14S6

THEORY AND PRACTICE OF ENGINEERING.

Book II.

Fig. 2703.

value of this no general rule can be given, but it must scarcely exceed the width of thearch ; and if particular circumstances require a greater radius, the joints of the voussoirsmust not be directed to the centre of the arch, but towards a nearer point.

These two first conditions fulfilled, the number of arcs of a circle of which the archshall be composed must be determined; it must not be less than 3, nor exceed 11. Thefollowing rules for describing these arches are usually adopted by engineers.

Flat Arches described tcith three Arcs of a Circle .— The length of the radii of the three arcscomposing the flat arch not being determined, when the two diameters of the curve arefixed, another element must be introduced. We will suppose first that the three arcsare all of GO° ; that is to say, each equal to the sixth part of the circumference.

The half of the span of the arch A C is equal to a, and its height C D = 5; the centres of thetwo arches described at the springing will necessarily be situated in the two points F, G, inthe line A B, and the centre of the third arch in a point E in the line D C prolonged;one of these centres being found, the two others will be also.

Call x the distance from the point C, in themiddle of A B, to the centre G. The triangle o

F G E being equilateral, we have E G equal F G,and, consequently,

CE+CD=FG+GB;or V 3 . x + b =x + a ;

and by resolving this equation,

a — b a/3 4- 1 . ..

"=V3^1 = -2- (a - 6 >

This value is constructed by setting off from thepoint C the line CK~a — b, and after havingformed on it the equilateral triangle CIIK, carry-ing from L to G the height Eli of the triangle.

The position of the centres of the three archesmay also be obtained by tracing the quarter of thecircle A I H (fig. 2704.) and the equilateral triangleAC I; then drawing D3I parallel to HI, andM F E parallel to I C.

Eet us now suppose that one of our conditions isthat the greater and lesser arc differ from eachother as little as possible. Call the half of thespan of the arch AC —a, and its height C D = 6;the radius A F of the arc of the springing y, andx that of the arc of the summit, the centre ofwhich is in E ; we shall have by reason of the rec-tangular triangle C F E,

(x-y) 9 = (*-&)* + (a — y) 9 .

In resolving this equation with relation to x, wefind

\ (b* + a*)-ay

* h ~y

We shall then have for the expression oi the re-lation of the two radii,

x _ ^ (b 2 + a g ) —yV ~~ by-f

By making its differential equal zero, and find-ing the value of y, we find, after abridging, b 9 + a 9 = c 9 ,be

c + (a —by

putting this value in that of x, we deduce from it,a c

x = c ~(^Zb)'

These two expressions are formed by cutting off from the line A D (fig. 2705.) the distanceD G. equal to a — £», and raising on the middle of the remaining part A G the perpendicularFI E ; the points E, F, where it meets the two diameters of the curve, will be the centressought. This method gives a greater difference between the lengths of the two arcs thanthe preceding, which appears preferable. When the relation of the two diameters is notless than one-third, the difference of the radii of the arcs of which the flat arch is composedis not sufficient for the passage from one to the other to be marked, and produces a dis-agreeable effect, and it must be described with three arcs of a circle ; but when the archis flatter, a greater number must be employed.

Fig. 2704.

Fig. 2705.