formly spaced along its length (Fig. 1) and calculate the weight of each segment of the arch together with the superincumbent load. Let these applied forces in the left half of the arch be P,, P,, P,...P,, and in the right half P’,, P’,, P’,...P’n; their horizontal distances from the crown joint a,, a,...and @’,, a’; .-, respectively. The axial points of the cross-sections are referred by coordinates (x, y) to a system of axes passing through the crown of the axis of the arch. The axial thrust and moment for any section inclined at the angle ¢n from the normal are then calculated as follows: In the left half of the arch: Nn=H - cos dbm— V- sindm+(P,+Pot ... + Pm) sin om Mn—M.—H-: Ym—V- Im + P, (2m—4@,) + P.-(%m— G2) +11 + Pa. (fun— Qm) In the right half of the arch: (511. N'n—=H -cos¢atV ‘sindmt (P',+Po+. +. +P'm) ‘sindm M’ p=M — HH ym tV tm + P's (am— a’,) + P's (2m — a’ s) +... + P'm (Sm — Om) With the notation previously employed, the approximate expression for the work of aan at the arch is w=t(-as+t Hence, with the assumption of oe eee and rigidity of the end sections, we obtain the following equations of con- dition expressing the Theorem of Least Work: pee M aM Y¥ aN fi an 8 Bye ail eS Md gs_o, pi T dal, dw MW aM, N dN fae Na ar s+ fz: av ¢s=9, where the integrals represent a summation over the entire length of the span. If the segments of the arch are assumed of uniform length ae the axis of the arch, and if a mean cross-section A,,A.,.., and a mean moment of inertia J,, Z,, 284 ARCHES AND SuSPENSION BRIDGES. are introduced for each segment, then Simpson’s Rule may be applied for the above summation. The substitution of the values of Mf and N, as well as of their differential coefficients, from the above expressions (511) yields the following three equations of condition: a,+bH-+c M,=—0 agate H+d My=0 fF... . cece eee (512. a, te V=0 Introducing the abbreviations of notation (P, + P+... + Pn): sin dn =Pm for the left,and P’m for the right half of the arch, P (am = 4,) + P2(tm — A.) +... +Pm(tn— @m)=Mn for the left and M’, for the right half of the arch, the coefficients in (512) are given by: a= — 5 [42 (M+ MY) +22 (M4M,) +44 (M, +M’,)+...+%(M+M’s)|+3[ 42% (P+?) +228 (P,4 P’,) +4 (P,+P%, ) +...+S2(P,+P%)| a, = 5 [ECM AM) +7 LEM) +7 (M,+M’,) +7 (Ma+M’s) | = =| 47 (M,—M’,) +24 (M,—M’,) +4 Caen *(513 b= [42 +2%44%+.. +2] + [= so Ltt | hoot | wey ole re +2] d=[+e+EtEet tz] on [ag tage net ee] + [28 420s fasting 1 4 wre) APPENDIX. MASONRY AND CONCRETE ARCHES. 285 With the values of H, M, and V calculated from equations (512), equs. (511) are to be used for determining the axial thrust and moment at the other sections of the arch and the re- sulting normal stresses. For a loading symmetrical about the crown, we have M’ = M and P’ —P; hence V = 0, so that only H and M, remain to be figured. This condition obtains for the dead load when the roadway is horizontal or symmetrically inclined about the crown of the arch. If we investigate only the three cases of loading mentioned above, then H and M, need to be-figured but once for the dead load and then for the full live load. For the load over the half-span, we need only find the value of V from the third of equs. (512) using M’ —0, P’ — 0, since H and M, for this case of loading are just one-half of the corresponding values for the full-span load. The stresses calculated for the dead and live loads separately are then to be added algebraically. Approximate Formulae. We assume an arch with parabolic axis, having a symmetrical form and carrying a symmetrical, continuous loading which amounts to q per unit length at the crown, q, per unit length at the ends, and varies between these values as a parabolic function so that at distance 2 a from the crown, g=got+ (q1 — qo) = - We also assume that /.cosé—I, l iol : is a constant, and substitute FA, for the second summation in the expression O for the coefficient b, along with which the terms in the expression for a, re- ferring to the axial thrust are neglected. Again replacing the summations by definite integrals, we readily obtain Lge ie eg A Be M=~3593K5XT Ty 216 30 Th Lf Spa. ages a7, Fe ZL, 45 I, ; Introducing the symbol f’ =f é + ZAP » the solution of equs. (512) Ao yields. — b6gatn, F i 7 ya { tere Fee) oa 7 f—s«d0 24 siee aoa (bids and, for the moment at the ends of the span, di aa | Stor f= ee) a: 7 7 10 12 These formulae are also applicable to the simplified, approximate design of flat segmental arches. With 9,=q, they reduce to the formulae for a load uniformly distributed over the entire span. For a uniform load of p per linear foot over one-half of the span, H and Mo take one-half the values given for a full-span load and, in addition, there appears at the crown sec- tion a vertical shear (directed upward on the loaded side) of V= =e pl. 286 ARCHES AND SUSPENSION RripGEs. b.) Procedure with the aid of influence lines. If it is de- sired to investigate various cases of loading, especially the criti- cal loadings for individual sections, it becomes necessary to construct the influence lines for three unknowns of the external forces. For this purpose we may apply the procedure developed in § 20 for the arch with fixed ends, the simplifications there given for parabolic arches with constant moment of inertia being also applicable to flat segmental arches of approximately uniform cross-section.* For arbitrary forms of arches (e. g., basket-handle and equilibrium curves) with variable moments of inertia, there is to be recommended the general graphic method in which the influence lines of the quantities H, X, and X, are obtained as funicular polygons. For any given loading, Fig. 2 Influence Lines — bythe exact formulae as By the formulae for the parabolic arch these quantities are obtained in the familiar manner by sum- ming the ordinates of their influence lines multiplied by the re- spective loads. With the aid of these quantities, we may then either draw the line of pressures or simply caleulate the moments about the core-points of any given section. If a, and y, are the coordinates of any core-point referred to the system of axes fixed by equ. (321), then, with the notation of § 20, by equ. (309), MM, = M,—H. y, —X,. 4, —X, ; from this value of the moment, the stress in the extreme fibers of the cross-section is computed in the ordinary manner. By the same equation, the influence line for M, may be derived from those for the quantities H, *On this approximate, simplified design see Th. Landsberg, ‘‘ Beitrag zur Theorie der Gewdlbe.’’ Zeitschrift d. Ver. deutscher Ing., 1901. APPENDIX. MASONRY AND CONCRETE ARCHES. 287 X, and X,, and, by its aid, the most unfavorable action of the loads may be accurately determined. It will always suffice, however, to use an approximate determination for the lengths of loading giving maximum stress and, for this purpose, we may employ the simplified relations established for the para- bolic arch (i. e., determining the limits of loading by equ. (381) or by means of the straight reaction locus and the hyperbolic reaction-envelope curve.) For the frequently occurring case of a circular arch of large rise-ratio, whose radial thickness increases from the crown to the ends in a uniform ratio with sec ¢, the following may be given in addition to the exact for- mulae of design. Adopting the determining quantities of the left end reaction, i. e., H, V’, and M, as the three unknowns, the loading consisting of a singie concentra- tion G, and with the notation indicatedin Fig. 2, we have the axial thrust and moment at any section given by Ps=V,-sin@+ H:cos¢— [a-sino] - Mx=M,4+V-r (sin bo— sing ) —H-r(cos¢ — cos dy) = [« -r (sin bd, — sing) lf. If do is the thickness at the crown, then, for any section, d= do seco 1 ‘ and J= Io. sec °O=7T = 5 do® .sec’g. Introducing the symbols for abbreviation, 5= Ayr 12 7 a=(14+5) sing —- sin*d 3 1 3 - (515. B= {Zot 5 sing cos -(F + cos" ) y= = sind (cos*é + zor+s) and the corresponding symbols aj, By, Yo, and a,, By, ‘1, for the points d= gq and ¢, respectively, we obtain the following expressions from equs, (182) (185) and (186) for the variation of the coordinates of any point in the axis of the arch: Ab- A =~ go [ (a— a) + ¥, rf (aa) cin he 4 (008% — costa) b+ Hr { (a— ay) c08 66 — 5: (B— Ba) } A a a a aoe ates (519 M=—e- d The coefficients b, c and d are determined by the equations (513). As is the length of an arch segment between two sec- tions. As a result of these disturbing influences, there may arise at certain points of the arch, particularly at the ends, tensile stresses which exceed the strength of the mortar and cause the formation of cracks. By giving the arch the necessary form and thickness, and by appropriate precautions in the erection, these undesirable effects should be prevented; but, if they do appear, the effective cross-section of the-arch at those points must be correspondingly reduced (Fig. 3), and this fact APPENDIX. MASONRY AND CONCRETE ARCHES. 291 must be taken into consideration in the execution of the de- sign. The position of the line of pressure, however, will not be materially altered by this circumstance. Fig. 3, ro With regard to the magnitude of the coefficient of elasticity E for masonry, the reader is referred to the remarks in Chap. II, fourth edition, of the ‘‘Handbuch des Briickenbaues.’’ The methods of design and formule given above may also be applied directly to arches of reinforced concrete. For these we have only to replace A and J in the design by their equiva- lent values which are obtained, in the familiar manner, by adding to the cross-section of the concrete that of the steel mul- ate Es tipled by n= =o Remarks on the Temperature Variation to be Assumed in Steel and Masonry Bridges. In evaluating the stresses caused by change of temperature in statically indeterminate structures, we are accustomed under our climatic conditions to assume a range of temperatures from — 25° to + 45° C., so that with a mean temperature of + 10° C. (= 50° F.), the thermal variations in metallic bridges amount to + 35° C. (= + 63° F.). In this assumption the fact is duly taken into account that the average temperature of a steel structure exposed to the direct rays of the sun is about 10° to 15° higher than that of the air in the shade. If the structure, however, is shielded in any portions from the direct solar rays, the resulting unequal heating must be considered. On the magnitude of these differ- ences of temperature, thorough observations were taken on the occasion of building two arch bridges at Lyons in 1886 and 1887.* For this purpose, a section 1 meter long of a plate arch having a box cross-section 900 mm. high and 800 mm. wide was used. In this test piece thermometers were inserted at eight points in holes filled with mercury and their readings taken several times a day for seven summer months. The test piece was provided with three coats of brown oil paint, closed at the end with boards and left resting freely with one long side turned toward the south. It was found that the difference of temperature between the warmest and coldest points of the *Ann. des ponts et chaussées. 1893, II., p. 438. 292 ARCHES AND SUSPENSION BRIDGES. steel on sunny days attained a value of 14° C. (25° F.) and that the average temperature of the steel rose to a value 15° C. (= 27° F.) above the temperature of the air in the shade. Surrounding the piece with a box of sheet iron effected a lower- ing of the maximum temperature by 7° and similarly it was found that a coating of light blue paint reduced the average temperature by about 5° C. (=9° F.). A temperature variation of + 35° C. (= + 68° F.) in the steel parts directly exposed to the sun and one of + 20° and — 385° C. (= + 36° and — 63° F.) in the parts shielded from the sun’s rays may therefore be properly assumed provided a temperature of -+10° C. (—50° F.) is taken for the initial unstressed condition. But even if the temperature has some other value at the moment that the structure is released to carry its own weight, it is possible to bring about such an initial condition of stress at the erection of statically indeterminate structures as to virtually fulfil the above assumption. In masonry bridges there cannot occur such large variations in temperature as in metallic structures. Although definite observations on this point are lacking, nevertheless it is fair to assume that massive masonry cannot attain the maximum temperature of the air in summer nor, on the other hand, can it cool down to the lowest air temperature even during pro- longed frosts. If we therefore base our calculations of the temperature stresses in masonry arch bridges on a range of tem- perature of + 20° C. (= =+ 36° F-.), all the demands of safety will be well met and, in the case of more massive masonry arches that are covered with earth, this range may even be reduced. y BIBLIOGRAPHY. 293 BIBLIOGRAPHY. Theory of Arches and Suspension Bridges. Navier. Rapport et mémoire sur les ponts suspendus. Paris, 1823. Navier. Résumé des lecons données a 1’école des ponts et chaussées, sur l’application de la mécanique 4 1’établissement des constructions et des machines. Paris 1826. Second edition, 1833, p. 277. Translated into German: first edition by G. Westphal, Hanover, 1851; second edition by G. 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Proceedings of the Ameri- ean Society of Civil Engineers, August. 1904, é INDEX. PacE Anchored ends, framed arch with............... 0c ccc cece cee ceeees 242 Approximate theory for suspension bridges.......................- 25 Arched TibS! cos ye: wewsig oa eee s Bes RE CAs eke eau Hanae aA eee 87 conditions for stability....... 0.0... 0. e eee 95 eriti¢al loading: oases aus ee ay Kea e ESAs dia See Seed wn bene HL OA 105 GETORMALIONS: sa ied ss danals A ntasd va an wae kteG one eles ce ge eens alk ee 98 external, forces: «sie va ee yee. Reh 624 REE Lucie Bees Sed ek seeds ope an 102, 110 graphic determination of fiker stresses................0000 00s 95 internal “stresses: se 'e4.4. 5g ore heeded aw segue aun ahi ae an oe 87, 95 laws Of LOWS cae au sare. acate ates won ama oa WS eos 107, 109 line of resistance eve. se riag eee eee Pee a Sak ee Bad ed dhe wee 95 MOPMAL “StHESSES 2 ails odie we asaya water AM Hee hate aoe Sian ee a 88, 107 reaction loctis: +04 sezenies de des Bee tae eds VE eaa sed eek ead aces 105 SHEATING StEESSCS: aK sow ay haw SAY Ree ee ees Gre aS aera 92, 109 tangent; Curves: 2vadeen dss euuda gee Se ses Sakedus ae eee ae FSG Sees 105 Without: HINGES: cing seeds Sees eben Ce MOR Latina ela ae te 147 Arches) gvogavasdeees sob tu she $55 Geeaes Hees Ga re oe SEE SE RES SEE 87, 209 combination with cable ........ 0... ccc cece eee eee 272 combination with straight triss 1... 0... . cece eee eee ee eee 270 connected to elastic piers........ 0... . eee eee eee ne 161 CONUNUOUS. wala cuss ca hes MEE a eee ye 187, 192 GEfOTMALIONS acka suede suse Goes Faas Fede Bee 114, 141, 175, 184 displacement of abutments............... 00. e eee eee eee eee 159 elastic: theory .< ess. seeteg eee era mates ede es SHUG twee ease es 279 CXACH THEOL) acsuajet a nae oes GRla eae Ae A Renae Wa nae Seace a 199 wéneral theory. «+ oy csexe gry aesses das sev ae ld chveee cy aoe ETERS 87 HoOriZOntall thrus os oscars sigue oa wawirdiee ayadierd. coe 120, 127, 129, 130 influence lines for stresses... 1.0... 00.0226 ccc c eee cee eeee 95 WASOHLY ANd) sccacwsudas sacked dane vedere regent eres ans 238 OUTICWIAE PONY BOD. iceyis es ists, auotdogh bs ie-ointear hldyelaia as ape nvalaibmeeaGeahe Avarecanens 10 conditions: for equilibrium: 5 cssisseccaerassweaeaiadauae mes sek 25 General. method Of deSiOM vi. ccs ica nwaa eae Viee wee eee eigen ace 8 GORDO eccrine Bini: ih SRE SEALS hae CARNE AEA nde he wasted woesarine gases amaleaeaed 240 Graphic methods, cantilever atell s ..0vecnena es ghees Sheets tear eeenraeneees 178 continuous stiffening truss.......... 0... cece eee ee eee 51 CORE. POINTS: cei sceteniaes HE anAM ere ReEeE Nee Kes ie 96 deflections of arched rib (Plate I)........ 0... cee eee eee 143 deflections of framed arch.......... 0.0... cece cece eee 236 fiber StTESSES: cae Ree ae BS See Ga ANNs OR aed Mette 97 Hat: parabolig debs icant caceakadn cat ware ada tintin 4 ame avers 158 franied arches: .es.yacwsse regard buwed eee Lewd es Bewlew 215, 223, 228 hingleless atcheS: ccc os scscaaiaaad densa Maus da wneud 4 Relais 152, 156 NASON ALCHES wxveseosnequsasarsdkh wats evan ndiawsesgedente 286 stiffened suspension bridges. ...-..-. 66s. ceee eee e eee eens 36, 48 two-hinged arched rib. ......... 6 eee eee eee ete eee 124, 133 two-hinged framed arch............ cece eee cece een eee eee 223 Graphs, moments and shears in arched ribs..............--05. 113, 138, 173 moments in cantilever arch. ... 10.2.0... cece eee eee eee 181 moments in stiffening trusS.... 22.0... esc e ec eeeeeeee eee 57 shears in stiffening truss........ 6... eee e eee eee eee eee eee 62 TT QU CFIC ci ccsiciss Casntba aes each Shee EEA RE COLNE WERRS DOG ae DAG Meena Gts 190 TWorizontal. CensiON: soneV axe sees eee cheba sR App ea a ata 1 stiffened suspension bridge, single span..........--...-.-++ 29, 30 three: SPans coceceserae eas eraser rsaaceaaegdesow nes 39, 40 special CaSCS ... 6... cee eee eee ete ee ete eens 45 Horizontal. Ghrust: