508 



BRIDGE. 



Theory. 



The pier in the first case, taking it at 5 feet, 

 5x74=370, and i?> + in will be 185+ 



will be 



185+900 or 



10S5 ; multiply this by 5, we have 5425, a little 

 more only than the overturning force, as the thick- 

 ness was taken at 5 feet, which is a little in excess. 

 The reader, if he chooses to go through the calcula- 

 tion fur himself, will find 4.97 agree exactly. 



In the second case, the pier =5.68 nearly, X 18 

 = 102.24., and .its half =51.12, which added to 900, 

 and multiplied by 5.68, gives 5402.3. A trifle in 

 excess, because 5.6S, like the former, is only an ap- 

 proximate number. 



The weight of the pier in this case making so 

 small a part of the whole resisting force, we may 

 readily believe, that its total immersion in water 

 would make no great addition to the requisite thick- 

 ness. Stone, when so immersed, loses about -J. of its 

 weight, being in specific gravity about 2-J times that 

 of water ; and, in the above example, were the whole 

 pier under water, it ought to be about a fiftieth part 

 thicker. 



We have hitherto supposed the arch equilibrated, 

 at least as far as is conveniently practicable, in which 

 case the horizontal thrust is represented by the rec- 

 tangle under the radius and thickness at crown. 

 But if the equilibration of the arch has not been at- 

 tended ta, we must consider whether any uncommon 

 weight about the shoulders may not produce, by the 

 help of friction, a thrust in the arch fully equivalent 

 to what would arise from a greater thickness at the 

 crown ; and our calculations are to be regulated ac- 

 cordingly. 



On the other hand, we have given the arch a 

 weight in the above example which is nearly that of 

 solidity. But in general the arch weighs much less. 

 The most common case, where the stability of the 

 pier is any way doubtful, is when it carries no more 

 than the ring of arch-stones, and before it is assisted 

 by the weight of the superincumbent backing. The 

 weight keeping the pier steady, is now much dimi- 

 nished ; while the horizontal thrust is unaltered ; 

 tor, if not propagated by weight, it is by means of 

 the friction of the sections propagated 'to the pier, 

 so as to act against it in the same manner as if com- 

 pleted. 



Now, as it is by no means likely that the arch 

 will be made thinner at the spring-courses than at 

 the crown, while any additional thickness of the 

 former is always in favour of the piers, we shall pro- 

 ceed upon the supposition, that a regular annulus, 

 or ring of stones, is laid on them everywhere of equal 

 thickness. Suppose this thickness, as before, to be 6 

 feet. In that case the semi-arch of the above dimensions 



measures 499.5, or 500 feet, and 



,J 2 1 + (- 



3a I 



T*= v 



600+ 



1500 



1500 



^=32.13 



4.18 



feet for the breadth of the pier. But it is 

 by no means likely that the arch would have 6 feet 

 thickness of crown in these circumstances ; 2, or at 

 most 3 feet, would, in all probability be thought 

 sufficient for a depth of keystone ; and a ring of 

 3 



arch-stones 2 feet deep will require a pier of 9 feet Theory 

 only. If we build up the pier behind the springing > -v- 

 for about 6 feet, this thickness may be reduced to 

 8 feet ; and it will be absolutely necessary to do so 

 in a case of this kind, to prevent the lower sections 

 of the arch from sliding away. 



The above example is taken for a semicircular 

 arch ; and though the reader must see, that the 

 thickness of the pier is in no certain proportion to 

 the span, it is nevertheless obvious, that those 

 writers who derive it from that, have hitherto erred 

 considerably in excess. It is usually stated at % for 

 semicircles ; but we see, that in the most unfavour- 

 able circumstances, it need not exceed - of the span, 

 and may often be made much less. This, however, 

 we state with limitation, referring to the height of 

 pier above given ; for were the pier much higher, it 

 must be made thicker ; if the pier be infinitely high, 

 the weight of the arch sinks into insignificance, and 

 the thickness, =^2*, which in the above arch 6 

 feet thick is =24 feet nearly, and in general, if 



the thickness at crown = - of radius, then - 



1 r * /2 



- = r *J ~ > that is, takingnhe span, ra=2tt, the 



thickness = s J ; whence this rule for the 



*> m 



thickness of a pier of infinite height. Find what 

 part the thickness at crown, is of the span, extract 

 the square root, and multiply it by the span for the 

 thickness ; or thus, multiply the diameter by the 

 thickness at crown, and extract the square root. 



One of the loftiest bridges with which we are ac- 

 quainted is that of Alcantara, over the Tagus, in 

 Spain. It is stated by Don Antonio Ponz, in his 

 Viage d' Espana, to consist of six arches, the two 

 largest 110 feet in span, the water at the lowest is 

 42 feet deep ; from the surface of which, to the be- 

 ginning of the springing of the middle arches, 87 j 

 and from thence to the upper surface, 76 ; which, 

 with the 4 feet and a half of parapet, make the 

 whole 205 feet and a half, (more correctly, 209i). 

 Taking then the thickness at crown as equivalent to 

 16 feet, and the diameter 110, the thickness for an 

 infinite height should be 42 feet. They are 38 in 

 thickness, and 129 feet high. Let us now try this 

 thickness by the general formula given in the earlier 

 part of this Section. 



The lower or immersed part 42 feet high, and 38 

 broad, is 1596 ; but of this are to be deducted on 

 account of the immersion, leaving for that part 958. 

 The pier from thence to the springing is 87 by 38, 

 or 3306. We must suppose such a pier built up be- 

 tween the arches to at least $ of the height, or about 

 20 feet ; but on account of a set off which appears 

 in the design, we shall suppose the breadth still 38 

 on an average, which makes 760, and the whole pier 

 5024, and its half is 2512. To this add | of the 

 semi-arch ; say $ X 55 X 16=660, and we have 3172. 

 By this number let us divide the product of the ho- 

 rizontal thrust and height of pier, that is, 16x55 

 X 129= 1 13520, and we find about 36 feet, very near 



