THK CIVIL ENGINKER AND ARCHITECTS JOURNAL. 



[March 



Let it also be supposed that the half arch removed had the effect 

 of exerting on the half arch represented certain pressures, of which 

 the resultant is i force P, acting at some point d in the crown. P will 

 be wholly horizontal, if the two halves of the arch be similar and 

 similarly circumst inci-d. For, if every part of their mutual pressure 

 were vertical, it must act upward on one half arch ; and in the oppo- 

 site direction, or downwards, on the other half arch. And this is 

 evidently inconsistent with the hyputlieais that both half arches are 

 similarly circumstanced. 



The other forces of the system are its pressures upon its abutment. 

 These pressures will have a single resultant, acting al some point, a: 

 and by equating the vertical and horizontal forces, it follows that the 

 components of this resultant are an upward vertical force equal to 

 the weight of the system (W), and a horizontal force equal to P, but 

 contrary in direction. 



Kow, the tendency of the weight W to turn the half arch about 

 the abutment may be considered the cause of the force P at d being 

 called into existence. As no horizontal forces but those mentioned 

 are supposed to act, the horizontal force P exists iu every part of 

 this arcli. 



Draw a c horizontal, and Q6 vertical from G the centre of gravity 

 nf the system : this latter line is that in which the weight of the 

 system acts. Draw also c d vertical. 



The vertical distance of the upper force P from a is equal to cd^ 

 and the horizontal distance of the force W from a is a 6. The mo- 

 ment of P about a is therefore P X cd, and that of W is W X a 6 : 

 these moments are equal to each other. 



ab 

 .■.W.ab=P.cd, orP=W. — . 



cd 



If, therefore, we knew the exact value of ab, cd, and the total 

 weiglitVV, we should be able to get P at once from the above simple 

 equation. But, in fact, we do not know the exact point a, which is 

 the point of application of the resultant of the forces at the spring- 

 ing ; neither have we ascertained the exact point c. All we know of 

 either point is that it is somewhere between the extrados and intradosi 

 at the springing and vertex respectively. ' 



It is not necessary, however, for our purpose to define either point 

 for the limits assigned for its position are generally small enough to 

 define P from the above equation with sufficient accuracy for pr ctical 

 purposes. By referring to the equation, it will be seen that the 

 greatest value of P is derived from giving a b its greatest, and cd its 

 least, possible value : and conversely, the least value of P is derived 

 from giving a 6 its least, and cd its greatest, possible value. The 

 real value of P lies betweeu these limits. 



1st. The greatest value of P is derived from giving ab its greatest 

 and ed its least value; which conditions are satisfied by supposing o 

 to be in the extrados at the springing, and d in the intrados at the 

 vertex. Hence, from the equation we get this rule — "The greatest 

 value of the horizontal thrust is derived from multiplying the tola 

 weight of the half arch by the horizontal distance of its centre o 

 gravity from the springing of the extrados, and dividing the product 

 by the leant height of the arch, or the height of the vertex of the 

 intrados above the springing of the extrados." 



2nd. The least value of P is derived from giving at its least, and 

 cd its greatest, value. Hence, a must be taken in the intrados, and d 

 in the extrados; and the corresponding rule will be — "Multiply the 

 otal weight of the half arch by the horizontal distance of its centre 

 of gravity from the springing of the intrados, and divide the product 

 by the greatest height of the arch, or the height of the vertex of the 

 extrados above the springing of the intrados." 



Having obtained these rules we may proceed to practical illustra- 

 tions of them. It may however be first observed that their accuracy 

 dependsentirely on the supposition that the half arch does not sustain 



Fig. 2. 



horizontal pressures from the effect of the loading : if it did, the hori- 

 zontal thrust P at d, would no longer be equal to the P at a, fig. 1. 



Let us take the case of a half arch, 

 a a' d d', supporting a mass of masonry 

 above it, as in the accompanying figure 

 2. This would be the case of a gate- 

 entrance, or arch beneath a tower. If 

 the tower be lofty compared with the 

 rise of the arch, we may suppose G the 

 centre of gravity of the half load cen- 

 trically situated. This hypothesis saves 

 the trouble of calculating the position 

 of the centre of gravity of that part o'^ 

 the masonry which is situated in the 

 spandril of the arch: the n^e/g/i/ how- 

 ever of that portion must be included in 

 the total load W ; taking G iu this posi- 

 tion, then the rest of the calculation is 

 very simple. Draw the vertical line 

 b b, indicating the direction in which 

 W acts. Then, as we have already 

 shown, the greatest and least values of H J 

 the horizontal or lateral thrust are re- ' - ' 

 spectively, 



Wx"-iandWx — 

 c d c d'. 



Suppose, for example, that the weight of the masonry W is 20 tons ; 

 that by measurement c d' the least rise of the arch= 8 feet, and a b= 

 4 feet, (the span of the arch being about four times as much), then we 

 get immediately the greatest value of the thrusts 20 tons X | or = 10 

 tons. And suppose it to be also found by measure that C d, the ex- 

 treme rise, IS 9 feet, and a' 6'=3^ feet, then the thrust = 20 tons X Si 

 -^ 9 = 7| tons. ' 



Now it will be observed that between the greatest and least value 

 of the thrust so obtained (7^ and 10 tons), there is a considerable dis- 

 crepancy, namely, upwards of two tons. But it is to be remembered, 

 that methods here explained, though founded on exact principles, 

 are merely approximative; and, moreover, we have purposely chosen 

 a much more unfavourable instance than will generally occur in prac- 

 tice. 



In making a section, as in the above figure, dividing the arch and 

 its loading into two parts, it is virtually assumed that the part of the 

 loading which is represented in the figure is not subject to forces 

 arising from its connection with the part supposed to be removed. 

 That, in general, this hypothesis is nearly correct, will be allowed 

 when it is considered that the masonry of the superstructure is laid 

 on gradually in successive horizontal courses, and that each stone is 

 supported in its place by those below it, and h^is no tendency to roll 

 over or move sideways. Where the workmanship is accurate, there 

 will be little strain between the two halves of the load, except from 

 " settling" or similar accidental causes, the nature of which precludes 

 specific calculation. 



The above method of determining the thrust cannot apply except 

 when the form of the structure is such as to allow of a tolerably 

 accurate guess as to the position of the centre of gravity of the load 

 sustained by the half arch. This can only be made where the specific 

 gravity of the loading is uniform, and the structure is so lofty com- 

 pared with the rise of the arch, that the mass contained in each span- 

 dril is too small to materially affect the centre of gravity. Where, 

 however, the rise of the arch is considerable compared with th ' 

 dimensions of the superstructure, we may resort to the following 

 method, which wi.l be found very convenient, and which will have 

 the advantage also of determining the thrust when the loading is 

 heterogeneous. 



