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 The stability of arches, 
 
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THE STABILITY OF ARCHES 
 
THE BROADWAY SERIES OF ENGINEERING HANDBOOKS 
 VOLUME XX 
 
 THE 
 
 STABILITY OF ARCHES 
 
 BY 
 
 ERNEST H. SPRAGUE, A.M.Inst.C.E. 
 
 ASSISTANT AT UNIVERSITY COLLEGE, LONDON; LECTURER AT THE 
 
 WESTMINSTER TECHNICAL INSTITUTE; FORMERLY PROFESSOR OF 
 
 ENGINEERING AT THE IMPERIAL CHINESE RAILWAY COLLEGE, 
 
 SHAN-HAI-KUAN 
 
 WITH FIVE FOLDING PLATES AND FIFTY-EIGHT 
 DIAGRAMS 
 
 LONDON 
 
 SCOTT, GREENWOOD & SON 
 
 8 BROADWAY, LUDGATE, E.G. 
 
 1916 
 
 [A^ rights reserved^ 
 
PREFACE 
 
 The present book is an attempt to put before 
 tbe reader the principles upon which the sta- 
 bility of an arch is determined. Much ingenuity 
 has been displayed in devising satisfactory 
 methods of investigation either for assuring the 
 stability of, or for estimating the actual maxi- 
 mum stresses in, an arch-ring. Most of these 
 depend ultimately on the elastic theory. In 
 using indirect methods for the purpose of sim- 
 plification there is always the danger of losing 
 sight of the degree of approximation attained, 
 and there is the added difficulty always insepar- 
 able from graphical methods of assuring the 
 necessary accuracy. This is particularly the 
 case in the graphical investigation of the stresses 
 in an arch-ring, and therefore particular atten- 
 tion has been given to the elastic theory, which 
 is our ultimate standard of reference. 
 
 The general deductions of this theory have 
 been confirmed by all recent experience, par- 
 ticularly for arches which are moderately flat ; 
 and as modern arches in masonry and concrete 
 are usually of this type, the elastic theory af- 
 fords the most satisfactory basis of investigation. 
 
VI PEEFACE 
 
 The author desires to acknowledge particu- 
 larly his indebtedness to Prof. Melan's " Theorie 
 des Gewolbes und des Eisenbetongewolbes im 
 besonderen," Handbuch fiir Eisenbetonbau, 
 Vol. I, and to " Leitfaden fiir das Bntwerfen 
 und die Berechnung gewolbter Briicken," by 
 G. Tolkmitt. 
 
 EENEST SPEAGUE, A.M.Inst.C.E. 
 
 University College, London. 
 May, 1916. 
 
CONTENTS 
 
 Preface ..... . . '" 
 
 CHAPTER I 
 
 PAOES 
 Introduction . . . . . . . . .1-10 
 
 Histovioal Details — Early Arches — Classification — 
 Terms Used — Particulars o£ Bridges Constructed 
 — Strength of Materials Used. 
 
 CHAPTER II 
 
 The Thbee-Pinned Arch H-36 
 
 Problem of the Arch — Critical and True Lines of 
 Pressure — The Linear Arch — Conditions Neces- 
 sary for its Determination — Three-Pinned Arch 
 -^Its Advantages— Construction of the Line of 
 Pressure, and Determination of the Thrust, Shear- 
 ing Foroe and Bending Moment— Eddy's Theorem 
 —Illustration of the Methods Used, and Compari- 
 son of the Results — Three-Pinned Spandrel-Braced 
 Arch — Example of ihe Determination of the 
 Stresses therein (i) by a Force Diagram, (ii) by 
 Influence Lines. 
 
 CHAPTER III 
 
 The Elastic Theory of the Arch 37-48 
 
 Austrian Experiments — Experiments with Polarized 
 Light — Horizontal and Vertical Displacements 
 Due to Bending, to Axial Thrust, and to Change 
 of Temperature — Fundamental Equations — Re- 
 marks on the Assumptions Made — Allowance for 
 Thrust and Change of Temperature. 
 
 CHAPTER IV 
 
 The Two-Hinged Arch 49-64 
 
 Advantages and Disadvantages — Determination of the 
 Line of Pressure^Horizontal Thrust, Graphically 
 and by Calculation — Method of Reaction Locus 
 Applied to a Parabolic Arch and to a Circular 
 Arch — Bending Moment by Influence Line 
 Method, 
 
vm CONTENTS 
 
 CHAPTER V 
 
 PAGES 
 
 The Hinseless Akch . . .... 65-82 
 
 Advantages and Disadvantages — General Expression 
 for Bending Moment — Example of Determination 
 of the Horizontal Thrust, Bending Moment and 
 Stresses — Method of Influence Lines— M. Mes- 
 nager's Method. 
 
 CHAPTER VI 
 
 Masonry and Conogete Arches . . . 83-102 
 
 Critical Line of Pressure — Conditions of Stability — 
 Formula for Stresses — Rule of the Middle Third — 
 Principle of Least Work — Position of True Line of 
 Pressure^Joints of Rupture — Curves of Minimum 
 and Maximum Thrust— Construction of the Critl. 
 oal Line of Pressure^Reduoed Load Curves 
 Method of Fictitious Joints^Oglio Bridge — Sirq^ 
 plified Treatment for Small Arches — .Asymmetric 
 Loading — Adjustment of the Arch to Suit the 
 Load — Adjustment of the Load to Suit the Arch — ■ 
 Stability of the Abutments — Intermediate Piers, 
 
 CHAPTER VII 
 
 Desion of Masonry and Concrete Arches . . 103-120 
 Problem to be Considered, and its Solution — Empiri- 
 cal FormulsB for Thickness of Arch-Ring — Form of 
 the Arch-Ring — Thickness of Abutments — ^Dimen- 
 sions of Piers — ^Tolkmitt's Method — ^Thickness of 
 the Arch-Ring — Best Form of Arch — Example. 
 
 CHAPTER VIII 
 
 Loads and Stresses 121-134 
 
 Effect of the Load — Weight and Strength of Arch 
 Materials — Dead Load — Live Load — Highway 
 Bridges — Railway Bridges — Equivalent Distributed 
 Load — Example — Stresses in the Arch-Ring — • 
 Position of the Critical Sections — Stresses by the 
 Elastic Theory — Example. 
 
 Appendix . . 135-138 
 
 Ordinates of a Circular Arc — Constructions for a Para- 
 bola — Tangent to a Parabola — Construction tor a 
 Flat Circular Arc. 
 
 Index 139-14X 
 
CHAPTBE I. 
 
 INTRODUCTORY. 
 
 1. Whether we regard the arch from an architec- 
 tural or from an engineering standpoint, that is to 
 say, from its artistic or its scientific side, the elegance 
 of its form and its combined lightness and strength 
 make it an object of the first interest to constructors. 
 The scientific aspect of the subject dates no further 
 back than about 200 years, but during that period it 
 has received its full share of consideration by mathe- 
 maticians and engineers. On the architectural and 
 practical side the use of the arch goes back to a re- 
 mote antiquity, although not so remote but that in- 
 dications of its origin are traceable. In the English 
 translation of the Bible the arch is only once men- 
 tioned (Bzek. XL. 16), and then only on account of 
 a mistranslation, we are informed. In Chaldea and 
 Egypt in early times we find only the simplest forms 
 of arch, and in many ruins of ancient cities, such as 
 Persepolis, no trace of it is found. In all probability 
 it had its origin and reached its highest development 
 in China ; for scattered over this vast country from 
 south to north are fine examples of arched bridges 
 of great antiquity. In the mountains which divide 
 Manchuria from China, fine arches exist in the Great 
 Wall built some 300 years B.C., which are still in 
 excellent preservation, and the bridge of arches which 
 1 
 
2 THE STABILITY OP ARCHES 
 
 Marco Polo speaks of crossing in the thirteenth cen- 
 tury serves its purpose equally well to-day ; whilst 
 all over China the arch has figured as a favourite 
 feature in architecture and landscape gardening, as 
 ■ may be seen in their pictures and pottery. 
 
 2. Owing to the tendency of the arch under nor- 
 mal conditions to sink at the crown and thereby to 
 cause spreading at the abutments, it requires strong 
 lateral support, in consequence of which the earliest 
 known examples at Nippur in Chaldea, about 4000 
 B.C., and at Dendera and other places in Egypt 
 about 3500 to 3000 B.C., were used below ground 
 level. These were of unburnt brick, and not more 
 
 r' 
 
 .^^^^ S^ 
 
 ^^ 
 
 i 
 
 Pig. 1. Pig. 2. 
 
 than 6 ft. span. The favourite construction in the 
 East at this early period was to build the arch, if we 
 may so call it, of horizontal slabs projecting one over 
 the other, as shown in Figs. 1 and 2, a fine example 
 of which is the Treasury of Atreus at Mycenae. Ex- 
 amples of this method of construction are found in 
 Egypt, Greece, Assyria, Italy, Mexico, India, China, 
 and elsewhere. Such structures, however, cannot 
 be regarded as true arches, because there is no hori- 
 zontal thrust, and this must be regarded as the 
 essential feature of the arch. Horizontal thrust, in 
 fact, appears at first to have been regarded with dis- 
 favour, for, as the Hindus say, . " the arch never 
 sleeps," meaning, of course, that it is always exerting 
 a thrust tending to its destruction. 
 
INTRODUCTOBT 
 
 3 
 
 Primitive arches of the form shown in Fig. 3 are 
 found at Deir-el-Bahri and in the round towers of 
 Ireland. Fig. i is an example from the Great Pyra- 
 
 FiG. 3. 
 
 Pig. 4. 
 
 mid ; whilst in Fig. 5, which occurs also in the 
 Pyramids, is shown a combination of both types, 
 
 Fig. 5. 
 
 whilst Fig. 6 shows another arrangement in the tomb 
 of the pyramid at Gizeh. In this we have an early 
 example of the semicircular arch, a favourite form 
 in ancient China, which appears to have been intro- 
 duced into Europe by the Etruscans, who were of 
 eastern origin. This people transmitted it to the 
 Eomans, who excelled in this class of construction, 
 of which they have left numerous examples, the 
 
4 THE STABILITY OF ABCHBS 
 
 Pont du Gard and the Aqueduct at Tarragona being 
 particularly fine specimens. 
 
 3. The scientific treatment of the subject began 
 about 1712 with Lahire. Previous to this we have 
 no evidence, and we can scarcely suppose that the 
 theory of construction in general had received much 
 attention, because the general condition of me- 
 chanical science was not such as to permit it. 
 Most of the bridges of ancient date, outside of mili- 
 tary bridges, were constructed by builders who be- 
 
 FiG. 6, 
 
 longed to a close corporation, and these expert 
 constructors moved from place to place constructing 
 bridges from their practical knowledge of the art, 
 without much consideration, we may suppose, of 
 the mechanical principles involved. Lahire (1712), 
 however, considered the equilibrium of the voussoirs, 
 which form the arch-ring, by treating them as a 
 system of frictionless blocks acted on by the load and 
 by the mutual reactions between their faces. Eytel- 
 
INTEODUCTORY 5 
 
 wein advanced the discussion by introducing the 
 frictional resistance of the surfaces; and Couplet 
 (1730), Coulomb (1773), Boistard (1822), Navier, and 
 others are associated with the early history of the 
 theory. M6ry first introduced the idea of the line 
 of pressure, whilst the names of Lam6, Clapeyron, 
 Eankine, Moseley, and many others of first-class re- 
 putation are associated with the later development 
 of the subject. 
 
 4. From an architectural standpoint arches may 
 be classified as i^— 
 
 (a) Semicircular or Norman arches. 
 
 (6) Segmental, that is, less than semicircular. 
 
 (c) Pointed or Gothic. 
 
 (d) Horseshoe or Etruscan. 
 
 (e) Elliptic or pseudo-elliptic. 
 
 5. The masonry arch consists of the arch-ring and 
 
 Spandrel 
 
 Pig. 7. 
 
 its abutments, which take the end thrust (Pig. 7), 
 and intermediate piers perform the same function in 
 the case of a series of arches. The arch-ring itself 
 
6 THE STABILITY OF ARCHES 
 
 consists of wedge-shaped stones or voussoirs care- 
 fully fitted, or of bricks bonded together. The cen- 
 tral voussoir at the crown of the arch is known as 
 the keystone, and those adjacent to the springings as 
 the imposts. The bearing surfaces are known as the 
 bed-joints, and these are perpendicular or nearly so 
 to the under surface or soffit of the arch, which sur- 
 face is also spoken of as the intrados ; whilst the 
 outer surface is called the extrados. The spandrels 
 or haunches are the lower portions of the extrados, 
 and the spandrel filling or backing is the material 
 between the extrados and the roadway. 
 
 6. Modern arches may be classified as : — 
 (a) Metal arches. 
 
 (&) Masonry and brickwork arches, 
 (c) Concrete and reinforced concrete arches. 
 
 The main difference in theory consists in the fact 
 that metal arches may be subject to considerable 
 tensile stress, whereas masonry and concrete arches 
 are designed with a view to completely avoiding any 
 tension in the arch-ring under all conditions of load- 
 ing, whilst reinforced concrete arches, though often 
 designed in the same way, may be permitted to suf- 
 fer tensile stress if properly reinforced, with a con- 
 sequent saving of material. 
 
 Increased knowledge of theory has naturally 
 tended to lighter construction and increased length 
 of spans. Few of the ancient bridges had long 
 spans. Those in China and the East are always 
 quite short. In England the longest span masonry 
 arch is over the Dee at Chester, the clear span being 
 203 ft., whilst in America the longest span is at 
 Washington, of 220 ft. The masonry arch near 
 
INTEODUCTOEY 7 
 
 Trezzo in Italy, built in 1380 and destroyed in 1416, 
 is therefore remarkable in having a span of 251 ft. 
 la 1906 a three-hinged arch of plain concrete wag 
 completed at Ulm in Germany of 215 ft. span, but 
 for long span bridges of this type armoured concrete 
 is generally preferred. Such an arch was con- 
 structed over the Tiber in Italy, having a span of 
 328 ft. with a rise of 32-8 ft. (See " Engineer," 16 
 May, 1913.) 
 
 7. For very large spang it is necessary to use steel 
 and iron. The table on next page, giving particulars 
 of various typical arches arranged in accordance with 
 their spans, will be found useful for the sake of 
 comparison. 
 
 8. The strength of masonry and concrete arches 
 loaded under similar conditions varies directly as the 
 strength of the material and the depth of the key- 
 stone, and inversely as the radius of curvature at the 
 crown. These considerations indicate the possibili- 
 ties at our disposal for increasing the span. The 
 following figures are approximately the crushing 
 strengths of the materials in common use. 
 
 Crushing Strengths of Arch Materials. 
 
 Good brickwork in Portland cement . 150 tons/ft. '^ 
 
 Cement concrete .... 100 to 180 
 
 Granite 
 Bessemer steel 
 Cast steel 
 Cast iron 
 
 1600 
 3000 
 3600 
 5000 
 
 From these figures we see that, under proper con- 
 ditions, cast iron should be a very suitable material 
 for arches, though it has in recent years been custo- 
 mary to regard it with suspicion on account of a few 
 
THE STABILITY OF ARCHES 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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INTRODUCTORY 
 
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10 
 
 THE STABILITY OF ARCHES 
 
 isolated failures. There seems no reason, however, 
 why cast iron , should not be used for structures in 
 compression, or subject to moderate tension, pro- 
 vided it is used in such a way as to avoid unknown 
 initial strains. 
 
 If we compare the strengths of different arches as 
 suggested above, with the help of these figures, we 
 find a great variation in their relative strengths. 
 Thus, for example, the figures in the last column 
 below represent the relative strengths of the bridges 
 named : — 
 
 Name. 
 
 Material. 
 
 Crushing 
 
 Strength = s. 
 
 
 3 II 
 
 ■sl 
 
 ad 
 R' 
 
 
 
 
 
 ■32 
 
 e3« 
 
 
 Maidenhead 
 
 Bridge 
 Caester Bridge 
 Plauen ,, 
 London „ 
 
 Brick in Cement 
 Sandstone 
 Hard Slate 
 Granite . 
 
 150 tons/fi;.2 
 
 400 „ 
 1670 „ 
 1600 
 
 5-25 
 
 4-0 
 
 4-11 
 
 4-75 
 
 169 
 140 
 214 
 162 
 
 4-66 
 11-4 
 38 
 47 
 
CHAPTBE II. 
 
 DISCUSSION OF THE PROBLEM. 
 
 Theee- Pinned Arch. 
 
 9. The problem to be solved in connection with the 
 arch may present itself in various ways. Firstly, the 
 form of the arch may be given and the material of 
 which it is constructed, the nature of the supports and 
 the load liable to come upon it, and it may be re- 
 quired to determine the degree of stability it pos- 
 sesses and the maximum stresses that may arise ; 
 secondly, the distribution of the load may be given, 
 and it is required to determine the form of arch best 
 adapted to the circumstances ; thirdly, the form of 
 the arch may be fixed, and the load distribution may 
 be arranged to suit it. 
 
 Two modes of procedure are in common use in 
 carrying out the investigation of the stability of 
 masonry and concrete arches. The simple and older 
 method which is best adapted to the treatment of 
 small spans is based upon the determination of the 
 critical line of pressure for the arch and the given 
 load upon it ; whilst the more modern and accurate 
 method aims at finding the true line of pressure by 
 means of the elastic theory. This method is univer- 
 sally applied to metal arches which are statically 
 indeterminate ; and though less reliable in the case 
 of masonry and concrete the results have been shown 
 (11) 
 
12 
 
 THE STABILITY OF AECSES 
 
 experimentally to be trustworthy within the limits of 
 accuracy required. 
 
 10. The Linear Arch.— Let ABCD (Fig. 8) be one 
 of the voussoirs of a masonry arch-ring, and let BFAD 
 be the load upon it. Then if Wj be the resultant of 
 their weights, and P, Q be the resultant reactions of 
 the adjacent voussoirs, then these three forces P, Q, 
 and Wj being in equilibrium, must pass through the 
 
 Pig. 8. 
 
 same point O and be proportional to the sides of a 
 triangle S, 1, 2 having its sides parallel to the three 
 forces, by the well-known theorem of the triangle of 
 forces. Thus the side S, 2 = g represents the pres- 
 sure Q on the face CD ; and this force reversed then 
 becomes the reaction on the adjacent block CDHG ; 
 it intersects the resultant load Wg on the second 
 voussoir and so determines the direction of the thrust 
 E on the next one, and so on. In this way, proceed- 
 ing from voussoir to voussoir, we obtain a link- 
 polygon for the given loads whose links represent 
 the lines of action of the forces acting at the voussoir 
 joints. This link-polygon is the line of pressure for 
 
DISCUSSION OF THE PROBLEM 13 
 
 the arch, and since a polygon of rigid jointed rods 
 coinciding with it would be in unstable equilibrium, 
 this polygon is often spoken of as the " linear arch," 
 this term being applied to the ideal arch for the 
 given load system, which has no bending moment 
 at any point along it. It is clear, however, that the 
 position of the linear arch is indeterminate unless 
 the magnitude direction and position of one of the 
 forces is known ; in other words, although any 
 number of polygons may be drawn for the given 
 loads, three independent conditions are necessary to 
 fix that particular one which is their true line of 
 pressure, and unless three such conditions are known 
 the problem of finding it is statically indeterminate, 
 that is to say, is impossible of solution by means of the 
 simple laws governing the statics of rigid bodies. If, 
 for example, we know three points through which the 
 line of pressure must pass, as in the case of a three- 
 pinned arch, when the friction of the pins is neglected, 
 then the problem can be easily solved ; but if one of 
 the ends is fixed, by doing away with the pin-joint, 
 then some other condition becomes necessary, such 
 as a knowledge of the direction of the thrust at the 
 fixed end ; and if both ends are fixed, by doing away 
 with the two end joints, then some other condition 
 must be given or assumed before the true line of 
 pressure can ,be found. In the case of a masonry or 
 concrete arch therefore, which is usually without 
 joints, no points at all on the line of pressure being 
 known, three other conditions are necessary for its 
 determination, and these are derived from the elastic 
 properties of the arch-ring. 
 
 11. For the above reasons, and because the exact 
 
14 THE STABILITY OF ARCHES 
 
 determination of the linear arch is both troublesome 
 and uncertain in such cases, it has often been pre- 
 ferred to make metal arches pin-jointed at the 
 crown and springings ; and even in the case of 
 masonry and concrete arches semi-rigid joints are 
 sometimes introduced at the crown and springings 
 to fix more definitely the position of the line of 
 pressure. Thus in the case of a bridge over the 
 Enz, near Hofen, plates of lead ^ in. thick were 
 placed over the centre third of the joints, thereby 
 ensuring tliat the line of pressure 
 lies within this space; and in 
 other cases granite rockers or 
 hinges of cast iron or steel have 
 been used, as shown in Fig. 9 ; 
 whereby, if we neglect the effect 
 of friction, the line of pressure is 
 constrained to pass through the 
 axis of the joint, and in any case 
 cannot pass far from it unless the joint seizes through 
 age and rust or excess of pressure. 
 
 12. The Three-pinned Arch. — If we assume that 
 the line of pressure passes through the axis of the 
 three pins, then its exact position becomes easy of 
 determination from simple statical considerations 
 alone ; and for this reason, the results being more 
 assured, arches of this type have been preferred, 
 where theory alone would recommend an arch-ring 
 without joints. Moreover, the effect of temperature 
 changes has no appreciable influence on the stresses 
 in this type of arch. 
 
 Method I. Determination of the Line of Pres- 
 sure. — Suppose A, B, C (Fig. 10) to be the three 
 
DISCUSSION OF THE PEOBLBM 15 
 
 pins of an arch, and Wj, W^, Wj, W^ to be any 
 system of vertical loads upon it. Set down the 
 loads to scale along a vector line 0, 1, 2, 3, 4, and 
 with any pole O construct a link-polygon Aabcde, 
 which may begin at A, and which ends on the 
 vertical through C at e. Then this will be an 
 equilibrium polygon for the loads, but will not be 
 the linear arch unless it passes through A, B, and C. 
 Join Ae, and draw 05 parallel to it. This deter- 
 
 PlG. 10. 
 
 mines the vertical forces at A and G, which will be 
 the same wherever the pole is taken. But if the 
 link-polygon is to pass through A and C, its closing 
 line must be parallel to AC, and therefore the pole 
 must lie on a line parallel to AG drawn through 5. 
 In order to make the polygon also pass through B 
 it must have a vertical depth at B equal to BE ; 
 
16 
 
 THE STABILITY OF ARCHES 
 
 hence its vertical depth B'E' must be increased so as 
 to make it equal to BE, and this may be effected by 
 decreasing the polar distance in the ratio B'E' : BE. 
 If then a new pole 0' be taken on the line through 
 5 parallel to AC and so that O'H' : OH = B'E' : BE, 
 then a new polygon starting from A should pass 
 through B and C, and this will be the linear arch 
 required ; that is to say, it will be the true line of 
 pressure. 
 
 Pig. H. 
 
 13. Method II. — This method has the advantage 
 that it may be applied as well to forces which are 
 not vertical. For example, suppose A, B, C (Fig. 11) 
 are the pins and W^, Wg, Wj, W^ are any loads. 
 Construct their vector figure 0, 1, 2, 3, 4 and with 
 any convenient pole Oj construct a link-polygon 
 
DISCUSSION OF THE PROBLEM 17 
 
 a, b, c, d for the loads between A and B, the first 
 and last links of which meet in e. Then a line Ej 
 through e parallel to 0, 2 represents the line of action 
 of the resultant, and if we draw through A and B 
 parallels to it, to cut the polygon in a and d, and 
 join ad, a line through Oj parallel to ad divides the 
 resultant 0, 2 at the point 5, into two parts which 
 represent the pressures on the pins ; but this polygon 
 will not be part of the linear arch unless it also 
 passes through A and B, and in order to do this it 
 must have its closing line ad to pass through A and 
 
 B. Join AB and through 5 draw 55' parallel to it ; 
 then if a new pole be taken anywhere on this 
 line, the new polygon starting through A will pass 
 through B also. Similarly, if we draw any link- 
 polygon for the loads between B and with any 
 pole Og, and find the pressures at B and C as before, 
 viz. 2, 6 and 6, 4, a line through 6 parallel to BC 
 determines the line on which the final pole must be 
 taken to make the polygon pass also through B and 
 
 C. Hence if we take the point as pole and start 
 a polygon from A, it should pass also through B 
 and C and be therefore the required linear arch. 
 
 14. Method III. The Beaction Locus. — This 
 method consists in finding the reactions due to each 
 individual load and from them determining the 
 resultant reactions by compounding the individual 
 components. One of these resultants being known 
 in magnitude, direction, and position, we have the 
 three necessary conditions for fixing the line of 
 pressure. Let A, B, C (Fig. 12) be the three pins 
 of the arch and let Wj, Wg, etc., be any loads on it. 
 Then if we first suppose all the loads removed 
 2 
 
18 
 
 THE STABILITY OP AECHES. 
 
 except Wj, the line of pressure for this load will 
 pass directly through B and C, because there is no 
 load between these points to deflect it. Produce 
 BC therefore to cut the line of action of the load 
 Wj in D. Then the reaction at A must also pass 
 through D, and AD is therefore its direction. A 
 triangle of forces 0, 1, 2 being now drawn, the re- 
 actions 1, 2 at B and 2, at A due to Wj are found. 
 Similarly, if we suppose all the loads except W^ 
 
 Fig. 12. 
 
 removed, the line of pressure will pass directly 
 through A and C. Hence if we produce the line 
 joining A to C to out Wj in B, the reaction at B 
 must pass through E, and if 1-3 represent "Wj, a 
 triangle of forces 1, 3, 4 will give the values of the 
 reactions at A and B due to ■W2. 
 
 If these are the only two loads we may complete 
 the parallelogram 2, 1, i, 0, and it will be seen that 
 3, O is the resultant reaction at B, and Oo is the 
 
DISCUSSION OF THE PEOBLEM 19 
 
 resultant reaction at A. If therefore we draw a 
 link-polygon with as pole, passing through A, it 
 should pass also through C and B and will be the 
 linear arch required. 
 
 In this way the reactions due to any number of 
 loads may be found for each in turn, and these may 
 then be compounded by means of a vector polygon 
 to give the resultant reactions at A and B. Either 
 of these being known the line of pressure can be 
 drawn without further difi&culty. 
 
 If we suppose the loads Wj and Wj to continually 
 change their position, the points D and E will move 
 along two straight lines which are the locus of the 
 points in which the reactions intersect as a load moves 
 across the arch. For this reason they are known 
 as the reaction locus for this particular case, and it 
 will be seen that as soon as the reaction locus is 
 known, it will be no more difficult to find the re- 
 actions in the case of a two-pinned arch than it is in 
 the present case, though the locus itself is not so 
 simply determined. 
 
 15. Determination of the Thrust, Shearing 
 Force, and Bending Moment at any Point by 
 means of the Line of Pressure. Eddy's Theorem. 
 — The effect on any section which is produced by the 
 external load is usually analysed into (1) a thrust, (2) 
 a shearing force, and (3) a bending moment. When 
 these are known we have all that is necessary for 
 the calculation of the stresses. Now all of these are 
 at once readily found at any section so soon as the 
 line of pressure is known. For suppose abed (Fig. 
 13) to be a portion of the line of pressure for an 
 arch and AB to be any section of the arch-ring, 
 
20 
 
 THE STABILITY OF AECHES 
 
 whose axis passes through the centroid of the sec- 
 tion C, and let the resultant thrust E acting along 
 cb cut the plane of the section in the load-point L. 
 Then the magnitude of E will be known from the 
 corresponding ray 0, 1 in the vector figure, and if 
 
 Pig. 13. 
 
 we resolve it into components le and eO normal and 
 parallel to the section AB, these components will be 
 the values of the normal thrust N on the section and 
 the shearing force S. Also because the resultant 
 passes through L at a distance CL from the axis of 
 the arch it causes a bending moment whose value is 
 E X r, where r = CP is the perpendicular distance 
 of the resultant from the point C. Draw CQ verti- 
 
DISCUSSION OF THE PEOBLEM 21 
 
 cal to cut E in Q. Then the triangles CPQ, TOl 
 are similar, and therefore 
 
 CQ : CP = 01 : H 
 or H . CQ = 01 . OP 
 
 But 01 measures the resultant E, and CP is its 
 lever-arm about C. Therefore 01 . CP is the bending 
 moment at the section and is equal to H . CQ ; that 
 is to say, the bending moment at any section is 
 measured by the product of the horizontal thrust H 
 measured in the scale of loads, into the vertical in- 
 tercept between the axis of the arch and the line of 
 pressure, measured in the scale of lengths. 
 
 This important theorem is known as Eddy's 
 Theorem, and is very convenient in the graphical 
 treatment of the arch because the intercepts between 
 the line of pressure and the arch-axis give a visual 
 perception of the relative values of the bending 
 moments at various points, similar to that which 
 the bending moment diagram does in the case of 
 beams under vertical loads. 
 
 16. Application of the Preceding Methods to 
 the Caseof a Three-Pinned Arch with Distributed 
 Loading. — Let ACB (Fig. 14) be a circular arch of 
 150 ft. span and 15 ft. rise, with pin joints at A, B, 
 and C, and suppose the dead load to be 26 cwt. per 
 foot run over the whole span, with an additional 
 live load of 18 cwt. per foot run on one- half the 
 span. It is required to find the thrust, shear, and 
 bending moment at any point. 
 
 Method I. By Means of the Line of Pressure. — 
 In a case of this kind the line of pressure may be 
 expected to lie so near the axis of the arch itself 
 
22 
 
 THE STABILITY OF AECHBS 
 
 under ordinary conditions that the vertical intercepts 
 between the two curves will be too small to ensure 
 reliable measurement. It will therefore be an ad- 
 vantage to distort the vertical scale of the drawing 
 
 by magaifying its vertical dimensions, say five times. 
 This should not be done graphically, because the 
 errors will be multiplied to the same extent, but the 
 ordinates should be calculated by the formula or by 
 
DISCUSSION OF THE PROBLEM 28 
 
 the Table oa page 135. In this way we get the el- 
 liptical curve ACjB which is a parallel projection of 
 the circular arc ACB. BCj and ACj are then joined 
 and produced to cut the resultants of the two loads 
 on the ribs ACj, BCj in D and B. This determines 
 the directions of the reactions at A and B due to 
 each of these loads taken separately, and if we now 
 set down 0.1= 97"5 tons and 1 . B = 165 tons the 
 magnitudes of the reactions due to these loads in- 
 dividually will be given by the triangles Pj . . 1 and 
 Pg . 1 . B whose sides are drawn parallel to the 
 directions of the forces, and if we complete the 
 parallelogram OPjlPj, Oo, OB will be the resultant 
 reactions at B and A. The polar distance scales 
 65 tons, but the actual horizontal thrust of the 
 arch will be five times greater than this, on account 
 of the vertical distortion, and will therefore be 
 65 X 5 = 325 tons. 
 
 Next draw FG parallel to O . 1 and join AF, GB. 
 These lines will be tangents to the line of pressure, 
 and since this curve will, for uniform loading, con- 
 sist of parabolic arcs, these arcs may be drawn in to 
 touch the tangents AF, FCi and C^G, GB by the 
 method described on page 138. 
 
 This determines the line of pressure, and the 
 bending moment at any point may now be found by 
 measuring the vertical ordinate between the line of 
 pressure and the arch- axis in feet and multiplying it 
 by the polar distance, viz. 65 tons. For example, 
 at I span from the left-hand abutment the ordinate 
 scales 4'6 ft. and the bending moment there is there- 
 fore 4-6 ft. X 65 tons = 299 tons ft. 
 
 The thrust in the arch at this point may be found 
 
24 
 
 THE STABILITY OF AECHES 
 
 by drawing the tangent to the curve, viz. HH^ and 
 Oh parallel to it. Its actual direction can be found 
 by making ak = ^ ah, when Ok will be the direction 
 required, and if we resolve Ok into components Ofc', 
 k'k parallel respectively to the tangent and radius of 
 the arc at M, these will give the values of the normal 
 thrust and of the shearing force at M, and in the 
 . same way the values of the thrust, shear, and bending 
 moment at any other point can be found, and curves 
 of thrust and shear constructed therefrom. 
 
 17. Method II. By Calculation. — Let Va, H (Fig. 
 15) be the vertical and horizontal components of the 
 
 148i tons. 
 
 Fig. 15. 
 
 reaction at A, and Vb, H be those at B. 
 
 Then Vb = 97-5 + ^ x 67-5 = lU-l tons, 
 and Va = 262^ - lU} 
 Taking moments about 0, we have 
 
 114| X 75 = H X 15 + 97-5 x 37-5 = 
 328i tons = the hor. thrust. 
 
 Further tan eA = -2=3-11 
 
 Vb^ 1141 
 H 328 
 
 tanfi^B = ~ = 
 
 ■■ -4514. 
 ^=•3486 
 
 .-. 61^ = 24° 18'; 
 
 6,. = 19° 13'. 
 
 Ea = H sec 6Ia = 3281 x 1-0972 = 359-9 tons. 
 Eb = H sec (9b = 3281 x 1-0590 = 347-4 tons, 
 
DISCUSSION OF THE PEOBLBM 
 
 25 
 
 To find the bending moment at the point M 
 (Fig. 16). 
 
 The bending moment at M = Va x 37-5 - H x y 
 - -^ X 165 X 18-75. 
 Now ^ = R cos <^ - R cos <^j = E (cos cf> - cos <^|,) 
 and 752 = R2 - (S - 15)^ = 30 E - 225 
 whence E = 195 ft. .-.00= 180 ft. 
 37-5 
 
 sin <f)-- 
 
 195 
 
 •1923. .-. <^ = ir 5'andcos<^=-9812. 
 
 Also cos ^0 =^ = '9231. 
 195 
 
 .-. 2/ = 195 (-9812 --9231) = 195 x -0581 = 11-33 ft. 
 
 Fig. 16. 
 
 .-. the bending moment at 
 M = 148|- X 37-5 - 328i x 11-33 - 82-5 x 18-75 
 = 5555 - 3718 - 1547 = 290 tons ft . 
 
 18. Method III. By Influence Lines. — In this 
 procedure the bending moment at any section due to 
 a unit load in any given position is found by draw- 
 
26 
 
 THE STABILITY OP ARCHES 
 
 ing an influence diagram for the particular section 
 considered. (See "Moving Loads by Influence 
 Lines " in this Series.) Thus, if we require to find 
 the bending moment at any section D (Fig. 17) an 
 influence line must be drawn for this particular sec- 
 tion to show the bending moment there for a unit 
 
 <i. B, 
 
 Pia. 17. 
 
 load placed anywhere on the span. Now the actual 
 bending moment at any section is the difference 
 between the positive bending moment due to the 
 vertical forces, less the negative bending moment 
 due to the horizontal thrust. But when a unit load 
 acts at any point distant x^ from the end A, the 
 bending moment at D due to the vertical forces is rep- 
 resented by a triangle AjD^B, for which 
 1 X aij (Z - a;,) 
 
 D,I),= 
 
 I 
 
 tons ft. 
 
 Also 
 
 I = 1 X X,, Yu = 
 
 •^1 
 T 
 
 and taking moments about C 
 
 H.2/„ 
 
 X, I 
 
 H = ^-J-, from which it appears 
 
DISCUSSION OF THE PROBLEM 27 
 
 that H varies as x^, and that the influence diagram 
 for H ig therefore a triangle whose maximum ordinate 
 
 is C,C.,. But when the load is at the centre a;, = — 
 
 and therefore the maximum value of H is — , 
 
 whilst the bending moment at any point x due to 
 the force H will be Hy. The ordinates of the 
 triangle AjC^Bj therefore when multiplied by y will 
 give the bending moment at the point D due to the 
 horizontal thrust, and the actual bending moment at 
 D will therefore be given by the vertical intercept 
 between the two triangles AjDjBj and AjC^Bj. 
 
 Thus in the present case we construct an influence 
 diagram for ^ span by setting up 
 
 CA =J^= ^^Q ^ ^^'^^ ■-= 28-325 tons, 
 ^ ^ iy, 4 X 15 
 
 the ratio being actually multiplied by the unit load 
 
 of 1 ton say. If we suppose CjCj drawn 2-8325 ins. 
 
 long the scale will be 1 in. = 10 tons. Join AjCj, 
 
 BjCj. Then the triangle A^CjBj is the influence 
 
 diagram for the horizontal thrust H, or the " H-line " 
 
 as it is called. Next set up 
 
 DjDj = ^1^^ ~ '^i) = Ix 112-5 = 28|. tons, 
 
 and join D^A^, D^Bj. Then AjD^Bj is the influence 
 diagram for the vertical forces. 
 
 The ordinate at any point of the shaded area will 
 now give the bending moment at D due to the unit 
 load at the given point. The bending moment at 
 D due to any given system of loads in any position 
 may now be readily found by multiplying the 
 ordinate under each load by the magnitude of the 
 
28 THE STABILITY OF AECHBS 
 
 load and summing the results. Also by a well-known 
 property of influence lines, if the load is uniformly 
 distributed the bending moment will be given by the 
 area of the influence diagram under the load multi- 
 plied by the load per unit length. 
 
 The maximum and minimum effects may therefore 
 be easily investigated in this way. 
 
 In the present case AjNi = 59'75 ft., DjDj = 
 13-96 ft., CA = 9-58 ft. 
 
 Hence the area AiDjN = i x 59-76 x 13-96 = 
 417-2 tons ft. ; 
 
 the area NC1C3 = i x 9-58 x 15-25 = 73-05 
 tons ft. ; 
 
 and the area Bfifi^ = ^ x 9-58 x 75 = 359 3 
 tons ft. 
 
 .-. the positive bending moment due to 
 2-2 tons p. ft. = 2-2 (417-2 - 73-05) = 757-1 tons ft., 
 
 and the negative bending moment due to 
 1-3 tons p. ft. = 1-3 x 359-3 = 467-1 „ „ 
 
 . -. the resultant bending moment = 290-0 tons ft. 
 
 thus checking exactly with the calculated result. 
 
 This will not, however, be the maximum bending 
 moment as is often assumed, because it will be seen 
 at once that the maximum value will occur when the 
 live load covers the left-hand portion of the arch 
 up to the point N only, in which case the bending 
 moment at ^ span will be greater than before by the 
 amount represented by the triangle NCiCj, viz. 
 7305 tons ft. The maximum bending moment at ^ 
 span is therefore 290 -f 73 = 363 tons ft. 
 
 19. Example of a Three-pinned Spandrel- 
 braced Arch. — -This type of arch is very economical 
 
stability of Arches.] 
 
 Fia. 18. 
 
 [To face p. 28. 
 
DISCUSSION OF THE PROBLEM 29 
 
 and convenient of construction in positions to which 
 it is adapted, and is consequently commonly used. 
 
 The maximum and minimum forces occurring in 
 each member are best found in one or other of the 
 following ways. First, a unit load is placed at any 
 one of the top joints, e.g. at Tg (Fig. 18), and the 
 reactions at B„, BqI due to this load are found by 
 constructing the triangle of forces hb^a, fig. 1. A 
 reciprocal figure or force diagram is then drawn in 
 the usual way for the whole structure, from which 
 the forces in each member may be scaled off. These 
 forces are tabulated, and when each is multiplied by 
 the actual dead or live load which occurs at any joint, 
 the force in any member which that load produces 
 will be known ; and it will therefore be easy to select 
 that combination of loads which produces either the 
 maximum or the minimum force in any particular 
 member. By the second method, an influence line 
 is drawn for the moment of resistance of each bar in 
 the structure, from which the maximum or minimum 
 force which can occur in it may be deduced vof the 
 usual way, by loading exclusively either the positive 
 or negative areas. 
 
 In order to compare the two methods, which may 
 be used mutually to check each other, we will con- 
 sider the case of the three-pinned arch shown in Fig. 
 18, which we will suppose to carry a uniform dead 
 load of 1-^ tons per foot and a live load of 1 ton per 
 foot, distributed along the upper chord by means of 
 a platform. The lower panel-points lie on a para- 
 bolic arc of 90 ft. span and 15 ft. rise. 
 
 Method I. — Suppose a 1 ton load placed in suc- 
 cession at each of the joints T(„ Tj, T2, T3 of the top 
 
30 
 
 THE STABILITY OF AECHBS 
 
 chord. Force diagrams are drawn for each case in 
 the usual way as shown in (i), (ii), and (iii) and the 
 forces in each member are scaled off and tabulated 
 as follows : — 
 
 1 
 
 2 
 
 3 
 
 i 
 
 5 
 
 6 
 
 7 
 
 
 Force due to Unit Load 
 
 Force due to Actual Dead toad I 
 
 Member. 
 
 
 at 
 
 
 of 2-2i Tons at 
 
 
 
 Ti 
 
 T2 
 
 T3 
 
 Ti 
 
 T2 
 
 Xs 
 
 ToTi . 
 
 + •73 
 
 + •IS 
 
 - ^41 
 
 + 16^4 
 
 + 4-05 
 
 - 9-25 
 
 T1T2 
 
 + ■525 
 
 + 1^07 
 
 - ^703 
 
 + 11^8 
 
 + 24^05 
 
 -15-8 
 
 t^t' . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 T^'Tji . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 T IT,' . 
 
 -•26 
 
 - ^505 
 
 - ^703 
 
 - 5^85 
 
 -11^35 
 
 -15-8 
 
 To^Tji . 
 
 -•14 
 
 - ^265 
 
 - ^41 
 
 - 3-16 
 
 - 5^95 
 
 - 9-25 
 
 B„B, . 
 
 + •575 
 
 + 1^135 
 
 + 1^72 
 
 + 12^9 
 
 + 25-3 
 
 + 38-7 
 
 B1B2 
 
 -•24 
 
 + ^87 
 
 + 2^03 
 
 - 5'4 
 
 -19-55 
 
 + 45-7 
 
 B2B3 
 
 -•025 
 
 - ^07 
 
 + 2^23 
 
 - •568 
 
 - 1^57 
 
 + 50-2 
 
 WSs' ■ 
 
 + •77 
 
 + 1^52 
 
 + 2-23 
 
 + 17-35 
 
 + 34^2 
 
 + 50-2 
 
 Bi^B/ . 
 
 + •68 
 
 + 1^355 
 
 + 2^03 
 
 + 15-3 ■ 
 
 + 30^5 
 
 + 45-7 
 
 B„iBii . 
 
 + •57 
 
 + 1^135 
 
 + 1^72 
 
 + 12^8 
 
 + 25-55 
 
 + 38-7 
 
 T„Bi . 
 
 -•92 
 
 - -22 
 
 + ^52 
 
 -20^7 
 
 - 4^95 
 
 + 11-7 
 
 T1B2 
 
 + •22 
 
 - •gs 
 
 + ^32 
 
 + 4^95 
 
 -220 
 
 + 7-2 
 
 T^B., 
 
 + •55 
 
 + 1^12 
 
 - ^74 
 
 + 12^4 
 
 + 25^2 
 
 -16-65 
 
 T2IB3 . 
 
 -•275 
 
 - ^53 
 
 - -74' 
 
 - 6^2 
 
 -11^9 
 
 - 16-65 
 
 T/B,i . 
 
 + •13 
 
 + ^258 
 
 + ^32 
 
 + 2^93 
 
 + 5^8 
 
 + 7-2 
 
 T,'W ■ 
 
 -•17 
 
 - ^335 
 
 + ^52 
 
 - 3^83 
 
 - 7^53 
 
 + 11-7 
 
 BflTy 
 
 + •56 
 
 + ^135 
 
 - ^315 
 
 + 12-6 
 
 + 3^04 
 
 - 7-1 
 
 BiTi 
 
 + •91 
 
 - ^41 
 
 - ^14 
 
 + 20 '5 
 
 + 9-2 
 
 - 3-15 
 
 B2T2 
 
 -•155 
 
 - ^675 
 
 + ^24 
 
 - 3-49 
 
 -15-6 
 
 + 5-4 
 
 B.,T3 . 
 
 
 
 
 
 + -5 
 
 
 
 
 
 + 11-25 
 
 B^'T.i . 
 
 + •088 
 
 + •iss 
 
 + ^242 
 
 + 1^98 
 
 + 3-54 
 
 + 5-4 
 
 Bi^T,! . 
 
 -•050 
 
 - •lO 
 
 - -U 
 
 - 1-125 
 
 - 2-25 
 
 - 3-15 
 
 B„-T„' . 
 
 -10 
 
 - -205 
 
 - ^32 
 
 - 2^25 
 
 - 4-62 
 
 - 7-1 
 
DISCUSSION OF THE PKOBLEM 
 
 31 
 
 8 
 
 9 
 
 10 
 
 u 
 
 12 
 
 13 
 
 14 
 
 I.. 
 
 Force due to Live Load of 
 16 Tons at 
 
 Total 
 Force due 
 to Dead 
 
 Total Force due to 
 Live Load. 
 
 Max. 
 
 Min. 
 
 Ti 
 
 Ta 
 
 Ts 
 
 Load. 
 Tons. 
 
 Tons. 
 
 Tons. 
 
 Tons. 
 
 Tons. 
 
 + 10-95 
 
 + 2-7 
 
 - 6-15 
 
 - 7-16 
 
 + 13-65 
 
 -18-38 
 
 + 6-49 
 
 - 25-5 
 
 + 7-88 
 
 + 16-05 
 
 -10-55 
 
 - 12-95 
 
 + 23-93 
 
 -32-60 
 
 + 10-98 
 
 - 45-5 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 - 3-9 
 
 - 7-6 
 
 - 10-55 
 
 - 12-95 
 
 + 23-93 
 
 - 32-60 
 
 + 10-98 
 
 - 45-5 
 
 - 2-1 
 
 - 3-98 
 
 - 6-15 
 
 - 7-16 
 
 + 13-65 
 
 -18-38 
 
 + 6-49 
 
 - 25-5 
 
 + 8-62 
 
 + 17-05 
 
 + 25-8 
 
 + 153-95 
 
 + 102-9 
 
 
 
 + 256-87 
 
 + 156-4 
 
 - 3-6 
 
 + 13-05 
 
 + 30-45 
 
 + 112-25 
 
 + 104-5 
 
 - 3-6 
 
 + 216-78 
 
 + 108-6 
 
 - -38 
 
 - 105 
 
 + 33-45 
 
 + ] 49-82 
 
 + 101-2 
 
 - 1-43 
 
 + 251-1 
 
 + 148-4 
 
 + 11-55 
 
 + 22-8 
 
 + 33-45 
 
 + 149-82 
 
 + 101-2 
 
 - 1-43 
 
 + 251-1 
 
 + 148-4 
 
 + 10-2 
 
 + 20-38 
 
 + 30-45 
 
 + 112-25 
 
 + 104-5 
 
 - 3-6 
 
 + 216-8 
 
 + 108-6 
 
 + 8-45 
 
 + 17-2 
 
 + 25-8 
 
 + 154-04 
 
 + 102-9 
 
 
 
 + 256-9 
 
 + 154-0 
 
 -13-8 
 
 - 3-3 
 
 + 7-8 
 
 - 13-61 
 
 + 15-6 
 
 -24-67 
 
 + 2-0 
 
 - 38-3 
 
 + 3-3 
 
 -14-7 
 
 + 4-8 
 
 + 6-08 
 
 + 18-7 
 
 -14-7 
 
 + 24-8 
 
 - 8-6 
 
 + 8-25 
 
 + 16-8 
 
 + 11-1 
 
 - 13-80 
 
 + 47-2 
 
 - 12-07 
 
 + 23-4 
 
 - 25-9 
 
 - 4125 
 
 - 7-95 
 
 + 11-1 
 
 - 13-80 
 
 + 47-2 
 
 - 12-07 
 
 + 23-8 
 
 - 25-9 
 
 + 1-95 
 
 + 3-87 
 
 + 4-8 
 
 + 6-08 
 
 + 18-7 
 
 -14-7 
 
 + 24-8 
 
 - 8-6 
 
 - 2-55 
 
 - 5-02 
 
 + 7-8 
 
 - 13-61 
 
 + 15-6 
 
 -24-67 
 
 + 20 
 
 - 38-3 
 
 + 8-4 
 
 + 2-02 
 
 - 4-73 
 
 - 5-43 
 
 + 10-4 
 
 -12-68 
 
 + ^-0 
 
 - 18-1 
 
 + 13-65 
 
 + rvl5 
 
 - 2-1 
 
 + 20-03 
 
 + 19-8 
 
 - 6-45 
 
 + 39-8 
 
 + 13-6 
 
 - 2-33 
 
 -10-1 
 
 + 3-6 
 
 - 2-86 
 
 .+ 10-8 
 
 - 12-43 
 
 + 7-9 
 
 - 15-3 
 
 
 
 
 
 + 7-5 
 
 + 22-50 
 
 + 15-0 
 
 
 
 + 37-5 
 
 + 22-5 
 
 + 1-32 
 
 + 2-32 
 
 + 3-6 
 
 - 2-86 
 
 + 10-8 
 
 - 12-43 
 
 + 7-9 
 
 - 15-3 
 
 - -75 
 
 - 1-5 
 
 - 2-1 
 
 + 20-03 
 
 + 19-8 
 
 - 6-45 
 
 + 39-8 
 
 + 13-6 
 
 - -15 
 
 - 3-07 
 
 - 4-73 
 
 - 5-43 
 
 + 10-4 
 
 -12 68 
 
 + 5-0 
 
 - 18-1 
 
 In the above Table, Col. 1 gives the name of each member of the arch. 
 Cols. 2, 3, and 4 show the force induced in each of these members by a 
 1-ton load situate at joints Tj, Tj, T, respectively, as measured on the 
 corresponding force diagram. Cols. 5, 6, and 7 show the forces actually 
 caused in the members by the actual dead load of 22J tons at the points Tj, 
 Tj, and Tg. Cols. 8, 9, and 10 show the same thing for the temporary 
 load of 15 tons at each joint. Col. 11 shows the total force in each mem- 
 ber due to the whole dead load. These values are found by adding the 
 forces produced by the loads at T,, Tj, and Tg in each bar to those pro- 
 duced by the loads acting on the opposite half of the arch, which is the 
 same thing as adding them to the forces produced by Tj, Tj, and Tg in the 
 corresponding members of the right-hand half. Cols. 12 and 13 give the 
 maximum and minimum forces which occur in the members, which are 
 found by adding all the positive forces for a maximum and all the nega- 
 tive forces for a minimum in each ease. Cols, 14 and 15 give the maxi- 
 mum and minimum forces due to the dead and live loads combined, 
 taking the worst combinations that can occur. 
 
32 
 
 THE STABILITY OF ARCHES 
 
 20. Method II. By Influence Lines. — In apply- 
 ing this method to find the maximum force in any 
 bar of a framed structure, as in the present case, an 
 influence line must be drawn for the bending moment 
 about the moment centre, from which the maximum 
 bending moment can be found and the maximum 
 force in the bar deduced. 
 
 Fig. 19. 
 
 For example, if we require to find the force in the 
 bar TjTj say, of the preceding example, an influence 
 line is drawn for the bending moment about the point 
 B2, which is the moment centre for that bar, because 
 if an imaginary section be taken through the three 
 members T^To, TjEj, and B^B^, by taking moments 
 about the point Bj, the unknown forces in the bars 
 
DISCUSSION OF THE PROBLEM 33 
 
 TjB.j, B^Ba are eliminated from the resulting equa- 
 tion, and the moment of the force in the bar TjT2 
 about the point B.,, will be in equilibrium with the 
 bending moment round that point due to the external 
 forces on the one side of the section ; and since the 
 force in T^Tj x T2B2 is equal to the bending 
 moment, the value of T^Ti is found by dividing this 
 bending moment by the length of the lever arm T^Bj. 
 
 It is therefore required, first of all, to construct 
 an influence diagram for the moment about B^, 
 showing how this moment changes as a unit load 
 moves over the arch. Now the actual moment at 
 any point is the difference between the positive 
 moments due to the vertical forces and the negative 
 moments due to the horizontal thrust of the arch, 
 and we have shown already, in § 18, how an influence 
 diagram may be constructed to show this. 
 
 Thus, in the present case, to find the force in the 
 bar T1T2, when a unit load is at Tj say, 15 ft. from 
 the L.H. end, we first construct the influence diagram 
 for the point Bg. Make 
 
 CiC„ = -^ = ^Q ^ ^f ^ = 19-8 tons ft. 
 ^ iy, 4 X 15 
 
 Using a scale of 1 in. = 10 tons, CjCg = 198 in. 
 
 Next make 
 
 DjDj = 3£jL^i)_ = ^Q ^ ^^ = 20 tons ft. = 2 ins. , 
 
 and complete the diagram. The intercept B^E at 
 15 ft. from the L.H. end then scales 35 tons ft., 
 and the force in TjT2 is therefore 
 3-5 3-5 
 
 T2Bij 6-75 ft. 
 
 ■52 tons. 
 
 Similarly, when the unit load is at T2, 30 ft. from 
 3 
 
34 THE STABILITY OF ABCHBS 
 
 the L.H. end, the intercept on the influence diagram 
 
 scales 7'0 tons ft., and the force in TjTg is then 
 
 7'0 
 
 - — - = 1-04 tons, and so on. 
 6'7o 
 
 Next to find the force in a lower chord member, 
 
 say BjBj. In this case the point Tj will be the 
 
 moment centre, and the influence line for H will 
 
 have its maximum ordinate = ^^ by the same 
 
 4y 
 
 reasoning as before, where y^ is the height of the 
 
 point Tj above the. springing-joint = 20 ft. The 
 
 maximum value of the bending moment due to H 
 
 will therefore be CO = — — — = 30 tons ft. Also 
 
 the ordinate for the bending moment due to the 
 vertical forces for the position Tj will be 
 
 B,B4 = ^^ ^ ^^ = 12-5 tons ft. 
 ^ * 90 
 
 If we plot these values and complete the diagram 
 
 we find from it that for a unit load at Tj the bending 
 
 moment about Tj is EjE^ =2-5 tons ft., and the force 
 
 in B,B„ = -i ^!^____ = - 0-23 tons ft., 
 
 lever arm of BjBa 
 
 the negative sign indicating tension, because the 
 
 moment about Tj is clockwise. 
 
 To find the force in a diagonal member. Suppose 
 we consider, for example, the diagonal T^Bj. The 
 moment centre for this member will be the point X 
 in which the bars TjT2 and B^Bg cut by the section 
 a/3 intersect. 
 
 Now if the unit load be on the left Tj, 
 
 M = Vb' . M - H2/2 . . (1) 
 and when it is on the right of Tj 
 
DISCUSSION OF THE PEOBLBM 35 
 
 M, = Vb (Z - «) - Hy, . . (2) 
 
 I - V 
 but if the unit load be distant v from T„ , Vb = — = — 
 
 
 
 and Vb' = j-. 
 
 .-. M, = I . M - H2/, . . (1') 
 and M, = ^ ~ ^ . (Z - u) - Hj/, . (2') 
 
 i 
 
 The first terms in these two equations are both 
 represented by straight lines, and if we put v = Z in 
 
 the expression — u, we get for the ordinate at the 
 extreme right the value u ; and if we put v = o in 
 
 the expression — - — {I - u) ^re get Z - m as the 
 I 
 
 value of the ordinate on the extreme left. If we 
 
 now draw in the lines GF^, FGj, and drop verticals 
 
 from Tj and T^ to meet them in ij and t^, the broken 
 
 line F<j, t^, G represents the influence diagram for the 
 
 bending moment at X due to the vertical forces 
 
 Iv 
 alone. If then we set up KL = —2^ as before, the 
 
 triangle FKG will be the influence line for the bend- 
 ing moment due to the horizontal thrust H. The 
 first of these bending moments will cause tension in 
 the bar TjBj and the second compression. The 
 effect of any load on the moment at X can now be 
 found from the influence diagram, and the force 
 which it causes in the bar TjB^ can be deduced by 
 dividing this moment by the lever arm. Thus, for 
 example, in the present case, tjtg scales S'd tons, 
 whilst the lever arm Xx is 15'4 ft., which makes the 
 
36 THE STABILITY OF AECHBS 
 
 tensile force in the bar TjBj = 0'22 ton when the 
 unit load is at Tj. Similarly, the force may be found 
 for any other position of the unit load, and, therefore, 
 for any system of loads in any position on the span. 
 
 Having thus found for any particular bar the force 
 caused in it when a load is placed at any point of the 
 span, the maximum force which will occur in it under 
 any arrangement of loading can easily be derived, 
 because when calculating the maximum tensile force 
 in any member, all the temporary loads which cause 
 compression, and therefore diminish the tensile force, 
 must be left out of consideration ; and similarly when 
 calculating the maximum compressive force all those 
 temporary loads which cause tension must be 
 omitted. 
 
 By comparing the results obtained above by this 
 method, it will be seen that they agree closely with 
 those obtained by Method I. 
 
CHAPTEE III. 
 
 THE ELASTIC THEORY OP THE ARCH. 
 
 21. When an arch has less than three joints, the 
 three conditions necessary to fix the position of the 
 line of pressure must be made up by a consideration 
 of the elastic properties of the arch-ring. It has 
 been objected to the elastic theory that the arch-ring 
 not being homogeneous throughout, especially when 
 joints exist, the theory cannot be properly applied ; 
 but this objection has little force in fact, because the 
 arch-ring in such cases is usually composed of dis- 
 tinct parts, like voussoirs, which are bounded by 
 joints sensibly normal to the axis, and there is no 
 theoretical objection to neighbouring parts having 
 different coefficients of elasticity, so long as the parts 
 themselves have uniform elastic properties. 
 
 The theory is, however, somewhat complex, and 
 involves assumptions and simplifications more or less 
 approximate, so that the results derived from its ap- 
 plication have until comparatively recent years been 
 rightly regarded with a certain amount of suspicion. 
 This distrust has been largely removed by the care- 
 ful and complete series of experiments carried out by 
 the Institution of Austrian Engineers and Architects, 
 the results of which were published about the year 
 1900. 
 
 These experiments were made on arches of 
 (37) 
 
38 THE STABILITY OF AECHES 
 
 masonry and brickwork and of concrete, plain and 
 armoured, having spans of from 1'35 to 23 m. (75"4 
 ft.) span, and 4-6 m. (15-1 ft.) rise. The large brick 
 and stone arches of 23 m. span were built in Port- 
 land cement mortar, 1 cement to 2 '6 sand. The 
 crown thickness was 0'6 m. (1 ft. llf ins.) and the 
 thickness at the springings was 1-14 m. (3 ft. 7i ins.). 
 The concrete arch was 0'7 m. (2 ft. 3^ ins.) thick 
 throughout, whilst the Monier arch was 0-35 m. 
 (1 ft. 1| ins.) at the crown and 0-6 m. (1 ft. 11| ins.) 
 at the springings. The longitudinal rods were 0'55 
 in. diameter, and the transverse rods 0'276 in. dia- 
 meter, the mesh being 2-^ ins. The width was 2 m. 
 (6-56 ft.). The loading was applied gradually from 
 one abutment to the crown, and the distortion of the 
 axis was carefully measured until the arches showed 
 signs of collapse. From these measurements the 
 values of E were calculated and were found to be as 
 follows : — 
 
 For masonry . . 60,400 kg./cm.a = 858,000 lb./in.2 
 „ brickwork . . 27,800 ,, = 393,000 
 
 „ plain concrete . 246,000 „ = 3,475,000 
 
 „ reinforced concrete 333,800 „ = 4,700,000 
 
 At a certain critical load cracks began to appear 
 in the arch-ring between J and ^ or between | and 
 I of the span, and at the supports, agreeing in 
 general with the position of the joints of rupture as 
 deduced from theory. 
 
 The breaking loads were as follows : — 
 
 Brickwork arch . . . 67"5 tons. 
 Masonry .... 74-02 ,, 
 Concrete . . . . 83-27 „ 
 Monier concrete . . . 146-12 ,, 
 
THE ELASTIC THEOEY OP THE ARCH 39 
 
 The breaking load for the masonry arches exceeded 
 the critical load by about 30 per cent, for the brick- 
 work arches by about 59 per cent, for the plain con- 
 crete by 31 per cent, and for the reinforced concrete 
 by 86 per cent. 
 
 The general result of the experiments was to show 
 that the elastic theory might be applied with confi- 
 dence to arches of masonry, brickwork, and concrete 
 of such proportions as are now common, and as a 
 direct outcome of them a stone arch of 213 ft. span 
 and 58'6 ft. rise was built over the Pruth in Austria. 
 A full account of these experiments is given in 
 Volume I, "Handbuch fiir Bisenbetonbau". 
 
 The general accuracy of the elastic theory of the 
 arch is also confirmed by experiments on models 
 under polarized light. Dr. E. G. Coker, who has 
 made many experiments on this subject, has kindly 
 directed my attention to a memoir of M. Mesnager 
 ("Determination complete sur un module r^duit des 
 tensions qui se produiront dans un ouvrage. An- 
 nales des Fonts et Chauss6es ") in which the author 
 describes a glass model of a bridge over the Ehone 
 which was constructed for the purpose of checking 
 the results as calculated in the usual way. This 
 model, examined under polarized light, gives an 
 ocular demonstration of the change of position of the 
 neutral axis of the arch-ring when the load is ap- 
 plied, which is quite in accordance with the deduc- 
 tions of theory, whilst the stresses obtained by the 
 optical method did not differ more than 14-7 per 
 cent, as a maximum, from the results of the elastic 
 method. 
 
 22. The Development of the Elastic Theory of 
 
40 
 
 THE STABILITY OF ARCHES 
 
 the Arch, Angular Distortion of a Rib due to 
 Bending. — Let a be the area of a horizontal strip of 
 the cross-section of an arch-ring (Fig. 20), distant u 
 from the axis CO through the centroid of the section, 
 and let AjBj, A^B^ be neighbouring sections, 8s 
 apart. Then if the arch-ring bend through an angle 
 
 Pig. 20. 
 
 S6 on the length Ss when the load is applied, and if 
 Sx is the change of length produced thereby on the 
 length Bs at distance u from the axis CO, then the 
 
 Sx 
 stress induced is s = B . »-, and the moment of re- 
 sistance of the. strip is therefore 
 
 sau = E . -~ . a . u. 
 Ss 
 
 The total moment of resistance of the cross-section 
 
 is therefore 
 
 E = sFl 
 or since hx = u . SO 
 
 \_ Ss _ 
 
 and since for any given cross-section of an arch-ring, 
 
 2|B.g-.a.« 
 
THE ELASTIC THEOEY OF THE AKCH 41 
 
 B, 8s and 86 may be regarded as constants without 
 sensible error, we get 
 
 E = E . ^ 2 [au^l 
 os 
 
 or since 2 [au'^] = I 
 
 E = EI . i^, 
 Ss' 
 
 and since for equilibrium the moment of resistance 
 
 E of the cross- section must be equal and opposite 
 
 to the bending moment M due to the external forces, 
 
 we get 
 
 M = EI.f^or8e = ^-«*- 
 
 A6 
 
 Ss EI 
 
 that is to say, the angle of flexure due to the bend- 
 
 ■ , M 
 mg moment is proportional to ^. 
 
 Hence, if a rib be fixed at one end and free at the 
 other, and a known system of loads be applied to 
 it, the bending moment at every point will be known 
 and the total angular displacement between its 
 extremities will be 
 
 '^] ■ ■ ■ ('» 
 
 23. Horizontal and Vertical Displacements due 
 to Bending. — Let C (Fig. 21) be the centre of 
 gravity of a small segment of rib, and suppose this 
 section to turn through an angle 86 on the length 8s, 
 when the load is applied, due to the elasticity of the 
 material, and let y be the vertical distance of the 
 free end B below C, and x the horizontal distance 
 between them. Due to the flexure of the element 
 8s, suppose B to move to B', and let 8a;, By be the 
 horizontal and vertical displacements of B due to 
 
42 
 
 THE STABILITY OP ARCHES 
 
 the change in angle 8^. Then since these displace- 
 ments are very small, BB' is sensibly perpendicular 
 to CB and 8a; is perpendicular to y. Hence in the 
 similar triangles CDB, BD'B' 
 
 8a; BB ~^ 
 
 i^CB^^ 
 
 and similarly -^ = „^, 
 ^ X CB' 
 
 Hence the horizontal displacement is Sa; 
 
 -y.he (2) 
 
 Fig. 21. 
 
 and the vertical displacement is 
 Introducing the value of 
 
 M.83 
 EI 
 M.Ss 
 
 Zy = x.^e . (3) 
 
 he = 
 
 we get 
 
 and 
 
 Sa; = 2/ 
 
 hy = X. 
 
 EI 
 
 M. h s 
 
 EI 
 
 (4) 
 (5) 
 
 These are the small horizontal and vertical com- 
 ponents of the actual displacement due to flexure 
 on a small length hs of the rib, and the total hori- 
 zontal and vertical displacements between the 
 extremities will be the sum of all these 
 
 r M . 8s- 
 or Aa; = S 2/ ■ " 
 
 and 
 
 \y 
 
 = Sp 
 
 EI 
 EI 
 
 (6) 
 (7) 
 
THE ELASTIC THEORY OF THE AECH 43 
 
 24. Horizontal and Vertical Displacements due 
 
 to Axial Thrust. — Let N be the normal thrust on 
 
 any cross-section of the rib, whose area is A. Then 
 
 if the resultant act along the axis, the stress on the 
 
 section will be uniform, and if we denote it by s„, then 
 
 N 
 s = X- The compressive strain on a small length 
 
 _8s ^ N 
 ^ AE 
 
 Ss will therefore be 
 
 E 
 
 8s, and if we de- 
 
 note compression by the negative sign, the 
 compression along the whole length will be 
 
 N , 
 
 total 
 
 As = 
 
 AE' 
 
 and the corresponding horizontal and vertical dis- 
 placements due to the thrust are therefore 
 
 • ■ (8) 
 
 Aa; 
 
 - 2 
 
 and 
 
 A2/= - 2 
 
 rN 
 Lae 
 
 8a; 
 
 "N 
 
 Lae 
 
 &y 
 
 (9) 
 
 25. Horizontal and Vertical Displacements due 
 to Change of Temperature, — Let u, be the co- 
 efficient of expansion of the material, and t the 
 number of degrees rise in temperature. Then the 
 expansion due to this rise on a length Ss will be 
 at . Ss and the corresponding horizontal and vertical 
 displacements will be at . 8x and at . Sy, whilst the 
 total displacements between the ends will be 
 
 AX = S, \at . 8x] . . (10) 
 
 and A2/ = 2 [at . iy] . . (11) 
 
 26. Total Horizontal and Vertical Displace- 
 ments! — Combining the values of ax from equations 
 (6), (8), and (10) and of aj/ from equations (7), (9), 
 and (11) we get 
 
44 THE TABILITY OF AECpES 
 
 total horizontal displacement 
 
 EI 
 and total vertical displacement 
 
 r^ ; 8a; + 2 [a^ . 8a;] . (12) 
 
 EI 
 
 m-^y 
 
 + 2 [a/! . Sy] . (13) 
 
 It is only in very fiat arches that the effect of the 
 thrust is appreciable, and as its inclusion complicates 
 the formulae it will be desirable in the first case to 
 neglect its effect, and to investigate later what allow- 
 ance should be made for it. Likewise the effect of 
 change of temperature may be left for future con- 
 sideration, whilst we discuss first of all the, stresses 
 due to bending which form by far the principal con- 
 tribution to the actual result. 
 
 27. Primary Equations of Condition for a 
 Hingeless Arch, — As we have already seen, the 
 problem of the arch requires three independent con- 
 ditions for its solution, and these conditions are, for 
 an arch without joints, derived necessarily from its 
 elastic properties. 
 
 If we assume that the abutments are .rigid, so that 
 when the load is applied the span remains unaltered, 
 then the total horizontal displacement as expressed 
 by equation (12) must be zero. Further, if the level of 
 the supports remains the same, then the total vertical 
 displacement as expressed by equation (13) must also 
 be zero. Finally, if the ends of the arch-ring are as- 
 sumed rigidly fixed in direction, so that however the 
 arch itself may bend under the load, the slope of its 
 axis at these ends is unaltered, then Afl in equation (1) 
 will also be zero, and therefore if we consider only 
 
THE ELASTIC THEORY OF THE ABCH 45 
 
 the bending moment due to the load, and consider 
 E as constant, we get from the above conditions 
 
 ^Mt/.Ss^Q . ■ . (14) 
 
 I 
 
 .Ma; 
 
 I 
 
 . (15) 
 
 and 2 — ■ - = . . . (16) 
 
 These are the fundamental equations of condition 
 upon which the solution of the arch problem depends, 
 whatever variations may be adopted in applying 
 them. 
 
 In the special case when the moment of inertia 
 of the arch-ring is constant, I will also disappear 
 from the equations, and if, as is often done, the axis 
 of the arch is divided into equal lengths 8s, Bs being 
 then also constant will vanish too, and the above 
 equations will then reduce to the simpler forms : — 
 2 [M.y\ = ; (14') ; % [Mx\ = ; (15)' ; and 2 [M] = 
 0; (16)'. 
 
 Sometimes also, when the moment of inertia of the 
 arch-ring is variable, with the object of simplifj'ing 
 the equations as much as possible, the lengths hs are 
 
 so chosen as to keep the' ratio — constant, in which 
 
 case the equations take the same simple forms as 
 those last derived. 
 
 These, however, are merely convenient variations 
 in the method of applying the fundamental equations. 
 
 28. Remarks on the Assumptions. — As regards 
 the first assumption that the span does not alter 
 when the load is applied, this is not exactly true, be- 
 cause when the centering of an arch is removed the 
 
46 THE STABILITY OF AECHES 
 
 material necessarily yields, if only on account of its 
 elasticity, and if no allowance were made for this the 
 alteration of form might seriously affect the stresses. 
 In arches up to about 150 ft. span, however, the 
 centering is usually built so that the rise is greater 
 than the final rise by about ^^^ of the span, or by 
 the amount of its probable deflection. The abutments 
 may also yield to a certain extent, and any increase 
 in the span will cause an increase in the horizontal 
 thrust amounting to about 17 tons for every -J^ ft., 
 according to Balet ("Analysis of Elastic Arches "), 
 whilst a rise in temperature of 15° F. will cause an 
 increase of about 3 tons in the thrust per foot of 
 width. In masonry and concrete fluctuation of tem- 
 perature is less than for metal arches and may be 
 taken as not more than 20° F. in this country, and 
 less when the average depth of filling exceeds about 
 30 ins. As regards the third assumption, the 
 fixure of the terminals is never absolutely rigid, be- 
 cause the elasticity of the abutments will permit a 
 slight rotation. This effect must, however, be very 
 slight in most cases and may safely be disregarded, 
 especially when the depth of the arch-ring increases 
 towards the abutments. 
 
 29. Allowance for the Thrust and Change of 
 Temperature. — It has been shown in sect. 26 (eq. 
 12), that the total horizontal displacement of the 
 free end of an arch rib is given by the expression 
 
 ihl Alii 
 
 and if we assume the span to remain unaltered, then 
 
 22/ . -"^ - ■S,^.Sx+'S,at.Sx=0 (equation 13). 
 ill AHi 
 
THE ELASTIC THEOEY OF THE ARCH 47 
 
 The value of a for stone and concrete varies from 
 •0000044 to -0000078 per degree Fahrenheit, and t 
 need not be taken higher than about 20° F. for high- 
 way bridges. The effect of the axial thrust N is to 
 shorten the span, and its effect may therefore be re- 
 garded as equivalent to a fall in temperature, and in- 
 cluded as such in the temperature effect. The terms 
 involving N are, however, always very small in com- 
 parison with those involving the bending moment, 
 and a small correction may be applied to allow for 
 them. Thus in the formula for H, sect. 31, viz., 
 
 H 
 
 
 the effect of the thrust may be sufficiently allowed 
 
 V I 
 
 for by adding to the denominator the term — . 
 
 E ■ n . A, 
 
 where - is the mean length of the horizontal seg- 
 n 
 
 ments into which the span is divided, A is the mean 
 
 area of the cross-sections, vis the distance of the 
 
 centre of curvature at the crown from the X axis, and 
 
 E is the radius of curvature at the crown, the value 
 
 — being usually nearly equal to unity for flat arches. 
 E 
 
 The maximum stresses due to change of temperature 
 usually arise at the crown and springings, and their 
 values may be estimated with sufficient accuracy by 
 means of the following formulae due to Melan : — 
 
 For the horizontal thrust due to change of tem- 
 perature 
 
48 THE STABILITY OF ARCHES 
 
 jj _ 45 Eatl„ 
 
 4 hh' 
 
 where I is a mean value of the moments of in- 
 ertia of the cross-sections and h, h' are the rise of 
 the arch and the rise of the Hne of resistance respec- 
 tively, a,nA h' = h ( 1 + -r . -psi) a'PPJ'oximately. 
 
 The bending moment at the crown due to change 
 of temperature then becomes 
 
 j^ _ 15 Ea< ■ I„ 
 
 and the extreme stresses due to change of tem- 
 perature are 
 
 (a) at the crown s, = — Bai A! ( l + 2 ^V 
 
 16 hh' \ tj 
 
 (b) at the springings Si == —Eat ^fl 4 _ 
 
CHAPTEE IV. 
 
 THE TWO-HINGED ARCH. 
 
 30. The two-pinned arch having joints at the abut- 
 ments, but continuous at the crown, is chiefly used 
 for steel bridges. No particular advantage appears 
 to be gained by employing pin joints at the abut- 
 ments only, because the determination of the forces 
 is a statically indeterminate problem, whilst the con- 
 struction is complicated by introducing pin joints ; 
 and for any except small arches the extra rigidity 
 gained by dispensing with the crown pin does not 
 appear to be sufficient to compensate for the other 
 disadvantages. > 
 
 The actual calculations are, however, less complex 
 than in the case of the hingeless arch, because only 
 one elastic condition is necessary to replace the loss 
 of one pin. This condition may be derived by as- 
 suming the span to remain unaltered by the loading. 
 We then have by equation 14 of last chapter 
 
 Now by Eddy's Theorem (sect. 15) if ACB (fig. 22) 
 be the axis of the arch, hinged at A and B, and ADB 
 be the line of pressure, whose polar distance is H, the 
 bending moment M at any point is H x CD, where 
 CD is the vertical intercept between the axis of the 
 arch and the line of pressure, measured in the hnear 
 (49) 4 
 
50 
 
 THE STABILITY OF ARCHES 
 
 scale of the drawing. If we introduce this value into 
 equation (14) above, we get 
 
 'B-.GB.y . gs" 
 
 I 
 
 
 
 (17) 
 
 and if we suppose the axis divided into small equal 
 
 Fig. 22. 
 
 lengths Ss, Ss being constant and the horizontal 
 thrust H being also constant, this reduces to 
 
 %m^] = 0. 
 
 But CD at any point = ED - EC = z - y where z 
 is the ordinate of the line of pressure and y that of 
 
 the arch-axis. Hence 
 
 we 
 
 get 2 
 
 or 2 
 
 I 
 
 = S 
 
 I 
 
 {« - y}y' 
 
 
 
 . (18) 
 
 Now if any link-polygon be drawn through A and 
 B, such as APB, for the given loads upon the arch, 
 its ordinates u will have a constant ratio to those of 
 the true linear arch ADB, which is merely an equili- 
 brium polygon for the same load system with a dif- 
 ferent polar distance. If then we write the ratio 
 
 — = r, z =r .u, and substituting this value of z in 
 
THE TWO-HINGED ARCH 
 
 51 
 
 equation (18) we get 
 
 m - 
 
 whence r 
 
 ru . y 
 
 = rS 
 
 I 
 
 [r 
 
 (19) 
 
 or when the cross-section of the arch-ring is rect- 
 angular, since I = -- where t is the thickness of 
 the ring, we get 
 
 m 
 
 •u-yl 
 . ^ J 
 
 (20) 
 
 The values of the numerator and denominator of 
 these expressions may obviously be found by measure- 
 ment from the drawing and subsequent calculation 
 and summation, and if the polar distance corres- 
 ponding to the trial-polygon AI'B be changed in the 
 
 inverse ratio -, the equilibrium -polygon drawn with 
 the new pole will be the true line of pressure ; or 
 
 1 ^m 
 
 true polar distance 
 assumed polar distance 
 
 
 (21) 
 
 or in the special case when the value of I is constant 
 
 1_ %[u.y] 
 
 It may be remarked that the determination of the 
 value of r fixes the value of the polar distance and 
 
52 THE STABILITY OF AECHES 
 
 therefore of the true horizontal thrust of the arch, 
 and when this is known there is no difficulty in 
 drawing the line of pressure or in calculating the 
 bending moments, etc. 
 
 31. Example. — It will be of interest to consider, 
 by way of comparison, the same arch and the same 
 loading as in the example of chapter II, except that 
 the arch-ring will now be made continuous at the 
 crown. This arch was 150 ft. span, . with a rise of 
 15 ft., loaded with 26 cwt. per foot, with 18 cwt. per 
 foot additional load covering the left-hand half. 
 
 Method I. Calculation of the Horizontal Thrust. — 
 If an arch of span I carry a uniform load . of w per 
 foot run, we have by the fundamental equation (14), 
 sect. 27, 
 
 My .Bs 
 
 = 0. 
 
 ^ I 
 
 Now M at any section is the difference between 
 the bending moment on a simply supported beam of 
 the same span and loaded in the same manner and 
 the bending moment due to the horizontal thrust H. 
 Hence if we denote by p, the bending moment on 
 the simply supported beam, M = /x - Ky. 
 
 Consequently the above equation becomes 
 
 5!, (f^ - ^y)y ■ ^s ^ Q 
 
 2/2. 8s 
 
 = HS 
 
 J. 
 
 whence H = — -j— ^approximately. 
 
THE TWO-HINGED AKCH 
 
 ds 
 
 63 
 
 or exactly H = 
 
 |«_ 
 
 2/2 . ds 
 
 B 
 
 or, if we assume the arch-ring to be of uniform cross- 
 section, 
 
 H =i'^£j^. 
 \y^ . ds 
 
 Fig. 23. 
 
 Now M- =^ ■ x{l - X) (Fig. 23) 
 
 and a; = - - E sin 0. 
 2 
 
 .•.^ = |(^-Rsine) (Z - |+E8ine 
 = |(^-E2sin2.). 
 Also 2/ = E(cos - cos a), sin a = — ^ and ds = B> . dO. 
 
 J-w /[2 _ ^2 gi^2 ^"N E(cos e - cos a) E . dO 
 
 •'■ H = -^ ^ 
 
 2 E2(cos $ - cos ay. 'R.de 
 
54 THE STABILITY OF ABCHES 
 
 which reduces to 
 
 j-(sin a-acos a)-— sin'o+— oos o(a - 4 sin 2a) 
 
 ij ^ j«_ ^ 5 f ^ 
 
 2K ■ i(o + i sin 2o) - sin 2a + a oos' o 
 
 Now in our case sin a = •3846176, a = 22° 
 37' 12" = -39476 radians, and E = 195 ft. 
 
 Inserting these values we get H = 187'55 w as 
 the value of the horizontal thrust. Now the hori- 
 zontal thrust due to a uniform load on one-half the 
 span is one- half of the horizontal thrust when the 
 whole span is covered. 
 
 Hence for a load of iv = 2-2 tons per ft. on the 
 left-hand half of the arch we get 
 
 TT 2-2 X 187-55 one Q 4- 
 
 Hj = = 206-3 tons, 
 
 A 
 
 and for a load of 1*3 tons per ft. on the right-hand 
 half of the arch we get 
 
 „ 1-3 X 187'55 ,010. 
 H, = = 121-9 tons. 
 
 ^ 2 
 
 Hence the horizontal thrust due to both is 
 H = 206-3 + 121-9 = 328-2 tons, 
 
 which is slightly greater than the horizontal thrust 
 when the arch was 3-pinned. 
 
 The bending moment at any point can now be 
 readily found from the equation M = /u, - Hy. 
 
 32. Method II. Graphically. — ^The arch was 
 drawn to a scale of 1 in. = 10 ft., and the span 
 having been divided into sixteen equal parts, the 
 loads on these were set off along a vector line as 
 shown in Plate I, to a scale of 1 in. = 20 tons, and 
 a pole Oj having been chosen in a convenient posi- 
 tion, a link-polygon, AFF . . . Bj, was constructed, 
 
pguTGONf 
 
 Stability of Arches.] 
 
 Plate i. 
 
 [To face page 55. 
 
THE TWO-HINGED ARCH 
 
 55 
 
 and the closing line 0-17 put in. A line O^a was 
 then drawn through Oj parallel to ABj, dividing the 
 vector line into two parts at a which represent the 
 vertical loads on the pins. The position of a will, of 
 course, be independent of the position of the pole 
 Oj, but if the link-polygon is to pass through B as 
 well as A, the pole must lie on the horizontal through 
 a, for it is only so that the closing line can be hori- 
 zontal. The true pole must therefore lie on the 
 horizontal through a, and the polar distance O^a must 
 at the same time be such as to satisfy the condition 
 (21') sect. 30. The axis of the arch was therefore 
 divided into equal lengths Ss, and verticals EP were 
 drawn through the mid-points of these. The lengths 
 of these verticals were then measured, for convenience, 
 in ^ in. units and the corresponding ordinates of the 
 arch with the same -^ in. scale. These lengths were 
 found to be as follows : — 
 
 Ordinate. 
 
 1 
 
 2 
 
 3 
 
 ' 
 
 5 
 
 6 
 
 7 
 
 8 
 
 ' y 
 
 0'38 
 
 1-02 
 
 1-58 
 
 205 
 
 2-42 
 
 2-70 
 
 2-88 
 
 2-98 
 
 ; y': 
 
 0-144 
 
 1-040 
 
 2-496 
 
 4-202 
 
 5-856 
 
 7-290 
 
 8-294 
 
 8-880 
 
 EF 
 
 1-28 
 
 3-62 
 
 5-57 
 
 7-18 
 
 8-48 
 
 9-82 
 
 9-82 
 
 9-93 
 
 , y.EF 
 
 0-46 
 
 3-69 
 
 8-80 
 
 14-72 
 
 20-52 
 
 25-16 
 
 28-28 
 
 29-59 
 
 i Ordinate . 
 
 " 
 
 10 
 
 11 
 
 1-2 
 
 13 
 
 14 
 
 16 
 
 16 
 
 y 
 
 2-98 
 
 2-88 
 
 2-70 
 
 2-42 
 
 2-05 
 
 1-58 
 
 1-02 
 
 0-38 
 
 11' 
 
 2[!/2] = 2 X 38-2 = 76-4 
 
 EF 
 
 9-60 9-05 8-25 7-22 5-98 4-52 2-88 105 
 
 2/.EF 
 
 28-61 26-06 22-27 17-47 12-26 7-14 2-94 0-38 
 
 
 2[j/ . EF] = 248-35 
 
 1 _ 248-35 
 f 76-4 
 
 3-25, 
 
56 THE STABILITY OF AECHES 
 
 Now the assumed polar distance was 5 ins. ; there- 
 fore the true polar distance is 
 
 5 ins. X 3-25 = 16'25 ins. = 325 tons. 
 
 If we draw a line through the pole Oj parallel to 
 the closing line ABj of the link-polygon, this deter- 
 mines the vertical reactions at the supports, and if 
 we take the true pole on the horizontal through the 
 point a so obtained, a new link-polygon starting from 
 one pin should pass through the other, and will be 
 the true line of pressure. 
 
 The values of the bending moments cannot as a 
 rule be found accurately from the bending moment 
 diagram, because the line of pressure lies very near 
 the axis of the arch ; but when the horizontal thrust 
 has been found, the bending moment at any point of 
 the arch-ring may be readily calculated by subtract- 
 ing from the bending moment which the load would 
 cause on a simply supported beam of the same span 
 the bending moment due to the horizontal thrust. 
 Thus the bending moment at ^ span in the present 
 case will be 
 TLy + \ wx^ - Nt,. X = 325 x 11-3+ ^ x 2-2 
 X 37-52 - 148| X 37-5 = 335-5 tons/ft. 
 
 The graphical determination of the thrust will be 
 found to be quite as reliable as when determined by 
 calculation, for on account of the fact that difference 
 terms are involved, the arithmetic must be more than 
 usually accurate to obtain a reliable result. 
 
 33. Method III. By means of the Reaction 
 Locus. — When a single load W rests at any point on 
 the arch, the reactions at the supports A and B must 
 intersect on the line of action of the load in some 
 
THE TWO-HINGED AECH 
 
 57 
 
 point C (Pig. 24) whose position may be found if the 
 horizontal thrust of the arch is known. For if nl be 
 the horizontal distance of W from the left-hand sup- 
 port, from the similar triangles CAD, cad 
 z _ cd _Va W 
 
 nl ad H 
 
 H 
 
 (1 - n) 
 
 whence z = -—— (1 - n). 
 H 
 
 Parabolic Arch, — If we suppose the axis of the 
 arch to be a parabola, and the moment of inertia of 
 the ring at any point to be proportional to the se- 
 cant of the inclination, then denoting the moment of 
 inertia of the arch-ring at the crown by I„, and its 
 
 value at any other section by I, we have I 
 
 J ds 
 
 ° ' dx 
 
 and H 
 
 = ^>^-y^^^t^=^^y.dx-,^f.dx. 
 
 i ' J I 
 
 Now when x is less than nl, 
 
 /i = Va . a; = W(l - n)x, 
 
58 
 
 THE STABILITY OF ARCHES 
 
 dx 
 
 and when x is greater than nl, 
 
 /n = Va . a; - W(a; - nl) = Wn(i - x). 
 Therefore 
 
 I jxy .dx = I W(l -n)x .y .dx + \ ^n{l - x)y . 
 
 Jo Jo J nl 
 
 where y = —jK-ii - ^) 
 
 V 
 
 = W(l - w) i^ r x\l - x)dx + ^^^ {' x{l - xf . dx 
 
 'Jo ^ Jn! 
 
 = W/.ZM1 - ^0 (1 + ^ - ..). 
 
 Also 
 
 1>^- 
 
 (ia; 
 
 A '^^^• 
 
 .-. H = I W^w (1 - «) (1 + w - n^) 
 and therefore z = h . 
 
 5(1 + n - n') 
 This is the equation to the reaction locus, and to 
 
 facilitate the plotting of it values of — are given be- 
 
 low for different values of ii up the centre of the 
 span : — 
 
 n 
 
 zlh 
 
 
 1-600 
 
 ■05 
 1-527 
 
 -10 
 1-468 
 
 ■15 -20 
 1-4151-379 
 
 •25 
 1-347 
 
 -30 
 1-322 
 
 ■35 
 1-304 
 
 -40 
 1-290 
 
 -45 
 1-283 
 
 ■50 
 1^280 
 
 Circular Arch, — The equation to the reaction locus 
 
 1 + BA; 
 
 for a circular arch is y 
 
 '■ . v„ which be- 
 
 1 - Aifc' 
 
 comes a very complex expression when the values of 
 A, B, and k are introduced in terms of the known 
 quantities. It may, however, be readily plotted 
 with the aid of the following tables (see Dubois' 
 
THE TWO-HINGED AECH 
 
 59 
 
 " Graphical Statics ") where a and /3 have the mean- 
 ings shown in Fig. 25 and k = ~, I being the 
 
 Fia. 25. 
 
 moment of inertia of the arch-ring assumed con- 
 stant, A its area, and r the radius of the circular arc. 
 
 Table I 
 
 giving values of A and Bfor given values 
 
 of a and $. 
 
 — 
 
 
 
 <i=0 
 
 a =10° 
 
 « = 20'' 
 
 a = 30° 
 
 = 40= 
 
 a = 50° 
 
 a = 60° 
 
 0=90° 
 
 
 1-20 
 
 1-19 
 
 1-17 
 
 1-14 
 
 1-08 
 
 1-00 
 
 0-88 
 
 
 
 
 
 0-2 
 
 1-21 
 
 1-20 
 
 1-18 
 
 1-15 
 
 1-10 
 
 1-01 
 
 0-90 
 
 
 
 0,2 
 
 
 0-4 
 
 1-24 
 
 1-24 
 
 1-21 
 
 1-18 
 
 1-13 
 
 1-05 
 
 0-94 
 
 
 
 h' 
 
 A 
 
 0-6 
 
 1-29 
 
 1-29 
 
 1-27 
 
 1-24 
 
 1-20 
 
 118 
 
 1-02 
 
 
 
 
 
 0-8 
 
 1-38 
 
 1-38 
 
 1-36 
 
 1-84 
 
 1-30 
 
 1-24 
 
 1-18 
 
 
 
 
 
 1-0 
 
 1-50 
 
 1-50 
 
 1-49 
 
 1-47 
 
 1-45 
 
 1-41 
 
 1-36 
 
 
 
 
 B 
 
 'a 
 
 0-234 
 
 0-238 
 
 0-221 
 
 0-203 
 
 0-178 
 
 0-146 
 
 0-107 
 
 
 
 a' 
 
 Example. — Suppose a = 40°. Then by Table I 
 B = 0-178 — , where a is the half-span and h is the 
 
 rise. 
 
60 THE STABILITY OF ABCHES 
 
 Table II, giving values of y„. 
 
 p 
 
 1=0 
 
 = 10° 
 
 = 20° 
 
 a = 30° 
 
 a = 40° 
 
 0=60° 
 
 = 60° 
 
 0=90' 
 
 
 
 1-280 
 
 1-282 
 
 1-288 
 
 1-300 
 
 1-316 
 
 1-340 
 
 1-376 
 
 1-571 
 
 0-2 
 
 1-290 
 
 1-292 
 
 1-298 
 
 1-309 
 
 1-327 
 
 1-348 
 
 1-380 
 
 1-571 
 
 0-4 
 
 1-322 
 
 1-324 
 
 1-329 
 
 1-340 
 
 1-354 
 
 1-374 
 
 1-403 
 
 1-571 
 
 n-fi 
 
 1-379 
 
 1-380 
 
 1-385 
 
 1-393 
 
 1-405 
 
 1-421 
 
 1-443 
 
 1-571 
 
 0-8 
 
 1-468 
 
 1-469 
 
 1-471 
 
 1-476 
 
 1-483 
 
 1-490 
 
 1-504 
 
 1-571 
 
 1-0 
 X a 
 
 1-600 
 
 1-600 
 
 1-599 
 
 1-587 
 
 1-594 
 
 1-591 
 
 1-588 
 
 1-571 
 
 X h 
 
 A = 1-08 — , 1-10 ^„ etc., for values of /3 = 0, 0-2 a, 
 
 etc., and by Table II. 
 
 y„ = 1 -316/1, l-327fe, etc., for corresponding values 
 
 of ^. 
 
 Thus when /8 = -4 x 40", for example 
 
 y = 
 
 1 + 0-178 ^.fc 
 1 - 1-13 g- . k 
 
 X 1-354/1. 
 
 Hence for an arch of 150 ft. span and 15 ft. rise, 
 a 
 
 ■' 1 - 1-13 X 52/c 
 
 and if the depth of the arch-ring is 18 ins. and the 
 radius of the arch-axis 195 fr., 
 
 1-5^ = 1 
 12 X 1952 
 
 k = 
 
 •• y 
 
 202911 
 
 202800 
 
 X 20-31 = 20-37 ft. 
 
 202772 
 In the case of flat arches, such as the one under 
 
stability of Arches.] 
 
 Plate ii. 
 
 [To face page 61. 
 
THE TWO-HINGED AECH 61 
 
 consideration, no sensible difference will be caused 
 by regarding the curve as parabolic. This has been 
 done in the present instance (Plate II). The reaction 
 locus was calculated from the tabular values above. 
 The span was divided into sixteen equal parts, and 
 verticals were drawn through the mid-points of each 
 of these to represent the lines of action of the loads, 
 as in the former solution. The directions of the re- 
 actions due to each load in turn were then determined 
 by joining the pins A and B to the points in which 
 each load intersects the reaction locus. A triangle 
 of forces was then drawn for each load, as shown, 
 from which the magnitudes of the reactions were 
 found. Their values at A were then plotted from 
 to 16, giving 0'16 as the resultant reaction at A. 
 The vertical loads 16-1', 1''2' etc., were next com- 
 pounded with this reaction, and a link-polygon was 
 drawn starting from A. This polygon passed through 
 B, as it should do, thereby checking the accuracy of 
 the drawing. It is therefore the true line of pressure. 
 The polar distance found in this way scaled 330 tons, 
 as against 3282 tons by calculation. 
 
 It will therefore be seen that either of the graphi- 
 cal methods may be relied upon to give a result 
 sufficiently near the truth for all practical purposes. 
 An absolute check should not in any case be expected, 
 because the assumptions are not exactly the same in 
 all three cases. 
 
 34. Construction of an Influence Line for the 
 Bending Moment at any Section of a Two-pinned 
 Parabolic Arch, — It has been shown above that the 
 horizontal thrust for a parabolic arch is 
 
 H = I wi w (1 - w) (1 + w - w''') 
 
62 
 
 THE STABILITY OF ARCHES 
 
 where n = - and x is the distance of the load W from 
 
 one end. 
 
 Hence for a unit load 
 
 h 
 
 . M (1 - w) (1 + W 
 
 If the values of H for different values of n be calcu- 
 lated from this formula and the results plotted at the 
 corresponding points, a fair curve drawn through 
 the points so obtained will give the horizontal thrust 
 when the load occupies any position on the span. 
 Now the actual bending moment M at any point is 
 equal to the bending moment /jl due to the same 
 load on a simply supported beam less the bending 
 moment H?/ due the horizontal thrust ; therefore 
 U = ^ - By (Pig. 26) 
 
 / 
 
 
 r — ^^ 
 
 /^ 
 
 ^ ^ 
 
 ^^^^^^^^^>\ 
 
 «A^ — J 
 
 ) 
 
 \ ' 
 
 --^ 
 
 Fia. 26. 
 
 or 
 
 where /* 
 
 x{l - «) 
 I 
 
 ¥=/^ _ H 
 
 y y 
 
 for a unit load. 
 
 If therefore we require to construct an influence 
 
 diagram for the bending moment at a particular 
 
 x(l - x) 
 section C, we set up CCi = -i— — ^ and join AC^, 
 
 BCj. An influence line ADB for the horizontal 
 thrust, or the H-line for the section, is then plotted 
 from the equation for the thrust given above, when 
 the shaded area between the two figures will be an in- 
 
THE TWO-HINGED ARCH 63 
 
 M 
 
 fluence diagram for values of — , from which there- 
 
 y 
 
 fore the values of M can be found directly. 
 
 By way of example and for the sake of comparison 
 with the 3-pinned arch already considered, an in- 
 fluence diagram will be drawn for a 2-pinned arch 
 of the same span and rise for a section C aX \ span. 
 
 The following table will facilitate the calculatipn 
 of the ordinates of the H line : — 
 
 Values of § . « (1 - n) (1 -|- n - n'') for Different Values 
 of 11. 
 
 n = 
 
 - n (1-m) {1 + n-n^) 
 
 05 
 •031 
 
 •10 
 ■061 
 
 15 
 •09C 
 
 20 
 116 
 
 25 -30 
 ■189-159 
 
 35 
 
 174 
 
 40 
 
 188 
 
 45 
 193 
 
 •50 
 •195 
 
 If we multiply each of these values by - = -— = 10, 
 
 in the present case we get the values of H due to 
 
 a unit load in the corresponding positions. These 
 
 values when plotted will therefore give the H line, 
 
 and the ordinate between this and the triangle ACjB 
 
 M 
 will give the values of — for a unit load in any position 
 
 on the span. The values of M at once follow. The 
 bending moment at the section for any system of 
 concentrated loads in any given position may there- 
 fore be easily found in the usual way by multiplying 
 each load into its ordinate on the diagram and sum- 
 ming the results with due regard to sign. At the 
 same time the influence area will give the value of 
 M for a distributed load. Now in the present instance 
 
 nn * 1 2;a - x) 37-5 x 112-5 „ .q 
 
 CG, at i span = -A-^ = ^gq ^ ,,.3 = 2-49. 
 
64 THE STABILITY OF ARCHES 
 
 The influence diagram will therefore be as shown in 
 Fig. 26, and the bending moment at ^ span will be 
 found by multiplying the area AC^E by 2 '2 tons per 
 ft., the negative area EDDj also by 2-2, tons per ft., 
 and the negative area BD,D by 1'3 tons per ft. 
 
CHAPTBE V. 
 
 THE HINGELESS ARCH. 
 
 35. An arch without hinges is not only less flexible 
 than an arch with joints, but tbe resistance which 
 the fixed ends offer to bending assist in relieving the 
 stresses in other portions of the arch, when other 
 conditions remain the same. Its main disadvantages 
 are that the calculation of the stresses requires great 
 accuracy, and the results obtained by graphical 
 methods are not to be trusted except as a check on 
 the calculations. At the same time the stresses due 
 to change of temperature may appreciably increase 
 the stresses due to the loading. If, however, the 
 arch can be erected in cold weather, the effect of 
 any rise in temperature will be to cause a reduction 
 in the maximum stresses, and the fixure of the ends 
 will therefore have a beneficial effect in this respect). 
 As the cost of construction is in general not greater 
 than in other kinds of arch, the advantages appear 
 to favour this type, so that the theoretical difficulties 
 appear to be the sole reason why the hingeless arch 
 has not been universally adopted for long spans, 
 where the difficulties of erection are the same in 
 any case. 
 
 36. General Expression for the Bending Mo- 
 ment at any Section of a Symmetrical Hingeless 
 (65) 5 
 
66 
 
 THE STABILITY OF ARCHES 
 
 Arch under Vertical Loading. — Let ACB (fig. 27) 
 be the axis of the arch, and AjCjBj be the hne of pres- 
 sure (at present unknown) for the load upon it. Then 
 by Eddy's Theorem, sect. 15, the bendingmoment M at 
 aay point is represented by the product of the hori- 
 zontal thrust H and the vertical intercept between 
 the arch-axis and the line of pressure at that point. 
 Thus at the point C, M = H x COi. 
 
 Pig. 27. 
 
 Let a horizontal line A2B2 be taken as the X-axis, 
 and its mid-point as origin, and let the vertical 
 through C cut it in Cg and the line A^Bj in D. 
 
 Then CCi = CiD - C^D - CCj. 
 .-. M = H(CiD - CjD -y) = B. .GJ) --H.[G^T> + y). 
 
 But H X CjD = ft. = the bending moment due 
 to the same system of loads on a simply supported 
 beam. 
 
 .-. M = ;a - H(C2D + y). 
 
 But CoD = z„ + 
 
 I 
 
 X where z„ is the mean 
 
 ordinate of the trapezium A^AjBgBj. 
 
 Then M 
 
 .-. M = yu - H^/ 
 Let Xj = H !A 
 
 B.Z,. - H 
 
 Zk 
 
 ■ .X. 
 
 I 
 
 ny - X^x 
 
 -^andX, = B..z„ 
 
 X,. 
 
THE HINGBLBSS AECH 67 
 
 Introducing this value of M into equations (14), 
 (15), and (16) of sect. 27, we get 
 
 2^.Ss-H2f . 
 
 5s-X,2f 
 
 .Ss -X,2|. 
 
 . 8s = . 
 
 . (U)' 
 
 2'q.ss-n:$^^ 
 
 .Ss-X,t^. 
 
 , Ss -Xj2^. 
 
 Ss = , 
 
 (15)" 
 
 2^.8s-H2| 
 
 . Ss - X, 2 1 . 
 
 Ss - X22j. 
 
 , Ss = . 
 
 (16)" 
 
 Now for an arch which is symmetrical about a 
 vertical axis S / • 8s = for symmetrical values of 
 X, since the terms have equal and opposite values on 
 each side of the centre. Also 2^ • 8s = for the 
 same reason ; and if we choose our origin so that 
 2- . 8s =, 0, the above equations reduce to 
 
 2^ . Ss - H 2^ . 8s = . . (14)"' 
 
 2^ . 8s - Xi 2 J . 8s = . . (15)"' 
 
 2^.Ss - X2 2-'j^..8s= . . (16)'" 
 
 Further, if we divide up the arch-axis so that the 
 
 Ss 
 ratio -J is kept constant, we can reduce these to the 
 
 form 
 
 %/x.y - B.%y'' = . . (14)"' 
 
 2m« - Xi2a;^ = . . (15)>^ 
 
 2/t - X, . M = . . (16)'^ 
 
 where n is the number of divisions into which the 
 arch-ring is divided. 
 
 Frorn these last equations we get 
 
H = 
 
 
 X, = 
 
 2a;2 
 
 ^2 = 
 
 2/^ 
 
 68 THE STABILITY OF ARCHES 
 
 . (22) 
 
 • ■ (23) 
 
 . . . (24) 
 
 37. In order to illustrate the application of these 
 formulae, we will apply them to the same example 
 as before, except that in the present case the ends 
 will be assumed as rigidly fixed, and to make the 
 example more generally useful, the arch-ring will be 
 supposed of variable depth. The material will be 
 taken as concrete weighing 150 Ib./ft.^, reinforce- 
 ment will be provided to the extent of 2 per cent of 
 the crown area, and this will be embedded sym- 
 metrically about the axis at 2 ins. from the outer 
 surfaces of the arch-ring. The thickness at the 
 crown will be taken as 2 ft. and at the springings as 
 3 ft., the extrados and intrados being circular arcs. 
 
 In order to divide the arch-ring into segments so 
 
 that the ratio -p may be constant, in accordance with 
 
 the condition assumed above, the values of the 
 moment of inertia were calculated by dividing the 
 half axis into five equal parts. The moment of 
 
 inertia of the concrete is I^ = — where b is taken 
 
 as 1 ft. and t is the thickness of the arch-ring at the 
 section. The moment of inertia of the steel is 
 I, = A, X (distance from axis)^, where A, is the total 
 area of the steel at the section. The total moment 
 of inertia of the cross-section is then I = I„ -f 15 I„ 
 taking the ratio of the elastic moduli as 16. The 
 
THE HINGELESS ARCH 
 
 69 
 
 values of I at six equidistant sections were estimated 
 to be as follows : — 
 
 Section. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 6 
 
 6 
 
 ft.'' 
 
 I 
 
 1-08 
 
 1-23 
 
 1-42 
 
 1'78 
 
 2-46 
 
 3-32 
 
 A horizontal base-line 0-7 (Fig. 28) was then drawn 
 equal in length to 1'6, and the values ^I were plotted 
 above and below it, to any convenient scale so as to 
 obtain a diagram representing the moments of inertia 
 
 Pig. 28. 
 at any point. Then -if parallel lines be drawn at 
 any slope as shown in Fig. 29 the ratio — will be 
 approximately constant throughout, and by trial and 
 
70 THE STABILITY OF AECHES 
 
 error it is easy to find a slope such as will divide 
 
 the arch-ring into a convenient number of parts, and 
 
 which will close up at the point 7. The values of 
 
 Bs so found are then set off along the axis of the 
 
 arch, and the loads Wj, Wg, etc., acting on these 
 
 lengths are assumed to act at their mid-points. 
 
 The corresponding values of 8a; measured along 
 
 the horizontal are given in the following table along 
 
 with other data required for solving the equations 
 
 (22), (23), and (24). Column 1 gives the section 
 
 considered ; column 2 gives the vertical ordinate y' 
 
 of the arch-axis at the section. By addition we 
 
 get ^2 [y] = 79'40 ft., and as n = 14 we have 
 
 79-40 
 
 — f- — = 11 '34 ft. as the mean value. Now in order 
 
 that the condition 2 j . 8s = 0, or in our case 
 
 Sy = 0, may be satisfied, the axis AjBj, from which 
 the values of y are measured, must be drawn at this 
 distance above the line AB. Column 3 gives the 
 values of 2/ = 2/' - 11"34. The algebraic total of 
 this column should be zero, but due to unavoidable 
 small errors we find it to be - 0'04. This error is 
 then distributed equally in column 4, thereby reduc- 
 ing the error of the total to + 0-002. Column 5 
 gives the values of x' corresponding to the values of 
 y', measured from the left-hand end ; column 6 gives 
 the horizontal lengths of the segments &x into which 
 the arch is divided, and column 7 gives the loads on 
 these segments, viz. for 2-2 tons per foot on the left half 
 and 1'3 tons per foot on the right half of the span. 
 These loads were next plotted on a vector line, and a 
 
THB HINGELESS AECH 
 
 71 
 
 
 
 W, 
 
 W3 O »« »0 iH O O 
 
 co 
 
 G<l 
 
 N 
 
 
 
 
 o 
 
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72 THE STABILITY OF ARCHES 
 
 bending moment diagram was drawn for them in the 
 usual way, as if they were on a simply-supported 
 beam, and the values of the bending moments /j, 
 at each section so found are given in column 8. 
 Column 9 gives the values of /J^y as obtained" from 
 columns 4 and 8 ; whilst column 10 gives the 
 squares of column 4. 
 
 Inserting the values of 2//.?/ and %y^ so found in 
 equation (22), we get 
 
 g^yy^ 63974 tons/ft.^ ^3 
 
 ^y'^ 210-92 ft.2 
 
 If we include the effect of the normal thrust on 
 the arch the denominator will be increased by an 
 amount which is approximately equal to 
 
 I v_ I 
 8s ■ R ■ n.A, 
 
 where v is the distance of the centre of curvature at the 
 
 crown from the horizontal axis of eo-ordinates = about 
 
 I 3 
 191 ft. in our case, and since ^ = = and A„ = 2'5 sq. 
 
 ft. say the value of the above term is approxi- 
 mately 6 tons. If we take this into account we get 
 H = 295 tons. 
 
 In order to find the values of x^ and x.^, and there- 
 by fix the position of the line of pressure and the 
 bending moments, the values oi /jx and x^ required 
 in equation (23) are found and tabulated below. 
 
THE HINGELESS ARCH 78 
 
 Table II, fm- the Calculations of Xj and Xj. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 Section. 
 
 Tons/Ft. 
 
 y in Ft. 
 
 /jj in Tons/Ft.2 
 
 j« in rt.2 
 
 1 
 2 
 3 
 4 
 5 
 6 
 7 
 
 7' 
 6' 
 
 5' 
 4' 
 3' 
 2' 
 V 
 
 1,740 
 2,940 
 3,900 
 4,270 
 4,430 
 4,970 
 4,980 
 
 4,840 
 4,620 
 
 4,330 
 3,880 
 3,280 
 2,400 
 1,000 
 
 + 66-40 
 + 50-90 
 + 39-15 
 + 29-10 
 + 20-15 
 + 11-80 
 + 3-90 
 
 - 3-90 
 -11-80 
 
 -20-15 
 -29-10 
 -39-15 
 -50-90 
 -66-40 
 
 + 115,500 
 + 149,700 
 + 152,700 
 + 124,200 
 + 89,280 
 + 58,650 
 + 19,420 
 
 - 18,880 
 54,520 
 
 - 87,260 
 -112,900 
 - 128,400 
 -122,200 
 
 - 66,400 
 
 4,409-0 
 
 2,590-8 
 
 1,532-7 
 
 846-8 
 
 406-0 
 
 189-2 
 
 15-2 
 
 9,939-7 
 2 
 
 19,879-4 
 
 
 2^ = 51,580 
 
 - 709,450 
 + 590,560 
 
 
 ■S,ii.x=- 18,890 
 
 
 From the above values we get by equation (23) 
 %ixx _ - 18890 _ .Qf;f^ f^. _ -cr «A - «B 
 
 -950 ft. = H 
 
 X — ^ 
 ^ ix- 19879 t 
 
 and by equation (24) 
 
 X„ = ^ = ^^^ = 3684 tons/ft.2 = H ,^„ = H ^^ + ^'^ _ 
 
 «A + «B 
 
 and 0A 
 
 2b = 
 
 2 X 3684 
 295 
 
 •95 X 160 
 
 295 
 
 Whence zp, = 12-73 ft. 
 and zb = 12-25 ft. 
 
 = 24-98 
 0-48. 
 
74 THE STABILITY OP ARCHES 
 
 The fixing couples at the supports can now be 
 found for 
 
 AAi = ^A - AA2 = 12-73 - 11-34 = 1-39 ft. 
 and BBj = «b - BB^ = 12-25 - 11-34 = 0-91 ft. 
 and Ma = H X AAj = 295 x 1-39 = 410 tons/ft. 
 and Mb = H X BBi = 295 x 0-91 = 268-5 „ 
 
 The value of the reaction at A can be found by 
 taking moments about B. Thus 
 Va X 150 + Mb - Ma - (165 x 112^ + 97-5 x 37i) = 
 whence Va = 149 tons. 
 
 The bending moment at any section can 'then be 
 found from the equation 
 
 -M = IX - Ky - B.Z, - B.^^ " ''^ - 
 
 = fj. - HU + 
 
 I 
 
 ZA - Zb 
 
 I 
 
 ..). 
 
 Thus at section 1 we get 
 M=1740 - 295 (- 7-934 + 12-49 + -0032 x 66-4) 
 
 = 1740 - 1406 = 334 tons/ft., 
 and at section 7, near the centre, 
 M = 4980 - 295 (3-606 + 12-49 + 0032 x 3 90) 
 = 4980 - 4760 = 230 tons/ft. 
 
 38. Treatment of the Hingeless Arch by means 
 of Influence Lines.^In order to investigate the 
 worst conditions of loading at any given section of 
 the arch-ring, the most convenient method is to 
 construct an influence line for the thrust, bending 
 moment, or other effect which it is desired to in- 
 vestigate. 
 
 We have seen in chapter V, sect. 34, that under suit- 
 able conditions ^~ .Ss= and that if the arch-axis be 
 
THE HINGELESS AECH 
 
 75 
 
 so divided as to make y- constant, then %[y] = 0, and 
 in equations 22, 23, and 24 of chapter v we have 
 
 H = |f (22), X, = ^ (23), and X, = ^ (24) 
 
 where Xj = H ^J^-^ and X^ = H . «„ (see sect. 36). 
 
 Having then divided up the axis of the arch- ring 
 so as to make -— constant in the way already ex- 
 plained (sect. 37), the position of the x axis is drawn in 
 as before, and this determines the values of the in- 
 tercepts y between the x axis and the arch-axis. 
 
 Graphical Evaluation of the Terms Syn?/, %y^, etc. — 
 Let ADjB (Fig. 29) represent the influence line 
 
 Pig. 29. 
 
 for bending moment for a unit load on a simply 
 supported beam AB. Then when a unit load Wj 
 acts in any position C, Wj . COj is the bending 
 moment /aj which it produces at a given section D. 
 Similarly if a load Wg act at E, Wj . EBj is the bend- 
 ing moment ft,^ which it produces at D, and so on for 
 other loads. Hence the total bending moment at D 
 is Wj . CCi + Ws . EBj + etc. = 5W . />. 
 
76 THE STABILITY OF ARCHES 
 
 Let AjCjBi, A1B2B1, AiPjBj be the bending mo- 
 ment diagrams for loads Wj, Wj, W3. 
 
 Then the bending moment diagram for the total 
 load is made up of the ordinates of these triangles ; 
 and therefore the ordinate DjDj is the sum of DjOi, 
 Djfe, and DjC. But D./l represents the bending mo- 
 ment at D due to the load Wj acting at C = 
 Wj . CCj ; similarly DJ) represents the bending mo- 
 ment at D due to Wj acting at E = Wj . BEj, and 
 so on. Hence the total ordinate D2D3 represents the 
 value of 2W/U., and therefore if the loads W are the 
 ordinates y, %fi.y is represented by the ordinate z — 
 D^Dg of the bending moment diagram drawn for 
 these loads, at the section D under consideration. 
 We therefore set off the values of y vertically along a 
 vector line (ii), Plate III, and with any pole O^ we 
 construct a link-polygon AjCj. Also setting off the 
 values of y horizontally as at (i), with a pole Oj at 
 the same polar distance as before, we draw a link- 
 polygon for these so-called loads. Then the inter- 
 cept wwj cut off on the X axis by the first and last 
 links gives the total mpment of the forces y about 
 this axis = 52/^. In the present case mm-^ = 2 x 
 5-8 ft. = 11-6 ft. Similarly if we suppose the same 
 points as before loaded with the values of x instead 
 of those of y, the ordinates of the link-polygon will 
 give the value of %nx. The so-called loads x are 
 therefore plotted along a vector Une in (iii), and any 
 pole O3 being taken, another link-polygon AjCj is 
 drawn, and its closing line AgCg. Then the ordinates 
 of this figure represent the values of %iix for any 
 section, whilst the intercept G^^ on the vertical axis 
 of the arch, between the first and last links produced 
 
THE HINGELBSS AECH 77 
 
 to cut it, represents the sum of the moments of the 
 X loads about that axis. Hence 2C3,. represents the 
 value of 2*^ for the whole arch, and therefore 
 
 X, = ^ = 4^ tons. 
 ^ ■S,x' 2C3TO 
 
 If therefore we suppose 2C^n to represent the unit 
 load to scale, the value of the ordinate z^ at any 
 point gives the value of X^ for a unit load at that 
 point, and is therefore the influence diagram for Xj. 
 
 Similarly since 
 
 H = ?/^ = ^2 
 
 if we suppose 2wtWj to represent the unit load to 
 scale, the value of the ordinate Zg at any point will 
 be the horizontal thrust for the unit load at that 
 point, and consequently the link-polygon, or more 
 accurately the curve which envelops it, is an in- 
 fluence line for the value of the horizontal thrust due 
 to the unit load in any position. It is therefore 
 spoken of as the H polygon. 
 
 Finally, if we set down the values of unity to any 
 scale along a vector line at (iv) and with any pole 
 O4 construct another link-polygon A4C4B4, the or- 
 dinates Z4 of this polygon will give the values of 
 21 X yu, for any position of the load. If we make 
 the polar distance = ^w = 07 where n is the number 
 
 of the unit loads, then X„ = — = \z^, and the 
 
 n 
 
 ordinates of the polygon measured to twice the linear 
 scale are the influence values for X^. 
 
 In this way we have influence diagrams represent- 
 ing the values of H, Xj, Xj. Now the bending 
 moment at any section of the arch is 
 
78 THE STABILITY OF ABCHBS 
 
 M = /x - (H2/ + Xio; + X2). 
 If therefore we adhere to the scale of the Xj diagram , 
 that is to say, twice the Hnear scale, and alter the 
 
 ordinates of the H diagram in the ratio -^ — and the 
 
 ordinates of the Xj diagram in the ratio js^—, where 
 
 y and x are the values taken at the point for which 
 the value of M is required, the algebraic sum of the 
 corrected ordinates will determine the influence 
 line for the values of Hy + X^a; + X2 at the section 
 considered. 
 
 In order to obtain an influence diagram for M, 
 therefore, we construct the influence line for the 
 values of fj. at the section considered. This diagram 
 is a triangle, and using the same vertical scale, viz. 
 twice the linear scale, the ordinate of this triangle 
 on the vertical axis of the arch, should measure 
 
 x' = -^ - X. The triangle being drawn in, the in- 
 
 fluence line for M will be the diagram between it 
 and the curve already found for the values of Hy + 
 X^x + Xj. 
 
 In Plate III the above procedure is illustrated. 
 Having drawn the curves AgCj, A3C3, and Afi^B^, 
 the ordinates of the first two polygons are altered in 
 
 the ratio -2— and — — by changing the polar distances 
 
 in the inverse ratios. This has been done for two 
 sections, viz. the section A at the extreme left, and 
 the section 6, for which the corresponding values 
 of X and y were measured. In this way the new 
 polygons Aja, AjC^, Ajfo, AjC were obtained. The 
 
Hnilr¥¥Tt 
 
 .^^. 'T 
 
 30 Co io 6« ^ So 55 iSo MO 
 
 Staiilify of Arclies. 
 
 Plate hi. 
 
 [Tofo,ce pnqe 79. 
 
THE HINGELESS AECH 79 
 
 ordinates of the Xj and X2 polygons were then added 
 and those of the H polygon subtracted, the result 
 being the curves A4C7B4, A4OHB4. In the first 
 case, since there is no bending moment at A for a 
 freely supported beam, the curve obtained is the in- 
 fluence diagram for the bending moment at A, and 
 its area, as measured by planimeter was found to be 
 + 222 sq. ins. for the left-hand half of the girder, 
 and - 2-17 sq. ins. for the right-hand half. There- 
 fore since the left-hand half carries a load of 2-2 tons 
 per foot and the right-hand half carries a load of 1'3 
 tons per foot, and remembering that the scale of the 
 drawing was 1 in. = 20 ft. and that the vertical scale 
 is twice the linear scale, the total bending moment at 
 A due to the whole load is (2-22 x 2-2 - 2'17 x 1-3) 
 X i^S tons/ft. = 412-6 tons/ft., which checks closely 
 with the result hitherto obtained by calculation (see 
 p. 74). 
 
 In the second case, dealing with the bending 
 moment at section 6, having obtained the resultant 
 curve for the values of Hy + Xja; -1- Xg, the triangle 
 ^i^^^i is drawn having its intercept on the vertical 
 axis = the distance of the section from the left-hand 
 end. The shaded area then represents the influence 
 diagram for M, and the algebraic sum of its areas, 
 which is here negative and represents - 154 tons/ft., 
 is the bending moment at section 5. 
 
 In this way influence lines for the bending moment 
 at any section may be drawn, and it is then easy to 
 see what distribution of loading will produce the 
 most severe moment, 'and to obtain the value of it. 
 
 Now the maximum stresses due to the bending 
 moment are always much in excess of the stresses 
 
80 THE STABILITY OF ARCHES 
 
 due to the axial thrust, and therefore the distribution 
 of load which determines the maximum bending 
 moment at any section may be taken as that which 
 produces maximum stress at the section. Also the 
 normal thrust N for this loading may be found ap- 
 proximately from the formula N = H seci^, where 
 1^ is the angle which the section makes with the 
 vertical. 
 
 39. By way of illustration suppose we require the 
 worst conditions of stress for the section 5. Having 
 constructed the influence diagram for the bending 
 moment at this section, we find that the areas of the 
 positive and negative parts are + 132, - 308, and 
 + 36 sq. ft., or + 66; - 154, and + 18 tons/ft. per 
 ton of load covering these segments. Hence the re- 
 sultant bending moment due to the dead load of 1'3 
 tons per ft. = (66 + 18 - 154) x 1-3 = - 91 tons/ 
 ft., and the maximum positive moment due to the 
 live load of '9 ton per foot covering the two extreme 
 segments is -9 (66 + 18) = 75-6 tons/ft., whilst the 
 maximum negative bending moment due to the live 
 load is '9 x 154 = 138-6 tons/ft. The actual greatest 
 and least values of the bending moment at the section 
 are therefore - 91 + 75-6 = - 15-4 tons/ft., and 
 - 91 - 138-6 = - 229-6 tons/ft. 
 
 The horizontal thrust H for these conditions of 
 loading are found from the H line to be 260 tons 
 and 376 tons respectively, whilst <f> = 16^°. 
 
 .-. N = 260 sec 15f and 376 sec 15^° resp. 
 = 270 and 390 tons. 
 
 M. Mesnager's Method. — This method is described 
 in " Engineering," 17 March, 1916. It affords an easy 
 means of calculating the stresses in a concrete arch 
 
THE HINGELESS AECH 
 
 81 
 
 which satisfies the conditions that its axis is paraboHc 
 and that its moment of inertia increases according 
 to the law I cos 6 -— I,,, where I„ is the moment of 
 inertia at the crown, and I is the moment of inertia 
 at any other section inclined 6 to the vertical. Then 
 if the thickness at the crown is /, the thickness at 
 
 any other point 
 
 If the arch is built in this 
 
 ^cos 6 
 way the bending moments at any point can easily 
 
 Fig. 30. 
 
 be found graphically, by the following construction, 
 a proof of which is given in the issue of " Engineering " 
 referred to above. Let ACB be the arch axis, and 
 let W be a load in any position. Make CE = i CD 
 and through E draw a horizontal FG. Make 
 6 
 
82 THE STABILITY OP ARCHES 
 
 CO = I CD and through O draw HK horizontal. 
 Also let E be the point in which the line of action of 
 W outs FG. Join EH, EK, and from draw OL, 
 OM parallel to EH, EK. Join EM, EL. Then the 
 intercept between these two lines and the arch axis 
 gives at every point the bending moment for the arch 
 produced by the load W on the same scale that the 
 triangle LEM represents the bending moment pro- 
 duced by the same load on a simply supported 
 girder. Thus ET represents the bending moment 
 on the arch-ring to the same scale that ES represents 
 the bending moment on a simply supported girder. 
 
 Since the ratio of all the vertical lines in the 
 figure will be the same whatever the scale of vertical 
 distortion, a diagram may be drawn to a large scale, 
 and this diagram can then be used for finding the 
 bending moments on an arch of the above type, 
 whatever its rise and span, the result being rigidly 
 accurate so long as the work done in compressing 
 the arch-ring is negligible in comparison with that 
 done in bending it, an assumption usually regarded 
 as sufficiently accurate for practical purposes. 
 
CHAPTEE VI. 
 
 MASONRY AND CONCRETE ARCHES. 
 
 40. Previous to the development of the elastic 
 theory, the investigation of the stability of masonry and 
 brickwork arches was always carried out by testing 
 whether an equilibrium polygon could be drawn to 
 lie within the middle-third of the arch-ring ; because 
 if this were possible it was argued that the arch 
 would be stable. This method of solution, however, 
 gives no clue to the actual stresses set up, because 
 the actual line of pressure is undetermined ; and 
 therefore where the greatest economy of material is 
 desirable and necessary, as in the case of very large 
 arches, the determination of the actual line of pres- 
 sure is required, and this is only possible by means 
 of the elastic theory of the arch ; but in the case of 
 small bridges, in which the thickness of the arch-ring 
 is of secondary importance, the older and less accu- 
 rate method may be used and is useful in checking 
 the results obtained by calculation. 
 
 41. It has already been explained that, treated 
 as a statical problem, no solution for a hingeless arch 
 is possible, because the three conditions necessary to 
 fix the position of the line of pressure are absent. It 
 was, however, realized by the early engineers that an 
 arch must necessarily be stable provided that :-^ 
 (83) 
 
• 84 THE STABILITY OF AECHBS 
 
 (1) There is no tension at any joint of the arch- 
 ring. 
 
 (2) That the crushing strength of the material is 
 not exceeded. 
 
 (3) That the Une of thrust nowhere makes an 
 angle with the normal at any bed-joint greater than 
 the least value of the friction angle. 
 
 These are the three fundamental conditions of 
 stability for an arch-ring. Now in order that the 
 first of these conditions may be 
 satisfied, it may easily be shown 
 -^ that, assuming a linear distribu- 
 M tion of stress, it is only necessary 
 for the line of thrust to lie within 
 certain critical limits, which for 
 a rectangular cross-section are 
 the middle-third points of the 
 ' arch -ring ; or if we allow for 
 
 "'■ ■ frictional resistance, it may be 
 
 taken, according to Scheffler, that provided the line 
 of pressure lies within the middle half the primary 
 condition of stability is satisfied. 
 
 For when the resultant thrust on any section of a 
 structure acts out of centre in the plane of the princi- 
 pal axis, the stress at any point distant y (Fig. 31) 
 from the axis XX through its centroid is given by 
 the formula 
 
 where e is the eccentricity of the resultant thrust, N 
 is the component of this thrust normal to the section, 
 and y is to be taken positive or negative according 
 
 
 
 A 
 
 1 
 
 1 
 
 1 
 
 I 
 _t_ 
 
 L J 
 
MASONRY AND CONCRETE ARCHES 85 
 
 as it is on the same side of the axis XX as the re- 
 sultant or not. Now for a rectangular section 
 
 
 ''" "12 
 
 
 in which case 
 
 the formula becomes 
 
 
 Therefore it the stress becomes zero 
 
 at the extreme 
 
 edge when y 
 
 = - - we get 
 
 = 1 - 6-f . 
 
 t 
 
 t 
 
 
 that is to say, when the resultant departs from the 
 axis of the section by one-sixth of its depth in either 
 direction the stress on the remote edge is zero, and if 
 more than this it becomes tensile. This is the well- 
 known "Eule of the Middle-Third," viz. that so long 
 as the resultant thrust lies within the middle - 
 third of the depth, there will be no tensile stress in 
 the arch-ring. 
 
 42. The question, however, recurs as to where 
 the resultant does actually lie when the shape of the 
 arch is known and the loading and other conditions 
 are given. In the absence of the elastic theory it is 
 necessary to rely upon the somewhat metaphysical 
 argument known as the " Principle of Least Work ''. 
 This important principle, which all experience tends 
 to justify, has been stated very clearly by Eankine 
 as follows : — 
 
 "If a body be in equilibrium under the action of 
 any system of forces, this system will consist of forces 
 
86 
 
 THE STABILITY OP AECHES 
 
 of two kinds, which may be described as active and 
 passive forces, standing to one another in the relation 
 of cause and effect, the passive forces being called into 
 play by the active forces ; and since these passive 
 forces will not increase after they have once pro- 
 duced equilibrium with the active forces, we conclude 
 that the passive forces will be the least which are 
 required to produce equilibrium." 
 
 In other words, we assume that Nature effects her 
 purpose with the least expenditure of mechanical 
 energy. It follows from this " Principle of Least 
 
 Pig. 32. 
 
 Work " that of all the equilibrium -polygons that can 
 b3 constructed for a system of loads on an arch, that 
 one is the true line of pressure which lies nearest to 
 the axis of the arch-ring. For if AA', BB' (Fig. 32) be 
 two neighbouring sections, unit distance apart, and s 
 is the stress at any distance y from the axis C'C, then 
 
 — = where s is the stress at unit distance 
 
 s" a 
 
 from the neutral axis NN, and since the strain-energy 
 
 in tension or compression per unit length is ^ volume 
 
 X— the total stram- energy developed between the 
 
MASONRY AND CONCBETB ARCHES 
 
 87 
 
 two sections will be given by the expressions 
 
 ^b . 8y . Sxs'^ fl + ^y 
 
 E 
 
 Sx.s- 
 ~2E" 
 
 b.Sy + 
 
 ^yb. 
 
 ^x . s} 
 
 A + + i i,^ I„ 
 
 
 b.hy 
 
 
 As„- . &x 
 2E 
 
 "l + 
 
 
 and this is least when a is greatest. But the posi- 
 tion of the load-point L and the position of the 
 neutral axis N are connected by the relation ON . CL 
 = k^, so that when a is greatest, the load-point 
 approaches nearest to the axis. Hence we derive 
 the following important conclusion, that the, result- 
 ant line of thrust will always endeavour to set itself 
 as near the axis of the arch-ring as possible, placing 
 itself in such a position that whilst remaining an 
 equilibrium-polygon for the given loads it will develop 
 the minimum amount of strain-energy in the arch- 
 ring. 
 
 We conclude therefore that if an equilibrium-poly- 
 gon can be drawn for the given load system that 
 will lie altogether within the middle-third of the 
 arch-ring, the primary condition of stability will be 
 satisfied, because the true line of pressure will lie 
 nearer to the axis than the one found, and therefore 
 the arch will be more stable than if this one were 
 the true line of thrust. 
 
 Now if we consider the manner in which an or- 
 dinary segmental arch usually yields under normal 
 conditions of loading, we find that cracks begin to 
 form along the sgfijt at the grown, and along tb§ 
 
88 
 
 THE STABILITY OF ARCHES 
 
 extrados towards the springings as shown at A, B, 
 and C. This indicates that the Hne of pressure has 
 passed outside the middle-third limits. In the criti- 
 cal position when the line of pressure just touches 
 the middle-third lines as at A, C, and B (Fig. 33), the 
 sections at which contact first occurs are known as 
 the joints of rupture, because the arch will first yield 
 
 ~7777777777777T7777777T7~77~' 
 
 TTTTTTTm-rrr^ 
 
 FiQ. 33. 
 
 at these points, and the equilibrium-polygon is then 
 known as the critical line of pressure. 
 
 Should the arch yield in the manner indicated 
 above, the critical line of pressure is known as the 
 curve of minimum thrust, because the horizontal 
 thrust of the arch is then the least that can occur 
 consistent with the first condition of stability. On 
 the other hand, if the arch is so shaped or so loaded 
 that it yields as shown in Fig. 34, the critical line 
 of pressure is known as the curve of maximum 
 thrust, because the horizontal thrust is then the 
 greatest that can occur within the same limitations. 
 
 43. Construction of the Critical Line of Pres- 
 sure. Reduced Load Curve. — In dealing with the 
 load upon an arch it is usual to consider a portion 
 
MASONRY AND CONCRETE ARCHES 
 
 89 
 
 of the arch contained between two longitudinal sec- 
 tions 1 ft. or 1 m. apart, as the case may be, and to 
 assume that the weights of the arch-ring and its 
 
 Pig. 34. 
 
 load act vertically ; and this assumption is on the 
 sate side, because if the arch- ring have a rise equal 
 to or less than half the span, any horizontal forces 
 due to earth pressure or other causes will increase 
 its stability. Usually an arch has a concrete jBlling 
 over the haunches, as shown in Fig. 35, upon which 
 
 Fig. 35. 
 
 rests the earth or other material. These materials 
 may be of much less specific weight than that of 
 the arch-ring. In such a case it is convenient to 
 construct a reduced load curve by diminishing the 
 ordinates of each material so as to obtain a diagram 
 which represents the amount of masonry haying the 
 
90 THE STABILITY OF AECHES 
 
 same weight. Thus, e.g. if the weight of the arch- 
 ring be 160 lb. /ft.*, that of the concrete backing 140 
 lb./ft.3, of the earth filling 100 lb./ft.^ and if there is 
 also a live load of 200 Ib./ft.^, the ordinates between 
 the line EB' and the extrados are reduced in the 
 ratio j^^ giving the curve eB' ; the ordinates above 
 BB'D are reduced in the ratio ^|^ and added to the 
 last, whilst the live load will be represented by 
 an additional height of masonry of -|^2. ft. on the 
 scale of the drawing. In this way we get the line 
 HH' which is the reduced load curve, such that 
 the area HH'CAB. below it represents a quantity 
 of masonry whose weight is equivalent to that of 
 the combined weights of the other materials. 
 
 44. Method of Fictitious Joints. — A further 
 simplification in the construction is introduced by 
 disregarding the actual voussoirs and having con- 
 structed the reduced load curve, dividing it into ver- 
 tical strips of equal width. If these widths are not 
 large, their resultant weights may without sensible 
 error be taken as acting along their centre lines. 
 For if we take the trouble to consider the load rest- 
 ing on each voussoir, as is often 
 done, and compound it with the 
 weight of the voussoir itself, the 
 difference in the results obtained 
 will be quite inappreciable, be- 
 cause the weight of the tri- 
 angular prism ceb (Fig. 36) is so 
 small compared with the pres- 
 sures on its faces that the line of pressure is not 
 sensibly deflected by it in passing from cb to ce, and 
 therefpre no appreciable error is made in regarding thg 
 
MASONRY AND CONCRETE ARCHES 
 
 91 
 
 load as consisting of segments bounded by vertical 
 planes, whose weights are assumed to act along the" 
 mid-ordinates, and if these planes are equidistant, the 
 weights contained by them are proportional to the 
 mid-ordinates under the reduced load curve, and 
 these may therefore be used as vectors to represent 
 them. 
 
 46. Construction of the Critical Line of Pres- 
 sure for iViinimum Thrust at the Crown, and for 
 
 Curve 
 
 Pig. 37. 
 
 Symmetrical Loading. — Having constructed the 
 reduced load diagram and divided it into vertical 
 strips of equal width, vectors O'l, 1'2, etc. (Fig. 37), 
 are set off proportional to the mid-ordinates, and a, 
 
92 THE STABILITY OF ARCHES 
 
 pole being taken on the horizontal through o, a link- 
 polygon ab to h is drawn, whose sides are produced 
 to cut the first liijk ab produced in the points 1', 2', 
 etc. Now if we consider the equilibrium of any por- 
 tion of the arch-ring, such as AA'B'B, and assume 
 the line of pressure to act at the upper middle-third 
 point C at the crown, then if B" were by chance the 
 point of rupture, the resultant thrust at this point 
 will just be tangential to the lower middle-third line 
 at B", and its direction must be such as to pass 
 through the point 3" in which the resultant of the 
 load intersects the horizontal thrust through C, ac- 
 cording to the principle that when a body is in equi- 
 librium under the action of three forces, the directions 
 of those forces must pass through the same point. 
 B"3" will therefore be the direction of the resultant 
 thrust at B". But if we find that B"3" produced 
 cuts the lower middle-third line, B" cannot be the 
 point of rupture, because this occurs at the point 
 where the line of pressure touches the middle-third 
 line. We therefore try the point D" in the same 
 way by joining D"4" and seeing if it cuts the lower 
 middle-third line, and so on until we arrive at a 
 point, say B", where the line B"5" first touches. 
 This will be the joint of rupture, and a line through 
 . the point on the vector figure parallel to B"5" will 
 determine the pole 0' which corresponds to the 
 critical line of pressure. If the link-polygon is now 
 redrawn with this pole, the result will be the critical 
 line of pressure for minimum thrust at the crown. 
 If, however, no such tangent can be found, it does 
 not necessarily follow that an equilibrium-polygon 
 cannot be drawn lying within the middle-third 
 
JB R I C K E>RID°iE OVER THE Oglip RivEf? J??^ CLEwr; SPfl"/ 
 
 ^CALE ' t'xtH a 2 f^lfiM-S- 
 
 Slabilili/ of Arclies.^ 
 
 Plate iv. 
 
 [ J'o face page 'J3. 
 
MASONRY AND CONCEETB AECHES 93 
 
 limits, for this may sometimes be effected by start- 
 ing at a point below C at the crown. 
 
 46. In order to make clear the above descrip- 
 tion • the line of pressure is drawn in Plate IV for 
 a bridge over the Oglio, in Italy. This bridge is 
 of brickwork and has a clear span of 21 m. and 
 a rise of 11-9 m. The arch is a circular arc, the 
 radius of the intrados being 24 4 m. and the radius 
 of the extrados 27 "5 m. with a crown thickness 
 of 1-4 m. The width is 7'5 m. The weight of 
 the brickwork is taken as 2000 kilo, per cub. 
 metre, the load up to DE being also brickwork of 
 approximately the same weight. 
 
 Eadius of DB = 54-5 m. From DE to the hori- 
 zontal PG the backing is gravel weighing 1600 kg. 
 per c. m. GE = 1-1 m. The greatest live load to be 
 carried is 8000 kg. per metre run. 
 
 The half arch was drawn to a scale of 1 in. = 
 2 m. and was then divided into twenty- two strips 
 each 1 m. wide, with another one of less width at 
 the end. The mid-ordinates were drawn in, as 
 shown. These ordinates are proportional to the 
 weights of the corresponding strips and represent 
 those weights to a scale of 1 in. = 2'" x T'S" x 
 2000 kg./cm.' = 30,000 kg. The live load being 8000 
 kg. per metre will be represented by ^Vinr = '^^'^ ^'^• 
 The ordinates of the gravel are reduced in the ratio 
 z^ws- = T' ^^^ these together with the live load or- 
 dinates are added to the brickwork, so that we get 
 the reduced load curve representing the equivalent 
 amount of brick. The mid-ordinates of the reduced 
 load diagram were then reduced i and plotted along 
 a vector line as shown, the scale being now 1 in. = 
 
94 THE STABILITY OF ARCHES 
 
 240,000 kg.. A pole O was arbitrarily chosen on the 
 horizontal through o, and a link-polygon abc . . . w 
 was drawn whose links were produced to cut the 
 first link in the points 1', 2', 6', 7', etc. These 
 points were then projected upon a horizontal drawn 
 through the upper middle-third point at the crown, 
 giving 1", 2", 6", 7", etc. These represent the 
 points through which the resultant pressures on each 
 section of the arch-ring must go on the assumption 
 that these resultants pass through the lower middle- 
 third point. Thus if we join the point 12" say, to 
 the lower middle- third point of section 12, viz. P, the 
 trial line 12" P will be the direction of the resultant 
 thrust on this assumption. But when produced it 
 passes outside the lower middle-third line, and there- 
 fore the joint of rupture cannot be at this section. 
 We find that as we approach the abutment the 
 trial lines become more nearly tangential to the 
 lower middle-third line, and in this example we find 
 that it is only when we reach the abutment that the 
 trial line becomes actually a tangent. We therefore 
 draw through 23 a line having this direction, and 
 this determines the pole Oj for the critical line of 
 pressure. Starting from the upper middle-third 
 point at the crown this line of pressure is drawn in as 
 shown. 
 
 The polar distance thus found measures the hori- 
 zontal thrust corresponding to the critical hne of 
 pressure to the vector scale of 1 in. = 240,000 kg., and 
 would in this case give the actual thrust of the arch 
 because no other link-polygon could be drawn within 
 the middle-third iines which would lie nearer to the 
 axis than the one drawn ; in other words, the critical 
 
MASONRY AND CONCRETE ARCHES 
 
 95 
 
 line of pressure will in this case be the true line of 
 pressure or approximately so. 
 
 47. Simplified Treatment of Small Segmental 
 Arches. — In the case of small segmental arches it 
 will usually be found, as in the preceding example, 
 that the joints of rupture lie so near the springings 
 that no error worth considering will be introduced if 
 we assume that the line of pressure is tangential 
 to the lower middle-third line at the springings 
 
 themselves. This greatly simplifies the investiga- 
 tion, because if we find the centre of gravity of the 
 reduced load-area, the horizontal thrust H acting 
 through the upper middle-third point C at the crown 
 (Fig. 38) will meet the vertical through the centre 
 of gravity in Q, and therefore the resultant reaction 
 E acting through the lower middle-third point at the 
 springing joint must have the direction EQ. A tri- 
 angle of forces abc will then determine the magni- 
 tudes of H and E when W is known. 
 
 Further, experience shows that an arch in process 
 of construction will support itself when extending 
 
96 
 
 THE STABILITY OF ARCHES 
 
 beyond the abutments to some distance, and it 
 may safely be taken that it may be trusted to do 
 so in any case as far as a joint AA' making an angle 
 of about 30° (Fig. 39) with the horizontal as shown 
 in Fig. 39. This part of the arch may therefore be 
 regarded as forming part of the 
 abutment itself, and it is only in 
 this way that we can explain the 
 stability of some arches which 
 must otherwise be regarded as 
 violating the primary condition 
 of stability. In this case the 
 springing joint may be regarded 
 as at AA'. 
 48. Construction of the Critical Line of Pres- 
 sure for an Arch, when the Load is Asym- 
 metric. — When the load upon an arch is asym- 
 metric, the difficulty of constructing the critical line 
 of pressure is greatly increased, because the highest 
 point of the curve will no longer be at the centre of 
 
 Fig. 39. 
 
 Fio. 40. 
 the arch as it must be for symmetrical loading. It 
 will, however, lie somewhere near the centre on the 
 side on which the load is greatest. A good way is 
 therefore to select a point C (Fig. 40) in a reasonable 
 position and two other points A and B on the lower and 
 upper middle-third lines respectively at points near the 
 springings where we may suppose the joints of rupture 
 
MASONRY AND CONCEETB AECHBS 97 
 
 are likely to occur. A link-polygon for the given loads 
 is then drawn to pasa through the points A, C, and 
 B as described in Chapter II, and if this lies with- 
 in the middle-third lines, the primary condition of 
 stability is satisfied. But if it is found to cut these 
 lines, then the point D where the polygon recedes 
 furthest from the axis should be noted, and a point D' 
 being taken on the middle-third line at this section 
 another link-polygon is drawn to pass through A, 
 D', and B. 
 
 In the same way it may be found after drawing 
 the first trial-polygon that the points A and B may 
 require to be chosen in a different position to enable 
 the link-polygon to lie within the middle- third, and 
 the trial-polygon will enable us to select the most 
 suitable positions ; but if no such polygon can be 
 drawn, then we must either (i) thicken the arch- 
 ring ; (ii) alter the form of the arch ; or (iii) alter 
 the distribution of the load. 
 
 49. Adjustment of the Arch to Suit the Load. — 
 Let Fig. 41 represent an arch which has been designed 
 to approximate to the form desired and which is pro- 
 portioned, as far as one can estimate roughly, to carry 
 the dead load, the proportions being those commonly 
 adopted. For the sake of convenience in drawing 
 suppose the vertical scale to be magnified four times. 
 The lower figure represents the arch and its load so 
 magnified, ac being the reduced load curve for the 
 dead load. The area under this curve divided into 
 five strips of equal width, and the loads, as repre- 
 sented by the mid-ordinates of these strips reduced 
 to \ size were set down along a vector line 0"1 . . . 5, 
 a? shown, A pole Oj was taken as the horizontal 
 7 
 
98 
 
 THE STABILITY OP ARCHES 
 
 through 5 at any convenient distance and a link- 
 polygon defgh was drawn, its first and last links in- 
 tersecting in I. A vertical through I then represents 
 the line of action of the resultant load on the semi- 
 arch. 
 
 Draw through the mid-point of the crown a hori- 
 zontal CL to represent the horizontal thrust at the 
 5 4 3 a I 
 
 Pig. 41. 
 crown, meet the resultant load in L and join the 
 centre A of the springing joint to L. Then AL re- 
 presents the reaction at the springing when the line 
 of pressure acts at its centre. Draw oOj parallel to 
 AL, and with Og as a new pole, construct another 
 link-polygon starting at A, which should pass 
 through G. This will represent the line of pressure 
 
MASONRY AND CONCEETB AECHBS 
 
 99 
 
 if we neglect any small alteration in the load due to 
 the modified form of the arch-ring. We now alter 
 the axis of the arch so as to make it conform to this 
 curve, and the new form of the arch-axis so obtained 
 is shown by a dotted line in the figure. 
 
 50. Adjustment of the Load to Suit the Arch. 
 — The third method of design consists in altering the 
 
 Fig. 42. 
 
 distribution of the load so as to make the line of 
 pressure coincide approximately with the assumed 
 form of the arch. This may be effected by altering 
 the specific weights of the backing materials or by 
 building openings in the spandrels when the load 
 over the haunches is too great or by means of 
 transverse arches, etc. In order to determine the 
 
100 THE STABILITY OP AECHES 
 
 distribution of load required to bring about the de- 
 sired result, the half-span (Fig. 42) is divided into 
 a convenient number of equal parts. Verticals 
 drawn through the points so obtained intersect the 
 arch-axis in the points a, b, c. At A, c, b, a, C draw 
 radial lines OA, Oc, etc. Perpendiculars to OA and 
 OC at A and C respectively intersect in Q, and there- 
 fore CQ, AQ are the directions of the line of pressure 
 at C and A. If therefore we take Q as a pole, the 
 lines Qo and Q^ will be the directions of the rays of 
 the vector figure corresponding to the first and last 
 links of the line of pressure, and if we draw lines 
 through Q parallel to the tangents at a, b, and c, 
 these lines will cut off lengths 01, 1-2, 2-3, 3-4 on 
 any vertical which are proportional to the loads on 
 the corresponding segments of the arch. 
 
 If, therefore, we set up ordinates dd', ee', etc., 
 equal to these lengths, at the mid-points of the 
 horizontal segnlents into which the half-span was 
 divided, and draw a fair curve d'e'fg' through the 
 point so obtained, this curve will represent the dis- 
 tribution of load which will make the line of pres- 
 sure coincide with the axis of the arch. If we 
 estimate the probable weight of the load and the 
 arch-ring for the segment adjacent to the crown, and 
 note that this weight is represented by the length 
 0-1 in the vector figure, the scale of the diagram 
 will be determined with sufficient accuracy, and the 
 magnitudes of the loads on the other segments will be 
 represented by the lengths 1'2, 2-3, 3'4 to the same 
 scale. 
 
 51. Abutments of an Arch, — As we have already 
 seen, if an ordinary flat arch yields under its load, 
 
MASONEY AND CONCRETE ARCHES 
 
 101 
 
 the line of pressure rises at the crown and sinks at 
 the springinga; but if the earth yield because the 
 horizontal pressure due to earth or otherwise forces 
 the abutment inward, the line of pressure sinks at 
 the crown and rises at the springings as indicated in 
 the sketch. In the first case therefore the line of 
 minimum thrust should be used, and in the second 
 case the line of maximum thrust. 
 
 In considering the stability of an abutment there- 
 fore, if we disregard the earth-pressure the line of 
 
 •t: 
 
 Pig. 43. 
 
 minimum thrust must be assumed as our critical line 
 of pressure, and the arch must be considered as fully 
 loaded, because the horizontal thrust is then greatest. 
 The resultant W (Fig. 43) of the load on the half arch 
 is then found, and a A of forces determines the values 
 of H and E. The weight W of the abutment is 
 next calculated and compounded with the previous 
 resultant E as shown, We thus get E' which must 
 
102 
 
 THE STABILITY OF AECHES 
 
 lie within the middle- third of the base, and satisfy 
 the other conditions of stability. 
 
 If this does not occur the width or slope of the 
 abutments must be modified until the required con- 
 ditions are fulfilled. 
 
 52. Intermediate Piers. — In the case of inter- 
 mediate piers the most unfavourable condition of 
 the load occurs when one span is fully loaded and 
 the adjacent span unloaded. In this case, the line 
 
 FiQ. 44. 
 
 of maximum thrust is assumed as the critical line of 
 pressure for the unloaded span, and the minimum 
 line of thrust for the loaded span. 
 
 The resultant thrusts Ej and E^ at the springings 
 (Fig. 44) intersect in a. The resultant of these is 
 compounded with the weight of the pier W3, and the 
 final resultant cuts the base in L. 
 
CHAPTBE VII. 
 
 DESIGN OF MASONRY AND CONCRETE ARCHES. 
 
 53. Wheeeas in the construction of metal arches it 
 is a matter of small importance what the exact form 
 of the arch may be, in the case of masonry and con- 
 crete it is desirable that the line of pressure, in the 
 worst case of loading, shall remain within or not 
 far outside the middle-third of the arch-ring, and in 
 order that this may be so with a minimum thickness 
 of arch-ring the axis of the arch should deviate as 
 little as possible from the line of pressure for the 
 load it carries. In this case the arch-ring will be 
 under pure thrust and the stresses will therefore 
 have their least value. Any deviation from the true 
 form necessitates a heavier arch, and, moreover, the 
 deformations due to the load are increased, the theory 
 becomes less correct, and the weight and cost of 
 material are greater than need be, together with the 
 expense of erection. Hence the problem is, to give 
 the arch such a form that its axis for a definite mean 
 load shall coincide with the line of pressure for that 
 load. The direct solution of this problem is difficult, 
 though it may be approximated to graphically by first 
 designing the arch for its assumed load by any of 
 the standard formulae given in the text-books, then 
 (103) 
 
104 THE STABILITY OP ARCHES 
 
 determining the line of pressure for this arch, and 
 afterwards modifying the form of the arch so as to 
 make it conform to the line of pressure ; or again 
 the load on the arch may be modified by openings or 
 arches in the spandrels or by changing the specific 
 weight of the filling material, as explained in the 
 last chapter. 
 
 54. Empirical Formulae for the Thickness of the 
 Arch-ring at the Crown. — The thickness of the 
 arch-ring will depend to some extent upon the load 
 which the arch is likely to carry, and in view of 
 this it is convenient to classify arches under four 
 groups : — 
 
 (1) Light arches which carry only their own weight 
 and a light load, such as arches which form a ceiling 
 without floors. 
 
 (2) Mean arches, used in buildings, which carry a 
 floor, as in the case of cellar vaults. 
 
 (3) Heavy arches for road bridges, tunnels, etc., 
 subjected to heavy loads but only slight impact and 
 vibration. 
 
 (4) Very heavy arches for railway bridges, subject 
 to heavy loading and strong vibration. 
 
 The thickness at the crown may then be calculated 
 approximately by the formula 
 
 where n is the number of the group to which the 
 arch belongs in the above classification, s is the ratio 
 of rise to span, and I is the span. Thus for an arch 
 carrying a floor and having a span of 3 ft. and with 
 s = 5^ we have w = 2 and 
 
DESIGN OP MASONET, ETC., ARCHES 105 
 
 a result rather too large for such small arches. 
 For n = 3,1 = 100 ft., s = | we get 
 
 t„ = (-3 + L\ (3-28 + 10) = 4-64 ft. 
 
 Trautwine's formulm for circular and elliptical 
 arches are as follows :— 
 
 for first-class stone : t„ = 0-25 ^E + 0-52 + 0-2 ; 
 for second-class work : — 
 
 K = 0-281 VE + 0-51 + 0-225 ; 
 for brickwork or fair rubble : — 
 
 t = 0-333 JB, + 0-52 -f- 0-267, 
 where E is the radius of curvature at the intrados 
 at the crown, I is the clear span, and the units are in 
 feet. 
 
 Low's formula : t, = 0-125 ^10{l - h) + 2IL 
 where h is the clear rise, and H is the height of the 
 surcharge above the extrados at the crown. 
 
 Bankine' s formulm : t„ = ^0-12E for a single arch 
 
 t„ = ^0-1 7E for a series of arches. 
 Perronet's formula for circular or elliptical arches : 
 
 t„ = l + 0035Z. 
 65. Form of the Arch-Ring.— Let CCi (Fig. 45) be 
 the line of resistance for a distributed load of w per foot 
 on the arch. At a distance x from the crown C the 
 
 downward load is I ivdx and the horizontal thrust 
 is H. Hence the slope of the curve at this point is 
 tan a = -j^ = == \wdx. 
 
 dy 1 (' 
 
106 
 
 THE STABILITY OP AECHBS 
 
 Consequently ^ = g. 
 
 If therefore E is the radius of curvature at the point 
 
 E ~ 5x2 H' 
 
 = - sec" a. 
 E 
 
 Hence H = wE cos" a 
 
 Now the normal pressure at Cj is N = H sec a, and 
 if the compressive stress is to be the same for all 
 
 FiQ. 45. 
 
 sections the thickness of the arch -ring t must be such 
 that 
 
 the pressure at the crown pressure at Cj 
 
 K " t 
 
 H H sec a 
 or — = 
 
 K t 
 
 .'. t = t„ sec a, or in other words, the projection of 
 any section on the vertical must be equal to the 
 crown thickness. 
 
 Further it may be shown that at the crown 
 E = r + t„ where r is the radius of curvature of the 
 intrados there. 
 
 According to the above investigation the proper 
 
DESIGN OF MASONEY, ETC., ARCHES 107 
 
 thickness of the arch-ring at any point may easily be 
 found geometrically. For if P be any point of the 
 intrados, and we set up PQ = t and draw QE hori- 
 zontal to cut the radius through P in E, PE will be 
 thickness required. It is usually suf&cient to de- 
 termine the thickness at one other point P' and to 
 draw a circular arch passing through both. The 
 position of this second point is at the joint of rupture 
 and this depends on the form of the arch. For semi- 
 
 FiQ. 46. 
 
 circular arches it may be taken at the section which 
 makes an angle of 30° with the horizontal (Fig. 46) 
 
 and corresponds at the intrados with the half-rise = 2- 
 
 The thickness is then t = <„ sec 60° = 2t„. 
 
 For elliptic or pseudo-elliptic arches the joint of 
 rupture at the haunches may be taken at the section 
 which is inclined at 45° or thereabout (Fig. 47), and 
 the thickness < = 1 -4 <„ ; or more conveniently, we 
 may take t = k .t„ where ^ is a co-efficient which 
 depends upon the ratio of rise to span, and whose 
 value may be taken as follows : — 
 
108 
 
 THE STABILITY OF AECHES 
 
 For a ratio i 
 
 1 
 i 
 1 
 s 
 
 k = 1-80 
 k = 1-60 
 k = 1-40 
 
 Fig. 47. 
 
 Croizette-Desnoyer's formulae for thickness at the 
 springings : — 
 
 for segmental arches. 
 
 s. 
 
 to 
 
 8. 
 
 i 
 
 4 
 
 A 
 
 A 
 
 1-40 
 
 1-24 
 1-60 
 
 1-15 
 
 I 
 
 1-10 
 
 t. 
 tl 
 
 1-80 
 
 1-40 
 
 for el] 
 
 for elliptic and pseudo-elliptic arches. 
 
 56. Thickness of the Abutments. — The thick- 
 ness of the abutments may be found from the 
 formula 
 
 (Fig. 48), where H is the height of the springings 
 above the foundation. 
 
 Thus for an arch of 100 ft. span, with s = |, 
 H = 15 ft., i. = 4-5 ft., 
 
 T = ("-a + i) (100 - 20 + 15 + 9) = 36-4 ft. 
 
DESIGN OF MASONRY, ETC., ARCHES 109 
 Trautwine's formula is a = •2E + O'lH + 2 
 
 Eankine states that a varies from ^ to ^ radius of 
 the intrados at the crown in existing structures. 
 
 German practice a = 1 + 0*04 (si + 4H). 
 
 Piers. — These are usually from 2\ to 3 times the 
 thickness of the arch- ring at the crown, except in 
 the case of abutment piers, which are constructed 
 when a series of arches are in line ; these are made 
 sufl&ciently strong to take the centre thrust from 
 either side in case of the collapse of one of the 
 spans. 
 
 Fig. 48. 
 
 The results of these formulae often differ widely 
 from one another ; they can therefore only be re- 
 garded as useful for the purpose of preliminary cal- 
 culations and the design should be verified by the 
 methods already described. 
 
 57. Tolkmitt's Investigation of the Form of 
 the Arch-Ring. — Tolkmitt (" Leitfaden fiir das 
 
110 
 
 THE STABILITY OF ARCHES 
 
 Entwerfen und die Berechnung gewolbter Bnicken," 
 1912) has dealt with the form of the arch-ring in a 
 more scientific way, and his method of procedure has 
 found considerable favour on the Continent. He be- 
 gins by assuming that the axis of the arch coincides 
 with the line of resistance for the dead load on it, and 
 then investigates the curve which the axis must as- 
 sume in order to satisfy this condition for a given 
 system of loading. Taking the centre of the crown 
 at the intrados as origin of co-ordinates, and ^, r; as 
 the co-ordinates of any point on the axis, and calling 
 A the area under the reduced load-curve = NMOS, 
 
 A = I 2 . df and tan «, 
 
 A 
 H 
 
 dr] 
 
 Fig. 49. 
 
 where a is the angle of inclination PQ8 of the normal 
 to the vertical. 
 
 Whence '^ = I . ^ = I . . (1) 
 
 When 2 is a function of the co-ordinates this equa- 
 tion can be solved, and its solution determines the 
 form of the arch-ring. 
 
 In the first place applying the result to a load 
 diagram whose upper boundary is horizontal, and 
 taking the intrados at the crown as origin of co-ordi- 
 nates, he shows that the form of the intrados may be 
 
DESIGN OF MASONET, ETC., ARCHES 111 
 expressed approximately by the equation (Fig. 49) 
 
 ^ ^ 2(1 - e) H - (i + e) x^ ' ^^^ 
 
 where « = ^t • • ■ • (3) 
 
 and z„ is the height of the reduced load surface above 
 the intrados. The radius of curvature of this curve 
 at the crown when a; = is 
 
 E„ = ? (1 - .) . . . (4) 
 
 and substituting the value of H from this equation in 
 (2) we get 
 
 ^ " 2R2„ - (1 + e) x^ ■ ■ ^^^ 
 
 With the help of (3) and (4) we get 
 
 B. = {-R + Q z, . . . (6) 
 and then by (3) 
 
 ^=ETi: • • • (^) 
 
 Nowletw = ^' = —^ . . (8) 
 
 The equation (5) may be written 
 
 y= 2mB-^^ ... (9) 
 
 The compressive stress at the crown for uniform 
 distribution of stress is 
 
 ^, = f . . . (16) 
 In designing an arch, assuming the thickness at 
 
112 THE STABILITY OF ARCHES 
 
 the crown, and the load there, then for a given span 
 we can calculate the proper form of the arch. Thus, 
 for example, suppose t„ - 2lt.,e„ = iit.,l = 60 ft. as 
 above, and s„ = 20,000 Ib./ft.^, the weight of the 
 material of the arch-ring being 120 Ib./ft.' 
 Then s„ in masonry units will be 166, 
 
 and by (4) E„ = - (1 - «) where H = sj, by (16) 
 and e = l^hj (3). 
 
 = 2(^1^ - l) = 81 ft., 
 and by (8) m - ^ ^ = 54-2. 
 
 ••• by (9) y = 
 
 IWV 
 
 54-2a;2 54-2a;2 
 
 2 X 54-2 X 81 - a;2 8780 - a;^' 
 and when a; = 30 ft., y = h= g_g^iL-_ = 6-2 ft. 
 
 For given values of I, h, and t,,, the radius of curva- 
 ture E at the crown decreases or increases accord- 
 ing as z„ becomes greater or smaller ; that is to say, 
 as the load increases the arch rises between the 
 crown and springings, and as the load decreases it 
 falls. It follows therefore that in designing an arch, 
 neither the fully-loaded nor the unloaded condition 
 should be used as the basis of calculation, but rather 
 a mean condition, viz. one in which half the moving 
 load is distributed uniformly over the arch ; for the 
 axis of the arch can, of course, be an equilibrium- 
 polygon for one condition of loading only, and the 
 
DESIGN OF MASONEY, ETC., AECHBS 113 
 
 deviations of this equilibrium- polygon from the ex- 
 treme cases above and below it v?ill be least if the 
 axis of the arch is made an equilibrium-polygon for 
 the mean load, where by the mean load is to be 
 understood 
 
 .„=*. + e + J 
 
 where w is the height of the moving load expressed 
 in masonry units, and e is the height of the dead 
 load over the extrados in masonry units. 
 
 Also at the supports y — h, when x = -. 
 
 A 
 
 Substituting these values in (5) and (9) we get 
 
 ^ = !:(?-»- 8--^ ett) • (^«) 
 
 and putting this last value of E into (9) 
 
 y = p h + m — , ■ ■ (12) 
 
 — • — x^ 
 
 ^ h 
 
 Formula (10) is a quadratic in R which on solution 
 
 gives 
 
 E = - <„ + ,4- 
 
 163 
 
 "0^ 1 8i£ 
 h 8 F 
 
 V(/ 
 
 8 l^ J V' ^^1 
 
 This oombined with (6) gives the horizontal thrust 
 
 ^-16 
 
 3 1 8<„g 
 
 A 8 12~ 
 
 + /f?.+i.^^f 4 32^^" 
 
 ■vc 
 
 *„ • 8 ■ P ) l-^ 
 
 8 
 
 (14) 
 
114 
 
 THE STABILITY OF AECHES 
 
 and since by (8) m = 
 
 i + e 
 
 we get m = 
 
 and by (3) e = 
 8 Hz 
 
 k2„ 
 
 (15) 
 
 H + 8«A 
 
 58. Example. — Given an arch having a clear span 
 of 60 ft. with a rise of 15 ft. and a crown thickness 
 of 2 ft. with a reduced load of 2 ft. over the crown, 
 it is required to find the horizontal thrust and the 
 proper form of the arch-ring. 
 
 Here«„ = 2ft. + 2ft. = 4 ft. ; f^ =- A = -267 
 
 h 15 
 
 Qt„z„ ^ 8 X 2 X 4 
 
 /2 
 
 602 
 
 = 018. 
 
 .-. by (14) 
 602 
 
 H = jg- [-267 + -125 + -018+ V(-410)2 + 4x -018] 
 
 = 225 [-410 + -490] = 202-6 ft.^ ; 
 that is to say, the horizontal thrust is equal to the 
 weight of 202'5 cub. ft. of the masonry of the arch-ring. 
 8x202-5x4 ^2^.3^^^ 
 
 Further by (15) m 
 and by (12) y 
 
 202-5 -1-8x2x4 
 24-3a;2 24-3a;2 
 
 602 
 4 
 
 15 + 24-3 
 15 
 
 — rr2 
 
 2358 - x^' 
 
 for X = 
 
 y = 
 
 5 
 0-26 
 
 10 
 1-08 
 
 15 
 2-56 
 
 20 
 4-96 
 
 25 
 12-62 
 
 30 ft. 
 15 ft. 
 
 The following values are the ordinates for a para- 
 oolic arc of the same rise and span : — 
 
 y = 
 
 0-42 
 
 1-67 
 
 3-75 
 
 6-67 ' 10-42 
 
 15 ft. 
 
DESIGN OP MASONRY, ETC., ARCHES 115. 
 
 „dby(n,B = 5«L0.(jl^.^) = .15 
 
 ft. 
 
 59. Determination of tlie Tiiici<ness of the 
 Arch-Ring. — The thickness of the arch-ring must 
 satisfy the two following conditions : first, the permis- 
 sible stress must not be exceeded, and second, it must 
 be possible to draw an equilibrium -polygon for the 
 given load which shall lie within the middle-third of 
 the arch- ring when the arch is loaded on one- half. 
 
 Since the actual position of the line of pressure is 
 not known, it is not advisable to base the first con- 
 dition upon the maximum stress at the edge of the 
 
 H , . , . 
 arch-ring, but rather on the value s„ = -rt which is 
 
 h 
 the mean stress at the crown. This stress then must 
 not be exceeded. Also in order that the second con- 
 
 _ wV^ 
 dition may be fulfilled t„ ~ -06 "g 
 
 Vw 
 
 Example 1. — An arch of 60 ft. span and 15 ft. 
 rise is built of material whose maximum permissible 
 stress is 2000 Ib./ft^. It carries a load of 200 Ib./ft.- 
 Find the least thickness of the arch- ring if its weight 
 is 150 lb./ft.s 
 
 HereZ = 60,. = ?00ft. = J.„=20000^^33^^ 
 150 6 150 
 
 ..= i.60^_J 
 
 133 
 
 = 1 = 1.,.. 
 
 Example 2.— Given that the load per foot run on 
 
•116 THE STABILITY OF ABCHES 
 
 an arch is w = 600 lb. per sq. ft. and the maximum 
 permissible stress s = 20,000 Ib./ft.^ If the span is 
 50 ft., find the proper rise and thickness of arch- 
 ring. Height of filling over crown = 3 ft. Weight 
 of filling material 120 Ib./ft.^ weight of arch-ring 
 150 lb./ft.3 
 
 -rr 20000 , OQ « J • 
 
 Here s„ = = la3 ft. expressed m masonry 
 
 i-0\j 
 
 units 
 
 ande = 3 X -^|5 = 2-4ft.,w = ^ = 4ft.,Z = 50ft. 
 150 150 
 
 ^.= ix50VA=2-lft. 
 
 Then z = 2-1 -f- 2-4 + I = 6-6 ft. 
 
 m = -^^ = -^ = 37-4. 
 
 y = 
 
 2mr - P' 
 .•. when X •= ^l = 25 ft. 
 
 y = h= 37-4 X 25^ ^9.55^^ 
 
 ^ 2 X 37-.4 X 41 - 625 
 
 60. Determination of the Best Form of Arch 
 when the Load-Curve has any Form. — On account 
 of openings in the spandrels, or the use of filling 
 materials having different specific weights, or other 
 causes, the upper boundary of the reduced load-area 
 is often not horizontal, as we have been assuming. 
 In the case of ordinary bridges for roads or railways 
 the deviation from a horizontal line is, however, 
 never very great, and therefore it is permiBsible to 
 
DESIGN OF MASONRY, ETC., ARCHES 
 
 117 
 
 assume as a first approximation that the upper 
 boundary is a horizontal line at a height z„ above 
 the intrados and to determine in this way the form 
 of the arch. The curve so found is not, however, 
 the correct curve, but serves as a basis for further 
 calculations. 
 Thus suppose ABODE (Fig. 50) to be the actual 
 
 reduced load-area in a given case. Then this area 
 may be regarded as arising from the case in which 
 the surface is horizontal as indicated by OF, by sup- 
 posing the part DEF subtracted, for the corrections 
 due to the change of form of the intrados may be 
 regarded as negligible. Let W be the weight of the 
 material represented by the area ABCF, SW be the 
 weight represented by DEF, and 8M be the moment 
 of 8W about A. Then if 8H and 8E are the corre- 
 sponding changes in the horizontal thrust H and in 
 the radius of curvature E, and if the actual values be 
 denoted by affixing a dash to the symbols used above, 
 we get for the actual load-area 
 W = W - 8W, E' = E - 8E, M' = M - 8M, 
 H' = H - 8H. 
 Then since the moment of the load-area may be 
 regarded as equal to the product of the horizontal 
 
118 THE STABILITY OF ARCHES 
 
 thrust into the rise of the line of pressure, and since 
 the points in which the line of pressure cuts the 
 vertical support reactions in either case may be taken 
 as coincident, we have the relation 
 H - 8H ^ M - 8M 
 H ~ M ■ 
 Also (E' + t,)z„ = H' and (E + K)z„ = H (eq. 6), 
 
 whence 8H = ^ . 8M . . (17) 
 
 M ^ ' 
 
 and 8E = ^ = g.^ . . (18) 
 
 The values of 8H, 8M and 8E are positive if the load, 
 area rises towards the supports, and negative when 
 it falls. 
 
 The procedure therefore is as follows : H and E are 
 first calculated on the supposition that the surface 
 of the reduced load-area is horizontal. M is then 
 found by the approximate formula 
 
 • (19) 
 
 M = - 
 8 
 
 h /™ J. h 
 ^' ■*■ h + m \W 15 
 
 8H and 8E are next calculated by means of formulae 
 (17) and (18). 
 
 Thence we deduce the correct horizontal thrust 
 H - 8H and the correct radius of curvature at the 
 crown E - 8E. The form of the arch may now be 
 improved by inserting for E its corrected value in 
 
 the equation y = _ — = and changing the value 
 
 2mE - x^ 
 
 of m so that the curve may pass through the sup- 
 ports as before ; that is to saj , make y = h when 
 X = il. 
 
 61. Example. — In Fig. 50 suppose |i = 30 ft., 
 h = 15 ft., ^„ = 2 ft., z„ = i ft., and the slope of 
 
DESIGN OF MASONRY, ETC., ARCHES 119 ■ 
 
 DE to be 1 in 10. CD = 6 ft., DF = 24 ft., EF 
 = 2-5 ft. 
 For a horizontal surface CF, by formula (14), 
 
 16 
 3600 
 16 
 
 z 1 8t„z l/z 1 8tzA\32tz, 
 
 4 1 8x2x4 
 15 8 3600 "^ 
 
 ^(•3935)2 
 
 + 
 
 32 X 2 X 4- 
 3600 
 
 = 225 [-3935 + -475] = 195 cub, ft. 
 Also by formula (15) 
 
 8ILz„ 8 X 195 X 4 
 
 m 
 
 B. + 8 t„z„ 195 + 8 X 2 X 4 
 and by (11) 
 
 3600 / 1 
 
 = 24-1 ft. 
 
 E 
 
 8\h m 
 
 8 
 
 i- + ^) = 48-6 ft. 
 15 24-V 
 
 Then by the figure 
 
 8W = ^^ ^ ^'^ = 30 ft.8, 
 
 and 8M = 8W X ^ = 240 ft.* 
 
 .-. by (19) M = 
 
 P 
 
 ^o + 
 
 h /m 
 
 3600 
 
 4 + 
 
 15 
 
 /24a 15 
 
 15 + 24-1 \ & 15 
 
 h + m\& "^ 15y 
 
 =■2666 ft.' 
 
 Hence by (17) 
 
 8H = ^ . 8M = -^ 
 M 2666 
 
 X 240 = 17-5 ft.3. 
 
 which is negative, because 8M is negative, 
 
 and by (18) 8E = g.^ = IZl^ ^ 4-4 ft. 
 
 ^ ^ ' M «r„ 4 
 
 The actual horizontal thrust of the arch is there- 
 fore 
 
 H' = H - 8H = 195 - 17-5 = 177-5 ft.^ 
 
120 THE STABILITY OF ARCHES 
 
 and the radius of curvature at the crown is 
 K = E - 8E = 48-6 - 4-4 = 44-2 ft. 
 
 Further, if in the equation y =-p, — = -„, we 
 
 2wiE - x^ 
 
 put y = h = 15 it. and x = ^l = 30 ft., and the 
 
 corrected value of E = 44-2 ft., 
 
 then 15 = — — where m' is the cor- 
 
 2m' X 44-2 - 900 
 
 rected value of m, from which we get m = 31 '5 ft. 
 The corrected form of the arch is therefore 
 
 y = 
 
 mx 
 
 31-5a;2 31-5a;2 
 
 2ot'E' - a;2 2 x 31-5 x 44-2 - x^ 2780 ■ 
 
CHAPTER VIII. 
 
 LOADS AND STRESSES. 
 
 62. The weight of masonry and concrete arches and 
 the dead load they carry is so much greater than in 
 the case of metal arches that the effect of impact and 
 vibration is not of much importance. Moreover, 
 the live load is relatively so small in bridges of any 
 size that if the line of pressure for the fully- loaded 
 arch coincide with the axis of the arch, it may safely 
 be assumed that under the worst conditions of load- 
 ing it will not pass beyond the critical limits. Also 
 concentrated loads are so much better distributed by 
 the filling, that they may be dealt with by reducing 
 them to an equivalent distributed load, and it is 
 further convenient to express the live load as well 
 as the filling in terms of an equivalent weight of 
 masonry whose density is that of the arch-ring. 
 
 Weight and Strength of Arch Materials. 
 
 
 Weight per 
 Cub. Ft. 
 
 Values o( s„ 
 
 
 Lb. 
 
 lb./ft.2 
 
 Masonry : Sandstone or Limestone . 
 
 135 to 150 
 
 40,000 to 60,000 
 
 Granite .... 
 
 165 
 
 100,000 
 
 Brick .... 
 
 120 
 
 30,000 to 40,000 
 
 Concrete .... 
 
 140 
 
 50,000 to 80,000 
 
 Reinforced Concrete 
 
 150 
 
 — 
 
 Gravel and Earth . 
 
 90 to 120 
 
 — 
 
 (121) 
 
122 THE STABILITY OF AECHBS 
 
 63. Dead Load. — The dead load consists of the 
 weight of the arch itself and the filling, etc., which 
 it carries. The following are the average weights 
 for the materials involved :— 
 
 The height of the filling over the extrados at the 
 crown may be taken 
 
 for railway arches as ft. 
 
 P 
 
 and for highway arches as — ft. 
 
 P 
 where p is the density of the material in lb. per cub. 
 foot. 
 
 Average Permanent Load on Boad Bridges. 
 
 Iron road bridges : — 
 
 Timber platform and ballast . 100 Ib./ft.M ,„. ,, ,,, „ 
 Cross girders . . . . 20 „ / = ^''" '"■'"• 
 
 Iron bridges with brick arches : — 
 
 Arches 48 lb."! 
 
 Concrete and asphalt . . . 42 „ I oi n ii, lu 2 
 
 Metalling 100 „ f = ^^" '"■'"• 
 
 Cross girders 20 „ J 
 
 64. Live Load.^ — The following formulae for the 
 live load, which may be assumed, are given in Melan's 
 " Plain and Eeinforced Concrete Arches'' (transla- 
 tion by D. B. Steinmann) : — 
 
 for very heavy vehls. iv = f 100 + 1^00?\3 + f. i^ f ^2 
 \ I J oe 
 
 „ heavy vehicles 10 = f 100 + -^^°° ^ ^ + ^ 
 
 I J 3e 
 light vehicles «; = ( 100 + ^^^^ ^^ + ^ 
 
 -n 
 
 e 
 
 .. railway bridges w = ('lOOO + ?0000\3 + e^ ^_ 
 
 When half the arch is loaded we must write |J 
 instead of I, 
 
LOADS AND STEBSSES 123 
 
 The load to be allowed for a dense crowd of people 
 may be taken as 100 lb. per sq. foot. 
 
 Highway Bridges. — For small spans heavy 
 
 wagons give the most severe loading, but for large 
 
 spans a crowd of people. The axle distances for 
 
 heavy wagons are about 10 ft., so that for spans up 
 
 to 20 ft. only one axle at a time comes upon the one 
 
 half of the bridge. In the case of a bridge 20 ft. 
 
 wide, which carries on one half an axle load of 6 tons, 
 
 if we consider the axle load as spread over 10 ft. width, 
 
 2x6 
 
 the average live load will be = -12 ton/ft.''' 
 
 10 X 10 
 
 Taking the heaviest freight wagon of 24 tons, 
 
 26 feet long, and a team of six horses weighing -35 
 
 ton each, extending over 36 ft., this gives upon a 
 
 length of 61 ft. and a width of 10 ft. an average 
 
 load of 
 
 24 + 6 X -35 „.„„ , • 
 
 — — — — = -0428 ton. 
 
 10 X 61 
 
 Without the horses, however, on a length of 25 ft. 
 
 24 
 we get = -096 ton/ft.^ 
 
 J.U X Ao 
 
 In dealing with concentrated loads on small spans 
 these are supposed distributed over an area drawn 
 at 36° to the vertical from the boundary of the load. 
 
 65. Railway Bridges. — In the case of locos the 
 axle spacings are as low as 6 ft., and although the 
 pressure is better distributed by means of rails, 
 sleepers, and ballast than in the case of roadway 
 bridges with a thin roadway, yet even in the case of 
 spans as small as 30 ft. the full weight of three 
 axles must be allowed on the half span. If the 
 maximum wheel-load for a 72-ton engine be taken 
 
124 
 
 THE STABILITY OF ARCHES 
 
 as 8 tons, this gives a total load of 48 tons distributed 
 over a width of say 12 ft. of archway (= length of 
 sleepers + 3 ft.), that is to say, over 15 x 12 = 
 180 sq. ft. of surface. 
 Hence we get an average of j^/^ = "267 ton/ft.^ 
 On longer spans, however, the equivalent distri- 
 buted load will diminish as the span increases up to 
 a certain point. Thus the following table gives the 
 equivalent distributed load for railway bridges of 
 varying span calculated for a maximum load of 
 66-ton engine and 1-ton tender with an increase of 
 2i per cent for future contingencies. 
 
 Span in feet . 
 
 Equivalent load 
 
 in tons per ft. 
 
 10 
 3-69 
 
 15 
 3-11 
 
 20 
 
 2-88 
 
 25 
 2-61 
 
 30 
 2-45 
 
 40 
 2-22 
 
 50 
 2-10 
 
 100 
 1-93 
 
 150 
 1-92 
 
 200 
 1-92 
 
 66. Example. — A highway bridge has an opening 
 of 100 ft. and a rise of 12 ft. It is constructed of 
 cement concrete weighing 135 Ib./ft.^, whose compres- 
 sive strength under test was 4000 Ib./in.^ The per- 
 missible mean compressive stress is taken as ^^ of 
 this value, and the height of the surcharge at the 
 crown is 2 ft. The densities of the concrete and 
 filling are 140 Ib./ft.^ and 100 Ib./ft.^ respectively. 
 Roadway level. 
 
 Here the height of filling at the crown reduced to 
 masonry units is 2 x if^ = 1'43 ft., and by formula 
 sect. 64, when the arch is loaded on one half we get 
 for the heavy loading 
 
 12000\ 3 4- 1-43 
 
 w =100-1- 
 
 Then by sect. 59 
 
 50 ; 3 X 1-43 
 
 = 350 lb./ft.2 
 
LOADS AND STEESSES 125 
 
 where s„ = 400 x 144 = 57600 Ib./ft.'' 
 
 = 1 X 100 /_?^ = 1-95 ft., or say 2 ft. 
 \ 57600 ■^ 
 
 The form of the arch, etc., may then be deduced 
 as in previous examples. 
 
 If we now put the value e = -01 ^ into the 
 
 H 
 
 formula 
 
 ' = '-('-?) 
 
 we get 
 
 . = .(l.-06-) 
 
 
 = s„ + -06 w{i\\ 
 
 or the maximum stress is greater than the mean 
 
 / /\ 2 
 
 stress by -06 M-)' 
 
 67. Stresses in the Arch-Ring. — If the pressure 
 upon the cross-section of an arch be uniformly distri- 
 buted over it, then the magnitude of this stress is 
 
 N 
 everywhere s„ = -v- where N is the normal thrust 
 
 and A is the area ; but if the resultant pressure does 
 not act along the axis, then regarding the arch as 
 elastic and capable of resisting tension, the stress in- 
 tensity at any distance y from the axis is given by 
 the formula 
 
 where e is the eccentricity ot the resultant thrust and 
 y is positive or negative according as it lies on the 
 same side of the axis as the resultant, or on the side 
 
126 THE STABILITY OF ARCHES 
 
 remote from it, and for a rectangular cross- section 
 
 k'~ = — , so that the stresses at the edges are 
 12 
 
 6e\ 
 
 - '{' - ?). 
 
 If, however, tensile stresses are not permissible, 
 then e must not bo greater than — for rectangular 
 
 cross-sections, where t is the thickness of the arch- 
 ring (see chap, vi., p. 86), that is to say, the re- 
 sultant thrust must lie within the middle-third of 
 the depth. So long as this is so, the value of s will 
 
 N 
 not exceed 2—, or the maximum stress will not ex- 
 A 
 
 ceed twice the stress when the distribution is uni- 
 form. 
 
 Now in the case of masonry and plain concrete 
 arches it is not usual to allow for any tensile stress, 
 so that if e is greater than ^t, the above formula can 
 no longer be used, but that part of the section in 
 which, according to the formula tensile stress would 
 occur, must be regarded as ineffective. In this case 
 the effective thickness AC of the section (Fig. 51) will 
 be less than its full thickness AB and will in fact be 
 AC = SAL = 3(1^ - e). 
 
 Hence the maximum stress will be 
 
 s - N _ 4N 
 
 3{it - e] 3{t - 2e)' 
 
 The maximum stress therefore increases very 
 quickly with e and becomes infinitely great when 
 e = -^t, whereas when tensile stress is permissible, 
 the maximum stress in this case will be only four 
 
LOADS AND STRESSES 
 
 127 
 
 times the mean stress. It is therefore very im- 
 portant that the resultant thrust shall not have a 
 large eccentricity. 
 
 The probable position of the critical sections at 
 vyhich the maximum stresses occur can only be 
 found by means of the elastic theory of the arch, 
 because, as we have shown, it is only in this way 
 that the actual position of the resultant thrust can 
 be fixed, although we can assert it will lie within 
 
 PiQ. 51. 
 
 the critical limits provided that an equilibrium can 
 be drawn for the given loading within those limits 
 (see chap. vi.). Hence a limiting value for the 
 maximum stress may be derived which is tolerably 
 reliable. Tolkmitt in his treatise already referred to 
 shows that for the half-span loaded with w tons per 
 foot, no line of pressure can be drawn which is closer 
 to the axis than 
 
 48' = 0-01 ^ at the crown 
 and U" = 0-0125 -=^ at the springmgs, 
 
 a 
 
128 THE STABILITY OF AECHBS • 
 
 where S' and 8" are the total Hmits of deviation from 
 the axis. The maximum value of ^8' lies at \l from 
 the crown, and since the thickness of the arch-ring 
 does not vary much within this distance from the 
 crown, the thickness at the crown may be made 
 
 <„ = 6 X -01 ^. 
 
 68. Stresses in the Arch-Ring by the Elastic 
 Theory. — When the steel reinforcement is asym- 
 metric with respect to the axis of the arch-ring, the 
 centroid of the cross-section is displaced towards the 
 side on which the steel is in excess ; or, where there 
 is only single reinforcement, towards the side on 
 which it occurs. It is only when the resultant thrust 
 passes through this point that the stress over the sec- 
 tion can be uniform. When this- does not occur, that 
 is to say, when the resultant of the load is eccentric, 
 the stress will vary over the section, and its value may 
 be found without difficulty by the ordinary formula 
 
 so long as there is no tensile stress, the moment of 
 inertia being that of the section reduced to concrete. 
 When, however, the resultant acts outside the 
 critical limits, so that tension occurs, the calculation 
 becomes more troublesome, for although the same 
 reasoning applies as in the case of beams, it is no 
 longer true that the neutral axis passes through the 
 centroid of the cross-section, its position depending 
 now on the value of the bending moment. Eefer- 
 ring to Fig. 52, and with the usual assumptions, we 
 obtain the following relations : — 
 
LOADS AND STRESSES 
 
 129 
 
 h'. - n 
 s, — m.c 
 
 n 
 
 • (1) 
 
 , n — a' 
 s, = m . c 
 
 • (2) 
 
 t-^' : M ^ ^ 
 
 K-c-'H 
 
 Ai».i». 
 
 M-f^- 
 
 
 K — »-ii 
 
 Fig. 52. 
 
 from (1) men + s,n = mch' or m(wic + s,) = vich'. 
 
 mch' 
 
 mc + s, , s, 
 
 m + _' 
 c 
 
 € = ''-!'.& . 
 
 .h' 
 
 C, = s/A/ = mcA,' . 
 T, = s. . A, = mcA, 
 
 n 
 
 - 
 
 a 
 
 
 ?f 
 
 
 h' 
 
 - 
 
 11 
 
 (3) 
 
 (4) 
 (5) 
 (6) 
 
 In the case of beams with no axial thrust, the re- 
 sultant axial force being zero, we have 
 
 C + C. - T. = 0, 
 but in the present case this becomes 
 
 C + C. - T. = P . . (7) 
 
 h' - n 
 
 en T_ , I, , n - a . 
 
 — .0 + mch., - mcA, 
 
 2 n 
 
 Vnb 
 12" 
 
 + wa; 
 
 , n - a 
 
 mh, . 
 
 n 
 
 h' - n 
 n 
 
 N 
 = N 
 
130 THE STABILITY OF AECBE8 
 
 c[n^.b + 2mA,'(w - a') - 2mA,{h' - n)] = 2Na;„. 
 
 . ^ ^ 2Na;„ ,qs 
 
 n^b + 2mA,'{n - a') - 2mA,{h' - n) 
 If the reinforcement is symmetrical about the axis 
 then A, = A/ and 
 
 ^ 2Nn ,g,^ 
 
 " n't + 2otA,(2» - h) ' ■ '^ > 
 
 In order to obtain the value of n, taking moments 
 about a point in the neutral axis, we get 
 
 N(w -e + r) = c.^--«+ A,'s/(m - a') + A,s,{h' - n) 
 
 = c. -— + mA, . c^ i-+wA,ci '— 
 
 "3 ■ '• " n '" n 
 
 .-. 3N(a;„ - e + r) = c[n^b + 3mA,' {n - a'f + 3mA, 
 {h' - nYl 
 _ 3Nw(re - e + r) ,n\ 
 
 °^ '^ " v?b + 3toA;(»i - a'Y + 3mA.(;i' - nf ' ^ ' 
 and therefore equating (8) and (9) we get 
 
 2 
 
 ri^b + 2mA,'(w - a') - 2mA,{h' - n) 
 
 = 3(a;„ - e + r) ,jq> 
 
 n^b + 3mA,'(« - aj + 3mA.{h' - n)"' ^ ' 
 Writing » = e - r and transforming, this becomes 
 
 n^ - 3v.n^ - ^. n[A,'{v - a') - A,{h' - v)\ 
 
 + ^[A 'a\v a) - A,h'{h' - v)] = . (11) 
 
 and if A, = A,', this becomes 
 o /Q , 9 ■ 6mA, ,„ , . , 6otA, 
 b 
 
 [a'{v - a') - h'Qi' - v)] = . . (11') 
 This equation determines the value of x„ and 
 
LOADS AND STEBSSES 
 
 131 
 
 then from (8) or (9) the value of c may be found arud 
 the stresses in the steel from (1) and (2). 
 
 Example (compare Melan's " Plain and Eeinforced 
 Concrete Arches "). — In an arch rib of 1 ft. depth, the 
 extreme moments per foot of width at the most 
 severely stressed section are Mj = 14880 Ib./ft. and 
 Mj = - 9720 lb. /ft. The axial thrusts corresponding 
 to the two cases of loading amount to 9500 lb. and 
 
 or 
 
 ^ 
 
 A. a' 
 
 I 
 
 E 
 
 I 
 
 IT 
 l~ 
 
 h. 
 
 } 
 
 H 
 
 * 
 
 // y It/t/JltMlltl/ 
 
 i • 
 
 t 
 
 
 Pig. 53. 
 
 8100 lb. respectively. If we take the amount of rein- 
 forcement at a' = 2'11 per cent and a" = 0'623 per 
 cent as obtained from Prof. Melan's. diagrams where 
 
 A A ' 
 
 a! = —land a" = — i, and assuming the steel is placed 
 h h 
 
 at O-l/i from the upper and lower faces of the arch- 
 ring (fig. 53), we have for the first case, viz. Mj = 
 14880 Ib./ft. by equation (8) 
 
132 THE STABILITY OF ARCHES 
 
 2P?l 
 
 c = 
 
 ■n?h + 2mA,'{n - a') - 2mA,(/i' - n) 
 where P = 9500 pounds. 
 
 6 = 1 ft. ; m = 15 ; a; = ^^ x 1 ; a' = -1 ; 
 
 A, = —xl;h'=-9 ft. 
 ' 100 
 
 and n is found from equation (11), viz. : — 
 
 „3 _ Sv.n^ -^. n\A;{v - a) - A.{h' - v)] 
 
 
 
 + ^ [A:a'(v - a') - A.h'{h' - v)] = 
 
 , , M, 14880 -i.Kcc 
 
 where v = e - r and r = ^=-i- = ^^^„ = 1 odd. 
 
 Pi 9500 
 
 «^A.g - a) - mA/g - a') 
 
 ^°^ S = bh + m(A + A/) 
 
 ^ 15 X -0211 (0-5 - 0-1) - 15 X -00623 (0-5 - 01) 
 1 X 1 + 15 (00623 + 4-0211) 
 
 ■1266 - 03738 -08922 
 
 1 + -410 1-410 
 
 = -0633 ft. 
 
 and e = ^ + cf = -5 + -0633 = -5633. 
 
 .w = -5633 - 1-566 = - 1-003. 
 .-. n" + 3-009w2 - 90w [-00623( - 1-003 - -1)--0211 
 (-9 + 1-003)] + 90 [-00623 x -1 (- 1-103) - -0211 
 
 X -9 X 1-903] = 0. 
 
 .-. m3 + 3-009W - 90« (- -00687 - -04015) + 90 
 
 ( - -000687 - -03613) = 
 
 •n? + 3-009n2 + 4-2318m - 3-3135 = 0. 
 
 Whence n = -533 ft. 
 
LOADS AND STEBSSBS 
 
 133 
 
 for n 
 = -5^ 
 •52 
 •51 
 •53 
 •54 
 
 2 X 9500 X -533 
 
 ■ " •5382 X 1 + 30 X -00623 (-538 - -1) - 30 x -0211 (-9 - -533) 
 
 12^?! =121^=76310 lb./ft.' 
 
 •28409 + -0809 - ^2323 •1327 
 
 = 630 lb./in.2 
 
 c' = c X - = 530 X ^ = 411 lb./m.2 
 e •563 
 
 and by (1) s. 
 
 h' - n 
 
 •367 
 
 15 X 530 
 
 7950 X '^^ = 5450 Ib./in.^ 
 •533 
 
 and by (2) s,' = mc 
 
 15 X 530 
 
 ■9 ~ 533 
 •533 
 
 ■533 - •I 
 
 •533 
 
 •433 
 
 = 7950 X ^- = 6460 Ib./in.^ 
 ■533 
 
 It will bs seen that this method of calculation is 
 very troublesome, and consequently many attempts 
 have been made to devise methods which will give 
 approximately the same results. None of these ap- 
 pear to be very satisfactory so far as agreement in the 
 results is concerned, though possibly sufficiently so, 
 considering the general uncertainty of the data. 
 Prof. Melan proceeds as follows : — 
 The moment of inertia of the cross-section 
 
 I = 5[(0^9 - x) (2-7 - x)a' + {x - 01) {x - 0^3)a"] 
 
134 TKE STABILITY OF AECHES 
 
 where 
 
 x= -15{a' + a") + 715-^ [a' + ay + 30 (0-9a' + O-la") 
 
 where x = —. 
 a 
 
 Then E„ = — . iPc = m . d^c = moment of resist- 
 
 X 
 
 ance of the section 
 
 or E, = —^ d'^ ■ =-= = -ptt; • — p- d^ .c = m' . d^c. 
 
 9-a; 15 09 -x 15c 
 
 The coefiBcients m and m', or the resisting mo- 
 ments per unit stress for d = i, depends only on a' 
 and a", and if these are plotted, the values of m and m 
 will be represented by curves. The chart so formed 
 (which is given in Melan's "Plain and Eeinforced 
 Concrete Arches," translated by D. B. Steinmann) is 
 very convenient for designing purposes and parti- 
 cularly when the sections are subject to reversed mo- 
 ments. In the example above the values of a' and 
 a'' there assumed were determined from the chart 
 for the assumed value of s = 13500 Ib./in.^ = 30 c. 
 The stresses obtained in this way were 
 in the upper surface of the concrete 450 Ib./in.^ 
 in the lower „ ,. ,, 295 „ 
 
 in the upper steel .... 13,500 ,, 
 in the lower „ . . . . 6,460 ,, 
 
 It is more usual, however, to make the reinforce- 
 ment the same both on the upper and lower sides. 
 
APPENDIX. 
 Calculation of the Ordinates of a Circular Arc. 
 
 CG = Jr^ - -r^ ; OF = Jr^ - {^ly. 
 .: CE= Jr^ - x' - Jr' - {ilf, 
 
 and if we make ~ = 1 
 
 2 
 
 CE = ^r' - x'^ - Jr'^ -1 
 and (r - Kf + 1^ = r-. 
 
 ■■■r = ^n^^ 
 
 and r^ = {(h'' + i-\ + ^. 
 
 Coefficients of '--for different values of 
 
 
 •1 
 
 •2 
 
 ■3 
 
 •1 
 •084 
 
 •5 
 
 •6 
 
 •7 
 
 ■8 
 
 ■9 
 
 10 
 
 for_ = 
 
 •1 
 
 •099 
 
 •096 
 
 ■091 
 
 ■075 
 
 ■064 
 
 ■051 
 
 ■036 
 
 ■019 
 
 
 
 I 
 
 •12 
 
 •119 
 
 •115 
 
 •110 
 
 •102 
 
 091 
 
 ■078 
 
 ■063 
 
 045 
 
 ■023 
 
 
 
 
 ■14 
 
 •138 
 
 •134 
 
 •128 
 
 •118 
 
 ■106 
 
 •092 
 
 ■074 
 
 ■053 
 
 ■028 
 
 
 
 
 •16 
 
 •158 
 
 •154 
 
 •146 
 
 •136 
 
 •121 
 
 •105 
 
 ■085 
 
 •061 
 
 ■033 
 
 
 
 
 •18 
 
 •178 
 
 ■174 
 
 •165 
 
 •154 
 
 •135 
 
 •120 
 
 ■098 
 
 •070 
 
 ■038 
 
 
 
 
 •20 
 
 •198 
 
 •192 
 
 •183 
 
 •169 
 
 •151 
 
 •130 
 
 ■104 
 
 ■074 
 
 ■039 
 
 
 
 (135) 
 
136 THE STABILITY OF AECHES 
 
 Example.— li I = 150 ft. and h = 15 ft., then 
 
 h_ 
 
 0-2. 
 
 
 
 
 r— — T— — -~^ 
 
 \^ E 
 
 ♦■Vx- -> 
 
 F / 
 
 G 
 
 M 
 
 / 
 
 Fig. 54. 
 Therefore the ordinates are 
 •192 X 150 = 28-8, and so on. 
 
 198 X 150 = 29-7, 
 
 Construction for a Parabola. 
 
 Method I. By Points. — Let A-5 (Fig. 55) be the 
 half- base and AO be the height of the parabola it is 
 
 required to construct, AO 
 being its axis. Complete 
 the rectangle A0B5 and 
 divide OB into any con- 
 venient number of equal 
 parts, such as 0-1, 1-2, 
 etc. Similarly divide B 5 
 into the same number of 
 equal parts at 1', 2', 3', 
 4'. Join 01', 02', etc. 
 Then the points in which 
 these lines cut the cor- 
 '°' ^' responding verticals de- 
 
 termiae points on the parabola. 
 
APPENDIX 
 
 137 
 
 Method II. — The following method avoids the ne- 
 cessity of dividing the sides of the rectangle into 
 equal parts, and is therefore quicker. Let CB (Fig. 
 
 Fig. 56. 
 
 56) be the base of the semi-parabola and BA its 
 
 height. Join AG. To obtain a point on the curve 
 
 draw any line Aa, and 
 
 draw ab horizontally to t r f s' ^' a' t 
 
 cut it in h. Erect a 
 
 vertical he to cut Aa in 
 
 c. Thus c is a point on 
 
 the parabola. 
 
 Method III. By 
 Tangents.— hei AT, BT 
 (Pig. 57) be any two 
 tangents, and let it be 
 required to construct a 
 parabola to touch these tangents at A and B. 
 Divide BT, ,AT into any convenient number of 
 equal parts, as shown, and join 11', 22', etc. These 
 
 Fig. 57. 
 
138 
 
 THE STABILITY OP ARCHES 
 
 lines will envelop the parabola, and piactically 
 define it. 
 
 To Draw a Tangent at any Point of a Para- 
 bola. — Let it be required to draw a tangent to the 
 parabola in Fig. 55 to touch it in any given point C. 
 Draw CD parallel to the base, and make OT = OD. 
 Join CT. Then CT is the tangent required. 
 
 Fig. 58. 
 
 To Construct a Fiat Circular Arc the Length of 
 Chord and Rise being given, — ^Let AB (Fig. 58) be 
 the chord or span and the rise. With O as centre 
 and radius OC describe a circle. Join AC, cutting this 
 circle in F, and make FD = FO. Join CD. Divide 
 CD, CO into as many parts as the points required on 
 the arc AC, and draw Al, A2, A3 to meet Bi, B5, 
 B6. The points o£ intersection are points on the arc 
 required. Similarly for the other half. 
 
INDEX. 
 
 Abutment, definition of, 5. 
 Abutments of arch, 100. 
 
 — ■ thickness of, 108. 
 
 Arc of circle, construction of, 
 
 138. 
 Arch, best form of, 116. 
 
 — circular, 58. 
 
 — concrete, 83. 
 
 — elliptic, 107. 
 
 — form of, 97. 
 
 — glass model of, 39. 
 
 — hingeless, 65. 
 
 — history of, 1. 
 
 — linear, 12. 
 
 — masonry, 83. 
 
 — materials, strength of, 7. 
 
 — parabolic, 57. 
 
 — theory of, 45. 
 
 — three-pinned, 11. 
 Arches, classification of, 5, 6. 
 
 — particulars of, 8, 9. 
 
 .— relative strength of, 10. 
 Arch-ring, definition of , 5. 
 Austrian experiments, 87. 
 
 B 
 
 Backing, 6. 
 
 Bed joints, 6. 
 
 Bending moment in aroh-ring, 
 
 19. 
 hingeless arch, 66, 
 
 79. 
 
 two-pinned arch, 61. 
 
 Brickwork, strength of ,;7, 121. 
 
 Cast iron, strength of, 7. 
 Cement concrete, strength of, 7. 
 Chester bridge, 6. 
 Circular arc, construction of, 
 
 138. 
 — ■ arch, 58. 
 
 Coefficient of expansion, 47. 
 Concrete, strength of, 121. 
 
 — arch, 83. 
 
 design of, 103. 
 
 Conditions of stability, 84. 
 
 Critical limits, 84. 
 
 Critical line of pressure, 83, 88, 
 
 96. 
 construction of, 
 
 91. 
 Crown, thickness of, 104. 
 
 D 
 
 Dead load, 122. 
 Displacement, angular, 41. 
 
 — due to axial thrust, 43. 
 
 — bending, 41. 
 
 temperature, 43, 46. 
 
 Distortion of rib, 40. 
 
 E, values of, 38. 
 Earth, weight of, 121. 
 Eddy's theorem, 19. 
 Elastic theory, 11, 37. 
 polarized light experi- 
 ments, 39. • 
 Elliptic arch, 107. 
 
 (139) 
 
140 
 
 THE STABILITY OF ABCHBS 
 
 Equations of condition for 
 hingelees arch, 44. 
 
 Equivalent distributed load, 
 124. 
 
 Expansion, coefficient of, 47. 
 
 Extrados, 6. 
 
 G 
 
 Glass model of arch, 39. 
 Granite, strength of, 121. 
 Gravel, weight of, 121. 
 
 H-line, 27. 
 Haunches, 6. 
 Highway bridges, 123. 
 HingelesB arch, equations for, 
 
 44, 45, 65. 
 Horizontal thrust for two- 
 pinned arch, 61. 
 
 — — by calculation, 52. 
 
 — — graphically, 54. 
 
 — — by reaction locus, 56.' 
 for hingeless arch, 72. 
 
 Impost, 6. 
 
 Influence area, 28. 
 
 — line method, 26, 62, 74. 
 
 Intrados, 6. 
 
 Iron cast, strength of, 7. 
 
 Joints of rupture, 38, 88. 
 
 K 
 
 Keystone, 6. 
 
 Lead joints, 14. 
 
 Least work, principle of, 85. 
 
 Limestone, strength of, 121. 
 
 Linear arch, 12. 
 
 Line of pressure, 12, 85. 
 
 determination of, 15. 
 
 Live load, 122. 
 Loads on arch, 121. 
 Load curve, 88. 
 
 form of, 99. 
 
 Low's formula for arch-ring, 
 105. 
 
 M 
 
 Masonry arches, 83. 
 
 • design of, 108. 
 
 Materials, strength of, 121. 
 — weight of, 121. 
 Mesnager's method, M., 80. 
 Middle-third rule, 85. 
 
 N 
 
 Neutral axis by polarized light, 
 
 89. 
 Normal thrust, 80. 
 
 Oglio bridge, 98. 
 
 Ordinates of circular arc, 135. 
 
 Parabola, construction of, 136. 
 
 — • tangent to, 138. 
 
 Parabolic arch, 57. 
 
 Perronet's formula for arch- 
 ring, 105. 
 
 Pier, definition of, 5. 
 
 Piers, 102, 109. 
 
 Polarized light, experiments 
 with, 39. 
 
 Principle of least work, 88. 
 
 B 
 
 Badius of curvature, 112. 
 Railway bridges, 123. 
 Hankine's formula for abut- 
 ment, 109. 
 
 arch-ring, 105. 
 
 Beaction locus, 17, 57. 
 Eeduced load curve, 88. 
 
INDEX 
 
 141 
 
 Reinforced concrete, weight of, 
 
 121. 
 Rupture, joint of, 38, 88. 
 
 s 
 
 Sandstone, strength of, 121. 
 
 — weight of, 121. 
 Shear in arch-ring, 19. 
 Soffit, definition of, 6. 
 Spandrel, definition of, 6. 
 
 — braced arch, 28, 29. 
 Span of arches, 7. 
 Springings, thickness of, 108. 
 Stability, conditions of, 84. 
 Statically indeterminate, de- 
 finition of, 13. 
 
 Strength of arches, 10. 
 
 — — cast iron, 7. 
 
 materials, 7, 121. 
 
 steel, 7. 
 
 Stress due to eccentricity, 84. 
 Stresses in arch, 121, 125. 
 
 Temperature effect, 14, 65. 
 
 — fluctuation of, 46. 
 Thickness of arch-ring, 115. 
 crown, 104. 
 
 springings, 108. 
 
 Three-pinned arch, 11. 
 
 example of, 21. 
 
 Thrust in arch-ring, 19, 46, 80. 
 
 parabolic arch, 58. 
 
 Tiber bridge, 7. 
 Tolkmitt's method, 109. 
 Trautwine's formulse for abut- 
 ments, 109. 
 
 — — — arch-ring, 105. 
 
 Voussoir, definition of, 6. 
 
 w 
 
 Weight of arch materials, 121. 
 
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