arr Thus the first equation becomes pa + pd=+m (c × x) + n (x − e), Or pa × pd=" + mx + mc + nx − ne. 2 Transferring the unknown quantities to the second side of the equation, we shall Multiply all the terms by 2, and divide by a, in order to get rid of xx, and we a 1394. This last equation is a formula for finding the thickness of all sorts of arches whose voussoirs are of equal depth, which we will now apply to fig. 573. The model was 36 inches and 3 lines in span. The arch consisted of two concentric circles, and it was divided into four equal parts, a vertical joint being in the middle, the two others being inclined at angles of 45 degrees. The piers whereon it was placed were 40 inches and 4 lines high, and on a very exact measurement the values were as follow: PE (a in the formula) was EH TI=KL KV (d in the formula) ML x AB (p in the formula) representing the thrust or 8·127 x 3 2pd=48.762 × 13.876 2MK × AB KH represented by 2mc (=5749 × 3 × 4·249) a = EP, the height of the pier being 40·333, bb aa 40.333 13.876 24.381 48.762 676 621 Substituting these values in the formula x =√2p+ +2.128-1.459; which gives a =5·8, or 5 inches 94 lines for the thickness of the piers to resist the thrust of the arch, supposing it to be perfectly executed. But, from the imperfection of the execution of the model, it was found that the piers required for resisting the thrust a thickness of 6 inches and 3 lines. 1395. When the piers of the model were made 7 inches thick the arch on its central joint was found capable of supporting a weight of three pounds, being equal to an addition of 8 superficial inches beyond that of the upper parts of the arch which are the cause of the thrust, and this makes the value of 2p in the formula 56 762 instead of 48.762, 787 629+124-828-86-458 and changes the equation to x=56·762+' +2:430-1.55; from 40.333 which we should obtain a = 7.366 inches, or 7 inches 3 lines, exhibiting a singular agree. ment between theory and practice. Rondelet gives another method of investigating the preceding problem, of which we do not think it necessary to say more than that it agrees with that just exhibited so singularly that the result is the same. It is dependent on the places of the centres of gravity, and therefore not so readily applicable in practice as that which has been just given. Second Experiment. 1396. Fig. 567., in a preceding page, is the model of an arch in freestone, which has been before considered. It is divided into nine equal voussoirs, whose depth to the extrados is 21 lines, and whose interior diameter is 9 inches. 1397. Having drawn the lines heretofore described, we shall find mL x AB expressed in the formula by 2ne, which is twice the vertical effort of the lower part of 2me, which indicated twice the vertical effort of the upper a, which represents the height of the piers, being 195, and 560.70 1121-40 45.60 5113.584 20109.60 6667.92 6.94 And all these values being substituted in the formula, will give 1398. Let the mean curve TKG of the arch (whatever its form) be traced as in figs. 573, 574., the secant FO perpendicularly to the curve of the arch, and through the point K, where the secant cuts the mean curve, having drawn the horizontal line IKL, and raised from the point B a vertical line meeting the horizontal IKL in the point i, draw iK from K to m, and the part mL from B to h, and the double thickness of the arch from B Let hn be divided into two equal parts at the point d, from which as a centre with a radius equal to half hn, describe the semi-circumference of a circle which will cut in E the horizontal line BA prolonged. The part BE will indicate the thickness to be given to the piers of the arches to enable them to resist the thrust. to n. 1399. The truth of the method above given depends upon the graphic solution of the following problem: To find the side BE of a square which shall be equal to a given surface mL x 2e; an expression which is equivalent to 2p, and we have already seen that x= √2p was a limit near enough; hence we may conclude that the thickness BE obtained by the geometrical method will be sufficiently near in all cases. Experiments on surmounted Arches. 1400. The interior curve of fig. 574. is that of a semi-ellipsis 81 lines high; it is divided into four parts by an upright joint in the crown and two others towards the middle of the haunches determined by the secant FO, perpendicular to the interior part of the curve. Having traced the mean circumference GKT, the horizontal IKL, and the vertical Bi, we shall find The effect of the thrust indicated by KL-iK=mL will be 2p therefore 175.5 d being 66.5, 2pd will be 351 x 66.5, which gives 351 O 23341.5 m, which is KM × AB, will be 19 x 9, which gives 171-0 c, that is, i K, being 174 lines, we have 2me=171 x 17 x 2, which b, which expresses the sum of the vertical efforts m+n, will be 769.50 lines, or a little more than 16 lines. The model of this arch would not however stand on piers less than 17 lines thick. In taking the root of double the thrust the result is 183 lines, as it is also by the geometrical method. Application to the Pointed Arch. 1401. The model which fig. 575. represents was of the same height and width as the last, and the voussoirs were all of equal thickness. Having laid down all the lines on the figure as before, we shall find iK of the formula to be x = √252 +63·8 +416 -6.45=12·46 lines for the thickness of the pier. In taking the square root of double the thrust the thickness comes out 15.88 lines, as it does by the geometrical method. Experiments showed that the least thickness of piers upon which the model would stand was 14 lines. Application to a surmounted Catenarean Arch. 1402. The lines are all as in the preceding examples (fig. 576.). The whole arch acts on the pier in the direction FT, which is resolved into the two forces Tƒ and Tm, and the formula, as before, is thus having found Bm=22}, we have the value of p=22} × 9=201; and 2p=402. bb aa 1403. This model was of the same dimensions as the preceding : b, which represents Tfx AB, will be 769-5; will be 6.41, and = 41·11. These values substituted in the formula give x= √402+41·11 −6·41 = 14.64 lines. ፩ 769-5 120 1404. Experiment determined that the pier ought not to be less than 16 lines, and the geometrical method made it 28. The following table shows the experiments on six different models. n E PSR Fig. 576. This table shows that, in practice, for surmounted arches, the limit x= √2p, or the thickness obtained for the construction by graphical means is more than sufficient, since it gives results greater than those that the experiments require, excepting only in the cassinoid; but even in the case of that curve the graphical construction comes nearer to experiment than the result of the first formula. 1405. It is moreover to be observed, that the pointed is the most advantageous form for surmounted arches composed of arcs of circles. We have had occasion to speak, in our First Book, of the boldness and elegance exhibited in this species of arches by the architects of the twelfth and thirteenth centuries; we shall merely add in this place that where roofs are required to be fire-proof, there is no form so advantageously capable of adoption as the pointed arch, nor one in which solidity and economy are so much united. 1406. Next to the pointed arch for such purpose comes the catenary (the graphical method of describing which will be found under its head, in the Glossary at the end of the work), and this is more especially useful when we consider that the voussoirs may all be of equal thickness. Application of the Method to surbased Arches, or those whose Rise is less than the Half Span. 1407. For the purpose of arriving at just conclusions relative to surbased arches, three models were made of the same thicknesses and diameters, with a rise of 35 lines, and in form elliptical, cassinoidal, and cycloidal. We however do not think it necessary, from the similarity of application of the rules, to give more than one example, which is that of a semi-ellipse (fig. 577.), in which, as before, the formula is H E F M The lines described in the foregoing examples being drawn, we have TI, represented by d, being 24.84, we have 2pd m, which is KM x AB, will be 14.66 × 9, which gives c, representing iK, being 8.5, 2mc b, which expresses the sum of the vertical efforts m + n (39′5 × 9) a, being always 120, b 355.5 a is 120 16543-44 +8.76-2.9625-22 lines, or a little less than 25 lines. 1408. In the model it was found that a thickness of 26 lines was necessary for the pier, and the lower voussoirs were connected with it by a cementing medium. Without which precaution the thickness of a pier required was little more than one tenth of the opening. Taking the square root of double the thrust, that is, of 666, we have 25-81, about the same dimension that the graphical construction gives. The experiments, as well as the application of the rules, require the following remarks for the use of the practical architect. 1409. I. The cassinoid, of the three curves just mentioned, is that which includes the greatest area, but it causes the greatest thrust. When the distance between the intrados and the extrados is equal in all parts, it will only stand, supposing the piers immoveable, as long as its thickness is less than one ninth part of the opening. 1410. II. The cycloid, which includes the smallest area, exerts the least thrust, but it can be usefully employed only when the proportion of the width to the height is as 22 to 7 in surbased arches, and in surmounted arches as 14 to 11. The smallest thickness with which these arches can be executed, so as to be capable of standing of themselves, is a little more than one eighteenth of the opening, as in the case of semicircular arches. 1411. III. The ellipsis, whose curvature is a mean between the first and second, serves equally well for all conditions of height, though it exerts more thrust than the last-mentioned and less than the cassinoid. 1412. It is here necessary to remark, that too thin an arch, whose voussoirs are equal in depth, may fall, even supposing the abutments immoveable, and especially when surbased: because, when once the parts are displaced, the force of the superior parts may lift up the lower parts without disturbing the abutments. Raking Arches. 1413. Let ACA' (fig. 578.) be the model of a raking arch of the same diameter and thickness as the preceding example, the voussoirs of equal thickness, and the piers of different heights, the lowest being 10 inches or 120 lines in height, and the other 14 inches or 174 lines. The tangent at the summit is supposed parallel to the raking lines that connect the springing. 1414. This arch being composed of two different ones, the mean circumference on each must be traced, and each has its separate set of lines, as in the preceding examples; E the horizontal line KL of the smaller arch is produced to meet the mean circumference of the other in S, and the interior line of its pier in g. PSR Fig. 578. PSR 1415. The part KLS represents the horizontal force of the part of the arch KGS, common to the two semi-arches ; so that if a joint be supposed at S, the part LK represents the effort acting against the lower part of the smaller arch, and LS that against the lower part of the larger arch. These parts resist the respective efforts as follows: the small arch with the force represented by iK, and the greater one with the force represented by gS. But as gS is greater than LS, transfer LS from g to f to obtain the difference fS, which will show how much LS must be increased to resist the effort of the larger half arch; that is, the effort of the smaller one should be equal to Lf; but as this last requires for sustaining itself that the larger one should act against it with an effort equal to KL, this will be the difference of the opposite effort, which causes the thrust against the lower part of the smaller arch and the pier from whence it springs. Hence, transferring fL from L to q, taking the half of iq and transferring it from L to h, the part hK multiplied by the thickness AB will be the expression for the thrust represented by p in the formula Having found hK=30 and AB=9, we have for the value of p 30 x 9=2741, and for that of 2p-549, d which represents IT, being 291, 2pd=16195. In 2mc; m, which represents MK x AB, will be 12 x 9=111, and 2m = 222. c, which represents iK, being 8, we have 2mc=222 x 8=1776. The height of the pier represented by a being 174, we have The vertical effort represented by b, or TF x AB, will be 41 × 9=375, =82.81 J= 549 +82-81 +464-216-23-08 for the thickness of the greater pier from which the smaller semi-arch springs. For the half of the greater arch, having produced the horizontal line IK'L', make K'r equal to VL', and bisect rL' in t; the line K't represents the effort of the smaller against the greater arch, which resists it with a force shown by 'K': thus making K'q' equal to 'K, the effort of the thrust will be indicated by q't x AB, whose value p in the formula will be x= √360+118 65 +51 48-7175-15.855 lines for the thickness of the smaller pier. |
