THEORY 
 
 OF 
 
 VflTTQQniD A DPUfiQ 
 
 VUUooUlt[ At[L)illlo , 
 
 BY 
 
 WM, GAIN, O.E., M, Am, Soc, O.E., 
 
 Professor of Mathematics and Engineering^ University 
 of North Carolina. 
 
 SECOND EDITION, REVISED AND ENLARGED. 
 
 NEW YOEK: 
 
 D. VAN NOSTRAND CO., PUBLISHEES, 
 23 MURK AY AND 27 WARREN STREETS. 
 
 1893. 
 

 
 GENERAL 
 
 COPYRIGHT. 1893. 
 D. VAN NOSTRAND COMPANY. 
 
 2 7 6 
 
TABLE OF CONTENTS. 
 
 CHAPTER I. 
 
 ARTICLE. PAGE. 
 
 1. Definitions 11 
 
 2. Portion of arch taken for investiga- 
 
 tion, Hypothesis 14 
 
 8. Segmeiital arch denned 15 
 
 SYMMETRICAL ARCHES. 
 
 4. Conditions of equilibrium of the arch 
 
 as a whole 17 
 
 r>. Conditions of equilibrium of the half 
 
 arch and line of resistance "2(> 
 
 6. Formulas for thrust at the crown 25 
 
 7. Scheffler's method of dividing up the 
 
 arch ring and spandrel and finding 
 
 loads in posit.oi: and magnitude '27 
 
 s. Example of a stone arch subjected 
 only to its own weight. Method of 
 tabulating loads ;i:id arms, and con- 
 structing trial lin j of resistance 30 
 
 Graphical method of finding horizon- 
 tal thrust. 
 
II. TABLE OF CONTENTS. 
 
 ARTICLE. PAGE . 
 
 9. Same arch subjected to symmetrical 
 
 live loads 30 
 
 Method of drawing trial line of resist- 
 ance, to pass through any two points 
 of half arch. 
 
 UN-SYMMETRICAL AECHES. 
 
 10. Formulas to ascertain position and 
 
 magnitude of thrust at crown 40 
 
 Example 1. Stone arch previously con- 
 sidered subjected to its own weight 
 and an eccentric load. Method of 
 drawing trial resistance line. 
 
 Examples 2, 3 and 4. 
 
 Effects of rolling load on piers in a 
 series of arches. 
 
 CHAPTER 1C. 
 
 11. A more correct method of ascertaining 
 
 loads and arms 53 
 
 12. A more convenient method of forming 
 
 diagrams of forces and drawing a 
 
 trial line of resistance 57 
 
 13. Usual method of drawing line of resis- 
 
 tance. Equilibrium Polygon 64 
 
TABLE OF CONTENTS. 1TI. 
 
 ARTICLE. PAGE. 
 
 13 (a). Method of constructing equilibrium 
 polygon when one abutment re- 
 action is given 68 
 
 14. Special Properties of the Equilibrium 
 
 Polygon 69 
 
 15. Example of an arch subjected to 
 
 wheel loads, by a graphical me- 
 thod, to pass an equilibrium poly- 
 gon through any three points. Pe- 
 culiar division of arch ring and cor- 
 responding computation of tables. 78 
 Examples. 
 
 CHAPTER III. 
 
 16. Properties of Curves of Resistance 89 
 
 17. Intersecting Curves of Resistance.. 1)1 
 
 18. Curve of Resistance corresponding to 
 
 minimum and maximum horizontal 
 thrusts, 92 
 
 19. Curves of minimum and maximum 
 
 horizontal thrusts for symmetrical 
 arches with symmetrical loads 94 
 
 20. Curve corresponding to both mini- 
 
 mum and maximum thrust at the 
 same time within assumed limits 9(J 
 
 21. Unit Stresses at any point of a joint. 
 Examples 97 
 
IV. TABLE OF CONTENTS. 
 
 ABTICLE. PAGE . 
 
 22. Three ways of expressing the moment 
 
 of the thrust on any joint, about the 
 centre of that joint 104 
 
 23. Method of failure of arches. 
 
 Woodbury's Tables 105 
 
 CHAPTER IV. 
 
 24. Historical note concerning authors who 
 
 have proposed to apply the theory of 
 
 the solid arch to voussoir arches 112 
 
 25. Precise statement of conditions to be 
 
 fulfilled in order that the theory of 
 the solid arch be applicable to the 
 voussoir arch lir> 
 
 26. General table for isolated loads for 
 
 arches whose rise is one-fifth the 
 
 span 117 
 
 General method of procedure. 
 
 27. Application to an arch of 100 feet 
 
 span , 1 25 
 
 28. Table of theoretical depth of keystone 
 
 for stone arches whose rise is one- 
 fifth of the span 136 
 
 29. Maximum intensity of unit pressure 
 
 to be allowed in stone or brick arches. 143 
 
TABLE OF CONTENTS. V. 
 
 AEKICLE. PAGE. 
 
 30. Formula for radius in terms of span 
 
 and rise 144 
 
 31. ' Empirical formulas for depth of key 
 
 by various authorities 145 
 
 CHAPTER V. 
 
 32. Fundamental equations concerning 
 
 solid arches ' ' fixed at the ends " 150 
 
 83. Lemma 154 
 
 34. Second general method (founded on 
 
 Eddy's Constructions for the solid 
 arch) for testing the strength and 
 stability of stone arches. Live load. 162 
 
 35. Example I. Arch of 12.5 ft. span 164 
 
 36. Location of line kk 166 
 
 37. Location of line m m 166 
 
 38. Reduction of ordinates of the type m b . 167 
 
 39. Proof that points c satisfy the con- 
 
 ditions for an "arch fixed at the 
 ends" 169 
 
 40. Location of the Poles and O and 
 
 finding the centres of pressure on 
 
 the joints 171 
 
 41 . Test of the accuracy of the work 1 72 
 
VI. TABLE OF CONTENTS. 
 
 ARTICLE. PAGE. 
 
 42. Example II. Arch of 25 ft. span 175 
 
 43. Example III. Arch of 50 ft. span 176 
 
 44. Example IV. Arch of 100ft. span... 177 
 
 45. Example V. Arch of 100ft. span sub- 
 
 jected only to its own weight. Inter- 
 esting results 177 
 
 46. No simple rule for testing the stability 
 
 of arches approximately 180 
 
 47. Second method not giving readily, the 
 
 the exact position of live load pro- 
 ducing the most hurtful effect 182 
 
 APPENDIX. 
 
 Experiments on wooden arches. 184 
 
PREFACE. 
 
 In the present edition of this work, the 
 method of drawing trial lines of resistance, 
 given in the lirst edition, is retained and 
 several new chapters added containing 
 discussions of new constructions, equilibri- 
 um polygons, properties of curves of re- 
 sistance, unit stresses, and in the last two 
 chapters, two independent developments 
 of the application of the theory pertaining 
 to solid arches "fixed at the ends" to 
 voussoir arches. The Appendix contains 
 the discussion of the experiments on wood- 
 en arches, at the limits of stability, given 
 in the former edition, though here the 
 matter is presented in a more condensed 
 form. 
 
 It will probably be admitted that the 
 theory* of solid arches, fixed at the ends, 
 applies directly to voussoir arches, when 
 no mortar is used in the joints and the 
 voussoirs fit perfectly between the skew- 
 
II. 
 
 backs when not under stress, the resist- 
 ance of the backing to distortion of the 
 arch under stress being neglected and all 
 loads being supposed to be transmitted 
 vertically to the arch ring. 
 
 If the line of resistance determined from 
 this theory, for the original joints, every- 
 where lies within the middle third of the 
 arch ring, no further trial, on the supposi- 
 tion of other bearing joints, has to be made, 
 so as to make the assumed and computed 
 joints agree. 
 
 The investigation in this treatise is lim- 
 ited to this case. 
 
 The theory is likewise approximately 
 applicable where thin layers of cement 
 mortar are interposed between some or all 
 of the arch stones that are allowed to 
 harden well before the centres are struck. 
 
 For such cases, the design of a number 
 of arches, for spans from to 160 feet, for 
 a rise of one- fifth the span, are given in 
 Chapter IV. and the attention of construct- 
 ors is particularly called to this table and 
 the comparison of results with certain em- 
 pirical formulas as shown in figure 25. 
 
III. 
 
 It has long been the opinion of the 
 author that the much heavier locomotive 
 loads of to-day require greater depths of 
 keystone than are given by many formulas 
 in current use, and he submits that the 
 results of Chapters IV. and V. effectually 
 establish this position and show the danger 
 of ignoring a theoretical treatment of the 
 subject even where only approximately 
 applicable, as in the case of arches as 
 actually built. The loads may not be 
 transmitted vertically and the spandrel 
 filling may resist spreading, and, in fact, 
 act partly as an arch ; but, it is better not 
 to count on these extra elements of stabil- 
 ity, except as neutralizing the dynamic 
 effect of the moving load alone, and the 
 middle third limit prescribed may be 
 looked upon in the light of introducing a 
 factor of safety against the effect of the 
 loads regarded as static. 
 
 The author has derived assistance in the 
 theory from Schemer's Theorie des Voutes ; 
 also from Prof. Greene's "Arches," and 
 Prof. Eddy's "New Constructions in 
 Graphical Statics. " 
 
IV. 
 
 He has endeavored to assign credit at 
 the proper places in the text to these and 
 other authorSo 
 
 CHAPEL HILL, May 3d. l<S ( .)o. 
 
THEORY OF VOUSSOIR ARCHES, 
 
 CHAPTER I. 
 
 1. The voussoir arch is composed of 
 blocks of stone, brick or other material, in 
 the shape of truncated wedges, called arch 
 stones or voussoirs, whose inferior surface, 
 known as the soffit, is usually cylindrical 
 and perpendicular to the radiating surfaces 
 of contact called the joints. Between the 
 stones, mortar, either common or prefer 
 ably hydraulic, is generally interposed. 
 
 The lowest voussoirs rest against the 
 abutments along the surfaces called the 
 skewbacks or springing joints and the 
 lower edge of these joints is called the 
 springing line, 
 
 Figure 1 represents a section of the arch 
 made by a plane perpendicular to a gene- 
 rating element of the cylindrical surface. 
 This plane intersects the soffit in a curved 
 line called the intrados and the exterior 
 
surface of the arch ring in a line called the 
 extrados. The intrados is generally an 
 arc of a circle or two arcs of circles as in 
 the gothic arch, an ellipse or a false ellipse 
 (basket-handle) composed of several arcs 
 of circles tangent to each other. 
 
 The crown of the arch is the highest 
 
 part of it, the crown joint is an imaginary 
 vertical joint through the crown. 
 
 The keystone or key is the highest arch 
 stone, extending generally either side of 
 the crown, so that there is no actual joint 
 at the crown. 
 
 The haunch or reins is the part of the 
 arch between the crown and skewback, and 
 the spandrel is the part of the structure 
 
13 
 
 above the extrados and limited by the road- 
 way or other superior surface. The arch 
 is limited at the ends or heads, by planes 
 called faces perpendicular to a generating 
 element of the soffit in right cylindrical 
 arches, which alone will be considered in 
 this treatise. The spandrel walls at the 
 faces are usually of a superior quality of 
 masonry, but the space between them is 
 often filled with inferior masonry, gravel 
 or earth, called the spandrel filling or 
 simply the backing. 
 
 In large arches this space is often 
 occupied by separate walls, running par- 
 allel with the roadway and connected by 
 flat stones or light arches, which support 
 the material of the roadway. 
 
 The span of the arch (Figure 1) is the 
 perpendicular distance between the spring- 
 ing lines, and the rise is the vertical dis- 
 tance from the plane of the springing lines 
 to the highest part of the intrados. 
 
 The axis or axes of the cylindrical sur- 
 face of the soffit will be called likewise the 
 axis or axes of the arch. 
 
 When the radial length of joint is the 
 
14 
 
 same throughout, the arch ring is said to 
 be of uniform section and its constant 
 depth, measured radially, is the depth of 
 keystone. 
 
 2. In investigating the strength and sta- 
 bility of right cylindrical arches, it is con- 
 venient to consider a slice of the arch and 
 load comprised between two vertical planes, 
 perpendicular to the axis or axes of the 
 arch, and one foot apart, and it will be 
 understood that all subsequent figures refer 
 to such a slice whether distinctly stated or 
 not. 
 
 As the spandrel filling between the end 
 walls offers less resistance to deformation 
 of the arch ring than the spandrel walls at 
 the faces, it is proper to consider the slice 
 to be taken in the interior, where the 
 backing is of the least resisting character, 
 in order that we may investigate the most 
 unfavorable case. 
 
 As the external forces, including the 
 weight of arch, will be considered sym- 
 metrical with respect to the vertical plane 
 midway between the faces of the slice, all 
 the forces may be considered as acting in 
 
15 
 
 the medial plane. The same evidently 
 obtains for the internal stresses at the 
 joints, so that we have simply to investi- 
 gate the equilibrium of forces in one 
 plane. 
 
 The hypothesis will be made that the 
 weight per cubic foot of the arch stones is 
 the same throughout ; similarly for the 
 spandrel filling. This is not exactly true, 
 but is certainly near enough, when stones 
 are selected from the same quarry, espec- 
 ially when the arch is investigated as to the 
 action of the very heavy eccentric loads of 
 to-day ; for any irregularity can be sup- 
 posed included in the hypothetical loading 
 (never exactly realized) with perfect 
 safety. 
 
 Other approximations will be carefully 
 noted as we proceed. 
 
 o. In constructing voussoir arches a 
 wooden frame or centre is built between 
 the abutments, whose upper convex surface 
 exactly coincides with the soffit of the arch 
 to be built. The lower arch stones are 
 laid first, and it is a matter of experience 
 that in a semi-circular arch the arch 
 
16 
 
 stones can be laid successively up to about 
 half the rise without any centering what- 
 soever. This corresponds to the joint 
 which makes an angle of 30 with the 
 horizontal, or 00 with the vertical. The 
 central portion of 12 i. has to be supported 
 by the centre, until finally the keystone 
 (which should exactly fit the remaining 
 space) is driven in and the centre removed. 
 
 If the spandrel wall is built up to the 3< 
 joint as solidly as the abutment, as should 
 always be done, the arch and spandrel be- 
 low this joint can be considered as rigid 
 and a part of the abutments; hence the 
 true arch is the central portion of 12i. 
 total amplitude. 
 
 On this account it is convenient to de- 
 fine a segmented arch as one whose total 
 amplitude, or angle between the planes of 
 its extreme joints, is equal to or less than 
 120. 
 
 The pressure of one arch stone against 
 another at the joints gives rise to molecular 
 stresses, represented by the little arrows in 
 figure 1. It is our object first to find the 
 resultants of these stresses on each joint. 
 
17 
 
 The thrust at the crown is thus the result- 
 ant of the stresses on one side of the 
 crown joint exerted against and balanced 
 by the stresses on the other side of the 
 joint. Similarly for any other joints. 
 
 SYMMETRICAL ARCHES. 
 
 4. We shall consider first an arch formed 
 of two branches AC, 130 (Figure 2), 
 symmetrical and placed in juxtaposition, 
 and comprised between two parallel ver- 
 tical planes perpendicular to the axis of 
 the arch, the arch being right cylindrical. 
 This arch, composed of voussoirs in the 
 shape of wedges, leans against two abut- 
 ments at its extremities A and B, and is 
 loaded not only with its own weight but 
 with any other weight whatsoever, distrib- 
 uted symmetrically on either side of the 
 crown C. The mass of the arch is sub- 
 ject to the laws of friction in its joints. 
 The adherence of the mortar, interposed 
 between the voussoirs, being difficult to 
 estimate will not be considered. As the 
 two half arches are symmetrical as to the 
 crown C, it is clear that the points of 
 
18 
 
 application, A and B of the reactions R a 
 and R 2 of the surfaces of support, will be 
 also symmetrical in relation to the vertical 
 
 passing through the crown, and that the 
 line AB will be horizontal, in whatsoever 
 manner the points A and B may vary 
 upon the surfaces of support. 
 
Now the weight of arch and load = P, 
 acting downwards through the crown C, 
 together with the reactions R r and R 2 
 acting upwards, form a system of forces 
 in equilibrium, and because H l and R 2 
 must meet P in a point D they are 
 equally inclined to P and hence are equal. 
 
 If we decompose the reactions, R : and 
 R 2 , into their horizontal and vertical com- 
 ponents Pj, Q 15 P 2 , Q 2 , we should have 
 P 1 = P 2 = the weight of half arch with its 
 load, and the thrust QjZuQ,,. 
 
 Let us consider now one of the two 
 halves, for example AC. Let EPj be the 
 vertical passing through the centre of 
 gravity of this half with its load; to hold 
 this mass in equilibrium it is necessary 
 that there exists at the crown a force 
 whose direction CE passes through the 
 point of intersection E of the vertical EP, 
 with the direction of the reaction R r 
 
 As the vertical component of R : equals 
 the weight P l acting through E and since 
 by the laws of statics, the algebraic sum 
 of the horizontal components of the forces 
 acting on the half arch AC, must equal 
 
zero, the thrust at the crown C of the 
 arch is necessarily equal to the second 
 component Q 1? of the reaction R. and must 
 be horizontal as it is. 
 
 From what preceeds we are allowed to 
 consider only a half-arch, leant against a 
 FIG. 3. 
 
 fixed surface at A, and solicited by a hori- 
 zontal force at C. 
 
 5. (Figure 3.) Let ab, oj)^ 2 /> 2 , be the 
 joints of an arch; P 19 P., the vertical 
 directions of the weights of the parts ab 
 b^a,\ <fbb 2 <? 2 ; including the loads on the 
 
parts bb l9 bb^ ; P 1? P, acting through 
 
 their common centres of gravity. 
 
 The horizontal force Q, at the crown, 
 combined with the reaction R 2 at the j oint 
 8 & 2 holds the part abb 9 a 9 in equilibrium 
 and similarly for the reactions on other 
 joints. 
 
 At the points where the direction of Q 
 cuts P,, P 2 , combine those forces with Q 
 as shown in the figure; the resultant of Q 
 and Pj cuts joints a l b 1 at Aj, which is 
 therefore the centre of pressure on that 
 joint. As the weight abb 2 a z with its load 
 equals P 2 and is the weight on joint a 2 & 2 , 
 the resultant of P 2 and Q will give the 
 force acting on ajb^ in direction, position 
 and magnitude ; it cuts 2 6 2 at A 2 , which 
 is therefore the centre of pressure on that 
 joint. 
 
 In the same way the resultants and cen- 
 tres of pressure on all the joints may be de- 
 termined. A broken line connecting these 
 centres of pressure on the various joints 
 will be called the line of the centres of 
 pressure or more briefly, the line of re- 
 sistanee. 
 
For voussoirs indefinitely small, it ap- 
 proaches indefinately a curved line and 
 will be called the curve of the centres of 
 pressure or curve of resistance. 
 
 That granted, in order that the arch 
 may remain in equilibrium, it is necessary : 
 
 (1.) That the points of intersection 
 C', A,, A 2 , fall in the interior of the 
 respective joints ab, a } b l9 ajb y If for any 
 joint this is not so; e. </., if the point A ] 
 was above b l9 the mass abb^ a l would then 
 turn around the edge 6 1? as an unresisted 
 couple would be formed. To explain : 
 suppose the resultant H l to pass outside 
 of joint a l b l y conceive two equal opposed 
 forces, each equal to R, to act at edge b,- 
 this does not disturb the equilibrium; 
 then R, the force acting through A, 
 (which is outside the joint) with its equal 
 but not directly opposed force at b l9 would 
 form the unresisted couple in question 
 which causes overturning. 
 
 (2.) That the directions c^A^ c 2 A 2 of 
 the pressures upon the joints do not make 
 angles with the normals to the respective 
 joints which exceed the angle of friction. 
 
If it was not so, sliding at the joints in 
 question would occur of the mass above or 
 below. 
 
 However, the friction of the materials 
 usually employed in construction is suffi- 
 ciently great to not give cause for fear 
 as regards sliding, generally. 
 
 It is very easy to alter the direction of 
 the joints should sliding be apprehended, 
 hence it will not be considered further. 
 
 The two former requirements refer to 
 stability^ supposing no crushing of the 
 material occurs. The last requirement 
 pertains to the strength of the masonry 
 at a mortar joint. 
 
 (3.) The resultant of the pressure on 
 any joint must not pass so near the edge 
 that crushing of the mortar or the stone 
 (or brick) may occur. Of course in prac- 
 tice a certain factor of safety would be 
 used, but as this subject will be treated in 
 full later, we shall only remark here that 
 reasons will be given further on why the 
 true line of resistance should not depart 
 at any joint more than one- sixth the depth 
 of joint from its centre, or in other words, 
 
24 
 
 should everywhere be confined to the mid- 
 dle third of the arch ring and in fact it is 
 best to use still narrower limits. 
 
 It is to be remarked that the foregoing 
 theory does not require horizontal resist- 
 ance in the spandrel, which is not gener- 
 ally built with the same care that is taken 
 in the construction of the arch stones, and 
 thus cannot generally be regarded as un- 
 yielding; hence when a line of resistance 
 such as C'Aj A., passes, somewhere, out of 
 the arch ring, a serious derangement of 
 the arch may occur, even though the 
 spandrel may -prevent its falling: hence it 
 appears to be a poor construction to build 
 such an arch in preference to an arch in 
 which the resultant pressures on the joints 
 everywhere keep within the limits pre- 
 scribed. This will be adverted to again. 
 
 If no line of resistance can be dniwn 
 within the prescribed limits, or crushing is 
 feared on any joint, the depth of the vous- 
 soirs must be increased on part or the 
 whole of the arch; or the profile may be 
 altered; or, finally, we may combine both 
 of these methods to secure stability. 
 
6. We shall now enter into detail to 
 show how lines of resistance can be drawn 
 through various points of the arch ring. 
 
 Let us consider as in Art. 2 a slice of the 
 arch 1 unit thick. In Fig. 4, let Q=the , 
 horizontal thrust at the point M of the ! 
 
 crown joint (compare Fig. 3 throughout) ; 
 </, its vertical distance below the apex C; 
 P= weight of arch and load C A on joint 
 at A; a horizontal distance from A, tlK- 
 centre of pressure on the joint, to the ver- 
 tical passing through the centre of gravity 
 of P; h vertical distance between C and 
 
A; b = vertical distance between M, the 
 point of application of Q at the crown, 
 and A. 
 
 If we consider another point A' of the 
 mrve of resistance, at a vertical distance 
 above A=e, we shall have an analogous 
 notation P', a', b'. 
 
 If we know the points M and A, we have, 
 taking moments about A, 
 
 P = bQ .-. Q = ............ (1). 
 
 If we know any two points as A, A', 
 =bQ = ('-f e\Qu also a' P = b' Q, 
 
 ~ 
 
 hence, Q = ? ~ ............. (2). 
 
 Having obtained from eq. (1) or eqs. (2) 
 and (o), Q and its point of application at 
 the crown, we find where the resultant 
 pressures cut each joint as in Art. 3. 
 
 If the first curve drawn passes outside 
 of the prescribed limits in one or more 
 places, take points on the limiting curves 
 
27 
 
 opposite the points of maximum departure, 
 and by eqs. (^) and (3), pass a curve 
 through two of these points. 
 
 If the.arch is stable at all, it will almost 
 invariably be found in practice that the 
 last curve so drawn will fulfill the required 
 conditions of remaining in the prescribed 
 limits. If not a third approximation may 
 be tried, and so on. 
 
 This is Dr. Scheffler's method of draw- 
 ing a line of resistance within prescribed 
 limits and in practice, after becoming 
 familiar with the leading cases, the first 
 trial is generally sufficient. 
 
 7. Let us proceed to show how to find 
 the centres of gravity of the weights abb^ 
 d^ abb^a^ (Fig. 3), as also the magnitudes 
 of those weights. If the arch is loaded 
 with any weights, reduce them to the 
 same specific gravity as that of the masonry 
 of the bridge supposed homogeneous, as 
 follows : Lay down these weights in their 
 exact positions on the arch, and alter the 
 vertical dimensions to conform to the 
 specific gravity of the stone. We shall 
 thus substitute blocks of stone, by scale, 
 
28 
 
 for the surcharge of earth, water, etc. 3 or 
 the rolling load. 
 
 We now divide the horizontal through 
 A (Fig. 5) into an appropriate number of 
 parts and through these points of division 
 
 draw verticals from the intrados to the 
 curve DE that limits above the reduced 
 load. 
 
 Regard each trapezoid DGG'D' as a 
 rectangle, and calculate its surface by 
 multiplying its horizontal width AA' by 
 
29 
 
 the mean vertical dg. Next regarding the 
 centre of gravity of each trapezoid as that 
 of the corresponding rectangle, we shall 
 find the centre of gravity of the trapezoid 
 DD'G'G, for .example, to be upon the 
 mean vertical dg, which equally divides 
 the horizontal AA'. Draw through C, the 
 joint CH; the weight DD'G'G will be 
 considered as resting on the joint CH, 
 which is in excess by the small triangle 
 CG'H, an error too small to be regarded 
 in flat arches. 
 
 Scheffler gives a graphical construction for correcting 
 the joints to correspond better to the weights, but it is very 
 tedious in practice and defaces the drawing with too many 
 many lines, so it is not given. In article 11 will be found a 
 method that will insure all desirable accuracy for any form 
 of arch. 
 
 The assumption that the whole weight 
 vertically over a voussoir acts on it is not 
 strictly true, whether the spandrel is of 
 earth or masonry ; for part of it is doubt- 
 less distributed in a different manner by 
 aid of the friction of the particles, or in 
 some cases from the spandrel itself acting 
 as a sort of arch. Again, any horizontal 
 spreading of the haunches of the arch is 
 
30 
 
 partially hindered by horizontal resistances 
 in the spandrels. These influences though 
 are on the side of safety, particularly when 
 the dynamic effects of moving loads are 
 considered. In the case of culverts the 
 earth exerts a thrust, as to which and the 
 treatment of culverts and tunnel arches, 
 see Science Series No. 42. 
 
 Assuming the loads over the arch to act 
 as above, we multiply the surface of each 
 trapezoid by the horizontal distance of its 
 centre of gravity from A ; the sum of these 
 moments divided by the sum of the trape- 
 zoid surfaces (which are also the volumes), 
 will give the horizontal distance from A 
 to the centre of gravity of the whole part 
 considered. This method will thus give us 
 
 the weights P 1? P 2 , (Fig. 3), as well as 
 
 the horizontal distance of their centres of 
 gravity from the crown. 
 
 8. First Example. Let us illustrate by 
 an example of a railroad bridge (Fig. 1.5) of 
 50 ft. span and 10 ft. rise, the arch being 
 a segment of a circle ; voussoirs 2.5 ft. deep ; 
 the spandrel walls, of the same specific 
 gravity as the voussoirs, rising 2. S3 ft. 
 
31 
 
 above the summit of the arch rin 5 . i'he 
 
 arch is 7 ft. thick, but we shall consider 
 but a vertical slice of it 1 ft, thick. 
 
32 
 
 In the following table the first column 
 gives the number of the joint from the 
 crown; the second (w) the width of the 
 horizontal divisions AA', A'A"- - of Fig. >"> ; 
 the third (v) the corresponding mean 
 heights dg - - -; the fourth (s), the prod- 
 uct of these dimensions, giving thus the 
 surface of each trapezoid. Column (c) 
 gives the distance of the centre of grav- 
 ity of each trapezoid from the crown ; 
 column (m) the product of (s) and (c). 
 Now we cumulate, going from the crown, 
 in the next two columns, these surfaces (s) 
 and products (s X c) ; column (S) being 
 formed by adding the surface of each 
 trapezoid to the total surface, just found, 
 which precedes it. The last quantity in 
 column (S) should = sum of col urn (s). 
 
 In the same way column (M) contains 
 the continued sum of column (m), and 
 hence its last number should equal the 
 sum of column (m). Dividing now the 
 numbers in column (M) by the corres- 
 ponding ones in column (S) we get, col- 
 umn (C), the horizontal distances of the 
 centre of gravity of each weight P l5 P 2 
 
, corresponding to joints 1, 2 - - -, from 
 the crown. 
 
 1 
 
 w 
 
 v 
 
 s 
 
 C 
 
 m 
 
 S 
 
 M 
 
 C 
 
 1 
 
 2 
 3 
 4 
 5 
 
 G 
 
 5 
 
 5 
 5 
 5 
 5 
 1.75 
 
 5.4 
 6.1 
 76 
 9.8 
 13.2 
 14.5 
 
 27. 
 30.5 
 38. 
 49. 
 66. 
 25.4 
 
 2.5 
 7.5 
 12.5 
 17.5 
 22.5 
 25.9 
 
 67.50 
 
 228.75 
 475. 
 857.50 
 148 5. 
 657.86 
 
 27. 
 57.5 
 95.5 
 144.5 
 210.5 
 235.9 
 
 67.50 
 296.25 
 771.25 
 1628.75 
 8113.75 
 3771.61 
 
 2.5 
 5.1 
 8 1 
 11. 
 14.7 
 16. 
 
 
 
 
 235.9 
 
 
 3771.61 
 
 
 
 
 The preceding table shows that the 
 surface (or volume, for a slice 1 ft. thick) 
 of the half arch with its load equals 235. ( .i 
 sq. ft.; its moment as to the crown is 
 3TTUJ1 and the distance of its centre of 
 gravity from the joint at the crown is 10 ft. 
 Let it be required to pass a curve of resist- 
 ance through the crown joint, K of its 
 depth from the summit of the arch and 
 through joint (> at }- 3 of its depth above its 
 lowest point. By measurement on the 
 drawing (Fig. 9) we finda=25.i> 16=1). (>, 
 6=11.2. We have also P=235.9 cubic 
 feet of stone; hence by formula (1), Art. 0, 
 
 235.9x96 
 ~lT2~ 
 
 = 202 cubic ft. stone 
 
which may be reduced to tons, when de- 
 sired, by multiplying by the weight in 
 tons of a cubic foot of stone. 
 
 If now at the points of intersection of 
 the horizontal through the point of appli- 
 cation of Q at the crown, with the verti- 
 cals passing through the centres of grav- 
 ity of the surfaces given in column (S), 
 [Pj, P 2 , - - -, of Fig. 3], we combine these 
 weights with Q, the points of intersection 
 of the resultants of Q with these weights 
 
 P,, Pjj, ,with the corresponding joints, 
 
 will be points in the curve of resistance 
 sought. 
 
 For example, to determine where the 
 line of resistance cuts joint 4, lay off the 
 distance in column (C), 11.3 horizontally 
 from the crown, then on a vertical lay off 
 upward from this point the corresponding 
 weight on joint 4, given in column (S) 
 144.5 ; drawing a horizontal line through 
 the last point found=Q = 20^, we get the 
 resultant by completing the triangle of 
 forces. 
 
 Producing this resultant to intersection 
 with joint 4, will give the centre of pres- 
 
sure on that joint. It will be advisable, 
 in practice, to prick off the centres of 
 gravity, taken from column (C), at one 
 operation and number each one with the 
 number of the corresponding joint to 
 avoid mistake. 
 
 On continuing this construction for 
 each joint, we shall find that the line of 
 resistance remains within the inner third 
 of the arch ring. 
 
 It may be remarked that the small tri- 
 angle mentioned is in excess only for the 
 joint in question; thus this error is not 
 carried on. 
 
 The ordinary method of constructing a 
 line of pressures is to combine any result- 
 ant with the next weight following, re- 
 garded as concentrated at its centre of 
 gravity. By this construction any small 
 error in draughting is carried on, whereas, 
 by the former method, it is confined only 
 to the joint where it occurs first. 
 
 With accurate instruments and care, 
 using a sufficiently large scale, this meth- 
 od should answer all the requirements of 
 accuracy, and will generally be found the 
 
shortest in the end; whereas, with many 
 joints, it is difficult to locate this curve 
 precisely by the ordinary method. 
 
 We have made use in the above example, 
 of formula (1) to compute the thrust at 
 the crown. This can preferably be found 
 without computation, as follows: at the 
 point H on the horizontal through the 
 crown, lay off, to the scale of force, verti- 
 cally upwards the weight of half arch and 
 load = #35.!.) and through the extremity of 
 the line, draw an indefinite horizontal. 
 The intersection of the latter with the line 
 drawn through the point <> first mentioned 
 and the assumed centre of pressure on 
 joint f>, will cut off a distance on this 
 horizontal, equal to Q to the scale of force. 
 This value of Q is then to be laid off in 
 constructing all the other triangles of 
 force. This method can likewise be fol- 
 lowed in the next example if preferred. 
 
 9. Second Example. (Fig. 7.) Suppose a load of two 
 40-ton engines, one on each side of the crown, over divis- 
 ions 2, 3, and 4, i. e., 15 ft. along the rails. We shall sup- 
 pose it to bear only on 6 ft. of the thickness of the viaduct. 
 Calling the weight of a cubic foot of stone = .07 ton and A, 
 the height of the block of stone 15 ft. long by 6 ft. wide that 
 is required to weigh as much as one engine; we have 
 6 X 15 X h .07 =40 .-./*= 6.3. 
 
We now form the following table which refers to Fig. 7, 
 which, as the arch and load is symmetrical, represents, as 
 before, only one-half the arch. 
 
 Joint 
 
 w 
 
 v 
 
 s 
 
 C 
 
 in 
 
 S 
 
 M 
 
 C 
 
 1 
 2 
 3 
 4 
 5 
 6 
 
 5 
 
 5 
 5 
 5 
 5 
 1.75 
 
 5.4 
 
 12.4 
 14. 
 16. 
 13.2 
 14.5 
 
 27 
 62 
 70 
 80 
 66 
 25 
 
 2.5 
 7.5 
 12.6 
 17.5 
 22,5 
 25.9 
 
 67 
 468 
 
 869 
 
 HI 18 
 1485 
 658 
 
 21 
 
 159 
 239 
 
 :; >5 
 330 
 
 67 
 532 
 1401 
 
 2809 
 4294 
 4952 
 
 2.5 
 6. 
 
 11.8 
 
 14.1 
 
 15. 
 
 
 
 
 330 
 
 
 4952 
 
 
 
 
 A line of resistance passing through the middle 01 the 
 crown, the point on the springing joint, as before, will be 
 found to be contained inside of the limiting curves, and is 
 drawn as in Fig. 7, taking care to lay off the centres of gravity 
 on the prolongation of Q. We find in this case a = 25. 6 15 
 = 10.6, P=330, 6=10.7. 
 
 . Q = 880 X 10.6 =327=c=23tone . 
 
 If it is desired to draw the curve corresponding to the 
 minimum of the thrust in the limits chosen (see Art. 19.) 
 we resort to equations (2; and (3). As the nearest approach 
 of the last line of pressures drawn to the outside limiting 
 curve, is at joint 2 ; pass a curve of resistance through the 
 point of intersection of that outside limiting curve with the 
 second joint and the previous point at the springing joint. 
 We find P= 330, a 10 6, <?=9.8 and from table 2, column 
 (S)Pi = S9; from column (c) and the drawing a! =9. 8 6 
 = 3.8. 
 
 From (2), Art. 6, 
 
 aP_ aiPl 3498-338 
 
 From (3) 
 
 =11.93 
 
 9.8 
 3498 
 
Laying off this latter distance, from the summit of the 
 arch ring, downwards, we draw the curve as before. It is 
 everywhere within the proper limits, and of course touches 
 the upper middle third limit at joint 2 and the lower 
 middle third limit at joint 6, as assumed. 
 
 If we suppose a weight of 13.3 tons to rest on division 3 
 on both sides of the crown, along 5 ft. of the length of the 
 rails, we shall find by forming a table and constructing the 
 line of resistance as in the last case above, that it passes 
 slightly below the upper limit at the crown, and is every- 
 where contained in the middle third of the arch ring. 
 
 A curve of resistance for a uniform load of 1.5 tons per 
 foot along the whole length of the bridge can be drawn to 
 follow very closely the curve drawn in the first example. 
 
 One or two more suppositions of isolated weights, sym- 
 metrically placed, were made, but in all cases it was found 
 that a curve of resistance could easily be drawn in the inner 
 third of the arch ring. The thrust is too small to fear crush- 
 ing, and the directions of the thrust are inclined to the nor- 
 mals of the arch joints at angles much smaller than the 
 angles of friction, hence sliding is not to be feared. 
 
 The curves of resistance drawn in the 
 preceding examples are not necessar- 
 ily the true ones, otherwise we should 
 at once conclude that thus far the arch 
 had stability. The true curve depends 
 upon the elastic yielding of the arch to the 
 weights acting on it and we shall see later 
 how the aid of a few principles from the 
 theory of elasticity will enable us to locate 
 it approximately. For the present, we 
 shall continue the subject by showing how 
 
any trial line of resistance can be made to 
 pass through any three points of an arch 
 ring, either unsymmetrical or unsyminet- 
 rically loaded. The method first given 
 below requires the solution of some equa- 
 tions: subsequently in Art. 15 a purely 
 graphical method is developed which will 
 doubtless generally be preferred. 
 
 UNSYMMETRICAL ARCHES. 
 
 10. In unsymmetrical arches, or arches 
 unsymmetrically loaded, there is a joint, 
 EF (Fig. 8), generally near the crown, at 
 which the thrust is horizontal. Where 
 the arch is only solicited by vertical forces, 
 by compounding them with this thrust, as 
 before, we find the resultants on every 
 joint, and it is evident in this case that the 
 horizontal thrust is the same all through 
 the arch. 
 
 It is more convenient however, to find 
 the inclined thrust at the crown and com- 
 bine the partial weights with it, to find 
 the resultant on each joint. 
 
 Problem : To find this inclined thrust 
 and its point of application : 
 
Let I, L, K be three points through 
 
 which to pass a curve of resistance (Fig. 
 
 (8), ACB a horizontal line drawn through 
 
 the highest point of the extrados, and let 
 
 there be: 
 
 Q, the horizontal thrust, i. e. the horizon- 
 tal component of any one whatsoever of 
 the pressures acting through I, L, K. ; 
 
 P, the vertical component of the pres- 
 sure S at the crown joint, which will be 
 considered positive, if it is directed up- 
 wards, as regards pressure from the right 
 part upon the left. 
 
 Let <?, be the vertical distance of the 
 point of application of S below the hori- 
 zontal ACB; (fv hi'y 9v h* ; 9v h*> the hori- 
 zontal and vertical co-ordinates of I, K 
 and L as to the point C as the origin ; 
 
 P 1 , P 11 , P 111 , the vertical components of 
 the pressures acting through I, K and L; 
 P 1? P 2 , P 3 , the weights of the segments C 
 , I, CK, CL, with their loads ; 
 
 #i> <3 2 > a s , the horizontal distances of 
 the centres of gravity of these segments 
 CI, CK, CL, from the points I, K and L 
 respectively. 
 
43 
 
 To abbreviate, let us put: 
 h 1 -q=b 1 h a -q=b 2 
 
 Observe that arch CI is in equilibrium 
 under the action of the left reaction tP l 
 acting up and Q acting to the right being 
 its components) the thrust S at the crown 
 (P acting up and Q acting to left being its 
 components) and the weight of arch CI 
 and load P l acting downwards. 
 
 Similarly, the part CK is in equilibrium 
 under the action of P'' (acting up) and Q 
 (acting to left) at K, the force P (acting 
 down) and Q (acting to the right) at the 
 crown and the weight P 2 . 
 
 Lastly, the part CL can be dissociated 
 from the rest and conceived to be in 
 equilibrium under the action of the reac- 
 tion at L acting upwards, the forces P 
 (acting up) and Q (acting to the left) at 
 the crown and the weight P 3 of CL and load. 
 
 Balancing vertical components for the 
 parts CI and CK respectively, we have, 
 P'-fP^P, ........................ (1) 
 
 P"_P = P t ....................... (2) 
 
44 
 
 Next taking moments about I, K and L 
 of the forces holding in equilibrium the 
 part CI, CK and CL respectively. 
 
 a,P 8 +g 8 P=b 8 Q ................. . ........... (4) 
 
 a 3 P 3 -g 3 P=b 3 Q ............................ (5) 
 
 If the third given point L of the curve 
 of pressures is upon the joint at the crown 
 C, the value of q is known, and we have : 
 g s = o, h 3 = q, P 8 = O. From eqs. (3) and 
 (4) we find 
 
 
 If L is not upon the joint at the summit, 
 we find * 
 
 T> 
 
 Q 
 
 .(10) 
 
 * Add-(4) and (3) and call the sum eq. (11); also subtract 
 (5) from (3.) Place the values of Q equal to each other in this 
 last eq. and eq. (11); reducing, bearing in mind that do-c? :j 
 =d lf e. 2 -e 3 = ci, &c., we find P as in eq. (8). Substitute 
 this value of P just found in eq. (11) and deduce Q. which 
 gives eq. (9). Eqs. (6) and (5) are only particular cases of 
 eqs. (8), (9) and (10) when P ;j = O. 
 
Example 1. Fig. 9 represents the same 
 viaduct, before considered in Art. 8, with 
 a load of 40 tons on 15 feet of length over 
 divisions "<?, 3 and 4, on one side of the 
 arch only. The table of Art. 8 refers to the 
 right half of the arch : the table of article 
 to the left side. 
 
 Let us pass a curve of resistance through 
 the middle of the crown and through a 
 point on each springing joint, y^ depth 
 joint above its lower edge. 
 
 We find from the drawing and tables. 
 
 9^=9%-^^ 1 = 2 :=10.75 
 
 P ] = 330, A = 15, a, = 10.6 
 
 P i = ^3(>, p 2 = l<>, a s = 9.6 
 
 From ((>) : 
 
 p= ,^q3*,P >=24 cubic ft< of gto 
 ffA-r^ffi 
 
 from (T): 
 
 Q ^l-^i?==^68 cubic ft. of stone; 
 
 From M, the middle of the crown joint, 
 lay off downwards MN=P, alsoNH=Q on 
 the horizontal through N; MH will then 
 represent tl on the crown joint 
 
47 
 
 in direction, position and magnitude; and 
 by combining it with the weight of each 
 artificial voussoir and load, on each side of 
 the crown, each acting through its centre 
 of gravity, we evidently obtain the result- 
 ants on the various joints in direction, posi- 
 tion and magnitude, and therefore can 
 trace the curve of pressures. For example, 
 to find the resultant on the third joint on 
 the left side of the arch : draw a horizon- 
 tal line through M and lay off on it the 
 distance of the centre of gravity of the 
 three first divisions, from M, which by 
 Table 2 ( Art.8) , column C, is found to be 8.8. 
 
 Draw a vertical through this point and 
 from its point of intersection with MH, lay 
 off upwards the weight 159 (column S) of 
 the three divisions in question. 
 
 From the upper extremity of this last 
 line draw a line || and equal to MH; com- 
 pleting the parallelogram of forces as per 
 figure, the point where the resultant cuts 
 joint 3 is the centre of pressure of that 
 joint, and the resultant is given in magni- 
 tude, position and direction by the diago- 
 nal. 
 
48., 
 
 The construction for the other joints is 
 the same. 
 
 The nearest approach of the curve of 
 pressures to the extraclos is on joint 1, of 
 the left side of the arch, where it is only 
 three-tenths (.3) of a foot (on a large 
 scale drawing it was found to be .35) from 
 the edge. The nearest approach to the 
 intrados is at joints 3, 4 and 5 on the right, 
 being only about .7 to .75 from the edges 
 at those joints. 
 
 Example 2 Draw a line of resistance for the bridge, 
 loaded as above, through the lower middle third limit at the 
 left springing, the upper middle third limit at the right 
 springing and at a point on the crown joint 1. 1 ft. above 
 the intrados. It passes above the middle third limit at joint 
 2 on the loaded side, its maximum departure, and just 
 touches the lower limit at joints 1 and 2 on the unloaded 
 side. 
 
 Example 3. By aid of formulas (8), (9) and (10), draw a 
 line of resistance through the lower middle third limit at 
 the left springing and the upper middle third limit at joint 
 
 2 under the load and also through the upper limit at the the 
 right springing joint. 
 
 The thrust at the crown will be found now to act 0.76ft. 
 below the centre of the joint, its horizontal component being 
 294 cu. ft. and its vertical component 18 cu. ft. The line 
 of resistance everywhere keeps within the middle third 
 limits except at joints 1 and 2 on the right where it passes 
 O.U and 0.12 respectively below the limit. 
 
 Example 4. A load of 13-3 tons was assumed on division, 
 
 3 on one side, and it was found that a curve of pressures 
 
49 
 
 nmld be drawn, for this . (-centric load, within the inner 
 third of the arch ring. 
 
 If the backing is raised higher, thus making the bridge 
 weigh more, a rolling load will have less effect upon it . 
 hence a less depth of keystone may be used. Other things 
 {he same, it is a simple question of economy, considering 
 the approaches, whether to increase the height of surcharge 
 above the arch ring, or the depth of the arch stones. 
 
 Fig. 10 shows the effect of rolling loads 
 in different positions, on the piers; the 
 middle bay not being loaded but with its 
 own weight, the end spans as per figure. 
 The resultants at the springing joints we 
 have before determined; combining the 
 two on any pier with the weight of pier, 
 according to the usual rule for three forces 
 not intersecting in one point, we obtain the 
 final resultants on the bases of the piers. 
 
 It is seen from the figure that the 40 
 tons on both sides produces a more hurtful 
 effect on the pier than a 40 ton load on 
 one side only. 
 
 By combining the weight of abutment 
 with the thrust on it, we find that the cen- 
 tre of pressure on the foundation course is 
 sufficiently within the limits for most cases 
 in practice. 
 
 The dotted line in the abutment gives 
 
50 
 
 the centres of pressure of all the forces 
 acting on each joint for the joints in ques- 
 tion. For example, to find where this 
 centre of pressure is on the springing 
 line, produced, we combine the inclined 
 resultant on the arch joint at the springing 
 with the weight of the abutment above 
 the springing line, acting through its cen- 
 tre of gravity. This resultant makes an 
 angle with the vertical of only 23, hence 
 sliding on the springing course is not to 
 be feared, if the abutment is solidly built. 
 
 The lines of resistance as drawn for the 
 three arches, piers and abutments are not 
 necessarily the true ones. 
 
 Further on will be given a method of 
 locating approximately, the true line of 
 resistance for a well built arch, with thin 
 mortar joints, between immovable abut- 
 ments. The abutments in the figure are 
 of such proportions as to be practically 
 immovable, as the centre of pressure on 
 the base is near its centre, but not so the 
 piers. The first pier on the right tends to 
 lean to the left, the second one to the 
 right, This tendency is resisted too by the 
 
central arch, which thus puts forth a 
 stronger horizontal thrust than assumed, 
 corresponding to a line of resistance pass- 
 ing nearer the intrados at the crown and 
 the extrados at the skewbacks, the maxi- 
 mum efforts being produced when the line 
 of resistance passes very near these curves 
 so that no crushing ensues. We are not 
 able to locate this line with our present 
 knowledge, but it is plain that this central 
 arch will put forth its maximum effort if 
 necessary, to prevent much motion inwards 
 of the tops of the piers, so that the centres 
 of pressure on their bases will not depart 
 as far from their centres as the figure 
 shows. As there will be some motion 
 however, it tends to cause the line of 
 resistance of the other arches to travel 
 down the skewbacks at the piers and to 
 move up the crown joints, from the slight 
 increase of span, thus giving rise to a less 
 horizontal thrust from those arches, which 
 again tends to correct the eccentricity of 
 the thrust on the piers. If preferred, the 
 lines of resistance can be redrawn in these 
 arches, corresponding to a minimum thrust 
 
(within reasonable limits) of the outer 
 arches and a maximum thrust of the cen- 
 tral arch, when the stability of the piers 
 will be more apparent. Experience incli- 
 , cates that piers of the proportions shown 
 are perfectly stable. 
 
CHAPTER II. 
 
 11. A MOKE CORRECT METHOD OF 
 MAKING OUT THE TABLE OF WEIGHTS 
 AND CENTRES OF (TRAVITY. 
 
 The method of finding the weights and 
 centres of gravity given in Art. 7, although 
 sufficiently correct for flat arches with a 
 small depth of key, is not so for thick 
 arches approaching the semicircular or 
 elliptical in form. The following is sug- 
 gested by the author as giving all desir- 
 able accuracy with but little more labor 
 than Scheffler's method. 
 
 The arch is preferably divided into a 
 number of equal voussoirs (see Fig. 11), 
 and the vertical lines drawn from the 
 upper ends of the joints to the reduced 
 contour of the surcharge, divides the latter 
 into trapezoids. As before, we draw the 
 medial dotted lines, which will be assumed 
 to pass through the centres of gravity of 
 the trapezoids, though the latter can be 
 found exactly by the usual graphical con- 
 
54 
 
 struction if desired. The area of a trape- 
 zoid width X mean height =: w v. On 
 multiplying the area by the distance from 
 the crown to the medial line of the trape- 
 zoid (c) we have the moment m = (w v) c 
 about the crown, for any trapezoid. 
 
 The quantities vr, v, s, c and m for Fig. 
 11 are entered in the table below, being the 
 upper numbers corresponding to the joint 
 given in the first vertical column. The 
 corresponding quantities for the voussoirs 
 are the lowest numbers of the horizontal 
 rows. 
 
 Calling r, in Fig. 11, the radius of the 
 extrados, r l9 that of the intrados and n the 
 proportion of the circumference included 
 by the voussoir, we have its content 
 
 7T(V' 2 T 2 ) 
 
 for a thickness of 1. Now 
 n 
 
 this is equal to the depth (r r^) X tne 
 
 (1 T f- V \~1 
 
 2 7t - -) hence meas- 
 n 'I /J 
 
 ure the middle length and depth on a 
 drawing, their product will give the 
 required volume of a voussoir (= 2.35 
 = 4.7 in this case). 
 
w 
 
 v 
 
 s 
 
 c 
 
 in 
 
 S 
 
 M 
 
 C 
 
 2.72 
 2.35 
 
 2.13 
 
 2. 
 
 5.79 
 4.7 
 
 1.38 
 1.18 
 
 7.99 
 5.55 
 
 10.49 
 
 13.54 
 
 1.29 
 
 2.27 
 2.35 
 
 3.16 
 2. 
 
 7.17 
 
 4.7 
 
 3.88 
 3.36 
 
 27.82 
 15.79 
 
 22.36 
 
 57.15 
 
 2.55 
 
 1.51 
 2.35 
 
 5. 
 
 2. 
 
 7.55 
 
 4.7 
 
 5.77 
 5.03 
 
 43.56 
 23.64 
 
 34.61 
 
 124.35 
 
 3.59 
 
 .05 
 2.35 
 
 7.2 
 2. 
 
 3.6 
 4.7 
 
 6.77 
 5.92 
 
 24.37 
 
 27.82 
 
 42.91 
 
 176.54 
 
 4.11 
 
Tlie centres of gravity of the voussoirs 
 will be assumed to lie on the (dotted) cen- 
 tre line of the* arch ring and midway 
 between the joints; the distances from 
 these points to the vertical through the 
 crown give the arms in column (c). The 
 volume (4.7) of a voussoir multiplied by its 
 c, gives the corresponding ID of the table. 
 
 This manner of considering the voussoirs 
 and surcharge separately is continued, until 
 in columns S and M the quantities referring 
 to the same joint are combined by the con- 
 tinued addition of the quantities in col- 
 ums (s) and (ni) respectively. 
 
 If the voussoirs are taken the same size, 
 there is really no necessity of entering 
 their dimensions; simply giving their com- 
 mon area in column (s). 
 
 When the voussoirs are taken small 
 enough this method gives all desirable 
 accuracy. 
 
 Concentrated loads on the arch are easily 
 included by introducing a third row of 
 numbers, for any voussoir affected, just 
 above those given for the trapezoids as 
 will be fully explained in a subsequent 
 article. 
 
12. A CONVENIENT METHOD OF DRAW- 
 ING A TKIAL LINE OF RESISTANCE IN AN 
 
 ARCH. 
 
 Let fig. l'-> represent a semi-circular arch 
 of Inn ft. span, :> ft. key and o ft. depth of 
 surcharge over the crown of the same 
 specific gravity as the voussoir. The live 
 load extends from the crown to the right 
 abutment and weighs .'),<< M> pounds per 
 foot of rails. If this bears on a width of 
 <] feet it is equivalent to a layer of stone 
 of the same density as the voussoir (15<> 
 Ibs. pr. cu. ft.) 3. 4 ft. high, as shown in 
 the figure. 
 
 The spandrel was divided up by verti- 
 cal lines, ."> ft. apart for 4" ft. from the 
 crown, then 2 ft. apart for the next 10 ft., 
 and 1 ft. apart for the remaining 3 feet. 
 The joints 1, 2, 3, . . . are then drawn 
 as in the figure. 
 
 o 
 
 The following is a condensed table of 
 loads (S in cubic feet) and distances from 
 the crown to their centres of gravity (C. 
 
58 
 
 LEFT HALF. 
 
 Joints 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 
 S 
 
 c 
 
 30 
 
 2.5 
 
 62 
 5.1 
 
 100 
 
 7.8 
 
 146 
 
 10.8 
 
 201 
 14. 
 
 272 
 17.4 
 
 362 
 21.1 
 
 475 
 25. 
 
 
 Joints 
 
 9 
 
 10 
 
 11 
 
 12 
 
 13 
 
 819 
 33.5 
 
 14 
 
 15 
 
 16 
 
 S 
 C 
 
 529 
 26.6 
 
 589 
 28.3 
 
 656 
 30. 
 
 732 
 31.7 
 
 869 
 34.5 
 
 925 
 35.5 
 
 1004 
 36.8 
 
 EIGHT HALF. 
 
 Joints 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 S 
 C 
 
 47 
 2.5 
 
 96 
 5. 
 
 151 
 
 7.7 
 
 214 
 10.5 
 
 286 
 13.6 
 
 374 
 16.8 
 
 481 
 20.2 
 
 611 
 23.9 
 
 Joints 
 
 9 
 
 10 
 
 11 
 
 12 
 
 13 
 
 14 
 
 15 
 
 16 
 
 S 
 
 672 
 
 739 
 
 812 
 
 895 
 
 989 
 
 1042 
 
 1101 
 
 1184 
 
 C 
 
 25.4 
 
 27. 
 
 28.6 
 
 30.3 
 
 32. 
 
 33. 
 
 33.9 
 
 35.2 
 
60 
 
 Let us pass a line of resistance one- 
 twelfth the depth of arch ring below 
 the centres of joints 8 and the crown 
 joint, or through a, c and m. By the 
 formula method of art. 10 we find by aid of 
 a drawing, &c., P = 32. S, Q = 444.H. 
 (It will be instructive for the reader to 
 test these values by the purely graphical 
 method of art. 15, which is generally to 
 be preferred). 
 
 From m lay off to the right horizontally, 
 i; then vertically upwards, 
 om represents the resultant 
 at the crown joint. 
 
 Lay off on om produced mo to the left 
 and equal to om. Through the points o 
 thus determined, draw_verticals and lay 
 off from o the distances 01, 02, - - , equal 
 to the values of S pertaining to joints 1,2, 
 
 , as taken from the tables pertaining 
 
 to the right and left sides respectively. 
 Straight lines from m to 1, 2, - - , repre- 
 sent the resultants of the thrust at crown 
 and load down to joints 1, 2, - - - , in 
 magnitude. Their positions are found as 
 follows : draw a horizontal through ra, and 
 
(51 
 
 lay off on it the numbers in column C; 
 the first table referring to the left half of 
 the arch, the last table to the right half. 
 
 From the points so found draw verti- 
 cal lines to intersection with wo, pro- 
 duced if necessary, which thus give the 
 points where the inclined thrust at in is 
 to be combined with the weight from the 
 crown to any joint, to find the resultant 
 on that joint; whose intersection with it 
 is thus the centre of pressure for that 
 joint. 
 
 Thus the weight from the crown to 
 joint 8 on the left, acts 25' to left of m\ 
 lay off 25' on the horizontal through m, 
 then drop a vertical to intersection b with 
 mo; then draw ~ba \\ mS of force diagram 
 for left of arch, to find a the center of 
 pressure for joint 8. Similarly d and c 
 are found for joint 8 on the right. These 
 should be the first constructions made to 
 test the values of P and Q found, which 
 correspond to the line of resistance pass- 
 ing through a, m and c. 
 
 The line of resistance thus drawn pass- 
 es below the middle third of the arch ring 
 
62 
 
 on the unloaded side, the following 
 amounts in feet: at joints 2, 3, 4, ."> and <), 
 .o, .4, .3, .2 and .1 respectively; it then 
 crosses the arch ring, passes above the mid- 
 dle third about joint 12, and cuts the 
 springing joint 4.5 feet outside of the arch 
 ring. 
 
 On the loaded side it passes above the 
 middle third 0.1 at joints 4 and 5; then 
 crosses the centre line and is just tangent 
 to the lower middle third limit at joint 
 10, below which it again crosses the arch 
 ring and passes into the abutment, cutting- 
 joint 10 about 3 feet outside of the arch 
 ring. 
 
 JZxercise. Draw a line of resistance 
 for the part a m c regarded as a segrnen- 
 tal arch, through the upper middle third 
 limit at joint 8 on the left, the lower 
 limit at joint 8 on the right and 1,25 ft. 
 above the intrados at the crown. 
 
 We should find P= -23.8, Q = 44'.U 
 and the line of resistance everywhere keeps 
 within the middle third, barely touching 
 the lower limit at joint 2 on the left and 
 passing 0.10 ft. inside of the upper limit at 
 
63 
 
 joint o on the right and corresponding 
 nearly to the maximum and minimum of 
 tlie thrust in the limits of the middle 
 third (see Art. 20). The span of the arch 
 amc is 75.45 feet, the rise 17.2 ft. or 
 between % and V 5 of the span. 
 
 If for a moment we regard the unaided 
 semi-circular arch first considered; since 
 the line of resistance (or in fact any line of 
 resistance that can be drawn inside the 
 arch ring of the upper portion) passes 
 outside the arch ring at the abutments, the 
 arch will fall, the parts S-10 rotating out- 
 wards about joints K> and the crown 
 descending. But with spandrels built of 
 solid masonry up to about joints 8 (called 
 the joints of rupture), the parts 8-16 can 
 be regarded as almost immovable and the 
 part a m c can be approximately treated 
 as a segimental arch on fixed abutments. 
 
 However, as the higher the abutments 
 the more their tops will yield to a hori- 
 zontal thrust, the depth of arch ring 
 determined for the segimental arch a m c, 
 regarded as resting on immovable abut- 
 ments, should be slightly increased to 
 
64 
 
 allow for the slight horizontal spreading 
 at a and c. This spreading is due partly 
 to the elastic yielding of the abutment 
 from 8 down to the foundation and partly 
 to the closing up the joints of the rather 
 fresh mortar in the vertical joints of the 
 spandrels when the centres are struck. 
 
 Some constructors, especially the French 
 engineers, increase the depth of arch ring 
 from the crown to the abutments so that 
 the true line of resistance shall not leave 
 the middle third (or other limits) any- 
 where. This is, of course, the best way 
 to build a semi-circular or elliptical arch, 
 the abutments being built with joints 
 inclined (about at right angles to the 
 thrust) and in fact treated as a part of the 
 arch in finding the true line of resistance. 
 
 13. THE USUAL METHOD OF DRAWING 
 A LINE OF RESISTANCE. EQUILIBRIUM 
 POLYGON. 
 
 In Fig. 13, representing half an arch, 
 suppose the thrust S at the crown is known 
 in position, direction and magnitude and 
 that the weights P 1? P 2 and P 3 of the sue- 
 
cessive voussoirs 01, 12, '23 and loads, have 
 been found and laid off in position as 
 shown. From some convenient point O, 
 draw a line parallel to in n Q9 the direction 
 of the thrust S at the crown acting at m, 
 
 and to the scale of force make the length 
 
 00 equal to S. 
 
 From the point o thus found, draw a 
 vertical line and on it lay off successively 
 
 01 = ?!, 12 = P a and 23 = P 8 ; also connect 
 the points 1, 2 and 3 with O by straight 
 
lines. These lines are called rays, O is the 
 pole and the figure is known as the /*'>/ w 
 diagram. The rays Ol, O2, O3, represent 
 the resultants on the joints 1, ^', '), respect- 
 ively in magnitude and direction. To 
 find their position on the arch, produce 
 the thrust S at m to intersection n } with 
 P. ; from this point, draw a line parallel 
 to ray Ol of the force diagram, intersect- 
 ing joint 1 at <7; produce this line on to 
 intersection <1 with P 2 , at which point 
 draw a parallel to ray O2, intersecting 
 joint 2 at 1> ; again produce the last line on 
 to intersection f with P 2 , at which point, 
 draw a parallel to ray O3 to intersection c 
 with joint 3. The points </, &, c, are the 
 centres of pressure on joints 1, '2, 3 and 
 a broken line connecting in with a, a with 
 6 and & with c (dotted line) is the line of 
 resistance. This method of combining each 
 resultant thrust on a j oint, with the weight 
 of next voussoir and load to find the result- 
 ant on the next lower joint, is open to the 
 objection that any error made is carried 
 on, whereas, by previous methods, any 
 error made in construction is confined to 
 
Jie joint in question. The polygon m ;/, 
 If is called an equilibrium polygon, and 
 we note that it does not coincide with the line 
 of resistance; also that it passes through 
 the centres of resistance a and &, but does 
 not pass through c ; hence an equilibrium 
 polygon may pass out of the arch ring 
 at certain points and yet the centres of 
 resistance on the joints be found in the 
 arch ring; so that it cannot be used alone 
 in testing the stability of an arch ring, 
 particularly for semi-cicular arches, though 
 it is generally sufficiently near the truth 
 for flat arcs. 
 
 If we produce /' c/, which gives the 
 direction of the pressure on joint 2 to 
 intersection n 2 with in u l produced, we 
 have n 2 , a point in the vertical through 
 the centre of gravity of the part of the 
 arch from the crown to joint '2. This is 
 true, because we know that we must com- 
 bine S at the crown, with the weight of 
 arch from O to ^ (Pj+P 2 ) at such a point 
 on the line of action of S as to give a 
 resultant on joint ^ that will coincide in 
 position and direction with the former 
 
C8 
 
 resultant d /'. The only point that satisfies 
 this condition is n g which proves the state- 
 ment. Similarly/' c produced to inter- 
 section with m n^ at ?* 3 gives a point in 
 the vertical passing through the centre of 
 gravity of first three voussoirs and load. 
 
 Conversely, if by previous methods we 
 compute the horizontal distances of n,, n 2j 
 n 3 , from the crown and thus fix the points 
 n 1? n 2 , n s in position, and then find the 
 resultants on joints 1, "2 and 3 by drawing 
 lines from n 1? n g and n s parallel to rays Ol, 
 O2, O3, to intersections a, b and c with 
 joints 1, 2 and 3, these resultants produced 
 to intersection will form sides of the 
 equilibrium ^polygon m n l df. 
 
 This principle will be utilized in apply- 
 ing the theory of the solid arch further on. 
 
 (13a.) Sometimes the position of the 
 thrust at the crown is not given, but the 
 thrust at the lower joint 3 (ray O3) is given, 
 passing through c. In this case, P,, P 2 and 
 P 3 having been laid off in position and the 
 force diagram constructed, draw from c a 
 line parallel to ray O3, to intersection f 
 with P 3 ; then a line parallel to ray O2, 
 
from f to intersection d with P 2 ; next a 
 line from d parallel to ray Ol, to intersec- 
 tion n a with P 1 ; lastly, draw from u l a line 
 parallel to ray Go = S, to intersection in 
 with crown joint; giving thus the same 
 equilibrium polygon m 7^ 1 df as before, and 
 the same line of resistance m a b c deter- 
 mined as above. 
 
 14. SPECIAL PROPERTIES OF THE EQUILI- 
 BRIUM POLYGON. 
 
 Let Fig. 14 represent an arch, or a por- 
 tion of an arch, of any kind, loaded in any 
 manner, the joints through A and B being 
 any two joints whatsoever, and the joint 
 through I being any intermediate joint. 
 
 Call W the weight of arch and load in- 
 cluded between joints A and B, W and 
 W" the weights of arch and load included 
 between joints A and I and joints I and B 
 respectively, 1' = horizontal distance from 
 the vertical through the centre of gravity 
 of W to point I. Assume the thrust upon , 
 the joint at I as having the direction Z I 
 H and at the point H, where it intersects 
 the vertical through the centre of gravity 
 
70 
 
 of W, draw H C for an assumed direction 
 of the thrust upon the joint at A, the point 
 
 C lying upon the vertical through A and 
 either above or below it, or in fact coin- 
 ciding with it as a special case. 
 
71 
 
 In the force diagram below let the ver- 
 tical Q P, to any scale, represent W, Q 
 G= W", and G P = W; then on draw- 
 ing a line through P, parallel to H C, and a 
 line through G parallel to H I, their inter- 
 section gives the pole O; so that drawing 
 through Z a parallel to Q O to intersection 
 with the vertical through B, we establish 
 the point D. Connect C arid D by a straight 
 line, called the closing line, the vertical 
 through I meeting it at E and the perpen- 
 dicular let fall from I upon it meeting it at 
 X Next, through the pole O, draw a line 
 O M parallel to C I) to intersection M with 
 Q P and call the length O M = T and M 
 P V. These lengths represent the two 
 components of the resultant P O which 
 acts along the line C H in true position. 
 The two resultants, whose intensities are P 
 O and O Q, acting up along the lines C H 
 and D Z respectively, support the weight 
 of the arch, or with the components of the 
 weight of the arch, form a system in equil- 
 ibrium. 
 
 At the point C decompose the left reac- 
 tion into two components, one vertical 
 
acting from C towards S at a horizontal 
 distance a from the vertical through I, and 
 the other acting from C towards D. 
 
 The intensities of these two components 
 are given by the lines PM= V and OM = T 
 in the force diagram, though they are 
 represented in the figure above to a dimin- 
 ished scale. 
 
 Now if we suppose the part of the arch 
 to the right of the joint through I to be 
 removed and its action on the left part to 
 be replaced by the resultant pressure it 
 exerts at I ( = O G of force diagram) act- 
 ing to the le% the left part of the arch, 
 extending from joint I to A, is evidently in 
 equilibrium under the action of this force, 
 the. weight W of this part AI with load 
 and the two components V and T at C. 
 
 Hence taking moments about I of th/jse 
 forces in equilibrium, we have 
 
 Va-WT = T. IX. 
 
 In the force diagram, the line ON H, 
 the pole distance, is the horizontal distance 
 from O to PQ. 
 
 Now the moment T. IX is equal is II. 
 
73 
 
 IE; since, if at the point E we decompose 
 T (acting along CD) into vertical and hor- 
 izontal components, the moment of the 
 former about I is zero and the moment of 
 the latter =H. IE, which is thus equal to 
 the moment of T about I = T. IX, whence, 
 Va-WT=H. IE. 
 
 Similarly if we conceive drawn another 
 equilibrium polygon A K J Y B, with pole 
 O', passing through the points A, J and B 
 in the verticals through C, I and D respect- 
 ively, we should have 
 
 Va-WT = H' JF, 
 
 where V is the component AR directed 
 vertically, T' the one along AB, of the 
 left reaction acting from A towards K 
 arid H 1 is the new pole distance. 
 
 The resultant whose line of action is 
 KJY, is the resultant on the joint passing 
 through I or the previous joint considered, 
 J being simply a point on that resultant 
 in the vertical through I, but not on the' 
 joint, unless the latter is vertical. It fol- 
 lows that W is unaltered and it remains 
 to be proved that V=V. 
 
74 
 
 We have, from the first equilibrium poly- 
 gon the system of forces V and T acting at 
 C, the weight of arch and load W acting I to 
 left of vertical B D, and lastly the resultant 
 reaction at D, all together constituting a 
 system of forces in equilibrium. Hence 
 taking moments about D, we have (AL 
 being horizontal) 
 
 V. AL=W1. 
 
 Similarly, from the second equilibrium 
 polygon, we see that the forces V, T', W 
 and the reaction at B, constitute a system 
 of forces in equilibrium; hence taking 
 moments about B, we have 
 
 V'.AL=W1 
 
 since the right members are equal in the 
 last two equations, it follows that V=V ; 
 whence the left members of the two 
 equations preceding the last two are equal 
 and we have 
 
 IE 
 
 H. IE=H'.JF .. H' = H ^ 
 j.b 
 
 or the pole distances vary inversely as the 
 ordinates IE and JF from I and J to 
 the closing lines CD and AB. 
 
75 
 
 The above conclusions equally hold if 
 the weights W, W" are replaced by their 
 components, Representing the weights of 
 successive voussoirs and loads, and the 
 equilibrium polygon is drawn for the 
 entire system, since the pressures on joints 
 A, I and B are not altered by the decom- 
 position ; whence the previous formulas all 
 hold and the principles in question are es- 
 tablished as before. 
 
 The reader who is acquainted with the 
 principles of the equilibrium polygon will 
 recognize that C H Z I) is the equilibrium 
 polygon for the vertical forces V ( = PM 
 at C, W at H, W" at Z and a vertical 
 force = MQ in intensity, at B. 
 
 Also the formula above, 
 
 Va-W'l' = H. IE, 
 
 proves that the moment of all the vertical 
 forces to the left of a point I is exactly 
 equal to the pole distance H measured to 
 the scale of force, multiplied by the verti- 
 cal distance from I to the line CD (known 
 as the closing line) measured to the scale 
 of distance. 
 
76 
 
 Similarly since V'=V, the same moment 
 is equal to H'.JF as shown above or the 
 principle is generally true for any equili- 
 brium polygon drawn as above. 
 
 As in drawing a line of resistance it is 
 generally most convenient to know the 
 thrust on the vertical joint U at the crown 
 in position, direction and magnitude, the 
 method of finding it for the polygon whose 
 pole is O', will now be indicated. At 
 the intersection of JY witli the vertical 
 through the centre of gravity of the por- 
 tion of the arch included between joints 
 U and I with its load, combine the thrust 
 at J ( = G O' in force diagram), acting from 
 left to right, with the weight of the por- 
 tion considered =WG on force diagram, 
 giving the thrust at the crown = O'W ini 
 magnitude and direction. We find it in 
 position by drawing through the above 
 intersection a line parallel to O'W as 
 as shown by the small arrow. From this 
 thrust at the crown we draw the line of 
 resistance right and left of the crown in 
 the usual manner. 
 
 If the joint through I is the vertical 
 
77 
 
 crown joint, then KJY is at once the line 
 of action of the thrust there, its magnitude 
 being equal to O'G in the force diagram. 
 
 Where the joint through I is near A, it 
 may happen, particularly when the tan- 
 gent to the centre line of the arch ring 
 near A is nearly vertical, that the vertical 
 through the centre of gravity of W or the 
 weight from joint I to joint A with load, 
 lies to the left of A. 
 
 On constructing the figure, however, it 
 will be found that all of the previous 
 equations hold, so that the above con- 
 clusions are generally true. 
 
 The principle proved above, enables us 
 to draw an equilibrium polygon through 
 any three points, as A, J and B of an arch. 
 To do this take I (Fig. 14) on a vertical 
 through J, draw the joint through I and 
 find the values and positions of W and 
 W". 
 
 Then, as detailed above, draw the trial 
 equilibrium polygon C H I Z D correspond- 
 ing to pole O, the points C and D being 
 in the verticles through A and B. 
 
 On drawing from O, OM || DC to inter- 
 
78 
 
 section M with QP and MO' || AB, a dis- 
 tance to the right whose horizontal pro- 
 jection is 
 
 H-H IE 
 JF, 
 
 the new pole O' is established. 
 
 The equilibrium polygon A K J Y B can 
 now be directly drawn, beginning at A, J 
 or B at pleasure, as it must pass through 
 these points. Thus beginning at J, we 
 draw KY || O'G to intersections K and Y 
 with W and W"; then lines parallel to 
 PO' and O'Q through K and Y respective- 
 ly should pass through A and B. 
 
 15. j&xample. 
 
 In Fig. 15 is snown an arch of 75 feet 
 span, 15 feet rise and 7.5 feet depth of 
 keystone, the top of the backing rising to 
 2 feet above the top of arch ring. The 
 specific gravity of this backing is supposed 
 to be 0.8 of the masonry of the arch ring, 
 and the heights above the arch ring are 
 reduced to eight- tenths of the original 
 on both sides of the arch, though it is 
 
80 
 
 only shown on the rignt half to avoid 
 confusion of lines. Cooper's class " extra 
 heavy A " locomotive is placed over the 
 left half in the position shown. The 
 weight on each pair of wheels will be 
 supposed to bear on the length and width 
 of a cross tie, and to be transmitted verti- 
 cally downwards. If we regard the cross 
 ties as 8 feet in length, the weight per 
 foot of length for drivers, is 15 -r-- 8 short 
 tons which is equivalent in weight to 20.8 
 cubic feet of stone, weighing .07 ton or 
 140 Ibs. per cubic foot. This supposes the 
 ring stones to be made of stone of this 
 specific gravity which corresponds to good 
 sandstone masonry. 
 
 The 'equivalent for the pilot and tender 
 wheels are 14.3 and 10.1 cubic feet respect- 
 ively. These weights, resting on the 
 successsive voussoirs through the span- 
 drels, are given in column s, Table II 
 (upper numbers), their distances from the 
 crown are given in column c and their 
 moments about the crown in column m. 
 The lower numbers pertaining to any 
 t, in columns s and m refer to the un- 
 
81 
 
 loaded half and are found from Table I by 
 summing up the corresponding quantities 
 for any joint which are made out as ex- 
 plained in Art. 11. 
 
 It will be observed that the division of 
 the arch ring is different from that hitherto 
 used. As we shall use this division in 
 a subsequent article, from considerations 
 pertaining to the theory of the solid arch, 
 we shall state that the centre line of the 
 arch ring is divided into 32 equal 
 parts (2.77 ft. long each), and then the 
 voussoir joints are taken in succession two 
 divisions apart, except for the two vous- 
 soirs next the springing and the two next 
 the crown where only one division is taken. 
 This division of the arch ring is easily made 
 with dividers. The volume of each small 
 voussoir is 2.77x7.5 = 20.8 and the large 
 ones have a volume = 2 X 20.8 ~ 41.6 cubic 
 feet. The horizontal distances from the 
 centre of each voussoir (taken on centre 
 line) to the crown are the lowest numbers 
 of column c, Table 1. The columns S, }.: 
 and C in either table are made out as 
 usual (see Art. 11). 
 
TABLE I. RIGHT HALF. 
 
 Joints. 
 
 \v 
 
 v 
 
 s 
 
 c 
 
 111 
 
 8 
 
 M 
 
 C 
 
 9 
 
 2.93 
 
 2.77 
 
 1.6 
 
 7.5 
 
 4.7 
 
 20.8 
 
 1.46 
 1.38 
 
 29 
 
 25.5 
 
 36 
 
 1.41 
 
 10 
 
 5.84 
 5.54 
 
 1.82 
 
 7.5 
 
 10.6 
 41.6 
 
 5.83 
 5.48 
 
 62 
 
 228 
 
 77.7 
 
 326 
 
 4.19 
 
 11 
 
 5.8 
 
 2.48 
 
 14.4 
 41.6 
 
 11.68 
 10.97 
 
 168 
 456 
 
 133.7 
 
 950 
 
 7.11 
 
 12 
 
 5.62 
 
 3.6 
 
 20.2 
 41.6 
 
 37.38 
 36.32 
 
 351 
 
 679 
 
 195.5 
 
 1980 
 
 10.13 
 
 13 
 
 5.5 
 
 5.12 
 
 28.2 
 41.6 
 
 22.93 
 21.53 
 
 647 
 896 
 
 265.3 
 
 3523 
 
 13.29 
 
 14 
 
 5.22 
 
 7.07 
 
 36.9 
 41.6 
 
 28.30 
 26.57 
 
 1044 
 
 llo.> 
 
 343.8 
 
 5672 
 
 16.49 
 
 15 
 
 4.9 
 
 9.4 
 
 46.1 
 41.6 
 
 33.36 
 31.33 
 
 1538 
 1303 
 
 431.5 
 
 8513 
 
 19.75 
 
 16 
 
 4.fi3 
 
 12.1 
 
 56.0 
 41.6 
 
 38.10 
 35.88 
 
 2134 
 1493 
 
 529.1 
 
 12140 
 
 22.95 
 
 17 
 
 2.22 
 
 14.4 
 
 32.0 
 20.8 
 
 41.57 
 38.03 
 
 1330 
 7 ( .1 
 
 581.9 
 
 14261 
 
 24. 30 
 
 
 
 
 581.9 
 
 
 14261 
 
 
 
 
83 
 
 TABLE II. LEFT HALF. 
 
 s 
 
 
 
 c 
 
 m 
 
 S 
 
 M 
 
 C 
 
 8 25.5 
 
 
 36 
 
 25.5 
 
 36 
 
 1.41 
 
 14.3 
 
 7 52.2 
 
 4.15 
 
 59 
 290 
 
 92.0 
 
 385 
 
 4.19 
 
 2(5.8 
 6 5(i. 
 
 12.25 
 
 328 
 624 
 
 174.8 
 
 1337 
 
 7.65 
 
 26.8 
 
 o 61.8 
 
 18. 
 
 482 
 11)30 
 
 263.4 
 
 2849 
 
 10.82 
 
 26.8 
 
 4: 
 
 22.5 
 
 603 
 1543 
 
 360.0 
 
 4995 
 
 13.87 
 
 26.8 
 3 
 
 27. 
 
 724 
 2149 
 
 465.3 
 
 7868 
 
 16.90 
 
 16.1 
 
 J 87.7 
 
 34.1 
 
 549 
 2841 
 
 569 . 1 
 
 11258 
 
 19.78 
 
 16.1 
 1 H7.6 
 
 38.9 
 
 626 
 3627 
 
 682.8 
 
 15511 
 
 22.72 
 
 52.8 
 
 
 2121 
 
 735. C, 
 
 17632 
 
 23.97 
 
 735.6 
 
 
 17632 
 
 
 
 
 All the computations in these and sim- 
 ilar subsequent tables were made with a 
 in. inch slide rule, which ensures suf- 
 ficient accuracy with but small mental 
 wear and tear. 
 
84 
 
 If it is desired to pass a line of resistance 
 through the points A and B at the spring- 
 ing joints and through J on the 5th joint, 
 we first draw the load line C 870 ... o for 
 the left half of arch and lay off on it from 
 Table II, column S, 08=25, 07 = 'X>, 
 etc.; also from column C lay off the 
 successive distances, on a horizontal 
 (dotted) line through the crown, to the 
 verticals through the centres of gravity of 
 the loads from the crown to any joint. 
 Similarly C' '.>, . . . , 17 is the load line for 
 the right half and Y Z is a vertical through 
 its centre of gravity. 
 
 We now assume the thrust at the crown 
 to act along same line HZ and draw rays 
 CO and C'O' parallel to it; at the point H 
 where this line meets the vertical through 
 the centre of gravity of the left half of 
 the arch and load, draw H A. Then draw 
 a ray from o in left load line, parallel to 
 II A to intersection with ray C O at O, 
 thus fixing the pole O. Make C' O' = C O 
 to fix the right pole, since C O gives the 
 magnitude of the thrust at the crown. 
 
 The thrust at crown meets Y Z at Z 5 
 
85 
 
 from which point draw a parallel to ray 
 O' IT to intersection D with the vertical 
 through B. Connect A, D and A, B and 
 mark the points E and F where they are 
 intersected by a vertical through J. 
 
 At the point where the trial crown thrust 
 meets the vertical through the centre of 
 gravity of arch and load from crown to 
 joint 5, 10.^ ft. to left of crown (Table 
 II) draw a parallel to ray Oo to intersection 
 I with the vertical through J. 
 
 Through the trial pole O (as in Art. 14) 
 draw O M || A D to intersection M with 
 load line ; then draw M P || A B a distance 
 to the right whose horizontal projection is, 
 
 T K 
 
 H' = H As the distances I E and J 
 
 J r . 
 
 F are rather short, double them and lay 
 of along same line C R through C, C L = 
 '2 x JF, CH = '2 x IE. Then if Sis the 
 intersection of a horizontal line through C 
 and a vertical through O, connect L and S 
 and draw R Q parallel to L S to intersec- 
 tion Q with C S produced, which construc- 
 tion gives H' = C Q. A vertical through 
 Q to intersect M P at P gives the new pole 
 
86 
 
 P. At the right, draw ray C' P ' parallel 
 and equal to a ray from P to C (not drawn) 
 and we have the new pole P' at the right. 
 
 We have only now to draw through 
 J a line parallel to ray Po to intersection 
 T with the line of action of the weight 
 from the crown to joint 5, to fix a point T 
 in the line of action of the new thrust at 
 the crown. 
 
 Through T draw a line parallel to PC, 
 and from the intersections K and Y of this 
 line with the verticals through H and Z 
 draw lines parallel to rays Go and O'-IT 
 respectively, which lines should pass 
 through A and B if the work has been 
 done correctly. 
 
 If this does not obtain the error is per- 
 haps largely due to not taking off the 
 lengths IE and JF (with dividers, never 
 with scale) with sufficient accuracy. At 
 any rate, the construction must be repeated 
 if necessary until the line of resistance will 
 pass through the three points A, B and J. 
 
 The centre of resistance on any joint is 
 found in a similar manner to that already 
 given for the springing joint B. The 
 
87 
 
 broken line connecting the centres of 
 resistance on all the joints is shown by 
 the clotted line to pass near the centre line 
 throughout, which is due to the great 
 depth of arch ring chosen in this case for 
 the sake of a clear figure. 
 
 When the point J is on the crown 
 joint, take I to coincide with it and draw 
 the line of action HZ of the trial thrust at 
 the crown through I = J. The construc- 
 tion then proceeds as before, only the 
 final thrust at the crown, K Y, is now 
 drawn through 1= J parallel to ray PC. 
 
 The construction is thereby simplified 
 for this case. 
 
 The graphical methods given above of 
 determining completely the thrust at the 
 crown for a line of pressures passing 
 through any three points in the arch ring 
 will probably be preferred to the analytical 
 methods of Art. 10. 
 
 The preceding article indicates the entire construction 
 for the case represented by Fig. 15. Let the reader repeat 
 this construction on a scale of 3 or 4 feet to the inch and 
 compare with the numerical values of the components of 
 the thrust at the crown found by the formula method. 
 
 As another example, in the arch shown by Fig. 9 (Art. 
 10) use the quantities in the Tables of Arts. 8 and 9 and 
 
88 
 
 pass a line of pressures through the lower middle third 
 limit at the left springing and through the upper middle 
 third limits at joint 2 on the left and at the springing joint 
 on the right. With a scale of 30 feet to the inch it is 
 found that the thrust at the crown acts 0.76 foot below 
 the centre of the joint, its horizontal component being 
 ; 294 cu. ft., vertical component 18 cu. ft. 
 
 At joints 1 and 2 on the right the centres of resistance 
 fell 0.14 and 0.12 respectively below the middle third limit. 
 
89 
 
 CHAPTER III. 
 
 CURVES OF RESISTANCE CORRESPONDING 
 TO THE MAXIMUM AND MINIMUM HOR- 
 IZONTAL THRUST. UNIT STRESSES. 
 
 16. PROPERTIES OF CURVES OF RESIST- 
 ANCE. 
 
 In Fig. 16, e, c a represents an unsymmet- 
 rical arch, or an arch acted on by forces 
 not symmetrical, vertical or inclined. 
 
 Let P resultant of the external forces 
 
 acting on the arch between a and c, not 
 including the reaction R at a. Then on 
 combining R=a& with P, we get the 
 centre of pressure c on the joint cc^ 
 Similarly we could proceed for other 
 points, b 9 d, e, of the curve of resistance, 
 corresponding to the resultant R=.o&7 
 acting through a in the direction ak. 
 
90 
 
 Let a l b c l d e^ be a second curve, cor- 
 responding to the reaction R' at a^ Now 
 if S is such a force, acting towards the 
 left, that when combined with R, it gives 
 R' as a resultant, we can find a point c l9 
 on joint cc,, of the new curve of resistance, 
 either by combining R' with P as before, 
 or by combining its components with P : 
 thus call the resultant of R and P, T; this 
 combined at I with S, gives a resultant 
 which cuts joint c c^ at t\, a point lying 
 between Id and c, Id being in the direction 
 of s produced. 
 
 By this construction it is seen that the 
 new curve of resistance, corresponding to 
 the reaction R' at a', passes through b and 
 d, the points where k I intersects the first 
 curve of resistance; for other joints, as ee iy 
 the new curve lies nearer kl than the first 
 curve; since when S acts to the left, the 
 combination of T, for any joint, with S, 
 gives a resultant acting between T and S, 
 which therefore cuts the joint nearer kl 
 than the first center of pressure. 
 
 The above supposes that neither R nor 
 S are vertical, but that both act to the 
 
91 
 
 left, whence the horizontal component of 
 R' exceeds that of R. The joints are, 
 moreover, not supposed inclined more than 
 !K) from the vertical, counting from the 
 top. 
 
 17. Prop. If two curves of resistance 
 cut each other, the curve which lies nearest 
 the straight line, which joins their common 
 points, corresponds to the greatest hori- 
 zontal thrust. 
 
 We have seen in the preceding article 
 that the two curves can only intersect on 
 the straight line k I (Fig. 10) as implied in 
 the proposition. 
 
 Now if, at any joint c c } , the centre of 
 pressure c l9 corresponding to the curve 
 a l b c l de l9 lies nearer kl, the straight line 
 joining b and d, than the curve abcde, 
 then we may suppose a force S, acting in 
 the direction kl, to be combined with T 
 at I, to effect it. The force S, thus found, 
 must therefore, when combined with R at 
 a give R'; since R and S produce, the 
 same effect as R'; so that all points of the 
 first curve can be found by combining R 
 with the resultant of the force P, up to 
 
the joint, and afterwards combining their 
 resultant with S. 
 
 The force S, acting to the left, increases 
 the horizontal component of the resultants 
 on each joint; hence the curve a^bc^de^ 
 corresponds to a greater horizontal thrust 
 than the curve abed e, as stated in the 
 proposition. 
 
 If the arch is symmetrical, the curves of 
 pressure are symmetrical with respect to 
 the crown, whence kl must be horizontal. 
 
 18. (1). If a cure e of resistance has two 
 points common to the intrados and an in- 
 termediate point common to the extrados, 
 it corresponds to the minimum horizontal 
 thrust. 
 
 For, suppose the curve a b c d e^ Fig. 16, 
 touches the extrados near c, the intrados 
 on both sides nearer the abutments. 
 
 Then any other curve of resistance, 
 a l b c l d e^ that remains in the arch ring, 
 must cut the first, only in points on the 
 straight line kl, joining any two points 
 of intersection. 
 
 Now the new curve, near the points of 
 contact of the first curve with the con- 
 
tour curves of the arch ring, must, if it 
 remains in the arch ring, pass nearer kl 
 than the first curve, whence, by Prop. 
 Art. 17, the first curve corresponds to a 
 less horizontal thrust. Q.E.D. 
 
 (2). If (f < <f <rre of resistance, a b c dc, 
 Fig. 17, has two points of' contact, b and 
 '7, with the extrados, and <m intermediate 
 point of contact c -with the intrados, it 
 corresponds to a iidninnwn horizontal 
 thrust, if b c d y in the vicinity of c, lies 
 
 between the intrados and the straight 
 line bd, 
 
 For any other curve, lying in the arch 
 ring, as the dotted curve, must lie nearer 
 the straight line kl, joining their points 
 of intersection, than the first, in the vicinity 
 of b c and d, and thus corresponds to a 
 greater horizontal thrust. This case of 
 
94 
 
 the minimum is rarely or never found 
 in practice. 
 
 (3). If) however , the intrados, in the 
 vicinity of c lies between the curve b c d, 
 Fig. 18, and the straight line bd> the 
 curve corresponds to <i imurinnuu hori- 
 zontal thrust ; since this curve lies nearer 
 Id than any other, as the dotted curve of 
 resistance. 
 
 It is seen that Id in Fig. 18 lies above 
 bd, whereas the reverse occurs in Fig. 17. 
 
 When a curve of resistance possesses 
 both the properties of the maximum and 
 minimum of the thrust, the arch is at the 
 limit of stability; as see all the figures 
 relating to the experiments in the Appen- 
 dix. The above principles were first stated 
 by Dr. Herman Schetfier. 
 
 I'.*. When the arch is symmetrical (and 
 symmetrically loaded), the curves of resist- 
 
95 
 
 ance are symmetrical with respect to the 
 vertical through the crown; hence for a 
 half arch and load we can state the follow- 
 ing propositions : 
 
 (1). When the point of contact with 
 the extrados is higher than the point of 
 contact with the intrados, the curve of 
 resistance corresponds to the minimum 
 horizontal thrust. 
 
 (:i). When the point of contact with 
 the extrados is lower than the point of con- 
 tact with the intr ados, the curve of resist- 
 ance corresponds to the maximum horizon- 
 tal thrust. It generally touches the intra- 
 dos at the crown joint. 
 
 The curve corresponding to the minimum 
 thrust generally touches the extrados at 
 the crown for usual loads and depths of 
 arch ring; but for thin arch rings with 
 little or no surcharge above the crown, 
 especially with gothic arches, the curve of 
 resistance passes below the crown and 
 touches the extrados at some lower point. 
 This likewise may happen, even for seg- 
 mental arches of usual depth, when a heavy 
 load is placed over the middle of either 
 
96 
 
 haunch, as shown in Art. 9 for the curves 
 drawn corresponding to the minimum hori- 
 zontal thrust within the middle third 
 limits. 
 
 For segmental arches, the lower point 
 of contact for either maximum or minimum 
 thrust is generally at the springing joint. 
 
 :><). The conclusions above hold equally 
 when we wish to find the maximum or 
 
 minimum thrust for curves contained with- 
 in the middle third or any other limits, 
 only we substitute the upper and lower 
 limiting curves for extrados and intrados 
 in the enunciations. 
 
 In Fig. 19 the dotted line represents a 
 curve of resistance corresponding to the 
 maximum and minimum of the thrust at 
 the same time, within the limits shown. 
 
 The part a b c corresponds to the max. and 
 
97 
 
 bed to the min. of the thrust within those 
 limits; for b lies above a straight line 
 drawn from a to c, Art. 18 (3), and c lies 
 between b and d, Art. 18 (1). 
 
 In an arch by itself, if but one curve of 
 resistance can be drawn within its contour 
 curves, thus corresponding at once to the 
 maximum and minimum thrust, the arch 
 is at the limit of stability. 
 
 UNIT STRESSES AT ANY POINT OF A JOINT. 
 
 21. In Figure 3 of Article 5, the result- 
 ants of the molecular stresses on the joints 
 ab, a t b 1? a 2 b 2 , are Q, R 15 R 2 , respectively. 
 To find these stresses at any point of a 
 joint as a 2 b 2 . The resultant R 2 on the 
 joint a., b 2 meets it at A 2 , whose distance 
 from the nearest edge ( a 2 A 2 in this case) we 
 shall call d. The joint is rectangular in 
 shape ; its width perpendicular to the plane 
 of the paper being unity, and its radial 
 length a 2 b 2 we shall call h. 
 
 As R 2 is generally inclined to the nor- 
 mal to the joint, resolve it at A 2 into two 
 components, the first, which call R, being 
 normal to the joint and the other acting 
 
98 
 
 parallel with the joint. The last compo- 
 nent is a shearing force and undoubtedly 
 lessens the resistance of the joint, but 
 because its influence is difficult to estimate 
 it is generally neglected. 
 
 It will first be assumed that the mortar 
 in the joint can withstand both tension and 
 compression, in which case R 2 can fall out- 
 side the joint a 2 b 2 (Fig. :>), a distance d 
 
 ^r* B 
 
 r/. kt 
 
 ""FE^ 
 
 > 
 
 
 
 ^ ( C7 
 
 ' 
 
 
 / i P* 
 
 + Pv A - R 
 
 
 I : 
 
 
 f 
 
 1 
 
 e 
 
 iH r ^C^ 
 
 
 ! 
 
 , 
 
 D 
 
 
 
 
 
 d 
 
 
 R _ 
 
 . i 
 
 measured along the line of the joint pro- 
 duced, and still stability be assured if the 
 mortar possesses sufficient strength. Simi- 
 larly for concrete arches. In Fig. *?<>, let 
 B D represent the joint a 2 b 2 of Fig. o and 
 at the point where the resultant on this 
 joint cuts the joint or the joint produced, 
 resolve it into shearing and normal compo- 
 
99 
 
 nents ; neglecting the former, we have the 
 normal component R, acting through a, a 
 distance d -f- J h from the centre E of the 
 joint (Fig. 20). 
 
 At the point E conceive two opposed 
 forces -{-R, R, equal and parallel to R to 
 act. This does not destroy equilibrium. 
 To avoid confusion + R and R are drawn 
 through A, a point in the normal to BD at 
 E, but it is understood that A is supposed 
 to coincide with E. The force R with the 
 force R forms a right-handed couple RR, 
 that can be replaced by the equal couple 
 pp or the forces, ,'.... c', c, equal and 
 opposed to the uniformly increasing ten- 
 sible resistances from E to B and the 
 compressive resistances from E to D, E 
 lying in the centre of gravity of the cross- 
 section. 
 
 From the theory of flexure, calling t = c 
 the stress per square unit at the extreme 
 edge B or D, we have the moment of the 
 couple RR = M = R (d + % h) = V 6 ch 2 ; 
 whence, 
 
 
100 
 
 The remaining force + R at E (A), acting 
 at the centre of gravity of the joint corres- 
 ponds to a uniform compression, 
 
 h 
 
 over the whole joint (shown by the little 
 arrows to the right, though really acting 
 along B D). 
 
 As the forces given by (1) and (2) act 
 at the same time, their algebraic sum or 
 " effect " (see figure) gives the actual 
 stresses along joint B D, which are thus 
 seen to be uniformly increasing from a 
 neutral axis, not passing through E. The 
 tension at B = t rand compression at D 
 
 As in voussoir arches the resultant rarely 
 or never passes outside the arch ring, let 
 us suppose hereafter that it cuts the joint 
 to which it refers, in its interior a distance 
 d from the nearest edge ; then we have, 
 
 _6R(^h-d) 
 
 h 2 
 R 
 
101 
 whence, 
 
 ,-,=,(,--).,... 
 .+,=,(,-)....< 
 
 As this theory supposes that the limit of 
 elasticity has been nowhere exceeded, the 
 stretch or shortening of the "fibres" is 
 proportional to the stress and therefore to 
 the distance from the neutral axis; hence 
 
 a plane section before strain remains a 
 plane section after strain as in the ordinary 
 theory of beams. 
 
 It is seen from (3) that t r is positive, 
 or tension is exerted at B, for d < K h ; 
 for d == K h, t r = o or there is no 
 
 stress at B and the stress at D = c -(- r = 2 
 
 T> 
 
 r- or double the average (see Fig. 21)* 
 
102 
 
 lastly, for d > K h, the stress on the joint is 
 compressive throughout (Fig. 22). 
 
 Hence when the resultant cuts the joint 
 within its middle third there are only com- 
 pressive forces exerted on the joint; when 
 it passes outside the middle third, tensile 
 
 Fig. 22 
 
 forces are brought into play if the mortar 
 can supply them. 
 
 In the last case, if the mortar cannot 
 resist tensile forces, the normal component 
 of the resultant will be decomposed into 
 stresses, over a length of joint 3d, pro- 
 
103 
 
 portional to the ordinates of a triangle and 
 the part of the joint beyond this length 
 (od) will open. 
 
 This is plain, because we have just found 
 that for d = } 3 h, that od h, was under 
 compression, the stress at the farthest edge' 
 from the resultant being nothing. 
 
 The stress at the most compressed edge 
 is likewise double the average on the part 
 of the joint under compression .*. it is 
 
 :3 - '- for all cases where the resultant lies 
 -} d 
 
 in an outer third of the arch ring and 
 the mortar cannot resist tension. 
 
 The last formula and formula (4) suffice 
 to give the maximum compression per 
 square unit for the cases to which they 
 refer, where the voussoirs are each in one 
 block. In case the arch ring is made of 
 several rolls as in brickwork or when we 
 meet with joints transverse to the radial 
 joints as we proceed along the latter, the 
 above theory is only approximately true. 
 
 Example 1. In an arch of 5 feet radial length of joint 
 at the springing, the resultant has a normal component of 
 73.36 tons and it acts 1 ft. below the centre of the joint. 
 
104: 
 
 (1). What is the stress at the intrados if the mortar 
 cannot resist tension ? 
 
 Answer, " xS ',''-= 32.6 tons pr. sq. ft. 
 
 (2). What is the stress at the intrados and extrados 
 if the mortar can resist tension? 
 
 Here B=73.3G. h=5ft... d =1.5 ft.; hence by eqs. (:J) 
 and (4), 
 
 Stress at extrados=2/ r i~^iH)!5i= 2.0. 
 V 5/5 
 
 Stress at intrados =2(2 ' J ^-^ ) = 32. 2 
 tons per square foot. 
 
 Example 2. In the example above, if the resultant acts 
 0.5 ft. above the centre of the joint, find the unit stivsst-s 
 at the intrados and extrados. 
 
 X'2. Iii the theory of the solid arch, 
 which will be presently referred to, it is 
 necessary to know the moment M of the 
 resultant on any joint with respect to the 
 centre of that joint. This may be ex- 
 pressed in three different ways: first, by 
 the product of the resultant by the perpen- 
 dicular from the centre of the joint upon 
 it; second, by R ( }., h d ) as above, since 
 the shearing component has no moment, 
 and lastly by multiplying the horizontal 
 component of the resultant on any joint by 
 the vertical distance from the centre of 
 the joint to where a vertical through this 
 
105 
 
 centre meets the resultant. This is plain, 
 since at the intersection of the vertical 
 with the resultant decompose the latter 
 into vertical and horizontal components. 
 The moment of the former about the 
 centre of the joint is zero; hence the 1 
 moment of the latter (the constant hori- 
 zontal thrust of the arch ) is equal to that 
 of the resultant itself. 
 
 i-). METHOD OF FAILURE OF ARCHES. 
 
 In the appendix will be found an ac- 
 count of a number of experiments on small 
 wooden arches at the limit of stability 
 with their corresponding resistance lines, 
 which, of course, correspond to the maxi- 
 mum and minimum thrust at the same 
 time, within certain limits. (See art. '2 '.) 
 
 In the fourth experiment, irith a yield- 
 UKJ pier, the top of the pier and the 
 haunches of the arch moved outwards and 
 the crown descended. In this case the 
 limiting line of resistance (for the slightly 
 deformed arch) touches the extrados at the 
 crown, the intrados at the haunches and 
 
106 
 
 c at the outside edges of the piers. This is the 
 method of failure of any symmetrical arch 
 witli yielding abutments. 
 
 With rigid abutments the method of 
 failure for gothic or sequential arches, with 
 a load at the crown is given by experi- 
 ments one and eleven, and for an eccentric- 
 load by experiments sixteen and seventeen. 
 The figures show the lines of resistance-. 
 The .arches rotated about those edges 
 \vhich the lines of resistance approached 
 nearest. As the arches were slightly de- 
 formed at the instant before rupture, if the 
 resistance lines had been drawn for the de- 
 formed arch, they would necessarily have 
 passed through the edges of the joints, as 
 no crushing there was experienced for 
 these very light wooden arches. 
 
 When arches having rigid abutments 
 are built with too thin an arch ring, it is 
 found that the arch fails by the crown 
 rising and the haunches falling inward. 
 TJie line of the centres of pressure passes 
 through the intrados at the crown and 
 
 o 
 
 abutment joints and touches the extrados 
 ;it the haunches, the segments of the arch 
 
107 
 
 rotating about these edges; the upper 
 segment turns upwards about the extrados 
 edge at the haunches and the lower 
 segment falls inwardly, rotating about the 
 springs. The spandrels, if any, must 
 crack over the springs, haunches and 
 crown; the lower segments fall inwardly 
 into the void between the abutments and 
 are speedily followed by the upper part 
 with the spandrels. 
 
 There are doubtless some arches in exist- 
 
 | ence now, perilously near the kind of 
 
 I rupture just outlined. Woodbury, in his 
 treatise on the ar,ch, has devoted consider- 
 
 ' able attention to this latter mode of failure. 
 He supposes the arch unloaded, the span- 
 
 sdrel filling to be of the same density as the 
 voussoirs and to be carried up level with 
 
 I the crown and the arch ring to be uniform 
 
 mi thickness. He found that a semi-circu- 
 lar arch whose depth was 1 / 53 of the span, 
 
 'could just contain one line of resistance 
 and no more within the limits of the arch 
 ring down to the lower joints of rupture, 
 and that when the span was 2 ( J times the 
 
 (depth of key, only one line of resistance 
 
108 
 
 could be inscribed within the middle third 
 drawn to the lower joints of rupture. The 
 lower joints of rupture made angles with 
 the vertical of 07 in the first case and 
 about fJ'-i in the second. 
 
 If we call s = span, h = rise, and k=. 
 depth key, the following table gives 
 Woodbury's limits of k in terms of the 
 span for segmented arches of various 
 ratios of Tito*; the first set of values 
 k -=- s giving the ratios of depth of key to 
 span which permits only one line of resist- 
 ance to be drawn within the limits of the 
 arch ring; the second set of values, the 
 
 h 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 s 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 10 
 
 
 
 
 
 
 
 1 
 
 k 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 
 
 8 
 
 47 
 
 50 
 
 M 
 
 00 
 
 65 
 
 78 
 
 k 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 s 
 
 25 
 
 Jf> 
 
 21 
 
 30 
 
 33 
 
 39 
 
 ratios of k to s in order that only one line 
 of resistance can be inscribed within the 
 middle third of the arch ring. If the 
 spandrel is carried above the crown, these 
 
109 
 
 ratios will become less; but, if after the 
 centres are struck, the spandrels are 
 brought to a level with the top of key- 
 stone, the last ratios should certainly never 
 become less or joints will open. In fact, if 
 these values are attained the construction 
 for the solid arch will give a line of resist- 
 ance passing slightly outside of the middle 
 third and thus bringing tensile stresses on 
 fresh mortar at some of the joints. Proper- 
 ly, the spandrels should be built up pro- 
 gressively from key to abutment, so that 
 the height at the key is attained before 
 that at the abutment. As it will be well 
 for the reader to test some of these values, 
 it may be mentioned that for the first set 
 of values above, the line of resistance 
 touched the intrados at the crown, the ex- 
 trados at the haunches, and the intrados at 
 the springing. The curves limited to the 
 middle third touched the lower limit at 
 the crown and springing and the upper 
 limit about the haunches. 
 
 Exercise. When the span is 100 feet, 
 rise 20 feet, and depth of key 4 feet 
 ( V 25 space), and the spandrel rises to a 
 
110 
 
 level from the top of key, construct the 
 single line of resistance. 
 
 The method of failure of segmental 
 arches with rigid abutments and an eccen- 
 tric load over the haunch of the left half 
 may be illustrated by a reference to fig. 
 19. Conceive the arch ring to diminish in 
 depth so that finally but one curve of re- 
 sistance can be drawn therein. It will be 
 found to touch the extrados to the left of 
 the crown and at the right springing 
 joint; it will touch the intrados at the left 
 springing joint and a little to the right 
 of the crown. 
 
 In a large arch, crushing at the edges is 
 experienced before this minimum depth of 
 arch ring is attained. In any case the 
 arch will sink at the joints under the load, 
 the joints at the intrados opening, whereas 
 the arch ring rises a little to the right of 
 the crown, since the pressure there is near- 
 ly all concentrated at the lower edge. At 
 the left springing, the lower part of the 
 arch rotates downwards about the lower 
 edge and at the right springing about the 
 upper edge of the joint. As a consequence, 
 
Ill 
 
 the arch divides into three parts; the left 
 part falling inwards, the middle portion 
 rising at the right but falling at the left 
 end and the right segment rotating up- 
 wards about the upper edge of the right 
 springing joint. 
 
 In any kind of an arch, loaded in any 
 manner, the method of failure is easily ar- 
 rived at by simply studying the line of 
 resistance pertaining to the case. 
 
112 
 
 CHAPTER IV. 
 
 LENE OF RESISTANCE DETERMINED AS IN 
 A SOLID ARCH. METHOD OF ISOLATED 
 LOADS FOR SEGMENTAL ARCHES. COM- 
 PUTATION OF DEPTH OF KEYSTONE. 
 
 '^4. A great many approximate solutions 
 have been proposed for the voussoir arch, 
 but none satisfactory. The true line of re- 
 sistance in an arch depends primarily upon 
 its elasticity, and likewise upon the care 
 with which the stones are cut and fitted, 
 the thickness and yielding of the mortar 
 joints, the settlement and time of striking 
 of the centres if the mortar joints have not 
 hardened, and finally the yielding of the 
 piers or abutments. So many of these in- 
 fluences cannot be exactly estimated that 
 the author has hesitated about applying 
 the theory of the solid arch " fixed at the 
 ends" to the voussoir arch, particularly on 
 account of its complexity, though 111 Van 
 Nostrand's Magazine for January and No- 
 vember, 1871), he claimed that the theory 
 
113 
 
 of the solid arch was the most exact solu- 
 tion for the cases assumed, and a graphical 
 treatment was given in the last named ar- 
 ticle, to which reference will be made fur- 
 ther on. 
 
 From Prof. Swain's article in Van Nos- 
 trand's Magazine for October, 1880, it is 
 to be inferred that the application of the 
 theory of elasticity to the stone arch had 
 already been considered by a few authors 
 mentioned. In 187!) Winkler published 
 his notable theorem (given in the article 
 last mentioned) ; also Castigliano applied 
 the theory of the solid arch, after the 
 method of " least work," to stone arches. In 
 the same year Prof. Greene, of the Uni- 
 versity of Michigan, published a more 
 practical treatment, founded on the ana- 
 lytical theory of the circular arch, using 
 the method of isolated loads. 
 
 In this chapter a method similar to Prof. 
 Greene's is used, the tables, however, be- 
 ing obtained by aid of Winkler's tables, for 
 the solid arch " fixed at the ends." The 
 xp<r/i of the centre line of the arch ring 
 was divided into equal parts, and the quan- 
 
114 
 
 titles in the tables found for isolated loads 
 at the points of division by interpolation 
 (by aid of diagrams to a large scale), from 
 Winkler's constants, which refer to equal 
 angular divisions, though a few direct com- 
 putations were made for loads very near 
 the abutment. Winkler's theory and tables 
 for the solid arch are given in Du Bois's 
 " Graphical Statics," and the preliminary 
 computations of c x and c 2 were made out as 
 explained in " Theory of feolid and Braced 
 Arches" by the writer, p. 00, the deforma- 
 tion of the arch ring, due to bending mo- 
 ments, being alone considered. The gen- 
 eral table given in this chapter refers only 
 to arches whose rise is one fifth of the span. 
 The method of equal horizontal divisions 
 adopted here offers great practical advan- 
 tages, and enables one with small labor, 
 comparatively, to investigate the strength 
 and stability of a given stone arch. The 
 position of the lice load causing maximum 
 departures of the centres of pressure from 
 the centres of the joints, is more accurately 
 ascertained than hitherto, and some unex- 
 pected conclusions were established. 
 
115 
 
 25. The theory of the solid arch " fixed 
 at the ends," is strictly applicable to a 
 solid arch of stone, iron or other material 
 perfectly fitted, when not under stress, to 
 rigid abutments, the theory requiring three 
 conditions to be satisfied, viz.: (1) the tan- 
 gents to the centre line at the springs are 
 fixed in position, (2) span invariable, and 
 (o) the vertical displacement of one spring- 
 ing above the other equals zero; all having 
 reference to the deformation of the arch 
 ring due to the stresses caused by its own 
 weight and any load that may be ap- 
 plied. 
 
 The theory is evidently rigidly applica- 
 ble to a voussoir arch, with no mortar in 
 the joints, provided the voussoirs are cut so 
 perfectly that the arch fits accurately be- 
 tween the rigid abutments when not under 
 stress. When thin cement mortar is used 
 in the joints and allowed to harden before 
 the centres are struck, the conditions are 
 but little altered ; but for bad fitting stones 
 and thick mortar joints, not sufficiently 
 hardened, neither this theory nor any other 
 
116 
 
 is exactly applicable, though it may be re- 
 garded as some guide in the relative di- 
 mensioning of arches of various spans. 
 
 It must be carefully observed, too, that, 
 if in a voussoir arch, without mortar, fitting 
 perfectly when not under stress between 
 the skewbacks, the centre of pressure on 
 any joint, as determined by the theory of 
 the solid arch, falls without the middle 
 third, the joint will only bear on a length 
 equal to three times the distance from the 
 centre of pressure to the nearest edge of 
 the joint (Art. 21). In this case only the 
 bearing surface of the joint must be inclu- 
 ded in the formulas for fixing the resist- 
 ance line. Hence a new determination has 
 to be made on this basis and so on, until 
 the assumed and computed bearing joints 
 agree. This method, although possible, is 
 very tedious, so that the theory is only a 
 practicable one when the arch ring is 
 such that the true line of the centres of 
 pressure lies everywhere within the middle 
 third. 
 
 Only such arches are examined in this 
 work, besides space does not permit the 
 
117 
 
 consideration of any arch rings except 
 those of uniform cross-section. 
 
 20. In Fig. *23 9 the circular curve, of 
 rise equal to one-fifth of its chord or span, 
 represents the centre line of an arch ring 
 of constant section. The span is 5 units 
 in length, the rise one unit. Each half 
 span is divided into 10 equal parts and 
 vertical lines drawn through the points of 
 division. Where the successive lines cut 
 the curve will be designated as points <>, 
 1, '2, :3, . . . , 10 and 1', >', .r, . . . 10' 
 for the left and right halves respectively, 
 the left springing point being called lo, 
 the right springing 10', the crown and 
 the numbers increasing regularly along 
 the arc from the crown to the two spring- 
 joints respectively. 
 
 If a single load W is placed on the arch, 
 supposed to be without weight, its equili- 
 brium polygon will consist of two straight 
 lines. When W is directly over a point 
 of the arch previously fixed, as 4, the 
 table gives at once c^ y and c a in terms of 
 h = rise of centre line of arch considered, 
 where c l = vertical distance from point 10 
 
118 
 
119 
 
 on arc to side of equilibrium polygon, c 2 
 the same for point 10' on arc and y ver- 
 tical distance above the crown to apex D 
 of equilibrium polygon. 
 
 Thus if W acts at 4 on arch, the rise of 
 whose centre line (above the centre of the 
 springing joints) is 10, the span 50, we 
 have from table, that the resultant at left 
 abutment acts .10 X 10 or l.G below 
 centre of joint, the resultant at right 
 springing, acts -f-. 290 X 10 or 2.90 above 
 centre of joint, and the apex D lies .21.0 
 X 10 = 2. 10 above the highest point of 
 centre line of arch ring, plus coefficients 
 corresponding to points above the arc and 
 minus below, as just indicated. On laying 
 off W on a vertical line just to left of arch 
 as shown, and drawing lines from the ex- 
 tremities parallel to the sides of the equi- 
 librium polygon passing through D, the 
 intersection gives the pole of the force 
 diagram. The length of the horizontal 
 line from the pole to the line representing 
 W, gives H = horizontal thrust due to W 
 alone, and the line divides W into the two 
 vertical components of the reactions V t 
 
120 
 
 tf 
 
 PQ 
 
 W 
 
 II 
 
 a" 
 
 w 
 
 +-f M 1 ! II 
 
 - x x Ci c; H 
 
 Ct 71 *- l^ CC b- X t^ tr: i 
 ' < ^C O t ^ r: Tl C : 
 COiCOOa>^HQOOCOb"' 
 
 7i cc -f ^i r: o i~ O O : 
 
 4 I f I I 
 
121 
 
 and V 2 at left and right springings re- 
 spectively. For W = 1 (to a large scale) 
 we can thus find graphically the co- 
 efficients in columns H, V, and Y 2 , but it 
 is better to compute them from easily 
 derived formulas: 
 
 V,::^H ;.V,= ^H; 
 
 X '> - X 
 
 H = wWAD + Bm 
 \ x 5 x/ 
 
 We have here, A D = 1 -{- y (cj, 
 
 paying attention to the sign of c,. Since 
 x is known, H can be at once computed 
 and afterwards Vj and V 2 . The coefficients 
 thus found are to be multiplied by W in 
 any application, as indicated in the table. 
 The moment about any point of the 
 centre line of the arch ring, for any weight 
 W is equal to the H corresponding, multi- 
 plied by the vertical distance from the 
 point to the equilibrium polygon corres- 
 ponding to W (see Art. 22). The alge- 
 braic sum of such moments due to any 
 number of weights, gives the total moment 
 
122 
 
 at the point and the sum of the IPs gives 
 the total horizontal thrust clue to the 
 weights. If the moment at the point, due 
 to the weight of the arch is found and 
 added to the preceding moment and its H 
 likewise added to the previous sum of the 
 IPs, then the quotient of the total mo- 
 ment divided by the total II, gives the 
 vertical distance from the point on the 
 centre line of the arch ring to the equili- 
 brium polygon corresponding to the weight 
 of arch and loading considered. This is 
 the method to be used in fixing any 
 point of the equilibrium polygon after 
 the theory of the solid arch " fixed at the 
 ends". When the moment is plus, the 
 equilibrium polygon is above the centre 
 line of the arch ring at the point; when 
 minus, below. 
 
 For convenience, the values of M, = 
 Hc 1 and M 2 = Hc 2 ( the moments at the 
 left and right springing joints) as well as 
 Mj-j-Mg are given in the general table. 
 The coefficient for a weight at the crown 
 ( point of arc) is only half of M t -f- M 2 
 for reasons that will appear directly. 
 
123 
 
 Let fig. 24 represent any arch whose 
 rise is one-fifth the span; the rise of ite 
 centre line is also one-fifth its chord from 
 centre to centre of springing joints. Di- 
 vide the half chord of the centre line into 
 ten equal parts and erect verticals at the 
 points of division. Reduce the length of 
 
124 
 
 the part of these verticals comprised be- 
 tween the extrados of the arch and the 
 horizontal roadway line to eight tenths of 
 each. This is best done graphically. 
 Thus if e g is laid off equal to 10 and ef 
 equal to 8 to any scale, and we lay off any 
 length from e along e g and from its ex 
 tremity draw a parallel to g f to intersec- 
 tion with 6/', the distance from this inter- 
 section to e gives the reduced length. The 
 material above the arch ring to the reduced 
 contour can then be regarded as weighing 
 the same per cubic foot (.07 ton) as the 
 arch ring. On drawing dotted verticles 
 half way between the first verticals, the 
 area of the trapezoids comprised between 
 any two successive dotted verticals will be 
 equal to the width multiplied by the length 
 of the full vertical between them, and it 
 will be regarded as a force acting along 
 the medial (or full) vertical. In fact, for 
 a length of arch equal to unity, this area 
 is also the volume of a prism having for a 
 base this area, and by multiplying by .07 it 
 can be reduced to tons. 
 
 In all the computations below the por- 
 
125 
 
 tion of the arch from either springing joint 
 to the first dotted vertical will be neglected, 
 as its influence is very small in fixing the 
 true equilibrium curve. The weight at the 
 crown is that due to the portion comprised 
 between the adjacent dotted verticals on 
 either side. This division of the arch is 
 different from that hitherto used. 
 
 27. Let us proceed now to the consider- 
 ation of an arch of 100 ft. span. The rise 
 is 20 ft. and the depth of the key 5 ft., the 
 horizontal roadway rising 2 ft. over the 
 crown. The first column in the table be- 
 low gives the joint of the arch at which 
 the weight is concentrated. The " depth" 
 of a " trapezoid " ( column 2) multiplied 
 by the constant " width " 5.2 ( column 3 ) 
 gives the area = volume == W, expressed 
 in cubic feet. We have only to multiply 
 these values, as well as those given in col- 
 umns H and M, by .07 to reduce to tons 
 when desired. 
 
 The coefficients of columns H and M, 
 -f- M 2 are taken from the general table 
 above. On multiplying these by the suc- 
 cessive values of W we get the horizontal 
 
126 
 
 thrusts and moments given in columns H 
 and M t of the table. Any thrust in col- 
 umn H is that due to the load at the point 
 corresponding, hence the sum gives the 
 total thrust due to loads 0, 1, 2, - - - , !. 
 On adding to this the same sum, less that 
 due to the load at the crown, we get the 
 thrust due to the weight of the entire arch 
 = 51*4.0 cubic feet, since the loads at equal 
 distances from the crown are the same. 
 Similarly M, for load at -f- M t for load 
 at O 1 , is the same as Mj for load at 
 <> + M 2 for load at 6 = ( M 4 + M 2 ) 
 
 Table for 100 ft. span, 20 ft. rise, 5 ft. depth of Key. 
 
 "S 
 o 
 
 "ft 
 
 <& 
 
 g 
 
 3 
 
 W 
 
 H 
 Coeff 
 
 H 
 
 M^Mo 
 Coeffs." 
 
 MI 
 
 PL| 
 
 ft 
 
 
 
 
 
 
 
 
 
 
 
 cu. ft. 
 
 
 cu. ft. 
 
 
 
 
 
 6.6 
 
 5.2 
 
 34.3 
 
 1.1639 
 
 39.92 
 
 -f-. 1769 
 
 + 6.07 
 
 1 
 
 6.8 
 
 
 35.4 
 
 1.1512 
 
 40.75 
 
 + .3489 
 
 + 12.35 
 
 2 
 
 7.2 
 
 
 37.4 
 
 1.0864 
 
 40.63 
 
 + .2944 
 
 + 11 01 
 
 3 
 
 8. 
 
 
 41.6 
 
 .9807 
 
 40.80 
 
 + .2089 
 
 + 8.69 
 
 4 
 
 9.1 
 
 
 47.3 
 
 .8473 
 
 40.08 
 
 + .1152 
 
 + 5.45 
 
 5 
 
 10.5 
 
 
 54.6 
 
 .6867 
 
 37.49 
 
 + .0075 
 
 + 0.41 
 
 6 
 
 12.2 
 
 
 63.4 
 
 .5098 
 
 32.32 
 
 .0973 
 
 6.17 
 
 7 
 
 14.4 
 
 
 74.9 
 
 .3327 
 
 24.92 
 
 .1780 
 
 13.33 
 
 8 
 
 17.2 
 
 
 89.4 
 
 .1706 
 
 15.25 
 
 .2143 
 
 19.16 
 
 9 
 
 20.3 
 
 
 105.6 
 
 .0497 
 
 5.25 
 
 .1688 
 
 17.83 
 
 583.9 317.41 12.51 
 
 549.6 277.49 20 
 
 Weight of arch = 11 33. 5 H=594.9 M 1= 250.2 
 
127 
 
 for load at 6. The same principle holds 
 for any two loads at equal distances from 
 the crown. The coefficient for the load at 
 the crown was not doubled, as there is no 
 other load corresponding to it. 
 
 On adding up the figures in the last 
 column and multiplying by h = 2() 1 we 
 find the total moment 250.2, and on 
 dividing this by the total H= 504.0 we 
 find that the resultant at the left springing 
 passes 0.42 foot below the centre of the 
 joint. The same holds at the right spring- 
 ing on account of symmetry. The equili- 
 brium polygon can now be drawn, as the 
 vertical component = : J weight of arch 
 500.75 and H = 51)4.0 are given as well as 
 Cj= .42. Construct the force diagram by 
 drawing Hand from its left extremity ; layoff 
 successively on a vertical downwards half 
 the load at crown and the loads at 1, 2, 
 - - - , (see fig. 24). We draw the equil- 
 ibrium polygon from a point .42 below the 
 centre of the left springing joint, as ex- 
 plained in Art. (13a). It is very near the 
 centre line and is shown approximately in 
 fig. 24. Its vertical distance from points 
 
128 
 
 l>, o, and o' on the arc are respectively 
 +.27, +.2, .25 and 1 ; so that multiply- 
 ing by H = 5 ( J5, the moment at these 
 points are -f- 101, -\- 120, 141) and 00 
 respectively. We have previously seen 
 (Art. 13) that the resultant along in n, say 
 of fig. 24, is strictly that pertaining to the 
 joint between m and n at a, the true re- 
 sistance curce passing slightly below rn\ 
 still for purposes of comparison below it 
 is near enough to consider the line m n 
 to represent the line of action of the result- 
 ant acting on the joint passing through 
 the point where the vertical through m 
 cuts the centre line, particularly as we shall 
 find that the maximum departure of the 
 line of resistance from the centre of the 
 joints, when the live load is considered, is 
 nearly always at a springing joint where 
 no error is made. Further, as the result- 
 ants on the upper joints are nearly perpen- 
 dicular to them for usual loads the inter- 
 section of a perpendicular from m on the 
 joint corresponding can be regarded as 
 the centre of pressure, for purposes of 
 comparison below. 
 
We are now prepared to consider the ad- 
 ditional influence of the live load, which in 
 all the subsequent examples in this chapter 
 will be taken as a locomotive load of Oooo 
 pounds per foot of track, 20 feet in length 
 or slightly greater or less, corresponding to 
 the horizontal divisions of the arch, 
 followed by a tender load of 2400 Ibs. per 
 ft. 30 ft. long, about, and this followed by 
 another locomotive load as before. This 
 about corresponds to Cooper's class extra 
 heavy A, without the pilot wheel. For 
 cross ties S ft. long, these loads for a slice 
 of the arch I foot thick, are 750 and 30o 
 Ibs. per ft. respectively. As the horizontal 
 divisions of the arch are 5.2 ft. each, the 
 locomotive load on each division = 750 x 
 5.2 Ibs., or the weight of 27.8 cubic ft. of 
 stone weighing 140 Ibs. per cubic ft.; the 
 tender load on each division is 11.1 cubic 
 ft. stone. The first locomotive will be as- 
 sumed to cover 4 divisions of 5.2 ft. each 
 or 2o.8 ft. ; the tender 6 divisions or 31.2 ft. 
 As M and H both vary, for any point as 
 we shift the live load, it is only by trial 
 
130 
 
 that the maximum value of M^-H=c, 
 can be found for the point considered. 
 
 Thus consider point 5 of centre line of 
 arch ring where, for dead load only we 
 have found M = -f-120 and H = 5 ( ,>4. ( ,. 
 On a large scale drawing similar to fig. 23 
 (except that loads at all points 1, 2, 3 - - - 
 are considered), on measuring the vertical 
 ordinates from point 5 of arc to the equili- 
 brium polygons corresponding to weights 
 W=27.8 at 4, 5, 0, 7, and adding the results 
 ( = .778) we have the total moment due 
 to locomotive load at 4, 5, 6, 7, = .778 X h 
 W = .778 X 20 X 27.8 =: -f 432.0. That 
 due to tender load at 8, 9, is similarly = 
 .061 X 20 X 11.1 = + 13.5 ; adding moment 
 = -\- 120 due to dead load, the total 
 moment at point 5 = -f- 500.1. The hori- 
 zontal thrust due to loads 27. 8 each at 4, 5 ? 
 0, 7, is found by adding the coefficients in 
 column H of general table for points 4, 5, 
 0, 7, and multiplying by 27.8 .-. 2.37G5 X 
 27.8 = 00.1. Similarly for tender load at 
 points 8, t), the thrust is .2203x11.1 = 2.4. 
 Adding the thrust due to weight of arch, 
 -VJ4.1) to the sum of these two and we have 
 
131 
 
 the total horizontal thrust = 663.4. On 
 dividing-}- 5 6 6.1 by this, we find that 
 the equilibrium polygon due to dead load, 
 locomotive load at 4, 5, (>, 7, and tender 
 load at 8, 9, passes 0.89 foot above the 
 centre of the joint through point 5 on arc, 
 whence on drawing a normal to the joint 
 through the point found, the resist- 
 ant is found to pass 0.8 above 
 centra measured along the joint. The 
 live load may now be moved to right 
 or left one or more divisions, but for no 
 other position is the resultant on joint 5 so 
 far from the centre as is found by a com- 
 putation similar to the above. A similar 
 investigation for point 6 with locomotive 
 loads at 4, 5, 0, 7 and tender loads at 8, 9, 
 gives M = +520, H = W>4, so that c = + 
 .7S or less than for point 5. From fig. 23 
 it is evident that for points 5 or 6 the live 
 load should not extend as far as point 2 
 from the left, as the moment due to a load 
 at 2 or to the right is negative. Trial 
 shows that for a maximum c at 6 the load 
 should not extend to o, but from 10 to 4 in- 
 clusive as just given. 
 
132 
 
 At the crown the max. departure is 
 caused by loads, say from 4 to 10 and 4 1 to 
 lo 1 ; but it is not so great as elsewhere, and 
 as the loading is unusual it will not be 
 further considered. 
 
 At point 3 1 of arc, for loc. loads at o, 1, 
 2, 3, tender at 4 to 9 inclusive, c - 3I'i) 
 H- 74(1 = . 5 or less than for any other 
 point hitherto examined. This value will 
 doubtless be increased slightly by moving 
 live load one or two divisions to the left. 
 
 Consider next the right springing joint. 
 M 2 and consequently c 2 , for dead load is 
 minus. This obtains for spans of about 
 35 ft. and upwards, for rise = l / 5 span, so 
 that the live load to left of crown giving a 
 positive moment at K) 1 acts against the 
 dead load; hence we should not expect to 
 find c 2 for such spans as great as c x at the 
 left springing, where the moments due to 
 both live and dead loads have the same 
 (minus) sign, but only trial can determine. 
 For spans less than about 35 ft., max. c 2 , 
 may be greater than max. c 13 as in fact was 
 found to be the case for an arch of 25 ft. 
 span. 
 
133 
 
 For this 100 feet span, M 2 -f- H was com- 
 puted for the front of the loc. load at 2', 1', 
 and 1 successively, and found to be great- 
 est when the loc. loads were at 0, 1, 2, 3 
 and tender loads at 4, 5, G, 7, 8, 1). The 
 computation proceeds as before, the sum of 
 the coefficients in columns M 2 and H of 
 the general table, for the points above be- 
 ing multiplied by hW and W respectively, 
 thus : 
 
 -f .923 X 20 X 27.8 = -f 513.2 
 + .853 X 20 x 11.1 = + 189.4 
 Dead Load M't = 250.2 
 
 Total, M 2 = + 452,4 
 
 4.382 X 27.8 = 121.8 
 2.597 X 11.1 = 28.8 
 Dead load thrust = 594.9 
 
 Total, H = 745.5 
 
 c 2 == + 452,4 - 745.5 = + .61 
 
 The computation for max. c t proceeds 
 after the same principle. The maximum 
 G I corresponds to loc. loads at 5, 0, 7, 8 and 
 tender load at 9. The moment coefficients 
 are taken from column M, in general table. 
 
134 
 
 - 1.0648 X 20 X 27.8 = 592.0 
 
 - .1888 X 20 X 11.1 =- 41.9 
 
 250.2 
 
 Total, M, = - 884.1 
 
 1.6998 X 27.8 = 47.25 
 .0497 X 11.1 = .55 
 504.90 
 
 Total, H = 642.70 
 c t = -884.1 -=- 642.7 = 1.370, or the re- 
 sultant passes 1.38 ft. below the centre of 
 the left springing joint for this position of 
 the live load. 
 
 To draw the left resultant in position 
 we next compute the vertical component 
 V a . Add up the coefficients in column V,, 
 of general table for points 5, 6, 7, 8 and 
 multiply by 27.8; also multiply (for tender 
 load at point 9) .9918 by 11.1; the sum 
 added to the half weight of arch gives the 
 total V x 
 
 3.0295 X 27.8= 100.S 
 
 .9918 X 11.1 = 11.0 
 
 % weight arch = 5 00. 7 
 
 Total, V, = 078.5 
 
135 
 
 From the point, 1.38 ft. below the centre 
 of the left springing joint, lay off verti- 
 cally upwards V\ = 678.5, to any scale, and 
 from the upper extremity of this line draw 
 a horizontal equal in length (to the scale 
 of VJ to total H = 042.7 to fix the pole 
 of the force diagram. A line from the 
 pole to the lower extremity of Y t gives 
 the magnitude and direction of the result- 
 ant on the left springing joint. It cuts 
 this joint .95 ft. below its centre For ac- 
 curacy the left springing joint should be 
 drawn to a large scale (anywhere along 
 the radius), and this construction made to 
 the large scale as this is found to be the 
 joint where the centre of pressure is far- 
 thest from the centre. 
 
 As the resultant passes 0.13 ft. outside 
 of the middle third the depth of key must 
 be increased. From this and some other 
 examples it was found that if the depth of 
 key was increased by 3 to 4 times the 
 departure, the line of resistance for the new 
 arch would lie inside the middle third lim- 
 it, just touching it at the critical joint, the 
 left springing in this instance. 
 
In this case the depth of key was in- 
 creased by 4 X 0.13 = 0.52 ft., or say 0.5 
 ft., so that the new key was 5.5 feet. A 
 new construction and computation for this 
 arch showed that the resultant on the left 
 springing joint passed .02 ft. inside the 
 lower middle third limit or practically 
 touched it. 
 
 If desired, the equilibrium polygon for 
 the entire arch can now be drawn for the 
 last loading considered. As before we com- 
 pute V t , H, c t , and it is best to compute 
 c 2 as a check on the construction which can 
 be made as explained in Art. loa. It is 
 much better, however, to find by computa- 
 tion the position of the centre of gravity 
 of the left half of arch and load c^nd deter- 
 mine by construction at once the centre of 
 pressure at the crown joint, from which 
 points the polygon can be drawn more ac- 
 curately towards either abutment. 
 
 28. On investigating arches of various 
 spans in the manner indicated above, it 
 was found that the springing joints were 
 the only ones necessary to examine, except 
 in the case of a 12.5 ft. span, 2.5 ft. rise 
 
137 
 
 and 2.2 ft. depth of key, where a single 
 concentrated load of 40,000 pounds over 
 point 6 was found to cause the resultant 
 on the joint through to reach the upper 
 middle third lin it. The load in no other 
 position gave as great a departure on any 
 other joint. In this case the values of c, 
 V, and H were computed, and from the 
 force diagram resulting the true direction 
 of the resultant on joint in position and 
 magnitude was obtained. 
 
 In any arch, the resultant on the critical 
 joint having been found in position and 
 magnitude, the normal component can be 
 scaled off and the maximum intensity of 
 stress at the most compressed edge found 
 as in Art. 21. 
 
 The following table gives the final results 
 of a series of constructions to determine 
 the depth of key for stone arches of rise 
 one-fifth the span, so that the line of 
 resistance should everywhere be contained 
 within the middle third of the arch ring 
 of uniform cross- section and just touch it 
 at the critical joints. 
 
 The arch ring was supposed to weigh 
 
138 
 
 140 Ibs. per cubic foot, the material above 
 it 112 Ibs. per cu. ft., and the loading 
 as given above, viz. : 6,000 Ibs. per foot of 
 track locomotive load for about 20 feet 
 (depending on the length of the horizontal 
 divisions of the arch), followed by a tender 
 load of 2,400 Ibs. per foot on about 30 feet, 
 and this followed by a second locomotive 
 and tender of the same weights, with the 
 one exception of the 12.5 ft. span, where a 
 load of 40,000 Ibs. on two drivers was 
 alone assumed. These loads were supposed 
 to bear equally on 8 feet cross ties and to 
 be transmitted vertically to the arch. 
 
 The actual lengths of locomotive and 
 tender loads measured along the rails for 
 the different spans was as follows : 
 
 Span. 
 
 Length loc. load. 
 
 Do. tender load. 
 
 25, 
 
 12.00, 
 
 none, 
 
 50, 
 
 15.60, 
 
 none, 
 
 75, 
 
 19.50, 
 
 none, 
 
 100, 
 
 20.80, 
 
 5.2, 
 
 125, 
 
 19.38, 
 
 12.92, 
 
 150, 
 
 23.25, 
 
 15.50. 
 
 It is only in the case of the 150 ft. span 
 
139 
 
 that the locomotive length appreciably 
 exceeded 20 ft., but the weight of arch 
 was so great, compared with that of the 
 load, that the error in finding the depth 
 of key was small. 
 
 TABLE GIVING THEORETICAL DEPTH OF KEY. 
 
 1 
 
 Rise 
 
 Key 
 
 Loo 1 Load 
 at 
 
 Tender 
 at 
 
 Maximum 
 Intensity 
 of Stress 
 
 Feet 
 
 Feet 
 
 Feet 
 
 
 
 Tons pr. eq. ft. 
 
 12.5 
 
 2.5 
 
 2.2 
 
 6 
 
 none 
 
 
 25. 
 
 5.0 
 
 2.G 
 
 0.1. ...9 
 
 * 
 
 9. 
 
 50. 
 
 10.0 
 
 3.5 
 
 4. ...9 
 
 " 
 
 14. 
 
 7n. 
 
 15.0 
 
 4.5 
 
 5 .... 9 
 
 * 
 
 22. 
 
 100. 
 
 20.0 
 
 5.5 
 
 5 .... 8 
 
 9 
 
 25. 
 
 125. 
 
 25.0 
 
 6.25 
 
 5. 6. 7 
 
 8,9 
 
 HO 
 
 150. 
 
 30.0 
 
 7. 5, 6. 7 
 
 8. 9 
 
 36. 
 
 As mentioned above, the joint of rupture 
 where the departure of the resistance line 
 from the centre of joint was greatest, was 
 found to be the left springing, except for 
 the 25 ft. span, where it was the right 
 springing and the 12.5 ft. span, where 6 
 was the critical joint. 
 
 The above values are plotted to scale ' 
 Fig. 25, being shown by the small circles. 
 A line through these circles is nearly 
 straight, but not sufficiently so for accu- 
 racy. The following table gives these 
 
and interpolated values for every 5 feet 
 for use of constructors: 
 
 DEPTH Or KEY !N FEZT 
 
 ~T 
 
 o,|~Z 
 
141 
 
 DEPTH OF KEY FOB ARCH OF UNIFORM SECTION AND RISE 
 \ SPAN, ALL DIMENSIONS BEING IN FEET. 
 
 Span 
 
 Key 
 
 Span 
 
 Key 
 
 Span 
 
 Key 
 
 5 
 
 1.96 
 
 55 
 
 3.7 
 
 110 
 
 5.80 
 
 10 
 
 2.12 
 
 60 
 
 3.9 
 
 115 
 
 5.95 
 
 12.5 
 
 2.20 
 
 65 
 
 4.1 
 
 120 
 
 6.10 
 
 15 
 
 2.27 
 
 70 
 
 4.3 
 
 125 
 
 6.25 
 
 20 
 
 2.43 
 
 76 
 
 4.5 
 
 130 
 
 6.40 
 
 25 
 
 2.60 
 
 80 
 
 4.7 
 
 135 
 
 6.55 
 
 30 
 
 2.77 
 
 85 
 
 4.9 
 
 140 
 
 6.70 
 
 35 2.95 
 
 90 
 
 5.1 
 
 145 
 
 6.85 
 
 40 3.13 
 
 95 
 
 5.3 
 
 150 
 
 7.00 
 
 45 3.30 
 
 100 
 
 5.5 
 
 155 
 
 7.15 
 
 50 tt.5<) 
 
 106 
 
 5.65 
 
 160 
 
 7.30 
 
 On fig. 25 are plotted for comparison the 
 depths of key purposed by Dejardin (line 
 D), Scheffler (line S, by interpolation from 
 his tables for rise = V. 4 and */ 6 span), Croi- 
 zette-Demoyers (line C D) and Trautwine 
 (line T) (see Art. 31). 
 
 The depths of key, as computed, are in 
 excess of most of the values given, all of 
 which refer to materials of only average 
 strength (second class masonry for the 
 Trautwine line). This excess was to be 
 expected, for most of the old formulas are 
 founded on the successful practice of the 
 past, and cannot therefore be expected to 
 give results corresponding to the very much 
 heavier locomotive loads of to-day, though 
 
142 
 
 'some of them may be a rude sort of a guide 
 in the design of common road bridges. 
 
 The French authors quoted, after find- 
 ing the depth of key (as plotted above), 
 then increase the radial length of joint to- 
 wards the abutment, making it vary as the 
 secant of the inclination to the vertical. 
 
 The above table is for an avch of uni- 
 form section. The same material can be 
 better disposed, perhaps, by making the 
 depth of keyless and increasing the length 
 of joint as we approach the abutment; but 
 the theoretical treatment of this case falls 
 under that of the arch of variable cross- 
 section, and it has to be omitted for want 
 of space. 
 
 The table gives the depth of key for im- 
 movable piers or abutments and arch 
 stones that fit perfectly between the skew- 
 backs, when laid on the centres (without 
 mortar) and supposed not under stress. 
 Where the abutments or pier* are yielding^ 
 either from not having a rock foundation 
 or from too small a width (particularly 
 in a series of arches), or where mortar 
 in the joints is used which is not hard 
 
143 
 
 when the centres are struck, or in any case 
 when the mortar joints are not thin and 
 hard, so that the arch cannot be regarded 
 as practically homogeneous throughout, 
 then an increase should be given to the 
 depths above, according to the best judg- 
 ment of the engineer. 
 
 The effect of temperature on the arch 
 
 been omitted though its effects are 
 
 large (see Science Series, No. 48). The 
 
 lynamic effect of the load is supposed to 
 
 In' taken up by the spandrels. 
 
 The arch ring has been supposed above 
 TO be of sandstone and weigh 140 Ibs. per 
 ruble foot. If it weighed 100 Ibs. the 
 same depth of key would correspond to 
 live loads 8 / 7 of those assumed, or a less 
 depth of key would suffice for same loads. 
 
 ^.). MAXIMUM INTENSITY OF STRESS AL- 
 LOWABLE IN STONE AND BRICK ARCHES. 
 
 The average pressure on a joint is equal 
 to the normal thrust divided by the area 
 of the joint, and this reaches a maximum 
 in existing bridges (according to Scheffler, 
 
144 
 
 for masonry weighing 150 Ibs. per cu. ft.) 
 from 17 tons per square ft. at the crown to 
 62 tons per square ft. at the springing. In 
 such arches the resultants on the joints acl 
 outside the middle third; but even if they 
 acted at the middle third limits, the inten- 
 sity of pressure at the most compressed 
 edge would be double the above or .34 and 
 124 tons per square ft. at crown and 
 springing respectively. Scheffler recom- 
 mends not exceeding average pressures at 
 the crown and springing, corresponding to 
 columns of the same material as the vous- 
 soirs 207 and 308 ft. high, or say 15.5 to *^> 
 tons per square ft. for stone weighing 150 
 Ibs. per cubic ft. 
 
 This rule seerns safe (for cement joints, not 
 common mortar) and corresponds for stone 
 weighing 150 Ibs. per cu. ft. or fair lime- 
 stone, to intensities of pressure at the most 
 compressed edges of 31 tons per sq. ft. at 
 the crown and 40 tons at the springing. 
 
 30. Existing structures that have done good service, :IH 
 well as arches which have failed, afford the data from 
 which the many empirical formulas for depth of keystone 
 have been derived. These formulas are not based on theory 
 but on successful practice and are valuable in their way, 
 
145 
 
 but very unsatisfactory iii some respects. Thus they differ 
 very greatly in their results, some giving double the depth 
 of Keystone for certain spans as others, and besides they 
 rarely make a distinction between a common road bridge 
 and one intended for the heaviest modern locomotives to 
 pass over at great speed. 
 
 In fact these formulas generally represent the practice 
 of the past, mainly for light road bridges (with a few excep- 
 tions) and can serve a useful purpose in the design of such 
 bridges; but most of them are evidently inadequate for the 
 heavy moving loads of to-day on railroads. 
 
 As preliminary to writing some of the best known of 
 the formulas, it will be convenient to give simple formulas 
 for expressing the radius in terms of the span, for ease of 
 reduction, as most of the formulas are expressed in terms 
 of the radius. 
 
 If we call s =span in feet, h = rise in feet of a circular 
 arch, r being the radius of the intrados, we have the w -11- 
 kuown relative 
 
 _ h 1 29 
 
 For = , r s 
 s 5 40 
 
 , -s 
 
 I, " 1.8 . 
 
 io 
 
 I'l 24 
 
 31. The following formulas, by French engineers, for k = 
 depth of keystone in feet, are taken from DuBosque ("Pouts 
 et Viaducts en Maconnerie") after reducing to English 
 
146 
 
 equivalents. They are intended to apply to best bricks or 
 to stone not so hard as granite, stone of "medium resist- 
 ance" as BuBosque styles it. 
 
 Perronefs formula is intended for every kind of arch, 
 semi-circular, segmental, elliptical or basket handle, and is 
 as follows : 
 
 k = l 4 .0347s .... (2). 
 
 A more recent author, DC jar din, gives the following for 
 circular arches : 
 
 = , k = l 4 0.1 r = l .'i.-, 
 
 =4- k=l+.05r=l J> 
 
 _=_ ,k = 14.035r=14 .037s 
 
 s s 
 
 = , k = 1 4 .02 r = 1 4 .026 s | 
 s 10 
 
 and for ei ptical or basket handled arches 
 
 =i,k = 14 .07 r .... (4) 
 s 3 
 
 in which r equals the radius of curvature at the crow r n. 
 
 Another French authority, M. Croizette Desnoyers, gives 
 the following for segmenta arches, the first also applying 
 to semi-circular arches. 
 
 v t\ ^JL, 97 \/ ). r 
 6 
 
 = 0.5 4. 27 \/ 2 r. 
 
 I 
 
 , , I 
 
 ,5 4 .253 V '2 r =.54 .327 V s, | 
 
 .5 
 
 .235\/2r = .54.342v/s, \ ....(5) 
 
 = , k =.5 4. 217 v 7 2 r : 
 s 10 
 
 .5 4 .35 V/S, 
 
147 
 
 He likewise adapts the first formula of (5) to elliptical 
 or false elliptical arches of small rise, by considering r to 
 represent the radius in feet of an arc of a circle of same 
 span and rise. 
 
 Dupuit gives a much smaller depth of key by the fol- 
 lowing formulas : 
 
 = , k = o.:j v 
 
 s 2 
 
 h 1 > " (fi) 
 
 < ,k =0.27 v/ B= V .073 s | 
 
 J 
 
 These call for good granite laid with care. 
 
 To all the above formulas must be added .02 //, u-Jiere H 
 is the height of the surcharge above the crown, reduced if 
 necessary to the density of earth. 
 
 This is plainly a very ru It: and i u-x.kot way of allowing 
 for an extra surcharge of earth, M:r- i ;v moleraie amount, 
 must add materially to the slahilit'/, the io;;d being fixed and 
 symmetrical with respect to the crown and thus giving a 
 line of resistance much nearer the centre line of the arch 
 ring than an eccentric rolling load, and not calling for an 
 extra section on account of the dynamic effect of the live 
 load. A:tually a smaller ar /h ring can be used up to a * 
 certain height with the same security, especially consider- 
 ing that the active (or passive) pressure of the earth around 
 the arch almost ensures stability (when crushing is not to 
 be feared) even for thin arch riu^s. 
 
 In any case the arch should be examined, first leaving 
 out the horizontal pressure of the earth, which generally 
 only adds to the stability, and afterwards considering it. 
 
 After finding the depth of keystone by preceding 
 formulas, the Europeans generally increase the (radial) 
 depth of arch ring from the crown to the springing. If 
 we call d the angle that any joint makes with the vertical, 
 and I the radial length of any joint for seymental archcx, 
 the following formula for this length is frequently used, 
 
 1 = k sec d (7). 
 
 Du Bosque prefers to find the radial length of the joint 
 
148 
 
 at the springing joints and crown by (7) and other formu- 
 las above, and draw an arc of circle through the upper 
 ends of those joints to define the extrados. 
 
 For semi-circular, elliptical or basket handled arches 
 the rule is to measure up from the springing line, half the 
 rise and draw a horizontal to intersection with the intrados ; 
 the joints drawn there normal to the intrados called " the 
 joints of rupture," must have a length equal to the depth 
 of keystone multiplied by a coefficient, which is 
 
 2 for semi-circles 
 
 1.8 " ellipses, &c., rise = span. 
 
 1.4 " - = 1 .. 
 5 
 
 As before, a circle is drawn through the upper ends of 
 the " joints of rupture," and the crown joint for the ex- 
 trados down to the joint of rupture and there tangents are 
 drawn to this circle to limit the masonry down to the 
 abutment.* 
 
 The above lengths of joints at crown and elsewhere are 
 in excess of average English and American practice, which 
 may be partly due to the poorer qualities of stone found in 
 France. We shall now give+some English and American 
 formulas. 
 
 Rankings formulas. Find the longest radius of curva- 
 turer of the arch ; then the depth of key for a single arch, 
 including tunnel arches in rock or conglomerate, is 
 
 k= y 0.12 r ( 8 )- 
 
 For an arch of a series the coefficient of r under the 
 radical should be 0.17; fora tunnel arch in gravel or firm 
 earth 0.27, and in wet clay or quicksand 0.48. The defect 
 
 * See Van Nostrand's Magazine for December, 1883, for 
 illustrations in article by E. Sherman Gould, C. ~E. 
 
149 
 
 in this formula is the lack of a constant term to give a 
 proper depth of key for arches of small span. It gives 
 smaller values even than Dupuit's formula for the very 
 best materials. 
 
 Trautwine's formula, for first-class masonry, 
 
 k =0.2 + 14 Vr-f0.5s .... (9) 
 
 has the constant term, but is wrong in principle, in that 
 for the same span the depth of key increases with r or as 
 the rise diminishes ; whereas the flatter the arch, for the 
 same span, the less should be the key (where crushing is not 
 in question) since a line of resistance can be more easily 
 inscribed within the same limits in a flat arch than in one 
 of greater rise when the key is the same. This last defect 
 characterizes all the formulas given above, but those of 
 Dejardin and Dupuit. 
 
 Trautwine increases the depth given above >e for second 
 class masonry, and about % for brick on fair rubble. 
 
150 
 
 CHAPTER V. 
 
 PRINCIPLES AFFECTING SOLID ARCHES 
 FIXED AT THE ENDS. LEMMA. 
 
 SECOND GENERAL METHOD OF LOCATING 
 THE TRUE LINE OF RESISTANCE. 
 
 32 Fundamental Equations of Solid 
 Arches "fixed at the ends" 
 
 Space forbids deducing the fundamental 
 equations of solid arches, but the reader is 
 referred to the author's " Theory of Solid 
 and Braced Elastic Arches," pages 21 to 
 3G, for a simple development of the theory. 
 The following are the three equations 
 which must be satisfied in order that a 
 line of resistance may be the true one: 
 
El 
 
 151 
 
 (1), 
 
 My - 
 KI =0 
 
 The first indicates that the end tangents 
 to the centre line of the arch ring are fixed 
 in direction; the second, that the deflec- 
 tion of one end of the arch below the 
 other is zero ; and the third, that the span 
 is invariable. 
 
 The centre line of the arch ring is sup- 
 posed divided into a great number of parts, 
 each equal to AS; M represents the mo- 
 ment of the resultant about the centre of 
 the joint traversing the middle of the 
 corresponding AS, E is the modulus of 
 elasticity for the corresponding voussoir 
 AS long, and I represents the moment of 
 inertia of a plane joint traversing the 
 centre of AS, about a horizontal axis 
 passing through the centre of the joint. 
 The origin of co-ordinates is taken at the 
 centre of the left springing joint; x is the 
 
152 
 
 horizontal and y the vertical distance from 
 this origin to the centre of the correspond- 
 ing AS. The summation extends over the 
 entire arch ring. 
 
 If we call H the uniform horizontal 
 thrust of the arch for vertical loading, and 
 v the vertical distance from the centre of 
 any joint traversing the middle of AS, to 
 the resultant acting on that joint, we have 
 by Art. 22, M = Hv. 
 
 As by the graphical method it would be 
 impracticable to divide the centre line of 
 the arch ring into a very great number of 
 parts, we must content ourselves with divid- 
 ing it into a certain number of parts of 
 appreciable length and find the M, E, I, x 
 and y for the middle of each part, which 
 gives a fairly good average and is suffici- 
 ently correct in practice. 
 
 If we regard the modulus E as constant 
 throughout the arch ring, it may be 
 droppe-i from the equations. 
 
 Similarly, replacing M by Hv, H may 
 be dropped as well as AS. 
 
153 
 
 Therefore, for an arch ring of constant 
 cross-section where I is constant, the three 
 conditions (1), (2) ard (3), reduce very 
 simply to 
 
 2(v) = ---- (4), 
 2 (vx) = .... (5), 
 .... (li). 
 
 In the case of the voussoir arch, if the 
 curve of the centres of pressure, as deter- 
 mined in position by the above equations 
 in a manner to be shown, lies everywhere 
 in the middle third of the arch ring, there 
 will be no tension exerted on any joint, so 
 that the theory of the solid arch exactly 
 applies when there is no mortar in the 
 joints and the stones are cut to fit perfectly. 
 
 If, however, the centre of pressure on 
 any joint without mortar lies outside the 
 middle third, only a part of this joint is 
 under compression (Art. 21), so that on 
 substituting the I for that part in eqs. (1), 
 (-0 and (o) for an arch of variable cross 
 section, a nearer approximation can be 
 made by another trial and so on. Event- 
 ually the assumed and computed values of 
 
154 
 
 I will practically agree when the line of 
 resistance can be regarded as fixed. In 
 case there is mortar in the joint that can 
 supply all needed tensile resistance, the 
 line of resistance can pass without the mid- 
 dle third without any change in the equa- 
 tions, as the voussoir arch, save for a dif- 
 ferent modulus for the thin mortar joints, 
 is subjected to the same deformation as the 
 corresponding solid arch, so that its line of 
 resistance is nearly identical. As in well 
 designed bridges the line of resistance 
 should nowhere pass out of the middle 
 third, for any loading, to avoid the possi- 
 bility of joints opening with the accom- 
 panying infiltration of water, as well as to 
 provide a factor of safety, the tentative 
 method above will rarely be needed, and 
 the line of resistance can be at once found 
 from (4), (5J and ((>) for the arch of con- 
 stant cross section. 
 
 33. LEMMA. 
 
 In the constructions needed for establish- 
 ing the true line of resistance, according to 
 
the theory of the solid arch, we shall have 
 to solve the following problem : 
 
 In fig. 20, having given a series of fixed 
 points b T , b 2 , b 3 , ..., 1> 8 , and having drawn 
 parallel ordinates 1> 3 v 3 , 1> 4 v 4 , ..., through 
 them, intersecting a line m, m in the points 
 m T , ni 8 , ..., it is required to draw in, m 8 so 
 as to satisfy the two conditions. 
 
 2 (m b) = o, ~ (mb . x) o 
 
 where m b represents the ordinate from 
 m l m g to any one of the points b, ordinates 
 above m t m 8 being counted positive, those 
 below negative, and x represents the dis- 
 tance (abscissa) from an assumed origin O 
 
156 
 
 to the ordinate m b to which it refers. If 
 we give x the subscript of the ordinate to 
 which it refers, then the above conditions 
 can be written in full, for this figure, 
 
 m 8 b 3 + m 4 b 4 + m 5 b 5 + m e b 6 (m 1 b, + m 2 
 b,4-m 7 b 7 +ni 8 b 8 )=o; 
 
 (m 8 b 8 .x 8 +m 4 b 4 .x : f : m 5 b 5 . x 5 + m 6 b e . 
 
 Xg) (nij bj . x x + m 2 b 2 . x 2 + m 7 b 7 . x 7 + 
 
 m, b s . x s ) = o. 
 
 Now draw a straight line from b 1 (= 
 vj to b s (= v 8 ) and designate where it in- 
 tersects the ordinates by v 2 , v 3 , . . . ; any or- 
 dinates as m 5 b 5 can be written 
 
 or generally, 
 
 mb vb vm ; 
 
 which gives m b plus when above m x m s 
 and minus below, as is imperative. Sub- 
 stituting in the equations above we have 
 
 2 (vb -vm) = o, 2 (vb vm) x = o. 
 .-. 2(Jb) = 2 (vm), 2(v) . x) = 2 (vm . x), 
 
157 
 
 If we call x the abscissa of the result- 
 ant of the lines of the type vb treated as 
 forces and x' the abscissa of the resultant 
 of the lines vm treated as forces, then, since 
 the moment of the resultant is equal to the 
 sum of the moments of the components, the 
 last eq. above can be written, 
 
 x 2 (vb) = x' u I; .Tin) ; 
 
 whence, in view of the relation, 3 (vb) = 
 2 (vm), we have x ' = x . This establishes 
 the proposition, that when the line n^ m 8 
 has been determined correctly, the result- 
 ant R of the lines vb, treated as forces, 
 must equal and coincide with the resultant 
 of the lines vm treated as forces. 
 
 We can quickly, to any convenient scale, 
 find the value of R = v 2 b, + V 3 m~ - 
 +v 7 b 7 . Its position can be found by tak- 
 ing moments, most conveniently, about an 
 ordinate AB through the centre of the liiu- 
 
 b, b, 
 
 If the lines v b, by pairs, are equidistant 
 
158 
 
 from AB, as happens in all the applica- 
 tions that follow, call the distances from 
 this ( dotted ) medial ordinate ( AB) to 
 the ordinate through b j? b 2 , b 3 and b 4 , 
 z,, z 2 , z 3 and z 4 respectively. Then 
 treating left handed moments as positive, 
 right handed as negative, we have the al- 
 gebraic sum of the moments about the me- 
 dial ordinate, equal to ( v 7 b 7 v 2 b 2 ) z 2 + 
 ( v 6 b e v 8 .b s ) z 3 -f(v 5 b 5 v 4 b 4 ) z 4 ; and 
 on dividing this by R (as found above) we 
 have the distance from the medial or- 
 dinate to R which can then be laid off in 
 position, as shown in the figure. 
 
 The differences as (v 7 b 7 v 2 b a ) can 
 readily be found by taking the distance v 2 
 b 2 in dividers and laying it off from v 7 
 along v 7 b 7 . The difference between the 
 two lines is to be measured to the same 
 scale as the ordinates v b in finding the 
 value of R above. (This method is to be 
 generally used in similar cases). 
 
 We next draw a trial line n l n g and di- 
 vide ordinates as v 5 n. (n 5 being the inter- 
 section of nj n s with the ordinate v 5 b 5 ) in- 
 
159 
 
 to two sets by a line drawn from v t (=b t ) 
 
 to n 8 . 
 
 The resultant T of the sum of the ordi- 
 nates from the line v, v 8 to v 1 n g can be 
 found in magnitude, by adding up the or- 
 dinates, and in position by taking mo- 
 ments about A as just explained. Lay it 
 off the computed distance to the left 
 of A. 
 
 Now the position of T is not changed 
 when YJ n 8 assumes its true position v, m 8 
 (m 1 m 8 being regarded as the true line to 
 satisfy the original conditions), since all 
 the ordinates in the triangular space v t v g 
 n s are altered in the same ratio. T is thus 
 fixed in position no matter where n 8 may 
 be placed on the line v s n 8 . 
 
 It follows, because of this property 
 and since the ordinates, by pairs, are 
 equidistant from AB, that the resultant 
 T' of the ordinates intercepted between 
 Vj n 8 and n t n 8 is at the same distance to 
 the right of A that T is to the left. Then 
 if u l is afterwards shifted to m 19 T' is un- 
 changed in position, since all ordinates are 
 altered in the same ratio. Finally, if n 8 is 
 
160 
 
 shifted to m 8 , m, remaining stationary, the 
 position and value of T" remain un- 
 changed. Hence lay off T' in position as 
 far to the right of A as T is to the left, and 
 get its trial value by adding up the ordi- 
 nates included between v l n s and n : 
 
 iV 
 
 From what has been proved above it is 
 plain that if n : n 8 has been drawn correct- 
 ly, the resultant of T and T' or of the or- 
 dinates between v 1 v g and n t n 8 must coin- 
 cide with and be equal to R; hence calling 
 I and I 1 the distances from T and T' 
 respectively to R, we have 
 
 If the trial, T representing the sum of 
 the ordinates from v x v 8 to v, n g , is not 
 equal to the true value of T just found, re- 
 duce the distance v g n to v g m 8 in the ratio 
 of the true T to the trial T just 
 found. 
 
 Change v 1 u 1 to v : m 1 in the ratio of the 
 true T' (given by formula above) to trial 
 T' = sum of ordinates from v l n,, to n l n a . 
 
161 
 
 The reduction in both cases is best 
 effected graphically. 
 
 As the sum of similar ordinates in the 
 triangle v l n 8 m l is the same as for the tri- 
 angle Y! m s nij and their resultant has the 
 same position, it is evident that m, m 8 is 
 the true closing line (as it is called) to sat- 
 isfy the conditions. 
 
 2 (mb) = o, 2 (mb . x) = o. 
 
 It is easy to see if the first condition is 
 fulfilled by taking the successive lengths, 
 m 3 b s , m 4 b 4 , m 5 b 5 , m 6 b 6 in dividers and 
 adding up along a straight line. Similarly 
 add the lengths, m, b,, m 2 b 2 , m 7 b 7 , m 8 b 8 , 
 along the same straight line. If the two to- 
 tal lengths agree, the condition 2 (mb) = o 
 is satisfied. 
 
 When the ordinates vb are equal at 
 equal distances from the medial line AB, 
 R must coincide with AB. Now the re- 
 sultant of 2 (vm) cannot pass through the 
 centre unless m l m 8 is drawn parallel to 
 v, v 8 , in which case the lines vm will all be 
 of equal length throughout. Their num- 
 ber, in the present instance, is 8, so that by 
 
162 
 
 the first condition 2 (vb) 2 (vn 
 have 
 
 v,m 1 = v g m 
 
 8 
 which at once determines the line m, ni g . 
 
 It is equally correct and shorter to take 
 the sum of the ordinates vb to one side of 
 AB and divide by 4 when the total num- 
 ber of ordinates is 8. 
 
 34. We shall now proceed to design a 
 series of stone (or brick) arch bridges, whose 
 rise is one-fifth the span^ so that the line of 
 resistance for the position of the rolling 
 load tried shall just be contained within 
 the middle third limits of the arch ring, 
 and the intensity of pressure on any edge 
 of a voussoire joint shall not exceed say oO 
 tons per square foot for the best brick, or 
 50 tons for good granite or sandstone. 
 
 The live load assumed is known in Coop- 
 er's Specifications as c< Class extra heary 
 A". We give below the distances in feet 
 from the front pilot wheel to each pair of 
 wheels in turn and on the same line, the 
 weight of the pair of wheels in tons of 2000 
 pounds : 
 
L63 
 
 Pair of Pilot Wheels f net 8 tons. 
 
 " Driver 
 
 ' 8 -! 
 
 " 15 
 
 " 
 
 13.83 
 
 ' 15 
 
 
 
 18.33 
 
 " 15 
 
 " 
 
 ' 22 H3 
 
 " 15 
 
 " Tender 
 
 < 20.92 
 
 " 9 
 
 
 
 * -40.42 
 
 " - 9 
 
 <t < 
 
 ' 45.25 
 
 " 9 
 
 " Pilot 
 
 54.25 
 
 " 8 ' 
 
 The position of the pilot wheel of the sec- 
 ond locomotive is given last, from which all 
 the wheels of the second locomotive can 
 be located when desired. For short spans 
 the above load may be used or 40 tons 
 equally distributed upon two pairs of dri- 
 vers, seven feet centre to centre, whichever 
 produces the most hurtful effect. 
 
 The arch ring will be supposed of uni- 
 form section throughout and composed of 
 brick or sandstone weighing 140 Ibs. (.(/! 
 tons) per cubic foot. The material above 
 the arch ring up to the level of the road- 
 way will be supposed to have a specific 
 gravity eight-tenths of that of the arch 
 ring. 
 
 The load on a pair of wheels is carried 
 through the rails and cross- ties to the bal- 
 
164 
 
 last. If these cross-ties are 8 feet in length 
 we shall suppose the load uniformly dis- 
 tributed along this length, so that for one 
 foot length of cross tie the load will be but 
 one-eighth the load on the whole length of 
 the cross tie. The load for one foot in 
 length of tie is then reduced to cubic feet 
 of masonry by dividing by 0.07. 
 
 35. Example ./(see plate, fig. 27). The 
 figure represents an arch of 12.5 ft. span, 
 2.5 rise, 2.2 ft. depth of keystone, and ra- 
 dius 9.06 ft., with a surcharge rising 2 ft. 
 above the crown to the level roadway and 
 loaded 3 ft. to the left of the crown with 
 20 tons on 8 ft. cross ties, equivalent to 
 35.7 cubic ft. of masonry of the same spe- 
 cific gravity as the arch ring (0.7 ton per 
 cubic foot) on 1 ft. length. The dotted 
 medial lines of the trapezoids are 0.< v . of 
 the same lines extended to the roadway, 
 thus giving the "reduced contour" shown. 
 It is evident, for this small span, that the 
 alternative load of 80,000 pounds equally 
 distributed on the two pairs of drivers 7 
 feet apart, must produce the most hurtful 
 effect. As only one pair of drivers can get 
 
165 
 
 on the half arch they are placed as shown, 
 not quite */4 8 P an to left of crown. The 
 centre line of the arch ring is divided into 
 'M equal parts, the dotted joints are drawn 
 as shown (only a few of these are drawn 
 near the left springing), two divisions apart 
 and radial lines (full lines) are drawn mid- 
 way between them. The portion of the 
 arch ring between a springing joint and 
 the iirst clotted joint constitutes one artifi- 
 cial voussoir, and the portion between any 
 two consecutive dotted joints likewise con- 
 stitutes a voussoir. 
 
 As the theory of the solid arch requires 
 that we find the moment for each voussoir 
 about the middle of its centre line, or where 
 the full radial lines cross it, at the points 
 a l5 a 2 , ... , the construction of Art. 15 ap- 
 plies, or we take the division of the arch 
 included between the full radial lines in 
 tabulating quantities, and find the result- 
 ants acting on these full line joints as in 
 Article 15. 
 
 The condensed tables of loads and arms 
 (made out as explained in Art. 15) are as 
 follows: 
 
166 
 
 LEFT SIDE. 
 
 Joint 8 
 
 7 
 5.7 
 
 6 
 
 5 
 
 4 
 
 3 
 
 2 
 
 1 
 
 
 
 vS 
 
 1.9 
 
 9.7 
 
 49.5 
 
 53.9 
 
 58.6 
 
 63.4 
 
 68.5 
 
 71.2 
 
 c 
 
 .24 
 
 .76 
 
 1.26 
 
 2.66 
 
 2.77 
 
 2.94 
 
 3.15 
 
 3 . 42 
 
 3.55 
 
 RIGHT SIDE. 
 
 Joint 
 
 9 
 
 10 
 
 11 
 
 12 
 
 13 
 
 14 
 
 15 
 
 16 
 
 
 S 
 C 
 
 1.9 
 
 .24 
 
 5.7 
 
 9.7 
 
 13.8 
 
 18,2 
 
 22.9 
 
 2.82 
 
 27.7 
 
 32.8 
 
 35.5 
 
 .76 
 
 1.26 
 
 3.78 
 
 2.31 
 
 3.36 
 
 .3.87 
 
 4.13 
 
 The loads S are laid off on either side of 
 the arch, the arms either side of the crown, 
 and for an assumed thrust, shown by the 
 upper inclined line through the crown, the 
 various resultants on the joints 1 to L6 are 
 found as in Art. 15. These resultants in- 
 tersect the verticals through a ly a 2 , . . ., a, G 
 at the points b 1? b 2 , . . . , b 16 . 
 
 36. To locate the line k k', measure the 
 ordinates from a straight line joining a t 
 and a 16 , to the points a 1? a 2 . . .,.a 8 ; add, di- 
 vide by 8 and lay off the distance verti- 
 cally from aj to k and from ^ , to k' and 
 draw the line k k'. 
 
 37. Regard kk 1 as the trial closing line 
 for points b; connect ^ and b 16 , also k and 
 
167 
 
 b 16 by straight lines and find R and the 
 trial T and T', exactly as explained in Art. 
 33, in position and magnitude. The posi- 
 tions are given on the figure. Trial T = 
 23.35, trial T'= 14.1, R = 38.78; 
 
 .-. true T = 38.78 14| = 22.81; 
 5.22 
 
 true T' = 38.78 f^| = 15.97. 
 5.22 
 
 ( 
 
 Hence we must lay off b x m = b, k ,^~r- 
 
 ,).)> 
 
 and b 16 m' = b 16 k/ u 'j ' 
 
 This is best done by the ratio lines as 
 shown, or the distances may be computed 
 and laid off to scale. The line mm' is 
 thus the true closing line of Art. 33, for 
 points b t b : , . . . b lti . 
 
 3S. The ordinates y,, y 2 , . . ., y 8 (from a 
 a 17 toa 15 a,, . . ., aj are next scaled off; also 
 the ordinates from k k' to a^ a,, . . ., a., 
 called k a 15 ka. : , . . ., k a 8 , and we find the 
 value of 2 (k a.y) 2 (k a 4 . y 4 + k a 5 . y. 
 + k a, . y 6 + k a 7 . y 7 4- k a s . y 8 k a x . y! - 
 ka, . y, ka 8 y.) = 11.50. 
 
168 
 
 Next, the ordinates from m m' to the 
 points b : , b,, . . ., b 16 , called mb^ mb 2 , . . ., 
 in b 16 are scaled off and 2 (nib . y) is found. 
 In the present instance the complete ex- 
 pression for this is (m b 8 + m b 9 ) y s + (m b 7 
 + m b 10 ) y 7 + (m b e + m b n ) y 6 + (m b. + 
 m b J2 ) y 5 + (mb 4 + mb i3 ) y 4 (ml> 8 + mb l4 ) 
 y 3 (m b 2 + in b 15 ) y 2 (m b t -f m b l6 ) j l 
 = lX.*-v ; S, ordinates above mm' being treat- 
 ed as positive, those below negative. 
 
 We have now only to reduce the ordi- 
 nates nib in the ratio of 11.50 to lJS.^8 (by 
 the proper ratio lines), and lay off the re- 
 duced lengths. from the line kk' vertically 
 up or down, according to the sign of m b 
 to find all the points c,, c,, . . ., c lG in the 
 true equilibrium polygon for the arch. The 
 ordinate from m in' to the point where the 
 trial thrust meets the crown is likewise re- 
 duced in the same ratio and laid off from 
 k k' to fix the true centre of pressure on 
 the crown joint, .05 ft. below the centre of 
 the joint. 
 
 The reader familiar with Prof. H. T. 
 Eddy's " Constructions in Graphical 
 Statics " will recognize that the above pro- 
 
169 
 
 cedure is founded upon his beautiful con- 
 structions for the solid arch fixed at the 
 ends. 
 
 30. It will now be shown that the points 
 <?, located as above, are points in the equi- 
 librium polygon, directly over the centre 
 of the artificial voussoirs, which satisfy the 
 three conditions for an arch fixed at the 
 ends (Art. 32) 
 > (ac) = o, 2 (ac . x) = o, 2 (ac . y). = o. 
 
 Referring to Art. 3*5 it is seen that the 
 line k k' was located in a manner satisfy- 
 ing the conditions, 
 
 2 (k a) = o, 2 (k a . x) o . . . (A), 
 
 as shown in Art. 33. 
 
 Lines of the type k a in these formulas 
 refer to vertical ordinates measured from 
 kk' to a t , a 2 , ... a l6 . Similarly mb repre- 
 sents a vertical ordinate from line m m' to 
 bj or b L , etc.; ordinates above kk' or mm 1 
 being regarded as plus, those below minus. 
 
 By the method used in Art. 37 or Art, 
 33 the line m m 1 was located to satisfy the 
 conditions, 
 
 2 (mb) = o, 2 (mb .x) = o. 
 
170 
 
 The ordinates m b were DOW all changed 
 in the same ratio, which does not affect the 
 position of mm 1 . The altered ordinates 
 were next laid off from k k', the new value 
 of in b being equal to k c in the figure. 
 
 The conditions just given are thus satis- 
 fied by the k c' s, 
 
 /. -2tkc) =o, ^(kc.x) = o...(B). 
 
 Also by the construction of Art. 08, since 
 every m b has been altered in the ratio 
 11.50 to 18.28 to the corresponding k c, 2 
 
 as we see by reference to the equations of 
 Art. 38. 
 
 If the right member of the last equation 
 is transferred to the left member, since, kc 
 - k a = a c, we have, 2 (a c . y) = o. On 
 subtracting eqs. (A) from (B) and writing 
 the equation just found in the group, we 
 have 
 
 2 (a c) = o, 2 (a c . x) = o, 2 (a c . y), 
 
 or the points c satisfy the conditions for an 
 arch " fixed at the ends." 
 
171 
 
 40. As the closing line m m' has been 
 shifted to kk' and the ordinates m b al- 
 tered in the ratio 11.50 to 18.28, by the 
 theory of equilibrium polygons, we draw 
 from the old pole O (on the left) a parallel 
 to m m 1 (in its first position) to intersection 
 J with the load line, then a horizontal to 
 the right a distance = old pole distance X 
 
 lo.ivO -r- i / i ic 
 
 - to r the position of the true pole for 
 11. oO 
 
 the left force diagram. This is easily ef- 
 fected graphically by laying oif J L and J 
 M in the ratio of 11.5 to 18.28 and draw- 
 ing M P parallel to L I, I being the point 
 where a vertical through O intersects the 
 horizontal through J. 
 
 The new pole may likewise be found by 
 the method of Art. 14, by drawing through 
 O a parallel to a line connecting b t and b l6 
 to intersection with load line, then from 
 this point, a parallel to a line connecting 
 the points Ci and c l6 , previously found, a 
 distance to the right whose horizontal pro- 
 
 18.28 
 lection = old pole distance X 7 
 
 11. oO. 
 
 To fix the new pole P' on the right, 
 
172 
 
 draw PC" equal and parallel to ray P C 
 on left. 
 
 The points c can be tested by drawing 
 the new resultants on the joints, having 
 given the new poles and the position of 
 the thrust at the crown. These resultants 
 produced to intersection with the respective 
 joints from a to a l7 give the centres of 
 pressure on the corresponding joints. 
 
 The centres of pressure all lie within the 
 middle third of the arch ring, except at 
 joint G where the thrust passes exactly l / Q 
 depth from centre. The intensity at the 
 upper edge of joint is therefore double 
 the mean. The normal component of this 
 thrust is the weight of 03.2 cu. ft. of stone 
 = 03.2 X .07 = 4.4 tons; the mean pres- 
 sure is thus 4.4 -r- depth joint '2.2 ft. = 2 
 tons and the intensity at upper edge is 
 therefore 4 tons per square foot. 
 
 41. To test the accuracy with which the 
 work has been done, we measure a c at the 
 points a x to a l6 , counting distances above a 
 plus, below minus; then find the co-ordi- 
 nates x, y, of each point & l to a 16 regarding 
 a as the origin, x being measured horizon- 
 
173 
 
 tally (along a a 17 ) to the ordinate through 
 any point a and y vertically above a a 1T 
 to point a as previously stated, and finally 
 from the products as shown in the table: 
 
 Joint 
 
 ac 
 
 y ac . y 
 
 x ac.x 
 
 4 
 
 - .13 
 
 1.89 
 
 .246 
 
 2.78 .36 
 
 5 
 
 -|- .40 
 
 2.26 
 
 .904 
 
 3.69 1.48 
 
 6 
 
 
 2.50 
 
 .950 
 
 4.60 
 
 1.75 
 
 7 
 
 
 2.70 
 
 .486 
 
 5.57 1.03 
 
 8 
 
 + .06 
 
 2 . 80 
 
 .168 
 
 6.52 .39 
 
 15 
 
 
 .92 
 
 .074 
 
 12.90 
 
 1.03 
 
 16 
 
 -f .30 
 
 .32 
 
 J'l'G 
 
 13.66 4.09 
 
 
 {-1.53 
 
 
 +2.924 
 
 J-10.13 
 
 1 
 
 .27 
 
 .32 .086 
 
 .34 .92 
 
 2 
 
 .21 
 
 .92 
 
 .193 
 
 1.12 
 
 .23 
 
 3 
 
 .07 
 
 1.45 
 
 .101 
 
 1.92 
 
 .13 
 
 9 
 
 - .10 
 
 2.80 
 
 .280 
 
 7.50 
 
 .75 
 
 1<) 
 
 - .18 
 
 '1. To 
 
 .486 
 
 8.48 1.53 
 
 11 
 
 .20 
 
 2.50 
 
 .500 
 
 9.41 
 
 1.88 
 
 12 
 
 - .18 
 
 2. -26 
 
 .406 
 
 10.34 
 
 1.86 
 
 13 
 
 - .13 
 
 1.89 
 
 .246 
 
 11.23 
 
 1.46 
 
 14 
 
 .00 
 
 1.45 
 
 .130 
 
 12.10 
 
 1.09 
 
 -1.43 2.428 9.85 
 
 We have here the ratios, 
 
 ( ac) 
 
 ~ 1.43 ' 2 ( 
 
 (+ ac . x) 
 
 ac . y) 
 10.13 
 
 2.924, 
 
 2.428 ' 
 
 2 ( ac.x) ~ 9.85' 
 These ratios should strictly be unity, so 
 that the conditions of an arch "fixed at 
 the ends," 
 
174 
 
 2 (ac) = o, 2 (ac . y) = o, 2 (ac . x) = o, 
 may be satisfied. The results, however, 
 are very close, as will be apparent on sup- 
 posing the points c all to be lowered one 
 hundredth of a foot only, when 2 (+ a c) 
 will become -f- 1.47 and 2 ( ac), 1. ")*^, 
 the minus sums now being the greater. The 
 changes will evidently be equally great in 
 the other sums, so that we have accidentally 
 here determined the centres of resistance 
 within about .01 foot, even on this small 
 scale. We may readily rest content though^ 
 with half a tenth of a foot error on each 
 joint owing to unavoidable errors of con- 
 struction. See the next example where 
 these errors are as pronounced as for any 
 arch examined and yet a shifting of the 
 points c by half a tenth of a foot is about 
 all that is necessary to satisfy the condi- 
 tions. 
 
 We conclude for the arch just examined, 
 for the given position of the live load, that 
 the line of resistence just touches the mid- 
 dle third limit at one point only, and that 
 it possesses the proper margin of safety 
 both as to strength and stability. 
 
175 
 
 42. Example II. A segmental stone arch of 25 ft. span, 5 
 ft. rise, radius 18.12 ft. and uniform depth of arch ring 2.5 
 ft., was next examined. 
 
 The depth of spandrel filling over the crown was 2 ft. 
 and the live i'oad-consisted of the last three driving wheels 
 of the locomotive above specified on the left half of the 
 arch, no load on right half. 
 
 The loads are given precisely as follows : 15 tons 3.8 ft. 
 from crown, 15 tons 8.3 ft. and 15 tons 12.8 ft. from crown. 
 
 The division of arch ring and the construction gener- 
 ally was exactly like that just given for Example I, so that 
 it is not necessary to enter into it. The true thrust at the 
 crown, after the theory of the solid arch, was found to act 
 .06 ft. below the centre of the crown joint and its horizontal 
 component was 120.25 cu. ft. = 8.4175 tons, the vertical com- 
 ponent 10'. 5 cu. ft. =0.735 ton. 
 
 If we call for brevity, the distance ac=v, we find in 
 this instance, 
 
 _2_(_j v ) _ UK). 2 (+vx) _25.64 2 (+vy}_ _6.32_ 
 "YPv) ""^53 ' 2 ( vx) "18^4 ' 3T(^-~vy)~ "BUT 
 
 If we conceive the thrust at the crown lowered 0.1 ft. 
 but maintaining its same direction and magnitude, the 
 points c will all be lowered 0.1 foot, and the new ratios will 
 be as follows : 
 
 _? lr.l> _ 126 s (+ vx) . _ 16 -^ 8 2(4-vy) _ -2.96 
 
 2 (v) ~2.30 ' "S ( vx) ~~ 28~75 ' 2 ( vy) ~~ 8.02 
 
 Here the minus t-rnis exceed the plus terms so much 
 that it is plain that the thrust has been lowered too much ; 
 in fact (neglecting any possible tilting) it is evidentsthat the 
 true position of points c lies between the first and last posi- 
 tions and is nearer the former than the latter; so that we 
 can safely say that the first series of points is certainly 
 within 0.05 foot of the correct position. 
 
 As the discrepancy above was as great as that found in 
 any case examined, it was thought worth while to show that 
 
176 
 
 the extreme limit of error in any rase was negligence and 
 that the construction affords practically exact results. 
 
 On constructing the centres of pressure on all the 
 joints, it was found that they nowhere leave the middle 
 third except at the right springing joint (17) where the cen- 
 tre of pressure was 0.08 ft. above the middle third limit. 
 Hence it was thought best to increase the depth of keystone 
 three times this amount, or 0.25 ft., making the radial depth 
 of arch ring uniformly 2.75 f*et. 
 
 This is so near the former value, 2.5 ft., that it was not 
 thought worth while to test it by another construction. 
 It will be assumed to satisfy all conditions. 
 
 The intensity of thrust at the upper edge of the right 
 springing joint (for the arch ring 2.5 ft. deep) is found to be 
 9.2 tons per square foot, at the lower edge of the left spring- 
 ing joint 8.7 tons. 
 
 43. Example III. Segmental stone arch of 50 feet span, 
 10 ft. rise, radius 36.25 ft., and height of surcharge above 
 crown 2 feet. A construction for a depth of keystone of 3 
 feet showed, for the loading to be given, that the line of 
 resistance passed outside the middle third ; hence, for a 
 second trial, a depth of arch ring of 3.5 feet was assumed. 
 The loading omitted the pilot wheel and consisted of the 
 eight drivers on the left half of the arch, viz. : 15 tons 4.25 
 feet from crown ; 15 tons 10 feet ; 15 tons 14.25 feet ; and 
 15 tons 19 feet, all to left of crown of arch. 
 
 The line of resistance nowhere passed outside the 
 middle third of the arch ring. At the springing joints the 
 resultants touch the lower middle third limit on the 
 loaded side ; and the upper limit on the unloaded side ; at 
 the crown the thrust passes through the centre of the 
 crown joint. The arch thus satisfies all the conditions 
 of stability. The horizontal component of thrust at 
 crown = 279.3 cu. ft. = 19.55 tons, and th.3 vertical com- 
 ponent = 21 cu. ft. = 1.47 tons. The normal component 
 of the thrust at the left springing = 413.5 cu. ft. = 28.95 
 tons, giving an intensity at the intrados of 
 
28 95 
 
 2 =16.5 tons, 
 
 o.o 
 
 which is within the proper limit. 
 
 This arch then satisfies all conditions for this loading. 
 
 44. Example IV. Stone arch of 100 ft. span, 20 ft. rise, 
 radius 72.5 ft., depth of keystone 5 ft., and height of sur- 
 charge above crown 2 ft. The Ir adiug consisted of one 
 locomotive and tender, as specified in Art. 34, covering the 
 left half of arch, the right half being unloaded. The 
 pilot wheel was placed 8 feet to left of crown, the other 
 wheels following in order at the distances given in Art. 35. 
 so that the last tender wheel was barely on the arch. 
 
 The line of resistance was ev< rywhere contained within 
 the middle third, except at the left springing joint where 
 it passed 0.17 ft. below the limit; hence to satisfy the 
 middle third limit the arch ring should be increased in 
 depth, say 3X 0.17= .51ft., making the depth 5.5 ft. 
 
 The thrust at the crown joint fell 0.4 ft. below centre 
 for the arch of 5 ft. key, its horizontal component being 
 689.7 cu. ft. =- 48.279 tons, and the vertical component 26.5 
 cu. ft. =1.855 ton. The intensity of thrust at lower edge 
 = 14.3 tons per sq. foot. 
 
 Tin? thrust at the left springing acted 1 foot below the 
 centre of the joint, its normal component being 1048 cu. ft. 
 = 73.36 tons. It acts as a uniformly increasing stress 
 over a depth of joint = 3x1.5=4.5 ft. ; hence the aver- 
 age stress is 73. 36 -f- 4. 5= 16.3 and the intensity at lower 
 edge is double this, or 32.6 tons per square foot. For a 5.5 
 ft. key the intensity is much less. 
 
 45. Example V. The arch of Ex. IV. of 100 ft. span, 20 
 ft. rise, and 5 ft. depth of key, subjected only to its own 
 weight. 
 
 Here we need consider only the right half of the arch, 
 since the thrust at the crown, from consideration of sym- 
 metry, must be horizontal as well as the line m m' (using 
 the designation given on plate for another span). 
 
 Hence for an assumed horizontal thrust at the crown, 
 
178 
 
 having found points such as b and drawn the line k k' i 
 before, we determine line m m' for points 6 exactly as \ 
 found kk' for points a ; or it is generally shorter to add up 
 with dividers the ordinates to points b above a trial line as 
 kk', and subtract from the sum the length we obtain by 
 adding up ordinates to points b below k k'. The difference 
 divided by 8 gives the amount the horizontal m m' is above 
 or below k k' to satisfy the condition 2 (,iub) = o. The 
 points b meant above are b () , b 1() , .... b lg . As a test the 
 sum of the ordinates from m m' to points b above should 
 exactly equal the sum to points b below m m'. 
 
 The sum of the products 2 (ka . y) is made out exactly 
 as before. Its value for the half arch in this case is 302. 
 Similarly find 2 ^nib . y) = mb . y () - mb ]() . y u) + mb n . y n 
 f mb^ . y 12 + mbjg . y lg - (mbtt, ' y u + mbl6 . J u mb M 
 . y ) = 3S8, the ordinates rub being measured from m m' up 
 (-f ) or down ( ) to points 6. 
 
 On diminishing all the ordiuates tub in the ratio of Su:( 
 to 388 and laying them off from k k', we find the points c 
 through which the resultants on the joints pass. 
 
 The point of thrust at the crown is found as usual, and 
 the new pole is found by laying off on a horizontal through 
 
 3SS 
 C' a distance equal to old pole distance ^ : . The points 
 
 c can now be tested and the resultants produced to intersec- 
 tion with all the joints. 
 
 It is interesting to compare the new curve of the 
 centres of pressure for the bridge unloaded, with that found 
 previously for the load on the left half. Thus reiuembi r- 
 ing that the depth of arch ring is 5 ft., one-sixth of which 
 is 0.83 ft. from the centre to curves defining middle third 
 limits, the following tables give the distance measured atony 
 any joint from its centre to the centre of pressure of the 
 joint, plus distances being measured upwards, minus dis- 
 tances downwards. The upper numbers, under the joint 
 numbers, correspond to the arch unloaded and the lower 
 numbers to the arch loaded. 
 
171) 
 
 KIGKT SIDE. 
 
 C i own 
 
 9 
 
 10 
 
 11 
 
 
 
 14 
 
 i' 
 
 
 17 
 
 .27 
 
 3 ' 
 
 -.20 
 
 () 
 
 K20 
 
 . if 
 
 4-. 32 
 
 (I 
 
 _-,, 
 
 .4(1 
 
 
 .57 
 
 .62 
 
 
 -.15 
 
 -f.05 
 
 +.10 
 
 .! 
 
 .i:. 
 
 LEFT SIDE. 
 
 
 7 
 
 G 
 
 .j 
 
 4 
 
 3 
 
 2 
 
 1 
 
 
 -.30 
 
 .'20 
 +.18 
 
 II 
 +.W) 
 
 +.20 
 
 +_75 
 
 +.35 
 
 
 
 
 + 13 
 
 .22 
 
 .65 
 
 - .4(1 
 1.00 
 
 It will be observed at joints 0, 5 and 10, where the ceu 
 hvs of pressure for bridge loaded are farthest from centre 
 of joints, that when the live load comes on the centres of 
 pressure remain on the same side of the centre line of the 
 arch ring as for bridge unloaded. At many other joints as 
 17 the reverse obtains. Hence, if the arch was not circular, 
 but of such a figure that, for bridge unloaded, its centre 
 line would be the focus of the centres of pressure on the 
 joints, the departure of the line of resistance for bridge 
 loaded at joints 0, 5 and 10 would not be so great as above, 
 though at joint 17 (which is often a critical joint), and at 
 some other points, it would beigreater. Therefore such a 
 design would often permit of smaller arch rings, such that 
 the line of resistance for the bridge loaded in any way could 
 still be inscribed in the middle third. However, in some of 
 the bridges examined (Exs. II. and III.) the centre of pres- 
 sure on joint 17, for bridge loaded, passed through the upper 
 middle third limit, so that if an arch having its centre line, 
 the line of resistance for arch unloaded, was used here of 
 the same depth as before, the centre of pressure on joint 17 
 would leave the middle third, and the arch would not be as 
 stable as before. 
 
 As we cannot tell, without a special investigation of 
 this kind, which design will prove the most economical, it 
 is well to hold on to the segmental circular arch until the 
 others, for ail kinds of loading, are proved the most ecouom 
 
180 
 
 ical, particularly as it is much easier to construct, and the 
 e conomy, if any, in replacing it by the other, must be small, 
 Writers generally, in advocating the catenarian curves, have 
 not properly considered the preponderating influence of 
 heavy eccentric loading. 
 
 40. An examination of the lines of re- 
 sistance in all the preceding examples fails 
 to indicate any simple approximate rule for 
 constructing them without recurring to the 
 theory of the solid arch. I have stated 
 elsewhere, partly on the strength of a few 
 constructions after the theory of the solid 
 arch, for uniform loads or comparatively 
 light eccentric loads, that " it seems highly 
 probable that the actual line of resistance 
 is confined within such limiting curves, ap- 
 proximately equidistant from the centre 
 line of the arch ring that only one line of 
 resistance can be drawn therein." From 
 the constructions above this rule is found 
 to indicate very roughly about the posi- 
 tion, but it is not precise enough in prac- 
 tice. Thus for the 100 ft. span above un- 
 loaded, the true curve passes .4 below the 
 C2ntre line (measured, not vertically, but 
 along the joint) at the springs, .35 above at 
 joints 4 and 13 and .37 below at the crown ; 
 
181 
 
 but the divergences are much greater at the 
 joints of rupture (0, 4 or o, 9 and 17) for 
 the arch heavily loaded on one side, as we 
 see from the table, and the same thing is 
 shown on the plate for the 12.5 feet span. 
 Hence we cannot state precisely that if 
 a line of resistance can be inscribed in the 
 middle third, the true line of resistance 
 will be found in the middle third. It is in 
 fact plain from the above constructions 
 that if only one line of resistance can be 
 inscribed in the middle third, the true line 
 will pass outside of it at certain points; for 
 the first line touches the curves limiting 
 the middle third at all the joints of rup- 
 ture, as it corresponds to both the maxi- 
 mum and minimum of the thrust within 
 those limits, whereas the true curve does 
 not at all the joints, hence it cannot agree 
 with the former and hence must lie outside 
 the middle third limits, since by assump- 
 tion only one line of resistance can be 
 drawn therein. 
 
 We can appropriately quote here Wink- 
 ler's notable theorem published in 1879, 
 
which is practically true for segmental 
 arches of constant cross-section. The the- 
 ory is given in full in an article by Prof. 
 Geo. F. Swain on the stone arch, in Van 
 Nostrand's Magazine for October, 18SO. 
 Winkler's theorem is as follows: 
 That line of resistance is approximately 
 the true one which lies nearest the centre 
 li)te of the arch ring ax deter rained by the 
 method of least squares. 
 
 This remarkable theorem is easily dem- 
 onstrated by aid of the theory of elasticity, 
 and while it is no aid practically in pre- 
 cisely fixing the true line of resistance, yet 
 the conclusions are valuable as confirming, 
 in a general way, the preceding construc- 
 tions. 
 
 47. The method of finding the true re- 
 sistance line given in this chapter is per- 
 fectly general and applies to any form of 
 arch of constant cross- section. The defect 
 in the method practically is that the most 
 hurtful position of the live load cannot be 
 readily ascertained. It is true that for sin- 
 gle loads the method of this chapter will 
 give the quantities c,, y and c 2 as defined 
 
in Chap. IV.,* from which we may make 
 out a table and proceed as in the preceding 
 chapter. This method is very long and is 
 rarely needed, as circular arches are gener- 
 ally built; and for these the quantities c 1? y 
 and c, can be readily found from existing 
 tables and formulas. 
 
 The positions of live loads assumed in 
 this chapter simply followed the roui^h 
 rule of putting the heaviest part of the 
 load over the middle of the haunches. The 
 constructions resulting were all made be- 
 fore the method of the preceding chapter 
 was developed. It is gratifying to note 
 thai the conclusions as to depth of key, 
 etc., are almost identical with those of 
 Chap. IV., where the most hurtful position 
 of the live load was carefully ascertained. 
 
 * S^H such constructions in "Theory of Solid and 
 Braet-d Elastic Arches, '' p. 79. 
 
184 
 
 APPENDIX. 
 
 The writer, in 1874, performed some experiments on 
 light wooden arches at the limit of stability, which tend to 
 confirm theory and are instructive in many ways. They 
 will be given in full below. The experiments were made 
 with great care ; the voussoirs being accurately cut, the 
 span kept invariable and horizontal, piers vertical, and 
 the weight applied very gently and without shock. 
 
 To avoid mistake the following momenclature will be 
 adopted : 
 
 Depth of a voussoir is the dimension in the direction of 
 the radius of the intrados _[_ to the axis of the arch. 
 
 Thickness of an arch is the dimension i! to the axis of 
 the arch. 
 
 ' Width of a pier is its horizontal dimension J_ to the axis 
 of the arch. 
 
 Height is measured vertically. 
 
 The dimensions will all be given in iuchi. s. 
 
 A Gothic arch (Fig. 28) of 14 in. span, and 12.12 in. 
 rise, was cut out of a poplar (tulip tree) plank, 3.65 in. 
 thick, consisting of 8 voussoirs, each 3. Go thick, i> deep, 
 and 4.U8 along their centre line from middle to middle 
 of joint : each vousscir weighing .52 Ib. Quite a 
 number of voussoirs were cut out of the same layers of 
 fibres and those selected that weighed exactly the same : 
 the voussoir to be tried being hung to one end of a delicate 
 balance beam, with a voussoir of the standard weight at the 
 other end. The two voussoirs at the crown not being cut 
 out of the same layers of fibres as the others, were shaved 
 oft' about the middle of the extrados (not touching the joints) 
 so as to weigh exactly 1 voussoir of the standard weight and 
 their centres of gravity were found experimentally, and 
 found to be at exactly similar points in both voussoirs, so 
 that the entire arch was symmetrical as to the crown. 
 
 The centres of gravity of the other voussoirs are taken 
 on the arc of a circle passing through the middle of the 
 joints, and for >my voussoir, equidistant from the joints 
 bounding that voussoir. For voussoirs whose sides are 
 
185 
 
 little inclined this is sufficiently near the truth, and by 
 dividing the arch ring into a sufficient number of artificial 
 roussoirs the result may be made as accurate as we 
 please. Still as no wood is homogeneous the results can 
 only be regarded as approximate as compared with the 
 hypothetical homogeneous arch, still sufficiently near to 
 establish the laws heretofore demonstrated. 
 
 When this arch was set up the joints apparently 
 fitted perfectly, and on placing a drawing-board by the 
 side of the arch and tracing off its contour curves, it was 
 found to be a perfect Gothic whose arcs, composing the 
 contour curves were correct arcs of circles described from 
 the springing points opposite. 
 
 A number of rectangular wooden bricks of exactly 1 
 voussoir in weight, of various sizes, were also cut out, as 
 well as half bricks, quarter bricks, etc., and some solid 
 rectangular piers of various dimensions. 
 
 A voussoir is taken as the unit of weight. 
 
 In experiments where weights were placed upon the 
 top of the arch, an assistant added brick after brick, 
 carefully balancing the load at the top on either side by 
 the fingers until the arch reached its balancing point ; 
 i. c., the point where it stood with the weight, but fell with 
 a slight jarring. 
 
 The two bottom voussoirs were, when necessary, kept 
 from sliding by two fastening tacks being driven into the 
 board on which the ur.'li rested, pressing against the arch 
 .03 above the springing line, or so little that it may be dis- 
 regarded. The board was carefully levelled at every exper- 
 iment by a spirit level, and the span kept invariably at 
 14 in. 
 
 There was little or no -vibration in the room. 
 
 Ftrst Experiment. With 8.2 voussoirs on the summit 
 of the arch it stood, though fell with 8.3 voussoirs on the 
 summit ; rotating on joints 2 on iutrado al edge, join's 4 
 at the extrados and at the upper edge of the crown joint, 
 the arch being forced out at the haunches and falling at 
 the crown. (See Fig. 28.) 
 
The following table gives in its first column the number 
 of joint from the crown; columns, the elementary weights 
 (4.1 voussoir being the weight on the summit that goes to 
 each abutment, the weight of each voussoir being taken as 
 unity) ; column in gives tho horizontal distance from the 
 crown to the centre of gravity of each voussoir with its 
 load, if any, which, in this case, is also the moment in 
 reference to the vertical through the crown of each voussoir. 
 Columns 8, M, and C have been before -xplained: 
 
 
 s 
 
 m 
 
 s 
 
 M 
 
 
 1 
 
 4.1 
 
 0.00 
 i Tn 
 
 4.1 
 ^ i 
 
 0,00 
 
 1 r/i 
 
 . < ih 
 
 00 
 
 1. 
 
 1. 
 1. 
 
 1. 
 
 8.1 
 
 1.70 
 
 4.68 
 6.79 
 7.88 
 
 5.1 
 
 G.I 
 7.1 
 8.1 
 
 1.70 
 
 6.38 
 13.17 
 21.05 
 
 .33 
 1.04 
 1.85 
 
is; 
 
 Try a line of resistance, passing 0.1 from the upper 
 edge of crown joint and 0.1 from the extrados edge of 
 the joint at the springing. It is found to cut joint 2 at 
 0.1 from the intrados. 
 
 From joint to joint 2 the line of pressures corres- 
 ponds to the minimum of the trust ; from joint 2 to joint 
 4, to the maximum within limiting curves 0.1 from in- 
 trados and extrados respectively. (Art. 20.) 
 
 Sliding would have occurred on joint 4, as the resultant 
 on that joint made an angle of 21 deg. with the normal, 
 but for the tacks before mentioned. 
 
 The diagrams for this and all the following experiments 
 were drawn to a scale of one-third the natural size, except 
 in the case of some of the pier experiments. 
 
 It may pertinently be "enquired, why at the limit of 
 stability, the centres of pressure should not be found at 
 the very edges of the joints in place of being ().". inch 
 from those edges ? The answer is simple : We have seen 
 in Art. 21 that when the centre of pressure on a joint leaves 
 the middle third, the joint begins to open, and this open- 
 ing is quite perceptible when this centre of pressure is 
 very near the edge. This opening of the joints causes a 
 deformation of the arch ring, so that the figure just before 
 rotation occurred is not that assumed in the drawing. If 
 the deformation had been known at the instant of rupture, so 
 that the true figure could have been drawn, then the line of 
 resistance would have passed through the very edges of the 
 joints 0, 2 and 4, as they alone were bearing at the time. 
 No attempt was made to find the deformed figure ; in fact, 
 it varied so rapidly just before rupture that it would have 
 been impossible to have found it. Similar remarks and 
 explanations apply to all the subsequent experiments. 
 
 Second Experiment. With the two voussoirs at the 
 crown in one solid piece, the arch could not give by rota- 
 tion, as the lower edge of crown joint could not open. 
 With a sufficient pressure on the crown, there was sliding 
 along joints 1, the coefficient of friction being small for 
 these wooden bl cks. 
 
188 
 
 We evidently have here a sufficient reason for making 
 the keystone in one solid piece. 
 
 Third Experiment. On placing a knife edge against a 
 notch .03 deep, cut into the bottom voussoir, 0.4 above the 
 springing line, on each side, the arch balanced with 11.1 
 voussoirs on the summit. The line of resistance must now 
 pass through the knife edges, and it will be found on 
 constructing a diagram that it will pass about 0.1 fromedpcs 
 at joints and 2, as before. 
 
 Fig. 29 
 
 Fourth Experiment (Fig. 29.) The same arch stood, 
 being very nearly on the balancing point, on solid piers In. 
 high. 1.9 wide, and 3.65 thick, each pier weighing 2.3 vous- 
 soirs, the intrados at the springing being at the inner edge 
 of pier. The piers were made vertical by a spirit level, and 
 their tops were upon the same level in every experiment 
 given. 
 
189 
 
 in the following table the pier is included opposite 
 joint 5 of the first column : 
 
 
 s 
 
 c 
 
 m S 
 
 M 
 
 C 
 
 I 
 
 I 
 
 1 
 1 
 
 2.3 
 
 1.7 
 
 4.68 
 6.79 
 
 7.88 
 7.95 
 
 1.70 1 
 4.f>8 2 
 6.79 3 
 7.88 4 
 18.28 0.3 
 
 1.70 
 6.38 
 13.17 
 21.05 
 39.33 
 
 1.7 
 3.19 
 4.39 
 5.26 
 6.24 
 
 
 6.3 
 
 
 39.33 
 
 
 
 A line of resistance 0.1 from edges of joints o and 8, 
 cuts the base of the pier 0.2 from its outer edge. 
 
 Fifth Experiment. With piers 40.47 in. high, 3.65 wide, 
 and 1.9 thick, weighing 10.1 voussoirs each, with the intra- 
 dos of arch at springing on a line with inner edge of pier, 
 the same arch balanced. The pier was built of a solid block 
 22 in. high and 5 bricks placed on top, one above the other 
 to make up the 40.47 in height. 
 
 A line of resistance drawn .1 from summit and .1 from 
 intrados at joint 3 passes .5 from outer edge of pier, or 
 about 1-7 width of pier. 
 
 If the figure of the deformed arch could have been 
 drawn at the instant of rupture the line would have 
 passed through the very edges. 
 
 Sixth Experiment. The pier of Exp. 4 (Fig. 29) was 
 moved outward (from the axis of the arch) > o that when 
 its inner edge was .1 from the springing, it stood with no 
 weight on the summit ; when it was .4 from edge, it stood 
 with .5 vs., fell with .6 vs. ; .5 from edge, balanced with .7r> 
 .vs. : .6 from edge balanced with .75 vs. ; .7 from edge 
 balanced with .37 vs. ; 1.0 from edge balanced with .12 vs. 
 
 On constructing the table and diagram as above for the 
 load .75 vs., we find the centre of pressure on joint 4, .63 
 from the inner edge, or slightly over the extreme limit 
 above, as should be the case. 
 
 Seventh Experiment. The same arch stood easily with 
 .75 vs. on the summit, on solid piers, 22. high, 3.65 wide, 
 and 1.9 thick, each weighing 5.1 vs. ; the arch fell with the 
 addition of .12 vs. more. 
 
190 
 
 On constructing this figure it will be found that the 
 line of centres of pressure, assumed 0.1 from edges of 
 joints and 3 as before, passes .63 from inner edge of 
 springing jc hit (us was stated above) and cuts the base of 
 pier .39 from its outer edge or about 1-9 the width of pier. 
 
 Enjhth Ejrparhnent. On moving this pier back as in the 
 (3th KJ-I>, : 
 
 0.47 the arch balanc* d with 1.12 vs. 
 
 0.53 " " " " 1.25 " 
 
 .5'.) " " " " 1.25 " 
 
 .63 " " " " 1.25 " 
 
 .7 " " " " 1.12 " 
 
 1. " " 1.IM) * 
 
 On constructing the line of resistance for a weight of 
 1.25 at the apex, passing 0.1 from the edge of joints and 3 
 as before, it will be found that the centre of pressure on 
 joint 4 is .7 from the edge, again slightly over the extreme 
 limit .63 found by experiment. 
 
 It is evident from an inspection of the arches in churches 
 that constructors were well aware that a higher pier might 
 be used when its inner edge was moved back a certain dis- 
 tance from the springing, which is equivalent to what we 
 have established above. 
 
 Ninth Experiment. With the pier used in E-JCI>. 4. and 
 the same arch, excepting that the two voussoirs at the 
 crown were in one piece, the arch and pier just balanced as 
 in Exp. 4. In fact the arch and pier can easily rotate on the 
 third joint and the outer edge of pier. 
 
 Tenth Experiment. --The same arch with piers 1.98 wide, 
 7.5 high and thickness of arch, each weighing 2 vs., stood 
 easily when a cylindrical pin .03 in diameter was placed 
 at the lower edge of crown joint. This joint bore at no other 
 point, hence the line of resistance passes through the pin. 
 Assuming it to pass .1 from the edge of joint 3. the con- 
 struction will show that it cuts the springing joint .6 from 
 inner edge and the base of pier .15 from its outer edge. 
 
 The experiments that we have just considered very 
 
191 
 
 clearly indicate the fallacy of that theory which supposes 
 that if a line of resistance passes outside the inntr third of 
 the arch ring, that it must fall. On the contrary, in every 
 case of the stability of the arches previously given, it is 
 impossible to draw a line of resistance everywhere contained 
 within the inner third of the arch ring. 
 
 Eleventh Experiment. Fig. oU. With this same Gothic 
 arch a segniental circular arch was now made of 24.24 in. 
 span and 7 in. rise ; the voussoirs being as before 2. deep 
 and :>.G5 thick. 
 
 With 7.C vs. on the summit, this arch balanced; the 
 weight being placed on a small stick resting on the trammit. 
 With a greater weight the rotation occurred on joints 0, 2 
 and 4, the crown falling. 
 
 1 
 
 2 
 3 
 4 
 
 s. 
 
 m. 
 
 s. 
 
 M. 
 
 C. 
 
 3.8 
 1. 
 1. 
 1. 
 1. 
 
 0.00 
 2.03 
 5.90 
 
 '.. us 
 12.11 
 
 3.8 
 4.8 
 
 5.8 
 (5.8 
 
 7.s 
 
 0.00 
 2.03 
 7 <>3 
 17.31 
 29.42 
 
 .00 
 .4-2 
 137 
 2.55 
 3.77 
 
 Fio. 30 
 
192 
 
 On trial it was found that the true line of resistance 
 passes .15 from the edges at joints 0, 4 and 2 ; giving the 
 characteristics of both a maximum and a minimum thrust. 
 
 The ends of this arch required fastening tacks thrust 
 into the board and pressing against voussoirs 4, .03 above 
 the springing as in the first exp., with the Gothic, to pre- 
 vent sliding. The thrust on joint 4 made an angle of 50 w 
 with the normal to that joint. 
 
 Twelfth Experiment. With this arch resting on piers 
 3 63 wide, 5.8 high and 2. thick, each weighing 1.5 vs., the 
 inner edge of pier being on a line with the springing, the 
 arch balanced with .5 vs. on the summit. 
 
 We find, by constructing a line of resistance passing 
 .15 from summit and the intrados at the third joint, that 
 it cuts the base of pier .24 from its outer edge. 
 
 Thirteenth Experiment. To form some idea of the 
 action of mortar of different degrees of hardness, pieces of 
 cloth .07 thick when not pressed, and .04 thick when 
 pressed between two flat surfaces by the hands were put 
 between the joints of the Gothic arch (Fig. 28), each piece 
 weighing .015 voussoir. 
 
 The span was then altered until the joints were all 
 close, when it was found to be 14.57, the rise to the apex 
 being 14.55. On placing a drawing-board by the side of 
 this arch and tracing its contour curves, they were found 
 to be very nearly arcs of circles, though not with their 
 centres at the springing points. To locate them ; measure 
 horizontally from the springing points .32 towards the 
 middle of the span, and then vertically downwards 0.1 to 
 the centres, from which the arch may be drawn. . 
 
 This arch balanced with 4.0 vs. at apex; fell with 4.05 
 vs. The limiting lines to the curve of resistance was found 
 to be distant .3 = 1-7 depth of joint from the contour 
 curves, at its nearest approach to them. 
 
 This arch spread outwards upon the application of the 
 weights, joint 2 being the point of rupture at the haunches ; 
 hence it is evident that if there had been a solid spandrel, 
 or in this case, simply the pressure of the hands, to resist 
 
193 
 
 this spreading, that the arch would not have fallen. The 
 spandrel would have supplied horizontal forces in addition 
 to the vertical^ones due to its weight. 
 
 If the spandrel were not solidly built, at least up to 
 joint '2, there would necessarily be derangement of the 
 arch. 
 
 The curves of resistance were drawn in all the fore- 
 going experiments, not taking into consideration the last 
 mentioned derangement of the arch, which would have 
 caused this curve to pass nearer the edges or exactly 
 through them. 
 
 In fact, in most of the experiments, just before rotating, 
 the edges alone seemed to be bearing. In the case of the 
 simple Gothic, without cloth joints, when a sufficient 
 weight was applied at the summit, the joint there and 
 joint 2 opened sensibly before the balancing weight was 
 put on. The segmeiital arch flew out at the second joints, 
 falling at the crown, only opening when near the balancing 
 point. 
 
 Isolated weights applied at the summit do not occur 
 in practice, and it is hardly probable that a well-built 
 viaduct, whose intrados is a segment of a circle with thin 
 joints, will spread appreciably after the mortar has well 
 set ; and this is necessarily a stronger form of arch than 
 the semi-circular, elliptical, or hydrostatic, where; the 
 spandrel thrust is generally required to cause stability. 
 
 If the latter profiles are desired, let the depth of the 
 voussoirs be increased towards the abutment, so as to 
 keep the line of resistance within the proper limits of the 
 arch ring, when the constructor will be assured of stability. 
 
 It certainly seems singular, that engineers should ever 
 rci-niti'iHdul an aivh like the hydrostatic, which necessarily 
 ri -quires a very effective spandrel thrust to keep the arch 
 from tumbling down. 
 
 The spandrels must in such cases be built with the same 
 'are used with the arch stones, thus increasing the expense, 
 while really losing in stnngth. 
 
 Fourteenth Experiment. In the joints of the same 
 
194 
 
 Gothic* arch, pieces of soft woolen cloth .15 thick when 
 not pressed, and .1 when pressed hard between two bri -ks 
 by the hands, were next inserted, each piece of cl:>tli 
 weighing .027 voussoir. The span, when the joints w r> 
 close, was found to be 15 in. ; rise to apex, 14.63, The 
 centres for describing the contour curves were 1.07 in. from 
 the springing points measured horizontally towards the 
 middle of span. 
 
 This arch balanced with 2.3 vs. on the apex. 
 
 Assuming this arch to preserve its figure, the curve of 
 resistance passes about one-fourth of the depth of joint 
 from the edges at its nearest approach to them. 
 
 This experiment gives us some idea of the effect of 
 thick plastic mortar joints or of uncentreing an arch with 
 fresh mortar joints. 
 
 Fifteenth Experiment. A Gothic arch of about half the 
 dimensions of the first given in Exp. 1 was cut out, really 
 lief ore the arch we have just been considering. 
 
 It was not found to be symmetrical as to weight, one- 
 half weighing 1-32 of the whole arch more then the. other 
 half. Still as arches in practice are unsymnu -tri'-al as to 
 weight at least it will be interesting to know, that assum- 
 ing this arch to be symmetrical, the curve of pressures 
 passes .075 from the edges at joints of rupture, mora- 
 lly with weights at the apex. 
 
 All the preceding experiments were repeated with t;:is 
 arch and the same laws approximate^ established. 
 
 In the experiment with the cloth joints the cloth was 
 .05 thick not pressed ; .04 when pressed hard b 
 The curve of resistance was found to pass .1 from tne 
 edges at the joints of rupture, with a weight on the ,-i>ex, 
 and nearly so in the pier experiment with no weight on the 
 apex. 
 
 Sixteenth Experiment. The Gothic arch given by Fig. 
 28 will now be considered with an uusymmetrieal loud. A 
 stout needle was thrust into the second voussoir from the 
 crown on the right side, in the, direction of a vertical 
 through its centre of gravity, as represent* d in Fit:. 31. With 
 
L95 
 
 a weight of 3.3 vs. on the top of the needle, the arch 
 balanct-d, opening at summit and lower end of joint 1 on 
 the right. The voussoir to which the weight was added 
 would have slid if urns had not been thrust into the edges 
 of its joints, thus supplying n force analogous to friction, 
 though not interfering at all with rotation. 
 
 We now form the following tabl s : the first being 
 condensed from the one referring to Lxp. :;. 
 
 LKFT .-IDE. 
 
 
 S 
 
 C 
 
 1.70 
 3.19 
 4.39 
 5.26 
 
 1 
 
 2 
 3 
 4 
 
 1 
 2 
 
 3 
 4 
 
196 
 
 RIGHT BIDE. 
 
 
 s 
 
 C 
 
 M 
 
 8 
 
 M 
 
 
 1 
 
 2 
 3 
 
 4 
 
 1. 
 
 4.3 
 1. 
 1. 
 
 1.7 
 
 4,68 
 6,79 
 
 7.88 
 
 1.70 
 20.12 
 6.79 
 
 7.88 
 
 1. 
 5.3 
 6.3 
 7.3 
 
 1.70 
 21.82 
 28.61 
 36.49 
 
 1.70 
 4.12 
 4.54 
 5. 
 
 7.3 
 
 36.49 
 
 A line of resistance can be drawn, as shown in the 
 figure, passing .18 from the extrados at joints 4 on left, and 
 1 on right and .18 from the intrados at joints and 3 on 
 the right. 
 
 The lower edge of the crown joint was imperfect, 
 being the only imperfect edge in the arch, and this may 
 account for the line of resistance retreating farther in 
 the arch than for a load in the summit as before r.>n- 
 sidered. 
 
 The thrust on joint 1, on the right, was inclined at 
 an angle of 15 to the normal to that joint, which accounts 
 for the sliding, as the joints were planed and across the 
 grain. 
 
 Seventeenth E.rperimciii. The segmented arch. Fig. :-:o, 
 was next tried with the eccentric load. 
 
 A short needle was thrust in voussoir 2 on the left, in 
 the direction of the vertical through its centre of gravity, as 
 shown in Fig. 32 ; the arch balanced with 5.4 voussoirs on 
 the top of this needle. 
 
 We form the following tables : 
 
 UIGHT SIDE. 
 
 
 s 
 
 m 
 
 S 
 
 M , C 
 
 .1 
 
 2 
 3 
 4 
 
 1 
 1 
 1 
 1 
 
 2.03 
 5.90 
 9.38 
 12.11 
 
 1 
 2 
 3 
 4 
 
 2.03 
 7.93 
 17.31 
 'J9.42 
 
 2.03 
 
 3.96 
 5.77 
 1 7.37 
 
198 
 
 LEFT SIDE. 
 
 
 S 
 
 M S 
 
 M C 
 
 2 
 3 
 4 
 
 1. 
 
 6.4 
 1. 
 1. 
 
 2.03 
 37.76 
 9.38 
 12.11 
 
 1. 
 7.4 
 8.4 
 9.4 
 
 2. 03 
 39.79 
 49.17 
 61.28 
 
 2.03 
 5.38 
 5.85 
 6.52 
 
 9.4 61,28 
 
 The voussoir on which the weight was placed would 
 have slid along its joints but for pins being thrust into its 
 edges in a manner that did not interfere with rotation. 
 
 A line of resistance was drawn that passes .15 from the 
 iutrados at joint 2 on the right and .2 distant from the 
 edges at joints 4, 1 and 4; hence the true curve will prob- 
 ably puss about .18 from these edges. This is nearly (.03 
 difference) what we obtained, for the limits from the 
 of the line of resistance in the llth Exp., Fig. '3(\. The 
 thrust on joint 1 on the left is inclined 16 .5 to the normal 
 to the joint, nearly wh<vi: we found before. The sliding in 
 this and the last experiment only occurred just before the 
 balancing weight was applied ; the line of pressures travel- 
 ling down the crown joint as the weight was increased, un- 
 til finally the direction of the pressure on joint 1 exceeded 
 the complement of the angle of friction. 
 
 Eighteenth Experiment. Figure 33 represents two rat- 
 ters 9.92 in length, 1.9 width (dimension in plane of paper) 
 and 3.60 thick, leaning against each other at the top and 
 against piers 7.7 high, 1.98 wide and 3.(> thick at their bot- 
 tom edge, which is moved back 0.6 from the edge of the pier. 
 The horizontal distance between the vertical piers is in in., 
 so that the feet of the rafters are 11.2 apart. Each rafter 
 weighed 2.3 vs. ; each pier 2. vs. The rafters and piers just 
 balanced in this position. 
 
 Reasoning as in Art. 2, we see that the thrust at the upper 
 edges of contact of the rafters is horizontal ; hence draw a 
 vertical line through the centre of gravity of the rafter 
 equal to its weight ; the resultant on the lower edge of the 
 rafter passes through this edge, and combined with the 
 
199 
 
 Fig. 33 
 
 weight of the pier acting through its centre of gravity, gives 
 the resultant thrust on the base of the pier. In thh- 
 it strikes twenty-two hundredths (.'22) from its outer edire. 
 
 This experiment was performed to ascertain whether 
 the resultant on the ba-e could ever be drawn through the 
 outer edge of base for the original figure. It seemed prob- 
 able, as the centres of pressure at the apex and top of the 
 pier were absolutely fixed, and there was only one real 
 
200 
 
 joint at the base of the pier ; but we see, even from this 
 case, that the joint opened sufficiently to deform the origi- 
 nal figure, so that the resultant cannot be drawn exactly 
 through the outer edge for the original figure. This should 
 offer a valuable hint to experimenters and constructors, not 
 
201 
 
 to look for the stability in similar structures that the the- 
 ory of ' ' rigid ' ' or incompressible bodies would give, espec- 
 ially structures composed of a great number of blocks with- 
 out cementing material. 
 
 Nineteenth Experiment. Fig. (34), represents a r-fter and 
 pier of the preceding experiment ; the rafter leaning against 
 a vertical rough plastered wall by its edge, the lower edge 
 resting on the pier 1.03 back from its inner edge. This was 
 the balancing position. 
 
 After several trials, assuming as we found in the pre- 
 ceding experiment, that the resultant strikes .22 from the 
 outer edge of the base of pier, it was found that the direc- 
 tion of the thrust against the wal? was inclined about 35 to 
 the horizontal, which is about what we should imagine the 
 angle of friction of the edge on the wall to be. If the thrust 
 at the upper edge be assumed horizontal as is usual, it will 
 be found that the final resultant passes outside the base of 
 pier; hence, such an assumption is false. The construc- 
 tion (Fig. 34), will also show that .32 v. of the rafter is sus- 
 tained by the wall, 1.98 v. being supported by the pier : i. c. 
 about one seventh of 'the weight of the rafter is upheld by 
 the friction of the plastered wall. 
 
 On leaning a half arch against a wall, it was found to 
 balance on higher piers than when the other half was placed 
 against it, 
 
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